# RealClimate

1. So how did they eventually discover the Beer-Lambert-Bouguet law? And pressure broadening? Inquiring minds want to know!

Comment by Barton Paul Levenson — 26 Jun 2007 @ 7:50 AM

2. Sweet!

I had run across some of this elsewhere, but you are doing a better job at it by far. Anyway, weak absoption is a linear function of the column abundance before saturation at the peaks, strong is with saturation and works according to the square root, and very strong follows the Beer-Lambert. But I would assume that these are still approximations – and that the actual function applicable at all column abundancies would be would be a function of column abundance and ppm.

Comment by Timothy Chase — 26 Jun 2007 @ 8:40 AM

3. PS to my earlier post

In fact, it would seem that one could numerically derive a formula for sensitivity based upon the spectra for both carbon dioxide and water vapor spectra if given the function for water expansion, the shape of the land masses and the total amount of water. As this has not been done, I assume I am missing more than a little.

Comment by Timothy Chase — 26 Jun 2007 @ 9:03 AM

4. Hey Ray or Spenser;

Could you clarify the statement, “If Koch and Angstrom had examined the changes over the range between a 10cm and 1 meter tube, they probably would have been able to determine the correct law for increase of absorption with amount, despite the primitive instruments available at the time.”. It seems confusing as it would seem that the obvious explanation is simply due to the inverse square law. You assistance in seeing your point would be most welcome…

Dave Cooke

Comment by L. David Cooke — 26 Jun 2007 @ 9:22 AM

5. Is there a way to calculate the forcing where increasing concentrations of CO2 causing a feedback of increasing H2O evaporation yield an amplified forcing from the combination of CO2 and water vapor?

[Response: Yes. This is done all the time in simple one-column models of the atmosphere, such as pioneered by Manabe. Since you don’t have the dynamics that control water vapor in such models, you replace all that with a n assumption that the relative humidity remains fixed as climate warms. Thus, more water vapor goes into the atmosphere, leading to more greenhouse effect, and amplifying the initial effect of CO2. Dave Archer’s online model has an option to hold the relative humidity fixed as you increase the temperature of the atmosphere, which allows you to explore this feedback. –raypierre]

Comment by Tim Jones — 26 Jun 2007 @ 10:21 AM

6. raypierre and Spencer Weart

Thank you very much. It works well for me.

Comment by Allan Ames — 26 Jun 2007 @ 10:46 AM

7. Answer to my question in #3:

1. With climate sensitivity to CO2, you are also taking into account the albedo – and climate sensitivity is in the long-run.
2. Once you take into account the albedo and over the long-run, you are also dealing with ice melt;
3. You have to include convection as well as the distribution of the gases;
4. You have to deal with how the system evolves over time, which means convection both within the atmosphere and the ocean;
5. You have to take into account weather variability and non-laminar flow;
6. You have to take into account storms, including hurricanes – and we have difficulty modeling those.

In the case of hurricanes, our calculations based simply upon an understanding of convection and surface temperature would suggest that wind speeds would be considerably lower than they are, and it appears that water spray in essence lubricates the storm, permitting it to achieve higher speeds.

But once you took all of this into account, you might as well throw in the carbon cycle, vegetation (albedo), soil (albedo), cloud formation (albedo, etc.), evaporation from soil, the spectra for the sun and its luminosity, solar cycles, and once you do that you pretty much have the entire climate model, including how it evolves over time – or there abouts. And at that point, you really aren’t talking about climate sensitivity any more since climate sensitivity would hold the level of carbon dioxide in the atmosphere constant.

Geez! No wonder you guys need supercomputers and fairly big grids.

Comment by Timothy Chase — 26 Jun 2007 @ 10:54 AM

8. I’ve heard the pressure broadening being questioned.
This explains a mechanism:
which is impacts during the emission process.

But the wikipedia explanation is for emission frequency broadening and not absorption frequency broadening which is relevant in our case?

Comment by mz — 26 Jun 2007 @ 12:00 PM

9. At root pressure broadening is a reflection of the fact that a molecule’s quantum levels change when you bring another molecule close to it, within a few nanometers as a practical matter. At low densities, such interactions, or collisions only involve the radiator and its single collision partner. This is called the impact approximation, e.g. that collisions only occur between pairs of molecules. At high densities the models assume that the radiator is in a cage of other molecules, which may or may not have some short range order (this is equivalent to a liquid). In either case you have to consider the distribution of the change in the quantum state energies over time as the molecules move relative to each other.

Changing temperature changes the number of collisions per second, and to a limited extent the time over which the collision takes place (temperature broadening). Unlike lifetime broadening, which is the same for every radiator, pressure and temperature broadening are heterogeneous (each molecule sees a different environment).

Comment by Eli Rabett — 26 Jun 2007 @ 3:08 PM

10. Please include above that the Hitran database was started in the late 1960s. This is not very clear above. It lends some perspective.

Thanks,

[Response: I’m a little unclear myself as to the precise relationship between the “AFGL tapes” referred to in Goody and Yung, and the Hitran database itself, other than that they seem to contain similar information organized in a similar way. Of course, Hitran, now, is far more comprehensive than the AFGL tapes are likely to have been. Can you suggest a convenient reference giving some history of the Hitran database and how it came to Harvard? –raypierre]

Comment by Richard Ordway — 26 Jun 2007 @ 5:37 PM

11. Does the broadening work similarly for absorption and emission?

Comment by mz — 26 Jun 2007 @ 6:06 PM

12. Thanks heaps for that explanation. I indeed got it wrong when trying to explain the CO2 saturation fallacy. At least I will now be a bit better equipped the next time it comes up.

RealClimate continues to be the standard reference for laypeople like myself.

Comment by Ender — 26 Jun 2007 @ 6:48 PM

13. RE #9 Eli Rabett

The cage – that would be due to van der Waals forces – inverse cube.

Lifetime broadening – due to the uncertainty principle as it applies to time and energy, the spontaneous decay from a higher energy state to a lower one – Lorentzian spreading.

Collision broadening – reduces the average lifetime of the excited state, and after numerous collisions results in what is called homogenous broadening. I would assume this is what you refer to as “pressure broadening.”

Doppler broadening – due to different molecules moving at different velocities. I would assume this isn’t significant for what we are dealing with here. Or would this be what you are refering to as “temperature broadening”?

Comment by Timothy Chase — 26 Jun 2007 @ 8:10 PM

14. Hi, Ray. I appreciate this good explanation of the greenhouse effect. I am trying to understand how radiation of longwave energy depends on the temperature of the atmosphere. This article says:

Roughly speaking, in the part of the spectrum where the atmosphere is optically thick, the radiation to space occurs at the temperature of the high, cold parts of the atmosphere. That’s practically zero compared to the radiation flux at temperatures comparable to the surface temperature; in the part of the spectrum which is optically thin, the planet radiates at near the surface temperature. Increasing CO2 then increases the width of the spectral region where the atmosphere is optically thick, which replaces more of the high-intensity surface radiation with low-intensity upper-atmosphere radiation, and thus reduces the rate of radiation loss to space.

My understanding is that a greenhouse gas is not a black body, so the Stefan-Boltzmann law does not apply. I have also been told that re-radiation of absorbed radiation is in quantum units. Can you please explain why the colder greenhouse gases at the top of the troposphere radiate less than warmer gas further down?

Is it correct to say that if there was no lapse rate, ie. the atmosphere was the same temperature all the way up, there would effectively be no greenhouse effect because it would radiate at the same temperature as the surface of the Earth? Or that if temperature increased going up the greenhouse effect would be negative, meaning it would cause cooling?

[Response: To answer your last question first, it is basically true that if there were no lapse rate there would be no greenhouse effect. If there weren’t any turbulent transfers at the ground, you could technically get a greenhouse effect from an isothermal atmosphere if there’s a temperature jump between the ground and the atmosphere, but turbulent transfers in practice limit this effect. (In this statement, I’m putting aside exotic cases like the scattering greenhouse effect on Early Mars from dry ice clouds, which can indeed work in the absence of a lapse rate). As for the earlier part of your query, even though greenhouse gases are not blackbodies, Kirchoff’s law still applies. That means that the emission is the product of an emissivity/absorptivity coefficient (that’s where your quantum effects come in) and the Planck function. The Planck function goes down with temperature, for any give frequency, and there you are. –raypierre]

Comment by Blair Dowden — 26 Jun 2007 @ 9:10 PM

15. raypierre

thanks for your response to #5. it obviates my semi objection in the “Saturated Gassy Argument” post comments completely. We really need simple answers like that (keeping relative humiditiy constant) now that are based on fundamentals.

I shudder to think what would happen if getting people to understand GHGs involved having them understand Pauling’s resonance model.

Comment by Marion Delgado — 27 Jun 2007 @ 1:31 AM

16. mz @ 8 and 11:

Yes, the broadening effect works for both emission and absorption. The Wikipedia article you pointed to is about spectral lines, and at other points in the article they note that “spectral lines” applies to both absorption and emission. (I’m guessing whoever wrote the text in the broadening sections is more used to thinking about emission spectra, and so got lazy and kept referring only to emission.)

Comment by Peter Erwin — 27 Jun 2007 @ 4:36 AM

17. When the product of the absorption factor times the amount of CO2 encountered equals one, then the amount of light is reduced by a factor of 1/e, i.e. 1/2.71282…

Minor quibble: I think that should be written as “reduced by a factor of e” (i.e. Flux_final = Flux_initial / e). “Reduced by a factor of X” usually means “divided by X”, so “reduced by a factor of 1/e” implies “divided by (1/e)”, which isn’t what you want to say…

[Response: If you think this bit of stylistics is confusing, you should have seen us trying to figure out whether “auf” in Angstrom’s paper (as in CO2 reduced “auf 2/3”) means “by 2/3” or “to 2/3”; fortunately we were able to double-check with Stefan. Prepositions are semantically tricky, even for native speakers. That’s why, given my druthers, I prefer equations to words. –raypierre]

Comment by Peter Erwin — 27 Jun 2007 @ 4:44 AM

18. Ray,

I agree with you that Angstrom was wrong in saying that increasing the amount of CO2 in the atmosphere would not raise the surface temperature, but his experiment is pretty convincing with a 50% increase in the amount of CO2 hardly altering the absorption.

The current models have the greenhouse gases radiating in their bands with the same intensity that they absorb, but the greenhouse gases lose most of that excitation by collisions to other air molecules. It defies the Law of the Conversation of Energy for the lowest layer of the atmosphere to warm the air, as it does every day, and also radiate back to Earth the same energy it has absorbed. Shouldn’t we be looking for another way in which CO2 can affect the climate, such as that suggested by Callendar.

IMHO, what really happens is that the radiation is absorbed in the first 30 m of the atmosphere, with a rate of absorption that decreases exponentially with height, just as the Beer-Lambert law predicts. If you then double the CO2, you will half the height of this layer, with the result that the warming from the CO2 will be concentrated in a smaller volume causing more heating. The heated air will rise, preventing the CO2 from re-emiting the energy it has absorbed to the surface.

However, the maximum absorption is at the surface of the Earth, (e.g. bottom mm. or 0.1″.) Therefore it will cause more evaporation, leading to more water vapour which is also a greenhouse gas. Moreover, it will also cause more melting of ice, and this will change the planetary albedo. In this way, higher CO2 levels lead to snowline rising in altitude and latitude, and so is how CO2 levels are linked to ice ages.

But, just as the absorption from the surface is saturated, so is the radiation emitted at the top of the atmosphere. Increasing CO2 cannot emit more radiation. In order for the planet’s climate to stabilise, if it loses ice albedo, then it will warm until there is an increase in albedo due to the formation of new clouds.

When the Arctic sea ice melts, then the change in albedo will be so great that it could cause a radical change in the atmospheric circulation before it settles into a new regime where there is adequate cloud cover to produce climate stability.

Stephen Weart described how the US Air Force saved the day by finding that the absorption bands were thinner at high altitudes. That sounds almost like a 1950’s cowboy movie with the US Cavalry coming over the horizon to rescue the heroine. However, I suspect that “The Day After Tomorrow” is closer to the truth, with a lone scientist who knows that disaster will strike, but no one will listen to him!

Comment by Alastair McDonald — 27 Jun 2007 @ 10:41 AM

19. Alastair, what’s special about the bottom millimiter or 100 mils? Why not the bottom micron, or even the bottom angstrom. Indeed, let us go all Zeno-esque and show that Earth’s surface cannot radiate at all. Only one thing wrong–reality isn’t cooperating, and indeed IR radiation in the CO2 band does make it to the stratosphere, and even escapes to space. The usual response when evidence conflicts with theory is to modify the theory accordingly.

[Response: There’s nothing wrong with the theory. The theory, as embodied in radiation models, is in extremely good agreement with all available observations. I don’t think Alastair has managed to understand either my post or Spencer’s. I don’t understand how Alastair could still be talking about all the absorption happening so low down, if he had understood the wavelength dependence of absorption in the graphs I showed. It’s very frustrating. I worked hard to make this stuff perfectly clear, but evidently I have not succeeded. –raypierre]

Comment by Ray Ladbury — 27 Jun 2007 @ 11:02 AM

20. raypierre (inline comment to #19) wrote:

I don’t think Alastair has managed to understand either my post or Spencer’s. I don’t understand how Alastair could still be talking about all the absorption happening so low down, if he had understood the wavelength dependence of absorption in the graphs I showed. It’s very frustrating. I worked hard to make this stuff perfectly clear, but evidently I have not succeeded.

The important thing is that the radiation at the lower levels has to make its way up through higher levels of the atmosphere before it is able to escape into space. It will be absorbed and re-emitted numerous times, and each time it could be re-emitted either towards space or torwards the ground – but ultimately it has to escape. But to escape, it must pass through the level where the atmosphere is dry and the effects of carbon dioxide dominate.

I am not sure why some have difficulty with this, but it is an easy principle to keep ahold of once you grasp it.

Comment by Timothy Chase — 27 Jun 2007 @ 11:28 AM

21. Alistair:

Aha! This is exactly the response I feared someone would have to this. First, Koch’s experiment was poorly designed and badly interpreted. Second, it ignored positive feedbacks. We are not dealing with an experiment with the correct measurements of the tube, carefully recorded, and with relative humidity kept constant.

The correct answer is, therefore, that Koch’s experiment is completely unconvincing because he didn’t account for differences between surface level and the upper radiant levels, didn’t measure precisely because he didn’t understand the importance of small differences in absorption over time – which you have because excess C02 lingers, and didn’t understand the importance of small differences in C02 and small but persistent increases in temperature because he didn’t account for water vapor’s response to temperature and pressure changes.

As a piece of the overall puzzle this is a tremendous post. Maybe my favorite on Real Climate so far.

Comment by Marion Delgado — 27 Jun 2007 @ 11:28 AM

22. Re #14: Thanks, raypierre, for your response. I don’t understand how the Planck function, which is about black body radiation, affects a greenhouse gas. Is it basically the same, except there is only one frequency of radiation instead of the black body distribution? If a cold greenhouse gas absorbs a photon emitted from the ground, and emits at a lower intensity, where does the rest of the energy go?

If the greenhouse depends on the the temperature of the gas that radiates into space, that implies all the other events of absorbing and re-radiating longwave radiation are of no ultimate consequence for determining the temperature of the Earth. But when a greenhouse gas aborbs radiation, it is either transformed into kinetic energy (ie. warms the local air) or is re-radiated. Some of this re-radiated energy should reach the ground and warm it. This process appears to be independent of the temperature of the air. I am having a hard time connecting this with your description. There are still a few pieces of this puzzle that I am missing.

[Response: The radiation is not independent of temperature, since the emitting molecules and the photons are interacting with each other and forming a kind of equilibrium. If you accept Kirchoff’s law for the moment, that tells you that the emission is the absorption coefficient times the very same Planck function B(nu,T) that governs blackbody emission. The statistical mechanics behind this is very tricky, because it’s not a true thermodynamic equilibrium. A very turgid derivation is given in Goody and Yung. If you like, you can just take Kirchoff’s law as a well-established experimental fact. That’s indeed how early investigators hit upon blackbody radiation to begin with. They weren’t measuring actual blackbodies, but rather were measuring the ratio of emission to absorptivity of various stuff, and finding that the ratios fell on a universal curve that depends on frequency and temperature but not on the stuff. That curve is what we would now call the Planck function, and it increases monotonically with temperature at any given frequency. Where I’m sympathetic with Alastair is that people take Kirchoff’s Law far too much for granted. It’s extremely well confirmed experimentally, but its conceptual foundations, though basically put on a sound footing by Einstein, are very, very subtle. They have bedazzled such greats as the mathematician David Hilbert. –raypierre]

Comment by Blair Dowden — 27 Jun 2007 @ 11:43 AM

23. Alastair McDonald (#18) wrote:

But, just as the absorption from the surface is saturated, so is the radiation emitted at the top of the atmosphere. Increasing CO2 cannot emit more radiation. In order for the planet’s climate to stabilise, if it loses ice albedo, then it will warm until there is an increase in albedo due to the formation of new clouds.

Adiabatic expansion increases the distance between CO2 molecules giving them more time to relax. That is why the center of the band isn’t saturated. Additionally there is always some spreading, even if it is only due to the lifetime broadening which results from the uncertainty principle as applied to time and energy.

Comment by Timothy Chase — 27 Jun 2007 @ 11:52 AM

24. This is a very nice article. I really enjoy seeing abstract concepts in physics explained so that most laymen can grasp the essentials of what’s being explored. It’s been decades since I was a physics undergrad, but I managed to understand the essentials fairly well.

Thanks.

Comment by JIm Frank — 27 Jun 2007 @ 1:35 PM

25. Raypierre and Alastair–that’s precisely the theory–Alastair’s theory–that I’m trying to get Alastair to see doesn’t hold water.
Alastair, I don’t mean to be dismissive. I just can’t see where you get the idea that absorption–which is isotropic–and emission–also isotropic–gives rise to a one-way opacity to IR radiation in the lower atmosphere. While the CO2 absorption lines are far from saturated in the upper atmosphere, they are still flattened a bit in the peak–that right there tells me that the IR getting through is not insubstantial.

Comment by Ray Ladbury — 27 Jun 2007 @ 2:50 PM

26. Marion,

I sympathise strongly with Ray. I too work hard to make my ideas clear, but seem to have little success :-(

It is highly unlikely that Dr Koch failed to keep his relative humidity constant. He was using pure CO2. See: http://docs.lib.noaa.gov/rescue/mwr/029/mwr-029-06-0268a.pdf Moreover, the saturation of greenhouse gases had already been discovered by Tyndall. See:

The Bakerian Lecture:
On the Absorption and Radiation of Heat by Gases and Vapours, and on the Physical Connexion of Radiation, Absorption, and Conduction
John Tyndall
Philosophical Transactions of the Royal Society of London, Vol. 151, 1861 (1861), pp. 1-36

Ray claims that his diagrams of bandwidth broadening show that the radiation is cascaded through the troposphere, but to me that seems an unreasonable conclusion to draw. It is true that bandwidth broadening will widen the lines, but for it to make much difference Ray has had to postulate a 400% increase in CO2. At present with less than 40% we are seeing global warming.

Moreover, when a line is broadened it happens at all altitudes. The effect is that the bandwidth at the bottom level then becomes the width one level up, and the width at one level up is then at level two. In other words, if we ignore the bottom level nothing has changed. Therefore the bottom level is where the real change is happening, and so the main increase in absorption will there, QED.

But more importantly, carbon dioxide only makes up about 400 parts per million of the atmosphere. Doubling the amount of CO2 will not double the number of collisions a CO2 molecule receives. In fact it will make practically no difference at all, since here we are considering pressure broadening, and the global surface atmospheric pressure can only change if the mass of the atmosphere changes. Adding another 400 ppm of CO2 to the atmosphere will hardly change the mass of the atmosphere!

Moreover, the low concentration of CO2 in the atmosphere also shows why the current models are wrong. When the CO2 is excited, some of that energy is converted to heat. But the heat goes mainly into N2 and O2 molecules which do not re-radiate the heat. They convect. Therefore the energy of radiation absorbed by the greenhouse gases is not re-radiated back to Earth. It is carried high into the atmosphere by convection, and then is radiated from below the tropopause either by exciting CO2 molecules or from the tops of clouds.

Although the radiation to space is from the same region in both schemes, in my scheme the surface global warming is driven by effect at the surface, not ten miles away by a belt of air with a density of only 1% of that at the surface.

Comment by Alastair McDonald — 27 Jun 2007 @ 4:02 PM

27. As I explained in my talk in Stockholm on 12 september 2006, Knut Ångström measured CO2 absorption on Mount Teide on Tenerife using solar radiation (with the Wien peak in the visible light), which means that the near infrared CO2 absorption bands hardly have any contribution to the absorption.
http://home.casema.nl/errenwijlens/co2/angstrommodern1.gif

He used a measurement instrument with two tubes: one with air and one with CO2, and measured the differerence in absorption. Of course he couldn’t measure a difference with this crude method.
http://home.casema.nl/errenwijlens/co2/angstromfig1.gif

That’s why Knut Ångström claimed the CO2 spectrum was saturated.

[Response: It’s true that most of Angstrom’s paper (which is available from the link we gave in Part I) dealt with absorption of solar radiation. There is, however, the almost throwaway reference to Herr J. Koch, who did indeed do the experiment for thermal IR. It was Koch’s experiment that was picked up by Monthly Weather Review and used to argue for saturation. –raypierre]

Comment by Hans Erren — 27 Jun 2007 @ 5:01 PM

28. “…the atmosphere still wouldn’t be saturated even if we increased the CO2 to ten thousand times the present level.”

Doesn’t this mean that the amount energy absorbed is very, very small for a given increase in concentration for these bands?

Shouldn’t we be concerned with the bands that are more readily absorbed, you know, that give the molecule the characterstics that make them green house gasses?

Comment by aaron — 27 Jun 2007 @ 6:17 PM

When a gas is in LTE then the absorption and emission is isotropic, but when Milne came up with the idea of LTE he wrote “This permits us to see in a general way why the state of local thermodynamic equilibrium in the interior of a star breaks down as we approach the surface.” My point is that plantary atmospheres have two surfaces; interior and exterior. Where the surface of the Earth is radiating into the atmosphere, the radiation is unidirectional. The same is true at the top of the atmosphere where the outgoing longwave radiation to space is also unidirectional.

BTW. Don’t confuse region emitting from the TOA with the non-LTE region. The non-LTE region does not emit net long wave radiation. It just gets hotter!

Comment by Alastair McDonald — 27 Jun 2007 @ 6:22 PM

30. Alastair McDonald (#26) wrote:

Moreover, when a line is broadened it happens at all altitudes. The effect is that the bandwidth at the bottom level then becomes the width one level up, and the width at one level up is then at level two. In other words, if we ignore the bottom level nothing has changed. Therefore the bottom level is where the real change is happening, and so the main increase in absorption will there, QED.

It is at the bottom level that water vapor dominates. It absorbs virtually all of the longwave which is available at that level. Therefore adding carbon dioxide at that level will have a miniscule effect. However, the stratosphere is a completely different matter. The stratosphere is very dry, and for this reason CO2 dominates in the stratosphere.

Moreover, the low concentration of CO2 in the atmosphere also shows why the current models are wrong. When the CO2 is excited, some of that energy is converted to heat. But the heat goes mainly into N2 and O2 molecules which do not re-radiate the heat. They convect. Therefore the energy of radiation absorbed by the greenhouse gases is not re-radiated back to Earth. It is carried high into the atmosphere by convection, and then is radiated from below the tropopause either by exciting CO2 molecules or from the tops of clouds.

How do individual molecules “convect”? As I understand it, thermal energy is random kinetic energy and as such will be isotropic. What can convect is an air mass. And whatever thermal energy might be lost due to some collisions can also be gained by other collisions? Can it not? And it wouldn’t have to be the same CO2 molecule losing energy as gains it. In addition, the greater the altitude, the lower the air density, the fewer the collisions.

Although the radiation to space is from the same region in both schemes, in my scheme the surface global warming is driven by effect at the surface, not ten miles away by a belt of air with a density of only 1% of that at the surface.

But the extensive set of observations of upwelling and downwelling radiation does not fit into your scheme. It fits mainstream theory. And the troposphere is responsible for only three quarters of the mass of the atmosphere – not ninety-nine percent. The rest of the atmosphere is above it. Beyond this, why is distance such a large factor in your view – if the stratosphere is in all directions?

Perhaps if you showed climatologists a little more patience and took time to better understand why they believe the things they do – no matter how silly they may seem – you would stand a better chance of showing them the errors of their ways. But your current approach hardly seems productive.

Comment by Timothy Chase — 27 Jun 2007 @ 7:13 PM

31. First, Alistair, we have to be clear what we’re discussing. I was not talking about Angstrom’s solar radiation measurements, even though you mentioned Angstrom’s experiment being conclusive, since it’s Koch’s experiment, as stated above, that was mainly used to argue for saturation. The post is called what Angstrom didn’t know, after all, so that must be the point of confusion. When you wrote “he used pure C02” that explained it to me. But in fairness, how do you square that one (tube of air, tube of pure C02) with “I agree with you that Angstrom was wrong in saying that increasing the amount of CO2 in the atmosphere would not raise the surface temperature, but his experiment is pretty convincing with a 50% increase in the amount of CO2 hardly altering the absorption.”? I think pure C02 is a little more than a 50% increase in the amount. Angstrom’s comparison was a point of data, not necessarily “crude” but also not really an experiment. It established what it established.

I believe what we mean by keeping relative humidity constant is just that – an experiment where you mimic the actual response of water vapor by keeping the atmosphere in the tube at the same relative humidity.

The thing that makes Koch relevant is his measurement of IR. Had raypierre been around then, I believe he would have recommended that Koch: measure very, very precisely and do numerous trials; use a longer tube.

And I wanted to add that a key factor for both angstrom and koch low-balling the importance of C02 concentration was not just that they didn’t fully envision conditions where energy radiates away, but that they weren’t modeling positive water vapor feedback (the magnitude of which but not the existence is still being hashed out). Water vapor is important because it’s an amplifier. C02 is important because it’s a tiny but persistent ratchet. Even if they had been right, and raypierre is showing nicely why they weren’t, the only change would be the scale and time frame of human-assisted warming, not the fact of it.

Comment by Marion Delgado — 27 Jun 2007 @ 9:07 PM

Given what Timothy Chase said just above, I would put your “doesn’t hold water” right up there with tamino’s observation that “urban islands of liberalism” skewed the data, masking the truth of Gavin’s “sunspots cause Republicans” theory.

More seriously, I think this issue just needs more diagrams.

Comment by Marion Delgado — 27 Jun 2007 @ 9:26 PM

33. Aaron (#28) wrote:

“…the atmosphere still wouldn’t be saturated even if we increased the CO2 to ten thousand times the present level.”

Doesn’t this mean that the amount energy absorbed is very, very small for a given increase in concentration for these bands?

Each doubling of carbon dioxide raises the temperature by roughly 2.9 degrees Kelvin. But this additional carbon dioxide isn’t absorbing energy which is somehow left over and hasn’t been absorbed before – its absorbing the same energy more times. The temperature of the system rises as the result of each parcel of energy staying within the system for a longer period of time – which means that the more times it is absorbed and re-emitted, the higher the temperature will be.

However, that wasn’t quite the issue that Ray was considering I believe. He was thinking of the issue of spreading or “broadening.” Higher temperatures and pressures will result in the broadening of the band so that it is able to absorb over a wider range of wavelengths. And even in the case of an isolated molecule, there will be some spreading – “lifetime broadening” as the result of the uncertainty principle as it applies to time and energy. (This was covered in the essay.) But there are other points which could be brought in, even with respect to the centerline.

Shouldn’t we be concerned with the bands that are more readily absorbed, you know, that give the molecule the characterstics that make them green house gasses?

I don’t know.

What do you have in mind?

It might help, though, if you checked to see whether or not the essay already dealt with it or at least some aspects of it.

Comment by Timothy Chase — 27 Jun 2007 @ 9:59 PM

34. Is there a good link(s) to publicly available literature (no charge preferrable) that shows the agreement between calculated and observed atmospheric IR emission spectra looking up or down or both, or at least some observed spectra that can then be modeled at the Archer site? The spectra I have calculated there seem quite reasonable, although I’m not sure his use of a constant tropospheric lapse rate, especially in the tropics, is completely justified. But then it is a radiation only model.

Comment by DeWitt Payne — 27 Jun 2007 @ 10:14 PM

35. RE #30 Timothy,

You wrote “But the extensive set of observations of upwelling and downwelling radiation does not fit into your scheme. It fits mainstream theory.”

I know of no set of observations of upwelling or downwelling radiation taken within the atmosphere. If you know of any I would be extremely grateful if you could tell me where to find them.

You also wrote “And the troposphere is responsible for only three quarters of the mass of the atmosphere – not ninety-nine percent. The rest of the atmosphere is above it. Beyond this, why is distance such a large factor in your view – if the stratosphere is in all directions?”

You are quite correct that the troposphere forms only 70% of the atmosphere, but my point was that the density of the air at 10 miles is only 1/100 of that at the surface. Of course the tropopause is at 10 km rather than 10 miles so I was exagerating. Moreover I believe that the height from which the radiation is emitted to space is at only 6km. That still means that the density of the air in that region will be much less than at the surface. In effect you have the tail wagging the dog! You can get some idea of the air density from this web page which shows air pressure, which is effectively the same thing.

http://www.metoffice.gov.uk/education/secondary/teachers/atmosphere.html#main

Comment by Alastair McDonald — 28 Jun 2007 @ 5:47 AM

36. [[Moreover, the low concentration of CO2 in the atmosphere also shows why the current models are wrong. When the CO2 is excited, some of that energy is converted to heat. But the heat goes mainly into N2 and O2 molecules which do not re-radiate the heat. They convect. Therefore the energy of radiation absorbed by the greenhouse gases is not re-radiated back to Earth. It is carried high into the atmosphere by convection, and then is radiated from below the tropopause either by exciting CO2 molecules or from the tops of clouds. ]]

Alastair, I think your mistake is in thinking that because the nitrogen and oxygen don’t radiate, there’s no radiation from a given level at all. The nitrogen and oxygen will collide with the carbon dioxide and the carbon dioxide will radiate. The net effect is that just as much is radiated as is received, once conduction and convection are accounted for.

Comment by Barton Paul Levenson — 28 Jun 2007 @ 6:44 AM

37. In #33 Timothy says:

Each doubling of carbon dioxide raises the temperature by roughly 2.9 degrees Kelvin. But this additional carbon dioxide isn’t absorbing energy which is somehow left over and hasn’t been absorbed before – its absorbing the same energy more times. The temperature of the system rises as the result of each parcel of energy staying within the system for a longer period of time – which means that the more times it is absorbed and re-emitted, the higher the temperature will be.

First, I think it should be 1.2 K, because the 2.9 K is after various feedbacks which are a separate issue. But the point I want to make is that the model presented here (and many other places), of warming the atmosphere each time longwave radiation is absorbed and re-emitted, seems to me to be in conflict with radiation balance model presented by Raypierre, where the key factor is the temperature of the greenhouse gas molecules that radiate into space, usually high in the atmosphere.

If both of these ideas are valid, I would like to understand the connection between them.

[Response: It’s not the “residence time” of energy in the system that counts for warming. In the end, what counts is the rate of escape out the top, for a given surface temperature. The greenhouse effect lowers that rate (for any given surface temperature) by replacing high temperature radiation from the ground with lower temperature radiation from aloft. That happens partly through “new” absorption of radiation that more or less used to escape directly from the surface, as well as absorption and re-emission of radiation that used to get absorbed and re-emitted at lower layers, but now (at higher CO2) does so at higher layers. It just confuses the matter to try to think of the heating the way Timothy is. For example, in the troposphere, the net infrared absorption-emission is a cooling effect, which balances convective heating. Does that mean that CO2 is cooling the troposphere and that it would get warmer if you took it out? No, of course not. There are dangers in thinking about local budgets of that sort. –raypierre]

Comment by Blair Dowden — 28 Jun 2007 @ 7:09 AM

38. So what I’m getting is: additional solar absorption is negligible (but exists), GHGs reabsorb heat in the IR bandwith radiated from the earth causing a slight warming because it takes longer to reach an equilibrium.

What is magnitude of this effect, as describe I would assume it were very slight. Most of what my flawed view of the GHG effect that gave it credibility was that it absorbed more solar energy which would require the atmosphere to heat until it radiated more energy.

The mechanism described above is missing something, mainly a link to temperature increase. I don’t see how it can cause an imbalance of the magnitude described. How much can increased CO2 concentration really slow reemission?

[Response: As far as warming of the surface goes, you can forget about solar absorption in the atmosphere. It’s there, but it’s a sideshow, and the greenhouse effect would work perfectly well in an atmosphere that was completely transparent to solar radiation. The effect has nothing to do with changes in the time it takes to reach equilibrium. Increasing CO2 increases the temperature even once you reach equilibrium. CO2, and other greenhouse gases, work on the “energy loss” side of the equation — infrared loss to space — not on the “energy gain” side (absorption of solar radiation). It’s just like putting on a blanket. Putting on a blanket doesn’t make you feel warmer by generating new heat. It makes you feel warmer by reducing the rate at which you lose heat to your environment. Same thing with greenhouse gases. The very same thing. I don’t know why you’d have the intuition that this is “a pretty small” effect. How can you have any intuition about it without having done the numbers? I don’t think this is the sort of thing to which the word “intuition” applies. The quantitative effects of CO2 on absorption are just too far outside what can be gleaned from common experience. One can have prejudices or preconceptions — like the prejudice that most early scientists had that humans were just too small to be able to change something as big as the global climate. That’s a prejudice that Lindzen and many other misguided people still have today. I hasten to add that I’m not by any means tarring you with the same brush, or indeed with any brush. I do want to make the point, though, that one should distinguish between a prejudice and an intuition. –raypierre]

Comment by aaron — 28 Jun 2007 @ 7:31 AM

39. Re: #35 (Alastair McDonald)

my point was that the density of the air at 10 miles is only 1/100 of that at the surface

The density of air at 10 miles is about 1/10 that at the surface, not 1/100. The very graph you link to shows this.

Comment by tamino — 28 Jun 2007 @ 8:23 AM

40. Blair Dowden (#37) wrote:

First, I think it should be 1.2 K, because the 2.9 K is after various feedbacks which are a separate issue.

Of course – I was writing in too abbreviated a manner. Actually I waivered on whether to use on figure or the other while trying to keep things short. Too short, it seems. But there has been at least one time where I brought up the amplification carbon dioxides’ effect by water vapor, and if I didn’ explain that the feedback had feedback, that became an issue for someone else – and I was trying to avoid that aspect of it. As Raypierre says, “there are many wrinkles.” Or as I like to think of it, “The way that can be spoken…”

But the point I want to make is that the model presented here (and many other places), of warming the atmosphere each time longwave radiation is absorbed and re-emitted, seems to me to be in conflict with radiation balance model presented by Raypierre.

And that is appreciated.

Honestly, I wasn’t thinking of the radiation heating the atmosphere each time it got absorbed and re-emitted – since its not that kind of radiation. But I was thinking in terms of a residence time while at the same time realizing that this could only be an approximation of sorts. It seemed an appropriate response to the apparent view that each photon gets absorbed only once – and that was what heated up the atmopshere. But you are right: the fundamental principle is that of radiation balance, and I probably should have brought it back to that.

Comment by Timothy Chase — 28 Jun 2007 @ 9:10 AM

41. Yeah, got all that. But a blanket doesn’t raise your body temperature and it won’t give you a fever. If you’ve lost heat, it will help you get back up to temperature faster, but it doesn’t actually warm you.

You can tar away, I don’t mind. I fall into that group (though I’m not a scientist, I just like the topic), but I’m not trying to pick a fight (I believe we do affect the climate, I just see no evidence that we are in any dangerous way). I’m just trying to learn.

I want to believe.

[Response: Really, no tar implied. You did well to spot the imperfection in the analogy. What makes the blanket/person analogy imperfect is that the absorption of solar energy by the earth remains nearly constant as you increase the CO2 (put on the blanket), whereas for a mammal’s body can adjust its metabolism. In practice, putting on the blanket allows you to maintain body temperature with less expenditure of calories. The blanket analogy does work properly for a person on the verge of hypothermia, since then their metabolism is going flat out to try to maintain body temperature but can’t do it, so core temperature drops, leading to bad consequences; putting on the blanket raises body core temperature back to survivable values. In my experience, the blanket analogy is the most accessible way to get across that CO2 warms the planet by making it harder for the planet to lose heat — by putting on a layer of insulation. If anybody can suggest a better analogy that gets around the metabolism problem, though, I’d be very grateful. –raypierre]

Comment by aaron — 28 Jun 2007 @ 9:51 AM

42. Alastair McDonald (#35) wrote:

You wrote “But the extensive set of observations of upwelling and downwelling radiation does not fit into your scheme. It fits mainstream theory.”

I know of no set of observations of upwelling or downwelling radiation taken within the atmosphere. If you know of any I would be extremely grateful if you could tell me where to find them.

Unfortunately, they don’t have an extensive library all in one place on the web, at least that I am aware of as of yet, and much of the material which is there is by subscription. However, performing a search for either “‘downwelling radiation’ altitudes” or “‘upwelling radiation’ altitudes” on Google will bring up quite a few hits. 629 and 941 respectively. Much of the data we get is at the surface, other satellite, but a fair amount appears to be from planes. I see twenty- and sixty-second intervals being mentioned, above and below cloud platforms, polarization, etc.

There is obviously a great deal of data out there, whether one is talking about measurements of radiation, the size of thermokarst lakes in Canada, the levels of carbon dioxide at thousands of different stations, measurements on the flow of water and nutrients in the ocean using various markers, etc. We aren’t speaking of a discipline which lacks data, which consists of only of arm-chair theorizing or which is likely to be overturned by the same.

Comment by Timothy Chase — 28 Jun 2007 @ 10:23 AM

43. raypierre (inline to #38) wrote:

It just confuses the matter to try to think of the heating the way Timothy is. For example, in the troposphere, the net infrared absorption-emission is a cooling effect, which balances convective heating. Does that mean that CO2 is cooling the troposphere and that it would get warmer if you took it out? No, of course not. There are dangers in thinking about local budgets of that sort.

You overestimate me, sir.

Honestly, I wasn’t even thinking in terms of local budgets, but how, for the system as a whole, the level of energy within that system may remain at a higher level than it would without the greenhouse effect. But in any case, “residence time” (which is more or less what I was getting at) would be a concept more apt to confuse than enlighten – although it does help at least initially to realize that the energy stays within the system for a longer period of time. In essence, greenhouse gases result in a longer queue. The energy going in is balanced by the energy going out – once the equilibrium is reached, but the amount within the system at any given moment is higher than it would be without the greenhouse gasses, hence the higher average temperature.

Basically the “blanket” model.

However, the problem (as I understand it) with local radiation budgets which you brought up isn’t simply that we are treating the local conditions in isolation from the global system, but that we would be looking at the radiation budget without considering convection, evaporation, condensation, etc. No part should be regarded in isolation from the rest of the system, but more importantly, no aspect should be treated in isolation form the others.

And so what we will generally look for is a simple framework for analysis through which we consider each relevant aspect where this framework can be applied anywhere within the system – at least for the purpose of understanding what is going on at specific points within the sytem. But in terms of being able to calculate the trajectory of the system as a whole, given all of the interactions, one will have to perform for the whole, taking into account all of those interactions. No single element exists in isolation from all the rest.

Nothing particularly profound, but for someone only a little more lost than myself, it might help.

Comment by Timothy Chase — 28 Jun 2007 @ 11:51 AM

44. Raypierre, in your response to #37 you said “…in the troposphere, the net infrared absorption-emission is a cooling effect, which balances convective heating.” I wonder if you could expand on how that works. I suspect that the idea of the greenhouse effect heating the air is not just simplistic but also misleading.

Comment by Blair Dowden — 28 Jun 2007 @ 12:07 PM

45. Alastair McDonald had asked where you can get the empirical data for downwelling and upwelling longwave radiation (#35) since I had said that such data is extensive (#30). I suggested a couple Google searches (#42), but what would appear to be one of the better and more organized repositories would be “The Atmospheric Radiation Measurement (ARM) Program.”

At the following two links, they describe in some detail the data they have, where the measurements are being taken (via satellites, ground-based, or planes, etc.), although they want you to pay to get the actual data.

The website itself will give the visitor a better idea of what kind of data is out there and how extensive it is.

Comment by Timothy Chase — 28 Jun 2007 @ 1:24 PM

46. Blair Dowden (#44) wrote:

Raypierre, in your response to #37 you said “…in the troposphere, the net infrared absorption-emission is a cooling effect, which balances convective heating.” I wonder if you could expand on how that works. I suspect that the idea of the greenhouse effect heating the air is not just simplistic but also misleading.

The surface and lower troposphere loses much of its heat due to moist air convection, but given that the stratosphere is extremely dry, the loss of heat in the upper troposophere must be due to something else: outgoing longwave radiation, I would presume.

Something along these lines is suggested here:

7 Dec 2004
Why does the stratosphere cool when the troposphere warms?
gavin
[edit URL]

Of course I would also be interested…

[Response: Actually, that was a good example of the weeds one can get stuck in on this subject. A better discussion is found here: http://www.realclimate.org/index.php/archives/2006/11/the-sky-is-falling/ – gavin]

Comment by Timothy Chase — 28 Jun 2007 @ 2:02 PM

47. Re #45 Timothy, Thanks for those two links.

If anyone else happens to know where I can get the information for free please let me know. The problem with paying, is that I don’t really know what I am getting until it is paid for :-(

FWIIW I suspect that the data does not exist because when the experimenters measured radiation they found that it did not match the models and so concluded that their data was wrong and did not publish!

At least Christy and Spencer were brave enough to publish :-)

[Response: Alastair, that’s really going too far. There is a figure in Goody and Yung showing the old Tiros top of atmosphere spectra that confirms the standard picture of radiative transfer. There are airborne FTIR observations looking down, shown in Liou’s textbook. There are Mars Global Surveyor TES spectra that confirm the models (I put one of those in Chapter 4 of my book, in the real gas section). The CKD paper on the water vapor continuum verifies the radiative transfer calculation against field observations. Looking upward from the ground, you can find even more data. There’s all of Dan Lubin’s upward looking FTIR data, both in the tropics (the CEPEX experiment) and in the Arctic. If it sometimes seems like there’s a paucity of downward looking spectral data covering the whole thermal IR, it’s because there really aren’t any serious scientific issues left to settle. Most spectrometers being flown look at narrow spectral windows with very high resolution, because that’s where the good science is right now. –raypierre]

Comment by Alastair McDonald — 28 Jun 2007 @ 5:07 PM

48. Alastair McDonald (#47) wrote:

Re #45 Timothy, Thanks for those two links.

If anyone else happens to know where I can get the information for free please let me know. The problem with paying, is that I don’t really know what I am getting until it is paid for :-(

Well, I usually can’t afford a new pair of socks…

FWIIW I suspect that the data does not exist because when the experimenters measured radiation they found that it did not match the models and so concluded that their data was wrong and did not publish!

At least Christy and Spencer were brave enough to publish :-)

Well, I dug around a little more, and I found something that might be just the sort of comparison you are looking for — although you need to keep in mind that this is for spectra resulting from the3 influences of clouds and aerosols – which are considerably more difficult to model than clear sky. Additionally, these are from a few years ago: 2000-1.

Climate Forcing by Clouds and Aerosols: Two Years of Field Studies

I will see if I can find some more stuff along these lines in the next few days.

Comment by Timothy Chase — 28 Jun 2007 @ 6:33 PM

49. Alistair McDonald – “FWIIW I suspect that the data does not exist because when the experimenters measured radiation they found that it did not match the models and so concluded that their data was wrong and did not publish!”

OR the published data is correct. I guess if you really want to believe in something you will invent all sorts of things. William of Occum said something long ago about multiplying entities and you are multiplying them a bit here. Unless you have measurements of your own that refute the published ones I think we can all assume that they are correct within current knowledge.

Comment by Ender — 28 Jun 2007 @ 7:16 PM

50. More comparisons of modeled vs measured spectra may be found here…

CAVE Publication List
http://www-cave.larc.nasa.gov/cave/pages/bibliog.html

This one contains a few:

Introduction to an Online Coupled Ocean-Atmosphere Radiative Transfer Model (2002)
Zhonghai Jin, Thomas Charlock, and Ken Rutledge
http://www-cave.larc.nasa.gov/cave/pdfs/Jin.AGU02.pdf

Comment by Timothy Chase — 28 Jun 2007 @ 8:20 PM

51. aaron (#41) said:

Yeah, got all that. But a blanket doesn’t raise your body temperature and it won’t give you a fever. If you’ve lost heat, it will help you get back up to temperature faster, but it doesn’t actually warm you.

This is a bad analogy. What you said is true because warm-blooded animals can control their temperature through evapotranspiration and changing the heat conductance (or resistance if you prefer) of their skin and/or fur.

If you take a cold-blooded animal, like a snake, throw a blanket over it, and shine a heat lamp on it, its temperature will increase more than if it didn’t have a blanket.

[Response: The blanket analogy might still be a useful communication tool if one said simply that the blanket makes you feel warmer, not because it generates energy, but because it insulates you and reduces the rate at which you lose heat. For a mammal, “feel warmer” means that your metabolism doesn’t need to struggle to maintain survivable body temperature. “feel warmer” here doesn’t mean an increase in temperature, but it still conveys the idea that the insulation is the important thing, just like the “insulation” of the Earth by CO2 is the main factor in the greenhouse effect. There are many levels of explanation, and what I’m groping for here in this analogy is the basic one-liner that conveys the essence of the greenhouse effect without doing too much violence to the real physics.

Actually, perhaps the blanket analogy isn’t as bad as all that. True, because your metabolism adjusts, a blanket doesn’t increase your body core temperature, but it does increase your skin temperature and the temperature of the air adjacent to your skin. That’s a big part of why you feel warmer. –raypierre]

Comment by Jason Patton — 28 Jun 2007 @ 10:11 PM

52. #41,51
Although the famous scene in ‘Goldfinger’ is an urban myth, there’s a grain of truth to it.
Anyone wrapped in a space blanket during a heatwave would have pretty poor odds of survival, since heat stroke sets in at a body temperature of 104 F. Go past that point and you will reach thermal equilibrium with your environment forever.

Comment by Alex Nichols — 29 Jun 2007 @ 3:59 AM

53. Will the snake increase to a higher temperature with the blanket? (I suppose it depends how windy it is, but it’s not too windy outside of the earth.)

This suggests to me that the effect would warm the atmosphere from the surface out, toward the current surface temperature, until emission is restored and the budget balances. The surface shouldn’t get hotter (not much atleast).

And how how strong is the effect, are we talking about a blanket, or a sheet?

I don’t know if I’m capable of understanding, or even have the time, but I’d like to get into the real physics.

Comment by aaron — 29 Jun 2007 @ 4:54 AM

54. Alastair wrote: “FWIIW I suspect that the data does not exist because when the experimenters measured radiation they found that it did not match the models and so concluded that their data was wrong and did not publish!”
Danger, Alastair! Danger! You are crossing the line into anti-science nutjob territory here. Remember what Isaac Asimov said,
“The most exciting phrase to hear in science, the one that heralds new discoveries, is not ‘Eureka!’, but ‘That’s funny …'”

The goal of experimental science is not to confirm a model, but rather to stretch the model with data until it breaks and the theorists have to fix it. This is their goal. Moreover, there is the issue of accountability: When experimentalists take data, the funding agency expects a report. If no data are forthcoming, there had better be a good explanation. All you do when you make statements like that is demonstrate that you don’t understand how science actually works.

Comment by ray ladbury — 29 Jun 2007 @ 8:14 AM

55. aaron (#53) wrote:

I don’t know if I’m capable of understanding, or even have the time, but I’d like to get into the real physics.

For those who really wish to step into the cathedral of climatology, a basic understanding of the greenhouse effect is all that is required.

The following will give you that:

10 Apr 2007
Learning from a simple model
http://www.realclimate.org/index.php/archives/2007/04/learning-from-a-simple-model/

How far you go in exploring climatology after that is up to you, but this website is a really good place to a great deal of it.

Comment by Timothy Chase — 29 Jun 2007 @ 9:25 AM

56. Re #54.

Ray (Ladbury), have you noticed a funny thing on Ray’s (Pierrehumbert) Figure 4.6: Some representative OLR spectra for Mars, observed by the Thermal Emission Spectrometer on Mars Global Surveyor at various times of day. from page 103 (110) of his book?

No matter what the time of day, the CO2 emissions remain the same, with a brightness temperature of around 200 K.

Comment by Alastair McDonald — 29 Jun 2007 @ 10:05 AM

57. Alastair, I’m afraid I don’t see that in the figure. Yes the variability is les than at other points in the spectrum, but isn’t that what would be expected for a ghg?

Comment by ray ladbury — 29 Jun 2007 @ 10:46 AM

58. Alastair McDonald (#56) wrote:

Ray (Ladbury), have you noticed a funny thing on Ray’s (Pierrehumbert) Figure 4.6: Some representative OLR spectra for Mars, observed by the Thermal Emission Spectrometer on Mars Global Surveyor at various times of day. from page 103 (110) of his book?

No matter what the time of day, the CO2 emissions remain the same, with a brightness temperature of around 200 K.

Pg. 110.

The red is the afternoon, the violet at sunset, and the blue at night.

There is a big difference between them, peaks of 8 x 10^-6, ~5.75 x 10^-6 and 3 x 10^-6 respectively. You were misreading it, Alastair.

Face it: climatology is a great deal more advanced than you thought. Climatology is a branch of physics, a very well developed one at that.

The links I provided above show it, at least with respect to our understanding of spectra.

Climate Forcing by Clouds and Aerosols: Two Years of Field Studies

CAVE Publication List (numerous peer-reviewed articles, many of which are directly relevant to this fairly specialized topic)
http://www-cave.larc.nasa.gov/cave/pages/bibliog.html

Sample article from list above:

Introduction to an Online Coupled Ocean-Atmosphere Radiative Transfer Model (2002)
Zhonghai Jin, Thomas Charlock, and Ken Rutledge
http://www-cave.larc.nasa.gov/cave/pdfs/Jin.AGU02.pdf

It is time to quit insisting that you are the only one that really understands how the greenhouse effect works and start learning about it. Put aware your fantasy and start learning about reality. You know some things already, but you could learn a great deal more.

You are bright enough.

[Response: By the way, the Figure 4.6 that’s currently there in the book is just a placeholder pulled from the TES website without modification. Eventually, I’ll re-plot it in consistent form using original data from one of the thousands of TES shots. The night-time curve is just a distraction for the things I’m discussing in Chapter 4, so I’ll eliminate that, but for the stuff you guys are talking about, thinking about why the night-time curve looks the way it does is quite interesting. For educational purposes, it would be nice to have an instrument like TES orbiting the Earth, but so far as I know there isn’t one. A similar instrument was sent to Venus, but unfortunately the sensor got stuck in its blackbody calibration position. –raypierre]

Comment by Timothy Chase — 29 Jun 2007 @ 10:51 AM

59. Re Rays response [Response: Alastair, that’s really going too far. There is a figure in Goody and Yung showing the old Tiros top of atmosphere spectra that confirms the standard picture of radiative transfer.]

Ray,

With my model the predicted OLR spectrum at the TOA is the same as yours, and what is measured experimentally. That sort of confirms my picture of radiative transfer :-) What I am saying is that if you measured the CO2 band OLR spectrum at 1000 feet it would look similar to that at the TOA.

I don’t have a copy of Liou’s book (to hand), but on Slide 4 of John Harries’s notes there are two OLR spectra for the tropics and the Arctic. It is funny that both show the same brightness temperature for CO2. It is almost as if CO2 has a fixed brightness temperature that is independent of atmospheric temperature.

I think that you are probably correct when you write “If it sometimes seems like there’s a paucity of downward looking spectral data covering the whole thermal IR, it’s because there really aren’t any serious scientific issues left to settle.” Everyone I speak to thinks that the problem of OLR is well and truly solved. But in A Busy Week for Water Vapor Philipona et al. found that the water vapour near the surface was increasing. That implies that CO2 is acting there, not high in the tropopause as you argue. There, you wrote “In equilibrium, the Earth must lose as much energy out the top of its atmosphere as it gains by absorption of Solar energy” which is true, but you also wrote t “Planets only have one way of losing energy, which is by infrared radiation to space, often called ‘Outgoing Longwave Radiation,’ or OLR.” However, there is another way. Planets can reach equilibrium by producing clouds which reflect the solar energy away, and so match incoming solar radiation to a fixed or even runaway OLR. This is what has happened on Venus, and is the cause of the dust storms on Mars. On Earth it is not dust storms which cool the atmosphere, it is El Nino and hurricanes!

Comment by Alastair McDonald — 29 Jun 2007 @ 11:36 AM

60. Alastair McDonald (#59) wrote:

I think that you are probably correct when you write “If it sometimes seems like there’s a paucity of downward looking spectral data covering the whole thermal IR, it’s because there really aren’t any serious scientific issues left to settle.” Everyone I speak to thinks that the problem of OLR is well and truly solved. But in A Busy Week for Water Vapor Philipona et al. found that the water vapour near the surface was increasing. That implies that CO2 is acting there, not high in the tropopause as you argue.

The articles I pointed to have charts for mid-atmosphere and they demonstrate the role of the upper atmosphere. The peaks you said looked the same in Raypierre’ book had peaks of 8 x 10^-6, ~5.75 x 10^-6 and 3 x 10^-6 at 20, 22.5 and >25 mu-m respectively.

You have it wrong and the climatologists have it right. You see what you want to see and always look for an easy way to dismiss the rest before really looking at it.

Comment by Timothy Chase — 29 Jun 2007 @ 12:04 PM

61. Alastair, please. Raypierre wrote in _A Busy Week …_
> the water vapor feedback discussed in Philipona et al. is not the same water vapor feedback usually discussed in
> connection with global warming. It is instead a surface water vapor feedback which adds additional surface warming
> on top of the usual things we talk about. The effect is already incorporated in the climate models used in IPCC
> forecasts, but the new observational study will be useful as a reality-check.

Comment by Hank Roberts — 29 Jun 2007 @ 2:31 PM

62. [[But in A Busy Week for Water Vapor Philipona et al. found that the water vapour near the surface was increasing. That implies that CO2 is acting there, not high in the tropopause as you argue. ]]

It doesn’t imply that at all. Changes in water vapor happen most strongly near the surface because the surface is where the source of the water vapor is — the oceans. And water vapor has a very shallow scale height compared to dry air, 2 km versus 8 km.

Comment by Barton Paul Levenson — 29 Jun 2007 @ 4:12 PM

63. Is it oversimple to say ‘the atmosphere warms from the bottom (incident sunlight warms the earth), warm earth warms the lower atmosphere; the atmosphere warms itself by mixing and radiation; the atmosphere cools from the top by radiation to space’?

Comment by Hank Roberts — 29 Jun 2007 @ 11:44 PM

64. I believe that after several readings of the two papers(I’m “optically thick” in another sense), I finally grasp the proper response to those who argue fallaciously that additional CO2 would have little or no effect because the absorbtion band in the infra red is near saturation so,(they conclude)there is little absorption of radiation emitted from the surface.
This is a false conclusion because (a) adding CO2 to the atmosphere will cause the thermal radiation emitted to space to originate from higher and colder levels than previously, with less radiation emitted. This disturbs the energy balance until the atmosphere near the surface heats up so that the atmosphere at high levels can radiate enough energy to restore the balance; and (b)adding more CO2 in effect adds more red M&Ms to the conveyor belt by enlarging the width of the spectrum in which CO2 can be absorbed. For the example given by Raypierre,at 1CO2 the band width is almost 3.5 microns (13.5 to 17), while at 4CO2 the band width is about 4.5 microns(13 to 17.5).
If I still don’t have a grasp its through no fault of the authors. Rather that the electromagnetic spectrum and it’s properties are far afield of my own background and experience.But I feel strongly that we’re at a crossroads regarding the future of our planet and it behooves those of us,in the general public to try to comprehend as much as possible.

Comment by Lawrence Brown — 30 Jun 2007 @ 1:05 PM

65. By the way I try constantly to pretend I believed the opposite of my actual position (what used to be called being on the alarmist side, but now I think is simply the precautionary side and unfortunately may soon be the mainstream), and respond accordingly.

Most denialism is (our) time-wasting deliberate repetition of debunked points for sheer PR purposes (and even to gain mass in search engine results). But some of it, I have to say, is the kind of abuse most scientific ideas should take.

In a similar way, the paper suggesting “most published results are false” I found nonsensical, but useful, for instance.

Comment by Marion Delgado — 30 Jun 2007 @ 7:01 PM

66. Now even if I read Knut Ångström very carefully, he only claims:
“Unter keinen Umstunden dorfte die durch die Kohlensure bewirkte Absorption der Erdstrahlung 16 Proc. Ubersteigen, und die Grusse dieser Absorption Ã¤ndert sich quantitativ mit dem Kohlensuregehalt sehr wenig, solange numlich derselbe nicht weniger als 20 Proc. des vorhandenen betrugt.”

“Under no circumstances will the absorption by CO2 of [IR] terrestrial radiation be larger than 16%, and the amount of this absorption changes quantitatively very little with the CO2 concentration, as long as CO2 concentration remains below 20%.”

IMHO Knut Ångström doesn’t claim saturation in the infrared here, he just contrasts it with the 60% absorption that Arrhenius claimed in 1896.

Furthermore I don’t think that pressure is that effective on absorption if expressed in atm cm (which is in fact 1000 dobson units). Koch measured two absorptions: 10% at 30 atm cm and 9.6% at 20 atm cm, which Ångström extrapolated to 16% for 250 atm cm.
Arhhenius tried to refute this, but in fact the 16% value was confirmed in 1901, by measurements in Berlin. Strangely enough this huge reduction in absorptive power by CO2 (from 60% to 16%) didn’t lead to a change in the climate sensitivity values by Arrhenius.

See the graph: black dots Ångström (1900) red dots Arrhenius (101)
http://home.casema.nl/errenwijlens/co2/angstrom1900/angstromarrhenius.gif

Comment by Hans Erren — 30 Jun 2007 @ 7:20 PM

67. Re: Blanket Analogy

Two things…

First. If its already HOT and you cover yourself with a nice wool blanket even a thin one, you won’t be able to lose heat fast enough and you’ll experience the wonderful symptoms of Heat Stroke, the proverbially “flu-like symptoms” of fatigue, nausea, vomiting, delirium, fever, death… So, I think the blanket analogy works quite well.

Second. A better analogy might be attic/wall insulation in a home. You still need a furnace to warm the house – but if you have better insulation the furnace needs to provide less heat to maintain a nice comfy temperature. But in our case, our furnace (sun) isn’t linked to a thermostat here on Earth and so doesn’t reduce its output despite our improved insulation – and so our house gets hotter. See above…

Comment by Robin Johnson — 1 Jul 2007 @ 12:32 AM

68. To the best of my knowledge, no one has yet answered the question posed at the beginning of this paper, that when an infrared flashlight of one watt is flashed at sea level, what fraction would be received by an astromanut above the atmosphere. Maybe because the answer is obvious to most, but I’ll take a crack at it anyway.
Using the graph of ‘Transmission Through a Tube of CO2’ shown in this paper, at 300 ppm concentration of CO2 and a temperature of 20C,and sea level pressure, a 1.0 meter tube filled with pure CO2 would be needed to simulate one atmosphere of equal cross section, when pressure broadening and the fact that pure CO2 collisions are different from the,mainly nitrogen and oxygen composition of the air. This would be equivalent to a column of air from the surface to the top of the atmosphere. The amount of light getting through would by about .66 watt. At 2xCO2, a length of tube of 5 meters would be needed to simulate the atmosphere and the amount of light received is about .64 watt. The amount drops to .6 watt at 4xC02.
These numbers are a lot less than the 4 watt deficit given in the first paper, from the 240 watts/m^2 entering as given in Gavin’s paper ‘Learning from a simple model’. This is because(I hope) of the fact that the watt originates from sea level and does not undergo the cooling and mixing that takes place in the greenhouse process. If I’m still in the dark(pun intended)its back to the drawing board.

Comment by Lawrence Brown — 1 Jul 2007 @ 4:57 PM

69. Laurence, what is the frequency and bandwidth of the flashlight? Is it cloudy? You need that information.

Comment by Eli Rabett — 1 Jul 2007 @ 9:44 PM

70. I really liked your detailed analysis in Part 2 of how the absorption of infrared depends on CO2 concentration. Enjoyed Part 1 as well.

Comment by Thinker — 2 Jul 2007 @ 2:13 AM

71. Thank you, Eli. I only left out three critical elements?! Not too good. I’ll have to make an assumption on frequency( wavelength) and, if allowed, assume that the light shines with equal intensity throughout the bandwidth. Assuming cloudlessness is pretty idealistic and doesn’t simulate the real world, but I thought it was an implied in the initial question at the beginning of the paper. Anyhow, I appreciate knowing that these factors have to be taken into account.

Comment by Lawrence Brown — 2 Jul 2007 @ 12:55 PM

72. How does the mesosphere and thermosphere above and with (much) higher temperatures than the stratosphere play into all of this? Is it that the density/pressure is so low as to be near non-existant and the so few molecules with high kinetic energy just don’t count other than in pristine theory?? If so, what would be a more scientific way of describing it? If not, why isn’t their radiation into space at a temp similar to or even greater than the surface?

Comment by Rod B — 2 Jul 2007 @ 1:52 PM

73. 67. Those are both good. I’d refine the first one though. Instead of “if it’s already HOT” change it to “if you (or another person) are already hot”. If it’s hot out, the blanket won’t cause you to overheat, it might even keep you from being overheated.

Comment by aaron — 2 Jul 2007 @ 8:38 PM

74. Rod B. I think you answered your own question. The international space station orbits in the upper part of the thermosphere–not much there.

Comment by ray ladbury — 2 Jul 2007 @ 9:04 PM

75. re 33: “…the more times each parcel of energy is absorbed and re-emitted, the higher the temperature will be.”

I just can’t get this. Take two molecules that keep exchanging a photon. Why does the temperature (average of the two) increase? Or change at all? Also, for the most part, the absorption of IR radiation by a molecule ends up as energy stored in its internal bonds or electron levels, neither of which increases the temperature of that molecule…does it?? What am I missing here?

Comment by Rod B — 3 Jul 2007 @ 12:20 PM

76. Rod B (#75) wrote:

I just can’t get this. Take two molecules that keep exchanging a photon. Why does the temperature (average of the two) increase? Or change at all? Also, for the most part, the absorption of IR radiation by a molecule ends up as energy stored in its internal bonds or electron levels, neither of which increases the temperature of that molecule…does it?? What am I missing here?

Its not that you have one photon bouncing back and forth between two molecules, its that the energy is slowed down before it finally escapes to space while energy keeps coming in at the same rate. At some point the rate at which energy leaves the system (the photons leaving the Earth and its atmosphere) has to be balanced with the energy which is entering the system (the new photons just keep coming in at the same rate). The only way to do this is to raise the temperature of the Earth to the point that its increased temperature raises the rate of energy loss enough to compensate form the decreasing of the rate of energy loss due to the greenhouse gases.

The more energy that is in the system at any given time, the higher the temperature at the surface and the higher the temperature in the lower atmosphere. Greenhouse gases require a higher level of energy to be in the system to achieve a balance between the rate at which energy enters the system and the rate at which energy leaves the system. More energy essentially means more photons at any given time where each photon will tend to take longer to get out.

A bit of an oversimplification (Ray will probably be pulling out his wooly hair when he reads this), but close enough.

[Response: It’s actually not so bad, so I guess I guess I get to keep my hair (wooly or not):) Fourier himself used an analogy with a bucket (actually he used “vase” not “seau”) with a hole near the bottom. Put it under a faucet and the bucket will fill up until the pressure forces water out the hole at the same rate it’s coming in. The amount of water in the bucket is analogous to temperature, the faucet in is analogous to sunlight (fixed supply) and the leak through the hole is the loss to space by infrared. Though he didn’t take the analogy further, in modern terms we’d say that if you decrease the size of the hole you need more water in the bucket in order to force water out at a rate that balances the input. You can sort of think of the bucket analogy as applying to photons, I suppose, but then you have to think harder to get the connection with temperature. –raypierre]

Comment by Timothy Chase — 3 Jul 2007 @ 2:06 PM

77. If equal numbers of water molecules and carbon dioxide molecules are put into the tube, who gets most of the IR photons? The water molecules by far because water molecules have a permenant electric dipole and carbon dioxide molecules do not. In fact carbon dioxide is a fairly weak absorber of IR energy as compared to water.

I think you guys are suffering from severe carbon dioxide anoxia.
What you guys are claiming is that the recent trend in global warming is due to the increase of the concentration of carbon dioxide from 0.033% by volume in 1980 to present value of 0.038% by volume. That claim, in the words of the late Senator Everett M. Dirksen, “Ole Golden Throat”, “is just so much hogwash”. Google, Gobal Warming: A closer Look at the numbers” Monte Hieb is mine safety engineer, knows a lot about gases and gives the correct computation re role of the water vapor and carbon dioxide, etc in the greenhouse effect. My advice: don’t challenge an engineer, they are never wrong.

The reason the last 10 years out of 12 have been the warmest on record is that period has been dominated by a strong El Nino cycle which has now come to an end. A La Nina cycle has started and the climate is going to cool down big time.

[Response: Far be it for me to start the whole engineers vs. scientists thing again, but Monte Hieb’s calculations are, to quote Senator Dirksen, “hogwash” and are almost as wrong as it is possible to be! – see: http://www.realclimate.org/index.php/archives/2006/01/calculating-the-greenhouse-effect/ . I hope that the rigour he applies to his mine safety calculations is significantly higher. – gavin]

Comment by Harold Pierce Jr — 3 Jul 2007 @ 3:13 PM

78. Gavin: You are just flat out wrong about carbon dioxide’s role in climate. At ca. 0.038% by volume carbon dioxide does diddly squat. Your article on “Water Vapor: Forcing or Feedback” is nonsense. Water is always present in real air and it does not matter how it gets there. Water molecules are light, agile, swift, frisky and pesky. Carbon dioxide molecules are slugs and dumbells. I’m an organic chemist and have been duking it out with water molecules at the bench for 40 years.

Here are agents that move enormous amounts of water into the atmosphere over land that is independent of carbon dioxide: Microbes, insects, plants, and animals. All surfaces have adsorbed and absorbed water, and large volumes are exchanged with the atmosphere that depends on only pressure and temperature. Climate and weather are strongly influenced by heating air over the of the land and humidity of the air over that land.

Gavin You just don’t know the personalities of these molecules.

[Response: Errr…. You might want to think about getting out of the lab a little more often…. – gavin]

Comment by Harold Pierce Jr — 3 Jul 2007 @ 4:25 PM

79. Harold, how many water molecules have you duked it out with in the stratosphere? Pretty dry up there last data I saw. And the CO2 band is a lot less saturated up there. Being an organic chemist does not mean you are an expert in atmospheric physics. I’m a physicist with >20 years experience. Here, I am a student. Join us and learn.

Comment by ray ladbury — 3 Jul 2007 @ 4:45 PM

80. Take Timothy’s text:

“the photons leaving the Earth and its atmosphere) has to be balanced with the energy which is entering the system (the new photons just keep coming in at the same rate). The only way to do this is to raise the temperature of the Earth to the point that its increased temperature raises the rate of energy loss enough to compensate form the decreasing of the rate of energy loss due to the greenhouse gases.”

Now _why_ to do this, the only way is to raise the temperature of the Earth —

Photons coming in: solar output, energy transferred in a whole lot of ways, but what heats the surface of the Earth is photons.

Photons come in specific energies, not a sliding scale, but stepwise. About the same number of each energy, on average, all the time, comes in from the Sun.

All of them that hit the ground basically end up as heat energy. The ground gets warmer. The ground warms the air.

The ground is at a temperature — an energy level — that’s going to be emitted in the infrared range. The photons coming off the ground in the infrared are in the band (the set of energies) that, when they hit a greenhouse gas molecule, get grabbed and turned into another kind of energy (the bond vibrates? stretches? the angle changes?) and that then causes another photon in the same infrared range to be emitted (or the molecule bumps another one and transfers the energy by bumping it a bit harder while it’s vibrating).

So we get energy in, over the Sun’s range of output. If there weren’t any greenhouse gases in our atmosphere, the ground would be radiating heat right into space, and be a lot colder. As more greenhouse gases accumulate, they’re floating around in the atmosphere and as they pick up energy from photons in the infrared (or by being bumped) they can emit infrared photons, which go in any random direction. Most of those hit another greenhouse gas molecule or hit Earth, lather rinse repeat.

Some don’t hit anything on their way off the planet — those remove heat energy. Nothing else does. Only outgoing infrared.

So —- more heat energy in the system bouncing around the more greenhouse gas molecules there are, and at the very top of the atomosphere, while there are a few more molecules of CO2 (and almost no water), they aren’t magically picking up the extra energy that’s bouncing around below.

The whole atmosphere’s expanded as it’s being heated from the bottom up. So at the top of the atmosphere the molecules get lifted up, and cool as they get further apart. The top of the atmosphere cools, because the lower atmosphere has heated and lifted up the top of the total atmosphere.

What’s happening with the energy leaving the planet? It was at a level matching solar input (in total, emitted as infrared energies). It still is, and the greenhouse gases at the top got lifted up, and they cooled (higher, thinner, fewer collisions).

Where’s the outgoing energy now, since we recently added a lot of greenhouse gas? It’s bouncing around in the Earth/Atmosphere, and in the oceans. Heat is penetrating the ground, as the drilling work has detected.

Over the next few centuries, all the heat bouncing around is going to mean there will be on average more faster (hotter) molecules zipping to the top of the atmosphere, whacking a greenhouse gas molecule harder, bumping it up to a higher energy than it had there in the high thin cold air (which can go into its bonds, and its bond energy can punch out a higher energy, hotter, shorter wavelength photon that before the GHG era).

The molecules at the new higher top of the expanded upper atmosphere eventually warm up to where the total energy being emitted into space is again equal to the total coming from the sun.

Still too many polysyllables, can someone make this simpler?

Comment by Hank Roberts — 3 Jul 2007 @ 6:43 PM

81. re 76: Sounds like trivia but it might help to understand what’s going on. My contention (question) is that the IR radiation absorbed by atmospheric gasses does NOT raise the temperature of the gasses/atmosphere, though I understand how sequential emission-absorption-re-emission eventually raises the temp of the earth surface. I also understand the temp of the atmosphere changes for a slew of other reasons, one of which might be the collision of gas molecules and the coincident exchange/conversion of bond energy to kinetic energy (=temp) — but not from IR absorption per se.

Comment by Rod B — 3 Jul 2007 @ 10:09 PM

82. re 80: “Still too many polysyllables, can someone make this simpler?”

I’m dubious over some of the details (especially upper atmos stuff), but overall I think your description is quite good!

Comment by Rod B — 3 Jul 2007 @ 10:27 PM

83. Rod B in #81

My contention (question) is that the IR radiation absorbed by atmospheric gasses does NOT raise the temperature of the gasses/atmosphere…

If you think of temperature as being a function only of the kinetic energy of the gas molecules, then absorption of a photon does not immediately raise the gas temperature. However, IIRC, you have to consider the various internal modes of the molecules as additional degrees of freedom over which to distribute energy. That’s why the heat capacity of a diatomic gas is higher than the heat capacity of a monatomic gas and the heat capacity of a polyatomic gas is higher yet. So a more general definition of temperature which takes into account the total energy of the system would say that the temperature does indeed go up when a photon is absorbed. I think. It’s been a long time since I had thermo, but I did look up gas heat capacities the other day. Google is your friend.

Comment by DeWitt Payne — 4 Jul 2007 @ 12:14 AM

84. Rod B (#81) wrote:

re 76: Sounds like trivia but it might help to understand what’s going on. My contention (question) is that the IR radiation absorbed by atmospheric gasses does NOT raise the temperature of the gasses/atmosphere, though I understand how sequential emission-absorption-re-emission eventually raises the temp of the earth surface.

Well, there is some warming do to the conversion of captured photons into kinetic energy, but it can go the other way as well, and this really isn’t where the main action is as far as heating the atmosphere goes, or so I have picked up while here. Evidently, much of the warming of the atmosphere is largely do to water vapor as the result of evaporation.

But I am still try to sort out the details.

I will be going over the two posts, the comments, etc.. Plus I have started on Ray’s book and have some other material. One thing I definitely want to get right.

Comment by Timothy Chase — 4 Jul 2007 @ 12:23 AM

85. Hank Roberts thanks, thats supercalifragilisticexpialidocious!

There have been a variety of predictions of the mismatch between CO2 equilibrium temperature and present day temperature based around standardised futures a la IPCC, but is anybody actually keeping track of the sensibly expected emissions looking forwards say five to ten years?

With 140+ coal fired power plants lighting their boilers this year alone and no doubt even more predicted for the next few years we are not going to hit any sort of flattening of CO2 emissions for a while. Not until something very messy hits the fan in a way that will reverse this trend. India, China and co are just hitting their straps and no amount of proselytising is going to stop that any time soon â�� at least not until we are all suffering a lot.

So if we look at say the next 10 to 20 years of vigorous anthropogenic CO2 addition, plus some reasonably likely carbon from drying forests and tundra, what is the likely CO2 in 2020 and 2030, and what is the equilibrium temperature for that awful state?

I think THATS the answer we need to be waving around on our placards.

Comment by Nigel Williams — 4 Jul 2007 @ 3:05 AM

86. #85 Nigel

I suspect what you’re looking for is the IPCC Special Report on Emissions Scenarios http://www.grida.no/climate/ipcc/emission/ and probably Chapter 10 of AR4 (http://ipcc-wg1.ucar.edu/wg1/wg1-report.html)… Can’t give you a summary, since I haven’t read them myself.

Comment by SomeBeans — 4 Jul 2007 @ 3:27 AM

87. [[My contention (question) is that the IR radiation absorbed by atmospheric gasses does NOT raise the temperature of the gasses/atmosphere,]]

Then your contention is wrong. Does blacktop heat up in the sun? Same process as the atmosphere heating up in IR light.

When an IR photon is absorbed by a carbon dioxide molecule, it raises that molecule’s energy. It is most likely to lose that energy in collision with a neighboring molecule. The neighboring molecule will move a little faster. Faster molecular motion = higher temperature.

Comment by Barton Paul Levenson — 4 Jul 2007 @ 6:52 AM

88. RE #78 Retired five years ago. My idea of a green enviroment is the hall or the cardroom at the rock in Richmond, BC. This place is fantastic and has 25 tables going 24/7/356.

I get plenty of exercise doing yardwork.

Comment by Harold Pierce Jr — 4 Jul 2007 @ 8:40 AM

89. re 83: Thanks, DeWitt. My understanding is that velocity is the only thing that determines temperature (mv^2 ala 0.5mv^2 = 1.5kT (the 1.5 factor might be off — I’m too lazy right now to look up a confirmation)). You can add heat energy without increasing temp (evaporation, e.g.) and raise temp without adding heat energy (air conditioning compressors, e.g.). Would you concur? Or point me in the right direction?

Comment by Rod B — 4 Jul 2007 @ 9:27 AM

90. re 87 (Barton): “…Then your contention is wrong. Does blacktop heat up in the sun? Same process as the atmosphere heating up in IR light.”

But aren’t the processes different? Sunlight hitting almost any solid or liquid material, if not reflected, will excite the motion of the entirew molecule and this, and only this, determines temperature. Because of the strangeness of material science, sunlight hitting a gas molecule does nothing (with some exceptions, but mostly ionization or molecular breakup). However, IR radiation, again inexplicably, does interact with some gas molecules, but through absorption of the E-M energy by the (predominately) intra-molecular bonds with no temperature increase. An entirely different process.

“…When an IR photon is absorbed by a carbon dioxide molecule, it raises that molecule’s energy. It is most likely to lose that energy in collision with a neighboring molecule. The neighboring molecule will move a little faster. Faster molecular motion = higher temperature…”

I agree with this, but note it is the collidee not the absorber that gets hotter. BTW, I think the most likely loss/transfer of energy comes from emission, not collision, but this is just a quibble.

Comment by Rod B — 4 Jul 2007 @ 9:53 AM

91. Re: #89 (Rod B)

Velocity is not the only thing that determines temperature. Temperature is (roughly speaking) proportional to the average energy per mode in the system. The modes can include velocity, internal vibration, potential energy, rotation, even magnetic field strength (one of the ways used to achieve ultra-low temperatures in a laboratory is “magnetic relaxation”).

The exact definition of temperature is: the rate of change of energy with respect to entropy, when volume and particle number are held constant. Not a very insightful definition for this discussion … but precise.

Comment by tamino — 4 Jul 2007 @ 10:42 AM

92. Rod,
Temperature has a precise thermodynamic definition as Tamino indicated. Consider a monoatomic gas. It has 3 degrees of freedom, and indeed its energy is 3kT/2. Now consider a diatomic gas–we’ve added new degrees of freedom, including rotation and vibration. Unlike kinetic energy, which is not quantized, these new energy modes are quantized, although the bands tend to be broad, and deform in response to surrounding molecules. Although these are INTERNAL to the molecule, they are motion. Indeed, if the atoms were part of a solid, they would share teh vibration with all their neighbors. Thus, I believe that in a collision, the energy may thermalize and add to the kinetic energy of the atoms. So, I don’t think it’s 100% correct to view the vibrational bands of CO2 as frozen out and isolated from the other types of energy. Indeed, given that a photon carries momentum, even in a radiative porcess, there is some change in momentum (albeit small) and therefore energy of the absorbing or emitting molecule.
In any case, the energy is not escaping, and it has to have some effect.

Comment by ray ladbury — 4 Jul 2007 @ 11:50 AM

93. Also, when you evaporate water, your molecules of water in vapor form are moving faster, and those in liquid form are moving slower; when you freeze water, the molecules are moving much slower yet. You can say there’s no temperature change on some large average total basis — but any glider pilot knows to look for places where nice puffy white cumulus clouds are forming, because heat is transferred to the surrounding air as the molecules of water vapor glom together into droplets. The air rises faster as the heat of condensation is released.

Now I”m hoping for some wild idea from the people who specialize in tuning microwave pumping to favor particular chemical reaction paths. Is here any way to do something analogous with favoring particular vibration/bond energy paths, somehow find something to tune in the upper atmosphere to make it more probable that an already common interaction will kick out an infrared photon?

Bypass the whole middle atmosphere, run the radio transmitters on solar power and tickle the CO2 or whatever else is handy= at the top of the atmosphere into turning a bit more of its energy into infrared, going out where we need it to go?

Just daydreaming.

Comment by Hank Roberts — 4 Jul 2007 @ 3:12 PM

94. Rod B in #89

I really can’t explain in any more detail than tamino did (#91) why you are wrong about only molecular translational kinetic energy determining temperature without the equivalent of a semester course in thermodynamics and statistical mechanics. I will add some numbers, though. For molar heat capacity in J/mole K at constant pressure, argon is 20.8, nitrogen is 28.8 and CO2 is 37.84. The atomic weight of argon and the molecular weight of CO2 are similar (40 and 44), so it’s not just a function of mass.

Hank Roberts in #93 said:

Now I”m hoping for some wild idea from the people who specialize in tuning microwave pumping to favor particular chemical reaction paths. Is there any way to do something analogous with favoring particular vibration/bond energy paths,

Over thirty years ago some of the people in the lab where I worked built a CO2 laser with the idea of trying to influence specific reaction rates by activating selected molecular bonds. The idea became dead in the water when it was found that the absorbed energy redistributed internally much (like orders of magnitude IIRC) faster than the rate of any chemical reaction. So you can’t in general, as far as I know, change the path of a chemical reaction by exciting a particular chemical bond in a molecule consisting of more than two atoms. To stimulate IR emission, you need a population inversion as in a laser, again IIRC.

Comment by DeWitt Payne — 4 Jul 2007 @ 4:39 PM

95. Is there any net energy transferred to the surrounding air molecules when carbon dioxide absorbs infrared radiation? This Wikipedia article suggests any energy absorbed is subsequently re-radiated.

Carbon dioxide molecules can absorb IR radiation. Collisions will immediately transfer this energy to heating the surrounding gas. On the other hand, other CO2 molecules will be vibrationally excited by collisions. Roughly 5% of CO2 molecules are vibrationally excited at room temperature and it is this 5% that radiates.

Five percent over what time period? And how is this relationship affected by changing temperature, pressure, or greenhouse gas concentrations? My feeling is that direct warming of the air is not a factor in the greenhouse effect, but I am not sure about that.

Comment by Blair Dowden — 4 Jul 2007 @ 7:45 PM

96. I’ve found a new toy for on line spectral modeling and am using it to illustrate the saturation fallacy. The Spectral Calculator allows separation of the various parameters being discussed here, such as temperature, pressure and composition. The first post is on temperature. The approach is to simplify by isolating one factor at a time.

Comment by Eli Rabett — 4 Jul 2007 @ 11:53 PM

97. Eli Rabett (#94) wrote:

I’ve found a new toy for on line spectral modeling and am using it to illustrate the saturation fallacy. The Spectral Calculator allows separation of the various parameters being discussed here, such as temperature, pressure and composition. The first post is on temperature. The approach is to simplify by isolating one factor at a time.

I would strongly recommend checking it out.

Eli is kind enough to walk you through using the tool and how it may be used. I was able to follow just fine – and then ended up playing around with it a bit. There are more features if you subscribe, and the prices are a little high for my shoe-string budget, but the free features are more than enough to insure that I will be returning more than a few times.

Comment by Timothy Chase — 5 Jul 2007 @ 1:24 AM

98. Blair Dowden (#95) wrote:

And how is this relationship affected by changing temperature, pressure, or greenhouse gas concentrations? My feeling is that direct warming of the air is not a factor in the greenhouse effect, but I am not sure about that.

You are probably right that ir heating of the atmosphere isn’t a major mechanism, but I suspect that warmer air will result in additional emissions where kinetic energy is transformed into longwave emissions. In Raypierre’s inline to your comment #37, he stated that “infrared absorbtion-emission has a cooling effect, which balances convective heating” where the latter is moist air convective heating due to evaporation. If I am not mistaken, this implies that some of the energy which is resulting in emission isn’t from absorption, but is the result of convective heating – and therefore molecular collisions.

Comment by Timothy Chase — 5 Jul 2007 @ 2:23 AM

99. Timothy, I think that you are right that there’s not a lot of atmospheric heating. It is mainly that some of the IR photons eventually make it back to the ground and are reabsorbed, further heating the surface. But the basic issue is that the amount of radiation escaping decreases, so the system must heat up until a new equilibrium (energy out=energy in) is established.

Comment by Ray Ladbury — 5 Jul 2007 @ 9:21 AM

100. My understanding of the process is that when Sunlight is absorbed by the Earth’s surface,the surface warms and emits radiation in the mostly infra red part of the spectrum.The infra red is strongly absorbed by C02 and H2O.The molecules absorb this energy transmit part of this into motion and collide with the predominant N2 and 02 gases of the atmosphere, increasing their velocity.This increase in motion of all the molecules raises the temperature of the air .
If we treat the atmosphere as layers through which the infra red radiation passes on its journey into space,half of the flow of the photons is up and half down.Layers close to the ground are warmer because they receive the direct radiation from the ground and also the re-transmitted radiation from the warmer layers above. This results in a thermal structure where the warmest air is at low altitudes and gets progressively cooler in the upper regions.This is oversimplified since I haven’t included convection but it’s basically why the thermal structure is what it is.
Since as the above paper shows, the air is not near saturation, adding more CO2 means more IR radiation is absorbed, more heating takes place and the cycle continues with more intensity. The differential temperature between the surface and the top to the troposphere is now greater and more layers have to be warmed at the top for the same amount of energy to escape. There are less polysyllables here but is it right?

Comment by Lawrence Brown — 5 Jul 2007 @ 9:58 AM

101. Re #100: Lawrence, if you read the posts just before yours (they may not have been there when you posted) you will see that while greenhouse gases absorb infrared radiation and warm the air, they also absorb kinetic energy from other molecules and radiate infrared. When this happens, there is less kinetic energy, thus local cooling. In fact, it appears there is more radiative cooling than warming.

You are right that the infrared radiation goes in all directions. This has little effect on the temperature structure of the atmosphere, which would be much the same without any greenhouse gases at all. It does warm the ground, which warms the air above by convection. Convection is how most energy moves through the atmosphere.

According to the radiation balance model, adding more greenhouse gas raises the average altitude at which they radiate into space. At higher altitude the atmosphere is cooler. A cooler greenhouse gas radiates less energy into space, so less energy is lost from the Earth system, so it gets warmer.

Comment by Blair Dowden — 5 Jul 2007 @ 11:53 AM

102. Blair, In the troposphere, convection is much more important than radiation for warming of the atmosphere. In the stratosphere, you can exchange energy, but warming by radiation is probably not all that significant. The main thing is that the CO2 in the stratosphere prevents IR from escaping and sends some of it back to the troposphere.

Comment by Ray Ladbury — 5 Jul 2007 @ 1:20 PM

103. re 89, 90, 91: Well, blow me down. To prove Tamino and Ray wrong I dug out my Feynman lectures. Unfortunately Feynman beat hell out of me. It’s as you say — maybe even worse. It seems the more atoms in a molecule the more true kinetic energy is stored in the internal bonds, and at three atoms/molecule the internal KE is TWICE that of the whole molecule’s traveling through space. I could not have been more wrong! Worse than that — you might want to check my math on the following (I don’t want to show it here because it might be misleading to the unwary). I calculated (theoretically) the kinetic energy of one CO2 molecule at 250 degreesK to be 1.56×10-20 joules, 1/3 “external” and 2/3 internal (Damn, again!), with the whole molecule zipping around “externally” at 375m/sec. Then I gave it one photon at 14.7um, calculating its energy at 1.35×10-20 joules. Absorbed into the internal bonds, one photon raised the molecules temp to 469 degrees — almost double. Does that sound right to you all??? I must go regroup!!

re 93 (Hank): I’m a little gun shy, but I think you are wrong. A molecule of water vapor at 100 degreesC has more internal “heat” energy than a molecule of water at 100degreesC. Doesn’t it?? Or does it somehow immediately (??) dissipate?

Comment by Rod B — 5 Jul 2007 @ 1:52 PM

104. >93, 94, thanks deWitt for the history; I was thinking of microwave pumping favored bond energy to increase a reaction product, thus: http://scholar.google.com/scholar?q=microwave+pump+reaction+path

and wondering if some microwave pumping could nudge molecules over the hump to emit an infrared photon—faster than a chemical reaction, but I guess you’re saying there’s no ‘tunable’ configuration that can be favored long enough to favor emitting an infrared photon?

Comment by Hank Roberts — 5 Jul 2007 @ 1:53 PM

Timothy, I think that you are right that there’s not a lot of atmospheric heating. It is mainly that some of the IR photons eventually make it back to the ground and are reabsorbed, further heating the surface. But the basic issue is that the amount of radiation escaping decreases, so the system must heat up until a new equilibrium (energy out=energy in) is established.

Well, I guess the best way for me to put it is that the greenhouse effect due to carbon dioxide in the stratosphere raises the temperature at ground level and sea level, resulting in more evaporation. This evaporation leads to increased water vapor, and water vapor results in moist air convection and an increase in the greenhouse effect. But the direct result of the greenhouse effect in the atmosphere in a cooling in which counterbalances the warming due to moist air convection. Finally, the cooling of the atmosphere due to the greenhouse effect involves the transformation of thermal/kinetic energy into long-wave infrared radiation, and while the direct trasformation of infrared radiation into thermal/kinetic energy in the atmosphere is certainly possible, within the atmosphere, the transformation of thermal/kinetic energy into infrared predominates.

Comment by Timothy Chase — 5 Jul 2007 @ 2:12 PM

106. Edit on #105

“The best way for me to put it…” got me to look at that last sentence. It should end:

“… and while the direct transformation of infrared radiation into thermal/kinetic energy in the atmosphere certainly occurs, within the atmosphere the transformation of thermal/kinetic energy into infrared is dominant.”

Comment by Timothy Chase — 5 Jul 2007 @ 2:36 PM

107. For this series, around 81 to 102, and Alastair in particular. What seems to me to need emphasis is the notion of a radiation field in which the molecules reside, and which is a consequence of being at a finite temperature. With molecules at temperature T and an isotropic field of thermal energy also at T, molecules are at equilibrium, joyfully emitting and absorbing at more or less the same rate. (For CO2 at the beach on a warm day, these radiations are relatively far up the chain of vibrational energy states.)

With flashlights or warm bodies like Earth, there may be additional fields of higher temperature, in which case there will be net absorption by the molecule in the direction of the gradient at the frequencies at which the molecule can exchange energy. Energy absorbed by the molecule will be subsequently re-radiated isotropically, thereby raising the temperature of its neighbors, but depleting the gradient at the same frequencies as were absorbed. So, yes, the atmosphere can be heated by absorption of energy emitted at the surface. It can also be cooled, depending.

Since the molecule can only come into equlibrium at the frequencies at which it emits and absorbs, it will usually be out of equlibrium at other frequencies. When the radiant field is of higher temperature than the molecule, the frequencies at which the molecule absorbs will appear less bright along the direction of the gradient. Some might call this a sort of saturation; some might not.

Cautionâ��speculation begins here. Now, Alastair, do you buy that when the radiant field is of lower temperature than the molecule, as can happen when local thermal equilibrium begins to fail, the molecular frequencies can appear to be at a higher temperature than the general field? And perhaps when we see particular temperatures for certain molecules as you have observed, we are looking at the point where they crossed out of LTE?

Comment by Allan Ames — 5 Jul 2007 @ 2:50 PM

108. Rod B (#103) wrote:

re 89, 90, 91: Well, blow me down. To prove Tamino and Ray wrong I dug out my Feynman lectures…

I have yet to break-out the calculator – although I used a spreadsheet a while ago.

[… he says, ending with an oh-so-charming and sheepish grin.]

re 93 (Hank): I’m a little gun shy, but I think you are wrong. A molecule of water vapor at 100 degreesC has more internal “heat” energy than a molecule of water at 100degreesC. Doesn’t it?? Or does it somehow immediately (??) dissipate?

I would be inclined to think that a water molecule which has just broken free of the liquid at the boiling point has more kinetic energy, but then again, the internal energy is quantized, and therefore the additional kinetic energy is probably due velocity, although there is probably an increased likelihood of excited internal states.

Comment by Timothy Chase — 5 Jul 2007 @ 3:06 PM

109. Rod wrote:
> A molecule of water vapor at 100 degreesC has more internal
> “heat” energy than a molecule of water at 100degreesC. Doesn’t it??

Nope. Unless “internal” is your point. Temperature is all the energy involved with the molecule; water vapor has its energy in the one molecule, motion, vibration, rotation between the atoms (it isn’t hydrogen bonded in a favorable position to other water vapor; when it does, it’s changing from water vapor to water, condensing into mist, energy going into the bonds that hold one molecule to the next).

Consider — you probably know the warning not to reheat a cup of water in a microwave, after you’ve brought it to a boil once. That’s because the first time you heat it to boiling, all the little gas bubbles in the cracks and crevices on the inside of the cup get dislodged. Those form points where the hot water easily turns to vapor, starting the boiling at a lot of little spots. You can see the same thing inside a glass kettle on a stove, strings of bubbles rising from particular points all over the inside surface as the water is just reaching the boiling point.

But if you bring a cup of water to a boil in a microwave, don’t open the door, let the water cool off, and do it again, and again — you are boiling off or physically removing all the little discontinuities.

What happens? It becomes possible to superheat the water because there aren’t those little points where the bubbles of vapor can easily form.

Result?

Kids, do NOT try this at home.

The caution is: If you’ve managed to reheat the water under these conditions, no banging on the countertop, no dirt falling into the cup from the top of the microwave chamber, clean cup, no nuclei to promote bubbles to form more easily at any point —- it’s possible the water will — all of it — be heated up to 100 degrees C, 212 fahrenheit, or even slightly higher.

And you open the door and the vibration or breeze causes a whole lot of that water to turn to vapor and you get a faceful of boiling water and steam.

Like I said, kids, do NOT try this at home. Look up “superheated” ….

Comment by Hank Roberts — 5 Jul 2007 @ 3:53 PM

110. Re #102 and #105: Ray and Timothy, I want to question the statement that carbon dioxide in the stratosphere is important to the greenhouse effect for two reasons: First, there is not much carbon dioxide there. Second, the temperature in the stratosphere rises with altitude, so the CO2 will radiate at a higher temperature, thus more heat will be lost compared to the upper troposphere.

I understood that greenhouse gases near the top of the troposphere were more significant.

Comment by Blair Dowden — 5 Jul 2007 @ 4:24 PM

111. Blair Dowden (#110) wrote:

Re #102 and #105: Ray and Timothy, I want to question the statement that carbon dioxide in the stratosphere is important to the greenhouse effect for two reasons:

Not a problem.

First, there is not much carbon dioxide there.

Temperature increase due to carbon dioxide is roughly proportional to the log of the concentration. While there isn’t much carbon dioxide in the stratosphere, this also implies that one doesn’t have to add much carbon dioxide to the atmosphere in order to double its concentration which directly results in an increase in temperature of approximately 1.2 degrees Kelvin and indirectly in an increase in temperature of approximately 2.9 degrees when one takes into account successive rounds of amplification due to increased water vapor in the troposphere.

Second, the temperature in the stratosphere rises with altitude, so the CO2 will radiate at a higher temperature, thus more heat will be lost compared to the upper troposphere.

When it radiates, roughly half of the radiation will be downwelling and half will be upwelling. Depending upon where the molecule is located, the photon will have a slight tendency to eventually reach either space or the ground as the result of a stochastic process of emission and re-absorption. However, upwelling radiation will tend to have slightly longer legs than downwelling radiation due to decreasing atmospheric density.

Comment by Timothy Chase — 5 Jul 2007 @ 4:46 PM

112. Blair, factor in the _time_ that has to elapse.
The temperature of the stratosphere is higher toward the top.
That doesn’t mean that when something causes it to LIFT UP slightly, it gets hotter. Quite the opposite — lift it up, it expands and gets cooler. That’s the prediction for when the planet warms up.

Surface warms.
Lower atmosphere warms (and because it’s warmed, it expands, and it can only expand upward not downward)
Upper atmosphere is being lifted higher from the surface
—- because it’s lifted, it expands and cools

A cool molecule emits its infrared photon at — a cooler temperature.
So the heat leaving the planet isn’t keeping up with the heat being held in the atmosphere/earth/ocean.

That takes a few hundred years — til the upper atmosphere, still out there a bit farther away, gets warmed up again and is emitting heat at the same rate as it originally was, the same rate as the incoming energy.

Meanwhile down in the lower atmosphere/earth/ocean system, it’s gotten hotter over those same several centuries and stayed hotter for quite a few centuries, until the CO2 gets captured in minerals and ocean sediments.

At the top you have the same number of CO2 molecules (or slightly more, over time). They’re the molecules

Comment by Hank Roberts — 5 Jul 2007 @ 5:03 PM

113. Blair, out of curiosity, where do you get your information that the stratosphere is depleted in CO2? The measurements I’ve been able to find suggest that the lower stratosphere (up to ~35 km) is only a few ppmv less than the troposphere, and that it is increasing at a rate that is not incomensurate with the increase in the troposphere. Most of the measurements I’ve seen date from the 70s and 80s. Do you have more recent measurements?

Comment by ray ladbury — 5 Jul 2007 @ 8:44 PM

114. re 107: a quicky question/clarification: the radiated E-M field has no temperature…., does it??

Comment by Rod B — 5 Jul 2007 @ 9:12 PM

115. Timothy, your statements in #111 are correct but do not address the issue of where most of greenhouse warming occurs. It may be worse than I said – carbon dioxide in the stratosphere radiates more energy than it absorbs, and thus causes the stratosphere to cool. This is independent of any effects from the troposphere. What I don’t know is whether or not this results in a (small) net cooling of the Earth.

I still think greenhouse warming take place mainly in the upper troposphere.

Hank, I am afraid you lost me with your comment. Photons move fast, so the greenhouse effect is almost instantaneous. Convection takes a little longer, maybe a few days.

Comment by Blair Dowden — 5 Jul 2007 @ 9:34 PM

116. a little public musing re my post 108 et al: It still doesn’t smell right. It says that E-M radiation is the direct primary and major cause of atmosphere temperature increases. Yet post after post here have implied otherwise, e.g. “a little bit….” Feynman’s equations also imply that any added energy to a molecule will get distributed between the internal kinetic energy and the overt kinetic energy with a definite ratio unique to each size (number of atoms) of molecule. Is this right?????

One other thought for mulling before I go try to find out. Is my contention of heat changing without changing the temperature only “valid” with averages ala Maxwell-Boltzman distribution? It sticks in my mind that the molecule that evaporates is the odd one out with much higher kinetic energy than the average speeding near the surface and luckily breaking free. It takes a much higher (but how much higher????) than average heat and temp with it, simultaneously adding temp and heat to the vapor and removing heat and temp from the liquid base. Any thoughts?

Still thinking aloud, fishing for responses: does that mean that evaporation from the sea surface cools the sea and warms the atmosphere? AND, possibly, the excited evaporated H2O molecule, with its kinetic energy distributed between its internal bonds and its overt velocity in a 2:1 ratio, can emit IR radiation, even if it had not absorbed any, with E-M energy taken from its bond energy???

Comment by Rod B — 5 Jul 2007 @ 10:12 PM

117. An afterthought that’s driving be batty: Go anywhere on the web or to any standard physics textbook and 99 times out of 100 (I’m guessing) it will say that 1.5kT = kinetic energy = 0.5mv2 of the overt total molecule. Can anyone find where it says the total kinetic energy of a gas molecule = K.E. = 1.5(r-1)kT where r is the number of atoms in the molecule? Or am I just blind?????

Comment by Rod B — 5 Jul 2007 @ 10:41 PM

118. > Photons move fast, so the greenhouse effect is almost instantaneous. Convection takes a little longer, maybe a few days
Blair, look up “mean free path” and the simulator Eli Rabett has been talking about. Photons move til the next interaction, which happens very fast in a gas. Yes, speed of light — but only til the next interaction. Else the heat would all rush off the planet every night. We had someone else here repeatiing basically these same beliefs a few weeks back, who finally left I guess. Where are you getting them, one of the other “climate science site” pages?

Comment by Hank Roberts — 5 Jul 2007 @ 10:50 PM

119. Blair Dowden (#115) wrote:

Timothy, your statements in #111 are correct but do not address the issue of where most of greenhouse warming occurs.

At the surface – moist air convection is what is primarily responsible for heating the troposphere.

It may be worse than I said – carbon dioxide in the stratosphere radiates more energy than it absorbs, and thus causes the stratosphere to cool.

Solar radiation includes UV rays which will heat the stratosphere as the result of ozone. This is the reason why a depleted ozone layer results in the cooling of the stratosphere over Antarctica, resulting in a higher temperature differential between it and the surface, increasing winds, results in an upwelling of nutrient-rich water from below, increasing the release of methane and carbon dioxide, intensifying the greenhouse effect and lofting moisture into the stratosphere which results in the slower replentishment of the ozone layer. Then there is terrestrial radiation. Then there are the aerosols.

Each plays a role.

But the important thing is actually fairly basic: if the system is in equilibrium, then all of the effects must cancel one another, at the surface, in the troposphere and in the stratosphere.

However, all of this assumes an equilibrium, and I see that Hank Roberts has done non-equilibrium in #112. Good thing too – since I have little doubt that he knows more about it than I do.

Comment by Timothy Chase — 6 Jul 2007 @ 2:35 AM

120. RE: #111 Carbon dioxide doesn’t emit any IR photons. All the IR energy that a carbon dioxide molecule absorbs is removed by collisions with nitrogen and oxygen molecules. Most of the greenhouse effect is due to water molecules near the surface of the earth and above and in fact, carbon dioxide plays little or no role in the greenhouse effect. At a concentration of 0.038% by volume there is too little molecular muscles to do anything except standby and watch water molecules, greedy little IR energy hogs, do all the work. The so-called feed back process is just a fantansy thought up by climatologists drinkin’ Thunderbird and smokin’ cheap Mexican pot on the patio at the Hotel California.

Air pressure is far more important than carbon dioxide on effecting the amount of water vapor in the air because the heat of vaporation of both solid and liquid water is primarily a function of pressure. For a non- associated liquid the heat of vaporation of solely a function pressure. Water is an associated liquid and the heat of vaporization also depends on a limited extent on temperature. This why high pressure cell are generally warm and dry in summer and cold and dry in winter. Low pressure have more moisture due to the lower atmosphere pressure.

Comment by Harold Pierce Jr — 6 Jul 2007 @ 4:49 AM

121. Harold, for someone who doesn’t understand the science, you have awfully strong opinions.

Comment by Ray Ladbury — 6 Jul 2007 @ 7:18 AM

122. [[You are right that the infrared radiation goes in all directions. This has little effect on the temperature structure of the atmosphere, which would be much the same without any greenhouse gases at all.]]

This is incorrect. Radiative transfer is a major process in the atmosphere. Layers of air are heated by the sunlight they absorb, and much more by the infrared they absorb both from the ground and from other layers of air.

Comment by Barton Paul Levenson — 6 Jul 2007 @ 7:31 AM

123. [[Absorbed into the internal bonds, one photon raised the molecules temp to 469 degrees — almost double. Does that sound right to you all??? I must go regroup!!]]

Temperature isn’t a concept that applies to one molecule. You need a group of them.

Comment by Barton Paul Levenson — 6 Jul 2007 @ 7:33 AM

124. [[RE: #111 Carbon dioxide doesn’t emit any IR photons.]]

Yes it does.

[[ All the IR energy that a carbon dioxide molecule absorbs is removed by collisions with nitrogen and oxygen molecules. Most of the greenhouse effect is due to water molecules near the surface of the earth and above and in fact, carbon dioxide plays little or no role in the greenhouse effect.]]

Wrong. Carbon dioxide provides about 26% of Earth’s greenhouse effect versus about 60% for water vapor.

[[ At a concentration of 0.038% by volume there is too little molecular muscles to do anything except standby and watch water molecules, greedy little IR energy hogs, do all the work.]]

It’s the absolute amount that matters, not the concentration. Most of the atmosphere, nitrogen, oxygen and argon, is not radiatively active. (Oxygen is a bit radiatively active in the UV). There are about 3 x 1015 kilograms of carbon dioxide in the air. That’s about 5.88 kilograms per square meter of Earth’s surface. That’s clearly enough to make a difference.

[[ The so-called feed back process is just a fantansy thought up by climatologists drinkin’ Thunderbird and smokin’ cheap Mexican pot on the patio at the Hotel California. ]]

You seem to have been indulging in a little of that yourself. I’d recommend reading up on some basic climatology instead. Try googling for “Clausius-Clapeyron law” and trying to understand why that implies a temperature feedback effect for carbon dioxide warming.

Comment by Barton Paul Levenson — 6 Jul 2007 @ 7:42 AM

125. The Hadley Center of the British Met Office gives a clear graph of the emission of IR radiation at higher altitudes due to increasing levels of CO2 in the atmosphere.

The graph is followed by an explanation of present day conditions and from a hypothetical higher cooler level. The explanation is straightforward.The IR emission can’t emit as much at this cooler layer and the atomosphere must warm up until the rate of IR emission returns to the original rate. The surface, in the example will produce a warming of 10K over the present. day,meaning that increasing the present emissions from an average height of about 5.5 kilometers, to 7 kilometers raises the surface by this much- 10K or 10C

Comment by Lawrence Brown — 6 Jul 2007 @ 9:06 AM

126. re 119 Chase: While the climate people may have their own definition of “equilibrium”, from the thermodynamic standpoint, if the atmosphere were at equilibrium with the radiation fields, then upwelling and downwelling radiation fields would be the same (definition of equilibrium, as distinct from steady state). We are at the bottom of distinctly non-equilibrium energy transport system, trying to explain it with equilibrium concepts.

My position is that the GHG effect is purely kinetic, not at all thermodynamic.

Comment by Allan Ames — 6 Jul 2007 @ 11:49 AM

127. re 114, Rod B: Does a radiated field have a temperature? Sort of. It is common to compare measured intensity to that of a black body and produce a “brightness temperature”. Searching on this term and AIRS should get some spectra.

You have too many mixed questions to answer, but KE = 1/2 M V^2. period.

Comment by Allan Ames — 6 Jul 2007 @ 11:56 AM

128. Rod, the total energy of a molecule in thermal equilibrium with its surroundings at temperature T is 3/2 nkT where N is the number of atoms (where did they get r from? The kinetic energy is 3/2 kT. That means that the internal energy is 3/2 (n-1) kT. Your formula was for internal energy, not kinetic.

Comment by Eli Rabett — 6 Jul 2007 @ 12:18 PM

129. RE #107 where Allan Ames wrote

“Now, Alastair, do you buy that when the radiant field is of lower temperature than the molecule, as can happen when local thermal equilibrium begins to fail, the molecular frequencies can appear to be at a higher temperature than the general field? And perhaps when we see particular temperatures for certain molecules as you have observed, we are looking at the point where they crossed out of LTE?”

I do buy that if you mean that when the brightness temperature (radiant field) is lower than the kinetic temperature of the air molecules, then the air is warmer than the radiant field produced by the surface of the Earth, and the system will no longer be in LTE. This happens every clear night when the surface cools and radiates less black body radiation towards the air than the air radiates towards it. The reverse process happens during the day. when the surface is warmed by solar radiation and the blackbody radiation field from it is warmer than the air containing the greenhouse gas molecules that are absorbing it. This means that the air next to the Earth’s surface is only in LTE twice per day, when the two temperatures cross.

And Re 114 where RodB wrote “the radiated E-M field has no temperature…., does it?”

It does have a temperature called the brightness temperature. This temperature is the Planckian temperature which is calculated by assuming that the radiator is a black body. When the Planckian (brightness) temperature and and the Maxwellian (kinetic) temperature correspond, then the gas is held to be in local thermodynamic equilibrium.

Comment by Alastair McDonald — 6 Jul 2007 @ 12:43 PM

130. >Hank Roberts has done non-equilibrium in #112. …

Tim, I’m sure I don’t know more about it — I noticed Blair forgot that after a change in CO2, Earth’s heat takes time to reach equilibrium. I pointed it out and tried as usual for simple words. That took me a year to understand even as a notion!

Read me cautiously! At _best_ I’m useful here as a copy editor. Mostly I’m trying to find simpler, clearer words, knowing the explanations without the math are more poetry than physics. I long ago worked as a ranger=naturalist cave guide; that set my level. Someone asked, always, on every single cave tour: “How many miles of undiscovered cave are there here?”

Blair — this is helpful, it’s a new online simulator, try this one: http://www.seed.slb.com/en/scictr/watch/climate_change/challenge.htm

Eli, you might like this one too. It’s based on Tom Fiddaman’s work, link therewith.
Here’s one screenshot. http://farm2.static.flickr.com/1439/740529390_35e55d5dba.jpg?v=0

Gavin, I nominate this simulator for your ‘Start Here’ and links lists.
More people watch than read; this one ought to be in firmware in all the video-pods (grin).

Comment by Hank Roberts — 6 Jul 2007 @ 12:43 PM

131. re “Temperature isn’t a concept that applies to one molecule….”

While it may not be practical, I beg to differ. Theoretically (and really) you can have one molecule of a gas (or anything else) that has a mass of a few amu’s and is tooling around space at a velocity and therefore has kinetic energy (0.5mv2) and therefore has temperature ala 1.5kT.

Comment by Rod B — 6 Jul 2007 @ 1:58 PM

132. re 127: “Does a radiated field have a temperature? Sort of. It is common to compare measured intensity to that of a black body and produce a “brightness temperature”.

I understand the concept, and maybe it’s a quibble, but the brightness temperature refers to the physical body that’s putting out the temperature-less E-M radiation.

“You have too many mixed questions to answer, but KE = 1/2 M V^2. period.”

Guilty; I’m trying to sort it out. I’m also trying to reconcile Feynman’s definition that K.E. = 3/2(r)kT where r is the #modes or degrees of freedom or (roughly) atoms/molecule. He doesn’t discount 1/2 m V^2, just says you have to include all masses and velocities of the molecule.

Comment by Rod B — 6 Jul 2007 @ 2:35 PM

133. Re #132 RobB where you wrote:
I understand the concept, and maybe it’s a quibble, but the brightness temperature refers to the physical body that’s putting out the temperature-less E-M radiation.

It’s not a quibble if you are correct, so I used Google and found these:
http://en.wikipedia.org/wiki/Brightness_temperature
http://scienceworld.wolfram.com/physics/BrightnessTemperature.html
The brightness temperature is a value calculated using the wavelength and intensity of the E-M radiation from the body. For a blackbody, the temperature calculated at all frequencies will be the same, and will be equal to the body’s temperature found in the normal way.

For a solid or liquid the normal temperature is not due to the kinetic energy of the molecule. It is the energy of the molecular vibrations. IMHO, this makes it rather surprising that gases emit lines with the same intensity as the continuous radiation from a solid or liquid.

Comment by Alastair McDonald — 6 Jul 2007 @ 3:04 PM

134. Thanks, Eli. Feynman’s use of “r” in the formula seemed odd to me too.
I think your distinction of “internal” and “kinetic” is easier to visually, though Feynman (and others) refer to it all as kinetic, by virtue of the actual kinetic energy from atom vibration and rotation internal to the molecule. In any case what I’m trying to verify is — does that internal energy affect the temperature? I.e. does absorption of E-M radiation per se, which I’m assuming goes into the “internal” energy, increase the temperature of said molecule? And by my calculations (but don’t bet the farm!), a BIG temp increase.

Comment by Rod B — 6 Jul 2007 @ 3:10 PM

135. Alastair (134), et al: not sure if you agree or not. In any case to reassert, representing something with a “temperature” to maybe aid the comprehension is not the same as someting actually having temperature. Varying eneregy fields of Electrostatic and Magnetic flux do not have temperature. They come from things with temperature (and one can relate the things temperature with the peak intensity of the radiation accurately), and they can change the temperature of other things. But no temp for them….

Comment by Rod B — 6 Jul 2007 @ 3:26 PM

136. re 132 Rod B: Google on specific heat of gases, and you find that the more modes of movement, the higher Cv and Cp. The numbers tells you more about the molecule than vice versa. Note Cp,Cv vary with T as more modes get activated at energies within the thermal continuum. Check out ammonia. There are several good expositions on the web on “heat capacity of gases.”

For the GH effect you need energy exchange at IR frequencies, not particularly Cp, but such exchange says there is a mode which would raise Cp so there is some connection.

Comment by Allan Ames — 6 Jul 2007 @ 3:28 PM

137. Temperature is the difference from absolute zero. A molecule twitching and tapdancing after ingesting a photon is warmer than the same molecule beforehand, when it was sitting comatose and close to motionless.

Comment by Hank Roberts — 6 Jul 2007 @ 3:33 PM

138. re 133 Alsatair +1 Rod B: For me the part to get ones head around is “thermal equilibrium”, which is where (and only where) temperature is defined. At a finite temp. there will be a radiation field, modified by emissivity, through which matter exchanges energy with other matter on a continuing basis.

And yes, there will be local temperature fluctuations because quanta are lumps.

Comment by Allan Ames — 6 Jul 2007 @ 3:41 PM

139. re 129 Alastair: As I said in 126, I think it is pretty clear that the system is mostly not in LTE. Have you explored the math of a non-LTE system?

Comment by Allan Ames — 6 Jul 2007 @ 3:57 PM

140. Re #139

I got caught out by non-LTE, which is in the state of the thermosphere. It is not the same as not LTE. I have resorted to calling not LTE local thermal unequilibrium or LTU.

In non-LTE, the plasma becomes very excited because there are too few collisions to relax it. In LTU the radiation field is not isotropic. It exists at the base of the atmosphere where the absorption exceeds the emission, and at the top of the atmosphere the emission exceeds the absorption. The layer at the base of the atmosphere where this applies is only 30 m deep. The layer at the top of the atmosphere where this applies starts at a height of 6 km ie it includes half the the tropsophere, all the stratosphere and all the mesosphere.

Comment by Alastair McDonald — 6 Jul 2007 @ 5:06 PM

141. A little thermodynamics. First, temperature is a macroscopic quantity. It does not apply to a single molecule. Look at the definition T=partial S/partial E. Now think of a single molecule–does it even have an entropy (apart from due to its chemical identity, I mean; remember we’re holding chemical potential, etc. constant). In point of fact the average kinetic energy will be 3/2 kT, so a temperature of a single molecule doesn’t make sense.
OK, now the question of the temperature of a radiation field. Well, the cosmic microwave background has a temperature (~3 Kelvins), and that is how we know anything about the Big Bang. That’s as real as anything.
Now on the question of the proportionality between energy and temperature. There’s nothing magical about 3kT/2. If you comfine your gas to 2 dimensions (as in an electron gas in a high-electron mobility transistor, or an adsorbed film on a surface, KE~kT). Likewise as the number of degrees of freedom of the molecule goes up, the proportionality constant (the specific heat) between energy and kT goes up. You can actually see this as you heat a gas up from a very low temperature. Initially, the temperature is too low to excite the quantized rotational and vibrational energy levels, so E~3kT/2. As T increases, you start to see more rotational and vibrational energy levels excited, and the specific heat actually increases. That is, you have to add more energy to get the same temperature rise. Now I’m sure this is review for most of you, but hopefully it refreshes memories of thermo or chemistry (at least if they are pleasant).

Comment by ray ladbury — 6 Jul 2007 @ 5:09 PM

142. Ray (141): a quicky on part of your post — and I’m definitely not being argumentative over detail trivia, but I think this concept is important for understanding (at least by me!) the physics of absorption/emission and, getting really basic here, temperature. First, while maybe just inadvertant, I think S = Q/T (or E/T) (fill in the partials).

So while a single molecule will have a specific and precise kinetic energy greater than zero per Maxwell-Boltzman probability, it won’t have a temperature of its own??? Or when that single molecule bumps another or the wall it won’t impart some ot its KE (and in turn temperature)??? Makes no sense. The universal gas laws are derived starting with a sole lonely molecule — is that just convention???

My understanding is the 3 degreesK is specifically the temperature of the ubiquitous leftover soup of the Big Bang, not the common usage CMB radiation. Q: what is the temperature of the incoming solar radiation at, say, 1,000,000km from the earth???

I’ll respond to the last half of your post (and that of others) later; it served up good food for thought. Thanks.

Comment by Rod B — 6 Jul 2007 @ 10:03 PM

143. Re #141 That does not affect the simple fact that the Earth system loses heat to space from the top of the atmosphere. Therefore the upper layers of the atmosphere must be losing heat, and are not in LTE. To compensate for the loss at the top of the atmosphere the air must be gaining heat somewhere, and basically the atmosphere gains heat at the Earth’s surface by absorbing radiation. This is the greenhouse effect shown by Horace de Saussure. The air at the base of the atmosphere is gaining heat, and so it is not in LTE either.

The current models see the whole of the troposphere as being in LTE and being homogeneous, ignoring the discontinuity at the top of the boundary layer. That is wrong.

Now that science has replaced religion as the custodian of the truth about how the world works, it is difficult for me to break down your belief that now it is the scientists who are preaching the word of God, and every thing they say is true.

The reason people are having trouble understanding the current theory is because it is wrong. The fact that we know that CO2 does cause climate change does not mean that the current theories are correct.

Comment by Alastair McDonald — 7 Jul 2007 @ 1:52 AM

144. Ray, I would like to continue your thermodynamics lesson from “As T increases, you start to see more rotational and vibrational energy levels excited, and the specific heat actually increases. That is, you have to add more energy to get the same temperature rise.” If a greenhouse gas is present, vibrational energy levels can cause infrared radiation to be emitted, reducing the energy level of the gas. This means you need to add even more energy (eg. by IR absorption) to get the same temperature rise. Apparently, in the stratosphere carbon dioxide radiates out more energy than it absorbs, causing cooling.

I am trying to get back to my old question: does the greenhouse effect heat the air directly in the lower troposphere? When water vapor or CO2 absorb IR radiation, it is translated into kinetic energy, ie. heats the air. The increased temperature causes IR to be emitted, reducing the temperature. This cannot balance, or it would be impossible for air with greenhouse gas to warm up. So there must be some direct warming.

Air also warms via convection from the surface, so it might actually emit more IR than it gets from its greenhouse gases (as in the stratosphere), but this does not mean greenhouse gases cause a net loss.

If the lower troposphere is being warmed directly, how much of a role does the upper troposphere play, where radiation is escaping into space? I am having a hard time putting these two processes together.

Comment by Blair Dowden — 7 Jul 2007 @ 7:06 AM

145. Alastair, First the philosophical: I think there is sufficient truth in the Universe that both science and religion can be custodians should they choose–not to mention philosophy, literature, humor…

Now the science–if the energy in=energy out (or nearly so), you have LTE. And it is nowhere necessary for the models to assume homogeneity–although the troposphere is fairly homogeneous, and homogeneity is a reasonable local approximation throughout. The reason people are having difficulty with the science is because it is subtle.

Perhaps it would help both you and us if you would lay out all your problems with current theory in a systematic fashion. As I’ve been pointing out in the other thread–you have to know the likely effect of an error (even a systematic one) before you can gauge its significance.

Rod B. You are correct and thanks for the correction. It’s been a long time since stat mech. If you are looking at a single molecule in a macroscopic container, it will certainly interact with the walls. However, keep in mind that in a solid, motion is constrained (you no longer have the 3/2 kT for kinetic energy). The molecule can collide with the walls elastically (not losing energy) or inelastically (losing energy). If the collisions are elastic, we can approximate the walls as objects–and not worry about the molecular structure. For inelastic collisions, the molecule either excites an oscillation in the solid or it gains energy from the solid. However, the oscillations in the solid are quantized (as with any oscillator), so we have to describe this process quantum mechanically, and it’s easiest to do so in terms of phonons–quanta of vibration or sound. It doesn’t make sense to look at the molecules in the solid as individual molecules–their energetics within the solid is completely different than when they are on their own.
WRT solar radiation–it makes sense to use the Stefan-Boltzmann law. Solar energy density decreases as the square of the distance from the Sun. Temperature scales as the 4th root of the energy density, so temperature decreases roughly as the square root of distance.

Comment by ray ladbury — 7 Jul 2007 @ 7:16 AM

146. Re 143 So do you suggest two regions, one where the Schwarzchild works, the other where we need formal field to mode coupling, or what? How close to what you want are the various LBL models?

Comment by Allan Ames — 7 Jul 2007 @ 8:35 AM

Alastair is trying to get away to focus on writing his paper proving all climatologists are wrong. He’s said so repeatedly.

As long as we keep trolling for him here, baiting him to go on posting, distracting him from his work, he won’t ever get his work together in his own website and paper.

Going into his ideas off-topic in designated RC discussion threads does distract everyone from the topics at RC as well.

It’d be a kindness to everyone not to question Alastair further here, eh? Let him get his work together in one place so it’s coherent.

Asking questions in topics at RC just tempts him beyond the limit of his self control to repeat his ideas without writing them up in a scientific paper.
And until he does, they’re just frequently asserted beliefs, not science.

Comment by Hank Roberts — 7 Jul 2007 @ 10:12 AM

148. re 145: We are mostly in agreement. One minor continuing quibble (or maybe not…): it’s still the individual molecule vibrating within the solid or liquid that produces (or is the basis for), in part, the Planck radiation…., I think…

And — temperature of the emitting body, not of the actual radiation, scales to the fourth root of the radiated energy density. I don’t think the “temperature of the E-M radiation flux” halfway between the Sun and Earth is proportional (roughly) to 5000/(5.6×1021m2)…

Comment by Rod B — 7 Jul 2007 @ 10:33 AM

149. Blair, As I said, it’s been a long time since stat mech, so I’m struggling with this as well. In part, we are falling victim to the limitations of language and visualization–necessary for understanding, but it only gives a piece of the story. For instance, we are talking about temperature AND about how an individual molecule will behave. However, temperature is a concept that only applies to fairly large aggregates of molecules. It is a concept that was developed when heat was still thought of as a substance (caloric), and that is why Boltzmann sought to make thermodynamics by deriving it from the physics of large numbers of molecules.
I think where we are getting confused is when we look at all “warming” as “temperature”. In reality what we are doing is adding energy to a system (or rather keeping it from escaping). Thermodynamics says when you do that the system has to heat up. But we’ve already seen there are other places the energy goes than kinetic energy–vibrational motion, rotational motion, even gravitational potential energy. Now when a molecule has a higher energy than surrounding molecules, there are a lot more ways for it to lose energy than to gain energy, so it will tend to thermalize–(that is move back to an energy close to 1/2 kt multiplied by the number of degrees of freedom). It can do this in a number of ways. It can fall from a great height, colliding with molecules to thermalize all the way down. It can lose vibrational energy via a collision or other interactions with surrounding molecules. It can emit a photon to relax back to its ground state. Some of these processes will heat the surrounding air. Some (e.g. photon emission) will not. However, half the photons emitted are now moving downward, so the molecules below, now have a flux of IR going downward and a flux going upward. More of them will absorb IR–and again, half the radiation goes down and half up. And so on, until IR photons are incident on the ground–or air near enough to the ground–we further warm the ground.
The key thing to remember is that you are adding energy to the system the system. That energy can warm things up. It can excite vibrational and rotational motion in molecules. It can melt ice or evaporate water. It can cause the atmospher to expand outward. It can warm water and land. However the important thing to remember is that you are increasing the energy of the system. Increasing the energy of a system always makes it less predictable. Does that help at all.

Comment by ray ladbury — 7 Jul 2007 @ 12:00 PM

150. Re: #121 Absolutely. This “Fightin’ Illini” ( B.Sc.(Hon), 1967. U. of I, U-C) and Anteater (Ph.D., UC Irvine 1972) does indeed has strong opinions, and this ace organic chemist does not take cheap shots from anybody.

Comment by Harold Pierce Jr — 7 Jul 2007 @ 12:51 PM

151. Rod, technically, it’s not a blackbody spectrum as such, but if you place a black body at that point, it will heat up to roughly that temperature.

Comment by ray ladbury — 7 Jul 2007 @ 1:10 PM

152. Harold, OK, so arrogating to yourself an aura of expertise based on your experience in a field related to climate studies by only the slimmest of threads–that’s OK. Calling you on it is a cheap shot. Whatever, dude, just trying to understand the rules of engagement here.

Comment by ray ladbury — 7 Jul 2007 @ 1:50 PM

153. Re: #150 (Harold Pierce Jr)

Your post #120 is so full of the most basic falsehoods that it’s impossible to take your claim of expertise seriously. It seems to me that Ray Ladbury was pretty easy on you.

Comment by tamino — 7 Jul 2007 @ 4:44 PM

154. Re 153: I’m a slave to my generous nature.

Comment by ray ladbury — 7 Jul 2007 @ 5:31 PM

155. tamino (#153) wrote:

Re: #150 (Harold Pierce Jr)

Your post #120 is so full of the most basic falsehoods that it’s impossible to take your claim of expertise seriously. It seems to me that Ray Ladbury was pretty easy on you.

I suppose I should really thank him…

I have been working on a page devoted to climate change fallacies as part of something a bit larger, and the material he wrote looks like good source material! What do you think: should credit him as a source?

Comment by Timothy Chase — 7 Jul 2007 @ 7:51 PM

156. re 151: absolutely true, Ray. […but you can’t seem to admit that electrostatic and magnetic flux (radiation) itself has no temperature…]

Comment by Rod B — 7 Jul 2007 @ 10:24 PM

157. local thermodynamic equilibriumâ��(Abbreviated LTE.) A condition under which matter emits radiation based on its intrinsic properties and its temperature, uninfluenced by the magnitude of any incident radiation.
LTE occurs when the radiant energy absorbed by a molecule is distributed across other molecules by collisions before it is reradiated by emission. LTE is needed for Planck’s law and Kirchhoff’s law to apply, and is typically satisfied at atmospheric pressures higher than about 0.05 mb. Laser radiation is an example of non-LTE emission.

Comment by Eli Rabett — 7 Jul 2007 @ 10:31 PM

158. Eli Rabett (#154) wrote:

LTE occurs when the radiant energy absorbed by a molecule is distributed across other molecules by collisions before it is reradiated by emission. LTE is needed for Planck’s law and Kirchhoff’s law to apply, and is typically satisfied at atmospheric pressures higher than about 0.05 mb.

Thank you, Eli. That answers a number of questions.

Actually, I had done a little reading earlier today, and I ran across the fact that a local thermodynamic equilibrium requires the local kinetic temperature to be equal to the Planckian temperature.

Local Thermodynamic Equilibrium — from Eric Weisstein’s World of Physics
http://scienceworld.wolfram.com/physics/LocalThermodynamicEquilibrium.html

Given what you have just said, a few pieces have started to fall together for me. If the matter corresponding to the kinetic temperature and the radiation corresponding to the Planckian temperature are strongly interacting, in all likelihood we have a local thermodynamic equilibrium.

I had been wondering how moist air convection would enter into this, whether it would result in a violation of the local thermodynamic equilibrium – but given what you have just stated, it would appear to be irrelevant since the local thermodynamic equilibrium would be a very good approximation on a small timescale (which is all that would be required for it to be considered as being applicable within a given environment) and convection would become important only over much greater time scales.

The same link above includes a link to a piece on Kirchoff’s law which argues that it is still “applicable” even under non-equilibrium conditions – if one performs a suitable averaging over all wavelengths. Of course, it is strictly applicable, that is, under LTE, it is applicable at each individual wavelength considered independently of the other wavelengths. Incidently, if I remember correctly, Gavin (?) mentioned at one point that it was by means of treating each wavelength independently of the rest that Kirchoff was first able to derive the law in the first place. This would seem to be a fairly weak derivation – if one were under the assumption that local thermodynamic equilibria were few and far between rather than realizing that they were the rule, not the exception.

Comment by Timothy Chase — 8 Jul 2007 @ 12:29 AM

159. Rod B.,
The concept of temperature in a radiation field is useful–photons interact after all and so have a thermodynamics of their own. BTW, you do realize that it’s possible to have a negative temperature (e.g. a population inversion), and if you do, it corresponds to a higher energy state than a positive temperature. Of course, left to itself, such a negative temperature state will quickly decay to a normal Arrhenius distribution, but you need to remember that all thermodynamic concepts are somewhat subtle–not to mention widely applicable. Do you remember the results from the last run at the Relativistic Heavy Ion Collider at Brookhaven? They were talking about the temperature of a quark-gluon plasma created by slamming uranium atoms together. That surely goes beyond anything Boltamann envisioned.

Comment by ray ladbury — 8 Jul 2007 @ 6:46 AM

160. Re #149: Ray, your comments on thermodynamic equilibrium did help me to re-think about how the atmosphere behaves. I might have found my missing link. This is where I am now:

When we follow the radiative process of the greenhouse effect, we see the Earth radiates infrared radiation, which is absorbed by greenhouse gases. This gets converted into kinetic energy, which means the air is warmer. The kinetic energy in turn excites the greenhouse gas molecules which causes them to emit infrared radiation in all directions. Some of this (ie. 324 watts per square meter) reaches the ground and warms the surface of the Earth.

So the air and the surface are being warmed by this process. But most of it takes place in the lower atmosphere, where there is ten times more water vapor than carbon dioxide, and water vapor absorbs over a wider spectrum. It seems that carbon dioxide can only contribute a few percent to the greenhouse effect.

But Raypierre makes a startling statement: If there was no temperature gradient in the atmosphere there would be no greenhouse effect. Lets construct that world and see what happens, by confining all the greenhouse gases to the lower atmosphere. All the same absorption and re-radiation of longwave energy will take place as it does now. The surface will receive the same 324 W/m2 of infrared radiation. The only difference is that those greenhouse gas molecules that radiate into space will do so at a higher temperature, the same as that at the surface. The higher temperature means more energy is being lost.

Where does that energy come from? It has to come from the surface of the Earth, and get there by convection. That would mean a lot more convection than what we have today. It appears that in this world the greenhouse gases do not lead to any warming, all they do is make the atmosphere a lot more violent. Instead of heat energy we get wind energy.

If we let our greenhouse gases seep higher into the atmosphere, they get cooler and radiate at a lower temperature. Less energy is lost, and there is less convection from the surface to compensate. More heat energy is retained, and the Earth gets warmer.

Relatively more carbon dioxide makes it to the upper atmosphere than water vapor, which condenses at lower temperatures. Therefore carbon dioxide plays a larger role in raising the altitude at which the atmosphere radiates into space. This explains why it appears to punch above its weight, contributing about 20 percent of the greenhouse effect instead of 2 or 3 percent.

Does this make sense?

Comment by Blair Dowden — 8 Jul 2007 @ 9:46 AM

161. Blair (160), a quicky while I digest the rest of your post: If the molecule heats up with IR absorption, why wouldn’t it cool down with later IR emission??

Comment by Rod B — 8 Jul 2007 @ 2:27 PM

162. Blair,
Let’s think of it this way. First how does energy get where it’s going. Well, the surface warms because sunlight hits it. Water evaporates and the surface heats the air. Water vapor and air rise, transporting lots of energy to the mid to upper troposphere–this is the dominant process for transporting energy to levels up to the stratosphere. Energy in the stratosphere gets there mainly from incoming UV sunlight being absorbed by O3 molecules, but there’s a small contribution from outgoing longwave radiation–OLR–(infrared or IR). OK, so how does the upper troposphere and stratosphere get rid of energy. Well, it can go back to the surface as wind or storms, but that keeps the energy in the system. The only way the climate loses energy is by OLR. Now OLR is not a very efficient energy transport mechanism. You only get it two ways: A collision excites a rotational or vibrational mode that relaxes via OLR or an OLR photon from below excites a mode that then decays via OLR. Collisional excitation is not common at atmospheric temperatures (only a few percent of molecules have such energies at these temperatures), but there are still lots and lots of molecules, so it happens. Only half the photons emitted on average will still be outbound. Moreover, if the atmosphere is sufficiently dense, there’s a good chance that the excited molecule will decay collisionally rather than radiatively. So, basically, near the surface, where the atmosphere is dense, there’s virtually no chance that an IR photon in the sensitive band of a greenhouse molecule will escape.
OK. Let’s look at Earth from a satellite. Where are the photons we see coming from. Looking at a spectral range far from any absorption lines, we’re seeing photons coming right from the surface. Now what about a photon in the absorption region of water (too broad to be a “line”)–any chance it came from the lower troposphere? Nope. It would be gobbled up again immediately–it had to come from the cloud tops. above which there’s virtually no water. What about CO2’s absorption lines–well CO2 is present and well mixed into the lower stratosphere at least, so a photon in this band is unlikely to escape before another CO2 molecule gobbles it up. And so the OLR in this band has to be coming from very high up, where the CO2 concentration is feeble enough that the probability of the photon being absorbed is small. But it’s very cold at the top of the troposphere/bottom of the stratosphere, so far fewer molecules will be excited collisionally, and the OLR in the CO2 band will be weak here. So the IR radiation in the CO2 band is an LTE process.
To summarize: The probability of an IR photon from the surface in the absoprtion region of a ghg escaping the atmosphere is virtually nil. It will be absorbed and a lower flux (decreased by 1/2 by the direction of the photon and by nonradiative relaxation) will be emitted, consistent with the temperature of the gas where this takes place. The process will repeat until the density of the greenhouse gas is sufficiently low above that the photon has a chance of escaping without being reabsorbed. Thus the intensity in this band will look like it comes from a source with the temperature of the region where the ghg peters out. Add more CO2 and the region where this density occurs rises and gets cooler so the emission in the band is more feeble. Less radiation overall escapes, so the planet warms until the OLR again equals the incoming radiation.
Does this make sense, Blair?

Comment by ray ladbury — 8 Jul 2007 @ 3:02 PM

163. A quick question. I have been having a hard time finding a good reference that characterizes the CO2 content vs altitude in the stratosphere. Spencer states that CO2 is well mixed well into the stratosphere. I’ve found a couple of references that say that at the troposphere/stratosphere boundary the CO2 content drops by 5-7 ppmv, but I have one guy contending that at the stratosphere boundary the CO2 content drops to 52 ppmv and one contending there is zero CO2 in the stratosphere. The references I’ve found on line are abstracts from rather old Nature articles. Anyone got anything better?

Comment by ray ladbury — 8 Jul 2007 @ 5:20 PM

164. Re #162: On a macro scale, energy is absorbed near the surface, and lost in the upper atmosphere. Convection is the connection between them. Am I right to say that this is the process that drives the convection?

How about this: The greenhouse effect drives convection in the atmosphere by separating the region of heat gain near the surface from the region of heat radiation into space higher in the atmosphere. The lapse rate (fall in temperature with altitude), not primarily caused by the greenhouse effect, is responsible for greenhouse warming, because a colder greenhouse gas radiates less energy. A smaller lapse rate would mean more energy lost from the upper atmosphere, leading to more turbulence and less warming. A higher lapse rate means less energy is lost, thus more warming and less energy differential to drive turbulence.

On a micro level, I want to examine this statement:

Collisional excitation is not common at atmospheric temperatures (only a few percent of molecules have such energies at these temperatures), but there are still lots and lots of molecules, so it happens. Only half the photons emitted on average will still be outbound. Moreover, if the atmosphere is sufficiently dense, there’s a good chance that the excited molecule will decay collisionally rather than radiatively.

The first sentance seems meaningless to me without a time scale. I expect to see something like “5 percent collisional excitation per second at 15 deg. C.”

Are you also saying that in a denser atmosphere there is less collisional excitation? Implying it goes up with temperature but down with pressure?

Comment by Blair Dowden — 9 Jul 2007 @ 6:09 AM

165. To clarify what I said above, the main driver in the atmosphere is the temperature difference between the equator and poles. This is a horizontal force, as opposed to the vertical force caused by the greenhouse effect.

The difference in solar radiation reaching the equator and polar regions is roughly the same as the 235 watts per square meter that need to be transported up into the atmosphere to be dumped into space, but the distance is smaller. I wonder how these forces compare in magnitude?

Comment by Blair Dowden — 9 Jul 2007 @ 6:38 AM

166. [[Blair (160), a quicky while I digest the rest of your post: If the molecule heats up with IR absorption, why wouldn’t it cool down with later IR emission?? ]]

Again, individual molecules don’t have a temperature, groups of molecules do. But extrapolating the question to a whole layer of atmosphere, you can have radiative balance at many different possible temperatures. The more IR a layer is absorbing, the higher the temperature at which it will achieve radiative equilibrium. (Or in the real atmosphere, radiative-convective equilibrium.)

Comment by Barton Paul Levenson — 9 Jul 2007 @ 7:18 AM

167. Barton, I think I have a word problem —- when they cool a single atom, what is it being measured in this work if not temperature? The words make it hard for me to grok. Maybe others are having the same language trouble as well. It’s likely the ‘physics for poets’ problem — using words where only math really explains what’s happening. When they cool a single atom, if it doesn’t have a temperature, what does it have less of after the procedure?

P Maunz, T Puppe, I Schuster, N Syassen, et alâ�¦ – Nature, 2004 – nature.com
… Cavity cooling of a single atom ….. Cooling of a single atom in an optical trap inside a resonator.

Comment by Hank Roberts — 9 Jul 2007 @ 9:14 AM

168. Hello:

I have a climate-change skeptic who has asked me to detail his misconceptions based on his inability to understand why man-made CO2 is the cause of global warming and how warming can be stopped by limiting CO2 emissions. I don’t want to take up or detract from the threads of both Parts I and II, which I have read and will reread to try to understand the arguments, but I’m not an atmospheric physicist/chemist and can’t adequately respond to his questions. I assume my email address can be “seen” by the moderator, and if he/she would be willing to discuss this with me “off-line”, I’d be much obliged. Thanks.

Comment by Taber Allison — 9 Jul 2007 @ 10:48 AM

169. Hank, In a single-atom trap, what they are doing is decreasing the kinetic energy of a single atom to the point where it can be confined with a laser and magnetic field. It’s kinetic energy, not temperature, per se. This is done to look at excitation/relaxation or radioactive decay properties of that single atom.

In a Bose-Einstein condensation experiment, they use similar techniques to cool an assemblage of atoms to the point where their quantum nature as bosons becomes apparent. Here, it is adequateto talk temperature.

Comment by Ray Ladbury — 9 Jul 2007 @ 11:28 AM

170. Thanks Ray; I nitpick at this because I think “cooling” and “warming” make people think of “temperature” — it’s a word problem, that the word doesn’t apply to cooling a single atom, or to what changeswhen a single molecule absorbs or emits an infrared photon, yet when all the kinetic energy that can be removed is removed, we’re still approaching “absolute zero” and, again, “zero” makes people think it’s a temperature.

Just foraging for simpler terms that still explain what’s meant.

Comment by Hank Roberts — 9 Jul 2007 @ 12:40 PM

171. I thought we (well, many of us) had concluded/agreed that an individual gas molecule has kinetic energy and temperature, which is pretty much defined by kinetic energy. Now I’m reading again that single molecules have no temperature. Can we get this figured out? How about a vote, maybe leading to a consensus?

Comment by Rod B — 9 Jul 2007 @ 5:29 PM

172. re 157, 158: If the Planckian and Maxwellian components are in equlibrium, would we not expect that a thermometer dressed in black would read the same as one in white?
Or does “local” mean “narrow region of frequency”?

Comment by Allan Ames — 10 Jul 2007 @ 9:16 AM

173. Rod, “temperature” is a word problem. I think it’s clear that temperature is an average, in macroscopic work.
Microscopic and submicroscopic work is different. Excerpts below from various definitions:

Temperature is a measure of the average kinetic energy of the particles in a sample of matter.
en.wikipedia.org/wiki/Temperature

In thermodynamics, the integrating factor of the differential equation referred to as the first law of thermodynamics.
In statistical mechanics, a measure of translational molecular kinetic energy (with three degrees of freedom).
http://www.novalynx.com/glossary-t.html

… the measurement of how fast the molecules are moving back and forth.
http://www.uwsp.edu/cnr/wcee/keep/Mod1/Unitall/definitions.htm

temperature is what you measure with a thermometer (this is kind of an operational definition).
… the temperature of a system tells how much the internal energy of the system grows upon a given increase of entropy
http://www.maxwellian.demon.co.uk/faq/glossary.html

Temperature refers to the temperature of the ambient air excluding direct heating of the sensor by solar radiation.
http://rps.uvi.edu/WRRI/glossary.html Glossary of Meteorological Terms

Comment by Hank Roberts — 10 Jul 2007 @ 9:31 AM

174. #168 Tabor:

Have a look at this:
http://gristmill.grist.org/skeptics
for clarification on common skeptic views. (RealClimate isn’t well suited to this sort of work ;-) )

Comment by SomeBeans — 10 Jul 2007 @ 10:34 AM

175. I wanted to provide more detail on my previous request. I have a scientific background (Ph. D. in plant ecology) and a research background in paleoecology. So, I understand climate history, but lack training in atmospheric chemistry and physics. I work for a non-profit conservation organization and have been having conversations with a donor about climate change. He is insistent that CO2 does not matter for current warming (he also argues for saturation, which is amply covered here). Here is an excerpt of how he describes the greenhouse effect:

“CO2 is well mixed with the other gases and is present from ground level to high altitudes. No radiation from the earth is reflected back. At altitude the air is colder, so the radiation is absorbed by a cold gas. Gore states that there is enough CO2 to absorb all the radiation in the opaque regions of the CO2 and none escapes – it just warms the CO2. There is also a radiative transfer between the warm earth and the cold gas which warms the gas and cools the earth, but this may be small.”

He cannot follow “Gore’s scheme” – Gore “says that adding CO2 will make the earth’s radiation warm the CO2 at lower altitudes, thus causing a slight warming. This in turn allows the atmosphere to hold more water vapor. This causes major global warming. Here is why I disagree. The earth’s radiation goes equally in all directions, so only the nearly vertical rays are absorbed at altitude, and the lower portion gets the most heat. Adding CO2 should not change this.”

My skeptic goes on to say that “CO2 makes up about .0003 of the atmosphere. Heating this one degree raises the temperature of the atmosphere by .0003 degrees..”

Having read through both parts I and II and read the responses, I have little trouble seeing the errors in his thinking. Take the last paragraph – I have read elsewhere, CO2 contributes app. 25% or more of the warming effect (with a range around that number) – most of the gas in the atmosphere N2 and O2 are not radiative gases and don’t contribute to greenhouse warming. So, the thinking that greenhouse warming is directly and simply proportional to CO2 relative concentration not to mention that the greenhouse effect is a consequence of “warming: CO2 are off base.

He also is wrong in stating that no radiation from the earth is reflected back. As was described in Part I, some of the absorbed IR is re-radiated in all directions. And, his description of Gore’s “scheme” is off.

This thread describes in detail how the greenhouse effect works, and how adding CO2 will lead to more warming, but I wanted to make sure I’m not missing anything, or that my friend is not raising something that hasn’t been covered here.

Thanks.

Comment by Taber Allison — 10 Jul 2007 @ 10:37 AM

176. #174

I’ve checked that out, and provided more detail in my next post, realizing that my post was not specific enough to warrant comment. The grist site does not address the issues that are being discussed here, as far as I can tell, for example, that site has not posted anything on saturation.

Comment by Taber Allison — 10 Jul 2007 @ 10:46 AM

177. Hank, I realize this is starting to sound like an arcane philosophical debate, but I think the concept is important. I didn’t see any of your sources in 173 that prohibited the state of just one molecule. Though they all implied/stated the predominately practical situation of moles of molecules. E.g. “the average kinetic energy of the particles in a sample of matter.”: how large a sample? 1023molecules? 10,000? Two? Hey, how about one? Average kinetic energy comes from a bunch of individual molecules all having their own individual kinetic energy (per Boltzman distribution for gasses). Now I recognize that its pretty hard to get a Boltzman distribution average when you get down to 2 or 3 molecules. But those 2 or 3 (or ONE) still have kinetic energy and hence temperature.

This seems important to me to get my hands around the process of radiation absorption/emission, energy transfer (to what in the sub-molecular or even sub-atomic level), variations of temperature for all of these processes, etc. I need to understand how it works at the molecular/atomic level — and that means ONE. That’s really the only way (for me at least) to understand the physics. Start with one electron, neutron, atom, molecule, photon, unit charge, etc. and go from there. If one molecule doesn’t get hotter, none do. The common accepted usage of the term not withstanding.

Comment by Rod B — 10 Jul 2007 @ 11:59 AM

178. Taber, if you can find out your donor’s source for his beliefs, it may be helpful. I tried some searches using phrases from your description and didn’t get a hit. In general it’s a familiar denial statement, but knowing more particularly if he’s relying on a PR site or a political source or one of the sites that claims to be science info may help clarify what he’s thinking and why.

Comment by Hank Roberts — 10 Jul 2007 @ 12:36 PM

179. Re #163

Ray,

This is more recent than your old Nature articles. It hasn’t been published yet! Figure 3 shows the vertical (and latitudinal) distribution of CO2.

Atmos. Chem. Phys. Discuss., 7, 9973-10017, 2007
http://www.atmos-chem-phys-discuss.net/7/9973/2007/
MIPAS reference atmospheres and comparisons to V4.61/V4.62 MIPAS level 2 geophysical data sets
J. J. Remedios1, R. J. Leigh1, A. M. Waterfall2, D. P. Moore1, H. Sembhi1, I. Parkes1, J. Greenhough1, M.P. Chipperfield3, and D. Hauglustaine
http://www.atmos-chem-phys-discuss.net/7/9973/2007/acpd-7-9973-2007.html

I think that the sudden drop in concentration above 80 km is due to the CO2 molecule being heavier than other air molecules. See http://www.atmos-chem-phys-discuss.net/7/9973/2007/acpd-7-9973-2007.pdf

[Response: I don’t think so – I think it is related to the photolytic destruction of CO2 in the mesosphere. (i.e. http://www.cosis.net/abstracts/EGU04/05569/EGU04-J-05569.pdf ) – gavin]

Comment by Alastair McDonald — 10 Jul 2007 @ 1:59 PM

180. Touche Gavin!

But over 70 km is in the ignorosphere, where the atmosphere begins to become non-uniform, with the scale heights of chemical species differing by their molecular weights. See http://en.wikipedia.org/wiki/Mesosphere

Comment by Alastair McDonald — 10 Jul 2007 @ 4:38 PM

181. Rod, some of the cites define temperature as an “average” — can you get an average temperature for a single molecule? Perhaps with multiple measurements.

Other cites define temperature in terms of the solution to the thermodynamic equation, the sum of kinetic energy.

This is a poetry question, not a physics argument:
http://www.newscientist.com/blog/shortsharpscience/

Tuesday, July 10, 2007 The poetry of science
What do poets and scientists have in common?

“… the need to communicate new and unfamiliar concepts â�� for which appropriate words and metaphors may not already exist â�� without resorting to lazy clichÃ©s that may distort the meaning….so much science now happens at a level we canâ��t see, so science depends more and more on metaphor and analogy to communicate itself.”

Clearly when people above say a molecule doesn’t have a temperature they mean they can’t calculate an average with one item. Clearly other writers use the word temperature to mean the kinetic energy in the one item.

Recall the 11th Commandment: “You do too know what I mean!” Doesn’t it apply to this discussion?

Comment by Hank Roberts — 10 Jul 2007 @ 4:59 PM

182. Taber Allison — I’m probably more of an amateur at this than you, but to me it appears that you have the essentials, and haven’t missed anything important for your purposes.

Comment by David B. Benson — 10 Jul 2007 @ 5:21 PM

183. Hank, it’s more fundamental than that. Remember the precise physical definition of temperature–partial derivative of energy wrt entropy (holding # of particles and volume constant). You can’t define the entropy of a single particle, so the definition breaks down when you have a single particle. There is more to temperature than kinetic energy.

Comment by ray ladbury — 10 Jul 2007 @ 7:06 PM

184. Re #175: Taber, most of the statements made by your skeptic are wrong, including some of those attributed to Gore. But there is one main misconception behind it all. The simplest model used to explain the greenhouse effect is that it acts like a blanket to keep the Earth warm. And that is fine if you are willing to take the scientist’s word for it and have no inclination to explore further.

But as soon as you look at the actual composition of the atmosphere the blanket model falls apart. As I am sure you realize from reading here, it is a lot more complicated than that. Your head is probably spinning from all the detail. Mine is.

Basically, the greenhouse does not work simply by absorbing heat and warming the air. If it did, your skeptic would be right that carbon dioxide can only play minor role and doubling it would not make much difference.

Instead, it works by changing the temperature at which the Earth loses energy into space. So while a great number of greenhouse gas molecules may absorb and re-radiate energy, the one that matters is the one that radiates it into space. That is more likely to happen in the upper atmosphere, where it is colder. The lower the temperature, the less energy is lost, and the more the Earth warms.

Cold air can hold less water vapor, so it decreases with altitude. So while on average there is ten times as much water vapor, there is relatively a lot more carbon dioxide in the upper atmosphere where it matters.

There are probably gaps in this explanation that you do not quite follow. If so, then ask. That is what this forum is about. I am still having trouble connecting the heat gain at the surface and lower atmosphere with the heat loss in the upper atmosphere. I am sure the answer is convection, but the details are not clear.

Comment by Blair Dowden — 10 Jul 2007 @ 7:42 PM

185. Alastair (and Gavin), thanks for the references–just what I needed. They guy on the other end of this debate is using science that is about 50 years out of data–i’t like he spent the last 50 years in a coma. Damn, I forgot to ask him if he knew who Milli Vanilli (sp) was. ;-)

Comment by ray ladbury — 10 Jul 2007 @ 8:39 PM

186. This horse is still kicking. It’s not obvious to me that I can’t get the average of one thing, unless I’m constrained by an arbitrary (for good reason maybe) construct of the definition. One house sold in all of Podunk County last year; what was the average sale price of houses in Podunk last year? None??? I’ll have to think about it.

So I can’t have temp with one molecule, but I can with TWO??? What if I start with one (no temp) and then add another from a distant unconnected place which had/has no temp (’cause it was/is just one), do I now have two molecules with no temp? Gets me back to the earlier post — if I have just one molecule and it has no temperature, does it have no kinetic energy? And if it crashes elastically into a wall, no energy/temp is transferred??? I certainly understand the usefulness of the “average” concept. But taken to the extreme of “one molecule can not have temp” creates absurdity after absurdity. And I want to know exactly what happens, say, when one photon is absorbed by one CO2 molecule. Does its energy and temperature not increase (from nothing?!?!) if its up there by itself? Does it increase if other molecules are in the vicinity (BTW, how near do the others have to be to count??)? It’s starting to sound a lot like the magical enigma of particle/quantum physics…

Comment by Rod B — 10 Jul 2007 @ 9:41 PM

187. This is why the Buddha said the first thing is the rectification of names (wry grin).
“Hotter” — can mean for one molecule, an increase in energy? But “hotter” not in temperature beacuse for temperature you need more than one molecule; “hotter” as in, er, livelier, more energetic, more spread out, whatever. Stirred up.

Two molecules, the one that got stirred up can pass some of that excitedness by bumping it, or by passing a photon from the one to the other. Have we got “temperature” with two molecules?

I’m banging on the words because Rod’s having trouble seeing how one molecule can absorb a photon, get more excited, then bump into another molecule and pass th at excitation on. Let’s try it for the fifth grader level and see if there are words that can be helpful.

We know it _happens_ so we know it’s real, but real means mathematics; finding words is up to us.

Comment by Hank Roberts — 10 Jul 2007 @ 9:56 PM

188. Let me introduce the idea of an ensemble. Take your one molecule. Put it in a box (this turns out to be important if you actually do the exercise) whose walls are at temperature T. Now imagine you have an infinite number of such boxes, so that the molecule is moving at all possible velocities subject to the constraint that the walls are at temperature T. You will find that the translational, vibrational and rotational average energy distributions are specified by the temperature T.

Comment by Eli Rabett — 10 Jul 2007 @ 10:38 PM

189. and remember: http://lemoncustard.com/media/rf8.gif

Comment by Hank Roberts — 11 Jul 2007 @ 12:35 AM

190. Rod B (#186) wrote:

But taken to the extreme of “one molecule can not have temp” creates absurdity after absurdity

For me the idea of a single molecule having a temperature is sounds like that Zen Koan question of,

“What is the sound of one hand clapping?”

I think for pretty much the same reason. When you are talking about temperature, you are speaking of a distribution of velocities, but if there is only one molecule, there can be no distribution or spread.

It makes about as much sense as the fact that you can’t have someone be a parent without their being the parent of someone. That with only one straight line, you can’t have something which is crooked. That to make a right angle with straight lines, you need more than one straight line. Temperature seems to be a quality when you first think of it, at least with human-scale objects, and structure might seem to be the same sort of thing, at least when one considers a single large object. But the structure of a thing consists of the relationships which exist among its parts, their positions and how they are connected.

Consider me with my four limbs and counting my neck and head as a fifth. Now rotate each in the same way so that my left arm is where my left leg used to be, and my left leg is where my right leg used to be. Now that is something that seems especially absurd.

The relationships matter. But to have a relationship requires two or more things.

“Distribution” sounds like a spatial concept, and that is roughly how they handle the “velocity distribution” in “phase space.”

Comment by Timothy Chase — 11 Jul 2007 @ 12:49 AM

191. #177 There’s no doubt that Boltzmann defined his thermodynamic laws on the basis of statistical mechanics, as was Einstein’s study of Brownian motion in 1905. If you start looking at the internal energies of molecules, you need to look at how energy is bound up in the bonds between atoms, if you start looking at individual atoms you need to start looking at spin and quantum electron states. Sooner or later, you’ll end up with the Standard model of QCD and perhaps have to get into debates about String theory!
However, I think this is all going in the wrong direction.
The point is that earth can only receive and lose heat via electromagnetic radiation at its boundary layer. The “messenger” is always the photon.
So it’s more important to look at the top of atmosphere radiation balance than go delving into the debates in fundamental high energy physics.

Comment by Alex Nichols — 11 Jul 2007 @ 2:40 AM

192. Re #187

Actually a molecule has four temperatures: kinetic, electronic, vibrational, and rotational. The average of the kinetic temperatures of all the molecules in a volume of gas is the Maxwellian temperature of the gas, i.e. its temperature measured with a thermometer. According to the Equipartition Theorem, when the gas is in thermodynamic equilibrium all four average temperatures will be the same. In other words the average kinetic, electronic, vibrational, and rotational temperatures will all be equal. But any two molecules in the gas will have different kinetic temperatures, and their other temperature will not necessarily be equal to their kinetic temperatures.

For instance, if a CO2 molecule absorbs a photon, then its vibrational temperature will increase, but its kinetic temperature will initially remain unchanged. If it is relaxed by a collision with an air molecule, then the air molecule will have a higher kinetic temperature, and so the air will have a higher Maxwellian temperature. In other words, the air warms (but not by much!)

Comment by Alastair McDonald — 11 Jul 2007 @ 4:24 AM

193. [[My skeptic goes on to say that “CO2 makes up about .0003 of the atmosphere. Heating this one degree raises the temperature of the atmosphere by .0003 degrees..”]]

The increased energy of the greenhouse gas is almost immediately distributed throughout local air by collisions. You don’t have CO2 heating up by 1 degree and everything else staying the same temperature. The whole atmosphere heats up.

Comment by Barton Paul Levenson — 11 Jul 2007 @ 7:22 AM

194. Rod B.,
First, the concept of an average doesn’t make sense if you are talking about averaging over a sample size of one. There, you are dealing with exact measurements. For a sample size of 2, you can come up with a meaningful average, but the concept of the average is not very useful, since two states may have the same average properties, but very different energy distributions for the two molecules. And so on. Remember in statistics class, the place where you could start approximating various distributions (Poisson, binomial…) as Normal was sample size of ~30. The larger the ensemble the harder it becomes to specify the system in terms of the energies of the indivicual molecules, but the more meaningful the average properties start to be.
Second thing to keep in mind. The concept of temperature dates from the very early history of thermodynamics–when heat was considered a kind of fluid. Temperature was the gradient that said which direction heat flowed. As thermodynamics became more formal, the temperature came to mean something very specific in the relation between energy and entropy–specifically it was the partial derivative (or slope) in the change in energy for a change in entropy (other variables held constant). The relationship between average energy (kinetic, potential, etc.) and temperature comes out of Maxwell’s (or more generally the Arrhenius) distribution, which relates temperature and average energy. Also, keep in mind that it’s not just kinetic energy, we’re talking about here–that only holds for billiard-ball molecules with no internal degrees of freedom. And don’t forget that tempearature applies to solids as well, where it relates to the energy levels of the oscillations between atoms in the solid (and that’s elastic potential and kinetic, not just kinetic).
So what you are dealing with is a concept that has 3 different meanings in 3 different models, all of which are related but just a little different.
Eli is right–you can think of things in terms of ensembles if that helps, but it’s the box-molecule system that really has the temperature in that case. When speaking of a single particle–or even 2 particles, it just makes much more sense to speak in terms of the indivudual energies of the two particles.

Comment by Ray Ladbury — 11 Jul 2007 @ 7:49 AM

195. So there’s a common problem in explanation here —– a CO2 molecule can absorb an infrared photon, then pass the energy along to other gases, warming the atmosphere. Even though there are relatively few, they are transferring the energy.

This latter argument seems like “the straw holds only an ounce of water, there’s no way you could drink a quart of water through it — or pump a hundred gallons through it.”

Comment by Hank Roberts — 11 Jul 2007 @ 9:32 AM

196. I am a retired research veterinarian,ignorant of physics, and hope that I may receive a little help from correspondents to this site. To facilitate this, I would like to put the following proposition which others may rubbish at will, thus providing illumination to me:

For a stable temperature balance at ground level, it is generally argued that influent energy (from the sun) should equal effluent energy (by convection/conduction and by infra red radiation). This appears to ignore planetary sequestration of energy and its subsequent release into the atmosphere. Until recent times, some surplus energy was being locked away “in the cellar”(fossil fuels). We’ve now found the key and are gorging on the goodies we’ve discovered. We have developed technology which has enabled a massive population increase and has been accompanied by unprecedended release of heat into the atmosphere. Would this alone not be sufficient to destabilise the climate, leaving aside possible indirect greehouse effects?

I find it hard to believe that this possibility has not been previously considered. It may be so facile that it was dismissed years ago, explaining my not having heard anything of the possibility in recent debates on the subject. In any event, I would appreciate enlightenment.

If there is any merit at all in my hypothesis, the direct effect on climate may or may not summate with the indirect greenhouse effects of CO2. (I remain to be convinced of the magnitude of the latter effects despite having been impressed with Raypierre’s arguments on absence of waveband saturation in qualititave terms.)

Of course, in the unlikely event that I have made a valid case, replacement of fossil fuel with a non carbon source of energy (such as nuclear)in equal amount wouldn’t have a mitigating effect on climate change.

Comment by Douglas Wise — 11 Jul 2007 @ 9:37 AM

197. Douglas, Welcome. Earth recieves 1.740Ã�10^17 W from the Sun–that’s 24/7/365.25. Compare that to global energy consumption of 1.5 x 10^13 W, and you can see we’ve still got quite a lot of catching up to do. Your basic point, though is valid–energy consumption cannot increase exponentially indefinitely.

Comment by Ray Ladbury — 11 Jul 2007 @ 10:13 AM

198. Mr. Wise, that’s been considered. The total heat produced by all human energy use (let alone the total energy stored in fossil fuels over the longer term) is extremely tiny compared to the total energy reaching the planet from the Sun. Photosynthesis just doesn’t store sunshine all that efficiently!

“… Superimposing the steeply climbing curve for fossil fuel upon the gradually rising curve for biomass energy, we see an increase from roughly 960 million metric tons annual oil-equivalent consumed in 1900 to roughly 10,600 million in the late 1990s. This constitutes approximately an eleven-fold increase in overall human energy use during the 20th century.
“…. In 1750, humankind consumed roughly 260 million oil-equivalent tons of energy. By 1900, the annual total had grown to about 900 million. This amounts to something less than a 4 fold increase in 150 years. The corresponding increase for
the 100 years between 1900 and 2000 was three times greater. If we were to graph this increase on the millennium scale employed in Figure 6.1, the line of increase would show an almost vertical rise.”

It’s a lot, but not compared to sunshine: “… in 1970 the total energy consumed in the U.S. was equal to the energy of sunlight received by only 0.15% of the land area of the continental U.S.” http://adsabs.harvard.edu/abs/1974STIA…7512425W

Comment by Hank Roberts — 11 Jul 2007 @ 10:23 AM

199. #196 Doug Wise

I’m not certain of the answer to your question but I suspect that the answer is that the heat directly realised by humanity is small compared to the heat arriving from the sun but the effect of the carbon dioxide realised is more significant.

The appropriate analogy is probably along the lines that burning your blanket gets you a small amount of heat in the short term but keeping it in an uncombusted form will keep you warm night after night!

Comment by SomeBeans — 11 Jul 2007 @ 11:00 AM

200. #196 We consume around 13 terawatts of energy per year, but the amount of solar energy incident on the planet is estimated at around 600 terawatts.
giving a ratio of roughly 2.16%.

But it gets worse, because the 13 terawatts includes hydroelectric and nuclear power production.
So human thermal energy production is probably less than 1% of incident solar radiation.

The other way of looking at it is, what was the climate like when C02 levels were much higher e.g. in the early Carboniferous before sequestration of the coal seams?

Pretty damn hot with average temperatures of around 22C compared to about 12 C today.

Later on in the middle carboniferous, conditions were not unlike today and the CO2 levels were also similar.
Cool geological periods with high C02 levels coincide with continental drift and super continents allowing ice cap formation.

Comment by Alex Nichols — 11 Jul 2007 @ 11:49 AM

201. Re #196: Doug, a unit of fossil fuel is only burned once, but the carbon dioxide produced remains in the atmosphere for tens or hundreds of years contributing the to greenhouse effect. The direct heat from combustion is negligible, so it is ignored.

Comment by Blair Dowden — 11 Jul 2007 @ 11:53 AM

202. Something I’ve been wondering about, and this seemed as good a place as

I’ve heard it claimed a few times (not by particularly authoritative
sources) that because CO2 higher in the atmosphere has a greater effect
on greenhouse warming (true?), emissions from aircraft have more of an
effect (per molecule) than emissions from cars, etc. Other than in the
very short term, is this true, or does the CO2 get mixed fast enough that
the difference is negligible?

Comment by Jeremy — 11 Jul 2007 @ 4:21 PM

203. Jeremy,
It’s been looked at, although the concern is not really CO2 which stay’s pretty well mixed well into the stratosphere (See Alastair’s link in response (#179) to my query above), but soot, other aerosols, water vapor, uncombusted fuel and so on could be a concern.

Comment by ray ladbury — 11 Jul 2007 @ 5:31 PM

204. Many thanks to those of you who so clearly and courteously addressed the concern I raised in posting 196 – one less thing for me to be sceptical about! Before raising another, I would like to mention in passing that the two graphs in the first reference suggested to me by Hank Roberts (198)would be sufficient in themselves to convince most sceptics that we constitute a threat to the planet, even if not through the medium of CO2.

Back to the CO2 story. Can any of you take the trouble to put me right over a few other matters? I have had bad experiences of models in my own field. Perhaps climate models are more robust – they’re certainly more sophistated and complex. Nevertheless, I note that one of your lead authors has also expressed misgivings on the subject (see “The Lure of Solar Forcing”) so, perhaps, I may be allowed to question the basis for assumptions in the greenhouse gas models? I will raise a few questions which have largely been provoked by Raypierre’s lead article to this Section and a graph in Wikipedia headed, “Radiation Transmitted by the Atmosphere” in a Section entitled, “Greenhouse Gas”.

1) It is sometimes stated that, without greenhouse gasses, we’d be 33 deg C colder (all to do with equilibrium, blackbody radiation and an albedo of 0.3 as I vaguely understand it). However, why does the Royal Society suggest that we’d only be 20-30 deg C colder and why the broad range? It doesn’t suggest a great deal of certainty with the present situation and thus doesn’t inspire a huge amount of confidence in predictions of future scenarios.
2) My concern primarily rests with the subject of wavelength saturation. Clearly, reducing atmospheric CO2 would make us colder. However , will increasing it make us hotter and, if so, by how much? I am not sure whether the answers solely emanate from modellers and, if they do, whether it would be possible to test their findings by direct experimentation.
3) It is my understanding that 60% of heat escaping the earth does so by convection to the top of the atmosphere (wherever that is – top of clouds, top of stratosphere, troposphere/stratosphere interface?)where there is conversion to OLR and escape to space. This process is essentially unaffected by greenhouse gasses. The remaining 40% of escaping heat leaves the earth’s surface as IRR but only 30% of this 40% (12%) ever gets out, the rest being absorbed by greenhouse gasses. Is my understanding correct even if naively expressed, partly correct or hopelessly wrong?
4) The Wikipedia graph shows practically all escaping IR being routed through a waveband frequency of 8.5 to 15 microns (assertion based on “eyeballing” the data only). Below 8.5, water vapour alone is sufficient to absorb all IR. Above about 15 microns, overlapping absorption bands of CO2 and water vapour prevent escape. Is this correct?
5) The Hitran spectroscopic archive relating to CO2 was used to generate the graph shown by Raypierre in the lead article to this Section. A second graph was generated to show what would happen if atmospheric CO2 quadrupled. This translated to about 10% extra absorption – about 3% at 13 microns and 7% at 17 microns.
6) My problem is that I don’t know whether the Hitran data already incorporate the effects of water vapour over the 10-22 micron range shown in the graph because I don’t know how the graph was generated. If they don’t, then the 7% potential extra absorption by CO2 around the 17 micron wavelength will be irrelevant except in very dry air while the 3% around the 13 micron area will, in reality, be much less, given water vapour. My judgement here is based on the Wikipedia graph.
7) This leads me to the conclusion that, if CO2 and water vapour at current levels are considered together, addition of more atmospheric CO2 will only provide very few of Raypierre’s extra red M and Ms. In other words, it is possible that the upside of potential warming is much more constrained than has been suggested.
8) Because I have no access to numbers (not bright enough to use them if I had) and no knowledge of the assumptions of climate models, I’m probably being ridiculously naive. However, I hope you can understand the logic behind my current scepticsm.
9) Raypierre invokes a virtual female experimenter (why female – is he a romantic as his name might suggest or merely being politically correct) with a tube of CO2 and a blazing torch of IR in the 10-22 micron bandwidth. It seems a pity that she couldn’t have had a torch blasting out a true blackbody spectrum and a tube with both CO2 and water vapour in it. Better still, would it be possible actually to carry out a genuine experiment (experimenter of either sex acceptable) to allow a test of model assumptions with respect to increasing CO2 against a background of an otherwise fixed current atmosphere?
10) If surface temperatures rise, won’t increasing convection rate act as a negative feedback?

If any of you out there have the patience to give me one more push, my scepticism may finally fall away and I’ll be able to concentrate on being totally depressed by contemplating the consequences of your projections!

Comment by Douglas Wise — 13 Jul 2007 @ 1:47 AM

205. Can’t answer all of your questions, but I’ll try the first two:

[[1) It is sometimes stated that, without greenhouse gasses, we’d be 33 deg C colder (all to do with equilibrium, blackbody radiation and an albedo of 0.3 as I vaguely understand it). However, why does the Royal Society suggest that we’d only be 20-30 deg C colder and why the broad range? It doesn’t suggest a great deal of certainty with the present situation and thus doesn’t inspire a huge amount of confidence in predictions of future scenarios.]]

The 255 K figure for Earth’s equilibrium temperature is based on its current albedo of about 0.3. The uncertain figures are probably based on the fact that we don’t know what Earth’s albedo would be without the atmosphere. The current albedo is mostly from clouds.

[[2) My concern primarily rests with the subject of wavelength saturation. Clearly, reducing atmospheric CO2 would make us colder. However , will increasing it make us hotter and, if so, by how much? I am not sure whether the answers solely emanate from modellers and, if they do, whether it would be possible to test their findings by direct experimentation.]]

Radiative forcing by CO2 (at least for concentrations similar to those at present) appears to be logarithmic:

RF = 5.35 ln (C/C0)

where RF is in watts per square meter and concentration C and original concentration C0 are in parts per million by volume (ppmv). For doubling CO2, this gives RF = 3.7 watts per square meter. For a climate sensitivity of 0.75 K per watt per square meter, this would bring about a 2.8 K temperature increase.

Comment by Barton Paul Levenson — 14 Jul 2007 @ 6:54 AM

206. Douglas, Barton does a good job on the treatment of CO2 forcing. I thought I’d chime in and correct a few misimpressions. First, convection is indeed the dominant mechanism for transporting heat–in the troposphere. However, it does not remove energy from the climate system–the energy is still in the atmosphere. The only way you lose energy is via longwave radiation that exits the atmosphere entirely (outgoing longwave radiation or OLR). Water vapor is important for many reasons. In the troposphere it is the dominant greenhouse gas, but it also drives much of the energy transport, since the heat of condensation puts a lot of energy high in the troposphere. Very high in the troposphere and into the lower stratosphere, the situation is very different. Here there is very little water vapor and much less convection (especially in the stratosphere–as the name implies).
So the question is how does the energy get from the upper troposphere where convection manages to transport it to space–and it’s here that adding CO2 makes a difference. In the upper troposphere and stratosphere, where most of the water vapor has condensed out of the atmosphere, CO2 is the dominant greenhouse gas. This is where the second blanket analogy comes from. Increase the weight of the blanket, and you are going to get warmer, because more energy will be re-radiated back into the troposphere. Does this help?

Comment by ray ladbury — 14 Jul 2007 @ 8:14 AM

207. Consider that heat escaping into space has to have some form, and we call that infrared photons.

With zero greenhouse gases, the infrared photons produced when sunshine heats the surface of the planet get radiated out into space (see the link I posted earlier on measuring the temperature of the Moon during an eclipse, sorting out the heat from the surface regolith vs. the heat from the underlying rock, noting the measurement was done through the upper part of Earth’s atmosphere so correcting for water vapor here.)

With some greenhouse gases, the infrared photons coming from the heated planet get involved in the atmosphere, the heat moves among various modes, most of which can be transferred to other gases by sideswiping and bouncing off them (and no, I haven’t a clue about the quantum mechanics of those interactions, but I’m sure I’m using poetry here not math — since we’re talking about atoms and small molecules, we’re talking probabilities not freeway or billiards interactions).

The other processes make gases move in bulk; weather.

At the top of the troposphere, the bulk motion from local heating quits rising in the temperate zones; in the tropics, and over very large volcanos, the atmosphere can billow up into the cold dry stratosphere; aircraft and rockets can deposit water up there as well. Note the attention given to the increase in number and location of “polar stratospheric clouds” which are being seen further away from the poles. We’ve changed that part of the atmosphere, not for the better.

At the top of the _stratosphere_ though, we’re faced with the question of how the heat energy escapes into space.

It can’t escape very much by conduction; nothing to touch. It can’t escape much by convection; nowhere to go, except for the very small light atoms. Hydrogen can get off the planet carrying some heat as velocity, but it’s trivial.

What can escape? Infrared photons.

Where do they come from? Some made it all the way from the ground, which “glows” in the infrared range. We see satellite infrared photos, and we can see both the water vapor and the ground as brighter where they’re hotter.

Some come from water vapor emitting infrared — photons that were emitted from where the water vapor is, almost all in the troposphere, and that happened to be headed upward and continued without interacting and left the planet.

Some come from CO2, which is well mixed in the lower atmosphere and which doesn’t freeze out so isn’t removed from the stratosphere like water vapor is. Most of those infrared photons are energy that has been working its way up through the atmosphere. Remember the atmosphere is always glowing in the infrared band —- curse of the infrared astronomers, the sky is _bright_ day and night for them.

I don’t know how long it takes an infrared photon on average to go from the ground to space; but the physics predicted that when the ground got warmer as CO2 increased, the stratosphere would get colder — the lower atmosphere warms up, the gases expand, the whole atmosphere gets a bit bigger, pushing the top of it a bit higher, and as it gets higher it expands and cools.

At the top, that cooler CO2 is getting bumped and jostled by other molecules, and it’s taking it a bit more interaction to accumulate enough energy that it can cough out an infrared photon. Which can go up, down, or sideways …..

Comment by Hank Roberts — 14 Jul 2007 @ 9:14 AM

208. Last questions of a skeptic …
I’ve been following this website for six months, learning about climate change. I also bought several books on Amazon by Houghton and Hartmann and have checked out the AIP website. I have learned a lot, and every objection I had has been answered.

These two articles, which get to the basic spectroscopic underpinnings of the model, are excellent. I wish they had been out there when I got first got motivated to see what all the fuss was about. Most of what is available is either way too dumbed down (which only raises educated non-specialist’s skepticism) or are the original papers, themselves (which are too expensive, and too terse to understand) . Writing an article at a level to explain things in a technically correct way is very difficult without writing and entire textbook. (Yes I have downloaded Raypierre’s book and am working through it). The authors have done a great service. Thanks.

However, I still have a few remaining questions. As I understand things, the best analogy of what is happening with CO2 absorption is it is like a Venetian blind slowly closing. Yes, the line centers are totally opaque, pushing the height where radiation at those frequencies can escape so high and cold they don’t contribute much at all to the total radiation budget. This just puts more burden on th frequencies in between, in the skirts of the lines, which much radiate more power, thus requiring a higher near surface temperature.

My questions, then, are:

1) How accurate is the model of a Lorentzian for a collisoned broadened lineshape far down in the wings?
What are the basic asumptions? Yes it works for the width at half-maximum (or 1/e^2 width), but what about way way down, far from line center, in the case of very high optical density?

2) Have any direct measurements been down in the far wings on these lines (since Koch 100 yrs ago)? The spectral plots I see in these articles (and in Houghton’s book) are calculated extrapolations of fits based on two parameters: full width at half maximum (FWHM) and oscillator strength.

3) I expect the short answer to 2)is : yes, the HITRAN database. But what is the HITRAN database for? Was is designed and measured to be used for chemical identification? If so, center wavelength (CW), line strength, and a FWHM may be OK to identifiy peaks. But a full spectrum for broadband radiation models may need more measurements at widely varying concentrations to see the actual shape of the wings. Far in the wings, details of the statistics of the phase interruptions which cause the spectal broadening may be important. What if the actual absoption differs from the Lorentzian by 20%, (or 50%)? What would that mean for the temperature predictions? I was a little surprised that one author seems not to know much about the origins of HITRAN. Is this uncertainty widespread among the climate modelling community? (Hey, I realize there is an awful lot of stuff to know …)

I fully expect to find out this is a non-issue
(as all my other ideas have turned out to be!).
I probably expect that the phase kicks from collisons happen over a timescale on the order of the center frequency, and therefore the deviations would appear so far fom line center whee the absorption is so ridiculously low it is not in the picture. The phase modulations from passing molecules may be slower, though. I have to ask these questions because the consequenes for reducing CO2 by the necessary 70% (!) are very serious. Before people start writing checks with their mouths about what emission caps should be, they better be sure they can pay up when the time comes. And that we are very very sure of the physics that we CAN check.

Thanks for any thoughts on this stuff.

Comment by Dave Dougherty — 14 Jul 2007 @ 5:58 PM

209. Re #208

Dave,

The wings of the lines are still being investigated. In April I was talking to a scientist who was working on the effects of the mixing ratios from the wings of two different absorbing gases.

You can find the story of HITRAN here at http://in-cites.com/papers/LaurenceRothman.html but note that he says that water vapour could block out the infrared radiation from a jet engine through absorption. Just think how quickly the weaker emissions of infra-red from Earth’s surface are blocked out by the thicker lower atmosphere.

In the 15 micron band cited by Ray Pierrehumbert, the optical dense lines do not get more than 30 m above the surface of the Earth. They certainly do not reach the top of the atmosphere, by definition. In that band, pretty well the only radiation that reaches the TOA is in the gaps between the lines where no absorption takes place. When you average the gaps and lines you get the band absorption. But the absorption mainly happens near the surface. It does not cascade up the atmosphere driven by falling temperatures. For CO2, Planck’s function is irrelevant, as is re-emission to the surface. Only water vapour and clouds acts in that way. But the rate of evaporation, and so the concentration of water vapour, depends amongst other things, on the concentration of CO2, and so the gross greenhouse effect does depend on CO2.

Because CO2 only acts near the surface it also affects the ice cover. The latest high levels of CO2 have led to the melting of the Arctic sea ice. See how the ice melt is almost a month earlier than previous years at http://www.abmcdonald.freeserve.co.uk/north.htm

When the ice has gone, the CO2 will act on the sea water and the global climate will return to the balmy days of the Eemian or even perhaps the Eocene.

Comment by Alastair McDonald — 15 Jul 2007 @ 5:25 PM

210. [[In the 15 micron band cited by Ray Pierrehumbert, the optical dense lines do not get more than 30 m above the surface of the Earth. They certainly do not reach the top of the atmosphere, by definition.]]

Gee, and I get 7,800 meters. Are we talking about the same thing? If you’re talking about exactly the line at 14.99 microns, I think you’ll find that very little IR energy from the Earth, or at least a very tiny fraction of it, is emitted between 14.98 and 15.00 microns.

If you use the band from 13 to 17 microns, you’ve got lines all through there, and a certain absorption coefficient, and the 99% depression only occurs at 7.8 km. Not 30 meters.

Comment by Barton Paul Levenson — 15 Jul 2007 @ 8:21 PM

211. Alistair,

I agree about the extremly high optical densities near the line centers. That is what makes the wing shape so important. If it is just being investigated now, it seems to me that this whole field is way way out on a limb. If the lineshape were a square box function, with infinitely steep sidewalls, they GW would have the same saturation dependence as you get for a smeared out gray body model. (Which is wrong I have been told.)

I need to find the papers for the radiatve forcing vs. CO2 concentration expressions in the IPCC report. Several different complicated expressions by different authors were included. If you plot them, they are not so different. Though they are functions of log(CO2 conc.), when you plot them over the range of doubling CO2, you are not way off from a simple linear increase. The reason for this dependence, I thought, had to do with the shape of the wings, which give a much softer (but more pernicious since it can go up and up) saturation (with concentration) behavior to the forcing. I have a suspicion that even if I pay \$100 for them, I still won’t get a straight answer. I’d bet they never checked whether the HITRAN linshapes represent reality at all. They just cranked away on their *line by line* codes on their computers.

Raypierre’s plot of the transmission for broadband light vs propagation distance is essentially the same phenomenon. Note that he is just calculating the trsnmission assuming the lines are lorentzian.

I have found the website for the HITRAN:
http://cfa-www.harvard.edu/hitran

In the original 1973 paper, the authors directly state that *the precise lineshape is a matter of some uncertainty … but it customary to start from a Lorentz shape. … All lineshapes in HITRAN are assumed to be Lorentzian* Later it states that some paper question the validity of the 1/(v-v0)^2 dependence in the far wings. I will try to find these old (1964 and 1969) papers I guess.

The whole thing is just a table of fits for line center wavelength, FWHM, and oscillator strength. Reference was made to fits using a third order polynomial. That would definitely be a fit just to the top of the lineshape. They 2004 paper on the site references this paper by Rothman as the latest word on the lineshapes:

L. S. Rothman, R. L. Hawkins, R. B. Wattson, and R. R Gamache, Energy levels, intensities, and
linewidths of atmospheric carbon dioxide bands, JQSR 48, 537-566 (1992).

I would pay for it, but I can’t seem to get access to pay the \$30 from Science Direct. ( I is such a scam that taxpayer funded research is handed over for a song to these academic publishing houses. This stuff should be free … dammit.)

What we need to see is the raw spectral data for a few of the lines, and see how they fit it in the wings. I really think this database is for applications like chemical identification or order of magnitude visibility estimates of targets (missile plumes) through the atmosphere.

This really bugs me. This stuff should be measured and re-measured with the latest modern detectors and sources (quantum cascade lasers etc.) Relying on data from the 1970s is ludicrous for such an important and subtle problem CO2 induced GW seems to be. This is the only part of the whole GW problem that can be measured with decent accuracy! (The CO2 conc. is the only other one.) It sound to me like the modelers just took this database on faith and ran much too far with it. I have seen no references to the need for further experimental determination of detailed lineshapes anywhere.

Even if Lorentzian fits turn out to be OK, it will be like being right the same way a stopped clock gives the correct time twice a day. Not very confidence inspiring about the underpinings of the rest of the GW story.

If I have to stomach being lectured by Cameron Diaz and Ed Begley Jr. on TV, then the researchers better get out the glass tubes and re-measure, or at least re-fit, some of these lines. If Lorentzians are OK in the wings, then tutorials should address why this is so, since the HITRAN data seems to be fit only at line center.

Comment by Dave Dougherty — 15 Jul 2007 @ 8:55 PM

212. > For CO2, Planck’s function is irrelevant, as is re-emission to the surface. Only water vapour and clouds acts in that way.

Comment by Hank Roberts — 15 Jul 2007 @ 10:41 PM

213. I’m still trying to grasp what happens with absorption/emission on a molecular scale to molecular energy, and have gotten mixed messages here. Does the absorbed IR radiant energy all go to the intramolecular bonding energy? and does this get distributed over the translational stretching, bending, or molecular rotation following some process that is energy and/or molecular makeup dependent? Does this bond energy appear as “regular” exo-molecular (is that a word??) kinetic energy (translation velocity)? Or, can the molecule transfer some of its bond energy to “regular” kinetic energy. What I’m getting at, does bond energy, directly or indirectly, tend to increase a molecule’s “pudding”? Pudding being a molecular quality that if it were evidenced in a lot of molecules would be temperature. Can the bond energy transfer as kinetic with a collision with another molecule. Finally when a molecule emits radiant energy does it all come from the bonds or can “regular” kinetic energy provide some? Finally (really), do electron energy levels play any part in infrared radiation absorption/emission?

Anyone care to help?

Comment by Rod B — 16 Jul 2007 @ 9:30 PM

214. Re #213

Rod, One would think that chemistry is simple, with atoms made of protons, neutrons and electrons. That explains atomic number, isotopes, atomic weight, electrical conduction, and the chemical composition of molecules. However, when you go into quantum mechanics it all becomes horribly complicated.
by

There are not just one, but four ways in which a gas molecule can be “heated.” (Solids and liquids have a different way, which is inter-molecular vibration which generates blackbody radiation.) Those four ways are the kinetic energy of the molecule, its electronic excitation, its vibrational state, and its rotational state. When one of the molecule’s electron gets excited, either by the absorption of a photon or collision with a free high energy electron, then the molecular electron can return to its ground state by emitting a photon, or the energy can be dissipated into kinetic, rotational and vibrational energy. The vibrational energy can emit a photon or be dissipated into kinetic and rotational energy. The rotational energy can be emitted as a photon or be dissipated into kinetic energy.

In general, electronic excitation (absorption) and relaxation (emission) involve visible and ultra violet (UV) photons. Vibrational excitation and relaxation involve infra red photons, and rotational excitation and relaxation involves microwave photons. This means that solar radiation is neded for electronic excitation, whereas terrestrial blackbody radiation only causes vibrational and rotational excitation. A further complication is that the high energy UV radiation from the Sun can break molecular bonds and provide the energy for new ones. This is how the ozone layer is formed.

In order for a molecule to absorb or emit a wave of electromagnetic radiation it must oscillate so that there is a change in the balance of its electromagnetic field. Carbon dioxide is a symmetrical linear molecule so its rotations do not generate a field. However, the water vapour molecule is bent so its rotations do produce a field and it can be rotationally excited by radiation of the appropriate frequency.

In general, an excited molecule can lose its energy without emitting a photon. “The most common mode of return is by thermal decay involving radiation-less transitions.” [Atkins, P.W. “Molecular Quantum Mechanics 2nd edn.” 1983] Thus the idea that the greenhouse effect works by the excited gases re-emitting back to the surface is nonsense!

Comment by Alastair McDonald — 17 Jul 2007 @ 9:15 AM

215. Alastair, please. Posting these statements overthrowing contemporary physics isn’t fair to new readers who may believe you.

Be fair: warn new readers that there are counterexamples in hardware to what you believe. I refute it thus:

“… lasers (e.g. dye lasers, carbon dioxide lasers) where the laser medium consists of complete molecules, and energy states correspond to vibrational and rotational modes of oscillation of the molecules….”

http://en.wikipedia.org/wiki/Carbon_dioxide_laser
“…. the laser is achieved by the following sequence:
1. Electron impact excites vibrational motion of the nitrogen. Because nitrogen is a homonuclear molecule, it cannot lose this energy by photon emission, and its excited vibrational levels are therefore metastable and live for a long time.
2. Collisional energy transfer between the nitrogen and the carbon dioxide molecule causes vibrational excitation of the carbon dioxide, with sufficient efficiency to lead to the desired population inversion necessary for laser operation.

“the laser transitions are actually on vibration-rotation bands of a linear triatomic molecule [O=C=O], the rotational structure of the P and R bands can be selected …..”

The math Dr. Weart describes being done in the 1950s led to the carbon dioxide laser before it led to our understanding of how the energy flows in the atmosphere. It’s the same physics though.

Comment by Hank Roberts — 17 Jul 2007 @ 10:31 AM

216. Thanks, Alastair. A couple of follow-ups: Do gasses not emit blackbody/Planck-type radiation? How about the Sun, or is its gas so compressed, even at the “surface”, that it acts like a liquid? Does bond energy change in liquids and solids? Do bond energy or electron levels affect heat (temperature) levels? If the predominate relaxation of gasses is other than re-emission of IR radiation, does that mean the temperature increase in the earth’s surface mainly comes from conduction or convection of heated greenhouse gasses down to the surface? Sounds a little shaky (but what do I know??!!) Does this apply to CO2 (and maybe other GH gasses???) but not H2O as I think you said above? Or is it that molecule #1 absorbs the IR and mainly relaxes by transferring its absorbed energy to molecule #2 as kinetic energy, then molecule #2 can relax by emitting radiation, ala “local thermodynamic equilibrium” mentioned above by Eli??? If molecule #2 can easily emit, why doesn’t molecule #1??

As I understand it, only a molecule with a dipole moment can absorb radiation and convert it directly to rotational energy. But other molecules (with other restrictions) can absorb IR directly into vibrational bond energy. Is this right and what you said in #214?

Given that electron energy levels are considerably less (in number) and higher (in level) than bond energy levels, is that why electron excitation is very discrete and comes generally from higher energy visible/UV/X-ray radiation?, while gas bonds can absorb a much wider range of photon energy (hf) levels and in the infrared?

Comment by Rod B — 17 Jul 2007 @ 12:04 PM

217. Alastair, on speculation, I figured the problem may be that CO2 is not a homonuclear molecule (more than one kind of element).

So, I tried a search for: non-homonuclear triatomic molecule

That search (either Google or Scholar) finds this: http://www.aiaa.org/content.cfm?pageid=406&gTable=mtgpaper&gID=67474
Abstract: http://pdf.aiaa.org/preview/1979/PV1979_1041.pdf

That may help. It’s early but it’s on the point you’re wrestling with.

Comment by Hank Roberts — 17 Jul 2007 @ 12:27 PM

218. Re #210

Barton, Tour results do not agree with Messerole, C.A., Mulcahy, F.M., Lutz, J., and Yousif, H.A. (1997) â��CO2 Absorption of IR Radiated by the Earthâ�� Journal of Chemical Education, 74 p. 316-7 who calculate that transmission of CO2 in the 2390 to 2275 cm^-1 band is less than 1% at a height of 25 m. I would be interested to see your calculations.

Regards, Alastair.

Comment by Alastair McDonald — 18 Jul 2007 @ 8:37 AM

219. Re #211

Dave,

Note that the number of lines makes the careful treatment of the wings both impractical and irrelevant.

Comment by Alastair McDonald — 18 Jul 2007 @ 9:28 AM

220. Re #212

Hank,

I thought that this document stated that water vapour emits at a low level of continuum radiation. I don’t think there is any doubt that clouds do!

Comment by Alastair McDonald — 18 Jul 2007 @ 9:32 AM

221. Re #215

Hank,

The physics that I am attacking is not contemporary. It is based on the 19th century concept of Kirchhoff’s Law which can be paraphrased as emitted radiation equals absorbed radiation on a per wavelength basis, or more formally as: “The emissivity of a medium is the ratio of the emitted energy to the Planck function at the temperature of the medium for a given wavelength. If the absorptivity is defined as the amount of energy absorbed at that wavelength divided by the appropriate Planck function, Kirchhoff’s law states that the two quantities are equal.” [Morcrette, J-J. “Radiation Transfer” ECMWF, 2000]

The laser is based on the concept of stimulated emission first proposed by Einstein in 1917. Stimulated emission occurs when the absorption of one photon results in two being emitted. This is in direct conflict with Kirchhoff’s law. Moreover, the initial CO2 molecules are not excited by radiation, but rather by the collision with an excited Nitrogen molecule. Nor does the CO2 emit a photon with the same energy that it absorbed from the N2. Moreover, it requires the aid of collision with a Helium molecule to return to its ground state.

So, it is nothing like Kirchhoff’s law at all :-(

Comment by Alastair McDonald — 18 Jul 2007 @ 10:02 AM

222. Hi Rod,

Blackbody radiation is emitted as a continuous spectrum whereas gases emit line radiation. In the same paper in which Einstein proposed stimulated emission, he proved that Kirchhoffâ��s Law applied to gases, and that they emit according to Planckâ��s function. For the proof, he used the Equipartition Theorem/ Boltzmann Distibution, but that does not apply to the vibrational energies of greenhouse gases at room temperature. They are â��frozen outâ�� due to quantum effects. He had already got the specific heats wrong and had been corrected by Debye in 1907. That was before Einstein was famous. Now of course, no-one would dare to correct Einstein, so if he did make a second error then it is not surprising that it has persisted.

The Sun emits continuous radiation that follows Planckâ��s function except for the Fraunhofer lines created by the gases of the chemical elements. It is argued that in the Sun these lines are so broadened by temperature and pressure that they are smeared out into blackbody radiation. I believe that your suggestion that the gases are so compressed that they behave like liquids has also been proposed. Inspection of the surface of the Sun tends to suggest that it is a liquid, but you have to be pretty brave or foolish to claim that it is molten!

When the energy of the molecules in a solid exceed a threshold they become liquid. This requires latent heat. Increase the energy further and it becomes the kinetic energy of the gas molecules. The temperature of the surface of the Earth is 99.9% due to the absorption of solar radiation, but if the air is cooler than the surface then it will be heated by conduction. When the air is warmer it will be cooled by conduction. Warming and cooling by convection is restricted to the atmosphere itself, (and the oceans.)

But, what I am saying is that most of the heating of the air at the surface is due to the greenhouse effect of the Earthâ��s blackbody radiation being absorbed by the greenhouse gases. That causes convection during the day only. H2O is more complicated because it also causes convection by being less dense than air, and it carries latent heat up into the atmosphere. Thus water provides a further method of cooling the surface by evaporation and warming the atmosphere by condensation.

I think LTE is a red herring. It only applies to stellar bodies where there is substantial emission from the gases and no other form of heating. There is a region called non-LTE where the solar radiation heats the gases and they are unable to cool because of a lack of collisons. However as far as outgoing longwave radiation is concerned, in the Earthâ��s atmosphere, the greenhouse gas molecules are de-excited by collisions before they can re-emit their absorbed energy, (but I may revise that view.) The point is that molecule 1 is a greenhouse gas molecule which can re-emit but it loses it energy to an air molecule, which cannot re-emit the energy. Molecule 2 will share its kinetic energy with other air molecules. In theory one of those will collide with Molecule 1 and re-excite it, but before it can emit it will be hit by another air molecule.

The radiation can be absorbed as rotational energy, vibrational energy, or electronic energy depending on it frequency of the photon. For rotational absorption you need a polar molecule, such as the bent water vapour molecule O / H O, in order to have and oscillating field. A rotating symmetrical molecule O = C = O does not create a field because it is neutral. But both polar and non-polar molecules can produce oscillating fields by stretching and bending, as can ozone O = O = O a triatomic homonuclear molecule.

Electronic levels are not fewer, just require more energy â�� higher frequency photons. Look up Hydrogen lines and you will find that there are several sets for the simplest of molecules.

Comment by Alastair McDonald — 18 Jul 2007 @ 11:49 AM

223. BTW, Alastair, I thought your referenced links in #210,220 offered a very good explanation of the basics of radiation transfer at the molecular level.

Comment by Rod B — 18 Jul 2007 @ 11:54 AM

224. There are continua associated with water and CO2 emission at very low levels.

Comment by Eli Rabett — 18 Jul 2007 @ 11:55 AM

225. Re.

222. Alistair. You must be aware that just about everyone thinks your stand on LTE is a bit odd. Including the US air force…please see http://www.dodsbir.net/Sitis/view_pdf.asp?id=DothH04.pdf

Comment by David donovan — 18 Jul 2007 @ 1:51 PM

226. Re #217.

Hi Hank,

That is an intriguing link you posted, but I cannot get to the full paper, nor to the paper to which they refer by Simmons and Lindquist. I have found the technical report on the NASA server which has lots of goodies. Try searching for carbon dioxide. It brings up Moon forming collisions and a Hadean Earth!

The trouble is the report only has absorption coefficients for 300 K but what I need is 200 K thru 300 K. Anyway, as they say, back to the drawing board :-)

Comment by Alastair McDonald — 18 Jul 2007 @ 1:52 PM

227. Re #205

I don’t think that the US Air Force has heard my views on LTE yet. I only came to the conclusion that it was irrelevant to OLR today!

Comment by Alastair McDonald — 18 Jul 2007 @ 2:19 PM

228. Alastair, thanks. This is great! But, a couple more followups. I would have guessed that electron energy levels number in the many hundreds and bonding energy levels, with their numerous combinations, in the many thousands, and which is one of the bases for “spreading”. Not correct?

If the GH gasses re-emit very little (in any direction) what causes the ultimate emission of about 240 W/m2 or so into outer space? Radiation making it directly from the surface to 60km or so seems screwy (most IR radiation is (first?) absorbed within 100m). Is it blackbody-type radiation? [I couldn’t tell from #222, 1st para. if gasses do or do not emit radiation ala Planck. As an aside, how thick/deep can a surface be and still be a radiating surface?]

most of the heating of the air at the surface is due to the greenhouse effect of the Earth’s blackbody radiation being absorbed by the greenhouse gases. That causes convection during the day only.” Since the Earth radiates 24 hrs/day, why “during the day only”??

Comment by Rod B — 18 Jul 2007 @ 3:09 PM

229. I have a request: would people responding to previous comments please reference the poster and time as well as the number? or at least quote enough of the text so as to make an unambiquous determination ? the numbers seem to change and it is difficult to scroll back and forth for visually and physically challenged readers.

I also notice that the php/sql engine has some difficulty at times, or i would suggest implementing a threaded comment system a la Usenet, but that might be too much to ask.

i suppose i could volunteer to code such an extension…

Comment by sidd — 18 Jul 2007 @ 4:49 PM

230. Re #228

I will push the boat out a little here because I am not sure if the following is true but …

Gaseous elements when vapourised tend to be atomic, or even plasmic and so can have no vibrational or rotational excitation levels. However, with just one electron hydrogen has 12 lines. But those electronic levels are never stimulated by infra-red photons, so they are irrelevant to the greenhouse effect. They are relevant to the interior of stars, and they are what Einstein considered. Milne, who came up with the idea of LTE was also thinking about electronic excitation and astrophysics. That is why LTE is not relevant to the Earth’s atmosphere.

The main greenhouse gases are H2O and CO2. Nitrogen and Oxygen do not absorb or emit infra-red radiation but do get heated by collisions with the excited greenhouse gas molecules. If they collide with a greenhouse gas at the same time as a photon then the the absorption line of the greenhouse gas is broadened. Obviously, high in the atmosphere where the air is thinner there are less collisions and less broadening.

Although the CO2 cannot absorb photons which produce pure rotation, the energy levels of the vibratiosn can be modfied by rotations. So a vibration line becomes a vibration band ie a series of lines for the different rotations that can be associated with the vibration. There are also several pure vibration lines due there being different isotopic versions of the greenhouse gas molecules. For instance 12C 16O 16O, 13C 16O 16O, 14C 16O 16O, 12C 16O 17O, etc. Then each of these has a full series of rotational variations.

I am arguing that the emissions to space come from between the lines and from between the bands. But I am also saying that there is a lot of radiation to space from the tops of clouds near the tropopause. That radiation is blackbody and there is no water vapour above it to prevent it escaping to space. The heat for that radiation is provided by the condensation of water vapour which was carried there by convection not by radiation.

You ask why is there convection] during the day only? A good question :-)

Without the Sun or the back radiation from the atmosphere which I have demolished then in a clear sky state the surface cools very quickly, and since the power of the radiation depends on the fourth power of the temperature, there is a big drop in the heating of the lower air which is also being cooled by conduction. There is also a little back radiation from water vapour cooling the air.

The link for hydrogen lines should be http://iapetus.phy.umist.ac.uk/Teaching/IntroAstro/Hydrogen.html

Comment by Alastair McDonald — 18 Jul 2007 @ 4:56 PM

231. So, Alastair, someone on Alpha (the International Space Station) or using an infrared imaging satellite could aim an infrared film camera at the horizon — from there they can clearly see the top of the troposphere and the clear stratosphere above it — and you’d predict that, viewed in the infrared, the stratosphere seen edge-on would not be emitting any infrared at all, it’d be black on the picture, and the clouds would show up as very bright in the infrared?

Is that a fair summary? If so you’ve got a falsifiable prediction.

Possibly it’s already been done and you can find an image somewhere.

Comment by Hank Roberts — 18 Jul 2007 @ 5:21 PM

232. re 230,231 (Alastair and Hank): I ran across a doc that kinda confirms the infrared radiation into space comes directly from the surface (between the lines so to speak, or, more accurately, mostly through the “window”), and (primarily) the clouds. Very little presumably comes from re-emitting greenhouse gasses. I have to go contemplate.

Comment by Rod B — 19 Jul 2007 @ 10:30 AM

233. Alastair McDonald (#222) wrote:

I think LTE is a red herring. It only applies to stellar bodies where there is substantial emission from the gases and no other form of heating. There is a region called non-LTE where the solar radiation heats the gases and they are unable to cool because of a lack of collisons.

The common view is that as long as the electromagnetic field and matter are strongly interacting, there is a local thermodynamic equilibrium. This is a view that was expressed by Eli Rabett back in #153:

LTE occurs when the radiant energy absorbed by a molecule is distributed across other molecules by collisions before it is reradiated by emission. LTE is needed for Planck’s law and Kirchhoff’s law to apply, and is typically satisfied at atmospheric pressures higher than about 0.05 mb.

But of course you go on to explain why you don’t think that the radiation and matter strongly interact:

However as far as outgoing longwave radiation is concerned, in the Earth’s atmosphere, the greenhouse gas molecules are de-excited by collisions before they can re-emit their absorbed energy, (but I may revise that view.) The point is that molecule 1 is a greenhouse gas molecule which can re-emit but it loses it energy to an air molecule, which cannot re-emit the energy. Molecule 2 will share its kinetic energy with other air molecules. In theory one of those will collide with Molecule 1 and re-excite it, but before it can emit it will be hit by another air molecule.

Interesting approach, Alastair. Greenhouse gas molecules (such as carbon dioxide) can always be de-excited by collisions before they re-emit, but they can never be excited to re-emission by collisions. The energy only goes in one direction.

This is the “law” if you will which in your view prevents the electromagnetic field and matter from strongly interacting.

In the common view, nature would have no preference for de-excitation versus excitation. Thus the Planckian temperature of the electromagnetic field will follow like the tail the dog which is the Maxwellian temperature of matter – resulting in Local Thermodynamic Equilibrium (LTE). But you maintain that the only place where this sort of strong interaction can take place is in stars. But why would such strong interaction take place there? Couldn’t we simply impose the same sort of “filter,” declaring that molecules always get de-excited by collision before re-emission even under those circumstances? If not, why not?

Comment by Timothy Chase — 19 Jul 2007 @ 10:59 AM

Comment by Hank Roberts — 19 Jul 2007 @ 11:41 AM

235. I would like to wade in on Alastair’s side generally, though perhaps not in all details. In particular, I would assure Timothy Chase that radiation and absorption go together. If there is one, there is the other, though other processes may compete to make it difficult to measure.

I also say that isolated CO2 molecules inside a hot sphere will have detectable radiation — this is what it means to be in complete thermal equilibrium, as opposed to LTE which can have either higher or lower thermal radiation compared to the vib-rot-tran components.

BTW, please note that the Einstein coefficients are kinetic, and apply with or without equilibrium.

Comment by Allan Ames — 19 Jul 2007 @ 12:00 PM

236. Allan Ames (#235) wrote:

I would like to wade in on Alastair’s side generally, though perhaps not in all details. In particular, I would assure Timothy Chase that radiation and absorption go together. If there is one, there is the other, though other processes may compete to make it difficult to measure.

Well, Alastair believes that he has some sort of integrated alternative theory of the greenhouse effect. All of the elements are strongly-coupled. However, the inability of carbon dioxide to absorb and re-emit except when it is close to the ground plays an essential role in his view – at least as far as I understand it.

In the mainstream view, the strong interaction between the greenhouse gases and the atmosphere (which exists in all parts of the atmosphere except the uppermost parts of the stratosphere where pressure is too low) is what results in the coupling of the electromagnetic field of thermal radiation and the atmosphere itself. The atmosphere is opaque to thermal radiation, and molecules can be both excited to re-emission and de-excited so that they fail to re-emit. The electromagnetic field and the atmosphere are strongly interacting. Such strong interaction implies strong coupling – and the achievement of a local thermodynamic equilibrium in which the Planckian temperature of the longwave radiation and the Maxwellian temperature of the atmosphere are equal.

Moreover, since greenhouse gases can absorb and re-emit in all the lower layers of the atmosphere (that is, beneath the highest layers of the stratosphere) and re-emit in all directions, they can absorb and re-emit between neighboring layers, and higher layers will indirectly interact with the surface via the intermediary layers. As such, in the mainstream view, water vapor dominates in the lower levels of the troposphere, but in higher layers which are especially dry, carbon dioxide dominates and plays the role of a greenhouse gas, the effects of which is to regulate the amount of water vapor which is in the lower atmosphere through water vapor feedback – since any given parcel of carbon dioxide stays in the atmosphere much longer than an equivilent parcel of water vapor.

But in Alastair’s view, it is precisely this strong interaction which prevents the electromagnetic field and greenhouse gases from becoming strongly coupled – because energy which might be re-emitted by greenhouse gas molecules is always lost in collisions, never gained. In his view, it is only within stars that a local thermodynamic equilibrium will be achieved. Assuming Alastair is correct about strong interaction preventing strong coupling in planetary atmospheres, I have to wonder whether the achievement of a local thermodynamic equilibrium would be even “less likely” in a stellar atmosphere.

Comment by Timothy Chase — 19 Jul 2007 @ 2:03 PM

237. PS to Allan Ames…

Anyway, I believe that at a certain level, talk about a local thermodynamic equilibrium is still just an approximation. For example, there is the electromagnetic field of the sun – solar radiation – then the electromagnetic field of the atmosphere – its thermal radiation. We can speak of these as separate fields and we can distinguish between them, their temperatures, their spectra, but it isn’t exactly as if each and every photon bears a label designating it as belonging to one field or the other.

Obviously solar radiation isn’t in local thermodynamic equilibrium (LTE) with the atmosphere, and thus the whole of the electromagnetic radiation within any given part of the atmosphere won’t be in LTE with the atmosphere. But the thermal radiation of the earth will be, for the most part – and it is something that we can distinguish from solar radiation – particularly at night. The atmosphere is transparent to visible light (mostly), and solar radiation is principally in the visible part of the spectrum. The atmosphere is opaque to the earth’s thermal radiation (for the most part), and so it makes sense to treat this thermal radiation as being in local thermodynamic equilibrium with the atmosphere. But a thermometer wrapped in black linen is something else – that’s why we speak of albedos.

Now is this good enough? Well, yes – as an approximation. If you want to get into the really detailed equations, that is possible, too. Hank Roberts pointed to a source which shows how the general direction. But while Einstein is more accurate, most people stick to Newton – because it is easier to work with the equations and they don’t need that kind of accuracy. The same works here, I believe. As for the greenhouse effect itself, the central issue is fairly basic, in my view at least: greenhouse gas molecules can gain or lose energy in collisions, and consequently, they can and do absorb and re-emit radiation throughout most of the atmosphere.

Comment by Timothy Chase — 19 Jul 2007 @ 2:27 PM

238. Re #224 where Eli Rabett wrote “There are continua associated with water and CO2 emission at very low levels.”

I have now found the reference to contiuum radiation that I was looking for. It is chapter 9 of “Principles of remote sensing of atmospheric parameters from space” February 1998 By R. Rizzi and updated by R. Saunders which can be found at
http://www.ecmwf.int/newsevents/training/rcourse_notes/DATA_ASSIMILATION/REMOTE_SENSING/Remote_sensing10.html

It is a little confused because it says first that the continuum is mainly due to H2O, perhaps due to its dimers. Then states that CO2, N2 and O2 also produce a continuum emission, but as far as I know these gases do not produce dimers.

The full document is worth reading if you want blackbody radiation, vibro-rotational bands, and line shapes and broadening explained with more authority than me. OTOH I do not endorse his view that atmospheric gases obey Kirchhoff’s Law!

Re #231 Hank,

Thanks for that suggestion of a limb spectrum. I am just trying to remember where I saw such a diagram recently, but I cannot find it. An accurate < 1/cm downward spectrum for clear skies would also work. The problem is finding one.

Comment by Alastair McDonald — 19 Jul 2007 @ 4:17 PM

239. re 236, Timothy: I’m asking a quicky question without having read the entire post; couldn’t wait.

…the achievement of a local thermodynamic equilibrium in which the Planckian temperature of the longwave radiation and the Maxwellian temperature of the atmosphere are equal.

Maybe I missed a turn somewhere, but isn’t Planck temperature determined by Maxwell temperature and always equal to it, at least on the “surface” of the gasses??

[ps I trust I’m not reopening the debate over the “temperature” of the actual E-M radiation…!]

Comment by Rod B — 19 Jul 2007 @ 9:21 PM

240. re 237, by Timothy: a quicky quibble and a clarification:

The photons do too bear a label by virtue of the hf energy level.

I halfway buy the “close enough” (for the girls I go with [;-)) argument, but am bothered by its conclusive nature. While just an partially educated hunch, my basis for being a skeptic rests greatly on the molecular/atomic/sub-atomic very precise physics that goes on between radiation and gas molecules. A physics that I gather is still based a bunch on assumptions and reasonable scientific guesses.

Comment by Rod B — 19 Jul 2007 @ 9:41 PM

241. It appears to me that Hank Roberts’s posts # 66 and 67 under Friday roundup belong in this thread.

Comment by Rick Brown — 19 Jul 2007 @ 11:10 PM

242. Rod B (#239) wrote:

Maybe I missed a turn somewhere, but isn’t Planck temperature determined by Maxwell temperature and always equal to it, at least on the “surface” of the gasses??

Well, at this point I will gladly defer to the American Meteorological Society:

local thermodynamic equilibriumâ??(Abbreviated LTE.) A condition under which matter emits radiation based on its intrinsic properties and its temperature, uninfluenced by the magnitude of any incident radiation.

LTE occurs when the radiant energy absorbed by a molecule is distributed across other molecules by collisions before it is reradiated by emission. LTE is needed for Planck’s law and Kirchhoff’s law to apply, and is typically satisfied at atmospheric pressures higher than about 0.05 mb. Laser radiation is an example of non-LTE emission.

http://amsglossary.allenpress.com/glossary/search?id=local-thermodynamic-equilibrium1

If there isn’t enough interaction, the local thermodynamic equilibrium breaks down. But I was unclear as to why exactly this happens. But the example of a laser is a good clue. If the particles are colliding infrequently, then the electromagnetic field which arises from absorption and re-emission will tend to destroy the Boltzmann distribution of velocities which exist between those particles.

Thermal Radiation in the Upper Atmosphere
A. R. Curtis, R. M. Goody
Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, Vol. 236, No. 1205 (Aug. 2, 1956), pp. 193

The derivation of Planck’s and Kirchoff’s laws presuppose the persistance of a Boltzmann distribution – but if the latter breaks down, the other two go out the window – which is why a gas laser is able to emit radiation at a precise frequency, etc., the radiation overwhelms the dis-order that would normally exist within matter and which would therefore be imposed upon the electromagnetic field. In essence, a laser is an instance of the tail wagging the dog.

Rod B (#240) wrote:

The photons do too bear a label by virtue of the hf energy level.

I was thinking the same thing, too.

You have the wavelength or frequency, the directionality in the case of solar radiation (unless there is scattering), etc.. But an individual photon could belong to one or the other far wing of the Planckian distribution. The sun, afterall, emits some longwave radiation. Similarly, there will be some photons emitted within the visible part of the spectrum even by objects at room temperature.

I halfway buy the “close enough” (for the girls I go with [;-)) argument, but am bothered by its conclusive nature. While just an partially educated hunch, my basis for being a skeptic rests greatly on the molecular/atomic/sub-atomic very precise physics that goes on between radiation and gas molecules. A physics that I gather is still based a bunch on assumptions and reasonable scientific guesses.

Actually the “close enough” is more or less implicit in the phrase “local thermodynamic equilibrium” itself, assuming the LTE is a stable one. If it is to persist, there must be interaction between the local neighborhood and the outside world. Sunlight being absorbed by the ground and re-emitted into the atmosphere as thermal radiation is a good example of this.

“… A physics that I gather is still based a bunch on assumptions and reasonable scientific guesses.”

Well, one could argue that we know of radiation only because we know of matter. Then again, one could make the opposite argument. Seems all pretty mysterious to me. Then again, as a former philosophy major, I came to think that knowledge itself was pretty mysterious. It took me ten years to arrive at a satisfactory definition of it, or rather, definitions of the terms in which it is defined.

At least on the surface, it seems to me that there is a fairly big divide between philosophy and empirical science. Descartes could go round in circles over whether what he was staring at was actually his hand or not or whether he could trust an argument which seemed to suggest that he existed at the time he held it firmly in mind. Empirical science, on the otherhand, attempts to ground itself in evidence and testability. None of the arguments are deductively certain except insofar as we aren’t really sure whether they apply only to our mental constructs. Euclid is a good example of this.

But what empirical science has going for it is justification from many different lines of inquiry – where the tentative conclusions which it arrives at are tested again and again – and improved upon as needed. The justification received by a conclusion which is justified by multiple lines of inquiry can be far greater than that which it receives from any one line of inquiry considered in isolation.

In contrast, when you have a single individual sitting in a small warm room staring at his hand by the light of a fire, there isn’t that much evidence for him to work with, certainly not in any systematic fashion. If he is especially inclined towards radical doubt, he is pretty much limited to that single line of inquiry which he happens to be thinking about at the moment.

After a while, it can get to be a pretty strange brew – the result of the tail of thought wagging the evidence on which it would otherwise be based.

Comment by Timothy Chase — 20 Jul 2007 @ 1:12 AM

243. In #236 Tim wrote Well, Alastair believes that he has some sort of integrated alternative theory of the greenhouse effect. All of the elements are strongly-coupled. However, the inability of carbon dioxide to absorb and re-emit except when it is close to the ground plays an essential role in his view – at least as far as I understand it.

It is not correct that I am arguing that CO2 is only able to to absorb near the ground. It can absorb at all altitudes, and when clouds emit blackbody radiation, the CO2 lines will be absorbed in the regions around them. However, in most of the atmosphere there is no radiation in the frequencies at which CO2 absorbs because it has already been absorbed by CO2 near the surface of the Earth (or around the clouds.)

OTOH, I am postulating that CO2 does not emit at any altitude because it is deactivated by collisions before it can do so. However, that does raise an interesting point. In the region where Goody & Jung (G&Y) say non-LTE exists i.e. where the time between collisions is greater than relaxation times, then CO2 emissions will occur. In other words, greenhouse gas emissions will come from the G&Y non-LTE region of atmosphere. Since this non-LTE region is at the top of the atmosphere, there is no surprise that is where emissions to space originate.

In #237 Tim wrote Assuming Alastair is correct about strong interaction preventing strong coupling in planetary atmospheres, I have to wonder whether the achievement of a local thermodynamic equilibrium would be even “less likely” in a stellar atmosphere.

The radiation from the surface of the Sun and from stars is blackbody radiation caused by the vibration of the atoms. The relative movement of the positively charged nuclei of the atoms generates a continuum oscillating electromagnetic field. Therefore there is no delay between a molecule becoming excited and its relaxation. (That delay only occurs in the much cooler planetary atmosphere where line radiation is generated by molecular vibrations, not atomic vibrations.) In the region of the Sun which is generating blackbody radiation, obviously the Maxwellian and Planckian temperatures are equal, and so that region is in LTE. Above that region is the Chromosphere where the Fraunhofer lines are produced.

So for strict LTE you need part of the the atmosphere to be a blackbody, which in the case of the Earth’s atmosphere it is not!

Comment by Alastair McDonald — 20 Jul 2007 @ 5:16 AM

244. [[Barton, Tour results do not agree with Messerole, C.A., Mulcahy, F.M., Lutz, J., and Yousif, H.A. (1997) “CO2 Absorption of IR Radiated by the Earth” Journal of Chemical Education, 74 p. 316-7 who calculate that transmission of CO2 in the 2390 to 2275 cm^-1 band is less than 1% at a height of 25 m. I would be interested to see your calculations.

Regards, Alastair. ]]

Certainly. Transmissivity in a column of gas is

T = exp(-k c L)

where k is absorption coefficient, c greenhouse gas concentration, and L path length. The k c L term is, of course, the optical thickness. We can use Beer’s Law here because for infrared light in Earth’s atmosphere, scattering is negligible.

Essenhigh (2001) gives a mean absorption coefficient from 13 to 17 microns of 1.48 reciprocal meter atmospheres. For a carbon dioxide concentration of 0.04% of an atmosphere, the path length needed for transmissivity to drop to 1% is about 7,800 meters.

For a blackbody at 288 degrees K (the Earth’s surface, although it’s actually a graybody with a mean emissivity around 0.95), 8.5% of the radiation is from 13 to 17 microns, using the Planck distribution.

The discrepancy may be from considering different bandwidths. If you calculate it from, say, 14.98 to 15.00 microns, of course the needed path length will be much smaller. Your miscalculation may be in assuming that the infrared radiation from the Earth is only in the exact bands that carbon dioxide absorbs. There is plenty that gets through. There are CO2 absorption line wings all over the spectrum, but they are only highly concentrated at a few points, like the 4.3 and 15 micron bands.

Comment by Barton Paul Levenson — 20 Jul 2007 @ 6:07 AM

245. Re #244

Barton.

I should have remembered that you were using Essenhigh’s values :-( They are from Hottel, H. C. and Egbert. R. B. Trans. Am. Soc. Mech. Engrs. 1941, 63, 297. See
http://pubs.acs.org/subscribe/journals/ci/31/i12/html/12box.html#refb

That document goes back to 1941 when it was believed that greenhouse gases absorbed bands of radiation. We now know that those bands are in fact groups of lines. The absorption only happens to the lines. When these lines are totally absorbed and not re-emitted no more absorption can occur. If these lines are totally absorbed in the first 50 m say, and on average they are 30% of the band width, then the absorption at 50 m will be 30% and the transmission 70%. However and 100 m, the absorption will not double to 60% nor will the transmission be 40%. They will still be 30% and 70% respectively.

Your calculation of a 1% transmissity at 7,800 does not happen, otherwise there would be no radiation emitted at the top of the atmosphere in the 13 to 17 micron band at all, but we know that the brightness temperature there is 215 K. I am arguing that 215K is the average of lines with a brightness temperature of 0K and others with a brightness temperature of 300K.

Comment by Alastair McDonald — 20 Jul 2007 @ 10:07 AM

246. Alastair McDonald (#245) wrote:

I should have remembered that you were using Essenhigh’s values They are from Hottel, H. C. and Egbert. R. B. Trans. Am. Soc. Mech. Engrs. 1941, 63, 297. See
http://pubs.acs.org/subscribe/journals/ci/31/i12/html/12box.html#refb

That document goes back to 1941 when it was believed that greenhouse gases absorbed bands of radiation.

Well now, that changes everything!

Everyone knows that when a new theory comes along and supplants an older theory, it replaces it whole. Perfect example: Einstein’s gravitational theory and Newton’s gravitational theory. Not even the geometry of space remains the same, and absolute time? Well, we know what happened to that.

Then again, there is that correspondence principle. Looking at Schwarzchild’s solution, the only way that Schwarzchild was able to derive the last constant left in his equations was by reference to how how curved spacetime approached the flat employed by Newton if one followed a straight line far enough.

We now know that those bands are in fact groups of lines. The absorption only happens to the lines. When these lines are totally absorbed and not re-emitted no more absorption can occur. If these lines are totally absorbed in the first 50 m say, and on average they are 30% of the band width, then the absorption at 50 m will be 30% and the transmission 70%. However and 100 m, the absorption will not double to 60% nor will the transmission be 40%. They will still be 30% and 70% respectively.

The bands are groups of lines, but the bands and the lines broaden as the result of pressure and temperature. Likewise, there is lifetime broadening of a single line simply as the result of the uncertainty principle as it applies to energy and time.

Like you I enjoy physics.

I am looking forward to learning more about it. The neat thing about it is that it seems there is always more to learn.

Comment by Timothy Chase — 20 Jul 2007 @ 11:27 AM

247. RE: 237 Timothy Chase

There are two issues. As regards the physics, since we are talking about the transfer of many joules of energy through a kilogram/sq.m. each day, a small error in the transport parameters can lead to large errors in temperature. So, we need the most accurate portrayal we can put together.

Then there is the further question of how to embed the physics in a broader climate model in a way that can be calculated in reasonable time. Have you seen a detailed analysis of what heat flow (and error) the best LBL models predict?

If Rod_B is still wondering about the temperature of single atoms, in non-LTE, temperature is the number that gives you the correct vibration population distribution, which presumes that a single atom can have a temperature.

I cannot find the post, but the reference to SAMM2 is very good. http://www.dodsbir.net/Sitis/view_pdf.asp?id=DothH04.pdf

A useful reference on LTE is http://edoc.ub.uni-muenchen.de/archive/00001056/01/Goussev_Oleg.pdf

It seems consistent to say that LTE is equilibration in the mechanical modes, but not necessarily in or with the radiation modes.

I still mostly agree with Alastair. And lines get narrower as the altitude goes up until you are in the nonLTE region.

Comment by Allan Ames — 20 Jul 2007 @ 1:54 PM

248. Allan Ames (#247) wrote:

RE: 237 Timothy Chase

There are two issues. As regards the physics, since we are talking about the transfer of many joules of energy through a kilogram/sq.m. each day, a small error in the transport parameters can lead to large errors in temperature. So, we need the most accurate portrayal we can put together.

I agree – accuracy is a good thing. Then again, it also helps to keep in mind that there are negative feedback processes involved in the establishment of equilibria. Not all errors grow large – and some errors average out. Tamino had a good essay on this a while back. Much of climatology is based upon this sort of principle.

Then there is the further question of how to embed the physics in a broader climate model in a way that can be calculated in reasonable time. Have you seen a detailed analysis of what heat flow (and error) the best LBL models predict?

Not offhand. Then again I am a philosophy major turned coder.

If Rod_B is still wondering about the temperature of single atoms, in non-LTE, temperature is the number that gives you the correct vibration population distribution, which presumes that a single atom can have a temperature.

I thought that “population” normally refered to more than one.

I cannot find the post, but the reference to SAMM2 is very good.
http://www.dodsbir.net/Sitis/view_pdf.asp?id=DothH04.pdf

That looks like something worth reading.

I might recommend the following:

(3) [M*]/[M] = J/A.

This situation is described as “radiative equilibrium”. At altitudes greater than about 150 km the ratio of the densities of the CO2 vibrational levels 01101 and 00101 is determined solely by the upwelling earthshine. The frequency of collisions is too small to have any impact and radiative equilibrium prevails.

(pg 5)

USERS’ MANUAL FOR SAMM2,
SHARC-4 and MODTRAN4 MERGED
H. Dothe, J. W. Duff, J. H. Gruninger, P. K. Acharya, A. Berk
AFRL-VS-HA-TR-2004-1145
Environmental Research Papers, No. 1145
http://www.dodsbir.net/Sitis/view_pdf.asp?id=DothH04.pdf

A useful reference on LTE is
http://edoc.ub.uni-muenchen.de/archive/00001056/01/Goussev_Oleg.pdf

I found the following to be of interest:

In recent years the observations of infrared molecular emissions from the middle and upper layers of the Earth’s atmosphere have attracted particular attention since trends in this altitude region are significant for understanding of global changes of the atmosphere. Simultaneously a number of sophisticated ground based and space infrared experiments provided unique data about the upper planetary atmospheres. However, an essential problem here is that in many cases the interpretation of these data requires accounting for deviations from the local thermodynamic equilibrium (non-LTE). The non-LTE effects are crucial in the middle and upper Earth’s atmosphere for the majority of infrared molecular bands and influence strongly the radiation emerging from these layers. In some cases the observations of these emissions from space are influenced by the non-LTE effects down to the lower stratosphere.

(pg 1)

Non-LTE diagnostics of the infrared observations of the planetary atmosphere
Oleg Goussev an der Fakultat für Physik der Ludwig–Maximilians–Universitat
Munchen 2002
http://edoc.ub.uni-muenchen.de/archive/00001056/01/Goussev_Oleg.pdf

I didn’t know, for example, that non-LTE effects become important in the middle of the Earth’s atmosphere and the above, but it suggests that LTE is a good approximation below this – as does the passage from the earlier paper.

I also found this interesting:

2.4 Partial LTE

The gas is said to be in a non-LTE state when the level populations do deviate from the Boltzmann law while the molecular velocity distribution remains Maxwellian. One may describe this state with the same probability formalism as was used above for the case of LTE by introducing at each point a set of excitation temperatures instead of the one parameter T. The complete LTE or non-LTE are hardly ever encountered in planetary atmospheres. In reality the gas is mainly in a state that might be called partial LTE in which the velocity distribution of molecules is Maxwellian but only some groups of levels obey the Boltzmann relation (Eq. 2.13, (p. 5)). If one considers a certain volume of gas and traces its state as the density decreases, then generally one will observe a series of the partial LTE states, in which ever fewer levels follow the Boltzmann relation. Two such states are discussed, which are of special importance.

(pg 7)

Non-LTE diagnostics of the infrared observations of the planetary atmosphere
Oleg Goussev an der Fakultat für Physik der Ludwig–Maximilians–Universitat
Munchen 2002
http://edoc.ub.uni-muenchen.de/archive/00001056/01/Goussev_Oleg.pdf

Across some parts of the spectra, (partial-)LTE will still hold up – as suggested in the first passage I cited.

It seems consistent to say that LTE is equilibration in the mechanical modes, but not necessarily in or with the radiation modes.

It seems fair to say that LTE is a good approximation in a variety of circumstances. However, as I pointed out in what you were responding to, one may perform more exact calculations which dispenses with this sort of approximation all together. But it will still be an approximation to one degree or another.

I still mostly agree with Alastair.

So you have stated.

But you have never actually said what it is that you agree with. That LTE doesn’t hold at all in planetary atmospheres? That the greenhouse effect occurs only in the lowest part of the atmosphere? That greenhouse gases are incapable of re-emission?

Actually I believe you already acknowledged that you disagree with the last of these, and as it is quite central to his “theory.”

In any case, it would help if you would be more specific – for the sake of accuracy if nothing else.

But as the papers you have cited would suggest, LTE is a fair approximation throughout much of the atmosphere in one band or another. Therefore Kirchoff’s law is applicable in many contexts – and greenhouse gases are able to perform their role as greenhouse gases throughout much of the atmosphere.

Given the emphasis which you place on accuracy, I would expect you to prefer a theories which recognize the fact that the greenhouse effect occurs throughout much of the atmosphere over a theory which holds that it only occurs near groundlevel.

And lines get narrower as the altitude goes up until you are in the nonLTE region.

In the piece that you were responding to I acknowledged the fact that LTE breaks down at the upper reaches. But it seems that I have learned a little more.

Thank you for suggesting the articles.

Comment by Timothy Chase — 20 Jul 2007 @ 4:53 PM

249. PS to 248

What is most interesting about the second paper which Allan Ames pointed out:

Non-LTE diagnostics of the infrared observations of the planetary atmosphere
Oleg Goussev an der Fakultat für Physik der Ludwig–Maximilians–Universitat
Munchen 2002
http://edoc.ub.uni-muenchen.de/archive/00001056/01/Goussev_Oleg.pdf

… is the existence of “non-Local Thermodynamic Equilibria,” where opacity and emissivity are functions of the frequency which are not strictly equal. Kirchoff’s Law is violated – but its violation is a matter of degree and specific to particular bands. Interestingly, the effect is partly dependent upon direction. In the lower atmosphere (including the layers where carbon dioxide is most important), the effects appear to be fairly minor. Judging from the literature I have seen so far, the effects (or deviations from Kirchoff’s Law) appear to be most important in plasma physics and mid to upper levels of the atmosphere. In essence, non-local thermodyanmic equilibria simply adds a new level of detail and precision in terms of our account of the atmosphere. Likewise, we have been taking this sort thing into account in some climate models at least as far back as 1999.

For those who might be interested, I think it is worth checking out. But in any case, it is probably best to learn Euclid before moving on to Riemann (that particular analogy seems especially apt) – just as you should learn differential calculus prior to integral. (I came in through the backdoor in both cases, but then in the former it was youth and in the latter ignorance. Oh well…)

Comment by Timothy Chase — 20 Jul 2007 @ 9:11 PM

250. Timothy (re 242), I didn’t see anything in the post that contradicted what I said in 239 and 240. In any case I agree with you. Now to read what comes later….

Comment by Rod B — 21 Jul 2007 @ 10:37 AM

251. Timothy (246) says “the bands and the lines broaden as the result of pressure and temperature. Likewise, there is lifetime broadening of a single line simply as the result of the uncertainty principle as it applies to energy and time.

Do the lines actually broaden (increase their miniscule bandwidth) or become more available for absorption?
Why wouldn’t the uncertainty principle just as easily/often narrow the lines. (And I trust we’re not counting on Heisenberg for AGW validation [;-)….)

Comment by Rod B — 21 Jul 2007 @ 11:04 AM

252. The lines are what’s observed in the instrument, photographed — the “line” is not the actual specific amount of energy transferred in any particular event.
It’s a sum over time of observations.

Same issue as with the question about specifying “the” altitude at which radiation escapes to space.

Comment by Hank Roberts — 21 Jul 2007 @ 11:29 AM

253. Rod B (#250) wrote:

Timothy (re 242), I didn’t see anything in the post that contradicted what I said in 239 and 240. In any case I agree with you. Now to read what comes later….

Typically I don’t assume that someone is agreeing with me or disagreeing with me: I am simply trying to address the question that has been posed – to the extent that I understand the issues which are involved. Obviously there was a little more involved in 242, but in large part that was meant to be somewhat entertaining for you and others and perhaps even a little playful for those who wanted to examine it a little more closely.

But even when I disagree with people, while I might get annoyed, perhaps even strongly annoyed, I at least try not to take it personally. Afterall, whether we realize it or not, there is always more that binds us than separates us. As I look at the world, the “us vs them” which many view the world in terms of is at worse temporary and ultimately unnecessary. While I am not Buddhist, I would regard it as a “veil of maya,” an illusion which we should try and help one another escape from – if possible. But there are oftentimes other demands which for one reason or another must take precedence – at least for the time being.

Rod B (#239) wrote:

Maybe I missed a turn somewhere, but isn’t Planck temperature determined by Maxwell temperature and always equal to it, at least on the “surface” of the gasses??

[Thinking for a moment…]

There will be typically be some difference between the two if the system is not in local thermodynamic equilibrium, and the more distant the system is from local thermodynamic equilibrium the greater the difference may be. As I understand it, a large part of the reason lies in the fact that radiation gets re-emitted before collisions can take place between the molecules which would otherwise equalize the two temperatures. Instead, the temperature of the re-emitted radiation will tend to reflect the temperature of the incident radiation.

Rod B (#240) wrote:

The photons do too bear a label by virtue of the hf energy level.

I halfway buy the “close enough” (for the girls I go with [;-)) argument, but am bothered by its conclusive nature. While just an partially educated hunch, my basis for being a skeptic rests greatly on the molecular/atomic/sub-atomic very precise physics that goes on between radiation and gas molecules. A physics that I gather is still based a bunch on assumptions and reasonable scientific guesses.

Well, regarding the photons “bearing labels,” as I understand it, the distribution of solar radiation will be a particular shape with a peak determined by the temperature of the sun and the same will be true of the blackbody radiation emitted by the earth’s atmosphere and surface. But in both cases, the distribution is probablistic, such that if you were to select a particular photon at random, some frequencies would be far more probable than others, but it could be at any given frequency. Likewise, no matter what direction a particular photon has, it could be a photon which has or has not be re-emitted from the surface or the atmosphere – simply because of the existence of scattering and the possibility (no matter how small it might be) that it has either been or not been absorbed by at least one molecule along its way.

You wrote,

“… but am bothered by its conclusive nature.

“While just an partially educated hunch, my basis for being a skeptic rests greatly on the molecular/atomic/sub-atomic very precise physics that goes on between radiation and gas molecules. A physics that I gather is still based a bunch on assumptions and reasonable scientific guesses.”

Well, in the final analysis, no knowledge is conclusive, at least not in any Cartesian sense.

At that level, one could argue that any given item of knowledge is merely a reasonable guess, scientific or otherwise. However, physics is capable of a degree of justification rarely encountered in everyday life – because of the many different lines of inquiry and wide body of evidence upon which a given scientific conclusion may be based. Granted, some of the tentative conclusions won’t achieve that level of justification – our discovery of the world is an ongoing process. But even the more tentative conclusions aren’t what we would ordinarily call “reasonable guesses” in as much as they are part of a systematic, scientific process.

Moreover, I would argue that much of our knowledge regarding the interaction of matter and electromagnetic fields is quite exacting and has achieved a great deal of justification. No doubt there are softer areas around the boundaries, but I suspect that such areas are far smaller than what many might suppose.

Comment by Timothy Chase — 21 Jul 2007 @ 12:54 PM

254. Re 248 Chase Not necessarily speaking for Alastair, I assert that, for all GHG (except some H2O spectra), we need to think in terms of lines, not bands, and we need to recognize that many of the interactions between the mechanical and radiative modes occur at the edge of, or out of, equilibrium, and that this failure is worst at the top and bottom of the atmosphere (and probably in violent storms). I have not found a LBL radiative transfer (RT) model that is self consistent in terms of temperature and the various coming and goings of quanta, but then I too have a lot to learn especially about all the historical work on RT.

Comment by Allan Ames — 21 Jul 2007 @ 12:59 PM

255. RE 251 Rod B Lines represent the absorption or emission probabilities versus frequency of particular transitions, characterized by a functional form and frequency. As the function broadens through radiationless interactions it usually flattens so as to preserve its integral over frequency, the oscillator strength.

Comment by Allan Ames — 21 Jul 2007 @ 1:13 PM

256. Allan Ames (#254) wrote:

Re 248 Chase Not necessarily speaking for Alastair, I assert that, for all GHG (except some H2O spectra), we need to think in terms of lines, not bands, and we need to recognize that many of the interactions between the mechanical and radiative modes occur at the edge of, or out of, equilibrium, and that this failure is worst at the top and bottom of the atmosphere (and probably in violent storms). I have not found a LBL radiative transfer (RT) model that is self consistent in terms of temperature and the various coming and goings of quanta, but then I too have a lot to learn especially about all the historical work on RT.

I agree.

Personally I think that the material you brought up is fascinating and I hope that I will be able to give it the attention that will be required to understand it. Likewise, assuming the motivation, I believe that Alastair stands roughly the same chance as myself to understand it. I hope he will make the effort, in part because this acknowledges that he is right – to a degree.

At the same time, I believe that understanding things in terms of LTE is useful, much like Gavin’s “Simple Model,” although it obviously much more accurate. At the same time, taking into account the nonlocal equilibrium effects will be an improvement – to the extent that they are not already taken into account.

At the same time, it should be acknowledged that the author of the second paper admits that what he is doing is an approximation which does not take into account all of the effects (such as scattering). As such it is also an approximation. But some approximations are nevertheless better than others.

But it should also be acknowledged that any calculations which are performed will generally be at the expense of other calculations. There are always tradeoffs. For example, to the extent that the equations which he gives are more complicated than those assumed in an LTE model, when they are implimented, we may need to reduce the resolution of the model in order to perform the calculations within a reasonable amount of time. Or alternatively, we may need to keep certain other simplifying assumptions involving convection or humidity as we increase computer power – even though we know that they are only approximations. But what is most important is the net accuracy of the projections and the rate at which the calculations are performed. As such this should be the standard by which we judge the various tradeoffs.

Comment by Timothy Chase — 21 Jul 2007 @ 4:49 PM

257. re Timothy (253) It makes sense that equilibrium exists in order for Planck temperature to equal Maxwell’s temp. But, your explanation is of apples, not oranges. I equate Planck temp (actually radiation) with his black/graybody radiation with the temp calculating the total radiation and the peak wavelength. This has nothing to do with re-emission and relaxation of bond energy ala greenhouse gasses…. Do it??

My photon labeling was far too coarse, and you’re correct. Most photons you could classify with a pretty good probability, but no certainty. Others one could miss terribly.

I wasn’t meaning to be philosophical with “we don’t know for sure”. Quantum physics says I might not be in Texas the next minute. Though I may have come across as too flippant (I’m a flippant kinda guy!), one of my areas of skepticism rests in the old argument: does adding more CO2 really significantly increase the absorption of the Earth’s radiation. And while the physics and science is predominately good and accurate, I’m not convinced that some of the physics and modeling assumptions and averages are known well enough or are close enough. It sounds like minutia, but this process of line/band spreading and adjusting of the global averages from it seems a highly leveraged process to me. While most here would strongly refute my contention, I am not convinced we know exactly (are “more accurately”, if you will) how it works. Though the understanding and classification of molecular bonding energy levels (in the thousands) seems quite detailed (no way for me to know how accurate, but I have no reason to dispute it), the absorption physics is too fuzzy for me. It still is line absorption, not “band” absorption and some say that the granularity is very small, e.g. absorbtivity (??) going for near 1 to near 0 in less than 0.01um. It would be a massive undertaking and likely days of supercomputer time to determine absorption factors and numerically integrate the marginal absorption over the entire IR with varying pressures and temperatures… and then project that globally and annually for 100 years. Maybe the averages and assumptions ― “5.4ln[C/C0] is about right” ― are O.K. But I’m skeptical. Teeny-tiny tweaks could have a multiplied leveraged effect on the global situation — and warming.

Comment by Rod B — 21 Jul 2007 @ 5:53 PM

258. Rod B (257) wrote:

re Timothy (253) It makes sense that equilibrium exists in order for Planck temperature to equal Maxwell’s temp. But, your explanation is of apples, not oranges. I equate Planck temp (actually radiation) with his black/graybody radiation with the temp calculating the total radiation and the peak wavelength. This has nothing to do with re-emission and relaxation of bond energy ala greenhouse gasses…. Do it??

I would say that a blackbody absorbs and re-emits radiation at all parts of the spectra – although the peak is determined by the temperature. Greenhouse gases are realistic bodies which absorb and re-emit only in certain parts of the spectra.

I am going off of a discussion in A Saturated Gassy Argument. Like you, I believed that the blackbody curve had little to do with absorption and re-emission by gases.

In response to something written by DeWitt Payne:

“…since gases do not emit any blackbody radiation.”

I don’t think this is strictly accurate. For any part of the spectrum where absorptivity/emissivity is close to 1 (saturated?), the shape of the emission spectrum in that wavelength range will (I think) be the blackbody spectrum for that temperature. In the Archer Modtran calculator, if you set the altitude to zero, look up and vary the surface offset temperature (tropical atmosphere, other settings default) from 0 to 30 degrees (to increase the specific humidity) the IR emission spectrum looks more and more like a blackbody spectrum and is practically indistinguishable at an offset temperature of 30 degrees. Eli or anyone, comments?
DeWitt Payne, 166

At the same time, I don’t quite see how gases could become blackbodies even within a given range of wavelengths.

Timothy Chase, 170

It’s my understanding that anything that isn’t perfectly transparent or perfectly reflective, i.e. an emissivity/absorptivity greater than zero, is a blackbody. When the emissivity is less than one, the emission spectrum will show fine structure as the emissivity of each line varies with wavelength.

DeWitt Payne, 173

Bart Paul Levenson then corrected Payne:

No. A blackbody has an emissivity/absorptivity of 1. A body with 0 180

However, a more extensive discussion of the same topic took place between Ray Ladbury and Ray Pierrehumbert in 154.

Anyway, I will respond to the rest of your post, but it probably helps to break things up.

Comment by Timothy Chase — 21 Jul 2007 @ 8:07 PM

259. PS to 258

I had cut off the bit by Bart Paul Levenson.

The full quote is from the discussion in A Saturated Gassy Arguement is:

No. A blackbody has an emissivity/absorptivity of 1. A body with 0 < e < 1 is a graybody. A body with e varying with wavelength is a realistic body.

Hopefully that will help my post make a little more sense.

In essence, the atmosphere is a realistic body which absorbs and re-emits radiation by means of greenhouse gases in the infrared where the radiation is “realistic body” radiation. As part of the atmosphere, all greenhouse gases have the same temperature, but different gases absorb and emit radiation in different parts of the spectra.

This is covered in more detail in the exchange between Ray Ladbury and Ray Pierrehumbert in the post #154 from the discussion following A Saturated Gassy Argument.

Comment by Timothy Chase — 21 Jul 2007 @ 8:59 PM

260. [[Why wouldn’t the uncertainty principle just as easily/often narrow the lines. ]]

Because the uncertainty is in the position of the lines, not their width. The broadening is a secondary effect, not a primary reality.

Comment by Barton Paul Levenson — 22 Jul 2007 @ 6:50 AM

261. [[does adding more CO2 really significantly increase the absorption of the Earth’s radiation.]]

Yes.

Comment by Barton Paul Levenson — 22 Jul 2007 @ 7:06 AM

262. Sources of Uncertainty

Rod B (#257) wrote:

I wasn’t meaning to be philosophical with “we don’t know for sure”. Quantum physics says I might not be in Texas the next minute. Though I may have come across as too flippant (I’m a flippant kinda guy!), …

Not a problem. But I try to be accurate, and I have the deepest respect for physics – although once you get to things like string theory I am not so sure. I guess I would have to say that the jury is still out.

… one of my areas of skepticism rests in the old argument: does adding more CO2 really significantly increase the absorption of the Earth’s radiation.

[Quick Note: I am probably going to keep my response to just the quotes above – as it has already gotten rather long.]

Well, we know that it should – and by a little over a degree Celsius, directly. However, some of the calculations that would have to be performed to “derive” a number like climate sensitivity from well-established physics and the exact geometry of the land, oceans, etc. would obviously be far too complex. Not that they actually use this particular “constant” anyway, not for the models themselves, instead it is something which will fall out of the calculations of a given model. Likewise, the “adjustment” of global averages in terms of temperature isn’t something that one would plug into the models. But I don’t know enough to say how they handle the initial temperatures when they begin a given run.

What follows are some of the sources of uncertainty which I can see, but it should be kept in mind that this is coming from someone who is not an expert, but merely on the outside looking in.

There are numerous points at which they have to make “simplifying assumptions,” for example, in calculating fluid behavior, but these are no doubt fairly good approximations. But I believe that one of the bigger sources of uncertainty is the matter of resolution, that is, how coarse is the grid? How many layers do you divide the atmosphere into?

The stratosphere is divided into roughly a handful at this point – not that this will have that much of an effect upon the behavior of the system. The troposphere more like fifteen. How many layers do you divide the ocean into? This varies. And not all of the grid is spatial, either. For example, they no doubt have a grid of sorts for the treatment of spectra. I believe that is what the “radiation codes” are about.

To make the calculations “exact” one would have to have a continuous grid in space and time. But even then it wouldn’t be exact insofar as you wouldn’t know exactly how matter is layed out, for example, in terms of the atmospheric constituents. Then of course calculating the behavior of ice is not something that one would do in terms of the fundamental equations of quantum mechanics, not for some time to come. And something similar would no doubt come into play with the radiation codes as well. How do the various atmospheric constituents interact?

Though the understanding and classification of molecular bonding energy levels (in the thousands) seems quite detailed (no way for me to know how accurate, but I have no reason to dispute it), the absorption physics is too fuzzy for me. It still is line absorption, not “band” absorption and some say that the granularity is very small, e.g. absorbtivity (??) going for near 1 to near 0 in less than 0.01um.

Well, the fundamental physics is known very exactly, but how molecules interact with one another even when they are molecules of the same exact makeup would be beyond our ability to solve exactly. Physicists have done something like this for standing water, analyzing its behavior reductionistically in terms of quantum mechanics such that the actual properties of water fall out of the equations, but even then there must be a grid of some sort and supercomputer power. Nevertheless, quite an accomplishment – and something which I find completely mind-boggling.

But the climate calculations involving ice doesn’t even begin to approach this in terms of its level of complexity. In fact, as the climate models have been done so far, all ice is the same. It melts uniformly – although Hansen has been doing calculations involving the effects of carbon pollution on the albedo of ice – and has found that in terms of current trends in the arctic this is as important as the increased levels of thermal radiation being absorbed by ice. But then you want to incorporate the nonlinear response of ice to its environment – and they are only beginning to work on this.

Likewise, Kirchoff’sLlaw is an approximation of sorts – a fair approximation, but an approximation nevertheless. In actuality climate modelers have already moved beyond Kirchoff’s Law in their calculations. Optics involving nonlocal thermodynamic equilibria. In fact, someone has been able to treate the problem in terms of exact integrals – but found it necessary to avoid dealing with certain phenomena such as scattering.

To look at a completely different level which we wouldn’t even begin to calculate in terms of physics, there is the behavior of organic matter (chum experiments in which one attempts to determine how much methane and carbon dioxide will be release with different degrees of mixing and at different temperatures. With this sort of thing they have to follow a well-defined recipe so that their experiments are reproducible..

But how about life? How many species of plants do you incorporate? On land? In the ocean? How many distinct levels of density do you assume? How do you model their responses? They are already dealing with these issues – and how one has to limit the calculations in terms of resolution. If I remember correctly, the number of species roughly a handful – although no doubt this will change. But then what about economies? Actually combined economy/climate model calculations are already being performed.

Given all this complexity, the calculations seem staggering. And currently we have computers capable of performing trillions of calculations per second, so this is doable at a certain level of approximation. Moreover, many of the “errors” with respect to one set of calculations will tend to be cancelled out by others – in the same way that the absolute level of error in predicting something which follows a bell distribution will grow as the square root of the population.

But more accuracy is always preferable. Nevertheless, anytime you try and improve upon a particular set of calculations, you have to keep in mind the fact that if the calculations are more intensive there, then certain other calculations will have to have some of their accuracy sacrificed, or the calculations will take longer or you will have to invest more in computers.

However, they do not tweak their calculations to make them fit the observed behavior of the climate system at a particular place or at a particular time. The equations are generalized. They may tweak global parameters – but their are very few of these to tweak. If they are to improve the modeling of climate behavior, the only thing they can do is improve the modeling in terms of the generalized equations, improve the initial data, or improve the resolution in one way or another.

As far as objectively judging the accuracy of their models in projecting future trends, there are a number of things they can do. They can compare their modeled behavior to actual empirical measurements and observations. They may perform multiple runs with slightly varied initial conditions to get a sense of how sensitive the system is to nonlinear behavior. They may improve the resolution for specific sets of runs. Or they can compare the trends projected by one model with those of another. In fact they do all of these.

But as for the accuracy of their calculations over a hundred years, Gavin and other climatologists with making calculations that far into the future – although we undoubtedly get a good sense of the trends which will be involved. Actually the biggest source of uncertainty lies in population, economies in terms of their behavior and response, and the extent to which we are able to control our emissions. These sources of uncertainty are in fact fairly negligible – for the next forty years. What we do today and for the next two or three decades will have very little impact upon what happens over this period of time.

But after that things begin to diverge – and more dramatically over time – largely due to what we do today and for the next several decades.

Comment by Timothy Chase — 22 Jul 2007 @ 3:19 PM

263. I’m almost getting there. There are still some details that seem poorly defined. First, I assume Planck radiation, commonly but technically incorrectly called “blackbody radiation”, is a function of wavelength and temperature of the radiating body with the emission by wavelength a smooth continuous curve (with the oddly skewed bell shape) over the entire spectrum but practically within a band with edges determined by the temperature. That same body will absorb the exact same radiation profile. Graybodies are those with emissivity less than one (and greater than zero) which means the total radiation (and absorption) is less than that of a blackbody, but such radiation curve is similar in shape, continuity and edges. For all practical purposes the emissivity does not vary by wavelength but is constant over the spectrum. This is commonly accepted for solids and liquids, but there is some uncertainty about gasses — I’ve seen many yesses and no’s. Gasses certainly absorb Planck-type radiation, like that emanating from the surface of the Earth in a smooth continuous function based on the temperature of the surface, which is derived from the 1/2mv2 kinetic energy of the ionized/charged molecules, atoms, and electron within the surface.

But gasses do not absorb radiation in a smooth continuous function, but rather in a highly discontinuous function strongly determined by the wavelength of a particular discrete extremely narrow band of this radiation. Secondly, it does not go into the 1/2mv2 kinetic energy of the gas molecule (and in turn increase its “pudding” which as I’ve said is “temperature” if it happens to a bunch of molecules — sorry!), but rather into the molecular bond energy (translation or bending oscillation) or the molecular rotational energy, or, in rare examples at shorter wavelengths, the electron energy level. Best guess for now is that this energy absorption does not increase the molecules’ temperature, though there have been conflicting opinions stated here. If it does not increase the temperature of the mole of molecules, it can’t be absorbed Planck-type radiation — though before it got absorbed and transferred it was. Does anybody accept or refute this? OTOH maybe it counts as Planck because it could later (soon though) turn into temperature with collisions and transferring bond energy from molecule #1 to kinetic energy in molecule #2. Does anybody know for sure??

Following the above thought then, if the molecule re-emits the earlier absorbed energy (again taking it from its bonds or rotation), the emission can’t (???) be Planck-type. Or is it? If not, I have no idea what type it is.

Going one further, assuming the above, can not the gasses still emit Planck-type radiation, separate from its bonding energy, by virtue of the gas’s kinetic energy from those molecules that have been ionized or have dipole moments (???? don’t know about that…) and based on the temperature of the atmosphere’s “surface”, wherever that might be? And this radiation will be a continuous function, not a pile of discontinuous discrete lines, and tend to cool the molecules.

I understand that some of the visible spectrum is absorbed by clouds (liquid) and that some of that is re-emitted toward Earth. Is this absorption wavelength specific or is it Planck-type coming and going??

Any thoughts?

ps one of the things that is confusing things is the generic and common use of “blackbody” to mean “Planck-type”, which comes in both blackbody (e = 1.0) and graybody (e not have the same temperature… do they???

A different quicky: Barton (260) says: “…the uncertainty [principle] is in the position of the lines, not their width…

Got ya. Makes sense. I assume the uncertainty could move the line left or right, in any case having the effect of broadening the original line. Thanks.

Comment by Rod B — 22 Jul 2007 @ 4:36 PM

264. repeating my ps paragraph from 263, which got clobbered by my attampt to say “e [less than sign] 1.0”

ps one of the things that is confusing things is the generic and common use of “blackbody” to mean “Planck-type”, which comes in both blackbody (e = 1.0) and graybody (e less than 1.0) forms. I suggest we adopt the popular but technically incorrect term of blackbody, and when we mean a true blackbody we call it a “blackbody with an e of 1.0”. But somethin’. I could buy Barton’s “realistic” term, though I don’t accept the e varying with wavelength part (with the possible exception of gasses ala the pontificating above.)

Comment by Rod B — 22 Jul 2007 @ 4:49 PM

265. Rod B (#264) wrote:

ps one of the things that is confusing things is the generic and common use of “blackbody” to mean “Planck-type”, which comes in both blackbody (e = 1.0) and graybody (e less than 1.0) forms. I suggest we adopt the popular but technically incorrect term of blackbody, and when we mean a true blackbody we call it a “blackbody with an e of 1.0”. But somethin’. I could buy Barton’s “realistic” term, though I don’t accept the e varying with wavelength part (with the possible exception of gasses ala the pontificating above.)

I will have to look at the earlier post a little later – currently I have some laundry going and a big project of sorts which I am trying to get a bit done on during the weekends. (Last week was a little busy for me, too – several twelve hour days at work.)

However, as far as realistic bodies go, here is a somewhat oversimplified example, but consider a blue book…

The book absorbs all visible light except for the blue light that it scatters. Now its emissivity will be high in those other parts of the visible spectrum, so you might expect it to glow in those. Except of course it is room temperature, and the vast majority of light which it re-emits will be in the infrared part of the spectrum. The blackbody curve simply becomes too shallow by the time you reach the visible part of the spectrum. So it absorbs the other parts of the visible light without emitting in them.

However, if you have bodies which are good absorbers of infrared light in certain parts of the electromagnetic spectra, they should also be equally good emitters in the same parts of those spectra. And the same principle applies, more or less, to gases. Anyway, this is how I understand it. There will be more variation in the case of gases, for example, due to the effects of nonlocal thermodynamic equilibria, but this becomes important only at certain pressures and temperatures for various gases. So long as nonlocal thermodynamic equilibria are not an issue, one should be able to calculate an emissivity at each wavelength which will result from the mixture of gases, where all of the gases are at the same temperature, and thereby treate the atmosphere at a given temperature and pressure as a realistic body.

*

We live in a world that is stranger than we can see. The land glows brightly in the infrared, appearing white to eyes specially adapted to it. Above the land at somewhat longer wavelengths the atmosphere is like a glowing fog, strangely-colored, with more distant objects shading to different colors than those nearby due to their different colors being faded and washed out by the vapors inbetween.

Comment by Timothy Chase — 22 Jul 2007 @ 7:23 PM

266. [[But gasses do not absorb radiation in a smooth continuous function, but rather in a highly discontinuous function strongly determined by the wavelength of a particular discrete extremely narrow band of this radiation. Secondly, it does not go into the 1/2mv2 kinetic energy of the gas molecule (and in turn increase its “pudding” which as I’ve said is “temperature” if it happens to a bunch of molecules — sorry!), but rather into the molecular bond energy (translation or bending oscillation) or the molecular rotational energy, or, in rare examples at shorter wavelengths, the electron energy level. ]]

Almost all (> 99%) of increases in a molecule’s level of energy will be quickly collided out and distributed among neighboring molecules.

Comment by Barton Paul Levenson — 23 Jul 2007 @ 6:32 AM

267. Re #266 Barton,

do you have a reference for “Almost all (> 99%) of increases in a molecule’s level of energy will be quickly collided out and distributed among neighboring molecules.”? I would very much like to see the calculations, or the theory behind that.

Cheers, Alastair.

Comment by Alastair McDonald — 23 Jul 2007 @ 8:18 AM

268. re #265 (Timothy) You raise some interesting things to contemplate. Is it accurate to say a true blackbody (at least liquids and solids for now) will absorb exactly as it emits, but does not emit exactly as it absorbs, ala the blue book (black book??) which reflects/scatters visible blue but emits no visible wavelengths despite it absorption of them.

2) The blue book is clearly absorbing with an absorption coefficient dependent on wavelength — if its reflecting blue it can’t be absorbing it. Does this put a chink in my assertion that absorption coefficients are independent of wavelength (again talking of only liquids and gasses)? Would it also apply to emissivity (if we’re talking of something less than a pure blackbody where a = e)?

3) I would think a gas emission would be at exactly the same wavelength as its absorption (ignoring the “spreading for the sake of discussion): it absorbs discrete wavelengths based on its internal discrete energy levels, and can emit by relaxing those same energy levels — much like the light absorbed and emitted via electron level energy changes. delta Energy = hf, coming and going. Then I’m still contending (asking?) that this discrete emission for a gas is not Planck radiation, by definition, which emitted radiation is determined only by temperature stemming from a Maxwell distribution of kinetic energy. Also I would think the gas’ discrete specific radiation is independent of temperature, other than its potential goes away if the molecule loses the internal bonding energy through collisions — which might increase temperatures.

Comment by Rod B — 23 Jul 2007 @ 9:37 AM

269. re #266 Barton says “Almost all (> 99%) of increases in a molecule’s level of energy will be quickly collided out and distributed among neighboring molecules.”

This would be in agreement with (one of) my contention. Though you are also saying that the molecule’s re-emission is theoretical and highly unlikely, as opposed to transferring its energy through molecular collisions, which can increase the atmospheric temperature. Do you concur with the last part — increasing temp?

Comment by Rod B — 23 Jul 2007 @ 9:53 AM

270. Alistair, Re #219

I do not think the wings of the lines are irrelevant. The whole point of the original article was the soft saturation of the absorption in raypierres plot of absorption with distance. The wings of the lines cause this, mainly.

In #230 you say: *I am arguing that the emissions to space come from between the lines and from between the bands.*

Exactly, that is why the wings are important. As they creep in with concentration increase, it is like a set of blinds closing. The line centers are opaque. this forces the blackbody trying to radiate between them to adjust to higher temperature to maintain power balance.

The other contribution to the soft absorption satuation tail, in principle, are weaker lines, at the edge of the band, which come into prominence as the concentration (optical density) is increased. However, these do not contribute as much as the main center lines to increase in integrated absorption with concentration of CO2.

DO the following calculation. Assume two lines, 1 and 2 with identical lorentzian lineshapes, and cross-sections sigma1 and sigma2. sigma2 = R sigma1 with R

Comment by Dave Dougherty — 23 Jul 2007 @ 11:44 AM

271. Re #270 seems like my post got cut off …

DO the following calculation. Assume two lines, 1 and 2 with identical lorentzian lineshapes, and cross-sections sigma1 and sigma2. sigma2 = R sigma1 with R

[Response: Don’t know what is going on here, but be careful with < signs (use & l t ; instead….) – gavin]

Comment by Dave D — 23 Jul 2007 @ 2:54 PM

272. The “less than” followed by a “greater than” sign get interpreted as HTML and I think it just assumes you’re trying to code the end of a paragraph and truncates. Might check your /View/Character Encoding and see what’s being done to what you type by the too helpful software.

Try “view source” on this page — when I do that I see Gavin’s successful “less than” sign as code, and his explanation how to write it as different code.

I don’t know if I can even paste back what Gavin wrote, if there’s nothing following this line, try ‘view source’ for this as well:
be careful with < signs (use & l t ; instead…

Comment by Hank Roberts — 23 Jul 2007 @ 3:17 PM

273. Re #270 & 271. How about you showing the calculations? I doubt I could do them anyway.

Comment by Alastair McDonald — 23 Jul 2007 @ 3:37 PM

274. Re #270 and #271 Sorry to crud up the thread. Hope this is better.

Do the following calculation: Assume two lines, 1 and 2 with identical lorentzian lineshapes, and cross-sections sigma1 and sigma2. sigma2 = R sigma1 with R less than 1.

We are interested in the derivative, with concentration (or OD at line center) of the integrated absorption for each line. Form the ratio and call it Q.

Plot Q vs. OD= sigma1*n*L for various values of R. I swept OD from 0.1 to 500 ( 10km/25m = 400) for R rangine from 1 to 1e-5.

For R=1, Q=1, of course. For OD very small ( say 0.1) then Q equals R. As the OD increases, Q increases, but never gets to 1. This means that a strong line alweays contributes more to the incremental absorption than a weak line at any concentration.

The exact amount depends on the details of the far wing shape of the lines ( about 10 to 20 times broadening parameter (HWHM)at the very large OD’s for main lines of CO2 at 15um. this is why knowing the wings is so important. Plot the function we are integrating. It is basically the lineshape with the center stomped out by the exp(-sigma*n*L* s(f)) factor (s(f) being the lineshape).

Because the weak lines at the edge of the branch( those at about 50% transmission since we are only concerned with ~doubling CO2) are fewer in number than the main lines in the center of the branch, their contribution will be further reduced.

I still think checking the exact lineshapes, rather than just assuming Lorentzian’s as HITRAN does, is important. We have this part of the problem under our control. It just takes time and spectrometers. Why introduce error right at the begining of the GW problem if we don’t have to. This directly impacts the Radiative forcing curves vs. CO2 concentration from the IPCC report which is important for deciding if there is any time left to do anything, and how high temps may go if we don’t.

There emay be a lot of lines, but the problem is surely much much smaller than the human genome project, and is maybe just as important.

Comment by Dave D — 23 Jul 2007 @ 4:10 PM

275. Re #273

I wrote some short Matlab files. If anyone wants to check what I did, I can send them.

Here is psuesdo Mathematica language of what I’m talking about:

The integrated tranmisions up through atmosphere for each line. (Yes n is a function of L, not important here, just total OD=sigma*n*L).

n=CO2 concentration
L= thickness of gas slab we are looking through
sigma = absortption cross section for each line

T1(f) = Integral( df exp(- sigma1 * n* L *s(f)) )
T2(f) = Integral( df exp(- sigma2 * n* L *s(f)) )

Taking s(f) as lorentzian: s(f) = 1/( (f/G)^2 +1).
Take G=1 to normalize frequency to broadening parameter.
OK since we are only interested in ratio, below.

dT1/dn = Integral( df (-sigma1 * s(f)*L) exp(- sigma1 * n* L *s(f)) )

dT2/dn = Integral( df (-sigma2 * s(f)*L) exp(- R*sigma1 * n* L *s(f)) )

These integrals are just the lineshape with the center suppressed by the tranmission. Only the wings contribute when sigma*n*L is big.

A1(2) = 1-T1(2) for absorption.

Q = (dT2/dn) / (dT1/dn).

Plot Q vs. n or (OD =sigma*n*L in range 0.1 to 500),
for variuos R’s like R=1 (equal line strengths) down to R=1e-5 ( a very weak line).

Comment by Dave D — 23 Jul 2007 @ 4:24 PM

276. Dave,

Thanks for letting us see those calculations. It helps in understanding your objections. However, they are not really valid. Let me explain why.

First, the Lorenzian line shape is only a first approximation. Calculation of the real wave shape is even more complicated than that. Secondly, the real wave shapes are being calculated and used in the models. The HITRAN data base only provides the line positions. When I criticise the radiation models I am told that I am wrong because of the care that is taken with these lines, but they are irrelvant to my objections.

On page 99 of Goody & Yung (the planetary radiation bible) they write:

If the composition is held constant, all of the ni [number density of the ith species] are proportional to the total pressure and (3.51) gives the important result, common to all impact theories, that the line width is proportional to the pressure [my emphasis]

Of course, doubling the CO2 concentration does not maintain a constant composition, but the change in pressure as a result of increasing CO2 from 280 ppm to 560 ppm is trivial. An increase in atmospheric pressure of 0.0280% is not going to cause the changes in global climate which are happening now. That is less than 1 mb!

The main greenhouse gas on Earth is water vapour. The way that CO2 influences the water vapour concentration and so the clouds (and the surface ice) is the answer to how CO2 affects the global climate. The venetian blind fine adjustment that you describe does happen, but it is trivial when compared with the velvet curtain effect produced by clouds.

The CO2 lines are saturated – optically thick – and this saturation happens within 30 m of the surface. Double the concentration and the saturation happens in the top 15 m. The air at the surface of the Earth gets warmer, there is more evaporation and less ice cover.

All that fancy maths with the Hamiltonians they use is irrelevant. It won’t stop the Arctic sea ice melting. The Arctic will become an ocean again with a maritime climate rather than the pseudo continental climate it has at present. The effects of that will spread to the Southern Hemisphere through the tele-connections that are only now being discovered.

The summer melt of the Arctic sea-ice is proceeding at an unprecedented rate. You have all been warned!

Comment by Alastair McDonald — 24 Jul 2007 @ 5:17 AM

277. [[Re #266 Barton,

do you have a reference for “Almost all (> 99%) of increases in a molecule’s level of energy will be quickly collided out and distributed among neighboring molecules.”? I would very much like to see the calculations, or the theory behind that.]]

“Under atmospheric conditions of interest, the time between collisions [is] usually much shorter than the natural transition lifetime.”

Hanel, R.A. 2003. Exploration of the Solar System by Infrared Remote Sensing. Cambridge, UK: Cambridge Univ. Press. p. 100.

For the amount of energy dE transferred, we can calculate the probability:

P(dE) = exp(-dE / (k T))

where k is the Boltzmann constant and the T the “bath temperature.” Now note that there are approximately 2600 molecules of nitrogen and oxygen for every molecule of carbon dioxide, and you can conclude that thermalization will dominate over excitation — radiative as well as collisional.

Comment by Barton Paul Levenson — 24 Jul 2007 @ 7:22 AM

278. [[re #266 Barton says “Almost all (> 99%) of increases in a molecule’s level of energy will be quickly collided out and distributed among neighboring molecules.”

This would be in agreement with (one of) my contention. Though you are also saying that the molecule’s re-emission is theoretical and highly unlikely, as opposed to transferring its energy through molecular collisions, which can increase the atmospheric temperature. Do you concur with the last part — increasing temp?]]

Certainly the infrared energy absorbed by the atmosphere results in raising its temperature. But radiative loss of that increased temperature mostly has to go through the greenhouse gases, since nitrogen and oxygen aren’t very radiatively active. A simple model would have a layer of atmosphere absoring infrared and then “reradiating it,” but in reality the process is a bit more complicated. The simple model gets the basic idea across, however, and for a sophisticated enough mathematical treatment can even get it quantitatively right.

Comment by Barton Paul Levenson — 24 Jul 2007 @ 7:25 AM

279. Thanks, Barton (278), but I’m confused. First off I was talking about an excited (by IR radiation) molecule colliding with another (any kind) and transferring its delta energy to it as kinetic and thereby increasing the temperature of the local atmosphere. Does this happen? Secondly, I thought you said the relaxation by re-emission of the IR very seldom happens because the excited molecule will, by magnitudes, be likely to lose its energy via collision long before it re-emits. Is this accurate?

Third, you say “Certainly the infrared energy absorbed by the atmosphere results in raising its temperature. …” Does the absorption and transfer of IR energy to the bond and rotational internal energy of the molecule raise its temperature? (For discussion, ignore the silly “one molecule can’t have temperature” debate.) I’ve pursued that a number of times here to either no avail or varying answers. I think the crux (cruxes??) of the question is: 1) is it only 1/2mv2 kinetic energy that determines temperature (as opposed to internal potential, chemical, etc.)? 2) Do the vibrating bonds or rotating molecule (1/2Iw(omega)2???) constitute temperature affecting kinetic energy?

Comment by Rod B — 24 Jul 2007 @ 11:00 AM

280. a ps to my #279 to Barton: an omitted (typo) but most important word: it should have read “Thanks, but I’m confused. …”

Comment by Rod B — 24 Jul 2007 @ 11:05 AM

281. Re #279 Rod B

I have already given you the answer to your question about temperatures in #214. I will try to explain again without repeating myself.

Heat is a form of energy. For solids and liquids the energy is contained in the form of vibration of the atoms. If you pass an electric current through a solid it warms up by absorbing the energy and vibrating more. The increase in vibrations as a liquid heats up causes it to expand, and we measure temperature by seeing how much the liquid mercury has expanded, in a normal thermometer.

Electromagnetic radiation is also a form of energy, and when the sun shines on a surface, the surface gets warmer because because it is absorbing the solar energy. The solar electromagnetic radiation is absorbed by the surface and causes atomic vibrations.

When the heat is great enough the atoms will have enough energy to break away from the surface and form a gas. In other words the substance has evaporated. In that state the atoms are formed into molecules with kinetic energy. They are are moving about at random. When they hit other molecules they share their kinetic energy in the collisions. These molecules can be other gas molecules or the glass of a common thermometer. Thus the thermometer is registering the average kinetic energy of the gas molecules.

All the gas molecules are traveling at different speeds, and Maxwell worked out a mathematical function which describes the distribution of those speeds. So the temperature of a gas measured with a thermometer is called its Maxwellian temperature.

Kirchhoff found that the radiation entering a solid equals the radiation leaving it, provided its temperature is not changing. This is of course just a matter of the conservation of energy. The energy entering a body must equal that leaving if none is being stored. What Kirchhoff also realised was that if you have two plates facing each other, both at the same temperatures, then there must be a function which describes the radiation at each wavelength. This led to Planck’s function, quantum theory, and the measurement of temperature by matching it to Planck’s function. The temperature of a body measured using it radiation is called its blackbody, or Planckian temperature. For most solids the Maxwellian temperature and the Planckian temperature are equal.

The question then is, does the Maxwellian temperature of the Earth’s atmosphere equal its Planckian temperature? If so it is in local thermodynamic equilibrium (LTE.)

HTH,

Cheers, Alastair.

Comment by Alastair McDonald — 24 Jul 2007 @ 3:51 PM

282. Re #276
This thread is almost dead, so no point is arguing much further.

The lorentzain expression is the only theorectical lineshape I am aware of. It arises from very simple assumptions about the statsitics ofthe collisions, and the collision duration compared to the center frequency of the line. Any thing else would have to be experimentally determined.

When I first got interested in AGW I *discovered* for myself the gray-body saturation problem. In simple gray-body calculations ( to get the 33-35K greenhouse effect) you can come to the conclusion that, having increased CO@ by 30%, we are more that halfway to saturation. That is, by adding just a bit more, we make the IR abosrption completely opaque, thus reaching maximum effect. If we measure 0.6C temp rise, maybe the max would be only double that.

I wrote as much to Gavin, and he sent me a reference to this paper, which set me straight on how things really work:

S. A. Clough, M. J. Iacono, Journal of Geographical Research, vol. 100, No D8, pg 16519-16535, 1995.

This is a line-by-line calculation of radiatve cooling rates. Right on page 2 of the paper, the authors say they are using data from HITRAN for line strengths and line widths. The mention they are doing clever numerical things to speed calculation, while only introdcuing 0.5% error which they say is small compare to the 5% error in the HITRAN data.

I am saying, possibly … maybe, the wings are not what HITRAN says, and therefore the error in the line-by-line calculation could be much bigger.

You are right about how strong the CO2 absorption is, but you can always detune from line center, between the lines, to a point where the transmission is much lower, and the aborption length is much much longer than 15 to 30m. The whole line does not saturate (in this sense) at the same time. In between is where the thermal blackbody radiation from the lower atmosphere and surface has to try leak out.

I bet that CO2 and H2O lines have never been measured accurate so far doen in the wings. Why would a chemist do it? The line positions are most important to them after all. Broadening is a *dirt* effect, unless you are interested in GW.

An experiment with broadband light would do all the integrals automatically, but you have to be able to measure the absolute transmission levels very carefully.
Measuring lineshapes is easier because it is a relative measurement (assuming the strnegth is already known). But there are so many lines to resolve, and you already have to do many air pressures, and CO2 concentrations to get the wings right.

I am not suggesting that the CO2 broadening is proportional to the CO2 concentration. CO2 is a trace gas. The broadening I use in independent of n(CO2) is my calculation, of course. It does say, however, that a strong line always dominates a weak line with the same linewidth at any pressure altitude.

I understand ( and believe from Manabe and Weatherall) the H2O feeedaback. I do agree that clouds are not accurately taken into account. Wasn’t there a bar graph in one of the IPCC reports which had about 10 different models that could’nt agree on the sign, much less the magnitude of the cloud feedback? Is there any prdiction on the change in dewpoint spread with increased CO2 ( with no cloud effects)?

I do agree that many times *experts* try to change the topic on you when you are trying to get a straight answer on a narrow, well-posed problem. This is because they really have’nt thought about the foundations on the field they work in. 99% chance you eventually get someone who finally knows the answer, and will tell you that your are wrong. But maybe there is something that was overlooked … Thats how I learn anyway.

Last, I don’t know what the conclusion would be if the wings were higher or lower (which I suspect) than Lorentzian. Would it be good or bad news for increasing temps? If saturation is slower, we may have more time to burn, but the final temp could be much higher and more dangerous. I don’t know.

Comment by Dave Dougherty — 25 Jul 2007 @ 1:41 AM

283. > what HITRAN says
I looked it up. Here’s the reference I posted in the Part I thread on that question

Comment by Hank Roberts — 25 Jul 2007 @ 9:45 AM

284. re my 279 and Alastair’s 214 & 281: a quick clarification: I assume you mean molecular vibration of a solid/liquid. Or does the incoming solar radiation exite the intramolecular atomic bonds also. And if so do those vibrating intramolecular bonds also increase the temperature? Which is just like one of my questions re gasses, which your posts came close to answering, but didn’t quite make it (or I just couldn’t see it…if so , sorry).
I’ll be more simple and specific: Start with a mole of CO2 molecules, darting about and crashing into each other (ignore the walls) and distributing their kinetic energy according to Maxwell/Boltzman distribution, which in turn dictates the measurable temperature of the mole. Now blast it with a pile of infrared radiation, some of which will be absorbed by a bunch of the molecules and be seen as internal bond vibrational (lateral or bending) or molecular rotational energy only — no molecular kinetic or electron level energy changes (for now). The QUESTION is: does the temperature of the gas increase?

If so, do both types — internal vibrational and rotation — increase the temperature, or just one? Or, instead, might the temperature of the mole increase only after collisions when one molecule’s internal bond energy transfers to another (CO2) as molecular kinetic energy. Or can maybe the IR radiation transfer directly to molecular kinetic energy of CO2 molecules, which would increase the sample’s temperature?

Another aside clarification: I’m under the impression the excitation of electron energy levels does NOT increase the temperature of a gas. Right of wrong?? How about if the electron is blasted free ala ionization and now one has charged atoms/molecules and electrons speeding around…(and presumably giving off Planckian radiation)?

Comment by Rod B — 25 Jul 2007 @ 12:48 PM

285. I think you’re back on the definition of temperature. The answer to your question just depends on which definition you’re using — and whether it’s applicable to a single atom or molecule and over what span of time.

“Temperature” isn’t a Platonic object. It’s how something behaves — on average, during some elapsed time.

Blast your imagined mole of CO2 with infrared photons, and some photons are absorbed.

Now — how would you _know_ the temperature of the CO2?

— Put a thermometer into it? Oops, the CO2 molecules have banged against the thermometer, violating your hypothetical requirement.

— Measure its infrared radiation? Oops, the CO2 molecules have emitted photons, violating your hypothetical requirement.

It will be warmer, when you measure it.

Comment by Hank Roberts — 25 Jul 2007 @ 2:06 PM

286. [[The lorentzain expression is the only theorectical lineshape I am aware of. It arises from very simple assumptions about the statsitics ofthe collisions, and the collision duration compared to the center frequency of the line. Any thing else would have to be experimentally determined.]]

There is also the “Doppler line shape,” which is important at high altitudes, and the “Voigt line shape,” which covers both Lorentz and Doppler broadening. See if you can find a copy of Goody and Yung’s “Atmospheric Radiation” (1989) in your local university physics library. They have an extensive discussion of line shapes.

Comment by Barton Paul Levenson — 26 Jul 2007 @ 7:05 AM

287. Re #283 where Dave Dougherty Says:

This thread is almost dead, so no point is arguing much further.

Well, I am finding your perspective interesting, although it seems you are unwilling to consider these issues from mine :-(

In #277 Hank has found an interesting and relevant paper: Rothman et al. (2005) entitled History and future of the molecular spectroscopic databases. It makes several points which agree with things both of us have written. For instance on line shapes they write:

Historically, the line shape used by most of the line-by-line codes was either the Lorentz function (ignoring translational effects), the Doppler function (ignoring collisional effects), or the Voigt profile which is the convolution of the two.

I was not sure what you mean by the gray-body saturation problem. The gray-body concept was used as an approximation for the absorption of greenhouse gases by averaging the effect over the blackbody spectrum. This was abandoned for a band model, since averaging the absorption of a band which changes with height together with a window which has a fixed absorption of zero with height is invalid. Discovery that the bands were themselves composed of lines and gaps meant that the band models had to be abandoned and the LBL (line-by-line) approach is now used. However, although line broadening may remove the gaps, it will not remove the window. Therefore, and here I think I am agreeing with you, line broadening will not cause a major increase in the greenhouse effect.

However, you seem to be looking for reasons why the anthropogenic increase in greenhouse gas concentrations will not lead to severe changes in climate. I am looking to see what caused the dramatic climate changes in the geological past, and how it was that they seem to have been associated with changes in carbon dioxide (and methane.)

Returning to Rothman et al., they do agree with you again when they write:

3.3. The far wing problem

All the models discussed above have been developed within the sudden impact approximation. Consequently, they cannot
provide an accurate description of the far-wing absorption, where the effects of the collision durations are no longer negligible[20]. The importance of an accurate description of the far wings of the strong H2O rotational and vibrational bands for atmospheric modeling is a very well-known example.

And with my contention that this work has now advanced into the overlap of wings:

Here too, the databases have been vastly improved and for instance, the 2004 HITRAN compilation [12] provides flexible tools for taking into account line mixing in CO2 Q branches. However a number of effects remain unaccounted for and need to be included in the near future.

You wrote:

I bet that CO2 and H2O lines have never been measured accurate so far down in the wings. Why would a chemist do it? The line positions are most important to them after all. Broadening is a *dirt* effect, unless you are interested in GW.

But there is an interest in GW and I think Rothman et al. have shown that these issues are being addressed. OTOH, the big error is that the models all assume that the emissions match the absorptions for those lines (Kirchhoff’s Law.) In fact the re-emissions will be affected by collisions, but two chemistry books I have found which mentioned that problem did not discuss it writing that it was uninteresting!

The reason the modelers cannot get the clouds right is because their radiation model is wrong, but try telling the experts that and they will disagree, treating you as a poor fool, and explaining that the clouds are very complicated. The fact that their radiation model is very simple, radiation in equals radiation out at micro and macroscopic levels, seems to pass them by. The Manabe and Weatherall model you mention ignores feedbacks from latent heat and clouds and uses one column to describe a circular flowing system that requires a minimum to two columns. Worse, the lapse rate is effectively fixed at the value supplied as a parameter.

The odds of being right and the expert being wrong must be much lower than 1 in 100. I would rate it below one in a million. But that does not mean it cannot happen. People do get hit by lightening despite its improbability. Anyway, it is too late for me to get my ideas out. With levels of CO2 at their current values we are already getting floods, droughts and wildfires in America, Europe, Asia, and Australia. The scientists still do not know what causes abrupt climate change, but are too proud to listen to the thoughts of someone outside the box. We’re all doomed!

Comment by Alastair McDonald — 26 Jul 2007 @ 7:11 AM

288. Re 287 and the final para: The odds of being right and the expert being wrong must be much lower than 1 in 100. I would rate it below one in a million. But that does not mean it cannot happen. People do get hit by lightening despite its improbability

The difference here is that while the chances of a particular individual being hit by lightening are extremely slim, given the population of the world i’m guessing (and guessing is right since have v limited knowledge of stats) it’s not too unlikely someone of the 6 billion plus will be struck sometime. Since Alistair is trying to prove the experts wrong in one aspect of science rather than all areas the comparison should be in nominating a particular person who wil struck beforehand which is surely a completely different magnitude of probability?

Comment by kevin rutherford — 26 Jul 2007 @ 8:23 AM

289. Re #284 where Rod B Says:

Re my 279 and Alastair’s 214 & 281: a quick clarification: I assume you mean molecular vibration of a solid/liquid.

Typically solids consist of crystal latices of atoms. Molecules are more a feature of gases. The heat energy of a solid is due to the vibrations between the positively charged nuclei of the atoms. It is their unconstrained movement which causes the continuum spectrum of blackbody radiation. Gases have fixed oscilations (vibrations and rotations) that their molecules can make and so it is a spectrum of lines. A solid, think of an iron bar, is just a mass of atoms vibrating in all directions.

Liquids fit somewhere in between solids and gases.

Your mole of gas will have a temperature. Put a thermometer in the gas and it will register a value that matches the average kinetic energy of the molecules. So it has a kinetic or Maxwellian temperature. Its Planckian temperature or brightness temperature is zero because we are assuming that it is not emitting any radadiation. Now blast it with a pile of infrared radiation … no molecular kinetic or electron level energy changes (for now) Its Maxwellian temperature is unchanged, and its Planckian temperature is unchanged because it is still not emitting any radiation. But it has aquired vibrational and rotational energy so now it has a vibrational temperature and a rotational temperature. Its electronic temperature is still zero.

But go back to the mole of gas before it was blasted. The molecules are colliding and that will cause them to vibrate and rotate. So in fact the gas will already have had a vibrational temperature and a rotational temperature. But those excited states are not permanent and will result in emissions, so the gas would have had a Planckian temperature. So if you take a mass of a greenhouse gas without any walls around it, it will slowly radiate away its heat (kinetic energy) via excitation of its vibrations and rotations and their emission of photons. This is what happens at the top of the atmosphere.

The blast of IR would have increased both the vibrational and the rotational temperatures, but as those excited states relax back to their equilibrium values by collisions then the gas kinetic temperature will rise. If they relax by re-emitting their energy, then the Planckian temperature will be temporarily raised.

So, as you wrote: the temperature of the mole increases only after collisions when one molecule’s internal bond energy transfers to another (CO2) as molecular kinetic energy. It can also transfer to other air molecules such as O2 and N2 in the Earth’s atrmosphere, and the kinetic energy will be shared by and with other CO2 molecules in the atmospheres of Mars and Venus..

The electronic excitation behaves just like the vibrational and rotational energies. The molecule/atom/ion will have an electronic temperature/energy which will be shared with its vibrational and then rotational states until it is degraded into thermal (kinetic) energy. I would imagine ions only give off photons when they collide, or are hit by electrons.

Finally, I would avoid the term Planckian radiation for blackbody or cavity radiation. Planck’s function describes the intensity for each wavelength of the continuous radiation from a blackbody emitting at a specificed temperature. If you then use the intensity and wavelength of an emission from a gas you can determine a Planckian or brightness temperature, but the gas is emitting line radiation not blackbody continuum radiation.

Although I have possibly raised more questions than I have answered I HTH,

Cheers, Alastair.

Comment by Alastair McDonald — 26 Jul 2007 @ 10:22 AM

290. Alastair, rock, shale, dirt, etc are just “crystal latices of atoms”??? Not molecular?

Comment by Rod B — 26 Jul 2007 @ 12:27 PM

“… One way to know if what you have is a crystal or not is to break it. Having a crystal form is what is known as a ‘physical property’.”

Comment by Hank Roberts — 26 Jul 2007 @ 12:46 PM

292. Re #290-1 A grain of sand is just a crystal of silica, and even dust or a speck of mud consists of millions of atoms in fixed proportions, not just one molecule. The kinetic theory of gases and Avagadro’s Hypothesis all depend on gaseous molecules. And of course,minerals break down into ions not molecules when dissolved in water.

However, I should have written: Inorganic solids consists of a crystal latices. Here is a picture of rock http://en.wikipedia.org/wiki/Image:Gabbro_pmg_ss_2006.jpg seen in a thin section 30 microns thick. It is just a load of crystals.

Wikipedia says “In general, the type of chemical bonds which hold matter together differ between the states of matter. For a gas, the chemical bonds are not strong enough to hold atoms or molecules together, and thus a gas is a collection of independent, unbonded molecules which interact mainly by collision. In a liquid, Van der Waals’ forces or ionic interactions between molecules are strong enough to keep molecules in contact, but not strong enough to fix a particular structure, and the molecules can continually move with respect to each other. In a solid, metallic, covalent or ionic bonds provide cohesion between molecules, and the positions of atoms are fixed relative to each other over long time ranges. This being said, however, there is a great variety in the types of intermolecular bonds in the different materials classes: ceramics, metals, semiconductors or polymers, and each material or compound may be different.” http://en.wikipedia.org/wiki/State_of_matter

So in a solid the position of the atoms are fixed relative to each other and it is the vibration of their nuclei which causes the blackbody radiation.

Comment by Alastair McDonald — 26 Jul 2007 @ 1:41 PM

“In a solid the positions of the atoms are fixed relative to each other _over_long_time_ranges_ [by] … a great variety in the types of intermolecular bonds …” [Emphasis added]

Remember those bonds? Follow cites forward from the deep past, like this one: http://prola.aps.org/abstract/PRB/v4/i6/p2029_1

You don’t need to invoke “the vibration of their nuclei” (whatever that may mean).

Comment by Hank Roberts — 26 Jul 2007 @ 2:41 PM

294. OK I’ve read it. It talks about scattering phonons in a glass not emitting photons from an opaque solid. And it says “Such a mean free path can be quantitatively explained by approximating the glassy structure with that of a crystal in which every atom is displaced from its lattice site.”

What I am saying is that blackbody radiation can be quantitatively explained by approximating the opaque solid with that of a crystal in which every atomic nuclei is vibrating at random within its lattice site.

Or you can view it as that the molecules are held so rigidly they cannot rotate or vibrate and the blackbody radiation is a result of the internal vibrations of the atoms.

But if you have a better description of how blackbody type radiation (i.e. continuous radiation) is produced by solids then let us have it.

Comment by Alastair McDonald — 26 Jul 2007 @ 5:03 PM

Comment by Hank Roberts — 26 Jul 2007 @ 5:45 PM

296. “Lightning,” people, “lightning,” not “lightening.” Lightening is what happens when something’s weight decreases.

Comment by Barton Paul Levenson — 26 Jul 2007 @ 6:16 PM

297. Well if you are prepared to take the humble opinion of someone who signs him/herself “xez” then go ahead.

However xez writes: “The pressure and doppler broadening effects are why we see a basically continuous spectrum from the solar photosphere even though it’s mostly ionized Hydrogen/Helium plasma; the pressure and temperature are so high the lines are very broad, and basically it has become a blackbody radiation source at the equivalent temperature.” That is an explanation I have seen for the Sun’s radiation, but how come the Sun’s spectrum also has lines if the lines have been smeared into continuum radiation?

Moreover it does not explain the black body radiation emitted from the surface of the Earth, such as that emitted by snow, nor any other terrestrial thermal emissions. There is no high temperature to produce a Doppler effect here, or high pressure to produce collisional broadening.

You will have to show a bit more understanding of the science, to get any more responses out of me.

Comment by Alastair McDonald — 26 Jul 2007 @ 6:33 PM

298. Alastair McDonald (#297) wrote:

Well if you are prepared to take the humble opinion of someone who signs him/herself “xez” then go ahead.

Well, I happen to put a lota stock in what a certain waskely wabett has to say. Then again he generally speaks of himself in the third person. People who speak of themselves in the third person are usually very important. Xez doesn’t speak of himself in the third person, so he might not be as important.

xez sez:

That is an explanation I have seen for the Sun’s radiation, but how come the Sun’s spectrum also has lines if the lines have been smeared into continuum radiation?

xez sez:

The solar corona is cooler and at lower density, so from it we do see more pronounced sharp spectral lines.

Alastair wrote:

Moreover it does not explain the black body radiation emitted from the surface of the Earth, such as that emitted by snow, nor any other terrestrial thermal emissions. There is no high temperature to produce a Doppler effect here, or high pressure to produce collisional broadening.

xez sez:

This is simply because there are effects like the doppler effect and raman scattering and so on that cause spectral line broadening as a result of higher ‘pressure’ and velocity within the material so that an element as a low density plasma will emit very narrow lines, the same element as a solid or high pressure gas tends to emit a much broader spectrum since the line-width becomes wider.

The operative word is like.

There are eight different types of local spectral broadening I can see:

Maybe xez is talking about one or more of those things.

Comment by Timothy Chase — 26 Jul 2007 @ 8:58 PM

299. alistair:

tamino, simon donner and eli have been hashing some of this out on open mind, maribo and rabett run. I think you should see what you make of that.

Comment by Marion Delgado — 26 Jul 2007 @ 9:29 PM

300. 1. Because there are other atoms in the sun besides hydrogen and helium

2. There is no Doppler shift from emissions from a solid if you are standing on the solid. OTOH, there is if someone is pegging it at you at a zillion km/h

Comment by Eli Rabett — 26 Jul 2007 @ 9:39 PM

301. Re #287

Hey Alistair,

That’s good stuff. I will read that article. Rothman is the guy in charge of HITRAN, and was an author on the paper I was looking for. Are they taking any new measurements of the lines?

Re #286 yeah. I should have said ‘theoretical collision-broadened lineshape’

Its interesting that this is being addressed. But, hey, 2004-2005 is pretty late to finally be addressing such a basic issue. Lots (decades) of modeling has gone on without anyone questioning the fundamental absorption calculations the whole problem rests upon, it seems.

I looked back at the plot of the absorption spectrum in the article. I agree that at the band center, the aborption in between the lines only come down to a factor of 100 ( Absorption Factor = OD?). Thus there is no additonal aborption occuring here with incresing CO2.

At the band edge, however, lines in the range of OD 1 to 10 with dominate the differential absorption over the contribution of the lines below OD=1, in the linear range.

Comment by Dave Dougherty — 26 Jul 2007 @ 11:33 PM

302. Re #300 No one is pegging it a me at a zillion km/h

Comment by Alastair McDonald — 27 Jul 2007 @ 3:40 AM

303. Alastair, just saying, Ockham.

> every atomic nuclei is vibrating at random within its lattice site. Or you can view it as that the molecules are
> held so rigidly they cannot rotate or vibrate and the blackbody radiation is a result of the internal vibrations of the atoms.

How could you test either of these? What test could distinguish this from the physics homework help site explanation?

Would it be finding an infrared picture of the Earth that showed no airglow in the infrared above the troposphere, as you said earlier? I’d think surely someone would have noticed if that were true, in all these years. But you said it should be true.

Is there any other way of falsifying your idea?

Comment by Hank Roberts — 27 Jul 2007 @ 10:47 AM

304. Alastair says:

Your mole of gas will have a temperature. Put a thermometer in the gas and it will register a value that matches the average kinetic energy of the molecules. So it has a kinetic or Maxwellian temperature. Its Planckian temperature or brightness temperature is zero because we are assuming that it is not emitting any radiation.

Would you care to explain why all gases do not radiate? True in the case of He or N2 the radiation rate would be infinitesimally slow, but for others such as CO2, or H2O not so. Of course if you let the radiation out of the volume in which the gas was enclosed the gas would cool unless you put energy in some other way.

Comment by Eli Rabett — 27 Jul 2007 @ 10:02 PM

305. I’m working on a summary of my contentions, questions, responses in this thread. The following is one piece of that summary that I’d like check its correctness. There has already been some agreement and some disagreement.

Contention: the energy in intermolecular bonding (consisting of lateral harmonic oscillations or bending oscillations, commonly called “vibrations”) or molecular rotation is in fact kinetic in nature but does NOT, by itself, add to the temperature of the molecules. The temperature of a gas is determined from only the kinetic energy evidenced by the “Center of Mass” translation movement (velocity) of the whole molecule(s). The temperature is simply measured through kinetic energy transfer of the molecule(s) crashing into the thermometer bulb; the internal bond vibration and rotational kinetic energies have no effect on that collision or transfer, so only the molecule(s)’ translational kinetic energy heats and expands the mercury. One other indication: as heat energy is added to a molecule(s) the temperature increases linearly per a certain heat coefficient — all heat goes into translation K.E. and the bond & rotational energy capacities are “frozen out” and do not increase. As the temperature continues to rise and the rotational and the bond levels open up and start accepting energy, the coefficient increases and greater heat energy must be now added to get the same molecular temperature increase.

Infrared radiative energy (photon) absorbed by a CO2 molecule goes only into the vibrational or rotational energy and has no effect, per se, on the temperature of the gas. Then (a few hundred nanoseconds (??) later), if a vibrational or rotational relaxation is by emission of another photon, there will never be a temperature change from this activity. The relaxation might (most likely) also come about by energy transfer to its own translational kinetic energy — shifting within the molecule, or to another’s through a collision (don’t know if the latter is direct…???), either of which now, after leaving the “internal” energy, does increase the temperature of the gas.

I know this all sounds like Physics 101, but the different information “out there” is amazing and I’m just trying to get it right. Thanks for the help and indulgence.

Comment by Rod B — 28 Jul 2007 @ 1:22 PM

306. Eli is sorry Rod, you really are not understanding what is happening.

Let us start from the beginning. An object with N atoms has 3N degrees of freedom. These can be described as the motion of each atom in space, which is three dimensional (x,y,z). An molecule with two atoms has six degrees of freedom, a molecule with three atoms has nine. However, since the atoms are linked to each other in a fixed consideration which makes other possible ways of describing the motion of the molecule better.

The most common one is to assign three coordinates (degrees of freedom, or dof for short) to the motion of the center of mass of the molecule. This energy associated with this would be the kinetic energy of the molecule moving through space. That leaves 3N-3 dof. Three (or two for linear molecules) dof are assigned to the rotation of the molecule moving through space. You can associate each of these with the rotation of the molecule perpendicular to these axes, x, y, and z. For a linear molecule, the z axis corresponds to the axis connecting the nuclei. This axis does not have an associated dof. That leaves 3N-6 (3N-5) dof for vibrations.

At thermal equilibrium (or local thermodynamic equilibrium) statistical mechanics and experiment shows us that each dof can be characterized by the same temperature T. The energy in each dof for all practical purposes is the same 1/2 kT (there are fine points here associated with the partitioning of energy in vibrational dof for which the vibrational quantum is much greater than kT and the distribution of energy among all of the molecules of the same type. That is best understood at the level of a physical chemistry/modern physics textbook, so not here).

Energy can be transferred to and from the thermometer bulb from rotational, translational and vibrational dof. In a thermalized situation they are all at the same temperature, tant pis.

Since temperature is a property of the entire collection of molecules again, there is another way of describing this, e.g. all possible arrangements of energy and everything else among a set of N molecules at temperature T called a canonical ensemble.

The issue of when dof are isolated from each other depends only on the collision frequency. If the collision frequency is orders of magnitude higher than energy loss by radiation, etc. from the system, the system will always be in local thermodynamic equilibrium defined by a temperature T.

Comment by Eli Rabett — 28 Jul 2007 @ 2:43 PM

307. Glad you are back, Rob.

Incidently, Eli’s response is “news” to me – but it really shouldn’t be. He is basically describing what is called phase space and analyzing thermal energy in terms of it. “Basic” statistical mechanics – whether its classical or quantum. The boson/fermion distinction gives it a twist when one goes from classical to quantum, but it is otherwise largely the same. I learned about phase space in some detail a long time ago in my own personal studies, but I haven’t really spent much time with it since.

As for his reasoning with regard to the equality of temperature with regard to each dimension of freedom, it makes perfect sense and would seem to be derivable from the equipartition theorem – which would also seem to require that each gas within the atmosphere assume the same temperature within the local thermodynamic equilibrium.

Comment by Timothy Chase — 28 Jul 2007 @ 5:04 PM

308. Rod,

Sorry for get the name wrong. I’ve actually drawn a blank on my own name once while trying to write a check. As for “the disagreements,” if it is simply a disagreement between an amateur climatology enthusiast such as myself and someone with a PhD who has been teaching for several decades, I would say that it is a virtual certainty that it is the latter who knows what he is talking about and not the former.

Comment by Timothy Chase — 28 Jul 2007 @ 5:16 PM

309. Re #303-304 etc.

Hank,

If you want to use Occam’s Razor on the problem of the origin of black-body radiation (BBR) then you should start with the original idea by Boltzmann, that BBR is produced by atomic oscillators with three degrees of freedom. He used that idea to explain Stefan’s experimental results. And that was the underlying idea that Planck used, but he had to quantumise the modes in order to get the curve to fit with the full experimental curve. In 1905 Einstein proved that atoms exist using Brownian motion, and that quanta exist using the photo-electric effect. The same year Boltzmann committed suicide because no-one would believe his idea that atoms do exist. (I know the feeling!)

We now know that an atom is made of a positively charged nucleus surrounded by a cloud of negatively charged electrons. BBR is easily explained by the nucleus oscillating in all direction, thereby creating an electromagnetic field, assuming that the net negative charge from the electron cloud remains unchanged. Trying to re-invent the wheel by imagining that a bar of iron glows red hot, due to the smearing of radiation caused by the rotation and vibration of non-existent molecules, goes completely against Occham’s Razor.

Eli, I did not write “all gases do not radiate.” I wrote “we are assuming that it is not emitting any radiation.”

Rod and I were discussing a hypothetical case where simplifications were made in order to understand the general principles. It is ridiculous to suggest that all gases do not radiate. We know that if they are subject to high enough temperatures and they have evaporated, the atoms of all elements will emit radiation due to quantum jumps by their electrons. However, we also know that at atmospheric temperatures no element emit at that type of line radiation, since they never receive collisions with enough energy to excite their electrons. At STP that type of emission is “frozen out.”

Rod,

A gaseous molecule has four forms of energy which affect its temperature: Translational (commonly called kinetic), Electronic (which is the level of excitation of the electrons above the ground state), Vibrational (a mixture of kinetic and potential energy), and Rotational (as a result of angular momentum.) The bond energy of a molecule is its chemical energy, and only comes into play during chemical reactions such as hydrogen burning in oxygen.

Eli is correct that when heat is added it is shared amongst all those four forms of energy. However, as I have explained above, at the temperatures found i the Earth’s atmosphere, the share going into electronic energy is negligaile. The electronic energy is frozen out, by no collision at that temperature is large enough to cause electronic excitement.

But not only that, the same is true for vibrational excitement. That too is frozen out! This means that when a CO2 molecule is excited by radiation it will lose its excitement due to collisons, but it will never receive a collision which is great enough to re-excite it.

In other words, Gavin’s simple model with the greenhouse gases re-radiating back to Earth is completely wrong. Greenhouse gases do not act like that. They absorb radiation which is shared with the the kinetic energy of the other molecules raising the temperature of the air. The air is heated directly by the radiation, not indirectly via conduction from the Earth’s surface.

In #308 Timothy wrote:

“… if it is simply a disagreement between an amateur climatology enthusiast such as myself and someone with a PhD who has been teaching for several decades, I would say that it is a virtual certainty that it is the latter who knows what he is talking about and not the former.”

As Benjamin Franklin wrote “In this world nothing can be said to be certain, except death and taxes.”

Cheers, Alastair.

Comment by Alastair McDonald — 29 Jul 2007 @ 8:39 AM

310. Re #301,

Dave,

It was Hank who found that paper not me. He is the one to thank.

The reason that it is only now that these issues are being raised is that it is only now that it is possible to handle those problems theoretically. In general, science operates by producing theories then using experiments to test them. Although it was possible to do the calculations for diatomic molecules, it was not until the 1970 that the theory of triatomic molecules, such as CO2 and H2O could be tackled. But even now the variations in collisions in the atmosphere make the task too vast to complete. Moreover, it was only during and after the nineties that the threat from climate change became obvious, and that urgency was needed.

But what do you think of my ideas?

Comment by Alastair McDonald — 29 Jul 2007 @ 8:50 AM

311. Alastair McDonald (#309) wrote:

Eli is correct that when heat is added it is shared amongst all those four forms of energy. However, as I have explained above, at the temperatures found i the Earth’s atmosphere, the share going into electronic energy is negligaile. The electronic energy is frozen out, by no collision at that temperature is large enough to cause electronic excitement.

Alastair,

We aren’t talking quantum states in electron orbitals but quantum states in molecular rovibration.

Alastair McDonald (#309) wrote:

In #308 Timothy wrote:

“… if it is simply a disagreement between an amateur climatology enthusiast such as myself and someone with a PhD who has been teaching for several decades, I would say that it is a virtual certainty that it is the latter who knows what he is talking about and not the former.”

As Benjamin Franklin wrote “In this world nothing can be said to be certain, except death and taxes.”

See above.

I rest my case.

Comment by Timothy Chase — 29 Jul 2007 @ 12:24 PM

312. Alastair McDonald (#310) wrote:

But what do you think of my ideas?

Alastair,

I don’t like saying this, but you tend to attract a certain element which will often have ulterior motives for fawning over your ideas. Your somewhat exaggerated estimation of your abilities (an imperfection which I no doubt suffer from on occasion) combined with your desire for such flattery is at times counterproductive.

Just something I’ve noticed on various occasions.

Comment by Timothy Chase — 29 Jul 2007 @ 12:47 PM

313. Re #310

No need to wait for theory. Ab-intio calculations always lag behind the experiments, in condensed matter physics anyway. (And this experiment is so simple!). This should have cried out for more measurements. You’d need that data anyway to verify your inevitable approximations, so why bother with theory? Even if you just measured a couple of lines, you could go with that for all of them, since the assumption that all lines are lorentzian is the default. What do you assume for collisions besides a Poisson distributed random varible process? As far as the dynamics of the collisons themselves go, it like calculating a trajectory for a falling leaf. You know the underlying laws, but so what? You work a lifetime for one leaf, then you need to integrate over all possible initial conditions. Why not just go measure the leaf distribution under the tree if that’s what you need?

I don’t like it when people stake out a position and gamble whether its right or not. (not just in this field, I get it a lot at my job too.) If they’re wrong, big deal, science is complicated … move on. If they’re right (like a stopped clock twice a day) then god almighty they are a genius an you are an idiot for raising questions. You have to try to tear arguments down which are near and dear to your heart and think objectively.

Comment by Dave Dougherty — 31 Jul 2007 @ 12:26 AM

314. A couple of (final??) questions: Alastair, et al: Is the likelihood of GHG radiation emission really very small compared to collision-like energy transfer within the gas (except possibly at very low density high altitudes)? If so, how does/can the surface, as opposed the atmosphere, possibly get heated via GHGs???

I still can’t understand why absorption/emission cycle is a net cooling effect in the troposphere. The molecule absorbs a photon and heats up. It then emits a photon and cools down. ??? Is it because the emitted photon has more energy which was picked up from some other heat source??? But, it would seem the total emitted energy would be less since much of the absorbed radiation energy will transfer to O2 and N2 via collision and never (damn near) get emitted. Also I would think an emitted photon on the average has the same discrete quantisized energy as one absorbed, though the number emitted doesn’t necessarily have to equal the number absorbed…

How is PV= NRT reconciled with the Total internal energy determining the temperature? Is it because the vibration and rotation energies are in fact kinetic and affect the pressure of a volume of gas just like the translation kinetic energy does??

BTW, do not believe my post #305 and a bunch just like it. They’re all wrong. Internal vibration and rotation energies do, in part, determine temperature.

Comment by Rod B — 1 Aug 2007 @ 10:25 AM

315. Some tidbits that may help:

(fairly old, interesting early info toward an explanation of separation between wet and dry layers of the atmosphere)

Molecular Geometry and 3D Structures of Molecules (Powerpoint)
classroom.sdmesa.edu/rfremland/chem%20152/molgeomf04.ppt

Every cloud has an invisible halo (this year’s news)
Science – Clouds are bigger than they look, according to new measurements by atmospheric scientists in Israel and the United States. They say that clouds are surrounded by a ‘twilight zone’ of diffuse particles, invisible to the naked eye, extending for tens of kilometers around the cloud’s visible portion.
(Science magazine; no longer available free, searching may find a copy somewhere)

Comment by Hank Roberts — 1 Aug 2007 @ 1:45 PM

316. Hank Roberts (#315) wrote:

Every cloud has an invisible halo (this year’s news)
Science – Clouds are bigger than they look, according to new measurements by atmospheric scientists in Israel and the United States. They say that clouds are surrounded by a ‘twilight zone’ of diffuse particles, invisible to the naked eye, extending for tens of kilometers around the cloud’s visible portion.
(Science magazine; no longer available free, searching may find a copy somewhere)

People should check out the following:

http://climate.gsfc.nasa.gov/publications.php

On the twilight zone between clouds and aerosols
Ilan Koren, Lorraine A. Remer, Yoram J. Kaufman, Yinon Rudich, and J. Vanderlei Martins
GEOPHYSICAL RESEARCH LETTERS, VOL. 34, L08805, doi:10.1029/2007GL029253, 2007

… and a great many more technical articles being made freely available by NASA in climatology.

Comment by Timothy Chase — 1 Aug 2007 @ 9:05 PM

317. Re #314 Where Rod wrote:

“BTW, do not believe my post #305 and a bunch just like it. They’re all wrong. Internal vibration and rotation energies do, in part, determine temperature.”

There is a good description of this at http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/eqpar.html
and an explanation of Kinetic Temperatures at:
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c1
In fact thr whole site is very good.

To answer your first question: the surface gets heated by solar radiation during the day, and by blackbody radiation from clouds at night. Without the clouds, even hot deserts get very cold at night.

However, water vapour, unlike CO2, can condense in the atmosphere. It is continuously condensing and then evaporating again, just as it does in clouds. See Hank’s #315 and Timothy’s #316. Therefore, water vapour does partly emit blackbody radiation, so performing in a way that is similar what happens in the GCMs. However, that is probably just a second order effect.

The first order effects are that CO2 and H20 absorb all the radiation from the surface in their bands, and do not re-radiate any of it back to earth. But for the troposphere to stay at a stable temperature then the air has to lose that absorbed heat, which it does by emitting blackbody radiation from clouds into the stratosphere, where there is no water vapour to absorb it, and the CO2 is too thin to absorb in the wings of the lines.

CO2 and O3 in the stratosphere absorbs the blackbody radiation from the clouds and infrared from the Sun, which causes it to be warmer than the troposphere, but how it loses its heat I have still to work out. Presumably, by radiating from the wings of the lines where the radiation cannot be absorbed at higher altitudes since the lines are thinner there.

To be honest, I still have some work to do, to get it all sorted out :-(

Comment by Alastair McDonald — 2 Aug 2007 @ 7:00 AM

318. For terrestrial conditions, the HITRAN and GEISA molecular databases indeed provide the best starting point for modelling absorption by trace gases. An excellent tool for calculating and visualizing the absorption by atmospheric gases is the Spectral Calculator (www.spectralcalc.com), also described by Eli Rabbet. High resolution spectra are available, and the raw data can be extracted. Here the HITRAN and GEISA databases can be perused and downloaded.

It should be noted that one must be careful to include other effects when needed. These include line coupling (most notably for carbon dioxide and methane), and continuum absorption from molecular oxygen and water vapor. Mie and Rayleigh scattering from aerosols and molecules respectively needs to be accounted for as well, in the appropriate wavelength regions. At very high altitudes, above ~90 km, the assumption of local thermodynamic equilibrium no longer applies, and care must be taken to account for this, although this effect is negligible for most Earth climate studies.

As a side note, if very high temperatures are being examined (e.g. the atmosphere of Venus), the HITEMP line catalog provides good data for water vapor, carbon dioxide, and carbon monoxide. These can be obtained at http://www.spectralcalc.com as well.

Comment by Martin — 2 Aug 2007 @ 8:25 AM

319. Re Martin’s #318,

If you use http://www.spectracalc.com to calculate the transmittance for the band bounded by wave numbers 620 1/cm to 720 1/cm for a Gas Cell of CO2 at STP and a cell length of 30 m (3000 cm) the absorption is 100%.

This does raise the question as to the partial pressure of the CO2 in this experiment. Does anyone know? Is it 1 bar or is it 0.35 mb?

Comment by Alastair McDonald — 2 Aug 2007 @ 9:04 AM

New people come here to RC to read, as it says, “Climate Science from Climate Scientists”

People who are _not_ climatologists, like, er, ahem, me, and you — who post a lot here — do get mistaken for real climatologists. I’m just another reader here trying to understand, sometimes by rephrasing to see if I can find better words. I still actually don’t know if you’re a working scientist.

For new readers coming in, who see what we write, it’s awfully confusing if we don’t, early and often, make clear we’re not climatologsts _and_ make clear what’s posting personal opinions, versus what info has cites to real scientists’ work.

People need to know which ideas are fervently repeated beliefs and which are footnotes to the literature, eh? A kindness.

Comment by Hank Roberts — 2 Aug 2007 @ 9:57 AM

321. Aha, Alastair. I think this (maybe??!!) answers another question with differing answers. The absorption/emission of energy into/from the rotational and vibrational energies in gas molecules is NOT classic blackbody/Planck absorption/emission (even if what is first absorbed is blackbody radiation from the earth’s surface.) It does not stem from the same physical process that generates blackbody radiation. E.g. blackbody is predominately continuous over a spectrum, the other is highly discrete and quantized. (I suppose for simple calculations and maybe understanding molecular absorption/emission can be considered as blackbody with some really crazy absorption coefficients.) One hole in my thinking, however: It seems the motions associated with rotational and vibration energy would partly satisfy the basis for blackbody radiation — jiggling of charged particles…????)

I understand H2O clouds emitting blackbody radiation back to earth. But this doesn’t seem enough for the required earth warming (just a guess) — for instance, doesn’t the H2O molecule give up a potful of heat energy to the rest of the atmosphere when it condenses to liquid clouds? Can’t gasses, like vapor, O2, N2, CO2, also emit blackbody radiation by virtue of the average temperature?

Comment by Rod B — 2 Aug 2007 @ 12:24 PM

322. Alastair McDonald (#317) wrote:

The first order effects are that CO2 and H20 absorb all the radiation from the surface in their bands, and do not re-radiate any of it back to earth. But for the troposphere to stay at a stable temperature then the air has to lose that absorbed heat, which it does by emitting blackbody radiation from clouds into the stratosphere, where there is no water vapour to absorb it, and the CO2 is too thin to absorb in the wings of the lines.

Actually it appears they radiate just fine, Alastair.

The first figure on this page (figure 23) shows the observation of CO2 reemission in the 15 μ from the mid-troposphere as observed by NIMBUS-4 IRIS (graph from a textbook published in 1976)

3.1 Physical basis of remote sounding
3 RADIATIVE TRANSFER AND REMOTE SOUNDING
http://ceos.cnes.fr:8100/cdrom-00/ceos1/science/dg/dg20.htm

Now here is something a bit older. It shows calculations by the military of infrared radiation of non-LTE infrared radiation for CO2:

SHARC, A Model for Calculating Atmospheric and Infrared Radiation Under Non-Equilibrium Conditions
January 24, 1994

Figure 2. Calculated Vibrational Temperatures for C02 States for a Solar Zenith Angle of 820.
(50-150 km)

NOTE: You will see that the temperature for all the displayed vibrational states approach each other at 50 km, indicating LTE conditions at 50 km and below.

The following are two more recent presentations regarding the measurement of CO2 from thermal spectra:

SPIE’s 4th Inter. Asia-Pacific Env. Remo. Sensing Symp.
Honolulu, HI, USA, 5655-17, 2004.

Imasu, R. and Y. Ota, A method for retrieving columnar CO2 concentration from thermal infrared radiation spectrum
Atmospheric Science from Space using Fourier Transform Spectrometry

Not available on the web, but the science is there, Alastair.

I realize that with all the time you spending theorizing, you would seem not to have any time to look up the actual science yourself. If you need me to, I can look up more when I get the chance.

Comment by Timothy Chase — 2 Aug 2007 @ 1:53 PM

323. Something else for Alastair…

An online article:

Abstract. The near-infrared nadir spectra measured by SCIAMACHY on-board ENVISAT contain information on the vertical columns of important atmospheric trace gases such as carbon monoxide (CO), methane (CH4), and carbon dioxide (CO2). The scientific algorithm WFM-DOAS has been used to retrieve this information. For CH4 and CO2 also column averaged mixing ratios (XCH4 and XCO2) have been determined by simultaneous measurements of the dry air mass. All available spectra of the year 2003 have been processed.

Carbon monoxide, methane and carbon dioxide columns retrieved from SCIAMACHY by WFM-DOAS: year 2003 initial data set
M. Buchwitz, et al
Atmos. Chem. Phys., 5, 3313–3329, 2005
European Geosciences Union
http://www.atmos-chem-phys.net/5/3313/2005/acp-5-3313-2005.pdf

An article to look up offline:

Abstract: The co-adsorption structures of CO and oxygen on Pd(100) and the angular distribution of reactive CO[2] desorption were studied with angle-resolved thermal desorption and low-energy electron diffraction. The CO[2] formation induced by heating the co-adlayer was extended to lower temperatures with an increase in the coverages, yielding five peaks; P[1]-CO[2] (∼400 K), P[2]-CO[2] (∼350 K), P3-CO[2] (∼290 K), P[4]-CO[2] (∼240 K) and P[5]-CO[2] (∼150 K)

Reaction diagram of carbon monoxide and oxygen on palladium (100) and angular distribution of reactive carbon dioxide desorption
OHNO, et al
1992, vol. 273, no3, pp. 291-300
Surface science (Surf. sci.)

Here are sounder measurements of thermal emissions in CO2 sensitive parts of the spectra from 1997:

http://cimss.ssec.wisc.edu/goes/sndprf/18mar4pl.gif
from
APPLICATION OF GOES-8/9 SOUNDINGS TO WEATHER FORECASTING AND NOWCASTING
W. Paul Menzel, et al
Bulletin of the American Meteorological Society
Volume 79, Number 10, October 1998
http://cimss.ssec.wisc.edu/goes/sndprf/sndprf.htm

Then there is Mars with its lower temperatures and pressures. For those who are interested, here are the results of some measurements…

http://starryskies.com/solar_system/mars/spirit/atmosphere01.html

Comment by Timothy Chase — 2 Aug 2007 @ 2:18 PM

324. Alastair, your references (Hyperphysics) in 317 throws a monkey wrench into my (our??) conclusions. It states the rotational and vibration energy plays NO PART in determining the temperature and is NOT INVOLVED with energy/heat transfer from collisions. What gives? It has one phrase that says the temperature “… as commonly measured… ” which is a slippery odd caveat, but I have no idea what it means. It implies that rotation and vibration energy levels increase the specific heat… but not temperature. ???!! I think Hyperphysics is wrong, but am confused out of any certainty.

Comment by Rod B — 2 Aug 2007 @ 2:47 PM

325. The following page has java animations of satellite data for thermal emissions from various gases, including co2 at 14.1, 13.4 4.45 μm bands in the upper- and lower-level atmosphere temperatures at various times of day.

The CIMSS Realtime GOES Page
http://cimss.ssec.wisc.edu/goes/realtime/grtmain.html

Here the elevation of land being determined by co2 absorption strength:

Figure 20. The 2-μm CO2 absorption strength (A) can be converted to topographic elevation (B). The derived elevations matches the USGS Digital Elevation Model (DEM) (C). The CO2 absorption strength image (A) is brighter for increasing strength. Because the atmospheric path length is smaller with increasing elevation, the absorption strength decreases, becoming darker in the image. The DEMs (B, C) are brighter for increasing elevation, thus are inversely correlated with the CO2 strength in (A).
http://speclab.cr.usgs.gov/PAPERS/tetracorder/FIGURES/fig20.dem.abc.gif

Figure 21. The 2-μm CO2 absorption strength versus USGS DEM elevation shows a linear trend with an excellent least squares correlation coefficient.
http://speclab.cr.usgs.gov/PAPERS/tetracorder/FIGURES/fig21.co2_graph.gif

Imaging Spectroscopy:
Earth and Planetary Remote Sensing with the USGS Tetracorder and Expert Systems
Roger N. Clark, et al
Journal of Geophysical Research, 2003.
http://speclab.cr.usgs.gov/PAPERS/tetracorder/

Comment by Timothy Chase — 2 Aug 2007 @ 2:48 PM

326. Rod B (#324) wrote:

Alastair, your references (Hyperphysics) in 317 throws a monkey wrench into my (our??) conclusions. It states the rotational and vibration energy plays NO PART in determining the temperature and is NOT INVOLVED with energy/heat transfer from collisions.

Actually, R Nave is explaining that he is making use of an oversimplification which treates molecules as point masses, and that you need to deal with rovibration if you want to properly deal with the specific heats of gases:

It is important to note that the average kinetic energy used here is limited to the translational kinetic energy of the molecules. That is, they are treated as point masses and no account is made of internal degrees of freedom such as molecular rotation and vibration. This distinction becomes quite important when you deal with subjects like the specific heats of gases. When you try to assess specific heat, you must account for all the energy possessed by the molecules, and the temperature as ordinarily measured does not account for molecular rotation and vibration.

Kinetic Temperature
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html

In otherwords, the author is admitting that the approach he is taking is inadequate for dealing with specific heat as it does not take into account the internal degrees of freedom due to rovibration.

Comment by Timothy Chase — 2 Aug 2007 @ 3:04 PM

327. Timothy (326): “……and the temperature as ordinarily measured does not account for molecular rotation and vibration.” (my emphasis)(from Hyperphysics). I made specific reference to this phrase in my post: what the hell is it supposed to mean? That the real temperature is different from what we measure???… because rotation/vibration is temperature we can’t measure???

Comment by Rod B — 2 Aug 2007 @ 4:18 PM

328. Rod B (#327) wrote:

Timothy (326): “……and the temperature as ordinarily measured does not account for molecular rotation and vibration.” (my emphasis)(from Hyperphysics). I made specific reference to this phrase in my post: what the hell is it supposed to mean? That the real temperature is different from what we measure???… because rotation/vibration is temperature we can’t measure???

It means that the oversimplified introductory “high school”-level approach which treates a molecule as a point mass in order to calculate kinetic energy the same way that you would that of large object undergoing simple linear motion does not work for calculating specific heat. Molecules cannot be treated as point masses if one is to account for their specific heats. Calculations of specific heat require you to recognize the fact that such molecules are not point masses but extended objects where their kinetic energy include rotation and vibration.

The measuring of specific heat and of thermal spectra are measures involving temperature which take these factors into account. But they are indirect – then again any measurement of molecules as molecules or as ensembles of molecules is indirect.

Comment by Timothy Chase — 2 Aug 2007 @ 4:44 PM

329. Hank Roberts (#320) wrote:

New people come here to RC to read, as it says, “Climate Science from Climate Scientists”

People who are _not_ climatologists, like, er, ahem, me, and you — who post a lot here — do get mistaken for real climatologists. I’m just another reader here trying to understand, sometimes by rephrasing to see if I can find better words. I still actually don’t know if you’re a working scientist.

Alastair appears to be the captain of the Voluntary Observational Ship “Gypsum King” which has won the VOS award for its contributions to marine weather observations in 1999. This is an achievement, and it should be acknowledged that he has made valuable contributions to meteorology, but it does not make him a climatologist. He is an amateur climatology enthusiast, like you and me who clearly has no expertise in the reemission spectra of greenhouse gases and the state of the field, including its solid grounding in empirical scientific observation. He should not pretend to be otherwise.

But he could learn a fair amount – if he chose to. In my estimation at least, he has the intelligence.

Comment by Timothy Chase — 2 Aug 2007 @ 4:52 PM

330. Rod, just a quickie re #324

The temperature that you measure with a thermometer is the kinetic or Maxwellian temperature. But you measure the temperature of the Sun or stars using a bolometer which registers the radiation. This is the brightness or Planckian Temperature. For a blackbody the two temperatures Maxwellian (kinetic) and Planckian (brightness) are equal. But for a gas, you do not have a continuum spectra. You have lines, and each line has a different intensity with peaks which do not match any blackbody spectra. But each peak can be allocated a brightness temperature by finding the temperature of the Planck function which passes through the peak. These lines are due to rotational and vibrational relaxation. On the other hand the gas still has a Maxwellian temperature that you can measure with a thermometer.

If you look up specific heat you will find that to raise the temperature of the gas you have to add kinetic and internal energy, but it is only the kinetic energy that is measured with a thermometer. Using the specific heat of CO2 it should be possible to work out the internal energy, rotational and vibrational excitation by subtracting the kinetic energy.

That’s all I have time to write at the moment.

Comment by Alastair McDonald — 2 Aug 2007 @ 5:03 PM

331. > the real temperature is … what we measure

Temperature is, first, a definition. Pick one:

All of them have to do with something that can be measured — an average property of a large number of molecules.

Rod, it’s how the universe behaves, regardless of what you believe should be ‘true’ or ‘real’ about any given atom. You can’t tell.

Why can’t you find the temperature of one atom?

“The more precisely
the POSITION is determined,
the less precisely
the MOMENTUM is known”

Comment by Hank Roberts — 2 Aug 2007 @ 5:05 PM

332. CORRECTION to #329

The award was in 1997.

Comment by Timothy Chase — 2 Aug 2007 @ 5:08 PM

333. And to be clear — bonds are something electrons do, but they’re something that happens, not something we can completely describe.

When energy is transferred, it’s not like the electron is a little solid lump orbiting a nucleus, and being bumped by an incoming photon into a higher orbit then dropping back down by emitting another photon to get rid of the energy, or changing its orbit to put the energy into a twisted or rotated bond angle. That’s poetry.

Comment by Hank Roberts — 2 Aug 2007 @ 5:10 PM

334. Hank Roberts (#333) wrote:

When energy is transferred, it’s not like the electron is a little solid lump orbiting a nucleus, and being bumped by an incoming photon into a higher orbit then dropping back down by emitting another photon to get rid of the energy, or changing its orbit to put the energy into a twisted or rotated bond angle. That’s poetry.

Its not orbiting, spinning or bending like a classical object, but something which exists in an orbital, like a cute little Schroedinger kitten doing its Einstein-Podolsky-Rosen thing with photons which stray to close in their wavy little ways. Coherences and decoherences all enter the picture, and it is all quite quantized and probablistic with all sorts of mind-bending, metaphysical mischief.

… or at least so I would gather.

Comment by Timothy Chase — 2 Aug 2007 @ 6:21 PM

335. [[Can’t gasses, like vapor, O2, N2, CO2, also emit blackbody radiation by virtue of the average temperature?]]

Only if their temperature is very high — in the thousands of degrees. At terrestrial temperatures, most gases radiate only at their absorption/emission lines.

Comment by Barton Paul Levenson — 3 Aug 2007 @ 7:04 AM

336. “Here, kitty, kitty, kitty…”
–attributed to Erwin Schroedinger

Comment by Barton Paul Levenson — 3 Aug 2007 @ 7:09 AM

337. Dang, this is getting messy. First, little stuff: 1) Barton, stuff does not have to be a certain temperature to emit “blackbody” radiation. Any temperature greater than absolute zero will do — witness the cosmic background radiation. Though the emission is material dependent in an elusive sort of way. 2) Timothy, I couldn’t tell if your reference (http://ceos.cnes.fr:8100/cdrom-00/ceos1/science/dg/dg20.htm) proved the emission from CO2 and H2O or just assumed it. But maybe it’s just me… It also stated that the emission from GHG molecules is blackbody type “per Planck’s function”, and implied it is not quantized as it is in the absorption. Is this correct? 3) Hank (for fun), the Bohr model works just fine here so no need to bolox it up with uncertainties and wave probabilities. Like the Copenhagen Convention (was it “convention”), if it works its true. And it describes LEDs pretty good. 4) Alastair, isn’t the use of a thermometer or bolometer a matter of convenience? If you could get there, have a thingy that doesn’t melt and a really good asbestos suit, could you measure the Sun’s temp with a thermometer?

Second, the big stuff: What I’m trying to determine (and have been up and down the hill a number of times [;-) ): Does a molecule emit radiation at all? Does it emit with the same discrete energy level (frequency) that it absorbed into its rotational and vibrational energies? Is that Planck functioned “blackbody”-type radiation, or something else? Or, might a bunch of gas emit both standard blackbody radiation and discrete radiation lines from relaxing rovational levels?

What I’m also trying to determine: Can (and does it sometimes) radiant energy absorbed by a molecule into its internal vibration and/or rotation energy transfer a la equipartition to its translation energy? Or another molecule’s translation energy? If so does it transfer in discrete packets as it was absorbed? If discrete packets, how can it equalize the equipartition balance? (Or maybe that’s just a tendency that might or not be realized…???)

What I’m really trying to determine: What is the temperature of a mole of gas (say CO2) that has a boltzman distribution of translation kinetic energy and every molecule is in its ground vibration and rotation states? Now lets add a bunch of absorbed radiation energy so that all of it goes into the molecules’ now excited rotation and vibration energy states — what’s the temperature now?
Side question: what happens to the temperature of the mole if electron energy levels are increased?
Side question: is not vibration and rotation true kinetic energy, albeit quantized, by virtue of the 1/2mv2 of the jiggling atoms and the Iw2 of the rotating molecule?

Comment by Rod B — 3 Aug 2007 @ 11:41 AM

338. RodB wrote:
> the Bohr model works just fine here so no need to bolox it up with uncertainties and wave probabilities. …
> What I’m also trying to determine: Can (and does it sometimes) radiant energy absorbed by a molecule into its internal
> vibration and/or rotation energy transfer a la equipartition to its translation energy?

Uh …. I think you’ll have to get beyond the Bohr model, Rod. Perhaps it’d help to ask for a pointer, say to the MIT Open Coursework, from someone with expertise in the area of radiation physics. Else you’re just inviting amateur and enthusiast opinions.

Comment by Hank Roberts — 3 Aug 2007 @ 12:47 PM

339. This may help:
Physics Help and Math Help – Physics Forums — Physics —- Quantum Physics —– thermal radiation in QM

Comment by Hank Roberts — 3 Aug 2007 @ 1:23 PM

340. Rod B. (#337) wrote:

Timothy, I couldn’t tell if your reference (http://ceos.cnes.fr:8100/cdrom-00/ceos1/science/dg/dg20.htm) proved the emission from CO2 and H2O or just assumed it.

I can’t resolve the server at the moment, but it is satellite imaging – and we know the spectra from the labs. Fingerprints, just like at a crime scene.

It also stated that the emission from GHG molecules is blackbody type “per Planck’s function”, and implied it is not quantized as it is in the absorption. Is this correct?

The difference between the thermal radiation of solids, liquids and gases is a matter of degree, not kind. Solids can approximate blackbody radiation, but blackbody radiation itself is an idealization. Individual gasses at low pressures and temperatures have well-defined lines and bands, but at higher pressures and temperatures both broaden with lines bleeding into lines and bands bleeding into bands. Dusts, crystals and alloys have relatively discrete spectra. However, impurities, clustering, ions and pressure broaden these spectra. And atmospheres are composed of multiple gasses where each gas is at the same temperature. Individually, the spectra of any one gas in the atmosphere is a poor approximation for a blackbody emitter, but taken together, the atmosphere does much better.

And this applies to both emission and absorption. Assuming LTE, a good emitter at a given wavelength will be an equally good absorber at the same wavelength.

Anyway, I might try to deal with what you threw to others, but I am at work right now, so I should keep it short.

Comment by Timothy Chase — 3 Aug 2007 @ 2:52 PM

341. I suppose this is relevant for something Angstrom didn’t know.

“Over a century ago, the physicist Kirchoff recognized … three fundamental types of spectra ….

[Kirchhoff’s three types of spectra. a) A continuous spectrum (blackbody spectrum) is radiation produced by warm, dense material; b) an emission line spectrum (bright line spectrum) is radiation created by a cloud of thin gas; and c) an absorption line spectrum (dark line spectrum) results from light passing through a cloud of thin gas.]

“… These Kirchoff spectral types are comparable to Kepler’s Laws in the sense that they are only a description of observable phenomena. Like Newton, who later was to mathematically explain the laws of Kepler, other researchers have since provided a sounder basis of theory to explain these readily observable spectral types.

“The development of the theory of quantum mechanics led to an understanding of the relationship between matter and its emission or absorption of radiation. If atoms are far enough apart that they do not affect each other, then each chemical element can emit or absorb light only at specific wavelengths. ….The strength of emission or absorption depends on how many atoms of the particular element are present as well as the temperature of the material, thus permitting both temperature and the chemical composition of the material producing the spectrum to be determined.

“If atoms are progressively jammed closer together, the wavelengths of emission or absorption by any given atom will be slightly changed, thus some atoms will emit/absorb at slightly longer wavelengths and others will emit/absorb at slightly shorter wavelengths…… where the emitting or absorbing atoms are thinly dispersed (the spectral features will look very sharp) and where they are tightly packed together (the spectral features are broadened). In the extreme case of high density, the emission lines become completely blurred together and one observes a continuous spectrum.”

Comment by Hank Roberts — 3 Aug 2007 @ 7:58 PM

342. Timothy, I disagree. While neither radiation is pure and ideal and may share some characteristics, there is (at least?) one major distinction. So-called blackbody radiation is not quantisized, in the same sense molecular absorption/emission is. A few lines spreading out into small bands and a couple of bands spreading into a little wider bands, while mitigating the quantizing nature, does not come close to the more or less continuous spectrum over a very wide range for blackbody radiation. Plus, (I think…) the underlying physics that generates the radiations are different.

Comment by Rod B — 3 Aug 2007 @ 9:33 PM

343. Hank,

I was holding onto these until later, but…

The following includes satellite images of CO2 and h20 reemission in atmosphere (towards the bottom):

McIDAS RETROSPECTIVE: VOLUME ONE
http://www.ssec.wisc.edu/gallery/mcidas-greatest-hits/firstgoes8.html

*

The following shows calculated degrees of cooling per day*wavelength as a function of pressure (which decreases exponentially with altitude) and wavelength for co2, ozone and water vapor.

Line-by-line calculation of atmospheric fluxes and cooling rates 2
http://www.aer.com/scienceResearch/rc/m-proj/abstracts/rc.clrt2.html

You will notice that by infrared radiation, water vapor always has a local atmospheric net cooling effect, there is only a small window where CO2 warms at roughly 12 km – and it cools for most altitudes-wavelengths, and ozone has a strong local atmospheric net warming effect until about 20 mb where it begins to radiate more radiation than it absorbs. With the exception of ozone, the direct warming effect of these greenhouse gases is principally due to radiation being absorbed by the surface. The atmosphere is warmed primarily by moist air convection in the troposphere, giving way to diffusion above the tropopause – with ozone playing a secondary role throughout ~8-24 km. But all of this is specific to season and latitude.

*

The distribution of carbon dioxide in ppm at 8 km as determined by the Atmospheric Infrared Sounder (satellite):

NASA AIRS Mid-Tropospheric (8km) Carbon Dioxide
July 2003
http://www-airs.jpl.nasa.gov/Products/CarbonDioxide/

Its infrared emissions demonstrate that it is not quite completely mixed even at this altitude.

Comment by Timothy Chase — 4 Aug 2007 @ 12:05 AM

344. > So-called blackbody radiation is not quantisized, in the same sense molecular absorption/emission is. — RodB

Rod. Saying “quantisized” and adding “so-called” and “in the same sense” get you nowhere productive. What’s your source??

Google “quantisized” and you find references along the lines of this thread: http://forum.physorg.com/index.php?showtopic=4385&st=15 (thread has a link to New Scientist’s decent article on Heim).

Do you _need_ to believe there are two different kinds of physics operating, to make some further point about climatology here?

Try the Cliff’s Notes page I linked — I gave only a brief excerpt above. There’s a good bit more there.

Comment by Hank Roberts — 4 Aug 2007 @ 3:06 AM

345. Re ‘#337 Where Rod writes : 4) Alastair, isn’t the use of a thermometer or bolometer a matter of convenience? If you could get there, have a thingy that doesn’t melt and a really good asbestos suit, could you measure the Sun’s temp with a thermometer?

No! They are different instruments and they measure different things.

A thermometer measures the kinetic energy of the molecules of a gas, or the vibrational energy of the atoms of a liquid or solid. A bolometer measures the radiation produced by the gas, liquid or solid. For a blackbody radiator like the sun where the atoms are closely packed then the two are the same. But if you had a dense cloud of Oxygen in space with a kinetic temperature 100 F, then the thermometer would read 100 F but the bolometer would read almost zero.

What I am saying may be the answer to the solar coronal heating problem. The solar corona is absorbing radiation from the sun and becoming electronically excited. But the gas atoms cannot emit that radiation until they become highly excited. Hence the blometers give high temperature readings. But if you placed a thermometer there (shaded from the solar radiation) then the gas would register a low temperature because the kinetic energy of the gas is low. The same is true for the Earth’s thermosphere.

Comment by Alastair McDonald — 4 Aug 2007 @ 5:57 AM

346. Re #339 and #341

Hank,

You should not believe everything you read on the web!

I don’t trust (am sceptical of) what Gavin and Ray write, far less people unwilling to use their full names such as Tev and Cliff. Moreover, it seems to me that Hydrodynamic is not entirely convinced by TeV’s explanations. I am certainly not :-(

But Cliff’s notes (who is Cliff) do not seem any better. He states that Kirchhoff found two types of line, absorption and emission but in fact they are the same lines. The emission lines turn into absorption lines when the radiation intensity exceeds the saturation value! Seems to me this proves Angstrom right!

He is also using the smearing of emission lines as the cause of blackbody radiation, but the electronic emission lines are “frozen out” at low temperatures, so how can they possibly produce blackbody radiation at terrestrial temperatures?

Google Bremsstralung radiation. That is also continuous like blackbody radiation, and has nothing to do with the smearing of electron lines.

Comment by Alastair McDonald — 4 Aug 2007 @ 6:31 AM

347. Rod B., I think you are getting hung up on terminology–black-body vs. line radiation. First, a black body does not exist in nature–that is nothing has 100% emissivity at all wavelengths. You can come close to blackbody radaition by looking at radiation emitted from a small hole in a cavity. This means that the probability of radiation escaping before it comes into thermal equilibrium with the walls and contents of the container (as well as the radiation field itself) is very small. So what does that equilibrium mean? First, all radiation starts by being emitted by a molecule/atom. However, the surrounding material influences the wavelength of emission and interacts with and influences the energy of the resulting energy. Lines broaden and eventually you get a continuum or close to it.
OK, now take away the container. The surrounding gas still broadens the line emission, but there is no container absorbing and re-radiating (with its vibrating atoms shifting frequencies), so the radiation retains its essentially broadened line character. Moreover, there are competing processes (collisional relaxation, etc.), so if the gas is too cool to excite the vibrational states etc., you get less radiation back. The system still comes to equilibrium, and the result is that if you look only at the portions of the spectrum where the molecule CAN emit, the spectrum LOOKS like a black body at the prevailing temperature. That’s what we mean by a gray body–the emissivity isn’t uniform.
This is my impression–anyone please feel free to correct me if I’m wrong. So, really there is no difference between “blackbody” radiation and line emission except that the blackbody radiation has come into equilibrium with its surroundings. Since this has to happen for all real radiation, all radiation is both black body and line emission.

Comment by Ray Ladbury — 4 Aug 2007 @ 6:42 AM

348. Rod B., All radiation is quantized–it comes in photons with energy h*nu. The differences Kirchoff described are continuous vs. line. This is precisely what Planck learned–you can’t get a blackbody spectrum unless you quantize the radiation.

Comment by Ray Ladbury — 4 Aug 2007 @ 7:01 AM

349. Alastair — CliffNotes are a North American college study guide series, around for decades.

You “don’t trust people unwilling to use their full names such as … Cliff….(who is Cliff)”?
Good grief, Alastair.

You just read the little excerpt, didn’t you, not even the page to which I gave a link?
You can’t not know who Cliff was, if you bothered to read even the one page.

http://www.cliffsnotes.com/WileyCDA/Section/id-305430.html

I quoted a bit of the original history of the science from that page. Attacking history isn’t helpful.

Look, you have a notion. Test it. Describe an experiment that could falsify it.

You do not believe carbon dioxide can radiate infrared photons. You know you can falsify your theory by looking for brightness in the infrared above the top of the clouds. Find a picture of the Earth’s horizon taken from space, see if there’s a sharp cutoff to the atmosphere viewed in the infrared, or a fuzzy edge.

“Kirchoff spectral types are comparable to Kepler’s Laws in the sense that they are only a description of observable phenomena.”

At this point you’re pushing an idea only you believe, that you haven’t published anywhere, that you don’t have math for, that argues that contemporary physics has to be wrong and you’re right. And you’re collecting people who really want to believe you for their own reasons, because they seem to think you’re proving climate change isn’t possible. I know that’s not where you’re going with your idea.

It’s not good for you or your readers when you keep “explaining” here why the climatologists are wrong with vague words and no published paper. Turn in the other direction — come at your idea as a scientist, suggest an experiment that can distinguish your idea from the other ideas. Look for that picture of the horizon in the infrared, since it would show a different picture according to your idea than the picture believed in by the infrared astronomers and climate physicists.

Stratospheric CO2 glows, or not, in the infrared. Look for pictures!

Comment by Hank Roberts — 4 Aug 2007 @ 10:05 AM

350. Hank (344): Oh! cram your “sources”. I was fudging (confusingly I admit) because of the uncertainty/quantum factor where ALL energy, including that created by a locomotive is quantisized, as Ray points out. Not the same as the quanta of molecular and electronic absorption/emission.

Comment by Rod B — 4 Aug 2007 @ 10:20 AM

351. Ray, There is a difference between blackbody radiation and line radiation. An iron bar glows red when heated. The lines which are emitted by iron vapour are green.

Blackbody radiation is thermal radiation, which depends on temperature. Line radiation is produced by electronic relaxation of atoms, or vibrational and rotational relaxation of liquid and gas molecules. These three are all produced in a different manner. Blackbody radiation is produced in yet a different manner. “Thermal radiation is generated when heat from the movement of charged particles within atoms is converted to electromagnetic radiation.” See http://en.wikipedia.org/wiki/Thermal_radiation

There are other types of radiation too such as fluorescence, phosphorescence, and Bremsstrahlung, X-rays and Gamma-rays. They are all electromagnetic radiation but they are all produced in different ways and have different characteristics.

Comment by Alastair McDonald — 4 Aug 2007 @ 10:22 AM

352. Alastair McDonald (#345) wrote:

What I am saying may be the answer to the solar coronal heating problem. The solar corona is absorbing radiation from the sun and becoming electronically excited. But the gas atoms cannot emit that radiation until they become highly excited. Hence the blometers give high temperature readings. But if you placed a thermometer there (shaded from the solar radiation) then the gas would register a low temperature because the kinetic energy of the gas is low. The same is true for the Earth’s thermosphere.

Alastair, physicists seem to believe that they have a solution to the solar coronal heating problem – without throwing out everything we know about the interaction between matter and electromagnetic energy. If you read the Wikipedia article a little further you will see this. Electronic excitation in the corona is important given the fact that the corona is hot enough that matter exists as a plasma. With regard to temperature measurements in the corona, there would however be a difference inasmuch as the corona is no doubt distant from a local thermodynamic equilibrium, but not to the degree that you imagine.

Electronic excitation is not an issue in the earth’s atmosphere because neither the radiation no matter are hot enough that electronic excitation would be involved.

As has been pointed out to you on numerous occasions (without any acknowledgement on your part, either in terms of agreement or disagreement), the thermal infrared radiation of which we speak is not due to electronic excitation, that is, it is not due electrons jumping from one orbital to another within an atom. It is due to the molecules themselves assuming different quantum rotational and vibrational states, and at higher temperatures it would involve stretching.

Think about it: how do we know that carbon dioxide absorbs infrared at various parts of the infrared band? At a basic level, without recourse to quantum mechanics, we can measure it in a lab. Why do we claim to know the same thing with regard to its emission at the same wavelengths? Because we are able to deal with emission the same way: measurements in a lab.

The various quantum molecular states involve energies far lower than what is involved in the corona of the sun. The longwave radiation relevant to the greenhouse effect as it exists within our climate system isn’t frozen out except under rather specific circumstances – because it is within the range of the energies associated with the quantum molecular states.

This isn’t electronic, it is molecular.

You claim that carbon dioxide is unable to participate in the greenhouse effect except near the surface of the earth. You claim that the reason is that the thermal radiation gets frozen out higher up in the earth’s atmosphere, or on other occasions, you claim that the energy gets carried off by non-greenhouse gases such as oxygen.

With regard to the latter, what every energy may be lost as the result of collisions may be gained by collisions, and as long as the collision rate exceeds the emission rate by a couple of orders of magnitude, a local equilibrium will be established between thermal radiation and the thermal state of matter – at above ~20 mb. Over all frequencies and wavelengths? No, but over the relevant frequencies and wavelengths.

Now with to the former, your argument is that the energy gets frozen out because the temperature of the atmosphere is too low to achieve electronic excitation. But the same would no doubt hold at the surface. It is able to achieve the state of excitation because the excitation is molecular, not pertaining to electrons shifting from one quantum orbital state to another, but to molecules shifting from one quantum rovibrational state to another. Genuine scientists studying the interaction of infrared thermal radiation and various gases know this as a result of lab experiments. Genuine scientists also know this as the result of quantum statistical mechanics, both in terms of how quantum mechanics applies to molecules and the equipartion theorem applies to both radiation and matter.

However, we also know this as the result of satellite imaging of the atmosphere.

I pointed out as much in 343:

The distribution of carbon dioxide in ppm at 8 km as determined by the Atmospheric Infrared Sounder (satellite):

NASA AIRS Mid-Tropospheric (8km) Carbon Dioxide
July 2003
http://www-airs.jpl.nasa.gov/Products/CarbonDioxide/

Its infrared emissions demonstrate that it is not quite completely mixed even at this altitude.

*

When you argue against the established science, you are making ad hoc arguments which are typically grounded in a lack of understanding and armchair-theorizing. You are doing so without recourse to empirical observation, systematic empirical laboratory experimentation, and both the relevant conceptual and mathematical framework. You seem to think that you are justified in doing this in part because the people who you are speaking with don’t know everything there is to know.

Well, neither do you.

What I hold is that the whole of the scientific community knows a great deal more than any one individual, or for that matter, what any one individual would ever be capable of knowing. There is a division of cognitive labor which that exists within a community which is ultimately all engaged in the same thing: the empirical study of the unified nature of reality. For any given well-established scientific knowledge, there are generally mulitiple independent lines of investigation which provide it with justification, and a conclusion justified by multiple independent lines of investigation is generally justified to a far greater degree than it would be if it were justified by any one line of investigation considered in isolation.

*

Consider an ant colony consisting of 10,000,000 individuals.

Each individual has perhaps 10,000 brain cells. Individually an ant brain isn’t capable of very much. But an ant colony has as collectively as many brain cells as a human being. These brain cells are by no means as tightly integrated as a human brain. Nevertheless, collectively an ant colony is capable of learning at a level which far exceeds that which any one ant brain is capable of in isolation, such that an older, larger colony knows better than a younger smaller colony than to attack a colony larger than itself.

The same principle applies to human beings.

Both genetically and in terms of the innate capacity of my brain, there is very little difference between me and someone who lived 800,000 years ago in a clan consisting of 100 individuals. However, as a human being living in today’s modern society, I have the benefit of a legacy stretching back to perhaps 10,000 years, and I have the benefit of living in a community consisting of nearly 7,000,000,000 individuals. Collectively we are capable of knowing a great deal more than any one individual is capable of in isolation.

*

The same principle applies even in terms of the temporal integration of your understanding from one moment to the next.

Engaging in extreme cartesian doubt where one would be unwilling to accept anything which is not immediately “self-evident,” including whatever memories one might have of a moment ago, if one were systematic about it, one would not even be able to take recourse to even the most basic distinctions between dreaming and perception, identification and evaluation, cognition and emotion, or even the subject who is aware and the object which the subject is aware of. At a slightly higher level, one might choose to accept perception, but regard any evidence much beyond this, including memories of more than a few seconds ago as objects which were placed there to create the mere appearance of an enduring and ancient world which was created ten seconds or ten thousand years ago.

But we are capable of moving beyond such extreme skepticism because we are able to recognize the principle that a conclusion which receives justification from multiple independent lines of empirical investigation is generally justified to a far greater degree than it would be if it only received justification from any one line of investigation considered in isolation. We realize that any argument for such extreme skepticism must presuppose the very sort of knowledge which it would deny us recourse, and as such is ultimately (self-referentially) incoherent. Thus for example a radical skeptic might choose to undercut all claims to knowledge by arguing that all claims to knowledge might ultimately be grounded in nothing more than a dream.

But what does the skeptic mean by “dream”?

*

At a certain level, you are making the same sort of mistake as the young earth creationist or radical skeptic.

You arbitrarily accept some of the conclusions of science, those which fit your personal “theory,” but not others. But how do you know the conclusions of science which you accept? What justifies them? The conclusions of science exist within a vast network of human knowledge which is far greater than that which any one individual is capable of understanding in isolation from a human community.

Right here on this blog we form a community which exists within a far vaster community. We are capable of learning from each other because we participate in a dialogue.

*

Consider this: if you have one individual with only three insights, he is capable of making only three connections between any two insights.

But if you have two individuals with three insights each, by learning from one another, they will have six insights and they will be capable of making fifteen connections between any two insights. Between five individuals, the community is capable of making one hundred and five connections. But if a given individual refuses to learn from the others, if he refuses to participate in the dialogue, then he is once again capable of making only three connections.

In terms of the principle involved, this is the position you place yourself in by refusing to acknowledge and learn from the insights of others.

*

This is important enough to point out and to stress both for your sake and for the sake of the communities which you are a member of.

When you insist that you have some alternative to modern empirical science, you know just enough to sound authoritative to some. By insisting that your own personal and largely ad hoc theory is an alternative to well-established science, you find it necessary to cast doubt on well-established science. As such you are undercutting the purpose of this community and inadvertently serving to further the interests of those who claim that there is no threat to humanity posed by global warming because global warming does not exist. Whether you choose to recognize it or not, your ad hoc skepticism serves the denialists.

I would ask that you reconsider your approach, for the sake of your understanding, this community, the human community and out of the recognition of the threat posed by climate change which you acknowledge exists.

Comment by Timothy Chase — 4 Aug 2007 @ 1:16 PM

353. Rod/Ray/Hank/Timothy/Alastair: I have gained from your discussions and so would attempt to clarify some things as I see them. Heat radiation is that radiation exchanged between bodies at equilibrium whatever the nature of the transitions. The black body has an infinite continuum of allowable frequencies. A cavity can approximate a black body arbitrarily well provided any port approaches zero size, and there is at least one region which absorbs at all frequencies inside. Black body radiation describes heat exchanged between black bodies.
Line or narrow band emitters can be at equilibrium, but parts of the spectrum will be empty. Similarly, line or narrow band absorbers can be themselves at equilibrium while letting higher energy radiation pass through.

Interacting quantum oscillators with the same transition frequency create higher and lower energy transitions. A large number of identical oscillators form bands, so a collection of atoms can form an approximation to a black body. Isolated atoms can (mostly) only emit single quanta. Interacting ideal gas molecules at our pressures show line broadening from occasional collisions, but spend too much time separated from neighbors to form bands. The exception is water, which molecules have a relatively strong attraction and a permanent dipole that makes them interact over much greater distances than they would otherwise, thereby forming a continuum at moderate concentration.

Since temperature is only defined at equilibrium, we rapidly get into difficulties in non equilibrium situations, which is most of the time. We can talk about apparent temperature, calculated from the ratio of vibrational states, which could be different from that using the distribution of velocities, and all these could be different from the apparent temperature of radiation passing through from a black body. This is the LTE state. Different thermometers will read differently in non-equilibrium situations. Go buy a dozen and some paint and try it.

Coherence is the result when atoms with inverted population distributions are placed in a cavity and are “stimulated” to emit in concert with a standing mode in the cavity – continuous mode lasers. Coherent radiation is monochromatic, and vice versa. I think this is what RodB was talking about.

Finally, the Einstein coefficients do not require equilibrium, so we can write a general NLTE transfer equation if we have the numbers, and as far as I can find we do.

Comment by Allan Ames — 4 Aug 2007 @ 1:51 PM

354. Allan Ames (#353) wrote:

Isolated atoms can (mostly) only emit single quanta. Interacting ideal gas molecules at our pressures show line broadening from occasional collisions, but spend too much time separated from neighbors to form bands.

This is why we speak of higher temperatures and pressures. But notice – speaking of a band should not be taken to imply that the individual lines no longer exist as distinct entities but only that they consist of an identifiable natural grouping. This will exist independently of the pressure or temperature. But even in the case of the bleeding between lines or bands which I mentioned above, the lines and bands will be distinguishable – its just that the boundaries between the lines or the bands will not be as well-defined due to spreading.

Allan Ames continues:

The exception is water, which molecules have a relatively strong attraction and a permanent dipole that makes them interact over much greater distances than they would otherwise, thereby forming a continuum at moderate concentration.

Water has some interesting properties including a fairly impressive ability to amplify parity violation which may have been a significant factor in the origin of the chirality of life. However, there natural broadening of lines due to the uncertainty relationship which exists between time and energy, line broadening, band broadening, etc. All of these involve to varying degrees a blending where while a line or band remains distinct from another, they are no longer entirely separate. They never were.

Allan Ames continues:

Since temperature is only defined at equilibrium, we rapidly get into difficulties in non equilibrium situations, which is most of the time. We can talk about apparent temperature, calculated from the ratio of vibrational states, which could be different from that using the distribution of velocities, and all these could be different from the apparent temperature of radiation passing through from a black body. This is the LTE state. Different thermometers will read differently in non-equilibrium situations. Go buy a dozen and some paint and try it.

Under LTE conditions, temperatures measured by different approaches will be identical since LTE is a local thermodynamic equilibrium.

With respect to thermometers which are painted different colors, well, that is why God created albedos. Temperatures are measured in the shade. When you speak of a thermometer which is painted a different color registering a different temperature, you are speaking of a thermometer which has been placed in the sunlight and/or which hasn’t had the opportunity to achieve equilibrium with its environment.

This is in 237. I wrote:

But a thermometer wrapped in black linen is something else – that’s why we speak of albedos.

The question at this point is, “How do you carve things?” You could define temperature in non-standard ways so that you are measuring temperatures in the sunlight using different colored thermometers just like you could define a color which consists of either blue or red, but it would simply anything. Indeed, it would make things more complex. Or we might choose not to distinguish between the thermal radiation of the sun and the local thermal radiation of either the atmosphere or the surface, as I suggested in 237. But we make these distinctions because they simplify matters – much like the choice of coordinate systems when attempting to solve Einstein’s field equations in General Relativity. Put your thermometers in the shade and let them achieve an equilibrium with their surroundings.

Allan Ames continues:

Coherence is the result when atoms with inverted population distributions are placed in a cavity and are “stimulated” to emit in concert with a standing mode in the cavity – continuous mode lasers. Coherent radiation is monochromatic, and vice versa. I think this is what RodB was talking about.

Granted. Monochromatic, in phase, etc. But this is the result of a population inversion, not the sort of thing that we are dealing with here.

Allan Ames continues:

Finally, the Einstein coefficients do not require equilibrium, so we can write a general NLTE transfer equation if we have the numbers, and as far as I can find we do.

I believe you are thinking of the following paper:

Non-LTE diagnostics of the infrared observations of the planetary atmosphere
Oleg Goussev an der Fakultat für Physik der Ludwig–Maximilians–Universitat
Munchen 2002
http://edoc.ub.uni-muenchen.de/archive/00001056/01/Goussev_Oleg.pdf

It is well worth reading. I have said as much earlier in this thread in 249.

And you are right: non-local thermodynamic equilibria are susceptible to mathematical description to one degree or another, although as I have pointed out previously it is best for people to start with the study of local thermodynamic equilibria then move onto the non-local. Study Newton before Einstein and Galileo before Schroedinger. Likewise there is a cost due to complexity such that there will be trade-offs were incorporation of the non-local will be at the expense of other calculations or increased calculation time.

But in any case the results of the study of non-local thermodynamic equilibrium has been largely incorporated into the radiation codes which are now in use.

The capability to compute quantities for atmospheric layers, which are not in local thermodynamic equilibrium (Non-LTE option).

pg 1

An Update on Radiative Transfer Model Development at Atmospheric and Environmental Research, Inc.
J. S. Delamere, et al
Twelfth ARM Science Team Meeting Proceedings, St. Petersburg, Florida, April 8-12, 2002

We have known of these effects for quite some time now.

Here is one of the earlier abstracts I have been able to find from 1989, although there are earlier versions of this paper:

In this, possible processes which can give rise to non-local thermodynamic equilibrium (non-LTE) effects in stratospheric hf through direct photochemical excitation are outlined and an assessment of their magnitude is made. This calculation is done for both the first (v = 1) and second (v = 2) excited state of hf. available kinetic data relevant to the excitation and deactivation of vibrationally excited hf in the stratosphere are reviewed and the implications of uncertainties in our knowledge of them are summarized.

Nonlocal Thermodynamic Equilibrium Effects in Stratospheric Hh by Collisional Energy Transfer from Electronically Excited O(Sub 2) and Implications for Infrared Remote Sensing
Kaye, Jack A.
Report Date : 01 OCT 1989

Additionally, you will notice that the incorporation of non-LTE effects into radiation codes in use goes back at least as far as 1996.

User’s Manual for SHARC-3, Strategic High-Altitude Radiance Code.
Corporate Author: PHILLIPS LAB HANSCOM AFB MA
Personal Author(s): Sharma, R. D. | Gruninger, J. H. | Sundberg, R. L. | Bernstein, L. S. | Robertson, D. C.
Organization Type: F – AIR FORCE
Report Date: 21 MAY 1996
Report Number(s): PL-TR-96-2104-ERP-NO-1193 (PLTR962104ERPNO1193)

Abstract: The Strategic High-Altitude Radiance Code, SHARC-3, calculates atmospheric radiance and transmission over a 1 – 40 microns spectral region for arbitrary lines of sight (LOS) between 50 and 300 km altitude. It models radiation due to nonlocal thermodynamic equilibrium (NLTh) molecular emissions, the dominant sources at these altitudes. This third release of SHARC has been upgraded to model LOS’s across and near the atmospheric terminator. It interpolates between several atmospheric profiles, which represent variations in species concentration through that region. In addition the spectral distribution of sunlight which has passed through the lower atmosphere is approximated by a series of optical depth functions for that part of the solar path below 50 km altitude. Auxiliary atmospheric profile generator and slit function programs enhance the capabilities of the code. Data for the previously supported major radiators have been upgraded where necessary, and CH4 and minor isotopes of H2O have been added. As before, auroral regions can be embedded in quiescent atmospheres, and industry-standard kinetics and radiation transport modules have been employed. This manual outlines the scientific basis of the code, details steps for code installation and use as supplied, and indicates how the user may modify the supplied chemistry if desired.

You will notice that this is the very same radiation research that modern climatology was based on at the same time in the 1980s. Military research – as has been stressed on numerous occasions. Nothing new.

In any case, I have pointed out just recently the existence of NLTE in the atmosphere:

SHARC, A Model for Calculating Atmospheric and Infrared Radiation Under Non-Equilibrium Conditions
January 24, 1994

Figure 2. Calculated Vibrational Temperatures for C02 States for a Solar Zenith Angle of 820.
(50-150 km)

One take-away point if you look at the graph: the difference between local and non-local thermodynamic equilibria is a matter of degree not kind. At some level of accuracy or another there will be some non-local effects however insignificant. Lets face it: reality can get rather messy. But it will be a matter of degree. Viewing much of the atmosphere strictly from inside a framework of either LTE or partial-LTE works quite well under most circumstances less well in others. But even any

In any case, climate models have incorporated the very effects which you believe are important.

Finally, if at times it seems that I am unfamiliar with some of this, it is because I haven’t started looking into climatology until the past couple of months and my background in physics is largely informal. But the research is there, the knowledge is there. And if you want, you can look it up – as I have done.

Comment by Timothy Chase — 4 Aug 2007 @ 6:42 PM

355. CORRECTION to #354

Where I wrote:

Allan Ames continues:

“Finally, the Einstein coefficients do not require equilibrium, so we can write a general NLTE transfer equation if we have the numbers, and as far as I can find we do.”

… using blockquotes, that is the extent of what I quote at that point, and the rest should be outside of the quote.

We I continue with,

I believe you are thinking of the following paper:

Non-LTE diagnostics of the infrared observations of the planetary atmosphere
Oleg Goussev an der Fakultat für Physik der Ludwig–Maximilians–Universitat
Munchen 2002
http://edoc.ub.uni-muenchen.de/archive/00001056/01/Goussev_Oleg.pdf

… that should be outside of what I am quoting. Sorry for the confusion on my part.

Comment by Timothy Chase — 4 Aug 2007 @ 7:33 PM

356. Hank (344, again), I really couldn’t get much out of the discourse in PhysicsForum that you referenced. Maybe it’s me. But thanks for the effort and assistance.

Comment by Rod B — 5 Aug 2007 @ 2:12 PM

357. All, I scanned the latest posts. They seemed informative, but I didn’t detect a direct answer to my very simple(???) 3rd question of #337, which was (with some clarification): Assume the temperature of a mole of gas (say CO2) that has a boltzman distribution of translation kinetic energy and every molecule is in its ground vibration and rotation states. (Is that possible?) Now add infrared radiation so that a percentage of the molecules (any % you want) has energy absorbed into its rotation and vibration energy levels, and assume no other energy transfer ala equipartition, collision, radiant emission, or from any other source, and that it is given enough time to settle (though given my assumptions I don’t know what “settle” would be…). Did the temperature of the mole, as measured by a thermometer, go up? down? or stay the same? Put another way (but still the same question), did the induced kinetic energy of the atomic vibrations and the molecular rotation add to the kinetic energy of the translation molecular movement?

A second side question (sorry): Assuming no other energy transfer, would this mole of CO2, if sufficiently compressed (though that might mess me up as it is itself an energy transfer…) emit black/graybody radiation per Planck’s function. And if so where would that emitted energy come from: translation, vibration, rotation, or some combo (I’m assuming not from electron, chemical or nuclear, etc.)?

Although it doesn’t seem quite as critical to me as before, I think this (and related questions on molecular energy transfer) has an impact on the base process of global warming — what gets warmed and how. I guessed it would have been straight-forward, but it’s looking far from that, including a bunch not even well known.

I need to go back and study the posts. They look good, and I appreciate yourall’s assistance and input.

Comment by Rod B — 5 Aug 2007 @ 4:38 PM

358. > assume no other energy transfer …. Did the temperature of the mole, as measured by a thermometer, go up?

Not as long as your assumption holds true, while no energy was transferred to the thermometer — the thermometer won’t detect anything until energy is transferred to or from it.

Comment by Hank Roberts — 5 Aug 2007 @ 9:27 PM

359. What I am saying is that cavity radiation, blackbody radiation, and thermal radiation are all the same thing, but it is different from line radiation of which there are many types. I prefer to use the term blackbody because it describes the prefect absorber, which through Kirchhoff’s law of emission equalling absorption, emits with the shape of Planck’s function.

I agree that a blackbody is an ideal which does not really exist. However, it is a good approximation to reality and Planck’s function can be used when modeling the radiation emitted by the Earth’s surface etc. The reason Planck’s function is written as B(T) is because it is the Blackbody function.

which describes the ideal

Comment by Alastair McDonald — 6 Aug 2007 @ 5:22 AM

360. Re #357

That is a good question. And the answer is that if the increase in the vibrational energy and rotational energy does not change, then the kinetic temperature (measured with a thermometer) will remain the same because the translational energy has not changed.

In fact what happens is that the vibrational energy can be re-emitted, or it can be degraded into rotational energy or translational energy. Water vapour can emit the rotational energy, but for carbon dioxide the rotational energy can only degrade to translational energy i.e. kinetic energy.

The answer to the second question is that CO2 does not emit blackbody radiation. It only emit lines related to it vibrational energy. Nitrogen does not even emit lines due to vibrational energy, whereas water vapour radiates vibrational lines, rotational lines, and continuum (blackbody?) radiation.

The energy for the lines comes from the excitation which may have arisen in collisions, making its original source translational energy. In other words as the gas radiates it cools down (kinetic energy gets less so registers a lower reading on a thermometer.)

HTH.

Cheers, Alastair.

Comment by Alastair McDonald — 6 Aug 2007 @ 6:19 AM

361. Re #357

Rod, that is a good question. And the answer is that if the increase in the vibrational energy and rotational energy does not change, then the kinetic temperature (measured with a thermometer) will remain the same because the translational energy has not changed.

In fact what happens is that the vibrational energy can be re-emitted, or it can be degraded into rotational energy or translational energy. Water vapour can emit the rotational energy, but for carbon dioxide the rotational energy can only degrade to translational energy i.e. kinetic energy.

The answer to the second question is that CO2 does not emit blackbody radiation. It only emits lines related to its vibrational energy. Nitrogen cannot even emit lines from its vibrational energy, whereas water vapour radiates vibrational lines, rotational lines, and continuum (blackbody?) radiation.

The energy for the lines comes from excitation, which may have arisen from absorption, but also from collisions so the original source is translational energy. In other words as CO2 radiates, it cools down (the kinetic energy gets less so it registers a lower reading on a thermometer.)

If you consider the centre of a large mass of CO2, the radiation emitted by the molecules will equal that absorbed from other molecules. The same equality will happen with the collisions sharing out the translational energy.

The conventional idea is that the translational and vibrational energy will also reach a stable balance, and so the vibrational emissions will depend on the translational energy, ie the temperature of the gas. This leads to a state of local thermodynamic equilibrium (LTE.)

However, LTE only applies to the centre of a mass of gas. In the atmosphere, scientists are already aware that the top of the atmosphere is in non-LTE. Where they are going wrong is not realising that at the other edge, the base of the atmosphere, LTE does not apply either. Here there is a net inflow of energy: radiation, sensible and latent heat, whereas at the top of the atmosphere there is a net loss.

HTH.

Cheers, Alastair.

Comment by Alastair McDonald — 6 Aug 2007 @ 7:49 AM

362. > CO2 does not emit blackbody radiation. It only emit lines related to it vibrational energy.
So does this modify your idea, and you no longer think the stratosphere is dark in the infrared band?
It sounds like you now do believe that in the upper stratosphere the CO2 is radiating in the infrared (as observed by the IR astronomers and the climatologists), and perhaps your argument is about how quantum mechanics works but not about what’s observed? But it also sounds like this eliminates the possibility of falsifying your idea by looking for IR radiation in the stratosphere.

Any other way to distinguish your idea from the commonly held physics? Must be able to falsify it if there’s a real difference.

Comment by Hank Roberts — 6 Aug 2007 @ 9:19 AM

363. Before the thread is cut, may I make a final plea for answers to queries I raised in post 204. Some but not all of the questions were answered. I would still appreciate enlightenment on the following:

In starting this debate, Ray Pierrehumbert explained that, at certain wavelengths, there is already saturation by CO2 (atmosphere optically thick and little or no outgoing OLR radiation to space from these wavelengths). However, increasing atmospheric CO2 would bring contiguous wavelengths in the wings into play such that they, too, would become optically thick. My question relates to whether modellers are considering these wing areas in isolation with respect to CO2. Is it not the case that they are already fairly blocked by water vapour such that extra CO2 would have much less greenhouse effect than would have been the case had no water vapour been present. It is my understanding that the Hitran data relate to single gases and not to greenhouse gas mixtures. Have any experiments of the type described by Raypierre been conducted anywhere with such mixtures?

One other query. CO2 is heavier than air and I understood that it was mixed in the lower atmosphere as a result of convection. Why, then, does it not form a thin layer immediately above the influence of convection? From what I’ve read it seems to be invading the stratosphere.

Sorry to drag you’all away from quantum mechanics but I’m only a simple soul looking for simple answers that I may be able to understand. Thank you in advance.

Comment by Douglas Wise — 6 Aug 2007 @ 10:55 AM

364. re Alastair, 360-1: I need to be picky to insure I understand your point exactly. You say,

“…if the increase in the vibrational energy and rotational energy does not change, then the kinetic temperature (measured with a thermometer) will remain the same because the translational energy has not changed.”

I assume your “increase” is what I posed in my question — the blasting of my CO2 with some IR, and not anything that happens later. Then your answer says NO, the temperature of the gas does not increase. Which implies that the kinetic energy of the vibrations and rotations DO NOT go into determining the temperature, only translation does. Is this your contention?

An anomaly: later you talk of the molecules emitting IR radiation and “cooling the gas.” Problem: IR line emission comes from vibration and rotation energy; if that doesn’t go into temperature re above, how can the gas cool when the ‘rotation’ energy is decreased (through IR radiation emission)?

2) If vibration energy transfers to translation (among others) does it then indirectly affect the temperature?

3) my understanding is that CO2 picks up rotation only from its vibrational levels. (CO2 does not have a dipole moment so can not absorb into rotation levels directly from IR, but only after vibration absorption makes the molecule a quasi dipole. (I don’t know if after the initial absorption the molecule can now absorb into rotational energy directly or not…) This would imply that rotation can transfer to translation or emission but only after going through vibration first. I think this is what you said; is it?

4) Why is it that CO2, if properly dense and reasonably heated, can not emit “continuum” blackbody radiation, but water vapor can??

5) Is Hank (362) correct? Are you changing your stance that GHG don’t re-emit? Or is it that you contend they can emit but lose their vibration/rotation energy so quickly through transfer via collision or otherwise to translation, that they don’t?

Comment by Rod B — 6 Aug 2007 @ 11:07 AM

365. Hank Roberts (#362) wrote:

It sounds like you now do believe that in the upper stratosphere the CO2 is radiating in the infrared (as observed by the IR astronomers and the climatologists), …

Actually Alastair (#361) wrote:

The energy for the lines comes from excitation, which may have arisen from absorption, but also from collisions so the original source is translational energy. In other words as CO2 radiates, it cools down (the kinetic energy gets less so it registers a lower reading on a thermometer.)

In fact what happens is that the vibrational energy can be re-emitted, or it can be degraded into rotational energy or translational energy. Water vapour can emit the rotational energy, but for carbon dioxide the rotational energy can only degrade to translational energy i.e. kinetic energy.

… so CO2 radiates energy gained from collisions. As for Co2 not emitting rotational, I believe that is right as well.

Finally, with respect to the difference between line thermal radiation and blackbody thermal radiation, I think they are as different as black and white. But the thing of it is that there is a continuem of grays between black and white. And as I have suggested, a similar continuem appears to exist between the two archetypal forms of thermal radiation. However, I personally still have a fair number of holes to fill in on this.

Comment by Timothy Chase — 6 Aug 2007 @ 11:40 AM

366. Re 357 Rod B:

Some questions as posed are difficult to answer in a few lines, so we answer hopefully related questions. Sometimes there are just too many questions and we get writers cramp.

1) At any finite temperature some fraction will be in higher states.
2) Adding net IR – heat – to a system will increase the total heat in the system by E = N h nu.
3) Energy is partitioned among the various modes if equilibruim is reached. Assuming you add energy to one mode, interaction with other atoms is usually necessary to achieve equilibrium.
4) BB radiation? If compressed enough probably. What upwells from Venus?
5) The new material made from compressed gas would have its own modes, not necessarily directly related to those of the molecules from which it was made. The energy of compressing the gas would “squeeze out” most of the lower energy motions. To bring OCO molecules close enough to allow quantum mechanical interactions says you put a lot of PV energy into it. Energy levels in closely coupled systems become properties of the group, not any individual atoms. Part of what distinguishes solids from gases is that there are different modes of motion of the atoms or electrons in the phase. Remember that all energy levels are derived from interactions between positively charged nuclei and negative electrons. Sometimes these levels are derivable from simple mechanical models like vibrating springs, sometimes not, like the electronic levels which are purely quantum mechanical.

Comment by Allan Ames — 6 Aug 2007 @ 12:56 PM

367. re 361, 362 McDonald & Roberts:

I keep missing the point of your debate. I think I have it, than all of a sudden the same point comes back from the opposite direction. Could I ask if any of the following align? If not, what, or if so, what else is missing?

My understanding is that molecules which are IR absorbers in spectrophotometers interact with thermal radiation at their characteristic wavelengths, ww. This interaction can have at least three effects. One is to absorb ww photons and re-radiate (almost) ww photons. Another is to remove ww photons and increase kinetic energies. A third is to remove kinetic energy and radiate ww photons.

Comment by Allan Ames — 6 Aug 2007 @ 1:23 PM

368. I’m trying to sort out which of Alastair’s statements are from his unique theory and which are from generally accepted physics, myself.

Comment by Hank Roberts — 6 Aug 2007 @ 2:26 PM

369. re 361, 362 McDonald & Roberts

Allan Ames (#367) wrote:

My understanding is that molecules which are IR absorbers in spectrophotometers interact with thermal radiation at their characteristic wavelengths, ww. This interaction can have at least three effects. One is to absorb ww photons and re-radiate (almost) ww photons. Another is to remove ww photons and increase kinetic energies. A third is to remove kinetic energy and radiate ww photons.

If I understand things correctly, the three effects exist and are all important. That seems to be something everyone agrees upon at this point. Once a local thermodynamic equilibrium over certain parts of the longwave spectra exists, the amount of energy being lost to kinetic energy after absorption is equal to the amount of kinetic energy being lost to reemission. And above 20 mb, the collision rate greatly exceeds the rate at which radiation is reemitted, so as an approximation, one may regard this as a local thermodynamic equilibrium.

However, as both you and Alastair have pointed out, the atmosphere isn’t strictly a local thermodynamic equilibrium even above 20 mb. I don’t know all of the details as of yet, but there are partial local thermodynamic equilibria and non-local thermodynamic equilibria – at different wavelengths. You probably know more about this part of it than I do. In any case, the deviation from a local or partial thermodynamic equilibrium is a matter of degree, and as the pressure increases, the deviation will decrease until for nearly all parts of the relevant spectra it becomes negligible at 50 km and below.

Finally, unless I am mistaken, the only real disagreement at this point is over the extent to which blackbody thermal radiation and line thermal radiation are different. I at least don’t think that is especially important. We will probably sort it out later – and learn more in the process.

Comment by Timothy Chase — 6 Aug 2007 @ 2:37 PM

370. Douglas Wise (#363) wrote:

However, increasing atmospheric CO2 would bring contiguous wavelengths in the wings into play such that they, too, would become optically thick. My question relates to whether modellers are considering these wing areas in isolation with respect to CO2. Is it not the case that they are already fairly blocked by water vapour such that extra CO2 would have much less greenhouse effect than would have been the case had no water vapour been present.

Well, it will certainly be blocked where water vapor is dominant, but generally water vapor doesn’t make it much above 4 km. Interestingly, though, even at 4 km it isn’t that evenly distributed. I know of some mpgs which show this as the result of infrared sounder readings. Similar mpgs exist for aerosols and I believe even for carbon dioxide. I will have to check. However, as you point out, CO2 goes well above 4 km. We know this from both gas measurements and soundings. For example, here is an infrared image of CO2 longwave reemission at 8 km:

NASA AIRS Mid-Tropospheric (8km) Carbon Dioxide
July 2003
http://www-airs.jpl.nasa.gov/Products/CarbonDioxide/

… so while it would be blocked below 4 km, it will absorb upwelling (and downwelling) radiation from different altitudes, and there will be reemission – principally from the far wings at lower altitudes and closer to the peaks at higher altitudes.

Douglas Wise (#363) wrote:

It is my understanding that the Hitran data relate to single gases and not to greenhouse gas mixtures. Have any experiments of the type described by Raypierre been conducted anywhere with such mixtures?

At a certain level, this is actually what the essay is about. But they have been repeated under different conditions with different mixtures since. And then of course there are the measurements we take of the atmosphere itself.

Douglas Wise (#363) wrote:

One other query. CO2 is heavier than air and I understood that it was mixed in the lower atmosphere as a result of convection. Why, then, does it not form a thin layer immediately above the influence of convection? From what I’ve read it seems to be invading the stratosphere.

Good questions.

The answer seems to be due to diffusion – which is driven by thermal-kinetic motion. It is a slow process – which is part of the reason why there is such a lag between the the anthropogenic emission of carbon dioxide and the achievement of equilibrium in terms of the greenhouse effect.

Anyway, I will look up the sounder-vids a little later, but if anyone wants to find them a little sooner, I believe they are at the same site as the link.

Comment by Timothy Chase — 6 Aug 2007 @ 3:39 PM

371. CORRECTION to 369

If I understand things correctly, the three effects exist and are all important. That seems to be something everyone agrees upon at this point. Once a local thermodynamic equilibrium over certain parts of the longwave spectra exists, the amount of energy being lost to kinetic energy after absorption is equal to the amount of kinetic energy being lost to reemission.

This is false.

The amount of energy being lost by absorption to kinetic energy will (typically) be less than the amount of energy being lost by kinetic energy to reemission, at least in the case of both water vapor and carbon dioxide. What is equal is the capacity to absorb or reemit. However, energy gained by convection will add kinetic energy, and on the whole, this kinetic energy is lost to radiation – which cools the atmosphere.

In the case of ozone, it is more split, but with radiation being lost to kinetic in the lower part, and less kinetic being lost to radiation above, so than on the whole ozone tends to warm the atmosphere. The reason for this difference is, I believe, due to the fact that it is absorbing ultraviolet radiation from sunlight rather than longwave thermal radiation from the earth.

Comment by Timothy Chase — 6 Aug 2007 @ 4:06 PM

372. Timothy, CO2 can absorb into rotation but not emit from it????

Comment by Rod B — 6 Aug 2007 @ 9:28 PM

373. Allan (366), I appreciate the helpful explanation why a sufficiently compressed gas acn emit BB radiation. I assume it would follow that the top of the atmosphere’s radiation is line/spectral only, and not BlackBody — not near warm or dense enough. (GHG emission only; not including the atmospheric albedo and such)

I don’t accept (or understand) your statement, “…all energy levels are derived from interactions between positively charged nuclei and negative electrons.” What about straight translation energy, rotation, or even vibration, which is movement of the entire atom?

Comment by Rod B — 6 Aug 2007 @ 9:41 PM

374. Rod B (#372) wrote:

Timothy, CO2 can absorb into rotation but not emit from it????

As I understand it, since carbon dioxide is a linear molecule which is only transiently dipolar as the result of bending, and therefore it would neither be able to absorb or emit in a strictly rotational mode. However, given the fact that it is transiently dipolar, vibration may be coupled with rotation, with the combined existing in quantized energy states. Those states can both absorb and emit.

Comment by Timothy Chase — 6 Aug 2007 @ 10:18 PM

375. Rates of diffusion are illustrated with a nice applet here: http://chem.salve.edu/chemistry/diffusion.asp#model

I vaguely recall CO2 is well mixed after several years’ time (the peaks from seasonal growth in different regions can be tracked as they move around the world– the variations can be followed until they get diffused.

Comment by Hank Roberts — 6 Aug 2007 @ 10:26 PM

376. [[It is my understanding that the Hitran data relate to single gases and not to greenhouse gas mixtures. Have any experiments of the type described by Raypierre been conducted anywhere with such mixtures?]]

In brief, yes. Combustion engineers did a lot of it in the ’40s to understand what happened in various engines and industrial processes. Carl Sagan used their data in his first model of the Venus atmosphere (1960).

Hottel, H. C.; Egbert. R. B. 1941. Trans. Am. Soc. Mech. Engrs. 63, 297.

Comment by Barton Paul Levenson — 7 Aug 2007 @ 7:35 AM

377. [[One other query. CO2 is heavier than air and I understood that it was mixed in the lower atmosphere as a result of convection. Why, then, does it not form a thin layer immediately above the influence of convection?]]

Interesting question. I have no idea what the answer is. Possibly it’s related to the fact that CO2 is such a small fraction of the air by volume.

Comment by Barton Paul Levenson — 7 Aug 2007 @ 7:36 AM

378. Re #364 where Rod wrote “I assume your “increase” is what I posed in my question — the blasting of my CO2 with some IR, and not anything that happens later. Then your answer says NO, the temperature of the gas does not increase. Which implies that the kinetic energy of the vibrations and rotations DO NOT go into determining the temperature, only translation does. Is this your contention?”

The answer to that question is Yes. The vibrational and rotational energy do not go into determining the (Maxwellian) temperature. But the energy of vibrations and rotations are not considered ‘kinetic’ energy. They should be thought of as vibrational energy and rotational energy, or as internal energy. It is true that during a vibration the energy of an atom change from pure potential energy, when the atom reverses direction , and pure kinetic energy but that kinetic energy is not what we are considering. The kinetic energy that is measured is the translational movement of the molecule.

2) If vibration energy transfers to translation (among others) does it then indirectly affect the temperature?

Yes again. The vibrational energy and the translation energy are interchanged in collisions, and as such will reach a stable balance. Thus before your blast of radiation the molecules would have been rotationally and vibrationally excited as the result of collisions, and after the blast they would have returned to a stable state. In the new stable state, the excess excitation would have been shared with the translational energy to raise the temperature of the gas.

3) my understanding is that CO2 picks up rotation only from its vibrational levels. (CO2 does not have a dipole moment so can not absorb into rotation levels directly from IR, but only after vibration absorption makes the molecule a quasi dipole. (I don’t know if after the initial absorption the molecule can now absorb into rotational energy directly or not…) This would imply that rotation can transfer to translation or emission but only after going through vibration first. I think this is what you said; is it?

No, what I have not said is that you actually get mixtures of vibration and rotation. The reason there are so many lines making up a CO2 vibration band is because the energy level depends on the vibration and rotation taken together. So CO2 can absorb vibrotional lines but, unlike water vapour, not pure rotational lines.

Vibrational energy is greater than rotational energy, so if a molecule loses some of its vibrational energy but not all, then it can go from being vibrationally excited to rotationally excited.

4) Why is it that CO2, if properly dense and reasonably heated, can not emit “continuum” blackbody radiation, but water vapor can?

That is a problem that still puzzles scientists. It may be because H2O can form dimers, or it may be that water vapour is condensing into water which does emit BBR, then evaporating again.

5) Is Hank (362) correct? Are you changing your stance that GHG don’t re-emit? Or is it that you contend they can emit but lose their vibration/rotation energy so quickly through transfer via collision or otherwise to translation, that they don’t?

Yes, I can’t make up my mind how the spectrum seen from space is formed. Both for Mars and for the Earth, the CO2 band is at a fixed brightness temperature which is independent of the surface temperature.

But currently I think that at the base of the atmosphere CO2 does not re-emit because it it losing its energy through collisions too quickly, but at the top of he atmosphere the air is too thin and so CO2 does emit from the non-LTE region.

Comment by Alastair McDonald — 7 Aug 2007 @ 9:00 AM

379. re 372 Rod B: This is going way beyond what we need for the current discussion, but there have been traces of the issue in other comments. Translational motion (of the center of mass in the absence of a field) is not quantized. As for the rest, please do some searches on molecular structure & bonding. Look at the bonding or valence electrons. These electrons are contributed by the atoms in the molecule and shared within the molecule. What we think of as vibrations are alternate solutions for the distribution of electrons around the atoms with energies differing by an IR quantum. The notion of mechanical analogues for electronic distributions comes from molecular mechanics (MM), which finds such analogues empirically. MM is the work horse in statistical mechanics, particularly solution properties of drugs. Spectroscopically, the mechanical analogues are useful, and even quantitative, but do not tell the whole story though they cover most of the IR where we are concerned. Look up triplet state and flouescence to see more on molecular spectroscopy.

On the matter of line broadening, note that even though molecules have no net charge or dipole, there are local variations in charge. One O might have a touch more + than the other, say, or there could be a small gradient across an O. These local charge fluctuations exist in all molecules and allow iteractions of the electrons when the atoms are close enough. Look up electric quadropole, and note that there are sextapoles, octapoles, etc, which increasingly come into play as the molecules get closer and closer.

Remember, for RC purposes, there is still a water continuum to work out.

Comment by Allan Ames — 7 Aug 2007 @ 11:02 AM

380. re Tomothy Chase, various. I find Alastair’s thoughts provocative, and would like to understand well enough to agree or disagree.

As for the origin of spectral lines in CO2, the earlier references on non LTE discuss them in detail. I would note the symmetry between absorption and emission.

I have been dismissive of non-equlibrium thermo, and now confess to having had an epiphany, also known as a cessation of stupidity from the following (just read the first few sections): http://home.earthlink.net/~dckennedy/pubs/nonEqThermoRad.pdf

Comment by Allan Ames — 7 Aug 2007 @ 11:04 AM

381. Allan Ames (#380) wrote:

re Tomothy Chase, various. I find Alastair’s thoughts provocative, and would like to understand well enough to agree or disagree.

I would recommend studying studying the science. I myself will be getting into it in more detail in the next few weeks. However, I can already see that our understanding of absorption and reemission both in the labs and the atmosphere is fairly well developed. And we have been taking measurements of longwave emissions from CO2 at mid-troposphere for some time. At this point they are becoming rather detailed.

NASA AIRS Mid-Tropospheric (8km) Carbon Dioxide
July 2003
http://www-airs.jpl.nasa.gov/Products/CarbonDioxide/

I have been dismissive of non-equlibrium thermo, and now confess to having had an epiphany, also known as a cessation of stupidity from the following (just read the first few sections): http://home.earthlink.net/~dckennedy/pubs/nonEqThermoRad.pdf

I have been interest in nonequilibrium for a while for a number of reasons and I will be reading the article with considerable interest. However, I noticed he is dealing with entropy production principles. This shouldn’t be a problem with regions near local thermodynamic equilibria, but if you are dealing with far from equilibrium it becomes more problematic. From what I can tell, investigations in that region are still fairly tentative.

To get a sense of the state of the field, I would recommend:

Classification and discussion of macroscopic entropy production principles
arXiv:cond-mat/0604482v3 [cond-mat.stat-mech] 2 May 2007
Stijn Bruers
http://arxiv.org/PS_cache/cond-mat/pdf/0604/0604482v3.pdf

Comment by Timothy Chase — 7 Aug 2007 @ 12:20 PM

382. Thanks to those who replied to my queries (#363). If I may, I would like to push Timothy Chase (#370) – or anyone else – into providing further information.

As I understand it, wavelength saturation implies almost zero transmission such that almost no infra red photons of the wavelength in question can reach the top of the atmosphere and escape as OLR. (If photons of this wavelength are both absorbed and re-emitted rather than transmuted to another energy form, then I take it that the density of greenhouse gases in a vertical column is sufficient, where saturation occurs, to richochet them up and down to the extent that hardly any make it to the top).

It is suggested that, as atmospheric CO2 increases, wavelengths in the wings also become optically thick and add to the greenhouse effect. I can fully appreciate this argument in a theoretical sense and if one ignores water vapour . However, if the relevant wing areas are already saturated or nearly saturated by a combination of CO2 and water vapour, then, clearly, there will be less upside potential for extra greenhouse gas warming than would have been the case had no water vapour been present.

Timothy Chase correctly points out that atmospheric water vapour distribution is patchy. In the atmosphere over deserts, for example, one might expect that extra CO2 could have its full extra warming potential. Over oceans, however, the potential would not be maximised. Timothy states that water vapour typically remains below 4 km but that there will be CO2 above that level still capable of acting as a second blanket. This is where I get into difficulty. If there is saturation or near saturation from a greenhouse gas mixture in the first 4km of the atmosphere, it seems to me that the CO2 above would be largely unemployed (redundant). I accept that I started the previous sentence with an “if”.

The question that I would like an answer to is this: When climate modellers determined climate sensitivity for a doubling of CO2, did they take into account the possibility of saturation or near saturation by a combination of water vapour and CO2 in the overlapping wavelengths? Alternatively, in producing a formula for radiative forcing by CO2 which is applicable in the range 180-280 ppm, have they addressed the possibility that the formula would have to change as CO2 levels significantly exceed those experienced in the previous one million years and possible saturation by a combination of water vapour and CO2 kicked in?

Comment by Douglas Wise — 7 Aug 2007 @ 1:01 PM

383. re 381 Tomothy Chase. Thank you for the references. I can summarize my epiphany as the notion that both fermions and bosons are part of thermodynamics, but it does not go much deeper than the statement — more a matter of tolerance than understanding. It does allow me to accept that having gas and radiation at different apparent temperatures is a thermodynamically consistent state, though not one described by the Schwartzschild equation as far as I can see.

Comment by Allan Ames — 7 Aug 2007 @ 1:50 PM

384. Douglas Wise — note the heat in the lower atmosphere isn’t going away just because there’s a lot of water vapor, it’s still in the system and is going to keep bouncing around until it gets transferred to the upper atmosphere.

Where there is much less water vapor, at the top of the atmosphere level where the heat can escape, there’s no question of saturation.

Try a pure poetry analogy — bubbles rising from the bottom of a pool full of water move up and out. Say the bottom of the pool fills up with mud — the bubbles rising up will move much slower while rising up through the lower few feet of mud, but once they get above the mud they still rise up to the top.

Comment by Hank Roberts — 7 Aug 2007 @ 2:06 PM

385. re 382 Douglas Wise. Are you familiar with LBL model at the HITRAN site or the transfer calculators? I am not sure but these may help. I hope the GCM modelers will say they would like to reproduce what the LBL models do.

Comment by Allan Ames — 7 Aug 2007 @ 2:17 PM

386. re 384. Hank, I’m probably better at poetry than physics but you either haven’t really helped or Raypierre got this thread off to a false start as regards explaining saturation to laymen. I liked the poetry concerning the red M and M sweets (candies) and the two fat kids who ate the lot (saturation). Your mud analogy contradicts this. The mud merely delays the photons on their journey to space but, nevertheless, they eventually get out (not saturation). Seems to me you can’t both be right. (Raypierre said nothing came out of the end of his female experimenter’s tube.) Am I being too literal with the poetry?

re 385. Thanks Alan, you may well have given me the information I need. It’s getting late in the UK and there’s no good my looking at your links till tomorrow when the whisky loses its influence. My fear is that, even tomorrow, I may not understand what’s in them. However, I’ll try and then, if necessary, come back to you for a translation that a geriatric biologist might be able to grasp.

Comment by Douglas Wise — 7 Aug 2007 @ 3:34 PM

387. Douglas Wise (#382) wrote:

If there is saturation or near saturation from a greenhouse gas mixture in the first 4km of the atmosphere, it seems to me that the CO2 above would be largely unemployed (redundant). I accept that I started the previous sentence with an “if”.

Saturation for a given layer of atmosphere means essentially that the emissivity at a particular wavelength is 1, or alternatively, that it is opaque to that wavelength. All the radiation is absorbed, but it is also re-emitted. However, it is re-emitted isotropically, that is, in all directions.

Eventually, all of the thermal radiation which enters the system will leave the system just as the sunlight which is absorbed at the surface gets reemitted as thermal radiation. However, the big question is how much the temperature must rise before the amount of thermal radiation leaving the system is equal to the amount of thermal radiation which is entering the system?

Infrared radiation which is re-emitted towards the surface is radiation which is not immediately escaping to space. Likewise, radiation which is absorbed by the surface will raise the temperature of the surface, increasing the amount of thermal radiation it emits. The feedback of absorption and reemission between the surface and the atmosphere will continue until the temperature of the surface has increased sufficiently for the amount of thermal radiation leaving the system to equal the amount entering the system.

A good read on this is at:

Learning from a Simple Model
http://www.realclimate.org/index.php/archives/2007/04/learning-from-a-simple-model

However, you might also want to check out the spreadsheet at:

htt://www.editgrid.com/user/timothychase/Greenhouse

In any case, since the radiation gets re-emitted and must pass through the layer of the atmosphere which is extremely dry and where carbon dioxide is dominant, the carbon dioxide can and will absorb the radiation even if it has been absorbed before, re-emitting roughly half of what it absorbs back towards the surface – just as the water vapor below does.

The question that I would like an answer to is this: When climate modellers determined climate sensitivity for a doubling of CO2, did they take into account the possibility of saturation or near saturation by a combination of water vapour and CO2 in the overlapping wavelengths?

With line-by-line analysis, they take this into account throughout the atmosphere, although models typically analyze things in terms of perhaps fifteen layers of atmosphere. However, climate sensitivity has been most accurately determined not by the models, but by paleoclimate analysis over the past 400,000 years. It would appear to be roughly 2.8 degrees Celsius.

Comment by Timothy Chase — 7 Aug 2007 @ 3:45 PM

388. PS to 387

To play around with the spreadsheet, you can do “File->Export As->Excel” and change the figures once you have the spreadsheet on your computer. Like the “simple model” itself, it is extremely oversimplified, but at least it should give you some sense for how the feedback process underlying the greenhouse effect works.

Comment by Timothy Chase — 7 Aug 2007 @ 4:00 PM

389. Alastair, good post (378). One thing still haunts me. I will assert (and might be wrong…) that vibrational energy is kinetic. It consists of particles of mass with velocity. The fact that it oscillates should not make any difference (other than affecting the average velocity. Translation oscillates of a sort, too (hits another molecule head on, comes to a stop, shoots out in another direction). Also, it is vibration that is the primary generator of temperature in liquids and solids. What you say sounds sensible; but I just can’t resolve this (for one) quandary.

Allan (379), maybe it’s just me (so sorry if I’m slogging down the thread), but I don’t think the thrust of this thread, which I’m (very slooooowly) working toward, can be comprehended without a pretty solid and detailed knowledge of inter- and intramolecular energy absorption, emission, storage, transfer. And it’s not totally clear to me, at least.

I did not follow the line about electron distribution being the alternate description of vibration energy. The vibration is of the atoms oscillating with another atom. Is it the electrons moving with the nucleus )or not???) that your referring to? Neither did I follow “just a touch more + charge”; O can get one more proton (then it’s no longer O) or lose one electron to ionization, e.g. Is the latter a “touch more +”?

A quick glance at your reference (380) looked promising. Thanks.

Comment by Rod B — 7 Aug 2007 @ 6:14 PM

390. Re #389

Rod, can I put it this way “When we talk about the kinetic energy of a molecule, we are referring to the the translational energy of the whole molecule, not any kinetic energy of its parts. We tend to use the terms ‘kinetic energy’ and ‘translational energy’ interchangeably.

But free molecules with translational energy only exist as gases, so the heat energy of liquids and solids is not translational energy. It is in the form of molecular and atomic vibrations.

It is important to note that the average kinetic energy used here is limited to the translational kinetic energy of the molecules. That is, they are treated as point masses and no account is made of internal degrees of freedom such as molecular rotation and vibration.

HTH,

Cheers, Alastair.

Comment by Alastair McDonald — 8 Aug 2007 @ 2:46 AM

391. Re #382 where Douglas asked:

When climate modellers determined climate sensitivity for a doubling of CO2, did they take into account the possibility of saturation or near saturation by a combination of water vapour and CO2 in the overlapping wavelengths?

The climate modellers are denying that saturation occurs. See Raypierre’s essay that heads this thread. They then claim that the increase in CO2 will cause a warming which will be amplified by the subsequent increase in water vapour. They do not know the amplification, so climate sensitivity is unknown. However they guess that for doubling CO2 temperatures will rise by 3 C.

FWIIW, I am saying that CO2 is saturated, but increasing CO2 will still increase global temperatures. That is because the saturated absorption happens closer to the ground and so it melts the ice at high altitudes and latitudes. The resulting change in albedo means that the surface absorbs more solar heat, and so there is a large rise in temperature, much more than the 3 C they predict.

Comment by Alastair McDonald — 8 Aug 2007 @ 7:47 AM

392. re 389 Rod B: In the overall discussion, it is a small point that molecules are not mechanical systems, though they sometimes act the same way. Check out “quantum oscillator” and “coupled quantum oscillator” and the “correspondence principle”. It is worthy to discuss mostly to note that interactions between molecular energy state can occur in many ways, with a variety of distance dependences. We are fortunate that these can be reasonably well described with Lorentzian average curves with pressure or concentration dependent widths, but we should keep in mind that the Lorentzian is an approximation, and each interaction will have its own statistics.

Comment by Allan Ames — 8 Aug 2007 @ 8:49 AM

393. re 391 Alastair: If one could run a GCM with LBL calculation in, say, 1KM intervals near the surface, with NLTE inclusion and consistent handling of energy exchanges, as would this accomplish what you wish? (For me this would be a good start, but just a start.)

Comment by Allan Ames — 8 Aug 2007 @ 8:58 AM

394. Alastair McDonald (#391) wrote:

The climate modellers are denying that saturation occurs. See Raypierre’s essay that heads this thread.

They are claiming that saturation occurs – at particular wavelengthes, but that there is both line broadening and band broadening as the result of higher temperatures and pressures – effects that are easily measured in the laboratories.

They then claim that the increase in CO2 will cause a warming which will be amplified by the subsequent increase in water vapour.

We know that it will be amplified by water vapor in large part due to the Clausius-Clapeyron relation. However, other factors are involved, in part due to amplification as the result of the albedo flip (melting ice will expose more dark soil and ocean which will increase the absorption of sunlight and thus thermal energy within the system), the distribution of the continents, etc..

They do not know the amplification, so climate sensitivity is unknown. However they guess that for doubling CO2 temperatures will rise by 3 C.

We know that the climate sensitivity given the current distribution of the continents is roughly 2.8 degrees as the result of of the paleoclimate record for the past 400,000 years.

FWIIW, I am saying that CO2 is saturated, but increasing CO2 will still increase global temperatures.

We know that CO2 is not saturated at most of the relevant wavelengths at higher altitudes as the result of laboratory experiments at different pressures and temperatures, the laws of quantum mechanics, and satellite imaging of radiation being reemitted at different altitudes which has the signature of carbon dioxide. The last of the involves our ability to measure emissions at more than 2000 different wavelengths.

For more info, including videos based on the data we are getting from the satellite measurements, please see:

AIRS – Multimedia: Videos: Animations
http://airs.jpl.nasa.gov/Multimedia/VideosAnimations/

Incidently AIRS stands for atmospheric infrared sounder – the name of the satellite which is getting the data which makes possible the imaging. In essence, CAT scans for the atmosphere, rendering visible that which is invisible to our eyes – given the limitations of human vision.

That is because the saturated absorption happens closer to the ground and so it melts the ice at high altitudes and latitudes.

If its saturated at groundlevel, then this is precisely where CO2 will nott be effective. Moreover, absorption by water vapor at the same wavelengths will completely swap the effects of carbon dioxide. Pesky lab experiments again.

RealClimate » A Saturated Gassy Argument
http://www.realclimate.org/index.php/archives/2007/06/a-saturated-gassy-argument/

The resulting change in albedo means that the surface absorbs more solar heat, and so there is a large rise in temperature, much more than the 3 C they predict.

… and the paleoclimate record says otherwise. 400,000 years of it.

Alastair, you could do much better than this.

You could learn the science rather than fantasy role-playing the unappreciated super genius. Think about it: something genuinely productive, maybe even helping us out. But what you just did is counterproductive.

Comment by Timothy Chase — 8 Aug 2007 @ 9:21 AM

395. Alastair writes:
> I can’t make up my mind how the spectrum seen from space is formed. Both for Mars and for the Earth,
> the CO2 band is at a fixed brightness temperature which is independent of the surface temperature.

Alastair, aren’t you here contradicting the basic idea of the whole process, that when CO2 increases, Earth warms up, and over several centuries Earth comes back to radiative equilibrium eventually by increasing the outgoing energy from the top of the atmosphere?

Or did you mean something different?

Comment by Hank Roberts — 8 Aug 2007 @ 10:39 AM

396. Timothy, a quibble, maybe. Do (can?) models do a line by line analysis?? The lines must number in the tens of thousands, all with different absorption constants that vary by molecule and by temperature. Then do 15 atmospheric layers, then do it in 100km by side sections, then project it with all other effects and forcings by some time interval through 50-100 years… and where most integrations have (ought?) to be done numerically. I’m no model nor supercomputer expert, but that sounds like a humongous number of flop calculations.

Comment by Rod B — 8 Aug 2007 @ 11:21 AM

397. Alastair, I’m sorry for being a tiresome tenacious guy, and I know I’m being picky. Everything gets close but still kinda dances around the question. You say (390),…. “When we talk about the kinetic energy of a molecule, we are referring to the translational energy of the whole molecule, not any kinetic energy of its parts. We tend to use the terms ‘kinetic energy’ and ‘translational energy’ interchangeably….” (emphasis mine). This sounds like it could be for convenience of analyzing, e.g. It comes close but it does not quite say, “Only a gas’ translation kinetic energy determines temperature. (Period) With a gas we use [not “tend to”] the terms kinetic energy and translation interchangeably because they are identical.

Even your source quote (which I have seen), … “It is important to note that the average kinetic energy used here is limited to the translational kinetic energy of the molecules. That is, they are treated as point masses and no account is made of internal degrees of freedom such as molecular rotation and vibration.” precisely says (implies) “for this analysis we will only look at the translation part of kinetic energy…” It does not, again, state definitively that translation energy in a gas is the only kinetic energy that determines the temperature.

I’m beginning to be convinced that the above is in fact true. I haven’t done the math yet, but playing with 3/2kT (translation only) going to 5/2kT to 3kT to 7/2kT (added vibration/rotation energies), it’s not obvious how adding energy to virotation can increase the T. (Though it gets messy when one changes the n of nkT and also changes the masses and average velocities of the equivalent 1/2mv2). And maybe the temperature of solids ands liquids coming only from vibration I can write off as just another mystery of physics (really, not a smart-a comment). Given this, then absorbed radiation can increase the temperature of a gas only after it gets transferred from virotation to translation, either in the same molecule or another via a collision. Secondly, emission decreases the temperature of a gas only if the source energy of virotation was just transferred from some molecule’s translation. Radiation being absorbed and immediately re-emitted (with whatever probability that happens) without any intervening energy transfer does not affect the temperature of a gas what-so-ever. Is this correct?

Comment by Rod B — 8 Aug 2007 @ 12:18 PM

398. Rod B (#396) wrote:

Timothy, a quibble, maybe. Do (can?) models do a line by line analysis?? The lines must number in the tens of thousands, all with different absorption constants that vary by molecule and by temperature. Then do 15 atmospheric layers, then do it in 100km by side sections, then project it with all other effects and forcings by some time interval through 50-100 years… and where most integrations have (ought?) to be done numerically. I’m no model nor supercomputer expert, but that sounds like a humongous number of flop calculations.

They were doing a trillion bits per second a few years back (2002) with the NEC Earth Simulator. I don’t know what they are doing now – but you know how computers have been advancing. They keep improving the resolution, and where they find that it makes a significant difference compared to previous calculations they improve it some more. Likewise, the algorithms are streamlined, made as efficient as they can make them – so that every flop counts. But yes, its a lot of calculations – and I believe it takes weeks to do a single simulation, although someone will hopefully correct me if I am wrong about that.

Comment by Timothy Chase — 8 Aug 2007 @ 12:24 PM

399. Timothy (et al) says, “We know that the climate sensitivity given the current distribution of the continents is roughly 2.8 degrees as the result of of the paleoclimate record for the past 400,000 years.” I understand this but, for what it’s worth, it’s one of the thing that makes me vaguely nervous and enhances skepticism. Sounds like measuring with an eyeball glance, marking with a paintbrush, and cutting with a diamond cutter.

Comment by Rod B — 8 Aug 2007 @ 12:30 PM

400. PS to 398

If I remember correctly the temporal resolution is at the level of a complete calculation for every simulated minute of every simulated day. The one exception to this is in the polar regions where due to wind speeds a higher temporal resolution is required.

Comment by Timothy Chase — 8 Aug 2007 @ 12:30 PM

401. Correction to 398 regarding the NEC Earth Simulator

32 teraflops – where a teraflop is a trillion floating-point calculations per second – and that was back in 1998. Ancient history as far as computers are concerned – given the doubling of chip speed every six months.

***** For Immediate Use January 21, 1998

Development of the World’s Fastest Ultra Computer for Resolving Environmental Problems on the Earth
~ NEC Awarded a Contract for Basic Design of Ultra-fast Parallel Processing Computer for Earth Simulator ~

NEC announced today to develop the world fastest ultra computer with the maximum performance of over 32 tera FLOPS (Note 1) as a part of the Earth Simulator Program promoted by Science and Technology Agency in Japan….

http://www.nec.co.jp/press/en/9801/2101.html

Comment by Timothy Chase — 8 Aug 2007 @ 12:38 PM

402. Rod, “temperature” is not a Platonic ideal object.

It’s a choice from a list of definitions.

You need to pick one. That means pointing to which dictionary you are using. Once people know what meaning you are choosing for the word, it’s possible to try to talk about the physics.

For most definitions of “temperature” it’s an average.

Note also that most definitions of “velocity” are: direction and speed.

Comment by Hank Roberts — 8 Aug 2007 @ 1:39 PM

403. The speed of computer instructions is not linearly related to finishing an application. As the speeds have increased (and the supercomputer speed has not come anywhere near doubling every six months since 1998) the number of OPS or FLOPS to complete the same process has also gone up significantly. For instance much of the speed is garnered through multiprocessing and the instructions just to coordinate and communicate within the multiprocessor goes up exponentially.

But, you might be right — maybe they can run a detailed comprehensive model in a few weeks. Though “streamlining algorithms” could also include taking some gross shotcut estimations — ‘let’s not numerically integrate every line; let’s just assume the answer is… 4.7…. [;-}

Comment by Rod B — 8 Aug 2007 @ 3:11 PM

404. Hank, when you say doubling CO2 will heat the earth by 2.4 degrees, which of the many alternate temperatures are you referring to??

Comment by Rod B — 8 Aug 2007 @ 3:20 PM

405. I wouldn’t say something like that. It’d be uninformative at best.

I’d point to a source in a science site that would include a definition of the term used. Same as I’m suggesting you do.

Comment by Hank Roberts — 8 Aug 2007 @ 3:52 PM

406. Rod, you know by now that you’re banging on a question that’s written about extensively, and the answer depends on how you set up the question.

Forty years ago:
http://prola.aps.org/abstract/PR/v150/i1/p153_1

Utterly spooky, talking about condensation, from the same issue:

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TVG-4JCBJS4-C&_user=10&_coverDate=09%2F15%2F2006&_rdoc=40&_fmt=summary&_orig=browse&_srch=doc-info(%23toc%235534%232006%23996309997%23627528%23FLA%23display%23Volume)&_cdi=5534&_sort=d&_docanchor=&view=c&_ct=61&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=c3e9f252ab9d8e19b6d3afe057b39718

Comment by Hank Roberts — 8 Aug 2007 @ 4:47 PM

407. Rod B (#404) wrote:

The speed of computer instructions is not linearly related to finishing an application. As the speeds have increased (and the supercomputer speed has not come anywhere near doubling every six months since 1998)…

They tend to use just about the best chips on the market at the time that they are constructed, but they aren’t constructed every six months. It would be interesting to find out what the most recent machine is and its specs. Of course, I can’t wait for the qubits, but that kind of tech still looks like it might take a while.

… the number of OPS or FLOPS to complete the same process has also gone up significantly. For instance much of the speed is garnered through multiprocessing and the instructions just to coordinate and communicate within the multiprocessor goes up exponentially.

Wild guess, but I would expect something logarithmic rather than exponential. We are talking about addresses where the length of the address would go up logarithmically with the number of addresses. Probably 2 to the power of something for the number of processors, that sort of thing. And oddly enough, Google has discovered that they can get away with doing what they do by stringing together a bunch of PCs.

But, you might be right — maybe they can run a detailed comprehensive model in a few weeks. Though “streamlining algorithms” could also include taking some gross shotcut estimations — ‘let’s not numerically integrate every line; let’s just assume the answer is… 4.7…. [;-}

True, but we have every reason to expect them to compare algorithms to test for accuracy and the effects upon the results from a given scenario – much like testing a new supercomputer with the calculation of pi out to so many decimal places.

In any case, these guys seem pretty detail oriented – if you check Gavin’s paper on the latest model.

Comment by Timothy Chase — 8 Aug 2007 @ 4:51 PM

408. Thanks to those who attempted to address my concerns expressed in posts #382 and #386. I will need time to study the links I have been given (and, possibly, professional help).

I was reassured to read the statement from Timothy Chase (#387)to the effect that modellers take account of the absorptive effects of water vapour/CO2 mixtures by using line by line analysis in up to 15 layers of atmosphere although the statement was subsequently queried by Rod B (#396).

However, I was confused by another of Timothy’s statements: He stated that “saturation from a given layer of atmosphere means that emissivity at a particular wavelength is 1, or alternatively, that it is opaque to that wavelength.” This seems to correspond to Raypierre’s definition: “Saturation refers to the condition where increasing the amount of CO2 fails to increase the absorption, because the CO2 was already absorbing essentially all there is to absorb at the wavelengths where it absorbs at all.” However, Timothy goes on to state that all the radiation absorbed will be re-emitted isotropically, that is, in all directions, thus explaining why higher level CO2 will not be irrelevant. He now appears to be singing from the same hymn sheet as Hank Roberts (#384) with his mud analogy.

Raypierre’s original essay appears to contradict this. He states that if you shine infrared of a particular wavelength up from the surface and there is sufficient CO2 and/or water vapour to saturate it by the time it has reached 4km, “essentially” none will get through. This must mean that higher level CO2 will be redundant with respect to that wavelength.

Am I being stupid or are Hank and Timothy contradicting Raypierre? Surely opaque is opaque is opaque? Alternatively, why describe a given wavelength as saturated if infrared of the same wavelength is re-emitted (albeit in haphazard directions). If the latter occurs, some of it will be up and thus higher level CO2 will obviously not be redundant at that wavelength. I am beginning to think that the answer may lie in Raypierre’s use of the word “essentially”, possibly implying that a tiny amount might get through the saturated layer. If this is the case, then higher level CO2 would be expected to have a tiny effect.

Please note that I am not disagreeing that increasing CO2 will cause more absorption “in the wings” thus reducing the bandwidth of infrared that can escape (with consequent global warming). Am I wrong to be thinking “horizontally” rather than “vertically”?

Comment by Douglas Wise — 8 Aug 2007 @ 5:15 PM

409. > Am I being stupid or are Hank and Timothy contradicting Raypierre?
Neither, I think. Remember that the wide range of sunlight once it reaches the ground becomes heat. Heat keeps the lower atmosphere moving. Eventually that motion is going to carry the warmth above the lower atmosphere, to the point where a photon can get away into the stratosphere.

Perhaps it’s time to urge Raypierre to help us out?

Comment by Hank Roberts — 8 Aug 2007 @ 6:04 PM

410. Douglas Wise (#408) wrote:

However, I was confused by another of Timothy’s statements: He stated that “saturation from a given layer of atmosphere means that emissivity at a particular wavelength is 1, or alternatively, that it is opaque to that wavelength.” This seems to correspond to Raypierre’s definition: “Saturation refers to the condition where increasing the amount of CO2 fails to increase the absorption, because the CO2 was already absorbing essentially all there is to absorb at the wavelengths where it absorbs at all.” However, Timothy goes on to state that all the radiation absorbed will be re-emitted isotropically, that is, in all directions, thus explaining why higher level CO2 will not be irrelevant. He now appears to be singing from the same hymn sheet as Hank Roberts (#384) with his mud analogy.

I will give it a shot, although if Raypierre wishes to step in and correct me, no problem…

Thing of it is, we are dealing with several issues at the same time.

Now what Raypierre is dealing with primarily is why adding carbon dioxide at a given layer does not result in saturation. While the peak becomes saturated, the wings are not saturated due to spreading where the spreading is both line spreading and band spreading due to pressure and temperature. Its not all at the same wavelength – which is why we get the logarithmic relationship between the partial pressure and temperature. That is, for every doubling the temperature goes up so many degrees. This argument holds whether you are dealing with water vapor or carbon dioxide.

However, it does not work for carbon dioxide at the same layer as were water vapor is dominant, because for all relevant wavelengths water vapor dominates and swamps out the effects of any additional carbon dioxide since carbon dioxide exists in much smaller quantities.

Now this is the point at which the absorption and reemission comes in.

The lower atmosphere is already opaque to any additional carbon dioxide at all relevant frequencies – unless you are talking about it being brought up to about the same level percentage wise as water vapor. Not bloody likely.

However, the longwave radiation which is absorbed and re-emitted at the layer where water vapor is dominant gets re-emitted and reaches the layer which is especially dry and where carbon dioxide is dominant. This layer is only partially opaque to longwave radiation – principally at the peak. Doubling the partial pressure of carbon dioxide at this level will increase the amount of longwave radiation which – rather than escaping to space heads in the direction of the surface. Some of this will be absorbed, and the direct effect will be (due to the greenhouse feedback between the atmosphere and the surface which I explained earlier) will be to raise the temperature of the surface by 1.2 degrees Celsius.

However, this increases the partial pressure of water vapor – another feedback. Then there are some additional feedbacks involving the albedo effect, the expansion of the atmosphere, etc., so that after all of the feedbacks the net result once the entire system reestablishes an equilibrium will that a doubling of the partial pressure of carbon dioxide implies an increase in temperature of 2.8 degrees.

So raising the partial pressure of carbon dioxide won’t increase the greenhouse effect if it is added at the same layer where water vapor is dominant (the lower troposphere) because that part of the atmosphere is already opaque – being dominated by water vapor. However, raising the partial pressure of carbon dioxide in the upper troposphere will increase the greenhouse effect with regard to a particular wavelength so long as that wavelength is not saturated in that upper layer. That was Hank’s point – as he was responding to your query regarding a specific wavelength.

But even if that particular wavelength (the peak) is saturated in the dry upper troposphere, the wings won’t be. That is one of Raypierre’s points. But then another one of Raypierre’s points is that the atmosphere expands due to the warming of the atmosphere which diminishes the density and degree of saturation of the uppermost layers where carbon dioxide plays an active role in the greenhouse effect. However, it takes a while for the uppermost layers to warm up after the adiabatic expansion and consequent cooling of those layers – where the process of warming up takes place by means of the relatively inefficient process of thermal diffusion.

These layers will have to warm up to the point at which they can more effectively absorb (due spreading) and reemit longwave radiation before they can fully participate in the greenhouse effect. This is the main reason why there is such a lag time before the climate system is able to achieve equilibrium again. And this I believe was another one of Raypierre’s points – although it came up in the comments, I believe.

I hope this helped!

Comment by Timothy Chase — 8 Aug 2007 @ 6:56 PM

411. Timothy (407), certainly not log related, though maybe 2N as opposed to exponential. Address length has little to do with it; coordinating, sync-ing, timing, and controlling many processors, each running a tidbit of the appl, does. Google’s appl, as complex as it is, is not the same as modeling.

but we have every reason to expect them to compare algorithms to test for accuracy…” You can’t really test for accuracy without running the algorithm. We inherently know the value of pi to a jillion places, and it will never differ. We don’t “know” (at least with similar precision) what to check our shortcut against in global climate modeling.

Comment by Rod B — 8 Aug 2007 @ 11:05 PM

412. A perhaps useful quote I’ve borrowed below — this is taken from one of the nonfiction columns in the online magazine Jim Baen’s Universe may help. I’ll claim fair use for this bit; attribution and information about the author follows. Full text by subscription at http://baens-universe.com/articles/The_Perpetual_Electron

People keep trying to describe what’s happening as Newtonian classical physics.

——— begin excerpt ———-

“… trying to imply and/or apply a classical idea of a ball oriented about an axis spinning and there being motion involved really has nothing to do [with] a quantum particle. Classical reality is completely meaningless and different and does not apply.” …
“Saying that electrons are fundamental particles is not sufficient to lay their behavior to rest. They have properties, and all properties are caused by something. There is no level below which we can stop searching for causes. We may not be able to find the causes, but they are there, waiting for us. Waiting for the day when we improve the power or subtlety of our equipment or the power or subtlety of our intellectual models.”
…” it is very likely that an electron is simply a manifestation of an actual piece of spacetime that cannot be reduced to anything else unless it is converted to a photon which would be another manifestation of spacetime.” … “we know diddly about the universe and will continue to peel back the layers …. Electrons are just manifestations of spacetime that emit certain unusual properties that when described via macroscopic ideas such as spinning balls confuses the living hell out of most people.” …“We must really and truly put our minds outside the classical ideas box and start thinking along the path of the weirdness that is quantum physics.”

—————-end quote; continuing excerpt identifying the author quoted ———————
— Doctor Travis S. Taylor is a scientist and engineer, as well as an author of fiction and non-fiction. His primary occupation is as a contractor working on projects for NASA, the Department of Defense and the intelligence community.
His current projects include advanced propulsion systems, ideas to help win the global war on terrorism, engineering future combat systems and helping to design the crew exploration vehicle for NASA’s return to the moon. He has a doctorate in optical science and engineering, a bachelor’s in electrical engineering, and three master’s degrees: one in astronomy, one in physics and one in aerospace engineering. He’s known as Doc Travis to readers of his hard science fiction novels. ….

Comment by Hank Roberts — 8 Aug 2007 @ 11:20 PM

413. Hank, written about extensively, but not with exactness. Maybe some of these papers use shorthand that is recognized by their professional circle, but it’s not by me. For instance, one of your sources (BTW, I couldn’t access two of them) said something like ‘temperature from kinetic energy being equivalent to the temp from the “total” energy; Is “total” vibration, rotation, electron, chemical and nuclear? (I’m just using this as a typical example; they might have well defined it in the whole specific article.) I’ve seen credible sources that said rotational and vibration energy directly affects temperature; and equally credible sources that said the opposite. We’ve had this in this thread! And I’m wanting to avoid the obscure and mysterious. Shirley you don’t think that if only I ask the question in a certain way that we are not experiencing global warming, aka temperature increases….

Comment by Rod B — 8 Aug 2007 @ 11:31 PM

414. [[Raypierre’s original essay appears to contradict this. He states that if you shine infrared of a particular wavelength up from the surface and there is sufficient CO2 and/or water vapour to saturate it by the time it has reached 4km, “essentially” none will get through. This must mean that higher level CO2 will be redundant with respect to that wavelength.

Am I being stupid or are Hank and Timothy contradicting Raypierre? Surely opaque is opaque is opaque?]]

The radiation from the ground doesn’t get through, but the greenhouse gases that have absorbed it radiate, and the IR from them goes up higher, and so on, and so on. All the layers affect all the other layers, even if not directly. (Though again, as I have been saying, there is IR that gets through the lowest layer. There is IR from the surface that goes all the way to space, though not a great amount of it — about 40 watts per square meter.)

Comment by Barton Paul Levenson — 9 Aug 2007 @ 7:05 AM

415. Re #397 Where Rod B Says:

‘ … absorbed radiation can increase the temperature of a gas only after it gets transferred from virotation to translation, either in the same molecule or another via a collision. Secondly, emission decreases the temperature of a gas only if the source energy of virotation was just transferred from some molecule’s translation. Radiation being absorbed and immediately re-emitted (with whatever probability that happens) without any intervening energy transfer does not affect the temperature of a gas what-so-ever. Is this correct?’

Yes! It is correct.

However, I think it may help if I explain the difference between heat and temperature. Heat is energy. For solids it is the energy of atomic vibrations. These are three dimensional, causing the contents of the bulb of a mercury thermometer to expand and increase the height of the column that is used to register temperature. But that temperature is what a mercury thermometer indicates. Let me call that the ‘Real’ temperature.

For all solids, the atomic vibrations of the positively charged nucleii and the negative electron clouds generates a continuum radiation which increases in intensity and average frequency with temperature. This radiation obeys the Stefan-Boltzmann Law http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html , and can also be used to measure the temperature. This temperature is called the Planckian temperature since it is based on Planck’s function for blackbody radiatiors. If the solid is not an ideal blackbody, then its Planckian temperature and its ‘Real’ temperature may not be the same. NB The blackbody radiation is sometimes called thermal radiation since it depends on temperature. But the temperature measured that way is called the blackbody or Planckian temperature, not the thermal temperature.

If you place a monatomic gas such as Helium in a heated container, then the gas molecules (atoms) will collide with the walls of the container and share their kinetic energy with the atomic vibrations of the container. Eventually thermodynamic equilibrium will be reached where a mercury thermometer will register the same temperature whether placed in the gas or on the container. This temperature is called the kinetic or Maxwellian temperature since it is due to the kinetic energy of the molecules, and corresponds to what I called the Real temperature of solids. At room temperature helium cannot emit vibrational or rotational photons since it consists of neutral atoms. Diatomic homogeneous molecules such as N2 and O2 have a similar problem with emitting photons since they too are neutral. Those gases have a zero Planckian temperature since they do not radiate.

The diatomic heterogeneous molecules such as carbon monoxide do absorb and emit by stretching.

Triatomic molecules, both homogeneous O3 and heterogeneous H20 & CO2, can absorb and emit vibrational energy

Polar triatomic molecules such as H2O can also absorb rotational energy. So even at this simple level it gets quite complicated.

The separate energies contained in vibrations and rotations cannot be measured directly with a thermometer, but the radiation emitted can, which leads to a Planckian or brightness temperatures for each line. A gas is then assumed to have four temperatures, due to its four types of energy: translational, electronic, vibrational, and rotational.

According to the Equipartition theorem http://en.wikipedia.org/wiki/Equipartition_theorem (ET) these four temperature should correspond, but the ET is a classical theory and does not apply to the quantum mechanics of low temperature gases such as the greenhouse gases on Earth.

That is where Raypierre and I fall out.

Comment by Alastair McDonald — 9 Aug 2007 @ 5:41 PM

416. Thanks, Hank (#409), Timothy (#410) and Barton (#414) for answering my queries in post #408. As you are all more or less singing from the same hymn sheet, I deduce that you must be right (or all instructed by a bad teacher!). In other words, ground effluent photons of wavelengths that match the absorption bands of greenhouse gases can be blocked at a low level of the atmosphere if there are enough of those gases present. This is defined as saturation. However, at the top of this low level there is re-emission of apparently the eqivalent number of photons of the same wavelength, carrying the same amount of energy, albeit directed in random directions.

If this is correct, why confuse a simple layman like myself by talking about saturation which implies that the photons of a given wavelength have had all their energy taken from them and been effectively finished off for good. Surely, scattering would be a more appropriate word to use.

As soon as one accepts that photons are just pinging about like pinballs in areas of so-called saturation but that , eventually, they all get out, one can readily understand why high altitude CO2 can be thought of as a second blanket.

However, if this is all there is to it and, if greenhouse gases are at a constant level such that one has eqilibrium, one would expect infrared to be emerging from the top of a saturated column of atmosphere at the same rate that it is going in. Under such circumstances, one would measure 100% transmission and zero saturation. This, therefore, must be wrong.

Perhaps the photons don’t get re-emitted in quite the way I envisage. Perhaps most end up hitting the surface of the earth, changing wavelength and trying an alternative escape route.

Hank mentioned convection. Perhaps there may be an explanation here but I can’t quite figure what it is.

It is clear that I have no understanding of physics and I appreciate that it must therefore be hard to explain to me what is really going on. It was Raypierre’s tube which blocked egress of infrared that doesn’t seem to tie in with your explanations. It seems you are all suggesting that there should be re-emission from the top of the tube such that, if I waited for an equilibrium situation, egress should match input.

Help!

Comment by Douglas Wise — 10 Aug 2007 @ 1:08 PM

417. Alastair McDonald (#415) wrote:

According to the Equipartition theorem http://en.wikipedia.org/wiki/Equipartition_theorem (ET) these four temperature should correspond, but the ET is a classical theory and does not apply to the quantum mechanics of low temperature gases such as the greenhouse gases on Earth.

That is where Raypierre and I fall out.

Actually I believe I was the only one who argued for the equality of the different temperatures on the basis of the equipartition theorem – although I knew that it was an approximation in partial- and non- local thermodynamic equilibria. (Actually, I don’t believe Rayspierre has ever really mentioned equipartition on RealClimate – or for that matter distinguished between the different kinds of temperatures.) However, I didn’t know that the equipartition theorem broke down quite so badly at the quantum level, at least with respect to specific heat. Thank you for bringing this to my attention.

Anyway, outside of rocket nozzles, cryogenics and lasers, there doesn’t seem to be much research devoted to rotational temperatures, or in the case of carbon dioxide, rovibrational temperatures. The literature will distinguish between different vibrational temperatures for carbon dioxide above 50 km, but as of yet I don’t see anything related to other temperatures as the result of non-vibrational or rotational-vibrational molecular states of greenhouse gases.

Perhaps this is something we could both look for in the time ahead. A bit esoteric, but since it seems to interest both of us, I think it would be worth digging into. Likewise we could both dig into partial- and non-local thermodynamic equilibria – as this is another interest we would seem to share.

Comment by Timothy Chase — 10 Aug 2007 @ 1:35 PM

418. Douglas Wise (#416) wrote:

However, at the top of this low level there is re-emission of apparently the eqivalent number of photons of the same wavelength, carrying the same amount of energy, albeit directed in random directions.

If this is correct, why confuse a simple layman like myself by talking about saturation which implies that the photons of a given wavelength have had all their energy taken from them and been effectively finished off for good. Surely, scattering would be a more appropriate word to use.

As soon as one accepts that photons are just pinging about like pinballs in areas of so-called saturation but that , eventually, they all get out, one can readily understand why high altitude CO2 can be thought of as a second blanket.

Well, it really is a different process from scattering. I myself wondered the same thing a few weeks ago, though. But for example, the temperature of the radiation absorbed and re-emitted tends to be the same as the temperature of the atmosphere itself so long as the atmosphere is (roughly) in local thermodynamic equilibrium – which for the vibrational states of carbon dioxide is from 50 km to below.

This is because energy tends to be equally distributed between the translational and vibrational for greenhouse gases. This is due to the fact that collisions are happening at a rate that is roughly two orders of magnitude or more above the rate of relaxation for the molecular states of excitation. And because it is actually absorption and re-emission, it is inelastic.

Energy is exchanged and a thermal equalibrium tends to be achieved between the thermal radiation and the atmosphere. The inelasticity of the interaction is likewise important in terms of the reradiation being isotropic, and the fact that the radiation is emitted becomes important once we get to moist air convection.

Additionally, as others have pointed out, the photon which is absorbed is actually finished off. When a photon is emitted, it will generally be after a large number of collisions, and will probably be emitted by a different molecule altogether.

In fact, the reemission is reemission of radiation, not of the same photon, and it is reemission by the gas, not by a given molecule since the molecule which emits is different from the molecule which absorbs. Alastair helped clarify this issue not too long ago.

As soon as one accepts that photons are just pinging about like pinballs in areas of so-called saturation but that , eventually, they all get out, one can readily understand why high altitude CO2 can be thought of as a second blanket.

The pinging about like pinballs is a great analogy. In this, the carbon dioxide acts like bumpers. The more bumpers, the longer pinballs will remain in play, and assuming they enter at a constant rate, the more pinballs which will be in play assuming one has a larger number of bumpers. It is a good analogy, but as suggested above, it has its limitations.

However, if this is all there is to it and, if greenhouse gases are at a constant level such that one has eqilibrium, one would expect infrared to be emerging from the top of a saturated column of atmosphere at the same rate that it is going in. Under such circumstances, one would measure 100% transmission and zero saturation. This, therefore, must be wrong.

Well, there are a couple of problems here actually.

The thermal radiation is entering at the bottom of the column, not the top. Solar radiation comes in through the top, but if it is scattered either at the clouds or the surface rather than being absorbed at the surface, then it isn’t part of the climate system’s thermal radiation. To become thermal radiation which is part of the climate system, it has to be absorbed by the surface and radiated as thermal radiation.

Secondly, 100% transmission would imply that no radiation is being reradiated towards the ground – but approximately half of the reradiated thermal radiation is downwelling, half upwelling. And downwelling radiation gets the chance to be absorbed by the surface, raising the temperature of the surface before being reradiated. This is what leads to the feedback.

Hank mentioned convection. Perhaps there may be an explanation here but I can’t quite figure what it is.

Well, convection is a major part of the equation. At nearly all levels of the atmosphere, carbon dioxide and water vapor have a local direct cooling effect. They raise the temperature, but they raise the temperature primarily at the surface. The lower atmosphere is warmed principally by moist air convection and the upper atmosphere is warmed primarily by the slow process of the diffusion of thermal energy.

The one major exception to this is the case of ozone – which is able to absorb radiation in the ultraviolet part of the spectrum. Since it tends to absorb most of its energy from solar radiation, in the lower part of the atmosphere where it is active, it tends to have a net warming effect.

It is clear that I have no understanding of physics and I appreciate that it must therefore be hard to explain to me what is really going on.

I knew less a couple of months ago – so you are not doing that bad…

It was Raypierre’s tube which blocked egress of infrared that doesn’t seem to tie in with your explanations. It seems you are all suggesting that there should be re-emission from the top of the tube such that, if I waited for an equilibrium situation, egress should match input.

The thermal energy entering at the bottom and the thermal energy leaving at the top will match – once the new equilibrium is established. The rate at which energy enters and the rate at which energy leaves will be equal at equilibrium. But these rates become equal only because the temperature at the surface is high enough that it is able to radiate more radiation and thus compensate for the increased opacity of the atmosphere.

Comment by Timothy Chase — 10 Aug 2007 @ 2:50 PM

419. I found this passage recently in a piece from the Energy Information Administration of the US Department of Energy

http://www.eia.doe.gov/cneaf/alternate/page/environment/appd_a.html

“What happens after the GHG molecules absorb infrared radiation? The hot molecules release their energy, usually at lower energy (longer wavelength) radiation than the energy previously absorbed. The molecules cannot absorb energy emitted by other molecules of their own kind. Methane molecules, for example, cannot absorb radiation emitted by other methane molecules. This constraint limits how often GHG molecules can absorb emitted infrared radiation. Frequency of absorption also depends on how long the hot GHG molecules take to emit or otherwise release the excess energy.”

Surely, this cannot be correct?

Comment by AEBanner — 12 Aug 2007 @ 5:52 PM

420. Re #419: AEBanner — I am not a physicist. However, I found a paper on-line in Proc. N. A. S. , vol. 12, 1927 by Joseph W. Ellis giving the longest absorption band at 7.67 microns, with presumeably the next being a 6.58 micron band. A paper from the Proc. Royal Society (?) in 1965 gives the only emission band for methane of import in the upper atmosphere as the 7.65 micron band.

Perhaps more usefully for you, I found that there exists a book entitled “The Physics and Chemistry of the Upper Atmosphere” by MH Rees.

Comment by David B. Benson — 12 Aug 2007 @ 6:41 PM

421. re 419. This is a BOMB that I trust is a DUD. The words blow a battleship sized hole into the absorption / re-emission process, don’t they?? Also: (Individual) Molecules are hot?? Molecules emit only at different frequencies than they absorb?? Am I reading this wrong? Or, as AEBanner implies, is DOE out to lunch? — which would be a helluva note!

Comment by Rod B — 13 Aug 2007 @ 11:02 AM

422. Re #419

I have e-mailed DoE for confirmation of this paragraph, and I really hope they reply.

By “hot” molecules, I assume they simply mean molecules which have absorbed photons and are therefore in an excited state.

Comment by AEBanner — 13 Aug 2007 @ 2:13 PM

423. Re #419 What the DoE are saying is quite correct. Let’s face it, if I had mentioned this earlier you would not have believed me.

What happens is that when a photon is absorbed is that the excited state first partly relaxes via a radiationless transition. See http://www.garfield.library.upenn.edu/classics1981/A1981LY47300001.pdf
Then it can emit the remaining energy as a photon, but the emitted photon will be at a longer wavelength than the absorbed one because it has less energy.

The climate modelers like Gavin are still working with Kirchhoff’s Law based on a paper by Einstein in 1917, but radiationless transfers have been known about since 1968, four years after the first edition of Goody’s bible Atmospheric Radiation was published!

Comment by Alastair McDonald — 13 Aug 2007 @ 3:54 PM

424. 422 AEBanner and 421 Rod B Yes, hot just means a photon has been absorbed and that the energy has not yet been lost to collision or radition. And Alastair is probably right about the models, though I have not been able to tell for sure. Gavin or someone needs to verify exactly what state they are in for us. The current basis for such things is NLTE radiative transfer theory — http://www.ssec.wisc.edu/library/coursefiles/05_rte.pdf would be a start. Its fig 5.1 is quite interesting.
You can backtrack this thread to various discussions on LTE, which will have some other recent references. Or search for line by line transfer models or radiative transfer models. The problem for GCM’s is that LBL models can suck up a whole computer by themselves, so various levels of approximation are used in the GCM’s and these vary substantially as to accuracy. OCO and CH4 do not work the way a lot of people think they do, or even the way many models do, which I believe is one of Alastair’s points. But they do assuredly bring about interactions between the radiation fields and the kinetic motion, just not the way some people expect.

Comment by Allan Ames — 13 Aug 2007 @ 5:23 PM

425. WOW you guys are still at this.!! I repeat my comment from Part I, before I even knew that part II existed.

Re 278 (from Part I, or 419 form here). The first 3 DOE sentences are valid. The rest is true but irrelevant since it is not constraining on the total process.
When the CO2 absorbs the energy & returns it to the air mostly by collisions but a little by re-emission, (once a hot CO2 collides once, then it loses enough energy that it canNOT reemit a photon of the specific wavelength that can be absorbed by another CO2) the surrounding air reabsorbs the energy within microseconds and within centimeters (at ground level) and since the air’s temperature is unchanged (conservation of energy) then the air will reemit IR energy per Wein & Stefan-Boltzmann in the wavelength range that CAN be absorbed or reabsorbed by the CO2, continuously, forever. This is HOW CO2 transports energy to higher elevations & out to space. Since the air & CO2 density is lower at higher elevations. a photon travels further going up than it does going down, hence energy is transported out to space. (in a zig-zag process of many absorbtions, at one microsecond per absorbtion, you can get a million absorbtions in a second!!)

Another comment on saturation. When CO2 absorbs a (typical IR) photon, the molecule goes to ~900K. (so said Eli years ago!) THEREFORE, the only way all the CO2 can be saturated is WHEN all the air is at ~900K. Otherwise the CO2 will collide with the air and cool back down to unsaturated conditions. OR looking at it in another way, IF all the CO2 DID stay saturated, then CO2 could not absorb photons, and the IR radiated photons would go directly to space faster, and hence cool down the air. We will not survive too well at ~900K. The saturation concept is a waste of energy!

A side comment- the concept that CO2 “TRAPS” photons is totally absurd. There is NO way in the world, that a ~900K CO2 molecule can exist for more than a few microseconds aurrounded by 288K air molecules. Besides if CO2 traps the energy & doesn’t give it back, then how does the energy warm the air? CO2 “catches and releases” energy. It just slows the transport to space process down, which is why GHGs cause warming.

My next question is since the air is now warmer, why doesn’t the Stefan-Boltzmann equation, that says that the energy transported out (either as a black body or gray – it doesn’t matter) is proportional to T^4, simply transport out MORE energy faster at the warmer temperature, to return the air to the original equilibrium temperature? ie Mother Natures natural compensation process for the greenhouse effect?

Comment by John Dodds — 13 Aug 2007 @ 5:27 PM

426. re 419 – 425: This is getting out of control! I’ll tentatively accept most of this, but a couple of things do not compute. 1) If the (numerous) discrete rotational and/or vibrational energy levels are precise for every (say) CO2 molecule, then the molecule can absorb or emit, or can transfer to translation energy at only one of those discrete levels. In the event of re-emission (assume at a lower frequency than the original absorption), it is not anywhere near obvious why another CO2 molecule can not absorb this photon as it has an energy level that exactly matches the re-emission. Can anyone explain why not? Moreover it seems much less likely that a different molecule from CO2 will absorb a CO2 molecules emission…

2) How do we come back around to a gas of normal density now emitting blackbody radiation ala Stefan-Boltzmann? For the umpteenth time, does it or not?

A follow up query: Does IR energy absorbed into the rotational or vibration levels have to relax almost immediately? Does a molecule never re-emit directly, as opposed to very seldom? Wouldn’t the average CO2 molecule absorb a number of photons near simuntaneously? Do/can collisions transfer energy from translation to rotation/vibration?

One last quibble/clarification for John. I assume you mean that when a molecule absorbs a typical IR photon, and then transfers its energy from rotation/vibration to translation, the temperature goes up ~900K. Am I correct? (And put your helmet on — the one-molecule-cannot-have-temperature contigent might be ready to fire…)

Comment by Rod B — 14 Aug 2007 @ 1:19 AM

427. Re #426

If you refer to #423 you will see that Alistair McDonald explained that an excited CO2 molecule can lose some energy by means of a “radiationless transition”, and he gave a reference to this effect which apparently has been known since 1968. Therefore, the energy of a new photon when emitted is less than the energy of the photon which initially excited the molecule. However, the energy of the new photon is then too small to excite another molecule, and its energy is presumably transferred into kinetic energy of other molecules, or the photon goes to ground, or escapes into space.

This seems to be OK, except that I simply do not understand how a “radiationless transition” works, and I should be very grateful for further explanation.

Re relaxation time for an excited CO2 molecule, I have been informed that 100 picoseconds (10^-10 sec) would probably be a top end figure, and it would more likely be 10ps.

Comment by AEBanner — 14 Aug 2007 @ 2:51 PM

428. re #427, et al: But there are thousands of discrete energy levels in a CO2 molecule. I also assume that a radiationless transistion/transfer does not have to transistion the full energy of whichever level, but can transistion say half of it (to translation) and half of it to a lower rotational level. [note, translation energy levels are far more numerous, near continuous, and almost do not exhibit similar quanta — other than via Planck’s constant.] This new 2nd rotational level then might re-emit but at a quanta that is 100% of its energy, which is half the energy of the original absorbed level, and thence at a lower frequency. But the energy quanta of this second level exactly matches a potentially available level in another — (actually every other, CO2 molecule. So why can’t the next molecule absorb that emitted energy quanta? Secondly, that particular quanta/level is less likely to exist in a non-CO2 molecule.

Comment by Rod B — 14 Aug 2007 @ 9:08 PM

429. re #427-II. If the relaxation interval is ~10psec, doesn’t that mean that the rotation and vibration levels absorb energy, but effectively never have any energy? If that’s correct how can the heat capacity (“n”kT) ever vary?

Comment by Rod B — 14 Aug 2007 @ 9:20 PM

Ok, clearly Angstrom is spinning by now …

A general formula for the rate of resonant transfer of energy between two electric multipole moments of arbitrary order using molecular quantum electrodynamics

A. Salam
Department of Chemistry, Wake Forest University, Winston-Salem, North Carolina 27109

(Received 14 July 2004; accepted 19 October 2004; published online 12 January 2005)

A general expression is derived for the matrix element for the resonant transfer of energy between an initially excited donor species and an acceptor moiety in the ground state, with each entity possessing an electric multipole moment of arbitrary order. In the quantum electrodynamical framework employed, the coupling between the pair is mediated by the exchange of a single virtual photon. The probability amplitude found from second-order perturbation theory is a product of the electric moments located at each center and the resonant multipole-multipole interaction tensor. Using the Fermi golden rule, a general formula for the rate of energy transfer is obtained. As an illustration of the efficacy of the theory developed, rates of excitation energy exchange are calculated for systems interacting through dipole-quadrupole, dipole-octupole, quadrupole-quadrupole, and the familiar dipole-dipole coupling. For each of the cases examined, the near- and far-zone limits of the migration rate are calculated from the result valid for all donor-acceptor separations beyond wave function overlap. Expression of the octupole contribution to the transfer rate in terms of its irreducible components of weights 1 and 3 leads to new features. The octupole weight-1 term is found to contribute only when the interaction is retarded, while the dipole-octupole weight-1 contribution appears as a higher-order correction term to the dipole-dipole rate. Order of magnitude estimates are given for the contributions of dipole-quadrupole and dipole-octupole terms relative to the leading dipole-dipole rate for near-, intermediate-, and far-zone separations to further understand the role played by higher multipole moments in the transfer of excitation and the mechanism dominating the process. ©2005 American Institute of Physics.

Comment by Hank Roberts — 14 Aug 2007 @ 11:05 PM

431. Allen Ames (#424) wrote:

The current basis for such things is NLTE radiative transfer theory — http://www.ssec.wisc.edu/library/coursefiles/05_rte.pdf would be a start. Its fig 5.1 is quite interesting.

Thank you, Allen. This looks fun!

*

AEBanner,

One point to keep in mind: if a molecule can lose energy during a collision, it can also gain energy in a collision. This is a large part of what goes on, actually, as what tends to raise the temperature of the lower atmosophere isn’t the absorption of radiation but moist air convection. The energy from this tends to be radiated both to space and to the surface by greenhouse gases. As such greenhouse gases have a net direct tendency to cool the atmosphere and tend to warm the atmosphere primarily by means of re-emitted radiation warming the surface which promotes moist air convection.

In any case, Allen, Alastair and I take an interest in how reality breaks from local thermodynamic equilibrium, but they may be a little farther than myself at this point.

Comment by Timothy Chase — 14 Aug 2007 @ 11:39 PM

432. Douglas asked in 382: “When climate modellers determined climate sensitivity for a doubling of CO2, did they take into account the possibility of …” and “… have they addressed the possibility that the formula would have to change as CO2 levels significantly exceed those experienced in the previous one million years …”

If you use the Find function for the words “climate sensitivity” in the topic I think you’ll get some help here; you’re thinking that climate sensitivity is something modelers come up with from physics calculations, I gather.

This may help clarify where the modelers get the estimates for climate sensitivity: http://julesandjames.blogspot.com/2006/03/climate-sensitivity-is-3c.html

On the second question, one million years isn’t much:
http://www.globalwarmingart.com/wiki/Image:Carbon_Dioxide_400kyr_Rev_png
if you’re willing to go four hundred million years, you can get some big numbers: http://www.globalwarmingart.com/wiki/Image:Phanerozoic_Carbon_Dioxide_png

Note the climatologists never did agree on whether to have their temperature scales run left-right or right-left; the zero point on one of those charts is on the right, and on the other one it’s on the left. Up is always more and down is always less, fortunately.

Comment by Hank Roberts — 15 Aug 2007 @ 12:17 AM

433. Re #428

You make a good point, Rod B, and I’ve wondered about that myself. I can only suggest that the process you described depends upon the absorption cross sections of the molecule for photons of different energies, and so is a matter of probability. The absorption lines adjacent to the 14.99 micron line for CO2 are somewhere between one and two orders down, if I remember correctly. So the chance of the transition happening twice ( in any “chain” ) would be between a factor of 10^2 and 10^4 down.

Hopefully, someone who really knows will provide a proper reply, because I believe this to be of great significance.

Comment by AEBanner — 15 Aug 2007 @ 6:50 AM

434. John Dodds (#425) wrote:
When the CO2 absorbs the energy & returns it to the air mostly by collisions but a little by re-emission, (once a hot CO2 collides once, then it loses enough energy that it canNOT reemit a photon of the specific wavelength that can be absorbed by another CO2) the surrounding air reabsorbs the energy within microseconds and within centimeters (at ground level) and since the air’s temperature is unchanged (conservation of energy) then the air will reemit IR energy per Wein & Stefan-Boltzmann in the wavelength range that CAN be absorbed or reabsorbed by the CO2, continuously, forever. This is HOW CO2 transports energy to higher elevations & out to space. Since the air & CO2 density is lower at higher elevations. a photon travels further going up than it does going down, hence energy is transported out to space. (in a zig-zag process of many absorbtions, at one microsecond per absorbtion, you can get a million absorbtions in a second!!)

Generally it doesn’t even make sense to speak of a photon being absorbed and reemitted. This is actually something which Alastair has pointed out on some occasions. The collisions and consequent transfers of energy simply occurs too often. I remember his running the equations once, and the figure was that collisions would occur at roughly a thousand times the rate of emission. Instead what we know as the result of quantum mechanics, experiments in the labs and infrared imaging of the atmosphere is that when radiation gets absorbed by the gas, it raises the temperature of the gas resulting in the reemission of energy which takes essentially a random path – if one were to track parcels of it.

Eventually some of that radiation will reach the surface and some will reach space. That which reaches the surface will warm the surface, either resulting in moist air convection which raises the tends to raise the temperature of the atmosphere or is reemitted in the form of thermal radiation. Eventually the energy gets out, and in fact at some point an equilibrium will be established where thermal radiation being emitted by the surface and thermal radiation leaving the atmosphere will be equal.

*

Incidently, Alastair brought up something called radiationless transfer at one point, but that is a very specialized phenomena which I believe has nothing to do with the atmosphere. But climatologists would know as they work with radiation physicists who can model this sort of thing: they have even incorporated the fairly esoteric non-local thermodynamic equilibria. But I wouldn’t worry about that for a little while. Throughout most of the atmosphere it isn’t that relevant, and where it does occur it is a matter of gradual deviation from what are called local thermodynamic equilibria.

*

Another comment on saturation. When CO2 absorbs a (typical IR) photon, the molecule goes to ~900K. (so said Eli years ago!) THEREFORE, the only way all the CO2 can be saturated is WHEN all the air is at ~900K. Otherwise the CO2 will collide with the air and cool back down to unsaturated conditions. OR looking at it in another way, IF all the CO2 DID stay saturated, then CO2 could not absorb photons, and the IR radiated photons would go directly to space faster, and hence cool down the air. We will not survive too well at ~900K. The saturation concept is a waste of energy!

When it cools by collision, it “heats up” other molecules by collision. At some point some of the molecules which recieve energy by collision will be greenhouse gas molecules which reemit.

A side comment- the concept that CO2 “TRAPS” photons is totally absurd. There is NO way in the world, that a ~900K CO2 molecule can exist for more than a few microseconds aurrounded by 288K air molecules. Besides if CO2 traps the energy & doesn’t give it back, then how does the energy warm the air? CO2 “catches and releases” energy. It just slows the transport to space process down, which is why GHGs cause warming.

Actually, throughout most of the atmosphere the radiation has the net tendency of cooling the atmosphere. But some of the radiation which is emitted by greenhouse gases will reach the surface, warming it and resulting in more thermal radiation or in moist air convection and some will escape to space. The moist air convection is what tends to raise the temperature of the atmosphere.

My next question is since the air is now warmer, why doesn’t the Stefan-Boltzmann equation, that says that the energy transported out (either as a black body or gray – it doesn’t matter) is proportional to T^4, simply transport out MORE energy faster at the warmer temperature, to return the air to the original equilibrium temperature? ie Mother Natures natural compensation process for the greenhouse effect?

Well, we aren’t talking about black body or gray body as the emissivity will differ in different parts of the spectra, meaning that the capacity to absorb and reemit will be different for different parts of the spectra. But with greenhouse gases being involved, a new equilibria is reached at a higher temperature roughly according to the T^4 where the thermal radiation entering the system from the surface will be equal to the thermal radiation escaping to space.

In essence, the higher temperature is the compensation for the presence of greenhouse gases in the atmosphere – as this is what makes it possible for the amount of thermal radiation entering the system to equal the thermal radiation which is leaving the system. Or to put it in terms of the problem you posed, the equilibria at the higher temperature will be reached before the even higher temperature which would be necessary to reemit radiation at a high enough rate to bring the temperature back down to the original temperature.

Comment by Timothy Chase — 15 Aug 2007 @ 8:33 AM

435. Hank (430), are you describing a transfer between quantized states, which for all practical purposes exist only in the rotation and vibration energy levels? If it’s from a quantized level to a non-quantized level (translation) then it would seem your analysis ala resonance doesn’t apply. (But the reverse, from translation to quantized levels, I’m not sure…) Between quantized states between different molecules, while following your rules of allowability, isn’t this transfer always with photons, i.e. emission/absorption of radiation?

Timothy (434), a couple of nits, though like bacteria can be terribly important: You say radiation absorption raises the temperature of the gas. I though we finally concluded that this is not correct, as absorption is only into rotation or vibration levels, which, until that absorbed energy gets transferred to translation, has no effect on temperature. Likewise emission per se from a greenhouse gas does not cool anything, other than emitted virotional energy which came from a transfer of energy from translation. And that transfer, not the emission, causes the cooling.

Isn’t radiationless transfer the transfer of energy from a virotional to another vibration/rotation level within the same molecule or from virotional to translation within the same or, maybe??, to a different molecule, none of which involves photons? Isn’t this important because the exchange between vibration/rotation levels and translation allows radiation to, indirectly, cause temperature changes in the gas. Alastair?

[Is there an industry accepted word that combines “vibration” and “rotation”??]

Comment by Rod B — 15 Aug 2007 @ 12:14 PM

436. I had stated in #434:

When it cools by collision, it “heats up” other molecules by collision. At some point some of the molecules which recieve energy by collision will be greenhouse gas molecules which reemit.

Rod B (#435) wrote:

Timothy (434), a couple of nits, though like bacteria can be terribly important: You say radiation absorption raises the temperature of the gas. I though we finally concluded that this is not correcrt, as absorption is absorbed only into rotation or vibration levels, which, until that absorbed energy gets transferred to translation, has no effect on temperature. Likewise emission per se from a greenhouse gas does not cool anything, other than the emitted energy from a rovational level might have come from a transfer for translation. And that transfer, not the emission, causes the cooling.

No problem.

When molecules have vibrational, rotational or vibrorational energy, they may also have translational energy which makes it possible for them to collide. When they collide, they may lose or gain these other forms of energy. Now I said “heats up” basically as a shorthand for any of these forms of energy since each also corresponds to a temperature. For example, depending upon the type of vibration (stretching, bending, etc) a given molecule of carbon dioxide may have several different vibrational temperatures at the same time – or rather, the aggregate of such molecules will, that is, the gas as a whole. Normally these are equal, that is, below 50 km, but above 50 km collisions will happen less often and thus they will tend to diverge. And since these are temperatures, emission actually does cool the gas.

Now one of the questions that is puzzling me is whether this is the case with rotational (which does not apply to carbon dioxide) and vibrorotational (which does) temperatures as well: are they equal to the vibrational and translational temperatures below 50 km? I honestly don’t know the answer as of yet. However, I suspect that if I stay here long enough I will have the answers to this and a great many other questions.

Comment by Timothy Chase — 15 Aug 2007 @ 12:36 PM

437. Rod, the text in 430 is not my analysis.

You wrote
> isn’t this transfer always with photons

Radiation transfer articles in the physics journals are written almost entirely in math, that I make no pretense to handle.

This may be why Spencer Weart’s AIP History section on radiation physics ends with the 1980s, because it was at that point good enough to be useful for climate work.

What matters for this issue is that the heat does move and the models describe how it moves well enough to be useful.

Comment by Hank Roberts — 15 Aug 2007 @ 12:52 PM

438. re 423 Alastair: I think the original radiationless transfer is Forster Transfer (with an umlauted o) (~Furster)
http://www.everything2.com/index.pl?node_id=625957
Such transfer can take place when the spectra overlap and the molecules are close, as between dye molecules adsorbed to a surface. It can be inter- or intra- molecular, like chlorophyll. Radiationless transfer is also what happens in a solid, where the spectral overlap is nearly total, and energy belongs to the whole solid, not just an atom. The relevance here is that close coupling between many molecules gives rise to bands, not lines, and many coupled molecules can have a continuum where energy is shared, just like a solid. This makes blackbody radiative properties. As far as I know right now, only OHO has enough heat of association that multimers are at all likely. Also see http://en.wikipedia.org/wiki/Dark_quencher

As for LTE and photons, the references back around #250 are worth a look if you have not done so recently. A lot of current discussion is covered in them.

Note to Douglas Wise: I see a big difference between the physics as expressed by radiation transfer theory, and what seems to be in the heads of a lot of people who think they know about GW. I sympathize because the older band theories support much misconception, and the pre quantum theories are still around.

Hank, Timothy: Rod B certainly has come a long way, has he not?

Comment by Allan Ames — 15 Aug 2007 @ 4:39 PM

439. > a long way
Definitely a long way into territory I’d never imagined (grin). I continue to hope Raypierre will come tell us where we’re getting to!

“It’s a long way from Isaac Newton,
It’s a long way to us …”

With apologies to http://www.ediacara.org/songs.html

Comment by Hank Roberts — 15 Aug 2007 @ 4:49 PM

440. I agree with Timothy and Allan that radiationless transfers are just a red herring, and are probably irrelevant to the greenhouse effect.

Comment by Alastair McDonald — 15 Aug 2007 @ 5:35 PM

441. Allen Ames (#438) wrote:

Hank, Timothy: Rod B certainly has come a long way, has he not?

I don’t mean to embarass him, but I pretty much always knew that he was/is special. And frankly I think a fair number of us have learned a great deal. Personally, that is one of my two biggest priorities. I am not sure what comes in first, though.

Comment by Timothy Chase — 15 Aug 2007 @ 5:39 PM

442. Rod 423,

Yes, when a molecule absorbs energy into a rotational etc mode IT HAS to gain energy. It HAS to heat up. Otherwise conservation of energy does not work.
If it gains energy it will be hotter, & one way or another (mostly collisions (as Eli said years ago, as Timothy said above), the energy will be returned to the rest of the air.
If you visualize a CO2 absorbing the smallest possible photon of energy, then it can only either reemit, of hit and lose some energy to the an other molecule (mostly N2 & O2.) If it hits then it cannot reemit.

Comment by John Dodds — 15 Aug 2007 @ 6:07 PM

443. Timothy (436), I think that rotation and vibration “temperature” should be properly called characteristic temperature to distinguish it from “real” temperature. (Much like the “characteristic temperature” of the electro-magnetic energy wave coming of a (black)body that has “real temperature”. If you add a pile of energy to vibration/rotation it does not increase the “real” temperature of the gas, as measured with a thermometer. It took me a long time to assimilate this and change my earlier opposite belief — thanks to a lot of helpful pounding form RC posters. Only translation (zipping around) energy affects the temperature of a gas. Please tell me I’m correct.

I also understand that energy can transfer among the three gasseous energy states (and can exchange rotation/vibration with translation from another molecule) and it works toward some equitable distribution within the molecule among the three states (equipartition). Secondly, with very few and insignificant exceptions, longwave radiation is directly absorbed and/or emitted into/from vibration or rotation energy.

I haven’t absorbed all of the posts yet, but I’ll jump to the question that’s really bothering me. There is a credible assertion going around that greenhouse gasses hardly ever gets to re-emit absorbed photons, except at very high altitudes. Question: where does the roughly 325 watts/square meter of downwelling longwave radiation that the earth has to absorb come from???

Comment by Rod B — 15 Aug 2007 @ 6:31 PM

444. ROD B We are all grappling to some degree with what is meant by “temperature”. For sure in RT theory the LTE temperature presumes equilibrium between kinetic energy and the analogs of harmonic motion. If these were in equilibrium with the external radiation field it would be a “colored body” equilibrium, depending on the emitters/absorbers present. (There are very few black gases.) As LTE begins to fail for lack of collisions, the last one left is, I think, the kinetic, since it holds the most energy. To change kinetic energy requires a collision of some kind, but (I think) the rot-vib’s have some internal conversions.

If the various energy components are not in equilibrium, it is reasonable to talk about average energy and so average temperature. “Apparent” might depend on how you were looking at the system. The emission from a set of nearly regular vibration levels will have an apparent temperature that depends on the separation and collisions driving the excitation. You may have noticed the OCO emission at 15 microns which seems to always have a temperature around 260. This is a sort of temperature (NLTE maybe). The only unambiguous definition seems to be, as said earlier, the proportionality between delta heat/delta entropy.

As you will no doubt hear, the primary source of it all is the sun.

Comment by Allan Ames — 15 Aug 2007 @ 8:26 PM

445. Don’t think “re-emit absorbed photons” — I had trouble with this. It’s not a solid entity that’s caught and thrown like a football.

The photon is a bookkeeping entry — an amount of energy transfer — not an entity; there’s no photon that exists after the molecule captured the energy — the energy then is in some other form. The molecule is doing a lot of other energy transactions all along. Those may include an outgoing photon, carrying away some energy.

Comment by Hank Roberts — 15 Aug 2007 @ 9:46 PM

446. Re 443
Types of temperature -Baloney!- er excuse me – a bright rosy red herring!
Conservation of energy is the key. Energy is either in a photon or in matter (or in a mag field) Going from the ground to space it goes from in matter (heat, translational or rotational or potential or binding energy in the atom or whatever, to photon. Conservation of energy requires one or the other & it can’t get lost on the way. You can’t NOT count it. (unless you are repealing E=mc^2) Since there is no nuclear fission or fusion in the air, AND since there ARE mechanisms to tranlate rotational etc to translational energy (eg a “hotter” CO2 with photon takes more space (higher energy state, vibrates in a larger space etc) than one without the photon- the larger hotter CO2 will perturb the nearest cooler molecules, to move energy to cooler molecules (collisions) then the energy is going to find its lowest state (ie equilibrium),- warm CO2 will cool down to warm the air. OR don’t you believe that the GHE warms the air? If the hot CO2 does not lose its eneregy to the air then how does the air warm up?
This is NOT even physics. It is plain ordinary common sense.
Sorry, but you guys are getting ridiculous.

Comment by John Dodds — 15 Aug 2007 @ 10:37 PM

447. This is kind of a continuous clarification…. John (442), I have to disagree (though it has been only a couple of weeks since I vociferously argued your point!) Adding energy, even “heat” energy, does not necessarily affect temperature. Otherwise there would be no differing specific heats (the amount of energy to raise one mass unit one degreeK: deltaQ = C*deltaT), which there are with gasses. Specifically the specific heat of a gas increases as vibration and rotation energies start to exceed their ground level. Nor, I guess, would you have things like heat of vaporization or fusion. I agree that rotation and vibration can be transferred to translation, and changes in translation energy, by definition changes temperature.

Allen, I agree the entropy equation (~ Q=ST) is valid, but I’m not comfortable with it as the ultimate say — entropy itself is not exactly unambiguous! Plus, I understand this can proven to be exactly as temperature defined by the Center-of-Mass kinetic energy. And, I agree the definition, or which “temperature” are we talking about, has to be clear. That’s why I call it “real” temperature as opposed to “characteristic” temperature, which is not real temperature but a characteristic that has some relevance to some related temperature. Talking of the “temperature” of an “E-M wave” or a “vibration level”, while physically accurate in the right contex, is not helpful, most pointedly because this entire blog is devoted to assessing REAL temperature.

Comment by Rod B — 15 Aug 2007 @ 10:49 PM

448. John (446), you’re being too loose with your terminology. The absorption of a photon by a GHG into its vibration/rotation energies DOES NOT increase the temperature of the gas. The transfer of (some of) that vibration/rotation energy to translation energy DOES increase the temperature, maybe a microsecond or so down the line.

Comment by Rod B — 15 Aug 2007 @ 11:07 PM

449. Rod Re where does the 325 watts come from?
The models say that the stratosphere (15-50+Km) gets cooler by transferring energy to the troposhere (0-15Km). (see Hansen et al 2005 figure 1.e)
So then why doesn’t the SBL (hotter air rises faster! hotter (gray)objects radiate more energy) just move the extra energy back to the stratosphere to cancel out the GHE & return to equilibrium? or doesn’t equilibrium work. Funny, every day the earth goes from cooling at night to warming when the sun comes up. IT IS REQUIRED that it pass thru equilibrium at the TOA twice every day! and since the energy gets all the way to the ground then it is required that the air be at equilibrium at all elevations every day also. (it takes under a second for a photon to go from ground to space via multiple CO2 absorbtions – so the SBL maintains AIR equilibrium VERY VERY quickly!) The cooler stratosphere & warmer ground is impossible if we are at equilibrium. BUT THAT temperature reality is confirmed by measurements. So HOW????

You might try adding energy to the ground from energy stored in the ocean/land per Hansen Nazarenko 2005 (which would also warm the stratosphere since we are adding energy that is not from the sun & not included in the models,) AND you might try subtracting energy from the stratosphere (because the earth mag field which has decreased for the last 100+ years no longer transfers as much energy to the charged particles in the stratosphere to send them towards the poles?- also not included in the models!

OR you might try continuing to think about it. You are doing well- Just don’t believe everything you hear – question everything. especially from the experts here! including Gavin’s step ONE of six, that the GHE is unchallengeable. (according to me- the GHE works, but the SBL just cancels it out almost immediately so there is no long term CO2 effect. CO2 can’t cause warming! The GHE doesn’t add energy. Arrhenius failed to account for the (gray body) SBL.-ie that hotter air radiates more energy faster to compensate for the slow down in transporting energy due to the increased absorbtions.)

Comment by John Dodds — 16 Aug 2007 @ 12:16 AM

450. John, I can’t recall or figure out what “SBL” stands for…

Comment by Rod B — 16 Aug 2007 @ 10:23 AM

451. re #450 SBL stands for ‘Stably stratified planetary boundary layer’. I suspect that John is confusing it with the PBL (planetary boundary layer) which convects, unlike the SBL which, as its name suggests, is stable, and is just another name for the PBL at night. See: http://en.wikipedia.org/wiki/Planetary_boundary_layer.

BTW, TOA is the top of the atmosphere which is roughly in equilibrium 24 hours a day. It is the BOA (Bottom of the Atmophere) where equilibrium only happens twice a day, when in the morning the surface is warmed by the sun to match air temperature, and in the evening when the surface cools below the air temperature.

Comment by Alastair McDonald — 16 Aug 2007 @ 1:37 PM

452. John Dodds writes:
> (it takes under a second for a photon to go from ground
> to space via multiple CO2 absorbtions

And it takes a good sprinter under a second to run 20 meters.

But it may take a drunken man half the day to get that far from the lamppost, to where he falls off the curb and ends his staggering.

Photons are emitted in any direction at random, and with more CO2 in the atmosphere — that’s a “drunkard’s walk” from their starting point.

The decreasing density of the atmosphere might be like having the drunk staggering about on a slope, rather than on the flat — he may go a bit farther in the easier direction with each stagger.

The timer ends when the drunk falls off the curb — when a photon goes out of the atmosphere into space.

There’s no way for the photons to know which way is up, and to prefer that direction, and to be able to choose it.

Comment by Hank Roberts — 16 Aug 2007 @ 2:18 PM

453. Re #451

You are getting a little too good at explaining this sort of thing, I hope you. I believe I actually understood what you had to say – and it was just too easy…

Comment by Timothy Chase — 16 Aug 2007 @ 2:30 PM

454. Re my #419
I found this passage recently in a piece from the Energy Information Administration of the US Department of Energy

http://www.eia.doe.gov/cneaf/alternate/page/environment/appd_a.html

“What happens after the GHG molecules absorb infrared radiation? The hot molecules release their energy, usually at lower energy (longer wavelength) radiation than the energy previously absorbed. The molecules cannot absorb energy emitted by other molecules of their own kind. Methane molecules, for example, cannot absorb radiation emitted by other methane molecules. This constraint limits how often GHG molecules can absorb emitted infrared radiation. Frequency of absorption also depends on how long the hot GHG molecules take to emit or otherwise release the excess energy.”

Surely, this cannot be correct?

Re Timothy Chase #434

“Incidently, Alastair brought up something called radiationless transfer at one point, but that is a very specialized phenomena which I believe has nothing to do with the atmosphere. But climatologists would know as they work with radiation physicists who can model this sort of thing: they have even incorporated the fairly esoteric non-local thermodynamic equilibria. But I wouldn’t worry about that for a little while. Throughout most of the atmosphere it isn’t that relevant, and where it does occur it is a matter of gradual deviation from what are called local thermodynamic equilibria.”

Re Alistair Mcdonald #440
“I agree with Timothy and Allan that radiationless transfers are just a red herring, and are probably irrelevant to the greenhouse effect.”

In my #419, I quoted a paragraph directly from the US Department of Energy which was dealing specifically with the GHG effect and “radiationless transfers” of energy.

So what am I to believe?

Comment by AEBanner — 16 Aug 2007 @ 5:21 PM

455. John, that document you’re referring to is dated 1994.
1994 seems like yesterday to me, but that info is quite old.
Or I am. Perhaps both.

Are you asking about their use of the term “quenching” to cover all the interactions that transfer heat between gas molecules without emitting photons?

I doubt that’s the same thing that the quantum physics people are talking about.

Comment by Hank Roberts — 16 Aug 2007 @ 6:25 PM

456. Re #453 & 454

TC wrote “I believe I actually understood what you had to say…”
If you are willing to believe what I have to say then you will find it much easier to understand than if you are determined to disbelieve it.

I should have written that spontaneous radiationless transitions are a red herring. Collision induced radiationless collisions are at the heart of the greenhouse effect, just as the US Department of Energy implied.

What seems to be the case is that, just as radioactive atoms have a half life, so do (vibrationally and rotationally) excited molecules. A laser works by electrically exciting a molecule (e.b. CO2) and allowing it to naturally relax quickly to a new longer half life state. Lots of molecules can be in that long life state and when one molecule relaxes it causes a chain reaction. The new relaxed state has a short half-life and soon relaxes to the ground state.

Because the half life of a vibrationally excited CO2 molecule is very much longer than the time between collisions of molecules at STP, then the absorbed vibrational energy becomes molecular translational (kinetic) energy, viz. a gaseous temperature rise.

This is new! But at STP there are very few collisions (none) which can induce vibrational excitation, but collisions (and spontaneous radiationless relaxation) can chip away at the vibrational energy, turning it first into rotational energy and then into translational (kinetic) energy.

HTH,

Cheers, Alastair.

Comment by Alastair McDonald — 16 Aug 2007 @ 6:27 PM

457. When are you guys going to get over this endless and fairly nonsensical discussion of radiative energy transfer in the atmosphere, and move on to the next step, radiative-convective energy transfer in the atmosphere?

That’s where you include the motion of air molecules in the atmosphere in your model, instead of pretending that the atmosphere is made up of fixed-position molecules.

Kerry Emanuel’s lectures (biased towards tropical cyclones, but still very applicable) are available online, and everyone here seems to be stuck on this one:

http://wind.mit.edu/~emanuel/geosys/node2.html

The conclusion of that lecture is this:
“So clearly, in radiative equilibrium, the surface air is not in thermal equilibrium with the surface. This creates the potential for convective heat transport away from the surface. This will be the subject of the next lecture.”

Compare this to Alastair’s statement: “I agree with Timothy and Allan that radiationless transfers are just a red herring, and are probably irrelevant to the greenhouse effect.”

Ah – let’s just point out here that convection is indeed an example of ‘radiationless energy transfer’ between point A (surface) and point B (upper atmosphere). It’s interesting to watch someone tie themselves into knots verbally, but I would say it’s time to move on to the next lecture in the series:

http://wind.mit.edu/~emanuel/geosys/node3.html

Unless of course, you wish to point out any logical or mathematical errors in the first lecture? It seems to me that the endless claims by Alastair that ‘scientists don’t understand radiative transfer in the atmosphere’ are just a backhanded attack on climate models.

After all, if we believe Alastair’s claims, then the radiative-convective models can’t be right either, nor can the global atmosphere-ocean coupled models, as they all rely on radiative transfer calculations as well.

Perhaps the really relevant link for this discussion is the following one:

DeSmogBlog: Clearing the PR Pollution that Clouds Climate Science

Comment by Ike Solem — 16 Aug 2007 @ 7:28 PM

458. Cutting through all of the above for the moment, and given what has been claimed in various posts, I still do not know where the 300+ watts/square meter of infrared radiative downwelling from the atmosphere/clouds to the surface (where it is absorbed) comes from. Odd question, I know, but the forest got lost in the trees somewhere. Anybody know the answer?

And to raise another old chestnut, likely related to the above question: I’ve looked at 2-3 college texts, including Ike’s reference in #457, and all use Planck’s blackbody function in calculating the energy flux through the atmosphere and between construct slices of the atmosphere. This says the atmosphere (gas) radiates energy in a near continuous spectrum based on the slice of atmosphere’s temperature ala Planck. But, the consensus in these threads (though not near unanimous) is that gasses do not emit so-called blackbody radiation. What gives?

Comment by Rod B — 18 Aug 2007 @ 10:18 AM

459. Re Alastair #456

The whole is greater than the sum of its parts.

Please accept the compliment in the way in which it was offered.

Comment by Timothy Chase — 18 Aug 2007 @ 12:15 PM

460. Rod, all of the gases in the atmosphere, looked at taken together in a snapshot, can be described as radiating approximately like a theoretical black body.

“Black body” is not a kind of radiation. It’s an appearance, of all the individual events overlapping.

Comment by Hank Roberts — 18 Aug 2007 @ 12:57 PM

461. As I said before, Ned Ludd would truly enjoy this discussion.

From the findings of the Intergovernmental Panel on Climate Change (IPCC) working group
“Our ability to quantify the human influence on global climate is currently limited because the expected signal is still emerging from the noise of natural variability, and because there are uncertainties in key factors”.

In other words, we know what we want to find. We haven’t found it yet. However, we know it is somewhere in the margins of error of our measurements.

[Response: Cites are always interesting. This quote is from the second assessment report (published in 1995). Try the latest version (2007) if you want something a little more relevant. – gavin]

Comment by Ken Newell — 20 Aug 2007 @ 12:58 PM

462. Can you describe the Natural Variability filter you constructed in 200 words or less?

Comment by Ken Newell — 20 Aug 2007 @ 1:52 PM

463. Please can anyone tell me if a GHG molecule can be raised into an excited vibrational or rotational state by collision with a nitrogen or an oxygen molecule?

Also, a web ref would be gratefully received.

Comment by AEBanner — 23 Aug 2007 @ 6:37 AM

464. re 463 AEBanner:

Comment by Allan Ames — 23 Aug 2007 @ 10:15 AM

465. AEB- from much above, any increase in energy is going to very quickly be spread across all the modes.

I wonder if a chaotic pendulum is a decent analogy for the way energy put into a molecule spreads?

(for a better analogy, build a chaotic pendulum that also has a solar panel and a capacitor and laser ….)

Comment by Hank Roberts — 23 Aug 2007 @ 10:22 AM

466. Re 463

As I understand it, for a CO2 molecule to be excited into the v2 bending vibrational mode of 667 cm^-1, it would have to receive energy equivalent to the average kinetic energy of a molecule at 369C.

I believe someone else has already done this calculation but I cannot find it.

For the v3 asymmetric stretching mode of 4.25 cm^-1 the temperature needed rises to 2950C.

These are the only two modes at which CO2 can absorb or emit radiation.

Water vapour can emit rotational energy after colliding with a molecule with a temperature of 27 K (kelvin).

Here’s a link but probably not what you want

st-socrates.berkeley.edu/~budker/Physics138/Alyssa%20Atwood%20Atm%20Spec5.ppt

Comment by Alastair McDonald — 23 Aug 2007 @ 10:46 AM

467. Good question AEB (463); I second it…

Comment by Rod B — 23 Aug 2007 @ 10:46 AM

468. Re #464, 5, 6

Many thanks, gentlemen.

Comment by AEBanner — 23 Aug 2007 @ 3:29 PM

469. Alastair McDonald (#466) wrote:

As I understand it, for a CO2 molecule to be excited into the v2 bending vibrational mode of 667 cm^-1, it would have to receive energy equivalent to the average kinetic energy of a molecule at 369C.

… and as I understand it, this is why we can’t just look at the average velocity but but have to look at the Maxwell velocity distribution.

Comment by Timothy Chase — 23 Aug 2007 @ 4:42 PM

470. Re #469 Where Timothy Chase Says:

… this is why we can’t just look at the average velocity but but have to look at the Maxwell velocity distribution.

Yhat is true, but without looking at the Maxwell Distribution it is obvious that we can apply the Equipartition Theorem (Boltzmann’s Disrtibution) to the 27K rotational energy, but that the 2950C v3 asymmetric vibrational energy will be ‘frozen out’ at room temperatures. Nor can I imagine that many molecules in the Earth’s atmosphere have a translational temperature of 396 C.

Comment by Alastair McDonald — 23 Aug 2007 @ 5:53 PM

471. Re Temperature. The question of vibrational modes in complex molecules is resolved at two levels, I think Boltzman said “(thermodynamic) processes occur according to the probability of the process taking place” This is not rubbish but the meaning is a bit obscure. Consider the temperature of GHGs with different vibrational modes, isolated molecules that are vibrating will continue to vibrate until it can radiate its energy it has an energy related temperature. In general molecules are not isolated and collide with other molecules in a random way which is generally called “thermal”, this random collision process is considered to be “real” temperature by some. The fact is that thermal collisions result in energy exchange between the resonant vibrations (rotation, bending etc.) of the polar molecules; the process is called thermalization. As far as atmospheric thermodynamics is concerned, thermal processes are dying out in the tropopause because the pressure is too low, the gasses above the tropopause undergo other energy transfer processes, it is perhaps a mistake to call their energy temperature.

I am very encouraged by what I read here, I did not know that GHGs could not re-absorb their own radiation, but it is probably true at least to some extent. The absorption/emission lines of these GHG molecule are not quantum processes, they are more like the twanging of a guitar string, fading with time. As Rod’s (421) comment on 419(AEBanner) says “re 419. This is a BOMB that I trust is a DUD. The words blow a battleship sized hole into the absorption / re-emission process, don’t they?” Well it isn’t a dud and it isn’t the only hole in the GHG idea.

Another giant hole is the concept of “downwelling radiation”. It is quite impossible for a cold body to transfer energy by radiation to a hot body. The Earth’s surface is warmer than the atmosphere so the idea of 390W being radiated downwards from the atmosphere is basically nuts. What is even funnier, if you can find one of the radiative balance diagrams, you will see that this downwelling stuff is miraculously absorbed 100% without any reflection by what must be a very black body indeed!

The whole concept of GHDs “blocking” or “trapping” radiation relies a process that, of itself, would produce a temperature gradient in the atmosphere over and above the adiabatic lapse rate due to gravitation, mathematically this would be a diffusion process. Diffusion processes are very slow, like conduction (a diffusion process), convection is much, much quicker and more spectacular, (hurricanes are just a very enthusiastic convection process). In boring old thermodynamic terms, the probability of a convection process occurring is very much greater than a diffusion process. It interesting to note that the heat transfer in the radiation zone of the Sun is by radiative diffusion and it is very, very slow, above radiation zone is the convection zone, the two processes cannot take place together. What is the IPCC going to do when the rest of the world realises it is onto a loser?

Comment by Dermod O'Reilly — 23 Aug 2007 @ 6:14 PM

472. Alastair,

Individual molecules do not have temperatures. Populations/distributions have temperatures. And it does not matter what you can or cannot imagine if we have the satellite images, have performed the experiments and have a fairly solid understanding of the physics.

You have the ability to make a positive contribution here. I would rather you did. Your intimation regarding Raypierre’s chart was not. Neither is this.

Quit placing so much emphasis on your imagination and more on the science – please.

Comment by Timothy Chase — 23 Aug 2007 @ 7:48 PM

473. PS (to 471)

Anyway, if you look up the Maxwell velocity distribution, you will notice that it has a very long tail.

… and honestly, Alastair, I thought you had done a good job of explaining the equilibria in 451, and there have certainly been times when you have made positive contributions, seen things before others. But too often you try to undercut the science in order to feed your own ego and end up making things easier for those who would deny that anthropogenic global warming is taking place or that there is any science. I really don’t think that is your intention.

Maybe we should email:
timothychase at gmail dot com

Comment by Timothy Chase — 23 Aug 2007 @ 8:20 PM

474. PS to my PS

I might not be able to email you back right away – I have to meet someone, and it will take a little time, and I have been up since 4:30 AM.

In the meantime, take care.

Comment by Timothy Chase — 23 Aug 2007 @ 8:27 PM

475. #471 “Another giant hole is the concept of “downwelling radiation”. It is quite impossible for a cold body to transfer energy by radiation to a hot body.”

It’s not impossible. The 2nd Law of Thermodynamics is purely a statistical net effect, not a microscopic physical principle. In fact, the 2nd Law is also time-symmetric. It says that a decrease in entropy is possible, just extraordinarily unlikely.

How else could clouds can be detected from by surface observations using ground-based remote infra-red sensing?

The presence of greenhouse gases simply means that the pathway to space is longer and thus slower than otherwise would be the case. A proportion of the radiant energy will “bounce off” GHG molecules several times before leaving the atmosphere entirely.

Since we are talking about non-equilibrium thermodynamics, it’s the rate at which incoming heat from a low entropy source (the sun) is lost to space that ultimately determines the temperature of the Earth and other planets.

Comment by Alex Nichols — 24 Aug 2007 @ 3:52 AM

476. RE #472-4

I have no wish to get into a flame war on-line or off-line so I will decline the possibility of emailing you.

The temperature of a gas is based on the average kinetic energy of its molecules. If you take the average kinetic energy of one molecule, you can obtain its temperature, which will remain constant between collisions. During that time it msy or may not change its excitation level.

What what you wrote earlier, I gather that you have only been interested in the greenhouse effect for about a year. In that case it is not surprising that you are unaware that gases can have several temperatures. However, if you read the document to which Allan Ames pointed http://www.dodsbir.net/sitis/view_pdf.asp?id=DothH04.pdf
you find on page 7/8 (14-15) the sentence “Defining a vibrational temperature has proven to be a useful concept because it not only indicates departure from equilibrium but also the extent of this departure.” The idea of translational, electronic, vibrational, and rotational temperatures was not my idea. It is all part of the science. In fact CO2 (molecules) have two vibrational temperatures!

HTH,

Alastair.

Comment by Alastair McDonald — 24 Aug 2007 @ 4:17 AM

477. Re 475 radiative transfer is not a statistical process, the transfer of energy occurs for each individual molecule as if no other existed, this is the process for isolated molecule which I was at pains to point out, yes? It is the difference between the stratosphere and the troposphere, GHGs are themalized in the troposphere.

You remark “It’s not impossible. The 2nd Law of Thermodynamics is purely a statistical net effect, not a microscopic physical principle. In fact, the 2nd Law is also time-symmetric. It says that a decrease in entropy is possible, just extraordinarily unlikely.” This is an old chestnut about statistical mechanics, there are a lot of variations “what is the chance of boiling water turning to ice?” however you say “a statistical net effect” the chance of a single event taking place is not zero, it is as you say “just extraordinarily unlikely”. Even if it did occur it would be finished before you noticed, think about it! From the moment you start to need a meaningful result in connection with global temperature and catastrophy, you will not feel the need to take sub-microscopic probabilities into account,

I am fascinated by your point about detecting clouds in IR “How else could clouds can be detected from by surface observations using ground-based remote infra-red sensing?” because the same point occurred to me; I found this: http://en.wikipedia.org/wiki/Infrared check the paragraph “meteorology”.
Nice that you were interested in my posting.

Comment by Dermod O'Reilly — 24 Aug 2007 @ 7:42 AM

478. Dermod, you seem to have some imprecision in your statements. First, atoms do not ignore the presence of other atoms around them. Lines are broadened, collisional relaxation occurs, etc. And, yes, radiative transfer is indeed an inherently statistical and quantum process.
On the other hand, you are technically correct that the greenhouse mechanism is not really a source of energy, but rather a decrease in a sink (radiative transfer). In the models, though, this is, irrelevant, as a diminished sink looks like a sink + a source.

Personally, I think people are getting way too wrapped around the axle about what is blackbody or greybody or quantized… The fact of the matter is the spectrum of radiation emitted by Earth’s surface starts off looking very much like a blackbody spectrum at about 290 K, but by the time it escapes to space, there are big holes in that blackbody spectrum corresponding to the absorption lines of GHGs. That’s the relevant physics.

Comment by Ray Ladbury — 24 Aug 2007 @ 8:57 AM

479. Alastair McDonald (#476) wrote:

However, if you read the document to which Allan Ames pointed http://www.dodsbir.net/sitis/view_pdf.asp?id=DothH04.pdf
you find on page 7/8 (14-15) the sentence “Defining a vibrational temperature has proven to be a useful concept because it not only indicates departure from equilibrium but also the extent of this departure.” The idea of translational, electronic, vibrational, and rotational temperatures was not my idea. It is all part of the science. In fact CO2 (molecules) have two vibrational temperatures!

It is a fascinating article, Alastair. We live in a world of great beauty.

Comment by Timothy Chase — 24 Aug 2007 @ 11:13 AM

480. Re #478 Ray, with respect, I think you may be confused because you do not have a good grasp of what I was saying, probably my fault. Let me try again. Statistical mechanics is the description of a large ensemble of particles with or without electrical characteristics (polarity etc.), parameters such as entropy, kinetic energy, temperature, pressure etc. etc. are assigned to this ensemble. A major feature of this ensemble is that the particles have to be sufficiently close to each other to collide elasticly , no collisions, no thermal process. In addition polar molecules interact with EM waves (infrared photons)and because they are polar some or all of this energy is converted into mechanical energy that is stored in the various oscillatory modes (vibration etc.). The source of this photon is not to be specified, a molecule a million miles away, a free electron laser etc., sure there is some relationship between the source vector and the effective crossection of the molecule but this is not to be confused with the random collision process.

On quantization (a quantum is just “a defined amount”) it is better to stick to the atomic level where only certain orbits (energy levels) are permitted, this is not comparable to the vibrations of polar molecules, not much dfferent from mass/spring oscillations.

I am not an enthusiast for blackbody descriptions in connection with planets, Planck was only concerned with “arrays of radiating antennas” (electrons attached to atoms in a fixed arrangement i.e. a solid. So many things are not blackbodies, conductors, transparent material etc.; liquids and gases come nowhere near this.

I do think you are quite right about the importance of convection. All the modellers using absorption/radiation (A/R) seem not to realise that this is a diffusion process which is VERY SLOW, it cannot account for the difference of surface temp. and top of atmosphere radiation temp. Convection arises from gravity which also happens to account for the adiabatic lapse rate which in turn, if you look with open eyes, explains the difference between surface temp and top of atmosphere radiation temp. Convection moves the energy from where it arrives to where it leaves, why should anyone want to look for anything else?

Comment by Dermod O'Reilly — 24 Aug 2007 @ 11:56 AM

481. re 472,3,4, et al. Just my two cents worth. The single molecule having temperature discussion has been dormant here for a while, though never resolved; I suspect the positions of the statisticians and the “simple body” folks are intractable. I still think a single molecule zipping around can crash into something and cause a jump (albeit a teeny tiny one) in a thermometer. But, who knows?

I’m also still bothered by the multiple temperatures of a molecule, though starting to comprehend it. Is a “useful definition of vibrational temperature”, which sounds like an abstract construct, “real” temperature? As I’ve asked before, will the measured temperature of a mole of gas with translation and vibration/rotation energy be greater than a mole with just translation (theoretically)?

Comment by Rod B — 24 Aug 2007 @ 12:03 PM

482. #477 I wasn’t using the point about the statistical nature of the 2nd Law as an explanation for infra-red photons reaching the ground from the atmosphere, merely suggesting that there’s nothing inherent in the law that rules it out.
However, a few things I’ve seen recently claim it’s a “physical impossibility”.

As far as I can see, all you can really say about temperature and emissivity is that colder bodies emit less photons at longer wavelengths and hotter bodies emit more at shorter wavelengths, according to a 4th power Law.

Nothing’s colder than the CMB radiation, including the instruments used to detect it, yet they seem to be able to detect the presence of the photons it emits. (O.K, in this case, in the radio wavelength)

Perhaps someone could explain the ground-based cloud measurements in the infra-red too.

e.g:-

“Ground-Based Infrared Remote Sensing of Cloud Properties over the Antarctic”

ASHWIN MAHESH,* VON P. WALDEN,1 AND STEPHEN G. WARREN
Geophysics Program and Department of Atmospheric Sciences, University of Washington, Seattle, Washington (22 August 2000)”

Obviously the contrast with a cold sky is important, but they needed to adjust their radiometers to avoid the
667-700 cm2 wavelengths of the atmosphere which would otherwise have swamped their readings, so these are detectable from the surface too, although perhaps just from close to the instrument.

The ARM observatory also does a lot of work in this area.

Comment by Alex Nichols — 24 Aug 2007 @ 12:43 PM

483. Fascinating post (471) Dermod. As a scientific skeptic, it is appealing, but…. if the downwelling IR radiation is severely constrained, where does the 320-350 watts/m>sup>2 shown on various radiation budget diagrams come from? Or, if the actual downwelling is a mere fraction of those numbers, this is not just a debate over relative degrees — but most of the fundemental GHG theory just got canned. This seems a bridge way too far.

On the other hand, this is the area of my biggest question. Intuitively it’s hard to comprehend how the IR surface absorbed downwelling is almost as much as the surface emitted IR and almost twice the absorbed solar radiation. The explanations, while maybe proven correct, are extremely complex and call upon whole bunches of the most esoteric physics, and to a simple Iowa farm boy sound a little bit like the HPFM process..

A minor point: I have no difficulty seeing the earth as a near blackbody in the IR spectrum. The Earth absorbs about 85% of solar radiation reaching it, and most of the stuff that reflect solar rays absorb IR radiation.

Comment by Rod B — 24 Aug 2007 @ 12:49 PM

484. > single molecule zipping around can crash into something

Then, we are no longer dealing with a single molecule, and we have a measurable temperature.

Comment by Hank Roberts — 24 Aug 2007 @ 1:41 PM

485. Re 483 You have problems with radiation budgets? One of my problem is: which one to believe? I tend to use the same one as the IPCC : http://ipcc-wg1.ucar.edu/wg1/Report/AR4WG1_Pub_Ch01.pdf (page 4 of 36 Fig.1) that anyone should suggest that the surface could radiate 390W/m2 into an atmosphere is at most 50C cooler beggars belief. Battered by this absurd suggestion I am driven to silence when I see that this cold atmosphere radiates 324W/m2 into a surface that is warmer. There is nothing in this IPCC citation that would indicate where “CO2 caused” global warming is coming from.

Other models propose that radiation is absorbed and re-emitted, they also rely on downward radiation and they are no more believable.

[Response: Now downward radiation doesn’t exist? What pray have people been measuring then? – gavin]

Comment by Dermod O'Reilly — 24 Aug 2007 @ 3:00 PM

486. Re #484

True!

But don’t forget the saying accredited to Richard Feynman:
Anyone who claims to understand quantum mechanics is either lying or crazy.

Comment by Alastair McDonald — 24 Aug 2007 @ 3:29 PM

487. Rod B (#481) wrote:

I’m also still bothered by the multiple temperatures of a molecule, though starting to comprehend it. Is a “useful definition of vibrational temperature”, which sounds like an abstract construct, “real” temperature? As I’ve asked before, will the measured temperature of a mole of gas with translation and vibration/rotation energy be greater than a mole with just translation (theoretically)?

Well, I tend to think that my hand is an abstraction.

I speak of it as a separate object which nevertheless belongs to me and which I use to perform various tasks – such as typing – so that it has functions which I find useful. But it isn’t really separate, is it? It is a part of a whole which is potentially separable but not actually separate, and if it were separate, it wouldn’t really be my hand any more, would it? At least not in any normal functional sense, although I suppose I might still have uses for it. But there are also properties and attributes which aren’t actually separable “except in thought only.”

We can speak of the length of an object, but the length isn’t something which exists apart from the object which is long, and when we choose to call one “thing” length, another width and another height, it really all is a matter of perspective now, isn’t it? All of these aspects are real in the sense that they exist, but they exist as different aspects of the same entity. Concepts such as “temperature” permit you to look at things from a certain perspective, regarding them in a certain way, where you are able to bring some of those aspects from the background into the foreground as an object of consideration through a process of articulation.

So I guess the long and the short of it is that I regard all of those temperatures as abstractions in one sense or another, but the kinetic undoubtedly has a wider range of applicability and is what we will know first – given the nature of the human beast. As for the measured temperature of your mole, how do you propose to measure it, theoretically or otherwise?

Comment by Timothy Chase — 24 Aug 2007 @ 11:19 PM

488. 484 & 486: Well that’s taking the quantum mechanics postulate — it ain’t there ’till we measure/look at it — to fun but ridiculous extremes. I’m safe as I do not claim to comprehend quantum mechanics. I think some of the earlier quantum pioneers said roughly the same thing.

Comment by Rod B — 24 Aug 2007 @ 11:45 PM

489. Re #485

Dermond,

Basically you are correct. The air at the base of the astmophere is warmed by absorption of blackbody radiation from the surface of the Earth. Gavin’s scheme where the air re-emits the absorbed radiation and is then heated by conduction with the surface only applies to the continuum radiation produced by water vapour vapour and its evaporation which provides a means of conduction through latent heat.

The down-welling radiation that Gavin’s pals are measuring is mostly from clouds. The bit that they can’t find – sometimes called The Energy Balance Closure Problem or The Blue Skies Anomaly – is from CO2, because it does not radiate back to Earth, only to space where it cannot be absorbed.

Comment by Alastair McDonald — 25 Aug 2007 @ 4:15 AM

490. Alastair McDonald (#489) wrote:

The down-welling radiation that Gavin’s pals are measuring is mostly from clouds. The bit that they can’t find – sometimes called The Energy Balance Closure Problem or The Blue Skies Anomaly – is from CO2, because it does not radiate back to Earth, only to space where it cannot be absorbed.

The blue skies anomaly is a phenomena involving the absorption of solar radiation, not the climate’s thermal emissions.

Are you saying that CO2 doesn’t reradiate in the earth’s atmosphere?

That would be hard to square with satellite images at over 2000 wavelengths, limb measurements and those being performed by planes, not to mention those done in the labs – under conditions involving the same physics. I believe the climatologists and radiation physicists have a fairly solid understanding of the reemission by carbon dioxide – right down to the quantum physics which is involved.

Don’t you agree, captain?

Comment by Timothy Chase — 25 Aug 2007 @ 9:59 AM

491. Dermond (485), I think Planck radiation is dependent only on the temperature of the radiating body, not on any other body or stuff. That’s how a hot sun radiates a bunch into a pretty cold “ether”.

Timothy (487), don’t be silly [;-). The reference said it was “defining a useful” term. That sounds like an abstraction. “Temperature” (ought to) refers to the non-abstract (and let’s not get all bollixed up with Plato, Satre, et al) feeling/measurement we call hot or cold. It ain’t complicated. If someone wants to come up with a useful term for something that is neither “hot” or “cold” they shouldn’t use “temperature” but rather invent a term not so confusing, like “goowan”.

Does vibration/rotation energy make the mole of gas “hotter” or not? Measure it any way you wish (while accounting for the “hotness” change resulting from the measurement itself.)

Alastair (489), cloud down welling sounds plausible, but is there enough energy in liquid/solid water in the sky to account for the roughly 320 watts/m2 of down welling? Sounds like a long row to hoe.

Comment by Rod B — 25 Aug 2007 @ 11:07 AM

492. Rod B (#491) wrote:

Timothy (487), don’t be silly [;-). The reference said it was “defining a useful” term. That sounds like an abstraction. “Temperature” (ought to) refers to the non-abstract (and let’s not get all bollixed up with Plato, Satre, et al) feeling/measurement we call hot or cold. It ain’t complicated. If someone wants to come up with a useful term for something that is neither “hot” or “cold” they shouldn’t use “temperature” but rather invent a term not so confusing, like “goowan”.

Actually I was being quite serious – on a number of different levels.

With regard to the different temperatures, they are all aspects of the same thing, and therefore unified in their existence by the thing which “possesses” them – much like the dimensions of length, width and height. Not only are they related, but intimately so through various transformations of energy from one form to another in accordance with physical laws. On a number of occasions, Tamino has even pointed out that at a fundamental level in terms of statistical mechanics they may be defined in the same way.

In fact, I believe the following was the incantation of illumination that he cast:

Perhaps once one retires, as both you and Alastair have, you begin to feel that the world doesn’t have quite so much use for you anymore. That isn’t true – but lets set that aside for a moment.

Feeling this way, both of you would appear to enjoy playing all sort of mischief so as to insure that world knows you are still here. But I know. So do the rest of us. You may also feel that with the darkening of the twilight you have little investment in what occurs forty or a hundred years hence. But wouldn’t you like to be remembered – and to leave a benevolent legacy for those who come after you?

I know I would – and I do.

According to legend, King Arthur will return when England needs him most.

We have need of you now. Please find a way to make yourself useful. Seeing the sorts of games you play I know that you have untapped intelligence and skill. Having played the games you have, you would be quite adept at uncovering the games that others play. And having learned what you have of arts as diverse as diverse as ethics and physics and human psychology, you could cast a light so that others may follow.

We are now living in times of shadow and darkness – and the world has need of your good will.

You have the time. Please give this – and us – some consideration.

Comment by Timothy Chase — 25 Aug 2007 @ 2:19 PM

493. Gavin,

The Latex library at:

/usr/www/users/realc/latexrender/class.latexrender.php

… seems to have a bit of a problem. As this works with other LaTexs, I would assume that the problem has something to do with the changes which have been made, quite possibly in Unix directory structure, having inadvertently broken something.

[Response: I have no clue why it is no longer working. Directory structure is fine, but no output is being generated in the tmp file, hence the errors. I’ll think about this some more when I get a chance. Sorry… – gavin]

Comment by Timothy Chase — 25 Aug 2007 @ 2:36 PM

494. gavin (inline to #493)

Response: I have no clue why it is no longer working. Directory structure is fine, but no output is being generated in the tmp file, hence the errors. I’ll think about this some more when I get a chance. Sorry…

No worries – and my apologies.

Honestly, I regard your time as rather valuable and this is quite small in scheme of things. Like others no doubt, I am just grateful that you are willing to spend time on Real Climate, its essays and participate in the discussions.

Thank you.

Comment by Timothy Chase — 25 Aug 2007 @ 4:33 PM

495. Timothy, O.K. not silly, but awfully esoteric. I understand how a thing possesses its length, width, and height; but how does it possess its length, length, and length? The molecule has some quality called temperature that stems from different energy levels of vibration… and another temperature from rotation levels… and another from translation. It’s doubly confusing since the amount or energy in each degree of freedom is 1/2kT ! So, does a molecule actually have different real temperatures among its energy stores? At least until it equalizes the energy stores ala equipartition? And then are the temperatures equal?

Looking just at translation T = mv2 / 3k. When all degrees of freedom get energy to their fullest is the temperature of the molecule (shuddup you statistical guys!) now T = mv2 / 7 k, for example (using just the translation kinetic energy)? Or do you now also add the 1/2mv2‘s and (1/2)Iw2‘s of the vibrations and rotations and then divide by 7k to get T. What it T now? Is it still different? Or does the energy get added to vibration and rotation without a change of T?

Same question I’ve been asking for some time, with different answers depending on the day of the week. You’re (all) giving me an astute analysis of the progression and development of the game replete with an in-depth psychology affecting the players. I just want to know the score. And I do mean “all”; Not picking on Timothy by a long shot. No one on RC nor none of the web-based forums, university lectures, astute papers and lab treatises that I’ve found (so far) simply tell (and describe) the score, just recite stuff that really does sound like incantations. (Maybe it’s me…)

Tamino might be right, but I like my Dad’s theory of the great benefit of getting old: you no longer have to be nice to people!.

Comment by Rod B — 25 Aug 2007 @ 9:14 PM

496. Try asking what coursework or material you’d need to understand to understand their “incantations” — usually there is some science education at some level required and missing, when the answers seem indistinguishable from magic. Since you’ve said you want a Newtonian explanation and don’t want any quantum mechanics you may be asking people who are too young to answer your question in the terms you’re requiring. Seriously, everything I’ve seen looking into the answers, once I start reading the footnotes and following references, does get into quantum math.

Comment by Hank Roberts — 26 Aug 2007 @ 12:08 AM

497. Rod b. (#495) wrote:

Same question I’ve been asking for some time, with different answers depending on the day of the week. You’re (all) giving me an astute analysis of the progression and development of the game replete with an in-depth psychology affecting the players. I just want to know the score. And I do mean “all”; Not picking on Timothy by a long shot.

Rod, I am serious.

At the same time, I was also trying to be as polite as possible while calling attention to the obvious. Beyond a certain point, it became obvious that you weren’t directing questions to Alastair because you thought him an expert but because you wished to bait Alastair into arguing against well-established physics. You knew that there were people who have every right to claim a degree of expertise, and while directing some questions to them, you were directing others to him oftentimes in the same post for the purpose of stirring up trouble for your own entertainment.

To the extent that you did this, you were gaming him and us. But Alastair had his own game, so I suppose he at least should not complain. Each of you had a way of playing on sympathy and goodwill.

Honestly, I think that both of you could put your intelligence and talent to better use. This isn’t a game. The people here are trying to make a difference. You may or may not like them or identify with them. It shouldn’t matter – given what other people will be facing in the coming decades. And while I may seem naive to you, I am not so naive as to think that you will actually want to help. I am simply pointing out that it is an option.

Regardless, I strongly suspect that the old games are over.

Comment by Timothy Chase — 26 Aug 2007 @ 2:44 AM

498. Re #490.

Timothy, you are correct. The Blue Skies anomaly is for solar not terrestrial radiation. I spotted that after I had posted the message.

I am saying that the radiation to space in the 15 micron band is emitted from the Non LTE region at the top of the atmosphere. In lower regions of the atmosphere there is no radiation in that band because the time between collisions is shorter than the half life of the vibrational excited molecules.

This has been said before by Jack Barrett “The roles of carbon dioxide and water vapour in warming and cooling the Earth’s troposphere” Spectrochima Acta Vol. 51A, No. 3, pp. 415-417, 1995.

He argued that this meant increasing CO2 would not cause global warming, and so was not believed.

Comment by Alastair McDonald — 26 Aug 2007 @ 2:46 AM

499. Re #482 Remote infra-red sensing studies.

I made a mistake in the units I quoted from the paper, when transferring a snippet from pdf to text.

It should read ‘667-700 cm-1 wavelengths’, not cm2, which would be a very strange wavelength!

Part 1 of the paper is at:-
http://www.atmos.washington.edu/~sgwgroup/sgwReferences/maheshWaldenWarren2001I.pdf

There’s a technical description of an IR radiometer from the manufacturer here, which may help elucidate exactly what they are measuring with modern instruments: –

http://www.yesinc.com/products/data/tir570/index.html

Comment by Alex Nichols — 26 Aug 2007 @ 3:17 AM

500. Alastair McDonald (#498) wrote:

I am saying that the radiation to space in the 15 micron band is emitted from the Non LTE region at the top of the atmosphere. In lower regions of the atmosphere there is no radiation in that band because the time between collisions is shorter than the half life of the vibrational excited molecules.

This has been said before by Jack Barrett “The roles of carbon dioxide and water vapour in warming and cooling the Earth’s troposphere” Spectrochima Acta Vol. 51A, No. 3, pp. 415-417, 1995.

He argued that this meant increasing CO2 would not cause global warming, and so was not believed.

Yes – he has been on the list of “skeptics” for quite some time.

Have you considered the fact that if “the time between collisions is shorter than the half life of the vibrational excited molecules” prevented a molecule from gas from reemitting, then water vapor could not reemit at the surface?

At one level, the argument which you are making regarding collision rates and half-lifes has a certain logic to it.

If a molecule could reemit only at or after the half life but was interrupted each time, then carbon dioxide and water vapor could not reemit under the very same circumstances that physicists say is required to achieve local thermodynamic equilibria. But the term is “half life.” This means that it is probabilistic – and has no memory of how long the molecule has been in the excited state. It parallels subatomic particle decay. This sort of thing is fairly basic, and should be dealt with in any undergraduate level course which touches on quantum mechanics even at the most introductory level.

If the gas is a certain temperature, then a certain percentage of molecules will be in the excited state (which will largely be a function of the Maxwell velocity distribution, at least under local thermodynamic equilibria conditions), and it does not matter which molecules are in the excited state or how long they have been so. Over a given duration, a certain percentage of them will reemit as determined by a law of exponential decay. Thus it is not the molecule which reemits, but one molecule which absorbs, and then after so many collisions another which emits, so that it is only the gas, the population, which reemits by both absorbing and emitting the radiation.

However, even if it made theoretical sense in terms of the physics that somehow a high rate of collision would prevent reemission, there would still be the lab experiments that we have been performing for over a century, the satellite images of reemission at various parts of the spectra and at various altitudes, and the measurements of reemission performed by planes.

As it is, theories, lab experiments and empirical measurements of reemission the atmosphere are all in agreement. The science is about as solid as it gets.

Comment by Timothy Chase — 26 Aug 2007 @ 12:43 PM

501. This may help:
http://eo.ucar.edu/skymath/tmp2.html
_______ begin quote _______

Prior to the 19th century, it was believed that the sense of how hot or cold an object felt was determined by how much “heat” it contained. …

Joule conclusively showed that heat was a form of energy. As a result of the experiments of Rumford, Joule, and others, it was demonstrated (explicitly stated by Helmholtz in 1847), that the various forms of energy can be transformed one into another.

When heat is transformed into any other form of energy, or when other forms of energy are transformed into heat, the total amount of energy (heat plus other forms) in the system is constant.

This is the first law of thermodynamics, the conservation of energy.

The energy associated with motion is called Kinetic Energy and this kinetic approach to the behavior of ideal gases led to an interpretation of the concept of temperature on a microscopic scale….

Temperature becomes a quantity definable either in terms of macroscopic thermodynamic quantities such as heat and work, or, with equal validity and identical results, in terms of a quantity which characterized the energy distribution among the particles in a system. (Quinn, “Temperature”)

The nature of radiation has puzzled scientists for centuries. Maxwell proposed that this form of energy travels as a vibratory electric and magnetic disturbance through space in a direction perpendicular to those disturbances.

The great question at the turn of the century was to explain the way this total radiant energy emitted by a black body was spread out into the various frequencies or wavelengths of the radiation. Maxwell’s “classical” theory of electromagnetic oscillators failed to explain the observed brightness distribution. It was left to Max Planck to solve the dilemma by showing that the energy of the oscillators must be quantized, i.e. the energies can not take any value but must change in steps, the size of each step, or quantum, is proportional to the frequency of the oscillator and equal to hv, where h is the Planck constant.

————– end quote ——————-

Comment by Hank Roberts — 26 Aug 2007 @ 2:14 PM

502. Timothy, me thinks thou doth psychoanalyze too much. I’ve been asking a simple straight-forward (albeit theoretical) question: Does energy in the vibrational and/or rotational bands increase the temperature (as commonly defined) of the gas? Maybe it’s me, but the question doesn’t sound terribly sinister and encumbered with deep-seated dark motivations to me.

I don’t want to help to do what, exactly? Questions about the temperature of a gas (to which nobody seems to have an answer and seems important to GW theory) or further exploring the claim that IR down-welling is in effect nonexistent (which blows a pretty big hole in GW (not just AGW), at least until scientifically rationalized), what — show I’m not a serious player who just wants to jab folks? Sorry! If you don’t have the answer, that’s perfectly O.K. and carries no negative connotations what-so-ever, but it doesn’t make me the villain. I don’t have the answer either.

Hank, it seems to me that, since gas is all about T, V, P, and n, something directly affecting the temperature (again, as commonly defined) shouldn’t have to be dug out from obscure references and footnotes in papers. Or maybe the answer to my question is just not known for sure — I’ve seen answers both ways from presumably reliable sources.

Nor, Hank, do I fear quantum mechanics (though I’m a long way from expert knowledge), but it does seem to be used to obfuscate or avoid the question at hand at times, even if in a congenial way We could debate the “single molecule has temperature” thing all day without getting any closer to the question at hand.

Comment by Rod B — 26 Aug 2007 @ 2:33 PM

503. So, Rod, is this part clear?

“When heat is transformed into any other form of energy, or when other forms of energy are transformed into heat, the total amount of energy (heat plus other forms) in the system is constant.”

— “energy” is not “heat” is not “temperature” —

This also may help:

____________________________

The point of the rather tedious calculation above, finding exactly how many different modes of vibration, or degrees of freedom, there are in the ovenfull of radiation, was to construct an argument parallel to that applied successfully to understand the properties of gases. The first conclusion would be that there is kT of energy, on average in each wave mode (not ½kT, because the standing wave is like an oscillator with both kinetic and potential energy).

For low frequencies, this is exactly what is observed. The amount of radiation shining out of the oven at low frequencies is correctly predicted by counting modes, as above, and allocating kT of energy to each mode.

The Ultraviolet Catastrophe

The problem is that as we go to higher frequencies, there are more and more possible degrees of freedom. The oven should be radiating huge amounts of energy in the blue and ultraviolet. But it isn’t. For this system, the equipartition of energy isn’t working! ….
____________________________________

Comment by Hank Roberts — 26 Aug 2007 @ 2:55 PM

504. Rod B (#502) wrote:

Does energy in the vibrational and/or rotational bands increase the temperature (as commonly defined) of the gas? Maybe it’s me, but the question doesn’t sound terribly sinister and encumbered with deep-seated dark motivations to me.

In response to this question: no, the energy in the vibrational and rotational modes does not increase the translational temperature, whereby “commonly defined” implies translational – so longer as the energy in the vibrational and rotational remains in those modes. But it does get reemitted or lost through collision, in which case it will raise the translational temperature assuming one does not already have a local equilibrium.

Rod B (#502) wrote:

Timothy, me thinks thou doth psychoanalyze too much.

I disagree.

The very act of communication on the part of the individual who is listening or reading what someone else has to say is a matter of understanding their intentions – of identifying what it is they intended to communicate. Someone who does not understand this or who is simply disregarding the motivation of others even when those motives become readily apparent is walking blindfolded listening only to the sounds about them.

Someone who is genuinely interested in the promotion of dialogue will be aware of the the psychological elements. So for example, if you engage in internet debates long enough, you come to realize that after a certain point a large part of what continues to fuel a given debate between individuals will be a matter of saving face.

If I am arguing with someone that I care about, I will let them save face – if I consider the results acceptable. If I am arguing with someone that I do not particularly care for, I will let them save face – given the same conditions.

A great deal of argument is about self-assertion which we will tend to view as a form of strength, but there is also strength in self-restraint – and permitting individuals to save face is one of the ways in which one exercises self-restraint. However, I do not believe it is wise to let your act of good will be taken advantage of.

I believe that you are basically a good person. However, you can get rather mischievious. On various occasions, it seemed as if Alastair was going to quit offering his own alternate raditional “theory” – which he is no more qualified to do than either you or I, it seemed quite clear that the questions you were asking and even directing to him were intended to get the whole argument going again. Particularly the further things went along.

To be perfectly honest, I realized you were trying to stir up trouble some time ago, but I didn’t mind so much so long as it was occasional. Not that big of a deal. Up to a point, I could even understand the motivation. But it became a well-established pattern, and then the pattern became tiresome.

However, I was unable to bring it out into to the open so long as I was unwilling to pay the price in terms of how it would affect someone else. Their behavior justified that paying the price. So I decided it was time to bring it out into the open.

At the same time, I believe you had other reasons for being here. I strongly doubt it was all games.

However, when you stated (#495)…

Tamino might be right, but I like my Dad’s theory of the great benefit of getting old: you no longer have to be nice to people!.

… that did not reflect particularly well on you.

Rod B (#502) wrote:

I don’t want to help to do what, exactly?

I believe that you understand enough of the physics to realize what the implications are for future generations. And at the very least you could have ceased in encouraging Alastair to contradict the science.

Rod B (#502) wrote:

Questions about the temperature of a gas (to which nobody seems to have an answer and seems important to GW theory)…

Unless I am mistaken, all of the questions you asked regarding temperature were answered by Eli, Ray and Tamino – who have a fair amount of expertise in the area. Additionally, literature was pointed out.

Rod B (#502) wrote:

…or further exploring the claim that IR down-welling is in effect nonexistent (which blows a pretty big hole in GW (not just AGW), at least until scientifically rationalized), …

You know better – and you have for quite some time.

As I have pointed out to him on numerous occasions, we have infrared images of IR emissions by carbon dioxide at 8 km – and at various other altitudes and at various channels. Likewise, we are flying planes which take measurements demonstrating downwelling IR emissions by carbon dioxide at various altitudes.

Rod B (#502) wrote:

… what — show I’m not a serious player who just wants to jab folks? Sorry! If you don’t have the answer, that’s perfectly O.K. and carries no negative connotations what-so-ever, but it doesn’t make me the villain.

I can understand playfulness, but what you did in terms of your interaction with Alastair went well beyond this. So while I would not consider you a villian, I believe your efforts are misdirected and consciously so.

Comment by Timothy Chase — 26 Aug 2007 @ 4:46 PM

505. RE #500 Where Timothy wrote “As it is, theories, lab experiments and empirical measurements of reemission the atmosphere are all in agreement. The science is about as solid as it gets.”

Nowhere in that post is there any reference or citation of a theory, lab experiment, or empirical measurement. This whole thread is based on a the denial of an experiment that was undertaken.

I don’t know whether you have access to the book by Goody and Walker, but it admits that the simple model explained by Gavin does not work for Venus. If you do the sums with an albedo of 80% you will find that the surface temperature is just not feasible. So that theory is wrong!

If you are going to believe everything your are told, of course you will be told the models work. They will gloss over the Venus problem, the Tropical Lapse Rate Problem, and the Energy Balance Closure Problem, and even the Martian polar inversion layer problem (hardly recognised yet) but which Raypierre is going to “eliminate” from his book. See response to #58.

Comment by Alastair McDonald — 26 Aug 2007 @ 5:02 PM

506. Lectures based on Goody & Walker’s 1972 book Atmospheres:

“There are multiple different tables of composition – not surprisingly, our understanding of the composition of the atmospheres of Venus and Mars have developed since Goody & Walker wrote Atmospheres in 1972.”

“… But we just want a simple comparison of Earth, Venus and Mars -Isn’t there a simpler way of looking at this?
For a THIN atmosphere – e.g. Earth, Mars – we can use a simpler model. The book has this…

Comment by Hank Roberts — 26 Aug 2007 @ 5:39 PM

507. Rod, part of the problem is that the notion of temperature of a single molecule does not make sense. The fact of the matter is that if you are talking about temperature in the thermodynamic sense, then the presence of molecules is irrelevant, and if you are talking about temperature in the sense of stat mech, you inherently assume a large number of molecules at or near equilibrium.
The relationship between energy and temperature is not as simple as 0.5mv^2 per degree of freedom. For instance, I can add energy to a system of ice, water and water vapor at the triple point and not change the temperature.
Adding energy to the vibrational/rotational excited states will probably increase the gas temperature–but probably through the relaxation process. That is, most of the vibrationally excited atoms will relax via collisions with other molecules (CO2 or other). Some will do so radiatively. A very few of the molecules accelerated by collisions with excited molecules will excite vibrational modes in other CO2 molecules. There is a continual competition between different processes, and small changes (like changes in CO2) can have a big effect on the ultimate balance in that competition. Don’t expect this to be simple.

Comment by ray ladbury — 26 Aug 2007 @ 5:50 PM

508. Alastair, it’s not “denial” to repeat an experiment with better instruments, and to extend the conditions.

The original post is to “discuss Herr Koch’s experiment in the light of modern observations.The discussion here is based on CO2 absorption data found in the HITRAN spectroscopic archive.”

Look again: http://www.realclimate.org/images/TransLongPaths.jpg
That’s science — revisiting the experiment, considering more recent work — not denial.

Comment by Hank Roberts — 26 Aug 2007 @ 5:58 PM

509. Alastair McDonald (#505) wrote:

RE #500 Where Timothy wrote “As it is, theories, lab experiments and empirical measurements of reemission the atmosphere are all in agreement. The science is about as solid as it gets.”

Nowhere in that post is there any reference or citation of a theory, lab experiment, or empirical measurement. This whole thread is based on a the denial of an experiment that was undertaken.

Lets look at lab experiments first:

The discussion here is based on CO2 absorption data found in the HITRAN spectroscopic archive. This is the main infrared database used by atmospheric radiation modellers. This database is a legacy of the military work on infrared described in Part I , and descends from a spectroscopic archive compiled by the Air Force Geophysics Laboratory at Hanscom Field, MA (referred to in some early editions of radiative transfer textbooks as the “AFGL Tape”).

Part II: What Ångström didn’t know
26 June 2007
Ray Pierrehumbert

Then of course you can check the AIP history by Spencer Weart:

The Discovery of Global Warming
The Carbon Dioxide Greenhouse Effect
August 2007
by Spencer Weart
http://www.aip.org/history/climate/co2.htm

Spencer Weart provides the following historical reference:

Martin, P.E., and E.F. Baker (1932). “The Infrared Absorption Spectrum of Carbon Dioxide.” Physical Review 41: 291-303.

… reviewed in:

Smith, R.N., et al. (1968). Detection and Measurement of Infra-Red Radiation. Oxford: Clarendon.

But if you look up what Raypierre mentions, you find that they are performing laboratory measurements and that these measurements are amazingly accurate nowadays:

We require these things, because of the instrumental capabilities of NASA satellites, to have incredible accuracy. That means we want to know line positions to one part in 10 million or even better. That’s a tough demand. We want to know the intensities to better than two percent. That’s tough, too. There are so many sources of error. For example, the pressure and the temperature in the cells that we’re using to measure these quantities. We’re pushing the envelope.

An interview with: Dr. Laurence Rothman
September 2004
http://cfa-www.harvard.edu/hitran/

Barton Paul Levenson also points out that they have even been done with mixtures. (#376)

Now with respect to “empirical measurements of reemission the atmosphere,” I had provided the following earlier in this thread:

#381
NASA AIRS Mid-Tropospheric (8km) Carbon Dioxide
July 2003
http://www-airs.jpl.nasa.gov/Products/CarbonDioxide/

#384
AIRS – Multimedia: Videos: Animations
http://airs.jpl.nasa.gov/Multimedia/VideosAnimations/

Then of course there is the US military article which Allan Ames originally brought to our attention:

USERS’ MANUAL FOR SAMM2,
SHARC-4 and MODTRAN4 MERGED
H. Dothe, J. W. Duff, J. H. Gruninger, P. K. Acharya, A. Berk
AFRL-VS-HA-TR-2004-1145
Environmental Research Papers, No. 1145
http://www.dodsbir.net/Sitis/view_pdf.asp?id=DothH04.pdf

It contains CO2 thermal emission images of the atmosphere as well. And some time ago I pointed out the fact that measurements were even being taken mid-atmosphere by planes and refered you to where you could get the data. Theory? Try quantum mechanics, radiation transfer theory, molecular spectroscopy, etc. You will get plenty of it – at least if you go to a university library.

Alastair McDonald (#505) wrote:

Alas I don’t know whether you have access to the book by Goody and Walker, but it admits that the simple model explained by Gavin does not work for Venus. If you do the sums with an albedo of 80% you will find that the surface temperature is just not feasible. So that theory is wrong!

Well, the simple model is simple. Of course it doesn’t work for Venus. Its conceptual. And if you will notice, when Gavin brings it up, he points out that there are some very fundamental problems with attempting to apply it to earth – such as the fact that it doesn’t incorporate the role of moist air convection. In fact its atmosphere is infinitely thin. But the Goddard Institute of Space Studies began by modeling the climates of other planets. They have actually done a fairly good job with Venus, I understand.

Alastair McDonald (#505) wrote:

If you are going to believe everything your are told, …

No, but I tend to trust the experts and I tend to trust in the fallibilistic, self-correcting process of science. Are you suggesting that the experts are untrustworthy? If so, how many different fields? Have they organized a conspiracy? Would you recommend that I trust the opinions of a retired ship captain who has decided to set himself up as an alternate authority on account of his poorly thought-out pronouncements of armchair physics instead?

Alastair McDonald (#505) wrote:

… of course you will be told the models work. They will gloss over the Venus problem, the Tropical Lapse Rate Problem, and the Energy Balance Closure Problem, and even the Martian polar inversion layer problem (hardly recognised yet) but which Raypierre is going to “eliminate” from his book. See response to #58.

Told the models work? Don’t know if you’ve noticed, but the literature does a pretty good job of pointing out where they are weak. So does Gavin, I believe. Venus? Dealt with it. Tropical Lapse Rate? Don’t know that much about it, but I began digging earlier this weak. It should help in understanding the expansion of the Hadley cells which is bringing drought our way. The things are expanding a bit more as the result of global warming than we initially predicted. Energy Balance Closure Problem? Sounds like a book-keeping issue of checking to make sure that all of the totals add up. Minor, judging from what I’ve seen, but somewhat thorny as you have to take into account all the effects (e.g., turbulence, time of day) and their interaction.

Science doesn’t have all the answers, but just because we don’t know everything doesn’t mean that we know nothing. Science progresses. Incidentally, where are your field experiments?

Raypierre?

I pointed out that you had misread his diagram in #58 with respect to the peaks and #60 with respect to the wavelengths. But normally things are graphed in terms of wavenumber, so perhaps some of your paranoia directed at Raypierre was understandable. However, he would still like to use a better one.

*

I understand the fact that you have questions and problems regarding how it all works. I do too. I am not a physics major, just a philosophy major who is now a computer programmer. But to have questions and problems is not the same thing as having an “alternative theory” and it does not justify arguing as if the physics isn’t well-established, the experts don’t know what they are talking about and somehow you do. Your confusion, the fact that something does not as of yet make sense to you, does not justify your claim to expertise.

*

You have been across the board in terms of your positions.

You claimed that carbon dioxide could not reemit except at the surface. You claimed that it could not reemit in the majority of the atmosphere because local thermodynamic equilibria presumably do not exist in the earth’s atmosphere. So according to you non-LTE prevents reemission. Then you claimed that the top of the atmosphere is non-LTE, and therefore only it could reemit. Then you have claimed that it is the rate of collision being too high which prevents reemission (i.e., hot LTE conditions), but you have also claimed that those parts of the spectra in which reemission would occur get frozen out – meaning that the temperatures are too low for reemission. You have claimed that the emissivity of carbon dioxide will suddenly jump from logarithmic to linear at the surface as the result of higher levels of carbon dioxide – resulting in a catastrophic runaway effect. You have claimed that reemission was due to electrons leaving their excited states when within the context of our atmosphere, reemission is due to molecules leaving their excited states. You have even gone so far as to claim that it was the nucleus which reemits.

The fact that you have jumped from position to position with such speed is itself an admission that you do not know and that you do not understand. And yet you have continued to claim expertise at various points and had been unwilling to ever say what your actual educational and professional background is. I had to look it up.

*

Back in 476 you tried to reverse our positions, claiming for example that I was unaware of some of the more esoteric details of the relevant radiation physics (namely, non-translational temperatures). I didn’t have a problem with this (and indicated as much in 479) – so long as you were willing to quit arguing with what is both theoretically and empirically well-established. I didn’t have a problem with this if it helped you save face – and meant that we could focus on science. The fact is I like you, but I was also willing to let someone else save face rather recently because it seemed the most likely way to get him to go elsewhere.

However, despite all of the evidence and detailed explanations which you have received, despite your desire to avoid conflict and apparent willingness to accept the scientific literature as indicated in 476, you were back to the old game by 489. I’d been played.

Currently your attempts at deligitimizing the advanced state of our knowledge in this area are as bad or worse than what we get from those who would deny that there is such a thing as the greenhouse effect. But you come across as someone who believes that such an effect exists, simply not in the way that the scientific community understands it – and therefore make those who recognize the existence of the greenhouse effect appear confused – when the confusion is entirely of your own making. As such you are giving those who oppose the science a gift which they couldn’t possibly hope to achieve on their own.

There are other effects. For example, when you derail discussions of the actual science into discussions of your ill-informed and often incoherent views. But this is really less important.

*

All of this is so unnecessary.

You have demonstrated the ability to pick up the science, to make the connections, to understand – when you weren’t looking to pick arguments with the experts on anything thath didn’t immediately make sense to you. Honestly you could be quite an asset. You even have been at times. But within the context of the debates in this forum with those who would deny the science, you have often been their greatest ally.

Comment by Timothy Chase — 26 Aug 2007 @ 7:25 PM

510. Timothy, me still thinks thou doth analyze too much, and it’s getting tiresome. Sometimes when a student asks a question it’s beneficial to accept it at face value and attempt an answer. Yes, he probably has peer pressure to do stuff, difficulty with his parental relationships, a pile of non-specific teen anxiety and currently really wanting to show off for the cute little Jr. in the first row. But he might just want to know how to factor. Getting yourself all wrapped around the axle and tied up in your jock over secondary sideshows is not particularly helpful. Also, you think Alastair will go quietly into the night if I just stopped goading him??!! Shirley you jest! Or are you just PO’d because he didn’t accept your answers carte blanche, so want to blame me? (There I go! Psychoanalyzing — against all my recommendations [;-) )

I have to go study the rest of the recent post(s); looked like they have some good stuff.

Comment by Rod B — 26 Aug 2007 @ 9:51 PM

511. Rod B (#510) wrote (in response to #504, I believe):

Also, you think Alastair will go quietly into the night if I just stopped goading him??!! Shirley you jest! Or are you just PO’d because he didn’t accept your answers carte blanche, so want to blame me? (There I go! Psychoanalyzing — against all my recommendations [;-) )

Honestly I think both of you could be assets, questions and all. But are you admitting that you have “goaded him” as you have just put it? If so, it would appear that my analysis of you was correct. Would things have gone smoothly without your encouraging him? Probably not, but it would seem that things would have gone a great deal more smoothly, perhaps even on a personal level for him.

But the latter is just sheer speculation on my part. I don’t know enough about him to say. Although I do suspect that he is taking all of this more seriously than you are.

But not even the climatologists would simply expect you to simply take their word for anything. Quite the contrary.

Or are you just PO’d because he didn’t accept your answers carte blanche…

What I would prefer on his part is his acknowledgement that the images of reemission in the mid-troposphere exist, the measurements of absorption and reemission have been performed and that we have graphs of the resulting data. So far every time a link has been provided he has refused to acknowledge that such things were made available. Just stony silence. Then he would make the same claims that no such evidence exists or that such reemission does not take place. It has nothing to do with accepting my answers with or without question. What I would also prefer is his acknowledgement of the fact that he is not an expert any more than I am.

Rod B (#510) wrote:

I have to go study the rest of the recent post(s); looked like they have some good stuff.

Frankly there are things that I would much rather be doing than having to deal with this sort of thing – like getting into more of the science. There is a great deal I would like to understand.

Comment by Timothy Chase — 27 Aug 2007 @ 1:13 AM

512. Re #509

Timothy,

If you read Spencers Weart’s history, or Bill Bryson’s “A History of Nearly Everything”, you will find that science is littered with mistakes and wrong turns. For instance, either Arhhenius or Angstrom must have been wrong. So what makes you so sure that the current climate theories are correct? The fact that the RealClimate scientists are so confident is no guarantee. Don’t forget what Max Planck wrote:

A new scientific truth does not succeed, generally speaking, because the opponents are convinced and declare themselves educated, but rather because the opponents gradually die out and the following generation is from the start familiar with it.

The skeptics, such as Jack Barrett, are saying “The models are wrong so they are over-estimate the effects of AGW.” The scientists are saying “We know that changes in climate happened in the past due to CO2, therefore our models are correct.” But you, as a philosopher, must know that neither of these assertions is logically correct. There is the possibility that the models are wrong, and that they are underestimating the effects on the way, which is what I believe.

We know that the Dansgaarg-Oeschger events happened but the models cannot reproduce them, therefore the models are wrong. That statement is logically correct. We know that the models were projecting that the Arctic ice would not disappear until 2070. How likely is that now when you see how much the ice has retreated this summer. The models are wrong again. Yes, the models usually get the lapse rate correct, but that does not mean that the models are correct. They are just fitted to the US standard atmosphere! And the models of Venus are fitted to its lapse rate, because the simple models does not work.

What I am saying is that the science produced by the skeptics is correct. That applies to Jack Barret, Christy & Spencer (tropical lapse rate problem), and now Stephen Schwartz. It is the conclusions that they draw from their facts that is wrong.

You wrote “Honestly you could be quite an asset.” That is what I am trying to be, but I am just a lone voice crying in the wilderness. So long as you apply uncritical faith in the scientists of today, then, in the words of Private James Frazer “We’re doomed!”

Comment by Alastair McDonald — 27 Aug 2007 @ 5:21 AM

513. Re #491 “Dermond (485), I think Planck radiation is dependent only on the temperature of the radiating body, not on any other body or stuff. That’s how a hot sun radiates a bunch into a pretty cold “ether”.”

This is off beam (pun intended) Rod, sorry to put it like this, but if everyone else shares your view there is no way this matter is going to be resolved here. Thermal radiation is no different from any other heat transfer process in the following respect: high energy tends to flows to low; this usually translates as “heat goes from hot to cold”, but the first statement is closer to the 2nd law of thermodynamics than saying “heat”. I notice I have said you are “off beam” about radiative heat and I apologise for this and will try to point you (no more puns) in a suitable direction. The expression used frequently for expressing the radiated energy is, as you know:
Wm2= εσT>4
I have no quarrel with this but we are concerned with the transfer of energy between two places (entities, systems etc.) and the proper expression is:
Wm2= ε σ(T1>4 – T2>4)
because the source (T1) and the sink (T2) temperature play a role, the radiation coupling is reciprocal. When deepspace is involved T>4 is rather small but it should not be forgotten, it matters to somebody. From this expression you can see how the 2nd law of thermodynamics comes into play. Notice that this is for purely radiative transfer, no CO2 or anything else must get in the way. As usual things begin to get rough when you go deeper; since we are dealing with photons there are other considerations that are not immediately obvious, the relevant science concerns the interaction of electromagnetic energy and matter and is called quantum electrodynamics (QED), you probably have heard of this but were not sure how it applied; this article in Wiki. is mainly historical but will lead you however far you want to go http://en.wikipedia.org/wiki/Quantum_electrodynamics .

That word temperature is a big deceiver. I have been watching the debate about “does a molecule of gas have a temperature?” with interest because the basic “thing” is energy. Thermal energy is most happy when it is about atoms, molecules, and particles in general, banging into each other, this is what we think about when we say something is hot or cold. This is a subtle matter and in the world of thermodynamics there is an explanation. In this reference http://en.wikipedia.org/wiki/Talk:Temperature you will find extensive discussion about temperature, in summary it separates the internal energy of the molecules from the kinetic (i.e. translational not rotational etc.) energy. If the definition of temperature seems restrictive it is because this is sometimes necessary when choosing words to identify physical processes.

Comment by Dermod O'Reilly — 27 Aug 2007 @ 6:14 AM

514. Alastair, there is more in science than “right” and “wrong”. Indeed, these terms are not particularly useful in scientific discussions because they really never apply. Descriptions are rather “complete” or “incomplete”. A model can be incomplete and still quite usable–viz. classical mechanics. What it boils down to is evidence. There is zero evidence that our current models have serious problems, and there is plenty of evidence that they perform adequately. The fact that we cannot reconstruct events from the distant past is not that troubling, as the outputs of these models are critically dependent on initial conditions, which are poorly constrained. Where we know initial conditions (Mt. Pinatubo, post 9/11/2001, etc.) the models perform well.
Science progresses at its own pace, and that pace is driven by evidence. Classical mechanics was just as incomplete a description of nature pre Michelson-Morley as post, but it was not until we had the evidence that that incompleteness became important. For now, the models do just fine.

Comment by Ray Ladbury — 27 Aug 2007 @ 8:18 AM

515. Rod, you can always get Alastair going, because he knows Einstein was wrong:
> Blackbody radiation is emitted as a continuous spectrum whereas gases emit line radiation.

This is a guaranteed distraction to any topic at RC — get started in that direction and it goes only in that direction.

On other questions Alastair is often helpful. It’s most likely that current models do underestimate warming, the IPCC is a very conservative group and specifically doesn’t talk about how nature got where it did in past warming events that are known to have happened but aren’t yet showing up in the models.

If you don’t want to talk about how models work and what we may be missing in the way of forcings, bring up radiation transfer, that’s a great way to avoid dealing with what we know has happened in past big warming events that aren’t yet explained or modeled, like the PETM.

As far as I know there aren’t any problems with the models on the “too cold too fast” end —- those work out and seem to be understood. It’s the rapid global warming events that aren’t in the models yet.

Comment by Hank Roberts — 27 Aug 2007 @ 9:29 AM

516. RE #511 Timothy Chase

Quote
“What I would prefer on his part is his acknowledgement that the images of reemission in the mid-troposphere exist, the measurements of absorption and reemission have been performed and that we have graphs of the resulting data. So far every time a link has been provided he has refused to acknowledge that such things were made available. Just stony silence.”
Unqoute

Please would you provide details (and/or web refs) of the wavelengths of the emitted radiation in the mid-troposphere? It would be very interesting to compare such figures with the DoE’s statement(re my #419)that CO2 emits only at longer wavelengths than those at which it absorbs.

I believe this statement to be of fundamental importance to consideration of the enhanced GH effect, and so far in this thread it has largely met with “stony silence”.

The problem seems to be that when an excited CO2 molecule emits another photon, it relaxes to a lower energy level in a rotational mode, but the difference in energy is less than the energy of the initially absorbed photon, and so does not match the requirements for further re-absorption by another CO2 molecule. This seems to negate the basis of the multi-layer models of the atmosphere.

Another question is what happens to the energy from the initially absorbed photon remaining in the lower energy rotational mode? Presumably it finds its way by collisions into the atmosphere and so raises the temperature of the atmosphere? Correct, or not???
But this would have been occurring for ever, unless the atmosphere has a way of converting this thermal energy into radiation which can then escape into space.

Comment by AEBanner — 27 Aug 2007 @ 10:16 AM

517. Mr. Banner, the page you point to is from 1994. The statement you’ve been quoting is one of the few on that page that has no reference/cite/footnote.

I haven’t found that exact statement anywhere else. Maybe it’s outdated, maybe it applies only when both molecules are at the same temperature/pressure/energy level. Offhand it makes sense that on average emission is at a lower energy than absorbtion, as they say — some of the energy will go into other modes. But you could ask:
URL: http://www.eia.doe.gov/cneaf/alternate/page/environment/appd_a.html
they have a contact link at the bottom of the page

>wavelengths of the emitted radiation in the mid-troposphere

If you look at the links he provided, the photographs specify the elevation and wavelength used in the image.

> the atmosphere has a way of converting this thermal energy into radiation which can then escape into space

That’s the whole point, if you start from the top of this thread and read the original post.

Comment by Hank Roberts — 27 Aug 2007 @ 10:50 AM

518. re (511): Timothy, you said (504), “… it seemed quite clear that the questions you were asking and even directing to him [Alastair] were intended to get the whole argument going again…. ” You implied that I am goading Alastair; I never was nor claimed to be.

In 511 you say, “… What I would prefer on his part is his acknowledgement that the images of reemission in the mid-troposphere exist, the measurements of absorption and reemission have been performed and that we have graphs of the resulting data. So far every time a link has been provided he has refused to acknowledge that such things were made available. Just stony silence. Then he would make the same claims that no such evidence exists or that such reemission does not take place.”

Sounds like “…PO’d because he didn’t accept your answers carte blanche…” (my words) to me. You go on to say, “…It has nothing to do with accepting my answers with or without question….” which is directly opposite the words above.

Just though I’d try to clear some things up, as irrelevant as they are.

Comment by Rod B — 27 Aug 2007 @ 11:14 AM

519. Dermod (513), I agree with the post. The Sun radiates a pile of energy to the Earth, and the Earth radiates energy to the Sun — but a trivial amount based on their respective surface temperatures (to the 4th power). This also brings up the (your?) question how the colder atmosphere radiates so much back to the earth. Perhaps it’s not classic blackbody radiation?? Perhaps it’s all coming from the low altitude warm troposphere?? I might have misread your post 485; sorry.

Comment by Rod B — 27 Aug 2007 @ 11:49 AM

520. Alastair McDonald (#512) wrote:

Re #509

Timothy,

I notice that what you have written has no acknowledgement of carbon dioxide reemissions in the mid-troposphere and above, something which is demonstrated by the longwave images. Likewise, there is no admission of the data we have from planes or the results that we have obtained in the laboratories. As far as I can tell, you still hold that carbon dioxide can reemit radiation either at the top of the atmosphere, or depending upon your mood, at the surface. Do you acknowledge the images? Do you acknowledge the data? Do you still claim that climatologists and physicists specializing in radiation transfer have next to no understanding of the radiative properties of carbon dioxide? If so, what are they missing?

Alastair McDonald (#512) wrote:

If you read Spencers Weart’s history, or Bill Bryson’s “A History of Nearly Everything”, you will find that science is littered with mistakes and wrong turns. For instance, either Arhhenius or Angstrom must have been wrong. So what makes you so sure that the current climate theories are correct? The fact that the RealClimate scientists are so confident is no guarantee. Don’t forget what Max Planck wrote:

A new scientific truth does not succeed, generally speaking, because the opponents are convinced and declare themselves educated, but rather because the opponents gradually die out and the following generation is from the start familiar with it.

Are you claiming that you have a “new scientific truth” when as I have pointed out:

You have been across the board in terms of your positions.

You claimed that carbon dioxide could not reemit except at the surface. You claimed that it could not reemit in the majority of the atmosphere because local thermodynamic equilibria presumably do not exist in the earth’s atmosphere. So according to you non-LTE prevents reemission. Then you claimed that the top of the atmosphere is non-LTE, and therefore only it could reemit. Then you have claimed that it is the rate of collision being too high which prevents reemission (i.e., hot LTE conditions), but you have also claimed that those parts of the spectra in which reemission would occur get frozen out – meaning that the temperatures are too low for reemission. You have claimed that the emissivity of carbon dioxide will suddenly jump from logarithmic to linear at the surface as the result of higher levels of carbon dioxide – resulting in a catastrophic runaway effect. You have claimed that reemission was due to electrons leaving their excited states when within the context of our atmosphere, reemission is due to molecules leaving their excited states. You have even gone so far as to claim that it was the nucleus which reemits.

The fact that you have jumped from position to position with such speed is itself an admission that you do not know and that you do not understand. And yet you have continued to claim expertise at various points and had been unwilling to ever say what your actual educational and professional background is. I had to look it up.

…?

Are you claiming that your poorly thought-out ideas constitute just such a revolutionary breakthrough? What is your revolutionary idea today? Does it constitute a rigorous scientific theory? What data do you have? What empirical tests have you performed? Are you claiming that you have any background in radiation transfer? Quantum mechanics?

If so, what is this background?

And what specifically are you arguing against with regard to current, mainstream climate theory? That it is not the final theory? That it doesn’t as of yet include all the effects and mechanisms which determine the course of the climate? They make no such claim. Are you claiming that what they are missing is something involving the radiative properties of either carbon dioxide or water vapor? If so, what?

Alastair McDonald (#512) wrote:

The skeptics, such as Jack Barrett, are saying “The models are wrong so they are over-estimate the effects of AGW.” The scientists are saying “We know that changes in climate happened in the past due to CO2, therefore our models are correct.”

Climatologists would not claim that their models are complete. They would however claim to have a great deal of evidence and justification for their understanding of the role that CO2 plays in the evolution of the climate system.

Alastair McDonald (#512) wrote:

But you, as a philosopher, must know that neither of these assertions is logically correct. There is the possibility that the models are wrong, and that they are underestimating the effects on the way, which is what I believe.

I would not claim to be a philosopher.

I would claim that I have a background in philosophy, that I have written eighty page critiques of Descartes’ “Six Meditations on First Philosophy,” “The Critique of Pure Reason” and of early twentieth century empiricism. This does not make me a philosopher or even someone who has the same degree of formal education as those who have obtained a PhD in philosophy.

I would claim some formal background in the history of philosophy. I would claim some formal, graduate level background in ancient greek philosophy.

But this does not make me someone who can offer expert opinion in ancient greek philosophy which is equivilent to that of someone who professionally specializes in ancient greek philosophy.

I would claim some formal background in dialectics. But this does not make me someone who can offer expert opinion in dialectics which is equivilent to that of someone who professionally specializes in dialectics.

As someone who has some graduate-level formal background in epistemology, I also claim to have studied formal and informal logic, to have some background in the relationship between foundationalism and coherentialism, and to have some understanding of the nature of self-referential argumentation. I recognize the fact that neither strong foundationalism nor strong coherentialism is logically defensible. I recognize the fact that a radical skepticism based upon the denial that knowledge exists or upon radical doubt that knowledge exists is self-referentially incoherent. I would claim to have applied self-referential argumentation at a technical level to certain other problems in philosophy, but I would not claim that such results have the same status as that of someone who has obtained a PhD in the relevant fields. I recognize the necessity of admitting corrigible knowledge (i.e., knowledge which lacks cartesian certainty) as a form of knowledge, that lacking this, in logic, no claim to any knowledge is logically defensible.

But even as a whole, this does not constitute a systematic theory of knowledge.

I would claim some formal background in analytic philosophy at the graduate level and I would claim some formal background in the philosophy of science. I would claim that advanced scientific theories are a form of knowledge. I would claim that scientific knowledge is cummulative. I would claim that a conclusion justified by multiple independent lines of investigation is often justified to a far greater degree than it would be if it were justified only by one line of investigation considered in isolation. I would claim that the advanced theories of empirical science cannot be tested in isolation, that there exist interdependencies between various scientific theories.

But this does not make me someone who is in possession of a systematic philosophy of science or who can offer expert level opinion equivilent to that of someone who specializes in the philosophy of science.

*

Do you have a PhD in any field that is relevant to climatology? Do you have a PhD in any advanced field of physics? Can you offer expert level opinion in radiation transfer theory? Do you have any graduate level background in any of this? Do you have any formal background? Have you systematically studied any of these disciplines?

Alastair McDonald (#512) wrote:

What I am saying is that the science produced by the skeptics is correct. That applies to Jack Barret, Christy & Spencer (tropical lapse rate problem), and now Stephen Schwartz. It is the conclusions that they draw from their facts that is wrong.

What basis do you have for making such claims? Why is it that their claims are correct when you so cavalierly dismiss so much of mainstream climatology? Do you have any formal training which justifies your making such claims as any form of expert level opinion?

Alastair McDonald (#512) wrote:

You wrote “Honestly you could be quite an asset.” That is what I am trying to be, but I am just a lone voice crying in the wilderness.

How can you be an asset if you so cavalierly dismiss so much of mainstream climatology?

Alastair McDonald (#512) wrote:

So long as you apply uncritical faith in the scientists of today, then, in the words of Private James Frazer “We’re doomed!”

I recognize that given the current state of our knowledge, we do not understand all the mechanisms which are relevant to understanding the climate system as a whole. I recognize that our knowledge of the evolution of the earth’s climate system is incomplete. But I also recognize that current climate theory is an advanced branch of physics in an advanced state. If this is what you genuinely claim to be an “uncritical faith,” then I have literally no understanding of what you mean by the term.

Comment by Timothy Chase — 27 Aug 2007 @ 11:53 AM

521. Re #517
Thank you, Mr Roberts, for your response to my #516. The DoE paper I referred to was updated in 2002. I think the following ref provided by Allan Ames in #464 is of great significance.

http://www.dodsbir.net/sitis/view_pdf.asp?id=DothH04.pdf

It is a joint research paper by Spectral Sciences, Inc., the US Air Force, and Boston College. It is highly authoritative, and immense. Pages 9 and 10 seem to show that an excited Co2 molecule can relax in stages, so that the energy of a particular emitted photon is less than the initially absorbed photons, so making re-absorption impossible. This supports the DoE statement. No doubt someone will wish to correct me on this point.

Another item of interest is that a CO2 molecule can be raised into an excited state by “resonant energy transfer” from a nitrogen molecule.

Comment by AEBanner — 27 Aug 2007 @ 3:12 PM

522. Ah, that’s a good link, thanks Mr. Banner.

I won’t pretend to follow this. Still hoping to bait one of the real physicists to join in — I wonder if a photon from a lower, warmer CO2 moledule can be absorbed by a higher, cooler CO2 molecule, because I still don’t find a clear argument that it can’t happen. But I’ve just skimmed this huge piece. It is probably worth a read by someone able to handle the physics and explain it. Oh, please?

excerpt——

(3) J/A =[M*]/[M]
This situation is described as “radiative equilibrium”. At altitudes greater than about 150 km the ratio of the densities of the CO2 vibrational levels 01101 and 00001 is determined solely by the upwelling earthshine. The frequency of collisions is too small to have any impact and radiative equilibrium prevails. …

The arguments cited above apply to stable species for which normally only the lowest vibrational level is populated and which may be excited by earthshine, sunshine, and collisions. …

“We then have a situation where the translational and rotational degrees of freedom can be described by one temperature and the vibrational levels are described by an entirely different temperature. Of course two different vibrational manifolds in the same molecule may be described by two different vibrational temperatures…..”

—– end excerpt ——

Comment by Hank Roberts — 27 Aug 2007 @ 3:54 PM

523. For Alastair — do you still agree you can falsify your theory by getting photographs of the limb of the Earth in infrared? Do you agree such photographs would show a fuzzy edge at the stratosphere if CO2 is radiating, as the infrared astronomers believe, but would show a black sky above the top of the clouds/troposphere if CO2 doesn’t radiate as you believe?

If that adequately allows you to test your theory, the images would be available from this system, I expect.

The SEAWIFS satellite system builds true color images for display from “eight (8) different bands in the visible and near infrared part of the electromagnetic spectrum…. the sensor is designed to collect all eight bands at exactly the same time.”
http://oceancolor.gsfc.nasa.gov/SeaWiFS/TEACHERS/TrueColor/

If you think this will let you falsify your theory, you might ask for the various infrared bands that are used to make up some of their many images showing the limb of the planet.

They have lots of imagery taken tangent to the horizon, though the public website images I’ve seen all combine all 8 bands for true color imagery, like these:
http://www.navis.gr/space/greece/images/large_08.jpg
http://www.geog.ucsb.edu/~jeff/wallpaper2/tn_europe_oblq_view_aug14_2001_above_red_sea_6400km_altitude_seawifs_wall_jpg.jpg
http://www.geog.ucsb.edu/~jeff/wallpaper2/cyclone_dina_jan242002seawifs_wall.jpg

Comment by Hank Roberts — 27 Aug 2007 @ 8:05 PM

524. Re #523

Hank,

No I do not agree that my theory can be falsified with photographs of the limb of the Earth. I am arguing that a view from outside the atmosphere will be the same as those currently obtained. What I need is a spectrum taken from 100 or 1000 m.

I had a look at the SEAWIFS page, but it is only for visible and near IR. The band I am interested in is the CO2 15 micron band since that has a major effect on the spectrum of the Earth (and Mars.) It is in the far infrared part of the spectrum.

I have performed an experiment which would have falsified my theory, but it didn’t. Unfortunately the results were not good enough to prove my theory. A major problem that I have is, because I am saying the current theory is wrong, no one believes me, and so they are not interested in understanding what my theory is.

I have just been reading James Rodger Fleming’s (Spencer Weart’s rival) book “Historical Perspectives on Climate Change” about Wiiliam Ferrel’s views on the greenhouse effect. Fleming writes: “He [Ferrel] made it clear that one-half of the radiation absorbed in the [atmospheric] envelope would be again radiated back to the [Earth’s] body.”

Fleming goes on to quote Ferrel:

“It is seen, therefore, of what great advantage the atmosphere is in raising the mean temperature of the earth’s surface. This effect is similar to that of the glass covering of a conservatory of plants, which is a diathermanous medium which permits the heat of the sun to pass through with facility, but is almost completely impenetrable by the heat radiated from the air and all bodies within.”[Ferrel, 1884]

I am saying that Ferrel was wrong on both points. The radiation is absorbed by the air and not re-radiated. The glass of a conservatory is transparent to infrared radiation, and in fact it is the band of solar radiation in the UV to which it is opaque!

But then, as I sure Timothy will enjoy pointing out, I do not have the Ferrel Cells named after me so I must be wrong :-(

Comment by Alastair McDonald — 28 Aug 2007 @ 6:14 AM

525. Re #506 where Hank Roberts Says:
26 August 2007 at 5:39 PM

Lectures based on Goody & Walker’s 1972 book Atmospheres:

“There are multiple different tables of composition – not surprisingly, our understanding of the composition of the atmospheres of Venus and Mars have developed since Goody & Walker wrote Atmospheres in 1972.”

Since the composition is not what I was discussing I ignored this post, but now having read on I have discovered that if you go to:
there are links to PDFs of all the chapters of Goody and Walker.

On pages 14 & 15 of Chapter 3 (pages 57 & 58 of the book) you can find the section to which I was referring. Using the simple model gives an impossible optical depth for Venus.

“… But we just want a simple comparison of Earth, Venus and Mars -Isn’t there a simpler way of looking at this?
For a THIN atmosphere – e.g. Earth, Mars – we can use a simpler model. The book has this…

But if the models is wrong for Venus, it is wrong. It may give reasonable results for Earth and Mars but it is still wrong. And that is what I am saying – the current models give reasonable results for today’s conditions but they are wrong, and are not predicting the future correctly. Actually, they did not even predict today’s melting Arctic ice correctly. Ray Ladbury may be satisfied with that performance, but I am not.

Comment by Alastair McDonald — 28 Aug 2007 @ 8:05 AM

526. Alastair,
The physics of melting ice is much less well understood than is that of the atmosphere. If I were to look for the failure of prediction, that is where I would start. And as I have said before, “right” and “wrong” are not particularly useful adjectives when discussing models. Models are simplified approximations of the system under study–by their very nature, they leave out some physics. Models provide insight or they do not. They provide adequate predictions or they do not. If a model does not reproduce the trends seen in the actual system, you look at the physics it leaves out, starting with the least understood aspects. Since I am not a climate scientist, it does not matter what I am satisfied with–what matters is whether they can get meaningful science out of the models. They do.

Comment by Ray Ladbury — 28 Aug 2007 @ 9:34 AM

527. Alastair, nobody’s “satisfied” or people would have put down the tools and gone home.
“All models are wrong; the practical question is how wrong do they have to be to not be useful” — George Box

Those of us asking for better information are looking for information that can be supported. Yours isn’t yet.
To support your unique theory you need some way it can be falsified or it’s just handwaving.
You need at least one thing that Einstein predicted one way, you predict another way, and is observable by third parties.
Until there’s a difference that can be observed, there’s no difference that makes a difference.

How about those infrared photographs of the horizon from space?

Comment by Hank Roberts — 28 Aug 2007 @ 9:34 AM

528. Alastair, your response above appeared after I wrote mine. Care to describe your experiment? I don’t understand what you mean about it not working.

Comment by Hank Roberts — 28 Aug 2007 @ 10:59 AM

529. Hank, he uses Real Climate as a platform for his ideas.

It gives him access to a much wider audience than he would have anywhere else. The climatologists who created Real Climate intended it to be used as a platform for teaching climatology to a wide audience. When he is humored as you are doing now, it encourages him in his attempts to undermine the strongest aspects of climatology and to attack the integrity of these climatologists to the point at which it would appear that at certain point they are made uncomfortable participating. He goes so far as to state that they ignore any evidence which suggests that the theories which form a core aspect of their discipline are invalid. And he shows every sign of being willing to do so again in the future.

People who visit here expect the “skeptics.”

They do not expect someone who teaches an alternative greenhouse theory and who knows just enough technical terms to sound convincing – at least for a brief period. He has repeatedly denied the existence of the mountain of evidence which we have on radiation transfer even when we have linked to various pieces of it repeatedly and he continues to do so even now. He shows signs of a willingness to do so in other areas. He derails discussions of the actual science with great frequency and at great length and will do so indefinitely if he is treated with a courtesy that he does not deserve. As such he diminishes the quality of Real Climate, undermines its ability to attract new participants and reach a wider audience.

As a matter of honesty, integrity and honor, I would not claim to have an expertise which I am lacking.

When he attempted to bait me into claiming such expertise I was not tempted for even a small fraction of a second. Nevertheless it opened some old wounds. While latter last even I was able to check the posts at Real Climate and was pleased to see new participants, including at least one that would appear to be a genuine asset, I was unable to participate because I was extremely upset. I am getting over it and will in all likelihood have it under control by the time that some “skeptic” attempts to use it against me. Nevertheless it is a deeply personal issue for me as I devoted a considerable part of my life to the attainment of a level of expertise which I will in all likelihood never achieve.

I do not have a problem with Alastair participating.

However, I at least will have a personal problem with his participating as he does now. Whether he is here or not, I will continue to participate because I believe in Real Climate, consider it to be an extraordinarily generous gift to all those who learn of it and visit it. I will continue to participate because I believe in its mission. I will continue to participate because I believe it is critical that people understand the issues – including the threat of loss of life which lies in the decades ahead. So long as he continues with what he is currently doing, I will not address him directly unless I believe that the conditions especially warrant it – as when he attacked the integrity of one of the climatologists. However, I see little reason in holding back from pointing out that he has no credentials, is entirely lacking of the educational background for any of the claims which he is making, and refuses to acknowledge the evidence which exists.

This is where I stand. I hope that even if you disagree with me that you will at least respect my position.

Comment by Timothy Chase — 28 Aug 2007 @ 11:04 AM

530. Oh, I know, Timothy; I often ask Alastair to distinguish statements he makes about the nature of reality that are unique to his theory from those taught in physics classes. I’m sure I’ll continue to ask. It’s the extreme case of the “please give me a cite for why you believe …. and a way to check your sources” request. Good practice anyhow.
Or maybe it’s just recreational typing; when I think that I give up for a while. This is probably one of those times.

As Patrick Moynihan once said, everyone is entitled to their own beliefs, but not to their own facts.

Comment by Hank Roberts — 28 Aug 2007 @ 12:56 PM

531. Re #528 Hank, My experiment was very crude so I have been reluctant to report it however …

Near where I live there is The Bournemouth Eye – a tethered balloon. It rises to about 100 m. I aquired an infrared thermometer with a range of 7 to 18 microns and used it to measure the temperature of the surface regularly as the balloon rose and again as it descended. The results showed that the temperature measured fell quickly for the first 50 m above the ground and then remained the same.

Note, what I claim I was reading was not the temperature of the surface which did not change, nor the air temperature which would have fallen by 1F. What I was measuring was the infrared radiation emitted by the surface. The reason the temperature reading fell was because the radiation in the 15 micron band was being absorbed by the air and not being re-emitted, in conflict with Kirchhoff and Einstein.

Convinced?

Comment by Alastair McDonald — 28 Aug 2007 @ 2:17 PM

532. Re #509

Timothy,

Sorry that I have not replied fully to your posts earlier but they are rather long and I am a slow reader.

Keeping this one short I would just like to mention first that the AIRS maps show CO2 concentration not radiation.

Second, the SAMM2 report is regarding a model not data. However, it does tend to support my view about radiation in the 15 micron band when they write:

Since the 15 μm CO2 band is very strong, its transitions to the ground state and low lying excited states are severely self-absorbed. This fact has important consequences for atmospheric radiation which we will point out as we go along. Similarly, solar radiation has negligible influence on the 15 μm radiation because of the small number of solar photons at this wavelength. The intensity of 15 μm radiation therefore shows no diurnal variation [Wise et al., 1995].

The lack of diurnal variation sort of implies that the 15 micron band originates at the TOA well away from the surface where the solar effect would have an influence.

Their Figure 6 goes against what I am saying, but it is modeled not measured, and it seems that they have forgotten to include the “freezing out” of the vibrational excitation at room temperature (STP) and below.

Cheers, Alastair.

Comment by Alastair McDonald — 28 Aug 2007 @ 2:56 PM

533. Nope. I just find too many reports of people taking photographs of the atmosphere in the band where you say it can’t happen.

Cameras have long been available in this band, for example this one from Mars Viking:
http://ntrs.nasa.gov/search.jsp?R=81317&id=1&qs=Ne%3D20%26N%3D4294910007%26Ns%3DArchiveName%257C0

Comment by Hank Roberts — 28 Aug 2007 @ 3:17 PM

534. AEBanner (#521) wrote:

Pages 9 and 10 seem to show that an excited Co2 molecule can relax in stages, so that the energy of a particular emitted photon is less than the initially absorbed photons, so making re-absorption impossible. This supports the DoE statement. No doubt someone will wish to correct me on this point.

Actually from what I can tell you are right.

Transitions from higher to lower states of excitation. But what this means essentially is that the energy becomes split into seperate “parcels,” each of which may be directed towards the surface or towards space. With such partial relaxation this changes the details, no doubt affecting some the atmospheric chemistry (which is part of the subject of the paper) and rendering the “random ladder climb” analogy that I had employed less accurate, but qualitatively things remain much the same. If the radiation reaches the surface, it warms the surface. If it warms the surface, this implies more thermal radiation.

Three points…

The net direct effect of carbon dioxide is to lower the temperature. There is an exception in the lower range where it is active, but this would appear to relatively small. I can look up the diagram a little later. It includes carbon dioxide, water vapor and ozone, showing their local effects in terms of warming or cooling in degrees per day I believe. As I understand it, the atmosphere is directly warmed principally is moist air convection in the lower levels, then both turbulence and diffusion which help to distribute the heat to the upper levels. However, ozone has a fairly broad range over which it has the net effect of warming the atmosphere and its overall net effect is to warm the atmosphere.

Comment by Timothy Chase — 28 Aug 2007 @ 3:42 PM

535. Regarding AIRS

AIRS shows CO2 concentration by means of its infrared emissions.

Comment by Timothy Chase — 28 Aug 2007 @ 3:57 PM

536. RE Alastair (#532),

It helps to keep in mind the fact that the name AIRS is an acronym meaning Atmospheric InfraRed Sounder. Its images of gas concentrations at various altitudes is due to its measurements of infrared emissions. Therefore if it is detecting CO2 at 8 km, this is because CO2 is emitting longwave radiation at 8 km.

If you are unable to lookup this information for yourself, I will gladly look it up for you. However, I am afraid that it will still be up to you to read it. I suppose that would mean that my efforts are wasted.

Comment by Timothy Chase — 28 Aug 2007 @ 4:29 PM

537. Re #535

The standard model claims that the CO2 emissions depend on temperature but you are claiming it shows concentration. Who is correct?

To be honest, I am pretty sure it is not the 15 micron emissions that they are using to measure the CO2 concentration. It will be near IR from re-emission of radiation absorbed from the sun.

Comment by Alastair McDonald — 28 Aug 2007 @ 4:33 PM

538. Alastair McDonald (#537) wrote:

Re #535

The standard model claims that the CO2 emissions depend on temperature but you are claiming it shows concentration. Who is correct?

Both.

AIRS does over 2000 different channels. I have mentioned this before. The spectra of carbon dioxide (and other greenhouse gasses) are a function not simply of the gas but of the pressure and temperature. Take enough measurements and you can get a great deal of information out of it.

Comment by Timothy Chase — 28 Aug 2007 @ 5:24 PM

539. re re-emission (534, et al). A repeat of a question/contention I made earlier and tried to remake recently but my post didn’t pass muster (probably too long, redundant and boring, but other than that I thought it pretty good… Or maybe I just botched the posting…). The assertion, refuted by some, that “an excited CO2 molecule can relax in stages, so that the energy of a particular emitted photon is less than the initially absorbed photons, so making re-absorption impossible” makes no sense to me. At a basic level there are near thousands of quanta energy levels (especially in rotation) and a molecule can absorb or emit only at those levels. If a CO2 emits at level “Q”, level Q then exists in every CO2 molecule, and some CO2 molecule will have that level vacant and be able to absorb the re-emitted photonic radiation. It matters not a twit if the first molecule started it all by absorbing a photon at level “X” (all not withstanding other affecting factors.) Is this not correct?

Comment by Rod B — 28 Aug 2007 @ 8:18 PM

540. Alastair,

More curious than angry.

Three questions: how long, did Rod know, and most especially, why?

Seen this sort of thing before, but nice twist. That’s what threw me off.

Comment by Timothy Chase — 28 Aug 2007 @ 8:28 PM

541. re Timothy (520 et al), I’m disappointed with your attitude. It sounds like you want to assure that RC is a mutual admiration and cooperative cheerleading society — and idea that pops up periodically here. The fact that you believe very strongly in what you profess and even feel you have solid evidence to back up your assertions does not warrant a scientific forum casting out or shunning all who don’t readily subscribe to your dogma. The moderators of RC are not that stringent. It might be O.K., even helpful, to shun the extreme crackpots; but crackpot is not defined by disagreeing with you or challenging your well thought out precepts.

Comment by Rod B — 28 Aug 2007 @ 8:30 PM

542. Rod B (#541) wrote:

re Timothy (520 et al), I’m disappointed with your attitude. It sounds like you want to assure that RC is a mutual admiration and cooperative cheerleading society — and idea that pops up periodically here. The fact that you believe very strongly in what you profess and even feel you have solid evidence to back up your assertions does not warrant a scientific forum casting out or shunning all who don’t readily subscribe to your dogma. The moderators of RC are not that stringent. It might be O.K., even helpful, to shun the extreme crackpots; but crackpot is not defined by disagreeing with you or challenging your well thought out precepts.

Rod,

Hank Roberts (#530) gave us a fairly good quote just a little while back:

As Patrick Moynihan once said, everyone is entitled to their own beliefs, but not to their own facts.

We have argued a great deal with Alastair and shown him time and time again that the criticisms he has of the mainstream effect don’t work and “his” theory is unworkable. Frustrating, but… whatever.

However, the data that we linked to, thats pretty hard stuff. The reemissions in the mid-troposphere actually recorded via satellite. Then of course there is the data from planes. Couldn’t actually give it to him, but available for purchase nevertheless. And not too expensive. However, he never even acknowledged that we made the images and data available to him time and time again.

And I didn’t like the attack upon the integrity of one of the climatologists. What does your book of “How to Play Fair with the Misinformed” have to say about that?

So yes, if that were what was going on I would have a serious problem with it. Particularly if someone kept on derailing conversations regarding the actual science with his own “theories” which made no sense whatsoever and did everything to undercut the purpose of Real Climate – which is to educate the public about climatology. But I can also see how that sort of thing might appeal to you.

I would also have a serious problem with someone claiming to be an expert – particularly an unacknowledged genius – in a field that he knew very little about. That would be “crackpot” as I understand the term. And I also have a real problem with the anti-science attitude that is so prevailent in certain parts of our society.

Likewise, if I had to weigh someone’s hurt feelings against someone else’s loss of life, let alone a whole lot of other someones’ lives, let’s see, on the one hand, on the other hand,… not much of a choice for me there. I’m sorry if that is a problem for you. However, given BAU for a few more decades I suspect we are going to face some choices that will be a great deal more difficult. Choices I would rather not see anyone have to make.

But I will remind you that I encouraged him to find a more productive role – at numerous points.

Not just trying to jettison him but wanting to see him grow into something more than he was before. As I had written in an essay for a particular audience at one point,

“… while recognizing one’s mistakes may be experienced prospectively as a form of death, the act itself brings a form of rebirth and self-transcendence, giving one the courage to revise one’s beliefs when confronted with new evidence.”

This is a view that I take very much to heart. What we should fear are not our mistakes or even our having to admit our mistakes to ourselves or to others, but our inability to admit our mistakes.

But I figure the following might help as well – at least in terms of understanding where I am coming from….

My first obligation, wherever I go, whoever I meet, and whatever I do is to understand. If I do not understand, then I cannot know what the consequences of my actions will be. If I do not understand, I can become trapped in a prison of conceptual illusion. If I do not understand, then I and those who are not necessarily my enemies may become trapped in needless conflict.

Understanding must come first. At that level, I am neither for you nor against you. At that level, neither you nor anyone else even enters the equation. This is who I am. This is what I am. Take from this what you will, for at the same level, you must face the same issue and the same alternative every moment of your life, and at that level, neither I nor any other human being matters, either.

http://groups.yahoo.com/group/creation_evolution_debate/message/81123

I don’t expect everyone to agree, but I believe that everyone should be willing to admit their mistakes – particularly when reality stares them right in the face.

Yes, Rod, I am an idealist.

But I also like to think that I am a good student of human nature.

Comment by Timothy Chase — 28 Aug 2007 @ 11:25 PM

543. Re #540

Timothy,

my rate of reading has not speeded up. It is still the same at one screenful per day. Thnkfully, your posts have become considerably shorter, and so I can now include them within that ration.

However, when you write “Three questions: how long, did Rod know, and most especially, why?” your brevity has me defeated. To what are you referring?

However, without knowing that, I can still give you an answer to your second question: you will have to ask Rod that! But you may be doing Rod a misjustice. I suspect that he is, like you, searching for the truth. And he, like you, is handicapped by his preconceived ideas. He believes that global warming is all a silly scare. You seem to believe that scientists can make no mistakes. You are both wrong.

Comment by Alastair McDonald — 29 Aug 2007 @ 3:29 AM

544. Ray and Hank,

You both seem to be making the same point. That although the models are not exactly right that is because no model can be. It is only a simplification of the real world.

But what I am saying is that the model of terrestrial radiation being reflected (or, if you prefer re-emitted) back to the ground is wrong. The air at the surface is not heated by conduction with the ground. It is heated by the radiation from the ground. Either way you get the same result: the air in the atmosphere is heated from below and it convects.

But that is where the similarity between the two systems ends. However, it pointless describing the correct model further if you cannot accept that simple difference.

But I will add that you cannot prove that a transparent colourless gas is acting at the surface of the Earth by using a satellite 100 miles up, no matter how sophisticated its cameras might be.

Comment by Alastair McDonald — 29 Aug 2007 @ 3:44 AM

545. Gentlemen,

Like Rod B, I have been asking similar questions in all my web postings on this thread but, until recently, have been confused by the answers since several appear to be inconsistent with each other while others answer questions I didn’t ask. However, as the thread has progressed, I believe things are becoming clearer, thanks largely to posts from AEBanner (#454 and 517), Allan Ames (#464), Ike Solem (#457) and Ray Ladbury (#478). Given my propensity to grab the wrong end of the stick, however, I would like to make a series of statements and follow them with a list of questions. I would appreciate advice on the veracity of the former and answers to the latter.

Statements:
1) At the top of the atmosphere, their is escape of long wave radiation which, at equilibrium, matches in energy terms that entering by solar radiation less albedo.
2) There is an “atmospheric window” in the wavelength range of roughly 8.5-13.5 microns where there are no blocking greenhouse gases except for trophospheric ozone. This window allows egress of radiation directly from the surface. However, in energy terms, this represents only about 17% of OLR.
3) Radiant energy absorbed by the atmosphere will be re-emitted in all directions. It will be re-emitted at longer wavelengths than that at which it is absorbed, the re-emitting wavelengths being longer at lower temperatures and pressure. This implies that a species of molecule which absorbs at one wavelength cannot absorb a photon re-emitted by a molecule of its own kind.
4) Approximately two thirds of the energy absorbed by the surface emanates from energy re-emitted in the atmosphere and only one third is attributable to directly absorbed solar energy.
5) Radiative transfer is a much slower energy transport mechanism than convection such that the latter process gets most of the effluent energy to the top of the atmosphere.
6) Radiation leaving the surface has a near blackbody spectrum. OLR does not. There are gaps attributable to greenhouse gases.
7) There is little water vapour in the high atmosphere.

Questions:
1) Will the “atmospheric window” get narrower with increasing CO2? I can picture the possibility of a curtain hanging on the right (long wave) side of the window which may become partly drawn, thus obscuring part of it. I can see this to be particularly likely if energy is leaving a dry surface with little overlying water vapour but less likely over, say, oceans.
2) Ray Ladbury (#478) talks of gaps in the blackbody spectrum in the case of OLR. However, Step 2 of “The CO2 problem in 6 Easy Steps” (Real Climate Archive, August 6th) cites a graph showing IR spectra taken from space. This graph has no explanation. Rightly or wrongly, I interpret it to be showing that Ray’s gaps are attributable to CO2 and ozone but not to water vapour. Is this correct? If so, is it explained by the dry upper atmoshere.
3) How does thermal energy reaching the top of the atmosphere by convection become OLR and make good its escape to space?
4) Various correspondents suggest that infrared in the near-15 micron range can be demonstrated in the high atmosphere. Could this emanate from re-emissions of atmospherically- absorbed solar irradiation, given that re-emissions from lower level CO2 molecules couldn’t apparently be an explanation?

Comment by Douglas Wise — 29 Aug 2007 @ 4:56 AM

546. Re #539 Rod B

I can only suggest that the process you described depends upon the absorption cross sections of the molecule for photons of different energies, and so is a matter of probability. The absorption lines adjacent to the 14.99 micron line for CO2 are somewhere between one and two orders down, if I remember correctly. So the chance of the transition happening twice ( in any “chain” ) would be between a factor of 10^2 and 10^4 down.

Comment by AEBanner — 29 Aug 2007 @ 6:18 AM

547. Gavin,

Quick question. Did you decide not to post something I submitted? Not a problem – light of a new day I am not even sure I would want it to go up. But it helps to know. I know that I have lost an strictly science post or two recently which were probably due to a technical glitch.

Comment by Timothy Chase — 29 Aug 2007 @ 8:17 AM

548. Re #547,

Gavin,

Rod has lost a post, and now so has Timothy. I am finding that RealClimate is very slow to respond since it was reconfigured. Moreover now I am no longer able to preview my posts before sending them :-(

The slowness may be due to this thread being too long, or because the RealClimate is now so popular that the server is often overloaded. Are you aware of our difficulties and is it possible you could do something about them?

Regards, Alastair.

Comment by Alastair McDonald — 29 Aug 2007 @ 10:08 AM

549. Alastair, try turning the other direction — when you can’t cite sources for what you believe, describe a way to test it.

Comment by Hank Roberts — 29 Aug 2007 @ 10:34 AM

550. Re #539 and #546

The way the molecules behave is determined by selection rules. for instance see http://www.iop.org/EJ/article/0022-3700/18/22/016/jbv18i22p4481.pdf Selection rules for rotational excitation of polyatomic molecules by slow electron impact

But for an explanation of the principles of electronic and vibrational (vibronic) energy changes see http://en.wikipedia.org/wiki/Franck-Condon_principle

Comment by Alastair McDonald — 29 Aug 2007 @ 10:43 AM

551. Hank, re your 515 referring to Alastair’s 222, etc. Maybe this got settled and I missed it, but I’d like to review my understandings and ask if everyone or anyone agrees.

1) So-called blackbody radiation (that which follows Planck’s function) is fundamentally different from gas “line spectrum” emission, though both are affected by quantum stuff. One derives from heat creating temperature and getting applied to (mostly) near-free electrons. The roaming and accelerating electrons are responsible for blackbody radiation and follow some distribution (other than Boltzman, maybe?) that does emit a continuous spectrum, for all practical purposes. Gas radiation stems predominately from the quanta energy levels of atomic vibration within a molecule and molecular rotation. This does not produce a continuous spectrum but discrete wavelength emission (though there are certain limited bandwidths where it looks close from the packing of the lines and spreading.) There is some gas emission/absorption at higher frequencies that relate to electronic/ionization energy levels, but this too is discrete.

2) Gasses can emit some so-called blackbody radiation in addition to the “line emission” described above.

3) Equipartition is the tendency of a molecule to equitably distribute and transfer its carried energy among translation, vibration and rotation levels. This is one way that vibration, say, can lose or gain energy. Another is (maybe indirectly) via molecular collisions. Another is IR, mostly, radiation emission and absorption. I say “tendency” because equipartition is a little loosey, has some quanta factors and follows distributions that, in part, are a function of temperature, so is somewhat quantized and non-linear.

4) I would think that a cold CO2 molecule is more likely to (re)emit from vibration, and maybe even rotation to some extent, because equipartition’s dependence on temperature makes it easier for a hot molecule to keep its vibration levels filled.

Comment by Rod B — 29 Aug 2007 @ 11:42 AM

552. Hank,

I described one way of testing by using an infrared thermometer from a balloon, but you seem more interested in disproving my ideas than understanding them :-(

Assuming that you are living in sunny California, (you will have to adapt the following experiment if you live elsewhere) park you automobile in the middle of a parking at midday and place a thermometer inside it in a shaded spot. Then with the windows closed and uncovered, wait for two hours and read the thermometer.

Repeat the experiment the following day, but this time leave each window open with a gap of 1 cm (half an inch.) This should make little difference to the amount of glass surrounding the car. If it is true that it is the glass reflecting the infrared radiation back into the car which causes the warming, then both runs of the experiment should give the same temperature reading.

In fact it is the air that absorbs the radiation, and by allowing the hot air to escape the amount of heat retained within the car is greatly reduced, and so also is the temperature reading.

Comment by Alastair McDonald — 29 Aug 2007 @ 11:44 AM

553. Douglas Wise,
OK, here’s my take on your points and questions–and remember, I’m just a dumb physicist, not and expert on atmospheres. OK, first, the 15 micron absorption line for CO2 is rather broad, and its shape is influenced by the surrounding gas. Also, the lifetime of the vibrational state is quite long, so as long as gas density is at least moderate, the chances of relaxation by collision or other de-excitation process is fairly high. A re-radiated photon can excite another CO2 vibrational state, but there aren’t that many re-radiated photons. Think of it this way: A CO2 molecule in an excited vibrational state has a higher energy than the vast majority of its neighbors. Since it is colliding with and interacting with those neighbors, energy must flow from it (and from the vibrational mode) into its neighbors translational motions. If the gas were hotter, more energy would flow into the vibrational mode from the collisions.
1) As CO2 concentrations increase, the wings of the CO2 absorption line get broader, so, yes the window does get a very small bit narrower.
2) The deviations from a “blackbody” (or really gray-body) radiation curve are largely due to absorption by ghgs of outgoing LWR.
3)Heat transported by convection gets up to the upper troposphere and heats the air there. The warmed air collides, exciting rotational and a few vibrational states. These high-altitude molecules then radiate in the bands where they can radiate. OK, now think of yourself on the space station looking down measuring the light coming from Earth. The light you see in the “windows” comes mostly right from the warm surface, but the LWIR comes from above the cloudtops for the wavelengths corresponding to water absorption features, and for CO2, from even higher than that. And because the upper troposphere is cold, there are “holes” in the spectrum compared to what we’d expect if all the radiation came from the warmer surface. Those holes represent energy, which has gone into heating the atmosphere and ultimately the surface.
4)There’s not that much intensity for sunlight in long wavelengths.

Make sense?

Comment by Ray Ladbury — 29 Aug 2007 @ 12:14 PM

554. re 546, 539: AEBanner (546) says, “…the process you described depends upon the absorption cross sections of the molecule for photons of different energies, and so is a matter of probability. The absorption lines adjacent to the 14.99 micron line for CO2 are somewhere between one and two orders down, if I remember correctly. So the chance of the transition happening twice ( in any “chain” ) would be between a factor of 10^2 and 10^4 down.”

I agree with the theory but question your numbers (though not smart enough to explicitly re-calculate). The chance of molecule 2 absorbing is less after molecule 1 absorbed at the same discrete frequency because the energy “pool” in that frequency is lessened — for a while. This is just following standard absorption/optical depth theory. It is not lessened just because molecule 1 re-emitted a photon; molecule 2 sees only a “pool” of photons and has no way of knowing their origin, nor does it have a way of selecting a photon emitted from the surface over an exact replica that happened to come from another similar molecule. Your probabilities appear too similar to the repeated coin flipping example.

In any event (whichever probability is correct), this does not say that molecule 2 CAN NOT absorb a photon emitted from molecule 1 of the same species, which others are contending.

Comment by Rod B — 29 Aug 2007 @ 12:31 PM

555. AEBanner,

I figure you might like this. Here are images composed from data showing infrared emissions from CO2 in different parts of the atmosphere – plus a few additional links for finding more…

From #322
The first figure on this page (figure 23) shows the observation of CO2 reemission in the 15 μm from the mid-troposphere as observed by NIMBUS-4 IRIS (graph from a textbook published in 1976)

3.1 Physical basis of remote sounding
3 RADIATIVE TRANSFER AND REMOTE SOUNDING
http://ceos.cnes.fr:8100/cdrom-00/ceos1/science/dg/dg20.htm

*

From #323

Abstract. The near-infrared nadir spectra measured by SCIAMACHY on-board ENVISAT contain information on the vertical columns of important atmospheric trace gases such as carbon monoxide (CO), methane (CH4), and carbon dioxide (CO2). The scientific algorithm WFM-DOAS has been used to retrieve this information. For CH4 and CO2 also column averaged mixing ratios (XCH4 and XCO2) have been determined by simultaneous measurements of the dry air mass. All available spectra of the year 2003 have been processed.

Carbon monoxide, methane and carbon dioxide columns retrieved from SCIAMACHY by WFM-DOAS: year 2003 initial data set
M. Buchwitz, et al
Atmos. Chem. Phys., 5, 3313–3329, 2005
European Geosciences Union
http://www.atmos-chem-phys.net/5/3313/2005/acp-5-3313-2005.pdf

*

From #323
Here are sounder measurements of thermal emissions in CO2 sensitive parts of the spectra from 1997:

http://cimss.ssec.wisc.edu/goes/sndprf/18mar4pl.gif
from
APPLICATION OF GOES-8/9 SOUNDINGS TO WEATHER FORECASTING AND NOWCASTING
W. Paul Menzel, et al
Bulletin of the American Meteorological Society
Volume 79, Number 10, October 1998
http://cimss.ssec.wisc.edu/goes/sndprf/sndprf.htm

*

#325
The following page has java animations of satellite data for thermal emissions from various gases, including co2 at 14.1, 13.4 4.45 μm bands in the upper- and lower-level atmosphere temperatures at various times of day.

The CIMSS Realtime GOES Page
http://cimss.ssec.wisc.edu/goes/realtime/grtmain.html

*

From #323
Measurements from Mars with its lower temperatures and pressures:
http://starryskies.com/solar_system/mars/spirit/atmosphere01.html
The CIMSS Realtime GOES Page
http://cimss.ssec.wisc.edu/goes/realtime/grtmain.html

*

#325
Here the elevation of land being determined by co2 absorption strength:

Figure 20. The 2-μm CO2 absorption strength (A) can be converted to topographic elevation (B). The derived elevations matches the USGS Digital Elevation Model (DEM) (C). The CO2 absorption strength image (A) is brighter for increasing strength. Because the atmospheric path length is smaller with increasing elevation, the absorption strength decreases, becoming darker in the image. The DEMs (B, C) are brighter for increasing elevation, thus are inversely correlated with the CO2 strength in (A).
http://speclab.cr.usgs.gov/PAPERS/tetracorder/FIGURES/fig20.dem.abc.gif

Figure 21. The 2-μm CO2 absorption strength versus USGS DEM elevation shows a linear trend with an excellent least squares correlation coefficient.
http://speclab.cr.usgs.gov/PAPERS/tetracorder/FIGURES/fig21.co2_graph.gif

Imaging Spectroscopy:
Earth and Planetary Remote Sensing with the USGS Tetracorder and Expert Systems
Roger N. Clark, et al
Journal of Geophysical Research, 2003.
http://speclab.cr.usgs.gov/PAPERS/tetracorder/

From #343, #352

The distribution of carbon dioxide in ppm at 8 km as determined by the Atmospheric Infrared Sounder (satellite):

NASA AIRS Mid-Tropospheric (8km) Carbon Dioxide
July 2003
http://www-airs.jpl.nasa.gov/Products/CarbonDioxide/

Its infrared emissions demonstrate that it is not quite completely mixed even at this altitude.

Hank Roberts provided the following in #533

Cameras have long been available in this band, for example this one from Mars Viking:
http://ntrs.nasa.gov/search.jsp?R=81317&id=1&qs=Ne%3D20%26N%3D4294910007%26Ns%3DArchiveName%257C0

I have provided the following in #384 and #509:

AIRS – Multimedia: Videos: Animations
http://airs.jpl.nasa.gov/Multimedia/VideosAnimations/

Chuck Booth provided this on a different thread:

Visualization of the global distribution of greenhouse gases using satellite measurements, by Michael Buchwitz. The Encyclopedia of Earth. Posted July 31, 2007
http://www.eoearth.org/article/Visualization_of_the_global_distribution_of_greenhouse_gases_using_satellite_measurements

… and in response I gave him the following:

University of Bremen IUP/IFE SCIAMACHY WFM-DOAS: Main page

Line-by-line calculation of atmospheric fluxes and cooling rates 2
http://www.aer.com/scienceResearch/rc/m-proj/abstracts/rc.clrt2.html

Comment by Timothy Chase — 29 Aug 2007 @ 12:39 PM

556. Timothy, I said (541), “…The fact that you believe very strongly in what you profess and even feel you have solid evidence to back up your assertions does not warrant a scientific forum casting out or shunning all who don’t readily subscribe to your dogma…”

Frankly I think your rebuttal post (542) says pretty much the same thing, though not exactly, and with considerably more erudition! You say you’re saving lives while Alastair is asking dumb questions and (implied) ought to be canned.

No, I do not approve attacks on integrity. But claiming someone is wrong in his analysis, even if “him” is an eminent scientist, is not an ad hom attack on integrity.

History is filled with scientific (and other) advances stemming from folks of a different field. Questioning someone smarter than me, per se, does not make me a crackpot, even if I don’t have a union card. (That would mean I can maybe only question 2% of the population!!)

And you conclude by saying, in effect, you have no desire to jettison Alastair (or whoever)… as soon as he agrees with me.

I don’t have “a problem” with the attitude. I simply think it inappropriate and unhelpful.

Comment by Rod B — 29 Aug 2007 @ 12:56 PM

557. PS to #355

The last is a line-by-line calculation the net local direct cooling or heating effect of various greenhouse gases. Not quite the same thing as the thermal images/data, but I thought that I would include it.

Comment by Timothy Chase — 29 Aug 2007 @ 1:08 PM

558. Douglas (545), these are good fundemental questions and comments. I’m going to offer my 2 cents worth, though my credentials are likely less than almost everyone else.

At the top of the atmosphere, their is escape of long wave radiation which, at equilibrium, matches in energy terms that entering by solar radiation less albedo.

Absolutely. This is the primordial concept. The mother that gets it all going.

[paraphrased] There is an … window…where there are [few] blocking greenhouse gases that directly egress 17% of OLR.

True, though there are different estimates between 10-20% with 10-12% being most likely

re 3) GHG radiation is isotropic. I do not believe re-emission has to be at a longer wavelength, though it is more likely. [Others will disagree] I do believe CO2 absolutely can absorb a photon (re)emitted by another CO2 molecule, though the re-emission and 2nd absorption is likely at a different wavelength than the 1st absorption. See other posts. [Others disagree]

two thirds of the energy absorbed by the surface emanates from energy re-emitted in the atmosphere and only one third is attributable to directly absorbed solar energy.

This is right, though one or two folks disagree and are part of an ongoing set-to. Down-welling IR absorption is 140% of the earth surface plus atmospheric solar absorption.

5), 6), and 7) are all accurate

Will the “atmospheric window” get narrower with increasing CO2?

I believe not except on a limited and insignificant scale at the edges due to line spreading. Spreading won’t come close to closing the window with a relatively greatly larger bandwidth. Some here might disagree. (Actually, I don’t fully buy the spreading concept, but that’s my burden and you should ignore it.)

gaps in the blackbody spectrum in the case of OLR

I think the gaps clearly stem from the radiation that got absorbed by whatever and not re-emitted (or 1st emitted in some cases) as it tried to get to the top. I don’t think their is any inherent qualitative distinction between greenhouse gasses. Others more knowledgeable might disagree.

I have no credible comment to statements 3) and 4).

Comment by Rod B — 29 Aug 2007 @ 1:44 PM

559. Rod B. (#556) wrote:

Timothy, I said (541), “…The fact that you believe very strongly in what you profess and even feel you have solid evidence to back up your assertions does not warrant a scientific forum casting out or shunning all who don’t readily subscribe to your dogma…”

Frankly I think your rebuttal post (542) says pretty much the same thing, though not exactly, and with considerably more erudition! You say you’re saving lives while Alastair is asking dumb questions and (implied) ought to be canned.

Rod, check out #555.

You can see how many links we have provided to data showing the reemission of radiation from carbon dioxide at many places in the atmosphere. Actually a few are from water vapor, but not that many – and there were more links. I have also pointed out that we have plenty of data from other sources – most particularly that from planes.

The first time I remember Alastair having acknowledged just one of the links was #532, but even then he didn’t acknowledge that it was thermal emissions from carbon dioxide. And that was only after I pushed hard. The first time he actually acknowledged that they were thermal emissions was #537.

As I said, Hank Roberts (#530) gave us a fairly good quote just a little while back:

“As Patrick Moynihan once said, everyone is entitled to their own beliefs, but not to their own facts.”

We have plenty of data. Not talking theory. Not talking about anyone having to accept a given theory. Just the data. Just the facts. Don’t just take my conclusions because I say so, but at least accept the facts, the data, because it is there.

Rod B (#556) wrote:

No, I do not approve attacks on integrity. But claiming someone is wrong in his analysis, even if “him” is an eminent scientist, is not an ad hom attack on integrity.

This was the most recent attack on Raypierre’s integrity (#505):

If you are going to believe everything your are told, of course you will be told the models work. They will gloss over the Venus problem, the Tropical Lapse Rate Problem, and the Energy Balance Closure Problem, and even the Martian polar inversion layer problem (hardly recognised yet) but which Raypierre is going to “eliminate” from his book. See response to #58.

But there have also been the wild goose chases where we have been given links that have nothing to do with what he is trying to prove, quotes from material taken out of context which say something quite different from what he is trying to show, the extreme stubbornness with regard to his positions, the recycling of arguments long after they have been addressed multiple times, the rudeness, etc.. I had actually been wondering about his state of mind.

Not any longer.

It is almost as if you two have decided to play “good cop, bad cop.” He takes no offense at all, is perfectly polite and extremely helpful:

Alastair (#548) wrote:

Gavin,

Rod has lost a post, and now so has Timothy. I am finding that RealClimate is very slow to respond since it was reconfigured. Moreover now I am no longer able to preview my posts before sending them.

Thank you for the concern, Alastair. It is appreciated.

Alastair McDonald (#550) wrote:

But for an explanation of the principles of electronic and vibrational (vibronic) energy changes see http://en.wikipedia.org/wiki/Franck-Condon_principle

That is an extremely helpful link, Alastair! Makes a great deal of sense, too, given my understanding of quantum mechanics.

You see, Rod?

You take all the offense, he takes none at all. And for the longest time you have been feeding him questions so that he could go on and on about his shifting, incoherent theory. All of the sudden he goes from incoherent and offensive to hyper-polite and extremely lucid.

I am thinking mindgames. Gaslight. Hank suspected as much at various points as well, at least with respect to Alastair.

Seeing something like this before plenty of times among creationists who tried to act extremely dumb just to see how hard the pro-evolution group would work to respond to them. Their way of demonstrating how much brighter they were than the supposedly well-educated pro-evolution people.

Here I think the purpose is different. Haven’t really got a clue. But I can at least see the pattern. Fairly sophisticated, actually.

Comment by Timothy Chase — 29 Aug 2007 @ 2:05 PM

560. Re #454 where Douglas Wise Says:

Gentlemen,
I am not sure if this is also addressed to me since it seems that I am regarded as a non-person (N-P perhaps ) here, but it is an interesting post so here is my 2 cents worth.
Statements:
1) At the top of the atmosphere, their is escape of long wave radiation which, at equilibrium, matches in energy terms that entering by solar radiation less albedo.

If more energy enters the atmosphere than leaves it, then it will warm and vice versa. At the top of the atmosphere the only way energy can enter or leave is by radiation, but the balance need not be maintained by a change in outgoing longwave radiation. It can also be achieved by a change in albedo. (This is obviously true but is often ignored.)

2) There is an “atmospheric window” in the wavelength range of roughly 8.5-13.5 microns where there are no blocking greenhouse gases except for trophospheric ozone. This window allows egress of radiation directly from the surface. However, in energy terms, this represents only about 17% of OLR.

Clouds can and do block that window 30% of the time which may be the reason the window is only 10% wide. See energy balance diagram below.

3) Radiant energy absorbed by the atmosphere will be re-emitted in all directions. It will be re-emitted at longer wavelengths than that at which it is absorbed, the re-emitting wavelengths being longer at lower temperatures and pressure. This implies that a species of molecule which absorbs at one wavelength cannot absorb a photon re-emitted by a molecule of its own kind.

This is not correct. Absorbed radiation tends to be remitted within an angle of 40 degrees of its incidence. In general it cannot be re-emitted at a higher frequency that that at which it is absorbed because that implies it has acquired more energy. It can re-emit at the frequency at which it absorbs. That is how a CO2 laser works.

4) Approximately two thirds of the energy absorbed by the surface emanates from energy re-emitted in the atmosphere and only one third is attributable to directly absorbed solar energy.

That is what this energy diagram from the IPCC TAR shows: http://www.grida.no/climate/ipcc_tar/wg1/fig1-2.htm

5) Radiative transfer is a much slower energy transport mechanism than convection such that the latter process gets most of the effluent energy to the top of the atmosphere.

Radiative transfer through the window is at the speed of light, much faster than convection! Moreover, convection stops at the tropopause, the boundary between the troposphere and the stratosphere. But inspection of the TAR energy balance diagram reveals that the combined Thermals and Evapotranspiration which make up the convection are about 100 W m-2 whereas the net long wave radiation is only about 65 W m-2.

6) Radiation leaving the surface has a near blackbody spectrum. OLR does not. There are gaps attributable to greenhouse gases.

The OLR spectrum does not contain gaps as such. There are bands where the OLR is less than that at the surface. However, there are gaps between the lines which make up the bands. The deepest band is the CO2 15 micron band. The widest bands are caused by water vapour.

7) There is little water vapour in the high atmosphere.
True. Above the troposphere, the stratosphere is very dry although it is wetted by contrails from aircraft. Above that altitude, the noctilucent clouds which may be due to water vapour seem to be increasing in frequency. Of course some clouds are formed from ice crystals not water aerosols.

Questions:
1) Will the “atmospheric window” get narrower with increasing CO2? I can picture the possibility of a curtain hanging on the right (long wave) side of the window which may become partly drawn, thus obscuring part of it. I can see this to be particularly likely if energy is leaving a dry surface with little overlying water vapour but less likely over, say, oceans.

What will happen is that the lines will get broader. So they will tend to close like a venetian blind rather than a curtain. However the bands, which you can think of as a curtain, will not get much broader. But see the introduction to this item by Ray Pierrehumbert to get his take on it. It is quite complicated.

2) Ray Ladbury (#478) talks of gaps in the blackbody spectrum in the case of OLR. However, Step 2 of “The CO2 problem in 6 Easy Steps” (Real Climate Archive, August 6th) cites a graph showing IR spectra taken from space. This graph has no explanation. Rightly or wrongly, I interpret it to be showing that Ray’s gaps are attributable to CO2 and ozone but not to water vapour. Is this correct? If so, is it explained by the dry upper atmoshere.

Here is one example of the OLR spectrum http://www.john-daly.com/spectra.gif If there were no greenhouse gases it would look like the 300 K line. If you look to the left of the CO2 band from the wavenumber of 550 cm-1 you can see that the line is at around 275K. This 25 K drop in brightness temperature is caused by water vapour absorption. You can also see that by inspecting this diagram http://origins.jpl.nasa.gov/library/exnps/f4-3.gif It has increasing wave number running right to left whereas on the Guam diagram they run left to right. But it is also calibrated in wavelength so the 15 um (micron) CO2 band can be easily identified and correlated on both diagrams.

3) How does thermal energy reaching the top of the atmosphere by convection become OLR and make good its escape to space?

Although clouds appear white in visible light, they behave like blackbody radiators in the infrared. They are not gases when the form clouds. Clouds that form near the top of the troposphere can radiate to space without that OLR being absorbed in the water vapour bands.

4) Various correspondents suggest that infrared in the near-15 micron range can be demonstrated in the high atmosphere. Could this emanate from re-emissions of atmospherically- absorbed solar irradiation, given that re-emissions from lower level CO2 molecules couldn’t apparently be an explanation?

The GUAM IR spectra shows that there are emissions in the 15 micron band with a brightness temperature of around 215K. It is generally believed that this radiation originated at the surface and 85K of brightness temperature was lost by absorption. This report from the US Air Force http://www.dodsbir.net/sitis/view_pdf.asp?id=DothH04.pdf (first posted by David Donovan) points out that the band does not alter on a diurnal basis therefore it is not due to solar activity.

Well guys, what’s wrong with that?

Comment by Alastair McDonald — 29 Aug 2007 @ 3:59 PM

561. Re #555 Timothy Chase

Many thanks for this great list of references. It will take me quite a while to work through it.

Comment by AEBanner — 29 Aug 2007 @ 5:04 PM

562. Re Alastair (#560)

Alastair,

You are brilliant! Obviously a top-notch climatologist. Are you Gavin?

Comment by Timothy Chase — 29 Aug 2007 @ 5:20 PM

563. Re #554 Rod B

I simply offered a suggestion to try to explain your problem. Perhaps it is not adequate, but it was the best I could do.

Quote

Comment by AEBanner — 29 Aug 2007 @ 5:33 PM

564. My only other guesses at this point would be M, J and R. (First initials.)

Comment by Timothy Chase — 29 Aug 2007 @ 6:34 PM

565. Timothy (559, et al), you say you buried Alastair in evidence and data and yet he persists, so can/cane him. Exactly what I accused you of. If one is plain obdurate, he simply should be ignored, not banished — though proving being obdurate is somewhat subjective.

I’m not charging to Alastair’s personal defense; nor am I accusing you individually other than as a current example. I’m just countering what I see as a negative non-specific characteristic. I’ve done it before with wholly different players. I believe it valid. I disagree with a couple of Alastair’s main contentions (and have so stated), but have less than expert certainty and simply want to understand what he is saying before I discard it out of hand. And while I have a level of respect and affinity for Alastair, it is no different for what I have for nearly everyone posting here, including you. I don’t think Alastair needs me to run his defenses; if he does he is in a heap of hurt!

I don’t agree that the attack on Raypierre that you cite is a clear violation of “non ad hom” attacks on integrity; though I admit it’s pushing the envelope, and to that end I would disapprove.

Comment by Rod B — 29 Aug 2007 @ 8:21 PM

566. Alastair, re-emission is within 40 degrees of the angle of incidence??? How do it do that??

Comment by Rod B — 29 Aug 2007 @ 8:36 PM

567. Re #566 That is in the case of a laser which I saw some where. The photon will not necessarily hit the centre of the molecule, so the stimulated photon will not be emitted in exactly the same direction as the stimulating photon.

But if it is a spontaneous emission I expect that it will be in any direction. Since we are mainly considering spontaneous emission, it may be better to ignore that point.

I am not sure myself how the selection rules etc. work, so I am interested in what anyone else turns up. BTW here is a description of a CO2 laser: http://www.laserk.com/newsletters/whiteCO.html

Comment by Alastair McDonald — 29 Aug 2007 @ 11:29 PM

568. RE #554 Rod B

I don’t think you are looking at the problem the right way. If you refer to the following ref., particularly pages 9 & 10, you will find some help.

http://www.dodsbir.net/sitis/view_pdf.asp?id=DothH04.pdf

This is a joint research paper by Spectral Sciences, Inc., the US Air Force, and Boston College. It is highly authoritative, and immense. Pages 9 and 10 seem to show that an excited Co2 molecule can relax in stages, so that the energy of a particular emitted photon is less than the initially absorbed photons, so making re-absorption impossible. This supports the DoE statement.

Comment by AEBanner — 30 Aug 2007 @ 9:13 AM

569. Re 568 where AEBanner says:

Pages 9 and 10 seem to show that an excited Co2 molecule can relax in stages, so that the energy of a particular emitted photon is less than the initially absorbed photons, so making re-absorption impossible.

Yes but we are only interested in the v2 15 micron (667 cm-1) band of CO2 for outgoing longwave radiation. See the two spectra:
http://www.john-daly.com/spectra.gif
and
http://origins.jpl.nasa.gov/library/exnps/f4-3.gif

On pages 9 & 10 both the diagram and the table show the 15 um line (01101) dropping straight to the ground state (00001) thus it does not pass through any intermediate levels, and so the emission could and does re-excite another molecule to the 01101 state from 00001.

On page 10 of http://www.dodsbir.net/sitis/view_pdf.asp?id=DothH04.pdf it says:

Around 15 μm, the bands result from a change of one quantum of bending mode v2. These transitions are very prominent in the emission from the earth’s atmosphere, and the sharp Q-branch can be clearly seen in spectra of the earth taken at spectral resolutions of a few wavenumbers from a balloon, rocket or satellite [Stair et al., 1985; Wise et al., 1995]. The reason for this is very simple. The spectrum of a 300 K blackbody peaks near 10 μm. Therefore, there are large numbers of 15 μm photons in the radiation, termed earthshine, emitted by the earth’s atmosphere as well as by the earth itself. Since the 15 μm CO2 band is very strong, its transitions to the ground state and low lying excited states are severely self-absorbed. This fact has important consequences for atmospheric radiation which we will point out as we go along.

I have not found the important consequences further on yet. Does anyone else know what they are?

Comment by Alastair McDonald — 30 Aug 2007 @ 10:15 AM

570. No, it doesn’t say impossible. It may be accurate for identical temperature and pressure, it may be correct describing a laser. It doesn’t say whether, for example, a higher, colder molecule can absorb a photon emitted by a lower, warmer molecule. I’m not a physicist, logic is a weak reed in dealing with quantum mechanics, words are not much use. You need to get someone who actually can do the math to address this question, and perhaps someone who can do the experiment.
Stating a conclusion because some text “seems to show” one thing isn’t sound.

Comment by Hank Roberts — 30 Aug 2007 @ 10:33 AM

571. AEBanner,
Interesting reference. By what these guys say, collisional excitation and relaxation dominate over their radiative counterparts out to ~65 km–well into the stratosphere. Since the temperature profile is increasing with altitude in the stratosphere, I think this implies that LTE is applicable throughout the range where CO2 contributes to the greenhouse effect. True?

Comment by Ray Ladbury — 30 Aug 2007 @ 10:37 AM

572. Huh.

Rod B (#565) wrote:

If one is plain obdurate, he simply should be ignored, not banished.

Banished?

I would save that for trolls – although that obviously would be a matter for the moderators. In all the time that I have been on Real Climate, I have never considered Alastair to be a troll – as I understand the term.

Ignored?

I will agree with you there – if someone refuses to acknowledge any and all facts and evidence against their position and does so over a long enough period of time. But how much time is an individual decision, and if they genuinely change I would consider what is in the past to be simply in the past. I believe you might understand why.

My reasons for giving people a great deal of latitude, time and chances are also intimately tied to the necessity of judging their words and actions even if I were to always keep my judgements to myself. But I would prefer not to go into the details of that as I prefer people to discover things for themselves – including their personal ethics. I don’t mind sharing principles, but like ships, I prefer to give worldviews a wide berth. A story for another time, perhaps.

In any case, I haven’t had the chance to keep up with the discussions of science that have been taking place and there are a lot of interesting leads to follow up…

Comment by Timothy Chase — 30 Aug 2007 @ 10:43 AM

573. AEBanner (#561) wrote:

Well, I am just returning the favor, I believe. I am going to have to make that paper a higher priority than I have managed so far. But there are other things to follow up, too, and I have been a little distracted as of late.

Comment by Timothy Chase — 30 Aug 2007 @ 11:00 AM

574. AEBanner (568), I really didn’t see in your ref that re-emission then absorption by the same molecular species is prohibited. They didn’t talk of it (much?) as opposed to other, implied more interesting or significant, exchanges. Maybe it (CO2 to CO2 at certain levels) doesn’t happen much, for a host of reasons. But nothing I read refuted my base level logic: I agree CO2 can relax in stages (which is changing energy levels in steps). But the energy levels available in a CO2 molecule are fixed and invariant, though some more likely to filled than others, sometimes significantly. If a molecule can emit at a discrete energy level (hf), it can absorb at the same level. And to repeat, if a molecule can absorb a photon at energy A, it has no way of telling if that photon originated from earthshine, another molecule, or another CO2 molecule.

I inferred for the ref that it’s possible for a CO2 (or any other specie) molecule to have a greater tendency to emit at certain levels than, for complicated non-linear reasons, its tendency to absorb at the same level. But nothing says it can’t.

Comment by Rod B — 30 Aug 2007 @ 12:05 PM

575. Hank said, “…logic is a weak reed in dealing with quantum mechanics, words are not much use. You need to get someone who actually can do the math to address this question, and perhaps someone who can do the experiment.
Stating a conclusion because some text “seems to show” one thing isn’t sound….”

Unfortunately this is very true! More unfortunately we all have to play the cards we’d been dealt.

Comment by Rod B — 30 Aug 2007 @ 12:20 PM

Local thermodynamic equilibria, partial local thermodynamic equilibria, non-local thermodynamic equilibria, transitions between excited states, etc. Interesting stuff indeed!

According to the paper, for CO2, deviation from LTE begins around 30 km as the average duration between collisions for a given molecule dips below that of one one-millionth relative to half-life. For two vibrationals it starts becoming noticeable by 50 km. For the third around 75 km. The three vibrational transitions come from two to ground and one excited state to the other.

Comment by Timothy Chase — 30 Aug 2007 @ 1:11 PM

577. PS to #576

One point that the author of…

http://www.dodsbir.net/sitis/view_pdf.asp?id=DothH04.pdf

… makes is that the deviation from LTE is a gradual one. It is also worth noting that non-LTE has been incorporated into the GCMs for some time now.

Comment by Timothy Chase — 30 Aug 2007 @ 2:15 PM

578. Haven’t read the latest posts, but I have a tentative thought that says I have to partially withdraw my vociferous assertions re re-emission and secondary absorption. I have been saying that CO2 (can) drop energy levels in steps and that if it drops to level B and re-emits from that level, that same level B exists in every other CO2 molecule and therefore can be absorbed. This assumes that the dropped energy delta from A to B was transferred to translation or something. But, if the A to B delta energy drop is the energy released with a photon, that photon will not (or hardly ever…) match any fixed energy level in any CO2 molecule, and therefore can not be absorbed by CO2. [slap! Slap!] I don’t know the probability of each, but, as a rough rule, it looks like I was only half right, as were the folks I claimed were totally wrong. Any of this make any sense to anyone?

Comment by Rod B — 30 Aug 2007 @ 4:05 PM

579. There have been several comments pointing out that I may have drawn the wrong conclusions from
http://www.dodsbir.net/sitis/view_pdf.asp?id=DothH04.pdf

OK, but the following paragraph comes from the DOE
http://www.eia.doe.gov/cneaf/alternate/page/environment/appd_a.html

Quote
What happens after the GHG molecules absorb infrared radiation? The hot molecules release their energy, usually at lower energy (longer wavelength) radiation than the energy previously absorbed. The molecules cannot absorb energy emitted by other molecules of their own kind. Methane molecules, for example, cannot absorb radiation emitted by other methane molecules. This constraint limits how often GHG molecules can absorb emitted infrared radiation. Frequency of absorption also depends on how long the hot GHG molecules take to emit or otherwise release the excess energy.
Unquote

Comment by AEBanner — 30 Aug 2007 @ 4:15 PM

580. Yep, but each time you quote it, since I can’t find the statement anywhere else, and since there’s no footnote for it (and every other statement around it is footnoted), and we don’t know when that particular text was revised, I have to wonder whether anyone else anywhere else has ever said that. I’m sure it’s moving up in Google’s hit list a bit since you’ve been reposting it at RC. Seriously, if you do believe it’s important, tracking down the author or someone who can check it for you might be a useful thing to do.

Comment by Hank Roberts — 30 Aug 2007 @ 5:01 PM

581. Re #578

I am only saying that CO2 molecules can reabsorb their 15 um radiation. If you look at the spectra, then CH4 only plays a small part in the absorption. However, there are two very important H20 bands and I have no information on whether they are or are not self-absorbing.

It gets more and more complicated, but that is the way quantum mechanics is, so there is no way to avoid it :-(

HTH,

Cheers, Alastair.

Comment by Alastair McDonald — 30 Aug 2007 @ 5:04 PM

582. Re #579 HR

Yes, I agree.

I do believe it is important, and I have already e-mailed DoE for further information, but sadly without response.

Another question. Can CO2 in the atmosphere absorb a photon directly into the 01101 energy level, or does the absorbed energy have to go into a higher level first? (re Dothe’s Fig 2)

Comment by AEBanner — 30 Aug 2007 @ 5:29 PM

583. Re #578 Rod B

I think I follow your logic for transitions between higher energy levels, but I don’t think it applies for transitions directly to the ground state. Does it? Re last sentence in my #582.

Comment by AEBanner — 30 Aug 2007 @ 5:57 PM

584. re 582, 583: this might seem to simplistic, but if a photon from earth’s blackbody radiation or any gas’s re-emission (or from wherever) exists at the equivalent level 01101, I assert the CO2 molecule will (can) gobble it straight away (subject to the other factors that determine whether a photon is absorbed or not.)

Comment by Rod B — 30 Aug 2007 @ 10:44 PM

585. Re #584

OK, but what are the “other factors”?

Comment by AEBanner — 31 Aug 2007 @ 10:37 AM

586. Re #580 HR

I’m sorry. I seem to have upset you. I’ll try to refrain from quoting THAT PARAGRAPH again.

Comment by AEBanner — 31 Aug 2007 @ 10:40 AM

587. > what are the “other factors”?

Google has worked up a pretty good tool for parsing natural language sentences into useful searches, I was informed by a programmer recently.

Comment by Hank Roberts — 31 Aug 2007 @ 12:00 PM

588. Re #587

Hank, thanks!

No 1 on that search – “Multiple Photon Excitation” looks like a great hit!

The quantum effect on polyatomic molecules is very complicated and has only recently (in the last decade or two) been investigated. It will be interesting to read what they have to say.

Now having started to read it, I do not think it gives an easy answer to the greenhouse effect. There are very many more polyatomic molecules than CO2 and H2O. I will try to read it in the hope that I might learn something relevant to global warming.

Comment by Alastair McDonald — 31 Aug 2007 @ 4:41 PM

589. Don’t leap at the first hit — it has useful history about what the theorists believed before the laser developers started doing the “impossible” and how the theorists adapted to that. But Google doesn’t come with a “wise and useful hits first” option — I’d suggest using “search within results” at least, and consider using Scholar instead once you get some terms down that Google finds.

Comment by Hank Roberts — 31 Aug 2007 @ 5:26 PM

590. AEBanner (585), much is probably beyond my scope but there is a probability function, which in turn might be temperature and pressure affectsd, that helps in determining whether a photon is absorbed or not.

Comment by Rod B — 31 Aug 2007 @ 9:39 PM

591. Many thanks to Ray Ladbury (#553), Rod B (#558) and Alastair McDonald (#560) for answering my post of #545. I think my understanding is advancing, albeit in small steps. If I may, I’d like to raise a few further issues.

I asked to what extent one might expect the “atmospheric window” to shrink as a result of increasing atmospheric CO2. ( My analogy of the right hand curtain being partly pulled across was regarded as less pertinent than the venetian blind analogy. However, the latter seems quite inappropriate because the thread was initiated following Raypierre’s observation that the 13.5 to 17 micron band was already fully saturated. The “window” could thus only be partly obscured by increasing absorption by CO2 of radiation in 12.5 – 13.5 range). Ray said the “window” might get “a very small bit narrower” and Rod suggested it would be unaffected “except on a limited and insignificant scale”. Would it be possible for anyone to tell me what this small effect would be as a percentage of the total extra warming attributable to increasing atmospheric CO2. I take it that it would be a small percentage only, with the balance made up from blocking of radiation at higher altitudes. However, if I understood Raypierre’s essay correctly, it would take two molecules of CO2 high in the trophosphere to have the same blocking effect as one at low altitude. This leaves me to wonder whether it is remotely possible that extra CO2 won’t have quite as much warming effect as the modellers assume, albeit plenty for us to be concerned about.

I remain somewhat mystified as to the translation mechanism for non radiative thermal energy into radiative thermal energy. In explanation, Ray said “heat transported by convection gets to the upper troposphere and heats the air there. The warmed air collides, exciting rotational and a few vibrational states. These high altitude molecules then radiate in the bands where they can radiate.” Which molecules does he mean? I understood (rightly or wrongly) that oxygen and nitrogen neither absorbed nor emitted radiation. This is no doubt a facile question but how are radiating photons created from non radiative thermal energy? Suppose, for example, that there were no ghgs in the high atmosphere. I would expect a colder climate but would I be correct? Effluent thermal energy can only escape to space as OLR. How?

Finally, AEBanner’s citing of the DoE quote which suggests that energy emitted from one CO2 molecule can’t be absorbed by another. Rod B descibes this as a “bomb” which, if correct, would blow a hole in AGW theory. I don’t see that this need be the case. Alastair states that “Guam IR spectra (of OLR) shows that there are emissions in the 15 micron band with a brightness temperature of 215 K. It is generally believed that this radiation originated at the surface and 85 K of brightness temperature was lost by absorption”. Surely, this general belief can’t be correct if those believing are of the view that radiation in the 15 micron band goes directly from the surface to space through the “atmospheric window”. If clouds are black body radiators, they would emit radiation at this wavelength which wouldn’t be blocked by water vapour and only somewhat by CO2 because it is fairly unsaturated at this level. Furthermore, I assume that photons absorbed by water vapour in the 5-9 micron band might result in longer wavelength re-emissions in the 13-17 micron band. As far as I am concerned, the DoE statement helps explain the concept of saturation which was meaningless to me before, given that many were claiming that re-emission always occurred at the same wavelength as absorption.

Quite possibly, I’m still holding the stick by its wrong end and I hope that, if so, you’ll enlighten me.

Comment by Douglas Wise — 1 Sep 2007 @ 7:44 AM

592. I kind of doubt that multi-photon absorption has that much of an effect on the greenhouse effect. All the searching I tried on it kept on turning up lasers – although there was a declassified document on the atmospheric effects of a nuclear bomb blast released in 1995. I am hoping that the FBI isn’t tracking those who download the document. Then again, having hung around Lenny Flank for so long and being married to someone who protested the first gulf war…

Comment by Timothy Chase — 1 Sep 2007 @ 10:19 AM

593. RE #578 Rod B

I think you are on the right lines here. I believe that photon emission is more likely to be collision induced than spontaneous, and some of the energy of the photon about to be emitted could be transfered into kinetic energy of the colliding molecules. So the emitted photon would have less energy than the initially absorbed photon gave to the absorbing molecule. Therefore, the emitted photon might well not have the required energy to be absorbed by another molecule of the same kind.

Comment by AEBanner — 1 Sep 2007 @ 3:56 PM

594. re 591, “….Therefore, the emitted photon might well not have the required energy to be absorbed by another molecule of the same kind.”

Or, by chance, any kind of molecule?? It would seem a lower energy photon might have a shot at H2O rotation(?) absorption. But we really don’t have a clue, do we? Can’t the re-emitted photon be almost at any (usually lower) energy level?

Comment by Rod B — 1 Sep 2007 @ 11:05 PM

595. Rod B (#594) wrote:

re 591, “….Therefore, the emitted photon might well not have the required energy to be absorbed by another molecule of the same kind.”

Or, by chance, any kind of molecule?? It would seem a lower energy photon might have a shot at H2O rotation(?) absorption. But we really don’t have a clue, do we? Can’t the re-emitted photon be almost at any (usually lower) energy level?

If one were talking about a single molecule which is not encountering collisions, then the energy emitted would have to be equal to the energy absorbed by that molecule. However, it wouldn’t necessarily be the same photon. You may have transitions from one excited state to a less excited state in accordance with the principle that Alastair (#548) cited earlier:

But for an explanation of the principles of electronic and vibrational (vibronic) energy changes see http://en.wikipedia.org/wiki/Franck-Condon_principle

It is worth noting, however, that the Franck-Condon principle is only an approximation, although I suspect a fairly good one.

But when we speak of the 15 micron being stongly self-absorbing, I take this to mean that it does not tend to make these transitions to lower excited states. This becomes important particularly at higher altitudes where it strongly diverges from LTE and is largely responsible for the cooling of the mesosphere and the lower thermosphere. This becomes especially important as this influences the circulation of the mesosphere, gaining energy from both nitrogen and O2, but particularly from collisions with photodisassociated O, or at least this is what I get from the abstract for:

New estimation of the 15 micron CO2 band cooling rate of the lower thermosphere
Ogibalov, V., Shved, G., Khvorostovskaya, L., Potekhin, I., Uzyukova, T
34th COSPAR Scientific Assembly, The Second World Space Congress, held 10-19 October, 2002 in Houston, TX, USA., meeting

Incidentally, I would assume that the cooling off of O strongly influences the chemistry – inasmuch as it will permit O to reassociate.

*

As this example indicates and as we have been reminded numerous times before, you can’t treat the molecules as if they exist in isolation. Collisions will happen, and in fact it is by means of such collisions that the greenhouse gases will tend to acquire their energy.. One question which occurs to me is whether such collisions may result in a molecule being knocked from a lower state of excitation to a higher state. In any case, for that proportion of photons which reach the surface after reemission will naturally warm it, and for this reason much of the energy will still be reabsorbed by greenhouse gases, albeit indirectly, either through thermal reemission from the surface, or what is more likely, moist air convection.

*

One other interesting point:

Abstract: Published data from rocketborne cryogenic-IR-spectrometer observations of the atmosphere in limb-viewing geometry are compiled in graphs and analyzed. It is shown that, except in very strong atmospheric emission bands (15-micron CO2 and 5.3-micron NO), off-axis leakage is the dominant radiance source at tangent altitudes above 100 km. The data are consistent with a bidirectional reflectance distribution function of 0.002-0.004 at 1 deg or telescope degradation factors of about 35. The implications of these findings for spectral modeling are considered.

Evidence for off-axis leakage radiance in high-altitude IR rocketborne measurements
Smith, Donald R., USAF, Geophysics Laboratory, Hanscom AFB, MA
IN: Stray light and contamination in optical systems; Proceedings of the Meeting, San Diego, CA, Aug. 17-19, 1988 (A89-41504 17-74). Bellingham, WA, Society of Photo-Optical Instrumentation Engineers, 1989, p. 30-36.

This would further suggest that the weaker emission bands radiate in part due to solar radiation, whereas with these stronger emission bands it is almost entirely due to thermal radiation from the earth’s surface.

Comment by Timothy Chase — 2 Sep 2007 @ 11:16 AM

596. Since we’re into the wildly uneducated speculations here, I wonder if “vibration” or “rotation” is quantized or continuously variable. I realize this isn’t’ even poetry, rather doggerel, attempting to describe something in words on the scale of single molecules.

Is it possible one can “spin up” a rotation by firing various-energy photons at it, and as the rotating particles wiggle thir electromagnetic environment, that will then can shake out new emitted photons whenever it happens to hit the right frequency to match ….

My literal blue-sky wishful thought is for some idea that could, with only a catalytic trickle of energy, tune the gases at the top of the atmosphere to highly favor emission of infrared photons (or, heck, I suppose any other kind) at the height where they have a good chance of departing the planet.

Of course that suggests any sufficiently advanced civilization elsewhere could occupy a planet indistinguishable from a cool red dwarf star from the outside.

This is merely an attempt to post something whacko enough to interest the science-fiction-friendly Ray to rejoin us (grin).
Happy Labor Day …

Comment by Hank Roberts — 2 Sep 2007 @ 12:33 PM

597. A couple of the easy ones: the molecules in the upper atmosphere heated through convection are predominately N2 and O2. When one collides with CO2, say, it can transfer some of its translation energy to CO2 translation or, as I understand…, directly to the CO2’s vibration or rotation internal energy (dropping the real temperature of the air with the latter process.) Molecules can transfer energy among their translation, vibration and rotation modes — within the same molecule or inter-molecularly. Whether and when this is done is problematic, gets very complex and depends on quantum mechanical pixie dust, among other variables. Only rotation and vibration can absorb/emit LW electromagnetic radiation energy (photons). Rotation and vibration energy levels are quantized, much like electronic quantum levels, though more plentiful.

Comment by Rod B — 2 Sep 2007 @ 10:44 PM

598. Re #596 and 597

For quantised molecular vibration states see

http://www.fas.org/sgp/othergov/doe/lanl/pubs/00285799.pdf

and

http://www.umsl.edu/~orglab/documents/IR/IR2.html

Comment by AEBanner — 3 Sep 2007 @ 6:08 AM

599. Re #594 Rod B

Reference to Fig 2 in

http://www.dodsbir.net/sitis/view_pdf.asp?id=DothH04.pdf

and to the infrared data in HITRAN for CO2 shows that the lowest radiant energy transitions are in the 15 micron region. Inter-molecular collisions can transfer some energy from a photon about to be emitted from a CO2 molecule into kinetic energy of translation, and so the energy of the photon is reduced, so making re-absorption by another CO2 molecule very unlikely (impossible?). This is in line wih the DoE statement.

Perhaps the lower energy photon may be absorbed by a molecule of water,as suggested by Rod B in #594, but this could only occur at altitudes of less than about 4 Km.

Comment by AEBanner — 3 Sep 2007 @ 10:19 AM

600. > altitudes
There’s not _zero_ water in the stratosphere, there’s enough of an increase to delay the recovery of the ozone layer, for example:
http://www.giss.nasa.gov/research/news/20010417/

Comment by Hank Roberts — 3 Sep 2007 @ 11:43 AM

601. Douglas Wise (#591) wrote:

However, if I understood Raypierre’s essay correctly, it would take two molecules of CO2 high in the trophosphere to have the same blocking effect as one at low altitude. This leaves me to wonder whether it is remotely possible that extra CO2 won’t have quite as much warming effect as the modellers assume, albeit plenty for us to be concerned about.

Well, how closely we know how carbon dioxide radiates which have some effects upon the atmospheric calculations. While we have some difficulty following all of the effects, this is actually what is most accurately known as the result of the physics of radiative transfer, and it is quite mature. There are some aspects of the interaction between chemistry and radiative transfer which are perhaps less mature, but that is a different matter. However, as I have pointed out on numerous occasions, we have fairly accurate measurements of the thermal emissions and spectra from carbon dioxide and other greenhouse gases, so we would seem to have a fairly accurate understanding of the radiative transfer which is currently taking place within the atmosphere.

Now with regard to the warming effect within the atmosphere of doubling CO2, obviously something comes out of the models – a climate sensitivity which is not assumed but is instead a product of a given model itself. However, the climate sensitivites which drop out of the models are not our only way of knowing what the climate sensitivity is. We have the paleoclimate record from the past 400,000 years, and it shows rather consistently a climate sensitivity of about 2.8 degrees.

One point, though: climate sensitivity is undoubtedly partly dependent upon the distribution of the continents on the globe as it incorporates other feedbacks. So for example, if you have the continents distributed mainly around the equator, you can have a form of global runaway albedo effect. Once you have passed the tipping point on either side, the climate sensitivity to CO2 doubling will be much smaller. This is largely what made possible Snowball Earth in the remote past – although it was more likely a slushball. But the continents won’t be moving that much in the next 100,000 years, so I believe it is a safe bet that our climate sensitivity will be roughly the same as it has been for the past 400,000 years.

Douglas Wise (#591) wrote:

I remain somewhat mystified as to the translation mechanism for non radiative thermal energy into radiative thermal energy. In explanation, Ray said “heat transported by convection gets to the upper troposphere and heats the air there. The warmed air collides, exciting rotational and a few vibrational states. These high altitude molecules then radiate in the bands where they can radiate.”

Well, outside of the partial local thermodynamic equilibria and the non-local thermodynamic equilibria (both of which are fairly narrow phenomena – although as we have seen from some of the papers that have entered the discussion, not really that narrow, principally in the mesosphere and above – and this will influence the manner in which energy is dissipated in those layers, and thus atmospheric circulation at those levels which will influence atmospheric circulation in the stratosphere and troposphere), the condition of local thermodynamic equilibrium holds. Where there is a local thermodynamic equilibrium certain other principles hold.

For example, you will have a Maxwellian distribution of kinetic energy, collisions will take place much more frequently that the half-life of reemission (this is actually what determines whether a local thermodynamic equilibrium applies to a given wavenumber – with a million collisions per half-life being enough to more or less insure it), and so the radiative “Planck” temperature will be equal to the kinetic Maxwellian temperature. Under LTE we know that Kirchoff’s law applies, which means that the emission at a given wavelength is equal to the absorption.

As such, moist air convection increases the temperature of the troposphere which increases the amount of thermal radiation which it emits by increasing the Maxwellian temperature and thus the Planck temperature.

Douglas Wise (#591) wrote:

Which molecules does he mean? I understood (rightly or wrongly) that oxygen and nitrogen neither absorbed nor emitted radiation. This is no doubt a facile question but how are radiating photons created from non radiative thermal energy?

Under LTE the local temperature for any given gas will be equal to the local temperature of the atmosphere itself. Thus the collisions from oxygen and nitrogen are sufficient to insure that the Planck temperature of greenhouse gases have sufficient energy to radiate at the corresponding wavenumbers.

Douglas Wise (#591) wrote:

Suppose, for example, that there were no ghgs in the high atmosphere. I would expect a colder climate but would I be correct? Effluent thermal energy can only escape to space as OLR. How?

Under those conditions the radiation wouldn’t be absorbed by the atmosphere, and as such would be immediately radiated into space rather than having some proportion radiated back to and warming the surface.

Douglas Wise (#591) wrote:

Finally, AEBanner’s citing of the DoE quote which suggests that energy emitted from one CO2 molecule can’t be absorbed by another. Rod B descibes this as a “bomb” which, if correct, would blow a hole in AGW theory. I don’t see that this need be the case. Alastair states that “Guam IR spectra (of OLR) shows that there are emissions in the 15 micron band with a brightness temperature of 215 K. It is generally believed that this radiation originated at the surface and 85 K of brightness temperature was lost by absorption”.

Well, what Alastair is describing there is a result of the greenhouse effect – but a bit removed from the LTE that applies within a given part of the atmosphere. In the absence of an atmosphere, or alternatively, in the absence of greenhouse gases, the effective temperature of the earth would be equal to the surface temperature. However, greenhouse gases diminish the efficiency with which the earth is able to radiate thermal energy into space, and as such they diminish its effective temperature – that is, the temperature that the earth appears to have as viewed at a distance. This is in essence the reduction in brightness temperature which he is speaking of, although unlike the effective temperature, the brightness temperature will be different for different parts of the spectra, being dependent upon the emissivity of greenhouse gases for those parts of the spectra as well as their distribution in the atmosphere.

Douglas Wise (#591) wrote:

Surely, this general belief can’t be correct if those believing are of the view that radiation in the 15 micron band goes directly from the surface to space through the “atmospheric window”.

Well, I am a little less sure in this area, but given the fact that CO2 is described as being strongly self-absorbing in this part of the spectra, I am of the view that it doesn’t radiate directly into space at this wavelength. Strongly self-absorbing would seem to imply that it is able to reabsorb at that wavelength and does so quite well. Nevertheless, much of the radiation with which the upper atmosphere is able to emit at that level is due to carbon dioxide at that wavelength.

Douglas Wise (#591) wrote:

If clouds are black body radiators, they would emit radiation at this wavelength which wouldn’t be blocked by water vapour and only somewhat by CO2 because it is fairly unsaturated at this level. Furthermore, I assume that photons absorbed by water vapour in the 5-9 micron band might result in longer wavelength re-emissions in the 13-17 micron band. As far as I am concerned, the DoE statement helps explain the concept of saturation which was meaningless to me before, given that many were claiming that re-emission always occurred at the same wavelength as absorption.

It does sound like a saturation of sorts, doesn’t it? But not quite the same thing. The saturation of which raypierre speaks is a saturation which more or less translates into an opacity at a given wavelength. If the atmosphere is opaque at that wavelength, it doesn’t mean that it no longer emits radiation at that wavelength, but more or less that none of the radiation can get through without absorption and consequent reemission. In a certain sense, this is where the blinders analogy falls down but pinballs in a pinball machine where you are adding bumpers does a pretty good job. I myself however prefer simply to think in terms of the atmosphere becoming opaque at a given wavelength. As it becomes opaque, it becomes less transparent, and any well-defined image at that wavelength would be lost.

Less and less of the energy at that wavelength will make it to space without making a round trip to the ground given the “isotropic” reemission of radiation (although Alastair has pointed out that this is actually an approximation – which nevertheless helps in terms of getting a handle on the process). However, if the atmosphere is completely opaque at a given wavelength, what this will imply is that parcels of energy (think of them as photons) which leave the surface at that wavelength will stand a fifty-fifty chance of making it out of the atmosphere without making another trip back to the surface and further warming the surface.

I hope this helps.

A bit more of an overview of a number of topics, actually, whereas the discussion had become rather specialized as of late. Fascinating stuff, but a little of a diservice to those coming into the conversation later than other participants.

Comment by Timothy Chase — 3 Sep 2007 @ 12:52 PM

602. PS to #600 (response to Douglas Wise)

The first sentence should have been, “Well, how closely we know how carbon dioxide radiates will have some effects upon the atmospheric calculations.” From what I can tell, after that sentence I have done a good job of not mangling the sentences. But I was still tuning up at the first sentence, juggling different ways of expressing the same thought.

Hopefully Ray Ladbury, Raypierre or someone else more knowledgable than myself will respond to your comment, but I figure I have done a fair job.

Comment by Timothy Chase — 3 Sep 2007 @ 1:15 PM

603. I would like to thank Timothy Chase for his response (#601) to my post of #600.

I accept that climate sensitivity falls out of the models and is consistent with 400000 years of paleoclimate data. Radiative forcing, however, is presumably built into the models. As explained in the iniating essay to this thread, the formula is log based such that increasing atmospheric CO2 has progressively reducing effects. In the previous one million years, atmospheric CO2 concentrations appear to have cycled between 140 and 280ppmv. Raypierre suggests that, at low levels of the atmosphere (say up to 8km – see Barton Paul Levenson’s post#210) the 13.5-17 micron bands are saturated at 300ppmv and above and that further absorption could only take place in the wings. Am I wrong to think that the radiative forcing formula is a mathematical calculation of the extent to which the atmosphere at this level is approaching “optical thickness”? The formula looks quite simple, leading me to suppose that the fact that high altitude CO2 has only half the blocking effect of lower altitude molecules may have been ignored. Clearly, the paleoclimate record for the last 400000 years does not resolve this point because it relates to times when the main CO2 absorbing bands were in the range from unsaturated to approaching saturated.

Timothy’s reply to my question relating to the escape of non radiative thermal energy to space appears to require the presence of ghgs. We have been assured that, without ghgs, the planet surface would be 20-33 deg C colder because all incoming short wave radiation would immediately leave the atmosphere as outgoing longwave radiation. This does not address what would happen to energy leaving the surface in non radiative form. Wouldn’t it be trapped in the atmosphere, having no way of being converted to radiation? (I suspect you’ll tell me this is ridiculously hypothetical and that, in any event, oxygen would convert to ozone and solve my dilemma.) Would it be mad to think of ghgs as a sort of buffering system? It seems they are blankets keeping heat in but nevertheless are necessary to let non radiative energy out of the top of the atmosphere.

Comment by Douglas Wise — 4 Sep 2007 @ 4:43 AM

604. Re where Douglas Wise says :
It seems they are blankets keeping heat in but nevertheless are necessary to let non radiative energy out of the top of the atmosphere.

Yes, that is true. It follows that if the atmosphere is saturated wrt. terrestrial radiation (in the greenhouse bands) at the base of the atmosphere, it must also be saturated wrt outgoing long wave radiation at the top of the atmosphere.

The way I visualise it, is that each layer of atmosphere behaves like a sieve removing a certain proportion of greenhouse gas photons, say 10%. The next layer will 10% of those remaining and so on. If the atmosphere is deep enough then eventuall there will be a layer where there are no GG IR photons, and although CO2 is only 4 parts per thousand of the earth’s atmosphere, with the billions of molecules it contains the height of total absorption is soon reached.

Now think of the top of the atmosphere, where the the sieves get wider as atmopheric pressure gets less, but that is unimportant. Just as every photon leaving the surface is trapped on it way up, looking down from the top everywhere you would see a photon escaping. If you add CO2 to the atmophere all that happens is that the radiation you see is emitted from higher in the atmosphere, because the sieves are tighter but still have the same overall effect.

Of course this means that if the incoming solar radiation is greater than this fixed outgoing radiation, then the planet will get hotter. This is what has happened on Venus. It has heated up until the planet’s surface melted and produced huge clouds of sulphur dioxide limiting the radiation at the bottom of the atmosphere to match the OLR, by reflecting away the solar flux.

On Earth the surface warms until water clouds form, and on Mars the planet warms up until huge dust storms shade the planet allowing it to cool.

Of course that is not what Raypierre is saying, but I think it makes more sense :-)

Comment by Alastair McDonald — 4 Sep 2007 @ 8:29 AM

605. > Wouldn’t it be trapped in the atmosphere, having no
> way of being converted to radiation?

Comment by Hank Roberts — 4 Sep 2007 @ 12:27 PM

606. I happened to read the publication of ‘herr’ John Koch (as Ångström did and Raypierrre didn’t),

John Koch, Beiträge zur Kenntnis der Wärmeabsorption in Kohlensäure., Öfversigt af Kongl. Vetenskaps-Akademiens Förhandlinger, 1901. N:o 6 p 475-488

Here is his figure 1:
http://home.casema.nl/errenwijlens/co2/Koch_fig1.gif

Comment by Hans Erren — 4 Sep 2007 @ 5:27 PM

607. Douglas, I’m not sure what you mean by saying a high-altitude CO2 molecule will be less efficient at capturing an IR photon. Are you referring to the lower amount of line broadening due to the lower pressure? Or are you referring to the decreased IR flux at higher altitude? To first order, the absorption cross section of a CO2 molecule will be constant regardless of where it is.
As to the absorption of CO2 in the wings of its line, keep in mind that the tails of the line are rather thick, so this is not insignificant.

Comment by ray ladbury — 5 Sep 2007 @ 4:49 AM

608. Re #606

Hans,

Thanks for that info. I had not realised that Dr John Koch had published a paper (nor perhaps did Ray.) My only source was the summary by Very of Angstrom’s summary of the work by Koch. They refer to “unpublished researches of Dr. J. Koch.”

In the figure you uploaded, I have translated the X-axis as “Layer thickness in cm” and the Y=axis as “Absorption in per cent” Curve I is presumably with a pressure of 1.0 atm and Curve II with a pressure of 0.5 atm. Would it be possible for you to upload the complete paper, since being over 100 years old it is surely out of copyright?

Comment by Alastair McDonald — 5 Sep 2007 @ 6:06 AM

609. I appreciate the replies of Alastair McDonald (#604), Hank Roberts (#605) and Ray Ladbury (#607) to my post of #603.

Hank, I believe you were answering my question as to how thermal energy which leaves the surface as sensible and latent heat and quickly reaches the high troposphere gets converted to radiative energy so that it can escape as OLR. Your succinct answer was that “everything radiates”. I take it, therefore, that you are implying that oxygen and nitrogen molecules can spit out photons which can then journey into space. Insofar as I understood Timothy Chase’s response (#601) to the same query I raised in #591, he suggested greenhouse gases would be necessary intermediaries. He actualy said “Under LTE the local temperature for any given gas will be equal to the local temperature of the atmosphere itself. Thus the collisions from oxygen and nitrogen are sufficient to ensure that the Planck temperature of the greenhouse gases have sufficient energy to radiate at the corresponding wavenumbers.”

Alastair suggests that my statement that “It seems that they (ghgs) are blankets keeping heat in but nevertheless are necessary to let non radiative energy out of the top of the atmosphere” is correct.

Ray, you started me on this hare so I’d be grateful for clarification to your statement that “Heat transported by convection gets to the upper troposphere and heats the air there. The warmed air collides, exciting rotational and a few vibrational states. These high altitude molecules then radiate in the bands where they can radiate.” Do you mean any old molecule (including oxygen and nitrogen as Hank seems to be implying) or just ghg molecules as Timothy and Alastair seem to be saying. Please accept that I’m not trying to stir up trouble and there may be no contradictions between you all – it just seems to me to be so with my lack of understanding of physics.

Alastair, I have managed to glean from this thread that your views may not be mainstream. However, to my simple mind, they seem as plausible as others I have read. I believe that you consider that the lower troposhere is already more or less saturated by a combination of CO2 and water vapour (barring the “atmospheric window”). You then argue that extra CO2 will will lead to saturation at a lower altitude, resulting in a lot of ice melt and reducing albedo with consequences worse than forecast by establishment modellers. However, in your recent reply, you seem to suggest that extra surface heating will encourage more cloud. Doesn’t cloud contribute more to albedo than ice? Perhaps there is no need to feel so depressed! The energy balance chart you referred me to seemed to require an outflow of energy of 235 watts/sq m. If you add 102 for that reaching the high troposphere by convection, 67 of solar energy absorbed by the atmosphere and never reaching the surface, 40 from the “window” and 30 from cloud tops, you exceed 235. Obviously extra CO2 will block some of this but extra cloud would add to it. I understand that a doubling of CO2 would make a difference of 3.7 watts/sq m. This doesn’t seem a lot when set against all the other possible variables. However, like you, I believe we should be taking action as soon as possible to reduce CO2 emissions. As a closet Malthusian, I also think we should prioritise measures to stabilise and ultimately reduce the human population. Climate change isn’t the only catastrophic effect our species is producing.

Finally, Ray, you ask me to be more specific about my concerns relating to the radiative forcing formula for CO2. My comment that high altitude CO2 had less blocking effect than low altitude was related to pressure effects. In the original essay, Ray Pierrehumbert discussed the lengths of tubes necessary to block outgoing IR of the appropriate wavelengths. At higher pressures, less IR can slip between gaps. I suspect that the formula for radiative forcing may relate to degree of saturation at any one altitude and not factor in changes in absorptive efficiency with changing pressure (altitude). I accept that this may be regarded as an absolutely unjustified suspicion, coming as it does from a complete ignoramus in the field.

Comment by Douglas Wise — 5 Sep 2007 @ 2:19 PM

610. Douglas Wise (#603) wrote:

I would like to thank Timothy Chase for his response (#601) to my post of #600.

Not a problem.

Douglas Wise (#603) wrote:

I accept that climate sensitivity falls out of the models and is consistent with 400000 years of paleoclimate data. Radiative forcing, however, is presumably built into the models.

Nope. The radiative forcing due to a given gas will be in large part the result of its distribution in the atmsopheric column, the temperatures and pressures at specific altitudes, thermal radiation coming up from the surface, etc.. This is afterall what HTRAN is all about.

However, calculation of the radiative forcing is again a job for the line-by-line codes that take into account atmospheric profiles of temperature, water vapour and aerosols. The most up-to-date calculations for the trace gases are by Myhre et al (1998) and those are the ones used in IPCC TAR and AR4.

These calculations can be condensed into simplified fits to the data, such as the oft-used formula for CO2: RF = 5.35 ln(CO2/CO2_orig) (see Table 6.2 in IPCC TAR for the others). The logarithmic form comes from the fact that some particular lines are already saturated and that the increase in forcing depends on the ‘wings’ (see this post for more details).

6 August 2007
The CO2 problem in 6 easy steps
Gavin Schmidt
Filed under: Greenhouse gases Climate Science
http://www.realclimate.org/index.php/archives/2007/08/the-co2-problem-in-6-easy-steps/

They even take into account the non-LTE effects in today’s climate models. So you have the absorptivity/emissivity of the gases, e.g., the specific spectra of absorption for each partial pressure and temperature. The current Hadley model (Hadgem1) divides the atmosphere into 38 levels , the first 5 km of ocean into 40 levels, and uses columns that are 1 degree by 1 degree.

Models ‘key to climate forecasts’
By Dr Vicky Pope
http://news.bbc.co.uk/1/hi/sci/tech/6320515.stm

Calculations for most models are run in intervals of a few minutes of simulated time.

Douglas Wise (#603) wrote:

As explained in the iniating essay to this thread, the formula is log based such that increasing atmospheric CO2 has progressively reducing effects. In the previous one million years, atmospheric CO2 concentrations appear to have cycled between 140 and 280ppmv. Raypierre suggests that, at low levels of the atmosphere (say up to 8km – see Barton Paul Levenson’s post#210) the 13.5-17 micron bands are saturated at 300ppmv and above and that further absorption could only take place in the wings. Am I wrong to think that the radiative forcing formula is a mathematical calculation of the extent to which the atmosphere at this level is approaching “optical thickness”?

I believe this is correct.

Douglas Wise (#603) wrote:

The formula looks quite simple, leading me to suppose that the fact that high altitude CO2 has only half the blocking effect of lower altitude molecules may have been ignored.

As I said, I don’t believe we use use that formula – and most certainly not for the entire atmospheric column. We use HTRAN and thus line-by-line calculation.

Douglas Wise (#603) wrote:

Clearly, the paleoclimate record for the last 400000 years does not resolve this point because it relates to times when the main CO2 absorbing bands were in the range from unsaturated to approaching saturated.

But we do know of earlier eras with much higher CO2 levels, 3000 ppm or more. The PETM was roughly 55 MYA. And as I have indicated, the actual empirical measurements which are part of HTRAN are quite detailed – as are the calculations. The logarithmic relationship is an approximation, although a fairly good one.

Douglas Wise (#603) wrote:

Timothy’s reply to my question relating to the escape of non radiative thermal energy to space appears to require the presence of ghgs. We have been assured that, without ghgs, the planet surface would be 20-33 deg C colder because all incoming short wave radiation would immediately leave the atmosphere as outgoing longwave radiation. This does not address what would happen to energy leaving the surface in non radiative form. Wouldn’t it be trapped in the atmosphere, having no way of being converted to radiation?

Evaporation leads to condensation leads to clouds – which are good blackbody emitters in the infrared. But if you are not talking about evaporation, then you don’t have moist air convection and heat will diffuse into the atmosphere much more slowly. At that point, we are speaking of thermal diffusion which is a much slower process.

Douglas Wise (#603) wrote:

(I suspect you’ll tell me this is ridiculously hypothetical and that, in any event, oxygen would convert to ozone and solve my dilemma.) Would it be mad to think of ghgs as a sort of buffering system? It seems they are blankets keeping heat in but nevertheless are necessary to let non radiative energy out of the top of the atmosphere.

In the absence of water vapor and other greenhouse gases, heat would diffuse through the atmosphere much more slowly from the surface, but it will also diffuse to the surface so long as the atmosphere is in contact with the surface. As the surface will emit radiation, it would also absorb heat from the atmosphere. No runaway effect.

Comment by Timothy Chase — 6 Sep 2007 @ 9:37 AM

611. Douglas Wise (#609) wrote:

I appreciate the replies of Alastair McDonald (#604), Hank Roberts (#605) and Ray Ladbury (#607) to my post of #603.

Hank, I believe you were answering my question as to how thermal energy which leaves the surface as sensible and latent heat and quickly reaches the high troposphere gets converted to radiative energy so that it can escape as OLR. Your succinct answer was that “everything radiates”. I take it, therefore, that you are implying that oxygen and nitrogen molecules can spit out photons which can then journey into space.

I believe he was being a little too succinct.

Non-ghgs won’t radiate. If anyone were to make this claim, you should ask what would be the spectra? Given Hank’s earlier recognition of the relationship between the line radiation and blackbody (or realistic body radiation) I believe he will agree. But heat would enter and leave the atmosphere by means of the much slower process of thermal diffusion as the result of being in contact with the surface.

Douglas Wise (#609) wrote:

Insofar as I understood Timothy Chase’s response (#601) to the same query I raised in #591, he suggested greenhouse gases would be necessary intermediaries. He actualy said “Under LTE the local temperature for any given gas will be equal to the local temperature of the atmosphere itself. Thus the collisions from oxygen and nitrogen are sufficient to ensure that the Planck temperature of the greenhouse gases have sufficient energy to radiate at the corresponding wavenumbers.”

Thats correct. However, even under non-LTE in the upper atmosphere (for example, 15 microns for CO2), collisions will result in greenhouse gases emitting. But with carbon dioxide at 15 microns in the mesosphere and thermosphere, this will be more due to collisions with photodisassociated O (from ozone) than than absorption and reemission.

Douglas Wise (#609) wrote:

Alastair suggests that my statement that “It seems that they (ghgs) are blankets keeping heat in but nevertheless are necessary to let non radiative energy out of the top of the atmosphere” is correct.

They are not necessary given thermal diffusion – the surface will still absorb heat from the atmosphere while the surface emits thermal radiation.

Douglas Wise (#609) wrote:

Ray, you started me on this hare so I’d be grateful for clarification to your statement that “Heat transported by convection gets to the upper troposphere and heats the air there. The warmed air collides, exciting rotational and a few vibrational states. These high altitude molecules then radiate in the bands where they can radiate.” Do you mean any old molecule (including oxygen and nitrogen as Hank seems to be implying) or just ghg molecules as Timothy and Alastair seem to be saying.

Since he is speaking of rotational and vibrational states he is speaking of greenhouse gases.

Douglas Wise (#609) wrote:

Please accept that I’m not trying to stir up trouble and there may be no contradictions between you all – it just seems to me to be so with my lack of understanding of physics.

I believe Hank will agree with respect to non-ghgs not radiating, and I believe that Alastair will agree with regard to thermal diffusion. Likewise, I believe Ray will agree that what he is speaking of is principally moist air convection. The atmosphere is rather stable in its absence.

Douglas Wise (#609) wrote:

Alastair, I have managed to glean from this thread that your views may not be mainstream. However, to my simple mind, they seem as plausible as others I have read. I believe that you consider that the lower troposhere is already more or less saturated by a combination of CO2 and water vapour (barring the “atmospheric window”).

Doesn’t matter how saturated it is below. If the thermal radiation is to make it out of the atmosphere, it must pass through the higher layers.

Douglas Wise (#609) wrote:

You then argue that extra CO2 will will lead to saturation at a lower altitude, resulting in a lot of ice melt and reducing albedo with consequences worse than forecast by establishment modellers. However, in your recent reply, you seem to suggest that extra surface heating will encourage more cloud. Doesn’t cloud contribute more to albedo than ice? Perhaps there is no need to feel so depressed!

It clouds both increase the albedo and are good absorbers of longwave, and thus are essentially blackbodies in that part of the spectrum. The net radiative effect of clouds can be positive or negative – depending upon the type of cloud. As for “nonstandard views” in this area, we have the spectra. Emissions at different altitudes of infrared at various wavelengths which are essentially fingerprints of the gases which emit at those altitudes.

Please see #555 above. I have included 12 links to images and animations of data received by satellites showing exactly what is being emitted at different altitudes.

AIRS – Multimedia: Videos: Animations
http://airs.jpl.nasa.gov/Multimedia/VideosAnimations/

… for a number of the animations, including a somewhat detailed explanation of how those animations are created from satellite data gathered from over 2000 different channels within the spectra.

Douglas Wise (#609) wrote:

The energy balance chart you referred me to seemed to require an outflow of energy of 235 watts/sq m. If you add 102 for that reaching the high troposphere by convection, 67 of solar energy absorbed by the atmosphere and never reaching the surface, 40 from the “window” and 30 from cloud tops, you exceed 235. Obviously extra CO2 will block some of this but extra cloud would add to it.

Thermal energy from the atmosphere emits 195 to space, including 165 from greenhouse gases and 30 directly from the clouds (i.e., without absorption by greenhouse gases – given a window). 40 of what the surface emits is not absorbed by the atmosphere but goes directly to space. Convection doesn’t go to space, but has to have its energy radiated into space, one way or another.

For a complete diagram, please see:

The Energy Balance and Natural Climate Variations
http://www.cara.psu.edu/climate/climatechangeprimer-pr4.asp

Douglas Wise (#609) wrote:

I understand that a doubling of CO2 would make a difference of 3.7 watts/sq m. This doesn’t seem a lot when set against all the other possible variables. However, like you, I believe we should be taking action as soon as possible to reduce CO2 emissions. As a closet Malthusian, I also think we should prioritise measures to stabilise and ultimately reduce the human population. Climate change isn’t the only catastrophic effect our species is producing.

What othere variables? Solar variability?

It has been declining since 1960. And since as far back 1880 it would appear that greenhouse gases (taken together) have been the dominant positive forcing – even over solar variability.

Douglas Wise (#609) wrote:

Finally, Ray, you ask me to be more specific about my concerns relating to the radiative forcing formula for CO2. My comment that high altitude CO2 had less blocking effect than low altitude was related to pressure effects. In the original essay, Ray Pierrehumbert discussed the lengths of tubes necessary to block outgoing IR of the appropriate wavelengths. At higher pressures, less IR can slip between gaps. I suspect that the formula for radiative forcing may relate to degree of saturation at any one altitude and not factor in changes in absorptive efficiency with changing pressure (altitude). I accept that this may be regarded as an absolutely unjustified suspicion, coming as it does from a complete ignoramus in the field.

We don’t use the simplified logarithmic formula in the global climate models but do precise calculations which take into account the pressures and temperatures at different levels in the atmosphere – and which even take into account non-LTE behavior.

However, calculation of the radiative forcing is again a job for the line-by-line codes that take into account atmospheric profiles of temperature, water vapour and aerosols. The most up-to-date calculations for the trace gases are by Myhre et al (1998) and those are the ones used in IPCC TAR and AR4.

These calculations can be condensed into simplified fits to the data, such as the oft-used formula for CO2: RF = 5.35 ln(CO2/CO2_orig) (see Table 6.2 in IPCC TAR for the others). The logarithmic form comes from the fact that some particular lines are already saturated and that the increase in forcing depends on the ‘wings’ (see this post for more details).

6 August 2007
The CO2 problem in 6 easy steps
Gavin Schmidt
Filed under: Greenhouse gases Climate Science
http://www.realclimate.org/index.php/archives/2007/08/the-co2-problem-in-6-easy-steps/

Also see:

Line-by-line calculation of atmospheric fluxes and cooling rates 2
http://www.aer.com/scienceResearch/rc/m-proj/abstracts/rc.clrt2.html

… for an example of the results from a line-by-line calculation at different altitudes for water vapor, carbon dioxide and ozone.

Comment by Timothy Chase — 6 Sep 2007 @ 9:41 AM

No, what I meant is that there’s no molecule that’s an energy roach motel, there’s no molecule that is unable to interact by absorbing and emitting photons at some wavelength, not necessarily infrared.

Energy has to leave a planet either by radiation — or by boiling off physical atmosphere. It doesn’t have to leave as _infrared_ photons, per se. It simply has to leave.

Look at Ray Pierrehumbert’s piece on science fiction atmospheres, here: http://geosci.uchicago.edu/~rtp1/papers/BAMS_SFatm.pdf

Pierrehumbert RT 2005: Science Fiction Atmospheres. Bulletin of the American Meteorological Society, 86, 696-698.

He writes:

“… The planetary settings of many other science fiction stories pose interesting questions from the point of view of physics of climate. On Frank Herbert’s Dune, … A completely dry planet with
a habitable temperature range is no problem, at least if one only needs it to remain habitable for a few hundred million years. For example, [a hypothetical] Venus with a pure Nitrogen-Oxygen atmosphere would have a mean surface temperature of around 300K….”

Take the planet out and leave the ball of gas — it won’t suddenly be unable to get warmer or cooler.

Comment by Hank Roberts — 6 Sep 2007 @ 10:50 AM

613. Hank, Alastair and Ray,

In my response #611 to Douglas Wise (#609), I had to say that “I believe [you] would agree…” to various points. Whether you agree or disagree, it would be helpful if you said so, in the case of disagreements, said why, and said anything else you might like to say.

However, it should also be pointed out that none of the four of us are climatologists.

Ray Pierrehumbert (the author of the essay) is a climatologist. So is Gavin Schmidt – who I quoted as saying that climate models do not use the simple logarithmic formula but use line-by-line calculation of absorption and emission at given altitudes, taking into account pressure and temperature, and thus spreading. But Ray Ladbury is a radiation physicist, so he has some expertise in this area.

Comment by Timothy Chase — 6 Sep 2007 @ 12:25 PM

614. Douglas,
OK. An atom/molecule can only emit a photon when it transitions from a higher energy state to lower one. Since earthshine is strongly peaked in the IR, it can only affect a few molecules–the so-called greenhouse gases. O2 and N2 cannot under normal circumstances radiate because they have no rotational spectrum (no dipole magnetic moment). They can have a rotational spectrum when excited electronically (mostly at high altitude).
At low altitude most of the relaxation of the excited vibrational state of CO2 is collisional, and this can impart energy to other molecules. Note, however, that even though the energy of the vibrational state is high compared to the thermal energy, a small percentage of molecules will have energy high enough to excite this mode. This process is not significant until the density of CO2 is sufficiently low that the IR photon has a chance of escaping–that’s the effective height you are looking at when you look at the 15 micron line. That is my understanding, anyway. Also note that this is not a perfect line–it gets broadened due to pressure, etc. The basic thing to remember is that if there is energy in one mode, it usually finds a way to get shared in the other modes.

Comment by Ray Ladbury — 6 Sep 2007 @ 1:14 PM

615. In the thermosphere, the O atoms come from O2 (photodissociation and dissociative collisions).

CO2 at higher altitudes will have significantly less pressure broadening, thus the range of wavelengths at which a molecule can absorb will be smaller. OTOH the peak absorption will be higher. The net of this is that: The emission from below on a CO2 line will be broader, the part that is out on the wings will pass through higher altitudes because the absorption lines are narrower. I spent a lot of July going through this

Comment by Eli Rabett — 6 Sep 2007 @ 5:28 PM

616. I’m still somewhat confused.

Minor point: I thought O2 and N2 had vibration and rotation energy levels, but unable to fill them with IR absorption because they are non-polar; could only fill them via collision or translation transfer. Is this wrong?

I thought this thread kinda concluded that 1) Planck function blackbody-type radiation comes from a totally different molecular action than IR radiation, which comes from the quantized vibration and rotation energies of polarized molecules (GH gases) — meaning the two ain’t the same thing, and that 2) gases theoretically radiate Planck function blackbody-type radiation (in addition to some radiating quantized IR line radiation), but from a practical viewpoint it is quite restrictive — requiring certain pressures and having “holes” in the spectrum, e.g. Correct or no? [As an aside one of the bothersome things is all (at least that I’ve seen) of the system level development and analysis of the earth’s energy balance use Planck function radiation (T^4, e.g.) entirely between the atmosphere levels…]

My primary confusion (still) is, after all of the above discussion of GH gases not re-emitting much, not absorbing emissions from the same type of GH gas, convection/latent heat rising to heat the upper atmosphere to get IR radiation finally leaving the earth because its really thin at the TOA, etc., etc., where does the over 300 watts/square-meter of IR atmospheric downwelling (absorbed by the earth) come from?

Comment by Rod B — 6 Sep 2007 @ 7:09 PM

617. I had written in #611:

Thats correct. However, even under non-LTE in the upper atmosphere (for example, 15 microns for CO2), collisions will result in greenhouse gases emitting. But with carbon dioxide at 15 microns in the mesosphere and thermosphere, this will be more due to collisions with photodisassociated O (from ozone) than than absorption and reemission.

Eli Rabett (#615) wrote:

In the thermosphere, the O atoms come from O2 (photodissociation and dissociative collisions).

… and then the single O combines with other O2 to form O3. I knew ozone was involved somehow – but I obviously should have checked.

Eli Rabett (#615) wrote:

CO2 at higher altitudes will have significantly less pressure broadening, thus the range of wavelengths at which a molecule can absorb will be smaller. OTOH the peak absorption will be higher. The net of this is that: The emission from below on a CO2 line will be broader, the part that is out on the wings will pass through higher altitudes because the absorption lines are narrower. I spent a lot of July going through this.

For those who are interested, it is well worth checking out…

The graphs are non-logarithmic and give you a decidedly different view from the graph above, from my perspective, a decidedly complementary view. You get to see the effects of pressure and temperature in the absorption curve – reminds me a bit of the “stone thrown in a lake” look of the wavefunction of a bound electron, although not quite.

As the center of the curve becomes progressively saturated, the wings take on more of the absorption. Additionally, Eli introduces the reader to some software where those interested can explore things for themselves – and if you care to understand some of the mathematics behind this, there is some of that as well.

Comment by Timothy Chase — 6 Sep 2007 @ 8:11 PM

618. Re #613 where Timothy said “Ray Pierrehumbert (the author of the essay) is a climatologist. So is Gavin Schmidt – ”

If you look up Ray Pierrehumbert’s CV I think you will find that he is a physicist not a climatologist. I suspect that Gavin is a mathematician. Not bad professions, but not equipped to handling the chaotic behavior of weather and other earth system sciences, where what you expect does not happen, cf. GWB and Katrina, Rumsfeld and Iraq. Or what you don’t does cf. The Boxing Day Tsunami.

But the big mistake they are making was pointed out by Spencer Weart when he wrote:

Although people did not deny the facts head-on, many denied them more subtly by failing to revise their accustomed ways of thinking. “Geoscientists are just beginning to accept and adapt to the new paradigm of highly variable climate systems,” wrote the NAS committee. And beyond geoscientists, “this new paradigm has not yet penetrated the impacts community”–the economists and other specialists who try to calculate the consequences of climate change.

Ray and Gavin are not geoscientists. I am!

Comment by Alastair McDonald — 6 Sep 2007 @ 8:13 PM

619. Re #616 where Rob says

“Minor point: I thought O2 and N2 had vibration and rotation energy levels, but unable to fill them with IR absorption because they are non-polar; could only fill them via collision or translation transfer.”

What you say is roughly correct, but O2 does emit microwave radiation due to its rotation. Moreover N2 can transfer its vibrational energy to CO2 molecules, a feature used in CO2 lasers.

Theoretically gases do not radiate Planck blackbody radiation, but despite that water vapour does!

But theoretically, their radiation depends on their temperature which is their average translational kinetic energy. Acording to the Theorem of Equipartition (ToE) all modes of energy will be equally populated. Thus the average speed in the x direction will be the same as that in the y an z directions. Fine!

But Einstein then says that this means it will be shared equally with the electronic excitation, the vibrational energy, and the rotational energy as well. No collisions at room temperature can induce any electronic excitation so that temperature is absolute zero. It is frozen out. Rotational energy is low, so collisions can induce that and rotational temperature will equal translational (the kinetic and thermometer) temperature.

The vibrational temperature is the key [edit for appropriateness]

The energy of the translational collisions is not great enough for the electronic temperature to match the translational and rotational temperatures. So the electronic energy frozen is out, and so partly is the vibrational temperature.

But it is complicated :-(

Comment by Alastair McDonald — 6 Sep 2007 @ 8:47 PM

620. Rod B (#616) wrote:

Minor point: I thought O2 and N2 had vibration and rotation energy levels, but unable to fill them with IR absorption because they are non-polar; could only fill them via collision or translation transfer. Is this wrong?

I believe you are right. However, if they can’t absorb, then they can’t emit – because they are nonpolar, and the only way that they will be able to lose vibrational and rotational energy will be by way of collision.

Rod B (#616) wrote:

I thought this thread kinda concluded that 1) Planck function blackbody-type radiation comes from a totally different molecular action than IR radiation, which comes from the quantized vibration and rotation energies of polarized molecules (GH gases) — meaning the two ain’t the same thing, and that …

It is always quantized – the true blackbody is an idealization. But it is also always smeared, where the lines and bands are broadened, getting closer to that idealization. Clouds are much closer, but there are also some solids that are much closer to the quantization we find among gases.

Rod B (#616) wrote:

2) gases theoretically radiate Planck function blackbody-type radiation (in addition to some radiating quantized IR line radiation), but from a practical viewpoint it is quite restrictive — requiring certain pressures and having “holes” in the spectrum, e.g. Correct or no?

Dusts and alloys will have holes as well, but if you change the mix in the alloy so that the structure becomes less regular, those holes will tend fill up.

Rod B (#616) wrote:

My primary confusion (still) is, after all of the above discussion of GH gases not re-emitting much, not absorbing emissions from the same type of GH gas, convection/latent heat rising to heat the upper atmosphere to get IR radiation finally leaving the earth because its really thin at the TOA, etc., etc., where does the over 300 watts/square-meter of IR atmospheric downwelling (absorbed by the earth) come from?

I would strongly recommend checking out Alastair McDonald’s post #569. It gives you links to two graphs which show the absorption over the longwave due to different molecules. As the graphs show, when you have an atmosphere composed of multiple greenhouse gases, the atmosphere comes much closer to the blackbody idealization than it would if you had only one greenhouse gas by itself. He also points out that the 15 micron on CO2 are generally strongly self-absorbing and rarely involve transitions to intermediate states.

But I would also suggest that even where there are transitions to lower states of excitation, those which are closest to the ground state will also tend to be self-absorbing. Likewise, radiation which is downwelling will, upon reaching the surface will be reemitted with the same spectra as the thermal radiation of the surface, and not necessarily as photons of the same wavelengths as those photons which were downwelling.

Comment by Timothy Chase — 6 Sep 2007 @ 8:55 PM

621. First, a minor quibble with Tim’s characterization of my expertise on climate change. I am definitely not an expert, just a physicist trying to parse out some pretty involved physics.
Now to the issue of equipartition, etc. First, people are tending to ignore the fact that we are talking about huge numbers of molecules. Thus, even when we are talking about rare events, they will occur with some frequency. This includes collisional excitation. Even at altitudes of ~10-60 km, with temperatures around 240 K, ~1.8% of molecules have sufficient energy to excite the 15 micron vibrational line in CO2. So, I don’t think it’s really accurate to say that this mode is “frozen out”. Thus, the 15 micron radiation we see from space is actually coming from molecules high in the troposphere where densities are still high enough that collisions (and occasional Earthshine) excite the vibrational mode, and it is ‘thermal’ radiation as well as being quantized. Add in other molecules, and you get additional lines as well as broadening of all lines, and the spectrum of thermal radiation fills in, giving a better approximation of a blackbody specturm.

Comment by Ray Ladbury — 7 Sep 2007 @ 8:08 AM

622. Alastair says, “Theoretically gases do not radiate Planck blackbody radiation, but despite that water vapor does!”

Then how do you explain the college-level development of earth energy balance using 2-3-4+ atmospheric levels all radiating ala Planck function? How do you explain solar radiation?

Timothy says,

“I believe you are right. However, if they can’t absorb, then they can’t emit – because they are nonpolar, and the only way that they will be able to lose vibrational and rotational energy will be by way of collision.”

That I’ll accept, even with Alastair’s O2 rotation microwave exception (that proves the rule…)

“It is always quantized – the true blackbody is an idealization….”

But I think not at the same level. (Theoretical) blackbody radiation is quantized like everything is at some level of n*h granularity; vibration, rotation, and electronic energy levels are quantized at massively larger steps caused by wholly different molecular action.

“Dusts and alloys will have holes as well…”

I understand. I also think I understand (??) that gases at low density may have far more “holes” than Planck radiation spectrum

“……….. …where there are transitions to lower states of excitation, those which are closest to the ground state will also tend to be self-absorbing. Likewise, radiation which is downwelling will, upon reaching the surface will be reemitted with the same spectra as the thermal radiation of the surface…”

It’s still not adding up in my mind. I think the referenced Alastair links were looking at radiation heading for space (or not). There still has to be about 325 watts/meter-squared, on the average over time, emitted from atmospheric greenhouse gases or clouds and absorbed by the earth. This compares to the 390 watts or so emitted by the earth as Planck function blackbody radiation into the atmosphere; and the roughly 170 watts absorbed from solar radiation; and the roughly 200 watts finally being emitted into space from high in the atmosphere and clouds. It doesn’t seem, with the exception of a little from clouds (??), that this 325 watts of downwelling can stem from Planck/blackbody radiation of the atmosphere — unless maybe it’s predominately being emitted in the lower few meters. (Or, do you get a big pile of Planck/blackbody radiation from N2, O2, etc??) Nor does it seem that it can come from GHG relaxation in the lower troposphere, given the discussion on how CO2, say, is far more likely to lose its excitation to N2 and O2 via collisions than it is to re-emit. Nor does re-emission from the top of the atmosphere seem to add up: the downwelling from the top can’t be more than the 200 watts escaping upward (given isotropic GHG/cloud emission), and even then that 200 downwelling watts has to survive a really hard obstacle course to make it back to the earth’s surface.

I don’t know the answer. But so far, as Red Skelton would say, “It just don’t look right to me.”

Comment by Rod B — 7 Sep 2007 @ 10:15 PM

623. Rod B. (#620) wrote:

Alastair says, “Theoretically gases do not radiate Planck blackbody radiation, but despite that water vapor does!”

Then how do you explain the college-level development of earth energy balance using 2-3-4+ atmospheric levels all radiating ala Planck function? How do you explain solar radiation?

Planck blackbody radiation is always an idealization.

However, gases will come closer to approaching this idealization at higher partial pressures (due to line broadening) and when they have a larger number of quantum states. Since water vapor is electrically dipolar, its quantum states include those that are pure rotation. It exists at a far greater partial pressures, and as such will experience greater line broadening. Given these two aspects it will come a great deal closer to blackbody radiation in certain parts of the spectra. Solar radiation is a little different since at this point we are talking about ionization. Molecular vibration and rotation would be insufficient to get us into the visible part of the electromagnetic spectra.

However, we are still talking about energy causing less energetic quantum states to enter more energetic quantum states which decay back into the less energetic – with matter consequently reemitting radiation. As such, there really isn’t a fundamental difference between thermal radiation which arises due to ionization and thermal radiation which arises due to strictly molecular mechanisms.

Minor point: I thought O2 and N2 had vibration and rotation energy levels, but unable to fill them with IR absorption because they are non-polar; could only fill them via collision or translation transfer. Is this wrong?

I believe you are right. However, if they can’t absorb, then they can’t emit – because they are nonpolar, and the only way that they will be able to lose vibrational and rotational energy will be by way of collision.

Rod B. (#620) responded:

That I’ll accept, even with Alastair’s O2 rotation microwave exception (that proves the rule…)

Well, the “exception proves the rule” doesn’t work very well in physics. But we were talking about electric dipoles and infrared radiation. What Alastair brought up was a little different – as it involves a weak magnetic dipolar moment.

Pleas see:

3.2 Molecular Oxygen Absorption

Oxygen is an electrically non-polar molecule and would not be expected to interact with microwave radiation. However, the molecule does exhibit a magnetic dipole moment. This magnetic moment is very weak in comparison with usually encountered electric dipole moments, but due to the high percentage of oxygen in air and the long transmission paths, [this type of…] absorption become appreciable.

Doc #846: Technical Progress Report For The Month Of February 1964
Lunar Research Program
Contract No. NASw-593
Prepared For:
Lunar And Planetary Office
National Aeronautics And Space Administration, Pg 22
http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660083539_1966083539.pdf

Incidentally, this can come in handy in the search for life on other planets. If there is highly reactive oxygen in the atmosphere giving rise to this signiture, this suggests that there is a biosphere which has discovered photosynthesis – since oxygen won’t stay for very long at high levels without being constantly replentished by a biological source.

I thought this thread kinda concluded that 1) Planck function blackbody-type radiation comes from a totally different molecular action than IR radiation, which comes from the quantized vibration and rotation energies of polarized molecules (GH gases) — meaning the two ain’t the same thing, and that …

It is always quantized – the true blackbody is an idealization. But it is also always smeared, where the lines and bands are broadened, getting closer to that idealization. Clouds are much closer, but there are also some solids that are much closer to the quantization we find among gases.

But I should explain a little further. The thread had more or less concluded that blackbody radiation and line radiation are simply two opposite poles along a continueem.

This included myself. See for example #298, see in particular my post #340.

The difference between the thermal radiation of solids, liquids and gases is a matter of degree, not kind. Solids can approximate blackbody radiation, but blackbody radiation itself is an idealization. Individual gasses at low pressures and temperatures have well-defined lines and bands, but at higher pressures and temperatures both broaden with lines bleeding into lines and bands bleeding into bands. Dusts, crystals and alloys have relatively discrete spectra. However, impurities, clustering, ions and pressure broaden these spectra. And atmospheres are composed of multiple gasses where each gas is at the same temperature. Individually, the spectra of any one gas in the atmosphere is a poor approximation for a blackbody emitter, but taken together, the atmosphere does much better.

Rod B., I think you are getting hung up on terminology–black-body vs. line radiation. First, a black body does not exist in nature–that is nothing has 100% emissivity at all wavelengths. You can come close to blackbody radaition by looking at radiation emitted from a small hole in a cavity. This means that the probability of radiation escaping before it comes into thermal equilibrium with the walls and contents of the container (as well as the radiation field itself) is very small. So what does that equilibrium mean? First, all radiation starts by being emitted by a molecule/atom. However, the surrounding material influences the wavelength of emission and interacts with and influences the energy of the resulting energy. Lines broaden and eventually you get a continuum or close to it.

This is my impression–anyone please feel free to correct me if I’m wrong. So, really there is no difference between “blackbody” radiation and line emission except that the blackbody radiation has come into equilibrium with its surroundings. Since this has to happen for all real radiation, all radiation is both black body and line emission.

Rod B. (#620) responded:

But I think not at the same level. (Theoretical) blackbody radiation is quantized like everything is at some level of n*h granularity; vibration, rotation, and electronic energy levels are quantized at massively larger steps caused by wholly different molecular action.

The different is a matter of degree, not kind. With solids you have more interactions between molecules which make possible a larger number of lines, bands and a larger variety of mechanisms for line and band broadening, but it is still pretty much the same thing, just more of it.

Rod B. (#620) responded:

I understand. I also think I understand (??) that gases at low density may have far more “holes” than Planck radiation spectrum.

Think of it this way…

With the interaction between matter and thermal radiation, you have two extremes and a continuem between them. The extremes are idealizations. One is strictly continuous blackbody radiation at all parts of the spectra. Even in the case of the reflective cavity with a small hole, this cannot be achieved – since to achieve the idealization the hole would have to be infinitely small. The other extreme consists of a single line which is infinitely thin at zero pressure and zero temperature – but there is always natural broadening due to quantum uncertainty, if nothing else. Two unachievable idealizations with the actual reality always being inbetween.

……….. …where there are transitions to lower states of excitation, those which are closest to the ground state will also tend to be self-absorbing. Likewise, radiation which is downwelling will, upon reaching the surface will be reemitted with the same spectra as the thermal radiation of the surface…

Rod B. (#620) wrote:

It’s still not adding up in my mind.

It’s still not adding up in my mind. I think the referenced Alastair links were looking at radiation heading for space (or not). There still has to be about 325 watts/meter-squared, on the average over time, emitted from atmospheric greenhouse gases or clouds and absorbed by the earth….

It adds up, even though it is fairly complicated

The Energy Balance and Natural Climate Variations
http://www.cara.psu.edu/climate/climatechangeprimer-pr4.asp

Planck/blackbody radiation of the atmosphere — unless maybe it’s predominately being emitted in the lower few meters. (Or, do you get a big pile of Planck/blackbody radiation from N2, O2, etc??) Nor does it seem that it can come from GHG relaxation in the lower troposphere, given the discussion on how CO2, say, is far more likely to lose its excitation to N2 and O2 via collisions than it is to re-emit.

As has been pointed out on a variety of occasions, while CO2 or other greenhouse gases can lose energy due to collisions with N2 and O2, they can also gain energy due to collisions with N2 and O2.

As Ray Ladbury most recently stated in #619:

First, people are tending to ignore the fact that we are talking about huge numbers of molecules. Thus, even when we are talking about rare events, they will occur with some frequency. This includes collisional excitation. Even at altitudes of ~10-60 km, with temperatures around 240 K, ~1.8% of molecules have sufficient energy to excite the 15 micron vibrational line in CO2.

This is what makes possible LTE conditions in which the Planck temperature, brightness temperature of reemission lines, and Maxwellian temperature are the same. If it were true that higher quantum states decayed to lower quantum states only at the exact half-life with collisions always taking place before , then collisions would prevent any reemission. But half-lifes don’t work that way. Energy is constantly being transfered by collisions, but reemission takes place according to a law of exponential decay where the states have no memory of how long they have existed. As long as a certain percentage are in an excited state at any given time, reemission will be taking place at a certain rate.

Additionally, we have plenty of empirical evidence which demonstrates reemission throughout the atmosphere. Please see my post #555.

Rod B. (#620) wrote:

Nor does re-emission from the top of the atmosphere seem to add up: the downwelling from the top can’t be more than the 200 watts escaping upward (given isotropic GHG/cloud emission), and even then that 200 downwelling watts has to survive a really hard obstacle course to make it back to the earth’s surface.

Whereever it gets reemitted at whatever altitude reemission takes place, it has only two directions it can go: towards the surface or to space. And whereever it gets absorbed, assuming LTE, the same amount must be reemitted. Whatever gets lost to collisions will also be gained – and more will be gained in the troposphere due to heat transport through moist air convection in accordance with the tropospheric lapse rate.

Comment by Timothy Chase — 8 Sep 2007 @ 12:46 PM

624. Re where Rod B Says:
7 September 2007 at 10:15 PM

Alastair says, “Theoretically gases do not radiate Planck blackbody radiation, but despite that water vapor does!”

Then how do you explain the college-level development of earth energy balance using 2-3-4+ atmospheric levels all radiating ala Planck function? How do you explain solar radiation?

The college level courses AND all the text books are wrong. That’s why Timothy says “It’s still not adding up in my mind.” And it is why those ideas don’t add up in your mind. That is why you and Timothy are still questioning!

I presume when you say “solar radiation” you are referring to the fact that it is close to the spectrum of a blackbody at a temperature of 5800K. I explain that by saying that the pressure within the Sun is so great that the plasmas are so compressed that they produce atomic vibrations which cause blackbody radiation.

I could even argue that the surface of the sun is in fact liquid, as anyone can see, but then I would be called an idiot so I wont say that. (Don’t bother telling me that you think I am wrong because I already know that.)

However, how do you explain the Fraunhofer lines if gases produce blackbody radiation?

Sorry to be so agressive :-(

It’s because I have hit the bottle :-)

Cheers, Alastair.

Comment by Alastair McDonald — 8 Sep 2007 @ 5:52 PM

625. OK, let’s go back to the blanket analogy: what greenhouse gasses do is keep energy from escaping from the surface to space. How does the blanket work? By decreasing the temperature difference between you and the air immediately surrounding you, because the energy you lose is proportional to the difference of the 4th powers of your temperature and that of the environment. So as the blanket heats up, net energy transfer to the blanket from the air around you slows and as the air heats up, net energy loss from you slows. And once the inner surface of the blanket heats up to roughly your body temperature, why is the blanket a good insulator? Because energy transfer across it is slow.
In the case of the atmosphere, you have lots of layers, and CO2 impedes energy transit. As the atmosphere heats up, net energy loss from the surface decreases–so there doesn’t have to be any energy transfer to the ground, just less energy lost.

Now as to the transition from line emission to quasi-continuum of blackbody radiation, think about what happens as you go from an atom, to a gas of atoms, to a liquid and eventually to a solid. Spectral lines broaden and coalesce until with a solid you have energy bands–ranges of energy the electrons can have, not just lines. You can see the start of this process in a gas with the broadening of spectral lines by interactions with surrounding molecules. The lines are still lines, and the atoms can only radiate for the same energy transitions, but the energies are a little smeared out. And how many molecules can radiate at those lines depends on how many have enough thermal energy–so it looks like a slice of blackbody just at those wavelengths. As you get more different molecules or more interactions as in a solid, the spectrum looks more blackbody. Finally the closest to a blackbody spectrum you get is emission from a small hole in a cavity. This is because you have not only the interactions of the gas and the walls of the container, but the interactions of the radiation field itself that come to equilibrium.

Comment by ray ladbury — 8 Sep 2007 @ 8:38 PM

626. No, no, no, Timothy. Planck blackbody radiation is not “always an idealization”. A smooth spectrum output is an idealization (99% of the time). But Planck function blackbody radiation (loosely named to include whatever spectrum) is as real as it gets.

Couldn’t wait. Now to read the rest of the post(s).

Comment by Rod B — 8 Sep 2007 @ 8:48 PM

627. Timothy (621), you say,

“… However, gases will come closer to approaching this idealization at higher partial pressures (due to line broadening) and when they have a larger number of quantum states…”

You (and others) are still confusing Planck blackbody radiation with line emission of, say, greenhouse gases. The former is thermal radiation (maybe if they officially called it that a pile of semantic debate could be avoided), generated by its own molecular processes; the latter is not thermal radiation (strictly speaking), does not have anything close to a “continuous” spectrum, and is generated by a different molecular process. I understand water vapor “line” radiation has near continuous quanta in certain parts of the spectrum (a slippery phrase as the spectrum is the crux of the observable difference) and looks a little like blackbody [thermal] radiation. But it ain’t.

You conclude this part of your post with

As such, there really isn’t a fundamental difference between thermal radiation which arises due to ionization and thermal radiation which arises due to strictly molecular mechanisms.

That’s a paradox nearing an oxymoron. Or, put another way, that’s wrong! Bear in mind the one and only full spectrum independent variable in Planck’s formulation is Temperature (the real kind). No molecular bonding coefficients; no molecular shapes or polarization; no molecular moments of inertia; etc.

I agree “… the “exception proves the rule” doesn’t work very well in physics…” but it makes the point and sounds good [;-) . I like and appreciate your helpful explanation of O2 absorption/emission.

Now, back to your first and main (and wrong…???) point. Hanks reference in 339 is addressing thermal radiation versus quantum physics, but in any case is a cacophonic forum with various folks doing the same debate we are having. It is helpful in this regard: It is clear that there is no fully complete understanding (agreement, really) nor a scientific consensus on the specifics of thermal radiation ICW greenhouse gas emission. Yet this is the cornerstone of global warming. Doesn’t smell right.

Back to the details: again the two radiation types are dissimilar. Gases, liquids, and solids can do and do both. (there’s still a debate on the gases piece.) Solids, the standard for blackbody radiation, also exhibits line radiation ala spectrographic analysis.

People seem to get all hung up with the “blackbody”. It’s true “blackbody” is an idealization (99% as I said) but the radiation is not. The confusion stems from the fact that the matter experiencing “blackbody” radiation usually has an emissivity (or absorption factor) less than 1.0 (leading to “graybody”, and they also can vary with frequency (leading to “graybody II”). But whatever its variance, thermal (blackbody) radiation originated from an entirely different molecular/atomic activity than line emission. Ray felt (347) that since “blackbodies” do not exist (they’re an idealization), blackbody radiation must be the same as “line” radiation. Not true. I’m sorry they chose the term “blackbody” to mean blackbody, graybody, and, theoretically, whitebody, and everything in between including the variance by wavelength. But it is what it is.

Timothy continues,

“The different is a matter of degree, not kind. With solids you have more interactions between molecules which make possible a larger number of lines, bands and a larger variety of mechanisms for line and band broadening, but it is still pretty much the same thing, just more of it.”

For the sake of brevity since this is getting quite long — wrong!

Timothy, with Ray’s help, says, in part,

“….. But half-lifes don’t work that way. Energy is constantly being transfered by collisions, but reemission takes place according to a law of exponential decay where the states have no memory of how long they have existed. As long as a certain percentage are in an excited state at any given time, reemission will be taking place at a certain rate. ……”

Hey! We’re getting somewhere. I think this says I misread the statements regarding collisions beating emissions by millions-billions to one to mean re-emission is unlikely to happen. But it sounds like you’re saying the re-emission from greenhouse gases will happen. It just has to go through a million-billion-gazillion molecular energy transfers before. Does this mean that re-emission (granted different photon/different molecule) from CO2 will nearly equal its absorption — eventually?

I’m getting a long ways up the hill, but can’t yet make it over the top. By simplistic numbers there is 519 watts/meter-squared getting absorbed somehow into the atmosphere/clouds. Assuming it all eventually gets into a greenhouse gas and all more eventually gets re-emitted: half up and (eventually) out and half down and (eventually) absorbed — that’s 260 down, not the budget’s 324, and 260 up, not 195. Maybe I’m missing something simple, but it still doesn’t totally add up.

Comment by Rod B — 8 Sep 2007 @ 11:04 PM

628. Alastair (622): you said “Theoretically gases do not radiate Planck blackbody radiation, but despite that water vapor does!”

I said, “Then how do you explain the college-level development of earth energy balance using 2-3-4+ atmospheric levels all radiating ala Planck function [T^4]? How do you explain solar radiation?”

Then you said, “The college level courses AND all the text books are wrong…. ”

So I say, “Well! That’s a helluva note!!”

It sounds like you now agree gases can radiate ala Planck because the Sun does. Granted it under pretty high pressure; but so what?

I’d have trouble with the surface of the Sun being a liquid, but a good case can be made for much of the interior Hydrogen being liquid — a superfluid at that maybe, if I recall.

You say, “how do you explain the Fraunhofer lines if gases produce blackbody radiation? ”

There is no reason under the Sun (pardon the pun) why a gas (or liquid or solid) can not experience both Planck thermal (blackbody) radiation and line/Fraunhofer radiation simultaneously. A common trait is a molecule absorbing its Fraunhofer line wavelength from its home body’s Planck radiation so the spectrum of the body we see is the continuous spectrum with characteristic narrow blank lines.

Ray (623) you might be on to something that will help lead me out of my dilemma. But something needs clarifying. You say, “…As the atmosphere heats up, net energy loss from the surface decreases–so there doesn’t have to be any energy transfer to the ground, just less energy lost…. ” How does this reconcile with the Earth’s steady state Energy Budget — the old 390 watts surface radiation leaving and 324 watts from greenhouse gases and clouds being absorbed.

Comment by Rod B — 8 Sep 2007 @ 11:52 PM

629. Rod B (#634) wrote:

No, no, no, Timothy. Planck blackbody radiation is not “always an idealization”. A smooth spectrum output is an idealization (99% of the time). But Planck function blackbody radiation (loosely named to include whatever spectrum) is as real as it gets.

Rod, for an object to be a true blackbody, the absorptivity at all wavelengths would have to be exactly equal to 1. For an object to be a true grey body, the absorbativity would have to be less than 1, but a single constant for all wavelengths.

Anything else is what would be refered to as a realistic body. In this thread, see Bart Paul Levenson #180, Tim #258, Tim #265. In A Saturated Gassy Argument, most definitely check out the exchange between Ray Ladbury and raypierre 154.

Then check out what Ray Ladbury (#623) just said:

Finally the closest to a blackbody spectrum you get is emission from a small hole in a cavity. This is because you have not only the interactions of the gas and the walls of the container, but the interactions of the radiation field itself that come to equilibrium.

As I have pointed out, for this to result in a true black body, the hole in the cavity would have to be infinitely small. Anything else would fail to achieve equilibrium, however small the deviation. You yourself found it necessary to qualify your statement by means of the phrase “loosely named.”

Comment by Timothy Chase — 9 Sep 2007 @ 5:00 AM

You are wrong with both the points you are making.

The greenhouse effect is not like a blanket. A blanket is a solid which prevents the warm air escaping and being replaced by cold air. It has a low coefficient of conduction but mainly works by preventing convection. In the atmosphere greenhouse gases do not prevent convection. In fact they participate in it. What greenhouse gases do is absorb the thermal radiation from the surface and pass that energy to the other molecules in the air so warming it. The air at the bottom of the atmosphere is being continually warmed by the greenhouse gas absorption, not by conduction with a warm human body.

If blackbody radiation was produced by the smearing of emission lines then an iron bar would glow green hot. But it would not produce blackbody radiation at room temperature, because those emission lines are frozen out. And solid iron is a homogeneous non polar molecule so it cannot emit rotational or vibrational energy either.

Blackbody radiation is a type of continuum radiation just like Bremsstrahlung. Thermal blackbody radiation is produced by inter-atomic vibrations which only exist in solids and liquids. You really ought to consult your textbooks rather than making up science to fit your preconceptions.

Comment by Alastair McDonald — 9 Sep 2007 @ 6:03 AM

631. Timothy,

You are correct that a blackbody is an idealisation, but blackbody radiation to which Rod was referring is not. It is also known as cavity radiation, and probably more correctly as thermal radiation since its intensity depends on the body’s temperature. Note that I say body temperature because only solid and liquid bodies can emit it. Gases do not and vapours only emit it when they condense into liquids.

Note, the intensity of line radiation emitted by gases is a function of their concentration. Line width broadening does not change the intensity of a line, only its width. So thermal radiation only applies to blackbody radiation. However, there is a problem using thermal radiation as a term because on earth the thermal radiation is in the infrared. Since the line radiation absorbed and emitted by greenhouse gases is also in the infrared, people tend to think that they are the same thing. They are not as I have explained to Ray above.

Comment by Alastair McDonald — 9 Sep 2007 @ 9:56 AM

632. Alastair, ever hear of phonons? Electronic transitions are not the only energy states of importance–especially in a solid. Most of your problems occur because you have an incomplete understanding of the physics. In some solids, the charge carriers are not even electrons or holes, but rather bound states of electrons, holes and phonons with an electric charge a fraction of that of an electron. Blackbody radiation is not a continuum phenomenon–as evidenced by Planck’s need to keep the quantum of action nonzero to obtain the blackbody spectrum. Bremmstrahlung is much simpler–the interaction of a fast-moving single charge with a sea of charge. You can find a derivation of it in an elementary E&M text book. Try finding an elementary derivation of blackbody radiation.

Comment by ray ladbury — 9 Sep 2007 @ 11:00 AM

633. Timothy, you just can’t get past the fact that the term “blackbody” as popularly used in talking about Planck functions is an unfortunate misnomer. It doesn’t mean the purist (and more accurate) definition of a true blackbody. It means any body that radiates ala Planck function. I admit I probably confuse things by switching between “blackbody” and “thermal” (which I would prefer), but I’m still using “blackbody” in the popularly accepted sense. So please stop refutting my “blackbody” function description by claiming I’m not using a perfect true blackbody.

Alastair (628), I agree with (but only 92%) and like your explanation. Makes sense. My only area of disagreement is the “only liquids and solids” part. I’m fully aware of the ongoing debate as to whether gases emit Planck thermal radiation. But one has no difficulty finding texts, papers, and forum posts that say they do, and clearly does not have to “make it up”. (One can find as well stuff that says they don’t.)

Comment by Rod B — 9 Sep 2007 @ 12:29 PM

634. Alastair McDonald (#628) wrote:

What greenhouse gases do is absorb the thermal radiation from the surface and pass that energy to the other molecules in the air so warming it. The air at the bottom of the atmosphere is being continually warmed by the greenhouse gas absorption, not by conduction with a warm human body. In the atmosphere greenhouse gases do not prevent convection. In fact they participate in it. What greenhouse gases do is absorb the thermal radiation from the surface and pass that energy to the other molecules in the air so warming it. The air at the bottom of the atmosphere is being continually warmed by the greenhouse gas absorption, not by conduction with a warm human body.

Actually greenhouse gases are predominantly directly responsible for cooling the atmosphere. The exceptions are a small region in carbon dioxide and a wider range of altitudes belonging to ozone.

I pointed this out in #343:

The following shows calculated degrees of cooling per day*wavelength as a function of pressure (which decreases exponentially with altitude) and wavelength for co2, ozone and water vapor.

Line-by-line calculation of atmospheric fluxes and cooling rates 2
http://www.aer.com/scienceResearch/rc/m-proj/abstracts/rc.clrt2.html

You will notice that by infrared radiation, water vapor always has a local atmospheric net cooling effect, there is only a small window where CO2 warms at roughly 12 km – and it cools for most altitudes-wavelengths, and ozone has a strong local atmospheric net warming effect until about 20 mb where it begins to radiate more radiation than it absorbs. With the exception of ozone, the direct warming effect of these greenhouse gases is principally due to radiation being absorbed by the surface. The atmosphere is warmed primarily by moist air convection in the troposphere, giving way to diffusion above the tropopause – with ozone playing a secondary role throughout ~8-24 km. But all of this is specific to season and latitude.

What warms the atmosphere is primarily moist air convection. The greenhouse gases primarily warm the surface, giving rise to moist air convection.

Alastair McDonald (#628) wrote:

If blackbody radiation was produced by the smearing of emission lines then an iron bar would glow green hot. But it would not produce blackbody radiation at room temperature, because those emission lines are frozen out.

If its peak emissions were in the green part of the spectrum, one might say that it is glowing green hot – although if its peak emission were in the green part of the spectrum it would appear white to our eyes – just like stars.

Why aren’t there any green stars?

However, even at room temperature, iron glows in the green part of the spectra – but very, very dimly – in accordance with its brightness temperature.

Brightness temperature
http://en.wikipedia.org/wiki/Brightness_temperature

Alastair McDonald (#628) wrote:

And solid iron is a homogeneous non polar molecule so it cannot emit rotational or vibrational energy either.

Is ozone non-polar given the fact that it consists of three atoms of the same kind? (Yes.) Is carbon dioxide permanently polar? (No.) What about magnetic moments? (O2 has a small magnetic moment which is responsible for its ability to absorb and emit in the microwave region.) The more the different types of excited quantum states that are possible, the more the emission lines.

Any molecule consisting of only two atoms of the same kind cannot be electrically dipolar, although there may be magnetic moments giving rise to microwave emissions. However, all molecules consisting of three or more atoms will have quantized vibrational states – and even though carbon dioxide is not permanently dipolar, vibration results in it being temporarily dipolar such that it is capable of quantized rotational states.

Alastair McDonald (#628) wrote:

Bremsstrahlung radiation is caused by the acceleration of a charge, Alastair.

And in the strictest sense, blackbody radiation exists only as an idealization since absorptivity and emissivity are never equal to 1 for all parts of the spectra. All thermal radiation is realistic body radiation – where the absorptivity varies in different parts of the spectra.

Alastair McDonald (#628) wrote:

Thermal blackbody radiation is produced by inter-atomic vibrations which only exist in solids and liquids. You really ought to consult your textbooks rather than making up science to fit your preconceptions.

Is water vapor incapable of inter-atomic vibrations? Is CO2 incapable of such vibrations? How about ozone? As far as thermal radiation is concerned, liquids and solids and solids differ from gases in the number of different types of excited quantum states which are possible – given the proximity of the atoms to one another and their bonds. But even in the case of dusts and alloys the quantization is often nearly as sharp in certain parts of the spectra as that of the atmosphere.

Alastair, there are some aspects of your motivation which in all honesty I would prefer not to examine too closely – for my own sake. However, your motivation obviously has very little to do with science.

This forum is intended primarily so that people can learn about the science of climatology. So far you have contributed a great deal more to the obsfucation – and in the process contributed to those who would deny that it is even a science even though you are quite clearly capable of learning and making positive contributions. A good case in point is where you deny that carbon dioxide is capable of emitting except near the surface – where absorption in any part of the spectra is already saturated due to water vapor. But there are others. This post of yours is a good case in point. You have been deliberately obnoxious towards Ray Ladbury, a physicist – who was simply explaining some of the actual science.

I do have a few questions regarding your motivation, though.

Why have you chosen to present your “alternative science” here? Why not create your own forum? Your own blog? There are places where you could create your own wiki and blog with 100 megabytes for images and files – where there is no limit on the size of the wiki itself.

Is it because Real Climate appeals to an audience which you would be unable to attract on your own?

When both parties to a relationship benefit this may be described as “symbiosis.” When one party principally benefits to the detriment of the other, the term “symbiosis” no longer applies.

Comment by Timothy Chase — 9 Sep 2007 @ 12:53 PM

635. Many thanks to Timothy Chase for taking so much trouble in trying to explain things to me in his posts of #606 and #611. It must be irritating to have to explain things which are straightforward to you but are causing difficulties to a thicko such as myself. I had almost got to the stage of totally accepting your explanations while still being a bit muddled in my mind. You have certainly helped a great deal over certain matters but every explanation prompts me to further thoughts and questions. May I please ask for forbearance and ask for yet more help?

In #601 you wrote: “If the atmosphere is completely opaque at a given wavelength, what this will imply is that parcels of energy (think of them as photons)which leave the surface at that wavelength will stand a 50-50 chance of making it out of the atmosphere without making another trip back to the surface and warming the suface.” What this would imply to me and, judging from Raypierre’s original essay, is that you are actually defining 50% transmittance and not saturation. You referred me to an energy balance diagram which clearly shows that there is very substantially more downwelling radiation than radiation leaving the top of the atmosphere. Your statement seems to me to be incompatible with the data presented in the diagram. If you were correct, 50% would escape at first attempt, 25% at the next and so on – in other words a lot more than is the case according to the diagram.

In my ignorance, therefore, I am still defining opacity as indicating that nothing can get through the lower (possibly saturated or opaque) layers but, having read the thread carefully, I am suggesting exceptions to explain how the radiation escapes as OLR. You may think this extremely naive but I would be interested in your response, so here goes with my list of possible exceptions:
1) “The atmospheric window”- 40W/sq m from surface (not contentious except insofar as it might be somewhat, albeit slightly, narrowed by increasing CO2 – and nobody I have asked has come up with a figure for this).
2) Areas at either edge of the main CO2 absorption band where a combination of CO2 and water (with overlapping absorption spectra) would cause total saturation in the prescence of water vapour but not in a dry overlying atmosphere.
3) Photons in the embrace of CO2 molecules lifted above the water vapour layer by convection.
4) Possibly, photons in the embrace of water vapour released when the vapour condenses to water droplets in cloud. This may be Micky Mouse thinking – what actually does happen when water vapour becomes cloud?

Further sources of potential OLR are, of course, also available. Clouds, as black body radiators, apparently push 30 W/sqm through the atmospheric “window” while there is also 67W/sqm of solar energy absorbed by the atmosphere (not the surface) to work on. Some of the latter will be temporarily interfered with by CO2 but nothing like as much as at lower levels because of much larger gaps between lines and no complementary blocking by water vapour.

I am suggesting that most of the surface radiation remains at low levels of the atmosphere and a substantial part of OLR may never have got to the surface in the first place. I have come to this conclusion based upon my interpretation of discussion in RealClimate on global dimming but my logic might be flawed so please enlighten me if this is the case.

Apparently, aerosols cause global cooling, partly by reflecting solar radiation and partly by absorbing it in the atmosphere and preventing it from hitting the ground. I understand that carbon absorbs solar radiation but what it then does with it ,I’m not sure. I believe SO2 both reflects and absorbs. The shape of the latter molecule must resemble that of CO2 and I take it that it has a dipole moment and therefore rotates and vibrates with the best of them but fancies shorter wave radiation than the ghgs. The molecules presumably also, therefore, re-emit the photons absorbed. If these got to the surface, they wouldn’t cool us – they’d do the opposite. But clouds are black body radiators. Thus, one would expect that clouds might accept short wave radiation ,just as the surface does, and, instead of spitting it back out in its absorbed form, would emit a blackbody spectrum. Then, of course, water vapour also absorbs some solar radiation. If one has a layer of water vapour, energy hitting the top (from above) is much more likely to be bounced out upwards than it is to penetrate the layer and get to the surface, just as rising photons from the surface are much more likely to be returned there than to get through.

Timothy, when I suggested that 3.7 W/sqm (which rightly or wrongly, I understood to be the extra surface energy which would be retained by a doubling of atmospheric CO2 and would thus have to be got rid of at the top of the atmosphere to get back to equilibrium)) was small as a percentage of the total energy flux and when set against other possible variables, you asked me what variables I had in mind. You proceeded to assume that it was solar variability and went on to dismiss this. However, though I might have misled you, this was not what was in my mind. RealClimate’s posts on dimming suggest that, between the 1960s to 1990s, aerosols reduced surface absorption of solar energy by 7W/sqm, nearly twice the amount but in the other direction than is claimed for CO2 doubling. I fully accept that some of this could have been an albedo effect but, I suspect, not the majority. It is therefore clear that switching solar absorbtion from the surface to the high atmosphere has a cooling effect on the surface. I also think it likely that absorption of IR by high level CO2 may be less warming than absorption at low level.

Other variables include ozone at tropospheric and higher levels, effects of clouds and even man’s extra release of surface energy (though I have been led to believe that the last is insignificant).

I would just say that, as a biologist, 3.7 as a percentage of 235 or 390 (whichever you want to choose) is a very small percentage to rely upon to be biologically significant. I can, in this instance, see the logic of the science. It’s the somewhat arrogant precision given to the numbers that worry me. CO2 is undoubtedly a greenhouse gas and its increase is man made. While not a fan of the precautionary principle, I can see that, in this scenario, the stakes are as high as they could be and can thus sympathise with its application.

Comment by Douglas Wise — 9 Sep 2007 @ 3:41 PM

636. I (#621) wrote:

… However, gases will come closer to approaching this idealization at higher partial pressures (due to line broadening) and when they have a larger number of quantum states…

Rob B (#625) responded:

You (and others) are still confusing Planck blackbody radiation with line emission of, say, greenhouse gases.

A great many others are “victims of this delusion.” Typically they have a strong background in physics. (I would be an exception to that rule.)

Rob B (#625) wrote:

I understand water vapor “line” radiation has near continuous quanta in certain parts of the spectrum (a slippery phrase as the spectrum is the crux of the observable difference) and looks a little like blackbody [thermal] radiation. But it ain’t.

And why do you claim that reemission by water vapor isn’t thermal radiation?

Because you choose not to use the word “thermal”?

*

I (#621) wrote:

As such, there really isn’t a fundamental difference between thermal radiation which arises due to ionization and thermal radiation which arises due to strictly molecular mechanisms.

Actually I should have said “… between thermal radiation which arises due to jumps in electron orbitals (up to higher energy states at absorption and down to lower energy states during emission) and thermal radiation which arises due to strictly molecular mechanisms.”

Not thinking, not clearly…

… never mind.

Rob B (#625) wrote:

That’s a paradox nearing an oxymoron. Or, put another way, that’s wrong! Bear in mind the one and only full spectrum independent variable in Planck’s formulation is Temperature (the real kind). No molecular bonding coefficients; no molecular shapes or polarization; no molecular moments of inertia; etc.

No structure, no matter…

… only idealization.

Rob B (#625) wrote:

I agree “… the ‘exception proves the rule’ doesn’t work very well in physics…” but it makes the point and sounds good [;-) . I like and appreciate your helpful explanation of O2 absorption/emission.

No problem.

*

Rob B (#625) wrote:

Now, back to your first and main (and wrong…???) point. Hanks reference in 339 is addressing thermal radiation versus quantum physics, but in any case is a cacophonic forum with various folks doing the same debate we are having.

There are a couple of important differences.

TeV, although more knowledgable than the fellow he is helping, probably hasn’t obtained his master’s degree as of yet. The fellow he is speaking to clearly understands that TeV is more knowledgable. But despite their general lack of expertise, TeV clearly understands that the closest thing we have to blackbody radiation is quantized, with transitions between discrete states being the cause of both absorption and emission.

Here, on the otherhand, you have people like raypierre, Eli, Ray Ladbury, Bart Levenson…

People with advanced degrees in physics willing to educate those who are less knowledgeable than they are. raypierre has a doctorate, Eli has a doctorate, Ray has a doctorate. Then you have people such as Alastair and yourself who are intent on denying that there is any such thing as expert knowledge in physics. People who have not obtained even the most basic degrees in the area and who are unwilling to acknowledge the expertise of those who have obtained PhDs in relevant fields.

Interesting contrast.

*

Rob B (#625) wrote:

It is helpful in this regard: It is clear that there is no fully complete understanding (agreement, really) nor a scientific consensus on the specifics of thermal radiation ICW greenhouse gas emission. Yet this is the cornerstone of global warming. Doesn’t smell right.

When it comes to radiation transfer, I doubt that you would be able to find anyone with a relvant degree (which I presume would include graduate level quantum mechanics) who would fail to acknowledge that the closest thing we have to blackbody radiation in the real world arises from transitions between discrete quantum states.

Do you know of any?

I believe quantum mechanics would be a prerequisite as the result of the quantized nature of electromagnetic energy and the quantized nature of matter.

Don’t you?

Of course, if you wanted to you could deny that matter and energy are quantized. You do have that latitude – if you wish to take it. Or you could simply claim that there is no scientific consensus in the matter. In fact, you could claim that the world is less than 10,000 years old or that the world is flat if you really wanted to – or if you prefer – that there is “no consensus” since not absolutely everyone agrees, and therefore the only objective position is to take is that these are still open questions.

Will you? Which ones?

Why not?

I suppose you could just claim that you can’t stand the stench.

*

Rob B (#625) wrote:

People seem to get all hung up with the “blackbody”. It’s true “blackbody” is an idealization (99% as I said) but the radiation is not. The confusion stems from the fact that the matter experiencing “blackbody” radiation usually has an emissivity (or absorption factor) less than 1.0 (leading to “graybody”, and they also can vary with frequency (leading to “graybody II”).

When the absorptivity varies within the visible part of the spectrum, the object is typically colored. Would you call something colored grey? When the absorptivity varies over different wavelengths, the object is said to be a realistic body with a realistic body spectrum.

*

Rob B (#625) wrote:

But whatever its variance, thermal (blackbody) radiation originated from an entirely different molecular/atomic activity than line emission. Ray felt (347) that since “blackbodies” do not exist (they’re an idealization), blackbody radiation must be the same as “line” radiation. Not true. I’m sorry they chose the term “blackbody” to mean blackbody, graybody, and, theoretically, whitebody, and everything in between including the variance by wavelength. But it is what it is.

Ray is a physicist with a background in radiation transfer. What is your expertise?

Your entire argument at this point turns on the claim “thermal (blackbody) radiation originated from an entirely different molecular/atomic activity than line emission.” Alright, what is the activity?

Vibration and rotation, and in the case of extremely hot gases, liquids and solids, electrons jumping up to higher energy orbitals upon absorption then back down to lower energy orbitals upon emission. It is the same. The only real difference is a matter of degree, principally in terms of the number of weak and strong bonds, the number of different quantum states, the complexity of the broadening mechanisms and the degree of broadening.

*

Rob B (#625) wrote:

Timothy, with Ray’s help, says, in part,

“….. But half-lifes don’t work that way. Energy is constantly being transfered by collisions, but reemission takes place according to a law of exponential decay where the states have no memory of how long they have existed. As long as a certain percentage are in an excited state at any given time, reemission will be taking place at a certain rate. …”

Actually I said it with Alastair’s help.

Back in high school I had picked up the two-volume Quantum Mechanics by Claude Cohen-Tannoudji, Bernard Diu and Franck Laloe, an english translation of the french. This is what I taught myself from, informally. Then I lost it on a bus while I was in the navy visiting Honolulu. In any case, it had been a very long time and I was rather rusty – so certain pieces which should have fallen into place didn’t – until Alastair used the term “half-life” in relation to reemission.

Then I responded in #500:

If a molecule could reemit only at or after the half life but was interrupted each time, then carbon dioxide and water vapor could not reemit under the very same circumstances that physicists say is required to achieve local thermodynamic equilibria. But the term is “half life.” This means that it is probabilistic – and has no memory of how long the molecule has been in the excited state. It parallels subatomic particle decay. This sort of thing is fairly basic, and should be dealt with in any undergraduate level course which touches on quantum mechanics even at the most introductory level.

If the gas is a certain temperature, then a certain percentage of molecules will be in the excited state (which will largely be a function of the Maxwell velocity distribution, at least under local thermodynamic equilibria conditions), and it does not matter which molecules are in the excited state or how long they have been so. Over a given duration, a certain percentage of them will reemit as determined by a law of exponential decay. Thus it is not the molecule which reemits, but one molecule which absorbs, and then after so many collisions another which emits, so that it is only the gas, the population, which reemits by both absorbing and emitting the radiation.

So what I wrote was confirmed by Ray Ladbury but I had put it together independently of him.

As I said, it was Alastair who gave me the missing piece.

I learn from others. I put things together. I try to understand.

What are you doing?

Rob B (#625) wrote:

Hey! We’re getting somewhere. I think this says I misread the statements regarding collisions beating emissions by millions-billions to one to mean re-emission is unlikely to happen. But it sounds like you’re saying the re-emission from greenhouse gases will happen. It just has to go through a million-billion-gazillion molecular energy transfers before. Does this mean that re-emission (granted different photon/different molecule) from CO2 will nearly equal its absorption — eventually?

How long is the half-life of a given state?

Roughly that long.

Exponential decay with no memory of how long any given molecule has been in a particular state.

*

Rob B (#625) wrote:

I’m getting a long ways up the hill, but can’t yet make it over the top. By simplistic numbers there is 519 watts/meter-squared getting absorbed somehow into the atmosphere/clouds. Assuming it all eventually gets into a greenhouse gas and all more eventually gets re-emitted: half up and (eventually) out and half down and (eventually) absorbed — that’s 260 down, not the budget’s 324, and 260 up, not 195. Maybe I’m missing something simple, but it still doesn’t totally add up.

Well, let’s look at the diagram I pointed you to:

The Energy Balance and Natural Climate Variations
http://www.cara.psu.edu/climate/climatechangeprimer-pr4.asp

Sunlight coming in is 342 watts per square meter. Reflected solar radiation going out is 77 off of clouds and 30 off of the surface. Outgoing longwave is 235. 342 = 77 + 30 + 235.

What is coming in equals what is going out.

Ok then.

Let’s drop the reflected and just look at absorbed radiation. That’s the 235.

The radiation which is absorbed by the atmosphere before it reaches the surface is 67. That which is absorbed by the surface is 168. 235 = 67 + 168.

What is entering the climate system as thermal energy equals that which is absorbed by the atmosphere from space plus that which is absorbed by the surface from space.

All right.

Let’s look at the radiation being absorbed by the atmosphere. We have 67 coming from space, 24 coming from the surface in the form of thermals, 78 coming from evapotranspiration, then 350 coming from surface radiation – with 40 watts per square meter from the surface passing through the atmosphere without being absorbed. 67 + 24 + 78 +350 = 519. Now once these watts have been absorbed by the atmosphere, where do they go?

165 goes from greenhouse gases to space. 30 goes from the clouds to space. 324 is back radiation to the surface. 165 + 30 + 324 = 519 — which is the same amount of energy being absorbed by the atmosphere.

What is coming in equals what is going out.

Let’s see…

Now we need to look at the radiation being absorbed by the surface.

168 is coming directly from sunlight and another 324 is back radiation (reemission) from the atmosphere. That’s 492.

Now let’s look at what is leaving the surface after having been absorbed. 24 from thermals. 78 from evapotranspiration. 390 from surface radiation (i.e., thermal emissions from the surface). That is 24 + 78 + 390 = 492.

What is coming in equals what is going out.

Comment by Timothy Chase — 9 Sep 2007 @ 5:57 PM

637. Douglas,

I don’t consider it a bother when someone is genuinely seeking the truth. But even when this is not the case, discussions will oftentimes give me the opportunity to clarify things in my own mind. I knew basically nothing regarding the greenhouse effect back in May – and I am still learning, both in terms of my understanding and in how I express various ideas.

Douglas Wise (#633) wrote:

In #601 you wrote: “If the atmosphere is completely opaque at a given wavelength, what this will imply is that parcels of energy (think of them as photons)which leave the surface at that wavelength will stand a 50-50 chance of making it out of the atmosphere without making another trip back to the surface and warming the suface.”

That statement was wrong, but I wasn’t taking the time that I probably should have in responding to you. Actually what stands a fifty-fifty chance is whether upon emission a photon will be upwelling or downwelling. Upwelling takes it towards space, downwelling towards the surface. But most of the radiation emitted by the atmosphere near the surface will be reabsorbed by the surface – if for no other reason than the fact that the surface is much closer, and a random walk is more likely to make it to the surface than to space – given an initial low altitude.

In fact, the halfway point as far as radiant energy is concerned is roughly 6 km in altitude – although it has been climbing as has the boundary between the troposphere and the stratosphere known as the tropopause. This halfway point is what is refered to as the “effective radiating level” or “layer.” It has the “effective radiating temperature” which is the temperature that the earth would appear to have as seen from a distance given its thermal emissions to space.

*

Douglas Wise (#633) wrote:

… I am still defining opacity as indicating that nothing can get through the lower (possibly saturated or opaque) layers but, having read the thread carefully…

How would that work?

Your view would seem to imply that reemission never takes place. But if thermal radiation is absorbed, it will be reemitted.

Energy may be lost to collisions, but energy may also be gained by collisions – resulting in reemission. If it is absorbed and reemitted a fair number of times, eventually it will make it out of the atmosphere simply as the result of a process which is essentially a form of diffusion. But there are other added complications which will tend to make things move a little more quickly than this might suggest. You were here for part of it: transitions from higher to lower states of excitation – where at least some of the energy which gets emitted is unlikely to be reabsorbed.

The other effects you mention (e.g., the atmospheric window, moist air convection, the freeing of latent heat) will also undoubtedly make it easier for energy to make it to space, but they aren’t strictly necessary.

In essence, when adding carbon dioxide or other greenhouse gas to the atmosphere, by making the atmosphere more opaque to thermal radiation, we are disturbing an equilibrium in which the rate at which energy enters the climate system equals the rate at which energy leaves the climate system. We are doing this by increasing the resistance to the flow of thermal energy from the surface to space, lowering the rate at which this occurs. So long as the rate at which energy leaves the system is lower than the rate at which energy enters the system, the temperature of the climate system must increase.

As the temperature increases, the rate at which energy is radiated increases – to the fourth power of the temperature, I believe. As such, the climate system will achieve a new equilibrium despite the increased resistance to the flow of thermal energy, but only as the result of it having a higher temperature which compensates for the increased resistance.

*

Douglas Wise (#633) wrote:

Apparently, aerosols cause global cooling, partly by reflecting solar radiation and partly by absorbing it in the atmosphere and preventing it from hitting the ground. I understand that carbon absorbs solar radiation but what it then does with it ,I’m not sure.

Sunlight which is not absorbed but which is simply reflected back into space has in a very important sense never made it into the climate system. It isn’t thermal radiation, at least not in relation to the climate system itself. As such, some aerosols will cool the climate system (temporarily masking the effects of added greenhouse gases) by reducing the amount of sunlight which gets absorbed and converted to thermal energy.

When you speak of black carbon, since this has a lower albedo, this implies that more sunlight gets absorbed and added to the thermal energy within the system. Black absorbs light making things hotter. Sulfur dioxide primarily increases the albedo – although the effects may be different depending upon the source of the radiation. While clouds are “blackbody” radiators, this is true primarily in the longwave. If they were blackbody radiators at all frequencies, given the temperature of the atmosphere they would appear black, absorbing all visible light but reemitting almost entirely within the infrared.

Since they are typically a lighter shade than that, they can’t be blackbody radiators in the visible part of the spectrum. As such, while they increase the greenhouse effect with what thermal radiation already exists within the climate system, they also reduce the amount of thermal energy which enters the system by scattering visible light back into space. Two competing effects. Which one wins? Depends upon the type of cloud.

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Douglas Wise (#633) wrote:

RealClimate’s posts on dimming suggest that, between the 1960s to 1990s, aerosols reduced surface absorption of solar energy by 7W/sqm, nearly twice the amount but in the other direction than is claimed for CO2 doubling. I fully accept that some of this could have been an albedo effect but, I suspect, not the majority. It is therefore clear that switching solar absorbtion from the surface to the high atmosphere has a cooling effect on the surface. I also think it likely that absorption of IR by high level CO2 may be less warming than absorption at low level.

How would that work?

The peak emission for the sun is in the yellow part of the spectrum. Relative to this it actually emits fairly little infrared. Since greenhouse gases are more or less transparent to visible light, they do not absorb visible light. So for the energy to enter the system as thermal energy which gets radiated away by greenhouse gases, it has to be absorbed somewhere else. Generally the surface. Maybe black carbon before it has a chance to settle on sea-ice or some glacier. Perhaps a dark cloud. But generally the surface. Although I can think of one important exception: ozone absorbs ultraviolet radiation in sunlight – warming the upper atmosphere.

*

Anyway, don’t worry too much about being new to this.

As I’ve said, I am pretty much in the same boat. As such, I am quite grateful that there exist exceptionally bright and talented individuals like the climatologists here who know a great deal more than I do. In addition to everything else, it means that I have people I can learn a great deal from.

Comment by Timothy Chase — 10 Sep 2007 @ 12:04 AM

638. Perhaps, I should clear up something. First, I did not mean to imply that blackbody radiation results from spectral broadening. Blackbody radiation is the distribution that results from the radiation field being in thermal equilibrium. However, photons do not interact with eachother, so the only way the radiation field can come into equilibrium is by interacting with matter in its vicinity. Spectral broadening brings the radiation field closer to blackbody, because it means the emissivity is nonzero over a larger proportion of the spectrum.

Comment by Ray Ladbury — 10 Sep 2007 @ 7:53 AM

639. Ray,

I looked up phonons on Wikipedia and it seems to me that they are the atomic oscillations in the crystal lattice of a solid that are generating the blackbody radiation.

Spectral broadening does produde a continuum spectrum over the band, but does not correspond to a Planck curve. The 15 um band has a spike in the centre and troughs on both sides.

Rod,

It is true that many (all) climatology say that greenhouse gas emissions depend on temperature just like blackbody radiation. However any proper science (physics or chemistry) textbook does not make that mistake.

Timothy,

You are correct that ozone is a greenhouse gas, and so is carbon dioxide but iron is not. It radiates continuum radiation and the greenhouse gases radiate line radiation. They originate in two very different ways, as does yet another form of radiation Bremsstrahlung.

However, thanks for pointing out that “But half-lifes don’t work that way.” That has given me the clue I have been looking for. I may not have to pick everyones brains any longer :-)

Comment by Alastair McDonald — 10 Sep 2007 @ 9:14 AM

640. Alastair, Yes, I know what the 15 micron band looks like. The emissivity is nonzero in that band and in the other absorption bands and zero elsewhere for CO2. The thing is that pressure effects, etc. broaden the line–effectively extending the region where the emissivity is nonzero. What you will find is that if you convolute the emissivity with the blackbody spectrum for the proper temperature, you get the appropriate emission spectrum. This is why you get different effective temperatures for different ghg lines–the emissions are coming from different altitudes at different temperatures.

Comment by Ray Ladbury — 10 Sep 2007 @ 12:40 PM

641. For those who are interested: a quick cite for the thermal absorption spectra of pure crystals…

Infrared absorption spectra of crystals arise from vibrations of the atoms in the crystalline lattice which are associated with a corresponding periodic change in dipole moment. The vibration spectrum of any crystal lattice cannot yet be predicted theoretically but a good deal of progress has been made in certain special cases. The extent to which the infrared spectra of diamond, silicon and germanium can be interpreted in terms of current theories will be briefly reviewed, with special reference to the problem of the two types of diamond. The explanation of the ferroelectric properties of barium titanate in terms of the structure of the crystal is still uncertain. One theory (Jaynes) predicts an infrared absorption band at 10T. No band is present in the spectrum at that wavelength. In micas the positions of the hydrogens have not been determined. It may be possible to do so by infrared techniques. This problem will be discussed. Of considerable interest, also, is the use of infrared as a means of estimating or groups in various micas.

Abstract to: INFRA-RED STUDIES OF CRYSTALS
G. B. B. M. SUTHERLAND
UNIVERSITY OF MICHIGAN
Symposium Program 7th Annual FREQUENCY CONTROL REVIEW OF TECHNICAL PROGRESS
18 – 20 MAY 1953
http://www.ieee-uffc.org/fc_history/symposia2.pdf

Fairly old, but people don’t seem to find pure monoatomic crystals that interesting anymore and focus on the more puzzling multi-atomics and alloys.

Comment by Timothy Chase — 10 Sep 2007 @ 2:20 PM

642. #638

Ray,

Yes, the current idea is that the greenhouse gases emit depending on their temperature, but that is not what the physical chemistry books say.

The climatologist say that, since the 15 um band roughly matches the 215K blackbody curve, then that is the temperature of the layer where it is emitting. That is around the tropopause, but I though that we had agreed it came from higher in the atmosphere, i.e. the thermosphere.

Comment by Alastair McDonald — 11 Sep 2007 @ 9:46 AM

643. Thanks, Timothy, for your post #635 in response to mine (#633).

I think I’m moving on and understanding more. I also think that I probably expressed myself badly in the last post – probably due to the absence of a hen-pecking wife who was away judging gundog tests in Germany and hence unable to monitor my whisky intake.

I tried to suggest that I defined opacity (or saturation)of a particular waveband as indicating that nothing (in terms of that waveband) could get from the surface to the top of the atmosphere. You effectively (but more politely) suggested that this was rubbish and indicated that I must therefore be denying the existence of re-emission.

When I first read the essay initiating the thread, I accept that I assumed that photonic energy grabbed by a ghg molecule would be converted to non radiative thermal energy. I now accept that a chain of events occurs which leads to a variety of outcomes which involve local heating of the atmosphere, re-emissions of photons of longer wavelengths than those absorbed in the first place and a limited number of re-emissions of photons of the original wavelength.

Because of my acceptance of re-emission, I have a slightly different mental picture of CO2 waveband saturation which I will explain thus (I’m not allowing for convection because I think it might complicate matters): As CO2 concentration increases, gaps between lines in the central area of the band start absorbing relevant IR and, furthermore, the two edges of the bandwidth start coming into play as well. Re-emission occurs (half up and half down). As this is repeated for every ascending layer of atmosphere, an ever smaller proportion of photons will get through until the number is so vanishingly small as to be effectively zero (=100% absorbance or 0% transmittance). I believe that I have interpreted Raypierre correctly when I suggest that this is the situation that obtains for CO2 in the 13.5-17 micron part of its absorption band. However, I don’t think you agree with my interpretation and cannot quite understand why.

I accept that one can identify OLR in the 13.5-17 micron band and it is therefore necessary to explain where it’s coming from. I don’t necessarily see this as a problem. In attempting my explanation, may I use the concept of the “effective radiating layer” that you raised in your post? I believe OLR brightness temperatures differ for different IR wavelengths (depending upon which ghgs are interfering) and give an indication of the mean temperature (and thus mean altitude) at which emission has occurred. Alastair has previously given me a cite that demonstrates this (and Ray has confirmed it in post #638). Alastair says that anything coming direct from the surface to space should have a brightness temperature of 300K. The closest we get to this is approximately 290K for IR travelling through the atmospheric window. This is made up of 40/70 from the surface and 30/70 from cloud tops. OLR of wavelengths which are fully saturated by water vapour in the lower atmosphere show a brightness temperature of 275K. (4/7 x 300 + 3/7 x 275) roughly equals 290 – as one would expect.

Water vapour can become cloud and cloud is a blackbody radiator in the IR range. I would suggest that a good proportion of the total solar energy (67W/sqm) absorbed by the atmosphere ends up in cloud and will be emitted over a wide wavelength range unimpeded by water vapour. Its range will also include the 12-18 micron bandwidth which will be somewhat impeded by overlying CO2. At the top of the water vapour layer, even without conversion to cloud, IR of wavelengths normally absorbed by water vapour might find escape to space on re-emission easier. This would certainly be the case if the water vapour column was somewhat higher that the length required for saturation, implying that any downwelling radiation would experience much more interference than upwelling. It is thus plausible to rationalise that most OLR in the water vapour absorbing range is emitted at the wet/dry interface of the atmosphere at a brightness temperature of 275K.

The question remains as to where the OLR in the CO2 absorbing range is coming from, particularly that in the 13.5-17 micron part of the band. I have already suggested that some will come from cloud (indirectly via atmospheric solar absorption) and, though the atmosphere may be CO2 saturated from 13.5-17 microns from the ground up, its top slice won’t be. Some IR will therefore get through gaps. I would also like to suggest that photons absorbed in the 8-13.5 bandwidth might result in re-emissions of others in the 13.5-17 micron range. Finally, non radiative thermal energy high in the atmosphere, having arrived by convection, will stimulate CO2 to spit out some photons. It seems clear that all this must take place somewhere between the cloudtops and the top of the atmosphere, giving a brightness temperature of only 215K.

I think it might be tempting providence to write QED at this stage. In any event, I am not trying to suggesest that the modellers are wrong, only trying to rationalise the mechanism by which the effects they predict could become manifest.

I would like to deal with one other issue where you seemed to disagree with a statement I made. I suggested that switching solar energy absorption from the surface of the planet to the atmosphere would have a cooling effect (on the surface). You asked how that would work. I think it self evident but I also think you may have misunderstood what I was attempting to infer, partly because I somewhat misguidedly linked it in with g