Part II: What Ångström didn’t know
By raypierre , with the gratefully acknowledged assistance of Spencer Weart
In Part I the long struggle to get beyond the fallacious saturation argument was recounted in historical terms. In Part II, I will provide a more detailed analysis for the reader interested in the technical nitty-gritty of how the absorption of infrared really depends on CO2 concentration. At the end, I will 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. 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").
Suppose we were to sit at sea level and shine an infrared flashlight with an output of one Watt upward into the sky. If all the light from the beam were then collected by an orbiting astronaut with a sufficiently large lens, what fraction of a Watt would that be? The question of saturation amounts to the following question: How would that fraction change if we increased the amount of CO2 in the atmosphere? 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. Think of a conveyor belt with red, blue and green M&M candies going past. You have one fussy child sitting at the belt who only eats red M&M’s, and he can eat them fast enough to eat half of the M&M’s going past him. Thus, he reduces the M&M flux by half. If you put another equally fussy kid next to him who can eat at the same rate, she’ll eat all the remaining red M&M’s. Then, if you put a third kid in the line, it won’t result in any further decrease in the M&M flux, because all the M&M’s that they like to eat are already gone. (It will probably result in howls of disappointment, though!) You’d need an eater of green or blue M&M’s to make further reductions in the flux.
Ångström and his followers believed that the situation with CO2 and infrared was like the situation with the red M&M’s. To understand how wrong they were, we need to look at modern measurements of the rate of absorption of infrared light by CO2 . The rate of absorption is a very intricately varying function of the wavelength of the light. At any given wavelength, the amount of light surviving goes down like the exponential of the number of molecules of CO2 encountered by the beam of light. The rate of exponential decay is the absorption factor.
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… . For this, or larger, amounts of CO2,the atmosphere is optically thick at the corresponding wavelength. If you double the amount of CO2, you reduce the proportion of surviving light by an additional factor of 1/e, reducing the proportion surviving to about a tenth; if you instead halve the amount of CO2, the proportion surviving is the reciprocal of the square root of e , or about 60% , and the atmosphere is optically thin. Precisely where we draw the line between "thick" and "thin" is somewhat arbitrary, given that the absorption shades smoothly from small values to large values as the product of absorption factor with amount of CO2 increases.
The units of absorption factor depend on the units we use to measure the amount of CO2 in the column of the atmosphere encountered by the beam of light. Let’s measure our units relative to the amount of CO2 in an atmospheric column of base one square meter, present when the concentration of CO2 is 300 parts per million (about the pre-industrial value). In such units, an atmosphere with the present amount of CO2 is optically thick where the absorption coefficient is one or greater, and optically thin where the absorption coefficient is less than one. If we double the amount of CO2 in the atmosphere, then the absorption coefficient only needs to be 1/2 or greater in order to make the atmosphere optically thick.
The absorption factor, so defined, is given in the following figure, based on the thousands of measurements in the HITRAN spectroscopic archive. The "fuzz" on this graph is because the absorption actually takes the form of thousands of closely spaced partially overlapping spikes. If one were to zoom in on a very small portion of the wavelength axis, one would see the fuzz resolve into discrete spikes, like the pickets on a fence. At the coarse resolution of the graph, one only sees a dark band marking out the maximum and minimum values swept out by the spike. These absorption results were computed for typical laboratory conditions, at sea level pressure and a temperature of 20 Celsius. At lower pressures, the peaks of the spikes get higher and the valleys between them get deeper, leading to a broader "fuzzy band" on absorption curves like that shown below.
We see that for the pre-industrial CO2 concentration, it is only the wavelength range between about 13.5 and 17 microns (millionths of a meter) that can be considered to be saturated. Within this range, it is indeed true that adding more CO2 would not significantly increase the amount of absorption. All the red M&M’s are already eaten. But waiting in the wings, outside this wavelength region, there’s more goodies to be had. In fact, noting that the graph is on a logarithmic axis, the atmosphere still wouldn’t be saturated even if we increased the CO2 to ten thousand times the present level. What happens to the absorption if we quadruple the amount of CO2? That story is told in the next graph:
The horizontal blue lines give the threshold CO2 needed to make the atmosphere optically thick at 1x the preindustrial CO2 level and 4x that level. Quadrupling the CO2 makes the portions of the spectrum in the yellow bands optically thick, essentially adding new absorption there and reducing the transmission of infrared through the layer. One can relate this increase in the width of the optically thick region to the "thinning and cooling" argument determining infrared loss to space as follows. 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.
Now let’s use the absorption properties described above to determine what we’d see in a typical laboratory experiment. Imagine that our experimenter fills a tube with pure CO2 at a pressure of one atmosphere and a temperature of 20C. She then shines a beam of infrared light in one end of the tube. To keep things simple, let’s assume that the beam of light has uniform intensity at all wavelengths shown in the absorption graph. She then measures the amount of light coming out the other end of the tube, and divides it by the amount of light being shone in. The ratio is the transmission. How does the transmission change as we make the tube longer?
To put the results in perspective, it is useful to keep in mind that at a CO2 concentration of 300ppm, the amount of CO2 in a column of the Earth’s atmosphere having cross section area equal to that of the tube is equal to the amount of CO2 in a tube of pure CO2 of length 2.5 meters, if the tube is at sea level pressure and a temperature of 20C. Thus a two and a half meter tube of pure CO2 in lab conditions is, loosely speaking, like "one atmosphere" of greenhouse effect. The following graph shows how the proportion of light transmitted through the tube goes down as the tube is made longer.
The transmission decays extremely rapidly for short tubes (under a centimeter or so), because when light first encounters CO2, it’s the easy pickings near the peak of the absorption spectrum that are eaten up first. At larger tube lengths, because of shape of the curve of absorption vs. wavelength, the transmission decreases rather slowly with the amount of CO2. And it’s a good thing it does. You can show that if the transmission decayed exponentially, as it would if the absorption factor were independent of wavelength, then doubling CO2 would warm the Earth by about 50 degrees C instead of 2 to 4 degrees (which is plenty bad enough, once you factor in that warming is greater over land vs. ocean and at high Northern latitudes).
There are a few finer points we need to take into account in order to relate this experiment to the absorption by CO2 in the actual atmosphere. The first is the effect of pressure broadening. Because absorption lines become narrower as pressure goes down, and because more of the spectrum is "between" lines rather than "on" line centers, the absorption coefficient on the whole tends to go down linearly with pressure. Therefore, by computing (or measuring) the absorption at sea level pressure, we are overestimating the absorption of the CO2 actually in place in the higher, lower-pressure parts of the atmosphere. It turns out that when this is properly taken into account, you have to reduce the column length at sea level pressure by a factor of 2 to have the equivalent absorption effect of the same amount of CO2 in the real atmosphere. Thus, you’d measure absorption in a 1.25 meter column in the laboratory to get something more representative of the real atmosphere. The second effect comes from the fact that CO2 colliding with itself in a tube of pure CO2 broadens the lines about 30% more than does CO2 colliding with N2 or O2 in air, which results in an additional slight overestimate of the absorption in the laboratory experiment. Neither of these effects would significantly affect the impression of saturation obtained in a laboratory experiment, though. CO2 is not much less saturated for a 1 meter column than it is for a 2.5 meter column.
So what went wrong in the experiment of poor Herr Koch? There are two changes that need to be made in order to bring our calculations in line with Herr Koch’s experimental setup. First, he used a blackbody at 100C (basically, a pot of boiling water) as the source for his infrared radiation, and measured the transmission relative to the full blackbody emission of the source. By suitably weighting the incoming radiation, it is a simple matter to recompute the transmission through a tube in a way compatible to Koch’s definition. The second difference is that Herr Koch didn’t actually perform his experiment by varying the length of the tube. He did the control case at a pressure of 1 atmosphere in a tube of length 30cm. His reduced-CO2 case was not done with a shorter tube, but rather by keeping the same tube and reducing the pressure to 2/3 atmosphere (666mb, or 520 mm of Mercury in his units). Rather than displaying the absorption as a function of pressure, we have used modern results on pressure scaling to rephrase Herr Koch’s measurement in terms of what he would have seen if he had done the experiment with a shortened tube instead. This allows us to plot his experiment on a graph of transmission vs. tube length similar to what was shown above. The result is shown here:
Over the range of CO2 amounts covered in the experiment, one doesn’t actually expect much variation in the absorption — only about a percent. Herr Koch’s measurements are very close to the correct absorption for the 30cm control case, but he told his boss that the radiation that got through at lower pressure increased by no more than 0.4%. Well, he wouldn’t be the only lab assistant who was over-optimistic in reporting his accuracy. Even if the experiment had been done accurately, it’s unclear whether the investigators would have considered the one percent change in transmission "significant," since they already regarded their measured half percent change as "insignificant."
It seems that Ångström was all too eager to conclude that CO2 absorption was saturated based on the "insignificance" of the change, whereas the real problem was that they were looking at changes over a far too small range of CO2 amounts. If Koch and Ångström 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’s worth noting that Ångström’s erroneous conclusion regarding saturation did not arise from his failure to understand how pressure affects absorption lines. That would at least have been forgivable, since the phenomenon of pressure broadening was not to be discovered for many years to come. In reality, though Ångström would have come to the same erroneous conclusion even if the experiment had been done with the same amounts of CO2 at low pressure rather than at near-sea-level pressures. A calculation like that done above shows that, using the same amounts of CO2 in the high vs. low CO2 cases as in the original experiment, the magnitude of the absorption change the investigators were trying to measure is almost exactly the same — about 1 percent — regardless of whether the experiment is carried out at near 1000mb (sea level pressure) or near 100mb (the pressure about 16 km up in the atmosphere).
26 June 2007 at 7:50 AM
So how did they eventually discover the Beer-Lambert-Bouguet law? And pressure broadening? Inquiring minds want to know!
26 June 2007 at 8:40 AM
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.
26 June 2007 at 9:03 AM
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.
26 June 2007 at 9:22 AM
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
26 June 2007 at 10:21 AM
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]
26 June 2007 at 10:46 AM
raypierre and Spencer Weart
Thank you very much. It works well for me.
26 June 2007 at 10:54 AM
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.
26 June 2007 at 12:00 PM
I’ve heard the pressure broadening being questioned.
This explains a mechanism:
http://en.wikipedia.org/wiki/Spectral_line#Broadening_due_to_local_effects
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?
26 June 2007 at 3:08 PM
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).
26 June 2007 at 5:37 PM
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]
26 June 2007 at 6:06 PM
Does the broadening work similarly for absorption and emission?
26 June 2007 at 6:48 PM
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.
26 June 2007 at 8:10 PM
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”?
26 June 2007 at 9:10 PM
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:
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]
27 June 2007 at 1:31 AM
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.
27 June 2007 at 4:36 AM
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.)
27 June 2007 at 4:44 AM
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]
27 June 2007 at 10:41 AM
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!
27 June 2007 at 11:02 AM
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]
27 June 2007 at 11:28 AM
raypierre (inline comment to #19) wrote:
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.
27 June 2007 at 11:28 AM
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.
27 June 2007 at 11:43 AM
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]
27 June 2007 at 11:52 AM
Alastair McDonald (#18) wrote:
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.
27 June 2007 at 1:35 PM
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.
27 June 2007 at 2:50 PM
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.
27 June 2007 at 4:02 PM
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.
27 June 2007 at 5:01 PM
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]
27 June 2007 at 6:17 PM
“…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?
27 June 2007 at 6:22 PM
Re #25 Ray (Ladbury)
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!
27 June 2007 at 7:13 PM
Alastair McDonald (#26) wrote:
(emphasis added)
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.
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.
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.
27 June 2007 at 9:07 PM
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.
27 June 2007 at 9:26 PM
Ray ladbury:
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.
27 June 2007 at 9:59 PM
Aaron (#28) wrote:
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.
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.
27 June 2007 at 10:14 PM
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.
28 June 2007 at 5:47 AM
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
28 June 2007 at 6:44 AM
[[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.
28 June 2007 at 7:09 AM
In #33 Timothy says:
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]
28 June 2007 at 7:31 AM
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]
28 June 2007 at 8:23 AM
Re: #35 (Alastair McDonald)
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.
28 June 2007 at 9:10 AM
Blair Dowden (#37) wrote:
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…”
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.
28 June 2007 at 9:51 AM
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]
28 June 2007 at 10:23 AM
Alastair McDonald (#35) wrote:
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.
28 June 2007 at 11:51 AM
raypierre (inline to #38) wrote:
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.
28 June 2007 at 12:07 PM
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.
28 June 2007 at 1:24 PM
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.
Downwelling longwave irradiance
http://www.arm.gov/measurements/measurement.php?id=downlwirrad
Upwelling longwave irradiance
http://www.arm.gov/measurements/measurement.php?id=uplwirrad
The website itself will give the visitor a better idea of what kind of data is out there and how extensive it is.
28 June 2007 at 2:02 PM
Blair Dowden (#44) wrote:
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]
28 June 2007 at 5:07 PM
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]
28 June 2007 at 6:33 PM
Alastair McDonald (#47) wrote:
Well, I usually can’t afford a new pair of socks…
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
http://geo.arc.nasa.gov/sgp/radiation/rad1.html
I will see if I can find some more stuff along these lines in the next few days.
28 June 2007 at 7:16 PM
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.
28 June 2007 at 8:20 PM
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
28 June 2007 at 10:11 PM
aaron (#41) said:
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]
29 June 2007 at 3:59 AM
#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.
29 June 2007 at 4:54 AM
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.
29 June 2007 at 8:14 AM
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.
29 June 2007 at 9:25 AM
aaron (#53) wrote:
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.
29 June 2007 at 10:05 AM
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.
29 June 2007 at 10:46 AM
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?
29 June 2007 at 10:51 AM
Alastair McDonald (#56) wrote:
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
http://geo.arc.nasa.gov/sgp/radiation/rad1.html
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]
29 June 2007 at 11:36 AM
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!
29 June 2007 at 12:04 PM
Alastair McDonald (#59) wrote:
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.
29 June 2007 at 2:31 PM
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.
29 June 2007 at 4:12 PM
[[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.
29 June 2007 at 11:44 PM
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’?
30 June 2007 at 1:05 PM
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.
30 June 2007 at 7:01 PM
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.
30 June 2007 at 7:20 PM
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 ändert 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
1 July 2007 at 12:32 AM
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…
1 July 2007 at 4:57 PM
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.
1 July 2007 at 9:44 PM
Laurence, what is the frequency and bandwidth of the flashlight? Is it cloudy? You need that information.
2 July 2007 at 2:13 AM
I really liked your detailed analysis in Part 2 of how the absorption of infrared depends on CO2 concentration. Enjoyed Part 1 as well.
2 July 2007 at 12:55 PM
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.
2 July 2007 at 1:52 PM
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?
2 July 2007 at 8:38 PM
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.
2 July 2007 at 9:04 PM
Rod B. I think you answered your own question. The international space station orbits in the upper part of the thermosphere–not much there.
3 July 2007 at 12:20 PM
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?
3 July 2007 at 2:06 PM
Rod B (#75) wrote:
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]
3 July 2007 at 3:13 PM
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]
3 July 2007 at 4:25 PM
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]
3 July 2007 at 4:45 PM
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.
3 July 2007 at 6:43 PM
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?
3 July 2007 at 10:09 PM
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.
3 July 2007 at 10:27 PM
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!
4 July 2007 at 12:14 AM
Rod B in #81
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.
4 July 2007 at 12:23 AM
Rod B (#81) wrote:
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.
4 July 2007 at 3:05 AM
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 � 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.
4 July 2007 at 3:27 AM
#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.
4 July 2007 at 6:52 AM
[[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.
4 July 2007 at 8:40 AM
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.
4 July 2007 at 9:27 AM
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?
4 July 2007 at 9:53 AM
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.
4 July 2007 at 10:42 AM
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.
4 July 2007 at 11:50 AM
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.
4 July 2007 at 3:12 PM
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.
4 July 2007 at 4:39 PM
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:
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.
4 July 2007 at 7:45 PM
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.
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.
4 July 2007 at 11:53 PM
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.
5 July 2007 at 1:24 AM
Eli Rabett (#94) wrote:
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.
5 July 2007 at 2:23 AM
Blair Dowden (#95) wrote:
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.
5 July 2007 at 9:21 AM
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.
5 July 2007 at 9:58 AM
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?
5 July 2007 at 11:53 AM
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.
5 July 2007 at 1:20 PM
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.
5 July 2007 at 1:52 PM
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?
5 July 2007 at 1:53 PM
>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?
5 July 2007 at 2:12 PM
Ray Ladbury (#99) wrote:
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.
5 July 2007 at 2:36 PM
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.”
5 July 2007 at 2:50 PM
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.
Caution�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?
5 July 2007 at 3:06 PM
Rod B (#103) wrote:
I admire your ambition and tenacity!
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.]
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.
5 July 2007 at 3:53 PM
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” ….
5 July 2007 at 4:24 PM
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.
5 July 2007 at 4:46 PM
Blair Dowden (#110) wrote:
Not a problem.
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.
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.
5 July 2007 at 5:03 PM
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
5 July 2007 at 8:44 PM
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?
5 July 2007 at 9:12 PM
re 107: a quicky question/clarification: the radiated E-M field has no temperature…., does it??
5 July 2007 at 9:34 PM
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.
5 July 2007 at 10:12 PM
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???
I’m getting a headache!
5 July 2007 at 10:41 PM
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?????
5 July 2007 at 10:50 PM
> 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?
6 July 2007 at 2:35 AM
Blair Dowden (#115) wrote:
At the surface - moist air convection is what is primarily responsible for heating the troposphere.
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.
6 July 2007 at 4:49 AM
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.
6 July 2007 at 7:18 AM
Harold, for someone who doesn’t understand the science, you have awfully strong opinions.
6 July 2007 at 7:31 AM
[[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.
6 July 2007 at 7:33 AM
[[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.
6 July 2007 at 7:42 AM
[[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.
6 July 2007 at 9:06 AM
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.
http://www.metoffice.gov.uk/research/hadleycentre/pubs/brochures/2005/clim_green/slide07.pdf
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
6 July 2007 at 11:49 AM
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.
6 July 2007 at 11:56 AM
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.
6 July 2007 at 12:18 PM
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.
6 July 2007 at 12:43 PM
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.
6 July 2007 at 12:43 PM
>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).
6 July 2007 at 1:58 PM
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.
6 July 2007 at 2:35 PM
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.
6 July 2007 at 3:04 PM
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.
6 July 2007 at 3:10 PM
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.
6 July 2007 at 3:26 PM
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….
6 July 2007 at 3:28 PM
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.
6 July 2007 at 3:33 PM
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.
6 July 2007 at 3:41 PM
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.
6 July 2007 at 3:57 PM
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?
6 July 2007 at 5:06 PM
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.
6 July 2007 at 5:09 PM
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).
6 July 2007 at 10:03 PM
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.
7 July 2007 at 1:52 AM
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.
7 July 2007 at 7:06 AM
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.
7 July 2007 at 7:16 AM
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.
7 July 2007 at 8:35 AM
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?
7 July 2007 at 10:12 AM
Please wait.
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.
7 July 2007 at 10:33 AM
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)…
7 July 2007 at 12:00 PM
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.
7 July 2007 at 12:51 PM
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.
7 July 2007 at 1:10 PM
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.
7 July 2007 at 1:50 PM
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.
7 July 2007 at 4:44 PM
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.
7 July 2007 at 5:31 PM
Re 153: I’m a slave to my generous nature.
7 July 2007 at 7:51 PM
tamino (#153) wrote:
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?
7 July 2007 at 10:24 PM
re 151: absolutely true, Ray. […but you can’t seem to admit that electrostatic and magnetic flux (radiation) itself has no temperature…]
7 July 2007 at 10:31 PM
From the AMS http://amsglossary.allenpress.com/glossary/search?id=local-thermodynamic-equilibrium1
local thermodynamic equilibrium�(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.
8 July 2007 at 12:29 AM
Eli Rabett (#154) wrote:
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.
Please see:
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.
8 July 2007 at 6:46 AM
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.
8 July 2007 at 9:46 AM
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?
8 July 2007 at 2:27 PM
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??
8 July 2007 at 3:02 PM
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?
8 July 2007 at 5:20 PM
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?
9 July 2007 at 6:09 AM
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:
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?
9 July 2007 at 6:38 AM
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?
9 July 2007 at 7:18 AM
[[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.)
9 July 2007 at 9:14 AM
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 al� - Nature, 2004 - nature.com
… Cavity cooling of a single atom ….. Cooling of a single atom in an optical trap inside a resonator.
9 July 2007 at 10:48 AM
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.
9 July 2007 at 11:28 AM
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.
9 July 2007 at 12:40 PM
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.
9 July 2007 at 5:29 PM
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?
10 July 2007 at 9:16 AM
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”?
10 July 2007 at 9:31 AM
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).
www.novalynx.com/glossary-t.html
… the measurement of how fast the molecules are moving back and forth.
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
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
10 July 2007 at 10:34 AM
#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
10 July 2007 at 10:37 AM
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.
10 July 2007 at 10:46 AM
#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.
10 July 2007 at 11:59 AM
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.
10 July 2007 at 12:36 PM
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.
10 July 2007 at 1:59 PM
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
www.atmos-chem-phys-discuss.net/7/9973/2007/
© Author(s) 2007. This work is licensed under a Creative Commons License.
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. www.cosis.net/abstracts/EGU04/05569/EGU04-J-05569.pdf ) - gavin]
10 July 2007 at 4:38 PM
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
10 July 2007 at 4:59 PM
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 clichés 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?
10 July 2007 at 5:21 PM
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.
10 July 2007 at 7:06 PM
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.
10 July 2007 at 7:42 PM
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.
10 July 2007 at 8:39 PM
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.
10 July 2007 at 9:41 PM
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…
10 July 2007 at 9:56 PM
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.
10 July 2007 at 10:38 PM
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.
11 July 2007 at 12:35 AM
and remember: http://lemoncustard.com/media/rf8.gif
11 July 2007 at 12:49 AM
Rod B (#186) wrote:
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.”
11 July 2007 at 2:40 AM
#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.
11 July 2007 at 4:24 AM
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!)
11 July 2007 at 7:22 AM
[[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.
11 July 2007 at 7:49 AM
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.
11 July 2007 at 9:32 AM
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.”
11 July 2007 at 9:37 AM
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.
11 July 2007 at 10:13 AM
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.
11 July 2007 at 10:23 AM
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!
One source: http://www.nd.edu/~philinst/pdfs/CHAPTER6.pdf
“… 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
11 July 2007 at 11:00 AM
#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!
11 July 2007 at 11:49 AM
#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.
11 July 2007 at 11:53 AM
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.
11 July 2007 at 4:21 PM
Something I’ve been wondering about, and this seemed as good a place as
any to ask:
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?
11 July 2007 at 5:31 PM
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.
13 July 2007 at 1:47 AM
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.
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.
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.
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!
14 July 2007 at 6:54 AM
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.
14 July 2007 at 8:14 AM
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?
14 July 2007 at 9:14 AM
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 …..
14 July 2007 at 5:58 PM
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.
15 July 2007 at 5:25 PM
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.
15 July 2007 at 8:21 PM
[[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.
15 July 2007 at 8:55 PM
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.
15 July 2007 at 10:41 PM
> For CO2, Planck’s function is irrelevant, as is re-emission to the surface. Only water vapour and clouds acts in that way.
Cite, please?
16 July 2007 at 9:30 PM
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?
17 July 2007 at 9:15 AM
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!
17 July 2007 at 10:31 AM
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.
17 July 2007 at 12:04 PM
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?
17 July 2007 at 12:27 PM
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.
18 July 2007 at 8:37 AM
Re #210
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.
18 July 2007 at 9:28 AM
Re #211
Dave,
There is some information on line shapes in this document http://www.ecmwf.int/newsevents/training/rcourse_notes/PARAMETRIZATION/RADIATION_TRANSFER/Radiation_transfer4.html#999581
Note that the number of lines makes the careful treatment of the wings both impractical and irrelevant.
18 July 2007 at 9:32 AM
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!
18 July 2007 at 10:02 AM
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
18 July 2007 at 11:49 AM
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 Kirchhoff�s Law applied to gases, and that they emit according to Planck�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 �frozen out� 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 Planck�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 Earth�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 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.
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 � higher frequency photons. Look up http://iapetus.phy.umist.ac.uk/Teaching/IntroAstro/Hydrogen.html�> Hydrogen lines and you will find that there are several sets for the simplest of molecules.
Hope this answers your couple of questions
18 July 2007 at 11:54 AM
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.
18 July 2007 at 11:55 AM
There are continua associated with water and CO2 emission at very low levels.
18 July 2007 at 1:51 PM
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
18 July 2007 at 1:52 PM
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
18 July 2007 at 2:19 PM
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!
18 July 2007 at 3:09 PM
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”??
Your referenced link is faulty.
18 July 2007 at 4:49 PM
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…
18 July 2007 at 4:56 PM
Re #228
Glad you like it Rod.
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
18 July 2007 at 5:21 PM
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.
19 July 2007 at 10:30 AM
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.
19 July 2007 at 10:59 AM
Alastair McDonald (#222) wrote:
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:
But of course you go on to explain why you don’t think that the radiation and matter strongly interact:
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?
19 July 2007 at 11:41 AM
http://radix.net/~bobg/climate/halpern.on.radiation.html
19 July 2007 at 12:00 PM
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.
19 July 2007 at 2:03 PM
Allan Ames (#235) 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.
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.
19 July 2007 at 2:27 PM
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.
19 July 2007 at 4:17 PM
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
or downloaded as a PDF from http://www.ecmwf.int/newsevents/training/rcourse_notes/pdf_files/Remote_sensing.pdf
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.
19 July 2007 at 9:21 PM
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…!]
19 July 2007 at 9:41 PM
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.
19 July 2007 at 11:10 PM
It appears to me that Hank Roberts’s posts # 66 and 67 under Friday roundup belong in this thread.
20 July 2007 at 1:12 AM
Rod B (#239) wrote:
Well, at this point I will gladly defer to the American Meteorological Society:
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.
Please see:
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
http://links.jstor.org/sici?sici=0080-4630%2819560802%29236%3A1205%3C193%3ATRITUA%3E2.0.CO%3B2-C&size=LARGE&origin=JSTOR-enlargePage
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:
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.
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.
20 July 2007 at 5:16 AM
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!
20 July 2007 at 6:07 AM
[[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.
20 July 2007 at 10:07 AM
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.
20 July 2007 at 11:27 AM
Alastair McDonald (#245) wrote:
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.
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.
20 July 2007 at 1:54 PM
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.
20 July 2007 at 4:53 PM
Allan Ames (#247) wrote:
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.
Not offhand. Then again I am a philosophy major turned coder.
I thought that “population” normally refered to more than one.
That looks like something worth reading.
I might recommend the following:
I found the following to be of interest:
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:
Across some parts of the spectra, (partial-)LTE will still hold up - as suggested in the first passage I cited.
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.
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.
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.
20 July 2007 at 9:11 PM
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…)
21 July 2007 at 10:37 AM
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….
21 July 2007 at 11:04 AM
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 [;-)….)
21 July 2007 at 11:29 AM
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.
21 July 2007 at 12:54 PM
Rod B (#250) wrote:
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:
[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:
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.
21 July 2007 at 12:59 PM
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.
21 July 2007 at 1:13 PM
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.
21 July 2007 at 4:49 PM
Allan Ames (#254) wrote:
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.
21 July 2007 at 5:53 PM
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.
21 July 2007 at 8:07 PM
Rod B (257) wrote:
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:
… I had stated:
DeWitt Payne had responded:
Bart Paul Levenson then corrected Payne:
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.
21 July 2007 at 8:59 PM
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:
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.
22 July 2007 at 6:50 AM
[[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.
22 July 2007 at 7:06 AM
[[does adding more CO2 really significantly increase the absorption of the Earth’s radiation.]]
Yes.
22 July 2007 at 3:19 PM
Sources of Uncertainty
Rod B (#257) wrote:
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.
[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?
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.
22 July 2007 at 4:36 PM
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.
22 July 2007 at 4:49 PM
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.)
22 July 2007 at 7:23 PM
Rod B (#264) wrote:
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.
23 July 2007 at 6:32 AM
[[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.
23 July 2007 at 8:18 AM
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.
23 July 2007 at 9:37 AM
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.
23 July 2007 at 9:53 AM
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?
23 July 2007 at 11:44 AM
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
23 July 2007 at 2:54 PM
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]
23 July 2007 at 3:17 PM
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…
23 July 2007 at 3:37 PM
Re #270 & 271. How about you showing the calculations? I doubt I could do them anyway.
23 July 2007 at 4:10 PM
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.
23 July 2007 at 4:24 PM
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).
24 July 2007 at 5:17 AM
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!
24 July 2007 at 7:22 AM
[[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.
24 July 2007 at 7:25 AM
[[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.
24 July 2007 at 11:00 AM
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?
24 July 2007 at 11:05 AM
a ps to my #279 to Barton: an omitted (typo) but most important word: it should have read “Thanks, but I’m confused. …”
24 July 2007 at 3:51 PM
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.
25 July 2007 at 1:41 AM
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.
25 July 2007 at 9:45 AM
> what HITRAN says
I looked it up. Here’s the reference I posted in the Part I thread on that question
http://www.realclimate.org/?comments_popup=455#comment-38413
25 July 2007 at 12:48 PM
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)?
25 July 2007 at 2:06 PM
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.
26 July 2007 at 7:05 AM
[[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.
26 July 2007 at 7:11 AM
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!
26 July 2007 at 8:23 AM
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?
26 July 2007 at 10:22 AM
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.
26 July 2007 at 12:27 PM
Alastair, rock, shale, dirt, etc are just “crystal latices of atoms”??? Not molecular?
26 July 2007 at 12:46 PM
Alastair, please.
http://www.madsci.org/posts/archives/2000-08/966273615.Es.r.html
“… 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’.”
26 July 2007 at 1:41 PM
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.
26 July 2007 at 2:41 PM
Well, no. Read it:
“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).
26 July 2007 at 5:03 PM
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.
26 July 2007 at 5:45 PM
http://www.physicsforums.com/showthread.php?p=1383874
26 July 2007 at 6:16 PM
“Lightning,” people, “lightning,” not “lightening.” Lightening is what happens when something’s weight decreases.
26 July 2007 at 6:33 PM
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.
26 July 2007 at 8:58 PM
Alastair McDonald (#297) wrote:
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:
xez sez:
Alastair wrote:
xez sez:
The operative word is like.
There are eight different types of local spectral broadening I can see:
http://en.wikipedia.org/wiki/Spectral_line#Broadening_due_to_local_effects
Maybe xez is talking about one or more of those things.
26 July 2007 at 9:29 PM
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.
26 July 2007 at 9:39 PM
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
26 July 2007 at 11:33 PM
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.
27 July 2007 at 3:40 AM
Re #300 No one is pegging it a me at a zillion km/h
27 July 2007 at 10:47 AM
Alastair, just saying, Ockham.
Your explanation:
> 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?
27 July 2007 at 10:02 PM
Alastair says:
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.
28 July 2007 at 1:22 PM
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.
28 July 2007 at 2:43 PM
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.
28 July 2007 at 5:04 PM
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.
28 July 2007 at 5:16 PM
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.
29 July 2007 at 8:39 AM
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.
29 July 2007 at 8:50 AM
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?
29 July 2007 at 12:24 PM
Alastair McDonald (#309) wrote:
Alastair,
We aren’t talking quantum states in electron orbitals but quantum states in molecular rovibration.
Alastair McDonald (#309) wrote:
See above.
I rest my case.
29 July 2007 at 12:47 PM
Alastair McDonald (#310) wrote:
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.
31 July 2007 at 12:26 AM
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.
1 August 2007 at 10:25 AM
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.
1 August 2007 at 1:45 PM
Some tidbits that may help:
http://nasadaacs.eos.nasa.gov/articles/1996/1996_cloudsclear.html
(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)
http://news.bbc.co.uk/2/hi/science/nature/6926597.stm
http://news.bbc.co.uk/2/hi/science/nature/6421303.stm
1 August 2007 at 9:05 PM
Hank Roberts (#315) wrote:
People should check out the following:
http://climate.gsfc.nasa.gov/publications.php
One should be able to find this article:
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.
2 August 2007 at 7:00 AM
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
2 August 2007 at 8:25 AM
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 www.spectralcalc.com as well.
2 August 2007 at 9:04 AM
Re Martin’s #318,
If you use 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?
2 August 2007 at 9:57 AM
Alastair, _please_please_please_ identify which statements are your opinions.
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.
2 August 2007 at 12:24 PM
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?
2 August 2007 at 1:53 PM
Alastair McDonald (#317) wrote:
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)
Now here is something a bit older. It shows calculations by the military of infrared radiation of non-LTE infrared radiation for CO2:
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:
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.
2 August 2007 at 2:18 PM
Something else for Alastair…
An online article:
An article to look up offline:
Here are sounder measurements of thermal emissions in CO2 sensitive parts of the spectra from 1997:
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
2 August 2007 at 2:47 PM
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.
2 August 2007 at 2:48 PM
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.
Here the elevation of land being determined by co2 absorption strength:
2 August 2007 at 3:04 PM
Rod B (#324) wrote:
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:
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.
2 August 2007 at 4:18 PM
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???
2 August 2007 at 4:44 PM
Rod B (#327) wrote:
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.
2 August 2007 at 4:52 PM
Hank Roberts (#320) wrote:
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.
2 August 2007 at 5:03 PM
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.
2 August 2007 at 5:05 PM
> the real temperature is … what we measure
Temperature is, first, a definition. Pick one:
http://www.google.com/search?q=define%3Atemperature
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?
Click here: http://www.aip.org/history/heisenberg/
“The more precisely
the POSITION is determined,
the less precisely
the MOMENTUM is known”
2 August 2007 at 5:08 PM
CORRECTION to #329
The award was in 1997.
2 August 2007 at 5:10 PM
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.
2 August 2007 at 6:21 PM
Hank Roberts (#333) wrote:
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.
3 August 2007 at 7:04 AM
[[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.
3 August 2007 at 7:09 AM
“Here, kitty, kitty, kitty…”
–attributed to Erwin Schroedinger
3 August 2007 at 11:41 AM
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?
3 August 2007 at 12:47 PM
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.
3 August 2007 at 1:23 PM
This may help:
Physics Help and Math Help - Physics Forums — Physics —- Quantum Physics —– thermal radiation in QM
http://www.physicsforums.com/showthread.php?t=17102
The thread begins with the question asked there:
“… reading how all matter emit continous spectra of radiation, and then reading about atoms and how all matter emit discrete spectra of radiation. … how does thermal radiation … fit into QMs desciption of radiation ?”
3 August 2007 at 2:52 PM
Rod B. (#337) wrote:
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.
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.
3 August 2007 at 7:58 PM
I suppose this is relevant for something Angstrom didn’t know.
http://www.cliffsnotes.com/WileyCDA/CliffsReviewTopic/Electromagnetic-Radiation-Light-.topicArticleId-23583,articleId-23482.html
“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.”
3 August 2007 at 9:33 PM
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.
4 August 2007 at 12:05 AM
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.
Radiation & Climate: Major Projects
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.
4 August 2007 at 3:06 AM
> 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.
4 August 2007 at 5:57 AM
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.
4 August 2007 at 6:31 AM
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 not