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Part II: What Ångström didn’t know

Filed under: — group @ 26 June 2007

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).


644 Responses to “Part II: What Ångström didn’t know”

  1. 301
    Dave Dougherty says:

    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.

  2. 302

    Re #300 No one is pegging it a me at a zillion km/h

  3. 303
    Hank Roberts says:

    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?

  4. 304
    Eli Rabett says:

    Alastair says:

    Your mole of gas will have a temperature. Put a thermometer in the gas and it will register a value that matches the average kinetic energy of the molecules. So it has a kinetic or Maxwellian temperature. Its Planckian temperature or brightness temperature is zero because we are assuming that it is not emitting any radiation.

    Would you care to explain why all gases do not radiate? True in the case of He or N2 the radiation rate would be infinitesimally slow, but for others such as CO2, or H2O not so. Of course if you let the radiation out of the volume in which the gas was enclosed the gas would cool unless you put energy in some other way.

  5. 305
    Rod B says:

    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.

  6. 306
    Eli Rabett says:

    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.

  7. 307
    Timothy Chase says:

    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.

  8. 308
    Timothy Chase says:

    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.

  9. 309

    Re #303-304 etc.

    Hank,

    If you want to use Occam’s Razor on the problem of the origin of black-body radiation (BBR) then you should start with the original idea by Boltzmann, that BBR is produced by atomic oscillators with three degrees of freedom. He used that idea to explain Stefan’s experimental results. And that was the underlying idea that Planck used, but he had to quantumise the modes in order to get the curve to fit with the full experimental curve. In 1905 Einstein proved that atoms exist using Brownian motion, and that quanta exist using the photo-electric effect. The same year Boltzmann committed suicide because no-one would believe his idea that atoms do exist. (I know the feeling!)

    We now know that an atom is made of a positively charged nucleus surrounded by a cloud of negatively charged electrons. BBR is easily explained by the nucleus oscillating in all direction, thereby creating an electromagnetic field, assuming that the net negative charge from the electron cloud remains unchanged. Trying to re-invent the wheel by imagining that a bar of iron glows red hot, due to the smearing of radiation caused by the rotation and vibration of non-existent molecules, goes completely against Occham’s Razor.

    Eli, I did not write “all gases do not radiate.” I wrote “we are assuming that it is not emitting any radiation.”

    Rod and I were discussing a hypothetical case where simplifications were made in order to understand the general principles. It is ridiculous to suggest that all gases do not radiate. We know that if they are subject to high enough temperatures and they have evaporated, the atoms of all elements will emit radiation due to quantum jumps by their electrons. However, we also know that at atmospheric temperatures no element emit at that type of line radiation, since they never receive collisions with enough energy to excite their electrons. At STP that type of emission is “frozen out.”

    Rod,

    A gaseous molecule has four forms of energy which affect its temperature: Translational (commonly called kinetic), Electronic (which is the level of excitation of the electrons above the ground state), Vibrational (a mixture of kinetic and potential energy), and Rotational (as a result of angular momentum.) The bond energy of a molecule is its chemical energy, and only comes into play during chemical reactions such as hydrogen burning in oxygen.

    Eli is correct that when heat is added it is shared amongst all those four forms of energy. However, as I have explained above, at the temperatures found i the Earth’s atmosphere, the share going into electronic energy is negligaile. The electronic energy is frozen out, by no collision at that temperature is large enough to cause electronic excitement.

    But not only that, the same is true for vibrational excitement. That too is frozen out! This means that when a CO2 molecule is excited by radiation it will lose its excitement due to collisons, but it will never receive a collision which is great enough to re-excite it.

    In other words, Gavin’s simple model with the greenhouse gases re-radiating back to Earth is completely wrong. Greenhouse gases do not act like that. They absorb radiation which is shared with the the kinetic energy of the other molecules raising the temperature of the air. The air is heated directly by the radiation, not indirectly via conduction from the Earth’s surface.

    In #308 Timothy wrote:

    “… if it is simply a disagreement between an amateur climatology enthusiast such as myself and someone with a PhD who has been teaching for several decades, I would say that it is a virtual certainty that it is the latter who knows what he is talking about and not the former.”

    As Benjamin Franklin wrote “In this world nothing can be said to be certain, except death and taxes.”

    Cheers, Alastair.

  10. 310

    Re #301,

    Dave,

    It was Hank who found that paper not me. He is the one to thank.

    The reason that it is only now that these issues are being raised is that it is only now that it is possible to handle those problems theoretically. In general, science operates by producing theories then using experiments to test them. Although it was possible to do the calculations for diatomic molecules, it was not until the 1970 that the theory of triatomic molecules, such as CO2 and H2O could be tackled. But even now the variations in collisions in the atmosphere make the task too vast to complete. Moreover, it was only during and after the nineties that the threat from climate change became obvious, and that urgency was needed.

    But what do you think of my ideas?

  11. 311
    Timothy Chase says:

    Alastair McDonald (#309) wrote:

    Eli is correct that when heat is added it is shared amongst all those four forms of energy. However, as I have explained above, at the temperatures found i the Earth’s atmosphere, the share going into electronic energy is negligaile. The electronic energy is frozen out, by no collision at that temperature is large enough to cause electronic excitement.

    Alastair,

    We aren’t talking quantum states in electron orbitals but quantum states in molecular rovibration.

    Alastair McDonald (#309) wrote:

    In #308 Timothy wrote:

    “… if it is simply a disagreement between an amateur climatology enthusiast such as myself and someone with a PhD who has been teaching for several decades, I would say that it is a virtual certainty that it is the latter who knows what he is talking about and not the former.”

    As Benjamin Franklin wrote “In this world nothing can be said to be certain, except death and taxes.”

    See above.

    I rest my case.

  12. 312
    Timothy Chase says:

    Alastair McDonald (#310) wrote:

    But what do you think of my ideas?

    Alastair,

    I don’t like saying this, but you tend to attract a certain element which will often have ulterior motives for fawning over your ideas. Your somewhat exaggerated estimation of your abilities (an imperfection which I no doubt suffer from on occasion) combined with your desire for such flattery is at times counterproductive.

    Just something I’ve noticed on various occasions.

  13. 313
    Dave Dougherty says:

    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.

  14. 314
    Rod B says:

    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.

  15. 315
    Hank Roberts says:

    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

  16. 316
    Timothy Chase says:

    Hank Roberts (#315) wrote:

    Every cloud has an invisible halo (this year’s news)
    Science – Clouds are bigger than they look, according to new measurements by atmospheric scientists in Israel and the United States. They say that clouds are surrounded by a ‘twilight zone’ of diffuse particles, invisible to the naked eye, extending for tens of kilometers around the cloud’s visible portion.
    (Science magazine; no longer available free, searching may find a copy somewhere)

    People should check out the following:

    http://climate.gsfc.nasa.gov/publications.php

    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.

  17. 317

    Re #314 Where Rod wrote:

    “BTW, do not believe my post #305 and a bunch just like it. They’re all wrong. Internal vibration and rotation energies do, in part, determine temperature.”

    There is a good description of this at http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/eqpar.html
    and an explanation of Kinetic Temperatures at:
    http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c1
    In fact thr whole site is very good.

    To answer your first question: the surface gets heated by solar radiation during the day, and by blackbody radiation from clouds at night. Without the clouds, even hot deserts get very cold at night.

    However, water vapour, unlike CO2, can condense in the atmosphere. It is continuously condensing and then evaporating again, just as it does in clouds. See Hank’s #315 and Timothy’s #316. Therefore, water vapour does partly emit blackbody radiation, so performing in a way that is similar what happens in the GCMs. However, that is probably just a second order effect.

    The first order effects are that CO2 and H20 absorb all the radiation from the surface in their bands, and do not re-radiate any of it back to earth. But for the troposphere to stay at a stable temperature then the air has to lose that absorbed heat, which it does by emitting blackbody radiation from clouds into the stratosphere, where there is no water vapour to absorb it, and the CO2 is too thin to absorb in the wings of the lines.

    CO2 and O3 in the stratosphere absorbs the blackbody radiation from the clouds and infrared from the Sun, which causes it to be warmer than the troposphere, but how it loses its heat I have still to work out. Presumably, by radiating from the wings of the lines where the radiation cannot be absorbed at higher altitudes since the lines are thinner there.

    To be honest, I still have some work to do, to get it all sorted out :-(

  18. 318
    Martin says:

    For terrestrial conditions, the HITRAN and GEISA molecular databases indeed provide the best starting point for modelling absorption by trace gases. An excellent tool for calculating and visualizing the absorption by atmospheric gases is the Spectral Calculator (www.spectralcalc.com), also described by Eli Rabbet. High resolution spectra are available, and the raw data can be extracted. Here the HITRAN and GEISA databases can be perused and downloaded.

    It should be noted that one must be careful to include other effects when needed. These include line coupling (most notably for carbon dioxide and methane), and continuum absorption from molecular oxygen and water vapor. Mie and Rayleigh scattering from aerosols and molecules respectively needs to be accounted for as well, in the appropriate wavelength regions. At very high altitudes, above ~90 km, the assumption of local thermodynamic equilibrium no longer applies, and care must be taken to account for this, although this effect is negligible for most Earth climate studies.

    As a side note, if very high temperatures are being examined (e.g. the atmosphere of Venus), the HITEMP line catalog provides good data for water vapor, carbon dioxide, and carbon monoxide. These can be obtained at http://www.spectralcalc.com as well.

  19. 319

    Re Martin’s #318,

    If you use http://www.spectracalc.com to calculate the transmittance for the band bounded by wave numbers 620 1/cm to 720 1/cm for a Gas Cell of CO2 at STP and a cell length of 30 m (3000 cm) the absorption is 100%.

    This does raise the question as to the partial pressure of the CO2 in this experiment. Does anyone know? Is it 1 bar or is it 0.35 mb?

  20. 320
    Hank Roberts says:

    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.

  21. 321
    Rod B says:

    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?

  22. 322
    Timothy Chase says:

    Alastair McDonald (#317) wrote:

    The first order effects are that CO2 and H20 absorb all the radiation from the surface in their bands, and do not re-radiate any of it back to earth. But for the troposphere to stay at a stable temperature then the air has to lose that absorbed heat, which it does by emitting blackbody radiation from clouds into the stratosphere, where there is no water vapour to absorb it, and the CO2 is too thin to absorb in the wings of the lines.

    Actually it appears they radiate just fine, Alastair.

    The first figure on this page (figure 23) shows the observation of CO2 reemission in the 15 μ from the mid-troposphere as observed by NIMBUS-4 IRIS (graph from a textbook published in 1976)

    3.1 Physical basis of remote sounding
    3 RADIATIVE TRANSFER AND REMOTE SOUNDING
    http://ceos.cnes.fr:8100/cdrom-00/ceos1/science/dg/dg20.htm

    Now here is something a bit older. It shows calculations by the military of infrared radiation of non-LTE infrared radiation for CO2:

    AD-A275 207
    SHARC, A Model for Calculating Atmospheric and Infrared Radiation Under Non-Equilibrium Conditions
    January 24, 1994
    http://stinet.dtic.mil/cgi-bin/GetTRDoc?AD=ADA275207&Location=U2&doc=GetTRDoc.pdf

    Figure 2. Calculated Vibrational Temperatures for C02 States for a Solar Zenith Angle of 820.
    (50-150 km)

    NOTE: You will see that the temperature for all the displayed vibrational states approach each other at 50 km, indicating LTE conditions at 50 km and below.

    The following are two more recent presentations regarding the measurement of CO2 from thermal spectra:

    Imasu, R and Y. Ota, CO2 columnar amount retrieved from thermal infrared spectra observed by IMG/ADEOS,
    SPIE’s 4th Inter. Asia-Pacific Env. Remo. Sensing Symp.
    Honolulu, HI, USA, 5655-17, 2004.

    Imasu, R. and Y. Ota, A method for retrieving columnar CO2 concentration from thermal infrared radiation spectrum
    Atmospheric Science from Space using Fourier Transform Spectrometry
    Bad Wildbad, Germany, 2003.

    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.

  23. 323
    Timothy Chase says:

    Something else for Alastair…

    An online article:

    Abstract. The near-infrared nadir spectra measured by SCIAMACHY on-board ENVISAT contain information on the vertical columns of important atmospheric trace gases such as carbon monoxide (CO), methane (CH4), and carbon dioxide (CO2). The scientific algorithm WFM-DOAS has been used to retrieve this information. For CH4 and CO2 also column averaged mixing ratios (XCH4 and XCO2) have been determined by simultaneous measurements of the dry air mass. All available spectra of the year 2003 have been processed.

    Carbon monoxide, methane and carbon dioxide columns retrieved from SCIAMACHY by WFM-DOAS: year 2003 initial data set
    M. Buchwitz, et al
    Atmos. Chem. Phys., 5, 3313–3329, 2005
    European Geosciences Union
    http://www.atmos-chem-phys.net/5/3313/2005/acp-5-3313-2005.pdf

    An article to look up offline:

    Abstract: The co-adsorption structures of CO and oxygen on Pd(100) and the angular distribution of reactive CO[2] desorption were studied with angle-resolved thermal desorption and low-energy electron diffraction. The CO[2] formation induced by heating the co-adlayer was extended to lower temperatures with an increase in the coverages, yielding five peaks; P[1]-CO[2] (∼400 K), P[2]-CO[2] (∼350 K), P3-CO[2] (∼290 K), P[4]-CO[2] (∼240 K) and P[5]-CO[2] (∼150 K)

    Reaction diagram of carbon monoxide and oxygen on palladium (100) and angular distribution of reactive carbon dioxide desorption
    OHNO, et al
    1992, vol. 273, no3, pp. 291-300
    Surface science (Surf. sci.)

    Here are sounder measurements of thermal emissions in CO2 sensitive parts of the spectra from 1997:

    http://cimss.ssec.wisc.edu/goes/sndprf/18mar4pl.gif
    from
    APPLICATION OF GOES-8/9 SOUNDINGS TO WEATHER FORECASTING AND NOWCASTING
    W. Paul Menzel, et al
    Bulletin of the American Meteorological Society
    Volume 79, Number 10, October 1998
    http://cimss.ssec.wisc.edu/goes/sndprf/sndprf.htm

    Then there is Mars with its lower temperatures and pressures. For those who are interested, here are the results of some measurements…

    http://starryskies.com/solar_system/mars/spirit/atmosphere01.html

  24. 324
    Rod B says:

    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.

  25. 325
    Timothy Chase says:

    The following page has java animations of satellite data for thermal emissions from various gases, including co2 at 14.1, 13.4 4.45 μm bands in the upper- and lower-level atmosphere temperatures at various times of day.

    The CIMSS Realtime GOES Page
    http://cimss.ssec.wisc.edu/goes/realtime/grtmain.html

    Here the elevation of land being determined by co2 absorption strength:

    Figure 20. The 2-μm CO2 absorption strength (A) can be converted to topographic elevation (B). The derived elevations matches the USGS Digital Elevation Model (DEM) (C). The CO2 absorption strength image (A) is brighter for increasing strength. Because the atmospheric path length is smaller with increasing elevation, the absorption strength decreases, becoming darker in the image. The DEMs (B, C) are brighter for increasing elevation, thus are inversely correlated with the CO2 strength in (A).
    http://speclab.cr.usgs.gov/PAPERS/tetracorder/FIGURES/fig20.dem.abc.gif

    Figure 21. The 2-μm CO2 absorption strength versus USGS DEM elevation shows a linear trend with an excellent least squares correlation coefficient.
    http://speclab.cr.usgs.gov/PAPERS/tetracorder/FIGURES/fig21.co2_graph.gif

    Imaging Spectroscopy:
    Earth and Planetary Remote Sensing with the USGS Tetracorder and Expert Systems
    Roger N. Clark, et al
    Journal of Geophysical Research, 2003.
    http://speclab.cr.usgs.gov/PAPERS/tetracorder/

  26. 326
    Timothy Chase says:

    Rod B (#324) wrote:

    Alastair, your references (Hyperphysics) in 317 throws a monkey wrench into my (our??) conclusions. It states the rotational and vibration energy plays NO PART in determining the temperature and is NOT INVOLVED with energy/heat transfer from collisions.

    Actually, R Nave is explaining that he is making use of an oversimplification which treates molecules as point masses, and that you need to deal with rovibration if you want to properly deal with the specific heats of gases:

    It is important to note that the average kinetic energy used here is limited to the translational kinetic energy of the molecules. That is, they are treated as point masses and no account is made of internal degrees of freedom such as molecular rotation and vibration. This distinction becomes quite important when you deal with subjects like the specific heats of gases. When you try to assess specific heat, you must account for all the energy possessed by the molecules, and the temperature as ordinarily measured does not account for molecular rotation and vibration.

    Kinetic Temperature
    http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html

    In otherwords, the author is admitting that the approach he is taking is inadequate for dealing with specific heat as it does not take into account the internal degrees of freedom due to rovibration.

  27. 327
    Rod B says:

    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???

  28. 328
    Timothy Chase says:

    Rod B (#327) wrote:

    Timothy (326): “……and the temperature as ordinarily measured does not account for molecular rotation and vibration.” (my emphasis)(from Hyperphysics). I made specific reference to this phrase in my post: what the hell is it supposed to mean? That the real temperature is different from what we measure???… because rotation/vibration is temperature we can’t measure???

    It means that the oversimplified introductory “high school”-level approach which treates a molecule as a point mass in order to calculate kinetic energy the same way that you would that of large object undergoing simple linear motion does not work for calculating specific heat. Molecules cannot be treated as point masses if one is to account for their specific heats. Calculations of specific heat require you to recognize the fact that such molecules are not point masses but extended objects where their kinetic energy include rotation and vibration.

    The measuring of specific heat and of thermal spectra are measures involving temperature which take these factors into account. But they are indirect – then again any measurement of molecules as molecules or as ensembles of molecules is indirect.

  29. 329
    Timothy Chase says:

    Hank Roberts (#320) wrote:

    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.

    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.

  30. 330

    Rod, just a quickie re #324

    The temperature that you measure with a thermometer is the kinetic or Maxwellian temperature. But you measure the temperature of the Sun or stars using a bolometer which registers the radiation. This is the brightness or Planckian Temperature. For a blackbody the two temperatures Maxwellian (kinetic) and Planckian (brightness) are equal. But for a gas, you do not have a continuum spectra. You have lines, and each line has a different intensity with peaks which do not match any blackbody spectra. But each peak can be allocated a brightness temperature by finding the temperature of the Planck function which passes through the peak. These lines are due to rotational and vibrational relaxation. On the other hand the gas still has a Maxwellian temperature that you can measure with a thermometer.

    If you look up specific heat you will find that to raise the temperature of the gas you have to add kinetic and internal energy, but it is only the kinetic energy that is measured with a thermometer. Using the specific heat of CO2 it should be possible to work out the internal energy, rotational and vibrational excitation by subtracting the kinetic energy.

    That’s all I have time to write at the moment.

  31. 331
    Hank Roberts says:

    > 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”

  32. 332
    Timothy Chase says:

    CORRECTION to #329

    The award was in 1997.

  33. 333
    Hank Roberts says:

    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.

  34. 334
    Timothy Chase says:

    Hank Roberts (#333) wrote:

    When energy is transferred, it’s not like the electron is a little solid lump orbiting a nucleus, and being bumped by an incoming photon into a higher orbit then dropping back down by emitting another photon to get rid of the energy, or changing its orbit to put the energy into a twisted or rotated bond angle. That’s poetry.

    Its not orbiting, spinning or bending like a classical object, but something which exists in an orbital, like a cute little Schroedinger kitten doing its Einstein-Podolsky-Rosen thing with photons which stray to close in their wavy little ways. Coherences and decoherences all enter the picture, and it is all quite quantized and probablistic with all sorts of mind-bending, metaphysical mischief.

    … or at least so I would gather.

  35. 335

    [[Can’t gasses, like vapor, O2, N2, CO2, also emit blackbody radiation by virtue of the average temperature?]]

    Only if their temperature is very high — in the thousands of degrees. At terrestrial temperatures, most gases radiate only at their absorption/emission lines.

  36. 336

    “Here, kitty, kitty, kitty…”
    –attributed to Erwin Schroedinger

  37. 337
    Rod B says:

    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?

  38. 338
    Hank Roberts says:

    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.

  39. 339
    Hank Roberts says:

    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 ?”

  40. 340
    Timothy Chase says:

    Rod B. (#337) wrote:

    Timothy, I couldn’t tell if your reference (http://ceos.cnes.fr:8100/cdrom-00/ceos1/science/dg/dg20.htm) proved the emission from CO2 and H2O or just assumed it.

    I can’t resolve the server at the moment, but it is satellite imaging – and we know the spectra from the labs. Fingerprints, just like at a crime scene.

    It also stated that the emission from GHG molecules is blackbody type “per Planck’s function”, and implied it is not quantized as it is in the absorption. Is this correct?

    The difference between the thermal radiation of solids, liquids and gases is a matter of degree, not kind. Solids can approximate blackbody radiation, but blackbody radiation itself is an idealization. Individual gasses at low pressures and temperatures have well-defined lines and bands, but at higher pressures and temperatures both broaden with lines bleeding into lines and bands bleeding into bands. Dusts, crystals and alloys have relatively discrete spectra. However, impurities, clustering, ions and pressure broaden these spectra. And atmospheres are composed of multiple gasses where each gas is at the same temperature. Individually, the spectra of any one gas in the atmosphere is a poor approximation for a blackbody emitter, but taken together, the atmosphere does much better.

    And this applies to both emission and absorption. Assuming LTE, a good emitter at a given wavelength will be an equally good absorber at the same wavelength.

    Anyway, I might try to deal with what you threw to others, but I am at work right now, so I should keep it short.

  41. 341
    Hank Roberts says:

    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.”

  42. 342
    Rod B says:

    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.

  43. 343
    Timothy Chase says:

    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.

  44. 344
    Hank Roberts says:

    > 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.

  45. 345

    Re ‘#337 Where Rod writes : 4) Alastair, isn’t the use of a thermometer or bolometer a matter of convenience? If you could get there, have a thingy that doesn’t melt and a really good asbestos suit, could you measure the Sun’s temp with a thermometer?

    No! They are different instruments and they measure different things.

    A thermometer measures the kinetic energy of the molecules of a gas, or the vibrational energy of the atoms of a liquid or solid. A bolometer measures the radiation produced by the gas, liquid or solid. For a blackbody radiator like the sun where the atoms are closely packed then the two are the same. But if you had a dense cloud of Oxygen in space with a kinetic temperature 100 F, then the thermometer would read 100 F but the bolometer would read almost zero.

    What I am saying may be the answer to the solar coronal heating problem. The solar corona is absorbing radiation from the sun and becoming electronically excited. But the gas atoms cannot emit that radiation until they become highly excited. Hence the blometers give high temperature readings. But if you placed a thermometer there (shaded from the solar radiation) then the gas would register a low temperature because the kinetic energy of the gas is low. The same is true for the Earth’s thermosphere.

  46. 346

    Re #339 and #341

    Hank,

    You should not believe everything you read on the web!

    I don’t trust (am sceptical of) what Gavin and Ray write, far less people unwilling to use their full names such as Tev and Cliff. Moreover, it seems to me that Hydrodynamic is not entirely convinced by TeV’s explanations. I am certainly not :-(

    But Cliff’s notes (who is Cliff) do not seem any better. He states that Kirchhoff found two types of line, absorption and emission but in fact they are the same lines. The emission lines turn into absorption lines when the radiation intensity exceeds the saturation value! Seems to me this proves Angstrom right!

    He is also using the smearing of emission lines as the cause of blackbody radiation, but the electronic emission lines are “frozen out” at low temperatures, so how can they possibly produce blackbody radiation at terrestrial temperatures?

    Google Bremsstralung radiation. That is also continuous like blackbody radiation, and has nothing to do with the smearing of electron lines.

  47. 347
    Ray Ladbury says:

    Rod B., I think you are getting hung up on terminology–black-body vs. line radiation. First, a black body does not exist in nature–that is nothing has 100% emissivity at all wavelengths. You can come close to blackbody radaition by looking at radiation emitted from a small hole in a cavity. This means that the probability of radiation escaping before it comes into thermal equilibrium with the walls and contents of the container (as well as the radiation field itself) is very small. So what does that equilibrium mean? First, all radiation starts by being emitted by a molecule/atom. However, the surrounding material influences the wavelength of emission and interacts with and influences the energy of the resulting energy. Lines broaden and eventually you get a continuum or close to it.
    OK, now take away the container. The surrounding gas still broadens the line emission, but there is no container absorbing and re-radiating (with its vibrating atoms shifting frequencies), so the radiation retains its essentially broadened line character. Moreover, there are competing processes (collisional relaxation, etc.), so if the gas is too cool to excite the vibrational states etc., you get less radiation back. The system still comes to equilibrium, and the result is that if you look only at the portions of the spectrum where the molecule CAN emit, the spectrum LOOKS like a black body at the prevailing temperature. That’s what we mean by a gray body–the emissivity isn’t uniform.
    This is my impression–anyone please feel free to correct me if I’m wrong. So, really there is no difference between “blackbody” radiation and line emission except that the blackbody radiation has come into equilibrium with its surroundings. Since this has to happen for all real radiation, all radiation is both black body and line emission.

  48. 348
    Ray Ladbury says:

    Rod B., All radiation is quantized–it comes in photons with energy h*nu. The differences Kirchoff described are continuous vs. line. This is precisely what Planck learned–you can’t get a blackbody spectrum unless you quantize the radiation.

  49. 349
    Hank Roberts says:

    Alastair — CliffNotes are a North American college study guide series, around for decades.

    You “don’t trust people unwilling to use their full names such as … Cliff….(who is Cliff)”?
    Good grief, Alastair.

    You just read the little excerpt, didn’t you, not even the page to which I gave a link?
    You can’t not know who Cliff was, if you bothered to read even the one page.

    http://www.cliffsnotes.com/WileyCDA/Section/id-305430.html

    I quoted a bit of the original history of the science from that page. Attacking history isn’t helpful.

    Look, you have a notion. Test it. Describe an experiment that could falsify it.

    You do not believe carbon dioxide can radiate infrared photons. You know you can falsify your theory by looking for brightness in the infrared above the top of the clouds. Find a picture of the Earth’s horizon taken from space, see if there’s a sharp cutoff to the atmosphere viewed in the infrared, or a fuzzy edge.

    “Kirchoff spectral types are comparable to Kepler’s Laws in the sense that they are only a description of observable phenomena.”

    At this point you’re pushing an idea only you believe, that you haven’t published anywhere, that you don’t have math for, that argues that contemporary physics has to be wrong and you’re right. And you’re collecting people who really want to believe you for their own reasons, because they seem to think you’re proving climate change isn’t possible. I know that’s not where you’re going with your idea.

    It’s not good for you or your readers when you keep “explaining” here why the climatologists are wrong with vague words and no published paper. Turn in the other direction — come at your idea as a scientist, suggest an experiment that can distinguish your idea from the other ideas. Look for that picture of the horizon in the infrared, since it would show a different picture according to your idea than the picture believed in by the infrared astronomers and climate physicists.

    Stratospheric CO2 glows, or not, in the infrared. Look for pictures!

  50. 350
    Rod B says:

    Hank (344): Oh! cram your “sources”. I was fudging (confusingly I admit) because of the uncertainty/quantum factor where ALL energy, including that created by a locomotive is quantisized, as Ray points out. Not the same as the quanta of molecular and electronic absorption/emission.


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