How does atmospheric CO2 and other Greenhousgases (GHG) affect the incoming (from sun) and outgoing (from earth) radiation. I understand that at certain wavenumbers (or areas of wavenumbers) in the Infrared (IR) these molecules (amongst others) absorb an IR-photon, get a bit "warmer" (means perhaps rotate faster or similar) and after a while reemit another photon with less energy (because the enhanced movement costs a bit of energy) ... resulting in a further warming. The emitted photon either hits another photon or will be going to space or will go back to earth.

Now you have lots of molecules (N) and statistically these three mechanisms lead to a warming of the surface of the earth (let aside scattering etc.).

My question is: 1. is this picture correct 2. when yes how are the different mechanisms in terms of photonic energies


No, this vision is (mostly) incorrect. In more detail :


Any molecule having a non zero electrical dipolar momentum will interact with the electromagnetic radiation. In the QM vocabulary it will absorb and emit photons. The CO2 molecule has 3 vibration modes and 2 of them produce an electrical dipole. The vibration energies of these modes are quantized and have energy levels En. For the first excited mode E1-E0 = h.f provides the 2 photons frequencies that will be absorbed/emitted at strongest. Their wavelengths happen to be 4µ and 15 µ what puts them squarely in the IR spectrum where Earth radiates.

So the first part of your vision was correct. CO2 but mostly H20 strongly absorb and emit photons in infrared. Because of Kirchhoff's law, this activity is a zero sum game, e.g the CO2 molecules absorb exactly as much as they emit. There is therefore no net heating of the atmosphere by IR.


There is no net energy transfer by "thermalisation". Indeed a vibrationnally excited CO2 molecule may collide with an N2 molecule, decay from E1 to E0 and transfer E1-E0 to the kinetic energy of the N2 molecule. However the inverse process exists too - an N2 molecule transfers E1-E0 to an unexcited CO2 molecule making its vibrational energy going from E0 to E1. In steady state both rates are obviously equal and there is no "heating up". Also common sense tells us that if there were a net energy transfer, the N2 molecules' temperature would diverge and reach an ultrarelativistic plasma rather fast.

It follows that the collisional processes are also a zero sum game


The right vision is then that in a mixture of 1 GHG gaz (non zero dipole like H20) and 1 non GHG gaz (zero dipole like N2), the GHG gaz will absorb and emit in the infrared spectrum and it will absorb exactly as much as it emits. The non GHG gaz lets everything pass through. Beside the radiation process which involves only the GHG gaz, there are collisions that involve both the GHG and the non GHG gaz. The role of this process is to make sure that the radiatively active GHG gaz and the non radiatively active gaz stay both at the same temperature. In steady state there is then no net energy transfer by collision between both species.

The fact that a GHG - non GHG mixture will be warmer than the case with non GHG only is an effect of density of radiation. Indeed in a non GHG atmosphere the radiation energy density is constant from the bottom to the top because the radiation goes through with a constant rate. As the matter doesn't interact with radiation, it just adds its own kinetic energy to the overall energy density.

In the case of GHG - non GHG mixture, the absorption/emission processes have for effect to decrease the photons' mean free path and thus to increase the radiation energy density as compared to the non GHG case. The GHG matter interacts with radiation in this case and its equilibrium temperature will be higher than in the non GHG case because of the higher energy density. The collisions will then make sure that the non GHG gaz will be at the same temperature as the GHG gaz. The overall result is that the whole GHG atmosphere will be at a higher temperature than the non GHG atmosphere.

Important caveat.

In the above I was describing the radiative and collisional processes only. In the real atmosphere add convection, phase change (evaporation, condensation etc), albedo changes and conduction. There is no reason to believe that these energy transfer processes are not impacted by radiation. Therefore making the jump from a GHG-non GHG mixture to a real atmosphere involves an implicit "all other things being equal" hypothesis which allows a qualitative conclusion but would be wrong for a quantitative conclusion.

  • $\begingroup$ many thanks for this very fine explicit answer. Though being from education a spectroscopist :), this remark helped a lot. $\endgroup$ – kampmannpeine Sep 10 '19 at 11:18

To make things simple you want to compare two atmospheres that are identical, except that one has infinitessimally more of some IR absorber. In the second case, some IR photons that were traveling upwards are absorbed, and thermalized (i.e. the energy is transferred to the surrounding gas molecules, before being re-emitted). The IR photons are basically transferring energy, in this case from the surface (or perhaps a lower level of the atmosphere), to wherever they are absorbed. If the surface is warmer than the atmosphere where our absorber molecule resides, then that layer is on net absorbing more energy than it was in the unperturbed case, so that layer of the atmosphere will warm (with respect to the control case). Eventually once we let the atmosphere respond (but then the two atmospheres have different temperature profiles), we see that there is also more downgoing IR radiation below our absorbed level, so the layers of atmosphere below, and the ground will see a net warming effect.

To think of it in another way, the radiative impedence of the system (surface cum atmosphere) to space is higher with the addition of the greenhouse gases, so given a fixed energy input (absorbed sunlight) the equilibrium temperature will rise. Beyond that first order effect, the details of atmospheric structure and flow begin to change, and then the hard part of the problem begins.


$\mathrm{CO}_2$ actually absorbs some of the solar radiation as it has absorption bands at $\approx 1.5$ and $\approx 2.5 \mu m$, and there is lots of energy at those wavelenghts in solar radiation.

The system boundary must be @TOA where we measure TSI of $1370 \frac{W}{m^2}$, or where we find $960 \frac{W}{m^2}$ which is the number we get after the ill-defined factor albedo has taken it´s share.

$960W$ would be the density of flux at the surface. The surface emits $390\frac{W}{m^2}$ which means that from $960W$ of irradiation at the surface only $780W$ is a absorbed, this comes from what is absorbed in $1m^2$ is emitted by $2m^2$. Since only half of the sphere is irradiated by the sun.

From emitted $390\frac{W}{m^2}$ into the atmosphere we get a mean atmospheric temperature of $255K$, which is only $240\frac{W}{m^2}$.

From emissionspectrum at the TOA we can see the temperature of $\mathrm{CO}_2$ at the wavelenghts it absorbs, which is $\approx 200K$ or about $100\frac{W}{m^2}$.

From this we can see that the surface heats the atmosphere and the surface gets more than enough energy from the sun, but it is kind enough to share it with some air. What really happens is that the atmosphere is a porous extension of the surface, that is incapable of absorbing much of solar radiation, and that it is the result of the sun heating the earth surface.

Some gasses like water get excited from the heat, carrying energy upwards as gas. This rise of matter from the surface from added heat of the sun is very much like the excited atom. Energy lifts up the atmosphere and earth, as well as the other planets in the solar system, is an excited planet. The gasses in the atmosphere doesn´t heat the surface, they get heated from the surface. The greenhousetheory is solely based on the assumption that the laws of nature for temperature, stefan-boltzmann law, doesn't apply for the surface of the earth and that it is warmer than what laws of nature can explain. A very strange starting point for a theory.

But there are many ways to show that the surface gets more than enough energy to get to the temperature we experience on the surface. One way is measurements, another way is simple calculations of heat transfer. The temperature of $255K$ as the "blackbody temperature" is totally misunderstood in greenhouse theory. It is what the system would be at if it was equally warm all the way through with a surface that was infinitly thin.

There is no energy missing in solar radiation to heat the earth to a mean temperature of $288K$. Actually, if heat transfer is calculated step by step from TOA using $\frac{W}{m^3}$ instead of $\frac{W}{m^2}$, to account for that the mass of earth must be heated to an extent where it can radiate a density of $390\frac{W}{m^2}$, there is no need to use albedo. Since surface temperature is measured at a distance above the actual surface it is more correct to calculate by volume. Another reason is that we are inside the system, so surfaces are not actually surfaces, they are volumes inside a system. We are immersed in a bath of fluid consisting of the absorbing gasses. If it is done that way one does not lose a single Watt. The "blackbody temperature" will then be $342\frac{W}{m^2}$ or $278.78K$, and that is very close to the surface. If the system would be a blackbody, the whole system including the solid earth mass, would have a temperature of $279K$ at the top of atmosphere.

So, greenhousegasses interact with radiation by getting heated and spreading that heat from a 2-dimensional surface to a threedimensional volume, moving energy around the earth to the night side faster than earth rotates. Since the cooling of earth mainly happens at the dark side where the sun doesn´t shine, the atmosphere cools the surface more efficiently by moving faster than the earth spins. And by extending the twodimensional surface to threedimensional volume, dividing the energy over more matter at a much lower density in the gasses.

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    $\begingroup$ This desperately needs formatting for clarity. $\endgroup$ – Jon Custer Sep 1 '16 at 13:51
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    $\begingroup$ VLQ voters: I re-formatted the text, there is no need to VtD! $\endgroup$ – peterh - Reinstate Monica Sep 1 '16 at 22:11
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    $\begingroup$ @LuffarBamse Good answer, welcome in the site! I re-formatted your text a little bit. Below the textbox you write, you can realtime follow, how your text will actually show. Here is a documentation, how can also you do this. (It is not a problem, if you don't do this whole on the spot, but you will get more upvotes if you do.) $\endgroup$ – peterh - Reinstate Monica Sep 1 '16 at 22:17

'zero sum game' Thermalization spreads the absorbed photon energy so rapidly throughout the thousands of surrounding molecules so rapidly that there is virtually undetectable heating rate (increase in T2) and the energy returned to the GH molecule is tiny and will take a long time to statistically accumulate enough energy in any given GH molecule to re-emit a photon. Put another way average energy transfer is very small; sigma(T2^4-T1^4) is very small since T2 barely increases.

  • $\begingroup$ This has several physics errors or inaccuracies that should be cleaned up. Contemplate the behavior of an isolated GHG molecule - what happens without any thermalization? $\endgroup$ – Jon Custer Apr 20 '16 at 20:16

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