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An absorption spectrum from high in the atmosphere of infrared radiation emitted from the earth, shows that for the 15µm wavelength there is almost complete absorption. This is attributed to absorption by CO2 in the atmosphere at this wavelength.

However, the concentration of CO2 in the atmosphere is generally about 0.04%, or 400 parts per million. This says to me that for any cubic metre of volume, there would be a large space not occupied by CO2 molecules through which such radiation would pass uninhibited, and therefore ultimately simply pass out to space.

This seems contradictory to me. Could someone perhaps elaborate on what is happening here.

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  • $\begingroup$ @my2cts: No I don't think so, to me it seems OP has a slightly oversimplified view of reaction cross sections. Not a grave mistake, very common I think in fact. But I see OP has already correctly answered their own question. $\endgroup$
    – AtmosphericPrisonEscape
    Oct 19, 2019 at 17:00
  • $\begingroup$ Aha, I just read the OP's answer. Quite detailed so I withdraw my comment. $\endgroup$
    – my2cts
    Oct 19, 2019 at 17:07

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Let's say a 15µm photon has a nonzero probability of interacting with any $CO_2$ molecule that it passes within a distance of 1 wavelength. How many $CO_2$ molecules are there in a cylinder with radius 15µm and extending from the earth's surface out to space? More than $10^{16}$.

I think what bothers you is the idea that $CO_2$ can be so tremendously much more effective at absorbing this wavelength than $O_2$ or $N_2$. Else why would the ratio be relevant? It's a bit like wondering how someone can die from arsenic poisoning when only 0.04% of their meal was arsenic. It's a small fraction of the total, but that doesn't mean it's not important.

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  • $\begingroup$ The reason I came to this was I was looking at the question of CO2 saturation, apparently within metres of the earth's surface. $\endgroup$
    – Joe
    Oct 12, 2019 at 11:15
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I thought I might answer my own question.

The absorption of radiation by matter follows a decreasing exponential relationship.

I = Iº exp(−α n X) where

Iº is radiation flux at origin, I is radiation flux at distance X from origin, n is number of molecules per m³, α is absorption cross section m²/molecule.

This equation is derived here http://astrowww.phys.uvic.ca/~tatum/stellatm/atm5.pdf It is a version of the Beer-Lambert law.

The international standard atmosphere at sea-level is T = 288K, pressure = 101325 Pa.

For a relative volume of CO2 of 0.04%, this gives a partial pressure of CO2 = (101325 x 0.0004) Pa

Solving the ideal gas law pV = nRT using the partial pressure of CO2, and applying Avogadro’s number as the number of molecules per mole, gives 10.2 x 10^21 molecules of CO2 in 1 m³.

Now we can determine the distance to absorb a given proportion of incident radiation, by using the equation above, and solving for X.

We need to know the absorption cross section of CO2.

This is found here http://vpl.astro.washington.edu/spectra/co2.htm from the PNNL link cm2/molecule vs. wavenumbers.

I am interested in the 15µm wavelength (667 per cm), which looks to be about 4.5 x 10^-18 cm2/molecule (or 4.5 x 10^-22 m²/molecule)

So now, for 99% absorption, or I/Iº = 0.01, we can solve for X.

The result is almost complete absorption in 1 metre!

The result is sensitive to the order of magnitude of the absorption cross section. It compares to the sometimes quoted experimental finding of Heinz Hug of complete absorption at 10 metres.

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  • $\begingroup$ Your self-answer is correct, it essentially boils down to the absorption cross section, or also called the 'opacity'. $CO_2$ concentrations in air do not fill up the geometric cross section, because the gas is evenly distributed in each air volume. Instead the opacity is a measure of absorption probability that has dimensions as if it were a geometric cross section. We also recently had an extended discussion on "Why is Mars cold" over at astronomy.se, which gives a bit more details on how the opacity behaves astronomy.stackexchange.com/questions/33628/why-is-mars-cold/… $\endgroup$
    – AtmosphericPrisonEscape
    Oct 19, 2019 at 17:03
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Even at this low concentration, a cubic millimeter of air still contains about $10^{13}$ $CO_2$ molecules. Plenty of opportunities to absorb photons.

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  • $\begingroup$ The question is not whether there may be some absorption, even by a single molecule. The question is how can it be close to 100% of the radiation emitted from, say, a square metre of area. There are 2500 times more other molecules (N2, O2) in your cubic millimetre, occupying a similarly greater space. $\endgroup$
    – Joe
    Oct 11, 2019 at 20:00
  • $\begingroup$ @Joe Suppose a photon of a certain wavelength has a certain probability of being absorbed in a space containing $10^{13}$ $CO_2$ molecules per cubic mm. Why would you expect the probability to change depending on the presence or lack of other molecules that do not interact with that wavelength? $\endgroup$
    – Solomon Slow
    Oct 11, 2019 at 21:24
  • $\begingroup$ @Solomon Slow I don't. But I would have thought that the probability of any radiation wave encountering a CO2 molecule in that space would be 1 in 2500, or 0.0004. If someone gave me those odds on not winning the lottery, I'd be a happy person. The large number of molecules disguises the even larger amount of space between them. $\endgroup$
    – Joe
    Oct 12, 2019 at 11:09
  • $\begingroup$ @Joe, I don't think it works that way. For 15µm photons and $CO_2$ molecules, there's an invariant property called the absorption cross section I don't know exactly how that's defined, but if you know the cross section, and if you know the density of $CO_2$ molecules within a given volume, then I'm pretty sure that's all you need to calculate the probability of a photon being absorbed within that volume. If other molecules that do not absorb 15µm light also are present, I would not expect their presence to change the answer. $\endgroup$
    – Solomon Slow
    Oct 12, 2019 at 14:27
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The atmosphere weighs as much as 10 meters of water. Taking 0.04 % of that means 4 millimeter. A glass pane of 4 mm absorbs all IR.

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At 0.04% I get that the spacing between the CO2 molecules is about 10 nm. So to get through, the electromagnetic waves would have to get past a lot of these molecules. Each one does a bit of absorbing. It's not so surprising that the net effect is that they absorb pretty much all the radiation at some wavelengths.

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  • $\begingroup$ 10 nanometers seems really small... What is that based on? The mean free path at altitude is probably more like 10 microns. $\endgroup$
    – tpg2114
    Oct 11, 2019 at 22:06
  • $\begingroup$ @tpg2114 Mean free path is not the same as distance. The distance at 1 atm is a bit larger than 10 nm but not much larger. Density a bit smaller than one millionth of dry ice, distances a hundred times larger than in the solid. $\endgroup$
    – user137289
    Oct 12, 2019 at 6:00
  • $\begingroup$ @Pieter Is the wave amplitude of relevance here? I'm thinking about the scenario where the radiation wave is travelling perpendicularly out from the surface of the earth, along a radius of the earth. $\endgroup$
    – Joe
    Oct 12, 2019 at 11:25
  • $\begingroup$ @Pieter Also, the end to end length of a CO2 molecule seems to be about 0.3 nm, so the spacing would be about 30 times greater? $\endgroup$
    – Joe
    Oct 12, 2019 at 11:35
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In the nineteenth century it was already discovered that concentration times pathlength is an experimental invariant. So to compare lab results with atmosphere you have to multiply the concentration with the atmospheric path length. For the atmosphere I use a scale height of 8043 m, this gives the following absorption spectrum with a central saturated band: Absorption spectrum EPA with scale height 8043m

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