Methane’s absorption bands are centered at 3.2 and 7.2 microns -- far off the peak of the Planck spectrum for a 290 K blackbody near 16 microns. Moreover, its absorption bands overlap with the water absorption region. So why is methane considered a stronger greenhouse gas than carbon dioxide, which has an absorption band at 15 microns?

  • $\begingroup$ Were you watching Congressional/electional developments lately then you would know this is more of a political science question than one that physics/meteorology could ever answer. $\endgroup$ – hyportnex Feb 28 at 15:54
  • $\begingroup$ Bert, unlike CO2, methane is highly flammable. This means that it cannot build up in the atmosphere like CO2 can. My opinion is that methane is not all that important as a greenhouse gas, as its IR absorption band overlaps with water (as you stated), and there are obviously processes in nature that oxidize atmospheric methane. Since methane has been created and released into the atmosphere for untold millions of years, the atmosphere would contain many percent methane if this wasn't true. $\endgroup$ – David White Feb 28 at 17:49

I'm just guessing that it's due to temperature- and pressure-broadening of the line shapes for CH4 versus CO2. You have to integrate the radiative transfer equation over frequencies as well as over an atmospheric column. So the line shapes, as well as the lines, also play a crucial role determining the overall effect. They're not just $\delta$-function-like. But I'm not immediately googling the answer I'm suggesting here. Anybody finding a cite?

 E d i t

Re @BertBarrois's comment below, let me suggest a totally different mechanism that might contribute to Bert's "methane problem".

CO2 is well-mixed in the atmosphere, whereas I don't know whether or not methane is. For our purposes here, let's suppose that (like ozone) methane isn't well-mixed. Then we can suggest the following.

All absorbed radiation is eventually reradiated (though possibly in several steps, i.e., several photons, at lower frequency). Now, the outgoing thermal radiation we're discussing is mostly coming from the Earth's surface (although a non-negligible part comes from atmospheric reradiation). And that surface radiation is non-isotropic, i.e., it's all from the ground going up. Greenhouse blanketing arises because the reradiated radiation >>is<< emitted isotropically, meaning that some of the radiation originally on its way up-and-out, gets sent back down to the surface. And then the whole process repeats, so to speak, with the net result that the equilibrium temperature gets warmer.

And part of that equilibrium temperature dependence is the vertical atmospheric layer where the reradiation occurs; closer to the ground means warmer. So that may contribute to the lesser effect of well-mixed CO2 versus (what we're supposing is) CH4 absorber amounts closer to the ground.

  • $\begingroup$ I have considered methane's rotational fine structure, but I don't see how T- & P-broadening can explain things. Broadening a line doesn't increase absorption integrated over frequency unless it avoids saturation. And don't forget that 7.2 microns is already block by water vapor. $\endgroup$ – Bert Barrois Mar 1 at 17:56
  • $\begingroup$ @BertBarrois Like you say, my suggested mechanism might not be contributing to your "methane problem" (and like I originally said, I'm just guessing :). So see the edit I made to my answer, guessing/suggesting an entirely different mechanism. Maybe one or both, or neither, contributes. $\endgroup$ – John Forkosh Mar 2 at 10:52
  • $\begingroup$ The IPCC report assumes that both CH4 and CO2 are well-mixed, unlike O3 and H2O: Climate Change 2013: The Physical Science Basis . I’ve been looking into the relative oscillator strength of CH4 and CO2, but am still puzzled by the fact that CO2 intrudes into a transmission window, whereas CH4 seems redundant with H20 at 7.2 microns. Take a look at this . $\endgroup$ – Bert Barrois Mar 3 at 12:39
  • $\begingroup$ @BertBarrois Well, I'm out of guesses. But the answer has to involve one or more of CH4's: (a)spectral lines along with their strengths and line shapes, (b)absorber amount, (c)its distribution in the atmosphere. Right? Unlike water vapor, I don't suppose absorber amount changes very rapidly or with lat,lon. You looking at any other variables? $\endgroup$ – John Forkosh Mar 4 at 7:34
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    $\begingroup$ I think I'm onto an explanation, and your (c) is on the right track., but it involves the distribution of H2O rather than CH4. I may have to answer my own question. Stay tuned. $\endgroup$ – Bert Barrois Mar 4 at 13:18

My own answer …

The IPCC report rates the global warming potential of greenhouse gases in terms of the incremental effect of each additional kilogram on IR transmission. The context includes all important greenhouse gases at baseline concentrations.

Even if CO2 and CH4 were examined out-of-context, it is difficult to estimate their relative oscillator strength from first principles, and collision rates are a further wild card. Per QM, the power radiated by a vibrating diatomic molecule with fractional effective charges $\pm Q$ at the ends should scale as ${{Q}^{2}}{{\omega }^{3}}/m$, where $m$ denotes reduced mass and $\omega $ vibrational frequency. The latter parameters are known, but only a brave chemist would dare to guess at the fractional effective charges based on the polarity of covalent bonds.

The IPCC report’s fundamental metric is radiative forcing, which quantifies the incremental effect of additional greenhouse gases on IR transmission, not from the ~288-K surface through the entire atmosphere, but from the ~220-K tropopause through the stratosphere. Since the CO2 absorption band centered at 15 microns is almost completely blocked (“self-saturated”) under baseline assumptions, extra CO2 merely widens the shoulders of the band. The CH4 band centered at 7.2 microns is almost completely blocked by water vapor within the troposphere, but the stratosphere is very dry, so CH4 has ample headroom to make trouble there. Wikipedia has nice figures of the stratospheric transmission spectra.

The opacity of the lower stratosphere in blocked bands governs the temperature gradient needed to drive radiative transport, which in turn governs the altitude of the tropopause, because convection gives way to radiative transport wherever the adiabatic gradient exceeds the radiative gradient. Each extra kilometer of altitude could raise the surface temperature as much as 6.5 degC, according to the effective lapse rate of the Standard Atmosphere, and the theoretical adiabatic lapse rate in dry air is even higher, 9.8 degC/km.

CH4 may never be as important as CO2 because it is much less abundant -- currently about 2 vs 400 ppm -- but gram-for-gram, it makes a greater marginal contribution to radiative forcing.

  • $\begingroup$ @John Forkosh -- What do you think? $\endgroup$ – Bert Barrois Mar 5 at 14:50

Basically it’s because methane has a more complex physical structure than carbon dioxide. It has five atoms and four bonds compared to carbon dioxide which has three atoms and two bonds.

This means that it can vibrate in more ways than carbon dioxide and hence it absorbs thermal radiation more strongly.

This is why it’s GWP (Global Warming Potential) is so high (at 21 x that of carbon dioxide).

  • $\begingroup$ That can’t be it. Methane has 9 vibrational modes, but thanks to symmetry, only 4 distinct frequencies, and only the 2 IR-active bands that I mentioned. Since they sit atop a water absorption band, I don’t see how they could pack much punch. $\endgroup$ – Bert Barrois Mar 1 at 17:47
  • $\begingroup$ @Bert Barrois: Why should that make a difference? It’s the total number of vibrational modes that counts in terms of how much thermal radiation it absorbs. $\endgroup$ – Mozibur Ullah Mar 1 at 17:52
  • $\begingroup$ The number of degenerate modes increases the strength of a given band, but since 7.2 microns is thoroughly absorbed by water, how could a little extra absorption by methane even matter? $\endgroup$ – Bert Barrois Mar 1 at 18:03
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    $\begingroup$ @Bert Barron’s: I’d encourage you to ask additional questions to the community. $\endgroup$ – Mozibur Ullah Mar 1 at 18:08

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