If we can use the Stefan-Boltzman equation to quantify the radiative flux leaving a solid body due to emission of radiation and to do it properly for a solid body we need to include the emissivity term then why, when considering the radiation emitted by the gas CO2 in the atmosphere, do we not use the SB equation utilising the absolute temperature of the atmospheric gas and the emissivity of CO2? The emissivity of CO2 is, I think, something like 0.002 (depends on temperature/pressure).
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2 Answers
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We don't
Emissivity of CO$_2$
Your claim of low emissivity of CO$_2$ is not entirely true. First off, it matters what the wavelength of emission is when determining emissivity. Carbon dioxide has specific bands of high absorption (which correspond to high emissivity). These bands are crucial for climate modeling, so listing a numerical emissivity value without specifying the wavelength is not a useful measurement.
Secondly, emissivity of a gas isn't as simple as that of a solid surface because solid surfaces are more or less opaque to the IR radiation emitted in accordance with the Stefan-Boltzman equation. Therefore, the surface area of the 'blackbody' is what is important when calculation emission or absorption. For a gas, however, IR radiation will penetrate some distance through it.
The result is that you need to calculate emissivity of a gas as a function of its optical depth. Staley and Jurica, 1970 present calculations where they do that for carbon dioxide in the 15 $\mu$m absorption band (the most important one for climate models). Their results show that at 20 C, CO$_2$ has an emissivity of 0.0823 at an optical depth of 1 cm; 0.244 at an optical depth of 10 m; etc.
Spectrum matters
As I mentioned above, the emission spectra matters a lot. In the case of CO$_2$, there is a specific emission band which corresponds to the absorption band. This means that any radiation emitted by CO$_2$ is emitted at just the perfect wavelength to be reabsorbed by other molecules of CO$_2$.
CO$_2$ also differs from water vapor in that it remains dispersed through the atmosphere and doesn't coalesce into weather formations. Water vapor aggregates into clouds; the emissions from those clouds can be significant because there is much more water vapor emitting in a cloud than in the (drier) air above the cloud. CO$_2$ doesn't exhibit that behavior and thus its emissions are mostly trapped by other CO$_2$, whereas blackbody emission from cloud banks can be significant on the global scale.
What does get emitted to space and lost is heavily attenuated by the presence of other CO$_2$ molecules and is insignificant compared to emissions from the Earth's surface, which pass through the absorption band gaps in water vapor and CO$_2$.
Conclusion
I don't actually know what factors each specific climate model uses. There may be some that account for the emissivity of CO$_2$. It may be that all the models account for it; it has been over a decade since I contributed to those models. However, whether they account for it or not, it is not a significant factor in driving temperature changes the way the emissivity of the Earth's surface is.
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$\begingroup$ My 0.002 came from reading some old books on refrigeration and heat engineering, around the 1950's I think. There were various graphs of it with pressure, temperature, etc. I will see if I can find them again. I would consider that the figures from these sources takes into account the spectral variation in emissivity of the gas. It gives effectively the percentage of heat emitted by radiation (compared to the SB maximum possible) for a bulk sample of gas at a particular temperature. So on my understanding of the SB equation the emissivity can have only a range between 1 and zero. $\endgroup$– user7733May 3, 2017 at 7:30
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$\begingroup$ I need to have a look at your Staley Jurica reference to find out exactly what they mean by an emissivity >1 . $\endgroup$– user7733May 3, 2017 at 7:31
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$\begingroup$ @user7733 You are right about emissivity values. Staley and Jurica used percentages, which I just copied. I edited them. $\endgroup$ May 3, 2017 at 12:02
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$\begingroup$ Acknowledged. However it still looks odd to me if you don't consider the emissivity as related to the whole planck curve as the total energy is the integral of the SB equation (in its true for as Hz^-1 Sr^-1) where the area under the planck curve is the total energy at a particular K. $\endgroup$– user7733May 3, 2017 at 19:15
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$\begingroup$ In the edit I threw into my answer (to your question) here, I show that CO$_2$ has a limited emission spectrum. CO$_2$ simply does not emit IR radiation at all wavelengths ($\epsilon \approx 0$). This must be combined with the curve for blackbody emission at Earth's surface temps. The two graphs overlap only around 15 um, so that is the only significant wavelength. $\endgroup$ May 3, 2017 at 19:21
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When looking at its role in planetary temperature, it is not important. Use the s-b equation for mean temperature of the atmosphere, -18C(255K), and the surface temperature, then you see what the effect the air(it doesn´t matter what gases it holds) has on surface temperatures. Emissivity is not really relevant, as emission is dependent on temperature only. Absorption does not cause emission, temperature does. Many people seem to think that emissivity, especially in the case of co2, is something that makes it more possible for co2 to be a warming agent. Not the case. Mean temperature of the atmosphere is -18C, and mean emission from co2 is according to that. Heat transfer equation show how co2 does not add any heat, it is an extra heat sink on the way to the ultimate heat sink in space.
Just use Stefan-Boltzmann equation and temperature. Solar heat is absorbed according to the geometric difference between the blackbody and earth, two volumes with shells. $TSI/(4/3^2)$. And since only half the surface area is irradiated, $(1/2*TSI)/(4/3^2)=383W/m^2$ or 286.7 Kelvin. The effective temperature is found by heat transfer to the surface and inverse square law, $1/4*(TSI-((1/2*TSI)/(4/3^2)))=244W/m^2$, or 256 Kelvin. Tropopause temperature, $((1/2*TSI)/(4/3^2))-(1/4*(TSI-((1/2*TSI)/(4/3^2)))=138W/m^2$ or 222 Kelvin. As you can see, heat dominates everything, and just modifying the blackbody to observed differences, absorption in depth of a volume + irradiation of only half the surface, gives an accurate distribution of temperature. It clearly shows why it is a big problem that the greenhouse effect use averaged solar heat over the whole surface. And also, albedo is not a concept that should be used for heat. I think it came from optics in the beginning, and it shouldn´t be used for temperature. Emitted heat depends on temperature only.
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1$\begingroup$ I don't understand what you are talking about. Albedo should be 'used for heat' because a higher albedo reflects solar radiation back into space, thus changing the W/m$^2$ received by the Earth's systems. Emitted heat does not depend on temperature only, there are obviously several other factors in the S-B equation, emissivity being important. Emissivity is certainly relevant to the operation of the Earth's systems; with $\epsilon = 0$ the Earth would have no way to lose heat to space and would be heated by the sun into a plasma ball. Several of the things you say appear to be flat out wrong. $\endgroup$ May 5, 2017 at 13:11
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$\begingroup$ Albedo is an effect from temperature, not a cause. Clouds form from heat, snow attracts heat. Heat transfer happen according to difference in temperature, so most heat will transfer to polar areas, or colder areas in general. Emission is dependent on temperature only, haven´t you read thermal physics? The draper point show that all solids start glowing at the same temperature, therefore emissivity is not relevant for the relationship between temperature and emission. So, no, considering temperature, emissivity can be left out. Absorption is also dependent on temperature of the emitter. $\endgroup$ May 5, 2017 at 16:27
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$\begingroup$ The point is, if I can do what I did in my answer, emissivity clearly is not relevant. I basically show that earth emit like a blackbody if geometrical differences are included. And that invalidates your argument. $\endgroup$ May 5, 2017 at 16:31