<p>The graph is showing the thermal emission from the Earth's surface. As you said, the atmosphere is optically thick at most infrared wavelengths. But eventually, if you go high enough, the atmosphere becomes optically thin. The atmosphere at that altitude emits infrared at wavelengths corresponding to its temperature.</p>
<p>It's a little more complicated. Most of the infrared absorption is due to water vapor. If you were to look at the IR spectrum of water vapor, you'd see a forest of individual absorption lines. In between the lines, the altitude at which you reach optical depth equals 1 is lower, and in the core of a line it's at a much higher altitude. Since temperature changes with altitude, the spectrum gets quite complicated.</p>
<p>When I was a grad student in astrophysics, we made observations from NASA-operated airplanes equipped with telescopes (specifically the Kuiper Airborne Observatory and the earlier, smaller Lear jet). We would fly around 40,000 feet, which was high enough at mid-latitudes to be in the stratosphere, but more importantly, it was above almost all the water vapor. Hence we were able to observe at wavelengths not accessible to ground-based observatories (in our case, 20-40 microns). During one of my oral exams, I was asked just this question - why 40,000 feet?</p>
<p><a href="https://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water</a></p>

CC BY-SA 4.0
2022-12-10T23:24:49.953Z
2022-12-10T23:24:49.953Z
John McCarthy
<p>Due to the large optical thickness, the outgoing radiation is emitted at stratospheric temperatures (the black-bodies indicated by ~210K temperatures).</p>
<p>Keep in mind that the black-bodies in the graph are scaled by their total integral, i.e. some constant <span class="math-container">$\times T^4$</span>. Therefore, the emission surface at 210K emits at way lower fluxes than the balmy, room-temperature black-body on the surface, where the atmospheric window signal is coming from.</p>
<p>I strongly presume that this graph has a linear axis in the y-direction, which heavily suppresses the low flux values. It also depends whether the flux <span class="math-container">$F_{\lambda}$</span> with units <span class="math-container">$[F_{\lambda}] = J  s^{-1} m^{-2} A^{-1}$</span> or the energy density <span class="math-container">$\lambda F_{\lambda}$</span> with units <span class="math-container">$[\lambda F_{\lambda}] = J  s^{-1} m^{-2}$</span> is plotted, which can suppress long-wavelength features.</p>
<p>I guess the usual view of Earth's emission spectrum I would have can be seen in some plots <a href="https://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/" rel="noreferrer">here</a>, where the 15 micron suppression is clearly visible (as is the inversion inside the feature - a telltale of the temperature structure in the atmosphere).</p>
<p>So TLDR: My money is on some scaling issue, as this plot is supposed to be the simple one, for public consumption.</p>

CC BY-SA 4.0
2022-12-04T22:15:41.730Z
Looking at that: https://en.wikipedia.org/wiki/Outgoing_longwave_radiation#/media/File:Nimbus_4_IRIS_OLR_1970.png

Here the drop is much less as compared to my first graph and "only" 2/3 are absorbed. This is in rough agreement with Planck's law (https://ncc.nesdis.noaa.gov/data/planck.html) using $\lambda=$15$\mu$m and T1=290K and T2=210K.
MichaelW
CC BY-SA 4.0
2022-12-04T21:40:26.467Z
2022-12-04T21:40:26.467Z
AtmosphericPrisonEscape
<p>Having in mind the absorption of IR radiation by various atmospheric gases, why is there almost no outgoing radiation into space where absorption is strong?</p>
<p><a href="https://i.sstatic.net/7U7vg.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7U7vg.png" alt="Enter image description here" /></a></p>
<p>At the <span class="math-container">$CO_2$</span> 15um wavelength, there is almost no outgoing radiation. OK, it sounds logical, because <span class="math-container">$CO_2$</span> absorbs that range of radiation.</p>
<p>But since <span class="math-container">$CO_2$</span> is a good absorber, according to <a href="https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation" rel="nofollow noreferrer">Kirchhoff's law</a> it should be a good emitter at the same wavelengths.</p>
<p>The radiation equation in its simplest form says:</p>
<p><span class="math-container">$$I_\nu = I_\nu(0) e^{-\tau\nu} + I_\nu^B [1-e^{-\tau_\nu}]$$</span></p>
<p><a href="https://i.sstatic.net/0LLc3.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0LLc3.png" alt="Enter image description here" /></a></p>
<p>I understand that when the optical thickness is large, the outgoing radiation is dominated by the <span class="math-container">$I_\nu^B$</span>, the Boltzmann value for black-body radiation.</p>
<p>When the atmosphere is optically thick, we would have this situation and, therefore, according to that picture, there should be outgoing radiation also at those wavelengths. Even taking into account the lower temperature of the atmosphere can, in my opinion, not explain, why outgoing radiation is almost zero.</p>
<p>But why is this the case?</p>

<atmosphere><greenhouse-gases><atmospheric-radiation><radiation-balance>
Why doesn't earth radiate at wavelengths where there is strong absorption?
MichaelW
# Why doesn't earth radiate at wavelengths where there is strong absorption?

(Asked by: username_0 on 2022-12-04)

Having in mind the absorption of IR radiation by various atmospheric gases, why is there almost no outgoing radiation into space where absorption is strong?

[![Enter image description here](https://i.sstatic.net/7U7vg.png)](https://i.sstatic.net/7U7vg.png)

At the $CO\_2$ 15um wavelength, there is almost no outgoing radiation. OK, it sounds logical, because $CO\_2$ absorbs that range of radiation.

But since $CO\_2$ is a good absorber, according to [Kirchhoff's law](https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation) it should be a good emitter at the same wavelengths.

The radiation equation in its simplest form says:

$$I\_\nu = I\_\nu(0) e^{-\tau\nu} + I\_\nu^B [1-e^{-\tau\_\nu}]$$

[![Enter image description here](https://i.sstatic.net/0LLc3.png)](https://i.sstatic.net/0LLc3.png)

I understand that when the optical thickness is large, the outgoing radiation is dominated by the $I\_\nu^B$, the Boltzmann value for black-body radiation.

When the atmosphere is optically thick, we would have this situation and, therefore, according to that picture, there should be outgoing radiation also at those wavelengths. Even taking into account the lower temperature of the atmosphere can, in my opinion, not explain, why outgoing radiation is almost zero.

But why is this the case?

**Tags:** atmosphere, greenhouse-gases, atmospheric-radiation, radiation-balance


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## 2 answers

### Answer by username_2 on 2022-12-04. Score: 6 | Accepted answer

Due to the large optical thickness, the outgoing radiation is emitted at stratospheric temperatures (the black-bodies indicated by ~210K temperatures).

Keep in mind that the black-bodies in the graph are scaled by their total integral, i.e. some constant $\times T^4$. Therefore, the emission surface at 210K emits at way lower fluxes than the balmy, room-temperature black-body on the surface, where the atmospheric window signal is coming from.

I strongly presume that this graph has a linear axis in the y-direction, which heavily suppresses the low flux values. It also depends whether the flux $F\_{\lambda}$ with units $[F\_{\lambda}] = J s^{-1} m^{-2} A^{-1}$ or the energy density $\lambda F\_{\lambda}$ with units $[\lambda F\_{\lambda}] = J s^{-1} m^{-2}$ is plotted, which can suppress long-wavelength features.

I guess the usual view of Earth's emission spectrum I would have can be seen in some plots [here](https://wattsupwiththat.com/2011/03/10/visualizing-the-greenhouse-effect-emission-spectra/), where the 15 micron suppression is clearly visible (as is the inversion inside the feature - a telltale of the temperature structure in the atmosphere).

So TLDR: My money is on some scaling issue, as this plot is supposed to be the simple one, for public consumption.

> **Comments:**
> **username_0**: Looking at that: https://en.wikipedia.org/wiki/Outgoing_longwave_radiation#/media/File:Nimbus_4_IRIS_OLR_1970.png  Here the drop is much less as compared to my first graph and "only" 2/3 are absorbed. This is in rough agreement with Planck's law (https://ncc.nesdis.noaa.gov/data/planck.html) using $\lambda=$15$\mu$m and T1=290K and T2=210K.  

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### Answer by username_1 on 2022-12-10. Score: 0

The graph is showing the thermal emission from the Earth's surface. As you said, the atmosphere is optically thick at most infrared wavelengths. But eventually, if you go high enough, the atmosphere becomes optically thin. The atmosphere at that altitude emits infrared at wavelengths corresponding to its temperature.

It's a little more complicated. Most of the infrared absorption is due to water vapor. If you were to look at the IR spectrum of water vapor, you'd see a forest of individual absorption lines. In between the lines, the altitude at which you reach optical depth equals 1 is lower, and in the core of a line it's at a much higher altitude. Since temperature changes with altitude, the spectrum gets quite complicated.

When I was a grad student in astrophysics, we made observations from NASA-operated airplanes equipped with telescopes (specifically the Kuiper Airborne Observatory and the earlier, smaller Lear jet). We would fly around 40,000 feet, which was high enough at mid-latitudes to be in the stratosphere, but more importantly, it was above almost all the water vapor. Hence we were able to observe at wavelengths not accessible to ground-based observatories (in our case, 20-40 microns). During one of my oral exams, I was asked just this question - why 40,000 feet?

<https://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water>


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