METEO 300
Fundamentals of Atmospheric Science

6.11 What happens when we have both absorption and emission in the infrared?

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6.11 What happens when we have both absorption and emission in the infrared?

Absorption and scattering apply to infrared radiation, just as they do to UV, visible, and microwave radiation. However, all matter that is at Earth-like temperatures is radiating in the infrared but isn’t hot enough to radiate significantly in the visible.

Remember that the peak in the Planck distribution function for a body at T = 300 K is given by Wien's Law:

λ max = 2898 μm  K T = 2898 μm  K 300 K =10 μm MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaeq4UdW2damaaBaaaleaapeGaciyBaiaacggacaGG4baapaqabaGcpeGaeyypa0ZaaSaaa8aabaWdbiaaikdacaaI4aGaaGyoaiaaiIdacaaMc8UaeqiVd0MaaeyBaiaaykW7caqGGaGaae4saaWdaeaapeGaamivaaaacqGH9aqpdaWcaaWdaeaapeGaaGOmaiaaiIdacaaI5aGaaGioaiaaykW7cqaH8oqBcaqGTbGaaGPaVlaabccacaqGlbaapaqaa8qacaaIZaGaaGimaiaaicdacaqGGaGaae4saaaacqGH9aqpcaaIXaGaaGimaiaaykW7cqaH8oqBcaqGTbaaaa@5D95@

To examine how extinction and emission work together, consider the case of an absorbing atmosphere with no scattering at a fixed wavelength. Assume that there is an absorbing gas between us and some radiation source which has a radiance I. Then I will be reduced as it travels through the absorbing gas:

d I absorption  = −σ n I ds=− κ a  I ds κ a ≡absorption coefficient,  SI units: m −1   MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@84E7@

[6.14]

But, at the same time, the layer with width ds is emitting:

d I emission  =  κ a   P e  ds This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[6.15]

Normally, we would multiply Pe by the emissivity for that wavelength. But because the absorptivity = emissivity for any given wavelength and type of matter by Kirchhoff’s Law, we can replace the emissivity, ε, with the absorptivity, κads. Thus:

dI=d I absorption +d I emission =− κ a  I ds+ κ a   P e  ds= κ a  ( P e −I ) ds or dI ds = κ a  ( P e −I )  MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@8546@

[6.16]

This equation is called Schwarzschild’s Equation. Please watch the video below (2:29).

Schwarzschild's Equation
Click here for transcript of the Schwarzschild's Equation video.

We can see what happens as we change the infrared absorptivity, and thus the emissivity, for cases in which there are two sources of radiation, a solid object like earth, and the atmosphere, which both absorb and emit. Look at Schwarzschild's Equation. dI over ds is a change in radiance over distance. Kappa a is the atmospheric absorption coefficient in units of meters to the minus 1. P sub e is the Planck distribution function spectrum of radiance, in that same direction. And I is the radius of the source. By the way, the atmosphere's temperature is low enough that p sub e is significant in the infrared, but not in the visible, where it is very small. So in the visible, Schwarzschild's Equation becomes the same as Beer's Law of Absorption, because P sub e is essentially 0. Not so in the infrared. Note that if P sub e equals I, then d sub I over ds equals 0. That is, the radiation does not change with distance. Let's look at a system where there is a radiation source at a single wavelengths, and in between the source and the observer is an atmosphere. We have seen that the amount of absorption, and thus emissivity, depends strongly on the wavelength of the infrared radiation. So let's look at some different wavelengths effectively. Let's start with a case in which the atmosphere does not absorb at all. Thus, the absorptivity and emissivity are 0, and all the observer sees is the source, but nothing from the atmosphere. For a second case where absorptivity is very small, there is some absorption of the source radiation, but there is also some emissions from the atmosphere itself. For the third case, of stronger absorptivity, the absorption is stronger so that the source is harder to see, but the atmosphere is now emitting more. Finally, if the absorptivity and emissivity equal 1, then all the radiation from the source is absorbed, and all the observer sees is the atmosphere near him.

satellite image of the United States
NOAA GOES satellite measurement of water vapor using the 6.5–7.0 μm wavelength band, which is strongly absorbed by water vapor.
Credit: NOAA

Physical Meaning

At every infrared wavelength, the change in the irradiance is due to the absorption of irradiance coming from some source, F, and the emission of irradiance by the absorbing gas itself. If the source is relatively strong (i.e., it produces a spectral irradiance greater than the Planck distribution function spectral irradiance from the gas), then the change in F will decrease mainly according to the Beer–Lambert Law. However, if the source is relatively weak, then the irradiance that gets to us will be due mostly to the gas itself.

Let’s look at this problem another way. Suppose we have an infrared sensor and are pointing it at the Sun. The Sun radiates some energy in the infrared, but at some wavelengths the water vapor and carbon dioxide are also radiating and the sensor may be picking up their irradiance. So what is the infrared sensor seeing?

Break the atmosphere into very thin layers, so thin that the temperature and amount of water vapor and carbon dioxide is almost constant over the thickness of each layer. Start with the layer closest to the sensor. It radiates both down to the sensor but also up to the layers above it. Since this first layer is just above us, the sensor measures the infrared radiation from it.

The next layer may be at a different temperature, but it too is radiating both down and up. Some of the irradiance in this second layer might get absorbed in the first layer depending upon what the first layer’s absorptivity is. If this absorptivity is a lot, then very little of the infrared radiation from the second layer will get through the first layer, according to the Beer–Lambert Law, but at the same time, the emissivity of the first layer will be great, so the irradiance of the first layer will be close to that of the Planck distribution function spectral irradiance at the layer’s temperature. As a result, the sensor will see mostly the first layer and not the second. If, on the other hand, the first layer has low absorptivity and thus low emissivity, then irradiance from the first layer to the sensor will be weak and irradiance from the second layer will get to the sensor along with irradiance from the first layer.

If we add a third layer, it too will be radiating the same amount of energy both down and up. Some or all of its irradiance might be absorbed in the first and second layers, so depending on the absorptivity of the first two layers, the sensor may or may not see any irradiance from the third layer.

You can see how we can build a model of atmospheric radiation by stacking individual layers with distinct amounts of water vapor and carbon dioxide on top of each other, one at a time, and as we add each one, use the Beer–Lambert Law to determine how much radiation from one layer is absorbed by the layers below or above it.

Quiz 6-3: Absorbed in thought

  1. Find Practice Quiz 6-3 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
  2. When you feel you are ready, take Quiz 6-3. You will be allowed to take this quiz only once. Good luck!