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How much thermal radiation does a given volume of gas emit? Does it go according to Stefan-Boltzmann law, i.e. as

$P=A\varepsilon \sigma T^4$,

with $A$ as the cross-sectional area, $\varepsilon$ as emissivity, $\sigma$ as Stefan-Boltzmann constant? If so, what is the value of emissivity? My first guess is $\varepsilon=1$.

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2 Answers 2

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You need to be clear what you mean by "thermal emission". The phrase means (to an astrophysicist) that the source function of the gas depends only on temperature. That is a necessary but not sufficient condition for the emitting object to be treated as a blackbody.

As such, it isn't clear at all that you can use the Stefan's law in any form whatsoever and certainly not by inserting a constant emissivity. A good counter-example would be thermal bremsstrahlung, which certainly is "thermal emission" that comes from gases (e.g. from the solar corona) and looks nothing like blackbody radiation.

That said, of course you can always say the power emitted is equal to $\epsilon A \sigma T^4$, where $\epsilon$ is a flux-weighted "emissivity" - basically a fudge factor that means the total power emitted is corrected for the fact your gas is not a blackbody. But $\epsilon$ is not an intrinsic property of the gas - it depends on composition, pressure, density and geometry. In such cases, it is also not a given that $\epsilon$ is temperature independent(!) because the amount of emission from a gas does not necessarily depend on $T^4$ (e.g. the bremsstrahlung example above).

However, the good news is that the total emission from an "optically thin" gas (one which is transparent to the radiation it is emitting) will be proportional to its volume. Or more accurately, it will be proportional to its emission measure, usually defined as $n_e^2 V$, where $n_e$ is the electron number density.

On the contrary, blackbody radiation does not necessarily tell you anything about the volume of the emitting object, only the emitting area visible to the observer. An example of a gas where this works would be the gas that makes up the Sun. The emission from the Sun approximates to a blackbody and can tell us what the emitting surface area of the Sun is, but not its volume without knowing or assuming something about its geometry.

If your gas is somewhere between optically thick and optically thin, then the solution is much more complicated and a proper radiative transfer approach is required. There is no simple equation and the link to the volume of the gas will be weak, but geometry-dependent.

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  • $\begingroup$ Wikipedia uses the Stefan-Boltzmann law for the idealized greenhouse model. en.wikipedia.org/wiki/Idealized_greenhouse_model. Is that a bit too ïdealized" then ? $\endgroup$
    – Cornelis
    Commented Jul 25, 2020 at 16:09
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    $\begingroup$ @Conelisinspace It absolutely does not assume that the Earth's atmosphere emits as a blackbody. It assumes the Earth is a blackbody at infrared wavelengths and then assigns a (wavelength-dependent) emissivity to the atmosphere. Obviously, the Earth's atmosphere is not a blackbody at visible wavelengths. $\endgroup$
    – ProfRob
    Commented Jul 25, 2020 at 16:42
  • $\begingroup$ Well, the temperature $T_a$ of the atmosphere causes also infrared wavelengths and the Stefan law is used to represent the radiative flux density from that atmosphere. $\endgroup$
    – Cornelis
    Commented Jul 25, 2020 at 17:19
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    $\begingroup$ The article seems clear enough to me. @Conelisinspace $\epsilon$ is ued as a fudge factor and depends on how optically thick the atmosphere is. It is not calculated from first principles and it is not an intrinsic property of the gas. I'll put an additional explanation into my answer. $\endgroup$
    – ProfRob
    Commented Jul 25, 2020 at 17:31
  • $\begingroup$ O.k., thank you for the "fudge factor", I already thought "why aren't there emissivity tables for gases ?". $\endgroup$
    – Cornelis
    Commented Jul 25, 2020 at 18:21
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I emailed a meteorologist, and he explained me that Stefan-Boltzmann law cannot be used for gases, as it assumes that the body can emit in all frequencies, whereas gases emit only in the very same frequencies where they absorb.

Consequently, I guess that the total power would need to be calculated by integrating Planck's law times the gas-specific spectrum, i.e. as

$P(T)=\int B_{\nu}(\nu,T)S(\nu)d\nu$.

Well, this does not have units of power exactly, as $B_{\nu}(\nu,T)$ has units of $Wsr^{-1}m^{-2}Hz^{-1}$, and consequently $P(t)$ has units of $Wsr^{-1}m^{-2}$, but I guess one might call it a power density.

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  • $\begingroup$ If you want to calculate the power from a volume look at the radiative transfer equation. The plane parallel solution seems good for this purpose. $\endgroup$
    – boyfarrell
    Commented May 13, 2017 at 0:59
  • $\begingroup$ You mean the second equation on this page: en.wikipedia.org/wiki/Radiative_transfer? Nice equation and seems to make intuitive sense, but the emission coefficient $j_{\nu}$ and the absorption opacity $k_{\nu,s}$ are still gas specific (even if we ignore scattering). The guy I emailed to said that you could calculate the total power of emission from Planck's law, so I guess you have to multiply Planck's law by the spectral lines (looked up from somewhere). I have only his word for it, though. $\endgroup$
    – Amateur
    Commented May 14, 2017 at 14:46
  • $\begingroup$ Secondly, en.wikipedia.org/wiki/Radiative_transfer looks at things within the gas, where a beam gradually builds up by emission. What I am interested in is the total amount of radiation that comes out. I wonder if there's a simpler formula for that. $\endgroup$
    – Amateur
    Commented May 14, 2017 at 14:51
  • $\begingroup$ Yes. That's the equation. But look at the local thermodynamic equilibrium equation and the Eddington solution. For your problem. Assume the gas has a well defined temperature. Then use the absorptivity and Planck's law to calculate the emission. Then solve with Eddington method. $\endgroup$
    – boyfarrell
    Commented May 14, 2017 at 15:34
  • $\begingroup$ All I wonder, could you somehow use the absorptivity and Planck's law on a macro level, like in that integral I give above. Like say that for a body of gas this thick (at this pressure), the absorptivity spectrum is just $S(\nu)$, if you look from the surface. $\endgroup$
    – Amateur
    Commented May 16, 2017 at 12:31

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