Thermal radiation intensity

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Imagine there is a semi-infinite semi-transparent body--such as a glass--at some temperature T. No surfaces exist. What is the energy intensity of thermally emitted light from any point in the body? If you use an absorption or emission coefficient (as I suspect is correct), can you explain how to derive the correct units from the usual absorption coefficient, which is not dimensionless, having units of cm^-1?
 

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  • #2
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Imagine there is a semi-infinite semi-transparent body--such as a glass--at some temperature T. No surfaces exist. What is the energy intensity of thermally emitted light from any point in the body?
In case of homogeneous temperature the thermal radiation is identical to the corresponding black body radiation at any point within the body. For inhomogeneous temperature you need to integrate the equation of radiative transfer: http://en.wikipedia.org/wiki/Radiative_transfer
 
  • #3
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In case of homogeneous temperature the thermal radiation is identical to the corresponding black body radiation at any point within the body. For inhomogeneous temperature you need to integrate the equation of radiative transfer: http://en.wikipedia.org/wiki/Radiative_transfer
The solution to the equation of radiative transfer isn't actually my interest. The problem is that I can't solve that equation without knowing the power of thermal emission (the coefficient [itex]j_\nu[/itex] in the wiki). You need to use an emissivity/absorption coefficient (or something like that) in order to not violate the second law of thermodynamics in an inhomogeneous medium, but I don't understand how that works. Apparently you can multiply Planck's black body curve by an emissivity coefficient, but I have also been told that emissivity=absorption... but the former is dimensionless and the later has dimensions of cm^-1, so I don't get any of this.


So... what is the actual power of thermal emission, basically? If i know precisely what the absorption coefficient is, as a function of wavelength, at a point, can I calculate the thermal emission intensity from that point (I'm talking about the energy emitted from the point, not that transmitted through the point)?
 
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  • #4
Khashishi
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The optical depth is given by the attenuation coefficient times the path length.
We know that at an optical depth of infinity, the spectrum must become equal to Planck's law, which will be written (##B_{\lambda,T}##)
At an optical depth of 1, 1/e of the light from behind will shine through, so 1-1/e of the light must come from within this block of glass. Following this line of thinking, we can get the emissivity per unit length.

An exercise for you is to do the simple calculus. I think you should end up with
##\epsilon = B_{\lambda,T} \alpha##
where ##\alpha## is the attenuation coefficient per unit length.
 
  • #5
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The solution to the equation of radiative transfer isn't actually my interest.
But that's the only way to answer your question for inhomogeneous temperature.

Apparently you can multiply Planck's black body curve by an emissivity coefficient, but I have also been told that emissivity=absorption... but the former is dimensionless and the later has dimensions of cm^-1, so I don't get any of this.
You need to use the dimensionless coefficient of absorption as explained by Khashishi. In the one-dimensionalequation of radiative transfer and without scattering Kirchhoff's law of thermal radiation turns the time independent equation of radiative transfer into

[itex]I'_\lambda = \alpha _\lambda \left[ {B_\lambda \left( T \right) - I_\lambda } \right][/itex]

with constant Temperature this results in

[itex]I_\lambda = B_{\lambda ,T} - \left( {B_{\lambda ,T} - I_{\lambda ,0} } \right) \cdot \exp \left( { - \alpha _\lambda x} \right)[/itex]
 
  • #6
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The optical depth is given by the attenuation coefficient times the path length.
We know that at an optical depth of infinity, the spectrum must become equal to Planck's law, which will be written (##B_{\lambda,T}##)
At an optical depth of 1, 1/e of the light from behind will shine through, so 1-1/e of the light must come from within this block of glass. Following this line of thinking, we can get the emissivity per unit length.

An exercise for you is to do the simple calculus. I think you should end up with
##\epsilon = B_{\lambda,T} \alpha##
where ##\alpha## is the attenuation coefficient per unit length.
So, it seems that if I have an absorption coefficient of 2 cm-1, the thermal energy emitted from the surface of a 1 cm3 volume is [itex]B_{\lambda,T}[/itex]exp(-2/1), since this is the energy lost by absorption, and emissivity=absorptivity.

If that is true, ok, but I am still missing something: This isn't the total energy emitted within the volume, it is the energy emitted from it, or the energy emitted within minus the energy absorbed before it can escape. So, how much energy is emitted within the volume?
 
  • #7
Khashishi
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multiply the emissivity by the volume to get the total
 
  • #8
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if I did that, the energy per volume would decrease as a function of A/V, where A is area, and V is volume. Obviously it is actually proportional to volume. The point is that some energy emitted is lost to absorption before it escapes the volume. I don't need to know the energy escaping the volume, that is easily calculated, I need to know the total energy emitted inside the volume.
 
  • #9
Khashishi
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Apologies, when I said emissivity, I meant radiant power density (units of W/m^3). This is the ##\epsilon## in my equation above. Check the units. Multiply this by volume to get total power.
 
  • #10
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I am not understanding. In thermal equilibrium, the energy emitted per unit volume must be the same as the energy absorbed. However, I want to know ALL of the energy emitted in the volume, not just the emissions which are ultimately transmitted. Therefore, a perfectly absorbing medium will be infinitely emitting. This is because a finite amount of radiation will be absorbed in an infinitesimal length, therefore, emitted radiation is infinite.

Obviously I am missing something but I don't get it.

Edit: Just a clarification: clearly it cannot be true that, as Khashishi implies, the energy emitted is the difference of that incident (i) and that transmitted (i-1/e). It should be much more than that since some emitted in the sample is absorbed before it can escape. So, how much is emitted???
 
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  • #11
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Sorry I meant "the difference of that incident, i, and that transmitted, i(1-1/e)"
 

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