How to calculate thermal radiation intensity without surfaces?

In summary, the energy intensity of thermally emitted light from any point in a semi-infinite semi-transparent body is equivalent to the corresponding black body radiation at that point for homogeneous temperature. For inhomogeneous temperature, one must integrate the equation of radiative transfer. The power of thermal emission can be calculated using the attenuation coefficient and the path length. The emissivity per unit length can be obtained through the calculus of Kirchhoff's law of thermal radiation. The total energy emitted within the volume can be calculated by multiplying the radiant power density (or emissivity) by the volume. In thermal equilibrium, the energy emitted per unit volume is equal to the energy absorbed, but the total energy emitted in the volume includes emissions that are ultimately transmitted. Therefore
  • #1
Hypatio
151
1
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
Hypatio said:
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
DrStupid said:
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
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
Hypatio said:
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.

Hypatio said:
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
Khashishi said:
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
multiply the emissivity by the volume to get the total
 
  • #8
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
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
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
Sorry I meant "the difference of that incident, i, and that transmitted, i(1-1/e)"
 

1. What is thermal radiation intensity?

Thermal radiation intensity is the amount of electromagnetic radiation emitted by a surface due to its temperature. It is also known as thermal emission or thermal flux.

2. How is thermal radiation intensity measured?

Thermal radiation intensity is typically measured in watts per square meter (W/m²). This unit represents the amount of energy emitted by a surface per unit area per unit time.

3. What factors affect thermal radiation intensity?

The intensity of thermal radiation is affected by the temperature of the emitting surface, the surface area, and the type of material. Other factors include the distance between the emitting surface and the receiving surface, and the properties of the surrounding environment.

4. How does thermal radiation intensity relate to temperature?

According to the Stefan-Boltzmann law, the thermal radiation intensity is directly proportional to the fourth power of the temperature. This means that as the temperature increases, the intensity of thermal radiation also increases significantly.

5. What is the difference between thermal radiation intensity and thermal radiation spectrum?

Thermal radiation intensity refers to the amount of radiation emitted by a surface, while thermal radiation spectrum refers to the distribution of this radiation over different wavelengths. The spectrum is dependent on the temperature of the emitting surface, and it follows the blackbody radiation curve.

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