Blackbody Radiation and Complex Refractive Index

In summary, blackbody radiation refers to the electromagnetic radiation emitted by an idealized perfect blackbody, which absorbs all incident radiation and re-emits energy solely based on its temperature, as described by Planck's Law. The complex refractive index is a measure of how light propagates through a medium, incorporating both the real part (which describes the speed of light in the material) and the imaginary part (which accounts for absorption). The interplay between blackbody radiation and the complex refractive index is crucial for understanding energy transfer in materials, optical properties, and the behavior of light in different media.
  • #1
elad
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TL;DR Summary
Emitted blackbody radiation into a medium with complex refractive index
Hi.

If the refractive index of a medium equals one, the total emitted blackbody intensity inside a medium is sigma*T^4.
In general, if the refractive index of a is a real number, the total emitted blackbody intensity inside a medium is n^2*sigma*T^4.
Now, when the refractive index of the medium is complex (n + ik), what will be the expression of the total emitted blackbody intensity inside the medium?
Will it be n^2*sigma*T^4 as before, or maybe (n^2+k^2)*sigma*T^4?

Thanks.
 
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  • #2
If the medium has an imaginary component, this will cause the electric and magnetic fields to be out of phase, so the real component of intensity will be reduced. I suppose that n^2*sigma*T^4 would have a cos theta factor applied. In addition, an imaginary component of refractive index means that the medium has attenuation, so the intensity will reduce exponentially with distance.
 
  • #3
elad said:
TL;DR Summary: Emitted blackbody radiation into a medium with complex refractive index

Hi.

If the refractive index of a medium equals one, the total emitted blackbody intensity inside a medium is sigma*T^4.
In general, if the refractive index of a is a real number, the total emitted blackbody intensity inside a medium is n^2*sigma*T^4.
Now, when the refractive index of the medium is complex (n + ik), what will be the expression of the total emitted blackbody intensity inside the medium?
Will it be n^2*sigma*T^4 as before, or maybe (n^2+k^2)*sigma*T^4?

Thanks.
Emissivity of thermal radiation is indeed tied to the (imaginary part of the) refractive index but it is not a simple polynomial factor.

Edit: I do not have the right book at hand but check for example this: https://neutrium.net/heat-transfer/calculation-of-emissivity-for-metals/

Edit2:
elad said:
In general, if the refractive index of a is a real number, the total emitted blackbody intensity inside a medium is n^2*sigma*T^4.
actually where did you get this from?
 
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  • #4
tech99 said:
If the medium has an imaginary component, this will cause the electric and magnetic fields to be out of phase, so the real component of intensity will be reduced. I suppose that n^2*sigma*T^4 would have a cos theta factor applied. In addition, an imaginary component of refractive index means that the medium has attenuation, so the intensity will reduce exponentially with distance.

tech99 - what do you mean by theta and why cos?

pines-demon said:
actually where did you get this from?
pines-demon - I got the form n^2*sigma*T^4 from the book thermal radiation heat transfer (Howell and Siegel). According to the book:
"If the refractive index is constant with frequency, integrating equation 17.44 over all ni yields the local total emitted blackbody intensity inside a medium, Ib,m=(n^2)*Ib, where Ib in this relation is for n=1. "
Equation 17.44 is the blackbody spectral intensity emitted locally inside a medium with n~=1 and with n a function of frequency.
Actually, you get n^2*sigma*T^4 from integration of Planck's law over frequency, when n~=1.
 
  • #5
elad said:
pines-demon - I got the form n^2*sigma*T^4 from the book thermal radiation heat transfer (Howell and Siegel). According to the book:
"If the refractive index is constant with frequency, integrating equation 17.44 over all ni yields the local total emitted blackbody intensity inside a medium, Ib,m=(n^2)*Ib, where Ib in this relation is for n=1. "
Equation 17.44 is the blackbody spectral intensity emitted locally inside a medium with n~=1 and with n a function of frequency.
Actually, you get n^2*sigma*T^4 from integration of Planck's law over frequency, when n~=1.
Oh I see this is within the media! Never seen this, how is it derived that would give you a clue. As intensities are involved there might be some ##|\tilde{n}|^2=n^2+\kappa^2## instead of a simple square.
 
  • #6
elad said:
pines-demon - I got the form n^2*sigma*T^4 from the book thermal radiation heat transfer (Howell and Siegel). According to the book:
"If the refractive index is constant with frequency, integrating equation 17.44 over all ni yields the local total emitted blackbody intensity inside a medium, Ib,m=(n^2)*Ib, where Ib in this relation is for n=1. "
Is this expression only valid when there is direct contact between the radiation source and the medium or does it even hold if the source is surrounded by a layer of vacuum and the medium forms a shell around this layer?
 
  • #7
Philip Koeck said:
Is this expression only valid when there is direct contact between the radiation source and the medium or does it even hold if the source is surrounded by a layer of vacuum and the medium forms a shell around this layer?
It is only valid if the source (and detector) are within the media. Imagine a sort of hot object in glass, the object emits ##n^2 \sigma T^4## where ##n## is the index of glass. If there object has a different emissivity ##\epsilon## then it is ##n^2\epsilon \sigma T^4##. If there is vacuum somewhere, then that has to be taken into account and would depend on the geometry.

Edit: vacuum not glass
 
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