# Thermal energy intensity per unit volume

I am trying to figure out what the total thermal energy emitted by completely transparent matter. It seems Planck's law can't do this because the thickness or geometry of the mass is not given, only it's surface area. The spectral radiance of a black body is given by Planck's law:

$B_\nu=\frac{2h\nu^3}{c^2}\frac{1}{e^\frac{h\nu}{k_BT}-1}$

which has units of total power per unit area of the body, per unit solid angle, per unit frequency (W m-2 sr-1 Hz-1)

Now, how do I find the thermal energy emitted per unit volume? I thought I could just get rid of the unit solid angle by writing

$B_\nu=\frac{8\pi h\nu^3}{c^2}\frac{1}{e^\frac{h\nu}{k_BT}-1}$

But I am still left with units of W m-2 Hz-1. Also, does Planck's law even matter for a transparent body, because a black body is completely absorbing? How else would you calculate the true amount of radiated thermal energy?

Andrew Mason
Homework Helper
I am not sure what you mean by a completely transparent body radiating. Do you mean a body with an absorption coefficient of zero? If that was the case, it would not radiate at all.

Try relating the surface area to the volume. If you know the power per unit area and you know the geometry of the body you can work out the power per unit volume.

AM

Sorry, I understand Kirchoff's law now. However, I don't see how to make the emission coefficient volumetric--or why these coefficients are continually related to surfaces and not volumes, understanding how a black body behaves doesn't seem to tell you anything about the thermodynamics of semi-transparent media, because the third dimension should appear everywhere.

Also, I still don't understand the second suggestion. If I have a solid transparent square, and another transparent square with an empty cavity in it (larger surface area), both at the same temperature, I should be able to show that the solid square radiates more energy due to it's greater volume.

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