Understanding Energy Density of Photon Gas

[Gregg]

Homework Statement

$u(\omega) d\omega \propto \frac{(\hbar \omega) (\omega^2)}{e^{\hbar \omega \over k_B T}-1} d \omega$

Homework Equations

The Attempt at a Solution

$\hbar \omega$ is the energy of a photon

$\frac{1}{e^{\hbar \omega \over k_B T}-1}$and this is the density of states for bosons. So you have the energy of the photon and the density of states. Why is there an extra $\omega^2$ term? I can't work out what it represents. I thought that it could be a consequence of the $d\omega$ but I am unsure.

 

[vela]

If I recall correctly (which I may very well not be doing), $\omega^2$ comes from the phase space factor. Essentially, you need to account for all the different direction the photon can be moving.

 

[andrien]

the phase space factor is proportional to
d^3(k)=k^2dkd(cosθ)dβ,where k^2 can be written as ω^2/c^2.

 

[Dickfore]

The term $\left[\exp(\hbar \omega/k_{B} T) - 1\right]^{-1}$ is the average number of photons with energy $\hbar \omega$ according to the Bose-Einstein distribution. The density of states is proportional to $\omega^2$.