# Homework Help: What do the terms mean in the energy density for photon gas $u(\omega) d\omega$ ?

1. May 12, 2012

### Gregg

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

2. May 13, 2012

### vela

Staff Emeritus
Re: What do the terms mean in the energy density for photon gas $u(\omega) d\omega # If I recall correctly (which I may very well not be doing),$\omega^2$comes from the phase space factor. Essentially, you need to account for all the different direction the photon can be moving. 3. May 17, 2012 ### andrien Re: What do the terms mean in the energy density for photon gas$u(\omega) d\omega #

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.

4. May 17, 2012

### Dickfore

Re: What do the terms mean in the energy density for photon gas ##u(\omega) d\omega #

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 $\omega^2$.