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 $\omega^2$.

 

[paranormal]

The extra $\omega^2$ term in the equation is a result of considering the energy density per unit frequency interval rather than per unit energy interval. This is a common convention in thermodynamics and statistical mechanics, where it is more convenient to consider energy as a function of frequency rather than of energy itself. The $\omega^2$ term arises from the fact that the energy of a photon is proportional to its frequency, and thus the energy density per unit frequency interval must also be proportional to the square of the frequency. This term does not have any physical significance on its own, but is necessary for the equation to be dimensionally consistent.