How to Determine the Pressure of a Photon Gas in a Cubic Box?

[Ruddiger27]

1. The problem Statement
A cubic box of volume V=L^3 contains energy in the form of photons in equilibrium with the walls at temperature T. The allowed photons energies are determined by the standing waves formed by the electromagnetic field in the box. The photon energies are (h/2pi)Wi = (h/2p)cKi, where Ki is the wavevector of the ith standing wave.
I have to find the pressure of the gas given:

2. Homework Equations

a: p(w)dw = (V/(c^3)(pi^2))w^2 dw

b: photons have no mass, so the chemical potenial is zero

c: int( x^3 / (exp^x -1 ) = (pi^4)/15

3. Attempt at Solution

I'm having a mental block here. The partition function of this is
Z= sum ( exp(-(e)/kt)) , where e is the energy of the photons. I put the density equation (1) as equal to the total energy E divided by the volume V.

I'm pretty sure this is wrong though. Do I have to find the density from the above equation (1), then use this to get the energy of an individual photon? How does density relate to energy?

 

[quasar987]

You have the density of state function [p(w)], so the partition function will be found by evaluating an integral. The integral analogue of the sum is this:

\int_0^{\infty} (\exp(-(e(w))/kt)*(\mbox{number of states btw w and w+dw})*(\mbox{average number of photons in the state of energy e(w)})dw

I let you find what each blob is.

 

[Ruddiger27]

Thanks, got it now.