Energy of a photon gas: two ways to get it, two different answers

Pacopag
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Homework Statement


I've been asked to calculate the energy of a photon gas in terms of the temperature. Assume non-interacting.
I'll spare the details, unless someone would like to see them, because the calculations can be found in most textbooks. Here's the problem:
When I do it using the partition function with Hamiltonian H = pc, I get
E = 3NkT
where N is the number of photons, k is Boltzmann's constant, and T is temperature.
When I do it by finding the density of states, i get
E \propto VT^4
Both answers are consistent with stuff I already know:
The first expression agrees with the equipartition theorem, but the second expression is the stefan-boltzmann law. So what the heck is going on here?

Homework Equations


The Attempt at a Solution

 
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I think I found the answer on Wikipedia under "Photon Gas." It seems that the number of photons is not fixed as in an ideal gas, and that
N \propto VT^3.
That gives the consistency I'm looking for.
 
Pacopag said:
E \propto VT^4
...
is the stefan-boltzmann law.
Not exactly. Perhaps some integrated version of.

Pacopag said:
I think I found the answer on Wikipedia ...
EXCELLENT!

Pacopag said:
It seems that the number of photons is not fixed as in an ideal gas, ...
Why do you think that the number of photons in an ideal gas is fixed? Grand canonical ensemble.

Pacopag said:
N \propto VT^3. That gives the consistency I'm looking for.
Indeed, you are deriving this from stat mech, relativity, and (a bastardized version of) QM.
 
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