# Photon gas: thermodynamics

1. Nov 28, 2007

### ed321

1. The problem statement, all variables and given/known data
I’m struggling to reconcile two results about the behaviour of a photon gas, any help would be appreciated:
First of all the Gibbs free energy=0, which means that dG=0=Vdp-SdT
But also p=1/3 U/V and S=4/3 U/T which means p=1/4 ST/V. Now if we call the entropy per unit volume S/V=s, p=1/4sT and differentiating both sides with respect to p at constant s we get 1=1/4s dp/dT which doesn’t agree with the result from the Gibb’s function.

2. Relevant equations
(My expressions for G, p and S agree with wikipedia: http://en.wikipedia.org/wiki/Photon_gas)

3. The attempt at a solution
I think I must not be differentiating p=1/4sT correctly, but I can’t see what I need to do differently.

Thanks.

2. Nov 29, 2007

### ed321

Would this be more appropriate somewhere else?

3. Nov 29, 2007

### ozymandias

Hey ed,

I'll rephrase your question for you.
Why do you think you can transform

$$0 = V dp - SdT$$

into

$$\frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_s$$

(which is really what you're comparing your derivation to - correct me if I'm mistaken.)

--------
Assaf
http://www.physicallyincorrect.com/" [Broken]

Last edited by a moderator: May 3, 2017
4. Dec 1, 2007

### ed321

I thought $$\frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_s$$
followed straight away from $$0 = V dp - SdT$$ by "dividing" by dT at constant s. Maybe this step isn't valid since you can't always treat dT like a ordinary number, but I can't see any reason this would be the case this time?

$$dG = 0 = V dp - SdT$$ gives
$$\frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_G$$, not
$$\frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_s$$.