Does $PV\propto T$ apply to a photon gas? For an ultrarelativistic ideal gas, I know that


*

*$p=\frac{u}3$;

*$TV^3 =$ constant;

*$pV\propto T$.


For a photon gas, I know that the first two results apply as well. However, I am unsure if the third result also applies (or how to prove it). My biggest concern is that while for an ultrarelativistic gas, the proportionality constant $Nk_B$ makes physical sense, it does not for a photon gas as $N$ doesn’t make sense (chemical potential is zero). From what I recall, for the ideal gas, I employ result 3 to obtain result 2. How do I prove the relationship for the photon gas, and how do I tackle my concern?
 A: The confusion might come from the fact that the particle number is a function of temperature. 
Let us start with a photon dispersion $\epsilon_k=c k$ .
The grand canonical partition function for $\mu=0$ and two independent polarisations is then given by:
$$
Z_G = \Pi_k \frac{1}{(1-e^{-\beta \epsilon_k})^2} = e^{-\beta \Phi}
$$
From here we can directly calculate $PV$ with the grand canonical partition function $\Phi$:
$$
 - PV = -k_B T \, \log Z_G= 2 k_B T \sum_k \log(1-e^{-\beta \epsilon_k} )
$$
Taking now the continuum limit of the sum:
$$
 2 k_B T \sum_k \log(1-e^{-\beta \epsilon_k} ) = \frac{2k_B T  V}{h^3} \int dk^3 \log(1-e^{-\beta \epsilon_k} )
=  \frac{8 \pi  V}{(hc)^3 \beta^4} \int_0^\infty dx  \, x^2 \log(1-e^{-x} )
$$
The remaining integral is known to be $-\frac{\pi^4}{45}$. So we find:
$$
 PV =  \frac{\pi^2}{45} \frac{k_B^4}{(\hbar c)^3} V T^4
$$
To relate this to the particle number we also have to calculate the average particle number. Here we must consider the operator average, since the chemical potential is zero. 
$$
 <N>= 2\sum_k <n_k> = \frac{2 V}{h^3} \int dk^3 \frac{1}{e^{\beta \epsilon}-1}= \frac{V}{\pi^2 \beta^3 (\hbar c)^3} \int_0^\infty \frac{x^2}{e^{x}-1} 
= \frac{2 \zeta(3) \, V k_B^3}{\pi^2 (\hbar c)^3} V T^3
$$
The integral appearing in here is $2 \zeta(3)$. With this we can calculate now also the relation between $PV$ with $<N>$.
$$
 PV= \frac{\pi^4}{90 \zeta(3)} <N> k_B T
$$
A: Non conservation of the number of particles does not imply that it is not possible to introduce an average number of photons $\left<N\right>$. The equation of state can be cast in the form of
$$
P= constant \left<N\right>k_BT
$$
Notice that a similar result is valid for the classical perfect gas in the gran-canonical ensemble, even though chemical potential is not zero.
