# Does $PV\propto T$ apply to a photon gas?

For an ultrarelativistic ideal gas, I know that

1. $$p=\frac{u}3$$;
2. $$TV^3 =$$ constant;
3. $$pV\propto T$$.

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?

• I think that the article in wikipedia outlines the solution, and the answer is yes en.wikipedia.org/wiki/… – anna v Feb 17 at 5:16

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.

$$= 2\sum_k = \frac{2 V}{h^3} \int dk^3 \frac{1}{e^{\beta \epsilon}-1}= \frac{V}{\pi^2 \beta^3 (\hbar c)^3} \int_0^\infty \frac{x^2}{e^{x}-1} = \frac{2 \zeta(3) \, V k_B^3}{\pi^2 (\hbar c)^3} V T^3$$

The integral appearing in here is $$2 \zeta(3)$$. With this we can calculate now also the relation between $$PV$$ with $$$$.

$$PV= \frac{\pi^4}{90 \zeta(3)} k_B T$$

Non conservation of the number of particles does not imply that it is not possible to introduce an average number of photons $$\left$$. The equation of state can be cast in the form of $$P= constant \leftk_BT$$ Notice that a similar result is valid for the classical perfect gas in the gran-canonical ensemble, even though chemical potential is not zero.