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?

  • 2
    $\begingroup$ I think that the article in wikipedia outlines the solution, and the answer is yes en.wikipedia.org/wiki/… $\endgroup$ – 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.

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

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

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


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.


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