# Why is Equation of State of Photon gas different from the Equation of State of Boson gas?

We know that, for Bosons, the Equation of State is given by $$pV=\frac{2}{3}U$$; where $$p$$ is the Pressure, $$V$$ is the Volume and $$U$$ is the Internal Energy of the Boson Gas. (see Equation of State of Fermion & Boson gas)

But, for Photon gas, the equation of state is given by $$pV=\frac{1}{3}U$$; where $$p$$ is the Pressure, $$V$$ is the Volume and $$U$$ is the Internal Energy of the Photon Gas. (see Equation of State of Photon gas)

But, we know that Photon is Ideal Boson. Then, why is the Equation of State of Photon gas different from Equation of State of Boson gas?

• Please do not edit a question to ask a different question after it has received answers. If you have a new question, ask a new question. May 1 at 20:16

The equation $$p = 2u/3$$, where $$u$$ is kinetic energy density, is true for any non-relativistic, non-interacting, point-like particles, whether they be spin 1 bosons or spin 1/2 fermions. Your first source assumes this by using a non-relativistic formula for kinetic energy.

Similarly, the equation $$p = u/3$$ is true for ultrarelativistic (or massless) bosons or fermions.

The point of difference in your question is that photons are massless, but in most other circumstances (W, Z, Higgs) bosons are massive and non-relativistic.

Photons are indeed bosons (spin 1), and from what I can see in the book all the formulas are written for a classical system of bosons. Photons are massless relativistic particles,so when you try to perform the integration the total energy is not: $$$$E=\frac{p^{2}} {2m}$$$$ but its relativistic form: $$$$E=pc$$$$ As you can see the main difference in computing the integral is in the quadratic dependence of the momentum p in the classical case, which becomes "linear" in the second case. If you put this energy in the bose-einstein distribution and try to compute the integral you should find a dirrerent result.

Edit Integrals of distributions like B-E have as a solution the product of two special mathematical functions, the Euler Gamma function and the Zeta Riemann function:

$$$$\int_{0}^{\inf} \frac{x^{s-1} } {e^{x}-1} dx = \Gamma(s) \zeta(s)$$$$

• Another thing is that you need to take into account a factor of $2$ due to polarization Apr 30 at 12:39
• @OfekGillon: The degeneracy doesn't affect the relations between pressure, volume, and internal energy cited by the OP, though. Apr 30 at 13:19