0

The density of states for a photon gas is defined by, $$D(\epsilon)=\frac{g}{2\pi^2}\frac{\epsilon^2}{(\hbar c)^2} $$ where g is the number of independent internal states for a photon. The question is now, what is the value of g?

Since photons are spin 1 particles, it should be 3, given by the fact that, $2s+1=g$.

But that's not the case as I read that it is actually 2, due to the two polarization sttaes of the em waves. Why is this?

Gandalf73
  • 381
  • Related: https://physics.stackexchange.com/questions/154468/difference-between-spin-and-polarization-of-a-photon – Hossein May 14 '22 at 22:34
  • 3
    Probably the answer is related to the fact that the photon is massless. A massless particle can't have a properly defined spin, but it has an helicity, and the photon has 2 helicity states. Classically these correspond to the two possible polarizations of an electromagnetic wave. I am not very expert though – Matteo May 14 '22 at 22:40

0 Answers0