# The equation of state of radiation

by micomaco86572
 Sci Advisor HW Helper P: 1,275 Heuristic argument: P = N * F/A P = N * dp_x/dt / A P = N * dp_x / (L / v_x) / A And v^2 = v_x^2 + v_y^2 + v_z^2 = 3v_x^2 -> v = sqrt(3) v_x We can make a similar argument for the components of momentum to get an overall factor of 3. P = N/V * / 3 Then for a photon gas is the energy, so we have P = Energy per particle * Number / Volume / 3 = energy density / 3 Another way I have seen it derived, is to take the EM Stress tensor, show that it must be traceless, and compare that with the general identity (for a perfect fluid) $$T^{\mu}_{\mu} = -\rho + 3p$$
 P: 54 If there is a layer of photon gas surrounding a black hole‘s surface, will the pressure of this gas still be isotropic? In other words, the energy tensor is still $$\begin{display} T^{\mu}_{\nu}=\left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & -p & 0 & 0 \\ 0 & 0 & -p & 0 \\ 0 & 0 & 0 & -p \\ \end{array} \right) \end{display}$$ , isn't it? Or the g11 is not equal to g22,g33 any more?
 P: 54 Does somebody know that which book gives the detail derivation of this formula—the energy-momentum tensor of radiation, namely $$\begin{display} T^{\mu}_{\nu}=\left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & -p & 0 & 0 \\ 0 & 0 & -p & 0 \\ 0 & 0 & 0 & -p \\ \end{array} \right) \end{display}$$ ?