 Solution

For an ideal gas with constant specific-heat ratio $\gamma$, the specific internal energy is $u=c_{V}T$ and $P=\rho (c_P-c_V)T=(\gamma -1)\rho u$. HenceThe stellar virial theorem consequently reads $\ddot{I}/2=2\mathcal{T}+3(\gamma -1)U-3P_{S}V+\Omega$. For a static nonrotating star with negligible surface pressure, $\mathcal{T}=\ddot{I}=P_S=0$, so $\Omega =-3(\gamma -1)U$.

Fully ionized nonrelativistic ions and electrons form a monatomic ideal gas, with $\gamma =5/3$, provided electron degeneracy pressure and radiation pressure are negligible. ThusBeing steady alone does not eliminate rotation or other macroscopic kinetic energy; those are additional assumptions needed for this relation.