 Solution

For a gas-pressure-supported stellar core of fixed mass and composition, neglect boundary terms for the illustrative whole-core estimate. The virial theorem gives $E=U+\Omega =-U$, and the monatomic gas gives $U=3M_c{k}_B\overline{T}/(2\mu m_u)$. Thus the effective heat capacity along the equilibrium sequence isThis negative heat capacity of a virialized gravitational system means that net energy loss makes a self-gravitating core contract and heat. It is different from the positive local material specific heat capacity. On the main sequence, hydrogen burning supplies most of the radiated energy, so rapid contraction is arrested. Hydrogen depletion increases the mean molecular weight; slow structural readjustment usually raises the central temperature rather than making it fall.

After core hydrogen exhaustion, a massive stellar core contracts and heats more rapidly until a later nuclear fuel ignites. There is no general reversal to cooling simply upon leaving the main sequence. If radiation pressure becomes dominant, the internal energy becomes $U=U_{\mathrm{gas}}+U_{\mathrm{rad}}$ with $3\int PdV=2U_{\mathrm{gas}}+U_{\mathrm{rad}}$. The virial theorem therefore gives $E=-U_{\mathrm{gas}}$; the radiation-dominated limit is marginally bound and its effective adiabatic index approaches $4/3$.

Quasistatic contraction can still maintain virial equilibrium. It fails dynamically only when sufficient support is lost, for example if pair creation reduces the effective adiabatic index below $4/3$, or when an iron core undergoes electron capture and photodisintegration. A fixed-composition gas-dominated contracting core with $T\propto 1/R$ is already a counterexample to any assertion that leaving the main sequence alone reverses the heating or destroys virial equilibrium.