 Solution

Multiply hydrostatic equilibrium by $4\pi r^{3}dr$ and integrate from centre to surface. Integration by parts, neglecting external surface pressure, givesFor an ideal gas with a constant adiabatic index, $U=\int PdV/(\gamma -1)$. ConsequentlyA fully ionized, nonrelativistic monatomic gas has $\gamma =5/3$, provided gas rather than radiation or degeneracy dominates the thermal equation of state. It then obeys $\boxed{2U+\Omega =0}$ and $E=U+\Omega =-U=\Omega /2$.

Added nuclear heat initially raises pressure and causes expansion. The expansion makes gravitational energy less negative and can lower the temperature of the self-gravitating gas, reducing the temperature-sensitive fusion rate. Conversely, a small contraction can raise the core temperature and increase burning. This negative feedback explains stable main-sequence burning in an ordinary nondegenerate stellar core; a degenerate core need not respond the same way.

Before hydrogen ignition, a cloud loses energy by radiation and contracts. Its gravitational energy decreases; for the ideal-gas virial relation, roughly half the released binding energy heats the gas and half is radiated. Heating continues until nuclear reactions can balance the luminosity, explaining the Kelvin-Helmholtz phase of pre-main-sequence evolution.

The effective adiabatic response varies with ionization, molecular dissociation, radiation pressure and degeneracy. Partial ionization can absorb compression energy and lower the local first adiabatic exponent below $4/3$, encouraging compressibility, convection and pulsational instability. A local outer layer with $\gamma \le 4/3$ does not by itself prove global collapse: ordinary stars have ionization layers of that kind. The dynamical stability criterion concerns a suitable pressure-weighted global response and its coupling to the eigenmode; an extended star with an effectively global exponent below $4/3$ can be dynamically unstable.