= Solution
\b[False.] <Stellar evaporation> removes <stars> above the escape energy after <two-body relaxation> has redistributed their energies. The remaining self-gravitating cluster need not become less dense; losing mass is not the same as holding its radius fixed.
For a slowly evolving isolated cluster, apply the <virial theorem> at each approximate equilibrium: $E=T+W=W/2=-T$. <Stars> evaporating with nearly zero energy at infinity remove little total energy, so the bound cluster has nearly constant negative $E$. With <gravitational radius> $R_g$ defined by $W=-GM^2/R_g$,
$$
\boxed{R_g=\frac{GM^2}{2|E|}\propto M^2.}
$$
For homologous structure, its characteristic mean density therefore scales as
$$
\boxed{\bar\rho\propto\frac{M}{R_g^3}\propto M^{-5}.}
$$
It increases as mass is lost. Escaping <stars> carrying positive energy make the bound energy more negative and strengthen the contraction tendency. This is <virial contraction under stellar evaporation>, related to the negative heat capacity of bound gravitational systems. A tidally limited <globular cluster> has additional radius constraints, so this scaling is not universal, but the claimed inevitable density decrease is false even in the simplest isolated evaporation model.
Back to article page
