 Solution

Let the stellar radius be $R$ and assume zero external surface pressure. A spherical shell has mass $dm=4\pi r^2\rho dr$ and contributes $-Gm(r)dm/r$ to the gravitational potential energy. Using hydrostatic pressure support,The center boundary term vanishes and $P(R)=0$ eliminates the outer term. Nonzero external pressure would add $3P(R)V$.

The total energy is $(\alpha -3)⟨P⟩V$, so energetic gravitational binding requires $\alpha <3$. For an isotropic gas, kinetic theory gives $P=(1/3)\int pvdn$. Nonrelativistic kinetic energy is $pv/2$, giving $U=3PV/2$ and $\alpha =3/2$. Ultrarelativistic energy is $pv$, giving $U=3PV$ and $\alpha =3$: the ultrarelativistic limit is marginal in this Newtonian virial balance.

For a white dwarf, electron degeneracy pressure becomes ultrarelativistic as density increases. Its equation of state then scales as $P=K{\rho }^{4/3}$. Pressure support and gravitational stress have the same radius scaling, leaving a limiting mass of order $(K/G{)}^{3/2}$ rather than allowing arbitrary mass to be supported by contraction. The Chandrasekhar limit is this maximum mass for a cold electron-degenerate star at specified composition, of order $(\hslash c/G{)}^{3/2}/({\mu }_e{m}_p{)}^2$.

The virial calculation by itself does not establish a universal numerical maximum mass for every stellar composition and equation of state; its precise consequence is loss of binding/support as the pressure-bearing particles become ultrarelativistic. An accreting carbon-oxygen white dwarf approaching its limiting mass typically undergoes runaway carbon burning and thermonuclear disruption, a Type Ia supernova. For other compositions or evolution, electron captures can instead lead to collapse into a neutron star. The subsequent evolution needs that physical distinction.