Codex @codex  0

Write the mean interior density asSincethe assumed outward decrease of $\bar{\rho }$ implies $\rho \le \bar{\rho }$. In mass coordinates, hydrostatic equilibrium isFor every interior mass $m\le m_r$, monotonicity givesIntegrating from the centre and using ${\int }_0^{m_r}{m}^{-1/3}dm=3m_r^{2/3}/2$ proves

Let $\mathcal{R}$ be the gas constant per mole. At the centre,Eliminating $T_c$ gives the Eddington quartic relation in its central form,

At the surface, set $P=0$ and $m_r=M$ in the upper pressure bound. Cubing it givesThe function $(1-\beta )/{\beta }^4$ decreases strictly as $\beta$ increases on $0<\beta <1$. Define ${\beta }^{\ast }$ by equality in the resulting bound:Then ${\beta }_c\ge {\beta }^{\ast }$, orand rearrangement gives exactly

Stellar homology givesFor an $n=3$ polytrope, $P=K{\rho }^{4/3}$, so these scalings implyThe equation of state in part i givesat fixed chemical composition. Substitution into the homologous mass scale proves the Eddington quartic relation

When $s=0$, part i gives the polytropic index $n=3$. Stellar homology givesbecause gas pressure is the fraction $\beta$ of the pressure required by hydrostatic equilibrium. The Kramers opacity law therefore scales asWith constant $\eta$, the opacity relation from part i gives

For an $n=3$ polytrope, $M\propto (K/G{)}^{3/2}$. The explicit $K(\beta )$ of part i obeysso the Eddington quartic relation isat fixed composition. Substitution gives