Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 317 / 4

The gas-to-radiation pressure ratio givesHenceand either pressure component gives

The radiative-pressure equation isDividing by hydrostatic equilibrium and substituting
$\kappa ={\kappa }_0(M_r/L_r)(L/M)$ givesa constant. Since both pressures vanish at the surface,
$P_{\mathrm{rad}}=(1-\beta )P$ and $\beta$ is constant throughout the star. Part i therefore haswith spatially constant $K$.

Writing $\rho ={\rho }_c{\theta }^3$ andgives the $n=3$ Lane-Emden equationThe surface is the first zero $\theta ({\xi }_1)=0$. The $n=3$ equation has no elementary closed-form solution; among nonnegative indices, the standard analytic cases are $n=0,1,5$. This Eddington standard model approximates chemically homogeneous massive main-sequence stars with substantial radiation pressure.

At any radius,Both gas and radiation pressures decrease outwards, so
$0<d{P}_{\mathrm{rad}}/dP<1$. ThereforeEquality is the Eddington luminosity, where radiation supplies the entire outward force and hydrostatic gas support is lost.

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