Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 317 / 3 / ii

Measure optical depth inward, so $d\tau /dr=-\kappa \rho$ and $\tau =0$ at the exterior boundary. The printed integral is shorthand with inconsistent integration labels; this differential definition fixes its meaning. In a plane-parallel atmosphere in radiative equilibrium, $F=\sigma T_{\mathrm{eff}}^4$ is constant and $d{P}_{\mathrm{rad}}/d\tau =F/c$.

For the Eddington surface boundary condition, approximate the outgoing frequency-integrated intensity by a constant $I_0$ over the outward hemisphere, with no incoming radiation. The angular moments at the surface areIntegrating the moment equation then gives $P_{\mathrm{rad}}=F(\tau +2/3)/c$. The Eddington closure approximation, with local thermodynamic equilibrium and radiative equilibrium, identifies the radiation energy density with $a_{\mathrm{rad}}{T}^4$ and sets $P_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$. Using $a_{\mathrm{rad}}c=4\sigma$ givesThe photosphere is represented by $\tau =2/3$, where $T=T_{\mathrm{eff}}$; the exterior boundary instead has $T(0)=2^{-1/4}{T}_{\mathrm{eff}}$. The closure and angular boundary condition are approximations, rather than exact isotropy of radiation at the surface.