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

The outward radiative flux is $F=L/(4\pi r^2)$. A photon of energy $E$ carries momentum $E/c$, so the momentum transferred per unit volume and time to matter is $\kappa \rho F/c$, when $\kappa$ is the appropriate transport opacity. Momentum conservation makes the radiation stress decrease outward by this amount. In the thin atmosphere, where the radiation stress is treated by its scalar radiation pressure,This is the first moment of the radiative transfer equation. The grey atmosphere approximation uses a frequency-independent transport opacity; in an optically thick diffusion regime it becomes the Rosseland mean opacity. Near the surface the plane-parallel atmosphere approximation removes geometric stress-divergence corrections.

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.

Taking the logarithmic derivative of the grey atmosphere temperature law, $4\log T=\log (\tau +2/3)+\text{constant}$, gives

Take the gas pressure $P$ to vanish at the exterior boundary and the atmosphere to be thin enough that surface gravity of a star is constant. With constant electron-scattering opacity, negligible radiative acceleration, and hydrostatic equilibrium in optical depth,If the radiative force is retained, $dP/dr=-\rho (g-\kappa F/c)$ for gas pressure, and the same result holds with the constant positive effective gravity $g_{\mathrm{eff}}=g-\kappa F/c$. A nonzero exterior gas pressure instead gives $d\log P/d\tau =1/(\tau +\kappa P(0)/g_{\mathrm{eff}})$. Total gas-plus-radiation pressure has the nonzero radiative surface term from part (ii), so the advertised exact $1/\tau$ formula presumes gas pressure with zero exterior gas pressure.

The stellar adiabatic exponent ${\Gamma }_2$ is defined at fixed specific entropy and composition byCombining parts (iii) and (iv), the radiative temperature gradient of the constant-opacity grey atmosphere isFor $1<{\Gamma }_2<4/3$, the adiabatic temperature gradient is strictly below $1/4$, and the Schwarzschild criterion is eventually violated. If ${\Gamma }_2$ is constant, solving for the crossing givesAt ${\Gamma }_2=4/3$ the radiative gradient approaches marginality from below, so there is no finite crossing in this model. If ${\Gamma }_2$ varies, the conclusion requires its subcritical value to persist in the deep layers. Uniform composition is needed for this use of the Schwarzschild criterion; a composition gradient requires the Ledoux criterion.