Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 2

Put $\mu =\cos \theta$ and $S=j/\kappa$. Integrating the radiative transfer equation over solid angle givesThe first integral is the radiative flux $F$. A stationary atmosphere with no local energy source has $dF/d\tau =0$, soBy definition, the mean intensity is $J=(4\pi {)}^{-1}\int Id\Omega$. Thereforewhich is radiative equilibrium: the angle-integrated emission and absorption rates are equal.

For $I=A+C\mu$, the angular moments areandThus this angular form satisfies the Eddington closure approximation, $c{P}_r=4\pi J/3$, and

Part (i) gives $S=J=A$. Substitution in $\mu dI/d\tau =I-S$ yieldsMatching powers of $\mu$ gives $A^{\prime }=C$ and $C^{\prime }=0$, the latter being constant flux. Hence $A=C\tau +A_0$.

At the surface there is no incoming intensity. Applying this condition to the outgoing hemisphere in the moment approximation,so $A_0=F/(2\pi )=2C/3$. ConsequentlyNow $A=J=j/\kappa =\sigma T^4/\pi$ and the effective temperature is defined by $F=\sigma T_e^4$. It follows thatAt the top of the grey atmosphere,

Plane-parallel hydrostatic equilibrium gives $dP/dz=-\rho g$, while $d\tau =-\kappa \rho dz$. HenceFor $\kappa ={\kappa }_0{P}^{\alpha -1}{T}^{4-4\beta }$,The grey-atmosphere relation from part (ii) can be writtenIntegrating from the zero-pressure top, where $T=T_0$, gives

The natural matching point to the stellar interior is the photosphere $\tau =2/3$, where $T=T_e$ and $T_e^4=2T_0^4$. Multiplying the preceding expression by ${\kappa }_0{T}^{4-4\beta }/g$ then yields the stellar surface boundary conditionThe local relation $L_r=4\pi r^2\sigma T^4$ identifies this matching temperature with the local effective temperature.