 Solution

Write $u=\cos \theta$, $S=j/\kappa$ and use axial symmetry. The radiative flux is $F=2\pi {\int }_{-1}^{1}uIdu$. Integrating the radiative transfer equation over directions givesThere are no energy sources in the atmosphere, so radiative equilibrium makes $F$ constant. Consequently

For the linear angular approximation $I=A+Cu$, the radiation-field moments areThus it satisfies the Eddington closure approximation, andSince $F$ is constant, $C^{\prime }=0$. Substitution in the radiative transfer equation, using $S=A$, gives $u{A}^{\prime }=Cu$, hence $\boxed{A^{\prime }=C}$.

In this moment approximation, the inward hemispheric flux at the surface isSetting this boundary moment to zero gives $A_0=2C/3$. ThereforeCombining it with the thermal source function and $F=\sigma T_e^4$ gives the grey atmosphere temperature law:

The surface condition is an approximate angular-moment condition. If the affine intensity were interpreted as an exact nonnegative boundary radiation field, no incident radiation would require $I(0,u)=0$ for every $u<0$; a linear polynomial vanishing on that interval would have $A_0=C=0$ and no outward flux either. The derived formal profile indeed has $I(0,-1)=-F/(4\pi )<0$. This positivity limitation of a linear Eddington intensity is why one uses the Eddington surface boundary condition as a moment closure, not the affine ansatz as an exact emergent angular intensity. An outward direction-independent intensity with no inward intensity gives the same boundary moments.

For the gas-pressure-dominated atmosphere, combine $dP/dz=-\rho g$ with $d\tau /dz=-\kappa \rho$. Thus the hydrostatic equilibrium in optical depth is $dP/d\tau =g/\kappa$. Let $w=T^4$. The grey atmosphere law implies $dw/d\tau =3T_0^4/2$. With the specified power-law opacity,At the outer vacuum limit take $P=0$ and $T=T_0$. Since $\alpha ,\beta >0$, integration givesThis is the grey-atmosphere pressure with power-law opacity relation.

At the matching radius the local radiative flux is $L_r/(4\pi r^2)=\sigma T^4$. In the thin plane-parallel atmosphere approximation this equals $F=\sigma T_e^4$, so $T=T_e$ and $T^4=2T_0^4$. Multiplying the pressure relation by ${\kappa }_0{T}^{4-4\beta }/g$ yields the required surface pressure-opacity relationHere $P\kappa$ is a product, not a pressure subscript.