For $I=A+C\mu$, the angular moments are
$$
J=\frac1{4\pi}\int I\,d\Omega=A,
$$
$$
F=\int\mu I\,d\Omega=\frac{4\pi}{3}C,
$$
and
$$
P_r=\frac1c\int\mu^2I\,d\Omega
=\frac{4\pi}{3c}A=\frac{4\pi}{3c}J.
$$
Thus this angular form satisfies the <Eddington closure approximation>, $cP_r=4\pi J/3$, and
$$
\boxed{C=\frac{3F}{4\pi}}.
$$

Part (i) gives $S=J=A$. Substitution in $\mu\,dI/d\tau=I-S$ yields
$$
\mu(A'+C'\mu)=C\mu.
$$
Matching powers of $\mu$ gives $A'=C$ and $C'=0$, 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,
$$
F=2\pi\int_0^1(A_0+C\mu)\mu\,d\mu
=\pi A_0+\frac{2\pi C}{3},
$$
so $A_0=F/(2\pi)=2C/3$. Consequently
$$
A=\frac{3F}{4\pi}\left(\tau+\frac23\right).
$$
Now $A=J=j/\kappa=\sigma T^4/\pi$ and the <effective temperature> is defined by $F=\sigma T_e^4$. It follows that
$$
\boxed{T^4=\frac34T_e^4\left(\tau+\frac23\right)}.
$$
At the top of the <grey atmosphere>,
$$
\boxed{T_0^4=\frac12T_e^4},
\qquad
\boxed{T_0=2^{-1/4}T_e}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-317/2/ii/solution