Put $\mu=\cos\theta$ and $S=j/\kappa$. Integrating the <radiative transfer equation> over solid angle gives
$$
\frac d{d\tau}\int_{4\pi}\mu I\,d\Omega
=\int_{4\pi}I\,d\Omega-4\pi S.
$$
The first integral is the <radiative flux> $F$. A stationary atmosphere with no local energy source has $dF/d\tau=0$, so
$$
\boxed{4\pi\frac j\kappa
=\int_{4\pi}I\,d\Omega}.
$$
By definition, the <mean intensity> is $J=(4\pi)^{-1}\int I\,d\Omega$. Therefore
$$
\boxed{\frac j\kappa=J},
\qquad
\boxed{4\pi\frac j\kappa=4\pi J},
$$
which is radiative equilibrium: the angle-integrated emission and absorption rates are equal.

Solution

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

 Solution

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