Put $\mu=\cos\theta$. For $I_\nu=A_\nu+B_\nu\mu^3$, the <mean intensity> is
$$
J_\nu=\frac12\int_{-1}^{1}I_\nu\,d\mu=A_\nu,
$$
because the cubic term is odd. The K-integral, or second angular moment, is
$$
K_\nu=\frac12\int_{-1}^{1}\mu^2I_\nu\,d\mu
=\frac{A_\nu}{3},
$$
because the $\mu^5$ contribution is also odd. Hence
$$
\boxed{K_\nu=\frac13J_\nu}.
$$
This intensity obeys the <Eddington closure approximation> even though it is not isotropic.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 315 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-315/1/b/solution