Because $T=T(\Psi)$,
$$
\nabla T=\frac{dT}{d\Psi}\nabla\Psi.
$$
The outward normal radiative flux is therefore
$$
\mathbf F\mathbin\cdot\widehat{\mathbf n}
=-\frac{4acT^3}{3\kappa\rho}
\frac{dT}{d\Psi}g.
$$
Integrating over the equipotential gives
$$
L_\Psi=-\frac{4acT^3}{3\kappa\rho}
\frac{dT}{d\Psi}S\langle g\rangle.
$$
Part (iii) gives $dm/d\Psi=\rho S\langle g^{-1}\rangle$, and hence
$$
\frac{dT}{d\Psi}=\rho S\langle g^{-1}\rangle\frac{dT}{dm}.
$$
Substitution yields
$$
\boxed{L_\Psi=-\frac{4acT^3}{3\kappa}
S^2\langle g\rangle\langle g^{-1}\rangle
\frac{dT}{dm}}.
$$
This is the radiative equation in the <equipotential stellar-structure approximation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 4 / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-317/4/iv/solution