Deep in an optically thick <grey atmosphere>, write $I_\nu=B_\nu+\delta I_\nu$ and retain the first spatial-gradient correction in the transfer equation:
$$
\delta I_\nu\simeq-\frac{\mu}{\kappa\rho}
\frac{dB_\nu}{dz}.
$$
Angular and frequency integration then gives the <radiative diffusion> flux
$$
F=-\frac{16\sigma T^3}{3\kappa\rho}\frac{dT}{dz}.
$$
For a thin <plane-parallel atmosphere>, constant <Rosseland mean opacity> $\kappa$, negligible external irradiation, and radius nearly equal to $R$, radiative equilibrium gives $F=L/(4\pi R^2)$. Therefore
$$
\boxed{\frac{dT}{dz}
=-\frac{3\kappa\rho L}{64\pi\sigma R^2T^3}}.
$$
With hydrostatic balance $dP/dz=-\rho g$, the equivalent pressure form is
$$
\boxed{\frac{dT}{dP}=\frac{3\kappa L}{64\pi\sigma GM\,T^3}},
$$
where $g=GM/R^2$.

Solution

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

{c}
{title2=$\kappa_R$}
{wiki=Rosseland_mean_opacity}

The Rosseland mean opacity is the harmonic frequency average weighted by $\partial B_\nu/\partial T$. It controls radiative diffusion because flux preferentially travels through low-opacity frequencies.


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 Incoming links: Rosseland mean opacity