 Solution

Let $r$ be radial position, $\mathcal{F}=L/(4\pi r^2)$ the outward radiative energy flux, and $P_r=a{T}^4/3$ the radiation pressure. A thin layer absorbs a fraction $\kappa \rho dr$ of an incident directed beam. Since a photon carries momentum $E/c$, the outward force per unit volume deposited by the radiation is $\kappa \rho \mathcal{F}/c$. Balance it against the radiation pressure gradient:Substituting $d{P}_r/dr=(4a/3)T^{3}dT/dr$ yields the radiative diffusion equationThe sign expresses outward energy flux down the temperature gradient. The approximation assumes optically thick material and near local thermodynamic equilibrium, with an appropriate mean opacity.

The printed attenuation law describes the loss from an incident beam; it is not generally the radial derivative of the net luminosity, since absorption is accompanied by emission. For a concrete counterexample, in a steady envelope without local energy generation the net luminosity is constant even though $\kappa \rho >0$ and $dT/dr<0$. Applying $dL/dr=-\kappa \rho L$ to that net luminosity would contradict energy conservation. The local momentum deposition argument supplies the physically correct derivation.