Measure <optical depth> inward, so $d\tau/dr=-\kappa\rho$ and $\tau=0$ at the exterior boundary. The printed integral is shorthand with inconsistent integration labels; this differential definition fixes its meaning. In a <plane-parallel atmosphere> in radiative equilibrium, $F=\sigma T_{\rm eff}^4$ is constant and $dP_{\rm rad}/d\tau=F/c$.

For the <Eddington surface boundary condition>, approximate the outgoing frequency-integrated intensity by a constant $I_0$ over the outward hemisphere, with no incoming radiation. The angular moments at the surface are
$$
F=2\pi I_0\int_0^1\mu\,d\mu=\pi I_0,\qquad P_{\rm rad}(0)=\frac{2\pi I_0}{c}\int_0^1\mu^2\,d\mu=\boxed{\frac{2F}{3c}}.
$$
Integrating the moment equation then gives $P_{\rm rad}=F(\tau+2/3)/c$. The <Eddington closure approximation>, with <local thermodynamic equilibrium> and radiative equilibrium, identifies the radiation energy density with $a_{\rm rad}T^4$ and sets $P_{\rm rad}=a_{\rm rad}T^4/3$. Using $a_{\rm rad}c=4\sigma$ gives
$$
\boxed{T^4(\tau)=\frac34T_{\rm eff}^4\left(\tau+\frac23\right)=\frac12T_{\rm eff}^4\left(1+\frac32\tau\right).}
$$
The <photosphere> is represented by $\tau=2/3$, where $T=T_{\rm eff}$; the exterior boundary instead has $T(0)=2^{-1/4}T_{\rm eff}$. The closure and angular boundary condition are approximations, rather than exact isotropy of radiation at the surface.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 317 / 3 / ii / Solution

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{title2=$P_{\rm rad}(0)=2F/(3c)$}

For no incident radiation and outward intensity approximated as direction independent, the surface angular moments are $F=\pi I_0$ and $P_{\rm rad}(0)=2\pi I_0/(3c)$. Integrating the radiation momentum equation with the <Eddington closure approximation> yields the <grey atmosphere> law $T^4=3T_{\rm eff}^4(\tau+2/3)/4$. This is an approximate angular boundary condition, not isotropy across both hemispheres.

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 Incoming links: Eddington surface boundary condition