Plane-parallel <hydrostatic equilibrium> gives $dP/dz=-\rho g$, while $d\tau=-\kappa\rho dz$. Hence
$$
\frac{dP}{d\tau}=\frac g\kappa.
$$
For $\kappa=\kappa_0P^{\alpha-1}T^{4-4\beta}$,
$$
\frac{d(P^\alpha)}{d\tau}
=\frac{\alpha g}{\kappa_0}T^{4\beta-4}.
$$
The grey-atmosphere relation from part (ii) can be written
$$
T^4=T_0^4\left(1+\frac32\tau\right),
\qquad
d\tau=\frac{2,d(T^4)}{3T_0^4}.
$$
Integrating from the zero-pressure top, where $T=T_0$, gives
$$
\boxed{P^\alpha=
\frac{2\alpha g}{3\kappa_0\beta T_0^4}
\left(T^{4\beta}-T_0^{4\beta}\right)}.
$$

The natural matching point to the stellar interior is the <photosphere> $\tau=2/3$, where $T=T_e$ and $T_e^4=2T_0^4$. Multiplying the preceding expression by $\kappa_0T^{4-4\beta}/g$ then yields the <stellar surface boundary condition>
$$
\boxed{\frac{P\kappa}{g}
=\frac{4\alpha}{3\beta}\left(1-2^{-\beta}\right)}.
$$
The local relation $L_r=4\pi r^2\sigma T^4$ identifies this matching temperature with the local effective temperature.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 2 / iii / Solution

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The photosphere is the layer from which most observed stellar radiation escapes; in a grey atmosphere it is represented approximately by optical depth $\tau=2/3$.


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 Incoming links: Photosphere