Stellar surface boundary condition

A stellar surface boundary condition matches the interior pressure and temperature to an atmospheric solution, commonly near optical depth $\tau =2/3$ where a grey atmosphere has $T=T_{\mathrm{eff}}$.

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

In a plane-parallel atmosphere with $d\tau =-\kappa \rho dr$ and $dP/dr=-\rho g$, one has $dP/d\tau =g/\kappa$. If gravity and opacity vary slowly, the pressure at optical depth $\tau$ is approximately $P\simeq g\tau /\kappa$.

Take an ideal gas grey atmosphere with $T^4=T_e^4(3\tau +2)/4$, constant surface gravity of a star $g$, vanishing external pressure and $\kappa ={\kappa }_0{P}^{1/2}{T}^8$. Integrating hydrostatic equilibrium in optical depth gives $P^{3/2}=8g[1/2-(3\tau +2{)}^{-1}]/({\kappa }_0{T}_e^8)$. Its logarithmic temperature-pressure gradient is $9\tau /16$, so the Schwarzschild criterion for a monatomic gas is crossed at ${\tau }_c=32/45$. The radiative continuation beyond that depth is unstable and must be replaced by convective transport.

In a grey atmosphere with $T^4=T_0^4(1+3\tau /2)$ and $\kappa ={\kappa }_0{P}^{\alpha -1}{T}^{4-4\beta }$, hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. With $\alpha ,\beta >0$ and $P(0)=0$, integration gives the displayed pressure-temperature relation. At the matching surface $T=T_e$, the Eddington surface boundary condition gives $T_e^4=2T_0^4$, hence $P\kappa /g=4\alpha (1-2^{-\beta })/(3\beta )$.

For a plane-parallel atmosphere in radiative equilibrium, constant surface gravity of a star $g$ and opacity $\kappa ={\kappa }_0{P}^{\alpha -1}{T}^{4-4\beta }$, match the stellar structure interior where the radiative flux satisfies $F=\sigma T^4$. The grey atmosphere law then puts the surface at optical depth $\tau =2/3$ with $T^4=2T_0^4$. Integrating hydrostatic equilibrium in optical depth from $P(0)=0$ gives $P^{\alpha }=2\alpha g(T^{4\beta }-T_0^{4\beta })/(3\beta {\kappa }_0{T}_0^4)$. Multiplication by ${\kappa }_0{T}^{4-4\beta }/g$ proves the displayed boundary condition for $\alpha >0$, $\beta \ne 0$. At $\beta =0$ the integral is logarithmic and the continuous limit is $P\kappa /g=4\alpha \log 2/3$. The relation assumes negligible radiation pressure and describes the Eddington closure approximation, rather than exact angular transfer.

Assume an ideal-gas grey atmosphere with $P(0)=0$, constant surface gravity of a star $g$, $T^4=T_e^4(1+3\tau /2)/2$ and $\kappa ={\kappa }_0\rho T^{4\beta +1}$ for $\beta >1$. Using $\rho =\mu P/(\mathcal{R}T)$ in $dP/d\tau =g/\kappa$ and integrating givesDifferentiation yields $d\log T/d\log P=[(1+3\tau /2{)}^{\beta -1}-1]/[2(\beta -1)]$. Applying the Schwarzschild criterion with monatomic adiabatic gradient $2/5$ gives ${\tau }_b=\frac{2}{3}[((4\beta +1)/5{)}^{1/(\beta -1)}-1]$. Since ${\tau }_b$ depends only on $\beta$, matching to $P=K{T}^{5/2}$ fixes $K\propto g^{1/2}{T}_e^{-2\beta -5/2}$.

For a grey atmosphere with $T^4=T_e^4(2+3\tau )/4$, an ideal gas and $\kappa ={\kappa }_0\rho T^{17}$, hydrostatic equilibrium in optical depth integrates to $P^2=512\mathcal{R}g[1/8-(2+3\tau {)}^{-3}]/(9\mu {\kappa }_0{T}_e^{16})$, taking $P(0)=0$. The logarithmic radiative gradient is $[(2+3\tau {)}^3-8]/48$. Equating it to the monatomic adiabatic temperature gradient $2/5$ gives $(2+3{\tau }_b{)}^3=136/5$ and the displayed onset temperature. Beyond this point the radiative profile must be replaced by convective transport.