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A fully convective monatomic ideal gas is an adiabatic stellar polytrope with index $n=3/2$. Its polytropic mass-radius relation givesAt the base of a thin grey atmosphere, the stellar surface boundary condition and electron-scattering opacity giveThe ideal-gas equation evaluated on the convective adiabat givesCombining these relations,The Stefan–Boltzmann law then gives

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:Angular and frequency integration then gives the radiative diffusion fluxFor a thin plane-parallel atmosphere, constant Rosseland mean opacity $\kappa$, negligible external irradiation, and radius nearly equal to $R$, radiative equilibrium gives $F=L/(4\pi R^2)$. ThereforeWith hydrostatic balance $dP/dz=-\rho g$, the equivalent pressure form iswhere $g=GM/R^2$.

For $I=A+C\mu$, the angular moments areandThus this angular form satisfies the Eddington closure approximation, $c{P}_r=4\pi J/3$, and

Part (i) gives $S=J=A$. Substitution in $\mu dI/d\tau =I-S$ yieldsMatching powers of $\mu$ gives $A^{\prime }=C$ and $C^{\prime }=0$, the latter being constant flux. Hence $A=C\tau +A_0$.

At the surface there is no incoming intensity. Applying this condition to the outgoing hemisphere in the moment approximation,so $A_0=F/(2\pi )=2C/3$. ConsequentlyNow $A=J=j/\kappa =\sigma T^4/\pi$ and the effective temperature is defined by $F=\sigma T_e^4$. It follows thatAt the top of the grey atmosphere,

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.