 Solution

Write $\mathcal{R}=k/(\mu m_H)$ for the specific gas constant, with the printed $m_H$ denoting the atomic mass unit, and $a_{\mathrm{r}}=4\sigma /c$ for the radiation constant. In the low-mass envelope the enclosed mass is approximately $M$ and the outward luminosity is approximately $L$. The stellar hydrostatic equation, stellar radiative diffusion, and ideal gas equation giveThe Kramers' opacity law becomes $\kappa =({\kappa }_0/\mathcal{R})P{T}^{-9/2}$. Dividing the two differential equations eliminates radius and mass density:Integrating from a surface matching point $(P_s,T_s)$ givesDeep compared with the photosphere, the surface terms are negligible. The radiative-zero white-dwarf envelope therefore hasThis is the leading envelope approximation used in the later parts, rather than an exact atmospheric boundary condition at $T_{\mathrm{eff}}$.

At the base, temperature, mass density and pressure match the isothermal degenerate core. Let $K_{\mathrm{NR}}=K_1{\mu }_e^{-5/3}$, where ${\mu }_e$ is the mean molecular weight per electron. Equating the gas and electron degeneracy pressure givesHence $P_b={\mu }_e^{5/2}{\mathcal{R}}^{5/2}{K}_1^{-3/2}{T}_c^{5/2}$. Substituting this into the envelope relation at $T=T_c$ yieldsThe matching of a white-dwarf envelope to a degenerate core thus gives $L/M\propto T_c^{7/2}$, the envelope relation underlying Mestel's cooling law. It depends on pressure as well as mass density continuity at the interface; otherwise the supplied core equation would not determine the matching mass density.