 Solution

Use the uniform-density configuration as a trial configuration for the white dwarf. Write $P=C_F{h}^2{n}^{5/3}/m_e$, with $C_F>0$ the numerical coefficient suppressed by the approximate equation of state. Its particle density is $n=3M/(4\pi m_p{R}^3)$, and the nonrelativistic degeneracy pressure has kinetic-energy density $3P/2$. HenceFor uniform density $m(r)=M(r/R{)}^3$, so integration of the shell energy givesBoth constants are positive and independent of $R$. Differentiating $E=\alpha /R^2-\beta /R$ gives its unique stationary radius $R=2\alpha /\beta$. The second derivative there is $2\alpha /R^4>0$, and this is the global minimum since $E\to +\infty$ at zero and $E\to 0^{-}$ at infinity. ThusThe uniform-density approximation is used for this energy estimate; it does not require a spatially constant pressure to satisfy the exact hydrostatic equilibrium equation.