Let $\omega=-\xi_1^2\theta'(\xi_1)$ be the <Lane-Emden surface mass constant>. The <Lane-Emden mass formula> gives
$$
R=a\xi_1,\qquad M=4\pi a^3\rho_c\omega,
\qquad a^2=\frac{5K_{\rm NR}}{8\pi G}\rho_c^{-1/3}.
$$
Thus $R\propto\rho_c^{-1/6}$ while $M\propto\rho_c^{1/2}$ at fixed composition. Eliminating central <mass density> gives the <polytropic mass-radius relation>
$$
\boxed{M\propto R^{-3},\qquad s=-3.}
$$
More explicitly,
$$
R=\xi_1(4\pi\omega)^{1/3}\frac{5K_{\rm NR}}{8\pi G}M^{-1/3}.
$$
Increasing the mass of a <nonrelativistic white dwarf> decreases its radius. This scaling ceases to apply when relativistic <electron degeneracy pressure> becomes important.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 317 / 3 / ii / b / Solution

{c}
{title2=$Y_n$}

For a Lane-Emden function whose first zero is $\xi_1$,
$$
Y_n=-\xi_1^2\theta'(\xi_1)
=\int_0^{\xi_1}\xi^2\theta(\xi)^n\,d\xi.
$$


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