<Radiative diffusion in a star> can be written
$$
\frac{dP_{\rm rad}}{dr}
=-\frac{\kappa\rho L_r}{4\pi cr^2}.
$$
Dividing by <hydrostatic equilibrium> gives
$$
\frac{dP_{\rm rad}}{dP}
=\frac{\kappa L_r}{4\pi cGm_r}
=\frac{\nu L}{4\pi cGM},
$$
which is constant by assumption. Since both $P_{\rm rad}$ and $P$ vanish at the surface, integration gives $P_{\rm rad}=(1-\beta)P$ with constant $\beta$.

Eliminating $T$ between the gas and radiation equations of state now gives
$$
P=K\rho^{4/3},
\qquad
K=\left[\frac3a
\left(\frac{\mathcal R}{\mu}\right)^4
\frac{1-\beta}{\beta^4}\right]^{1/3}.
$$
Thus the star is an $n=3$ <stellar polytrope>. Put
$$
\rho=\rho_c\theta^3,
\qquad
r=\alpha\xi,
\qquad
\alpha^2=\frac K{\pi G}\rho_c^{-2/3}.
$$
The structure equations become the <Lane-Emden equation>
$$
\boxed{\frac1{\xi^2}\frac d{d\xi}
\left(\xi^2\frac{d\theta}{d\xi}\right)=-\theta^3},
\qquad
\boxed{\theta(0)=1,\quad\theta'(0)=0}.
$$
The surface is the first zero $\xi_1\simeq6.89685$. Writing
$$
\omega_3=-\xi_1^2\theta'(\xi_1)\simeq2.01824,
$$
the <Lane-Emden mass formula> gives
$$
M=4\pi\alpha^3\rho_c\omega_3
=\frac{4\omega_3}{\sqrt\pi}
\left(\frac KG\right)^{3/2}.
$$
Therefore
$$
\boxed{M=\lambda\frac{(1-\beta)^{1/2}}{\beta^2}},
$$
where
$$
\boxed{\lambda=
\frac{4\omega_3}{\sqrt\pi}
\left(\frac3a\right)^{1/2}
\left(\frac{\mathcal R}{\mu}\right)^2G^{-3/2}}.
$$
The additional radiative-equilibrium relation is $1-\beta=\nu L/(4\pi cGM)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 317 / 3 / ii / Solution

{c}

If $\rho=\rho_c\theta^n$, $r=\alpha\xi$, and $\xi_1$ is the first zero of the Lane-Emden function, then a spherical polytrope has
$$
M=4\pi\alpha^3\rho_c\left[-\xi_1^2\theta'(\xi_1)\right].
$$


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 Incoming links: Lane-Emden mass formula