A <stellar polytrope> of index $n$ obeys
$$
P=K\rho^{1+1/n},
\qquad
\rho=\rho_c\theta^n,
\qquad
r=\alpha\xi.
$$
Combining <hydrostatic equilibrium>, $dP/dr=-Gm\rho/r^2$, with mass conservation, $dm/dr=4\pi r^2\rho$, gives the <Lane-Emden equation>
$$
\boxed{\frac1{\xi^2}\frac d{d\xi}
\left(\xi^2\frac{d\theta}{d\xi}\right)=-\theta^n},
\qquad
\boxed{\alpha^2=\frac{(n+1)K}{4\pi G}
\rho_c^{1/n-1}}.
$$
Regularity and normalization at the stellar centre require
$$
\boxed{\theta(0)=1,\qquad\theta'(0)=0}.
$$
Integrating the Lane-Emden equation from the centre then gives the enclosed mass
$$
\boxed{m(r)=-4\pi\rho_c\alpha^3\xi^2\theta'(\xi)}.
$$

For $n=5$, direct substitution into the equation finds
$$
\boxed{\theta=(1+\xi^2/3)^{-1/2}},
$$
so $\beta=1/3$ and $\gamma=-1/2$. This profile has no finite first zero and hence has infinite radius, but its total mass converges:
$$
\boxed{M=4\pi\sqrt3\,\rho_c\alpha^3}.
$$
Its mean density over the full, infinite configuration is consequently zero.

For a perfect gas, $T\propto P/\rho\propto\rho^{1/5}$, so $T/T_c=\theta$. The luminosity from <CNO cycle> burning with $\epsilon=\epsilon_0\rho T^{11}$ is therefore
$$
L=4\pi\epsilon_0\alpha^3\rho_c^2T_c^{11}
\int_0^\infty\xi^2\theta^{21}\,d\xi
=A\epsilon_0\alpha^3\rho_c^2T_c^{11},
$$
where
$$
A=4\pi\int_0^\infty\xi^2(1+\xi^2/3)^{-21/2}d\xi
=6\pi\sqrt3\,B(3/2,9)\simeq1.03.
$$
Thus $A$ is of order unity. The model is physically poor because the $n=5$ polytrope has infinite radius and zero mean density, while strongly temperature-sensitive CNO burning changes the thermal gradient and commonly creates convection. Real cores also have evolving composition, non-polytropic opacity and energy transport, and boundaries supplied by the surrounding star.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 317 / 1 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-317/1/solution