Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 317 / 2 / Solution

The density exponent in $ϵ={ϵ}_0{\rho }^a{T}^b$ reflects the number of reacting pairs per unit mass; for ordinary two-body hydrogen burning, $a\simeq 1$. The temperature exponent is the local logarithmic sensitivity of the thermally averaged nuclear cross-section. Near $2\times 10^7\mathrm{K}$,are standard approximations.

Spherical regularity makes central scalar profiles even in $r$. Since density and temperature decrease outward in a normal stellar core, they have expansionswith $\alpha ,\beta >0$. The high power $T^b$ then suppresses burning rapidly away from the centre, especially for the CNO cycle.

The luminosity equation isExpanding and integrating givesThe luminosity generated by a thin shell scales locally as $r^2\exp [-((a+1)\beta +b\alpha )r^2]$. Its maximum therefore haswhen the temperature term dominates. Comparable central structures then give

Near the centre, mass conservation gives $m(r)=4\pi {\rho }_c{r}^3/3+O(r^5)$. Substitution into hydrostatic equilibrium and integration yieldsFor the perfect-gas law $P=\rho \mathcal{R}T/\mu$,Taking the two-body value $a=1$ in the earlier expression for $\gamma$ and eliminating $\alpha$ gives

Solved by gpt-5.6-sol high.