The <stellar structure equations> are
$$
\frac{dM_r}{dr}=4\pi r^2\rho,\qquad
\frac{dP}{dr}=-\frac{GM_r\rho}{r^2},
$$
$$
\frac{dL_r}{dr}=4\pi r^2\rho\epsilon,\qquad
\frac{dT}{dr}=-\frac{3\kappa\rho L_r}
{16\pi a_{\rm rad}cr^2T^3},
$$
with
$$
P=P_g+P_{\rm rad},\qquad
P_g=\frac{\rho k_BT}{\mu H},\qquad
P_{\rm rad}=\frac{a_{\rm rad}T^4}{3},
\qquad
\kappa=\kappa_0\rho T^{-7/2}.
$$
The last relation is the <Kramers opacity law>. Uniform energy release per unit mass means $\epsilon$ is constant, so integration of the mass and luminosity equations gives $L_r=\epsilon M_r$. At the surface,
$$
\boxed{\epsilon=\frac LM=M^{-1}L}.
$$
Thus $\epsilon=M^aL^b$ has
$$
\boxed{a=-1,\qquad b=1}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 317 / 1 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-317/1/i/solution