At $x=1/2$, define
$$
q=\frac{31}{96},
\qquad
T_{1/2}=q\frac{\mu m_uGM}{k_BR},
\qquad
\rho_{1/2}=\frac{3M}{2\pi R^3}.
$$
Differentiating the temperature profile gives
$$
-\left.\frac{dT}{dr}\right|_{R/2}
=\frac{29}{48R}\frac{\mu m_uGM}{k_BR}.
$$
Since the luminosity is already constant there, the <radiative diffusion in a star> equation yields
$$
L=\frac{29\pi a_{\rm rad}c}{36}
\frac{R}{\kappa_{1/2}\rho_{1/2}}
q^3\left(\frac{\mu m_uGM}{k_BR}\right)^4.
$$

For the <Kramers opacity law> $\kappa=\kappa_0\rho T^{-7/2}$, substitution gives
$$
\boxed{L=C_KM^{11/2}R^{-1/2}},
$$
where
$$
\boxed{
C_K=\frac{29\pi^3a_{\rm rad}c}{81\kappa_0}
q^{13/2}
\left(\frac{\mu m_uG}{k_B}\right)^{15/2}}.
$$
Thus $\alpha=11/2$ and $\beta=-1/2$ in the notation of the question.

For constant <electron-scattering opacity> $\kappa_{\rm es}$,
$$
\boxed{L=C_{\rm es}M^3},
$$
where
$$
\boxed{
C_{\rm es}=\frac{29\pi^2a_{\rm rad}c}{54\kappa_{\rm es}}
q^3\left(\frac{\mu m_uG}{k_B}\right)^4}.
$$
Hence $\delta=3$ and $\gamma=0$: the radius cancels, and the mass dependence is shallower than under Kramers opacity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 317 / 1 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-317/1/ii/solution