Write the <neutrino degeneracy parameter> as $\xi_\nu=\mu_\nu/T_\nu$. Requiring one degenerate species not to exceed today's <critical density> gives
$$
\frac{\xi_\nu^4T_\nu^4}{8\pi^2}
\leq6\times10^4T^4.
$$
Using $T/T_\nu=(11/4)^{1/3}$,
$$
\boxed{\xi_\nu\leq
\left(48\pi^2\times10^4\right)^{1/4}
\left(\frac{11}{4}\right)^{1/3}
\simeq65}.
$$
Several equally degenerate species would strengthen the bound by the fourth root of their number.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 310 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-310/2/d/solution