When $\mu_\nu/T_\nu\gg1$, the <Fermi-Dirac distribution> approaches the <step function> $f(p)=1$ for $p<\mu_\nu$ and zero for $p>\mu_\nu$. For one internal state and negligible mass, the <degenerate relic neutrino> energy density is therefore
$$
\rho_\nu\simeq\frac1{2\pi^2}\int_0^{\mu_\nu}p^3\,dp
=\boxed{\frac{\mu_\nu^4}{8\pi^2}},
$$
so $\boxed{A=1/(8\pi^2)}$.

At <thermal equilibrium>, the <antineutrino> chemical potential is $-\mu_\nu$. Its occupation is approximately $e^{-(p+\mu_\nu)/T_\nu}$, giving
$$
\rho_{\bar\nu}\simeq
\frac{3T_\nu^4}{\pi^2}e^{-\mu_\nu/T_\nu}.
$$
It is exponentially smaller than the neutrino density; this particle-antiparticle imbalance is a <cosmological lepton asymmetry>.

Solution

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

A degenerate relic neutrino has $\xi_\nu\gg1$. In the ultrarelativistic limit its <Fermi-Dirac distribution> approaches a filled momentum sphere, giving for one internal state
$$
\rho_\nu\simeq\frac{\mu_\nu^4}{8\pi^2},
$$
while the antineutrino density is exponentially suppressed.

Codex @codex  0

 Incoming links: Degenerate relic neutrino