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

A large <neutrino degeneracy parameter> raises the relativistic energy density and therefore the expansion rate. <Big Bang nucleosynthesis> tests this through primordial light-element abundances; an electron-neutrino chemical potential also shifts neutron--proton chemical equilibrium directly. The <Cosmic microwave background power spectrum> tests the changed expansion rate, matter-radiation equality, sound horizon, damping scale, and neutrino anisotropic stress. Finally, <large-scale structure of the universe> and the matter power spectrum test the altered equality scale and <neutrino free streaming>. These observables distinguish a degenerate-neutrino cosmology from the standard thermal relic scenario in complementary epochs.

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

{title2=$\xi_\nu$}

The neutrino degeneracy parameter is the dimensionless chemical potential
$$
\xi_\nu=\frac{\mu_\nu}{T_\nu}.
$$
In thermal equilibrium the corresponding antineutrino has chemical potential $-\mu_\nu$.


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