 Solution

When ${\mu }_{\nu }/T_{\nu }\gg 1$, 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 thereforeso $\boxed{A=1/(8{\pi }^2)}$.

At thermal equilibrium, the antineutrino chemical potential is $-{\mu }_{\nu }$. Its occupation is approximately $e^{-(p+{\mu }_{\nu })/T_{\nu }}$, givingIt is exponentially smaller than the neutrino density; this particle-antiparticle imbalance is a cosmological lepton asymmetry.