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Use $\hslash =c=k_B=1$ and count one populated helicity per neutrino or antineutrino. For massless particles, the Fermi-Dirac distribution givesWhen ${\mu }_{\nu }/T_{\nu }\gg 1$, the occupied states form an almost sharp Fermi sea. Replacing the distribution by $\Theta ({\mu }_{\nu }-p)$ givesThe relative finite-temperature correction is of order $(T_{\nu }/{\mu }_{\nu }{)}^2$.

Chemical equilibrium gives ${\mu }_{\bar{\nu }}=-{\mu }_{\nu }$. For positive large ${\mu }_{\nu }$, the corresponding antineutrino distribution is dilute, soIt is exponentially negligible. The neutrino degeneracy parameter ${\xi }_{\nu }={\mu }_{\nu }/T_{\nu }$ is conserved in the assumed adiabatic massless evolution: redshifting preserves the distribution with both $T_{\nu }$ and ${\mu }_{\nu }$ proportional to $a^{-1}$.

There is a sign qualification in the printed absolute-value formula. If ${\mu }_{\nu }<0$, the distribution of neutrinos themselves is exponentially suppressed, and the antineutrinos, with positive chemical potential, form the degenerate sea. In either case the degenerate neutrino and antineutrino energy density isThus the formula with $|{\mu }_{\nu }|$ describes the dominant member of the pair, or their leading total, rather than the named neutrino distribution for both signs. No extra factor of two is present: only one member is densely occupied. Additional populated internal states multiply the result by their degeneracy.

Bosons have symmetric many-particle states, so any number may occupy the same one-particle state. At low temperature, an ideal Bose gas therefore places a macroscopic population in its lowest one-particle state; below the critical temperature this is Bose-Einstein condensation.

Fermions have antisymmetric many-particle states and obey the Pauli exclusion principle, so each one-particle state can hold at most one fermion of each available internal label. At zero temperature, a noninteracting Fermi gas fills distinct states in increasing order of energy up to the Fermi energy. Its many-particle ground state is thus the filled Fermi sea.