Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 49 / 2

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.

Let $r_{\nu }=T_{\nu ,0}/T_0$, and retain the assumption that these neutrinos remain massless. Constancy of the neutrino degeneracy parameter gives $|{\mu }_{\nu ,0}|=|{\xi }_{\nu }|r_{\nu }{T}_0$. The condition on the critical density therefore yields the critical-density bound on massless neutrino degeneracyFor the standard instantaneous-neutrino decoupling approximation before electron-positron annihilation in cosmology, cosmological entropy conservation heats photons relative to the decoupled neutrinos. The electromagnetic entropy degrees of freedom change from $2+(7/8)4=11/2$ to $2$, givingThis numerical value requires that thermal-history assumption; the bound in terms of $r_{\nu }$ is the general answer if large degeneracy changes decoupling or subsequent entropy transfer. Setting $T_{\nu ,0}=T_0$ would instead give $26.2$. For $N$ equally degenerate flavors with the same temperature, the leading bound becomes $36.8N^{-1/4}$ under the standard temperature assumption. The neglected thermal corrections make a small change near this loose energy-density limit; this is not a detailed Big Bang nucleosynthesis bound.

Two effects must be distinguished in Big Bang nucleosynthesis. The positive extra neutrino energy density increases the Hubble parameter through the Friedmann equation. Holding weak rates fixed, faster expansion causes earlier neutron-proton cosmological weak freeze-out, leaving more neutrons, and leaves less time for neutron decay before nuclear reactions begin. Both tendencies increase the primordial helium mass fraction.

Electron-flavor degeneracy also changes the charged-current balance directly. Chemical equilibrium for $n+{\nu }_e\leftrightarrow p+e^{-}$ gives, with negligible electron chemical potential,This is the primordial helium response to electron-neutrino degeneracy. Positive ${\xi }_{{\nu }_e}$ favors conversion of neutrons to protons and reduces the neutron abundance; negative ${\xi }_{{\nu }_e}$ favors the opposite balance. When neutrons are the limiting ingredient and nearly all surviving neutrons enter helium-4, the neutron-limited helium synthesis estimate isIn this usual proton-rich regime, the direct electron-neutrino effect lowers $Y_p$ for positive degeneracy and raises it for negative degeneracy, other conditions held fixed. At sufficiently large negative degeneracy the equilibrium ratio can instead exceed one; protons then become limiting. The corresponding ideal complete-capture estimate is $Y_p\simeq 2\min (r,1)/(1+r)$, so the neutron-limited trend is not a universal monotonic law at arbitrary negative chemical potential. Degeneracy in the other flavors primarily affects helium through the expansion rate in this simplified discussion.

For very large degeneracy the weak reaction rates themselves also change, so a quantitative freeze-out calculation must include both the altered distributions and the altered expansion rate. The two effects can compete for a positive electron-neutrino asymmetry. The primordial helium mass fraction change is not determined by $|{\mu }_{\nu }|$ alone: its flavor and sign matter as well as its energy density.