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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.

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.

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.

There are two diagrams: the $t$-channel Z boson exchange from the weak neutral current, and a crossed $u$-channel W boson exchange from the weak charged current. The charged-current vertices pair the incoming neutrino $k$ with the outgoing Electron $p^{\prime }$ and the incoming Electron $p$ with the outgoing neutrino $k^{\prime }$. The exchanged momentum has square $u=(k-p^{\prime }{)}^2$. With charge flow from the first vertex to the second, the exchanged boson is $W^{+}$. The corresponding panels show both topologies; no photon or Higgs diagram survives the stated neutrino assumptions.

Only a $t$-channel Z boson connects the muon-neutrino line to the Electron line. Its endpoints are the weak neutral current vertices $Z{\nu }_{\mu }{\bar{\nu }}_{\mu }$ and $Z{e}^{-}{e}^{+}$. A W boson couples ${\nu }_{\mu }$ to a muon, so it cannot produce an additional elastic charged-current diagram with external electrons. There is no photon-neutrino vertex, and a strictly massless Standard Model neutrino has no Higgs Yukawa vertex. There is one diagram.

Flavour issue in the printed effective interaction. The displayed charged-current factor uses an Electron field even for the muon-neutrino term. Taken literally, this introduces an extra ${\nu }_{\mu }e$ charged-current interaction, contradicting the Standard Model diagrams in part (a). In the Standard Model that factor uses the charged lepton of the same flavour; for Electron scattering the charged-current contribution is therefore present only for ${\nu }_e$. The weak flavour selection in neutrino-electron scattering fixes the physically consistent assignment used below. The literal printed consequence is given afterwards, so the discrepancy is explicit.

At leading order in the Electron mass and in the four-fermion regime, define ${\Gamma }_L^{\alpha }={\gamma }^{\alpha }(1-{\gamma }^5)$. The weak neutral current yields the scattering amplitudeAn overall amplitude phase has no observable effect. The Fermi interaction is valid when the exchanged invariants are small compared with the squared weak-boson masses.

Average over the two initial Electron spin states, and sum over final spins. There is no averaging over a sterile right-handed incoming neutrino. Using the fermion spin sum, the averaged square isTo display the trace contraction, defineThe supplied gamma matrix trace identities give the two tensors $8(S(k^{\prime },k)+iE(k^{\prime },k))$ and $4[(c_V^2+c_A^2)S(p^{\prime },p)+2i{c}_V{c}_{A}E(p^{\prime },p)]$. Symmetric-antisymmetric mixed contractions vanish. Set $X=(k^{\prime }\cdot p^{\prime })(k\cdot p)$ and $Y=(k^{\prime }\cdot p)(k\cdot p^{\prime })$. The remaining contractions are $S(k^{\prime },k)S(p^{\prime },p)=2(X+Y)$ and $E(k^{\prime },k)E(p^{\prime },p)=-2(X-Y)$, using the stated negative epsilon-contraction sign. Their sum givesThe massless Mandelstam variables satisfy $s+t+u=0$, and the two products are $s^2/4$ and $u^2/4$, respectively. Therefore

In the centre-of-momentum frame, momentum conservation gives $E=\sqrt{s}/2$. If $z=\cos \theta$ is the angle cosine between $p$ and $p^{\prime }$, then $u=-s(1+z)/2$. The massless two-body Lorentz-invariant phase space is $d{\rho }_2=d\Omega /(32{\pi }^2)$, while the flux is $2s$. ThusIntegrate ${\int }_{-1}^{1}dz=2$ and ${\int }_{-1}^1(1+z{)}^{2}dz=8/3$:Consequently the intended coefficients areThis is the massless neutrino-electron contact cross-section.

For completeness, the literal printed muon-neutrino charged-current term would instead shift both couplings by one, exactly as in part (c):For arbitrary real $c_V,c_A$, its linear terms cannot be represented by the requested quadratic form with coupling-independent constants. The Standard Model result and the literal printed interaction are not the same problem.

For Electron-neutrino scattering the weak charged current adds an Electron left-current. The supplied Fierz rearrangement has a minus sign for matrix components; the extra minus from ordering the external fermions cancels it. Accordingly the charged contribution is constructive in the neutrino-current/Electron-current ordering used in part (b):Thus ${\mathcal{M}}_e$ is obtained from the physically consistent ${\mathcal{M}}_{\mu }$ by $c_V\mapsto c_V+1$, $c_A\mapsto c_A+1$. The right-Electron coupling $c_V-c_A$ is unchanged, while the left-Electron coupling $c_V+c_A$ increases by two. This is the charged-current shift in neutrino-electron scattering.

For the antineutrino, choose the external phase convention in whichThe reversed ordering of the antineutrino spinors implements crossing symmetry; the displayed overall sign can be changed by an external-state phase. Its neutrino trace has the opposite antisymmetric part, so the $s^2$ and $u^2$ weights exchange:The phase-space integration already performed in part (b) immediately givesEquivalently, with $a=c_V+1,b=c_A+1$, these are $G_F^{2}s(a^2+ab+b^2)/(3\pi )$ and $G_F^{2}s(a^2-ab+b^2)/(3\pi )$. This is the neutrino-antineutrino interchange of chiral scattering weights. It uses the contact limit, not the full resonant $s$-channel propagator.

If the printed flavour-mismatched interaction is retained literally, its ${\nu }_{\mu }e$ amplitude already equals the displayed ${\nu }_{e}e$ amplitude. There is then no further flavour shift between those two processes, although antineutrino crossing still exchanges the weights. The distinction identified in part (b) is therefore necessary for the Standard Model interpretation of all three processes.

Use the weak hypercharge normalization $Q=T_3+Y$. Write $U(x)\in SU(2{)}_L$ and the hypercharge transformation as $e^{i\beta (x)Y}$. A left-handed doublet transforms as ${\psi }_L\mapsto e^{i\beta Y_L}U{\psi }_L$, while a right-handed singlet transforms as ${\psi }_R\mapsto e^{i\beta Y_R}{\psi }_R$. The Higgs doublet transforms as $\varphi \mapsto e^{i\beta /2}U\varphi$.

The representations per fermion generation are:

There is no right-handed neutrino in the minimal Standard Model. If such a sterile singlet were added, its hypercharge would be zero. The two hypercharge columns give equivalent normalizations, not different physical assignments. This is the electroweak representation and hypercharge table.

Yukawa interactions use $\varphi$ for down-type quarks and charged leptons, but $\tilde{\varphi }=i{\sigma }^2{\varphi }^{\ast }$ for up-type quarks. The conjugate doublet transforms as $e^{-i\beta /2}U\tilde{\varphi }$, using the pseudoreality of the $SU(2)$ fundamental representation. For example, the hypercharge sums in the down, charged-lepton, and up terms areThe doublet indices contract between ${\bar{\psi }}_L$ and the scalar, and colour indices contract in the quark terms. This verifies gauge invariance.

The indices $i,j=1,2,3$ label fermion generations, specifying which left-handed family and right-handed family are coupled. They are not weak-isospin indices. Each fermion type has its own complex Yukawa matrix. After symmetry breaking these matrices determine masses; the mismatch of the two quark left-handed mass rotations produces the CKM matrix. Left fields are weak doublets, right fields are weak singlets, and the scalar is a hypercharge-$1/2$ doublet in the stated convention.

Absent in the minimal Standard Model. The single charged-lepton Yukawa matrix supplies both the fermion mass matrix and the Higgs coupling, so the same left/right rotations diagonalize both. There is no off-diagonal $h\tau \mu$ vertex. With massless neutrinos, the separate charged-lepton family numbers prevent this transition perturbatively. If neutrino masses and mixing are added, loop-induced charged-lepton flavour violation can occur but is extremely suppressed; a sizeable rate would require an additional source of flavour violation. Part (c) provides one such source.