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A thermally populated extra light neutrino species is relativistic during nucleosynthesis; its $0.001\mathrm{eV}$ mass is negligible at MeV temperatures. One neutrino-antineutrino species adds $(7/8)\times 2=7/4$ to $g_{\ast }$. Thus the expansion rate increases by $\sqrt{12.5/10.75}$ and the freeze-out temperature by $(12.5/10.75{)}^{1/6}$. Earlier freeze-out leaves more neutrons, while faster cooling leaves less time for decay. The helium-4 abundance increases. This assumes the new species was populated at the standard neutrino temperature; an unpopulated species would not contribute this extra density.

Put $s_W=\sin {\theta }_W$ and $c_W=\cos {\theta }_W$. The left-handed doublet haswhile the right-handed charged-lepton singlet has $T_R^3=0$ and $Q_R=-1$. There is no weakly coupled right-handed neutrino in the stated field content. For a fermion $f$, write $g_L^f=T_{3L}^f-Q_f{s}_W^2$ and $g_R^f=T_{3R}^f-Q_f{s}_W^2$, with $g_R^{\nu }=0$ here. Since ${\bar{f}}_L{\gamma }^{\mu }{f}_L=\bar{f}{\gamma }^{\mu }{P}_{L}f$ and similarly for $R$,Consequently the neutral-current vector and axial couplings in the specified overall normalization areThese coefficients accompany $g/c_W$. A convention with $g/(2c_W)$ uses twice these vector and axial coefficients; mixing the two conventions would make the widths wrong by a factor of four. For the neutrino, $c_V-c_A{\gamma }^5=\frac{1}{2}{P}_L$, so the chiral projection is already built into the vertex.

For either massless final pair with momenta $k_1,k_2$, $p=k_1+k_2$, the invariant amplitude is, up to an irrelevant overall phase,The spin-summed fermion tensor isThe symmetric part, using the four-gamma trace, isThe vector-axial interference is antisymmetric in $\mu ,\nu$, so it drops out of the symmetric $Z$ polarization sum. Also $p_{\mu }{T}_f^{\mu \nu }=0$ for massless external fermions, by their Dirac equations; hence the $p_{\mu }{p}_{\nu }/m_Z^2$ part contributes zero. Using $k_1\cdot k_2=m_Z^2/2$ givesThe massive-vector spin average supplies the factor $1/3$ for the three spin states of the massive $Z$. The massless two-body Lorentz-invariant phase space is $d{\Phi }_2=d\Omega /(32{\pi }^2)$, and so $\int d{\Phi }_2=1/(8\pi )$. The spin-averaged amplitude above is angle-independent in the rest frame, givingThe massless leptonic Z decay widths for one charged-lepton flavor and one active neutrino flavor respectively areThere is no color factor for leptons and no extra final-state identical-particle factor for a particle-antiparticle pair. The neutrino projectors already exclude sterile helicity states, so no additional factor of two is needed. These are tree-level massless widths; electroweak radiative corrections and charged-lepton masses have not been included. If desired, using $G_F=g^2/(4\sqrt{2}{m}_W^2)$ and $m_W=c_W{m}_Z$ converts the common prefactor to $G_F{m}_Z^3/(12\sqrt{2}\pi )$.

For a quantitative thermal-relic prediction, assume one ordinary weakly interacting neutrino flavor: one populated neutrino helicity and one antineutrino helicity, negligible chemical potential, early thermal equilibrium, and relativistic decoupling around the MeV epoch. Equal neutrino and antineutrino populations imply no new lepton asymmetry. The Fermi-Dirac distribution is maintained while the weak interaction rate exceeds the expansion rate. After neutrino decoupling, the distribution free streams, its momentum scale falls as $a^{-1}$, and its comoving number is conserved. A mass of $1\mathrm{keV}$ is negligible at Big Bang nucleosynthesis temperatures, but the relic later becomes nonrelativistic. Its energy density then changes from radiation-like $a^{-4}$ scaling to matter-like $a^{-3}$ scaling. Stability prevents its removal by decay; decoupling while relativistic prevents a subsequent equilibrium Boltzmann suppression of its number.

At weak freeze-out, photons, electrons, positrons and all four neutrino flavors are relativistic and approximately at the same temperature. The effective number of relativistic energy degrees of freedom changes fromThe expansion rate at fixed temperature is larger by $\sqrt{50/43}\simeq 1.078$. Equating the supplied weak interaction time with the supplied age gives $T_f^3=\sqrt{g_{\ast }}$ when $T_f$ is measured in MeV. ThereforeEarlier cosmological weak freeze-out at a higher temperature leaves more neutrons. With negligible chemical potentials, the neutron-to-proton ratio in equilibrium is $r_f=\exp (-Q/T_f)$, where $Q=m_n-m_p\simeq 1.293\mathrm{MeV}$. If almost all surviving neutrons enter helium-4, the primordial helium mass fraction before neutron decay is $Y_f=2r_f/(1+r_f)$. Literal use of the given rate normalizations yieldsEquivalently, the leading response is $\delta Y_f/Y_f=[Q/(T_f(1+r_f))]\delta T_f/T_f$. This is a roughly one-percentage-point absolute increase in the helium fraction, rather than a one-percent relative increase.

The deuterium bottleneck delays nuclear burning, allowing free-neutron decay after freeze-out. If its end temperature is $T_N$, and ${\tau }_n$ is the neutron mean life, a sharper instantaneous-freeze-out estimate conserves total baryon number and givesThis formula evolves the neutron fraction, not just the neutron-to-proton ratio, because each neutron decay also creates a proton. After electron annihilation, let $r_T=(4/11{)}^{1/3}$. The radiation counts relevant to the age at fixed $T_N$ areConsequently $t_N^{(4)}/t_N^{(3)}\simeq \sqrt{3.36/3.82}\simeq 0.939$. Less time is available for neutron decay, adding to the helium increase. As an explicit illustrative completion, $T_N=0.07\mathrm{MeV}$ and ${\tau }_n\simeq 880\mathrm{s}$ give an increase of approximately $0.012$ in this toy model. The precise helium fraction requires the nuclear-burning temperature, baryon abundance and properly normalized weak rates; the question supplies only approximate time laws. In particular, the large no-decay fractions in the table must not be interpreted as precise predictions for the observed primordial abundance. The robust effect is extra relativistic energy raises the primordial helium fraction, with an absolute change of order $10^{-2}$ for one thermally populated flavor.

Before electron-positron annihilation in cosmology, the pair has two spin states, just as photons do, and the fermionic number integral gives $n_{\nu +\bar{\nu }}/n_{\gamma }=3/4$ at equal temperature. The factor follows by expanding $(e^x+1{)}^{-1}$ into an alternating exponential series: its $x^2$ integral is $3/4$ of the bosonic one.

After neutrino decoupling, cosmological entropy conservation in the photon-electron plasma changes its relativistic entropy count from $2+(7/8)4=11/2$ to $2$. The decoupled neutrinos do not share this heating. HenceThese counts include both the new neutrino and its antineutrino; another factor of two would double count them. This is the number density of a relativistic neutrino relative to photons, evaluated after photon heating.

They are nonrelativistic today, so the relic-neutrino energy density is dominated by rest mass:The number follows from the supplied critical density, without a Hubble-unit convention. It is conditional on the thermal history stated above; a sterile, colder or nonthermally produced state need not have this abundance.

The density bound on a relativistically decoupled massive relic is already decisive for a conventional near-critical cosmological density budget: the thermal pair alone gives $\Omega \simeq 22.4$. The same additional relativistic pair increases the effective number of neutrino species by one during nucleosynthesis and raises the primordial helium mass fraction, so its expansion effect supplies another constraint. Thus a stable, fully thermalized MeV-decoupling neutrino pair of this mass is incompatible with the usual cosmological density budget.

This conclusion is not a prohibition on every particle with that mass. Earlier decoupling followed by entropy dilution, a reduced production abundance, or interactions that change the assumed thermal history can alter both tests. At fixed mass, requiring this pair to contribute no more than the supplied critical density would demand a number density at most $1/22.4$ of the standard value, or a thermal temperature at most $(1/22.4{)}^{1/3}\simeq 0.355$ of the standard neutrino temperature. No numerical abundance or exclusion independent of these assumptions follows from the mass and equal particle-antiparticle counts alone.

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.

A collisionless relativistic species redshifts with $T_{\nu }\propto a^{-1}$. The coupled plasma instead obeys cosmological entropy conservation, $g_{\ast s}{T}^3{a}^3=\mathrm{constant}$, and is heated relative to a decoupled species as massive particles annihilate. The effective number of relativistic energy degrees of freedom $g_{\ast }$ and effective number of relativistic entropy degrees of freedom $g_{\ast s}$ coincide while all species share a common relativistic temperature. Use $g_d:=g_{\ast s}(T_{\mathrm{dec}})\simeq 100$ for the Standard Model bath, excluding the already decoupled extra species.

Just before ordinary neutrino decoupling, the coupled bath contains two photon states, four electron/Positron states and six ordinary neutrino/antineutrino states. ThereforeThe extra-neutrino temperature relative to that bath is then $(43/(4g_d){)}^{1/3}$. After the ordinary neutrinos decouple, electron/Positron annihilation heats only the photon bath: its entropy count changes from $2+(7/8)4=11/2$ to $2$. Both neutrino populations miss that last heating, soThis is the two-stage entropy dilution of an early-decoupled neutrino. Numerically $T_{\nu ,0}\simeq 6.8\times 10^{-14}\mathrm{GeV}$. Relative to ordinary relic neutrinos it is cooler by $(43/(4g_d){)}^{1/3}\simeq 0.48$.

Let $g_{\nu }$ count the extra neutrino's populated helicity and antiparticle states. For one conventional neutrino plus antineutrino species $g_{\nu }=2$, the same state count as photons. The Fermi-Dirac distribution supplies the $3/4$ number-density factor, soThus a two-state species has about $2.9\%$ of the photon number density. Stating $g_{\nu }$ explicitly also covers a different choice of populated states.

The weak charged current creates a left-handed neutrino and a right-handed outgoing positron in the massless limit. In the parent's rest frame their momenta are opposite, so these two fixed helicities give total angular-momentum projection of magnitude one along their flight axis. A spin-one W boson can supply that projection, and its leptonic width remains nonzero:A spin-zero pion cannot produce the same helicity configuration. The forbidden component is rescued only by a charged-lepton mass insertion that reverses its helicity, giving helicity suppression.

This can also be seen directly from the pion decay constant. Lorentz symmetry gives a vacuum-to-pion axial current matrix element proportional to $f_{\pi }{p}_{\pi }^{\mu }$. Contracting $p_{\pi }=k_e+k_{\nu }$ with the leptonic current givesusing the massless neutrino equation and $(/k_e+m_e)v_e=0$. Hence ${\mathcal{M}}_{\pi }\propto m_e$ and ${\Gamma }_{\pi }\propto m_e^2\to 0$. The W boson polarization vector is not constrained to be proportional to the total momentum, so its amplitude has no analogous compulsory mass factor.

Both the left-handed electron and the electron neutrino participate in weak charged currents through $W^{\pm }$, and both have weak neutral currents mediated by the Z boson. Their electromagnetic interactions differ: the electron has electric charge $-e$ and couples to the photon, while the neutrino is electrically neutral and has no tree-level electromagnetic coupling.

In the minimal massless-neutrino Standard Model, the active neutrino has only a left-handed particle state and a right-handed antiparticle state. The electron has both chiralities: its right-handed component has electromagnetic interactions and a weak neutral current, but no weak charged current. Its nonzero mass comes from a Yukawa coupling to the Higgs field; the minimal massless-neutrino model contains no right-handed neutrino for an analogous Dirac mass term.

Observed neutrino oscillations establish nonzero neutrino mass-squared differences. Atmospheric muon-neutrino disappearance depends on travel distance and energy, and solar electron neutrinos convert into other active flavors; reactor and accelerator oscillations provide complementary evidence. In propagation, the relative phase of two mass eigenstates is $\Delta m_{ij}^{2}L/(2E)$. If all masses were zero there would be no such relative phase. Thus at least two neutrino mass eigenstates are nonzero in the three-neutrino description with two independent measured splittings. A flavor state is a superposition of mass eigenstates, so the evidence does not assign a single definite mass to a flavor label.

The two tree-level Feynman diagrams have t-channel W boson exchange. The antiquark process reverses the fermion-flow arrows on the lower line.

Each vertex contributes $g/(2\sqrt{2})$ multiplying a chiral current. At momentum transfer $|q^2|\ll M_W^2$, the gauge-boson propagator is approximated by its momentum-independent term $g_{\alpha \beta }/M_W^2$, up to the overall phase. The longitudinal term does not contribute to these conserved massless currents. Thus the coefficient is $g^2/(8M_W^2)=G_F/\sqrt{2}$, giving the Fermi interaction. For the quark and antiquark amplitudes, apart from irrelevant overall phases,andThe low-energy operator form is ${\mathcal{L}}_{\mathrm{eff}}=-(G_F/\sqrt{2})(\bar{e}{\gamma }^{\alpha }(1-{\gamma }^5){\nu }_e)(\bar{u}{\gamma }_{\alpha }(1-{\gamma }^5)d)+\text{h.c.}$; its external spinor matrix elements are the displayed products.

For an unpolarized initial parton, average over its two spin states. There is no factor $1/2$ for the incoming active neutrino, which has one allowed helicity. The initial color average cancels the final color sum, so there is no surviving color multiplicity in this parton cross section. DefineThe symmetric and antisymmetric parts have zero cross contraction. Their separate contractions are $128[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)]$ and $128[(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)]$. Thus the given Levi-Civita symbol identity yieldsIncluding the coefficient $G_F^2/2$ and the initial-parton spin average givesMassless kinematics and $q=p-p^{\prime }=k^{\prime }-k$ imply $p\cdot k=p^{\prime }\cdot k^{\prime }=s/2$. From $y_q=k\cdot q/(k\cdot p)$, $p^{\prime }\cdot k=p\cdot k^{\prime }=s(1-y_q)/2$. ThereforeTo reduce the supplied phase-space formula, use the massless center-of-mass flux $2s$ and $d{\Phi }_2=d\Omega /(32{\pi }^2)$. This gives $d\sigma /d\Omega =\overline{|\mathcal{M}{|}^2}/(64{\pi }^{2}s)$. In that frame $y_q=(1-\cos \vartheta )/2$, so $d\Omega /d{y}_q=4\pi$. HenceThe antiquark channel's angular factor is a consequence of the chiral current and the reversed lower-line spinor order, rather than a color factor.