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If a chiral fermion triplet uses one normalized linear combination of two neutral leptons, adding the perpendicular combination as a gauge singlet completes their canonical kinetic terms. The two combinations are related to the original fields by a real orthogonal matrix. Hence their summed kinetic quadratic form is exactly that of the two original fields, with no mixed terms. The singlet carries the same chirality; no right-handed component is created for a purely left-handed massless neutrino.

Use the current convention in the question, whose vertex prefactor is $g/(2\cos {\theta }_W)$. The neutral-current vector and axial couplings are thenFor a charged lepton, $T_3=-1/2$, $Q=-1$, givingFor an active neutrino, $T_3=1/2$, $Q=0$, so $c_V=c_A=1/2$. The right-handed weak singlet contribution is purely vectorlike in the charge term; combining it with the left-handed contribution gives precisely this Dirac-current form. These coefficients would be halved if the vertex prefactor were $g/\cos {\theta }_W$ instead.

Let $p=k+k^{\prime }$ be the Z boson momentum and $k,k^{\prime }$ the outgoing massless lepton momenta. The decay amplitude isUse the fermion spin sums $\sum u\bar{u}=/k$ and $\sum v\bar{v}=/k^{\prime }$. Anticommutation of ${\gamma }_5$ with the Dirac gamma matrices gives the symmetric part of the trace,The vector-axial cross term is antisymmetric in $\mu ,\nu$, and hence drops out of the massive vector polarization sum. The massless equations also give $p_{\mu }{L}^{\mu \nu }=0$: in the symmetric part, the first two contractions sum to $p^{\nu }k\cdot k^{\prime }$, canceling the last; the antisymmetric term vanishes because $p=k+k^{\prime }$. The longitudinal $p_{\mu }{p}_{\nu }/m_Z^2$ term therefore makes no contribution.

Since $2k\cdot k^{\prime }=m_Z^2$, contraction with $-g_{\mu \nu }$ gives $4(c_V^2+c_A^2)m_Z^2$. Thus the massless fermion Z decay spin sum, including all three initial polarizations, isTo use this fully summed expression for the decay width of one unpolarized Z boson, apply the massive-vector spin average, dividing by three. The width of each fixed polarization is the same after integration over all directions, so that procedure also gives the physical single-particle width. Omitting the division would sum the widths of three initial states.

For two massless distinguishable final particles, the two-body Lorentz-invariant phase space isConsequentlyThe tree-level electroweak symmetry breaking relation is $m_W=m_Z\cos {\theta }_W$. Combining it with the Fermi constant normalization givesFor an active neutrino, the vertex itself projects out inactive chiralities, so no additional helicity factor is needed. This is the tree-level massless leptonic Z decay width; lepton masses and radiative corrections have been excluded.

Use the electroweak hypercharge normalization $Q=T_3+Y$, rather than $Q=T_3+\hat{Y}/2$. In the latter convention all tabulated hypercharges would be doubled. The electroweak representation and hypercharge table for one minimal Standard Model generation isThe quarks are colour triplets, the leptons are colour singlets, and all listed right-handed fields are weak singlets. A sterile right-handed neutrino, if added, has representation $(1,1{)}_0$ and changes none of the anomaly traces.

On each left weak doublet, $T_3=\mathrm{diag}(1/2,-1/2)$. The electric charges are therefore $1/6\pm 1/2=(2/3,-1/3)$ for $Q_L$ and $-1/2\pm 1/2=(0,-1)$ for $L_L$. On each right singlet $T_3=0$, so its electric charge equals its hypercharge. These are the observed quark and lepton charge assignments.

For Standard Model anomaly cancellation, traces must include both components of each doublet and all three colours. Every right weak singlet has $T_3^2=0$, givingOn the left doublets, $T_3^2=I/4$, while ${\mathrm{tr}}_L{T}_3=0$ separately in each doublet. ConsequentlyThe linear trace isEquivalently, ${\mathrm{tr}}_{L}Q=3(2/3-1/3)+(0-1)=0$. This proves cancellation of the mixed weak-hypercharge gauge anomaly within one generation.

The one-generation hypercharge anomaly traces for the cubic condition areThus both required anomaly differences vanish generation by generation. The right-handed subtraction is important: counting all physical fields with the same chirality sign would not give the cubic cancellation.

The elements have several complementary origins: primordial light nuclei, stellar and explosive fusion products, and nuclei built by neutron capture. Chemical enrichment also requires their ejection, rather than merely their production inside a star. Nucleosynthesis and transport into the interstellar medium must therefore be considered together.

Hydrogen and much of the helium are primordial. During Big Bang nucleosynthesis, a cooling, expanding plasma assembled surviving neutrons and protons into deuterium, helium-3 and especially helium-4, with small abundances of lithium-7 or its beryllium-7 progenitor. Unstable nuclei at mass numbers five and eight, together with the rapidly falling density and short available time, obstructed sustained production of heavier nuclei. The carbon, oxygen and heavier material in later generations consequently require other sites. Cosmic rays fragment heavier interstellar nuclei and contribute importantly to the light elements lithium, beryllium and boron, which ordinary stellar burning tends to destroy.

Burning stages depend on initial stellar mass. Stellar nuclear fusion can release energy when the products have greater nuclear binding energy than the reactants. The resulting stellar thermostat links nuclear burning to hydrostatic equilibrium: when fuel exhaustion lowers heating, contraction raises the central temperature until another fuel can ignite, if the core becomes hot enough. Higher nuclear charges bring larger Coulomb barriers, so later charged-particle burning generally requires progressively higher temperatures. Whether the core reaches them depends on its mass, electron degeneracy pressure, mixing and envelope loss.

An approximately solar-composition $1M_{\odot }$ star first spends most of its life on the main sequence, converting hydrogen into helium mainly through the proton–proton chain. The net conversion is four protons into one helium-4 nucleus, with positrons and neutrinos; the small CNO cycle contribution redistributes CNO catalyst abundances but is not a net source of carbon at this stage. Exhaustion of central hydrogen leaves a helium-rich core. A surrounding hydrogen-burning shell continues to add helium, while the core contracts and the envelope expands into a red giant.

The low-mass helium core becomes supported substantially by electron degeneracy pressure. At roughly $10^8\mathrm{K}$, the Triple-alpha process ignites, first making transient beryllium-8 and then carbon-12 through the Hoyle state. Because degeneracy initially weakens the usual expansion response, this ignition is a helium flash. It heats and lifts the degeneracy of the core, leading to a stable phase of core helium burning. The Triple-alpha process makes carbon; competing carbon-12 alpha capture makes oxygen:After central helium exhaustion, the star develops a carbon-oxygen core, a helium-burning shell and a more external hydrogen-burning shell. On the Asymptotic giant branch, recurrent thermal pulses of an asymptotic-giant-branch star arise from unstable helium-shell burning. Stellar dredge-up can bring processed material toward the surface, with its efficiency depending on mass and composition. The eventual stellar wind removes the envelope; the hot exposed core can illuminate a planetary nebula, and the remnant becomes a carbon-oxygen white dwarf. A usual isolated $1M_{\odot }$ star does not ignite carbon burning. In particular, an element synthesized in its core need not all escape: much of the carbon and oxygen remains locked in the white dwarf.

An approximately $5M_{\odot }$ star burns central hydrogen predominantly through the CNO cycle, usually in a convective core. The CNO nuclei act as catalysts; the slow nitrogen-14 proton-capture step tends to accumulate nitrogen-14 while converting hydrogen into helium. After central fuel exhaustion, a hydrogen-burning shell surrounds the contracting helium core. Unlike the $1M_{\odot }$ case, core helium burning normally begins before strong electron degeneracy, without a classical helium flash. The same Triple-alpha process and carbon-12 alpha capture create a carbon-oxygen core.

The $5M_{\odot }$ star then enters the Asymptotic giant branch, with hydrogen and helium burning in shells. Repeated thermal pulses and stellar dredge-up can enrich its envelope with freshly made carbon and s-process nuclei. Hot-bottom burning can occur when the base of the convective envelope is sufficiently hot: the CNO cycle processes dredged-up carbon into nitrogen, so a massive AGB star need not become carbon-rich despite making carbon below its envelope. Helium-shell reactions on neon-22 can also provide neutrons for the s-process. For an ordinary $5M_{\odot }$ model, strong envelope loss generally precedes carbon ignition, leaving a carbon-oxygen white dwarf. The exact threshold for carbon ignition and the size of these yields depend on metallicity, mass loss and mixing; they are not universal integers fixed solely by initial mass.

A $20M_{\odot }$ star reaches much more advanced stages. It starts with convective-core CNO cycle hydrogen burning, followed by core helium burning. After central helium exhaustion, contraction can ignite carbon burning at temperatures approaching $10^9\mathrm{K}$. Carbon-fusion channels make products including neon, sodium and magnesium. Neon burning then uses photodisintegration of neon-20 to release oxygen-16 and an alpha particle, followed by capture of that alpha particle on another neon-20 nucleus; the combined network releases energy and produces magnesium-24. Oxygen burning makes silicon-group nuclei through channels such asAt still higher temperature, silicon burning rearranges nuclei by photodisintegration and captures of protons, neutrons and alpha particles. It is not principally direct fusion of two silicon nuclei. Networks approach quasi-equilibrium and then nuclear statistical equilibrium, producing an iron-group core whose precise mixture depends on the electron fraction in stellar matter. Outer shells retain different stages of burning, giving an approximately layered structure. Increasing stellar neutrino energy loss makes advanced burning stages much shorter than hydrogen burning.

Near the iron group, nuclear binding energy per nucleon is close to its maximum. Further fusion cannot provide the sustained energy source required to support the core. As the core grows, electron capture reduces electron-pressure support and photodisintegration absorbs energy, assisting gravitational collapse. A successful core-collapse supernova ejects some of the envelope and processed shells, and leaves a neutron star or black hole. The shock also causes explosive burning, changing the pre-existing abundances. A $20M_{\odot }$ initial mass does not uniquely fix the explosion or remnant: envelope loss, rotation, composition and fallback all matter, and some collapses fail to eject much material.

Ejection determines the enrichment of the interstellar medium. Low- and intermediate-mass stars return helium, CNO-processed matter, dredged-up carbon and s-process products through stellar winds and envelope expulsion. Massive-star winds can expose hydrogen- or helium-burning products before collapse. Successful core-collapse supernovae eject substantial oxygen and other alpha-capture products, together with some iron-group matter formed in hydrostatic and explosive burning. Material that falls back or remains in a stellar remnant is withheld from the interstellar medium. Subsequent mixing, cooling and star formation distribute the escaped material into new stars. The distinction between production and ejection is why an internal burning inventory alone is not a stellar yield.

Binary stars provide an additional major iron-group source. In a Type Ia supernova, a carbon-oxygen white dwarf in a binary undergoes thermonuclear disruption after an appropriate accretion or merger history. Explosive carbon and oxygen burning makes intermediate-mass and iron-group nuclei. Much initially synthesized nickel-56 decays through cobalt-56 into iron-56; this radioactive decay also powers the supernova light curve. Such explosions have delay times and yields different from those of massive-star collapse, and help explain why iron enrichment need not track oxygen enrichment exactly.

Neutron capture builds much of the material beyond iron. Charged-particle reactions encounter increasing Coulomb barriers as nuclear charge grows, but a neutral neutron does not encounter that barrier. Neutron capture raises mass number, and subsequent beta-minus decay can turn a neutron into a proton and raise atomic number. The relative capture and decay times select different paths through the nuclear chart.

In the slow neutron-capture process, most unstable intermediates decay before the next capture, so the path stays near stable nuclei. In Asymptotic giant branch stars, important neutron sources areThe carbon-13 source operates in suitable partially mixed layers, while the hotter helium-shell pulses can activate the neon-22 source. Captures on pre-existing seed nuclei produce elements such as strontium and barium and, under suitable conditions, lead. Stellar dredge-up and stellar winds return these products to the interstellar medium. Massive stars also have a weak s-process, notably using neon-22 during helium- and carbon-burning phases; it need not produce the same abundance range as the main AGB contribution.

In the rapid neutron-capture process, very high neutron densities permit many captures before typical beta-minus decays. The path moves to neutron-rich nuclei, which later decay toward stability, producing heavy nuclei including lanthanides and actinides. Neutron-rich ejecta from neutron star mergers provide an established site: a merger can eject dynamical material and disk outflows, and radioactive decay of the newly made nuclei powers a kilonova. Special neutron-rich outflows associated with some massive-star collapses are additional candidate sites. It would be unjustified to assume that every ordinary core-collapse supernova produces the full heavy r-process pattern; the required neutron richness and expansion history are restrictive.

The two neutron-capture processes do not make every isotope beyond iron. Some proton-rich isotopes bypass their main paths. Gamma-process nucleosynthesis, for example, uses photodisintegration of pre-existing heavy seed nuclei in hot explosive material. Other proton-rich reaction paths can matter in suitable environments. Thus “heavier than iron” describes a family of production mechanisms rather than a single further ordinary fusion stage.

Binaries change both stellar evolution and the delivery of elements. Roche-lobe overflow, wind transfer and mergers can strip a donor, rejuvenate a mass-gaining star, alter core and envelope masses, and change the timing and character of an eventual explosion. Binary stripping can remove the hydrogen envelope of a massive core-collapse supernova progenitor. Accretion onto a white dwarf can cause a classical nova, ejecting hydrogen-burning products and particular isotopes while ordinarily leaving the white dwarf intact; this differs from the destructive Type Ia supernova. Compact binary stars can lose orbital energy through gravitational radiation and merge, providing the neutron star merger route to heavy r-process material. Consequently chemical enrichment cannot be inferred entirely from isolated-star tracks, even when a simplified population calculation treats binary components as evolving independently.

Use $Q=T_3+Y$ and $T_a={\tau }_a/2$. For each generation $r=e,\mu ,\tau$, the leptons areBoth are color singlets. The conjugate antilepton fields transform in the conjugate representations and have opposite hypercharge; they are already included by the conjugate fields in the Lagrangian. The two components of $L_r$ have electric charges $0,-1$, and $R_r$ has charge $-1$. There is no right-handed neutrino representation in this field content. These assignments use the electroweak representation and hypercharge table in the same hypercharge normalization as the scalar doublet.

For a doublet of hypercharge $Y$, defineThe $SU(2)$ term is absent for a singlet. Under the local transformation $\psi \mapsto e^{iY\omega }U\psi$, the gauge fields transform so that $D_{\mu }\psi \mapsto e^{iY\omega }U{D}_{\mu }\psi$. Consequently the required gauge-covariant kinetic terms areThe fields of definite chirality are defined with $P_{L,R}=(1\mp {\gamma }_5)/2$. Since $\bar{L}$ has hypercharge $+1/2$, the Yukawa interactionis invariant: its hypercharges add to zero, and $\bar{L}\varphi$ contracts the two doublet indices. A bare charged-lepton mass violates these assignments. No renormalizable neutrino Yukawa term is available without a right-handed neutrino. A gauge-invariant scalar potential depending on ${\varphi }^{†}\varphi$ can select the nonzero vacuum, without altering these kinetic and Yukawa constructions.

There is an explicit local unitary gauge transformation whenever $r_{\varphi }=(|{\varphi }_1{|}^2+|{\varphi }_2{|}^2{)}^{1/2}\ne 0$:Thus, with the real field $\rho =\sqrt{2}{r}_{\varphi }-v$,This proves the polar coordinates for the electroweak Higgs doublet constructively. The gauge choice is local about the nonzero vacuum; it need not be nonsingular at zeros of the doublet.

The three identical-generation kinetic terms permit unitary flavor changes. A singular value decomposition gives $U_L^{†}{Y}_{\ell }{U}_R=\mathrm{diag}(y_e,y_{\mu },y_{\tau })$ with $y_r\ge 0$. Rotate both components of each left doublet together, and the right singlets independently. In this basis,The left/right cross terms combine into this Dirac mass. Since the neutrinos have no mass matrix, their simultaneous left rotation preserves a diagonal charged current. This is charged-lepton Yukawa matrix diagonalization; the neutrinos remain massless in the stated renormalizable model.

The scalar kinetic term givesIntroduce $s_W=g^{\prime }/\sqrt{g^2+g^{\prime 2}}$ and $c_W=g/\sqrt{g^2+g^{\prime 2}}$. The combinations $W^{+}=(A_1-i{A}_2)/\sqrt{2}$, $W^{-}=(W^{+}{)}^{\ast }$, $Z=c_W{A}_3-s_{W}B$ and $A_{\gamma }=s_W{A}_3+c_{W}B$ therefore diagonalize the quadratic terms:The factor $1/2$ applies to the real neutral field, not to the complex charged field. The massless photon corresponds to the unbroken generator $T_3+Y$, which annihilates the vacuum. This gives the electroweak doublet gauge-boson mass matrix and its diagonal spectrum.

Take metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and write $M=M_W$. The outgoing electron and antineutrino momenta are $k$ and $k^{\prime }$. Up to an irrelevant overall phase, the amplitude isA fermion spin sum givesAnticommuting ${\gamma }_5$ through the massless momentum and gamma matrix, and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$, reduces the trace to twice the four-gamma trace minus twice its ${\gamma }_5$ counterpart. In particular its symmetric part isThe epsilon-tensor part is antisymmetric and contributes neither to the unpolarized polarization sum nor to the real linear polarization used below. There is no extra factor for selecting the neutrino helicity: the chiral vertex has already done so.

The external massless equations imply $p_{\mu }{J}^{\mu }=0$ for $p=k+k^{\prime }$, so $p_{\mu }{T}^{\mu \nu }=0$. With $k\cdot k^{\prime }=M^2/2$, averaging over the three initial spin states givesIndeed $-g_{\mu \nu }{T}_S^{\mu \nu }=16k\cdot k^{\prime }=8M^2$. This is a massive-vector spin average, not an average over the final spins.

In the rest frame, eliminating the momentum delta function and integrating the remaining radial delta function $\delta (M-2|\mathbf{k}|)$ reduces the two-body Lorentz-invariant phase space to $d{\Phi }_2=d\Omega /(32{\pi }^2)$. ThereforeThis derives the massless leptonic W decay width without invoking a precomputed decay formula.

For the specified polarization there is no initial spin average. Put $\mathbf{k}=(M/2)\hat{\mathbf{n}}$, ${\mathbf{k}}^{\prime }=-(M/2)\hat{\mathbf{n}}$ and ${\hat{n}}_z=\cos \theta$. The contraction selectsIt follows thatHere $\int {\sin }^2\theta d\Omega =8\pi /3$. The total rate agrees with the spin-averaged result, but its normalized polar distribution is $(1/{\Gamma }_z)d{\Gamma }_z/d\cos \theta =3(1-{\cos }^2\theta )/4$ instead of the isotropic unpolarized distribution. This is polarized massless leptonic W decay.

In the specified dust model $H_0=2/(3t_0)$, so ${\rho }_{\mathrm{crit}}=m_{\mathrm{pl}}^2/(6\pi t_0^2)$ and $n_B={\Omega }_{B0}{\rho }_{\mathrm{crit}}/m_B$. Photons and the six neutrino/antineutrino helicity states giveUsing the supplied GeV estimates literally givesThe rounded time/temperature conversions are unusually crude: converting $10$ Gyr to $4.8\times 10^{41}{\mathrm{GeV}}^{-1}$ and $3$ K to $2.6\times 10^{-13}$ GeV instead gives $Y_B\sim 10^{-10}$. Either set gives the same order-of-magnitude mass conclusion below, but the numerical abundance should identify which supplied units are used.

For a stable relic, the relevant comparison is energy density, not merely particle number. The matter density left over after the baryons is at most nine times the baryon density here. ConsequentlyPut $x=(m_{Y\mathrm{nr}}/m_Y{)}^{1/3}$ and use $Y_Y\simeq C_Y{x}^{3/2}{e}^{-x}$ with $C_Y\simeq 10^{-3}$. The condition becomesFor the literal supplied numbers the right side is about $30.5$, giving $x\gtrsim 25.6$ and $m_Y\lesssim 40$–$50$ GeV. With the physically consistent conversions the estimate is about $60$ GeV. The intended rough conclusion is therefore a sufficiently suppressed branch with $m_Y$ no larger than order $10^2$ GeV, within the toy cosmological particle freeze-out law. Comparing the two number yields alone gives a similar rough scale but misses the mass weighting. This is a thermal relic density bound for a tunable fermion.

For $m_Y\gtrsim m_{Y\mathrm{nr}}$, relativistic decoupling would instead leave $Y_Y\simeq 135\zeta (3)g_Y/(8{\pi }^4{g}_{\ast s})\simeq 4\times 10^{-3}$, incompatible with the matter-density bound for such heavy stable particles. These are conditional bounds on a populated, stable thermal relic. A short lifetime or an initially suppressed nonthermal population can change the conclusion; the supplied information alone does not constrain every possible production and decay history.

Assume $\lambda >0$ and $v>0$. The minima of the scalar potential have $|\varphi |=v$. Choose the vacuum expectation value ${\varphi }_0=v{e}_3$. Its stabilizer subgroup consists exactly of matrices $\mathrm{diag}(R_2,1)$ with $R_2\in O(2)$, soThe continuous spontaneous symmetry breaking is $SO(3)\to SO(2)$ and has two broken generators. For the full orthogonal group, the cross-product notation requires the gauge connection to transform in the adjoint, as an axial vector: $R[A{]}_{\times }{R}^{-1}=[(\det R)RA{]}_{\times }$. The scalar field transforms as the ordinary vector $R\varphi$. This axial transformation of orthogonal gauge connections makes the stated covariant derivative transform correctly even for $\det R=-1$; assigning an ordinary-vector transformation to the connection would fail.

Near this nonzero vacuum, unitary gauge removes the two angular Goldstone bosons and gives $\varphi =(0,0,v+h)$. PutThese are the physical charged gauge bosons, neutral massless gauge boson, and radial scalar field. Directly substituting into the original gauge field strengths givesThe scalar field kinetic term has no radial-vector cross term, because ${\partial }_{\mu }\varphi$ and ${\mathbf{A}}_{\mu }\times \varphi$ lie in perpendicular internal directions. The complete physical gauge-scalar Lagrangian becomesFor each real vector the mass term is $m^2{A}_{\mu }{A}^{\mu }/2$, while for the complex pair it is $m_W^2{W}_{\mu }^{+}{W}^{-\mu }$. Hence the adjoint triplet Higgs spectrum isThe radial scalar also has $m_h^2=\lambda v^2$ in the question's potential normalization. The two lost scalar modes supply the longitudinal polarizations of the massive vectors through the Higgs mechanism.

For the lepton kinetic terms let $P_L=(1-{\gamma }^5)/2$, $P_R=(1+{\gamma }^5)/2$, and denote the left neutral triplet entry by $n_L=c_{\alpha }{\nu }_L+s_{\alpha }{N}_L$. The missing singlet is the perpendicular combinationIts overall sign is immaterial. The real orthogonal matrix $\left(\begin{array}{cc}{c}_{\alpha }& s_{\alpha }\\ -s_{\alpha }& c_{\alpha }\end{array}\right)$ is orthogonal, so the sum of the kinetic terms of $n_L$ and $L^S$ is ${\bar{\nu }}_{L}i/\partial {\nu }_L+{\bar{N}}_{L}i/\partial N_L$ with no mixed terms. Combining with the right triplet gives ordinary Dirac field kinetic terms for $E^{+},N,e$, and only a left-handed kinetic term for the massless neutrino. This is orthogonal completion of mixed neutral lepton kinetic terms. The heavy charged field is $E^{+}$, as printed in the PDF, not the TeX's $N^{+}$.

To compute the charge commutator, the canonical anticommutation relations implyThe quartic terms cancel on reordering the fermions. Thus well-defined normal-ordered integrated bilinears obey fermionic bilinear charge algebra, $[Q_C,Q_D]=Q_{[C,D]}$. Left/right mixed commutators vanish since $P_L{P}_R=0$.

In either triplet, writeThe displayed current has a factor two relative to the projected bilinears, since $1\mp {\gamma }^5=2P_{L,R}$. Consequently its charges are $T^{+}=2(Q_K^L+Q_K^R)$ and $T^{-}=(T^{+}{)}^{†}$. The neutral combinations have canonical anticommutators because $c_{\alpha }^2+s_{\alpha }^2=1$. Applying the matrix commutator giveswhere $Q_{\ell }=e\int :(E^{+†}{E}^{+}-e^{†}e):$ is the electromagnetic charge in the lepton sector. The neutrino, heavy neutral lepton and singlet contribute zero. This establishes triplet charged-current closure on electromagnetic charge, with the requested proportionality independent of normalization choices for the weak Noether charges. The PDF correctly writes $J^{-}=(J^{+}{)}^{†}$; the TeX repeats $J^{+}$ incorrectly. Equal-time algebra does not require conservation of the separate fermionic weak currents after symmetry breaking. For full gauge-theory charges, the corresponding charged-boson contributions must also be included.

Use the source convention $Q=T_3+Y$, with electric charge measured in units of the positron charge. Expanding the covariant derivative in the fermion kinetic term gives interaction $-\bar{f}{\gamma }_{\mu }(g{T}_3{A}_3^{\mu }+g^{\prime }Y{B}^{\mu })f$. Substituting the given neutral-field rotation, the coefficient of $Z^{\mu }$ isThe photon coefficient is $g{s}_W(T_3+Y)=eQ$, with $e=g{s}_W=g^{\prime }{c}_W$. For a Dirac field, $T_3$ acts only on its left-handed component. ThusThe weak neutral current interaction consequently has the requested form, withThe electroweak representation and hypercharge table gives the following values for one generation:

These neutral-current vector and axial couplings use the source's vertex prefactor $g/(2c_W)$. They are twice the coefficients used with prefactor $g/c_W$; the latter convention must not be mixed into this calculation. For the neutrino, the chiral projector removes the inactive right-handed component automatically.

Set $G=g/(2c_W)$ and suppress the species label. For outgoing momenta $k_1,k_2$, the amplitude is, up to an irrelevant phase,The final-state spin sum is the Dirac trace of $/k_1{\gamma }^{\mu }(c_v-c_a{\gamma }^5)/k_2{\gamma }^{\nu }(c_v-c_a{\gamma }^5)$. Anticommutation of ${\gamma }^5$ with the gamma matrices makes its symmetric partThe vector-axial cross term is proportional to the antisymmetric Levi-Civita symbol, so it contracts to zero with the symmetric massive vector polarization sum $-g_{\mu \nu }+p_{\mu }{p}_{\nu }/M_Z^2$. Since $k_1^2=k_2^2=0$, the symmetric tensor is transverse to $p$ as well. Only $-g_{\mu \nu }$ remains, andMultiplying by $G^2$ proves the massless fermion Z decay spin sum:

For the decay width, average over the three initial polarizations. Rotational invariance makes the integrated width the same for each initial spin state, so the massive-vector spin average is legitimate even for a single prepared polarization. In the rest frame, the delta functions in two-body Lorentz-invariant phase space leaveThis follows by setting ${\mathbf{k}}_2=-{\mathbf{k}}_1$ and integrating $\delta (M_Z-2|{\mathbf{k}}_1|)$; the radial integration supplies a factor $1/2$. HenceFor a quark species multiply by three colors, with strong-interaction radiative corrections when comparing with data. For one effectively massless active neutrino, $c_v=c_a=1/2$, so ${\Gamma }_{\nu }=g^2{M}_Z/(96\pi c_W^2)=G_F{M}_Z^3/(12\sqrt{2}\pi )$.

The total resonance width contains visible charged-lepton and hadron decays plus $N_{\nu }{\Gamma }_{\nu }$. Thus the measured invisible contribution determinesEarly resonance measurements supported three light active neutrino species, with experimental uncertainties; the OPAL analysis of its 1989 data explicitly compares the three- and four-species predictions. Under the question's assumption that every possible family has a massless neutrino with ordinary weak couplings, this constrains the number of such families. The inference does not exclude additional neutrinos heavier than $M_Z/2$ or gauge singlet sterile states with no ordinary $Z$ coupling. This is light-neutrino counting from the Z width.

Let $\epsilon =i{\sigma }_2=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, so ${\epsilon }^{-1}=-\epsilon$. Direct multiplication of the Pauli matrices gives $\epsilon {\sigma }_i^{\ast }\epsilon ={\sigma }_i$, equivalently $\epsilon {\sigma }_i^{\ast }{\epsilon }^{-1}=-{\sigma }_i$. For $U=\exp (i{\theta }_i{\sigma }_i/2)\in SU(2)$ this impliesTherefore ${\varphi }^c=\epsilon {\varphi }^{\ast }$ transforms as the same weak fundamental representation doublet as $\varphi$, while complex conjugation reverses its hypercharge to $-1/2$. This is the Higgs conjugate doublet. The Yukawa coupling product $\bar{L}{\varphi }^c{u}_R$ is an $SU(2)$ singlet, and its hypercharges add toIt is consequently gauge-invariant and can supply an up-quark mass.

In unitary gauge, ${\varphi }^c=((v+H)/\sqrt{2},0{)}^T$. Writing $y_u$ for the full arbitrary coefficient of the Yukawa coupling term, its physical form isA one-generation field rephasing makes the mass positive, and its Higgs boson coupling is $m_u/v$. The printed Yukawa expression has an additional trailing $m$ after $R^{+}$ which is never defined. If it denotes a constant, absorb it into $y_u=f^{+}m$; in the conventional dimensionless Yukawa notation without that factor, $y_u=f^{+}$. It does not affect the gauge-transformation argument.

For three families, use a singular value decomposition of the two mass matrices: $U_{uL}^{†}{M}_u{U}_{uR}=D_u$ and $U_{dL}^{†}{M}_d{U}_{dR}=D_d$, with positive diagonal entries. The original left-handed weak current then becomes a current between mass eigenstates withA complex $3\times 3$ matrix has eighteen real parameters, and $U^{†}U=I$ imposes nine independent real conditions, leaving nine. A generic unitary matrix can be built from three complex plane rotations and three diagonal phases: each plane rotation supplies one angle and one phase. Thus these are three rotation angles and six phases. Rephase each of the six Dirac quark mass eigenfields, applying the same phase to its two chiralities so the diagonal masses stay real. A common phase changes no entry of $V$, leaving five effective rephasings. Removing five of the six phases leaves three angles and one physical phase. This CKM parameter counting assumes generic nondegenerate masses; degeneracies give additional basis freedom.

Exactly massless neutrinos have an arbitrary unitary flavor rotation that changes no mass term. After diagonalizing the charged-lepton mass matrix, choose the neutrino basis with the same left-handed rotation. The apparent lepton charged-current mixing matrix becomes the identity. Thus mixing of exactly massless neutrinos is removable; it is not an analogue of physical CKM mixing.

For the requested CP symmetry calculation, write $j_{ij}^{\mu }={\bar{u}}_i{\gamma }^{\mu }(1-{\gamma }^5)d_j$. The supplied charge conjugation transformations, with the fermion-reordering sign, giveFor $\Gamma ={\gamma }^{\mu }(1-{\gamma }^5)$, use $C({\gamma }^5{)}^T{C}^{-1}={\gamma }^5$ and the given gamma matrix relation to obtainUnder parity, the current acquires the Lorentz parity matrix and $1+{\gamma }^5$ becomes $1-{\gamma }^5$. Under charge conjugation the charged vector also changes sign and becomes its conjugate. The two minus signs cancel, and the parity matrices cancel in the contracted operator. ConsequentlyWith $k=g/(2\sqrt{2})$, the interaction is $-k\sum (V_{ij}{O}_{ij}+V_{ij}^{\ast }{O}_{ij}^{†})$. Thereforewith the transformed density evaluated at $x_P$. This is the CP transformation of a charged quark current.