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A system identifying penetrating muons, often by tracking beyond an absorber that stops most other charged particles. Its efficiency and hadron punch-through background require calibration.

A muonic atom has a negative muon bound in the Coulomb potential of an atomic nucleus. Its much larger reduced mass than in an electronic chemical atom greatly reduces the orbital scale, increasing sensitivity of transition energies to the nuclear charge radius and nuclear charge distribution. Measuring muonic X-ray transitions therefore complements electron scattering. Strong relativistic and nuclear-size effects in heavy systems require a treatment beyond a point-nucleus nonrelativistic Coulomb spectrum.

The proper mean lifetime of a free muon, approximately $2.2\mu \mathrm{s}$. Independent particle decays give survival probability $\exp (-\Delta \tau /{\tau }_{\mu })$. In a frame where the Lorentz factor is constant, $\Delta \tau =\Delta t/\gamma$; interactions of a stopped negative muon in matter can add capture channels.

The rest mass of the muon, approximately $105.66\mathrm{MeV}$ in natural units. Keeping it in phase space is necessary in precision lepton universality tests.

The muon neutrino is the neutral lepton coupled to a muon by the weak charged current. It carries lepton number $+1$; its antineutrino has the opposite lepton number.

Accelerator-produced muons circulating in a magnetic storage ring allow a direct comparison between their measured laboratory mean lifetime and $\gamma$ times the rest lifetime. The 1977 experiment at $\gamma =29.33$ measured positive- and negative-muon lifetimes near $64.4\mu \mathrm{s}$, consistent with time dilation. This is a controlled complement to atmospheric survival measurements, with different systematic effects.^[ref]

The weak neutral current couples the Z boson to a fermion without changing its electric charge. Its vector and axial-vector coefficients depend on weak isospin, electric charge, and the weak mixing angle.

The neutral current transfers no electric charge at a fermion vertex; “neutral” does not mean that every participating particle is uncharged. For example, ${\nu }_{\mu }{e}^{-}\to {\nu }_{\mu }{e}^{-}$ transfers energy and momentum to an electron through Z boson exchange. By contrast, a charged current can convert a muon neutrino into a muon, as in ${\nu }_{\mu }n\to {\mu }^{-}p$. Both processes conserve total electric charge. A photon also carries no electric charge, but an active neutrino has no electromagnetic charge coupling: its neutral current is a distinct weak interaction.

The experimentally useful contrast on a nucleon target iswhere $N$ is a nucleon and $X$ is an allowed final state of hadrons. The charged current makes a charged lepton available for identification; the outgoing neutrino in a neutral current leaves no bubble-chamber track. A sample of hadronic events without an identified muon therefore poses an experimental question: how many could instead be neutron interactions, or charged current events whose muon was missed? The answer requires a detector-response and background analysis, not merely an inspection for an absent line.

How can the relative abundance of tracks and showers distinguish hypotheses? Consider an original teaching model, with chosen inputs throughout. Condition on a sample of $N$ selected events, each assigned exactly one of two labels: track or shower. Multitrack events and unclassified events lie outside this model. A shower label need not identify an electron; background can imitate it. Conversely, a true electron interaction need not receive a shower label.

Let $b$ be the background fraction among selected events. Among the selected signal events, let $f$ be the fraction whose associated charged lepton is an electron, with the remainder associated with a muon. These are fractions after event selection, rather than fractions of the incident neutrino particle flux. Define the conditional probabilitiesHere $a$ and $r$ describe classification within the selected sample. They are not measured historical detection efficiencies. The law of total probability gives the overall shower probabilityThis formula is the bridge between an interaction hypothesis and a visible label. It keeps beam contamination, background and mistaken classification conceptually distinct: genuine electron neutrino interactions belong in $f$, while non-signal events belong in $b$.

Assume independent random variables for the event labels, all with this same $q$. If $K$ counts shower labels, every specified pattern with $k$ showers has probability $q^k(1-q{)}^{N-k}$. Counting those patterns with a binomial coefficient yields the binomial distributionThe track count is $N-K$, so its expected value is $N(1-q)$. At fixed $N$, the two counts cannot fluctuate independently. The variance above measures sampling fluctuations under the stipulated model, not uncertainty in $a$, $r$, $b$ or $q_b$.

Choose $N=40$, $a=0.90$, $r=0.05$, $b=0.10$ and $q_b=0.40$. Compare a signal with only muon-associated interactions ($f=0$) against one with equal selected electron- and muon-associated interactions ($f=1/2$). These give $q=0.085$ and $q=0.4675$, respectively. Their expected shower counts are $3.4$ and $18.7$; their expected track counts are $36.6$ and $21.3$. A background-only model with the specified $q_b$ gives $16$ showers and $24$ tracks on average. These are expectations, so they need not be integers.

For a hypothetical observation $K=4$, the likelihood ratio of the two signal models isThe binomial coefficient cancels because the same hypothetical sample is used in both likelihoods. This ratio is not the probability that either hypothesis is true. A separate original check gives $\mathrm{Pr}(K\le 5)\simeq 0.880$, $4.57\times 10^{-6}$ and $1.44\times 10^{-4}$ for the muon-associated, equal-mixture and specified background-only models, respectively.

Why are the controls essential? If the background-only model may choose any $q_b$, taking $q_b=K/N$ can fit the observed shower fraction. Counts alone then fail to distinguish it from a signal model. Independent background measurements and calibrated particle identification constrain that freedom. Changing the decay path additionally tests where the candidate-producing particles originate. The power of the experiment comes from combining apparatus interventions with event classification; the small binomial distribution model isolates only the latter part of that reasoning.

In 1947, Cesare Lattes, Hugh Muirhead, Giuseppe Occhialini and Cecil Frank Powell reported charged particle tracks in which a stopping particle was followed by a secondary particle. Their paper treated the interpretation as provisional.^[ref] The eventual distinction between pions and muons made a striking use of photography: a tiny permanent record could distinguish particles by how they slowed and transformed.

The method already had a history. In their jointly authored 1937 report, Marietta Blau and Hertha Wambacher described photographic plates exposed for five months at Hafelekar, $2300\mathrm{m}$ above sea level.^[ref] Neither shared the 1950 Nobel Prize in Physics, awarded to Cecil Frank Powell.^[ref] This was a development of photographic particle detection, with earlier contributors, rather than its invention by one person.

In his own later account, Cesare Lattes recalls calibrating emulsion shrinkage and the particle range–energy relation with accelerator-produced particles. Giuseppe Occhialini took borax-treated and untreated plates to Pic du Midi for about a month. The untreated plates provided a control; the treated plates retained more events because latent image fading was reduced. Later, Cesare Lattes exposed plates at the higher Chacaltaya station in Bolivia and obtained more candidate events.^[ref] The Pic du Midi cosmic ray exposure was roughly $3\mathrm{km}$ above sea level.^[ref]

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The bundle filename says 2018, but the paper's cover establishes Trinity Term 2019, 24 April 2019, and formal code A15272W1. These are solutions to that 2019 paper. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$, and the Feynman i-epsilon prescription. Write $/p={\gamma }^{\mu }{p}_{\mu }$. In question 1, $D=4-ϵ$, $g_D={\mu }^{ϵ/2}g$, and the amputated gluon self-energy insertion is $i{\Pi }^{\mu \nu ,ab}$. This fixes the overall sign and factor of $i$ in the results below; defining the insertion as $-i\Pi$ instead reverses all corresponding definitions of $\Pi$. The quadratic Casimir operator in the Adjoint representation is $C_A=N_c$ for $SU(N_c)$, hence $C_A=3$ in QCD.

After the scalar field obtains a vacuum expectation value $h$, the fermion mass matrix in a canonically normalized basis is $M=h\lambda$. Physical masses are its singular values, not arbitrary entries or signed eigenvalues. ThusThe independent Yukawa coupling $b$ can fit the tau lepton mass relative to the doublet scale, but the electron and muon are predicted to have equal masses. The values supplied by the paper instead give $m_{\mu }/m_e\simeq 105.7/0.511\simeq 207$. Consequently exact unbroken $S_3$ with this representation assignment cannot reproduce all three masses. Complex phases of $a,b$, or changes of basis, cannot split the two equal singular values. Additional breaking of the symmetry or a different field content would be needed; the discrepancy cannot be removed merely by choosing different invariant Yukawa couplings.

Let $v_H=⟨H⟩$ be the vacuum expectation value in the convention $M=v_H\lambda$. For canonically normalized fermion kinetic terms, physical charged lepton masses are the nonnegative values in the singular value decomposition of $M$, equivalently the square roots of the eigenvalues of $M^{†}M$.

In case (i),The opposite algebraic eigenvalues $y$ and $-y$ of the doublet block do not represent different physical masses. The unavoidable twofold degeneracy conflicts with the three distinct nonzero measured charged lepton masses, for any assignment of the three families.

In case (ii),One can fit the tau lepton mass by choosing $|v_H{y}_3|$, but the electron and muon remain massless. Thus neither exact dihedral group assignment with this single singlet complex scalar field reproduces the observed spectrum. Additional symmetry-breaking Yukawa interactions or other sources of mass are required; a basis change cannot remove the degeneracy or change the matrix rank.

The middle panel shows the Feynman diagram for the leptonic W boson decay. With the charged-current Lagrangian density sign in the question, its vertex Feynman rule isAn overall field-phase convention can change the overall amplitude sign but cannot change the decay width. With the standard Dirac field current ordering, the amplitude isThe unusual order of barred fields in the printed description should be interpreted as this physical weak charged current matrix element.

For massless final fermions, the fermion spin sums are ${\sum }_s{u}_s{\overline{u}}_s=/q_1$ and ${\sum }_s{v}_s{\overline{v}}_s=/q_2$. The massive-vector spin average divides by three. Thuswith the chiral traceAnticommuting ${\gamma }_5$ through the two following gamma matrices and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$ gives its symmetric partIts remaining part is an antisymmetric epsilon tensor from the gamma matrix trace identities; it vanishes against the symmetric polarization tensor $K$. Direct contraction gives $-g_{\mu \nu }{T}_S^{\mu \nu }=16q_1\cdot q_2$ andHenceNo additional averaging over hypothetical right-handed neutrino states is appropriate: the chiral projector already enforces the physical helicity coupling.

On shell, $p=q_1+q_2$, $p^2=M_W^2$ and $q_i^2=0$, so $q_1\cdot q_2=M_W^2/2$ and $p\cdot q_i=M_W^2/2$. The squared matrix element reduces to $g^2{M}_W^2/3$. In the parent rest frame, integrate the spatial momentum delta first in two-body Lorentz-invariant phase space. The remaining radial integral isConsequently the massless leptonic W decay width isLepton universality gives $\Gamma (W^{-}\to {\mu }^{-}{\overline{\nu }}_{\mu })=\Gamma (W^{-}\to {\tau }^{-}{\overline{\nu }}_{\tau })={\Gamma }_{\ell }$ in the same massless approximation. The muon and tau lepton masses produce small corrections if retained.

With an identity CKM matrix, the allowed quark channels are $d\overline{u}$ and $s\overline{c}$. The corresponding third-generation channel $b\overline{t}$ is closed by the top threshold. Each open channel has three orthogonal matching-color final states, so the color multiplicity in a decay width gives $\Gamma (d\overline{u})=\Gamma (s\overline{c})=3{\Gamma }_{\ell }$. There is no initial color average because the W boson is colorless. The tree-level W branching fractions are therefore:

The rounded percentages sum to 99% because of rounding. The exact fractions sum to one, with total leptonic fraction $1/3$ and hadronic fraction $2/3$. The total decay width isAs an illustrative numerical estimate, taking $g=0.65$ and $M_W=80\mathrm{GeV}$ gives ${\Gamma }_{\ell }\simeq 0.224\mathrm{GeV}$, each open quark width $\simeq 0.672\mathrm{GeV}$ and ${\Gamma }_W\simeq 2.02\mathrm{GeV}$. These are tree-level model estimates with the stated illustrative inputs, not a fit to a measured width. Strong radiative corrections increase the inclusive quark widths.

The produced quark and antiquark radiate gluons, generating a parton shower. As their virtualities fall towards the strong-interaction scale, perturbation theory gives way to hadronization: confinement converts the colored partons into color-neutral hadrons, typically seen as two particle jets, with additional radiation sometimes producing resolved extra jets. Unstable hadrons subsequently decay; the primary quarks are not observed as free particles.

Including the CKM matrix, the charged current couples an accessible up-type flavour $u_i$ to every kinematically accessible down-type flavour $d_j$. ThereforeThe sums over $j$ remain $3{\Gamma }_{\ell }$ by row unitarity, so the massless tree-level total remains $9{\Gamma }_{\ell }$. Quark mixing redistributes the hadronic branching fractions; it does not create a new total color factor. If a family is above threshold, only its accessible channels enter, and the unitarity sum must not be applied to an incomplete set of final flavours.

For completeness, CKM parameter counting starts with $N^2$ real degrees of freedom of an $N\times N$ unitary matrix. This follows by counting $2N^2$ real entries and $N^2$ independent Hermitian constraints in $V^{†}V=I$. The real-orthogonal subclass has $N(N-1)/2$ mixing angles; the remaining $N(N+1)/2$ parameters can be viewed as phases. Rephasing each up-type and down-type mass eigenfield changes $V_{ij}$ by $e^{-i{\alpha }_i}{e}^{i{\beta }_j}$. Of the $2N$ phases, the common phase does nothing, so $2N-1$ are removable. ThusFor three generic nondegenerate families this is three angles and one CP-violating phase. Degenerate masses can allow additional changes of basis and reduce the count.

Use Minkowski metric signature $+---$ and natural units. With no lepton mixing, choose the weak charged current conventionIts conjugate convention exchanges $J$ and $J^{†}$ without changing the interaction. The relevant four-fermion interaction gives the muon decay amplitude, up to an overall phase,Here $G_F$ is the Fermi constant; the factors $1-{\gamma }_5$ are twice the left chiral projector, so there is no additional factor of two to insert.

For the final-spin sum let ${\rho }_{\mu }=(/p+m)(1+{\gamma }_5/s)/2$, where $m=m_{\mu }$. It already describes the specified initial polarization; there is no further initial-spin average. The two gamma matrix traces areThe electron-mass term vanishes in its chiral trace. In the muon trace, anticommutation with ${\gamma }_5$ givesFor the chiral fermion trace contraction, defineThe supplied trace conventions imply $L^{\alpha \beta }=8K^{\alpha \beta }(k,q)$ and $H_{\alpha \beta }=4K_{\alpha \beta }(q^{\prime },r)$, where $r=p-ms$. The factor four in the second trace follows from the polarization projector. The symmetric parts contract to $2[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)]$. The Levi-Civita symbol contraction, including $i^2=-1$, contributes $2[(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)]$. Mixed symmetric-antisymmetric contractions vanish. Thus $K^{\alpha \beta }(a,b)K_{\alpha \beta }(c,d)=4(a\cdot c)(b\cdot d)$, and

Now neglect $m_e$ as well. Put $Q=p-k$ and $c=\hat{\mathbf{k}}\cdot \mathbf{s}$. The given moment integral over two-neutrino relativistic two-body phase space yieldsIn the muon rest frame, $k\cdot Q=mE$, $k\cdot r=mE(1+c)$, $r\cdot Q=m^2-mE(1+c)$ and $Q^2=m^2-2mE$. Substitution givesCombining this with the Lorentz-invariant phase-space measure, whose one-particle denominator is $(2\pi {)}^{3}2E$, and $d^{3}k=E^{2}dEd\Omega$, gives the polarized muon decay distributionConsequently $C=-2$, $D=-2$, $E=1$. Integration over the electron direction removes the polarization term, and ${\int }_0^1{x}^2(3-2x)dx=1/2$, recovering $\Gamma =G_F^2{m}^5/(192{\pi }^3)$.

The spin vector is axial and the momentum direction is polar: under parity, $\mathbf{s}$ stays unchanged while $\hat{\mathbf{k}}$ reverses. Their scalar product therefore changes sign. Its nonzero coefficient demonstrates parity violation in polarized muon decay, arising from the left-chiral weak charged current. At the upper endpoint the negative muon's electron is preferentially emitted opposite the spin, providing a sign check on the result.

Take the muon mass to be $m_{\mu }$, and use the quantum electrodynamics convention ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ with $e>0$. Let $p_1,p_2$ be the incoming muon and antimuon momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. External particles are on shell: $p_1^2=p_2^2=m_{\mu }^2$ and $k_1^2=k_2^2=0$. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

The QED Feynman rules needed here are the vertex $-ie{\gamma }^{\mu }$, the Dirac propagatoran incoming muon spinor $u(p_1)$, an incoming antimuon adjoint $\bar{v}(p_2)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define $R(q)=(/q+m_{\mu })/(q^2-m_{\mu }^2+i0)$ and $/ϵ={\gamma }^{\mu }{ϵ}_{\mu }$.

With the convention that a diagram contributes $i\mathcal{M}$, the two muon-antimuon annihilation amplitudes areTheir denominators are respectively $t-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$, where $t=(p_1-k_1{)}^2$, $u=(p_1-k_2{)}^2$. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k\cdot ϵ=0$, and represents an equivalence class $ϵ\sim ϵ+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with ${ϵ}^0=0$ and ${ϵ}^{\ast }\cdot ϵ=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace ${ϵ}_1^{\ast }$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external Dirac equations giveAway from propagator poles, $R(q)(/q-m_{\mu })=(/q-m_{\mu })R(q)=I$; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under ${ϵ}_i\mapsto {ϵ}_i+\alpha k_i$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as $p,k$ and outgoing momenta as $p^{\prime },k^{\prime }$, with $p+k=p^{\prime }+k^{\prime }$. The outgoing muon contributes $\bar{u}(p^{\prime })$, the incoming photon contributes $ϵ(k)$ without conjugation, and the outgoing photon contributes ${ϵ}^{\prime \ast }(k^{\prime })$. The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are $p+k$ and $p-k^{\prime }$, with denominators $s-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing $p_2\mapsto -p^{\prime }$ and one outgoing photon momentum $k_1\mapsto -k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final photons require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.

The ideal two-shower event fraction is $B_e^2$, approximately $4\%$ in simple channel counting or $3.18\%$ using the measured tau lepton electronic branching fraction. The selected particle decay chain isThe electron and Positron each make a contained electromagnetic shower; muons penetrate, and generic hadrons do not deposit all their energy as a contained electromagnetic shower.

To estimate $B_e$, use lepton universality and neglect final masses and Quantum chromodynamics corrections. The electronic and muonic decay widths each carry weight one. The accessible inclusive hadronic channels have weights $3|V_{ud}{|}^2$ and $3|V_{us}{|}^2$, where three is the color multiplicity in a decay width. The charm quark is inaccessible in physical tau lepton hadronic decays. With $|V_{ud}{|}^2+|V_{us}{|}^2\simeq 1$, the total weight is $1+1+3=5$, givingThe 2021 measured value $B_e=0.1782$ instead gives $B_e^2=0.0317552$.^[ref] The difference reflects effects omitted by the simple free-channel estimate, especially the hadronic contribution to the total decay width.

There are two tau leptons per produced pair and two per selected pair, so the fraction of all produced tau leptons belonging to these two-shower events is also $B_e^2$. The unrestricted fraction of individual tau leptons decaying electronically is $B_e$; these are different selections. A real high-energy shower cut gives an additional acceptance factor: $N_{\mathrm{selected}}/N_{\mathrm{pairs}}=B_e^2{\epsilon }_{ee}$, which cannot be computed without the cut and detector response. No factor of two is needed for the unique $e^{+}{e}^{-}$ assignment.

The magnetic rigidity gives the momentum, the bending direction gives the sign of electric charge, and the calorimetric energy then gives the mass. For planar tracks perpendicular to a uniform magnetic field, the Lorentz force and circular motion giveTrace each track away from its production or decay vertex. With the magnetic field into the page, a positive track bends counterclockwise and a negative track bends clockwise. For example, a positive particle initially travelling left accelerates downwards. The known incoming negative pion fixes the sign convention experimentally. A radius alone does not distinguish electric charge signs.

For a calibrated total energy $E$, the relativistic energy-momentum relation givesIf the calorimeter instead supplies only kinetic energy $T$, solve $(T+m{c}^2{)}^2=p^2{c}^2+m^2{c}^4$ to obtainCompare with candidate pion, kaon, and proton masses, together with the electric charge sign and shower behavior. An electron makes an electromagnetic shower; a muon is penetrating; a hadron generally requires a hadronic calorimeter and its own calibration. If either energy escapes or the response is uncalibrated, simply substituting a deposited signal for $E$ need not identify the particle correctly. The absent figure prevents attaching identities to individual drawn tracks, but the requested identification procedure does not require its geometry.

The decay is allowed in the Standard Model. Begin with ordinary muon decay, ${\mu }^{-}\to {\nu }_{\mu }{W}^{-\ast }$, $W^{-\ast }\to e^{-}{\bar{\nu }}_e$. The outgoing electron can emit a virtual photon that converts to $e^{-}{e}^{+}$. The bottom panel of the figure in part (b) shows one representative diagram. Radiation from other charged lines and exchange of the identical final electrons complete the decay amplitude.

The final total electric charge is $-1$. Its electron-family lepton number is $1+1-1-1=0$, and its muon-family lepton number is $1$, matching the initial muon. The extra pair therefore violates no relevant Standard Model selection rule. It is also kinematically allowed since $m_{\mu }>3m_e$. The virtual pair costs two electromagnetic vertices: the decay amplitude is order $G_F{e}^2$, and the decay rate is suppressed by roughly two powers of the fine-structure constant, with additional phase space and logarithmic factors.

For electron-positron annihilation into a muon pair the gauge diagrams are s-channel photon and Z exchange. For electron-positron annihilation into electron neutrinos there is s-channel Z exchange and t-channel W exchange, with interference. For electron-positron annihilation into W-boson pairs there are s-channel photon and Z diagrams and a t-channel electron-neutrino diagram. All three gauge amplitudes are required for the proper high-energy cancellation. Electron-Yukawa-suppressed s-channel Higgs diagrams can also produce the massive muon or W pair; they vanish in the usual zero-electron-mass approximation used for the leading gauge comparison.

There is no energy-independent ordering. For unpolarized beams and inclusive Born cross sections the useful regimes are:

The Z-pole ratio follows $[\frac{1}{2}{]}^2/[(-\frac{1}{2}+{\sin }^2{\theta }_W{)}^2+{\sin }^4{\theta }_W]\simeq 1.99$. Treating every weak reaction as uniformly slower misses both the resonance and forward exchange.

Away from thresholds and narrow resonances, photon-mediated annihilation gives one colour copy proportional to $Q_q^2$ relative to the muon cross-section. The accessible flavours are $u,d,s,c$, each with three colours, soBottom production is closed. QCD radiative corrections raise the leading parton-model result by a relative term beginning at ${\alpha }_s/\pi$; the requested estimate does not require a supplied value of ${\alpha }_s$.