Lepton

A lepton is a Standard Model fermion without color charge. The charged leptons have electric charge $-1$ and their corresponding neutrinos have charge zero. There are three generations; their antiparticles carry opposite electric charge and lepton number. Charged leptons and neutrinos have lepton number $+1$, with the corresponding antiparticles carrying $-1$.

An antilepton is the antiparticle of a lepton, with opposite additive quantum numbers. A Positron is the charged antilepton corresponding to an electron.

A lepton with nonzero electric charge, such as an electron or muon; neutrinos are electrically neutral.

The tau is the charged lepton of the third fermion generation. In the massless approximation its charged-current W boson partial decay width equals those of the electron and muon, reflecting lepton universality.

The tau lepton has electric charge $-1$ in units of the positive elementary electric charge; ${\tau }^{+}$ is its antiparticle. Its weak charged current decays make the tau lepton useful for testing lepton universality. They also illustrate how particle detection can establish a short-lived parent through visible daughters and a consistent invisible recoil.

The visible angular distribution and laboratory energy in ${\tau }^{-}\to h^{-}{\nu }_{\tau }$ carry information about the parent's spin. For fixed hadronic invariant mass, $E_h=\gamma (E_h^{\ast }+\beta p_h^{\ast }\cos \vartheta )$. A spin-zero recoil has analyzing coefficient ${\alpha }_h=1$. For a spin-one recoil coupled through the V-A interaction, longitudinal and transverse partial rates have weights $m_{\tau }^2:2m_h^2$ and opposite slopes, giving ${\alpha }_h=(m_{\tau }^2-2m_h^2)/(m_{\tau }^2+2m_h^2)$. Thus the harder-helicity ordering can reverse when the recoil mass exceeds $m_{\tau }/\sqrt{2}$. A mean invariant mass without spin-dependent spectral weights cannot fix the sign for a mixed sample.

For $N$ tau-pair events and a tag mode with branching fraction $B$, the tagged-hemisphere count is $S=2NBϵ$ and the double-tag event count is $D=N{B}^2{ϵ}_{ii}$. Eliminating $B$ gives $N=S^2{ϵ}_{ii}/(4D{ϵ}^2)$. Efficiency correlations matter: ${ϵ}_{ii}$ need not equal ${ϵ}^2$. An inclusive branching-fraction fit must include unlisted decay categories.

Counterrotating electron and Positron beams at SPEAR collided inside the Mark I magnetic particle detector. Lead–scintillator scintillation counters identified electrons through large shower pulses; small pulses plus hits beyond the iron return identified muons.^[ref]

The selection required opposite-charge electron–muon tracks, no detected photons, momenta above $0.65\mathrm{GeV}/c$, and transverse noncoplanarity above $20^{\circ }$. Multitrack hadron samples calibrated momentum-dependent misidentification; collinear lepton pairs checked identification losses. Photon and charge control samples tested backgrounds. Acceptance corrections depended on the proposed signal.^[ref]

The 1975 claim was anomalous events without a conventional explanation; new-particle pair production was a possible interpretation.^[ref] Martin Lewis Perl's lecture describes subsequent evidence for a charged lepton.^[ref]

The particle identification problem matters as much as the apparent missing four-momentum in electron-positron collisions. A penetrating hadron can imitate a muon; a particle shower can imitate an electron; an undetected photon or a detector gap can imitate invisible recoil. An absence of additional recorded signals is conditional on detection efficiency. In a proposed signal model, the selected count has the schematic formwhere $\mathcal{L}$ is integrated collider luminosity, $\sigma$ a cross section, and $\epsilon$ the combined acceptance and detection efficiency. A single count cannot determine all three factors. A control sample must probe the relevant momentum and angular range; its abundant events do not automatically calibrate a differently distributed signal.

The beam-overlap factor $L$ relating a reaction's cross section $\sigma$ to its ideal production rate $dN/dt$. Integrated collider luminosity $\mathcal{L}=\int Ldt$ has units of inverse area and gives the expected production count $\mathcal{L}\sigma$ when $\sigma$ is constant.

An absolute collider luminosity calibration that scans the transverse separation of two colliding beams while measuring an interaction rate. Known bunch populations and the overlap integral determine the visible monitor cross section. Correlations between horizontal and vertical profiles require a joint overlap treatment.

The time integral of the incident-beam overlap flux, with inverse-area units. For a fixed production cross section, the mean produced event count is $N={\mathcal{L}}_{\mathrm{int}}\sigma$. A selected count includes efficiency and acceptance, often integrated over final-state kinematics. This is distinct from astrophysical radiative luminosity, which has power units.

In the modern weak charged current interpretation, one allowed assignment isThe electron antineutrino, tau neutrino, muon neutrino and antineutrino of the tau neutrino carry away energy and momentum. There are four undetected particles in this decay assignment. Interchanging which parent decays electronically gives the opposite electric charge assignment. Each parent conserves electric charge and, in this approximation, total lepton number; four-momentum conservation applies separately to production and to each particle decay.

If $B_e$ and $B_{\mu }$ denote the corresponding branching fractions, independent parent decays give an inclusive mixed-flavour fraction $2B_e{B}_{\mu }$. The factor two counts the two assignments to opposite parent charges. For chosen $B_e=B_{\mu }=1/5$, the mixed fraction is $2/25$, before acceptance and detection efficiency. Correlations between the parents' spins can affect angular selection even when the fully inclusive branching fractions factorize.

The neutrinos in that equation are a physical interpretation to test. Neither a deficit in visible energy nor two unusual tracks alone establishes their identities or uniquely counts them. Even a compatible invariant mass and an angular distribution require particle identification, background checks and agreement across further particle decay channels. The direct detection milestone for the tau neutrino was in 2000, separate from the charged tau lepton discovery.

Combine the two massless neutrinos in ${\tau }^{-}\to {\ell }^{-}{\bar{\nu }}_{\ell }{\nu }_{\tau }$ into $q$. In the parent rest frame, four-momentum conservation givesThe lower bound follows from the nonnegative Minkowski norm of the sum of two future-directed null four-vectors; the upper bound follows from $E_{\ell }^{\ast }\ge m_{\ell }$. Substituting gives the displayed energy bounds. They are kinematic: they do not specify the weak interaction decay amplitude or the probability density function inside the interval.

If the charged daughter is also massless, put $x=2E_{\ell }^{\ast }/m_{\tau }\in [0,1]$. A parent with Lorentz factor $\gamma$ and speed $\beta$ in natural units producesHere ${\theta }^{\ast }$ is the daughter angle relative to the boost direction in the parent rest frame. Two opposite parent momenta therefore need not produce two opposite visible daughter momenta: the invisible system can recoil differently in each particle decay.

Choose $m_{\tau }=2\mathrm{GeV}$ and $\sqrt{s}=6\mathrm{GeV}$ purely for illustration. Then each parent has $E_{\tau }=3\mathrm{GeV}$, $|{\mathbf{p}}_{\tau }|=\sqrt{5}\mathrm{GeV}$ and $\beta =\sqrt{5}/3$. Treat all six daughters as massless, including the charged electron and antimuon, and generate each three-body particle decay with constant squared decay amplitude. Choose the parent direction isotropically and the two decays independently; this omits the physical production angular law and spin correlations.

The Lorentz-invariant phase space factorizes through $t=q^2$ into a two-body decay to $\ell +q$, followed by a two-body decay of $q$:After angular integration the first factor contributes $1-t/m_{\tau }^2=x$, while the massless second factor is constant. Since $|dt|=m_{\tau }^{2}dx$, the normalized chosen density is $f(x)=2x$. Drawing $x=\sqrt{U}$ with $U$ uniform on $[0,1]$ implements this density. Isotropic directions, followed by a Lorentz boost for the invisible pair and another for the parent, conserve the complete four-momentum of each generated event.

Define transverse acoplanarity by $\alpha =\pi -\Delta \varphi$, where $\Delta \varphi \in [0,\pi ]$ is the angle between the two visible momentum projections perpendicular to the beam. A fully visible two-body event with zero initial transverse momentum has $\alpha =0$ and $M_{\mathrm{miss}}^2=0$. The toy tau lepton events spread away from that point. These observables are complementary: an angular imbalance asks about directions, while the missing invariant mass also uses the visible energies and longitudinal momenta.

A Lorentz-invariant phase space density supplies allowed configurations and their geometric measure. A physical weak interaction decay amplitude reweights them and correlates spins. Consequently the plotted density is a lesson about conservation and available configurations, not a predicted experimental rate or a fit to the discovery sample. The massless charged-daughter approximation is also unsuitable near the muon rest-energy threshold.

Use natural units $c=1$ and the Minkowski metric with signs $(+,-,-,-)$. In the ideal centre-of-momentum frame, assume a known initial four-momentum $P=(\sqrt{s},0)$, no unaccounted radiation and exact visible reconstruction. DefineThe missing four-momentum in electron-positron collisions is a subtraction of measured quantities. It is not itself a particle detector hit. With $M_{e\mu }^2=(p_e+p_{\mu }{)}^2$, expansion gives the useful identityFor any collection of future-directed massless invisible four-momenta $q_i=E_i(1,{\hat{\mathbf{n}}}_i)$,Thus a positive missing invariant mass excludes exactly one massless invisible particle under the stated assumptions. It does not exclude one massive invisible particle, identify a neutrino flavor, or determine the number of massless invisible particles beyond that lower bound. Several collinear massless particles can even have zero combined invariant mass. Detector resolution and unaccounted radiation invalidate a literal event-by-event interpretation of these ideal inequalities.

For an original exact example in natural units, choose total energy $6\mathrm{GeV}$ and visible massless four-momenta, in $\mathrm{GeV}$,Two possible invisible massless four-momenta are $(2;-1/2,-1/2,\sqrt{14}/2)$ and $(2;-1/2,-1/2,-\sqrt{14}/2)$. Splitting either into two positive collinear parts gives a three-particle interpretation with identical recoil. Alternatively, one invisible particle of mass $\sqrt{14}\mathrm{GeV}$ has that recoil. This example tests the ambiguity of the subtraction, without asserting any of these alternatives is a tau lepton decay.

The energy component of the difference between the known initial four-momentum and the measured visible final four-momenta. In equal-beam electron-Positron collisions it is $\sqrt{s}-{\sum }_i{E}_i$ in natural units. It can arise from escaping neutrinos, but also from detector losses; it is distinct from the missing invariant mass and missing transverse momentum.

The invariant mass of the inferred invisible system, obtained from its four-momentum. It is generally different from the mass of each invisible constituent.

The muon is the negatively charged second-generation lepton, with electric charge $-1$ and lepton number $+1$. Its antiparticle is the antimuon.

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.

A weak charged current decay described at energies much below the W boson mass by a four-fermion interaction with coefficient given by the Fermi constant. At leading order with negligible final-state masses, its width is $G_F^2{m}_{\mu }^5/(192{\pi }^3)$.

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.

A vertically incident particle annihilating with a stationary target into two equal-energy muons gives each half the incident longitudinal momentum, while their transverse momenta are opposite. At height $H$ above a locally flat surface, each proper flight time is $2H{m}_{\mu }{c}^2/[c(p_{\mathrm{in}}c)]$, since $\gamma v_z=p_z/m_{\mu }$. Independent exponential particle decays then give the displayed joint survival probability. Atmospheric energy loss, scattering and curvature are omitted; replacing both trajectories by vertical paths would miss the two-body angular correction.

For an unpolarized parent charged lepton undergoing a three-body leptonic V-A interaction decay, neglecting daughter masses and radiative corrections, the normalized charged-daughter energy distribution in the parent rest frame is $f(x)=2x^2(3-2x)$, where $x=2E_{\ell }^{\ast }/m_{\mathrm{parent}}$ lies in $[0,1]$. Direct integration gives ${\int }_0^{1}f(x)dx=1$ and ${\int }_0^{1}xf(x)dx=7/10$. A specified parent spin orientation also produces angular terms, so this one-variable law alone does not describe polarized samples.

For a negative muon with rest-frame spin vector $\mathbf{s}$, the leading massless-electron distribution is $d\Gamma /(dxd\Omega )=G_F^2{m}_{\mu }^5{x}^2[(3-2x)+(1-2x)\hat{\mathbf{k}}\cdot \mathbf{s}]/(384{\pi }^4)$, where $x=2E_e/m_{\mu }\in [0,1]$. Its final-spin sum is $64G_F^2(k\cdot q^{\prime })(q\cdot (p-m_{\mu }s))$. The spin vector is axial whereas the electron direction is polar, so their scalar product reverses under parity. Its occurrence demonstrates parity violation by the left-chiral weak charged current.

The antimuon is the antiparticle of the muon. It has electric charge $+1$ and lepton number $-1$.