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Assume unpolarized beams, negligible electron and muon masses, a single scattering s-channel spin-one mediator with equal left and right vector couplings, and leading-order QED. Z boson exchange and radiative corrections are omitted. The angular factors are the Wigner D-matrix elements $d_{m{m}^{\prime }}^1({\theta }^{\star })$, with $m,m^{\prime }=\pm 1$. Up to common phases and normalization they are $(1+\cos {\theta }^{\star })$ for equal pair-spin projections and $(1-\cos {\theta }^{\star })$ for opposite projections. The four orthogonal helicity channels add incoherently, givingWith $s$ the squared center-of-mass energy, the normalized differential scattering cross-sections areThe second expression includes the solid angle Jacobian: a distribution per angle is different from one per $\cos {\theta }^{\star }$. Integration gives $\sigma =4\pi {\alpha }^2/(3s)$.

Parity reverses all momenta but leaves the spin axial vector unchanged, so it interchanges $L\leftrightarrow R$. The $LR\to LR$ and $RL\to RL$ amplitudes have equal magnitudes, as do $LR\to RL$ and $RL\to LR$. Thus every configuration has the same rate as its parity transform, proving parity conservation. The summed distribution is also forward-backward symmetric. Strictly, parity reverses both the beam and outgoing momenta, leaving their relative angle unchanged; forward-backward reflection by itself is not the definition of parity.

Assume the same left and right couplings $g_l,g_r$ at the electron and muon vertices. This lepton universality assumption is necessary to describe both vertices using just two numbers. Define $a=|g_l{|}^2$, $b=|g_r{|}^2$ and $z=\cos {\theta }^{\star }$. The two same-chirality channels have squared coupling weights $a^2,b^2$ and angular factor $(1+z{)}^2$; the other two each have weight $ab$ and factor $(1-z{)}^2$. Hence, for a common positive normalization $C$,If $|g_l/g_r|\ne 1$, then $a\ne b$, so a forward-backward term survives. At the amplitude level, parity interchanges the $g_l^2$ and $g_r^2$ channels; their unequal squared magnitudes show directly that the interaction violates parity. Equivalently the chiral current contains both vector current and axial current terms, with coefficients $(g_r+g_l)/2$ and $(g_r-g_l)/2$; unequal magnitudes preclude a pure parity eigen-current.

The odd-in-$z$ term is a useful signature within these assumptions, but it should not itself be called a parity-odd scalar: the unpolarized relative angle is unchanged when all momenta are reversed. If the two species have independent couplings, the coefficient is instead proportional to $(a_e-b_e)(a_{\mu }-b_{\mu })$, and a parity-violating coupling at only one vertex need not produce an unpolarized forward-backward asymmetry.

There is one tree-level Feynman diagram: the distinguishable electron and muon exchange a photon in the $t$ channel. There is no identical-particle exchange diagram.

Multiplying the external Dirac spinors, the two Feynman vertices and the photon propagator givesThe overall phase of this scattering amplitude is convention dependent. Four external spin labels with two choices each give $2^4=16$ possible configurations before taking a limit. For $E\gg m,M$, the vector QED interaction preserves fermion chirality, which becomes helicity for each external particle. Thus the surviving configurations, written as $(h_e,h_{\mu }{)}_{\mathrm{in}}\to (h_e,h_{\mu }{)}_{\mathrm{out}}$, areHere $L,R$ specify helicity along each particle's own momentum. The other configurations require at least one mass-suppressed flip and vanish in the strictly massless limit.

Define the Mandelstam variables for the scattering labels by $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2=q^2$ and $u=(p_1-p_4{)}^2$. Dropping the electron and muon masses givesThese identities follow by expanding the corresponding squared four-momenta, using $p_i^2=0$ and four-momentum conservation. Substitution into the supplied spin-averaged matrix element yieldsThe $1/4$ initial spin average is already included in the given expression; no further averaging factor belongs here.

To implement crossing symmetry with the annihilation labels, replace the old scattering four-momenta by $(p_1,-p_3,-p_2,p_4)$. The scattering Mandelstam variables then become $(s_{\mathrm{old}},t_{\mathrm{old}},u_{\mathrm{old}})=(t,s,u)$, where now $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. ConsequentlyIn the centre-of-momentum frame, take $\theta$ between the incoming electron and outgoing positive muon, $p_3$. For negligible masses, $t=-s(1-\cos \theta )/2$ and $u=-s(1+\cos \theta )/2$. ThereforeDefining $\theta$ using the outgoing negative muon interchanges $t$ and $u$ and leaves the result unchanged.

Away from narrow resonances and quark thresholds, the QED annihilation cross section into each accessible quark flavour is its squared fractional electric charge times the muon-pair cross section, multiplied by the number $N_c$ of distinguishable color charges. Thus the inclusive hadronic ratio isWith $N_c=3$, the continuum predictions are $R_{uds}=2$, $R_{udsc}=10/3$ and $R_{udscb}=11/3$ as successive charm quark and bottom quark channels become accessible. The measured normalization and steps support three color charges. At leading order these hadronic events form two back-to-back jets from the quark and antiquark; quantitative comparisons include QCD corrections and avoid the nonperturbative threshold region.

The extra resolved jet is explained by $e^{+}{e}^{-}\to {\gamma }^{\ast }\to q\bar{q}g$, with gluon radiation from either the quark or antiquark. One possible color charge routing is initial $q_r{\bar{q}}_{\bar{r}}$ followed by $q_r\to q_b+g_{r\bar{b}}$; the final $q_b$, ${\bar{q}}_{\bar{r}}$ and $g_{r\bar{b}}$ combine to a net colour singlet.

The three-jet cross section contains one additional strong interaction vertex in the amplitude, so its ratio to the two-jet rate is of order ${\alpha }_s=g_s^2/(4\pi )$ times a resolution-dependent kinematic coefficient. A resolved fraction of order $15\%$ is consistent with perturbative QCD and a strong coupling constant much larger than the fine-structure constant. It is not an equality ${\alpha }_s=0.15$: the jet definition and cuts are needed. Photon radiation has a much smaller electromagnetic rate. The large hadronic third-jet rate supports gluon radiation; identifying the radiated object specifically as a colour-octet gluon also uses the QCD vertex and colour-flow description, not merely the existence of three visible jets.

A fission reactor supplies predominantly electron antineutrinos, with typical energies of a few MeV. Even if they oscillate to muon antineutrinos, producing a real muon through a charged current requires roughly the muon rest energy, over $100$ MeV for a free nucleon target. For example,Thus a muon spectrometer cannot observe this appearance channel from a reactor. Measure electron antineutrino disappearance instead, most commonly through inverse beta decay, ${\bar{\nu }}_{e}p\to e^{+}n$, using the prompt Positron and delayed neutron signal.

Treat the atmospheric proton as stationary in the Earth frame, the incident antiproton as vertically downward, and neglect atmospheric energy loss, scattering and the Earth's curvature initially. Take $m_p{c}^2=0.9383\mathrm{GeV}$. Conservation of energy and equal outgoing energies giveEqual energies imply equal momentum magnitudes. Their transverse momenta cancel, so the two muons have equal downward components $p_z=p_{\bar{p}}/2$, whereEach muon therefore makes an angle $\theta$ with the downward vertical satisfyingThe paths are not both vertical: replacing their path lengths by $20\mathrm{km}$ would miss this two-body kinematic correction. Since $v_z=p_z{c}^2/E_{\mu }$, the proper flight time over vertical height $H$ isFor equal-energy atmospheric muon-pair survival, independent exponential particle decays with mean proper muon lifetime ${\tau }_0=2.2\mu \mathrm{s}$, both survive with probabilityEquivalently each travels about $20.55\mathrm{km}$, with proper flight time about $0.404\mu \mathrm{s}$, and survives with probability about $0.832$. The result is decay-only survival to an ideal surface, not a prediction including air interactions or detector efficiency. For a spherical Earth of radius $R_E$, the exact path length is $\ell =(R_E+H)\cos \theta -\sqrt{R_E^2-(R_E+H{)}^2{\sin }^2\theta }$; inserting $P_{\mathrm{both}}=\exp [-2\ell /({\gamma }_{\mu }{v}_{\mu }{\tau }_0)]$ gives a small curvature correction.

In the pion rest frame, momentum conservation makes the muon and neutrino momenta opposite, with common magnitude $p$. For a massless neutrino, $E_{\nu }=pc$; energy conservation and the relativistic energy-momentum relation giveSubstitute $pc=m_{\pi }{c}^2-E_{\mu }$ into the second equation. The $E_{\mu }^2$ terms cancel, leavingFor completeness, the complementary relativistic two-body decay energy is $E_{\nu }=(m_{\pi }^2-m_{\mu }^2)c^2/(2m_{\pi })$. The result is a total energy, including the muon rest energy, rather than its kinetic energy.

Let $M$ denote an incident muon and $S$ a detector signal. The selected sample is conditioned on $S$. The law of total probability givesBy Bayes' theorem, the contamination, namely the conditional probability that a selected particle is not a muon, isFor example, per $10000$ incident particles the expected selected counts are $8550$ true muons and $100$ other particles. The denominator is the $8650$ selected particles; the rejected muons are outside this sample.

For a proper lifetime inferred from a delayed decay flash, the measured interval ends when the light returns to the detector, not when the muon decays. Put passage at $(t,x)=(0,0)$ in the detector frame. If decay occurs at $t_d$, its distance is $v{t}_d$, and the backwards photon takes another $v{t}_d/c$ to arrive. ThereforeThe muon's proper elapsed time isApplying time dilation directly to the 10 ns reception interval would transform the wrong pair of events.

In standard configuration the frames are inertial, their corresponding axes are parallel, S${}^{\prime }$ moves at $v$ along the positive $x$ axis of S, and their origins coincide at $t=t^{\prime }=0$. Here $(t,x,y,z)$ and $(t^{\prime },x^{\prime },y^{\prime },z^{\prime })$ label the same event, $c$ is the speed of light and $\gamma =(1-v^2/c^2{)}^{-1/2}$.

A clock fixed in S${}^{\prime }$ has $\Delta x^{\prime }=0$. The inverse time transformation givesso the moving clock accumulates less proper time between its two ticks than the elapsed S-coordinate time.

Without time dilation the flight time from the given altitude would be $66.7\mu \mathrm{s}$, about $30.3$ rest-frame mean lifetimes, giving survival fraction only $e^{-30.3}\simeq 6.8\times 10^{-14}$ even at speed $c$. Relativistic muons can instead reach the ground with appreciable survival. Momentum-selected decay measurements test this clock-rate dependence experimentally.^[ref]

A mean-lifetime survival energy estimate requires a mean flight time no greater than the dilated mean lifetime:This corresponds to survival probability at least about $e^{-1}$. There is no sharp minimum energy for an individual random decay. For a required probability $p$, use $\exp [-H/(\gamma v{\tau }_0)]\ge p$. Energy loss in the atmosphere is neglected in this estimate. In the muon's frame the atmospheric path is length-contracted to $H/\gamma \simeq 660\mathrm{m}$; the ground reaches it in about ${\tau }_0$, giving the same survival probability.

For the two-body decay, use the rest frame of A. Momentum conservation makes B and the photon oppositely directed with equal momentum magnitude $p$. Energy conservation and the energy–momentum relation giveSubstitution and cancellation of $E_B^2$ giveA nonzero emitted photon requires $m_A>m_B$.

Experimental evidence. Tests of special relativity check linked predictions about light propagation, clocks and four-momentum. Several independent kinds of experiment support those predictions.

The observed partial-drag coefficient agrees with this first-order prediction rather than simple Galilean addition of $u$. It tests a different consequence of the Lorentz transformation from a particle-lifetime measurement.^[ref]

Atmospheric muons and the two frame descriptions. Atmospheric cosmic rays create unstable particles whose decay chains produce muons. A stopped-muon experiment measures the time between a stopping signal and the subsequent decay electron, obtaining the proper mean lifetime ${\tau }_0\simeq 2.2\mu \mathrm{s}$. The associated light-travel distance is only $c{\tau }_0\simeq 660\mathrm{m}$, yet substantial numbers of energetic atmospheric muons reach detectors after travelling several kilometres.

For an ideal selected cohort produced height $h$ above a detector and moving downward at constant speed $v$, the flight time in the terrestrial frame is $h/v$. Time dilation predicts a laboratory mean lifetime $\gamma {\tau }_0$, hence survival fractionWithout time dilation, the predicted exponent would be $-h/(v{\tau }_0)$ and the surviving population much smaller. Altitude comparisons constitute a cosmic-ray muon time-dilation experiment: select comparable momenta or energies and account for detector acceptance, atmospheric energy loss and backgrounds before interpreting the count ratio.

For illustration, not as a claimed measurement, take $h=10\mathrm{km}$ and $v=0.998c$. Then $\gamma =15.82$, the laboratory flight time is $33.4\mu \mathrm{s}$, and the predicted surviving fraction is about $0.38$. Omitting time dilation would give only about $2.5\times 10^{-7}$ on the same idealized constant-speed model.

In the muon rest frame, its decay still follows $e^{-\Delta \tau /{\tau }_0}$. The atmosphere and ground move upward. Their separation, measured simultaneously in that frame, is shortened by length contraction:The same survival fraction follows. The terrestrial description uses a longer-lived moving particle; the muon-frame description uses a shorter atmospheric flight path. They describe the same production and detection events. This consistency illustrates both time dilation and length contraction, while the survival measurement is not two independent experimental tests merely because it can be described in two frames.

Energy-momentum invariant. For a particle with rest mass $m$,Since $E=m{c}^2+K$, substitution and cancellation of $m^2{c}^4$ giveThis relativistic energy-momentum relation reduces to $K\simeq p^2/(2m)$ when $K\ll m{c}^2$.

Laboratory beam energies. If both final muons are at rest in the laboratory, their total momentum is zero and their total energy is $2m_{\mu }{c}^2$. Four-momentum conservation therefore requires equal and opposite incoming momenta. Because the electron and Positron have equal rest masses, they then have equal total energies:Subtract each particle's own rest energy:Using the usual handbook rest energies gives approximately $105.1\mathrm{MeV}$ for each beam. These are the ideal kinematic threshold values. Just above threshold, the pair has small, equal and opposite momenta in this laboratory frame.

Positron momentum. Apply the relativistic energy-momentum relation to $E_{+}=m_{\mu }{c}^2$ and use its stipulated negative-$x$ direction:The electron has the opposite momentum, $p_{e,x}=+p_0$ with $p_0=c\sqrt{m_{\mu }^2-m_e^2}$.

Boost to the electron rest frame. The electron's velocity is $v=p_{e,x}{c}^2/E_{-}=p_0/m_{\mu }>0$. Thus $S^{\prime }$ moves in the positive-$x$ direction, and its Lorentz factor satisfiesThe positron has the same laboratory total energy as the electron, so $v{E}_{+}/c^2=v{m}_{\mu }=p_0$. The supplied Lorentz transformation then givesAs a sign check, the same boost gives $p_{e,x}^{\prime }=\gamma (p_0-v{E}_{-}/c^2)=0$, as required for the electron rest frame.

The charged meson decays proceed through a virtual $W^{+}$:The first quark vertex has CKM factor $V_{ud}$, while the second has $V_{us}$. In the simple Cabibbo description these are approximately $\cos {\theta }_C$ and $\sin {\theta }_C$, so the strange amplitude carries a suppression $\tan {\theta }_C$ relative to the nonstrange one, or ${\tan }^2{\theta }_C\simeq 0.053$ in their corresponding squared weak-coupling factors.

For a charged pseudoscalar and a negligibly massive neutrino, the leading leptonic energy width has the formin natural units. The helicity factor $m_{\mu }^2$ is common to these two muon channels, while the threshold factor is much less favorable for the pion. Using the usual masses, its value is about $0.182$ for the pion and $0.910$ for the charged kaon: the pion's threshold factor alone is only about one fifth as large. Including the explicit $m_P$ factor gives a kinematic ratio about $0.0564$. The pseudoscalar decay constants $f_P$ also matter, so Lorentz-invariant phase space and CKM factors alone do not determine the full lifetime ratio.^[ref]

For $K^{+}\to {\pi }^{+}{\pi }^0$, one tree topology has the spectator $u$ unchanged and $\bar{s}\to \bar{u}+W^{+\ast }$ followed by $W^{+\ast }\to u\bar{d}$. The spectator $u$ can combine with $\bar{d}$ to form ${\pi }^{+}$, and the $u\bar{u}$ component supplies ${\pi }^0$. This carries the strangeness-changing weak factor $V_{us}^{\ast }{V}_{ud}$. Hadronization, other quark rearrangements and strong final-state interactions determine the hadronic amplitude; the diagram is not a complete precision width calculation. The small pion–muon energy release suppresses the pion leptonic channel kinematically, whereas the strange weak vertex suppresses the kaon channels through Cabibbo suppression.

Both the muon and electron have the same electric charge magnitude. At equal kinetic energy $K$ per particle, their Lorentz factors are $\gamma =1+K/(m{c}^2)$, so the heavier muon has the smaller Lorentz factor. At equal total relativistic energy the same ordering follows from $\gamma =E/(m{c}^2)$, provided that energy is large enough to create either particle.

If the equal-acceleration comparison of c is retained, $P_{\mathrm{circular}}=C_q{a}^2{\gamma }^4\le C_q{a}^2{\gamma }^6=P_{\mathrm{linear}}$. Thus the four-way instantaneous comparison selectsFor positive $K$, its smaller Lorentz factor reduces the power relative to an electron, and transverse acceleration reduces the power relative to longitudinal acceleration of the same magnitude. This conclusion follows from the constraint actually stated in c.

For the customary engineering comparison at equal applied force, the result changes. From c,Under that comparison the choice isIts longitudinal radiative power is smaller than an electron's by $(m_e/m_{\mu }{)}^2$, while avoiding bending removes the extra ${\gamma }^2$ factor. At a fixed bend radius $R$, the circular loss is $C_q{\gamma }^4{v}^4/R^2$, giving the familiar $m^{-4}$ suppression at fixed ultrarelativistic energy and $R$.

An energy budget alone does not specify a unique optimization. The force, acceleration, bend radius, acceleration time, and length are otherwise unrestricted. For example, in a linear accelerator a constant small force $F$ gives radiative power proportional to $F^2$, and reaching a specified final three-momentum takes time proportional to $F^{-1}$; the total radiated energy therefore tends to zero as $F\to 0$. One cannot infer a unique accelerator choice without an additional constraint. The standard practical preference is the second boxed answer, with its stated comparison. Muon decay introduces a separate lifetime constraint if slow acceleration is contemplated, but it was not part of the specified radiative-loss calculation.

Cecil Frank Powell's 1950 lecture illustrates the successive decays of pions and muons through their tracks and explains why the nearly fixed secondary muon range, allowing for range straggling, supports a single neutral decay partner. He describes its identification as a neutrino tentatively. The neutral particle itself leaves no recorded track. He contrasts this decay with the nuclear reactions caused by stopped negative pions, connecting the latter with the nuclear force.^[ref]

Here is a modern special relativity derivation for the dominant leptonic pion decay, assuming a free pion at rest, a massless muon neutrino, and no emitted photon. Write $m_{\pi }$ and $m_{\mu }$ for the pion and muon masses. By conservation of momentum the daughter momenta are opposite with equal magnitude $p$. By conservation of energy and the energy–momentum relation,Substitute $E_{\mu }=m_{\pi }{c}^2-pc$, square, and cancel $p^2{c}^2$. This givesThe resulting $p$ is fixed, although the muon's direction is not set by the incoming stopping track. Thus the kink angle relative to that earlier track is not the angle between the two daughters: in the pion rest frame, the daughters are back to back. A neutral-partner arrow in a diagram represents inferred momentum, not an observed string of grains.

For an original calculation, adopt rounded input values $m_{\pi }{c}^2=139.57\mathrm{MeV}$ and $m_{\mu }{c}^2=105.66\mathrm{MeV}$. They give $pc=29.79\mathrm{MeV}$, $T_{\mu }=4.119\mathrm{MeV}$ and $E_{\nu }=29.79\mathrm{MeV}$. Their sum satisfies $T_{\mu }+E_{\nu }=(m_{\pi }-m_{\mu })c^2=33.91\mathrm{MeV}$. These are rounded calculation inputs and outputs, not a fit to the 1947 plates.

At fixed material and particle identity, a fixed initial kinetic energy suggests a narrow particle range distribution broadened by range straggling and measurement uncertainty. The synthetic example convolves a single chosen range with a normal distribution; a second example superposes two such chosen ranges. It explains why repeated comparable stopping distances can be informative, without pretending to extract a historical spectrum.

The decay equations do not identify the neutral partner as a neutrino on their own: a massless photon would satisfy the same energy and momentum equations. Nor does any isolated kink establish the pion's identity. The inference gains force from calibrated particle range, track grain density, repeated decay patterns and independent particle properties. Historical particle names should not obscure the distinction: a muon is a lepton, whereas a pion is a meson.

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.

A secondary cosmic ray here is a particle produced by an atmospheric interaction of a primary cosmic ray or its descendants. A muon at ground level need not have traveled from the original astronomical source.

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.

At SPEAR, counterrotating electron and Positron beams collided inside a magnetic particle detector. Finer steps in stored-beam energy resolved a sharp production peak. A magnetometer and measured orbit positions monitored the energy settings; Bhabha scattering normalized the cross sections. The paper attributed the observed width to beam-energy spread and radiation effects, reporting an upper limit on intrinsic width. Its muon identification system was not functioning: the non-electron two-body excess was not conclusively a muon pair.^[ref]

A scan must compare normalized production rates, since more selected events can result from more beam encounters or different detection efficiency. A proposed diagnostic is to revisit energies on either side of the peak while monitoring the energy calibration and event selection. Such neighboring settings test the local production background; their event totals alone are not an independent measurement of the intrinsic decay width.