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When an electron of Lorentz factor $\gamma >1$ annihilates with a stationary equal-mass positron, transverse four-momentum conservation makes the photons' transverse components opposite. Eliminating their positive frequencies gives the displayed relation for noncollinear photon angles to the incoming electron. The exactly collinear forward/backward configuration is physically allowed but makes the quotient $0/0$; its undivided cross-multiplied identity remains valid. The angle relation thus requires nondegeneracy or an appropriate limit, rather than an assertion that every allowed angle pair defines the quotient.

Alvarez describes a progression from measured coordinates to momenta and angles, then conservation-constrained fits testing alternative track identities. Computer programs made the repeated tests practical; an insufficiently constrained event could retain several hypotheses.^[ref]

A mathematical formulation explains what such a fit can establish. Let $y$ collect measured momenta and angles, with positive-definite covariance matrix $V$. For a chosen reaction hypothesis, adjust them to $x$ by minimizing a generalized least squares objective,The constraints $g$ express four-momentum conservation and any independently known daughter masses. For each proposed identity, calculate energy from the energy–momentum relation, $E_i^2=m_i^2{c}^4+c^2|{\mathbf{p}}_i{|}^2$. A missing neutral particle adds unknowns; a measured decay vertex can add geometric constraints. Compatibility depends on the measurement errors and the independent constraints, not just the smallest raw ${\chi }^2$ among unrelated fits.

A simple independent check uses a single momentum conservation constraint. Suppose two measured one-dimensional momenta are $0.32$ and $-0.28\mathrm{GeV}/c$, each with independent standard deviation $0.02\mathrm{GeV}/c$, and a stipulated parent at rest requires their sum to vanish. Minimizing the two equal-weight squared changes gives fitted values $0.30$ and $-0.30\mathrm{GeV}/c$, with ${\chi }^2=2$. A fit repairs a mismatch according to a hypothesis; it does not prove that hypothesis. This one-component illustration is not a complete reaction fit.

James H. Christenson, James Watson Cronin, Val Logsdon Fitch and René Turlay reported the result in 1964. At Brookhaven's AGS, 30 GeV protons struck beryllium; collimators selected a neutral particle beam at 30 degrees. A downstream helium decay volume reduced interactions.^[ref]

The flight distance suppressed the short-lived kaon component. Each arm of a two-arm magnetic spectrometer had spark chambers before and after its magnet. Cherenkov detectors and scintillation counters triggered photographic recording in coincidence.^[ref]

The helium sample yielded $45\pm 9$ signal events; $(2.0\pm 0.4)\times 10^{-3}$ was the ratio to all charged decay channels.^[ref] That denominator must be retained when comparing this result with a total branching fraction.

The evidence was kinematic. Later regenerated-beam quantum interference established that the short- and long-lived states reached the same two-pion final state, excluding alternatives such as distinct pion-like particles.^[ref] The 1964 observation established evidence for CP violation; it did not determine every parameter of the CKM matrix. Inferring time-reversal symmetry violation from CP violation invokes CPT symmetry, rather than a direct comparison of time-reversed transitions in that experiment.

Compton scattering redistributes photon energy and momentum in a collision with a charged subatomic particle, here an electron. For a free electron initially at rest in the chosen inertial frame, an incident photon of energy $E=hc/\lambda$ leaves with energy $E^{\prime }=hc/{\lambda }^{\prime }$ after deflection through $\theta$. The angle $\theta \in [0,\pi ]$ is between the incident and outgoing photon directions; it is not the electron recoil angle. Four-momentum conservation giveswhere $m_e$ is the electron rest mass and $T_e$ its final kinetic energy. The photon energy loss pays for the electron recoil: neither energy nor momentum disappears.

For an original worked example, stipulate $E=70\mathrm{keV}$, the rounded electron rest energy $m_e{c}^2=511\mathrm{keV}$, and the rounded Compton wavelength ${\lambda }_C=2.426\mathrm{pm}$; $\mathrm{keV}$ means $10^3$ electronvolts. Treat these rounded inputs as the example's constants. They give the following calculated values, with energies rounded to three decimal places:

Since $0\le 1-\cos \theta \le 2$, every outgoing photon has $0<E^{\prime }\le E$ and $T_e\ge 0$. At zero deflection there is no recoil; backscattering transfers the most energy. At fixed $\theta$, the absolute wavelength increase is independent of incident energy, while the fractional energy loss grows with $E/(m_e{c}^2)$. This makes Compton scattering useful for inferring energy transfers from directions in photon detectors.

The free-electron-at-rest assumption matters. An electron in a bound state has binding energy and an initial momentum distribution. Neglecting these is useful when the transferred energy is large compared with the relevant binding energy and the initial momentum is small compared with the resolved momentum transfer. Otherwise the outgoing photon energies can broaden or depart from the stationary-target formula; an initially moving electron can even give energy to the photon.

A material may also scatter X-rays through a coherent elastic scattering channel with an effectively unshifted component. When an intact chemical atom of mass $M$ takes the recoil, the corresponding Compton wavelength $h/(Mc)$ is much smaller than the electron value. The single free-electron curve does not predict that component's weight. More generally, energy versus angle is a kinematic constraint; angular wave intensity requires a differential scattering cross-section, together with target and detector information.

A display of three-body final states in two independent pairwise squared invariant masses. For a parent of mass $M$ decaying into daughters of masses $m_1,m_2,m_3$, four-momentum conservation gives $s_{12}+s_{13}+s_{23}=M^2+m_1^2+m_2^2+m_3^2$. A narrow intermediate two-body resonance fixes one pairwise invariant mass and appears as a band.

A parent of rest mass $M$ can decay to on-shell daughters of rest masses $m_i$ only if $M\ge {\sum }_i{m}_i$. In the parent rest frame, each daughter energy is at least its rest energy, so four-momentum conservation proves necessity. The excess energy controls the available Lorentz-invariant phase space. A two-body decay threshold is the two-daughter special case; spin and other selection rules can still forbid an energetically allowed channel.

A sum of particle four-momenta at specified events is a four-vector, but sums recomputed at simultaneous events in different inertial frames need not correspond to that same four-vector. For opposite equal electric charges initially at rest a distance $d$ apart, a longitudinal electric field switched on simultaneously in the laboratory produces zero total laboratory three-momentum. In a frame moving at $v>0$, the switch-on events are separated by ${\gamma }_{v}vd/c^2$; during that interval only one force acts. The late equal-time total three-momentum is $-2{\gamma }_{v}mv+{\gamma }_{v}eEvd/c^2$, while the Lorentz transformation of an equal-laboratory-time sum varies as the laboratory pair's energy grows. Closed-system four-momentum conservation requires including the field and external apparatus.

For a real scalar field with interaction $-\lambda {\varphi }^4/4!$, the first Dyson series term is $-i\lambda \int {\varphi }^4{d}^{4}x/4!$. Four labeled external legs can attach to the four fields in $4!$ ways, or in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)2!2!=24$ ways after separating incoming and outgoing legs. The factorial cancels and spacetime integration supplies the four-momentum conservation delta function. This gives the connected tree scattering amplitude $-\lambda$. The separate identical final-state symmetry factor belongs in the event phase-space integral and is not another vertex factor.

The fixed-target production threshold concerns the incident accelerator beam, whereas antiproton mass from momentum and velocity concerns an outgoing secondary beam. To see the distinction, consider a free proton hitting a stationary free proton, producingThe final electric charge is $+2e$ and final baryon number is $3-1=2$, as initially. Approximate all four final rest masses by the same proton rest mass $m$. With metric signature $(+---)$, define the squared centre-of-mass energy by $W^2=c^2(P_1+P_2{)}^2$. The incident proton has total energy $E=m{c}^2+K$, and the target four-momentum is $(mc,0)$. Four-momentum conservation and the mass shell relation giveAt threshold the four final particles have zero relative momentum in their centre-of-momentum frame, so $W_{\mathrm{thr}}=4m{c}^2$. HenceOnly $2m{c}^2$ of new rest energy is created. The larger incident kinetic energy is required because the final system must also carry the initial laboratory momentum. The incident total energy at threshold is $7m{c}^2$; neither this nor $K_{\mathrm{thr}}$ is the selected secondary momentum.

The 1955 authors explicitly qualified the stationary-nucleon threshold: target nucleon motion inside an atomic nucleus can lower the required incident energy.^[ref]

The kinematic reason is visible for two on-shell incoming nucleons:A target momentum directed against the beam increases $W^2$. Applying this illustration to a bound nucleon also requires binding energy and residual-nucleus recoil; the free stationary-target $6m{c}^2$ result is therefore not a universal sharp threshold for a copper target. A low observed secondary momentum is entirely compatible with a much larger accelerator kinetic energy.

For an incident particle of rest mass $m_1$ and total energy $E_1$ hitting a stationary target of mass $m_2>0$, four-momentum conservation gives $W^2=m_1^2{c}^4+m_2^2{c}^4+2m_2{c}^2{E}_1$. Applying the relativistic collision mass threshold gives the displayed necessary condition. The incoming energy must also satisfy $E_1\ge m_1{c}^2$; other conserved quantum numbers or reaction dynamics can impose further restrictions.

A bubble chamber using liquid hydrogen as both target and recording medium. For a beam interaction on a proton, the target's initial momentum can usually be approximated as zero on the scale of the beam momentum. This makes four-momentum conservation especially useful: the initial state does not include the unknown motion of a proton bound inside a heavier atomic nucleus. The approximation still neglects molecular binding and thermal motion.

With relativistically normalized external states, $\mathcal{M}$ is the reduced S-matrix element after removing the four-momentum conservation distribution. A relativistic scattering cross-section combines $|\mathcal{M}{|}^2$ with the Lorentz-invariant phase-space measure and the invariant flux factor. Identical-leg factorials in Feynman vertices and the identical final-state symmetry factor in the event integral have different roles.

In a head-on lepton-proton collision with the proton along $+z$, hadronic four-momentum conservation gives $y_h={\sum }_h(E_h-p_{zh})/(2E_e)$ and $Q_h^2=p_{T,h}^2/(1-y_h)$. With $s\simeq 4E_e{E}_p$, the reconstructed Bjorken scaling variable is $x_h=Q_h^2/(s{y}_h)$. This is useful for charged current events whose final neutrino escapes. Initial-state radiation and unmeasured forward hadrons spoil the ideal closure.

The on-shell measure $d^{3}p/(2E)$ is Lorentz invariant. The product measure together with the four-dimensional Dirac delta distribution enforces four-momentum conservation. For equal masses in two-body scattering, $d{\Phi }_2=\sqrt{1-4m^2/s}d\Omega /(32{\pi }^2)$.

A stationary particle of rest mass $m$ absorbing a photon of energy $E$ acquires total invariant massThis follows from four-momentum conservation, since the incoming photon has momentum magnitude $E/c$. If the system splits into two equal particles, the largest kinematically permitted daughter rest mass is $M/2$, attained when both are at rest in the centre-of-momentum frame. Their common laboratory speed at threshold is $cE/(m{c}^2+E)$. Above this daughter-mass threshold there is insufficient centre-of-momentum energy; below it, relative kinetic energy is available.

Under $z\mapsto (az+b)/(cz+d)$ with $ad-bc=1$, the coordinate measure contributes $|c{z}_i+d{|}^{-4}$ at insertion $i$. The Koba-Nielsen factor contributes $|c{z}_i+d{|}^{-{\alpha }^{\prime }{\sum }_{j\ne i}{p}_i\cdot p_j}$. Four-momentum conservation makes this exponent ${\alpha }^{\prime }{p}_i^2$, so the factors cancel precisely on the displayed tachyon mass shell. The correlator alone is covariant; its product with all the coordinate measures is invariant.

For a reconstructed electron–Positron pair, four-momentum conservation givesNeglecting the electron mass compared with the daughter energies, and calling the opening angle $\theta$, this becomesAs an original example, choose $E_{+}=2\mathrm{GeV}$, $E_{-}=3\mathrm{GeV}$ and $\theta =60^{\circ }$. Then $M_{ee}{c}^2=\sqrt{6}\mathrm{GeV}\simeq 2.449\mathrm{GeV}$, although the pair's laboratory energy is $5\mathrm{GeV}$. The pair carries net momentum. Neither its laboratory energy nor the original proton beam energy is its invariant mass; unobserved products carry additional four-momentum.

For equal, oppositely directed incoming beams with equal total energy $E_b$, their total momentum vanishes. Thus the incoming center-of-mass energy isStipulated $1.50\mathrm{GeV}$ beams therefore provide $3.00\mathrm{GeV}$ of incoming center-of-mass energy. This identity is exact for the stated equal-beam geometry. Radiation before annihilation can reduce the energy forming the final state; the apparatus also has a distribution of beam energies. A nominal machine setting is therefore not an exact event-by-event final-state invariant mass.

Put $Q_u=2/3$, $s=q_s^2$, $u=q_u^2$, and suppress the common colour factor ${\delta }_{ji}$. Define $i\mathcal{M}$ by the product of Feynman rules. With massless external Dirac spinors the two graphs giveThe massless Dirac equation gives $/p_{1}u(p_1)=0$ and $\bar{u}(p_3)/p_3=0$. ThusBy four-momentum conservation, $p_2=p_3-q_u$, soThe two terms cancel:This is the Ward identity for the incoming photon; replacing its photon polarization vector by its four-momentum produces zero. The outgoing photon obeys the analogous identity. Each exchange graph separately fails this test, so the cancellation requires their sum. Consequently the physical scattering amplitude is invariant under ${\epsilon }^{\mu }(p_2)\mapsto {\epsilon }^{\mu }(p_2)+c{p}_2^{\mu }$.

Use the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. The external on shell relations are $p_1^2=p_2^2=m_e^2$ and $p_3^2=p_4^2=m_t^2$. Expanding the three Mandelstam variables and using four-momentum conservation, $p_3+p_4=p_1+p_2$, gives

For the outgoing massless gauge bosons, take $k_1^2=k_2^2=0$. Work away from internal poles, with the Feynman i-epsilon prescription understood. The Dirac equations and four-momentum conservation giveMultiplying by the neighboring Dirac propagator numerator cancels its denominator. The two fermion-exchange diagrams therefore yield the matrix commutatorThe annihilation vector current obeys $q_{\lambda }{J}_c^{\lambda }=\bar{v}(/p_1+/p_2)t^{c}u=0$. Direct contraction of the three-boson tensor givesThusThe terms proportional to $J_c^{\mu }$ cancel, leaving the non-Abelian Ward identity for fermion annihilationThe open-index tensor is generally nonzero: neither ${ϵ}^{abc}$ nor $J_c\cdot k_2$ is constrained to vanish. For a concrete nonzero configuration, take centre-of-momentum frame kinematics $p_1=(E,0,0,p)$, $p_2=(E,0,0,-p)$, $p=\sqrt{E^2-m^2}$, $k_1=(E,E,0,0)$, $k_2=(E,-E,0,0)$. Standard Dirac spinors with two-component spin labels $\chi =(1,0{)}^T$ for $u$ and $\eta =(0,1{)}^T$ for $v$ give the spinor current $(0,2E,2iE,0)$ and its contraction with $k_2$ is $2E^2$. Choosing both colour spinors $(1,0{)}^T$ and $(a,b,c)=(1,2,3)$ multiplies this by $1/2$, still nonzero. This explicitly rules out an unconditional tensor identity $k_{2\nu }{\mathcal{M}}^{ab\mu \nu }=0$.

For a physical polarization vector ${\epsilon }_{1\mu }^{\ast }$, however, $k_1\cdot {\epsilon }_1^{\ast }=0$, and thereforeThis is the required decoupling of a pure-gauge polarization vector on leg 2 when the other leg is physical. Interchanging the legs gives the corresponding identity for leg 1. The role of the three-boson diagram is explicit: it cancels the noncommuting colour-current term left by the two fermion exchanges.