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In the center-of-momentum frame, four-momentum conservation gives two opposite photon momenta, each of energy $M_{\pi }/2$. Let $z=\cos {\theta }_{\ast }$, with the axis along the Lorentz boost from that frame to the laboratory. Reversing the axis specified in the source changes $z$ to $-z$ and has no effect on the following product. The Lorentz transformation of four-momentum givesWriting $s=\sin (A/2)$, b now impliesThe isotropy assumption is $dN=a|dz|$ on $-1\le z\le 1$. Each interior opening angle has two inverse branches, $z$ and $-z$, both of which must enter the change of variables for probability densities. The positive branch has derivativeConsequently the diphoton opening-angle distribution isIts support is $2\arcsin (1/\gamma )<A<\pi$ for $0<\beta <1$, and its integral is $2a$, the integral of the original angular probability density before normalization. Thus the normalized probability density is $(2a{)}^{-1}dN/dA$. Omitting one inverse branch would incorrectly halve the total count.

All daughter particles are massless, so the parent mass $M$ supplies the only kinematic scale at tree level. Each vertex has two derivatives, making its scattering amplitude scale as ${\lambda }_i{M}^2$. For the four-$b$ vertex, symmetrizing over the derivative-bearing legs yields a sum of pairwise four-momentum products, which is proportional to $M^2$ by four-momentum conservation.

For $n$ massless daughters, the Lorentz invariance Lorentz-invariant phase-space measure scales as $M^{2n-4}$: each $d^{3}q/(2E)$ contributes two powers and the four-dimensional Dirac delta removes four powers. Including the parent factor $1/(2M)$ gives $\Gamma \sim |{\lambda }_i{|}^2{M}^{2n-1}$. ThusThe unspecified factors are dimensionless angular integrals and identical-particle factors. Each result has mass dimension one, as a decay width must in natural units.

For a photon incident on a stationary electron, producing an additional electron and positron leaves three massive particles in the final state. Four-momentum conservation gives $W^2=m_e^2{c}^4+2m_e{c}^2{E}_{\gamma }$ for the centre-of-momentum energy. Its minimum is $3m_e{c}^2$, so the incident threshold is $4m_e{c}^2$. At threshold all final particles are at rest in their centre-of-momentum frame, which moves at $4c/5$ relative to the target laboratory frame. This differs from a heavy-target threshold because the electron target's recoil cannot be neglected.

The neutral Sigma baryon and Lambda baryon both have valence content $uds$, but different spin-isospin couplings. An electromagnetic transition changes those internal couplings without changing any quark flavour. In the simple ground-state quark model, the nonzero magnetic transition is associated with the up- and down-quark electromagnetic currents; the strange-quark spin operator alone cannot change the $ud$ pair from spin one to spin zero. Four-momentum conservation gives photon energy $E_{\gamma }=(M_{\Sigma }^2-M_{\Lambda }^2)/(2M_{\Sigma })$ in natural units.

A particle decay into three daughter particles subject to four-momentum conservation and the relativistic energy-momentum relation. The daughter energies generally vary continuously even in the parent's rest frame. Grouping two daughters into a recoil system of variable invariant mass turns the remaining kinematics into a relativistic two-body decay.

When two massive particles coalesce without additional outgoing radiation, four-momentum conservation makes the final four-momentum their sum. For collinear incident velocities, its rest mass is the positive square root of the displayed invariant, and its final velocity equals the center-of-momentum frame velocity $c^2{p}_{\mathrm{total}}/E_{\mathrm{total}}$. Relative kinetic energy can become internal rest energy, so the final rest mass is generally larger than the sum of the incident rest masses.

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.

Use natural units $c=1$ and assume open phase space, $M>m_1+m_2+m_3$. In the parent's rest frame, define $s_{23}=(p_2+p_3{)}^2$. Four-momentum conservation givesThe upper recoil-mass limit follows from $E_1\ge m_1$. For positive $m_2,m_3$, the lower recoil-mass limit follows in their pair rest frame and is attained when their relative momentum vanishes. If both recoil particles are massless, $s_{23}=2E_2{E}_3(1-\cos {\theta }_{23})$ reaches zero with nonzero parallel momenta. If exactly one is massless, $s_{23}$ approaches the nonzero daughter's squared mass only when the massless particle becomes soft. Thus that upper $E_1$ bound is a supremum for a genuine three-body event with all massless particles of positive energy. If $m_1=0$, the lower endpoint is likewise a soft limit. Multiplying the mass expressions for $E_1$ by $c^2$ restores energy units.

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.

In the rest frame of the parent particle, four-momentum conservation requires $M=E_1+E_2\ge m_1+m_2$. For two equal masses the daughter momentum magnitude is $\frac{1}{2}\sqrt{M^2-4m^2}$; the two-body decay phase space closes at $M=2m$. For a finite constant decay amplitude, the decay width tends to zero at this threshold.

A two-track invariant mass and a direction test answer different questions. In units $c=1$, assign each track the pion mass $m_{\pi }$ and reconstruct its momentum before magnetic deflection. With $p_{\pm }=|{\mathbf{p}}_{\pm }|$ and opening angle $\alpha$,This follows by expanding the squared sum of the two four-momenta. The angle $\theta$ between the reconstructed parent momentum $\mathbf{P}={\mathbf{p}}_{+}+{\mathbf{p}}_{-}$ and the neutral particle beam direction $\hat{\mathbf{n}}$ satisfiesA fully reconstructed two-body decay channel gives the parent mass and $\theta \simeq 0$, within experimental resolution. A three-body decay channel with an unseen neutrino generally misses both constraints; an accidental correct invariant mass alone is insufficient. Notice that the track opening angle $\alpha$ need not be small even when the parent direction angle $\theta$ vanishes.

The 1964 analysis selected $490<M_{\pi \pi }<510$ MeV and remeasured events with $\cos \theta >0.9995$, comparing adjacent mass bands.^[ref]

For an original dimensionless example, take $m_{\pi }=1$, $\hat{\mathbf{n}}=(0,0,1)$ and ${\mathbf{p}}_{\pm }=(\pm \sqrt{3},0,3)$. Each energy is $\sqrt{13}$; the opening angle is $60$ degrees since $\cos \alpha =(9-3)/(9+3)=1/2$. Yet $\mathbf{P}=(0,0,6)$ points exactly along the beam, and $M_{\pi \pi }^2=4(13)-36=16$. Thus a parent of mass $4$ passes both ideal tests. These chosen numbers illustrate four-momentum conservation, not detector performance.

If a fixed component of a Lorentz four-vector vanishes in every Lorentz frame, the vector is zero. For its time component, arbitrary infinitesimal boosts force every spatial component to vanish; for a spatial component, first use rotations and then a boost. Applied to the incoming-minus-outgoing total four-momentum, this shows that energy balance in every inertial frame implies full four-momentum conservation. The statement fails for general tensors: a nonzero antisymmetric tensor has zero diagonal components in all frames.