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Why add the daughter four-momenta? Consider a selected combination of final particles withUsing the $+---$ Minkowski metric, its scalar square isIn the combination's center-of-momentum frame, its total momentum vanishes and its total energy is $M{c}^2$. Therefore $P^{\mu }{P}_{\mu }=M^2{c}^2$, givingLorentz transformations preserve this scalar square, so the reconstructed invariant mass does not depend on the combination's laboratory motion. It includes relative kinetic energy and need not equal the sum of the daughter rest masses. A random pair has an invariant mass too; that number does not itself establish a common parent.

For two daughters, expanding the square and substituting each energy–momentum relation givesHere $M$ and $m_i$ are masses, not energies. Reporting $M$ in $\mathrm{GeV}/c^2$ and momenta in $\mathrm{GeV}/c$ is equivalent to calculating numerically with $M{c}^2$ and $c{\mathbf{p}}_i$ in $\mathrm{GeV}$. Setting $c=1$ in natural units suppresses these conversion factors; it does not change the measured quantity.

Alvarez's lecture illustrates invariant mass searches with $K\pi$ combinations and contrasts resonance lifetimes near $10^{-23}\mathrm{s}$ with the much longer lived states the chambers originally targeted. These resonances decayed before producing a resolvable parent track.^[ref]

Operationally, choose a reaction hypothesis, require an acceptable bubble-chamber kinematic fitting result, and form specified daughter combinations. Accumulate their invariant masses in a histogram across events. A particle resonance is inferred from the ensemble's dynamics; its short-lived parent is not directly photographed. Daughter tracks can be visible even when the production and decay positions of the parent cannot be separated. Wrong combinations create background, and several combinations from one event are statistically correlated.

A projectile of mass $m_1$ elastically scattering from a stationary target of mass $m_2<m_1$ has the displayed maximum laboratory angle. In $c=1$ units, let its incident energy be $E>m_1$ and $s=m_1^2+m_2^2+2m_{2}E$. Boosting the outgoing four-momentum from the center-of-momentum frame gives $\tan \theta =\sin \phi /[\Gamma (\cos \phi +B)]$, with $\Gamma =(E+m_2)/\sqrt{s}$ and $B=(m_1^2+m_{2}E)/[m_2(E+m_2)]>1$. The maximum at $\cos \phi =-1/B$ has $\tan {\theta }_{\max }=1/[\Gamma \sqrt{B^2-1}]=m_2/\sqrt{m_1^2-m_2^2}$. Equality is allowed; a strict upper bound would exclude a realizable outgoing direction. The angular cap is independent of incident energy.

The interaction $-g\varphi {\Psi }_i^{\ast }{\Psi }_i$ gives one cubic vertex $-ig$ joining a real-scalar line to a particle-antiparticle pair of either complex scalar species. The second-order term of the Dyson series isWick theorem contracts the incoming ${\Psi }_1{\overline{\Psi }}_1$ pair at one vertex, the outgoing ${\Psi }_2{\overline{\Psi }}_2$ pair at the other, and the two $\varphi$ fields with each other. The two assignments of $x,y$ cancel the factor $2!$. Thus there is one connected tree-level Feynman diagram, the $s$-channel exchangeWith standard relativistic external-state normalization,up to the physically irrelevant common sign convention for $\mathcal{M}$.

In the center-of-momentum frame, $q=(E_1,-\mathbf{p})$ and $p=(E_1,\mathbf{p})$, soPair creation is kinematically possible exactly when $s\ge 4M_2^2$. The threshold incident momentum is therefore

Let the inertial frame $S^{\prime }$ move at speed $v$ in the positive $x$ direction relative to $S$, with parallel axes and coincident origins at zero time. The inverse Lorentz transformation givesHere ${\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ and ${\mathbf{u}}^{\prime }=d{\mathbf{x}}^{\prime }/d{t}^{\prime }$ is the particle's velocity in $S^{\prime }$. Dividing spatial differentials by $dt$ gives the relativistic velocity addition lawFor the collision, set $c=1$ during the algebra. Let the incident projectile have energy $E>m_1$ and momentum $p=\sqrt{E^2-m_1^2}$. The target is initially at rest. The squared invariant mass, boost speed and Lorentz factor of the center-of-momentum frame areIn that frame the projectile has $p_{\ast }=m_{2}p/\sqrt{s}$ and $E_{\ast }=(m_1^2+m_{2}E)/\sqrt{s}$. An elastic collision changes the direction of its momentum, through an angle $\phi$, while preserving these magnitudes. Transforming the outgoing four-momentum back to the laboratory givesThe maximum occurs at $\cos \phi =-1/B$. Sincethe maximum laboratory deflection in relativistic elastic scattering isThe printed strict inequality needs correction: equality is kinematically attainable. For example, in $c=1$ units take $m_1=2$, $m_2=1$, and incident $E=3$. A projectile scattered through $30^{\circ }$ with final energy $16/7$ has momentum $(3\sqrt{5}/7,\sqrt{15}/7)$. The target then has energy $12/7$ and momentum $(4\sqrt{5}/7,-\sqrt{15}/7)$; both outgoing particles satisfy their mass shells and conserve total four-momentum. This gives $\sin \theta =1/2=m_2/m_1$ exactly. The inequality is strict only away from the maximizing direction.

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.