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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.

Continue with $\eta =-g$ and dot products in that auxiliary metric. Odd gamma traces vanish. The two current tensors areContracting givesLet $u=2M^2-s-t$. The on-shell products are $p_1\cdot k_1=p_2\cdot k_2=(M^2-t)/2$, $p_1\cdot k_2=p_2\cdot k_1=(M^2-u)/2$, and $k_1\cdot k_2=s/2$. Substitution yieldsThe result is already averaged over incoming colours and spins and summed over outgoing ones. For $M\to 0$, it becomes $4g_s^4(t^2+u^2)/(9s^2)$.

An additional check uses the centre-of-mass angle and velocity ${\beta }_t=\sqrt{1-4M^2/s}$:It tends to $4g_s^4/9$ at threshold and is positive throughout the physical region. Keeping both printed minus signs would instead give $4g_s^4[(M^2-t{)}^2+(M^2-u{)}^2-2M^{2}s]/(9s^2)$, displaying the incorrect mass contribution explicitly.

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

At leading order, neglect the transverse momentum of the incoming partons and of the produced on-shell boson. In the hadron centre-of-mass frame,Then $m_W^2=(p_{+}+p_{-}{)}^2=x_{+}{x}_{-}s$, andSolving these product and ratio equations givesOver $-2<y<2$, each momentum fraction ranges betweenThey are anticorrelated: increasing $y$ increases $x_{+}$ while decreasing $x_{-}$. Higher-order recoil changes the simple two-to-one kinematics.

The charged-current process iswith a $t$-channel $W^{-}$ emitted at the electron-to-neutrino vertex and absorbed at the antimuon-to-antineutrino vertex. It is not an $s$-channel Annihilation of the unlike-flavour charged leptons into one $W$.

At high momentum transfer the incident electron's left-chiral fraction tends to $4/5$. In the collision centre-of-mass frame, the incoming antimuon moves along $-z$ and its charged-current allowed right helicity has spin along $-z$. A rest-frame spin at Angle $\theta$ to $+z$ projects onto that spin direction with probability ${\sin }^2(\theta /2)$. Therefore, if ${\sigma }_0$ denotes the cross section for a pure left-helicity electron and fully selected antimuon spin,It vanishes at $\theta =0$ in the ultrarelativistic selection limit and is maximal at $\theta =\pi$. Finite-mass corrections can fill the suppressed orientation. The absolute normalization depends on beam energy and the exchanged-boson propagator; the spin-angle dependence is what is fixed here.

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.

The center-of-momentum frame moves with velocity $\mathbf{V}=c^2{\mathbf{p}}_{\mathrm{tot}}/E_{\mathrm{tot}}$. Using the total four-momentum in solution givesFor $0<\theta \le \pi$, its direction bisects the two photon directions, and its speed decreases from the limiting value $c$ to zero. At $\theta =\pi$, the lab already is the center-of-momentum frame. At $\theta =0$, the formal speed is $c$ and no inertial rest frame exists; this endpoint is a limit, not an allowable boost.

With the reversed electric field, reflection about $x=d/2$ exchanges the electron and Positron. They therefore meet there with equal and opposite momenta. Each travels $d/2$, and the work-energy theorem gives its kinetic energy:Their combined kinetic energy is $1.000\mathrm{MeV}$, while each total relativistic energy is $\mathcal{E}=0.511+0.500=1.011\mathrm{MeV}$. The laboratory is the center-of-momentum frame of the colliding pair.

Local four-momentum conservation in the electron-positron annihilation excludes one real photon: the incoming total four-momentum is timelike, with zero spatial component and positive energy, while a single photon's four-momentum is null and has $|\mathbf{p}|=\mathcal{E}/c>0$. Two photons can balance their momenta by travelling in opposite directions, soThe equal photon energies follow from opposite equal momenta, not merely from sharing the incoming energy arbitrarily. Here $hc\simeq 1.23984\mathrm{MeV}\mathrm{pm}$. This is the free local annihilation idealization implicit in the question; significant momentum transfer to the external apparatus would change the kinematic system being considered.

The laboratory remains the center-of-momentum frame. At the relativistic collision mass threshold, the two final W bosons can have arbitrarily small opposite momenta. The available relativistic energy is $2m_e{c}^2+eEd$, soEach incoming particle must acquire about $80.378489\mathrm{GeV}$ of kinetic energy, travelling about $16.08\mathrm{km}$. Equality is the ideal kinematic threshold; a nonzero production rate needs energy above it. This estimate neglects acceleration losses and treats the W bosons with the supplied nominal rest mass.

A linear particle accelerator avoids repeatedly bending high-energy electrons and Positrons. This greatly reduces the bending synchrotron radiation loss, which at fixed bending radius grows approximately as the fourth power of the relativistic energy for a given particle species. Its useful collision energy can therefore be increased without the severe circular electron collider radiation penalty. The price is long structures or high accelerating gradients: each bunch normally passes once, requiring substantial beam power, fresh bunches, and very tight focusing for high luminosity. A circular particle accelerator reuses its accelerating structures and circulates bunches through many collisions, but needs a large ring and substantial radiofrequency power to replace synchrotron radiation losses. Heavier particles such as protons radiate far less at a given energy and radius, although a proton's constituent structure makes the hard-collision energy less sharply specified than in an electron-positron annihilation experiment.

Use $c=1$ and define the total four-momentum $P^{\mu }=(E,\mathbf{P})={\sum }_i(E_i,{\mathbf{p}}_i)$. For an inertial frame moving with velocity $\mathbf{V}$ relative to the laboratory, the Lorentz boost givesChoose $\mathbf{V}$ parallel to $\mathbf{P}$ and require zero total three-momentum. ThenThis requires $E>|\mathbf{P}|$, so that the total four-momentum is timelike. Its invariant mass is $\sqrt{E^2-|\mathbf{P}{|}^2}$, the total relativistic energy in the center-of-momentum frame. If all particles are exactly parallel photons, the total four-momentum is null and no finite-velocity center-of-momentum frame exists.

Sign qualification. The minus sign printed in the source describes the laboratory's velocity in the center-of-momentum frame, or the parameter of an active Lorentz boost with a plus-sign spatial rule. It is not the velocity of the center-of-momentum frame relative to the laboratory under question 1's passive convention. For example, a single particle with $(E,P_x)=(5,3)$ has its rest frame moving at $V_x=3/5$. The printed $V_x=-3/5$ would give $P_x^{\prime }=(5/4)(3+3)=15/2$, which is not zero. This explicitly resolves the sign rather than assigning a rest frame moving away from the particle.

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.

From c, $\sin (A/2)\ge 1/\gamma$, with equality when $z=0$: the photons are emitted perpendicular to the Lorentz boost axis in the center-of-momentum frame. ThereforeAt the minimum, the map $z\mapsto A$ is stationary: writing $A=A_{\min }+\delta$, its leading behavior is $\delta =\beta z^2/\gamma +O(z^4)$. A nonzero interval of uniformly distributed $z$ is therefore compressed near $A_{\min }$, producing an integrable inverse-square-root singularity in the opening-angle probability density. No differentiation of that probability density is needed. For finite $\gamma >1$, the probability density vanishes at $A=\pi$. For $\gamma =1$ the center-of-momentum frame is the laboratory and every event has $A=\pi$: the probability distribution becomes a Dirac delta distribution, rather than the singular $\beta =0$ substitution in c. At large $\gamma$, $A_{\min }\sim 2/\gamma$ in radians.

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.