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Interpret $m$ as the rest mass of each initial ball and assume that no energy or momentum escapes the collision. Let $\gamma =(1-v^2/c^2{)}^{-1/2}$ be its Lorentz factor. The total initial energy and momentum areTheir conservation determines the final four-momentum. For a single lump, $P={\gamma }^{\prime }{m}^{\prime }{v}^{\prime }$ and $E={\gamma }^{\prime }{m}^{\prime }{c}^2$, henceThe final invariant mass follows from $m^{\prime 2}{c}^4=E^2-c^2{P}^2$:This is the rest mass of a coalescing relativistic collision. For $v>0$, $m^{\prime }>2m$: some kinetic energy becomes internal energy in the final rest frame. Setting $m^{\prime }=2m$ would discard that energy and contradict conservation. At small speeds $v^{\prime }\sim v/2$, while $v=0$ gives $m^{\prime }=2m$.

Fix the phi-fourth theory convention ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4/4!$, with mostly-minus Minkowski spacetime and $\hslash =c=1$. Define the scattering amplitude by stripping the overall momentum-conserving delta function from $S-1=i\mathcal{M}$. If instead the coupling is defined by ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4$, every vertex coupling below must be replaced by $24\lambda$; the question does not specify this factorial convention.

The momentum-space Feynman rules are a Feynman propagator $i/(k^2-m^2+i0)$ for each internal scalar line, a Feynman vertex $-i\lambda$ with four scalar legs, and four-momentum conservation at each vertex. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. For an amputated scattering amplitude there are no external propagators; external states have the usual relativistic normalization. Sum the connected diagrams and all inequivalent assignments of the labelled external momenta. The overall delta function is $(2\pi {)}^4{\delta }^4(\sum p_{\mathrm{in}}-\sum p_{\mathrm{out}})$.

These Feynman rules follow by expanding the Dyson series $T\exp (i\int d^{4}x{\mathcal{L}}_{\mathrm{int}})$ and applying the Wick theorem to the free-field correlation functions. A Wick contraction gives the free Feynman propagator; the $4!$ ways of contracting a vertex cancel its factorial denominator. The expansion's $1/V!$ cancels permutations of identical interaction insertions, while any remaining automorphisms give the Feynman-diagram symmetry factor. Fourier transformation produces momentum conservation, and the LSZ reduction formula amputates external propagators to obtain the scattering amplitude.

For six-point amplitudes in phi-fourth theory, an original pair of diagrams is:

In the tree-level Feynman diagram, let the internal four-momentum be $r=p_1+p_2-q_1=q_2+q_3+q_4$. The labelled diagram has symmetry factor one, so its contribution isThis is one channel. The full tree scattering amplitude sums the ten unordered partitions of the six labelled external legs into two groups of three, since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$.

For the one-loop triangle Feynman diagram, set $P=p_1+p_2$, $Q_{12}=q_1+q_2$, $Q_{34}=q_3+q_4$, so $P=Q_{12}+Q_{34}$. A consistent routing has denominatorsWith the external labels held fixed, no nontrivial graph automorphism remains, so again the symmetry factor is one. The contribution isThere are other labelled assignments and other one-loop topologies in the full scattering amplitude; the displayed expression belongs to the particular drawn diagram.

One can also express this scalar triangle Feynman integral using Feynman parameters. With $x+y+z=1$, completing the square in $x{D}_0+y{D}_1+z{D}_2$ givesThe identity $1/(D_0{D}_1{D}_2)=2{\int }_{x,y,z\ge 0}dxdydz\delta (1-x-y-z)/(x{D}_0+y{D}_1+z{D}_2{)}^3$ and the shifted momentum integral yieldThus ${\mathcal{M}}_{△}=-{\lambda }^3/(16{\pi }^2)$ times the displayed parameter integral. It is ultraviolet finite in four dimensions, although physical threshold singularities must retain the Feynman i-epsilon prescription.

For the fixed-target production threshold, the incoming four-momenta are $(E,\mathbf{p})$ and $(m,0)$, with $E=\sqrt{m^2+|\mathbf{p}{|}^2}$. Their invariant mass obeysIn the centre-of-momentum frame, four final particles have total energy at least $4m$, attained when all four are at rest in that frame. Therefore $s\ge 16m^2$, equivalently $E\ge 7m$, andThis is the kinematic threshold for $m>0$. Exactly at equality the final phase space has zero volume; a nonzero production rate requires a strict inequality.

For a massive particle of rest mass $m$, its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$, where the four-velocity is $U^{\mu }=\gamma (c,\mathbf{v})$, $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. With metric signature $(-,+,+,+)$, $P\cdot P=-m^2{c}^2$. A photon has four-momentum $(E/c,(E/c)\hat{\mathbf{n}})$, a future-pointing null vector. Four-momentum conservation states that the total incoming and outgoing four-momenta agree for an isolated interaction; this includes both energy and momentum conservation in every inertial frame.

Take the incoming photon direction as $x$. Its energy is $\hslash \omega =\mu m{c}^2$, so the total initial four-momentum isThe invariant mass available after photon absorption is therefore $M=m\sqrt{1+2\mu }$. In the centre-of-momentum frame, the two equal daughters have opposite momenta. Their total energy is at least their combined rest energy $2\alpha m{c}^2$, with equality precisely when both daughters have zero momentum in that frame. Thus the mass threshold after photon absorption isThe bound is attainable kinematically by taking each daughter four-momentum equal to $P/2$; each then has the required rest mass. It immediately gives ${\alpha }_{\max }\to 1/2$ as $\mu \to 0$.

At this threshold both daughters move with the centre-of-momentum frame. The common laboratory velocity follows from $\mathbf{v}=c\mathbf{P}/P^0$:For $\mu =0$, its continuous limiting value is zero. At smaller allowed daughter rest masses, some of the centre-of-momentum energy appears as their relative kinetic energy, so their four-momenta need not be $P/2$.

The labels X and Y cannot be uniquely assigned from the surviving text: their decay vertices and daughter tracks were contained in the missing figure. The available conservation laws nevertheless specify how the assignment would be obtained.

The initial ${\pi }^{-}p$ state has $(Q,B,S)=(0,1,0)$, with $S$ denoting strangeness. At a primary strong interaction vertex all three numbers must be reproduced. A final Lambda baryon or Sigma baryon has $B=1$, $S=-1$ and therefore needs accompanying $S=+1$, for example a neutral kaon $K^0=d\bar{s}$ or a positive kaon $K^{+}=u\bar{s}$, with the remaining electric charge balanced. At later weak interaction decay vertices, strangeness need not be conserved; it would be wrong to enforce it there.

A displaced $p{\pi }^{-}$ pair with the appropriate invariant mass identifies ${\Lambda }^0\to p{\pi }^{-}$. A displaced ${\pi }^{+}{\pi }^{-}$ pair can identify $K_S^0\to {\pi }^{+}{\pi }^{-}$ after an invariant mass and kinematic check. For a candidate radiative Sigma-to-Lambda transition, the neutral parent satisfiesin natural units. Its photon energy in the parent rest frame is $(m_{\Sigma }^2-m_{\Lambda }^2)/(2m_{\Sigma })$. With known charged four-momenta, each unobserved neutral momentum is the difference between the incoming and visible outgoing momenta at its vertex; its inferred direction must agree with the dashed segment.

This information cannot establish which dashed segment was called X. For a concrete ambiguity, bothobey the primary electric charge, baryon number, and strangeness constraints and can yield a neutral kaon decay plus a displaced Lambda baryon decay. Their distinct neutral segments and the extra photon would be resolved by the missing topology and kinematics. The supplied possible Sigma baryon decay alone is insufficient to label X and Y. A labelled answer is therefore a source gap, rather than a completed identification.

The conversion from a measured beam peak to a particle resonance mass is fixed, but the missing graph supplies neither the peak position nor its width. Let $T_R$ be the incoming pion kinetic energy at the peak and $\Delta T$ the full width at half maximum after allowing for instrumental resolution. With a stationary proton and natural units,Thus the peak invariant mass isThe width must also be transformed to the centre-of-mass frame. To first order about the peak,For a broad feature, transform the two half-height positions separately: if they are $T_{-}$ and $T_{+}$, then the corresponding width in invariant mass is $\sqrt{(m_p+m_{\pi }{)}^2+2m_p{T}_{+}}-\sqrt{(m_p+m_{\pi }{)}^2+2m_p{T}_{-}}$. An energy-dependent numerator or width can also displace the observed maximum from the resonance pole parameter.

The low-energy ${\pi }^{+}p$ channel is normally dominated by ${\Delta }^{++}(1232)$, a Delta baryon with $Q=2$, $B=1$ and $I=3/2$. As an external reference example, not a reading of the absent plot, $M=1232\mathrm{MeV}$ and $\Gamma =117\mathrm{MeV}$ give $T_R\simeq 190\mathrm{MeV}$, $\Delta T\simeq 154\mathrm{MeV}$ and $\tau \simeq 5.63\times 10^{-24}\mathrm{s}$.^[ref] The actual graphical estimates remain unavailable. In particular, using the laboratory width directly in $\tau =\hslash /\Delta T$ misses the factor $m_p/M$.

Neglect atomic binding energy and take the target electron at rest. Use natural units temporarily. The final state contains three electrons and one Positron, so its threshold invariant mass is $4m_e$. For incident total energy $E$,The corresponding incident kinetic energy is $6m_e{c}^2=3.066\mathrm{MeV}$. At threshold the products have no relative momentum in the centre-of-momentum frame; their common laboratory motion requires more energy than merely the pair's rest energy.

Conservation laws. Use natural units $c=1$ and write $p_i^{\mu }=(E_i,{\mathbf{p}}_i)$ with the $+---$ metric. In the parent's rest frame, which is the final-state centre-of-momentum frame, $P^{\mu }=(M,0)$. Four-momentum conservation and the relativistic energy-momentum relation giveA real massless daughter has strictly positive energy. The kinematic threshold is $M\ge {\sum }_i{m}_i$; below, no decay satisfying these conditions exists. The endpoint discussion assumes open phase space, $M>{\sum }_i{m}_i$, to avoid collapsed threshold configurations.

Two-body energies. Momentum conservation requires ${\mathbf{p}}_2=-{\mathbf{p}}_1$, so both daughters have the same momentum magnitude $p$. Their on-shell relations implyand subtraction eliminates $p$:Since $E_1+E_2=M$, solving the sum and difference equations gives the relativistic two-body decay resultAs a check, their common momentum magnitude iswhere $\kappa$ is the Källén function. It vanishes at the massive two-body threshold and is real on the allowed side. If the symbols $M,m_i$ denote masses in kilograms rather than rest energies, multiply the displayed energy formulas by $c^2$ and the displayed momentum formula by $c$.

Minimum energy in three-body decay. The on-shell inequality already gives $E_1\ge m_1$. For $m_1>0$, the bound is attainable by putting daughter 1 at rest and letting daughters 2 and 3 recoil back to back:The last equation has a solution because $M-m_1>m_2+m_3$; its left side increases continuously from $m_2+m_3$ to infinity. ThusIf $m_1=0$, the same formula describes the lower boundary $E_1\to 0^{+}$. It is an infimum, not an attained physical zero-energy massless particle: this is a soft-particle endpoint.

Maximum energy from the recoil mass. Group daughters 2 and 3 into a recoil system with four-momentum $Q=p_2+p_3=P-p_1$ and squared invariant mass $s_{23}=Q^2$. Directly in the parent rest frame,Maximizing $E_1$ therefore means minimizing the recoil invariant mass. This approach does not assume a recoil rest frame when that system is null.

If both recoil daughters have positive masses, use their pair rest frame. Their total energy there is at least $m_2+m_3$, with equality when both are at rest relative to one another. Equivalently, their future-directed four-velocities satisfy $p_2\cdot p_3\ge m_2{m}_3$, givingEquality means the two recoil daughters have a common velocity in the parent frame: their momenta are parallel and ${\mathbf{p}}_2/m_2={\mathbf{p}}_3/m_3$. Daughter 1 recoils in the opposite direction. HenceAt this endpoint their total energy and its sharing areThese explicitly exhibit a permitted configuration attaining the bound when both recoil masses are positive.

If exactly one recoil daughter is massless, say $m_3=0$ and $m_2>0$, work in the rest frame of daughter 2. For any positive photon-like daughter energy there,The lower recoil-mass bound is approached only as $E_3^{(2)}\to 0^{+}$. Thus the same general formula with $m_3=0$ is the supremum of $E_1$ for a genuine three-body event, rather than an attained maximum. Interchanging 2 and 3 covers the other mixed case. The distinction is important because the paper's category “at least one has mass” includes this soft-particle endpoint as well as the attainable two-massive-particle case.

If both recoil daughters are massless, instead computeIt can be exactly zero at nonzero $E_2,E_3$ when their momenta are parallel. Then $|\mathbf{Q}|=E_2+E_3$, and momentum conservation makes $|{\mathbf{p}}_1|=|\mathbf{Q}|$. Solving $E_1+|{\mathbf{p}}_1|=M$ and $E_1^2-|{\mathbf{p}}_1{|}^2=m_1^2$ givesBoth recoil daughters can have positive energies: set $E_2=x|{\mathbf{p}}_1|$ and $E_3=(1-x)|{\mathbf{p}}_1|$ for any $0<x<1$. This endpoint is attained without a soft daughter, even though their combined four-momentum is null and has no rest frame.

Combining the three-body decay energy endpoints, the closed kinematic range isFor masses in ordinary units, multiply both endpoint expressions by $c^2$. Open endpoints occur precisely in the soft limits described above. These bounds specify kinematic possibilities; the decay amplitude can still vanish at a boundary, so a possible endpoint need not have a nonzero event density.

The tau-neutrino charged-current threshold requires specifying the stationary target and the lightest allowed final state. For a free neutron, the lowest charged current channel is ${\nu }_{\tau }n\to {\tau }^{-}p$. At threshold both final particles are at rest relative to one another in the centre-of-momentum frame, soThe equal-nucleon-mass approximation gives $m_{\tau }+m_{\tau }^2/(2m_N)\simeq 3.46\mathrm{GeV}$. More generally $E_{\nu }^{\mathrm{thr}}=[(m_{\tau }+M_X{)}^2-M_T^2]/(2M_T)$ for a target of mass $M_T$ and a recoil system with minimum allowed invariant mass $M_X$. A stationary electron in ${\nu }_{\tau }{e}^{-}\to {\tau }^{-}{\nu }_e$ instead gives $(m_{\tau }^2-m_e^2)/(2m_e)\simeq 3.09\mathrm{TeV}$. The customary $3.5\mathrm{GeV}$ answer therefore presumes a nucleon target; nuclear thresholds and practical deep inelastic scattering thresholds differ.

At the Z boson pole, the accessible primary fermion pairs are $e^{+}{e}^{-}$, ${\mu }^{+}{\mu }^{-}$, ${\tau }^{+}{\tau }^{-}$, the three ${\nu }_i{\bar{\nu }}_i$ channels, and $u\bar{u},d\bar{d},s\bar{s},c\bar{c},b\bar{b}$. The top quark is too heavy. The charged channels also have photon exchange, and the electronic channel has a substantial nonresonant Bhabha scattering contribution. On-shell W boson pairs are above threshold; they are not primary $Z$ decays.

A tau lepton pair gives two approximately back-to-back narrow low-multiplicity cones, usually one or three charged tracks per cone, with overall charge zero and missing energy and momentum from neutrinos. Opposite hemispheres can show electron, muon or hadronic tags. A decay vertex or track impact-parameter information helps establish the short tau lepton flight. Reject Bhabha scattering and dimuon events by their nearly full visible beam energy, particle identification and little missing momentum; reject ordinary quark multijets by their broader high-multiplicity activity and larger visible invariant masses. Control two-photon backgrounds with visible transverse momentum, forward activity and event geometry. Selection efficiencies depend on decay mode and must be calibrated.

Use a high-multiplicity hadronic mode with only one invisible neutrino, and reconstruct the total visible four-momentum $p_h$. In the tau lepton rest frame,Fit the endpoint of the hadronic invariant mass distribution, preferably together with its laboratory energy boundary at known beam energy. The two-dimensional boundary uses more information than the single largest observed mass. Detector smearing, unseen neutral products, backgrounds, beam radiation and hadronic spectral modeling must be included. Absence of a displaced endpoint generally sets an upper bound rather than a positive neutrino mass measurement. Strictly the flavour state contains several mass eigenstates; an unresolved endpoint constrains the corresponding weighted mass spectrum.

Neglect radiation, take $E_{\tau }=M_Z/2\simeq 45.6\mathrm{GeV}$, $m_{\tau }\simeq 1.777\mathrm{GeV}$, and treat a fixed hadronic invariant mass $m_h=1.5\mathrm{GeV}$ as a single recoil system in ${\tau }^{-}\to h^{-}{\nu }_{\tau }$. With negligible neutrino mass,Thus $E_h^{\min ,\max }=\gamma (E_h^{\ast }\mp \beta p_h^{\ast })$, numerically approximately $32.50$ to $45.60\mathrm{GeV}$. In the ultrarelativistic limit the range is $E_{\tau }{m}_h^2/m_{\tau }^2\le E_h\le E_{\tau }$. A spread of hadronic invariant masses broadens these endpoints; an average mass alone is not a strict endpoint specification for the whole sample.

Let $P_{\tau }=+1$ denote a positive-helicity ${\tau }^{-}$, and measure $\vartheta$ from its laboratory flight direction in its rest frame. For a spin-zero hadronic recoil, the left-helicity tau neutrino and angular momentum conservation giveFor example a tau lepton with spin along $+z$ can emit a left-helicity neutrino along $-z$, carrying spin along $+z$, with the spin-zero recoil along $+z$. A positive-helicity tau lepton then gives a harder visible recoil. If $x_h=E_h/E_{\tau }$, $r=m_h^2/m_{\tau }^2$ and $\beta \simeq 1$, the more general tau hadronic polarimetry formula isFor spin-zero recoil ${\alpha }_h=1$. For a spin-one system produced by the V-A interaction, the longitudinal and transverse weights are $m_{\tau }^2:2m_h^2$, with opposite angular slopes; summing gives ${\alpha }_h=(m_{\tau }^2-2m_h^2)/(m_{\tau }^2+2m_h^2)$. At $m_h=1.5\mathrm{GeV}$ this is about $-0.175$, so the negative-helicity tau lepton has the higher mean total hadronic energy, about $39.43$ versus $38.67\mathrm{GeV}$, within that spin-one model. ^[ref]

The supplied mean invariant mass does not specify the recoil's spin composition or its helicity amplitudes. Consequently the last sign cannot be universal: positive helicity is harder for a spin-zero recoil, negative helicity is harder for the specified $1.5\mathrm{GeV}$ spin-one recoil, and a mixture requires its measured spectral weights. Polarization of an intermediate vector or axial-vector meson also changes the division of energy among its daughter hadrons; that information is not fixed by the total jet invariant mass alone.

Nuclear-field pair creation is $\gamma Z\to e^{-}{e}^{+}Z$. A virtual photon exchanged with the nucleus supplies recoil momentum, while the real photon and virtual photon attach to the electron line joining the outgoing pair. The figure shows a representative graph. The second attachment ordering must be included in the full amplitude. A real photon cannot create a free massive pair in vacuum: its invariant mass is zero, whereas $(p_{e^{-}}+p_{e^{+}}{)}^2\ge 4m_e^2$. Nuclear recoil resolves that kinematic obstruction and permits energy-momentum conservation.

Initial-state radiation of gluons gives the W boson nonzero transverse momentum. Its transverse Lorentz boost then changes the charged-lepton $p_T$, washing out the sharp $M_W/2$ edge and permitting events above it. The recoil must be modelled with Quantum chromodynamics and constrained by measured hadronic activity. Independently, the finite W boson decay width spreads its actual invariant mass, also smearing the fixed-mass endpoint. Either is a theoretical broadening mechanism; neither is a detector-resolution effect.

Choose $W^{+}\to {\mu }^{+}{\nu }_{\mu }$. Important backgrounds include Quantum chromodynamics production of heavy-flavour hadrons with semileptonic decays, in-flight pion or kaon decays, and hadrons penetrating the calorimeters. These generally have accompanying jet activity; semileptonic decays can contain real neutrinos, while jet mismeasurement can fake missing transverse momentum. Heavy-flavour tracks can originate at a displaced decay vertex. Require a high-$p_T$ isolated muon, agreement between inner tracking and the external muon system, and substantial missing transverse momentum inconsistent with detector noise or a mismeasured jet.

The genuine electroweak backgrounds include $Z\to {\mu }^{+}{\mu }^{-}$ with one muon lost and $W\to \tau \nu \to \mu \nu \nu \nu$. Veto a second reconstructed opposite-sign muon and use the dimuon invariant mass when both are present to reject the Z boson channel. The tau-mediated spectrum is softer and contains extra missing particles; template fits and control samples estimate its residual contribution. Detector acceptance determines how much of the lost-muon background survives.

In a relativistic two-body decay, four-momentum conservation in the parent's rest frame requires ${\mathbf{p}}_1^{\ast }=-{\mathbf{p}}_2^{\ast }$ andFor fixed daughter masses, this fixes a unique $p^{\ast }$, independent of direction. In the Lorentz-invariant phase space integral, the Dirac delta distribution enforcing conservation of energy therefore performs the radial momentum integration. Only the solid angle remains to be integrated; the resulting $p^{\ast }$ can be taken outside that integral, even if the matrix element depends strongly on direction.

For three or more final particles, four-momentum conservation generally leaves continuously variable daughter energies and momenta, so the decay rate includes integrations over them, or equivalently over independent invariant masses in a Dalitz plot. Integrating over a variable daughter invariant mass is also necessary when an unstable daughter is treated with a finite decay width rather than a fixed mass.

With the four-momenta in a, $q^2=-2E{E}^{\prime }(1-\cos \theta )$, $p_2\cdot q=m_p(E-E^{\prime })$, and $p_2\cdot p_1=m_{p}E$. Thus the measured beam energy $E$, scattered electron energy $E^{\prime }$ and scattering angle $\theta$ determineHere $Q^2$ measures the spacelike momentum transfer, $y$ is the fractional energy transfer, and the Bjorken scaling variable $x$ is the struck parton's longitudinal momentum fraction in the leading collinear description. The hadronic invariant mass satisfies $W^2=(p_2+q{)}^2=m_p^2+Q^2(1/x-1)$; large $Q^2$ and $W^2$ select deep inelastic scattering.

A beam spectrometer measures $E$ before a thin stationary hydrogen target. Tracking layers on both sides of a magnetic spectrometer measure the outgoing electron direction, charge and curvature; an electromagnetic calorimeter measures its energy. Matching the track to its particle shower, with downstream muon rejection where useful, separates the scattered electron from hadronic tracks. A downstream hadronic detector measures the hadronic activity and supports event-quality checks. The sketch in a shows these separate measurements.

The scattered electron provides a clean, single-particle reconstruction. Reconstructing the entire hadronic system requires summing many charged and neutral fragments, correcting unequal particle calorimeter response and losses through beam holes, and sometimes accounting for escaping neutrinos. Thus the leptonic definition is experimentally convenient. It is not a fundamental requirement of Lorentz invariance: exact reconstruction of $p_X$ gives the same $q=p_X-p_2$, and hadronic or combined reconstructions can be preferable in some kinematic regions.

Use a clean Z boson sample with final-state radiation, $Z\to e^{+}{e}^{-}\gamma$, and independently calibrated electron four-momenta. For measured $P=p_{e^{+}}+p_{e^{-}}$ and measured photon direction ${\hat{n}}_{\gamma }$, write $p_{\gamma }=E_{\gamma }(1,{\hat{n}}_{\gamma })$. The known Z boson mass suppliesComparing the reconstructed photon energy with this prediction constrains its scale versus energy, detector position and conversion status. In practice fit the $e^{+}{e}^{-}\gamma$ invariant mass distribution, allowing for the physical Z boson decay width, detector response and backgrounds, instead of assigning every event exactly $M_Z$. Select final-state-radiation topologies: an initial-state-radiation photon is not part of the Z boson decay and generally does not satisfy the same mass constraint. The calibration applies over the populated photon energy range; extrapolation to still higher energies needs a validated response model.

Tag an energetic initial-state radiation photon in $e^{+}{e}^{-}\to \gamma Z^{(\ast )}\to \gamma +\text{invisible}$, observing a single-photon signature with missing energy and missing momentum. At known squared center-of-mass energy $s$, the leading single-photon recoil invariant mass isAbove the Z boson mass, radiative-return events cluster at $m_{\mathrm{recoil}}\simeq M_Z$. At or near the pole the full off-shell recoil spectrum is fitted instead. Folding the Breit-Wigner resonance cross section with the calculable QED radiation kernel gives a predicted tagged rate proportional to ${\Gamma }_e{\Gamma }_{\mathrm{inv}}$ for fixed total decay width. A fit with the known luminosity and visible line-shape parameters therefore determines ${\Gamma }_{\mathrm{inv}}$ directly from invisible events, rather than subtracting visible widths. Comparing identically selected radiative-return events with visible Z boson decays can instead constrain ${\Gamma }_{\mathrm{inv}}/{\Gamma }_{\mathrm{vis}}$ while cancelling much of the luminosity normalization.

The detector needs efficient single-photon triggering, good photon energy and angular resolution, reliable timing, and nearly hermetic charged-particle and particle calorimeter coverage. Forward vetoes reject radiative Bhabha scattering with escaped electrons; additional-photon vetoes reject $\gamma \gamma$ events with a lost photon. Require sufficient transverse photon momentum that a recoiling visible system could not all disappear down a beam hole. Timing rejects cosmic rays and beam backgrounds. Detector inefficiencies and beam holes must be measured, since visible final states with lost particles imitate the signal. The irreducible $e^{+}{e}^{-}\to {\nu }_e{\bar{\nu }}_e\gamma$ contribution from $t$-channel W boson exchange and its interference must be included when isolating the Z boson invisible decay width.

In gluon-fusion Higgs production, two gluons couple to a closed top quark loop, and the Higgs boson couples to that massive quark through its Yukawa interaction. The right panel of the diagram in c shows one explicit colour-flow component: internal $t_r\to t_b$ at one gluon insertion, then $t_b\to t_r$ at the other; the colour-neutral Higgs boson insertion preserves the internal color charge. Include both loop orderings. The color charge factor isso the two incoming gluons are projected onto a colour singlet. The Yukawa coupling is $y_t=\sqrt{2}{m}_t/v$, and the $h\bar{t}t$ vertex has strength $m_t/v$. No direct tree-level $hgg$ mass coupling is needed. Integrating out a sufficiently heavy top quark generates an effective interaction proportional to $(h/v)G_{\mu \nu }^a{G}^{a\mu \nu }$, which is allowed by gauge invariance even though the gluon is massless.

There is also no kinematic obstacle: two massless particles with nonparallel momenta can have nonzero total invariant mass. In particular, head-on gluons have $\hat{s}=4E_1{E}_2$, providing the energy needed for the Higgs boson at the LHC.

The incident particle obeys the energy–momentum relation $E^2=p^2{c}^2+m_1^2{c}^4$. The stationary target contributes energy $m_2{c}^2$ and zero momentum. A single product inherits the total four-momentum, so its invariant mass satisfiesHenceHere $E$ is the incident total energy, including rest energy. In a one-particle final state, exact production of a specified mass fixes the incident energy; there is no adjustable final relative motion to absorb excess energy. A resonance with a finite decay width relaxes that ideal exact equality over its line shape.