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For physical four-momentum $P^{\mu }=md{x}^{\mu }/d\tau$, the covariant-component equation is $d{P}_{\mu }/d\tau =({\partial }_{\mu }{g}_{\nu \sigma })P^{\nu }{P}^{\sigma }/(2m)$. The equation without $1/m$ instead uses a tangent parametrization, or the affine parameter $\lambda =\tau /m$ when the tangent is the physical momentum.

A change $\lambda \mapsto A\lambda +B$ rescales a geodesic tangent and all its linear Killing constants by $1/A$. A null norm remains zero and cannot fix $A$, unlike proper-time normalization of a material particle. Ratios of the constants, such as the photon impact parameter $b=|h|/(ck)$ in Schwarzschild spacetime, are independent of the affine scale. A physically normalized four-momentum fixes the photon energy and hence selects a tangent scale.

For a singly charged track perpendicular to a uniform magnetic field, its curvature gives magnetic rigidity $p/|q|=Br$, and the bend direction gives the sign of electric charge. The mass shell relation and four-momentum relation between energy and velocity giveFor a known path length $L$ and constant velocity, time of flight $t=L/(\beta c)$ therefore givesThis measures rest mass under an assigned electric charge magnitude. Curvature alone measures $p/|q|$, so independent evidence for $|q|=e$ matters. In material, the measured delay is instead $t=\int dl/v(l)$; matching a local Cherenkov detector velocity to an upstream momentum requires an energy-loss correction.

As an original illustrative calculation, choose $p=1.20\mathrm{GeV}/c$ and $L=12.0\mathrm{m}$, with rounded input rest energies $m_p{c}^2=0.93827\mathrm{GeV}$ and $m_{\pi }{c}^2=0.13957\mathrm{GeV}$. Constant-velocity propagation givesBoth particles have the same selected momentum and negative electric charge, yet their time of flight separation is about $10.51\mathrm{ns}$. These are calculated values for a deliberately simplified example, not measured discovery events.

For example, the toy energy loss changes the central antiproton velocity from $0.7878c$ to $0.7745c$. Directly substituting the slower downstream value together with the upstream momentum would overestimate the rest mass. The two panels thus describe different stages of a deliberately specified model; their numerical consistency depends on propagating energy between stages. Likewise, the scattered-pion markers represent an erroneous C2 response, not actual slow pions; an upstream C1 signal still rejects them.

For future-directed particles with total four-momentum $P$ and metric signature $(-+++)$, the positive square root of $-P_{\mu }{P}^{\mu }$ is their energy in the frame where total spatial momentum vanishes. Its invariant definition remains valid at a collision in spacetime.

In an isolated relativistic collision, total four-momentum is conserved; its time component states conservation of total relativistic energy.

This is the relativistic form of conservation of energy, expressed through the conserved four-momentum.

A particle accelerator that brings counterpropagating electron and Positron beams into collision. The known initial four-momentum and tunable center-of-mass energy make resonance scans and missing-energy measurements possible.

The energy component of a transferred four-momentum in a specified frame. In fixed-target deep inelastic scattering the target rest frame gives $\nu =E-E^{\prime }$, where $E,E^{\prime }$ are incoming and scattered electron energies. It is frame dependent even though $Q^2=-q^2$ is Lorentz invariant.

Equal laboratory recoil angles constrain outgoing momenta, not equality of masses. In units $c=1$, take an incoming mass $m$ with energy $2m$ striking a stationary mass $2m$. The final momenta $m(13/8,\sqrt{3}/2,\sqrt{57}/8,0)$ and $m(19/8,\sqrt{3}/2,-\sqrt{57}/8,0)$ conserve four-momentum and have mass shells $m^2$ and $4m^2$. Both trajectories make angle $\arctan (\sqrt{19}/4)$ with the incident axis. This is a concrete counterexample to inferring equal masses from equal outgoing angles alone.

A source on a timelike circular Schwarzschild geodesic at radius $r>3GM/c^2$ emits light to a static observer at infinity on the orbital symmetry axis. Such a ray has zero conserved axial angular momentum. The locally measured photon energy is the contraction of its four-momentum with the observer's four-velocity. Comparing the emitter with the observer at infinity therefore gives ${\nu }_{\mathrm{em}}=U^t{\nu }_{\infty }$, where $U^t=[1-3GM/(c^{2}r){]}^{-1/2}$, proving the formula.

Equivalently the gravitational redshift factor $\sqrt{1-2GM/(c^{2}r)}$ multiplies the transverse Doppler effect factor $1/\gamma$ measured by a local static observer. For $R_S=2GM/c^2\ll r$,The coefficient contains both gravitational and orbital time dilation. It assumes an observer at infinity on the symmetry axis; other viewing directions generally introduce a first-order longitudinal Doppler effect.

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.

The four-force is the proper time derivative of four-momentum. For constant rest mass and coordinate three-force $\mathbf{F}=d\mathbf{p}/dt$, its components are $K^{\mu }=\gamma (\mathbf{F}\cdot \mathbf{v}/c,\mathbf{F})$ in the convention $p^{\mu }=(E/c,\mathbf{p})$. Thus it packages the power transfer and three-force into one Lorentz-covariant vector.

The generators of spacetime translations in a quantum field theory. With the mostly-minus Minkowski metric, $P\cdot a=H{a}^0-\mathbf{P}\cdot \mathbf{a}$. A momentum eigenstate satisfies $P^{\mu }|p⟩=p^{\mu }|p⟩$, where $p^{\mu }$ is its four-momentum. After subtracting the vacuum energy, a translation-invariant vacuum state has eigenvalue zero.

A four-vector is a vector whose four vector components transform under a Lorentz transformation as $U^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{U}^{\nu }$. Its individual vector components generally change between inertial frames; the Minkowski metric combines them into invariant scalar products. Listing four quantities is not enough to make a four-vector: their transformation law matters.

The most direct example is the displacement between two spacetime events, written $(c\Delta t,\Delta x,\Delta y,\Delta z)$. Multiplying the time difference by the speed of light gives all four vector components the same length units. A position four-vector describes an event relative to a chosen origin, while a displacement does not depend on a common translation of the origin. Four-momentum and four-velocity use the same transformation rule for different physical quantities.

The Fourier-space quadratic Maxwell kernel after adding $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$. For nonzero $\xi$ and non-null four-momentum, its transverse and longitudinal momentum projectors give $K=-k^2(P_T+{\xi }^{-1}{P}_L)$ and its inverse is $-(P_T+\xi P_L)/k^2$. The inversion of the gauge-fixed Maxwell kinetic operator supplies the photon propagator with its vacuum boundary prescription.

Each Killing vector field supplies a conserved quantity for freely moving test particles: it turns a geometric symmetry into a first integral of the geodesic equation. Let $\gamma (\lambda )$ be a geodesic with affine parameter $\lambda$, tangent vector $U^{\mu }=d{x}^{\mu }/d\lambda$, and ${\nabla }_{U}U=0$. Wherever a Killing vector field $K$ is defined along the geodesic,The statement applies to Riemannian manifolds and to timelike geodesics, null geodesics and spacelike geodesics in a Lorentzian manifold. It requires an affine parameter; for a massive test particle, proper time is one choice. See Tong, section 4.3.2, equation (4.32).

For a worked example in the Euclidean plane, take $\gamma (t)=(1+t,2-t)$ with affine parameter $t$ and velocity $U=(1,-1)$. The Euclidean Killing vector fields for translation and rotation giveThe rotational field along the geodesic is $K_{\mathrm{rot}}(\gamma (t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy mass $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of linear momentum, and the second is the signed $z$ component of angular momentum about the origin. Their numerical values use toy units. The geodesic is straight; the rotation flow orbits are circles.

This conservation law follows from metric compatibility and the Killing equation:The geodesic equation kills the acceleration term, and the symmetric product $U^{\mu }{U}^{\nu }$ selects the vanishing symmetric derivative. This tensor contraction is the whole proof outline; see Carroll, equation (5.43).

In a stationary spacetime, a chosen time-translation Killing vector field $K$ gives Killing energy. With metric signature $(-+++)$, $c=1$, and a massive test particle's four-momentum $p^{\mu }=m{U}^{\mu }$, the physical normalization isHere $\Phi$ generates the axial rotational symmetry when one exists. In coordinates with $K={\partial }_t$ and $\Phi ={\partial }_{\varphi }$, these are $E=-p_t$ and $L_z=p_{\varphi }$. Both are conserved quantities along the geodesic, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every spacetime has a global timelike Killing vector field, and a stationary field can become spacelike in an ergoregion of a stationary spacetime. Conservation on the field's domain does not imply a universal global positive energy; see Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges.

Set the inertial cylindrical angle to $\Phi =\varphi +\omega t$. The Minkowski metric becomes $d{s}^2=(c^2-{\omega }^2{r}^2)d{t}^2-2\omega r^{2}dtd\varphi -d{r}^2-r^{2}d{\varphi }^2-d{z}^2$. For a massive geodesic parametrized by proper time, its cyclic coordinates give conserved covariant four-momentum components $p_t=m[(c^2-{\omega }^2{r}^2)\dot{t}-\omega r^2\dot{\varphi }]$, $p_{\varphi }=-m{r}^2(\dot{\varphi }+\omega \dot{t})$ and $p_z=-m\dot{z}$. Their combination gives $\dot{t}=(p_t-\omega p_{\varphi })/(m{c}^2)$. The radial Euler-Lagrange equation gives the displayed coordinate acceleration, although the free worldline is straight in inertial coordinates. The conserved inertial energy is $p_t-\omega p_{\varphi }$, and the inertial axial angular momentum is $-p_{\varphi }$.

The inclusive product of hadronic-current matrix elements, summed over final states with a Lorentz-invariant phase-space measure. Electromagnetic current conservation makes both indices transverse to the exchanged four-momentum.

The bridge from detector readout to a particle resonance is four-momentum conservation. In units $c=1$, sum the measured final-state four-momenta:For two photons separated by angle $\alpha$, $|{\mathbf{p}}_i|=E_i$ gives the displayed diphoton formula. Two back-to-back photons of $60\mathrm{GeV}$ each have $m_{\gamma \gamma }=120\mathrm{GeV}$. A four-lepton candidate instead uses four four-momenta in the same sum. Collision debris outside the selected decay is not included. Since the partons carry unknown fractions of the incoming proton momenta, reconstructing the decay products is more informative than assigning the full beam energy to each candidate.

Suppose the intrinsic decay width is negligible compared with the stipulated detector spread. Let the reconstructed signal mass follow a normal distribution with mean $m_0=120\mathrm{GeV}$ and standard deviation ${\sigma }_m$. For a total expected selected signal $S=60$ and flat background density $b=8$ events per $10^9$ electronvolts, the expected count density isThis is a deliberately chosen response model, not a fit to either experiment. In the fixed window $m_0-a<m<m_0+a$ with $a=3\mathrm{GeV}$, direct integration givesThe expected signal counts are $57.2700$ for ${\sigma }_m=1.5\mathrm{GeV}$ and $40.9614$ for ${\sigma }_m=3\mathrm{GeV}$. Doubling the spread lowers the fraction in the same window without changing the total signal. Detector resolution can therefore change the visibility of a peak even with identical particle production.

As a separate counting example, stipulate an observed integer count $n=70$ and an exactly known background mean $B_w=48$. The Poisson distribution gives a likelihood $L(s)=e^{-(48+s)}(48+s{)}^{70}/70!$ for nonnegative signal mean $s$. Its maximum-likelihood estimator is $\hat{s}=22$, and the likelihood ratio statistic against $s=0$ isThis tests a specified window with a specified background, and is not either experiment's discovery significance. If the background mean is allowed to vary freely, background alone can choose $70$ and fit that count exactly. Control samples, a defensible mass-shape model, nuisance parameters and corrections for searching many masses are therefore essential to the experimental inference.

Conservation of four-momentum forbids an isolated massive initial particle from decaying into a single photon: the initial Minkowski norm is $m^2{c}^2>0$, while the final norm would vanish. This argument is independent of inertial frame. A second final particle or external recoil changes the problem because the total final four-momentum need not be null.

The neutron and Positron together have more rest mass than the initial proton, so a finite incident energy is necessary even before recoil is considered. In the proton's rest frame, use four-momentum conservation and a negligible neutrino mass. The initial squared center-of-mass energy is $m_p^2{c}^4+2m_p{c}^2{E}_{\nu }$; the least final squared center-of-mass energy is $(m_n+m_e{)}^2{c}^4$. Equating them gives the threshold above. At threshold the final particles share one center-of-mass velocity; they are not both at rest in the laboratory.

Writing $\Delta =m_n-m_p$, the result separates intoThe positive second term is the laboratory recoil correction. With rounded rest energies $m_p{c}^2=938.272\mathrm{MeV}$, $m_n{c}^2=939.565\mathrm{MeV}$ and $m_e{c}^2=0.511\mathrm{MeV}$, the calculated threshold is about $1.806\mathrm{MeV}$. These rounded inputs follow the recommended particle masses.^[ref]

As a further ideal energy balance, suppose the Positron stops and annihilates with an electron, and all prompt annihilation gamma rays are contained. Its prompt visible energy is $T_{e^{+}}+2m_e{c}^2$. Neglecting neutron recoil gives $E_{\mathrm{vis}}\simeq E_{\nu }-(m_n-m_p)c^2+m_e{c}^2\simeq E_{\nu }-0.782\mathrm{MeV}$. Escaping gamma rays, detector response and recoil change that measured sum. The delayed capture signal is a separate energy deposit, not an extra contribution to the incident electron antineutrino's prompt energy reconstruction.