Four-momentum

A massive particle has four-momentumFor a massless particle, $P^{\mu }=(E/c,\mathbf{p})$ with $E=c|\mathbf{p}|$ and $P^2=0$.

A fixed set of four-momenta at fixed events sums to a four-vector, whether or not the set is the whole physical system. Its Minkowski norm is invariant. Reapplying a frame-dependent particle-selection rule or sampling different events need not give the transformed same sum. Physical conservation across different hypersurfaces is a separate issue involving interactions and boundary flux.

Two particles with unequal rest energies $r,R$ and equal opposite velocities have total energy $\gamma (r+R)$ but nonzero total momentum. If they form a composite while emitting a photon transverse to their initial axis, the final composite rest energy $Q$ satisfies the displayed identity. It follows by squaring the residual four-momentum. Neglecting binding, put $Q=r+R$; the minimum incoming energy is $(r+R)[\chi +\sqrt{1+{\chi }^2}]$, where $\chi =\epsilon (r+R)/(4rR)$. An exact threshold requires the composite rest mass, not just the incoming constituent masses.

A sum of particle four-momenta at specified events is a four-vector, but sums recomputed at simultaneous events in different inertial frames need not correspond to that same four-vector. For opposite equal electric charges initially at rest a distance $d$ apart, a longitudinal electric field switched on simultaneously in the laboratory produces zero total laboratory three-momentum. In a frame moving at $v>0$, the switch-on events are separated by ${\gamma }_{v}vd/c^2$; during that interval only one force acts. The late equal-time total three-momentum is $-2{\gamma }_{v}mv+{\gamma }_{v}eEvd/c^2$, while the Lorentz transformation of an equal-laboratory-time sum varies as the laboratory pair's energy grows. Closed-system four-momentum conservation requires including the field and external apparatus.

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.

Three-momentum is the spatial part of a particle's four-momentum, $p^{\mu }=(E/c,\mathbf{p})$. Its components depend on the reference frame, while $p_{\mu }{p}^{\mu }=m^2{c}^2$ is a Lorentz invariant.

An observer with future-directed unit velocity $u^{\mu }$ measures particle energy $E=-u_{\mu }{p}^{\mu }$ and spatial physical momentum given by the orthogonal projection of its four-momentum in the displayed formula. With signature $(-+++)$ and $p_{\mu }{p}^{\mu }=-m^2$, its squared magnitude is $p_{\perp \mu }{p}_{\perp }^{\mu }=E^2-m^2$. Components in the observer's orthonormal spatial frame agree with $\gamma m\mathbf{v}$. They differ in general from coordinate canonical momentum components. For a comoving observer in flat FRW, a conserved conventional spatial comoving covector has locally measured magnitude divided by the scale factor.

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.

The mass shell is the locus of four-momenta satisfying the relativistic dispersion relation for a fixed mass $m$. With mostly-minus Minkowski metric, $p^2=m^2$; with mostly-plus Minkowski metric, $p^2=-m^2$. The positive-energy sheet represents particle four-momenta. For $m=0$ this is the light cone. The condition of being on shell refers more generally to satisfying the field equations.

In an isolated relativistic interaction, the sum of incoming four-momenta equals the sum of outgoing four-momenta.

In an isolated collinear completely inelastic capture, add incoming total energies and signed momenta. Their invariant fixes the composite rest mass, including internal excitation. If it later converts all energy to two opposite axial photons, the two conservation equations immediately give the displayed photon energies. The laboratory frequency ratio is $(E+Pc)/(E-Pc)$, not generally one; equality holds in the composite rest frame.

In energy units $m=m_{\mathrm{in}}{c}^2$ and $M=M_A{c}^2$, capture of a photon of energy $e$ by a massive particle into one resonance satisfies $E-P\cos \theta =(M^2-m^2)/(2e)=2E_0$, with $P=pc$. Head-on incidence gives the largest invariant mass at fixed $E$, so the threshold solves $E+P=2E_0$. Using $E^2-P^2=m^2$ gives the title expression. It is the minimum over incident angle; above that energy the exact angle is fixed by the invariant equation, not arbitrary. The soft-photon regime $2E_0\gg m$ excludes exactly co-propagating incidence.

When $e\ll m\ll E_0$, the incoming massive-particle energy is $E\simeq 2E_0/(1-\cos \theta )$. Momentum addition then gives $\tan \varphi \simeq [e/(2E_0)]\sin \theta (1-\cos \theta )$. Differentiating the angular factor gives its maximum at $\cos \theta =-1/2$. This is an asymptotic result for both the energy-angle constraint and final deflection; corrections are controlled by $m^2/E_0^2$ and $e/E_0$.

For a photon and a massive particle with zero total momentum, their momenta are opposite and equal. Conservation of total energy fixes the same momentum magnitude before and after elastic scattering, independently of direction. A right-angle photon turn therefore makes the recoiling particle move perpendicular to the original photon direction in that frame. Transform both velocity components, or the complete four-momentum, to obtain the laboratory recoil; a transverse component is reduced by the boost Lorentz factor when the particle has zero longitudinal velocity in the zero-momentum frame.

If the total incoming four-momentum is timelike and has centre-of-momentum energy $W$, production of particles with rest masses ${\mu }_i$ requires $W\ge {\sum }_i{\mu }_i{c}^2$. In the centre-of-momentum frame, total final energy is a sum of $\sqrt{{\mu }_i^2{c}^4+c^2|{\mathbf{p}}_i{|}^2}$, each at least its rest energy. For massive products equality occurs with all at rest in that frame. A single product requires equality of the entire invariant mass, not just an inequality; a massless product cannot have nonzero energy and be at rest.

For an incident particle of rest mass $m_1$ and total energy $E_1$ hitting a stationary target of mass $m_2>0$, four-momentum conservation gives $W^2=m_1^2{c}^4+m_2^2{c}^4+2m_2{c}^2{E}_1$. Applying the relativistic collision mass threshold gives the displayed necessary condition. The incoming energy must also satisfy $E_1\ge m_1{c}^2$; other conserved quantum numbers or reaction dynamics can impose further restrictions.

For a photon of energy $h\nu$ hitting a stationary target of mass $m_2$, the incoming invariant energy satisfies $W^2=m_2^2{c}^4+2m_2{c}^{2}h\nu$. The massive multi-product threshold is the displayed value when ${\sum }_i{\mu }_i>m_2$. If the algebraic bound is nonpositive, the lower limit of allowed positive frequencies can be zero. A single final product instead fixes the frequency by invariant-mass equality, and a final massless particle can make a nominal threshold an infimum rather than an attained value.

// Target: classical-mechanics.bigb

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.

A photon hitting a resting mass $m$ has squared total energy invariant $m^2{c}^4+2m{c}^{2}E$. Producing a larger mass $M$ and an outgoing photon requires this invariant to exceed $M^2{c}^4$ if the photon has positive energy. The displayed threshold is therefore the limiting infimum as the final photon becomes soft, not an attained physical state with a zero-energy photon. Above threshold the outgoing photon energy in the centre-of-momentum frame is positive.

An incident particle of rest mass $m$ and energy $E$ striking a stationary identical target has total squared invariant energy $W^2=2m{c}^2(E+m{c}^2)$. Producing $N\ge 2$ final particles of the same rest mass requires $W\ge Nm{c}^2$, giving the threshold. Equality means no final kinetic energy in the centre-of-momentum frame. From the Lorentz factor, the corresponding incident speed is at least $cN\sqrt{N^2-4}/(N^2-2)$.

When an electron of Lorentz factor $\gamma >1$ annihilates with a stationary equal-mass positron, transverse four-momentum conservation makes the photons' transverse components opposite. Eliminating their positive frequencies gives the displayed relation for noncollinear photon angles to the incoming electron. The exactly collinear forward/backward configuration is physically allowed but makes the quotient $0/0$; its undivided cross-multiplied identity remains valid. The angle relation thus requires nondegeneracy or an appropriate limit, rather than an assertion that every allowed angle pair defines the quotient.

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.

The rest energy of a particle of invariant mass $m$ is $m{c}^2$, the energy measured in its rest frame.

The center-of-momentum frame is an inertial frame in which the total spatial momentum of a system vanishes. For a two-particle initial state, the momenta are equal and opposite and the total invariant energy is $\sqrt{s}$.

The centre-of-momentum energy is the total energy measured in the centre-of-momentum frame. For a scattering process with total four-momentum $P$, it is the Lorentz-invariant quantity $\sqrt{P^2}=\sqrt{s}$ in the mostly-minus convention.

For a particle of rest mass $m$,

For a massive free particle with invariant rest mass $m$, speed $v<c$, and Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$, its total energy is $E=\gamma m{c}^2$. It includes the rest energy $m{c}^2$ and the relativistic kinetic energy $(\gamma -1)m{c}^2$. Here $c$ is the speed of light. At fixed total energy, a larger speed requires a smaller rest mass.

For an isolated decay of a parent of rest mass $M$ into products of rest masses $m_i$, the difference between total final and initial kinetic energies is $(M-{\sum }_i{m}_i)c^2$ in every inertial frame. Conservation of total energy and inclusion of all products are essential.

Rest mass is the invariant mass $m$ in the relativistic energy-momentum relation. It equals the energy measured in the particle's rest frame divided by $c^2$.

Adding two null photon four-momenta gives the displayed invariant mass, with opening angle $\theta$ measured in the same frame as the energies. A nonzero mass of the pair does not give either photon a rest mass. At fixed energies it vanishes for parallel photons and is greatest for opposite directions. A reconstructed mass peak must be assessed with background, resolution and identification controls before assigning a parent particle.

Logarithmic differentiation of the diphoton mass gives $dM/M=(d{E}_1/E_1+d{E}_2/E_2)/2$ at fixed angle. If $s_i$ are fractional energy standard deviations and $\rho$ their correlation coefficient, covariance propagation gives the displayed formula. A common multiplicative calibration error transfers directly to the reconstructed mass; independent errors combine differently. Angular errors, if present, add the corresponding derivative and covariance terms.

For sequential on-shell two-body decays $C\to B+t$ and $B\to A+s$, with massless $s,t$, evaluate their energies in the B rest frame. Their pair mass is $2E_s{E}_t(1-\cos \theta )$, so its squared invariant mass ranges from zero to the displayed endpoint. The mass ordering is $m_C>m_B>m_A>0$. spin correlations can alter the distribution inside these limits, but cannot move the kinematic endpoints while the same on-shell decay chain applies.

For the sequential decay defining the cascade-decay invariant-mass endpoint, momentum conservation gives the displayed eventwise identity. Thus $m_{At}^2$ has minimum $m_A^2{m}_C^2/m_B^2$ and maximum $m_A^2+m_C^2-m_B^2$, attained at the opposite relative angles from the $s,t$ pair endpoints. The final A particle is massive, so its lower pair-mass bound need not vanish.

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.

For $\mathbf{E}=(0,0,E)$, $\mathbf{B}=0$, and initial momentum $(p_0,0,0)$,With the initial position at the origin,

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.

A four-force preserving the particle's rest mass. For a massive particle, $P=mU$ and $U^2=c^2$ in the mostly-minus convention imply $F\cdot U=c^{2}dm/d\tau$. Thus constant mass is equivalent to orthogonality of the force and four-velocity. Energy transfer in a moving frame is allowed even though rest mass does not change.

A nonzero four-force cannot have all components constant in an inertial frame along an entire constant-mass trajectory. If $U=U_0+F\tau /m$, preserving $U^2=c^2$ for all proper times requires both $U_0\cdot F=0$ and $F^2=0$. The orthogonal complement of timelike $U_0$ is negative definite, forcing $F=0$. A constant laboratory three-force or a four-force of constant invariant magnitude is a different, consistent condition.

For three-momentum $p=\gamma mv$, force and acceleration satisfy

The associated four-force uses proper time instead of coordinate time and is a different, four-dimensional object.

The three-force measured in a particle's instantaneous rest frame. For a pure force, ${\mathbf{f}}_0=m_0{\mathbf{a}}_0$, where ${\mathbf{a}}_0$ is the rest-frame acceleration. In that frame the four-force has zero temporal component and spatial part ${\mathbf{f}}_0$; the four-force itself is a different object.

Under a constant force $F$ with $p(0)=0$, momentum is $p=Ft$ andThe speed approaches $c$ without reaching it at finite time.

For a rest-frame decay $1\to 2+3$,The decay is allowed exactly when $m_1\ge m_2+m_3$.

In the centre-of-momentum frame with total energy $\mathcal{E}$, four-momentum conservation for a final massive particle of rest mass $m>0$ and one photon gives equal and opposite three-momenta. The energy–momentum relation and $|{\mathbf{p}}_{\gamma }|=E_{\gamma }/c$ give the displayed energies, andThe photon has positive energy when $\mathcal{E}>m{c}^2$. Kinematics fixes the speed and the opposition of the two final directions, while an angular distribution requires information about the interaction.

An isotropic two-photon decay of a massive parent with laboratory Lorentz factor $\gamma >1$ has the displayed normalized opening-angle probability density, with $\beta =\sqrt{1-{\gamma }^{-2}}$ and $2\arcsin (1/\gamma )<A<\pi$. To derive it in units $c=1$, the laboratory photon energies are $E_{1,2}=\gamma M(1\pm \beta z)/2$, where $z$ is uniform on $[-1,1]$. The diphoton invariant mass gives ${\sin }^2(A/2)=1/[{\gamma }^2(1-{\beta }^2{z}^2)]$. Apply the change of variables for probability densities to both signs of $z$. Its minimum-angle peak is an inverse-square-root singularity. At $\gamma =1$, the probability distribution is concentrated at $A=\pi$.

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.

For $C\to B+t$ with massless $t$, use $p_C=p_B+p_t$ in the B rest frame. The mass-shell identity gives $E_t=(m_C^2-m_B^2)/(2m_B)$ and ${\mathbf{p}}_C={\mathbf{p}}_t$. The denominator differs from the familiar daughter-momentum formula in the parent's rest frame. Units here have $c=1$.

A stationary particle of rest mass $m$ absorbing a photon of energy $E$ acquires total invariant massThis follows from four-momentum conservation, since the incoming photon has momentum magnitude $E/c$. If the system splits into two equal particles, the largest kinematically permitted daughter rest mass is $M/2$, attained when both are at rest in the centre-of-momentum frame. Their common laboratory speed at threshold is $cE/(m{c}^2+E)$. Above this daughter-mass threshold there is insufficient centre-of-momentum energy; below it, relative kinetic energy is available.

The Källén function controls two-body relativistic kinematics. In the rest frame of total invariant mass $\sqrt{s}$, the magnitude of each final momentum is $\sqrt{\kappa (s,m_1^2,m_2^2)}/(2\sqrt{s})$. It also determines the invariant flux factor in a two-particle collision.

For two-to-two scattering $A+B\to C+D$,They satisfy $s+t+u=m_A^2+m_B^2+m_C^2+m_D^2$ in units $c=1$.

With a specified ordering of external legs, this crossed channel carries the difference between an incoming and an outgoing energy-momentum vector. A one-particle exchange pole requires that difference to be on the exchanged particle's mass shell. Ordered amplitudes in one spatial dimension can label the crossed invariant using the difference of the two rapidity-parameterized external momenta; the external-leg convention fixes which crossed invariant is named $t$.

The direct channel combines the two incoming four-momentum vectors. A one-particle pole in this channel has the intermediate particle's squared mass equal to $s$. Imaginary rapidity differences can satisfy this mass relation for a bound state even though there is no pole at real physical scattering rapidity.