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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.

Compton scattering redistributes photon energy and momentum in a collision with a charged subatomic particle, here an electron. For a free electron initially at rest in the chosen inertial frame, an incident photon of energy $E=hc/\lambda$ leaves with energy $E^{\prime }=hc/{\lambda }^{\prime }$ after deflection through $\theta$. The angle $\theta \in [0,\pi ]$ is between the incident and outgoing photon directions; it is not the electron recoil angle. Four-momentum conservation giveswhere $m_e$ is the electron rest mass and $T_e$ its final kinetic energy. The photon energy loss pays for the electron recoil: neither energy nor momentum disappears.

For an original worked example, stipulate $E=70\mathrm{keV}$, the rounded electron rest energy $m_e{c}^2=511\mathrm{keV}$, and the rounded Compton wavelength ${\lambda }_C=2.426\mathrm{pm}$; $\mathrm{keV}$ means $10^3$ electronvolts. Treat these rounded inputs as the example's constants. They give the following calculated values, with energies rounded to three decimal places:

Since $0\le 1-\cos \theta \le 2$, every outgoing photon has $0<E^{\prime }\le E$ and $T_e\ge 0$. At zero deflection there is no recoil; backscattering transfers the most energy. At fixed $\theta$, the absolute wavelength increase is independent of incident energy, while the fractional energy loss grows with $E/(m_e{c}^2)$. This makes Compton scattering useful for inferring energy transfers from directions in photon detectors.

The free-electron-at-rest assumption matters. An electron in a bound state has binding energy and an initial momentum distribution. Neglecting these is useful when the transferred energy is large compared with the relevant binding energy and the initial momentum is small compared with the resolved momentum transfer. Otherwise the outgoing photon energies can broaden or depart from the stationary-target formula; an initially moving electron can even give energy to the photon.

A material may also scatter X-rays through a coherent elastic scattering channel with an effectively unshifted component. When an intact chemical atom of mass $M$ takes the recoil, the corresponding Compton wavelength $h/(Mc)$ is much smaller than the electron value. The single free-electron curve does not predict that component's weight. More generally, energy versus angle is a kinematic constraint; angular wave intensity requires a differential scattering cross-section, together with target and detector information.

The Compton wavelength of a subatomic particle of rest mass $m>0$ is ${\lambda }_C=h/(mc)$, where $h$ is the Planck constant and $c$ the speed of light. It is the wavelength of a photon whose energy $hc/{\lambda }_C$ equals that subatomic particle's rest energy $m{c}^2$. It is a characteristic scale, rather than a measured geometric radius of the subatomic particle.

The reduced Compton wavelength is ${\bar{\lambda }}_C=\hslash /(mc)={\lambda }_C/(2\pi )$, with $\hslash =h/(2\pi )$. Keep that factor of $2\pi$: the Compton scattering wavelength shift below uses $h/(m_{e}c)$, not $\hslash /(m_{e}c)$. Here we choose the rounded electron value ${\lambda }_C=2.426\mathrm{pm}$, where $1\mathrm{pm}=10^{-12}\mathrm{m}$; the maximum single-collision wavelength increase is $2{\lambda }_C=4.852\mathrm{pm}$.

Why this scale? A photon of wavelength $\lambda$ has momentum magnitude $h/\lambda$. At $\lambda \sim {\lambda }_C$ that momentum is comparable with $mc$, so transfer of an appreciable fraction of it can make the recoiling subatomic particle relativistic. Even for longer wavelengths, Compton scattering can resolve the small energy transferred into electron kinetic energy.

In the 1923 note Waves and Quanta, Louis de Broglie proposed a periodic process associated with rest energy, remaining in phase with a wave along the moving particle's path. He distinguished the internal frequency slowed by time dilation from the wave frequency based on total energy.^[ref] The proposal developed into his 1924 thesis.^[ref] In the 1929 lecture The wave nature of the electron, he reconstructed the argument using a Lorentz transformation.^[ref]

Here is that quantum phase argument in modern notation. Set ${\omega }_0=m{c}^2/\hslash$ in the rest frame. A Lorentz transformation to a frame where the particle has velocity $v$ givesAlong $x=vt$, proper time is $\tau =t/\gamma$, soThis is the harmony: evaluating the spatial wave at the moving particle gives the slower clock phase. At fixed $x$, its angular frequency is instead $E/\hslash =\gamma {\omega }_0$. Since $k=p/\hslash$, the de Broglie wavelength follows as $\lambda =2\pi /k=h/p$. The proposed clock was a theoretical construction, not an observed mechanical vibration.

A parent of rest mass $M$ can decay to on-shell daughters of rest masses $m_i$ only if $M\ge {\sum }_i{m}_i$. In the parent rest frame, each daughter energy is at least its rest energy, so four-momentum conservation proves necessity. The excess energy controls the available Lorentz-invariant phase space. A two-body decay threshold is the two-daughter special case; spin and other selection rules can still forbid an energetically allowed channel.

At constant radiative efficiency of black-hole accretion $\eta$ and the Eddington luminosity, a black hole grows exponentially with the Salpeter time $t_S=\eta t_E/(1-\eta )$, where $t_E={\sigma }_{T}c/(4\pi G{m}_p)$. Its doubling time is $t_S\log 2$. The factor $1-\eta$ accounts for the supplied rest energy that escapes as radiation instead of increasing the hole's mass.

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 circular timelike orbit with conserved energy equal to rest energy at infinity. Substituting $E/(m{c}^2)=(1-2\mu /r)/\sqrt{1-3\mu /r}$ and solving $E=m{c}^2$ gives $r=4\mu$, with $\mu =GM/c^2$. It is unstable. Strictly bound circular orbits have $r>4\mu$ and $E<m{c}^2$; radial stability requires $r>6\mu$. This limiting orbit is not the smallest radius ever reached by a bound-energy plunging trajectory.

A free scalar Klein-Gordon equation with rest mass $m$ isHere $c$ is the speed of light and $\hslash$ the reduced Planck constant. In the static limit, comparison with the static Yukawa Green function identifiesThis $\lambda$ is the reduced Compton wavelength, rather than $h/(mc)$; the two differ by $2\pi$. It is an exponential decay length of this exchange field, not the geometric size of a nucleon or an exact radius of a bound state. This article uses $\lambda$ for a length; the 1935 paper used that letter for its inverse.

For a deliberately rounded worked example, stipulate $\hslash c=197.3\mathrm{MeV}\mathrm{fm}$ and $m{c}^2=140\mathrm{MeV}$. ThenA femtometre is $10^{-15}\mathrm{m}$. The chosen rest energy is a convenient pion-scale input; neither it nor the resulting Yukawa potential is asserted to be an exact measured nuclear interaction. Holding $\hslash c$ fixed, chosen rest energies $70$, $140$ and $280\mathrm{MeV}$ give lengths $2.819$, $1.409$ and $0.7046\mathrm{fm}$. Doubling the exchange rest mass halves the decay length.

Range alone does not guarantee binding: the potential energy must overcome the kinetic energy cost of localizing a wavefunction. The existing exponential variational bound for the Yukawa potential makes this precise. For the same toy $V$, with reduced mass $M$, choosing a normalized exponential wavefunction proportional to $e^{-\alpha r}$ with $\alpha =\kappa /2$ yieldsFor additional stipulated inputs $M{c}^2=470\mathrm{MeV}$ and $A=100\mathrm{MeV}\mathrm{fm}$, this gives $⟨H⟩\approx -3.66\mathrm{MeV}$. A negative value proves a bound state in this scalar model by the quantum variational principle; it is an upper bound on its ground-state energy, not a prediction for the deuteron. A nonnegative trial value would not prove the absence of binding.

Material constituents such as particles and their composites. In cosmology, the narrower phrase pressureless matter usually refers to nonrelativistic components whose rest energy dominates their random kinetic energy, giving negligible pressure and density proportional to the inverse cube of the scale factor. This usage distinguishes the matter contribution from relativistic radiation in cosmology and vacuum energy.

The time-translation Killing vector field of the Schwarzschild metric gives a conserved Killing energy per unit rest mass,For unit rest mass, this is the total energy relative to infinity, including rest energy. At release $dr/d\tau =0$, so the four-velocity normalization gives $dt/d\tau =f(r_0{)}^{-1/2}$ andSubstitution into the radial four-velocity normalization yieldsChoosing the inward square root and evaluating at the ground provesThis identity is exact within the assumed Schwarzschild metric, despite its familiar Newtonian mechanics form. The quantity $dr/d\tau$ is not exactly the locally measured speed. A Schwarzschild static observer measures $v_{\mathrm{loc}}=(dr/dt)/f=(dr/d\tau )/\sqrt{f(r_0)}$. Thus the distinction disappears in the stated weak-field approximation, giving $v_{\mathrm{loc}}^2\simeq 2gh$ for $h\ll R_E$.

Write the equation of state as $P=w\rho c^2$. Integrating the cosmological perfect-fluid continuity equation for constant $w$ gives $\rho \propto R^{-3(1+w)}$. With spatial flatness and the geometric cosmological constant set to zero, the Friedmann equation is $H^2=8\pi G\rho /3$.

In S1, $w=-1$ describes vacuum energy: $T_{\mu \nu }=-\rho c^2{g}_{\mu \nu }$ and $\rho$ is constant. Its negative pressure drives accelerated expansion, $\ddot{R}/R=8\pi G\rho /3=H_1^2$, and the expanding solution is $R\propto e^{H_{1}t}$. This is the ideal de Sitter spacetime approximation to cosmic inflation. An exactly constant vacuum component cannot end its phase unaided; a phase transition or reheating mechanism is required for the assumed change to S2.

In S2, $w=1/3$ describes an isotropic bath of ultrarelativistic particles, such as photons, with negligible rest mass relative to their energy. The number per physical volume falls as $R^{-3}$, while the energy per particle redshifts as $R^{-1}$, so $\rho \propto R^{-4}$. This radiation-dominated universe has $R\propto (t-t_{\ast }{)}^{1/2}$ on its expanding branch.

In S3, $w=0$ describes pressureless matter, or relativistic dust, with negligible random kinetic energy. Conserved particle number and approximately fixed rest energy give $\rho \propto R^{-3}$. A flat universe containing only this component is the Einstein-de Sitter universe, with $R\propto (t-t_{\ast }{)}^{2/3}$. The constants $t_{\ast }$ are integration offsets in the individual phases; they need not represent one common start time when phases are matched.

Use Minkowski spacetime coordinates $x^{\mu }=(ct,x,y,z)$ and the Minkowski metric of signature $(+,-,-,-)$. For a massive particle with three-velocity $\mathbf{v}=d\mathbf{x}/dt$ and $|\mathbf{v}|<c$, proper time satisfiesThus the four-velocity components areHere upper indices indicate components in the chosen coordinates; lowering with the Minkowski metric changes the signs of the spatial components. Its invariant product defined by the Minkowski metric is

The four-momentum is $p^{\mu }=(E/c,\mathbf{p})=m{U}^{\mu }$, so $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. The Taylor expansion of the Lorentz factor at small $|\mathbf{v}|/c$ isConsequently $\mathbf{p}=m\mathbf{v}+O(m|\mathbf{v}{|}^3/c^2)$, recovering ordinary momentum. At rest the four-momentum has energy $E_0=m{c}^2$, so subtracting this rest energy gives the kinetic energyThus the nonrelativistic limit preserves the rest-energy offset as well as the usual kinetic energy.

Let $p$ be the magnitude of the incident particle's momentum. The relativistic energy-momentum relation givesThe stationary particle contributes rest energy $m{c}^2$ and zero momentum. Conservation of four-momentum therefore gives equal photon energies $\epsilon =(E+m{c}^2)/2$. A photon has momentum magnitude $\epsilon /c$, so if the two momentum vectors meet at angle $\theta$,Substitution gives the angular relation for two-photon electron-positron annihilation, whose derivation depends only on the stipulated equal rest masses and equal photon energies:All energies and the angle here are in the laboratory inertial frame.

For perpendicular photon trajectories, $\cos \theta =0$ and hence $E=3m{c}^2$. Since $E=\gamma m{c}^2$ with Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$,For the small-speed limit, set $\beta =v/c\ge 0$ and $\delta =\pi -\theta$. The angle formula impliesUsing $\arcsin z=z+O(z^3)$ therefore gives $\delta =\beta +O({\beta }^3)$, soAt zero incident speed the momenta are equal and opposite, so $\theta =\pi$, consistent with the expansion.

The relativistic energy of a particle of rest mass $m$ is $E=\gamma m{c}^2$, where the Lorentz factor is $\gamma =(1-v^2/c^2{)}^{-1/2}$. Using the binomial series at small $v/c$ givesThe second term is the familiar Newtonian kinetic energy. The first is the rest energy, which remains when $v=0$. Thus the new physical idea is mass-energy equivalence: a body's mass represents energy $m{c}^2$ even at rest. The total energy is not merely energy due to motion.

The incident four-momentum has energy $E$ and spatial momentum $p=\gamma mv$, while the stationary target has energy $m{c}^2$ and zero spatial momentum. By the relativistic energy-momentum relation, $c^2{p}^2=E^2-m^2{c}^4$. For the fixed-target threshold for production of equal-mass particles, the squared total centre-of-momentum energy isIn the centre-of-momentum frame, each of the final $N$ particles has energy at least its rest energy $m{c}^2$. Four-momentum conservation therefore requires $W\ge Nm{c}^2$, and thusFor the physically relevant $N\ge 2$, this is equivalent to $\gamma \ge (N^2-2)/2$. The speed is an increasing function of $\gamma$, soConsequentlyEquality is the kinematic threshold at which all final particles are at rest in their common centre-of-momentum frame. It shows why a collision with a stationary target needs more incident energy than merely the additional rest energy: some laboratory energy must remain as collective motion. A one-particle final state of the specified mass, with no other outgoing energy, is impossible because its invariant energy would be $m{c}^2$, whereas the two-particle initial state has invariant energy at least $2m{c}^2$.

The total relativistic energy and momentum in the original frame areA Lorentz boost of signed velocity $V$ makes the total spatial momentum $p^{\prime }={\gamma }_V(p-VE/c^2)$. Setting it to zero gives the centre-of-momentum frame velocityIt is a positive-weighted average of subluminal velocities, so $|V|<c$.

In the rest mass of a coalescing relativistic collision, four-momentum conservation gives $P_3=P_1+P_2$ provided no energy or momentum escapes in additional radiation. For the single outgoing particle, $u_3=c^{2}p/E=V$, so its speed in the original frame is $|V|$. Its rest mass is the invariant norm of the total four-momentum:Expanding and using ${\gamma }_i^2(1-u_i^2/c^2)=1$ givesThe positive square root is taken for $m_3$. The factor multiplying $2m_1{m}_2$ is the relative Lorentz factor, at least one. Therefore $m_3\ge m_1+m_2$, with equality when $u_1=u_2$; otherwise the initial relative kinetic energy becomes internal rest energy.

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 $P/2$.

Power is energy per time. In geometrized units, energy has the dimension of length and time is converted to length using $c$, so the dimensionless geometrized luminosity isEquivalently, $c^5/G$ is the corresponding SI power unit. The lifetime for emitting the entire rest energy at the given constant luminosity isThe same result follows from $M_{\odot }^{(\mathrm{time})}/L_{\odot ,\mathrm{geom}}$. The stipulated full-mass radiation time is of order $10^{13}$ years; this is the constant-luminosity energy budget asked for, not a stellar-evolution calculation.

The trace-free mass quadrupole moment is diagonal:The same constant factors multiply its third derivatives, giving ${\stackrel{...}{Q}}_{ij}{\stackrel{...}{Q}}_{ij}=(2/3)({\stackrel{...}{I}}_{zz}{)}^2$. The quadrupole formula and the preceding infall result therefore giveFor infall from infinity, $P=m^5/(15z^5)$ and $|\dot{z}|=\sqrt{m/(2z)}$. To find the emitted energy, integrate power over time, using $dt=|dz|/|\dot{z}|$:The total initial mass is $2m$, so the radiated fraction is $1/840\simeq 1.19\times 10^{-3}$. The Sun would emit that fraction of its rest energy inunder the stipulated constant luminosity. This is the head-on quadrupole radiation from equal masses prediction with the prescribed stopping rule. At the endpoint $|\dot{z}|=1/2$ and $m/z=1/2$, so extending the slow-motion weak-field formula that far is an extrapolation, not a controlled strong-field prediction.