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Choose the $M$ ball's initial velocity as $+v$; the $3M$ ball then has velocity $-v$. Both have Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$. Their total relativistic energy and signed relativistic momentum areThe laboratory frame is not the centre-of-momentum frame, despite the equal and opposite speeds. Four-momentum conservation and the relativistic energy-momentum relation give the invariant mass $M_f$ of the combined incoming system:The relativistic collision mass threshold for a final object of rest mass $5M$ is $M_f\ge 5M$. At the threshold,At equality, the outgoing object has laboratory velocity $V=c^2{p}_i/E_i=-v/2$, directed towards the initially heavier ball's motion. Its nonzero final kinetic energy is why setting $4\gamma M{c}^2=5M{c}^2$ would give the wrong threshold.

For the literal rest mass of a coalescing relativistic collision, if there is exactly one outgoing object of exactly fixed rest mass $5M$ and no radiation or other products, four-momentum conservation selects the equality above, rather than allowing every larger $v$. Speeds above this threshold require a larger final invariant mass or additional outgoing energy carriers.

Choose particle 1's incoming velocity positive and particle 2's negative. The final particle is stationary, so four-momentum conservation requires zero total incoming momentum:It follows that ${\gamma }_1{\beta }_1=1/\sqrt{2}$. Squaring gives ${\beta }_1^2/(1-{\beta }_1^2)=1/2$, henceThe initial total energy becomes the final rest energy. ThereforeThe extra invariant mass is the incoming kinetic energy retained as internal energy of the merged particle: $(M-3m)c^2$. It does not violate conservation of energy or four-momentum conservation; the sum of the separate rest masses is not a conserved quantity in an inelastic collision.

Put $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$, the common Lorentz factor. The initial total relativistic energy and momentum areAssume that coalescence emits no additional radiation. By four-momentum conservation, the final rest mass $m_f$ is the rest mass of a coalescing relativistic collision:The final velocity is $V=c^2{P}_x/E=v/3$, the centre-of-momentum frame velocity.

For $v\ll c$,The leading term is the classical sum of masses. The first correction is the internal energy retained after sticking, divided by $c^2$: the initial classical kinetic energy is $3m{v}^2/4$, while the common-motion kinetic energy is $\frac{1}{2}(3m/2)(v/3{)}^2=m{v}^2/12$. Their difference is $2m{v}^2/3$, consistent with mass-energy equivalence.

For a particle of rest mass $m_0$ and velocity $\mathbf{u}$, the relativistic energy and momentum areFor a photon of wavelength $\lambda$ travelling along $\hat{\mathbf{n}}$,Here $h$ is Planck constant and $c$ is the speed of light.

Let $\epsilon$ and ${\epsilon }^{\prime }$ be the incident and scattered photon energies. The stationary electron initially has energy $m_e{c}^2$. By four-momentum conservation, its final energy and momentum areUse the electron's relativistic energy-momentum relation, rather than assuming its recoil is nonrelativistic:Cancelling common terms givesSince $\lambda =hc/\epsilon$, the Compton wavelength shift isThe word elastic here means that the final electron has the same rest mass as initially and the complete photon-electron system conserves total energy; the photon alone loses energy to recoil.

For backscattering, $1-\cos \theta =2$, and henceChoose the incident direction as $+x$, so that the scattered photon travels in the $-x$ direction. An observer moving with velocity $+\beta c\hat{\mathbf{x}}$ meets that photon head-on. The Lorentz transformation of four-momentum gives the observed energyThis is the approaching-observer relativistic Doppler effect. Set $D=\epsilon /{\epsilon }^{\prime }=2.511/0.511$ and solve:The observer must move opposite to the scattered photon, along the original incident direction, at approximately $0.9205c$. Moving with the scattered photon would redshift it instead.

Invariant and collision. For a particle of rest mass $m$, special relativity gives relativistic energy and momentumTherefore the relativistic energy-momentum relation isIt is the squared Minkowski norm of the four-momentum, multiplied by $c^2$, and hence is preserved by every Lorentz transformation. Explicitly a boost with speed $U$ along $x$ gives $E^{\prime }=\Gamma (E-U{p}_x)$, $p_x^{\prime }=\Gamma (p_x-UE/c^2)$, with $p_y^{\prime }=p_y$, $p_z^{\prime }=p_z$. Expanding $E^{\prime 2}-c^2{p}_x^{\prime 2}$ gives $E^2-c^2{p}_x^2$ because ${\Gamma }^2(1-U^2/c^2)=1$. Spatial rotations preserve $p^2$, completing the invariance argument for all inertial frames.

For the incoming moving particle, $v=3c/5$ and $\gamma =5/4$. The total laboratory energy and momentum before the perfectly inelastic collision areFour-momentum conservation identifies these with the final particle's energy and momentum. Its rest mass is thereforeThe result exceeds the sum $3m_0$ of the incoming rest masses; some initial kinetic energy has become internal rest energy. Since $p{c}^2/E=v$ for a massive particle,

Two-body decay. Work first in the parent's rest frame, denoted by a star. Four-momentum conservation requires equal and opposite daughter momenta. Their equal rest masses give equal energies:Thus their rest-frame velocities are $v_{\ast \pm }=\pm 3c/\sqrt{10}$. Applying the relativistic velocity-addition formula with boost $v^{\prime }=3c/13$ gives the two laboratory velocities:Their laboratory energies follow directly from the Lorentz transformation of four-momentum:The corresponding signed laboratory momenta areTheir sums recover $E_i$ and $p_i$, and each satisfies $E_{\pm }^2-p_{\pm }^2{c}^2=(m_0{c}^2/2{)}^2$. One daughter moves forwards and the other backwards in the laboratory; both are collinear with the original flight axis. Numerically, $v_{+}\simeq 0.967615c$, $v_{-}\simeq -0.919138c$, $E_{+}\simeq 1.980756m_0{c}^2$ and $E_{-}\simeq 1.269244m_0{c}^2$.

Take inertial axes parallel, with frame $S^{\prime }$ moving at $+v\hat{\mathbf{x}}$ relative to $S$ and origins coincident at zero time. The Lorentz factor is ${\gamma }_v=(1-v^2/c^2{)}^{-1/2}$; $\Delta x,\Delta y,\Delta t$ are separations of the same pair of events in $S$, and primed quantities are their separations in $S^{\prime }$. The transverse coordinates are unchanged. For two events on an inertial clock's worldline,The proper time $\Delta \tau$ is that clock's elapsed time; uniform speed $w$ gives $\Delta t={\gamma }_w\Delta \tau$. Coordinate intervals between spatially separated events must instead be Lorentz-transformed with their spatial separation included.

For the round trip in the station frame, the rendezvous distance is $u{t}_{\mathrm{out}}$, and the return has the same speed and distance. Thus $t_{\mathrm{round}}=2t_{\mathrm{out}}=10a/(\sqrt{7}c)$. Its departure and arrival both occur at the station, so $\Delta x=0$ for these endpoints:This uses the return speed as measured by the station. The elapsed time aboard A, apart from an ideal instantaneous turn, would be $\Delta t_{S,\mathrm{round}}/{\gamma }_u$.

For the crash, A's station-frame velocity components are $(3c/5,\sqrt{7}c/5)$, while B's are $(3c/5,0)$. Here ${\gamma }_u=5/3$ and ${\gamma }_v=5/4$. Four-momentum conservation, with no separately escaping radiation, gives the wreckage totals in the station frame:In B's original inertial frame the same totals areThe invariant wreckage mass is $m\sqrt{14/3}$, exceeding $2m$ because internal energy remains in the wreckage. Its momentum cannot be specified without naming a frame; both natural frames have been given.

Momentum identity. From the Lorentz factor, $v^2/c^2=1-{\gamma }^{-2}$. The relativistic three-momentum therefore satisfiesTogether with relativistic energy $E=\gamma m{c}^2$, this gives the energy–momentum relation $E^2-p^2{c}^2=m^2{c}^4$.

Parent in the laboratory. The rest energy is $m_0{c}^2=1\mathrm{GeV}$, while the total relativistic energy is $3\mathrm{GeV}$. ConsequentlyIts momentum magnitude is $p_{0}c=\sqrt{3^2-1^2}\mathrm{GeV}=\sqrt{8}\mathrm{GeV}$.

Daughters in the zero-momentum frame. The centre-of-momentum frame is initially the parent's rest frame. Conservation of momentum gives ${\mathbf{p}}_1^{\ast }=-{\mathbf{p}}_2^{\ast }$. Equal daughter rest masses then imply equal relativistic energies. Conservation of relativistic energy givesTheir speeds are thereforeThe two velocities are opposite, and $p_X^{\ast }c=\sqrt{(1/2{)}^2-(1/3{)}^2}\mathrm{GeV}=\sqrt{5}\mathrm{GeV}/6$. Their total kinetic energy is $m_0{c}^2-2m_X{c}^2=m_0{c}^2/3$.

Invariant of the complete system. Before the decay, in the centre-of-momentum frame,After the relativistic two-body decay,In the laboratory, four-momentum conservation fixes the summed daughter energy to $3\mathrm{GeV}$ and their summed momentum to the parent's ${\mathbf{p}}_0$. Thus the same calculation isThe invariant is formed after summing the daughter four-momenta; adding their separate squared rest energies would omit the relative-motion contribution to the invariant mass of the pair.

Experimental evidence. Tests of special relativity check linked predictions about light propagation, clocks and four-momentum. Several independent kinds of experiment support those predictions.

The observed partial-drag coefficient agrees with this first-order prediction rather than simple Galilean addition of $u$. It tests a different consequence of the Lorentz transformation from a particle-lifetime measurement.^[ref]

Atmospheric muons and the two frame descriptions. Atmospheric cosmic rays create unstable particles whose decay chains produce muons. A stopped-muon experiment measures the time between a stopping signal and the subsequent decay electron, obtaining the proper mean lifetime ${\tau }_0\simeq 2.2\mu \mathrm{s}$. The associated light-travel distance is only $c{\tau }_0\simeq 660\mathrm{m}$, yet substantial numbers of energetic atmospheric muons reach detectors after travelling several kilometres.

For an ideal selected cohort produced height $h$ above a detector and moving downward at constant speed $v$, the flight time in the terrestrial frame is $h/v$. Time dilation predicts a laboratory mean lifetime $\gamma {\tau }_0$, hence survival fractionWithout time dilation, the predicted exponent would be $-h/(v{\tau }_0)$ and the surviving population much smaller. Altitude comparisons constitute a cosmic-ray muon time-dilation experiment: select comparable momenta or energies and account for detector acceptance, atmospheric energy loss and backgrounds before interpreting the count ratio.

For illustration, not as a claimed measurement, take $h=10\mathrm{km}$ and $v=0.998c$. Then $\gamma =15.82$, the laboratory flight time is $33.4\mu \mathrm{s}$, and the predicted surviving fraction is about $0.38$. Omitting time dilation would give only about $2.5\times 10^{-7}$ on the same idealized constant-speed model.

In the muon rest frame, its decay still follows $e^{-\Delta \tau /{\tau }_0}$. The atmosphere and ground move upward. Their separation, measured simultaneously in that frame, is shortened by length contraction:The same survival fraction follows. The terrestrial description uses a longer-lived moving particle; the muon-frame description uses a shorter atmospheric flight path. They describe the same production and detection events. This consistency illustrates both time dilation and length contraction, while the survival measurement is not two independent experimental tests merely because it can be described in two frames.

Energy-momentum invariant. For a particle with rest mass $m$,Since $E=m{c}^2+K$, substitution and cancellation of $m^2{c}^4$ giveThis relativistic energy-momentum relation reduces to $K\simeq p^2/(2m)$ when $K\ll m{c}^2$.

Laboratory beam energies. If both final muons are at rest in the laboratory, their total momentum is zero and their total energy is $2m_{\mu }{c}^2$. Four-momentum conservation therefore requires equal and opposite incoming momenta. Because the electron and Positron have equal rest masses, they then have equal total energies:Subtract each particle's own rest energy:Using the usual handbook rest energies gives approximately $105.1\mathrm{MeV}$ for each beam. These are the ideal kinematic threshold values. Just above threshold, the pair has small, equal and opposite momenta in this laboratory frame.

Positron momentum. Apply the relativistic energy-momentum relation to $E_{+}=m_{\mu }{c}^2$ and use its stipulated negative-$x$ direction:The electron has the opposite momentum, $p_{e,x}=+p_0$ with $p_0=c\sqrt{m_{\mu }^2-m_e^2}$.

Boost to the electron rest frame. The electron's velocity is $v=p_{e,x}{c}^2/E_{-}=p_0/m_{\mu }>0$. Thus $S^{\prime }$ moves in the positive-$x$ direction, and its Lorentz factor satisfiesThe positron has the same laboratory total energy as the electron, so $v{E}_{+}/c^2=v{m}_{\mu }=p_0$. The supplied Lorentz transformation then givesAs a sign check, the same boost gives $p_{e,x}^{\prime }=\gamma (p_0-v{E}_{-}/c^2)=0$, as required for the electron rest frame.

Use energy units $m=m_p{c}^2$, $M=m_A{c}^2$, $e=E_{\gamma }$ and $P=pc=\sqrt{E^2-m^2}$. The soft-photon resonant capture threshold follows from four-momentum conservation:For fixed proton energy the left side is largest at head-on incidence. Its threshold is $E+P=2E_0$. Combining that with $(E-P)(E+P)=m^2$ givesEquality is allowed for a head-on photon. Here $e=k_{B}T=2.3482\times 10^{-4}\mathrm{eV}$, soA very soft photon needs an enormous proton Lorentz factor to reach the specified invariant mass. A true thermal photon distribution and a finite-width resonance would smear this monoenergetic threshold.

For $\theta =0$, $E-P=m^2/(E+P)\le m$, whereas $2E_0\gg m$, so production is impossible. For each nonzero angle, sufficiently large energy is possible, and $P\simeq E$ givesTo find the maximum deflection in soft-photon capture, note that the final particle's transverse momentum comes entirely from the photon:Writing $x=\cos \theta$, maximization gives $(1-x)(2x+1)=0$ and the interior maximum $x=-1/2$. Therefore, to the same extraordinarily accurate ultrarelativistic order,The neglected relative corrections are of order $m^2/E_0^2$ and $e/E_0$, both around $10^{-24}$. The final particle is effectively collinear with the proton despite a large photon incidence angle.

In the convention of iii, the invariant decay amplitude is ${\mathcal{M}}_{s_1{s}_2}=-g{\bar{u}}_{\chi }^{s_2}(p_2)u_{\Psi }^{s_1}(p_1)$. Including the common overall $i$ in the definition changes no squared decay amplitude. An unpolarized ensemble is an incoherent equal mixture of the two initial spin states; hence take a spin average over $s_1$ and a spin sum over the unobserved final $s_2$. Different final spin states are orthogonal outcomes, so their probabilities, rather than their decay amplitudes, add.

Using the Dirac adjoint and the fermion spin sum givesThe gamma-matrix trace identities used here are $\mathrm{tr}{\gamma }^{\mu }=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$. The mass-shell conditions $p_1^2=M^2$ and $p_2^2=p_3^2=m^2$, together with four-momentum conservation, implyThereforeIn the parent's rest frame this also says $E_2=E_3=M/2$.

The two complex scalar fields have scalar propagatorscontractions of two unconjugated fields or two conjugated fields vanish. Draw an arrow on each line in the direction of positive conserved charge flow: a solid line carries one unit and a dashed line two units. A particle arrow follows its physical momentum; an antiparticle arrow opposes it.

There are two conjugate three-leg interaction vertices, both with Feynman rule $-ig$: one has a charge-two arrow entering and two charge-one arrows leaving, and the other reverses every arrow. These describe the two conjugate terms ${\varphi }^2{\Phi }^{†}$ and ${\varphi }^{†2}\Phi$. The two identical $\varphi$ attachments cancel the $1/2$ in the Lagrangian density. All crossed assignments are obtained by moving external lines between initial and final states while maintaining the arrows. There is no interaction vertex with three equal-sign charges.

Each interaction vertex conserves four-momentum, supplying $(2\pi {)}^4{\delta }^4(\sum p_j)$ with all momenta entering. Integrate each independent loop momentum with $d^{4}k/(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. No minus sign is assigned to a closed scalar loop. These rules are for $i\mathcal{M}$ after amputating external lines and removing the overall four-momentum conservation factor.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$, four-momentum conservation gives total energy $E=(\gamma m_1+m_2)c^2$ and momentum $\mathbf{p}=\gamma m_1\mathbf{v}$. The compound system therefore hasHere $M^2{c}^4=E^2-c^2{p}^2$. The excess over $m_1+m_2$ is retained internal energy, assuming absorption without escaping radiation.

Use the $+---$ metric signature, with the time component of four-momentum equal to $E/c$. The incident particle has energy $\gamma m{c}^2$ and momentum of magnitude $mc\sqrt{{\gamma }^2-1}$. Adding the stationary target's four-momentum gives total laboratory quantitiesThe centre-of-momentum energy $W$ follows from the invariant Minkowski norm of the total four-momentum:At the relativistic collision mass threshold, the two final particles have zero relative velocity in their centre-of-momentum frame. Their total energy there is therefore $2\alpha m{c}^2$. Four-momentum conservation requires $W\ge 2\alpha m{c}^2$, soThe condition $\alpha >3/2$ ensures ${\gamma }_{\min }>1$. This is the minimum required Lorentz factor; larger values allow final kinetic energy in the centre-of-momentum frame. Equivalently, the incident laboratory kinetic energy at threshold is $({\alpha }^2-9/4)m{c}^2$.

The conserved initial four-momentum of a particle with positive rest mass satisfies $P_{\mu }{P}^{\mu }=m^2{c}^2>0$. A single photon instead has null four-momentum, $K_{\mu }{K}^{\mu }=0$. Four-momentum conservation for an isolated one-body decay would require $P^{\mu }=K^{\mu }$, contradicting these Lorentz invariants. Thus decay into a single photon is impossible in every inertial frame.

Equivalently, the incoming energy and momentum obey $E^2-c^2|\mathbf{p}{|}^2=m^2{c}^4$, whereas a single outgoing photon obeys $E=c|\mathbf{p}|$. Changing the laboratory speed cannot change this mismatch of the invariant mass shell.

The charged weak current changes one neutron down quark into an up quark:The other two valence quarks are spectators in this tree-level Feynman diagram. The intermediate W boson is highly virtual; the available neutron decay energy is far below its on-shell rest energy. Charge is conserved at each vertex, and the electron's lepton number is balanced by the electron antineutrino.

The reversed quark-level charged current is allowed as an interaction, but a free proton cannot decay as $p\to n+e^{+}+{\nu }_e$ because $m_p<m_n+m_e$. Four-momentum conservation leaves no accessible final-state Lorentz-invariant phase space. The absence of this free-proton beta-plus decay is therefore kinematic, not an absence of a $u\to d$ weak vertex. Binding-energy differences in a nucleus, or incoming energy in a collision, can change whether the corresponding conversion is possible. Exotic baryon-number-violating proton decay is a separate question and is not generated by this ordinary charged-current diagram.

For a vacuum electromagnetic wave, the four-wavevector is $K^{\mu }=(\omega /c,\mathbf{k})$ with $|\mathbf{k}|=\omega /c$ and $K^2=0$. Its phase can be writtenBecause $K$ and $X$ transform as Lorentz four-vectors, their contraction is a Lorentz scalar. This proves phase invariance, with the same phase convention in both frames.

For a hypothetical free emission $e\to e+\gamma$, four-momentum conservation gives $P=P^{\prime }+Q$. The initial and final electrons have $P^2=P^{\prime 2}=m_e^2{c}^2$, and $Q^2=0$. Squaring gives $P^{\prime }\cdot Q=0$. In the final electron's rest frame, however, $P^{\prime }\cdot Q=m_e{E}_{\gamma }>0$ for a nonzero future-directed photon. ThusAn external field, another body or a medium can exchange momentum and change this conclusion; those are different systems.

Three relevant observable effects are the relativistic Doppler effect in spectral-line shifts, including the transverse time-dilation contribution; relativistic aberration, observable as a change of apparent light direction with observer motion; and relativistic beaming of moving emitters, associated with strongly asymmetric jet brightness. Apparent superluminal motion supplies a related light-travel-time test: $v_{\mathrm{app}}/c=\beta \sin \vartheta /(1-\beta \cos \vartheta )$ can exceed one without a material speed exceeding $c$. These interpretations require an emission and source-motion model; a cosmological or gravitational redshift by itself is not a direct test of special-relativistic kinematics.

For Compton scattering from an initially stationary electron, let incoming and outgoing photon energies be $\epsilon$ and ${\epsilon }^{\prime }$ and their mutual angle be $\psi$. Four-momentum conservation gives $P+Q=P^{\prime }+Q^{\prime }$. Squaring $P^{\prime }=P+Q-Q^{\prime }$ and using both electron mass shells and both photon null norms givesConsequently the outgoing angular frequency and shift areEquivalently, the Compton wavelength shift is ${\lambda }^{\prime }-\lambda =h(1-\cos \psi )/(m_{e}c)$. The electron recoil is essential; treating its energy as unchanged would suppress the effect.

To prove the vanishing component theorem for a Lorentz four-vector, first suppose $Y^0=0$ in every inertial frame. An infinitesimal boost along spatial direction $i$ gives $Y^{\prime 0}=Y^0-\beta Y^i+O({\beta }^2)$. It too vanishes for every $\beta$, so each $Y^i=0$. If instead a specified spatial component vanishes in every frame, spatial rotations first imply that all three spatial components vanish in a chosen frame, and a boost in the specified direction then forces $Y^0=0$. ThereforeA physical consequence is that conservation of energy in every inertial frame implies full four-momentum conservation: apply the result to the difference between outgoing and incoming total four-momenta.

The statement fails for general tensors. A nonzero antisymmetric rank-two tensor has $F^{00}=0$ in every frame, because antisymmetry is preserved. For an off-diagonal component, the nonzero Minkowski metric is another counterexample: ${\eta }^{01}=0$ in all orthonormal Lorentz frames, while $\eta$ is certainly not zero. The transformation law can preserve structural zero components without forcing the entire tensor to vanish.

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.

By four-momentum conservation, $p_{\pi }=p_1+p_2$. With the $(-,+,+,+)$ Minkowski metric, the neutral pion has $p_{\pi }^2=-M_{\pi }^2$, while each photon has $p_i^2=0$ and $|{\mathbf{p}}_i|=E_i$. Taking the Minkowski norm givesThe half-angle identity therefore yieldsThis is a measurement of the diphoton invariant mass using two photon energies and their opening angle, independent of the neutral pion's laboratory velocity.