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Each on-shell particle satisfies $E_i^2-c^2|{\mathbf{p}}_i{|}^2=m_i^2{c}^4$. Thus for a selected subset $A$,It is Lorentz invariant when the same particle labels and physical states are compared, with each energy and momentum transformed as that particle's four-momentum. No requirement that the particles share a velocity is needed. Their energies can change under interactions while their individual fixed rest masses remain the same.

A frame-dependent rule for choosing “half” is more delicate: different subsets can have different sums of rest-mass squares. If every particle has the same rest mass and the number selected is fixed, however, the displayed sum is unchanged even if the selected identities change. In a massless plasma it is identically zero for every subset. The precise condition is constancy of the selected scalar sum, not a blanket claim that every frame-dependent selection must change it.

The covariance of a selected particle momentum sum follows because, for a fixed set of particle four-momenta, their sum $P_A^{\mu }={\sum }_{i\in A}{p}_i^{\mu }$ is a four-vector by linearity of the Lorentz transformation. Thereforeis invariant. The subset need not comprise the entire plasma, and its particles need not move together. This expression includes cross terms between different particles; it is not the sum of their individual rest-mass squares in a.

The comparison must use the same subset and the transformed four-momenta at the same chosen events. Selecting “the lowest-energy half” anew in each frame need not do that. For example, take four photons along $\pm x$, with energies $E_0,2E_0$ in each direction. The two lowest in the original frame are an opposite pair and have invariant energy squared $4E_0^2$. After a boost at $c/2$ along $+x$, both forward photons are redshifted enough to become the lowest pair; that new collinear pair has zero invariant mass. This does not contradict covariance, because the subset changed.

If energies are resampled on a different simultaneity slice in an interacting plasma, they can also refer to different events. A physical conserved subsystem total must then include exchanged field energy and any boundary flux as appropriate. Such operational conservation requirements are distinct from the exact algebraic covariance of a sum of fixed four-vectors.

The magnitude of the electric force is $f=e\mathcal{E}$, directed so that the electron gains energy. The work across the gap is $e\mathcal{E}L=50\mathrm{MeV}$. With the source precision $m{c}^2=0.511\mathrm{MeV}$,Reducing the field by 20 percent gives kinetic energy $40\mathrm{MeV}$ andA large fractional energy change therefore produces only a small speed change, since the original electron is already highly relativistic.

For the transit time, combine the distance formula from solution with ${\gamma }^2-1=[ft/(mc){]}^2$:This is slightly longer than $L/c$, since the electron starts from rest. For the reduced field it becomes $3.37798\times 10^{-8}\mathrm{s}$.

The initial momentum magnitude satisfies $p^2/(m^2{c}^2)={\gamma }_i^2-1=288$. Equality of longitudinal and transverse magnitudes therefore givesAn electric field along $z$ changes $p_z$, while the magnetic Lorentz force only rotates the transverse momentum. Its magnitude $p_{\perp }$ remains fixed. At the longitudinal turning point $p_z=0$, henceFor an electron initially moving toward positive $z$, the decelerating field points toward positive $z$ because its charge is $-e$. The electric work over $\Delta z=1\mathrm{m}$ givesThe radius is determined by transverse momentum rather than total energy:so the requested radius plot is horizontal. The cyclotron angular frequency changes as $e|B|/(\gamma m)$ and the helix pitch shrinks to zero. These statements neglect radiation, as stipulated.

The equal and opposite incoming momenta make the laboratory the centre-of-mass frame. The initial total four-momentum is timelike. One outgoing photon would have null four-momentum, so an isolated two-body annihilation cannot produce exactly one photon. Two photons are kinematically possible and must have opposite momenta and equal energies:The quoted $0.511\mathrm{MeV}$ is the rest energy $m_e{c}^2$, not a mass in SI units. The energy balance includes both rest energies as well as the $1\mathrm{GeV}$ total kinetic energy.

This minimum concerns the complete final state. A detector with incomplete acceptance could register only one of the two photons, or none; conservation laws do not impose a minimum detected count without an ideal-detection assumption. Additional outgoing photons are also possible.

The electron moves in the direction for which the electric field does positive work. Its kinetic energy gain is the magnitude $eEL=30\mathrm{MeV}$. HenceThis uses the relativistic energy-momentum relation, rather than the nonrelativistic kinetic-energy formula. Acceleration is opposite the field direction for a negatively charged electron, although the energy gain is positive along its actual displacement.

Put $\epsilon =\hslash \omega$. The total four-momentum of the perpendicular photons isAlthough each constituent is massless, the sum has $P_{\mathrm{tot}}^2=2{\epsilon }^2/c^2$. Its invariant mass is thereforeThe rest energy is the total energy in the frame where the two photon momenta cancel, not the sum of individual photon rest energies.

The centre-of-mass velocity follows from the total energy and momentum:A boost in this direction with ${\Gamma }_{\mathrm{CM}}=\sqrt{2}$ gives zero total spatial momentum and total energy $\sqrt{2}\hslash \omega$. The constituent photons remain moving at $c$ in that frame.

For the magnetic-dominated crossed-field orbit, choose $V=E/B$ along $x$, which is subluminal when $|E|<c|B|$. Then $E_y^{\prime }=0$ and $B_z^{\prime }=B/\Gamma$. The initial velocity in this frame is $(-V,0,0)$, and the constant particle Lorentz factor is ${\gamma }^{\prime }=\Gamma$ because a magnetic field does no work.

Define the signed relativistic cyclotron frequency $\Omega =q{B}_z^{\prime }/(\Gamma m_e)$, retaining $q=-e$. With $t^{\prime }$ as parameter, the exact primed trajectory isIt is a circle of radius $|V/\Omega |$ with centre $(0,V/\Omega )$; its initial velocity is $(-V,0,0)$ as required. Transforming back givesIn the laboratory this is a stretched cycloidal trajectory with cusps where the electron is instantaneously at rest. Its average drift is $⟨v_x⟩=V=E/B$, along $\mathbf{E}\times \mathbf{B}$ for the positive field orientation. Its energy oscillates as the electric field exchanges energy with it; it does not accelerate without bound. If $E=0$, the initially stationary electron remains stationary. At $|E|=c|B|$, neither of these finite boosts exists; that separate null-field limit is not one of the two strict cases requested.

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.

The asserted necessity of equal masses is false. Equal unsigned outgoing angles imply equal longitudinal momenta when the transverse momenta cancel; they do not imply equal final speeds or energies for unequal masses.

An explicit counterexample uses $c=1$, masses $M_1=m$ and $M_2=2m$, initial projectile energy $E_1=2m$, and a stationary target. The initial total energy is $4m$ and momentum is $(\sqrt{3}m,0,0)$. Choose final four-momentaThey sum to the incoming four-momentum. Their squared norms are respectively $m^2$ and $4m^2$, so the collision is elastic. Both trajectories make the same acute angle with positive $x$:Thus no proof of the printed claim is possible.

The general centre-of-mass frame construction shows why. Let $p^{\ast }$ be the outgoing momentum magnitude there, ${\theta }^{\ast }$ the projectile scattering angle, and $E_1^{\ast },E_2^{\ast }$ the two energies. After a boost ${\beta }_{\mathrm{CM}}$ along $x$, the equal-angle condition for two forward nonzero trajectories isWith $E_1$ the initial projectile energy, its solution isprovided the right side lies between minus and plus one and both final longitudinal momenta are positive. Equal masses are one allowed case, giving ${\theta }^{\ast }=\pi /2$; they are not necessary. For the counterexample the right side is $3/16$. Even equal masses do not force equal angles for every scattering direction: the symmetric outgoing configuration is an additional choice.

Let $\epsilon =E_{ph}$ and $\gamma =W/(M_0{c}^2)$. For a head-on photon and particle collision, boost to the particle's initial rest frame. The incident photon energy is ${\epsilon }_{\ast }=\gamma (1+\beta )\epsilon$. On the inverse-Compton branch $\epsilon <\gamma \beta M_0{c}^2$, the maximal outgoing lab energy corresponds to rest-frame backscattering into the particle's forward direction. The exact Compton scattering recoil formula gives ${\epsilon }_{\ast }^{\prime }={\epsilon }_{\ast }/[1+2{\epsilon }_{\ast }/(M_0{c}^2)]$. Boosting the backscattered photon back to the lab givesFor $\gamma \gg 1$ this becomesOnly in the additional Thomson scattering limit $4\gamma \epsilon \ll M_0{c}^2$ may the denominator be neglected, yielding the printed estimate $4[W/(M_0{c}^2){]}^2{E}_{ph}$. Ultrarelativistic motion alone does not justify it. For example $\gamma =1000$ and $\epsilon =M_0{c}^2$ would give $4\times 10^6{M}_0{c}^2$ from that estimate, exceeding the available total energy; the recoil-corrected maximum is about $999.75M_0{c}^2$.

For an arbitrarily hard incident photon, maximizing over direction needs a further qualification: if $\epsilon \ge \gamma \beta M_0{c}^2$, the largest outgoing photon energy is the unchanged $\epsilon$, rather than this amplified backscattering branch. The exact photon energy gain is $\Delta {\epsilon }_{\max }={\epsilon }_{\max }^{\prime }-\epsilon$. Calling the outgoing energy the gain is accurate only when amplification is large. These distinctions prevent the inverse Compton scattering estimate from violating conservation of energy outside its low-rest-frame-photon-energy regime.

For causal reachability on a spatial circle, put $P=2\pi R$ and use a representative $\bar{x}\in [-P/2,P/2)$ of the spatial circle. The lower-left panel of part c is a fundamental strip with its vertical edges identified. On the covering space the charged world line is still the solution of part b; on the quotient it isA jump between opposite strip edges is a coordinate seam, not a discontinuity of the physical world line. At its $n$th complete lap, $x=nP$, soFor $a=q{E}_x/m>0$, the speed increases on each return while remaining below $c$. Its kinetic energy increases byThere is no single-valued electrostatic potential energy $-q{E}_x\bar{x}$ on a circle: the nonzero electromotive force is $\oint \mathbf{E}\cdot d\mathbf{l}=E_{x}P$. A globally defined magnetic vector potential one-form $A_x=-E_{x}t$ instead makes the driving explicit. In an embedded circuit, Faraday's law realizes this electromotive force through changing magnetic flux. More generally an external apparatus maintaining the field supplies the increasing particle energy. In an isolated dynamical field-plus-particle system, backreaction would change the field. Thus this is not a cyclic machine extracting work from a conservative potential, and it does not violate conservation of energy.

A useful plot must keep three different times separate: source emission $t_e$, atomic absorption or re-emission $t$, and return reception $t_R$. For the cloud centroid,Let $q=f_0/f<1$. Resonant excitation requires $q{D}_{\mathrm{in}}(t)=1$. Since $D_{\mathrm{in}}$ decreases from $\gamma (1+\beta )$ to $\gamma (1-\beta )$, its zero-width resonance first becomes possible atAt equality the line centre is reached only at infinite approach distance, where illumination vanishes. A finite source start time further limits how closely that ideal threshold is approached. For $\beta >{\beta }_{\min }$ the line-centre event satisfiesThere is a second useful critical speed,for which excitation is centred on closest approach, $t_{\ast }=0$. For ${\beta }_{\min }<\beta <{\beta }_{\perp }$ it occurs on approach; for $\beta >{\beta }_{\perp }$ it occurs after closest approach. Even a receding ultra-relativistic atom can see the incident light blueshifted because $\gamma$ is large.

A photon emitted toward the origin near the atomic rest-frame frequency $f$ has return Doppler factorThe sign differs from that in $D_{\mathrm{in}}$ because the returning ray travels in the opposite direction. The returning line centre decreases from $\gamma (1+\beta )f$ on early approach to $\gamma (1-\beta )f$ on late recession, and is $f/\gamma$ at closest-approach emission. The origin receives that closest-approach radiation at $t_R=L/c$, not at zero. Prompt re-emission at exact resonance would giveIt is above $f_0$ below ${\beta }_{\perp }$, equal at ${\beta }_{\perp }$, and below $f_0$ above it. A finite fluorescence lifetime shifts emission later along the orbit and therefore usually toward a lower return frequency.

The following plots adopt a compact, optically thin cloud, weak excitation, isotropic incoherent spontaneous emission in the atomic rest frame, and negligible velocity dispersion. They show an intensity envelope as the observable amplitude; its rms electric-field amplitude is proportional to the square root of this intensity. An incoherent fluorescence signal has no fixed optical phase. The four rows include a nonrelativistic speed, both critical speeds, and an ultra-relativistic speed. All rows share one intensity normalization, so off-resonant signals remain visibly weak.

A model for fluorescence from a uniformly moving cloud generating the plotted envelopes iswhere $\mathcal{L}(\Delta )=[1+4{\Delta }^2/{\Gamma }^2{]}^{-1}$ is a Lorentz line shape of full width $\Gamma$. The factor $1-\beta s$ is the incident photon-flux correction per source-frame time. The population is an exponential memory of earlier absorption events, not an instantaneous resonance indicator. For a detector of fixed small area at the origin, its received bolometric intensity scales asTwo powers of $D_{\mathrm{out}}$ come from aberration of light, one from photon energy, and one from arrival-rate compression. Finite lifetime is dilated to $\gamma \tau$ in the source frame; over an interval with slowly changing geometry its observed decay time is $\tau /D_{\mathrm{out}}$. Thus approach can compress and amplify a pulse, while recession stretches and redshifts it. Illumination and collection each supply an inverse-square geometric factor in the prompt weak-excitation limit. Strong forward beaming can even make a detuned approaching precursor brighter than the later line-centre feature, as in the ultra-relativistic row; the intensity maximum need not occur at exact resonance.

Absolute amplitude and a unique detailed spectral line profile cannot be inferred from $f,\tau ,v$ alone: cloud size, atom number, source power, optical depth, angular emission pattern and saturation were not specified. Finite cloud size convolves these point-centroid curves with additional delays. The plots describe incoherent, frequency-redistributed emission near $f$; a coherent Rayleigh scattering component would instead retain the driving frequency in the atom frame and add a different spectral component. These qualifications do not change the kinematic resonance thresholds or return-time relations.

An invariant has the same value when the same physical event or state is described in different inertial frames. A conserved quantity has the same value as the physical system evolves, subject to its equations and isolation conditions. These are different comparisons. The invariant mass of a heated object can change with time, while an isolated system's conserved energy has different numerical values in boosted frames.

Under a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, a Lorentz four-vector transforms asA covector transforms by the inverse matrix, $W_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{W}_{\nu }$. A general tensor has one corresponding factor for each index; for example,Contraction with the Minkowski metric produces Lorentz scalars; it does not by itself make those scalars conserved.

Use $\hslash =c=1$ and the mostly-plus Minkowski metric. Then the source's Klein-Gordon equation is $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. For a plane wave $\varphi =C{e}^{-i\omega t+i\mathbf{k}\cdot \mathbf{x}}$, substitution givesThus a real propagating wave vector requires $|\omega |\ge m$, or $|E|\ge m{c}^2$ on restoring units. The positive-energy branch has a rest-energy threshold $m{c}^2$. At $|\omega |=m$, $\mathbf{k}=0$; frequencies below the mass threshold give evanescent spatial dependence rather than real-wave-vector propagation. Both signs of $\omega$ occur in the classical equation; a real scalar field combines complex-conjugate modes. Its group velocity is $\mathbf{k}/\omega$ and is subluminal, though the positive-branch phase speed exceeds one when $m>0$.

For a static spherically symmetric scalar field, the equation outside the source isSetting $\psi =r\varphi$ gives ${\psi }^{\prime \prime }-m^2\psi =0$, soDecay at infinity selects $B=0$, the Yukawa potential. It is singular at the origin and describes a point source distributionally; a source-free solution regular at the origin and decaying at infinity must be zero. For $m=0$, $\varphi =A/r+B$, and decay sets the constant $B$ to zero.

Write $\Gamma =(1-v^2/c^2{)}^{-1/2}$ for the Lorentz factor of the inertial frame $S^{\prime }$, distinguishing it from the particles' changing Lorentz factors. Apply the Lorentz transformationto each world line in a. Both particles initially have velocity $-v$ in $S^{\prime }$. The switching events have coordinatesThus the Positron starts accelerating earlier in $S^{\prime }$. Its velocity first becomes less negative, vanishes when its laboratory velocity reaches $v$, and then becomes positive. The electron continues at $-v$ until $t^{\prime }=0$, then accelerates toward $-c$. The relativistic velocity-addition formula makes this precise:The electric field parallel to the Lorentz boost has unchanged magnitude, but its switching surface is $t^{\prime }+v{x}^{\prime }/c^2=0$, rather than $t^{\prime }=0$ everywhere.

In the laboratory, the total three-momentum is zero for all $t$: the two Lorentz forces cancel at equal laboratory times. In $S^{\prime }$, each active force is still $qE$, since this is one-dimensional motion with a parallel electric field. Define $t_p^{\prime }=-\Gamma vd/c^2$. For $v>0$ the total three-momentum, evaluated at equal $t^{\prime }$, isThis follows by integrating $d{p}_p^{\prime }/d{t}^{\prime }=F$ from its switching event and $d{p}_e^{\prime }/d{t}^{\prime }=-F$ from its later switching event. During the interval between them only the Positron feels the prescribed electric field, so momentum conservation for the particle pair fails. Once both forces act, the total three-momentum is constant again.

The equal-time momentum of an externally driven particle pair illustrates a distinction. The sum of particle four-momenta at a fixed pair of events is always a four-vector. However, a fresh sum at equal time in each inertial frame generally uses different events. Here transforming an equal-$t$ laboratory sum would give $-2\Gamma v\sqrt{m^2{c}^4+F^2{c}^2{t}^2}/c^2$, which varies with $t$; this differs from the equal-$t^{\prime }$ answer above. For example, with $c=m=F=d=1$ and $v=0.6$, the late-time equal-$t^{\prime }$ sum is $-0.75$, whereas transforming the equal-$t=0$ sum gives $-1.5$. This is a concrete failure of treating the pair's equal-time sums as components of a single frame-independent four-vector.

A closed system's total four-momentum, including the electromagnetic field and the apparatus supplying it, is conserved and independent of the choice of spacelike integration surface under the usual isolated-system boundary conditions. The particles alone exchange energy and momentum with an external system, so neither that independence nor universal particle-only momentum conservation is guaranteed.

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.

The laboratory remains the center-of-momentum frame. At the relativistic collision mass threshold, the two final W bosons can have arbitrarily small opposite momenta. The available relativistic energy is $2m_e{c}^2+eEd$, soEach incoming particle must acquire about $80.378489\mathrm{GeV}$ of kinetic energy, travelling about $16.08\mathrm{km}$. Equality is the ideal kinematic threshold; a nonzero production rate needs energy above it. This estimate neglects acceleration losses and treats the W bosons with the supplied nominal rest mass.

A linear particle accelerator avoids repeatedly bending high-energy electrons and Positrons. This greatly reduces the bending synchrotron radiation loss, which at fixed bending radius grows approximately as the fourth power of the relativistic energy for a given particle species. Its useful collision energy can therefore be increased without the severe circular electron collider radiation penalty. The price is long structures or high accelerating gradients: each bunch normally passes once, requiring substantial beam power, fresh bunches, and very tight focusing for high luminosity. A circular particle accelerator reuses its accelerating structures and circulates bunches through many collisions, but needs a large ring and substantial radiofrequency power to replace synchrotron radiation losses. Heavier particles such as protons radiate far less at a given energy and radius, although a proton's constituent structure makes the hard-collision energy less sharply specified than in an electron-positron annihilation experiment.

Let $z^{\mu }(\tau )$ be the point electric charge's world line, with proper time $\tau$ and four-velocity $U^{\mu }=d{z}^{\mu }/d\tau =\gamma (c,\mathbf{u})$. For the field event $x^{\mu }$, choose the retarded source event: $R^{\mu }=x^{\mu }-z^{\mu }({\tau }_{\mathrm{ret}})$ is future-directed and null. Write $R^{\mu }=(R,R\mathbf{n})$, where $R=|\mathbf{x}-\mathbf{z}(t_{\mathrm{ret}})|=c(t-t_{\mathrm{ret}})$. In the instantaneous rest inertial frame of that source event,Consequently the covariant expression matching this Coulomb potential isThe denominator is a Lorentz scalar, so the right-hand side is a four-vector with the required instantaneous-rest components. This describes the retarded potential in Lorenz gauge, with no added free electromagnetic wave. Lorentz covariance alone does not establish the boundary conditions or exclude additional terms depending on four-acceleration. The physical input is the retarded point-source solution of Maxwell equations: the potential at the retarded event depends on the instantaneous four-current, while its derivatives generate the acceleration field. Equivalently, evaluating the retarded source integral gives a Jacobian $1/(1-\mathbf{n}\cdot \mathbf{\beta })$ from the light-cone constraint; this fixes that input for arbitrary smooth subluminal motion.^[ref]

In any inertial frame, put $\mathbf{\beta }=\mathbf{u}/c$ and $\kappa =1-\mathbf{n}\cdot \mathbf{\beta }$. Then $-R\cdot U=\gamma cR\kappa$, and the Liénard–Wiechert potential isDifferentiate using $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, including the position and time dependence of the retarded time. For example, $\partial t_{\mathrm{ret}}/\partial t=1/\kappa$ and $\nabla t_{\mathrm{ret}}=-\mathbf{n}/(c\kappa )$. The resulting Liénard-Wiechert field isHere the dot denotes a derivative with respect to source coordinate time. The first term is the velocity or non-radiative field: it falls as $R^{-2}$ and exists even without acceleration. The second is the acceleration or radiation field, falling as $R^{-1}$ and transverse to $\mathbf{n}$. Its Poynting flux falls as $R^{-2}$, so its outward power integrated over a large sphere stays finite as its radius tends to infinity. The first term's squared-field flux instead vanishes in that limit. This large-distance distinction defines radiation; a moving near field can carry local energy without constituting radiation escaping to infinity.