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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.

For a freely moving unstable particle of fixed speed $v=\beta c$ and proper mean lifetime ${\tau }_0$, the mean travelled distance before decay is ${\ell }_d=v\gamma {\tau }_0$. Survival beyond distance $L$ has probability $e^{-L/{\ell }_d}$, since elapsed proper time is $L/(\gamma v)$. For rest mass $m$ and three-momentum magnitude $p$, ${\ell }_d=p{\tau }_0/m$.

In the simple box-state counting used here, the final three-momentum lies in a spherical shell: $dn=C{p}^{2}dp$, where the common constant $C$ contains the normalization volume and angular factor. Momentum conservation fixes the other final particle's momentum. Hence the density of states isCombining parts (a) and (b) givesWith the paper's rounded masses,By Fermi's golden rule, equal squared matrix elements make the rate ratio equal to this density ratio. Thus the naive constant-matrix-element estimate is $R\simeq 0.287$, which favours the electronic decay by about $3.49$ to one.

This is the particular noncovariant state-counting convention in the question. A calculation with the Lorentz-invariant phase-space measure places factors $1/(2E)$ in the measure and correspondingly normalizes its decay amplitude differently. Only the complete product of phase space and squared matrix element is convention-independent; mixing the two conventions would give a wrong rate.

The laboratory is the parent's rest frame. In this relativistic two-body decay, conservation of momentum gives equal and opposite daughter momenta, of magnitude $p$. Their total relativistic energies satisfySubtracting the relativistic energy-momentum relations eliminates $p$:Solving the two linear equations givesSince three-momentum obeys $p{c}^2=Ev$, the daughter speeds areTheir directions are opposite; the initial state does not specify their common axis. The kinetic energies are $K_1=5m{c}^2/8$ and $K_2=3m{c}^2/8$. Their sum $m{c}^2$ is the loss of rest energy when the total daughter rest mass is $3m$ instead of $4m$.

The proper mean lifetime ${\tau }_0$ is the mean lifetime measured in the unstable particle's own rest frame, or equivalently the mean elapsed proper time before decay. In the exponential decay model, its survival probability to proper time $\tau$ is $e^{-\tau /{\tau }_0}$. For the daughter of rest mass $m$, the Lorentz factor is ${\gamma }_1=E_1/(m{c}^2)=13/8$, so time dilation gives a laboratory mean lifetime of ${\gamma }_1{\tau }_0=3.25\mu \mathrm{s}$.

At constant speed $v_1$, travelling a distance $L$ requires laboratory time $L/v_1$ and proper time $L/({\gamma }_1{v}_1)$. Its mean decay length isfor ${\tau }_0=2\times 10^{-6}\mathrm{s}$. HenceThe mean lifetime sets the scale of an exponential distribution; it is not a fixed deadline, so travelling beyond one mean decay length has a nonzero probability.

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.

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.

For uniform circular motion, the transverse Lorentz force supplies the change in direction of three-momentum:The magnetic rigidity relation gives $pc=0.299792458B[\mathrm{T}]r[\mathrm{m}]\mathrm{GeV}=2.99792458\mathrm{GeV}$. Using $m_e{c}^2=0.511\mathrm{MeV}$,The coordinate centripetal acceleration is $a=u^2/r$. For transverse acceleration, the magnitude of the four-acceleration, or proper acceleration, is $a_0={\gamma }^{2}a$. The supplied Larmor formula, in its proper acceleration form, therefore givesThe emitted power here is also the laboratory total emitted power: in the instantaneous rest inertial frame the dipole radiation pattern has zero net radiated three-momentum, so both emitted energy and elapsed coordinate time gain the same Lorentz factor on transformation. This does not assert that a directional detector flux is invariant.

Numerically $\mathcal{T}\simeq 1.048\times 10^{-5}\mathrm{s}$, $P\simeq 2.18\times 10^{-10}\mathrm{W}$ per electron, andThe synchrotron's radiofrequency accelerating system must replace this lost energy to keep $r$ and magnetic rigidity fixed over many revolutions. The fraction lost per turn is about $4.77\times 10^{-6}$, so a fixed-orbit calculation over one turn is self-consistent.^[ref]

Assume $E,B>0$ for the directions and parametrizations; the boost formulas retain their signed meaning for reversed fields. A Lorentz boost with velocity $w\hat{\mathbf{z}}$ along $\mathbf{E}\times \mathbf{B}$ transforms the transverse electric field and magnetic field asIn the electric-dominated case chooseThe inequality ensures $|w|<c$, so this is an inertial frame. The field-removing boosts for perpendicular electric and magnetic fields are governed by the electromagnetic field invariants $\mathbf{E}\cdot \mathbf{B}=0$ and $E^2-c^2{B}^2>0$ explain why a purely electric inertial frame exists and a purely magnetic one does not.

The electron is not initially at rest in this convenient inertial frame: its initial velocity is $-w\hat{\mathbf{z}}$ and its initial three-momentum is $p_z^{\prime }=-{\gamma }_{w}mw$. Thereafter $p_x^{\prime }=-e{E}^{\prime }{t}^{\prime }$, while $p_z^{\prime }$ stays constant. The electric field accelerates it toward $-x$; as its relativistic energy increases, its negative transverse velocity decreases in magnitude. Defining $a=e{E}^{\prime }/m>0$ and $s=a\tau /c$ gives a useful parametric world line:Here $\tau$ is the proper time. These formulas follow by integrating $U^{\prime 0}={\gamma }_{w}c\cosh s$, $U^{\prime x}=-{\gamma }_{w}c\sinh s$, and constant $U^{\prime z}=-{\gamma }_{w}w$. Its projection onto the $t^{\prime },x^{\prime }$ plane is a hyperbola, but the spatial trajectory includes the initial transverse motion and is not a straight line.

Transforming back with $z={\gamma }_w(z^{\prime }+w{t}^{\prime })$, $t={\gamma }_w(t^{\prime }+w{z}^{\prime }/c^2)$ givesThe laboratory trajectory starts from rest toward $-x$, then bends toward $+z$. It is unbounded in $-x$ and ultimately has $u_z\to w=c^{2}B/E$ and $u_x\to -c/{\gamma }_w$. Thus the late velocity has magnitude $c$ as a limit, consistent with the relativistic energy-momentum relation, rather than undergoing a closed cyclotron orbit.

Use $c=1$ and define the total four-momentum $P^{\mu }=(E,\mathbf{P})={\sum }_i(E_i,{\mathbf{p}}_i)$. For an inertial frame moving with velocity $\mathbf{V}$ relative to the laboratory, the Lorentz boost givesChoose $\mathbf{V}$ parallel to $\mathbf{P}$ and require zero total three-momentum. ThenThis requires $E>|\mathbf{P}|$, so that the total four-momentum is timelike. Its invariant mass is $\sqrt{E^2-|\mathbf{P}{|}^2}$, the total relativistic energy in the center-of-momentum frame. If all particles are exactly parallel photons, the total four-momentum is null and no finite-velocity center-of-momentum frame exists.

Sign qualification. The minus sign printed in the source describes the laboratory's velocity in the center-of-momentum frame, or the parameter of an active Lorentz boost with a plus-sign spatial rule. It is not the velocity of the center-of-momentum frame relative to the laboratory under question 1's passive convention. For example, a single particle with $(E,P_x)=(5,3)$ has its rest frame moving at $V_x=3/5$. The printed $V_x=-3/5$ would give $P_x^{\prime }=(5/4)(3+3)=15/2$, which is not zero. This explicitly resolves the sign rather than assigning a rest frame moving away from the particle.

For constant laboratory force $\mathbf{F}$ and initial rest, integration gives three-momentum $\mathbf{p}=\mathbf{F}t$. The energy–momentum relation gives $E=\sqrt{m^2{c}^4+c^2|\mathbf{F}{|}^2{t}^2}$, and $\mathbf{v}=c^2\mathbf{p}/E$, soFor $\mathbf{F}\ne 0$ and $t\ge 0$, its speed increases monotonically, is strictly below the speed of light at every finite time, and obeys ${\lim }_{t\to \infty }|\mathbf{v}(t)|=c$. Thus $c$ is the limiting supremum, rather than a speed attained at a finite time. The degenerate case $\mathbf{F}=0$ leaves the particle at rest and does not have this limit.

Both the muon and electron have the same electric charge magnitude. At equal kinetic energy $K$ per particle, their Lorentz factors are $\gamma =1+K/(m{c}^2)$, so the heavier muon has the smaller Lorentz factor. At equal total relativistic energy the same ordering follows from $\gamma =E/(m{c}^2)$, provided that energy is large enough to create either particle.

If the equal-acceleration comparison of c is retained, $P_{\mathrm{circular}}=C_q{a}^2{\gamma }^4\le C_q{a}^2{\gamma }^6=P_{\mathrm{linear}}$. Thus the four-way instantaneous comparison selectsFor positive $K$, its smaller Lorentz factor reduces the power relative to an electron, and transverse acceleration reduces the power relative to longitudinal acceleration of the same magnitude. This conclusion follows from the constraint actually stated in c.

For the customary engineering comparison at equal applied force, the result changes. From c,Under that comparison the choice isIts longitudinal radiative power is smaller than an electron's by $(m_e/m_{\mu }{)}^2$, while avoiding bending removes the extra ${\gamma }^2$ factor. At a fixed bend radius $R$, the circular loss is $C_q{\gamma }^4{v}^4/R^2$, giving the familiar $m^{-4}$ suppression at fixed ultrarelativistic energy and $R$.

An energy budget alone does not specify a unique optimization. The force, acceleration, bend radius, acceleration time, and length are otherwise unrestricted. For example, in a linear accelerator a constant small force $F$ gives radiative power proportional to $F^2$, and reaching a specified final three-momentum takes time proportional to $F^{-1}$; the total radiated energy therefore tends to zero as $F\to 0$. One cannot infer a unique accelerator choice without an additional constraint. The standard practical preference is the second boxed answer, with its stated comparison. Muon decay introduces a separate lifetime constraint if slow acceleration is contemplated, but it was not part of the specified radiative-loss calculation.

A radiation pattern specifies emitted power per solid angle as a function of direction. For a slowly moving point electric charge with acceleration $\mathbf{a}$, $dP/d\Omega =q^2|\mathbf{n}\times \mathbf{a}{|}^2/(16{\pi }^2{ϵ}_0{c}^3)$. Its integral over directions is the Larmor formula; its fore-aft symmetry implies zero net radiated three-momentum in the instantaneous rest inertial frame. Directional detector power includes the distinction between source emission time and arrival time.

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.