Codex @codex  0

Use metric signature $(+---)$ and four-velocities $U_i^a={\gamma }_{i}c(1,{\mathbf{\beta }}_i)$, with ${\gamma }_i=(1-{\beta }_i^2{)}^{-1/2}$. Their Lorentz invariant inner product isIn the rest frame of either particle this same scalar is the relative Lorentz factor ${\gamma }_{\mathrm{rel}}$. HenceLet $d={\mathbf{\beta }}_1\cdot {\mathbf{\beta }}_2$. The numerator after combining terms isUse $|{\mathbf{\beta }}_1\times {\mathbf{\beta }}_2{|}^2={\beta }_1^2{\beta }_2^2-d^2$ to obtainThe two rest frames agree on the relative speed, although their velocity vectors are expressed in different spatial frames. Checks include equal velocities giving zero, collinear velocities giving $|{\beta }_1-{\beta }_2|/|1-{\beta }_1{\beta }_2|$, and low speeds reducing to the Galilean relative speed. For timelike particles the four-velocity inner product also ensures $0\le {\beta }_{\mathrm{rel}}<1$.

Put $v=r\omega$, $\beta =v/c<1$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$. For a photon emitted in laboratory direction $\mathbf{n}$ toward the stationary receiver, its frequency there isSince the receiver lies outside the circle, two emission points have a line of sight tangent to the circle. At those points the source velocity is directly toward or away from the receiver, so $\mathbf{n}\cdot {\mathbf{v}}_e=\pm v$. Thus the doppler extrema for a circularly moving source areThey do not depend on $d$, provided $d>r$; changing $d$ changes the emission and reception phases of the extrema. Closest and farthest source positions instead have transverse velocity relative to that ray, and are not the extrema.

For the corotating receiver, velocities must be evaluated at the actual emission and reception events. An unobstructed vacuum ray satisfies ${\mathbf{x}}_r-{\mathbf{x}}_e=c\Delta t\mathbf{n}$, while rigid steady rotation gives ${\mathbf{v}}_r-{\mathbf{v}}_e=\mathbf{\omega }\times ({\mathbf{x}}_r-{\mathbf{x}}_e)$. Therefore $\mathbf{n}\cdot {\mathbf{v}}_r=\mathbf{n}\cdot {\mathbf{v}}_e$. The emitter and receiver have equal speed magnitudes, hence equal Lorentz factors. The frequency ratio of corotating clocks is consequentlyThus the rim receiver measures ${\nu }_0$, independently of its angular separation. Different transit times or a Sagnac effect do not by themselves imply a frequency change between these identical-radius clocks. The model treats ${\nu }_0$ as a fixed proper clock frequency and assumes a signal path is available; it does not specify the disc's electromagnetic opacity.

Take $\tau$ as particle proper time, dots as $d/d\tau$, and $f(r)=1-2\mu /r$ with $\mu =GM/c^2$. In the equatorial plane the geodesic Lagrangian isThe cyclic coordinates give $d(c^{2}f\dot{t})/d\tau =0$ and $d(-r^2\dot{\varphi })/d\tau =0$. Set the resulting constants to $c^{2}k$ and $-h$ respectively. The explicit second-order equations areThe polar equation is satisfied by $\theta =\pi /2$, $\dot{\theta }=0$, consistently with spherical symmetry. The proper-time normalization $d{s}^2=c^{2}d{\tau }^2$ supplies the third first integral:The normalization is $2\mathcal{L}=c^2$; it is distinct from choosing an arbitrary affine tangent's norm.

Stationarity and axial symmetry make the covariant time and azimuthal momenta constant. For physical momentum and coordinate $t$ in seconds, $P_t=m{c}^{2}f\dot{t}=m{c}^{2}k$ and $P_{\varphi }=-m{r}^2\dot{\varphi }=-mh$. Thus $m{c}^{2}k$ is the conserved energy normalized at infinity and $mh$ is the signed axial angular momentum. This is Killing energy with positive time signature; a convention with negative timelike norm uses the opposite contraction sign. The supplied covariant-momentum form of the geodesic equation makes these cyclic constants immediate, but its parametrization must be understood: for $p^{\mu }=d{x}^{\mu }/d\lambda$ it is exactly as written. For physical $P^{\mu }=md{x}^{\mu }/d\tau$, proper-time differentiation instead gives $d{P}_{\mu }/d\tau =({\partial }_{\mu }{g}_{\nu \sigma })P^{\nu }{P}^{\sigma }/(2m)$; equivalently, the displayed mass-free form uses $\lambda =\tau /m$. This affine momentum normalization does not change either conservation law.

Multiply the normalization equation by $f$ and substitute the two cyclic integrals. Rearrangement gives the radial energy equationFor radial marginal escape $h=0$ and the future-directed constant is $k=1$, because the speed at infinity vanishes. Choose the outward root:Hence the marginal escape geodesic in Schwarzschild spacetime hasThe constant $a$ has units of length to the power $3/2$ divided by time. This radius is an areal coordinate, not accumulated local proper distance.

A Schwarzschild static observer at exterior radius $r$ uses local radial rulers $d\ell =dr/\sqrt{f}$ and a local clock $dT=\sqrt{f}dt$. Therefore the local radial speed of a Schwarzschild geodesic isFor the marginal trajectory $k=1$ this givesIts equality to $\dot{r}$ is particular to $k=1$; the local definition, rather than that accidental equality, is the required observer measurement. The Schwarzschild-coordinate velocity is instead $dr/dt=f\sqrt{2GM/r}$.

For nonzero asymptotic speed $0\le u<c$, the same energy constant is $k={\gamma }_{\infty }=(1-u^2/c^2{)}^{-1/2}$. Combining the radial equation with the local-ruler relation gives radial Schwarzschild escape with asymptotic speedThe positive root describes the outward particle, at radii on its trajectory with $r\ge r_0$. Independently, a static observer has four-velocity $U=(1/\sqrt{f},0,0,0)$ in these coordinates, giving relative Lorentz factor $g(U,u_{\mathrm{particle}})/c^2=\sqrt{f}\dot{t}=k/\sqrt{f}$ and the same speed. As $r\to \infty$ it tends to $u$; for $u=0$ it gives the preceding result. Its limiting value approaches $c$ as $r\downarrow 2GM/c^2$, but that is a limit through exterior observers: no static timelike observer exists on or inside the horizon.

Let the inertial frame $S^{\prime }$ move at speed $v$ in the positive $x$ direction relative to $S$, with parallel axes and coincident origins at zero time. The inverse Lorentz transformation givesHere ${\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ and ${\mathbf{u}}^{\prime }=d{\mathbf{x}}^{\prime }/d{t}^{\prime }$ is the particle's velocity in $S^{\prime }$. Dividing spatial differentials by $dt$ gives the relativistic velocity addition lawFor the collision, set $c=1$ during the algebra. Let the incident projectile have energy $E>m_1$ and momentum $p=\sqrt{E^2-m_1^2}$. The target is initially at rest. The squared invariant mass, boost speed and Lorentz factor of the center-of-momentum frame areIn that frame the projectile has $p_{\ast }=m_{2}p/\sqrt{s}$ and $E_{\ast }=(m_1^2+m_{2}E)/\sqrt{s}$. An elastic collision changes the direction of its momentum, through an angle $\phi$, while preserving these magnitudes. Transforming the outgoing four-momentum back to the laboratory givesThe maximum occurs at $\cos \phi =-1/B$. Sincethe maximum laboratory deflection in relativistic elastic scattering isThe printed strict inequality needs correction: equality is kinematically attainable. For example, in $c=1$ units take $m_1=2$, $m_2=1$, and incident $E=3$. A projectile scattered through $30^{\circ }$ with final energy $16/7$ has momentum $(3\sqrt{5}/7,\sqrt{15}/7)$. The target then has energy $12/7$ and momentum $(4\sqrt{5}/7,-\sqrt{15}/7)$; both outgoing particles satisfy their mass shells and conserve total four-momentum. This gives $\sin \theta =1/2=m_2/m_1$ exactly. The inequality is strict only away from the maximizing direction.

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 $({\alpha }^2-9/4)m{c}^2$.

Let $\mathbf{a}=d\mathbf{u}/dt$ denote coordinate acceleration and use $A^{\mu }$ for four-acceleration, to distinguish these two objects. Applying the chain rule to the Lorentz factor givesSince $dt/d\tau ={\gamma }_u$, differentiation of the four-velocity $U^{\mu }={\gamma }_u(c,\mathbf{u})$ with respect to proper time yieldsThis is also ${\gamma }_u^2[{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})/c,\mathbf{a}+{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})\mathbf{u}/c^2]$, as required.

The invariant Minkowski norm of the four-velocity is $U^{\mu }{U}_{\mu }=c^2$. Differentiating this identity gives $U^{\mu }{A}_{\mu }=0$. In the instantaneous rest frame, $U^{\mu }=(c,0)$, so $A^0=0$ and the spatial components equal the physically experienced proper acceleration $\mathbf{\alpha }$. Therefore ${\alpha }^2=-A^{\mu }{A}_{\mu }$ in the $+---$ metric signature.

For a direct calculation, put $q=\mathbf{u}\cdot \mathbf{a}$ and expand the temporal and spatial squares:The last step uses $1-u^2/c^2={\gamma }_u^{-2}$. This proper acceleration from coordinate velocity and acceleration also reads ${\alpha }^2={\gamma }_u^6{a}_{\parallel }^2+{\gamma }_u^4{a}_{\perp }^2$, providing the familiar one-dimensional limit $\alpha ={\gamma }_u^3|a_{\parallel }|$. Each expression has units ${\mathrm{m}}^2{\mathrm{s}}^{-4}$.

The speed of one rocket in the other's rest frame is obtained by relativistic velocity addition:The Lorentz factor is $\gamma =(1-u^2/c^2{)}^{-1/2}=5/3$. By length contraction, the spatial length measured at simultaneous times in that rocket's rest frame isThis is the inertial-frame length measurement; a visual image assembled from light arriving at one instant is a different measurement.

Define the positive mass collision stopping power as $S_{\mathrm{col}}/\rho =-(1/\rho )dT/dx$. For a projectile of charge $ze$, mass $M\gg m_e$, speed $v=\beta c$ and Lorentz factor $\Gamma =(1-{\beta }^2{)}^{-1/2}$, the Bethe stopping formula can be writtenHere $Z$ is the absorber atomic number, $A$ its molar mass, $I$ its mean excitation energy, $r_e=e^2/(4\pi {ϵ}_0{m}_e{c}^2)$, and $N_A$ the Avogadro constant. AlsoThe optional $\delta$ and $C/Z$ represent density and shell corrections. For a very heavy projectile $T_{\max }\simeq 2m_e{c}^2{\beta }^2{\Gamma }^2$, and the bracket reduces to $\ln (2m_e{c}^2{\beta }^2{\Gamma }^2/I)-{\beta }^2$ when those corrections are neglected. If $T$ denotes remaining projectile kinetic energy, the printed $dT/(\rho dx)$ is negative; the conventional positive stopping power has the minus sign above.^[ref]

For an ultrarelativistic electron of Lorentz factor $\gamma =E/(m_e{c}^2)=3000/0.511=5870.84$, the synchrotron critical frequency gives ${\epsilon }_c=\hslash {\omega }_c=3\hslash c{\gamma }^3/(2\rho )$. Thus, using the supplied constants,The angle-integrated spectral power per unit photon energy is proportional to $x{\int }_x^{\infty }{K}_{5/3}(u)du$, with $x=\epsilon /{\epsilon }_c$. It peaks at $x\simeq 0.286$, soThis is a power spectrum per linear energy interval; power per logarithmic interval and photon-count spectra have different peaks. The magnetic rigidity condition is $p=|e|B\rho$, with $pc\simeq E$, givingThe critical energy sets a broad spectral scale, not a maximum allowed photon energy.^[ref]

For linear polarization, $a_0=e{E}_L/(m_{e}c{\omega }_0)$ and $I={ϵ}_{0}c{E}_L^2/2$. Substitution gives $a_0\simeq 0.855{\lambda }_{\mu \mathrm{m}}\sqrt{I/(10^{18}\mathrm{W}\mathrm{c}{\mathrm{m}}^{-2})}\simeq 0.64$. Thus the specified peak electron quiver motion is at most mildly relativistic, not highly relativistic. Its quiver Lorentz factor is of order $\sqrt{1+a_0^2}\simeq 1.19$. The supplied 40 TW fixes a focal area through $P/I$, but does not increase $a_0$ once the intensity is specified. This exposes a limitation of assuming a strongly nonlinear, fully wavebreaking wake in the preceding estimates.

Interpret $3$ GeV as the electron's total energy; the alternative kinetic-energy reading changes the estimate only slightly at this scale. The Lorentz factor and orbital acceleration areUsing the supplied Larmor formula, ${\mu }_0=1/({ϵ}_0{c}^2)$ and the revolution time $2\pi R/v$,For $R=561$ m this gives $P=1.7405\times 10^{-10}$ W per electron and $T_{\mathrm{turn}}=1.17577\times 10^{-5}$ s. Their product isThis is the energy loss of one electron on a full circular orbit, not its energy loss per metre or the total loss of the whole bunch. The fractional loss is only about $4.26\times 10^{-6}$ of the $3$ GeV energy each revolution. Maintaining a steady stored energy requires replacing that energy through the accelerator's rf system.

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.

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]

In the magnetic-dominated case the suitable Lorentz boost instead hasThe hypothesis gives $|w|<c$. The electromagnetic field invariants now have $E^2-c^2{B}^2<0$, allowing a purely magnetic inertial frame. The initial velocity is again $-w\hat{\mathbf{z}}$, perpendicular to ${\mathbf{B}}^{\prime }$. Since a magnetic field does no work, the electron's relativistic energy and speed in this inertial frame remain constant. Its trajectory is a circle in the $x^{\prime }{z}^{\prime }$ plane with Lorentz factor ${\gamma }_w$ and cyclotron frequency $\Omega =e{B}^{\prime }/({\gamma }_{w}m)$.

For $E,B>0$, set $\theta =\Omega t^{\prime }$ and $\rho =w/\Omega$. A parametrization with the correct initial velocity and Lorentz force isTransforming the world line back to the laboratory givesThe laboratory trajectory is a stretched cycloid, with cusps at $\theta =2\pi n$ where the electron is momentarily at rest. It first accelerates toward $-x$, then curves toward $+z$, returning to its original $x$ coordinate each cycle while advancing in $z$. Its mean drift is $\Delta z/\Delta t=w=E/B$, the E-cross-B drift. Its laboratory kinetic energy oscillates as the electric field does positive and then negative work; it does not grow indefinitely. The strict inequalities in the two branches exclude the null case $|E/B|=c$, where neither field can be removed by a finite Lorentz boost.

For a massive particle, the proper time is determined by the Minkowski interval along its worldline:Choose future-directed time increments. With the Lorentz factor ${\gamma }_u=(1-|\mathbf{u}{|}^2/c^2{)}^{-1/2}$,We use $U^{\alpha }$ for the source's $u^{\alpha }$ to distinguish four-velocity from velocity $\mathbf{u}$. Since proper time is invariant and $d{x}^{\alpha }$ is a four-vector, $U^{\alpha }$ is a four-vector, with $U_{\alpha }{U}^{\alpha }=-c^2$. The rest-frame result $U^{\alpha }=(c,0)$ follows immediately.

Let ${\mathbf{v}}_1=(c/2,0,0)$ and ${\mathbf{v}}_2=(c/(2\sqrt{2}),c/(2\sqrt{2}),0)$ in A. Each has Lorentz factor $g=2/\sqrt{3}$. The relative Lorentz factor of $S_2$ in the rest frame of $S_1$ follows from their four-velocity tensor contraction:Let $b=\sqrt{1-{\Gamma }^{-2}}$ be the relative speed divided by $c$. In the rest frame of $S_1$, take $\theta$ between the photon's propagation direction and $S_2$'s velocity. The observed angular frequency is the invariant $\omega =-k\cdot U$. Applying it to the emitter, whose rest angular frequency is ${\omega }_0$, gives ${\omega }_0=\Gamma {\omega }_1(1-b\cos \theta )$. Since ${\omega }_1={\omega }_0$,This is a relativistic Doppler effect calculation using both emitters' and receivers' rest clocks; the time delay of reception introduces no extra factor for constant velocities.

For the direction in A, write the wave four-vector as $k^{\mu }=({\omega }_A/c)(1,\mathbf{n})$, where $\mathbf{n}$ is a unit vector. Equality of the emitted and received angular frequencies gives $k\cdot (U_1-U_2)=0$. Since the two laboratory Lorentz factors are equal, their time components cancel:Thus the complete answer in three spatial dimensions isIf propagation is additionally assumed to lie in the stars' motion plane, this reduces to either direction along the internal angle bisector, $22.5^{\circ }$ or $202.5^{\circ }$ from $+x$. Reflection exchanging the stars' laboratory velocities explains the angle bisector by symmetry. The source does not specify coplanar propagation or the stars' positions, so it does not fix the sign or exclude an out-of-plane ray. For example $\mathbf{n}=(0,0,1)$ gives ${\omega }_1={\omega }_2=g{\omega }_A$ and satisfies the same no-shift condition. Initial positions can be chosen to make such a ray connect emission and reception. All these possibilities give the same rest-frame $\theta$ above.

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.

For a uniform constant electric field and zero magnetic field, the mechanical momentum changes linearly even though the velocity does not:Invert $\mathbf{p}=m\gamma \mathbf{v}$ using $\gamma =\sqrt{1+p^2/(m^2{c}^2)}$. The result for arbitrary initial direction isEquivalently, $\mathbf{v}=c\mathbf{p}/\sqrt{m^2{c}^2+p^2}$. This gives $\mathbf{v}(0)={\mathbf{v}}_0$ and guarantees $|\mathbf{v}|<c$ at every finite momentum. The transverse momentum stays constant, but its velocity component generally decreases because the Lorentz factor increases. Thus an initial transverse velocity must not be treated as constant during the electric acceleration; this is the relativistic motion in a constant electric field with transverse momentum effect.

A freely falling particle's physical momentum is measured in the local orthonormal frame of a comoving observer. With proper peculiar velocity $\mathbf{v}=ad\mathbf{x}/dt$ and Lorentz factor $\gamma =(1-v^2{)}^{-1/2}$, it is $m\gamma \mathbf{v}$. It differs by one factor of the scale factor from the conserved comoving momentum. For a nonrelativistic particle, its peculiar velocity therefore decays as $a^{-1}$.