The upper-left panel shows the three causal structure regions, with null boundaries included in the appropriate causal future and causal past. For the charged trajectory, use part b with $a=q{E}_x/m$, choosing the $x$ direction so that $a>0$. The electric field gives the constant three-force $q{E}_x$ and hence the shifted hyperbolaIt starts with a vertical tangent at the origin, stays strictly inside the future light cone, and asymptotes to $x=ct-c^2/a$. The other panels implement the coordinate and topology changes used below; they are original spacetime diagrams, not copies of source artwork.

Let $m$ denote the constant rest mass, set $x(0)=0$, and put $a=f_x/m$. The relativistic three-force is $f_x=d{p}_x/dt$, soIt follows thatIntegrating the velocity with the initial position fixed givesThese formulas allow either sign of $a$; their $a\to 0$ limit is rest at the origin. For $a\ne 0$, the trajectory obeysThis is hyperbolic motion. Its coordinate acceleration decreases while its proper acceleration has constant magnitude $|a|$. In particular, a constant three-force does not give constant coordinate acceleration at relativistic speed.

Use coordinates $X^{\mu }=(ct,\mathbf{x})$ and the mostly-plus Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(-1,1,1,1)$. Along a future-directed timelike world line, proper time satisfiesThe four-velocity and four-acceleration areThe Minkowski norm is fixed: $U^{\mu }{U}_{\mu }=-c^2$. Differentiating with respect to proper time, using the constant Minkowski metric, givesIn the instantaneous rest reference frame, $U^{\mu }=(c,0,0,0)$, so the orthogonality says $A^0=0$ there. This construction uses a timelike world line; a massless particle has no rest-frame proper time parameter.

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.

Use a neutral, infinitely long wire on the $x$ axis. Denote the total current by $I=\int j_{x}d{A}_{\perp }$, and write $r^2=y^2+z^2$. Outside the wire the magnetic field and electric field areThese follow from cylindrical symmetry and Ampère's law. The zero temporal component of the given total four-current specifies lab neutrality; the $\rho$ in $j_x=\rho v_0$ must refer to a carrier density, not that zero net charge density.

TODO: normalization from a literal volume current density requires the unspecified wire cross-section. The source does not supply its area or profile. The exterior field family below is fully determined in terms of $I$; it cannot be reduced to $j_x$ alone without that missing datum. If $j_x$ was intended as a cross-section-integrated line current instead, set $I=j_x$ and interpret its accompanying charge density as charge per length. No inside-wire profile is invented.

For a boost $u=v_0$ along positive $y$, let $\Gamma =(1-u^2/c^2{)}^{-1/2}$. Apply the Lorentz transformation of electromagnetic fields at corresponding events:Here the subscript labels the perpendicular-to-wire boost, not the direction of the transformed vector. In these formulas evaluate the original field at $y=\Gamma (y^{\prime }+u{t}^{\prime })$, $z=z^{\prime }$ and $x=x^{\prime }$. The wire moves in this frame, so the fields are time-dependent. The local current transforms to $J^{\prime 0}=0$, $J^{\prime x}=j_x$, $J^{\prime y}=J^{\prime z}=0$. A nonzero transformed electric field is possible even with zero net charge density: it is not a purely electrostatic field.

For a boost $u=v_0$ along positive $x$, the transverse coordinates are unchanged. The transformed fields areThe transformed line charge and current are ${\lambda }^{\prime }=-\Gamma uI/c^2$ and $I^{\prime }=\Gamma I$. These reproduce the radial electric field through Gauss's law. A boost comoving with the electrons removes their individual current contribution, but the positive background then moves and the total current is not zero.

For two identical wires, their lab interaction is attractive, with magnitude ${\mu }_0{I}^2/(2\pi d)$ per length. The total four-current is spacelike, $J^2=-j_x^2<0$, so no inertial frame makes it a pure charge density. Also $B^2-E^2/c^2=B^2>0$ forbids a frame in which the exterior field is purely electric. For an arbitrary boost along the wires, the electrical part is repulsive and the magnetic part attractive:The latter always dominates, and their sum remains $-{\mu }_0{I}^2/(2\pi d)$.

Even a more general boost cannot make the entire-wire force electrical alone. At the second wire choose $y=d,z=0$, so $\mathbf{B}=B{\mathbf{e}}_z$ and $\mathbf{j}=j{\mathbf{e}}_x$ with $jB>0$. The physical Lorentz force density is $f=(0,0,-jB,0)$. If the boosted magnetic force vanished, ${\mathbf{j}}^{\prime }\times {\mathbf{B}}^{\prime }=0$, then ${\mathbf{E}}^{\prime }\cdot {\mathbf{B}}^{\prime }=0$ would also give ${\mathbf{j}}^{\prime }\cdot {\mathbf{E}}^{\prime }=0$. Thus $f^{\prime 0}=0$, which from the boost of $f$ requires $V_y=0$. But for such a boost $f^{\prime y}=-jB<0$, whereas ${\rho }^{\prime }{E}_y^{\prime }={\Gamma }_V^2{V}_x^{2}jB/c^2\ge 0$. The electrical term alone cannot equal the actual force. Therefore there is no such frame for both complete wires, although the force on electrons considered separately can be electrical in their rest frame.

For $P^{\mu }=(E/c,\mathbf{p})$, directly contract the supplied vector:Thus $\boxed{U_W\cdot P=0}$ whenever its stated normalization is defined, namely $m_0>0$ and $|\mathbf{s}|>0$. This is the Pauli-Lubanski orthogonality property.

The description as a unit vector, however, needs an additional assumption. From the given components and the mass shell,It equals minus one only for spin parallel to momentum, including the zero-momentum case. A concrete counterexample is $\mathbf{s}=(s,0,0)$ and $\mathbf{p}=(0,m_{0}c,0)$: the proposed normalized vector has square minus two. Orthogonality is still true. A genuine Pauli-Lubanski pseudovector constructed by boosting a fixed rest-frame spin has invariant norm $-m_0^2{c}^2|{\mathbf{s}}_{\mathrm{rest}}{|}^2$; the source's generic spatial expression and denominator do not supply that normalization for arbitrary $\mathbf{s}$.

The electromagnetic four-potential is $A^{\mu }=(\Phi /c,\mathbf{A})$, and the vacuum four-wavevector is $K^{\mu }=(\omega /c,0,0,k)$ with $k=\omega /c>0$. Define $\psi =K_{\mu }{X}^{\mu }=\omega t-kz$ and $C=(\omega k{)}^{-1/2}$. The given transverse potential is $A_y=-C\sin \psi$. Its physical fields areso the Poynting vector points along positive $z$. For our signature, ${\partial }^{\mu }=(c^{-1}{\partial }_t,-\nabla )$, $F^{0i}=-E_i/c$, and $F^{ij}=-{ϵ}_{ijk}{B}_k$. ThereforeThe square-root prefactor is simply the source's chosen wave normalization; an overall dimensional amplitude could multiply every field.

The four-current is $J^{\mu }=(c\rho ,\mathbf{j})$, with $J_{\mu }=(c\rho ,-\mathbf{j})$. Charge conservation takes the equivalent formsLowering both indices of $F$ changes the sign of its time-space entries: $F_{0i}=E_i/c$, while $F_{ij}=-{ϵ}_{ijk}{B}_k$. The temporal component of ${\partial }^{\mu }{F}_{\mu \nu }$ is $\nabla \cdot \mathbf{E}/c={\mu }_{0}c\rho$, by Gauss's law. Its spatial components are $c^{-2}{\partial }_t\mathbf{E}-\nabla \times \mathbf{B}=-{\mu }_0\mathbf{j}$, by the Ampère-Maxwell equation. HenceFor the displayed free wave the current vanishes because $k=\omega /c$; the component derivation also proves the sourced identity. The two homogeneous Maxwell equations follow separately from expressing $F$ in terms of a potential.

Force-sign qualification. With these definitions the physical Lorentz force density is $f_{\mu }^{\mathrm{phys}}=F_{\mu \nu }{J}^{\nu }$, whereas the source prints the reversed contraction $J^{\nu }{F}_{\nu \mu }=-f_{\mu }^{\mathrm{phys}}$. ConsequentlyThe physical time component is the power transferred to matter per volume divided by $c$, as in Poynting's theorem. For example parallel positive $j_x$ and $E_x$ do positive work, but the literal printed contraction has negative time component. This sign cannot be repaired while simultaneously retaining the source's tensor definition and the Maxwell identity above; reverse the force contraction to describe the physical force.

For a boost along positive $x$, let $\gamma =(1-v_x^2/c^2{)}^{-1/2}$. Parallel field components are unchanged, while transverse components mix. The Lorentz transformation of electromagnetic fields givesThese signs correspond to $S^{\prime }$ moving with $+v_x$ relative to $S$, not the inverse boost. They can be obtained from $F^{\prime }=\Lambda F{\Lambda }^{\mathsf{T}}$ for the electromagnetic field tensor.

Take the stated static vector potential with zero scalar potential. In the laboratory,Let $S^{\prime }$ move along $z$ at the electron's reference velocity $v$. The inverse transformation is $z=\gamma (z^{\prime }+v{t}^{\prime })$. The Lorentz transformation of electromagnetic fields giveswith all other components zero. The sign follows from $\mathbf{v}\times \mathbf{B}=v{B}_x{\mathbf{e}}_y$.

Although an observer fixed in $S^{\prime }$ sees an oscillating field, this is not an exact source-free propagating electromagnetic wave. The phase covector has ${\omega }^{\prime }=-\gamma v{k}_u$, $k_z^{\prime }=\gamma k_u$, so its squared norm isIt is the boost of a static spatial grating, not a null vacuum wave. Equivalently, $E^{\prime }/B^{\prime }=v<c$ and $B^{\prime 2}-E^{\prime 2}/c^2=B^2$ is nonzero away from nodes; a vacuum plane wave would have vanishing electromagnetic field invariants. In the ultrarelativistic limit it can be a useful nearly wave-like approximation along the electron's path, but the exact distinction remains.

A charged electron cannot actually maintain the given constant velocity through this nonzero transverse field without an external constraint: the Lorentz force deflects it. The calculation therefore uses the prescribed straight reference trajectory, a weak-deflection approximation, or an instantaneously comoving inertial frame. A globally fixed electron rest frame is not available for its exact deflected motion.

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.

A boost into a frame moving with velocity $\mathbf{V}$ transforms the temporal component of the four-wavevector as $K^{\prime 0}=\Gamma (K^0-\mathbf{V}\cdot \mathbf{k}/c)$. Therefore the general relativistic Doppler effect iswhere $\vartheta$ is the laboratory angle between the boost and propagation directions. This is also the frequency measured from the phase rate along the observer's world line, ${\omega }^{\prime }=K\cdot U_{\mathrm{obs}}$. The frequency is an observer-dependent contraction of the invariant wave phase.

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.