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For identical particles of rest mass $m$ and electric charge $q$, the four-current is $j^{\mu }=(q/m)\rho u^{\mu }$. The matter stress-energy tensor obeys ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, with the opposite force density in the electromagnetic stress-energy tensor. Antisymmetry of the electromagnetic field tensor makes its contraction with two four-velocities vanish, so mass continuity still holds. Removing that term leaves the displayed covariant Lorentz force law. The matter tensor alone is not divergence-free when an electromagnetic force acts.

Charge invariance means that the total electric charge of a specified collection of particles is unchanged by a Lorentz transformation. It differs from conservation of electric charge, which compares the same isolated system at different times. For a convective charge distribution with rest density ${\rho }_0$, the four-current is $J^{\mu }={\rho }_0{U}^{\mu }$, where $U^{\mu }$ is its four-velocity; its measured charge density increases under a boost while the simultaneous spatial volume contracts, leaving total charge unchanged.

The four-divergence of a four-vector field is the Lorentz scalar contraction ${\partial }_{\mu }{V}^{\mu }$. With $x^{\mu }=(ct,\mathbf{x})$, it equals $c^{-1}{\partial }_t{V}^0+\nabla \cdot \mathbf{V}$. The four-current has zero four-divergence, expressing local charge conservation.

For an ideal infinite filament with zero line charge density and current $I$ in a laboratory frame, the line-integrated four-current is $(0,I)$ in temporal/axial components. An axial boost of speed $u$ gives the title density and current. The laboratory magnetic field ${\mu }_{0}I\hat{\mathbf{\phi }}/(2\pi r)$ therefore becomes both an electric field from ${\lambda }^{\prime }$ and a magnetic field from $I^{\prime }$. Charge neutrality is frame-dependent for a current-carrying system. A model built from oppositely charged moving and resting populations makes this result explicit; the charge of each fixed material particle set remains invariant.

Assume an infinite, stationary, uniform thin wire, and no added external homogeneous fields. Its four-current is $j^{\mu }=(\sigma ,0,0,J)\delta (x)\delta (y)$. A boost along the wire leaves the transverse delta functions unchanged and transforms the time and longitudinal components, givingThe densities are measured per unit simultaneous length in their respective frames; treating charge alone as invariant per unchanged coordinate length would miss the current contribution.

For a coaxial cylinder, Gauss's law gives $E_r(2\pi rL)={\mu }_0\sigma L$ in the stated $c=1$ units. For a circular loop, Ampère's law gives $B_{\phi }2\pi r={\mu }_{0}J$. Thus $E_r={\mu }_0\sigma /(2\pi r)$ and $B_{\phi }={\mu }_{0}J/(2\pi r)$. The same reasoning in the primed frame gives the same expressions with primed densities.

Since $\hat{z}\times \hat{\phi }=-\hat{r}$ and $\hat{z}\times \hat{r}=\hat{\phi }$, the Lorentz transformation of electromagnetic fields gives $E_r^{\prime }=\gamma (E_r-v{B}_{\phi })$ and $B_{\phi }^{\prime }=\gamma (B_{\phi }-v{E}_r)$. Substituting the laboratory fields reproduces the primed-density formulas exactly; parallel components vanish in both frames.

Use SI units, $x^a=(ct,\mathbf{x})$ and metric signature $(+,-,-,-)$. The four-current is $j^a=(c\rho ,\mathbf{j})$. Charge density and current transform together under Lorentz transformations, and the continuity equation becomesWith the four-potential $A^a=(\phi /c,\mathbf{A})$, the Lorenz gauge ${\partial }_a{A}^a=0$ converts Maxwell equations to $\square A^a={\mu }_0{j}^a$, where $\square =c^{-2}{\partial }_t^2-{\nabla }^2$. Choosing no incoming homogeneous radiation, the retarded electromagnetic potential isCurrent conservation ensures this solution obeys the Lorenz gauge; a homogeneous solution may be added when additional field boundary data are specified.

For a thin loop, let $R=|\mathbf{x}|$, $\hat{\mathbf{R}}=\mathbf{x}/R$ and $t_R=t-R/c$. ThenIn the small-loop magnetic-dipole approximation, expand $|\mathbf{x}-{\mathbf{x}}^{\prime }|=R-\hat{\mathbf{R}}\cdot {\mathbf{x}}^{\prime }+\cdots$. The constant term integrates to zero because $\oint d{\mathbf{\ell }}^{\prime }=0$. The first nonzero term isFor the oriented circular loop the last integral is $\pi r^2\mathbf{n}\times \hat{\mathbf{R}}$. Thus with magnetic dipole moment $\mathbf{m}(t)=\pi r^{2}I(t)\mathbf{n}$,The $R^{-2}$ term is the near dipole potential and the $R^{-1}$ term its radiation potential. This is the leading small-$r$ result at fixed $R,\omega$; besides $r/R\ll 1$, the expansion of the retarded current requires $\omega r/c\ll 1$.

If only $r/R\ll 1$ is imposed and the loop need not be small compared with a wavelength, keep the retarded phase around the loop. For fixed $r,\omega$ and $R\to \infty$, the leading far-field expression is insteadwhere $\cos \theta =\mathbf{n}\cdot \hat{\mathbf{R}}$ and $J_1$ is the Bessel function of the first kind. Integrating the sinusoidal phase on a circle gives this Bessel factor; $J_1(z)\sim z/2$ recovers the radiation part of the boxed dipole result.

Use SI units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. Write $x^{\mu }=(ct,x,y,z)$, ${\partial }_{\mu }=(c^{-1}{\partial }_t,\nabla )$ and the four-current $J^{\mu }=(c\rho ,\mathbf{J})$, with Greek indices from $0$ to $3$. The electromagnetic field tensor is the antisymmetric tensor whose components areIndices are raised and lowered with $\eta$. Define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$, with ${ϵ}^{0123}=+1$. Then the covariant Maxwell equations areHere $\rho$ and $\mathbf{J}$ are charge and current density, ${\mu }_0$ is the vacuum magnetic permeability, and ${ϵ}_0{\mu }_0{c}^2=1$. For example the $\nu =0$ equation gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, and the spatial equations give $\nabla \times \mathbf{B}-c^{-2}{\partial }_t\mathbf{E}={\mu }_0\mathbf{J}$. With the stated dual convention, ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$; the second tensor equation gives the two homogeneous Maxwell equations.

A material segment of the line with rest length $d{x}^{\prime }$ has laboratory length $dx=d{x}^{\prime }/\gamma$, while its charge $dQ$ is unchanged. ThusEquivalently the line-integrated four-current transforms from $(c{\lambda }^{\prime },0)$ to $(\gamma c{\lambda }^{\prime },\gamma u{\lambda }^{\prime })$. Gauss's and Ampère's laws in $S$ now give $E_r=\lambda /(2\pi {ϵ}_{0}r)$ and $B_{\phi }={\mu }_{0}I/(2\pi r)$, exactly the fields found in part (a). Charge density is frame-dependent even though the charge of the same set of particles is invariant.

In $S$, adding the stationary opposite line cancels charge density and electric field but contributes no current. HenceThe neutral current filament in a boosted frame is not neutral. Transform its line-integrated four-current:The same result follows by adding the original resting line $({\lambda }^{\prime },0)$ to the new negative line, whose primed density and current are $(-\gamma \lambda ,+\gamma \lambda u)$. In particular ${\lambda }^{\prime }-\gamma \lambda =-\gamma \lambda u^2/c^2$ because $\lambda =\gamma {\lambda }^{\prime }$.

Transforming the total fields givesFor positive $\lambda$, $\mathbf{u}\times \mathbf{B}$ points inward, explaining the electric-field sign. The primed Gauss and Ampère fields calculated from ${\lambda }_{\mathrm{tot}}^{\prime }$ and $I_{\mathrm{tot}}^{\prime }$ reproduce these expressions. Both a moving charge distribution and a moving current must be transformed; boosting just one constituent would give an inconsistent total.

Take $x^{\mu }=(ct,x,y,z)$ and metric signature $(+,-,-,-)$. The four-current, four-potential and four-gradient areThe first two transform as contravariant four-vectors under an inertial-frame Lorentz transformation, within a consistently transformed electromagnetic gauge. The lowered-index gradient is a covector. Their contraction is a Lorentz scalar:Indeed, $J^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{J}^{\nu }$ and ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\alpha }{}_{\mu }{\partial }_{\alpha }$, so the two matrices cancel in the contraction. Local charge conservation in one inertial frame implies local charge conservation in every inertial frame. Integrating over a region gives the charge change equal to minus its outward current flux; global conserved charge requires no flux through the system boundary.

Put wire 1 on the $z$ axis and wire 2 at $x=d$, with positive $x$ pointing away from wire 1. Let each wire have stationary positive linear charge density $+\lambda$ and moving electron line density $-\lambda$ in the lab. For equal currents, both electron populations have drift velocity $+v\hat{\mathbf{z}}$, so the signed conventional current is $I=-\lambda v$.

First consider wire 2's positive charges in their own rest frame, which is the lab. Wire 1 has zero total line charge there, so its external electrostatic field is zero. The positive-charge force per length is therefore $\boxed{f_{+}=0}$.

Next use the electron rest frame $S^{\prime }$ moving at $+v\hat{\mathbf{z}}$. Wire 1's ion line density becomes ${\lambda }_{+}^{\prime }=\gamma \lambda$. Its electron density becomes ${\lambda }_{-}^{\prime }=-\lambda /\gamma$: the electron rest density is lower than its lab density. The transformed total isEquivalently this follows from the four-current transformation ${\lambda }_1^{\prime }=\gamma ({\lambda }_{1,\mathrm{net}}-v{I}_1/c^2)$, with ${\lambda }_{1,\mathrm{net}}=0$ and $I_1=-\lambda v$. By the electrostatic field of a line charge,A wire-2 electron is at rest in $S^{\prime }$ and feels only the electric force, $F_{e,x}^{\prime }=-e{\lambda }_1^{\prime }/(2\pi {ϵ}_{0}d)$, toward wire 1.

Apply the supplied transverse relativistic force transformation with the electron's lab velocity $\mathbf{u}=\mathbf{v}$. Its denominator is $\gamma (1-v^2/c^2)=1/\gamma$, so $F_{e,x}^{\prime }=\gamma F_{e,x}$, or $F_{e,x}=F_{e,x}^{\prime }/\gamma$. There are $\lambda /e$ electrons per unit lab length. ThusAdding $f_{+}=0$ gives exactly the attractive magnetic result. This electrostatic-frame derivation of parallel-wire forces uses electrostatics in each test carrier's own rest frame; no magnetic force acts on that stationary test carrier there. The other wire can still carry current in that frame. Transform the force per electron first and then count electrons per lab length: transforming a force per unit length as if it were a particle force would introduce an incorrect extra contraction factor.

Use a boost along the wire axis when comparing frames. The current variable in a four-current must be current density, in $\mathrm{A}{\mathrm{m}}^{-2}$, rather than a scalar total current in amperes. For the wire-force calculation it is convenient to integrate charge and current over each cross-section and work with line quantities.

First lower the index of the four-current with the Minkowski metric: $J_{\beta }=g_{\beta \alpha }{J}^{\alpha }$. The displayed expression is thenThereforeIts one free lower index transforms with the inverse coordinate Jacobian. The contraction with the metric here lowers an index; it does not make the current a scalar because the index $\delta$ remains free.

Write $a_y=E_{A,y}^0$, $a_z=E_{A,z}^0$, $b_y=E_{B,y}^0$ and $b_z=E_{B,z}^0$, evaluated at the same transverse coordinates $(y,z)$. A superscript $0$ labels each filament's own rest frame, rather than a single common frame in all cases. The filaments are infinite and uniform along $x$, so the change of longitudinal coordinate under a Lorentz boost creates no extra dependence in these functions. After the acceleration transients have passed, the magnetic field of each filament vanishes in its own rest frame.

For a filament moving at $+v\hat{\mathbf{x}}$, use the inverse of part (a): its transverse electric field in the laboratory is $\gamma {\mathbf{E}}^0$ and its magnetic field is $\mathbf{v}\times \mathbf{E}/c^2$. Superposition for the two co-moving filaments therefore gives, in $O$,In the common co-moving frame $O^{\prime }$ both filaments are at rest, soFor example, $B_y^{\prime }=\gamma [-\gamma v(a_z+b_z)/c^2+v\gamma (a_z+b_z)/c^2]=0$, a useful sign check. The supplied $\eta$ is the proper linear charge density. Thus each laboratory density is $\gamma \eta$ and each laboratory electric current is $\gamma \eta v$, consistently with length contraction and the four-current transformation. In $O^{\prime }$ each density is $\eta$ and both currents vanish.

For either normalized two-spinor ${\chi }^s$, $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=p^{2}I$ givesThe Dirac equation alone does not fix $N$: a normalization convention is needed. In the usual covariant external-spinor convention ${\bar{u}}_r{u}_s=2m{\delta }_{rs}$, choose real positive $N=\sqrt{E+m}$. Then $u_r^{†}{u}_s=2E{\delta }_{rs}$ and ${\bar{u}}_s{\gamma }^{\mu }{u}_s=2p^{\mu }$. A second common convention is unit rest density, $\bar{u}u=1$, giving $N=\sqrt{(E+m)/(2m)}$ and density $E/m$. Both explain why the boosted density is not one particle per unit coordinate volume: the density is a four-current component and changes under a Lorentz transformation.

One may instead choose unit laboratory density, $N=\sqrt{(E+m)/(2E)}$, or box normalization $N=\sqrt{(E+m)/(2EV)}$ for one particle in volume $V$. An infinite plane wave is not normalizable to one particle over all space; a localized wave packet is required for that. Thus a unique $N$ cannot be inferred from the prompt without specifying which of these conventions is intended.

With Cartesian Minkowski spacetime coordinates $x^{\mu }=(ct,x,y,z)$, introduce the four-current $J^{\mu }=(c\rho ,\mathbf{J})$ and electromagnetic four-potential ${\mathcal{A}}^{\mu }=(\varphi /c,\mathbf{A})$. The antisymmetric electromagnetic field tensor $F^{\mu \nu }={\partial }^{\mu }{\mathcal{A}}^{\nu }-{\partial }^{\nu }{\mathcal{A}}^{\mu }$ has components $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}^{ijk}{B}_k$. Maxwell equations becomeThe time and space components reproduce Gauss's law and the Ampère-Maxwell equation; the dual equation gives absence of magnetic monopoles and Faraday's law of induction. Antisymmetry gives charge conservation ${\partial }_{\nu }{J}^{\nu }=0$. The relativistic Lorentz force is $d{p}^{\mu }/d\tau =q{F}^{\mu \nu }{u}_{\nu }$. Every displayed equation transforms tensorially under a Lorentz transformation; the electric field and magnetic field are different components of the same object.

Two independent algebraic electromagnetic field invariants areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations but changes sign under spatial reflection. For perpendicular fields, the sign of the first distinguishes a frame with purely magnetic field from one with purely electric field. Both invariants vanish for a plane electromagnetic wave with $\mathbf{E}\perp \mathbf{B}$ and $E=cB$, although the field itself is nonzero.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

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.

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.

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.