Fields of a Moving Point Charge

Derivation

Fields of a Moving Point Charge

D-169 Home PU-204 Threads fields · force · waves Depends on Retarded Potentials and Jefimenko's Equations

Statement

For a point charge \( q \) moving on an arbitrary prescribed trajectory \( \mathbf{w}(t) \) with velocity \( \mathbf{v} = \dot{\mathbf{w}} \) and acceleration \( \mathbf{a} = \dot{\mathbf{v}} \), the retarded solutions of Maxwell's equations give the Liénard–Wiechert potentials \( V = \frac{1}{4\pi\varepsilon_0}\frac{qc}{Rc - \mathbf{R}\cdot\mathbf{v}} \) and \( \mathbf{A} = \frac{\mathbf{v}}{c^2}V \), where \( \mathbf{R} = \mathbf{r} - \mathbf{w}(t_r) \) is drawn from the retarded position and every source quantity is evaluated at the retarded time \( t_r \) defined by \( R(t_r) = c(t - t_r) \). Differentiating them yields the exact field \( \mathbf{E} = \frac{q}{4\pi\varepsilon_0}\frac{R}{(\mathbf{R}\cdot\mathbf{u})^3}\left[(c^2 - v^2)\,\mathbf{u} + \mathbf{R}\times(\mathbf{u}\times\mathbf{a})\right] \) with \( \mathbf{u} \equiv c\hat{\mathbf{n}} - \mathbf{v} \), \( \hat{\mathbf{n}} = \mathbf{R}/R \), together with \( \mathbf{B} = \frac{1}{c}\hat{\mathbf{n}}\times\mathbf{E} \). The field separates cleanly into a velocity (near) field falling as \( 1/R^2 \) and an acceleration (radiation) field falling as \( 1/R \) — only the latter carries energy to infinity.

Why it matters

This is the complete classical answer to the question "what field does a moving charge make?" Every phenomenon of classical radiation — antennas, synchrotron light, bremsstrahlung, Thomson scattering, the blue of the sky — is contained in the acceleration term of this one formula. The \( 1/R \) fall-off of that term is the entire reason radiation exists: the Poynting flux \( \sim E^2 R^2 \) then survives as \( R \to \infty \), so accelerated charges irreversibly export energy. A charge in uniform motion, whose field falls as \( 1/R^2 \), radiates nothing.

The derivation also delivers two structural lessons. First, electromagnetic "news" travels at \( c \): the field here-and-now is dictated by what the charge was doing on the past light cone, not by where it is now. Second, the factor \( (1 - \hat{\mathbf{n}}\cdot\mathbf{v}/c)^{-1} \) — a purely geometric compression of the information arriving from an approaching source — is the seed of relativistic beaming, and makes synchrotron sources brilliantly forward-collimated.

Assumptions

The trajectory \( \mathbf{w}(t) \) is prescribed and the charge strictly subluminal, \( v < c \) at all times.If \( v \geq c \) (possible for the phase speed of a source pattern, or for a charge exceeding \( c/n \) in a medium) the retarded-time equation can have zero or several roots: the single-root construction fails and one obtains Cherenkov-type shock cones built from multiple retarded contributions.

Retarded boundary conditions — no incoming radiation from infinity.The wave equation equally admits advanced solutions evaluated at \( t + R/c \); choosing retarded solutions injects the thermodynamic arrow of time by hand. Dropping the choice gives the Wheeler–Feynman absorber picture, in which the observed asymmetry is traced to boundary conditions of the universe.

Lorenz gauge, \( \nabla\cdot\mathbf{A} + \frac{1}{c^2}\frac{\partial V}{\partial t} = 0 \), inherited from the retarded-potential prior result.In another gauge (e.g. Coulomb) the potentials look completely different — \( V \) becomes instantaneous — but \( \mathbf{E} \) and \( \mathbf{B} \) are unchanged; only the final field formulas are gauge-invariant statements.

The charge is structureless and pointlike, \( \rho(\mathbf{r}',t) = q\,\delta^3(\mathbf{r}' - \mathbf{w}(t)) \).A finite charge distribution smears the retarded time across its body and introduces form factors; conversely the strict point limit makes the self-field and self-energy divergent, the classical disease that renormalisation later inherits.

The trajectory is not corrected for the energy and momentum the charge itself radiates.Self-consistency requires adding the radiation-reaction force (Abraham–Lorentz–Dirac); ignoring it is excellent whenever the energy radiated per characteristic time is small compared with the particle's kinetic energy, but fails in extreme fields (modern petawatt lasers, classical atom collapse).

Derivation

\[ V(\mathbf{r},t) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho\!\left(\mathbf{r}',\, t - \tfrac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|}\, d^3 r', \qquad \mathbf{A}(\mathbf{r},t) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}\!\left(\mathbf{r}',\, t - \tfrac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|}\, d^3 r' \]

Starting point: the retarded potentials, the causal Lorenz-gauge solutions of the inhomogeneous wave equations (prior result: retarded potentials / Jefimenko). Each source element contributes with the delay light needs to cross \( |\mathbf{r}-\mathbf{r}'| \). A

\[ \rho(\mathbf{r}',t) = q\,\delta^3\!\big(\mathbf{r}' - \mathbf{w}(t)\big), \qquad \mathbf{J}(\mathbf{r}',t) = q\,\mathbf{v}(t)\,\delta^3\!\big(\mathbf{r}' - \mathbf{w}(t)\big) \]

Specialise the source to a single point charge on the trajectory \( \mathbf{w}(t) \). The current is charge times velocity concentrated on the worldline; this pair automatically satisfies continuity \( \partial_t \rho + \nabla\cdot\mathbf{J} = 0 \). A

\[ V(\mathbf{r},t) = \frac{q}{4\pi\varepsilon_0}\int\!\!\int \frac{\delta^3\!\big(\mathbf{r}' - \mathbf{w}(t')\big)\,\delta\!\left(t' - t + \tfrac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|}\, dt'\, d^3 r' \]

Insert \( 1 = \int \delta\big(t' - t + |\mathbf{r}-\mathbf{r}'|/c\big)\, dt' \) to convert the implicit retarded evaluation into an explicit integral over a source time \( t' \). This is the key move: it lets the spatial delta act at fixed \( t' \), where it is an honest three-dimensional delta. Substituting \( t_r' \) directly into \( \delta^3 \) is illegal because the retardation makes its argument depend on \( \mathbf{r}' \) nontrivially. B

\[ V(\mathbf{r},t) = \frac{q}{4\pi\varepsilon_0}\int \frac{\delta\!\left(t' - t + \tfrac{R(t')}{c}\right)}{R(t')}\, dt', \qquad R(t') \equiv |\mathbf{r} - \mathbf{w}(t')| \]

Perform the \( d^3r' \) integral: the spatial delta simply sets \( \mathbf{r}' = \mathbf{w}(t') \) everywhere. One scalar delta in time remains. B

\[ \frac{d}{dt'}\left[t' - t + \frac{R(t')}{c}\right] = 1 + \frac{1}{c}\frac{dR}{dt'} = 1 - \frac{\hat{\mathbf{n}}\cdot\mathbf{v}}{c} \equiv \kappa, \qquad \frac{dR}{dt'} = -\,\hat{\mathbf{n}}\cdot\mathbf{v} \]

Composition rule preparation: \( R^2 = \mathbf{R}\cdot\mathbf{R} \) with \( \mathbf{R} = \mathbf{r} - \mathbf{w}(t') \) gives \( R\,dR/dt' = -\mathbf{R}\cdot\mathbf{v} \), hence \( dR/dt' = -\hat{\mathbf{n}}\cdot\mathbf{v} \). Because \( v < c \), \( \kappa = 1 - \hat{\mathbf{n}}\cdot\mathbf{v}/c > 0 \) strictly: the argument of the delta is monotone in \( t' \), so it has exactly one root \( t_r \) — each field point sees exactly one retarded image of the charge. C

\[ \delta\!\left(t' - t + \frac{R(t')}{c}\right) = \frac{\delta(t' - t_r)}{\left|1 - \hat{\mathbf{n}}\cdot\mathbf{v}/c\right|_{t_r}} = \frac{\delta(t' - t_r)}{\kappa(t_r)} \]

Delta-function composition \( \delta(f(t')) = \sum_i \delta(t'-t_i)/|f'(t_i)| \), with the single root \( t_r \) established in step 5 and \( \kappa > 0 \) removing the absolute value. This Jacobian is pure geometry — the same compression factor that makes an approaching train occupy the crossing longer than its rest length suggests. C

\[ V(\mathbf{r},t) = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{R\,\kappa}\bigg|_{t_r} = \frac{1}{4\pi\varepsilon_0}\,\frac{qc}{Rc - \mathbf{R}\cdot\mathbf{v}}, \qquad \mathbf{A}(\mathbf{r},t) = \frac{\mu_0}{4\pi}\,\frac{qc\,\mathbf{v}}{Rc - \mathbf{R}\cdot\mathbf{v}} = \frac{\mathbf{v}}{c^2}\,V \]

Evaluate the remaining \( dt' \) integral with step 6; repeat steps 3–6 verbatim for \( \mathbf{A} \) (the extra factor \( \mathbf{v}(t') \) just rides along and is evaluated at \( t_r \)). These are the Liénard–Wiechert potentials. Note \( V \) is not \( q/4\pi\varepsilon_0 R \): the charge in effect "counts more than once" when approaching, by the factor \( 1/\kappa \). A

\[ \frac{\partial t_r}{\partial t} = \frac{1}{\kappa} = \frac{Rc}{Rc - \mathbf{R}\cdot\mathbf{v}}, \qquad \nabla t_r = -\,\frac{\hat{\mathbf{n}}}{c\,\kappa} = -\,\frac{\mathbf{R}}{Rc - \mathbf{R}\cdot\mathbf{v}} \]

Derivative bookkeeping before differentiating the potentials. Both follow from implicit differentiation of the light-cone condition \( R(t_r) = c(t - t_r) \): vary \( t \) at fixed \( \mathbf{r} \), or \( \mathbf{r} \) at fixed \( t \), and solve using \( dR/dt_r = -\hat{\mathbf{n}}\cdot\mathbf{v} \) and \( \nabla R|_{t_r\,\mathrm{fixed}} = \hat{\mathbf{n}} \). Every subsequent spatial or time derivative must be routed through these, because \( t_r(\mathbf{r},t) \) is itself a field. C

\[ \nabla V = \frac{qc}{4\pi\varepsilon_0}\,\frac{(Rc - \mathbf{R}\cdot\mathbf{v})\,\mathbf{v} - (c^2 - v^2 + \mathbf{R}\cdot\mathbf{a})\,\mathbf{R}}{(Rc - \mathbf{R}\cdot\mathbf{v})^3} \]

Gradient of step 7: \( \nabla V = -qc\,(4\pi\varepsilon_0)^{-1}(Rc - \mathbf{R}\cdot\mathbf{v})^{-2}\,\nabla(Rc - \mathbf{R}\cdot\mathbf{v}) \), expanding \( \nabla(Rc) = c\hat{\mathbf{n}} + c\,\mathbf{v}\,(\hat{\mathbf{n}}\cdot\nabla t_r)\)-type chain-rule terms and \( \nabla(\mathbf{R}\cdot\mathbf{v}) \) with \( \mathbf{R} \) and \( \mathbf{v} \) both carrying \( t_r \)-dependence via step 8. The acceleration enters here for the first time, through \( \dot{\mathbf{v}}\,\nabla t_r \). C

\[ \frac{\partial \mathbf{A}}{\partial t} = \frac{qc}{4\pi\varepsilon_0}\,\frac{(Rc - \mathbf{R}\cdot\mathbf{v})\left(-\mathbf{v} + \tfrac{R\,\mathbf{a}}{c}\right) + \tfrac{R}{c}\,(c^2 - v^2 + \mathbf{R}\cdot\mathbf{a})\,\mathbf{v}}{(Rc - \mathbf{R}\cdot\mathbf{v})^3} \]

Time derivative of \( \mathbf{A} = \mathbf{v}V/c^2 \): product rule gives \( \frac{1}{c^2}\big(\mathbf{a}\,\tfrac{\partial t_r}{\partial t}V + \mathbf{v}\,\tfrac{\partial V}{\partial t}\big) \), with \( \partial t_r/\partial t = 1/\kappa \) from step 8 and \( \partial V/\partial t \) computed by the same chain-rule discipline. C

\[ \mathbf{E} = -\nabla V - \frac{\partial \mathbf{A}}{\partial t} = \frac{q}{4\pi\varepsilon_0}\,\frac{R}{(\mathbf{R}\cdot\mathbf{u})^3}\left[(c^2 - v^2)\,\mathbf{u} + (\mathbf{R}\cdot\mathbf{a})\,\mathbf{u} - (\mathbf{R}\cdot\mathbf{u})\,\mathbf{a}\right], \qquad \mathbf{u} \equiv c\,\hat{\mathbf{n}} - \mathbf{v} \]

Add steps 9 and 10 with signs. The \( (Rc - \mathbf{R}\cdot\mathbf{v})\mathbf{v} \) terms cancel identically; the survivors organise around the combination \( \mathbf{R} - R\mathbf{v}/c = (R/c)\,\mathbf{u} \), and \( Rc - \mathbf{R}\cdot\mathbf{v} = \mathbf{R}\cdot\mathbf{u} \). Everything on the right is evaluated at \( t_r \). B

\[ \mathbf{E}(\mathbf{r},t) = \frac{q}{4\pi\varepsilon_0}\,\frac{R}{(\mathbf{R}\cdot\mathbf{u})^3}\Big[\underbrace{(c^2 - v^2)\,\mathbf{u}}_{\text{velocity field } \propto 1/R^2} \; + \; \underbrace{\mathbf{R}\times(\mathbf{u}\times\mathbf{a})}_{\text{acceleration field } \propto 1/R}\Big] \]

Package the acceleration terms with the BAC–CAB identity \( \mathbf{R}\times(\mathbf{u}\times\mathbf{a}) = \mathbf{u}\,(\mathbf{R}\cdot\mathbf{a}) - \mathbf{a}\,(\mathbf{R}\cdot\mathbf{u}) \). Power counting in \( R \): since \( \mathbf{R}\cdot\mathbf{u} \sim R \), the first term scales as \( R\cdot R^{-3} = R^{-2} \) (generalised Coulomb field, tied to the charge) while the second scales as \( R\cdot R \cdot R^{-3} = R^{-1} \) (radiation, detachable). B

\[ \mathbf{B} = \nabla\times\mathbf{A} = \frac{1}{c}\,\hat{\mathbf{n}}\times\mathbf{E} \]

Curl of \( \mathbf{A} = \mathbf{v}V/c^2 \) with the same chain-rule bookkeeping (step 8): every term in \( \nabla\times\mathbf{A} \) assembles into \( \hat{\mathbf{n}}\times \) the corresponding term of \( \mathbf{E} \). The magnetic field of a point charge is always transverse to the line of sight to the retarded position, for both field types. C

Result

\[ V = \frac{1}{4\pi\varepsilon_0}\frac{qc}{Rc - \mathbf{R}\cdot\mathbf{v}}, \quad \mathbf{A} = \frac{\mathbf{v}}{c^2}V, \qquad \mathbf{E} = \frac{q}{4\pi\varepsilon_0}\frac{R}{(\mathbf{R}\cdot\mathbf{u})^3}\left[(c^2\!-\!v^2)\,\mathbf{u} + \mathbf{R}\times(\mathbf{u}\times\mathbf{a})\right], \quad \mathbf{B} = \frac{1}{c}\hat{\mathbf{n}}\times\mathbf{E} \]

Reading. All source quantities \( \mathbf{R}, R, \hat{\mathbf{n}}, \mathbf{v}, \mathbf{a} \) are evaluated at the retarded time \( t_r \) fixed by \( R = c(t-t_r) \), and \( \mathbf{u} = c\hat{\mathbf{n}} - \mathbf{v} \). The field at any point is the sum of two physically distinct pieces. The velocity field \( \propto (c^2-v^2)\mathbf{u}/R^2 \) is the charge's deformed Coulomb field: it points along \( \mathbf{u} \), which is exactly the direction from the charge's linearly extrapolated present position, and it stays attached to the charge. The acceleration field \( \propto \mathbf{R}\times(\mathbf{u}\times\mathbf{a})/R \) exists only while the charge accelerates, is transverse (\( \perp \hat{\mathbf{n}} \) in the far zone), and, falling only as \( 1/R \), carries a Poynting flux \( \sim E^2R^2 \) that survives to infinity: this piece is electromagnetic radiation. The factor \( (\mathbf{R}\cdot\mathbf{u})^{-3} = (Rc\kappa)^{-3} \) with \( \kappa = 1 - \hat{\mathbf{n}}\cdot\mathbf{v}/c \) produces violent forward beaming as \( v \to c \).

Units check. \( [\,\mathbf{R}\cdot\mathbf{u}\,] = \mathrm{m}^2\,\mathrm{s}^{-1} \), so the prefactor \( R/(\mathbf{R}\cdot\mathbf{u})^3 \) carries \( \mathrm{m}\cdot\mathrm{m}^{-6}\mathrm{s}^{3} = \mathrm{m}^{-5}\,\mathrm{s}^{3} \). Velocity term: \( [(c^2-v^2)\mathbf{u}] = \mathrm{m}^3\,\mathrm{s}^{-3} \); acceleration term: \( [\mathbf{R}\times(\mathbf{u}\times\mathbf{a})] = \mathrm{m}\cdot(\mathrm{m}\,\mathrm{s}^{-1})(\mathrm{m}\,\mathrm{s}^{-2}) = \mathrm{m}^3\,\mathrm{s}^{-3} \) — the two terms agree, as they must. Total: \( \frac{q}{4\pi\varepsilon_0}\times \mathrm{m}^{-5}\mathrm{s}^3 \times \mathrm{m}^3\mathrm{s}^{-3} = \frac{q}{4\pi\varepsilon_0}\,\mathrm{m}^{-2} \), which is \( \mathrm{V}\,\mathrm{m}^{-1} \) — an electric field. For the potential, \( [qc/(Rc-\mathbf{R}\cdot\mathbf{v})] = \mathrm{C}\cdot(\mathrm{m\,s^{-1}})/(\mathrm{m^2\,s^{-1}}) = \mathrm{C\,m^{-1}} \), and \( \mathrm{C\,m^{-1}}/(4\pi\varepsilon_0) = \mathrm{V} \). Consistent.

Limiting cases

Static charge (\( \mathbf{v} = 0, \mathbf{a} = 0 \)): \( \mathbf{u} = c\hat{\mathbf{n}} \), \( \mathbf{R}\cdot\mathbf{u} = Rc \), and \( \mathbf{E} = \frac{q}{4\pi\varepsilon_0}\frac{\hat{\mathbf{n}}}{R^2} \), \( \mathbf{B} = 0 \) — Coulomb's law recovered exactly.
Slow uniform motion (\( v \ll c \), \( \mathbf{a}=0 \)): \( \mathbf{E} \approx \frac{q\hat{\mathbf{n}}}{4\pi\varepsilon_0 R^2} \) and \( \mathbf{B} = \frac{\mu_0}{4\pi}\frac{q\,\mathbf{v}\times\hat{\mathbf{n}}'}{R^2} \) — the Biot–Savart field of a moving charge, to first order in \( v/c \).
Uniform relativistic motion (\( \mathbf{a}=0 \), any \( v \)): the velocity field alone survives and, remarkably, points radially from the present position, flattened into a pancake: \( \mathbf{E} = \frac{q}{4\pi\varepsilon_0}\frac{(1-\beta^2)\,\hat{\mathbf{R}}_p}{R_p^2\,(1-\beta^2\sin^2\theta)^{3/2}} \) — identical to boosting the Coulomb field with the field-transformation rules, a strong cross-check.
Non-relativistic radiation (\( v \ll c \), \( \mathbf{a} \neq 0 \)): the acceleration field reduces to \( \mathbf{E}_{\mathrm{rad}} = \frac{q}{4\pi\varepsilon_0 c^2}\frac{\hat{\mathbf{n}}\times(\hat{\mathbf{n}}\times\mathbf{a})}{R} \), with \( |\mathbf{E}_{\mathrm{rad}}| \propto \sin\theta \) — the dipole pattern; integrating the Poynting flux gives the Larmor power \( P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3} \).
Far zone (\( R \to \infty \)): only the acceleration field survives, with \( \mathbf{E} \perp \mathbf{B} \perp \hat{\mathbf{n}} \) and \( |\mathbf{E}| = c|\mathbf{B}| \) — locally a plane wave, connecting to the free-wave solutions.

Breaks when

Effective source speed reaches \( c \). The whole construction hangs on \( \kappa = 1 - \hat{\mathbf{n}}\cdot\mathbf{v}/c > 0 \) guaranteeing one retarded time. For a charge exceeding the local light speed \( c/n \) in a dielectric, several retarded times contribute simultaneously and the fields pile up on a Mach-like cone — Cherenkov radiation — which the single-root Liénard–Wiechert formula cannot describe; as \( v \to c \) in vacuum the potentials diverge on the forward direction.
Radiation reaction becomes significant. The trajectory was prescribed, ignoring the momentum the emitted field carries away. When the energy radiated in a characteristic time approaches the particle's own energy scale (electrons in \( \gtrsim 10^{22}\,\mathrm{W\,cm^{-2}} \) laser fields, tight synchrotron orbits, the classical hydrogen atom), the fields react back on \( \mathbf{w}(t) \) and one must solve the coupled Abraham–Lorentz–Dirac problem, with its notorious runaway and pre-acceleration pathologies.
Quantum regime. When emitted photon energies \( \hbar\omega \) are comparable to the particle's kinetic energy, or fields vary on the scale of the Compton wavelength, the classical continuous-field description fails: photon recoil (Compton scattering), discrete emission, and vacuum polarisation require QED. The classical formula is the correspondence-limit envelope of the quantum emission rate.
Point-charge self-field. On the worldline itself \( R \to 0 \) and both field pieces diverge; self-energy is infinite. Any question about the force of the charge's own field on itself lies outside this derivation.

Failure modes

The "obvious" substitution. Writing \( V = \frac{q}{4\pi\varepsilon_0 R} \) at the retarded position — i.e. assuming \( \int \rho(\mathbf{r}', t_r')\, d^3r' = q \). It does not: the retardation ties \( t_r' \) to \( \mathbf{r}' \), and the delta-function Jacobian produces the extra \( 1/\kappa \). Missing this factor loses beaming, the Doppler asymmetry, and ultimately the correct radiation fields.
Differentiating at frozen retarded time. Computing \( \nabla V \) or \( \partial\mathbf{A}/\partial t \) while treating \( t_r \) as a constant. Every derivative must pass through \( \nabla t_r = -\hat{\mathbf{n}}/c\kappa \) and \( \partial t_r/\partial t = 1/\kappa \); forgetting them silently deletes all acceleration terms and "proves" that no charge radiates.
Present-time source data. Evaluating \( \mathbf{v} \), \( \mathbf{a} \), \( \mathbf{R} \) at the observation time \( t \) instead of \( t_r \). For any appreciably moving charge this misplaces the source and breaks causality; the error grows with \( R \).
Misreading the uniform-velocity field. Concluding that because the formula uses retarded quantities, the field of a uniformly moving charge points back to the retarded position. The algebra conspires so it points from the present position — the retarded data contain an exact linear extrapolation.
Using the radiation term where the near field dominates. Inside \( R \lesssim c^2/a \) (or within a wavelength of an oscillating source) the velocity field is the larger piece; antenna near-field measurements, capacitive coupling, and reactive impedance all live there.
Treating \( \mathbf{B} \) as an independent unknown. It never is: \( \mathbf{B} = \hat{\mathbf{n}}\times\mathbf{E}/c \) holds exactly, near zone and far zone alike.

Discussion

The deepest structural fact in the derivation is the factor \( \kappa^{-1} = (1 - \hat{\mathbf{n}}\cdot\mathbf{v}/c)^{-1} \), and it is worth internalising its origin. The potential at \( (\mathbf{r},t) \) integrates the charge density over the past light cone, and a charge moving toward the observer stays in near-tangency with that cone for longer — its contribution is smeared over a longer stretch of source time, so it "counts" more. This is the same geometry as the relativistic Doppler effect, and cubed in the field denominator it becomes relativistic beaming: for \( \gamma \gg 1 \) the radiated power is funnelled into a forward cone of half-angle \( \sim 1/\gamma \). Synchrotron light sources are engineering built directly on \( (\mathbf{R}\cdot\mathbf{u})^{-3} \).

The velocity/acceleration split answers the old question "does a uniformly moving charge radiate?" with a clean no, and says precisely why. The velocity field, however distorted, falls as \( 1/R^2 \); its Poynting flux through a large sphere vanishes as \( 1/R^2 \), so the energy in that field is convected along with the charge, never lost. Only the \( 1/R \) acceleration field delivers finite power to infinity. Radiation is therefore not a new substance but the far-surviving part of the one field — the part created whenever the charge's velocity changes and the field lines must "kink" to reconnect the old Coulomb pattern to the new one. The kink propagates outward at \( c \); its transverse field is the radiation.

Historically these potentials (Liénard 1898, Wiechert 1900) predate special relativity, yet they are exactly Lorentz-correct — a reminder that Maxwell's theory was relativistic before relativity was articulated. The uniform-motion limit reproduces, from pure retardation algebra, the same flattened field one gets by boosting the Coulomb field with the field-transformation rules; the agreement of two completely different calculations is one of the classic consistency checks of electrodynamics.

Covariantly, the whole result compresses to one line: \( A^\mu(x) = \frac{q}{4\pi\varepsilon_0 c}\,\frac{U^\mu}{U^\nu (x - w)_\nu}\Big|_{\tau_r} \), where \( U^\mu \) is the charge's four-velocity and \( \tau_r \) the proper time at which the worldline pierces the past light cone of \( x \). The denominator \( U^\nu(x-w)_\nu \) is the invariant behind \( Rc - \mathbf{R}\cdot\mathbf{v} = Rc\kappa \), which exposes \( \kappa \) as a frame-dependent shadow of a scalar. Differentiating covariantly gives \( F^{\mu\nu} \) with the same near-field/radiation split; the radiation part satisfies \( F^{\mu\nu}F_{\mu\nu} = 0 \) (null field), the algebraic signature of pure radiation. Dirac's decomposition \( \frac{1}{2}(F_{\mathrm{ret}} - F_{\mathrm{adv}}) \) isolates the finite self-force from the divergent self-Coulomb piece, the covariant route to radiation reaction.

Common misconceptions. (i) "The field points to where the charge was" — only the data are retarded; for uniform motion the field direction extrapolates exactly to the present position, and radiation fields point transversally, not back along \( \hat{\mathbf{n}} \) at all. (ii) "Potentials are just calculation aids here" — in this problem the fields are essentially impossible to guess directly; the potential route with careful retarded-time bookkeeping is the physics. (iii) "A charge moving at constant relativistic speed carries a stronger total charge" — the \( 1/\kappa \) enhancement of \( V \) fore and suppression aft rearrange the field; the total flux of \( \mathbf{E} \) through any enclosing surface remains \( q/\varepsilon_0 \) (Gauss's law is exact and instantaneous in form).

Worked examples

Example 1 — Field of a relativistic electron passing at closest approach. An electron travels in a straight line at \( v = 0.90\,c \). Find the electric field it produces at a point at perpendicular distance \( b = 1.0\,\mathrm{nm} \) from its track, at the instant of closest approach, and compare with the field along the direction of motion at the same distance.

\[ \mathbf{E} = \frac{q}{4\pi\varepsilon_0}\,\frac{(1-\beta^2)\,\hat{\mathbf{R}}_p}{R_p^{2}\left(1-\beta^2\sin^2\theta\right)^{3/2}} \]

For \( \mathbf{a} = 0 \) only the velocity field survives; converting retarded to present-position variables \( (R_p, \theta) \) gives this closed form (limiting case above), with \( \theta \) the angle between the track and the line to the field point. B

\[ \theta = 90^\circ:\quad E_\perp = \frac{q}{4\pi\varepsilon_0 b^2}\,\frac{1-\beta^2}{(1-\beta^2)^{3/2}} = \gamma\,\frac{q}{4\pi\varepsilon_0 b^2}; \qquad \theta = 0:\quad E_\parallel = \frac{q}{4\pi\varepsilon_0 b^2}\,(1-\beta^2) = \frac{1}{\gamma^2}\,\frac{q}{4\pi\varepsilon_0 b^2} \]

Evaluate the angular factor at the two extremes; symbols first. The transverse field is enhanced by \( \gamma \), the longitudinal field suppressed by \( \gamma^2 \) — the pancake. B

\[ \frac{q}{4\pi\varepsilon_0 b^2} = \frac{(8.988\times 10^9\,\mathrm{V\,m\,C^{-1}})(1.602\times 10^{-19}\,\mathrm{C})}{(1.0\times 10^{-9}\,\mathrm{m})^2} = 1.44\times 10^{9}\,\mathrm{V\,m^{-1}} \]

Static Coulomb reference value at \( b = 1\,\mathrm{nm} \). A

\[ \gamma = \frac{1}{\sqrt{1-0.90^2}} = \frac{1}{\sqrt{0.19}} = 2.294 \quad\Rightarrow\quad E_\perp = 2.294 \times 1.44\times 10^{9} = 3.3\times 10^{9}\,\mathrm{V\,m^{-1}}, \quad E_\parallel = \frac{1.44\times 10^{9}}{5.26} = 2.7\times 10^{8}\,\mathrm{V\,m^{-1}} \]

Insert \( \beta = 0.90 \), \( \gamma^2 = 5.26 \). A

\[ E_\perp \approx 3.3\times 10^{9}\,\mathrm{V\,m^{-1}}, \qquad E_\parallel \approx 2.7\times 10^{8}\,\mathrm{V\,m^{-1}} \quad (\text{ratio } \gamma^3 \approx 12) \]

Reading. The passing electron delivers its Coulomb field as a compressed transverse pulse, \( \gamma \) times stronger and \( \sim b/\gamma v \) briefer than a slow charge would — the physical basis of the equivalent-photon (Weizsäcker–Williams) picture of relativistic collisions.

Units check. \( \mathrm{V\,m\,C^{-1}} \times \mathrm{C} / \mathrm{m}^2 = \mathrm{V\,m^{-1}} \); \( \gamma \) is dimensionless.

Example 2 — Where does radiation take over, and how much power flows? An electron is momentarily at rest but accelerating at \( a = 1.0\times 10^{17}\,\mathrm{m\,s^{-2}} \) (typical of an electron in a strong RF cavity field). Find the distance at which the radiation field first exceeds the Coulomb field on the plane \( \theta = 90^\circ \), and the total radiated power.

\[ E_{\mathrm{vel}} = \frac{q}{4\pi\varepsilon_0 R^2}, \qquad E_{\mathrm{rad}} = \frac{q\,a\sin\theta}{4\pi\varepsilon_0 c^2 R} \]

With \( v = 0 \) at the retarded instant, \( \mathbf{u} = c\hat{\mathbf{n}} \), \( \kappa = 1 \): the velocity field is pure Coulomb and the acceleration field takes the non-relativistic dipole form. A

\[ \frac{E_{\mathrm{rad}}}{E_{\mathrm{vel}}}\bigg|_{\theta=90^\circ} = \frac{a R}{c^2} = 1 \quad\Rightarrow\quad R_c = \frac{c^2}{a} \]

Form the ratio symbolically and set it to unity; the crossover radius depends only on \( a \). B

\[ R_c = \frac{(2.998\times 10^{8}\,\mathrm{m\,s^{-1}})^2}{1.0\times 10^{17}\,\mathrm{m\,s^{-2}}} = \frac{8.99\times 10^{16}}{1.0\times 10^{17}}\,\mathrm{m} = 0.90\,\mathrm{m} \]

Numbers last. Inside a metre the electron's field is essentially Coulombic; beyond, the radiation term dominates. A

\[ P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3} = \frac{(1.602\times 10^{-19})^2\,(1.0\times 10^{17})^2}{6\pi\,(8.854\times 10^{-12})\,(2.998\times 10^{8})^3}\,\mathrm{W} = 5.7\times 10^{-20}\,\mathrm{W} \]

Integrate the far-zone Poynting flux \( S = \varepsilon_0 c E_{\mathrm{rad}}^2 \) over the sphere using \( \int \sin^2\theta\, d\Omega = 8\pi/3 \) — the Larmor formula — then insert numbers. B

\[ R_c = \frac{c^2}{a} \approx 0.90\,\mathrm{m}, \qquad P = \frac{q^2a^2}{6\pi\varepsilon_0 c^3} \approx 5.7\times 10^{-20}\,\mathrm{W} \]

Reading. Even a violent \( 10^{17}\,\mathrm{m\,s^{-2}} \) acceleration radiates a minuscule absolute power from a single electron — but \( 10^{10} \) electrons bunched coherently radiate \( N^2 \) times this, which is how antennas and free-electron lasers turn the same formula into watts.

Units check. \( \mathrm{C^2 (m\,s^{-2})^2 / (F\,m^{-1}\, m^3\,s^{-3})} = \mathrm{C^2\,m^2\,s^{-4}\cdot m^{-2}\,s^{3} / (C^2\,J^{-1})} = \mathrm{J\,s^{-1}} = \mathrm{W} \). Correct.

Problems

A proton moves at \( v = 0.50\,c \) directly toward an observation point; at the retarded instant its distance is \( R = 1.0\,\mathrm{\mu m} \). Compute the Liénard–Wiechert potential \( V \), compare it with the naive Coulomb value \( q/4\pi\varepsilon_0 R \), and find \( |\mathbf{A}| \).
Solution
Head-on approach means \( \hat{\mathbf{n}}\cdot\mathbf{v} = v \), so \( \kappa = 1 - v/c = 0.50 \). Naive Coulomb value: \( V_C = \frac{(8.988\times 10^9)(1.602\times 10^{-19})}{1.0\times 10^{-6}} = 1.44\times 10^{-3}\,\mathrm{V} \). Liénard–Wiechert: \( V = V_C/\kappa = 2.88\times 10^{-3}\,\mathrm{V} \) — exactly double, because the approaching charge's contribution is geometrically compressed onto the light cone. Vector potential: \( |\mathbf{A}| = vV/c^2 = \frac{(1.499\times 10^8)(2.88\times 10^{-3})}{8.988\times 10^{16}} = 4.8\times 10^{-12}\,\mathrm{V\,s\,m^{-1}} \) (i.e. \( \mathrm{T\,m} \)), directed along \( \mathbf{v} \).
For an ultrarelativistic charge with \( \beta = 0.99 \), compute the ratio of the scalar potential directly ahead of the charge to that directly behind (same retarded distance \( R \)). What does this asymmetry become in the field (which carries \( \kappa^{-3} \))?
Solution
Ahead: \( \kappa_+ = 1 - 0.99 = 0.01 \), so \( V_+ = V_C/0.01 = 100\,V_C \). Behind: \( \kappa_- = 1 + 0.99 = 1.99 \), so \( V_- = V_C/1.99 = 0.5025\,V_C \). Ratio \( V_+/V_- = 1.99/0.01 = 199 \). For the fields the corresponding kinematic weighting is \( \kappa^{-3} \): \( (\kappa_-/\kappa_+)^3 = 199^3 \approx 7.9\times 10^{6} \). This is the origin of relativistic beaming — at \( \gamma = 1/\sqrt{1-0.99^2} \approx 7.1 \), emission concentrates into a forward cone of half-angle \( \sim 1/\gamma \approx 8^\circ \).
An electron moves uniformly with \( \gamma = 10 \). At present-position distance \( R_p = 1.0\times 10^{-10}\,\mathrm{m} \), find the field transverse to the motion, the field along the motion, and their ratio.
Solution
Coulomb reference: \( E_C = \frac{(8.988\times 10^9)(1.602\times 10^{-19})}{(10^{-10})^2} = 1.44\times 10^{11}\,\mathrm{V\,m^{-1}} \). Transverse (\( \theta = 90^\circ \)): \( E_\perp = \gamma E_C = 1.44\times 10^{12}\,\mathrm{V\,m^{-1}} \). Longitudinal (\( \theta = 0 \)): \( E_\parallel = E_C/\gamma^2 = 1.44\times 10^{9}\,\mathrm{V\,m^{-1}} \). Ratio \( E_\perp/E_\parallel = \gamma^3 = 1000 \). The field is squeezed into a pancake of angular thickness \( \sim 1/\gamma \) about the transverse plane; the total Gauss flux is still \( q/\varepsilon_0 \).
A model antenna electron oscillates as \( x(t) = x_0\cos\omega t \) with \( f = 100\,\mathrm{MHz} \) and \( x_0 = 1.0\,\mathrm{mm} \). Verify the non-relativistic treatment is legitimate, then find the peak radiation field \( E_{\mathrm{rad}} \) and peak \( B_{\mathrm{rad}} \) at \( r = 10\,\mathrm{m} \) on the plane \( \theta = 90^\circ \).
Solution
\( \omega = 2\pi f = 6.283\times 10^{8}\,\mathrm{s^{-1}} \). Peak speed \( v_{\max} = \omega x_0 = 6.28\times 10^{5}\,\mathrm{m\,s^{-1}} = 2.1\times 10^{-3}\,c \) — safely non-relativistic (\( \kappa \approx 1 \)). Peak acceleration \( a_{\max} = \omega^2 x_0 = (6.283\times 10^8)^2 (1.0\times 10^{-3}) = 3.95\times 10^{14}\,\mathrm{m\,s^{-2}} \). Radiation field: \( E_{\mathrm{rad}} = \frac{q\,a_{\max}}{4\pi\varepsilon_0 c^2 r} = \frac{(8.988\times 10^{9})(1.602\times 10^{-19})(3.95\times 10^{14})}{(8.988\times 10^{16})(10)} = 6.3\times 10^{-13}\,\mathrm{V\,m^{-1}} \). Magnetic field: \( B_{\mathrm{rad}} = E_{\mathrm{rad}}/c = 2.1\times 10^{-21}\,\mathrm{T} \). (A real antenna gets to measurable fields by moving \( \sim 10^{18} \) conduction electrons coherently.)
The classical death of the atom: an electron in a circular "Bohr" orbit of radius \( r = 5.29\times 10^{-11}\,\mathrm{m} \) with speed \( v = 2.19\times 10^{6}\,\mathrm{m\,s^{-1}} \) is continuously accelerating, so by this derivation it must radiate. Compute its centripetal acceleration, the Larmor power, and estimate the time to radiate away its total binding energy \( 13.6\,\mathrm{eV} \). What does the answer force upon physics?
Solution
Acceleration: \( a = v^2/r = \frac{(2.19\times 10^6)^2}{5.29\times 10^{-11}} = 9.07\times 10^{22}\,\mathrm{m\,s^{-2}} \). Larmor power: \( P = \frac{q^2a^2}{6\pi\varepsilon_0 c^3} = \frac{(1.602\times 10^{-19})^2 (9.07\times 10^{22})^2}{6\pi (8.854\times 10^{-12})(2.998\times 10^8)^3} = 4.7\times 10^{-8}\,\mathrm{W} \). Energy scale: \( 13.6\,\mathrm{eV} = 2.18\times 10^{-18}\,\mathrm{J} \). Crude timescale: \( \tau \sim \frac{2.18\times 10^{-18}}{4.7\times 10^{-8}} \approx 4.6\times 10^{-11}\,\mathrm{s} \) (a careful integration of the inspiral, with \( P \) growing as \( r \) shrinks, gives \( \approx 1.6\times 10^{-11}\,\mathrm{s} \)). Classical electrodynamics predicts every atom collapses in tens of picoseconds. Atoms are stable; therefore the classical field-plus-trajectory picture must fail at atomic scale — this calculation is the quantitative doorway to quantum mechanics.