Field of a Uniformly Moving Charge

A proton coasting round the LHC at 7 TeV carries an electric field, and that field is not a sphere of influence. It is a disc. Fore and aft it is suppressed by $1/\gamma^2 \approx 1.8\times10^{-8}$ ; broadside it is enhanced by $\gamma \approx 7461$ ; and essentially all of it is squeezed into a sheet a fifth of a milliradian thick. A charge near the beamline feels almost nothing at all — and then, for under a femtosecond, feels a hammer blow. Nothing was added to electrostatics to produce this. It is the Coulomb field, seen from a moving frame:

\vec E = \frac{q}{4\pi\varepsilon_0}\,\frac{1-\beta^2}{\big(1-\beta^2\sin^2\theta\big)^{3/2}}\,\frac{\hat{\vec R}}{R^2}, \qquad \vec B = \frac{1}{c^2}\,\vec v\times\vec E .

Oliver Heaviside wrote that angular factor down in 1889 — sixteen years before Einstein, using nothing but Maxwell's equations. He noticed that its surfaces of constant potential are not spheres but oblate spheroids, flattened along the motion by exactly $1/\gamma$ , and he told his friend George FitzGerald. Within months FitzGerald had suggested that matter itself might be flattened by the same factor.

Key takeaways

Signature $(+,-,-,-)$ , $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass. $u^\mu = \gamma(c,\vec v)$ , $A^\mu = (\phi/c,\vec A)$ .
The boosted Coulomb field: $\boxed{\vec E = \dfrac{q}{4\pi\varepsilon_0}\dfrac{1-\beta^2}{(1-\beta^2\sin^2\theta)^{3/2}}\dfrac{\hat{\vec R}}{R^2}}$ with $\vec R$ from the charge's PRESENT position and $\theta$ the angle between $\vec R$ and $\vec v$ .
The magnetic field is not optional. $\vec B = \vec v\times\vec E/c^2$ falls out of the same boost — no new law, no new constant. The cleanest proof that magnetism is relativity (`aq-sr-magnetism-relativistic-effect`).
Fore-and-aft $1/\gamma^2$ , broadside $\times\gamma$ , so the contrast is $\gamma^3$ : at $\beta = 0.8$ the sideways field is 4.63 times the forward one; at $\beta = 0.9$ , 12.07 times.
It is not radiation. $\vec E^2 - c^2\vec B^2 > 0$ strictly and $\vec E\cdot\vec B = 0$ : electric-type in every frame, boostable back to pure electrostatics. Radiation is the null case, and it needs acceleration.
Key idea: the equipotentials are ellipsoids contracted by exactly $1/\gamma$ — and that is where length contraction came from. Heaviside found the oblate spheroid in 1888–89; FitzGerald, reading it, proposed that rulers contract by the same factor. The contraction hypothesis was reverse-engineered from this field.

The core insight

Take a Coulomb field. Boost it. Everything else on this page is bookkeeping. The charge $q$ is invariant, so the source does not change — only the field pattern does, and it changes exactly as a boost of the tensor $F^{\mu\nu}$ demands: the transverse components of $\vec E$ are multiplied by $\gamma$ , the longitudinal one is left alone, and $\vec B = \vec v\times\vec E/c^2$ is forced into existence, because "purely electric" is not a Lorentz-invariant condition. The one geometric fact that makes the answer look strange — the $(1-\beta^2\sin^2\theta)^{-3/2}$ — is nothing but the statement that the rest-frame distance to the source, re-expressed in lab coordinates, is stretched along the motion by $\gamma$ . The field of a moving charge is a Coulomb field wearing a Lorentz transformation.

Where this lives today

The pancake is the working geometry of every collider. Beam-beam interactions — the fields a passing bunch inflicts on the opposing one — are computed from equation (1) with $\gamma$ in the thousands, and the resulting kick is what limits luminosity. In ultraperipheral heavy-ion collisions two lead nuclei miss each other entirely and interact only through these squashed fields; because $Z = 82$ , the equivalent-photon flux is enhanced by $Z^2 \approx 6700$ . Light-by-light scattering, predicted in 1936 and observed at the LHC in 2017, was seen exactly this way: two Heaviside pancakes passing in the night. The same field explains why magnetism exists (`aq-sr-magnetism-relativistic-effect`): the $\vec B = \vec v\times\vec E/c^2$ of a drifting conduction electron, summed over a wire, is Ampère's law. And the transverse space-charge force inside an accelerator bunch is suppressed by $1/\gamma^2$ : the beam's electric self-repulsion is almost exactly cancelled by the magnetic attraction of the parallel currents. High-energy beams hold together because a boosted Coulomb field very nearly cancels itself.

The physical picture

Sit in the charge's rest frame $S'$ . You see the most boring field in physics: a spherically symmetric hedgehog of straight radial lines, with $\vec B' = 0$ . Now boost to the lab frame, in which the charge sails along the $x$ -axis at speed $v$ . Two things happen, and they are not the same thing.

The field lines stay straight and radial — but they bunch. They still point away from the charge, and (a fact we will prove, because it looks impossible) away from where the charge is now. What changes is their density: boosting multiplies the transverse components of $\vec E$ by $\gamma$ and leaves the longitudinal one alone, so lines crowd into the equatorial plane and thin out along the poles. Push $\beta$ toward 1 and the hedgehog is crushed into a pancake.

A magnetic field appears, and you did not put it there. In $S'$ there was none; in the lab there is $\vec B = \vec v\times\vec E/c^2$ , circling the line of motion — precisely the Biot–Savart field of the current $q\vec v$ . $\vec E$ and $\vec B$ are components of one object $F^{\mu\nu}$ (`aq-sr-electromagnetic-field-tensor`), and "this field is purely electric" is a statement about a frame, not about the field. Magnetism is what electrostatics looks like from a moving train. The one thing that does not happen is radiation. The pattern is rigid: it translates with the charge, carrying its energy along like a bow wave, and nothing escapes to infinity. To lose energy for good you need an acceleration, a $1/R$ field, and a genuinely null $F^{\mu\nu}$ — the subject of `aq-sr-lienard-wiechert-potentials`.

The core equations

The field, and the equation this page owns:

\boxed{\;\vec E(\vec r,t) \;=\; \frac{q}{4\pi\varepsilon_0}\;\frac{1-\beta^2}{\big(1-\beta^2\sin^2\theta\big)^{3/2}}\;\frac{\hat{\vec R}}{R^2}\;} \qquad (1)

where: $\vec R = \vec r - \vec v t$ runs from the charge's present position to the field point and $\theta$ is the angle between $\vec R$ and $\vec v$ . Physical reading: a Coulomb field with an angular weighting. The prefactor $1/\gamma^2$ suppresses everything uniformly; the denominator then claws it back, and more, near $\theta = 90°$ . The magnetic field, which is not an independent object:

\boxed{\;\vec B \;=\; \frac{1}{c^2}\,\vec v\times\vec E \;=\; \frac{\mu_0}{4\pi}\,\frac{q\,\vec v\times\hat{\vec R}}{R^2}\,\frac{1-\beta^2}{\big(1-\beta^2\sin^2\theta\big)^{3/2}}\;} \qquad (2)

where: $\mu_0 = 1/\varepsilon_0c^2$ . Physical reading: $|\vec B| = \beta\sin\theta\,|\vec E|/c$ — the magnetic field is always smaller than $E/c$ , by the factor $\beta\sin\theta < 1$ . That inequality is the field invariant of eq. (4), and it is why this field can never be mistaken for a light wave. ⚠ The sign is easy to get backwards — see the warning card. The potentials, from which (1) and (2) both follow:

\phi = \frac{\gamma q}{4\pi\varepsilon_0}\,\frac{1}{\sqrt{\gamma^2(x-vt)^2 + y^2 + z^2}} \;=\; \frac{q}{4\pi\varepsilon_0}\,\frac{1}{R\sqrt{1-\beta^2\sin^2\theta}}, \qquad \vec A = \frac{\vec v}{c^2}\,\phi \qquad (3)

where: the two forms of $\phi$ are algebraically identical. Physical reading: $A^\mu$ is just the rest-frame $A'^\mu = (\phi'/c,\vec 0)$ boosted — a timelike four-vector tipped over, which is why $\vec A \parallel \vec v$ and carries the tell-tale $1/c^2$ . The surfaces $\phi = \text{const}$ are oblate spheroids, squashed along the motion by $1/\gamma$ : the Heaviside ellipsoid. The invariants, which classify the field once and for all:

\vec E^2 - c^2\vec B^2 \;=\; \vec E^2\big(1 - \beta^2\sin^2\theta\big) \;>\; 0, \qquad \vec E\cdot\vec B = 0 \qquad (4)

where: both are Lorentz scalars (`aq-sr-electromagnetic-field-invariants`). Physical reading: strictly positive, in every frame, at every point — the field is electric-type, meaning some observer sees it as pure electrostatics. That observer is the charge. A radiation field has $\vec E^2 - c^2\vec B^2 = 0$ exactly, and no boost turns one into the other.

Deriving the boosted Coulomb field

Derivation AOwned result: boost the four-potential, differentiate, and let the field tell you what it is

The safest derivation uses no field-transformation law at all — because that law is the single easiest thing to get sign-wrong, and this article's base got it wrong. Instead we boost the potential, a four-vector, which transforms in the one way a four-vector can, and then differentiate. Whatever comes out is right by construction, and can be used to audit the transformation law rather than depend on it.

Step 1. Boost the four-potential. In the rest frame $A'^\mu = (\phi'/c,\vec 0)$ with $\phi' = q/4\pi\varepsilon_0r'$ — a static charge has no current, so there is nothing else in the room. In the lab, \phi = \gamma\phi', \qquad A_x = \gamma\,\frac{v}{c^2}\,\phi', \qquad A_y = A_z = 0, and since $x' = \gamma(x-vt)$ , $y'=y$ , $z'=z$ , we have $r'^2 = \gamma^2(x-vt)^2 + y^2 + z^2$ . Why: the boost tips a purely timelike $A'^\mu$ over, and the spilled time-component becomes $\vec A$ — the whole origin of the vector potential of a moving charge, complete with the $1/c^2$ that makes magnetism feeble.

Step 2. Differentiate. With $\vec E = -\nabla\phi - \partial_t\vec A$ , \vec E = \frac{\gamma q}{4\pi\varepsilon_0}\;\frac{\vec R}{\big[\gamma^2(x-vt)^2+y^2+z^2\big]^{3/2}}, \qquad \vec R = (x-vt,\;y,\;z). Why: look at the numerator. It is $\vec R$ — the vector from the charge's present position, not the retarded one. The $-\partial_t\vec A$ term is doing that work: on its own $-\nabla\phi$ would tilt the field, and the time-derivative of $\vec A$ cancels the tilt exactly.

Step 3. Read off the angular form. Since $x - vt = R\cos\theta$ and $y^2+z^2 = R^2\sin^2\theta$ , \gamma^2(x-vt)^2 + y^2 + z^2 = R^2\big(\gamma^2\cos^2\theta + \sin^2\theta\big) = \gamma^2R^2\big(1-\beta^2\sin^2\theta\big), because $\gamma^2\cos^2\theta + \sin^2\theta = \gamma^2 - \sin^2\theta(\gamma^2-1)$ . Substituting collapses the $\gamma$ 's to $\gamma/\gamma^3 = 1-\beta^2$ and delivers eq. (1). Why: that one line is the entire angular structure. The rest-frame distance $r'$ , written in lab coordinates, is $\gamma R\sqrt{1-\beta^2\sin^2\theta}$ — stretched along the motion. It sits inside an inverse-square law, so cubing it produces the $-3/2$ power. There is nothing else to the formula.

Step 4. Get $\vec B$ for free. $\vec A = \vec v\phi/c^2$ with $\vec v$ constant, so $\vec B = \nabla\times\vec A = \vec v\times(-\nabla\phi)/c^2$ ; and $-\nabla\phi = \vec E + (\vec v/c^2)\partial_t\phi$ , whose second piece is parallel to $\vec v$ and dies in the cross product. Hence \boxed{\;\vec B = \frac{1}{c^2}\,\vec v\times\vec E\;} exactly, with a plus sign. Why: this is a derivation of the sign, not a choice of convention. A positive charge moving along $+\hat x$ has $\vec E \parallel +\hat y$ at a point directly above it, and the formula gives $\vec B \parallel \hat x\times\hat y = +\hat z$ — out of the page, exactly where the right-hand rule and Biot–Savart put it. *Verified in verify_aq-sr-field-uniformly-moving-charge.py: $\phi$ and $\vec A$ are boosted symbolically, $\vec E$ and $\vec B$ obtained by `sympy` differentiation with no transformation law assumed, and then (a) $\vec E\times\vec R_{\text{present}} = \vec 0$ identically, (b) $\vec E$ matches eq. (1) at 50 digits, (c) $\vec B - \vec v\times\vec E/c^2 = \vec 0$ identically.*

Go deeper · Derivation B · 2 minOwned result: the equipotentials are ellipsoids contracted by 1/gamma — the Heaviside ellipsoid

Derivation BOwned result: the equipotentials are ellipsoids contracted by 1/gamma — the Heaviside ellipsoid

The angular factor is usually presented as an algebraic curiosity. It is not. It is a shape, and that shape is the reason anyone ever thought of length contraction. This derivation extracts it, and then shows that the same $1/\gamma$ governs the field's angular width — with a number attached.

Step 1. Find the level surfaces of $\phi$ . From eq. (3), $\phi$ depends on position only through $\gamma^2(x-vt)^2 + y^2 + z^2$ , so $\phi = \text{const}$ is the surface \frac{(x-vt)^2}{(a/\gamma)^2} + \frac{y^2 + z^2}{a^2} = 1 . Why: an oblate spheroid — equatorial radius $a$ , polar semi-axis $a/\gamma$ along the motion. In the rest frame it was a sphere of radius $a$ . The equipotential has been contracted along the motion by exactly $1/\gamma$ — the FitzGerald–Lorentz factor, with nothing assumed about rulers. As $\beta\to1$ the polar semi-axis goes to zero and the spheroid degenerates to a flat disc — which is exactly what Heaviside wrote in a footnote of the 1889 paper: "the surfaces of equilibrium are not spheres, but are concentric oblate spheroids", becoming "a flat circular disc" as the speed approaches that of light. He had the pancake in 1889. What he did not have was any reason to think matter shared its fate; that leap was FitzGerald's.

Step 2. Now measure the pancake. Write $\theta = 90° - \delta$ , so $\delta$ is the angle away from the equatorial plane. Then $\sin^2\theta \approx 1 - \delta^2$ , so 1 - \beta^2\sin^2\theta \;\approx\; \frac{1}{\gamma^2} + \beta^2\delta^2 \qquad\Longrightarrow\qquad \frac{E(\delta)}{E(0)} = \Big(1 + \gamma^2\beta^2\delta^2\Big)^{-3/2}. Why: the profile is a universal function of the single variable $\gamma\beta\delta$ — every uniformly moving charge in the universe has the same one. That is what "confined to a cone of half-angle $1/\gamma$ " actually means, and it is now sharp enough to put a number on.

Step 3. Put the number on it. The field falls to half its equatorial peak when $(1+\gamma^2\beta^2\delta^2)^{3/2} = 2$ , i.e. $\gamma\beta\delta = \sqrt{2^{2/3}-1} = 0.766421$ . So the pancake's full width at half maximum is \boxed{\;\Delta\theta_{\text{FWHM}} = \frac{2\sqrt{2^{2/3}-1}}{\gamma\beta} \;\xrightarrow[\;\beta\to1\;]{}\; \frac{1.5328}{\gamma}\;} Why: $1/\gamma$ was always an order-of-magnitude gesture; this is the coefficient. For the LHC's 7 TeV protons ( $\gamma = 7461$ ) the pancake is $\mathbf{0.2055}$ mrad thick, not "about 0.134 mrad" — and the same constant reappears, unchanged, in the time profile of the flyby pulse. *Verified in verify_aq-sr-field-uniformly-moving-charge.py: the level surfaces of $\phi$ are shown to satisfy the spheroid equation with axis ratio exactly $1/\gamma$ ; $\sqrt{2^{2/3}-1}$ is computed in `mpmath` at 50 digits and matched against the article's printed 0.766421 / 1.5328 / 0.2055 mrad.*

Go deeper · Derivation C · 2 minOwned result: the flyby impulse does NOT depend on gamma — and that is why virtual photons work

Derivation COwned result: the flyby impulse does NOT depend on gamma — and that is why virtual photons work

The peak field of a fast charge grows like $\gamma$ ; the time it is applied shrinks like $1/\gamma$ . The two cancel — exactly, not approximately. Everything from Rutherford scattering to the equivalent-photon method follows from that one fact, and it is the result that decides whether relativity matters for a scattering problem at all.

Step 1. Set up the flyby. The charge moves along $\hat x$ ; the target sits at perpendicular distance $b$ , so $R^2 = v^2t^2 + b^2$ and $\sin\theta = b/R$ . Feeding $\gamma^2R^2(1-\beta^2\sin^2\theta) = \gamma^2v^2t^2 + b^2$ into eq. (1), the transverse field $E_y = E\,(b/R)$ is E_y(t) = \frac{\gamma q}{4\pi\varepsilon_0}\;\frac{b}{\big(\gamma^2v^2t^2 + b^2\big)^{3/2}} . Why: eq. (1) in the only variable the target cares about. Its peak is at $t=0$ : $E_y^{\text{max}} = \gamma q/4\pi\varepsilon_0b^2$ — the Coulomb field, enhanced by $\gamma$ .

Step 2. Read the pulse width. The bracket halves the field when $\gamma^2v^2t^2 + b^2 = 2^{2/3}b^2$ , i.e. at $t = \pm\,0.766421\,b/\gamma v$ , so $\Delta t_{\text{FWHM}} = 1.5328\,b/\gamma v$ . Why: the same constant $2\sqrt{2^{2/3}-1}$ as the angular width — it is the same function of the same variable, because the pancake swept past you is the pulse. Peak $\propto\gamma$ , width $\propto1/\gamma$ : the product is $\gamma$ -free, and that is the whole result, visible before any integral is done.

Step 3. Do the integral exactly. $\displaystyle\int_{-\infty}^{\infty}\frac{\mathrm{d}t}{(a^2t^2+b^2)^{3/2}} = \frac{2}{a\,b^2}$ with $a = \gamma v$ , so \boxed{\;\int_{-\infty}^{\infty} E_y\,\mathrm{d}t = \frac{\gamma q b}{4\pi\varepsilon_0}\cdot\frac{2}{\gamma v\,b^2} = \frac{q}{2\pi\varepsilon_0\,b\,v}\;} and every $\gamma$ has vanished. Why: the transverse impulse is completely independent of the projectile's Lorentz factor. A 7 TeV proton and a 7 GeV proton, passing at the same $b$ , hand over the same transverse momentum — which is why non-relativistic small-angle Coulomb-scattering formulae survive at relativistic energies.

Step 4. Look at what the target sees. During the flyby $\vec E$ is nearly transverse, $\vec B$ perpendicular to it, and $|\vec B| = \beta\sin\theta|\vec E|/c \to |\vec E|/c$ . Why: those are the field relations of a plane light wave. At the peak the invariant is $\vec E^2 - c^2\vec B^2 = \vec E^2/\gamma^2$ — not zero, but as close to null as makes no difference. So one may replace the passing charge by a spectrum of equivalent photons: the Weizsäcker–Williams method, whose error is exactly that $1/\gamma^2$ . Eq. (4) is the thing that tells you so. *Verified in verify_aq-sr-field-uniformly-moving-charge.py: the impulse integral is done symbolically in `sympy` and shown to be independent of $\gamma$ and of $\beta$ except through $v$ ; the peak field, the FWHM and the invariant $\vec E^2 - c^2\vec B^2 = \vec E^2/\gamma^2$ at $\theta = 90°$ are each recomputed in `mpmath` at 50 digits.*

Limiting cases

Case	Result	What it says
$\beta\to0$	$\vec E \to q\hat{\vec R}/4\pi\varepsilon_0R^2$	Coulomb. The angular factor switches itself off
$\beta\ll1$ , keep $O(\beta)$	$\vec B \to \mu_0 q\,\vec v\times\hat{\vec R}/4\pi R^2$	Biot–Savart. Magnetism is the first-order relic of a boost
$\theta = 0,\pi$ (fore/aft)	$E_\parallel = E_{\text{Coul}}/\gamma^2$	suppressed by $\gamma^{-2}$ ; and $\vec B = 0$ on the axis
$\theta = \pi/2$ (broadside)	$E_\perp = \gamma\,E_{\text{Coul}}$	enhanced by $\gamma$ ; $\vec B$ maximal, $B = \beta E_\perp/c$
ratio of the two	$E_\perp/E_\parallel = \gamma^3$	4.63 at $\beta = 0.8$ ; 12.07 at $\beta = 0.9$
$\gamma\gg1$	$\Delta\theta_{\text{FWHM}} = 1.5328/\gamma$	the pancake — and that is the true coefficient, not 1
equipotentials	oblate spheroid, axis ratio $1/\gamma$	the Heaviside ellipsoid; a disc as $\beta\to1$
any $\beta<1$	$\vec E^2 - c^2\vec B^2 > 0$	never null. Never radiation
flyby impulse	$\int E_\perp\,\mathrm{d}t = q/2\pi\varepsilon_0 b v$	no $\gamma$ at all — peak $\times$ width cancels

Worked examples

1 · A charge at $\beta = 0.8$ — the whole field, in numbers

Take $q = 1.00\ \mathrm{nC}$ moving at $\beta = 0.8$ ( $\gamma = 5/3$ ). Evaluate its field 1.00 m away, fore and broadside, and get the magnetic field too.

Step 1 — the benchmark, and the two directions. At rest, $E_{\text{Coul}} = q/4\pi\varepsilon_0R^2 = \mathbf{8.98755\ V/m}$ . Fore and aft ( $\theta = 0$ ) the angular factor is 1, so only the prefactor acts: $E_\parallel = (1-\beta^2)E_{\text{Coul}} = 0.36\times8.98755 = \mathbf{3.23552\ V/m}$ , and $\vec B = 0$ there since $\vec v\parallel\vec E$ . The field ahead of a charge doing four-fifths of light speed is barely a third of its static value.

Step 2 — broadside ( $\theta = 90°$ ). Now $1-\beta^2\sin^2\theta = 1/\gamma^2$ and the angular factor is $\gamma^3$ : E_\perp = \frac{1-\beta^2}{(1-\beta^2)^{3/2}}\,E_{\text{Coul}} = \gamma\,E_{\text{Coul}} = \mathbf{14.9793\ V/m}. The contrast between broadside and forward is $\gamma^3 = 4.62963$ — check: $14.9793/3.23552 = 4.6296$ ✓.

Step 3 — the magnetic field, twice. $B = \beta E_\perp/c = 0.8\times14.9793/(2.99792\times10^8) = \mathbf{39.972\ nT}$ . Cross-check against Biot–Savart-with- $\gamma$ : $\mu_0 q v\gamma/4\pi R^2 = (10^{-7})(10^{-9})(2.39834\times10^8)(1.66667) = \mathbf{3.9972\times10^{-8}\ T}$ ✓ — the same number to five figures, by two independent routes.

Step 4 — is it radiation? $cB = 11.9834 < E_\perp = 14.9793$ , so \vec E^2 - c^2\vec B^2 = 14.9793^2 - 11.9834^2 = \mathbf{+80.776\ (V/m)^2} \;>\; 0, with $\vec E\cdot\vec B = 0$ . And eq. (4) demands this equal $E_\perp^2/\gamma^2 = 224.379/(5/3)^2 = \mathbf{80.776}$ — it does, to every digit carried. There is a frame in which this charge's field is pure electrostatics: the charge's own.

2 · A 7 TeV proton in the LHC — how thin is the pancake?

Step 1 — the anisotropy. $\gamma = E/m_pc^2 = 7000/0.938272 = \mathbf{7461}$ (5 s.f.: 7460.5). Broadside the field is $\gamma = 7461$ times Coulomb; fore and aft it is $1/\gamma^2 = \mathbf{1.797\times10^{-8}}$ times it. The ratio between the two directions is $\gamma^3 = 4.15\times10^{11}$ — a charge directly in front of the oncoming proton feels, for practical purposes, nothing.

Step 2 — the thickness. The naive quote is $1/\gamma = \mathbf{0.1340\ mrad}$ . The honest one, from the half-maximum condition of derivation 2, is \Delta\theta_{\text{FWHM}} = \frac{1.5328}{\gamma} = \mathbf{0.2055\ mrad} . The field of a 7 TeV proton is a disc two ten-thousandths of a radian thick.

Step 3 — the pulse, and the punchline. A target at $b = 1.00\ \mathrm{mm}$ sees a spike of duration $\Delta t = 1.5328\,b/\gamma v = \mathbf{6.85\times10^{-16}\ s}$ — under a femtosecond — peaking at $\gamma q/4\pi\varepsilon_0b^2 = \mathbf{10.7\ V/m}$ , against $1.44\ \mathrm{mV/m}$ if the same proton crawled past. And yet the impulse is $q/2\pi\varepsilon_0bv$ : exactly what the slow proton delivers. A field 7461 times larger, applied for 1/7461 of the time. That is the cancellation of derivation 3, in one machine.

Figure

THE HEDGEHOG BECOMES A PANCAKE — AND ITS EQUIPOTENTIALS ARE THE ELLIPSOID FITZGERALD READ. In the charge's rest frame the field is the most boring object in physics: a sphere of straight radial lines, equal in every direction. Boost it. The transverse components of E are multiplied by γ and the longitudinal one is untouched, so the field lines crowd into the equatorial plane and thin out fore and aft. Quantitatively: the field ahead is suppressed by 1/γ², the field broadside is enhanced by γ, and the contrast between them is γ³ — a factor of 4.63 at β = 0.8, and 4×10¹¹ for a 7 TeV LHC proton. The surfaces of constant potential tell the same story as a shape rather than a formula: they are no longer spheres but OBLATE SPHEROIDS, squashed along the direction of motion by exactly 1/γ, degenerating to a flat disc as β → 1. Heaviside computed those spheroids in 1888-89 and told his friend George FitzGerald. FitzGerald's leap was to ask whether MATTER, being held together by electromagnetic forces, might be squashed by the same factor — and the length-contraction hypothesis was born, reverse-engineered out of this picture, sixteen years before Einstein derived it from a postulate.

Figure

THE FLYBY: THE PEAK GROWS AS γ, THE PULSE SHRINKS AS 1/γ, AND THE IMPULSE DOES NOT MOVE. A target sitting at impact parameter b watches a charge go past. The transverse field it sees is E_y(t) = γqb / 4πε₀(γ²v²t² + b²)^{3/2}. At the moment of closest approach the field peaks at γq/4πε₀b² — the Coulomb value, multiplied by γ. But the pulse lasts only Δt = 1.5328 b/γv, the same universal constant 2√(2^{2/3} − 1) that sets the angular width of the pancake, because the pulse IS the pancake sweeping past. Multiply a peak that grows like γ by a width that shrinks like 1/γ and the γ cancels: the exact time-integral is ∫E_y dt = q / 2πε₀bv, with no γ anywhere in it. A 7 TeV proton and a 7 GeV proton passing at the same impact parameter deliver the SAME transverse kick. This is why non-relativistic small-angle Coulomb-scattering formulae keep working at relativistic energies — and, because during the flyby E ⊥ B with |B| → |E|/c, why the passing charge may be replaced by a spectrum of equivalent photons: the Weizsäcker-Williams method, whose error is exactly the invariant E² − c²B² = E²/γ².

Common trapThis article's base prints the field-transformation law with BOTH signs flipped — and applied to its own starting point it gives B = MINUS v × E / c².

The AllQFT base for this article states the transformation it is about to use as

\vec E_\perp = \gamma\big(\vec E\,'_\perp + \vec v\times\vec B\,'\big), \qquad \vec B_\perp = \gamma\Big(\vec B\,'_\perp - \tfrac{1}{c^2}\,\vec v\times\vec E\,'\Big) \qquad \textbf{— WRONG.}

The error is structural, not typographical. The forward law, taking the lab to a frame moving at $+\vec v$ , is $\vec E\,'_\perp = \gamma(\vec E + \vec v\times\vec B)_\perp$ , $\vec B\,'_\perp = \gamma(\vec B - \vec v\times\vec E/c^2)_\perp$ — exactly what the live `aq-sr-transformation-electric-magnetic-fields` prints, correctly. To invert it you must swap the primes and flip the sign of $\vec v$ . The base swapped the primes and forgot the velocity:

\vec E_\perp = \gamma\big(\vec E\,'_\perp - \vec v\times\vec B\,'\big), \qquad \vec B_\perp = \gamma\Big(\vec B\,'_\perp + \tfrac{1}{c^2}\,\vec v\times\vec E\,'\Big) \qquad \textbf{✓ correct inverse.}

Feed the base's version its own starting point and it self-destructs. With $\vec B\,' = 0$ it returns $\vec B = -\vec v\times\vec E/c^2$ — exactly minus the field the same base prints two cards later as $\vec B = +\vec v\times\vec E/c^2$ . Physically it puts the field of a positive charge moving along $+\hat x$ into the page above the charge, where the right-hand rule and Biot–Savart both put it out. The base contains, side by side, a transformation law and a result that differ by a sign. Derivation 1 settles which is right, using no transformation law at all — and it is not the law.

Common trapThe field points at the charge's PRESENT position. This is not action at a distance, and it stops being true the instant the charge accelerates.

Equation (1) points $\vec E$ along $\hat{\vec R}$ , where $\vec R$ runs from where the charge is now. That looks like a violation of causality, and every year it convinces somebody that electromagnetism is secretly instantaneous. It is not.

The potentials are still retarded, and the news still crawls at $c$ . What happens is an extrapolation: in the retarded geometry the combination $r_{\text{ret}}(\hat{\vec n} - \vec\beta)$ is the displacement from the present position, because in the light-travel time $r_{\text{ret}}/c$ the charge advanced by precisely $r_{\text{ret}}\vec\beta$ . The field is not seeing the future; it is dead-reckoning, and for unaccelerated motion dead-reckoning is exact.

Break the assumption and the illusion dies at once. Stop the charge at time $t_0$ , and an observer at distance $d$ goes on being pulled toward the phantom present position — the place the charge would have reached — until $t_0 + d/c$ , when the kink in the field lines finally arrives and tells it the truth. That kink is the radiation field. The present-position property is a property of uniform motion, not of electromagnetism, and it is precisely the term that survives when $\vec a = 0$ in `aq-sr-lienard-wiechert-potentials`.

Physics of the portrait

Origins · 2 minOliver Heaviside and how this was discovered

Oliver Heaviside

1850–1925 · the field of a moving charge · the Heaviside ellipsoid · the man who handed FitzGerald the contraction

Oliver Heaviside left school at sixteen, worked as a telegraph operator, went deaf, and spent most of his life in poverty and self-imposed isolation, doing mathematics no university had asked for. He is the reason Maxwell's twenty equations in twenty variables are the four vector equations we now call Maxwell's; he invented the operational calculus, coined "impedance" and "inductance", and predicted the ionosphere. And in the winter of 1888 he did the calculation on this page — the field of a charge in steady rectilinear motion at any speed, not the slow-motion approximation everyone else had. He printed it in The Electrician (a four-part series beginning 9 November 1888) and then in full in the Philosophical Magazine (Ser. 5, 27 (167), 324–339, 1889; the paper is dated 27 December 1888): c\,E = \frac{q}{r^2}\,\frac{1 - u^2/v^2}{\big(1 - u^2\sin^2\theta/v^2\big)^{3/2}} in his own notation ( $c$ = permittivity, $v$ = the speed of light, $u$ = the charge's speed). That is equation (1) of this article, sixteen years before Einstein. In a footnote he recorded what the level surfaces do: "the surfaces of equilibrium are not spheres, but are concentric oblate spheroids", degenerating to "a flat circular disc" as $u\to v$ . The Heaviside ellipsoid. The same papers predicted what we now call Cherenkov radiation — a charge outrunning light in a dielectric — half a century before Cherenkov measured it. And he showed the moving charge's field to his friend George Francis FitzGerald, who asked the question Heaviside had not: if the electrostatic field that binds a body together is squashed by $1/\gamma$ , might the body be squashed by $1/\gamma$ too? FitzGerald's answer, a short letter to Science (13 (328), 390, 1889), is the length-contraction hypothesis. It was reverse-engineered out of Heaviside's ellipsoid.

History

1888

— Heaviside computes the field of a charge in steady motion at any speed, in a four-part series in The Electrician beginning 9 November, finding the fore-aft suppression and the broadside enhancement.

1889

— Heaviside publishes the full solution (Phil. Mag. Ser. 5, 27 (167), 324-339; the paper is dated 27 December 1888), printing the angular factor (1 - u²sin²θ/v²)^(-3/2) and noting in a footnote that the surfaces of equilibrium are concentric oblate spheroids — the Heaviside ellipsoid.

1889

— FitzGerald, in correspondence with Heaviside about that result, proposes in a letter to Science (13 (328), 390) that material bodies contract along their motion by the same factor.

1892

— Lorentz arrives independently at the contraction hypothesis.

1905

— Einstein derives length contraction from the two postulates, with no ether; Heaviside's field becomes simply a boosted Coulomb field.

1908

— Minkowski recasts the calculation as a single boost of the tensor F^μν, and the ellipsoid becomes a statement about geometry rather than about electricity.

Going further

Texts. Griffiths, Introduction to Electrodynamics, 4th ed., §10.3.2 and Example 12.13 · Purcell & Morin, Electricity and Magnetism, 3rd ed., ch. 5 · Jackson, Classical Electrodynamics, 3rd ed., §11.10 · Rindler, Introduction to Special Relativity, 2nd ed., §38.

Provenance. Signature $(+,-,-,-)$ , hence $p\cdot p = +m^2c^2$ ; $m$ is the invariant rest mass. $A^\mu = (\phi/c,\vec A)$ , $u^\mu = \gamma(c,\vec v)$ . This article's base prints the inverse field-transformation law with both cross-product signs flipped, which — applied to its own rest-frame starting point $(\vec E\,', \vec B\,' = 0)$ — yields $\vec B = -\vec v\times\vec E/c^2$ , contradicting the base's own later equation cards and reversing Biot–Savart. The corrected inverse is derived, machine-asserted, and cross-checked against a derivation that uses no transformation law at all (boost $A^\mu$ , then differentiate). Heaviside's 1889 formula and his "concentric oblate spheroids" footnote are quoted from the primary text (Phil. Mag. Ser. 5, 27 (167), 324–339). The exact publication day of FitzGerald's 1889 Science letter is not asserted here: it could not be confirmed to the standard the rest of this page is held to. The volume, issue and page are. All numbers are recomputed in `mpmath` at 50 digits against this article's printed digits.

Scope. This page owns the relativistic account: the boost of $A^\mu$ , the angular factor as a statement about the rest-frame distance, the Heaviside ellipsoid, the induced $\vec B$ with its sign, the invariant classification, and the $\gamma$ -free impulse. Retarded-time machinery belongs to `aq-sr-lienard-wiechert-potentials`; the E/B transformation law to `aq-sr-transformation-electric-magnetic-fields`; the current-carrying wire to `aq-sr-magnetism-relativistic-effect`. The live counterpart `aq-em-fields-moving-point-charge` (EMAG.RAD.FMOVQ.0611) covers the same field and calls it, aptly, the Heaviside pancake.

Builds on. `aq-sr-transformation-electric-magnetic-fields` · `aq-sr-four-potential` · `aq-sr-electromagnetic-field-tensor` · `aq-sr-charge-invariance-continuity`. Leads to. `aq-sr-magnetism-relativistic-effect` · `aq-sr-electromagnetic-field-invariants` · `aq-sr-covariant-lorentz-force` · `aq-sr-dual-field-tensor`.

If the field pattern is squashed by 1/γ along the motion, why is the FORWARD field weaker rather than stronger?

Because two effects are fighting and you have counted only one. Squashing the pattern longitudinally does crowd the lines — but the transformation of the components boosts only the transverse ones, by $\gamma$ , leaving the longitudinal one alone. The second effect wins everywhere except in the equatorial plane, and the net result is $1/\gamma^2$ fore and aft, $\times\gamma$ broadside. Field lines are a misleading picture here; the potentials and the invariants are not.

Does the moving charge's magnetic field mean it is radiating? It has E and B, perpendicular, like a light wave.

No — and the invariants say so with no room for argument. A light wave has $\vec E^2 - c^2\vec B^2 = 0$ exactly: a null field. The moving charge has $\vec E\cdot\vec B = 0$ (true, they are perpendicular) but $\vec E^2 - c^2\vec B^2 = \vec E^2(1-\beta^2\sin^2\theta) > 0$ strictly, for every $\beta<1$ and every $\theta$ . Positive, not zero — so there exists a frame in which $\vec B$ vanishes altogether, and it is the charge's own. You cannot boost away a light wave. You can always boost away this. As $\beta\to1$ the invariant shrinks to $\vec E^2/\gamma^2$ : arbitrarily close to null, which is the licence for the equivalent-photon approximation and the reason its error is $1/\gamma^2$ rather than zero.

Why does the impulse not depend on γ? Surely a faster projectile hits harder.

It hits harder and for less time, and the two cancel exactly: peak $\gamma q/4\pi\varepsilon_0b^2$ , duration $\propto b/\gamma v$ , product $\gamma$ -free — and the exact integral confirms it, $\int E_\perp\,\mathrm{d}t = q/2\pi\varepsilon_0 bv$ . It is why the Rutherford cross-section keeps its non-relativistic impact-parameter form for fast projectiles, and why the transverse beam-beam kick does not run away with energy. Relativity does not always amplify. Here it rearranges, and hands the total straight back.

Heaviside had the right field in 1889. Why is that not special relativity, sixteen years early?

Because a formula is not a theory. Heaviside had one correct solution of Maxwell's equations for one system, obtained inside an ether framework he never abandoned, and he read the flattening as something the ether did to the field. He did not have the relativity principle, the invariance of $c$ as a postulate, the transformation of time, or the claim that the same $\gamma$ governs clocks, rulers and momentum. FitzGerald's leap — that matter might contract too — was an inspired guess bolted onto a mechanical picture, not a deduction. What Einstein supplied was not the answer but the reason: a postulate from which Heaviside's ellipsoid, FitzGerald's contraction and a great deal else all follow at once, with the ether removed.

Is θ measured from the present position or the retarded position?

The present one: in eq. (1), $\vec R$ runs from where the charge is now. If you work instead in the retarded geometry — as the Liénard–Wiechert form does — you have a different vector $\hat{\vec n}$ and a different denominator $(1-\hat{\vec n}\cdot\vec\beta)^3$ , and the two are linked exactly by (1 - \hat{\vec n}\cdot\vec\beta)\,r_{\text{ret}} = R\sqrt{1-\beta^2\sin^2\theta}. Same field, two coordinate systems. The present-position form exists only because the motion is uniform; the retarded form survives when it is not.

Does the field carry momentum? Does a uniformly moving charge drag something along with it?

Yes, and yes. $\vec S = \vec E\times\vec B/\mu_0$ is non-zero everywhere off the axis, so the field carries momentum density $\vec S/c^2$ along with the charge — the bow wave. But no net energy leaves: $|\vec S| \propto 1/R^4$ while a sphere's area grows only as $R^2$ , so the flux through it dies as $R\to\infty$ . The energy sloshes, and travels, and never escapes. (Trying to read that co-moving field energy as an "electromagnetic mass" of the electron is the origin of the notorious $4/3$ problem — resolved by treating the stress-energy tensor properly, `aq-sr-electromagnetic-stress-energy-tensor`, and emphatically not a reason to reintroduce a velocity-dependent mass.)

Field of a Uniformly Moving Charge

Where this lives today

The physical picture

The core equations

Deriving the boosted Coulomb field

Limiting cases

Worked examples

1 · A charge at $\beta = 0.8$ — the whole field, in numbers

2 · A 7 TeV proton in the LHC — how thin is the pancake?

Physics of the portrait

Going further

Before this

Where this leads

Field of a Uniformly Moving Charge

Where this lives today

The physical picture

The core equations

Deriving the boosted Coulomb field

Limiting cases

Worked examples

1 · A charge at β=0.8\beta = 0.8β=0.8 — the whole field, in numbers

2 · A 7 TeV proton in the LHC — how thin is the pancake?

Physics of the portrait

Going further

Before this

Where this leads

1 · A charge at $\beta = 0.8$ — the whole field, in numbers