Magnetism as a Relativistic Effect

The wire argument is usually filed under pedagogy, but the physics it exposes is load-bearing wherever the carriers are not slow. In an accelerator, a bunch of $10^{11}$ protons at $\gamma \sim 7000$ is a line charge whose Coulomb field, boosted, becomes an intense magnetic field — and the two forces very nearly cancel. The electric field of the bunch pushes the protons apart; the magnetic field of the same bunch, which is nothing but that electric field seen from the lab, pulls them together; and the net space-charge defocusing goes like $1/\gamma^2$. That is the wire calculation run backwards, and it is why high-energy beams are so much easier to hold together than low-energy ones. Space-charge codes compute it from the field transformation, not from Biot–Savart, because at $\gamma \sim 10^3$ the "magnetic" and "electric" labels have stopped meaning anything frame-independent. The same reasoning drains the mystery out of the unipolar generator and the homopolar disc, machines that caused a century of confusion precisely because observers disagreed about whether the electromotive force was electric or magnetic. There is no fact of the matter. There is only a fact about the force. And the philosophical prize is large. Relativity plus Coulomb's law implies magnetism. You do not get to postulate the magnetic field independently; given special relativity and a static interaction between invariant charges, it is forced. Relativity is not a correction to electromagnetism but its explanation — which is why the electromagnetic sector of the Standard Model contains a single gauge field, not two. What is invariant is the field tensor of The Electromagnetic Field Tensor and the two scalars of Electromagnetic Field Invariants; everything else is a choice of observer.

Set up the wire the way Purcell did. In the laboratory frame $S$ the wire lies along $\hat x$: a lattice of positive ions at rest with line charge density $+\lambda_0$, and a sea of conduction electrons drifting at $-v_d\hat x$ (electrons are negative, so drifting backwards makes conventional current flow forwards) with line charge density $-\lambda_0$. The densities are equal and opposite: the wire is exactly neutral, and the current is $I = \lambda_0 v_d$ along $+\hat x$. Now send a test charge $q$ past the wire at velocity $u\hat x$, at perpendicular distance $r$. In the lab there is no electric field, so all the force on $q$ must be magnetic: $\vec F = q\,\vec u\times\vec B$. Go to the test charge's rest frame $S'$. Here $q$ is at rest, so it cannot feel a magnetic force at all — $q\vec v\times\vec B$ with $\vec v = 0$ is exactly zero. Whatever force it feels must be electric. If the wire were still neutral there would be no force and $q$ would not accelerate — contradicting the lab, where it plainly does. So the wire cannot be neutral in $S'$. And it is not. In $S'$ the ions stream backwards at $-u$, so their lattice contracts and their density goes up. The electrons, already moving against $u$, move faster in $S'$ than in the lab, so their density goes up too. Both rise — but not by the same factor, and the residue is a genuine net line charge. It is negative when $u$ points along the current, so it attracts a positive test charge: precisely the direction of the lab magnetic force, and precisely why parallel currents attract.

Everything rests on one asymmetry between charge and charge density.

where: $\lambda_{\rm pr}$ — the proper line charge density, measured where the lattice is at rest; $\gamma_w = (1-w^2/c^2)^{-1/2}$ with $w$ the lattice's speed in the frame of interest. Physical reading: the charge on a segment does not change, but the segment's length does. Contract the length by $1/\gamma_w$ and the charge per unit length rises by $\gamma_w$. Charge invariance is what makes this work — a separate, load-bearing fact, established in Charge Invariance and Continuity. Boosting the lab-neutral wire into the test charge's frame gives the induced density. This is the article's owned result, and it is exact:

where: $u$ — the test charge's lab speed, parallel to the wire; $\gamma_u = (1-u^2/c^2)^{-1/2}$; $I = \lambda_0 v_d$ — the lab current. Physical reading: every $\gamma_{v_d}$ cancels. The electrons' proper density carries a $1/\gamma_{v_d}$ and their boosted Lorentz factor carries a $\gamma_{v_d}$; the two annihilate exactly. Nothing is truncated at order $\beta^2$ — the result holds at any $u$ and any $v_d$, right up to $c$. The base article's claim that this is a leading-order statement is simply wrong. The residue is an ordinary line charge, so it makes an ordinary Coulomb field. Distances perpendicular to the boost are unchanged, so $r' = r$:

where: $B_\phi$ — the lab magnetic field of the wire, $\mu_0\varepsilon_0 = 1/c^2$. Physical reading: the pure electric field the moving observer measures is $\gamma_u u$ times the pure magnetic field the lab measures — the transverse field-transformation law, whose general form belongs to Transformation of Electric and Magnetic Fields, here reproduced by hand from length contraction alone. Finally the two frames are reconciled by a real transformation, not by a hand-wave:

where: the general transverse-force law is $F_y = F'_y/[\gamma(1 + u v'_x/c^2)]$, collapsing to $F_\perp = F'_\perp/\gamma_u$ for a particle momentarily at rest in $S'$. Physical reading: the minus sign is an attraction — a positive charge moving with the current is pulled toward the wire, the standard right-hand-rule result and the reason parallel currents attract. This is the magnetic force, derived without ever assuming a magnetic force exists.

1 · A real copper wire, and the imbalance you would have to measure

A copper wire of cross-section $A = 1.00\ \mathrm{mm^2}$ carries $I = 1.00\ \mathrm{A}$. Copper has $n = 8.5\times10^{28}$ conduction electrons per cubic metre. A charge $q = 1.00\ \mathrm{\mu C}$ passes at $u = 100\ \mathrm{m\,s^{-1}}$, at $r = 1.00\ \mathrm{cm}$. Find the drift speed, the induced line charge, the fractional charge imbalance, and the force — and show that it is the ordinary magnetic force.

Step 1 — identify. Neutral wire, test charge moving parallel to it: equations (2)–(4) apply verbatim. The only inputs are $n$, $A$, $I$, $u$, $r$.

Step 2 — the charge in the wire, and the drift speed. Each lattice carries $\lambda_0 = n e A = (8.5\times10^{28})(1.602177\times10^{-19})(1.00\times10^{-6}) = \mathbf{1.36185\times10^{4}\ \mathrm{C\,m^{-1}}}$ — 13 619 coulombs per metre, where a lightning stroke moves a few coulombs in total. Then $v_d = I/\lambda_0 = 1.00/1.36185\times10^{4} = \mathbf{7.343\times10^{-5}\ \mathrm{m\,s^{-1}}}$, so $\beta_d = v_d/c = \mathbf{2.449\times10^{-13}}$. The electrons crawl at 73 micrometres per second.

Step 3 — the induced line charge. $\gamma_u = 1.000\,000$ to twelve figures at $100\ \mathrm{m\,s^{-1}}$, so $$\lambda'_{\rm net} = -\frac{\gamma_u u I}{c^2} = -\frac{(1.000000)(100)(1.00)}{8.98755\times10^{16}} = \mathbf{-1.1127\times10^{-15}\ \mathrm{C\,m^{-1}}}.$$

Step 4 — the fractional imbalance. Compare it with the charge actually there: $$\frac{|\lambda'_{\rm net}|}{\lambda_0} = \frac{1.1127\times10^{-15}}{1.36185\times10^{4}} = \mathbf{8.17\times10^{-20}}.$$ In the test charge's frame the two lattices agree to nineteen decimal places — and that residual disagreement is the entire magnetic force.

Step 5 — the boundary number. $E'_r = \lambda'_{\rm net}/2\pi\varepsilon_0 r = \mathbf{-2.00\times10^{-3}\ \mathrm{V\,m^{-1}}}$, so $F'_\perp = qE'_r = \mathbf{-2.00\times10^{-9}\ \mathrm{N}}$, and with $\gamma_u = 1$ the lab force is the same, toward the wire. Check against magnetism: $B_\phi = \mu_0 I/2\pi r = \mathbf{2.00\times10^{-5}\ T} = 20.0\ \mathrm{\mu T}$ (about 40% of the Earth's field — which is why Ørsted's compass swung), and $q u B_\phi = (1.00\times10^{-6})(100)(2.00\times10^{-5}) = \mathbf{2.00\times10^{-9}\ \mathrm{N}}$. Identical. The relativistic route and the magnetic route agree to every figure carried.

2 · The same wire at $u = 0.6c$, where the $\gamma$ is not decoration

Now send the same $1.00\ \mathrm{\mu C}$ charge past the same wire at $u = 0.6c$. Here $\gamma_u = 1.25$ and the transverse-force factor does visible work — this is the case that exposes whether you actually did Step 4 of the second derivation or merely asserted it.

Step 1 — identify. Nothing changes except $u$. $\gamma_u = 1/\sqrt{1-0.36} = 1/0.8 = \mathbf{1.25}$, and $u = 0.6c = 1.79875\times10^{8}\ \mathrm{m\,s^{-1}}$.

Step 2 — the induced charge, now $\gamma$-enhanced. $$\lambda'_{\rm net} = -\frac{\gamma_u u I}{c^2} = -\frac{(1.25)(1.79875\times10^{8})(1.00)}{8.98755\times10^{16}} = \mathbf{-2.5017\times10^{-9}\ \mathrm{C\,m^{-1}}}.$$ Six orders of magnitude larger than in Example 1 — yet still a fractional imbalance of only $1.84\times10^{-13}$.

Step 3 — the field and force in $S'$. $E'_r = \lambda'_{\rm net}/2\pi\varepsilon_0 r = \mathbf{-4496.9\ \mathrm{V\,m^{-1}}}$, giving $F'_\perp = qE'_r = \mathbf{-4.4969\times10^{-3}\ \mathrm{N}}$. Cross-check against (3): $-\gamma_u u B_\phi = -(1.25)(1.79875\times10^{8})(2.00\times10^{-5}) = \mathbf{-4496.9\ \mathrm{V\,m^{-1}}}$ — the same, as it must be.

Step 4 — transform the force back, properly. $$F_\perp = \frac{F'_\perp}{\gamma_u} = \frac{-4.4969\times10^{-3}}{1.25} = \mathbf{-3.5975\times10^{-3}\ \mathrm{N}}.$$

Step 5 — the boundary number, and what would have gone wrong. The lab magnetic force is $q u B_\phi = (1.00\times10^{-6})(1.79875\times10^{8})(2.00\times10^{-5}) = \mathbf{3.5975\times10^{-3}\ \mathrm{N}}$ toward the wire — agreeing exactly with the transformed electrostatic force. Had we skipped the transverse-force law and declared $F_\perp = F'_\perp$, we would have predicted $4.4969\times10^{-3}\ \mathrm{N}$ — 25% too large, and wrong in a way that grows without bound as $u\to c$: at $u = 0.99c$ the error is a factor of $7.1$. This is why the factor must be transformed, not waved.

Magnetism can be a $(v/c)^2$ effect and still bend a compass needle because of a single number: the drift velocity, absurdly small, dragging an absurdly large charge.

Texts. Purcell & Morin, Electricity and Magnetism, 3rd ed., ch. 5 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.3 · Rindler, Introduction to Special Relativity, 2nd ed., ch. 7 · Jackson, Classical Electrodynamics, 3rd ed., ch. 11 · Page, Am. J. Sci. 34, 57 (1912).

Provenance. Signature $(+,-,-,-)$; $p\cdot p = +m^2c^2$; $m$ is the invariant rest mass, and no velocity-dependent mass appears anywhere. Base-defect repairs (logged): (i) the base's boxed $\lambda'_{\rm net}$ contains a literal English phrase inside the formula and an uncancelled $(\gamma_u/v_d)\cdot v_d$ — discarded and re-derived. (ii) The base calls the result "leading order"; it is exact, proved symbolically with $u$ and $v_d$ both free. (iii) The base dismisses the transverse-force factor as one that "cancels when comparing physical accelerations"; it is replaced by the actual transformation law $F_\perp = F'_\perp/\gamma_u$, whose omission over-predicts the force by 25% at $u = 0.6c$. All numbers here are recomputed in verify_aq-sr-magnetism-relativistic-effect.py, which parses this article's own printed digits.

Scope. The general field-transformation law is not derived here — it belongs to Transformation of Electric and Magnetic Fields, and the wire calculation instead reconstructs one component of it from length contraction alone. Charge invariance itself is established in Charge Invariance and Continuity, and the four-current in The Four-Current Density.

Builds on. Charge Invariance and Continuity · The Four-Current Density · Transformation of Electric and Magnetic Fields. Leads to. The Electromagnetic Field Tensor · The Covariant Lorentz Force Law · Covariant Maxwell Equations. Cross-track. The electrodynamics course carries a parallel Magnetism as a Relativistic Effect (aq-em); special relativity is the canonical home for the derivation, and the electrodynamics article recalls it.