---
title: "Covariant Electromagnetism"
book: "University Physics — Year 3"
subject: physics
language: en
chapter: 6
exercises: 12
source: https://one-course.com/books/physics/5/en/chapter/6-covariant-electromagnetism
---

# Chapter 6 — Covariant Electromagnetism

A copper wire carries ten amperes. Its electrons drift at a tenth of a millimetre per second — slower than a snail — and the wire is electrically neutral to fantastic precision. Yet a charge moving alongside feels a measurable pull. Viewed from that charge’s own frame, the “magnetic” explanation evaporates: the charge is at rest, and a resting charge feels no magnetic force at all. What pulls it, in that frame, is an *electric* field — conjured by relativity itself, because [length contraction](https://one-course.com/books/physics/5/en/chapter/4-relativistic-kinematics#prop-b3-relativistic-kinematics-contraction) unbalances, by one part in $10^{24}$, the two streams of charge in the wire. Magnetism is electricity seen from a moving seat. This chapter rewrites the electromagnetism of the Year 2 volume in the language of the previous three: [four-vectors](#def-b3-covariant-electromagnetism-fourvector) for charge and potential, one antisymmetric tensor holding $\vect E$ and $\vect B$ together, Maxwell’s four equations collapsing into two lines that read the same in every [inertial frame](https://one-course.com/books/physics/5/en/chapter/4-relativistic-kinematics#thm-b3-relativistic-kinematics-postulates). Beyond elegance, the payoff is working physics: how fields transform, why a fast charge’s field flattens into a pancake, and why the light of a relativistic electron sweeps forward like a headlight — the principle of the synchrotron light sources that X-ray proteins and batteries today.

## 6.1 Four-vectors for charge and current

**Definition 6.1 (Four-vectors).**

A *four-vector* is a quadruple $a^\mu = (a^0, \vect a)$, $\mu =
0, 1, 2, 3$, whose components mix under a boost exactly as $(ct,
\vect r)$ do:

$$
a'^0 = \gamma(a^0 - \beta a^1) , \qquad
a'^1 = \gamma(a^1 - \beta a^0) , \qquad
a'^{2,3} = a^{2,3}
$$

for a boost at $\beta c$ along $x$. Any two four-vectors give the invariant scalar product $a\cdot b = a^0b^0 - \vect a\cdot\vect b$, the same number in every frame — the pattern behind $c^2t^2 - x^2$ and $E^2 - p^2c^2$. Position $(ct, \vect r)$ and [four-momentum](https://one-course.com/books/physics/5/en/chapter/5-relativistic-dynamics#def-b3-relativistic-dynamics-four-momentum) $(E/c, \vect p)$ are the two we have; electromagnetism now supplies three more: current, potential, and wave-vector.

**Proposition 6.2 (The four-current and charge conservation).**

Charge density $\rho$ and current density $\vect\jmath$ form the four-current

$$
J^\mu = (\rho c,\ \vect\jmath\,) :
$$

a cloud of charge of proper density $\rho_0$ moving at $\vect u$ has $J^\mu = \rho_0\gamma_u(c, \vect u)$ — density grows by $\gamma_u$ because the cloud’s volume contracts, while the charge itself is invariant (a decisive experimental fact: atoms with fast inner electrons stay exactly neutral). Charge conservation is the invariant statement

$$
\frac{\partial(\rho c)}{\partial(ct)} +
\operatorname{div}\vect\jmath = 0 ,
$$

the continuity equation of the Year 2 volume, now visibly the same law for all observers.

**Partial proof.** The moving cloud: a box of proper volume $V_0$ holds charge $\rho_0
V_0$; in the lab the box is contracted to $V_0/\gamma_u$, so $\rho =
\gamma_u\rho_0$, and $\vect\jmath = \rho\vect u$ by definition of a current density. That $(\rho c, \vect\jmath)$ then transforms as a [four-vector](#def-b3-covariant-electromagnetism-fourvector) follows because it equals $\rho_0/c$ times the four-velocity $\gamma_u(c, \vect u)$, itself [four-vector](#def-b3-covariant-electromagnetism-fourvector) by construction. Invariance of charge is an experimental input. ∎

## 6.2 One tensor for both fields

**Definition 6.3 (Four-potential and field tensor).**

The potentials of the Year 2 volume assemble into the four-potential

$$
A^\mu = \Big(\frac{V}{c},\ \vect A\Big) ,
$$

and the measurable fields $\vect E = -\vect\nabla V -
\partial_t\vect A$, $\vect B = \operatorname{\vect{curl}}\vect A$ are the six independent components of one antisymmetric array, the *electromagnetic field tensor*

$$
F^{\mu\nu} =
\begin{pmatrix}
0 & -E_x/c & -E_y/c & -E_z/c\\
E_x/c & 0 & -B_z & B_y\\
E_y/c & B_z & 0 & -B_x\\
E_z/c & -B_y & B_x & 0
\end{pmatrix} .
$$

$\vect E$ and $\vect B$ are not two fields but six faces of one object; which face an observer calls “electric” depends on the observer’s motion.

**Theorem 6.4 (Maxwell, covariantly).**

The four Maxwell equations of the Year 2 volume are the component form of two frame-independent statements: the sourced pair (Gauss and Ampère–Maxwell) is

$$
\sum_\mu \partial_\mu F^{\mu\nu} = \mu_0\,J^\nu ,
$$

and the sourceless pair (no monopoles, Faraday) is the identity guaranteeing that $F$ derives from a [four-potential](#def-b3-covariant-electromagnetism-tensor). The Lorentz force is $\dd p^\mu/\dd\tau = qF^{\mu\nu}u_\nu$ — force law, magnetic included, in one line. Because both sides of each equation transform identically, *Maxwell’s theory needs no correction to be relativistic*: it was relativity’s first finished piece, born 1865.

**Partial proof.** Expand the $\nu = 0$ component: $\sum_i\partial_i(E_i/c) =
\mu_0\rho c$, i.e. $\operatorname{div}\vect E = \rho/\varepsilon_0$ (using $c^2 = 1/\mu_0\varepsilon_0$). The $\nu = 1$ component collects $-\partial_t(E_x/c^2) + (\partial_yB_z - \partial_zB_y)
= \mu_0 j_x$: the $x$ component of Ampère–Maxwell. The other components repeat the pattern; the sourceless pair is the equality of crossed second derivatives of $A^\mu$ (checked in [Exercise 6.3](#exo-b3-covariant-electromagnetism-3)). That $F^{\mu\nu}$ transforms as a (two-index) tensor — each index like a [four-vector](#def-b3-covariant-electromagnetism-fourvector) — is admitted; its consequences are the next proposition. ∎

## 6.3 How the fields transform

**Proposition 6.5 (Field transformation).**

For a boost at velocity $\vect v = v\,\vect e_x$: the components *along* the motion are untouched, the transverse ones mix,

$$
E'_x = E_x , \qquad
E'_y = \gamma(E_y - vB_z) , \qquad
E'_z = \gamma(E_z + vB_y) ,
$$

$$
B'_x = B_x , \qquad
B'_y = \gamma\Big(B_y + \frac{v}{c^2}E_z\Big) , \qquad
B'_z = \gamma\Big(B_z - \frac{v}{c^2}E_y\Big) .
$$

Compactly, transverse to the boost: $\vect E' = \gamma(\vect E +
\vect v\wedge\vect B)_\perp$ and $\vect B' = \gamma(\vect B - \vect
v\wedge\vect E/c^2)_\perp$. A pure $\vect B$ in one frame is $\vect
E$ and $\vect B$ in another: the fields are one phenomenon.

**Proof.** *Admitted at this level.* ∎

**Example 6.6 (Magnetism from a neutral wire).**

A neutral wire carries current: positive lattice at rest, electrons drifting. For a test charge moving alongside, boost to its frame: the two charge streams contract *differently* (they have different velocities), the wire acquires the net line charge $\lambda' =
-\gamma vI/c^2$, and its radial electric field pulls the charge with exactly the force the lab called $q\vect v\wedge\vect B$. Magnetism is what the second-order term $v u/c^2$ of relativity looks like when $10^{28}$ elementary charges per metre conspire: each effect is fantastically small, but the wire is neutral to even greater precision, so the tiny imbalance is the whole story ([Problem 6.1](#pb-b3-covariant-electromagnetism-1)).

![One force, two accountings. Lab: the neutral wire’s current makes B, the moving charge feels q v B. Charge’s frame: the positive lattice now moves and contracts, the electrons (slower there) spread out; the wire is net charged and its E' does the pulling. Same experiment, same outcome.](https://one-course.com/images/onecourse/chapters/physics-5/b3-covariant-electromagnetism/fig-acdaaf9ea586.svg)

*One force, two accountings. Lab: the neutral wire’s current makes $\vect B$, the moving charge feels $q\vect v\wedge\vect B$. Charge’s frame: the positive lattice now moves and contracts, the electrons (slower there) spread out; the wire is net charged and its $\vect E'$ does the pulling. Same experiment, same outcome.*

**Example 6.7 (The pancaked field of a fast charge).**

Transform the Coulomb field of a charge into the frame where it moves at $\gamma \gg 1$: the longitudinal field is unchanged while the transverse one is multiplied by $\gamma$. The once-spherical field flattens into a disc of opening angle $\sim 1/\gamma$ around the plane through the charge — accompanied, transverse to the motion, by a magnetic ring $B' = vE'/c^2$. To a stationary observer the passage of an LHC proton ($\gamma \approx 7000$) is a sub-picosecond slap of nearly-crossed $\vect E$ and $\vect B$: almost a pulse of light — the reason fast beams talk to matter the way photons do.

![The electric field of a point charge, at rest and at high speed: boosting multiplies the transverse components by and leaves the longitudinal ones alone, squeezing the field into a disc perpendicular to the motion.](https://one-course.com/images/onecourse/chapters/physics-5/b3-covariant-electromagnetism/fig-813ed69ef47c.svg)

*The electric field of a point charge, at rest and at high speed: boosting multiplies the transverse components by $\gamma$ and leaves the longitudinal ones alone, squeezing the field into a disc perpendicular to the motion.*

## 6.4 Invariants, light, and the headlight effect

**Proposition 6.8 (The two field invariants).**

From $F^{\mu\nu}$ one can build exactly two independent scalars:

$$
E^2 - c^2B^2
\qquad\text{and}\qquad
\vect E\cdot\vect B ,
$$

the same in every [inertial frame](https://one-course.com/books/physics/5/en/chapter/4-relativistic-kinematics#thm-b3-relativistic-kinematics-postulates). Consequences: if $E > cB$ somewhere (with $\vect E\cdot\vect B = 0$), some frame sees a pure electric field there; if $E < cB$, some frame sees pure magnetic; and a plane light wave, with $E = cB$ and $\vect E\perp\vect B$, has *both* invariants zero — it is light in every frame: no boost can turn it into a static field, only redshift it. One cannot catch up with a light wave; Einstein’s teenage question answers itself in two invariants.

**Partial proof.** Check invariance under the standard boost by direct substitution of [Proposition 6.5](#prop-b3-covariant-electromagnetism-transform) ([Exercise 6.6](#exo-b3-covariant-electromagnetism-6)); that no third independent invariant exists is admitted. For the wave: $E = cB$ and orthogonality give both zero; a frame with a static field would need a nonzero invariant. ∎

**Proposition 6.9 (Four-wave-vector, Doppler and aberration).**

A plane wave’s frequency and direction form the null [four-vector](#def-b3-covariant-electromagnetism-fourvector) $k^\mu = (\omega/c, \vect k)$, $\|\vect k\| = \omega/c$. Transforming it gives at once the relativistic Doppler formula of [Chapter 4](https://one-course.com/books/physics/5/en/chapter/4-relativistic-kinematics#ch-b3-relativistic-kinematics) and the *aberration* of directions:

$$
\cos\theta' = \frac{\cos\theta - \beta}{1 - \beta\cos\theta} .
$$

Read backwards, a source radiating isotropically in its rest frame beams, in the lab, half its light into the forward cone $\theta
\lesssim 1/\gamma$: the *[headlight effect](#prop-b3-covariant-electromagnetism-wavevector)*. An electron circling at $\gamma \sim 10^4$ in a storage ring sweeps a $0.1\,\mathrm{mrad}$ searchlight of X-rays around the ring — the synchrotron light that fills protein-crystallography beamlines.

**Partial proof.** The phase $\omega t - \vect k\cdot\vect r = k^\mu x_\mu$ counts wave crests passing events — an invariant — so $k^\mu$ must transform as a [four-vector](#def-b3-covariant-electromagnetism-fourvector). Apply the boost to $(\omega/c, k\cos\theta,
k\sin\theta, 0)$: the time component gives Doppler, the ratio of spatial components the aberration formula. Setting $\theta' =
90^\circ$ (the sideways ray of the source frame): $\cos\theta =
\beta$, i.e. $\theta \approx 1/\gamma$ for $\gamma \gg 1$ — half the sphere folds into the forward cone. ∎

![Aberration folds the radiation of a fast source into a forward cone of half-angle 1/: the headlight effect, and the working principle of synchrotron light sources.](https://one-course.com/images/onecourse/chapters/physics-5/b3-covariant-electromagnetism/fig-0abf01c97f87.svg)

*Aberration folds the radiation of a fast source into a forward cone of half-angle $\sim 1/\gamma$: the [headlight effect](#prop-b3-covariant-electromagnetism-wavevector), and the working principle of synchrotron light sources.*

**Method 6.10 (Working covariantly).**

(1) Identify the [four-vectors](#def-b3-covariant-electromagnetism-fourvector) in play ($x^\mu$, $P^\mu$, $J^\mu$, $A^\mu$, $k^\mu$) and prefer their invariant products to components. (2) To transform fields, split into components along and transverse to the boost and apply [Proposition 6.5](#prop-b3-covariant-electromagnetism-transform). (3) Check the two invariants before and after — the fastest error detector in the subject. (4) When a magnetic problem looks mysterious, ride with the charge: in its frame only $\vect E'$ acts. (5) Trust Maxwell: the equations never need relativistic “corrections”, only relativistic reading.

![An aurora: solar-wind charges steered by the Earth’s magnetic field into the polar atmosphere. What one observer calls magnetic steering, another calls electric acceleration — the fields mix under the transformations of this chapter.](https://one-course.com/images/onecourse/chapters/physics-5/b3-covariant-electromagnetism/img-f5d604651824.jpg)

*An aurora: solar-wind charges steered by the Earth’s magnetic field into the polar atmosphere. What one observer calls magnetic steering, another calls electric acceleration — the fields mix under the transformations of this chapter.*

## 6.5 Exercises

**Exercise 6.1 ★.**

(a) Boost $a^\mu = (5, 3, 0, 0)$ (units of some $a_0$) by $\beta =
0.6$: compute $a'^\mu$ and check $a\cdot a$ is unchanged. (b) Show that the sum of two [four-vectors](#def-b3-covariant-electromagnetism-fourvector) is a [four-vector](#def-b3-covariant-electromagnetism-fourvector). (c) Is $(c,
\vect v)$ of a particle a [four-vector](#def-b3-covariant-electromagnetism-fourvector)? And $\gamma(c, \vect v)$? (d) Why is “the electric field” alone not part of any [four-vector](#def-b3-covariant-electromagnetism-fourvector)?

**Solution of Exercise 6.1.**

(a) $\gamma = 1.25$: $a'^0 = 1.25(5 - 1.8) = 4.0$, $a'^1 = 1.25(3 -
3) = 0$; $a\cdot a = 25 - 9 = 16 = 16 - 0$. (b) The transformation is linear, so it distributes over sums. (c) $(c, \vect v)$: no — its “time” component is the same for all particles while the mixing demands otherwise; $\gamma(c, \vect v) = \dd x^\mu/\dd\tau$: yes, a [four-vector](#def-b3-covariant-electromagnetism-fourvector) divided by the invariant $\dd\tau$. (d) $\vect E$’s three components mix with $\vect B$’s, not with any scalar: the fields fill a two-index tensor, not a [four-vector](#def-b3-covariant-electromagnetism-fourvector).

**Exercise 6.2 ★.**

A copper wire of section $1.0\,\mathrm{mm}^{2}$ carries $10\,\mathrm{A}$; conduction-electron density $n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}$. (a) Compute the drift speed. (b) Write $J^\mu$ for the electron fluid and for the lattice. (c) Check the continuity equation for each. (d) The wire is neutral: what is $J^\mu$ for the whole wire, and which component survives?

**Solution of Exercise 6.2.**

(a) $u = I/nSe = 10/(8.5 \times 10^{28} \times 10^{-6} \times
1.6 \times 10^{-19}) = 7.4 \times 10^{-4}\,\mathrm{m}/\mathrm{s}$. (b) Lattice: $(nec, \vect 0)$; electrons: $(-nec, -ne\vect u)$ — their current density $\rho_-\vect u = -ne\vect u$ points against their drift, and that *is* the conventional current’s direction. (c) Both are static and uniform: every term vanishes. (d) Total: $(0, \vect\jmath)$ with $j = I/S = 10^{7}\,\mathrm{A}/\mathrm{m}^{2}$ — a pure current with no charge, the configuration that makes the wire’s relativity subtle.

**Exercise 6.3 ★.**

(a) From the matrix of $F^{\mu\nu}$, read off which components give $E_y$ and $B_x$. (b) Write $F^{\mu\nu}$ for a pure uniform field $\vect B = B\vect e_z$, and for the field of a plane wave ($\vect E = E\vect e_y$, $\vect B = (E/c)\vect e_z$). (c) Verify on components that $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu
A^\mu$ reproduces $\vect B = \operatorname{\vect{curl}}\vect A$ for $\mu\nu = 12$. (d) Why must $F$ be antisymmetric for the force $qF^{\mu\nu}u_\nu$ to do no work in the particle’s own frame?

**Solution of Exercise 6.3.**

(a) $E_y$: $F^{20} = E_y/c$; $B_x$: $F^{32} = B_x$. (b) For $\vect B
= B\vect e_z$, only $F^{12} = -B$ and $F^{21} = +B$; for the wave, $F^{20} = E/c$, $F^{12} = -E/c$ (and antisymmetric partners). (c) $F^{12} = \partial^1A^2 - \partial^2A^1 = -\partial_xA_y +
\partial_yA_x = -(\operatorname{\vect{curl}}\vect A)_z = -B_z$: matches the matrix. (d) The particle’s rate of energy change is $\propto F^{0\nu}u_\nu$; in the rest frame $u_\nu = (c, \vect 0)$ and antisymmetry makes $F^{00} = 0$: a force that can never work on a particle at rest — the defining property of the magnetic part.

**Exercise 6.4 ★.**

A parallel-plate capacitor at rest holds $\vect E = E_0\vect e_y$, no $\vect B$. Give $\vect E'$ and $\vect B'$ in a frame moving (a) along $\vect e_y$ (perpendicular to the plates); (b) along $\vect
e_x$ (parallel to the plates), and interpret the appearing $\vect B'$ as the field of the moving surface charges; (c) check the invariants in case (b); (d) in case (b), the plates also contract: which quantities ($\sigma$? $E'$? the plate separation?) change, and consistently with what?

**Solution of Exercise 6.4.**

(a) Boost along the field: longitudinal components unchanged, $\vect
E' = E_0\vect e_y$, $\vect B' = \vect 0$. (b) Boost along $\vect
e_x$: $E'_y = \gamma E_0$, $B'_z = -\gamma vE_0/c^2$ — the plates now stream as surface currents $\pm\sigma'v$, and two opposite current sheets enclose exactly such a field. (c) $E'^2 - c^2B'^2 =
\gamma^2E_0^2(1 - \beta^2) = E_0^2$; $\vect E'\cdot\vect B' = 0$: both preserved. (d) The plates contract along $x$, so $\sigma' =
\gamma\sigma$, consistent with $E' = \sigma'/\varepsilon_0 = \gamma
E_0$; the separation, transverse to the boost, is untouched.

**Exercise 6.5 ★★.**

Motional EMF unified. A conducting rod slides at $\vect v$ on rails across a uniform $\vect B$. (a) Lab account: which force drives the electrons along the rod, and what EMF results? (b) Rod-frame account: what field drives them, and where does it come from in [Proposition 6.5](#prop-b3-covariant-electromagnetism-transform)? (c) Show the two EMFs agree at order $v/c$. (d) Faraday’s flux rule of the Year 1 volume covered both “moving circuit” and “changing field” cases with one formula: what does relativity say about why that unification had to work?

**Solution of Exercise 6.5.**

(a) The magnetic force $-e\vect v\wedge\vect B$ pushes electrons along the rod: EMF $= vBL$. (b) In the rod’s frame the rod is at rest — no magnetic force on stationary charges — but the transformation delivers $\vect E' = \gamma\,\vect v\wedge\vect B$: an honest electric field does the driving. (c) $E'L = \gamma vBL
\approx vBL$ at order $v/c$. (d) It had to work because “motional” and “transformer” EMFs are one phenomenon read in two frames: the flux rule is covariance wearing 1831 clothes — Einstein’s 1905 paper opens with exactly this magnet-and-conductor asymmetry.

**Exercise 6.6 ★★.**

(a) Using [Proposition 6.5](#prop-b3-covariant-electromagnetism-transform), verify by direct computation that $E'^2 - c^2B'^2 = E^2 - c^2B^2$ for a boost along $x$. (b) Verify $\vect E'\cdot\vect B' = \vect
E\cdot\vect B$. (c) A region holds $\vect E$ and $\vect B$ perpendicular with $E = 2cB$: find the frame with a pure electric field (direction and speed). (d) Why can no frame make the field of a plane light wave purely electric or purely magnetic?

**Solution of Exercise 6.6.**

(a) The $x$ components are untouched; for the transverse ones, $E_y'^2 + E_z'^2 - c^2(B_y'^2 + B_z'^2) = \gamma^2[(E_y - vB_z)^2 +
(E_z + vB_y)^2 - c^2(B_y + vE_z/c^2)^2 - c^2(B_z - vE_y/c^2)^2]$; the cross terms cancel and the squares collect $\gamma^2(1 -
\beta^2) = 1$ times the untransformed combination. (b) Same bookkeeping on $E_xB_x + E_yB_y + E_zB_z$. (c) $E^2 - c^2B^2 =
3c^2B^2 > 0$: boost along $\vect E\wedge\vect B$ at $v = c^2B/E =
c/2$; there $B' = 0$ and the surviving field is $E' = E/\gamma =
\sqrt3\,cB$ — which squares to the invariant, as it must. (d) A light wave has both invariants zero: any frame must reproduce $E' = cB' \perp$, never a pure field.

**Exercise 6.7 ★★.**

Crossed fields. In the lab, $\vect E = E\vect e_y$ and $\vect B =
B\vect e_z$ with $E < cB$. (a) Show the frame moving at $\vect v_{\text{d}} = (E/B)\,\vect e_x$ sees a pure magnetic field. (b) Describe the motion of a charge released at rest, seen from that frame and back in the lab (a cycloid drifting at $v_{\text{d}}$). (c) The velocity filter of the Year 1 volume passed particles of speed $E/B$ undeflected: re-derive that in one line from (a). (d) What happens, qualitatively, when $E > cB$ — and why is the drift-frame trick then impossible?

**Solution of Exercise 6.7.**

(a) With $v_{\text{d}} = E/B$: $E'_y = \gamma(E - v_{\text{d}}B) =
0$; $B'_z = \gamma(B - v_{\text{d}}E/c^2) = B/\gamma$: pure, slightly weakened magnetic field. (b) There: a circle at $qB'/\gamma m$; back in the lab: that circle plus the uniform drift — a cycloid creeping at $E/B$ perpendicular to both fields, the trajectory of charges in a magnetron. (c) A particle moving at exactly $v_{\text{d}}$ is at rest in the drift frame, where the only field is magnetic and it feels nothing: undeflected. (d) For $E > cB$ the required drift exceeds $c$: no such frame; instead a pure-$E$ frame exists (previous exercise) and the charge is accelerated without bound along the field.

**Exercise 6.8 ★★.**

The four-wave-vector $k^\mu = (\omega/c, \vect k)$. (a) Show that requiring the phase to be invariant forces $k^\mu$ to transform as a [four-vector](#def-b3-covariant-electromagnetism-fourvector). (b) Derive the longitudinal Doppler formula from its time component. (c) Derive the aberration formula. (d) Starlight aberration: the Earth orbits at $29.8\,\mathrm{km}/\mathrm{s}$; through what angle does a star’s apparent position sweep over a year? (Bradley measured $20.5''$ in 1728 — the first direct proof that the Earth moves.)

**Solution of Exercise 6.8.**

(a) The phase $\omega t - \vect k\cdot\vect r$ counts crest-crossing events, an invariant number; it equals $k^\mu x_\mu$, and invariance of the product for all $x^\mu$ forces $k^\mu$ to transform four-vectorially. (b) $\omega'/c = \gamma(\omega/c -
\beta k_x)$ with $k_x = (\omega/c)\cos\theta$; for $\theta = 0$, $\omega' = \omega\sqrt{(1-\beta)/(1+\beta)}$. (c) $\cos\theta' =
k'_x/(\omega'/c)$: the stated formula. (d) $\beta = 10^{-4}$: $20.5''$; over a year the apparent position sweeps an ellipse of that angular semi-axis — Bradley’s aberration.

**Exercise 6.9 ★★.**

A charge $q$ moves at constant $\vect v = v\vect e_x$, $\gamma \gg
1$. (a) By transforming Coulomb’s field, show that in the transverse plane through the charge the field is boosted to $\gamma
q/4\pi\varepsilon_0b^2$ at distance $b$, while straight ahead and behind it is crushed by $1/\gamma^2$. (b) Show a stationary observer at distance $b$ sees a field pulse of duration $\sim b/\gamma v$. (c) For an LHC proton passing at $b = 1\,\mathrm{cm}$: peak field and pulse duration. (d) In what precise sense is this pulse “almost light”? (Check $E' \approx cB'$ and the invariants.)

**Solution of Exercise 6.9.**

(a) In the rest frame, Coulomb; boosting multiplies transverse components by $\gamma$ (at the transverse plane, $E'_\perp = \gamma
q/4\pi\varepsilon_0b^2$) while the field along the motion, evaluated ahead or behind at lab distance $r$, maps to a rest-frame distance $\gamma r$: reduced by $1/\gamma^2$. (b) The pancake of angular width $1/\gamma$ sweeps past at $v$: duration $\sim (b/\gamma)/v$. (c) $E' = 7250 \times 1.6 \times 10^{-19} \times 9 \times 10^{9}/10^{-4} \approx
0.1\,\mathrm{V}/\mathrm{m}$, lasting $2b/\gamma c \approx 9\,\mathrm{fs}$. (d) $B' = vE'/c^2 \approx E'/c$ and both invariants are $O(1/\gamma^2)$: locally indistinguishable from a light pulse — the basis of the “equivalent photon” description of fast-charge collisions.

**Exercise 6.10 ★★★.**

Chasing a light wave. A plane wave has $\vect E = E_0\vect e_y
\cos(\omega(t - x/c))$, $\vect B = (E_0/c)\vect e_z\cos(\omega(t -
x/c))$. An observer chases it at $\beta$. (a) Transform the fields: show $E'_0 = E_0\sqrt{(1-\beta)/(1+\beta)}$ and $B'_0 = E'_0/c$. (b) Show the frequency transforms by the same factor: the wave stays a wave, redshifted, with $E' = cB'$ always. (c) What becomes of the wave’s energy density (proportional to $E^2$) and of the number of photons? (d) Conclude: what would “riding alongside a light beam” require, and which invariant forbids it?

**Solution of Exercise 6.10.**

(a) $E'_y = \gamma(E - vB) = \gamma E_0(1 - \beta)\cos(\cdots) =
E_0\sqrt{(1-\beta)/(1+\beta)}\cos(\cdots)$, and $B'_z = E'_y/c$ by the same algebra. (b) The phase transforms with the same Doppler factor: same null wave, redder and weaker. (c) Energy density falls as the Doppler factor squared; the photon number is unchanged — each photon’s $h\nu$ carries the whole factor. (d) “Riding alongside” means the factor $\to 0$ with $\beta \to 1$: the wave never becomes static because its invariants are zero — there is no frame in which light stands still, which is where this book’s relativity began.

**Exercise 6.11 ★★★.**

Synchrotron light and the death of circular electron machines. An ultrarelativistic charge on a circle of radius $r$ radiates the power $P = q^2c\gamma^4/6\pi\varepsilon_0r^2$ (admitted — Larmor’s formula of the Year 2 volume, boosted). (a) Show the energy lost per turn is $\Delta E = q^2\gamma^4/3\varepsilon_0r$. (b) LEP: electrons at $100\,\mathrm{GeV}$ on $r = 3.1\,\mathrm{km}$: compute $\gamma$ and $\Delta E$ per turn — what fraction of the beam energy is re-bought every lap? (c) LHC: protons at $6.8\,\mathrm{TeV}$, same tunnel: compute $\Delta E$ per turn and compare. (d) Explain, with the $\gamma^4 =
(E/mc^2)^4$ factor, why the electrons’ successor is a *linear* collider or a much larger ring, and why proton rings survive.

**Solution of Exercise 6.11.**

(a) Per turn, $\Delta E = P \times 2\pi r/c$: the stated result. (b) $\gamma = 10^{11}/5.11 \times 10^{5} = 1.96 \times 10^{5}$: $\Delta E \approx
2.9\,\mathrm{GeV}$ per turn — three per cent of the beam energy re-injected every lap; LEP’s klystrons were the largest radio transmitter on Earth. (c) Protons: $\gamma = 7250$, $\gamma^4$ smaller by $(m_{\text{p}}/m_{\text{e}})^4 \approx 10^{13}$: about $5\,\mathrm{keV}$ per turn — negligible. (d) $\gamma^4 = (E/mc^2)^4$: at equal energy the electron radiates $10^{13}$ times more; hence linear colliders (radiate once, not per turn) or gigantic rings for electrons, while proton rings scale happily to $27\,\mathrm{km}$.

**Exercise 6.12 ★★★.**

The relativistic cyclotron. A charge in a uniform $\vect B$, at relativistic speed. (a) From $\dd\vect p/\dd t = q\vect v\wedge\vect
B$ with $E$ constant, show the orbit is a circle traversed at $\omega = qB/\gamma m$: the cyclotron frequency drops with energy. (b) A classical cyclotron pushes at fixed $\omega_0 = qB/m$: show that after $N$ turns with $\gamma - 1$ growing linearly to its final value $\Gamma$, the accumulated phase slip reaches a quarter RF period when $\Gamma \approx 1/2N$; for $N = 100$ turns, what proton kinetic energy is that, and how does it compare with the historical ceiling of classical cyclotrons (some $20\,\mathrm{MeV}$, bought with extra margin and voltage)? (c) Two cures exist: ramp the frequency (synchrocyclotron) or shape $B(r)$ to grow as $\gamma$ (isochronous cyclotron): explain each in one sentence. (d) The PSI isochronous cyclotron delivers $590\,\mathrm{MeV}$ protons: by what factor does its field at the rim exceed the central field?

**Solution of Exercise 6.12.**

(a) $|\vect p|$ is constant (no work); $\dot{\vect p} = q\vect
v\wedge\vect B$ turns $\vect p$ at rate $qvB/p = qB/\gamma m$. (b) Slip per turn $2\pi(\gamma - 1)$; with $\gamma - 1$ growing linearly to $\Gamma$, total slip $\approx \pi N\Gamma$; a quarter period is $\pi/2$: $\Gamma \approx 1/2N$. For $N = 100$: $\Gamma =
5 \times 10^{-3}$, $E_k \approx 4.7\,\mathrm{MeV}$ — the right scale; real machines stretched it to $\sim20\,\mathrm{MeV}$ with high dee voltages (fewer turns). (c) Synchrocyclotron: sweep the RF downward during each pulse to follow $qB/\gamma m$. Isochronous cyclotron: let $B(r)$ grow $\propto\gamma(r)$ so the ratio never changes and the beam stays continuous. (d) $\gamma = 1 + 590/938 = 1.63$: the rim field must exceed the central field by that factor.

## 6.6 Problem: Magnetism at a snail’s pace

**Problem 6.1.**

Weekend problem — how a $10^{-24}$ imbalance runs every motor

Electrons drift through a lamp cord more slowly than honey creeps, and $v^2/c^2$ for that drift is $10^{-24}$ — yet the magnetic force it produces lifts cars in scrapyards. This problem does the full two-frame accounting for a straight wire and a moving charge, and finds relativity hiding in every electromagnet. Data: copper wire, section $S = 1.0\,\mathrm{mm}^{2}$, current $I = 10\,\mathrm{A}$, conduction-electron density $n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}$; test charge $q = 1\,\text{µ}\mathrm{C}$ at distance $d = 1.0\,\mathrm{cm}$, moving parallel to the wire at $v = 10^{5}\,\mathrm{m}/\mathrm{s}$ (an ion-beam speed, to keep numbers visible); $\mu_0 = 4\pi\times10^{-7}\,\mathrm{H}/\mathrm{m}$.

**Part I — The lab-frame account.**

1. Compute the electrons’ drift speed $u$ , and $u^2/c^2$ .
2. Model the wire as two superposed line charges: the lattice $+\lambda$ at rest and the electrons $-\lambda$ drifting at $u$ , with $I = \lambda u$ . Compute $\lambda$ .
3. Compute $B$ at the charge’s position and the magnetic force $F = qvB$ on it — direction included, for $v$ parallel to the conventional current $I$ (recall: parallel currents attract).
4. Why is there no *electric* force in the lab?
5. The same wire’s electric field if it carried a net charge of just one electron in excess per metre: compare the force it would exert on $q$ with the magnetic force of question 3, and conclude how precisely “neutral” must be measured before magnetism can be attributed.
6. A $1\,\text{µ}\mathrm{C}$ charge at $10^{5}\,\mathrm{m}/\mathrm{s}$ : is the force of question 3 measurable? (Compare with the weight of a grain of sand, $\sim10^{-6}\,\mathrm{N}$ .)

**Part II — Changing seats.** Boost to the frame of the test charge (speed $v$ along the wire).

7. In that frame, what are the velocities of the lattice and of the electrons (which drift opposite to $I$ , hence gain speed in this boost — compose them properly)?
8. Integrated over the wire’s section, the four-current per unit length is $(c\lambda_{\text{tot}}, I, 0, 0)$ with $\lambda_{\text{tot}} = 0$ here. Transform it and show the wire’s line charge in the new frame is $$\lambda' = -\gamma_v\,\frac{vI}{c^2} .$$
9. Evaluate $\lambda'$ , in coulombs per metre and in electron charges per metre. Which stream got denser, and why (contract each stream separately if you prefer)?
10. Compute the electric field $E' = \lambda'/2\pi\varepsilon_0d$ at the charge, and the electric force $qE'$ on it.
11. Compare $qE'$ with the lab’s $qvB$ from Part I, and explain the residual factor $\gamma_v \approx 1 + 5.6 \times 10^{-8}$ (which [four-vector](#def-b3-covariant-electromagnetism-fourvector) ’s transformation law relates the forces?).
12. In this frame there is also a magnetic field (the wire still carries current): why does it exert no force here?

**Part III — The moral of the numbers.**

13. The relative imbalance $|\lambda'|/\lambda$ : compute it and write it as $\gamma_v vu/c^2$ .
14. How can an effect of order $10^{-19}$ of the wire’s charge produce a macroscopic force? (What enormous number does it multiply?)
15. Two parallel wires with $10\,\mathrm{A}$ each at $1\,\mathrm{cm}$ : compute the force per metre, and check it against the definition-grade value $2 \times 10^{-7}\,I_1I_2/d$ newtons per metre.
16. An electromagnet is ten thousand turns of this story: estimate the field of a $10^{4}\,$ -turn coil per metre carrying $10\,\mathrm{A}$ (solenoid formula of the Year 1 volume), and the force per square centimetre it exerts on iron ( $\sim B^2/2\mu_0$ ).
17. If relativity were switched off ( $c \to \infty$ in the transformation), what would remain of $\lambda'$ , of $B$ ’s force, of motors and scrapyard magnets?
18. Why does the test charge at *rest* near the wire feel nothing, even though the electrons stream past it? (Which cancellation protects it, and to what accuracy?)

**Part IV — Beyond the wire.**

19. The Year 2 volume derived magnetic fields from Ampère’s law with currents as given sources. What does this chapter add to that account — what *is* the magnetic field, seen from this problem?
20. Iron magnets have no battery: what plays the role of the current in permanent magnetism (one sentence; the honest microscopic answer is quantum and waits five chapters)?
21. A single electron beam in vacuum (no lattice): does a co-moving observer see it attract or repel itself more than a lab observer does? Reconcile the two frames’ accounts of a beam’s self-pinching.
22. Make the last point quantitative: show that two parallel like-charged beams moving together at $\beta$ repel with a net force (electric repulsion minus magnetic attraction) reduced by exactly $1/\gamma^2$ from its rest value — and say which frame’s account makes the factor obvious.
23. The energy for a lamp arrives at nearly $c$ while its electrons drift a metre per hour: what actually carries the energy along the cord (recall the Poynting vector of the Year 2 volume)?
24. Estimate $vu/c^2$ for the electrons of this problem and a pedestrian test charge ( $v = 1\,\mathrm{m}/\mathrm{s}$ ): even there, the force is first-order measurable with a compass needle — who demonstrated current deflecting a compass, and in what year?
25. Summarise the named result: a neutral $10\,\mathrm{A}$ wire, watched from a seat moving at $10^{5}\,\mathrm{m}/\mathrm{s}$ , carries $-10^{-11}\,\mathrm{C}/\mathrm{m}$ of relativistic charge — and that $10^{-24}$ -order bookkeeping error of [length contraction](https://one-course.com/books/physics/5/en/chapter/4-relativistic-kinematics#prop-b3-relativistic-kinematics-contraction) , multiplied by $10^{28}$ electrons, is the entire magnetic force of the lab.

**Solution of Problem 6.1.**

**1.** $u = I/nSe = 7.4 \times 10^{-4}\,\mathrm{m}/\mathrm{s}$; $u^2/c^2 =
6 \times 10^{-24}$. **2.** $\lambda = I/u = nSe = 1.36 \times 10^{4}\,\mathrm{C}/\mathrm{m}$ — fourteen kilocoulombs per metre in each stream. **3.** $B = \mu_0I/2\pi d = 2.0 \times 10^{-4}\,\mathrm{T}$; $F = qvB =
10^{-6} \times 10^{5} \times 2 \times 10^{-4} = 2 \times 10^{-5}\,\mathrm{N}$, directed *toward* the wire (parallel currents attract). **4.** The two line charges cancel exactly: $\lambda_{\text{tot}}
= 0$, no field to zeroth order. **5.** One excess electron per metre: $E = 2k\lambda_1/d =
2.9 \times 10^{-7}\,\mathrm{V}/\mathrm{m}$, force $2.9 \times 10^{-13}\,\mathrm{N}$ — $10^8$ times smaller than the magnetic force. The wire could hide a hundred million stray electrons per metre before electrostatics rivalled magnetism: attributing the force to $\vect B$ is safe. **6.** $2 \times 10^{-5}\,\mathrm{N}$ is twenty sand-grain weights: easily measurable. **7.** Lattice: $-v$. Electrons (drifting at $u$ opposite $I$, i.e. opposite the boost): speed $(u + v)/(1 + uv/c^2)$ — faster than the lattice. **8.** $\lambda' = \gamma_v(\lambda_{\text{tot}} - vI/c^2) =
-\gamma_v vI/c^2$. **9.** $\lambda' = -10^{5} \times 10/9 \times 10^{16} =
-1.1 \times 10^{-11}\,\mathrm{C}/\mathrm{m}$: about $7 \times 10^7$ excess electrons per metre. The electron stream, faster in this frame, is the denser one — each stream contracts by its own $\gamma$. **10.** $E' = 2k|\lambda'|/d = 2 \times 9 \times 10^{9} \times
1.1 \times 10^{-11}/0.01 = 20\,\mathrm{V}/\mathrm{m}$, pointing at the wire; force $qE' = 2 \times 10^{-5}\,\mathrm{N}$, attractive. **11.** Identical to $qvB$ up to the factor $\gamma_v = 1 +
5.6 \times 10^{-8}$: exactly the transformation law of a transverse force (the four-force), $F_{\text{rest}} = \gamma F_{\text{lab}}$. **12.** The wire still carries a (larger) current, hence $\vect
B' \neq 0$ — but our charge is at rest here, and a magnetic field grips only moving charges. **13.** $|\lambda'|/\lambda = \gamma_v vu/c^2 = 8.2 \times 10^{-16}$. **14.** It multiplies $\lambda/e \approx 8.5 \times 10^{22}$ elementary charges per metre: $8.2 \times 10^{-16} \times 8.5 \times 10^{22}
\approx 7 \times 10^{7}$ electrons per metre — macroscopic. Matter is so enormously charged that its neutrality is a razor’s edge; relativity tips the razor. **15.** $F/L = \mu_0I_1I_2/2\pi d = 2 \times 10^{-7} \times 100/0.01 =
2 \times 10^{-3}\,\mathrm{N}/\mathrm{m}$ — the formula that once *defined* the ampere. **16.** $B = \mu_0nI = 4\pi\times10^{-7} \times 10^{4} \times
10 = 0.13\,\mathrm{T}$; magnetic pressure $B^2/2\mu_0 \approx
6300\,\mathrm{Pa} \approx 0.6\,\mathrm{N}/\mathrm{cm}^{2}$: whole cars hang on integrated relativity. **17.** $c \to \infty$ kills $\lambda'$, and with it the force, the field’s magnetic part, motors, dynamos and scrapyard cranes: magnetism has no non-relativistic existence. **18.** At rest the charge sees the lab’s neutral wire: the two streams cancel to whatever precision matter is neutral — known experimentally to fantastic accuracy — and no force acts. **19.** That the “given sources” picture was one frame’s slice of a single tensor: $\vect B$ is the piece of the electromagnetic field that a given observer’s motion assigns to currents rather than charges. **20.** Electron spin: each electron is an elementary magnet (an intrinsic, quantum “current”), and iron is matter in which these align — [Chapter 12](https://one-course.com/books/physics/5/en/chapter/12-spin-and-two-level-systems#ch-b3-spin-two-level) and [Chapter 22](https://one-course.com/books/physics/5/en/chapter/22-electromagnetism-in-matter#ch-b3-electromagnetism-in-matter) take this up. **21.** In the beam’s rest frame the repulsion is pure Coulomb and *maximal*; in the lab, the parallel currents’ magnetic attraction nearly cancels it. No contradiction: the lab’s transverse force is the rest frame’s divided by $\gamma$, and slower dynamics ([time dilation](https://one-course.com/books/physics/5/en/chapter/4-relativistic-kinematics#prop-b3-relativistic-kinematics-dilation)) completes the account. **22.** Net lab force $= qE_{\text{lab}} - qvB_{\text{lab}} =
(1 - \beta^2)qE_{\text{lab}} = F_{\text{Coulomb}}/\gamma^2$ per the transformation; in the rest frame the factor is transparent: pure electrostatics, with the blow-up watched through dilated time. **23.** The Poynting vector: energy streams through the *fields* around the conductors at nearly $c$, the electrons merely marshalling it; the cord’s copper is a guide, not a pipe. **24.** $vu/c^2 \approx 8 \times 10^{-21}$ — and yet Ørsted saw his compass swing beside a wire in 1820: the multiplier of $10^{22}$ charges was already at work a century before anyone could name it. **25.** A neutral $10\,\mathrm{A}$ wire, viewed from $10^{5}\,\mathrm{m}/\mathrm{s}$, carries $-1.1 \times 10^{-11}\,\mathrm{C}/\mathrm{m}$: [length contraction](https://one-course.com/books/physics/5/en/chapter/4-relativistic-kinematics#prop-b3-relativistic-kinematics-contraction)’s $10^{-16}$ bookkeeping difference, multiplied by $10^{23}$ charges per metre, *is* the magnetic force — all of magnetism is relativity audited at a snail’s pace.