Part D · Minkowski · spine

Electromagnetism was relativistic all along: the field tensor

E and B are not two fields but two blocks of one antisymmetric tensor, and a boost stirs them into each other — which is why a neutral wire pulls on a charge moving beside it.

needs 27, 25~2½ h slides + 5 h problemsproblems: ./mr problems 28

Unit 25 found the crisis: Maxwell's equations are not Galilean-covariant, and the substitution that repairs them is the Lorentz boost. Unit 26 built Minkowski space and unit 27 put particles on it. What has not yet been done is the obvious thing — take Maxwell's equations, which were relativistic before anyone knew what that meant, and write them in objects the Lorentz group acts on. That is this unit.

The mathematics is one object: an antisymmetric $(0,2)$-tensor field $F$ on Minkowski space. Six independent components, which in any chosen inertial frame split three-and-three into what unit 23 called $\vec E$ and $\vec B$. The split is frame-dependent; $F$ is not. Everything else in the unit is bookkeeping with that object: raising and lowering with $\eta$, contracting with the four-velocity, taking a divergence, dualising with the Levi-Civita tensor, and building the stress-energy tensor out of it quadratically.

The physics is Purcell's: magnetism is electrostatics plus length contraction. The drift speed of conduction electrons is about $10^{-13}c$, so the relativistic correction to Coulomb's law is a part in $10^{26}$ — but a metre of copper carries $1.4\times10^{4}$ C of mobile charge, and the product is a force you can measure with a compass.

How to read this deck

One experiment runs through all five sections: a current-carrying wire and a charge moving parallel to it, done twice, in the lab and in the charge's rest frame, with every number carried to the digit where the two answers meet. Section 1 builds $F$ and shows its antisymmetry is forced. Section 2 derives the transformation rules and does the wire both ways. Section 3 writes Maxwell's four equations as two, and extracts the two invariants. Section 4 builds $T^{\mu\nu}$ and recovers Poynting's theorem from a single divergence. Section 5 is radiation, including the part classical physics never fixed.

Conventions, unchanged from unit 27: $x^\mu = (ct, x, y, z)$, $\eta = \mathrm{diag}(-1,1,1,1)$ in an inertial basis, Greek indices $0$ to $3$, Latin $1$ to $3$, Einstein summation as in unit 16. SI units throughout, with $\varepsilon_0\mu_0 = 1/c^2$. Proofs are on the slides where they are short and in these notes under Full proof where they are not; results imported from elsewhere are marked cited.

28 · overview

One wire, five sections

the experiment

A neutral wire carries $I = 1.00$ A. A charge $q = 1.00$ nC, $r = 2.00$ cm away, moves parallel to it at $v = 3.00\times10^{6}\ \mathrm{m\,s^{-1}}$.

In the lab the force is magnetic, $q\vec v\times\vec B$. In the charge's own frame the charge is at rest, so there is no magnetic force — yet it is still pulled in. What supplies the force?

where the force comes from, in four steps

1. Lab: the wire is neutral, so $\vec E = 0$; Ampère gives $B = 10.0\ \mu$T and a force $qvB = 3.000\times10^{-8}$ N towards the wire.

2. Charge's frame: lattice and electrons move at different speeds, so their charge per metre contracts by different factors. The wire carries $\lambda' = -\gamma Iv/c^2$.

3. Gauss turns $\lambda'$ into a field $E'$ at the charge; $qE'$ is the force, larger than step 1's by exactly $\gamma$.

4. That factor is how force transforms, $F_\perp = F'_\perp/\gamma$: one effect, two descriptions.

The plan. 1 one field, forced to be antisymmetric · 2 boosting, and the wire done twice · 3 Maxwell as two equations; two invariants · 4 $T^{\mu\nu}$, and Poynting · 5 radiation, and the runaway nobody fixed.

getting around

Words with a dotted underline are glossary terms: click one for its definition. G lists every term in the unit, N shows the notes, where the longer proofs live. A solid underline, like the wire's invariants, is a piece of the running example, shown on hover.

The answer to "something has to supply that force" is given away by the section titles, but it is worth sitting with the puzzle for a minute before reading on. The charge is at rest in its own frame. A force on a charge at rest is, by definition, $q\vec E$. So the wire — which is electrically neutral in the lab, to an accuracy no experiment has ever challenged — must carry a net charge in the charge's frame. Charge is not what is frame-dependent here: the total charge on a closed system is a Lorentz invariant (that is an experimental fact about atoms, and unit 34 will make it a theorem about a closed 3-form). What is frame-dependent is charge per unit length, because length is.

The four steps, and where each is done. Step 1 is Worked example 2 — the wire in the lab, with a drawing of the directions. Step 2 is Worked example 3 — the same wire, from the charge, which draws the two rows of charges in both frames (exaggerated, since the real imbalance is a part in $10^{15}$). Step 3 is its continuation, Worked example 3, continued — the net charge and its field. Step 4 is The two answers, to the last $\gamma$, which also gets the same number a third way, from the field transformation rules with no charge densities at all. The size of the effect — why something a part in $10^{15}$ is a force you can feel — is Purcell's moral.

Five sections, but the load is unevenly spread. Sections 1 and 3 are notation that has to become automatic: by unit 32 you will be moving indices around without thinking, and the practice starts here. Section 2 is the physics. Section 4 is the one that units 29 and 35 build directly on. Section 5 is short and ends in an admission of failure, which is rarer in a physics course than it ought to be.

section 1

One field, not two

What kind of object can push a charge around without changing its mass?

28 · one field

First, what a force is in spacetime

definition · four-force

For a worldline with four-momentum $p$, the four-force is $f = \mathrm{d}p/\mathrm{d}\tau$, a vector in $V$ at each event.

In a frame, $\mathrm{d}\tau = \mathrm{d}t/\gamma$ and $p = (E/c,\ \vec p)$, so

$$f^\mu = \gamma\Big(\tfrac1c\vec F\cdot\vec v,\ \ \vec F\Big),\qquad \vec F = \frac{\mathrm{d}\vec p}{\mathrm{d}t}.$$

Newton's three-force is the spatial part, divided by $\gamma$. The time part is the power, also divided by $\gamma$ — one object, not two.

constraint

For a particle of fixed mass, $\eta(p,p) = -m^2c^2$ is constant, so

$$\eta(f, u) = 0 .$$

One line: differentiate $\eta(p,p)$ along the worldline and use $p = mu$.

A four-force is never free: it must be $\eta$-orthogonal to the four-velocity at every event, or the particle's mass changes.

The one line, in full. $\frac{\mathrm d}{\mathrm d\tau}\eta(p,p) = 2\eta(\mathrm{d}p/\mathrm{d}\tau, p) = 2m\,\eta(f,u)$, using bilinearity and symmetry of $\eta$ exactly as one differentiates $|\vec r|^2$ in unit 03. If $m$ is constant the left side is zero, so $\eta(f,u) = 0$.

Where the $\gamma$s come from. $f^i = \mathrm{d}p^i/\mathrm{d}\tau = (\mathrm{d}t/\mathrm{d}\tau)\,\mathrm{d}p^i/\mathrm{d}t = \gamma F^i$; and $f^0 = \mathrm{d}(E/c)/\mathrm{d}\tau = \gamma\,\dot E/c = \gamma\,\vec F\cdot\vec v/c$, using the work–energy relation $\dot E = \vec F\cdot\vec v$, which unit 27 derived from the action. Check the constraint in components: $\eta(f,u) = -f^0u^0 + f^iu^i = -\gamma\frac{\vec F\cdot\vec v}{c}\gamma c + \gamma F^i\gamma v^i = 0$ ✓.

Dimensions. $[f] = [p]/[\tau] = \mathrm{kg\,m\,s^{-1}}/\mathrm{s} = \mathrm N$, and $[f^0] = [\vec F\cdot\vec v]/c = \mathrm{W/(m\,s^{-1})} = \mathrm N$ ✓ — all four components carry the same dimension, which is what lets them be components of one vector.

Warning. A force that changes the rest mass — a rocket burning fuel, a body being heated — is not covered by this constraint. Then $\eta(f,u) = -c^2\,\mathrm{d}m/\mathrm{d}\tau$, and the deficit is exactly the rest energy being added or removed. Unit 27's relativistic rocket is the worked case.

28 · one field

The constraint, drawn: $f$ is $u$ reflected in the light line

On a Minkowski diagram $\eta$-orthogonal is not a right angle. At each event of a worldline, $f$ is $u$ reflected in the light line through that event.

In one space dimension, $u = \gamma(c, v)$, and

$$\eta(f,u) = \gamma\big(-cf^0 + vf^1\big) = 0$$ $$\iff\quad \frac{f^0}{f^1} = \frac vc .$$

$u$ leans from the time axis by slope $v/c$; $f$ leans from the space axis by the same slope. Mirror images in the $45^\circ$ line.

two readings

At rest ($u$ vertical), $f$ is horizontal: $f^0 = 0$, a pure push, no power — the rest energy $mc^2$ cannot change.

As $v\to c$, $u$ and $f$ fold onto the light line together; neither ever crosses it.

28 · one field

The one guess, and what it forces

Unit 23's Lorentz force $\vec F = q(\vec E + \vec v\times\vec B)$ is linear in the velocity. Make the mildest possible relativistic guess: the four-force is linear in the four-velocity, with coefficients built from the field at that event and the charge,

$$f^\mu \;=\; q\,F^\mu{}_\nu\,u^\nu ,$$

for some array $F^\mu{}_\nu(x)$ of 16 numbers per event. Now impose the constraint $\eta(f,u)=0$ for every worldline through the event. Writing $F_{\mu\nu} := \eta_{\mu\alpha}F^\alpha{}_\nu$, the constraint reads $q\,F_{\mu\nu}u^\mu u^\nu = 0$.

theorem · antisymmetry is not assumed

If $F_{\mu\nu}u^\mu u^\nu = 0$ for every future-pointing timelike $u$ with $\eta(u,u)=-c^2$, then $F_{\mu\nu} = -F_{\nu\mu}$.

proof

Let $S_{\mu\nu} := \tfrac12(F_{\mu\nu} + F_{\nu\mu})$, the symmetric part; only it survives contraction with $u^\mu u^\nu$. $Q(w) := S_{\mu\nu}w^\mu w^\nu$ is a quadratic form vanishing on the open cone $\mathcal T$ (scaling $u$ scales $Q$ by $\lambda^2$, and every $w\in\mathcal T$ is some $\lambda u$). A polynomial vanishing on a nonempty open set vanishes identically, so $Q\equiv0$; polarisation gives $S = 0$. $\square$

A 2-d slice of $V$. $Q = 0$ on the hyperbola, hence on every ray through it, hence on the open shaded cone $\mathcal T$ — and a quadratic that vanishes on an open set vanishes everywhere.

Full proof, slowly. Write $F_{\mu\nu} = \eta_{\mu\alpha}F^\alpha{}_\nu$ and split it into symmetric and antisymmetric parts, $F_{\mu\nu} = F_{(\mu\nu)} + F_{[\mu\nu]}$, as unit 16 does for any bilinear map. The antisymmetric part contracts to zero with $u^\mu u^\nu$ automatically. So the hypothesis says $S(u,u) = 0$ for every unit-timelike $u$, where $S = F_{(\mu\nu)}$ is symmetric.

The set $\mathcal T$ of future-pointing timelike vectors is open in $V$ (it is the preimage of an open set under the continuous maps $w\mapsto\eta(w,w)$ and $w\mapsto w^0$). Every $w\in\mathcal T$ is $\lambda u$ for a unique $\lambda>0$ and a unit-timelike $u$, and $S(w,w) = \lambda^2S(u,u) = 0$. So the quadratic polynomial $w\mapsto S(w,w)$ in the four coordinates of $w$ vanishes on the open set $\mathcal T$. A polynomial vanishing on a nonempty open subset of $\mathbb R^4$ is the zero polynomial — differentiate it enough times at an interior point, or note that its restriction to any line through that point is a one-variable polynomial with infinitely many roots. Hence $S(w,w) = 0$ for all $w\in V$, and polarisation, $S(w,z) = \tfrac12[S(w+z,w+z) - S(w,w) - S(z,z)]$, gives $S \equiv 0$. $\square$

What the figure is for. The hypothesis is only about the hyperbola — physical four-velocities, all with $\eta(u,u) = -c^2$ — which is a thin set. The proof needs an open set, and gets it for free from homogeneity: a quadratic form that vanishes at $u$ vanishes along the whole ray through $u$. The rays through the hyperbola sweep out the shaded cone, which has interior. The slice shows two of the four dimensions; in $V$ the hyperbola is a three-dimensional hyperboloid and the cone is a solid cone, and the argument is word for word the same.

Why this matters more than it looks. Nobody put antisymmetry in. It came out of two things: that the four-force is linear in the four-velocity, and that mass is constant. The six independent components of an antisymmetric $4\times4$ array are then exactly the number of components in $(\vec E, \vec B)$ — three and three. That coincidence is the whole unit in one sentence.

The assumption that is doing work. Linearity in $u$ is a guess, not a theorem. It is the simplest thing consistent with the known low-speed law, and it is right, but a velocity-dependent force of a different shape is not logically excluded by relativity alone. What relativity excludes, given linearity, is a symmetric part.

28 · one field

The field tensor, defined

definition · the electromagnetic field tensor

A smooth assignment, to each event $x$ of Minkowski space, of an antisymmetric bilinear form $$F_x : V\times V \to \mathbb R,\qquad F_x(w,z) = -F_x(z,w),$$ that is, an antisymmetric $(0,2)$-tensor field. Its value on a pair of vectors is a number every frame agrees on.

This is unit 16's definition of a tensor, with $\mathbb R^3$ replaced by $V$ and the Euclidean inner product replaced by $\eta$. A tensor field on Minkowski space is just a tensor attached to every event, varying smoothly.

The physical content is one equation:

$$\boxed{\ \frac{\mathrm{d}p^\mu}{\mathrm{d}\tau} \;=\; q\,F^\mu{}_\nu\,u^\nu\ }$$

what it is not

It is not "six numbers that transform in a certain way". The transformation law is a theorem about the components in different bases, proved at the start of the next section. The object is the bilinear map. Unit 16 made this point for the inertia tensor; here, for the first time, it stops being a matter of taste, because $F_{\mu\nu}$ and $F^{\mu\nu}$ have different signs in half their entries.

Dimensions. $[f] = \mathrm N$ and $[q u] = \mathrm{C\,m\,s^{-1}}$, so $[F^\mu{}_\nu] = \mathrm{N\,s\,C^{-1}m^{-1}} = \mathrm{V\,s\,m^{-2}} = \mathrm T$. The field tensor carries the dimension of magnetic field; the electric components will therefore appear divided by $c$, and they do.

Smoothness. "Smooth" is more than is needed and less than is true. The derivations below use $C^2$, so that mixed partials commute by Clairaut (unit 03). Real fields of interest — a point charge's, a wave front's — are not $C^2$ everywhere, and unit 20's distributions are the repair; nothing in this unit depends on the repair, but the boundary of a charged conductor is a place where the honest object is a distribution and the jump conditions of unit 23 are its content.

An aside on names. "Field strength" and "field tensor" mean the same thing here; in gauge theory $F$ is the curvature of a connection, and unit 34 will make that literal. The name "Faraday tensor" is also used, and MTW's Gravitation writes it $\mathbf F$ and treats it as a 2-form from the start.

28 · one field

Its six components

Fix an inertial basis. $F$ has $16$ components $F_{\mu\nu} = F(e_\mu, e_\nu)$, antisymmetry leaves six, and the six are named:

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

Compactly: $F_{0i} = -E_i/c$ and $F_{ij} = \varepsilon_{ijk}B_k$, with unit 14's Levi-Civita symbol.

this is a naming, not a derivation

The assignment is fixed by demanding that $q F^\mu{}_\nu u^\nu$ reproduce $q(\vec E + \vec v\times\vec B)$ — which is Worked example 1, after the slides on raising indices. Until then, read it as six labels.

count

$\binom42 = 6$ independent entries. Three with a time index, three without. That is exactly $\vec E$ and $\vec B$, and it is why there are two fields in three dimensions and one in four.

Read the matrix carefully. The row index is $\mu$, the column index $\nu$, both running $0,1,2,3$. Antisymmetry is visible: the transpose is the negative. The $3\times3$ lower-right block is $\varepsilon_{ijk}B_k$: $F_{12} = \varepsilon_{123}B_3 = B_z$, $F_{13} = \varepsilon_{132}B_2 = -B_y$, $F_{23} = \varepsilon_{231}B_1 = B_x$ ✓.

Why $E/c$ and not $E$. Because $x^0 = ct$ carries a $c$, and because $[F] = \mathrm T$ as computed on the previous slide, while $[E] = \mathrm{V\,m^{-1}} = \mathrm{T\,m\,s^{-1}}$. Dividing by $c$ makes the dimensions of every entry agree, which they must: a tensor's components in an orthonormal basis all carry the same units. In Gaussian units $E$ and $B$ already share dimensions and no $c$ appears — one of the few places where Gaussian is genuinely tidier, and the reason older books look different here.

On conventions. Books differ by signs in this matrix, according to their signature ($-+++$ versus $+---$) and their sign for $F = \partial A - \partial A$. This deck follows Carroll's Spacetime and Geometry §1.8–1.9, which matches MTW; Jackson uses $+---$ and gets $F^{0i} = -E_i/c$. Nothing physical turns on it, but comparing two books without checking their conventions wastes an afternoon. Whenever you meet a new source, check one entry — the sign of $F_{0i}$ — and you can read the rest.

28 · one field

Raising and lowering, and why it now matters

$\eta$ is a nondegenerate bilinear form, so $w \mapsto \eta(w,\cdot)$ is an isomorphism $V\to V^*$ (unit 16's dual space). In components that is index lowering:

$$w_\mu = \eta_{\mu\nu}w^\nu,\qquad w^\mu = \eta^{\mu\nu}w_\nu,$$

with $\eta^{\mu\nu}$ the inverse matrix, numerically $\mathrm{diag}(-1,1,1,1)$ again. For a $(0,2)$-tensor, raise one index or both:

$$F^\mu{}_\nu = \eta^{\mu\alpha}F_{\alpha\nu},\qquad F^{\mu\nu} = \eta^{\mu\alpha}\eta^{\nu\beta}F_{\alpha\beta}.$$

in Euclidean space you could be sloppy. Not here

$\eta^{00} = -1$, so raising a $0$ index flips the sign:

$$F^{0i} = +E_i/c,\qquad F_{0i} = -E_i/c,$$

while $F^{ij} = F_{ij}$. In unit 16 the metric was $\delta_{ij}$ and up and down were the same numbers; that is why the distinction could be planted quietly in unit 04 and left. It cannot be left now.

Why the isomorphism is the right way to see it. A vector and a covector are different animals: one is an element of $V$, the other a linear functional on it. Nothing identifies them until a nondegenerate bilinear form is chosen, and then $w\mapsto\eta(w,\cdot)$ does it. Nondegeneracy is exactly what makes it injective, hence (equal dimensions) an isomorphism. The inverse map is "raising". This is the same construction as unit 04's gradient: $\mathrm{d}f$ is a covector always, and $\nabla f$ is its image under the inner product's isomorphism — which is why the gradient depends on the inner product and the differential does not.

The arithmetic, once. $F^{0i} = \eta^{0\alpha}F_{\alpha}{}^{i}$ — no, do it in one step from the fully-lowered components: $F^{0i} = \eta^{0\alpha}\eta^{i\beta}F_{\alpha\beta} = \eta^{00}\eta^{ii}F_{0i} = (-1)(+1)(-E_i/c) = +E_i/c$ (no sum on the repeated $i$ in $\eta^{ii}$ here; it is diagonal). Similarly $F^{ij} = \eta^{ii}\eta^{jj}F_{ij} = F_{ij}$, and for the mixed form $F^i{}_0 = \eta^{ii}F_{i0} = +E_i/c$ while $F^0{}_i = \eta^{00}F_{0i} = +E_i/c$ as well — equal here, but for a general tensor they would not be.

Abstract index notation. From here on, an expression like $F^{\mu\alpha}F^\nu{}_\alpha$ is read as a tensor, with $\mu,\nu$ naming its slots rather than denoting numbers; the repeated $\alpha$ is a contraction (unit 16), which is a basis-independent operation. The rules are three: every index appears once up and once down when summed; free indices match on both sides of an equation; and you may raise or lower any index on both sides at will. Fluency in this is the actual prerequisite for units 32 to 37, and it is bought only by doing the problem sheet.

28 · one field

Which entries flip: the sign pattern of $F^{\mu\nu}$

Raising both indices multiplies row $0$ by $\eta^{00} = -1$ and column $0$ by $-1$. The $(0,0)$ entry would be flipped twice, but it is zero; so exactly the six $\vec E$ entries change sign and the $\vec B$ block does not.

In matrices, $F^{\mu\nu} = \eta^{\mu\alpha}F_{\alpha\beta}\eta^{\beta\nu}$ is $\eta\,F_{\text{low}}\,\eta$: a diagonal matrix on the left scales rows, on the right scales columns.

Raise only the first index and only row $0$ flips: $F^\mu{}_\nu = \eta F_{\text{low}}$. Index position also fixes order: $F^\mu{}_\nu \ne F_\nu{}^\mu$, they differ by a sign.

Reading the figure. In matrix language $F^{\mu\nu} = \eta^{\mu\alpha}F_{\alpha\beta}\eta^{\beta\nu}$ is the product $\eta\,F_{\text{low}}\,\eta$ with $\eta = \mathrm{diag}(-1,1,1,1)$. Multiplying on the left by a diagonal matrix scales rows; on the right, columns. So the rule is mechanical: every entry with exactly one $0$ index flips. The same picture tells you what the mixed forms do — $F^\mu{}_\nu = \eta F_{\text{low}}$ flips only row $0$ — which is the whole content of the "ingredients" line in Worked example 1.

Check the last claim. $F_\nu{}^\mu = F_{\nu\alpha}\eta^{\alpha\mu}$, and antisymmetry gives $F_{\nu\alpha} = -F_{\alpha\nu}$, so $F_\nu{}^\mu = -\eta^{\mu\alpha}F_{\alpha\nu} = -F^\mu{}_\nu$ ✓. With a symmetric tensor the two would be equal, which is why the inertia tensor of unit 16 never forced the distinction.

28 · one field

Worked example 1 — the Lorentz force falls out

example · unpack $q F^\mu{}_\nu u^\nu$

Ingredients. $u^\nu = \gamma(c, \vec v)$ from unit 27, and the mixed components $F^\mu{}_\nu = \eta^{\mu\alpha}F_{\alpha\nu}$: $$F^i{}_0 = F_{i0} = \frac{E_i}{c},\quad F^i{}_j = F_{ij} = \varepsilon_{ijk}B_k,\quad F^0{}_j = -F_{0j} = \frac{E_j}{c},\quad F^0{}_0 = 0 .$$

Spatial part. $qF^i{}_\nu u^\nu = q\big(F^i{}_0u^0 + F^i{}_ju^j\big) = q\big(\tfrac{E_i}{c}\,\gamma c + \varepsilon_{ijk}B_k\,\gamma v^j\big) = \gamma q\big(E_i + \varepsilon_{ijk}v^jB_k\big) = \gamma q(\vec E + \vec v\times\vec B)_i .$

And $f^i = \gamma F^i_{\text{3-force}}$, so the $\gamma$s cancel: $\ \dfrac{\mathrm{d}\vec p}{\mathrm{d}t} = q(\vec E + \vec v\times\vec B)$ — unit 23's law, recovered.

Time part. $qF^0{}_\nu u^\nu = qF^0{}_ju^j = q\tfrac{E_j}{c}\gamma v^j = \tfrac\gamma c\,q\vec E\cdot\vec v$, and $f^0 = \gamma\dot E_{\text{en}}/c$, so $\ \dot E_{\text{en}} = q\vec E\cdot\vec v$: the magnetic field does no work, because $F_{ij}$ is antisymmetric and $v^iv^j$ is not.

The step worth pausing on is the last clause. The power delivered is $\vec F\cdot\vec v = q(\vec E + \vec v\times\vec B)\cdot\vec v$, and the magnetic block contributes $\varepsilon_{ijk}v^iv^jB_k$, which is zero because $\varepsilon_{ijk}$ is antisymmetric in $ij$ and $v^iv^j$ is symmetric — the contraction of a symmetric with an antisymmetric pair of indices vanishes, unit 16's lemma. "Magnetic forces do no work" is usually presented as a curiosity of the cross product; it is a one-line index identity, and in this formalism it is the same identity as the four-force constraint $\eta(f,u)=0$.

The ingredients line, checked. $F^i{}_0 = \eta^{i\alpha}F_{\alpha0}$; the only nonzero $\eta^{i\alpha}$ is $\eta^{ii}=+1$, so $F^i{}_0 = F_{i0} = -F_{0i} = +E_i/c$ ✓.

Dimensions. $[qF^i{}_\nu u^\nu] = \mathrm{C\cdot T\cdot m\,s^{-1}} = \mathrm{C\,V\,s\,m^{-2}\,m\,s^{-1}} = \mathrm{C\,V\,m^{-1}} = \mathrm N$ ✓.

Limit. $v\ll c$ gives $\gamma\to1$, $\tau\to t$, and the equation becomes exactly Newton's with the Lorentz force — no relativistic correction survives at first order in $v/c$ in the form of the law, which is why the law was found a century before the framework.

28 · one field

Where $F$ comes from: the four-potential

Unit 23 and unit 24 wrote $\vec B = \nabla\times\vec A$ and $\vec E = -\nabla\phi - \partial_t\vec A$. Assemble:

definition · the four-potential

$A^\mu = (\phi/c,\ \vec A)$, a vector field on Minkowski space; equivalently $A_\mu = (-\phi/c,\ \vec A)$. Then

$$F_{\mu\nu} \;=\; \partial_\mu A_\nu - \partial_\nu A_\mu ,\qquad \partial_\mu := \frac{\partial}{\partial x^\mu}.$$

Check one entry: $F_{0i} = \tfrac1c\partial_tA_i + \partial_i(\phi/c) = -E_i/c$ ✓, and $F_{ij} = \partial_iA_j - \partial_jA_i = \varepsilon_{ijk}B_k$ ✓.

gauge freedom, four-dimensionally

$A_\mu \to A_\mu + \partial_\mu\chi$ leaves $F$ unchanged, by Clairaut. Unit 23's two separate gauge transformations, $\vec A\to\vec A+\nabla\chi$ and $\phi\to\phi-\partial_t\chi$, are one.

$\partial_\mu$ carries a lower index: $\partial_\mu f$ are the components of the covector $\mathrm{d}f$, which is unit 04's point arriving with teeth.

Why $\partial_\mu$ is naturally down. Under a change of inertial basis $x'^\mu = \Lambda^\mu{}_\nu x^\nu$, the chain rule gives $\partial'_\mu = (\Lambda^{-1})^\nu{}_\mu\partial_\nu$ — the inverse-transpose rule, which is exactly how covector components transform (unit 16). So $\partial_\mu A_\nu$ is a $(0,2)$-tensor, and its antisymmetric part is $F$. The notation $\partial^\mu = \eta^{\mu\nu}\partial_\nu$ is legal and means $(-\frac1c\partial_t, \nabla)$ — note the sign on the time part, another place where sloppiness costs.

The sign in $A_\mu$. $A_0 = \eta_{00}A^0 = -\phi/c$. Every book that writes $A^\mu = (\phi/c,\vec A)$ with $-+++$ must have $A_0 = -\phi/c$; if you see $A_0 = +\phi/c$ the signature is $+---$.

Does every $F$ come from an $A$? On Minkowski space, yes, and the reason is unit 23's Poincaré lemma: a field satisfying the homogeneous Maxwell equations of section 3 is "closed", and on a contractible region closed implies exact. On a region with a hole the implication can fail — the magnetic monopole and the Aharonov–Bohm effect both live in that gap. Unit 34 states this properly as de Rham cohomology; for now, take $A$ as given on all of Minkowski space and note that existence of $A$ is precisely half of Maxwell's equations, which is why section 3 has only two equations to prove and not four.

Check the entry computation. $F_{0i} = \partial_0A_i - \partial_iA_0 = \frac1c\partial_tA_i - \partial_i(-\phi/c) = \frac1c(\partial_tA_i + \partial_i\phi) = -\frac1c E_i$, since $E_i = -\partial_i\phi - \partial_tA_i$ ✓.

section 2

Boosting a field

The wire, twice: magnetism in one frame, electrostatics in the other, the same force.

28 · boosting

The transformation law is a theorem

theorem · change of basis

Let $e'_\mu = e_\nu\,(\Lambda^{-1})^\nu{}_\mu$ be a second inertial basis, so that vector components obey $w'^\mu = \Lambda^\mu{}_\nu w^\nu$. Then the components of the $(0,2)$-tensor $F$ in the two bases are related by

$$F'_{\mu\nu} = (\Lambda^{-1})^\alpha{}_\mu (\Lambda^{-1})^\beta{}_\nu\,F_{\alpha\beta},\qquad\text{equivalently}\qquad F'^{\mu\nu} = \Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta\,F^{\alpha\beta}.$$

proof

$F'_{\mu\nu} = F(e'_\mu, e'_\nu) = F\big(e_\alpha(\Lambda^{-1})^\alpha{}_\mu,\ e_\beta(\Lambda^{-1})^\beta{}_\nu\big)$, and bilinearity pulls both scalars out. $\square$ The raised version follows by raising with $\eta$ and using $\Lambda^\top\eta\Lambda = \eta$.

Nothing here is about electromagnetism: it is unit 16's transformation law, in four dimensions, for a tensor that happens to be antisymmetric. What is about electromagnetism is which physical quantities sit in which slots.

Full proof of the raised version. $F'^{\mu\nu} = \eta^{\mu\rho}\eta^{\nu\sigma}F'_{\rho\sigma} = \eta^{\mu\rho}\eta^{\nu\sigma}(\Lambda^{-1})^\alpha{}_\rho(\Lambda^{-1})^\beta{}_\sigma F_{\alpha\beta}$. The defining property of a Lorentz transformation, $\eta_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu = \eta_{\mu\nu}$ (unit 26), rearranges to $\eta^{\mu\rho}(\Lambda^{-1})^\alpha{}_\rho = \Lambda^\mu{}_\beta\eta^{\beta\alpha}$, and substituting twice gives $F'^{\mu\nu} = \Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta F^{\alpha\beta}$. The moral: for a Lorentz transformation the "inverse" and the "transpose" differ only by where $\eta$ is applied, which is why one never has to keep track of inverses explicitly once the indices are in the right places.

Read the theorem as bookkeeping. It says nothing about the world. The content is in the previous section — that there is a tensor whose components are $\vec E$ and $\vec B$. Once that is established, how the components change under a boost is forced, and the next slide just does the matrix multiplication.

Historical note. Lorentz (1904) and Poincaré (1905) both had the field transformation rules before anyone had a field tensor; Lorentz found them by demanding that Maxwell's equations keep their form, which is unit 25's calculation done backwards. Minkowski's 1908 "Grundgleichungen" supplied the six-component antisymmetric object — his "space-time vector of the second kind", which Sommerfeld renamed the six-vector in 1910 — and Einstein's 1905 paper contains the rules in section 6 without the geometry. Getting the rules by matrix multiplication, as below, is the easy modern route; getting them by insisting the equations be covariant is the historical one, and the problem sheet asks for both.

28 · boosting

The mixing rules

Take $\Lambda$ the boost of unit 26 with speed $v$ along $\hat x$, and multiply out. Split into components along and across the boost:

theorem · field transformation rules $$\begin{aligned} E'_\parallel &= E_\parallel, & B'_\parallel &= B_\parallel,\\ \vec E{}'_\perp &= \gamma\big(\vec E + \vec v\times\vec B\big)_\perp, & \vec B{}'_\perp &= \gamma\Big(\vec B - \tfrac{1}{c^2}\vec v\times\vec E\Big)_\perp . \end{aligned}$$

Along $\hat x$: $E'_y = \gamma(E_y - vB_z)$, $E'_z = \gamma(E_z + vB_y)$, $B'_y = \gamma(B_y + vE_z/c^2)$, $B'_z = \gamma(B_z - vE_y/c^2)$.

checks

Dimensions. $[vB] = \mathrm{m\,s^{-1}T} = \mathrm{V\,m^{-1}}$ ✓; $[vE/c^2] = \mathrm{T}$ ✓.

Limit $v\to0$. Both reduce to $\vec E{}'=\vec E$, $\vec B{}'=\vec B$ ✓.

First order in $v/c$. $\vec E{}' = \vec E + \vec v\times\vec B$, $\vec B{}' = \vec B$ — exactly unit 25's Galilean field transformation, the rule its magnetic limit is built on, now seen as a truncation rather than a theory.

Full derivation of one entry. The boost matrix is $\Lambda^\mu{}_\nu = \begin{pmatrix}\gamma&-\gamma\beta&0&0\\-\gamma\beta&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$ with $\beta = v/c$ (unit 26; this is the transformation to a frame moving at $+v\hat x$). Then $F'^{02} = \Lambda^0{}_\alpha\Lambda^2{}_\beta F^{\alpha\beta} = \Lambda^0{}_0\Lambda^2{}_2F^{02} + \Lambda^0{}_1\Lambda^2{}_2F^{12} = \gamma\,\frac{E_y}{c} - \gamma\beta\,B_z$. Since $F'^{02} = E'_y/c$, multiply by $c$: $E'_y = \gamma(E_y - vB_z)$ ✓.

And $F'^{13} = \Lambda^1{}_\alpha\Lambda^3{}_\beta F^{\alpha\beta} = \Lambda^1{}_0\Lambda^3{}_3F^{03} + \Lambda^1{}_1\Lambda^3{}_3F^{13} = -\gamma\beta\frac{E_z}{c} + \gamma(-B_y)$, and $F'^{13} = -B'_y$, so $B'_y = \gamma(B_y + vE_z/c^2)$ ✓. The other four go the same way; two of them ($F'^{01} = F^{01}$, $F'^{23} = F^{23}$) are immediate because the boost acts as the identity on the $23$ block and $\Lambda^0{}_\alpha\Lambda^1{}_\beta F^{\alpha\beta}$ collapses by antisymmetry.

The pattern. Components along the boost are unchanged; components across it mix, each picking up the other field crossed with $\vec v$, with a relative sign and a $1/c^2$ on the magnetic side. That relative sign is what makes the invariant $B^2 - E^2/c^2$ work, and it is what stops the mixing from being a rotation.

A trap. These rules transform the field at the same event, evaluated at corresponding coordinates. $\vec E{}'(t',x')$ is built from $\vec E, \vec B$ at the event whose primed coordinates are $(t',x')$, not at the same numerical $(t,x)$. Forgetting this is the commonest error in problems where the field also depends on position — the wire is one, and worked examples 2 and 3 do the substitution explicitly.

28 · boosting

The mixing rules, plotted: a pure $\vec B$ seen on the move

A field that is purely magnetic in the lab ($\vec E = 0$, $\vec B = B\hat z$), seen by an observer moving at $\beta\hat x$. An electric field appears and both grow without bound, but the difference of their squares never moves.

Put $\vec E = 0$ into the rules of the previous slide:

$$B'_z = \gamma B,\qquad E'_y = -\gamma vB .$$

So $B'^2 - E'^2/c^2 = \gamma^2B^2(1 - \beta^2) = B^2$ at every speed: the flat dashed line.

the wire's charge

$\beta = 0.0100$ gives $E'_y = -\gamma vB = -30.0015\ \mathrm{V\,m^{-1}}$ — the dot at the far left, where both curves have barely moved. Worked example 3 gets the same number from charge densities.

As $\beta\to1$, $|E'|/cB' = \beta\to1$: a fast observer sees $\vec E\perp\vec B$ with $E\approx cB$, almost a light wave.

What to look for. Neither curve is the interesting one; their difference is. $\gamma$ and $\gamma\beta$ are $\cosh$ and $\sinh$ of the rapidity (unit 26), so $(cB', -E'_y)$ traces a hyperbola as the observer speeds up — the field components undergo a boost of their own in the $(E_y/c,\ B_z)$ plane. Worked example 4 draws that plane. The invariant $B^2 - E^2/c^2$ is the "length" of the pair under that boost, exactly as $c^2t^2 - x^2$ is for an event.

The limit, physically. A charge passing at nearly $c$ has, in the frame where it moves, a field squeezed into a thin pancake perpendicular to the motion, with $E\approx cB$ and $\vec E\perp\vec B$. That is why the field of an ultra-relativistic particle behaves like a pulse of radiation — the Weizsäcker–Williams idea of "equivalent photons". The plot shows the same thing from the other side: an observer racing past a static magnetic field sees something increasingly like light.

The numbers. $v/c = 3.00\times10^{6}/2.99792458\times10^{8} = 0.0100069$, $\gamma = 1.0000500730$, $\gamma\beta = 1.00074\times10^{-2}$, and $\gamma vB = 1.0000500730\times3.00\times10^{6}\times1.00\times10^{-5} = 30.0015\ \mathrm{V\,m^{-1}}$ ✓.

28 · boosting

Worked example 2 — the wire in the lab

example · the magnetic story

Setup. Wire along $\hat x$, $I = 1.00$ A in $+\hat x$; charge $q = 1.00$ nC at $(0, r, 0)$, $r = 2.00$ cm, velocity $v\hat x$ with $v = 3.00\times10^{6}\ \mathrm{m\,s^{-1}}$.

The field. The wire is neutral: the lattice's positive line density and the conduction electrons' negative one cancel exactly, so $\vec E = 0$. Ampère (unit 23) gives

$$B_z = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi\times10^{-7})(1.00)}{2\pi(0.0200)} = 10.0\ \mu\mathrm T .$$

The force. $\vec F = q\vec v\times\vec B = qvB_z\,(\hat x\times\hat z) = -qvB_z\,\hat y$, so

$$\begin{aligned}F_y &= -(1.00\times10^{-9})(3.00\times10^{6})\\ &\qquad\times(1.00\times10^{-5})\\ &= -3.000\times10^{-8}\ \mathrm N,\end{aligned}$$

towards the wire. A positive charge moving with the current is attracted — parallel currents attract.

The lab. Ions and electrons equally spaced, so no $\vec E$. Above the wire $\vec B$ points out of the page; $\hat x\times\hat z = -\hat y$ turns $q\vec v\times\vec B$ towards the wire.

How neutral is "neutral"? Experimentally, the charge of an atom is zero to better than $10^{-21}$ of the electron charge, and a current-carrying wire shows no external electric field to comparable precision. The cancellation is not an idealisation; it is one of the best-tested facts in physics, and it is what makes the next slide surprising.

The numbers behind the numbers. Copper has $n = 8.49\times10^{28}$ conduction electrons per cubic metre (one per atom). In a $1.00\ \mathrm{mm^2}$ wire the mobile charge per metre is $\lambda = neA = (8.49\times10^{28})(1.602177\times10^{-19})(1.00\times10^{-6}) = 1.3602\times10^{4}\ \mathrm{C\,m^{-1}}$ — thirteen thousand coulombs in every metre of ordinary lamp flex, on each sign. The drift speed needed to carry 1.00 A is then $u = I/\lambda = 7.352\times10^{-5}\ \mathrm{m\,s^{-1}}$: seven hundredths of a millimetre per second, so $u/c = 2.452\times10^{-13}$. Keep both numbers; the whole of Purcell's argument is their product.

Direction, carefully. The figure has $x$ to the right, $y$ up towards the charge, and $z$ out of the page, a right-handed set. With the current along $+\hat x$, the Biot–Savart field at $(0,r,0)$ is $\vec B = \frac{\mu_0I}{2\pi r}\,\hat I\times\hat r = \frac{\mu_0I}{2\pi r}\,\hat x\times\hat y = \frac{\mu_0I}{2\pi r}\hat z$. Then $\hat x\times\hat z = -\hat y$, giving a force towards the wire. If you got $+\hat y$, check that cross product; it is the one people reverse.

Check, dimensions. $[\mu_0 I/r] = \mathrm{N\,A^{-2}\cdot A\,m^{-1}} = \mathrm{N\,A^{-1}m^{-2}\cdot m} $, which reduces to tesla ✓. Check, limit. $v\to0$: the force vanishes, as it must — a charge at rest beside a neutral wire feels nothing.

28 · boosting

Worked example 3 — the same wire, from the charge

example · Purcell's argument, in full

Step 1 — proper densities. The lattice is at rest in the lab with density $+\lambda$, so its proper line density is $\lambda$. The electrons move at $-u\hat x$ with lab density $-\lambda$, so their proper density is $-\lambda/\gamma_u$ (their lab density is already contracted).

Step 2 — boost to the charge. In the charge's frame the lattice moves at $-v$, so $\lambda'_+ = \gamma_v\lambda$. The electrons' speed there follows from unit 26's velocity addition, $\gamma_{u'} = \gamma_u\gamma_v(1 + uv/c^2)$, so $\lambda'_- = -\gamma_v\lambda\,(1 + uv/c^2)$.

The same stretch of wire in both frames, exaggerated ($\gamma_v = 1.25$, $1 + uv/c^2 = 1.2$; really $1 + 5\times10^{-5}$ and $1 + 2.5\times10^{-15}$). In the lab the window holds as many electrons as ions. Seen from the charge both rows bunch up, the faster electrons more: the wire is negative.

Reading the figure. Charge cannot appear or vanish, so the only way the count in a fixed window can change is through spacing. The ions are at rest in the lab, so their lab spacing is their proper spacing, and from the charge's frame it shrinks by $\gamma_v$. The electrons' lab spacing is already contracted from their proper spacing by $\gamma_u$; in the charge's frame they move at $(u+v)/(1+uv/c^2)$, faster than the ions, so they are contracted more. The lab's exact balance is therefore broken in the charge's frame, and in the direction of more electrons. If the charge moved against the current, the electrons would be the slower row and the wire would turn positive — the force would push the charge away, which is the antiparallel-currents-repel rule.

Step 2, in full. A line of charge with proper density $\lambda_0$, moving at speed $w$, has density $\gamma_w\lambda_0$ in that frame: the charges are fixed in number per proper length, and the proper length is contracted by $1/\gamma_w$. Charge itself is invariant. For the electrons: in the lab their velocity is $-u$; boosting to a frame moving at $+v$ gives $u' = -(u+v)/(1+uv/c^2)$ by unit 26's addition law, and the identity $\gamma_{u'} = \gamma_u\gamma_v(1+uv/c^2)$ follows from it by one page of algebra — or, faster, by noting that $\gamma_{u'}c^2 = -\eta(U_{\text{elec}}, U_{\text{frame}})$ is an invariant pairing of two four-velocities, which evaluates to $\gamma_u\gamma_v(1+uv/c^2)$ in the lab. That is the argument worth remembering: $\gamma$ between two objects is an inner product of their four-velocities.

28 · boosting

Worked example 3, continued — the net charge and its field

example · Purcell's argument, concluded

Step 3 — the net. $\ \lambda' = \lambda'_+ + \lambda'_- = -\gamma_v\lambda\,\dfrac{uv}{c^2} = -\gamma_v\dfrac{Iv}{c^2} = -3.3381\times10^{-11}\ \mathrm{C\,m^{-1}} .$

Step 4 — the force. Gauss (unit 05) on a cylinder, with $r' = r$ because transverse lengths do not contract:

$$E'_y = \frac{\lambda'}{2\pi\varepsilon_0 r} = -30.0015\ \mathrm{V\,m^{-1}},$$ $$F'_y = qE'_y = -3.00015\times10^{-8}\ \mathrm N .$$

The wire is charged, the charge sits still, and the force is pure electrostatics.

The charge's frame, end on. The wire is negative, the charge is at rest, and the only force available is electric — pointing at the wire, as in the lab.

Step 3, the cancellation. $\gamma_v\lambda - \gamma_v\lambda(1+uv/c^2) = -\gamma_v\lambda uv/c^2$, and $\lambda u = I$. Note that the two enormous densities $\gamma_v\lambda = 1.3603\times10^{4}$ and $-1.3603\times10^{4}\ \mathrm{C\,m^{-1}}$ differ by a fraction $uv/c^2 = 2.454\times10^{-15}$ — fifteen significant figures of cancellation, leaving $3.34\times10^{-11}\ \mathrm{C\,m^{-1}}$. This is why the argument feels like a conjuring trick: the answer is the tiny residue of two huge numbers, and it is exactly the size of an everyday magnetic force.

Step 4, why $r' = r$. The boost is along $\hat x$ and the separation is along $\hat y$; unit 26's transverse scale factor is 1. Gauss's law in the primed frame is legitimate because in that frame the problem is electrostatics: the charge is at rest, and although the wire's charges move, the configuration is stationary, so the fields are static and unit 23's results apply verbatim.

Check the arithmetic. $Iv/c^2 = (1.00)(3.00\times10^{6})/(8.98755\times10^{16}) = 3.3380\times10^{-11}$; times $\gamma_v = 1.0000500730$ gives $3.3381\times10^{-11}\ \mathrm{C\,m^{-1}}$. Then $2\pi\varepsilon_0r = 2\pi(8.8541878\times10^{-12})(0.0200) = 1.11265\times10^{-12}$, and the quotient is $30.0015\ \mathrm{V\,m^{-1}}$ ✓.

28 · boosting

The two answers, to the last $\gamma$

Lab: $F_y = -3.000\times10^{-8}$ N. Charge's frame: $F'_y = -3.00015\times10^{-8}$ N. They differ by $\gamma_v = 1.0000500730$. That is not a discrepancy — it is the law.

theorem · transverse force

If a particle is instantaneously at rest in $S'$ and moves at $v\hat x$ in $S$, then

$$F_\perp = \frac{F'_\perp}{\gamma}.$$

proof

$p^y$ is unchanged by an $x$-boost, so $\mathrm{d}p_y = \mathrm{d}p'_y$; and $\mathrm{d}t = \gamma\,\mathrm{d}\tau = \gamma\,\mathrm{d}t'$. Divide. $\square$

the loop closes

$\dfrac{F'_y}{\gamma_v} = \dfrac{-3.00015\times10^{-8}}{1.0000500730} = -3.000\times10^{-8}\ \mathrm N$ ✓

And the same answer comes out of the mixing rules with no charge densities at all: with $\vec E = 0$ in the lab, $E'_y = \gamma(E_y - vB_z) = -\gamma vB_z$, which is $-30.0015\ \mathrm{V\,m^{-1}}$ ✓. Three routes, one number.

Why $p^y$ is unchanged. $p$ is a four-vector; under the boost $\Lambda$ along $\hat x$, the components $p^2$ and $p^3$ are untouched, because $\Lambda$ acts as the identity on that block. This is the same statement as "transverse lengths do not contract", for a different vector.

The three routes, named. (i) Lab magnetostatics: $q\vec v\times\vec B$. (ii) Rest-frame electrostatics: transform the sources (line densities), then apply Gauss. (iii) Transform the fields with the mixing rules. Route (iii) is one line and route (ii) is a page — but route (ii) is the one that explains, because it says where the charge came from. A formalism that gets the answer without telling you what happened is worth less than it looks; keep both.

The honest bookkeeping. Route (ii) and route (iii) agree exactly, both giving $E'_y = -\gamma vB_z$, because $-\gamma_v Iv/c^2/(2\pi\varepsilon_0 r) = -\gamma_v v\mu_0I/(2\pi r) = -\gamma_v vB_z$, using $1/(\varepsilon_0c^2) = \mu_0$. That identity is the whole content of the agreement, and it is worth doing on paper once.

A caution about "the force in the charge's frame is larger". Forces are not invariants and the comparison is not paradoxical, but it does mean that "the force" is an unhelpful thing to compare between frames. The four-force $f^\mu$ is the invariant object; its spatial part divided by $\gamma$ is what a Newtonian would call force, and the $\gamma$ is bookkeeping about whose clock the momentum change is measured against.

28 · boosting

Purcell's moral: magnetism is a $10^{-13}$ effect

The relativistic correction to the wire's charge balance is a fraction

$$\frac{uv}{c^2} = \frac{(7.352\times10^{-5})(3.00\times10^{6})}{8.98755\times10^{16}} = 2.454\times10^{-15},$$

with $u/c = 2.452\times10^{-13}$ carrying most of the smallness. But it multiplies $\lambda = 1.3602\times10^{4}\ \mathrm{C\,m^{-1}}$, and $10^{-15}\times10^{4}$ is $10^{-11}$ C per metre — enough to pull on a nanocoulomb with $3\times10^{-8}$ N.

Magnetism is what is left of Coulomb's law after a cancellation of fifteen figures fails to be exact.

One factor of $uv/c^2$ takes the charge per metre down almost fifteen decades.

do not over-claim

This argument explains the field of a steady current in a neutral conductor seen from a boosted frame. It does not derive magnetism from electrostatics in general: a permanent magnet's field comes from electron spin, which is quantum, and no boost turns an intrinsic magnetic moment into a charge density.

What is true in general is the tensor statement: $\vec E$ and $\vec B$ are blocks of one object, so no frame-independent line separates them.

Purcell's book. Electricity and Magnetism (Berkeley Physics Course vol. 2, 1965; 3rd edn with Morin, 2013) chapter 5 is the source, and it is the rare textbook chapter worth reading for pleasure. Purcell's pedagogical bet was to put relativity before magnetism, so that $\vec B$ arrives as a consequence rather than a second postulate. The bet is only partly winnable — see the warning — but the chapter is where most physicists first see that the two fields are one.

Reading the figure. The two dots are the same kind of quantity, charge per metre, so they can share an axis: $1.36\times10^{4}$ for the mobile electrons (and, equal and opposite, the lattice) and $3.34\times10^{-11}$ for what survives the boost. The gap is $\log_{10}(1/2.454\times10^{-15}) = 14.6$ decades, which is the "fifteen figures". Nothing else in the problem is anywhere near that small.

The amplification, stated as a ratio. Magnetic to electric force on a charge moving at $v$ beside a wire carrying drift speed $u$: the magnetic force is $\sim qvB$ and the Coulomb force one would get from the uncancelled electron line alone is $\sim q\lambda/2\pi\varepsilon_0r$; their ratio is $uv/c^2$. The reason magnetism is not negligible is not that the effect is big but that the thing it is a small fraction of — the raw Coulomb attraction of $1.4\times10^{4}$ C — is astronomically big. Two one-metre lengths of lamp flex, if you could strip the electrons out of them, would attract each other across a metre's gap with a force of $1.4\times10^{18}$ N.

An honest loose end. Purcell's argument as given treats the electrons as a continuous fluid of uniform proper density. Real conduction electrons have a thermal velocity distribution with $v_{\text{th}}\sim10^{6}\ \mathrm{m\,s^{-1}}$, enormously larger than the drift. The argument survives because only the mean velocity enters the current and the leading density correction is linear in the mean; the second-order corrections are real but are absorbed into the observed neutrality of the wire at rest.

section 3

Maxwell, in two lines

Four vector equations, eight scalar ones, and they are two tensor equations.

28 · maxwell

The four-current

definition · the four-current

$J^\mu = (c\rho,\ \vec J)$, a vector field on Minkowski space. For a single species of charge with proper density $\rho_0$ moving with four-velocity $u$: $J^\mu = \rho_0 u^\mu$.

The second form is the honest one: it manifestly builds a vector, and $\rho = \gamma\rho_0$, $\vec J = \rho\vec v$ drop out of it. Charge per unit volume is frame-dependent for the same reason charge per unit length was.

theorem · covariant continuity $$\partial_\mu J^\mu = 0 \iff \partial_t\rho + \nabla\cdot\vec J = 0 .$$

proof

$\partial_\mu J^\mu = \tfrac1c\partial_t(c\rho) + \partial_iJ^i$. $\square$

Unit 23's continuity equation, unit 22's mass version, and this are one statement: the divergence of a flux four-vector vanishes.

Why $c\rho$ and not $\rho$. Dimensions again: $[\vec J] = \mathrm{A\,m^{-2}} = \mathrm{C\,m^{-2}s^{-1}}$ while $[\rho] = \mathrm{C\,m^{-3}}$, and $c\rho$ fixes it ✓. Equivalently, $J^\mu = \rho_0u^\mu$ with $u^0 = \gamma c$ forces it.

Superposition. A conductor has two species, lattice and electrons, so $J^\mu = \rho_{0+}u_+^\mu + \rho_{0-}u_-^\mu$. Each term is a vector, so the sum is; and the lab-frame cancellation of the time components is the wire's neutrality. Nothing in the formalism requires a single fluid.

Charge invariance. $\int\rho\,\mathrm{d}^3x$ over a region containing all the charge is the same number in every frame, even though $\rho$ and the volume element each change. The clean proof needs the divergence theorem in four dimensions, which arrives in unit 34; for now it is cited, and it is the reason "the charge of the electron" is a constant of nature rather than a function of how fast the electron is going.

28 · maxwell

The wire's four-current is spacelike

In the lab the wire's $J$ has no time component. Resolved along a moving frame's axes ($\beta = 0.5$ here, for visibility) it acquires one, and it is negative: a charged wire.

In the lab, $\rho = 0$ and $\vec J = (I/A)\hat x = 1.00\times10^{6}\ \mathrm{A\,m^{-2}}\,\hat x$, so

$$J^\mu = (0,\ J_x,\ 0,\ 0),\qquad \eta(J,J) = J_x^2 > 0 .$$

Boost to the charge: $J'^0 = \gamma(J^0 - \beta J^1) = -\gamma\beta J_x$, so

$$\rho' = -\frac{\gamma vJ_x}{c^2} = -3.3381\times10^{-5}\ \mathrm{C\,m^{-3}} .$$

the wire again

Times the cross-section $A = 1.00\ \mathrm{mm^2}$: $\rho'A = -\gamma vI/c^2 = \lambda'$ ✓ — Worked example 3's net charge, with no counting of ions and electrons at all.

The running example's four-current. In the lab the wire has $\rho = 0$ and $\vec J = (I/A)\hat x = 1.00\times10^{6}\ \mathrm{A\,m^{-2}}\ \hat x$. So $J^\mu = (0, 10^{6}, 0, 0)$ and $\eta(J,J) = |\vec J|^2 > 0$: the four-current is spacelike. That is exactly why a frame exists in which the charge density vanishes — the lab is one — and why no frame exists in which the current vanishes. In the charge's frame $J'^0 = \gamma_v(0) - \gamma_v\beta|\vec J| \ne 0$: the wire is charged there, which is worked example 3 seen from the current side rather than the density side. P6 makes this a theorem.

Reading the figure. Components along tilted axes are read by drawing parallels: go out along the $x'$ axis, then parallel to the $ct'$ axis, until you reach the tip of $J$ (dashed). With $J = (0, J_x)$ the parallelogram gives $J'^1 = \gamma J_x$ and $J'^0 = -\gamma\beta J_x$, the second drawn as the arrow pointing down the $ct'$ axis. Since $J'^0 = c\rho'$, the sign says the moving observer sees a negative charge density, matching $\lambda' \lt 0$ for a charge moving with the current.

The arithmetic. $\gamma vJ_x/c^2 = 1.0000500730\times(3.00\times10^{6})(1.00\times10^{6})/8.98755\times10^{16} = 3.3381\times10^{-5}\ \mathrm{C\,m^{-3}}$; times $1.00\times10^{-6}\ \mathrm{m^2}$ gives $3.3381\times10^{-11}\ \mathrm{C\,m^{-1}}$ ✓. Dimensions: $[vJ/c^2] = \mathrm{m\,s^{-1}\,A\,m^{-2}/(m^2s^{-2})} = \mathrm{A\,s\,m^{-3}} = \mathrm{C\,m^{-3}}$ ✓.

28 · maxwell

The sourced pair, in one equation

theorem · the inhomogeneous Maxwell equations $$\boxed{\ \partial_\mu F^{\nu\mu} \;=\; \mu_0 J^\nu\ }$$

is, component by component, Gauss's law ($\nu = 0$) and the Ampère–Maxwell law ($\nu = i$).

$\nu = 0$: $\ \partial_iF^{0i} = \partial_i(E_i/c) = \nabla\cdot\vec E/c$, and $\mu_0J^0 = \mu_0c\rho$. So $\nabla\cdot\vec E = \mu_0c^2\rho = \rho/\varepsilon_0$ ✓

$\nu = i$: $\ \partial_0F^{i0} + \partial_jF^{ij} = -\tfrac1{c^2}\partial_tE_i + (\nabla\times\vec B)_i = \mu_0J_i$ ✓

bonus: continuity, free

$\partial_\nu\partial_\mu F^{\nu\mu} = 0$, because $\partial_\nu\partial_\mu$ is symmetric (Clairaut) and $F^{\nu\mu}$ antisymmetric. Hence $\partial_\nu J^\nu = 0$ automatically: charge conservation is a corollary of Maxwell, not an extra law. Unit 24 found the displacement current by insisting on this; here it is one line.

Index order matters. $\partial_\mu F^{\nu\mu} = +\mu_0J^\nu$ but $\partial_\mu F^{\mu\nu} = -\mu_0J^\nu$. With an antisymmetric tensor you must say which slot the divergence is taken in. A book that writes $\partial_\mu F^{\mu\nu} = \mu_0J^\nu$ has made a different choice somewhere — the signature, the sign of $F$, or the index order — so check all three before copying a formula out of it.

Full check of $\nu = i$. $\partial_\mu F^{i\mu} = \partial_0F^{i0} + \partial_jF^{ij}$. Now $F^{i0} = -F^{0i} = -E_i/c$ and $\partial_0 = \frac1c\partial_t$, so the first term is $-\frac{1}{c^2}\partial_tE_i$. And $F^{ij} = F_{ij} = \varepsilon_{ijk}B_k$, so $\partial_jF^{ij} = \varepsilon_{ijk}\partial_jB_k = (\nabla\times\vec B)_i$. Setting the sum equal to $\mu_0J_i$ and using $\mu_0\varepsilon_0 = 1/c^2$ gives $\nabla\times\vec B = \mu_0\vec J + \mu_0\varepsilon_0\partial_t\vec E$ ✓, unit 24's equation with the displacement current included.

Dimensions. $[\partial F] = \mathrm{T\,m^{-1}}$ and $[\mu_0J] = \mathrm{N\,A^{-2}\cdot A\,m^{-2}}$; $\mathrm{T} = \mathrm{N\,A^{-1}m^{-1}}$, so both sides are $\mathrm{N\,A^{-1}m^{-2}}$ ✓.

Limit. Statics: $\partial_t \to 0$ returns $\nabla\cdot\vec E = \rho/\varepsilon_0$ and $\nabla\times\vec B = \mu_0\vec J$, unit 23 exactly ✓.

What has been gained. Not brevity — four equations became two, and one of the two is free. What has been gained is that covariance is now manifest: both sides of $\partial_\mu F^{\nu\mu} = \mu_0J^\nu$ are components of a four-vector, so if the equation holds in one inertial frame it holds in all of them, with no calculation. Unit 25's laborious chain-rule check is replaced by an inspection of index structure. That is the entire practical value of tensor notation, and it is worth a lot.

28 · maxwell

The symbol, and the tensor

definition · the Levi-Civita symbol

$[\mu\nu\rho\sigma]$ is $+1$ for an even permutation of $(0,1,2,3)$, $-1$ for an odd one, $0$ otherwise. It is an array of numbers, declared to be the same in every basis — not a tensor.

definition · the Levi-Civita tensor

$\epsilon_{\mu\nu\rho\sigma} := \sqrt{|\det\eta|}\,[\mu\nu\rho\sigma] = [\mu\nu\rho\sigma]$ here, a $(0,4)$-tensor: the unique totally antisymmetric one with $\epsilon_{0123} = 1$.

theorem · when they coincide

Under a change of basis by $\Lambda$, the components of the tensor pick up $\det\Lambda$. For $\Lambda\in SO(1,3)$, $\det\Lambda = +1$, so the components are unchanged and the tensor's components equal the symbol in every proper Lorentz frame.

Under a reflection they differ by a sign: $\epsilon$ is a pseudotensor with respect to the full Lorentz group. Raised: $\epsilon^{0123} = \eta^{00}\eta^{11}\eta^{22}\eta^{33}\epsilon_{0123} = -1$.

Full proof of the theorem. Transform the array as if it were a $(0,4)$-tensor: $\epsilon'_{\mu\nu\rho\sigma} = (\Lambda^{-1})^\alpha{}_\mu(\Lambda^{-1})^\beta{}_\nu(\Lambda^{-1})^\kappa{}_\rho(\Lambda^{-1})^\delta{}_\sigma\epsilon_{\alpha\beta\kappa\delta}$. The right-hand side is totally antisymmetric in $\mu\nu\rho\sigma$ (it inherits the antisymmetry), and in four dimensions the space of totally antisymmetric $(0,4)$-tensors is one-dimensional, so $\epsilon' = \kappa\,\epsilon$ for some number $\kappa$. Evaluating at $(\mu\nu\rho\sigma) = (0123)$ and recognising the Leibniz formula for a determinant gives $\kappa = \det(\Lambda^{-1}) = 1/\det\Lambda$. For $\Lambda$ Lorentz, $\Lambda^\top\eta\Lambda = \eta$ gives $(\det\Lambda)^2 = 1$, so $\det\Lambda = \pm1$ and $\kappa = \det\Lambda$. $\square$

Why bother with the distinction. Because on a curved manifold (unit 32) $\sqrt{|\det g|}$ is not 1 and the two objects genuinely differ: the symbol keeps its values, the tensor acquires a factor. Meeting the distinction here, where the factor happens to be 1, is much easier than meeting it there.

Identities worth having. $\epsilon^{\mu\nu\rho\sigma}\epsilon_{\mu\nu\rho\sigma} = -24$; $\epsilon^{\mu\nu\rho\sigma}\epsilon_{\alpha\nu\rho\sigma} = -6\,\delta^\mu{}_\alpha$. The minus signs are the one lowered $0$ index; in Euclidean three dimensions the analogous identities ($\varepsilon_{ijk}\varepsilon_{ijk} = 6$) have no sign. P11 proves them.

Parity, physically. That $\epsilon$ flips under reflection is the mathematics behind unit 24's observation that $\vec B$ is an axial vector: $\vec B$ sits in the $ij$ block of $F$, and converting that block to a 3-vector uses $\varepsilon_{ijk}$. A mirror-image experiment has the same $F$ and the opposite "$\vec B$", which is bookkeeping, not physics — until the weak interaction, where it stops being bookkeeping.

28 · maxwell

The dual, and the other pair

definition · the dual field tensor $$\tilde F^{\mu\nu} := \tfrac12\,\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}.$$

Computing the components: $\tilde F^{0i} = -B_i$ and $\tilde F^{ij} = \varepsilon_{ijk}E_k/c$. So dualising is the substitution

$$\vec E/c \longmapsto -\vec B,\qquad \vec B\longmapsto \vec E/c,$$

a duality rotation by a quarter turn; $\tilde{\tilde F} = -F$.

theorem · the source-free pair $$\partial_\mu\tilde F^{\nu\mu} = 0 \iff \nabla\cdot\vec B = 0\ \text{ and }\ \partial_t\vec B + \nabla\times\vec E = 0 .$$

Equivalently, without the dual,

$$\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0,$$

the Bianchi identity. It is automatic once $F = \partial A - \partial A$: substitute and the six terms cancel in pairs by Clairaut.

count

Two tensor equations, eight scalar equations, exactly Maxwell's four vector laws. And half of them say only "$F$ comes from a potential".

Check the component claims. $\tilde F^{01} = \frac12\epsilon^{01\rho\sigma}F_{\rho\sigma} = \epsilon^{0123}F_{23} = (-1)(B_x) = -B_x$ ✓. $\tilde F^{12} = \frac12\epsilon^{12\rho\sigma}F_{\rho\sigma} = \epsilon^{1203}F_{03}$, the two surviving terms ($\rho\sigma = 03$ and $30$) being equal. The permutation $(1,2,0,3)$ has two inversions, hence is even, so $\epsilon^{1203} = \epsilon^{0123} = -1$, and $F_{03} = -E_z/c$, giving $\tilde F^{12} = +E_z/c$ ✓, which is $\varepsilon_{12k}E_k/c$ ✓.

Full proof of the Bianchi identity. $\partial_\lambda F_{\mu\nu} = \partial_\lambda\partial_\mu A_\nu - \partial_\lambda\partial_\nu A_\mu$, and cyclically. Adding the three gives six terms: $(\partial_\lambda\partial_\mu A_\nu - \partial_\mu\partial_\lambda A_\nu) + (\partial_\mu\partial_\nu A_\lambda - \partial_\nu\partial_\mu A_\lambda) + (\partial_\nu\partial_\lambda A_\mu - \partial_\lambda\partial_\nu A_\mu)$, and each bracket vanishes by Clairaut for $C^2$ fields. $\square$

Full check of the components of the identity. Take $(\lambda\mu\nu) = (123)$: $\partial_1F_{23} + \partial_2F_{31} + \partial_3F_{12} = \partial_xB_x + \partial_yB_y + \partial_zB_z = \nabla\cdot\vec B$ ✓. Take $(012)$: $\partial_0F_{12} + \partial_1F_{20} + \partial_2F_{01} = \frac1c\partial_tB_z + \partial_x(E_y/c) - \partial_y(E_x/c) = \frac1c[\partial_tB_z + (\nabla\times\vec E)_z]$ ✓. The four independent index triples give the four scalar equations: one from $(123)$ and three from $(0ij)$.

Magnetic monopoles. If one existed, the right-hand side would be $\mu_0 J^\nu_{\text{mag}}$ and the duality $E/c\leftrightarrow B$ would become a symmetry of the full theory. Dirac showed in 1931 that a single monopole anywhere would force electric charge to be quantised — the most attractive argument for something nobody has seen. That $\partial_\mu\tilde F^{\nu\mu} = 0$ is here a theorem (from $F = \partial A - \partial A$) and not an assumption is precisely why the potential formalism excludes monopoles.

28 · maxwell

The two invariants

theorem · every frame agrees on exactly two numbers $$\tfrac12F_{\mu\nu}F^{\mu\nu} = B^2 - \frac{E^2}{c^2} \;=:\;\mathcal I_1,\qquad \tfrac c4 F_{\mu\nu}\tilde F^{\mu\nu} = \vec E\cdot\vec B \;=:\;\mathcal I_2 .$$

proof of the first

$F_{\mu\nu}F^{\mu\nu} = 2F_{0i}F^{0i} + F_{ij}F^{ij} = 2\big(-\tfrac{E_i}{c}\big)\big(\tfrac{E_i}{c}\big) + \varepsilon_{ijk}\varepsilon_{ijl}B_kB_l = -\tfrac{2E^2}{c^2} + 2B^2 .$ $\square$

Both are scalars because every index is contracted; that is the whole argument, and it is unit 16's contraction lemma.

on the wire

Lab: $\mathcal I_1 = (10.0\,\mu\mathrm T)^2 = 1.0000\times10^{-10}\ \mathrm{T^2}$, $\mathcal I_2 = 0$.

Charge's frame: $B'^2 - E'^2/c^2 = \gamma^2B^2 - \gamma^2v^2B^2/c^2 = B^2$ — the $\gamma^2$ cancels against $1-v^2/c^2$ exactly, giving $1.0000\times10^{-10}\ \mathrm{T^2}$ ✓, and $\vec E{}'\perp\vec B{}'$ so $\mathcal I_2 = 0$ ✓.

Full proof of the second. $F_{\mu\nu}\tilde F^{\mu\nu} = \frac12\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}$. Of the 24 nonzero index assignments, group them by where the index $0$ sits; by the total antisymmetry of $\epsilon$ and of each $F$, all four positions contribute equally, so the sum is $4\,\epsilon^{0ijk}F_{0i}F_{jk}$ summed over $i,j,k$. With $\epsilon^{0ijk} = -\varepsilon_{ijk}$, $F_{0i} = -E_i/c$ and $F_{jk} = \varepsilon_{jkl}B_l$, this is $4\cdot\frac{E_i}{c}\varepsilon_{ijk}\varepsilon_{jkl}B_l = 4\cdot\frac{E_i}{c}\cdot2\delta_{il}B_l = \frac8c\vec E\cdot\vec B$. Halving gives $F\tilde F = \frac4c\vec E\cdot\vec B$, hence the stated normalisation. $\square$ (The identity used is $\varepsilon_{ijk}\varepsilon_{ljk} = 2\delta_{il}$, unit 14.)

Dimensions. $\mathcal I_1$ is in $\mathrm{T^2}$ ✓ (both terms: $[E/c] = \mathrm T$). $\mathcal I_2$ is in $\mathrm{V\,m^{-1}T}$, which is $\mu_0$ times an energy flux — since $\mu_0\vec S = \vec E\times\vec B$, the two carry the same dimension ✓.

Why exactly two. The Lorentz group acting on the six-dimensional space of antisymmetric forms has orbits of dimension at most four, so at least two independent invariants exist; and no more than two are algebraically independent because the complex combination $\vec Z = \vec E/c + i\vec B$ transforms under the complexified rotation group, whose only invariant of a complex 3-vector is the complex number $$\vec Z\cdot\vec Z = \frac{E^2}{c^2} - B^2 + \frac{2i}{c}\,\vec E\cdot\vec B = -\mathcal I_1 + \frac{2i}{c}\,\mathcal I_2 ,$$ one complex invariant, hence two real ones and no more. That complex 3-vector is the Riemann–Silberstein vector, and the observation is Minkowski's.

$\mathcal I_2$ is a pseudoscalar, not a scalar: it changes sign under a reflection, because $\epsilon$ does. Under the proper orthochronous group, which is all that physical boosts and rotations generate, it is invariant.

28 · maxwell

Worked example 4 — which field can you transform away?

example · reading the invariants

The rule. If $\mathcal I_2 = \vec E\cdot\vec B = 0$ and $\mathcal I_1 > 0$, a frame exists where the field is purely magnetic; if $\mathcal I_2 = 0$ and $\mathcal I_1 \lt 0$, purely electric; if $\mathcal I_1 = \mathcal I_2 = 0$ with $F\ne0$, neither, ever. If $\mathcal I_2\ne0$, both fields are nonzero in every frame.

Case A. $\vec E = 6.00\times10^{5}\,\hat y\ \mathrm{V\,m^{-1}}$, $\vec B = 1.00\times10^{-3}\,\hat z$ T. Then $\mathcal I_2 = 0$ and $\mathcal I_1 = 1.00\times10^{-6} - 4.0055\times10^{-6} = -3.0055\times10^{-6}\ \mathrm{T^2} \lt 0$: purely electric somewhere.

The boost. Kill $B'_z = \gamma(B_z - vE_y/c^2)$: take $v = c^2B_z/E_y = 1.4979\times10^{8}\ \mathrm{m\,s^{-1}}$, i.e. $\beta = cB/E = 0.499654$, $\gamma = 1.15443$. Then $E'_y = \gamma(E_y - vB_z) = \gamma E_y(1-\beta^2) = E_y/\gamma = 5.1974\times10^{5}\ \mathrm{V\,m^{-1}}$.

Check. $-E'^2/c^2 = -3.0055\times10^{-6}\ \mathrm{T^2}$ ✓ — the invariant is unchanged to five figures.

Why the rule is true. Suppose $\mathcal I_2 = 0$ and $\mathcal I_1 > 0$; then $\vec E\perp\vec B$ and $E \lt cB$. Boost with $\vec v = \vec E\times\vec B/|\vec B|^2$. Since $\vec E\perp\vec B$ we have $|\vec E\times\vec B| = EB$, so $|\vec v| = E/B$, which is below $c$ exactly when $E \lt cB$ ✓. In that frame the transverse rule gives $\vec E{}'_\perp = \gamma(\vec E + \vec v\times\vec B) = 0$ by construction, since $(\vec E\times\vec B)\times\vec B = -|\vec B|^2\vec E$ when $\vec E\perp\vec B$. For $\mathcal I_1 \lt 0$ use $\vec v = c^2\,\vec E\times\vec B/E^2$, of magnitude $c^2B/E \lt c$ ✓.

The arithmetic. $E/c = 6.00\times10^{5}/2.99792458\times10^{8} = 2.00138\times10^{-3}$ T, squared $4.00554\times10^{-6}$; minus from $B^2 = 1.00\times10^{-6}$ gives $-3.00554\times10^{-6}\ \mathrm{T^2}$. $\beta = cB/E = 0.4996541$, $\gamma = 1.154434$, $E'_y = 6.00\times10^{5}/1.154434 = 5.19735\times10^{5}\ \mathrm{V\,m^{-1}}$, and $(5.19735\times10^{5}/c)^2 = 3.00554\times10^{-6}$ ✓ to six figures.

28 · maxwell

The invariants, drawn: boosts slide along hyperbolae

$\vec E = E\hat y$, $\vec B = B\hat z$, boosts along $\hat x$. A boost moves the point $(E/c, B)$ along its hyperbola $B^2 - E^2/c^2 = \mathcal I_1$, never off it. Case A rides its branch down to the axis. A point on a dashed null line stays on it.

For $\vec E = E\hat y$, $\vec B = B\hat z$ and a boost along $\hat x$, the mixing rules read

$$\tfrac{E'}{c} = \gamma\big(\tfrac Ec - \beta B\big),$$ $$B' = \gamma\big(B - \beta\tfrac Ec\big):$$

a Lorentz boost of the pair itself. Its orbits are the hyperbolae of constant $\mathcal I_1$.

example · the rule, read off the picture

Case B — the wire. $\mathcal I_1 = +1.00\times10^{-10}\ \mathrm{T^2} \gt 0$, $\mathcal I_2 = 0$: purely magnetic somewhere, and the lab is that frame. No boost removes the magnetic field, which is why the charge's frame has both.

A branch that crosses an axis has a frame with one field zero there. The null lines cross neither axis away from the origin, so a light wave keeps both fields in every frame.

Reading the figure. Case A is plotted to scale in millitesla: $E/c = 2.0014$ mT, $B = 1.000$ mT, $\mathcal I_1 = -3.0055\ \mathrm{mT^2}$, and the boost at $\beta = 0.49965$ carries it to $(1.7337, 0)$, which is $E' = 5.1974\times10^{5}\ \mathrm{V\,m^{-1}}$ as on the previous slide. The upper branch is drawn through $(0, 1)$ mT to show its shape; the wire's own point is $(0, 0.0100)$ mT, the same curve scaled down a hundredfold, too close to the origin to see. Hyperbolae of constant $\mathcal I_1$ are to this plane what the calibration hyperbolae of unit 26 are to the $(x, ct)$ plane, and for the same reason: the transformation preserves a quadratic form of signature $(1,1)$.

The null case. $\mathcal I_1 = \mathcal I_2 = 0$ with $F\ne0$ means $E = cB$ and $\vec E\perp\vec B$ in every frame: a plane electromagnetic wave (unit 24). A boost can change its amplitude and frequency — that is the Doppler factor of unit 26 — but never its shape. Problem P4 asks you to prove there is no frame where a light wave is at rest, and the invariants prove it in one line, which is worth contrasting with the 16-year-old Einstein's version of the same question.

section 4

Where the energy is

One symmetric tensor, one vanishing divergence, and unit 24's two theorems fall out together.

28 · energy

The stress-energy tensor of the field

definition · the electromagnetic stress-energy tensor $$T^{\mu\nu} \;=\; \frac{1}{\mu_0}\Big(F^{\mu\alpha}F^{\nu}{}_{\alpha} \;-\; \tfrac14\,\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\Big).$$

Mathematically: up to scale, the only symmetric $(2,0)$-tensor quadratic in $F$, built with $\eta$ alone, that is traceless — proof in the notes. The first term is the obvious quadratic contraction; the second subtracts its trace.

Physically, the components are unit 24's:

$$T^{00} = u,\qquad T^{0i} = \frac{S_i}{c},\qquad T^{ij} = -\sigma_{ij},$$

with $u = \tfrac{\varepsilon_0}2E^2 + \tfrac1{2\mu_0}B^2$, $\vec S = \vec E\times\vec B/\mu_0$, and $\sigma$ the Maxwell stress tensor.

two properties, immediately

Symmetric: $F^{\mu\alpha}F^\nu{}_\alpha = \eta_{\alpha\beta}F^{\mu\alpha}F^{\nu\beta}$ is visibly symmetric in $\mu\nu$.

Traceless: $T^\mu{}_\mu = \frac1{\mu_0}(F^{\mu\alpha}F_{\mu\alpha} - \frac44F_{\alpha\beta}F^{\alpha\beta}) = 0$, because $\eta^\mu{}_\mu = 4$.

a sign to keep straight

$T^{ij} = -\sigma_{ij}$: unit 24's $\sigma$ is the force per area transmitted across a surface, while $T^{ij}$ is the flux of $i$-momentum in direction $j$. Momentum flowing out is momentum lost.

Full proof of the uniqueness claim. A $(2,0)$-tensor quadratic in $F$ and built with $\eta$ alone has every index either carried by an $F$ or contracted; the only ways to spend two free indices are $F^{\mu\alpha}F^\nu{}_\alpha$, $F^{\mu\alpha}F_\alpha{}^\nu = -F^{\mu\alpha}F^\nu{}_\alpha$, and $\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$. So the symmetric candidates span a two-dimensional space, $aF^{\mu\alpha}F^\nu{}_\alpha + b\,\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$; taking the trace gives $(a + 4b)F_{\alpha\beta}F^{\alpha\beta}$, so tracelessness forces $b = -a/4$ and fixes the combination up to the overall $a$. The physics fixes $a = 1/\mu_0$ by demanding $T^{00} = u$. $\square$ (A third quadratic, $F^{\mu\alpha}\tilde F^\nu{}_\alpha$, is excluded by "built with $\eta$ alone": it needs $\epsilon$. It is in any case not independent — it equals $\tfrac14\eta^{\mu\nu}F_{\alpha\beta}\tilde F^{\alpha\beta}$, a multiple of $\eta^{\mu\nu}$ and nothing new.)

Full computation of $T^{00}$. $F^{0\alpha}F^0{}_\alpha = F^{0i}F^0{}_i$ (the $\alpha=0$ term vanishes by antisymmetry), and $F^0{}_i = \eta_{ij}F^{0j} = F^{0i} = E_i/c$, so the product is $E^2/c^2$. With $\eta^{00} = -1$ and $F_{\alpha\beta}F^{\alpha\beta} = 2(B^2 - E^2/c^2)$, $$T^{00} = \frac1{\mu_0}\Big(\frac{E^2}{c^2} + \frac12\Big(B^2 - \frac{E^2}{c^2}\Big)\Big) = \frac{1}{\mu_0}\Big(\frac{E^2}{2c^2} + \frac{B^2}{2}\Big) = \frac{\varepsilon_0E^2}{2} + \frac{B^2}{2\mu_0} \checkmark$$ using $1/(\mu_0c^2) = \varepsilon_0$.

$T^{0i}$. $F^{0\alpha}F^i{}_\alpha = F^{0j}F^i{}_j = \frac{E_j}{c}\varepsilon_{ijk}B_k = \frac1c(\vec E\times\vec B)_i$, and $\eta^{0i}=0$ kills the trace term, so $T^{0i} = (\vec E\times\vec B)_i/(\mu_0c) = S_i/c$ ✓. Note $T^{i0} = T^{0i}$: energy flux divided by $c^2$ equals momentum density, $\vec g = \vec S/c^2$, which unit 24 asserted and which is here a consequence of the symmetry of a tensor. That is the cleanest known argument for $\vec g = \vec S/c^2$.

$T^{ij}$, in full. $F^{i\alpha}F^j{}_\alpha = F^{i0}F^j{}_0 + F^{ik}F^j{}_k = (-E_i/c)(E_j/c) + \varepsilon_{ikl}\varepsilon_{jkm}B_lB_m = -E_iE_j/c^2 + \delta_{ij}B^2 - B_iB_j$. Subtracting $\frac14\delta^{ij}\cdot2(B^2 - E^2/c^2)$ and dividing by $\mu_0$ gives $T^{ij} = -\varepsilon_0(E_iE_j - \frac12\delta_{ij}E^2) - \frac1{\mu_0}(B_iB_j - \frac12\delta_{ij}B^2) = -\sigma_{ij}$ ✓.

Tracelessness, physically. $T^\mu{}_\mu = 0$ says the field has no rest frame and no rest energy — it is the tensor statement that the photon is massless. Unit 22's perfect fluid has $T^\mu{}_\mu = -\rho c^2 + 3p$, so tracelessness is $p = \rho c^2/3$: the equation of state of radiation, derived here rather than assumed, and the one unit 39 needs for the radiation era.

28 · energy

One divergence, two conservation laws

theorem · the divergence identity $$\boxed{\ \partial_\nu T^{\mu\nu} \;=\; -\,F^{\mu}{}_{\nu}J^{\nu}\ }$$

The right-hand side is minus the Lorentz force density: $-F^\mu{}_\nu J^\nu$ has components $-(\vec J\cdot\vec E/c,\ \rho\vec E + \vec J\times\vec B)$.

Proof, in three moves. Expand the derivative; use $\partial_\nu F^{\nu\alpha} = -\mu_0J^\alpha$ on the first piece; and show the remaining two pieces cancel by the Bianchi identity. The full working is in the notes.

In vacuum, $J = 0$ and $\partial_\nu T^{\mu\nu} = 0$: energy and momentum are locally conserved, in exactly unit 22's sense — there a flux tensor for mass and momentum; here one with a row and a column more, and the extra row is energy.

$\partial_\nu$ contracts the column index, so each row of $T$ is one conservation law: row $0$ energy, rows $1$–$3$ the three components of momentum. Four laws, one equation; Worked example 5 reads each off.

Reading the figure. In each row, the $\nu = 0$ entry is a density and the $\nu = j$ entries are the flux of that density in direction $j$; $\partial_\nu$ of the row is $\frac1c\partial_t(\text{density}) + \nabla\cdot(\text{flux})$, the shape of every continuity equation since unit 22. Row $0$: density $u$, flux $\vec S/c$ (times $c$ overall). Row $i$: density $S_i/c$, which is $c\,g_i$, flux $-\sigma_{ij}$. The symmetry of $T$ is visible as the shared entries $S_i/c$ in row $0$ and column $0$.

Full proof

Write $\mu_0\partial_\nu T^{\mu\nu} = \partial_\nu\!\big(F^{\mu\alpha}F^\nu{}_\alpha\big) - \tfrac14\partial^\mu\!\big(F_{\alpha\beta}F^{\alpha\beta}\big)$, using $\eta^{\mu\nu}\partial_\nu = \partial^\mu$. Expand: $$\mu_0\partial_\nu T^{\mu\nu} = \big(\partial_\nu F^{\nu}{}_{\alpha}\big)F^{\mu\alpha} + F^{\nu\alpha}\partial_\nu F^{\mu}{}_{\alpha} - \tfrac12 F^{\alpha\beta}\partial^\mu F_{\alpha\beta}.$$

First term. $\partial_\nu F^\nu{}_\alpha = \eta_{\alpha\beta}\partial_\nu F^{\nu\beta} = -\eta_{\alpha\beta}\mu_0J^\beta = -\mu_0J_\alpha$ (the sign because Maxwell reads $\partial_\nu F^{\beta\nu} = \mu_0J^\beta$). So the first term is $-\mu_0F^{\mu\alpha}J_\alpha$, which is the claimed right-hand side times $\mu_0$.

The last two cancel. Put $X^\mu := F^{\nu\alpha}\partial_\nu F^\mu{}_\alpha$, and lower $\mu$: $X_\mu = F^{\nu\alpha}\partial_\nu F_{\mu\alpha}$. The Bianchi identity gives $\partial_\nu F_{\mu\alpha} = -\partial_\mu F_{\alpha\nu} - \partial_\alpha F_{\nu\mu}$, so $$X_\mu = -F^{\nu\alpha}\partial_\mu F_{\alpha\nu} - F^{\nu\alpha}\partial_\alpha F_{\nu\mu} = +F^{\nu\alpha}\partial_\mu F_{\nu\alpha} - F^{\nu\alpha}\partial_\alpha F_{\nu\mu}.$$ In the last term relabel $\nu\leftrightarrow\alpha$ and use antisymmetry twice: $-F^{\alpha\nu}\partial_\nu F_{\alpha\mu} = +F^{\nu\alpha}\partial_\nu F_{\alpha\mu} = -F^{\nu\alpha}\partial_\nu F_{\mu\alpha} = -X_\mu$. Hence $2X_\mu = F^{\nu\alpha}\partial_\mu F_{\nu\alpha}$, i.e. $X_\mu = \tfrac12F^{\alpha\beta}\partial_\mu F_{\alpha\beta}$ — exactly the third term with the opposite sign. They cancel. $\square$

Dimensions. $[T] = \mathrm{J\,m^{-3}}$, so $[\partial T] = \mathrm{J\,m^{-4}} = \mathrm{N\,m^{-3}}$; and $[F J] = \mathrm{T\cdot A\,m^{-2}} = \mathrm{N\,A^{-1}m^{-1}\cdot A\,m^{-2}} = \mathrm{N\,m^{-3}}$ ✓.

The physical reading. Field energy-momentum is not conserved by itself; it is conserved jointly with the charges', and the exchange rate is the Lorentz force density. Unit 29 will show that this is Noether's theorem for spacetime translations, and that the $T^{\mu\nu}$ written here is the symmetrised (Belinfante) version of the one Noether's theorem hands you.

28 · energy

Worked example 5 — reading off Poynting and the stress balance

example · the four components of one identity

$\mu = 0$. Left: $\partial_\nu T^{0\nu} = \partial_0T^{00} + \partial_iT^{0i} = \tfrac1c\big(\partial_tu + \nabla\cdot\vec S\big)$. Right: $-F^0{}_\nu J^\nu = -F^0{}_jJ^j = -\tfrac{E_j}{c}J_j$. Multiply by $c$:

$$\partial_tu + \nabla\cdot\vec S = -\vec J\cdot\vec E ,$$

which is Poynting's theorem (unit 24), letter for letter.

$\mu = i$. Left: $\partial_0T^{i0} + \partial_jT^{ij} = \partial_t\big(S_i/c^2\big) - \partial_j\sigma_{ij}$. Right: $-\big(F^i{}_0J^0 + F^i{}_jJ^j\big) = -\big(\rho E_i + (\vec J\times\vec B)_i\big)$. So

$$\partial_t\vec g - \operatorname{div}\sigma = -\big(\rho\vec E + \vec J\times\vec B\big),\qquad \vec g = \vec S/c^2,$$

which is unit 24's momentum balance, letter for letter.

What was gained. Unit 24 proved these separately, each by a page of vector identities. Here they are the four components of one tensor equation, and the symmetry $T^{0i} = T^{i0}$ — which unit 24 could only check — is why $\vec g = \vec S/c^2$.

Check the signs in the $\mu=i$ line. $F^i{}_0 = E_i/c$ and $J^0 = c\rho$, so the first piece is $-(E_i/c)(c\rho) = -\rho E_i$ ✓. $F^i{}_j = \varepsilon_{ijk}B_k$ and $J^j$, so the second is $-\varepsilon_{ijk}J^jB_k = -(\vec J\times\vec B)_i$ ✓.

A worked special case: the plane wave. For a wave travelling along $\hat x$ with $\vec E = E\hat y$, $\vec B = (E/c)\hat z$, both invariants vanish and $u = \varepsilon_0E^2$, $\vec S = \varepsilon_0cE^2\hat x = uc\,\hat x$, $T^{xx} = u$, all other $T^{ij}$ zero. The whole tensor is then $T^{\mu\nu} = u\,n^\mu n^\nu$ with $n^\mu = (1,1,0,0)$ null, which is as compact as a stress-energy tensor gets. Radiation pressure on a perfect absorber is $T^{xx} = u = S/c$: for the solar constant $1361\ \mathrm{W\,m^{-2}}$ that is $4.540\times10^{-6}$ Pa. P8 does this properly.

Where it goes. The pair (symmetric, divergence-free) is exactly what unit 35 needs on the right-hand side of the Einstein equations, because the left-hand side is symmetric and has identically vanishing divergence by the contracted Bianchi identity of unit 33. Everything in this slide is a rehearsal.

A caution. $\partial_\nu T^{\mu\nu} = 0$ in vacuum gives a local conservation law. Turning it into a conserved total requires integrating over a spacelike slice and discarding a boundary term, which is legitimate for fields falling off fast enough; in curved spacetime the same manipulation fails, and "the total energy of a gravitating system" becomes a genuinely subtle notion. Unit 35 flags it; unit 40 returns to it.

section 5

Radiation, honestly

One invariant gives the radiated power. The recoil from it breaks the theory.

28 · radiation

Larmor, made covariant

theorem · the Liénard formula $$P = \frac{\mu_0q^2}{6\pi c}\,a^\mu a_\mu = \frac{\mu_0q^2}{6\pi c}\,\gamma^6\Big(|\vec a|^2 - \frac{|\vec v\times\vec a|^2}{c^2}\Big).$$

Why. (i) $P$ is a Lorentz scalar: in the instantaneous rest frame the dipole pattern is $\sin^2\theta$ (unit 24), so the radiated four-momentum has no spatial part there, and $\mathrm{d}E = \gamma\,\mathrm{d}E_0$, $\mathrm{d}t = \gamma\,\mathrm{d}\tau$ divide out. (ii) Assuming $P$ depends only on $u$ and $a$ and is quadratic in $a$, the only available scalar is $a^\mu a_\mu$. (iii) The constant is fixed by the rest frame, where Larmor gives $\mu_0q^2|\vec a|^2/6\pi c$.

check

Limit. $v\to0$: $\gamma\to1$, $\vec v\times\vec a\to0$, and $P\to\mu_0q^2a^2/6\pi c$, Larmor ✓ (unit 24's form with $1/\varepsilon_0c^3 = \mu_0/c$). Dimensions in the notes.

Step (i), drawn. In the instantaneous rest frame the radiation goes out as $\sin^2\theta$ about $\vec a$: as much along $-\hat n$ as along $\hat n$, so the radiated momentum sums to zero and only energy leaves.

Step (i), in full. In the instantaneous rest frame the angular distribution of dipole radiation is $\propto\sin^2\theta$ about the acceleration (unit 24, cited), which is symmetric under $\hat n\to-\hat n$. So the momentum radiated in a proper time $\mathrm{d}\tau$ cancels: $\mathrm{d}P^\mu_{\text{rad}} = (\mathrm{d}E_0/c,\ \vec 0)$. In any other frame, $\mathrm{d}E = \gamma\,\mathrm{d}E_0$ (the time component of a four-vector whose spatial part vanishes) and $\mathrm{d}t = \gamma\,\mathrm{d}\tau$. Hence $\mathrm{d}E/\mathrm{d}t = \mathrm{d}E_0/\mathrm{d}\tau$: the power is the same number in every frame.

Step (ii) is an assumption, and is labelled as one. It excludes a dependence on $\mathrm{d}a/\mathrm{d}\tau$, which is not obviously illegitimate — the radiation-reaction force of The recoil, and where classical theory stops depends on exactly that. The honest statement is: Liénard's formula is derived in Jackson (Classical Electrodynamics, 3rd edn, §14.2–14.3) from the Liénard–Wiechert fields by integrating the Poynting flux over a sphere, and the argument here is a consistency check that reproduces it. It is cited in that sense. Problem P9 walks through the covariant route in detail.

Dimensions. $[\mu_0q^2a^2/c] = \mathrm{N\,A^{-2}\cdot C^2\cdot m^2s^{-4}/(m\,s^{-1})} = \mathrm{N\,A^{-2}A^2s^2\,m\,s^{-3}} = \mathrm{N\,m\,s^{-1}} = \mathrm W$ ✓ — a power, as it must be, and the $6\pi$ is a pure number. Same check for the recoil two slides on: $[\tau_0] = [\mu_0q^2/mc] = \mathrm{N\,s^2/(kg\,m\,s^{-1})} = \mathrm s$ ✓, so $m\tau_0\dot{\vec a}$ is a force.

Full expansion of $a\cdot a$. Differentiating $u^\mu = \gamma(c,\vec v)$ with respect to $\tau = t/\gamma$ and using $\dot\gamma = \gamma^3\vec v\cdot\vec a/c^2$ gives $a^\mu = \big(\gamma^4\vec v\cdot\vec a/c,\ \gamma^2\vec a + \gamma^4(\vec v\cdot\vec a)\vec v/c^2\big)$, whence $a\cdot a = \gamma^4|\vec a|^2 + \gamma^6(\vec v\cdot\vec a)^2/c^2$. Using $|\vec v\times\vec a|^2 = v^2a^2 - (\vec v\cdot\vec a)^2$ turns this into $\gamma^6(|\vec a|^2 - |\vec v\times\vec a|^2/c^2)$ ✓; the two forms are identities of each other.

28 · radiation

Synchrotron loss: the $\gamma^4$ that ended LEP

Energy lost per turn, $q^2\gamma^4/3\varepsilon_0R$, in LEP's ring for electrons and the LHC's for protons. Both lines have slope 4; at equal energy they sit $(m_p/m_e)^4 \approx 10^{13}$ apart. Above about 330 GeV an electron in LEP's tunnel would radiate more than its whole energy each turn.

circular motion

Circular motion, $\vec v\perp\vec a$: $a\cdot a = \gamma^4a^2$, so $P\propto\gamma^4$. Energy lost per turn is $\Delta E = q^2\gamma^4/3\varepsilon_0R$, with $R$ the ring's bending radius (not $\rho$, which in this unit is charge density).

Numbers. LEP at 104.5 GeV, $R = 3096$ m: $\gamma = 2.045\times10^{5}$, $\Delta E = 3.41$ GeV per turn. LHC protons at 6.8 TeV, $R = 2804$ m: $\gamma = 7247$, $\Delta E = 5.93$ keV. A factor $5.7\times10^{5}$, and the reason LEP's tunnel now holds the LHC.

Reading the figure. Both lines are $\Delta E = e^2\gamma^4/3\varepsilon_0R$ with $\gamma = E/mc^2$, computed for $R = 3096$ m (electrons) and $R = 2804$ m (protons). On log axes a power law is a straight line whose slope is the exponent, so both have slope 4; the vertical gap between them at a fixed energy is $4\log_{10}(m_p/m_e) = 13.1$ decades, less the small difference in radius. The dashed line is $\Delta E = E$. The electron line meets it near 327 GeV, a hard ceiling for a ring of that size no matter how much RF power is installed; the practical ceiling, set by the RF, was LEP's 104.5 GeV. The way out is a bigger ring ($\Delta E\propto1/R$, so only linearly) or no ring — a linear collider, which does not radiate this way at all.

The numbers, checked. $\Delta E = P\cdot(2\pi R/v)$ with $|\vec a| = v^2/R$, so $\Delta E \approx q^2\gamma^4/(3\varepsilon_0R)$ for $v\to c$. LEP2: $\gamma = 104.5\ \mathrm{GeV}/0.511\ \mathrm{MeV} = 2.0450\times10^{5}$, giving $\Delta E = 5.459\times10^{-10}\ \mathrm J = 3.407$ GeV per turn — three per cent of the beam energy, every turn, and the reason LEP could not be pushed much past 100 GeV per beam. LHC: $\gamma = 7247$, giving $9.507\times10^{-16}\ \mathrm J = 5.93$ keV, entirely negligible as a beam loss but a real heat load on a cryogenic beam pipe. Protons win by $(m_p/m_e)^4 = 1.1\times10^{13}$ at equal energy, reduced here to $5.7\times10^{5}$ because the energies differ.

28 · radiation

The recoil, and where classical theory stops

A radiating charge loses energy, so something must push back. Matching the work done over a period to the radiated energy gives the Abraham–Lorentz force

$$\vec F_{\text{rad}} = \frac{\mu_0q^2}{6\pi c}\,\dot{\vec a} = m\tau_0\,\dot{\vec a},\qquad \tau_0 = \frac{\mu_0q^2}{6\pi mc},$$

with $\tau_0 = 6.266\times10^{-24}$ s for an electron — the light-crossing time of $\tfrac23$ of the classical electron radius $r_e = 2.818\times10^{-15}$ m.

and now the trouble

$m\dot{\vec v} = m\tau_0\ddot{\vec v} + \vec F_{\text{ext}}$ is third order. With $\vec F_{\text{ext}} = 0$ it admits $\vec a(t) = \vec a_0e^{t/\tau_0}$: a free electron that accelerates itself forever — a runaway solution.

Discarding runaways by a boundary condition at $t=+\infty$ buys pre-acceleration: the charge starts moving a time $\sim\tau_0$ before the force arrives.

Neither is acceptable, and no fully satisfactory classical repair exists. This is stated, not solved.

Status, honestly. The Abraham–Lorentz force is not a theorem of the theory built in this unit; it is a patch, obtained by energy bookkeeping over a period of periodic motion and then asserted pointwise. Its covariant version, the Lorentz–Abraham–Dirac equation $m a^\mu = qF^\mu{}_\nu u^\nu + m\tau_0(\dot a^\mu + a^\alpha a_\alpha u^\mu/c^2)$ (Dirac 1938), has the same disease. The Landau–Lifshitz reduction replaces $\dot a$ by its zeroth-order value from the external force, giving a second-order equation with no runaways; it is the equation used in practice, it agrees with LAD to the order at which either can be trusted, and it is a reduction of order, not a derivation.

Why $\tau_0$ is the scale at which to stop believing. $c\tau_0 = 1.88\times10^{-15}$ m, comparable to $r_e$. At that scale the electron's own Coulomb energy exceeds $mc^2$, and quantum electrodynamics has long since taken over — the Compton wavelength $\hbar/mc = 3.9\times10^{-13}$ m is two hundred times larger. So the runaway is a symptom of extrapolating a classical point-charge theory past its domain, and the honest summary is that classical electrodynamics is an effective theory with a known, computable, and rather small domain of validity at the bottom end.

The self-energy, in one line. A uniformly charged shell of radius $a$ has field energy $\sim q^2/8\pi\varepsilon_0a$, which diverges as $a\to0$. Every difficulty on this slide traces to that divergence, and the modern name for the cure is renormalisation. Rohrlich's Classical Charged Particles is the standard book-length treatment; chapters 2, 6 and 9 need nothing this course has not taught, while the rest assumes quantum field theory.

Is it measurable? Radiation reaction is real and routinely important — in synchrotrons it is the damping that makes electron beams shrink, and in ultra-intense laser fields it dominates. What is not measurable is the pathology: no experiment has ever probed the $\tau_0$ timescale, so the runaways are a defect of the mathematics, not a prediction that has been refuted.

28 · radiation

Runaway and pre-acceleration, plotted

A force $F_0$ switched on at $t = 0$. Without radiation reaction the acceleration would jump straight from $0$ to $F_0/m$ (dotted). With it, either the charge runs away — shown with no force at all — or, if runaways are forbidden, it starts accelerating before the force arrives.

Write the equation as $\ \vec a - \tau_0\dot{\vec a} = \vec F_{\text{ext}}/m$. Its general solution is any particular one plus $\vec a_0e^{t/\tau_0}$, the runaway.

The only solution that stays bounded as $t\to+\infty$ is

$$\vec a(t) = \frac1{\tau_0}\int_t^\infty e^{(t-s)/\tau_0}\,\frac{\vec F_{\text{ext}}(s)}{m}\,\mathrm{d}s ,$$

an average of the force over the future, weighted over a time $\tau_0$.

for the step

$t > 0$: $a = F_0/m$. $\ t \lt 0$: $a = (F_0/m)\,e^{t/\tau_0}$ — nonzero before the force exists. For an electron the lead time is $\sim 6\times10^{-24}$ s.

Check the formula. Differentiate under the integral: $\dot{\vec a} = -\frac1{\tau_0}\vec F_{\text{ext}}(t)/m + \frac1{\tau_0}\vec a$, so $\vec a - \tau_0\dot{\vec a} = \vec F_{\text{ext}}/m$ ✓. Adding $\vec a_0e^{t/\tau_0}$ still solves the equation, but grows without bound; demanding boundedness as $t\to+\infty$ forces $\vec a_0 = 0$. That demand is the "boundary condition at $t = +\infty$" of the previous slide, and the integral shows its price: $\vec a(t)$ depends on the force at later times.

For the step. With $F_{\text{ext}} = F_0$ for $s > 0$ and $0$ before: if $t > 0$ the integral is $\frac{F_0}{m\tau_0}\int_t^\infty e^{(t-s)/\tau_0}\mathrm{d}s = F_0/m$; if $t \lt 0$ only $s > 0$ contributes and the integral is $\frac{F_0}{m\tau_0}e^{t/\tau_0}\int_0^\infty e^{-s/\tau_0}\mathrm{d}s = (F_0/m)e^{t/\tau_0}$. That is the solid curve. The runaway curve is drawn with $a_0 = 0.08\,F_0/m$, an arbitrary small start; any $a_0 \ne 0$ gives the same shape shifted in time.

What to look for. Every curve is smooth on the scale $\tau_0$, so on any timescale an experiment can reach — nanoseconds against $10^{-23}$ s — the pre-accelerating solution is indistinguishable from the step. The pathology is real in the mathematics and invisible in the laboratory, which is the honest summary of the previous slide.

28 · closing

What this bought

the wire, closed

Lab: $3.000\times10^{-8}$ N, magnetic.
Charge's frame: $3.00015\times10^{-8}$ N, electrostatic.
Ratio $\gamma_v$, exactly.
$\mathcal I_1 = 1.0000\times10^{-10}\ \mathrm{T^2}$ in both.

the objects

$F_{\mu\nu}$ antisymmetric, six components.
$J^\mu$, with continuity as a corollary.
$\tilde F$, $\mathcal I_1$, $\mathcal I_2$.
$T^{\mu\nu}$: symmetric, traceless, with $\partial_\nu T^{\mu\nu} = -F^\mu{}_\nu J^\nu$.

what uses it

29: $T^{\mu\nu}$ from Noether, honestly.
30: the equivalence principle.
32–33: the same index calculus with $\eta\to g$ and $\partial\to\nabla$.
38: a wave equation for a tensor, sourced by $T$.

The one habit to carry forward: write the tensor equation first, then choose a frame. Every calculation in this unit that was hard in three-vector notation — Poynting's theorem, the momentum balance, the field transformations — became a line of index bookkeeping, and every one that was easy stayed easy.

The bridge to units 32 onwards is narrower than it looks. In curved spacetime, $\eta_{\mu\nu}$ becomes $g_{\mu\nu}(x)$ and $\partial_\mu$ becomes $\nabla_\mu$; the field tensor is still $F_{\mu\nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu$ — and, because the connection is symmetric, that equals $\partial_\mu A_\nu - \partial_\nu A_\mu$ unchanged. Maxwell's equations in curved spacetime are the ones on these slides with one symbol replaced. That is not a coincidence; it is the point of writing them this way.

What was skipped. Radiation from an arbitrary worldline (the Liénard–Wiechert fields), the covariant treatment of media, and the Lagrangian that produces Maxwell's equations. The last is unit 29's opening move: $\mathcal L = -\frac1{4\mu_0}F_{\mu\nu}F^{\mu\nu} - A_\mu J^\mu$, and you now have every object in it.

Reading. Purcell & Morin chapter 5 for the wire, done at length and with pictures; Griffiths Introduction to Electrodynamics §12.3 for the field tensor at this level, skipping §12.1–12.2 if units 26 and 27 are solid; Carroll Spacetime and Geometry §1.8–1.9 for the conventions used here. Jackson chapter 11 is the thorough treatment but uses $+---$; convert one entry before reading it.

28 · checkpoint

Checkpoint

you can now

Given $F^{\mu\nu}$ and a boost, derive the $\vec E$ and $\vec B$ transformation rules and verify both invariants numerically; and show by explicit computation that $\partial_\nu T^{\mu\nu} = -F^\mu{}_\nu J^\nu$, reading Poynting's theorem off the time component and the momentum balance off the spatial ones.

it is false if

you cannot say which index of $F$ the divergence is taken in and why the sign depends on it; or you compute $\mathcal I_1$ in two frames and get two numbers; or the proof of the divergence identity stops at "the other terms cancel".

to check it

Close the deck. On paper: the six mixing rules from $F'^{\mu\nu} = \Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta F^{\alpha\beta}$; then $\mathcal I_1$ for the wire in both frames, to five figures; then $\partial_\nu T^{\mu\nu}$ in full, Bianchi step included. Then ./mr problems 28.