Part C · the analytic toolkit · spine

Maxwell: the displacement current, and light

One term, added because Ampère's law contradicted charge conservation, turns four field equations into a wave equation whose speed is computed from two bench-top constants — and it is the speed of light.

needs 23, 20, 18~3 h slides + 6 h problemsproblems: ./mr problems 24

Unit 23 ended with a warning. The Helmholtz theorem says the curl data of a field must be divergence-free, and Ampère's law $\nabla\times\vec B = \mu_0\vec J$ therefore requires $\nabla\cdot\vec J = 0$. Whenever charge piles up somewhere — on the plate of a capacitor, on an antenna, in any circuit carrying alternating current — that is false. This unit repairs the law, and the repair is forced: there is one term that does it, and it is Maxwell's displacement current. The consequence of that one term is that electric and magnetic fields propagate as waves, at a speed fixed by two constants measured with batteries, coils and a balance. That speed is the speed of light.

Two pieces of mathematics arrive with the physics. First, the wave operator in three dimensions and its retarded Green's function, computed by unit 20's contour methods: every Fourier mode of the field is unit 20's undamped oscillator, and the $i0$ prescription decides that light is emitted after the charge moves, not before. Second, gauge freedom: the potentials carry more information than the fields, and the extra is pure redundancy, which a good choice of gauge spends to make the equations simple.

How to read this deck

One object runs through it: a parallel-plate capacitor being charged by a steady current. Section 1 adds Faraday's law; section 2 finds that Ampère's law fails between the plates and repairs it; section 3 assembles the complete system and follows energy and momentum into the capacitor; section 4 derives light and computes its speed; section 5 writes the capacitor's potentials in two gauges; section 6 computes the retarded Green's function and asks how soon the rest of the world hears that the capacitor was switched on; section 7 asks whether it radiates.

Every number quoted was computed. Proofs are on the slides where they are short; the longer ones are in these notes under Full proof, and anything imported is labelled cited with where its proof lives. Terms with a dotted underline are glossary entries — click one, or press G. N opens the notes, R stacks the deck into one readable page.

24 · overview

One capacitor, one missing term, seven sections

Two discs 5 cm in radius, 1 mm apart, charged through a wire at 1 A. Is there a magnetic field between the plates, where no charge flows? Ampère's law gives two answers, depending on the surface chosen. Repairing it predicts light.

the route · from a contradiction to light

1. Through the flat $S_1$ the wire carries $I$; through the bag $S_2$ nothing does. Ampère's law needs both fluxes equal, i.e. $\nabla\cdot\vec J = 0$ — false on a charging plate.

2. Charge conservation fixes the one term that repairs it: $\nabla\times\vec B = \mu_0\vec J + \mu_0\varepsilon_0\,\partial_t\vec E$. In the gap, $\varepsilon_0\partial_t\vec E$ carries the current $I$ across.

3. In empty space, take the curl of Faraday's law and use the repaired law: $\nabla^2\vec E = \mu_0\varepsilon_0\,\partial_t^2\vec E$, a wave equation.

4. Its speed is $1/\sqrt{\varepsilon_0\mu_0} = 2.998\times10^8\ \mathrm{m\,s^{-1}}$, from two bench-top constants — the speed of light.

One loop $\gamma$, two surfaces it bounds. $S_1$ is pierced by the wire; $S_2$ slips between the plates, where no current flows.

The plan. 1 Faraday's law · 2 the repair, forced by continuity · 3 the complete system, its energy and momentum · 4 light: $c$, plane waves, polarisation · 5 potentials and gauge · 6 the retarded Green's function, by residues · 7 radiation, and a failure recorded.

The route, step by step. Step 1 is Ampère's law meets a charging capacitor and Where the inconsistency lives in section 2. Step 2 is Worked: the displacement current, from continuity alone, and Exactly one term shows that no other term would do. Step 3 is The vacuum wave equation in section 4; it uses Faraday's law, which section 1 supplies, and the identity $\nabla\times\nabla\times = \nabla\nabla\cdot{} - \nabla^2$ of unit 23. Step 4 is Worked: the speed of light from two bench-top constants, which also asks how good the 1860s measurements had to be.

Getting around. A dotted underline is a glossary term — click it, or press G. A solid underline like the capacitor above is the running example: hover it. N shows these notes, where the long proofs are.

The problem sheet follows the same order. P1–P3 are the displacement current and its consequences (dimensions and $c$; the capacitor driven at 50 Hz and at 1 GHz; the proof that the constraint equations propagate); P4–P6 are energy, momentum and plane waves (a resistor fed from the side, the force between two hemispheres by the stress tensor, sunlight and an ellipse of polarisation); P7–P8 are gauge, ending with a moving point charge in Lorenz and Coulomb gauge; P9–P11 are the wave operator's Green's functions — a second route to $G_\mathrm{ret}$, Kirchhoff's formula focusing a pulse, and the two-dimensional tail; P12–P13 are radiation, ending with the classical atom's collapse time.

Why this example. It is the configuration in which Ampère's law fails most visibly, it is small enough that every field in it can be computed exactly, and it touches every section: its $\vec B$ between the plates is the displacement current made visible; its energy enters through the rim, not along the wire; its plates attract with the pressure the stress tensor predicts; its potentials are the simplest non-trivial example of two gauges; and whether it radiates is a question the Larmor formula answers with a number.

Units and constants. SI throughout, as in unit 23: $\varepsilon_0 = 8.8541878128\times10^{-12}\ \mathrm{F\,m^{-1}}$, $\mu_0 = 1.25663706212\times10^{-6}\ \mathrm{N\,A^{-2}}$. The deck writes $c$ for $1/\sqrt{\varepsilon_0\mu_0}$ from section 4 on, once it has earned the name. Spatial coordinates near the capacitor are cylindrical, $(s, \phi, z)$, with the axis along the wire.

section 1

Faraday: a changing magnetic field curls the electric one

The one empirical law this unit adds.

24 · Faraday

Faraday's law

electromotive force

For a closed curve $\gamma = \partial S$, the work per unit charge done by the electric field once round it: $\mathcal E = \oint_{\partial S}\vec E\cdot\mathrm d\vec r$, in volts. For a static field it is $0$: $\vec E = -\nabla\varphi$ (unit 04).

Faraday's law · measured

For every fixed oriented surface $S$ with boundary $\partial S$, $$\oint_{\partial S}\vec E\cdot\mathrm d\vec r = -\frac{\mathrm d}{\mathrm dt}\int_S\vec B\cdot\mathrm d\vec A .$$

the minus sign is Lenz's law

The induced current opposes the change in flux. With a plus sign a growing flux would drive a current that increased it: a runaway, and a free source of energy.

The upward normal $\hat n$ orients $\partial S$ anticlockwise seen from above. The flux of $\vec B$ grows, so $\oint\vec E\cdot\mathrm d\vec r < 0$: the induced $\vec E$ runs clockwise, and a current along it would make a downward $\vec B$ inside the loop — opposing the growth.

What is measured and what is derived. Faraday's law enters here as Coulomb's law and Biot–Savart entered unit 23: as an empirical input. Faraday found it in 1831 by moving a magnet through a coil and by switching on the current in a second coil wound on the same iron ring; the galvanometer kicked when the current changed and not while it was steady. Everything else in this section is derived.

Reading the figure. The orientation convention is the one of Stokes' theorem (unit 04): the boundary is traversed so that the surface lies on the left when the normal points to your head. With $\hat n$ up that is anticlockwise from above. The flux $\int_S\vec B\cdot\mathrm d\vec A$ is positive and growing, so the law makes the circulation negative — the induced field, drawn in the second colour, runs the other way round. A wire laid along $\partial S$ would carry a current that way, and by the right-hand rule that current makes a field pointing down through $S$, against the change. That is all Lenz's law says.

Units. $[\partial_t B\cdot\text{area}] = \mathrm{T\,m^2\,s^{-1}} = \mathrm{V}$, since $\mathrm T = \mathrm{V\,s\,m^{-2}}$ (from $F = qvB$: $\mathrm N = \mathrm C\cdot\mathrm{m\,s^{-1}}\cdot\mathrm T$, and $\mathrm N\,\mathrm m = \mathrm{C\,V}$) ✓. A weber, $1\ \mathrm{T\,m^2}$, is a volt-second.

Moving loops. The law is stated for a fixed surface. For a loop that moves, the change of flux has a second source — the loop sweeping through a static $\vec B$ — and the EMF round the moving wire then comes from the Lorentz force $q\vec v\times\vec B$ of unit 23, not from $\vec E$. That the two cases give the same "flux rule" was, for Einstein, the first sign that something was wrong with the separation into electric and magnetic fields; unit 25 takes it up.

24 · Faraday

Faraday's law, pointwise

differential form

For $C^1$ fields, Faraday's law holds for every $S$ iff $\ \nabla\times\vec E = -\partial_t\vec B$.

proof

Stokes (unit 04) on the left, differentiation under the integral (unit 18) on the right: $\int_S(\nabla\times\vec E + \partial_t\vec B)\cdot\mathrm d\vec A = 0$ for every $S$. Take $S$ a small flat disc with normal $\hat n$ at a point; a continuous integrand whose flux through every such disc vanishes has $\hat n$-component zero (unit 05's localisation lemma, applied in the plane of the disc). $\square$

The capacitor, charged at steady current, has a $\vec B$ that does not change (section 2 computes it), so $\nabla\times\vec E = \vec 0$ there and the electric field between the plates is still a gradient.

Faraday adds nothing to the running example — which is why the answer section 2 finds is exact, not approximate. It matters as soon as the current varies: the closing slide measures by how much.

consistent with no monopoles

The divergence of a curl vanishes, so $\partial_t(\nabla\cdot\vec B) = -\nabla\cdot(\nabla\times\vec E) = 0$: a field with no sources at one instant has none later.

The localisation step, in full. Let $\vec F = \nabla\times\vec E + \partial_t\vec B$, continuous, and suppose $\vec F(\vec x_0)\cdot\hat n = \eta > 0$ for some unit $\hat n$. By continuity $\vec F\cdot\hat n > \eta/2$ on a small disc centred at $\vec x_0$ with normal $\hat n$, so the flux through that disc is positive, contradicting the hypothesis. Hence $\vec F\cdot\hat n = 0$ for every $\hat n$ at every point. The converse direction is Stokes read backwards.

Consistency with $\nabla\cdot\vec B = 0$. The divergence of a curl vanishes (unit 23), so Faraday's law implies $\partial_t(\nabla\cdot\vec B) = 0$: a field with no sources at one instant has none later. Faraday's law and the absence of monopoles are compatible, and the next slide asks whether Ampère's law is compatible with charge conservation in the same way.

24 · Faraday

Four equations, one of them suspect

What units 23 and this one have established, for fields that may now depend on time:

equation	source	status
$\nabla\cdot\vec E = \rho/\varepsilon_0$	Coulomb + Gauss	kept
$\nabla\cdot\vec B = 0$	Biot–Savart	kept
$\nabla\times\vec E = -\partial_t\vec B$	Faraday	new, measured
$\nabla\times\vec B = \mu_0\vec J$	Ampère	suspect

Two divergences and two curls: exactly what Helmholtz asked for in unit 23. The question is whether the curl data are divergence-free, as Helmholtz also demanded.

Faraday's curl data pass

$\nabla\cdot(-\partial_t\vec B) = -\partial_t(\nabla\cdot\vec B) = 0$ ✓.

Ampère's do not, in general

$\nabla\cdot(\mu_0\vec J) = -\mu_0\partial_t\rho$ by the continuity equation. Zero only when no charge accumulates anywhere — magnetostatics. The capacitor's plates accumulate charge at $1\ \mathrm{C\,s^{-1}}$.

Keeping Gauss's law for time-dependent fields is itself a hypothesis: that the field of a charge is determined by the charge now. It survives every test; unit 28 shows why it had to.

Why keep Gauss's law at all? In statics it was derived from Coulomb's law. For moving charges there is no Coulomb's law to derive it from, and taking it over is an assumption. What supports it: charge is measured by the flux of $\vec E$ through a surrounding surface, and the charge of an electron is the same whatever it is doing — a hydrogen atom and a helium atom are both neutral to about one part in $10^{20}$, although the electrons in them move at quite different speeds. That is the empirical content. Unit 28 shows it is forced once $\vec E$ and $\vec B$ are recognised as parts of one tensor field on spacetime.

Why keep $\nabla\cdot\vec B = 0$? The same kind of reason: no magnetic charge has been found, at rest or in motion. Faraday's law makes it self-consistent (previous slide).

The continuity equation was proved in unit 23's Lorentz-force slide from the integral statement of charge conservation by the localisation lemma. It is not a consequence of the field equations yet; by the end of the next section it will be.

section 2

Maxwell's repair

A term added on purely logical grounds, whose consequence is light.

24 · the repair

Ampère's law meets a charging capacitor

Integral form of Ampère, from unit 23: for a loop $\gamma$ and any surface $S$ with $\partial S = \gamma$,

$$\oint_\gamma\vec B\cdot\mathrm d\vec r = \mu_0\int_S\vec J\cdot\mathrm d\vec A .$$

Take $\gamma$ a circle of radius $s > a$ round the wire, and two surfaces it bounds:

$S_1$, the flat disc, pierced by the wire: right side $\mu_0I$.
$S_2$, the bag between the plates: no current crosses it, right side $0$.

Same left side, two different right sides. The law is not wrong about $\vec B$; it is inconsistent.

Side view. $S_1$ and $S_2$ share the loop $\gamma$; together they enclose the shaded region $V$, which holds the left plate and its growing charge $Q$. Both are oriented along the current; as part of $\partial V$, $S_1$'s outward normal points back along the wire.

Reading the figure. The loop is drawn edge-on, as an ellipse round the wire, and its radius $s$ is larger than the plates' radius $a$, so the bag $S_2$ can pass round the left plate's edge and through the gap without touching either plate. Any surface spanning $\gamma$ is either pierced by the wire or passes through the gap; these two are the representatives.

History. Maxwell met the problem in 1861–62, in On Physical Lines of Force, from a mechanical model of the aether in which elastic displacement of the medium carried a current; hence the name "displacement current", which survives although the model does not. The logical argument of Worked: the displacement current, from continuity alone — that the divergence of Ampère's law contradicts continuity — is how the result is now taught, and is the form Maxwell himself gave in the 1865 Dynamical Theory of the Electromagnetic Field, where the model was dropped.

24 · the repair

Where the inconsistency lives

two surfaces make a closed one

$S_1$ and $S_2$ together bound the region $V$ of the previous figure, containing one plate. With outward normal on $\partial V$, $$\int_{S_1}\vec J\cdot\mathrm d\vec A - \int_{S_2}\vec J\cdot\mathrm d\vec A = -\oint_{\partial V}\vec J\cdot\mathrm d\vec A = \frac{\mathrm dQ_V}{\mathrm dt}\ne0 .$$

Ampère's law needs the flux of $\vec J$ to depend only on the loop, and by the divergence theorem that holds iff $\nabla\cdot\vec J = 0$. On a charging plate, $\mathrm dQ/\mathrm dt = I = 1\ \mathrm A$.

a geometric fact, not a measurement

No experiment is needed to see the problem. Ampère's law in integral form is simply not a well-defined statement unless $\vec J$ has zero divergence.

Unit 23 already knew this: its Helmholtz slide showed that curl data must be divergence-free. The capacitor is the physical case in which that hypothesis fails.

The orientation, carefully. Orient both surfaces so that their normals point along the current, to the right in the figure, which matches the circulation of $\gamma$ by the right-hand rule. $\partial V$ consists of $S_1$ with normal reversed (pointing out of $V$, to the left) and $S_2$ with its own normal (pointing out of $V$, to the right, since $S_2$ bulges to the right and wraps the left plate). So $\oint_{\partial V}\vec J\cdot\mathrm d\vec A = -\int_{S_1} + \int_{S_2}$, and charge conservation for $V$ says this equals $-\mathrm dQ_V/\mathrm dt$. That is the displayed line.

What unit 23 already said. Its Biot–Savart proof used $\nabla\cdot\vec J = 0$ at exactly one step, to show $\nabla\cdot\vec A_0 = 0$; and its Helmholtz slide showed that curl data must be divergence-free. This slide is the physical case in which that hypothesis fails.

24 · the repair · example

Worked: the displacement current, from continuity alone

worked example 1 — take the divergence of Ampère's law, and repair it

Step 1 — the contradiction. The divergence of a curl is zero, so Ampère gives $0 = \nabla\cdot(\nabla\times\vec B) = \mu_0\nabla\cdot\vec J$. Continuity gives $\nabla\cdot\vec J = -\partial_t\rho$. Together: $\partial_t\rho = 0$ everywhere — false on the plates.

Step 2 — what the repair must do. Write $\nabla\times\vec B = \mu_0\vec J + \vec X$. Taking divergences,

$$\nabla\cdot\vec X = -\mu_0\nabla\cdot\vec J = \mu_0\partial_t\rho .$$

Step 3 — express $\partial_t\rho$ in fields. Gauss: $\rho = \varepsilon_0\nabla\cdot\vec E$, so $\mu_0\partial_t\rho = \nabla\cdot(\mu_0\varepsilon_0\partial_t\vec E)$. So $\vec X = \mu_0\varepsilon_0\partial_t\vec E$ works:

$$\nabla\times\vec B = \mu_0\vec J + \mu_0\varepsilon_0\,\partial_t\vec E .$$

Step 4 — the capacitor. Between the plates $E = Q/\varepsilon_0\pi a^2$ (unit 23's surface-charge jump), so the displacement current density is

$$\varepsilon_0\partial_tE = \frac{I}{\pi a^2} = \frac{1.00}{\pi(0.0500)^2} = 127.3\ \mathrm{A\,m^{-2}} .$$

Its flux through the plate area is exactly $I$: the current that stops at the plate continues across the gap as $\varepsilon_0\partial_t\vec E$.

Step 5 — both surfaces agree. Through $S_2$ the new term contributes $\mu_0\varepsilon_0\int\partial_t\vec E\cdot\mathrm d\vec A = \mu_0I$ ✓, the value $S_1$ gave from $\vec J$.

Checks. $[\varepsilon_0\partial_tE] = \mathrm{C\,V^{-1}m^{-1}}\cdot\mathrm{V\,m^{-1}s^{-1}} = \mathrm{A\,m^{-2}}$ ✓. Statics: $\partial_t\vec E = 0$ returns Ampère ✓. $\partial_tE = 1.438\times10^{13}\ \mathrm{V\,m^{-1}s^{-1}}$ — a large rate for a small effect.

Step 1, the one identity. $\nabla\cdot(\nabla\times\vec B) = 0$ for $C^2$ fields by Clairaut; it was proved in unit 23's notes to the Biot–Savart slide.

Step 4, the field between the plates. Ignoring fringing, the plates carry $\pm Q/\pi a^2$ per unit area and the field between them is $\sigma/\varepsilon_0$, by the jump condition across each plate (unit 23). Fringing is a correction of relative size about $d/a = 0.02$ and does not affect the flux argument, which uses only Gauss's law on a box round the left plate: whatever the shape of the field, its flux out of the plate's box is $Q/\varepsilon_0$, so $\varepsilon_0\frac{\mathrm d}{\mathrm dt}(\text{flux}) = I$ exactly.

Step 5, why it works for every surface. The repaired right-hand side has zero divergence by construction: $\nabla\cdot(\vec J + \varepsilon_0\partial_t\vec E) = \nabla\cdot\vec J + \partial_t\rho = 0$. So its flux through a surface depends only on the boundary loop, by the argument of the previous slide, and the law is well defined. The combination $\vec J + \varepsilon_0\partial_t\vec E$ is sometimes called the total current; it is the thing that is really conserved round a circuit containing a capacitor.

The number. $\pi a^2 = 7.854\times10^{-3}\ \mathrm{m^2}$, so $I/\pi a^2 = 127.32\ \mathrm{A\,m^{-2}}$ and $\partial_tE = 127.32/8.8542\times10^{-12} = 1.4380\times10^{13}\ \mathrm{V\,m^{-1}s^{-1}}$. Equivalently, with $C = \varepsilon_0\pi a^2/d = 69.54\ \mathrm{pF}$, the voltage climbs at $I/C = 1.438\times10^{10}$ volts per second: a kilovolt in $69.5$ ns.

Is it unique? Step 3 found a repair. Any other differs from it by a field with zero divergence, and the next slide shows that, under four stated hypotheses, that field must vanish.

24 · the repair

Exactly one term

theorem · uniqueness of the repair

Let $\vec X$ be added to Ampère's law so that it is consistent with continuity for all fields. Suppose $\vec X$ is (i) linear in $\vec E, \vec B$, (ii) local, with at most first derivatives, (iii) zero for static fields, (iv) consistent with mirror symmetry. Then $\vec X = \mu_0\varepsilon_0\partial_t\vec E$.

mirror symmetry · parity

Under $P:\vec x\mapsto-\vec x$: $\vec v\mapsto-\vec v$, $\vec F = m\vec a\mapsto-\vec F$. For $q(\vec E + \vec v\times\vec B)$ to flip, $\vec E\mapsto-\vec E$ but $\vec B\mapsto+\vec B$, because $(-\vec v)\times\vec B = -(\vec v\times\vec B)$.

proof

By (i)–(ii), $\vec X = \alpha\vec E + \beta\vec B + a\,\partial_t\vec E + b\,\partial_t\vec B + p\,\nabla\times\vec E + r\,\nabla\times\vec B$ — the only vectors linear in the fields with at most one derivative that commute with rotations.
(iii): static fields may have $\vec E, \vec B, \nabla\times\vec B\ne\vec 0$, so $\alpha = \beta = r = 0$. Faraday makes $\nabla\times\vec E = -\partial_t\vec B$, so $p$ merges into $b$.
(iv): the left side $\nabla\times\vec B$ flips sign under $P$ ($\nabla\mapsto-\nabla$, $\vec B$ fixed), as $\partial_t\vec E$ does; $\partial_t\vec B$ does not. So $b = 0$.
Consistency: $\nabla\cdot(a\partial_t\vec E) = a\,\partial_t\rho/\varepsilon_0$ must be $\mu_0\partial_t\rho$, so $a = \mu_0\varepsilon_0$. $\square$

Continuity fixed the divergence of $\vec X$; symmetry and locality fixed the rest. That is what "a term added on purely logical grounds" means.

Full proof of the list of candidates

A term linear in the fields and local with at most one derivative is a linear combination of the components of $\vec E$, $\vec B$ and their first partial derivatives in $t$ and $\vec x$, with constant coefficients (homogeneity of space and time). It must be a vector: if the coordinate axes are rotated, $\vec X$ must rotate as $\nabla\times\vec B$ does, or the law would single out a direction. The vectors that can be built linearly from a vector field $\vec E$ and its first derivatives, rotating correctly, are $\vec E$, $\partial_t\vec E$ and $\nabla\times\vec E$; the other rotation-covariant combinations of first derivatives are the scalar $\nabla\cdot\vec E$ (not a vector) and the symmetric traceless part of the derivative matrix (a different kind of object, with five components). This is the decomposition of the $3\times3$ matrix $\partial_iE_j$ into trace, antisymmetric part and symmetric traceless part — and the antisymmetric part of a $3\times3$ matrix is a vector, the curl, by unit 03's identification of antisymmetric matrices with $\mathbb R^3$ through the cross product. The same for $\vec B$. That is the six-term list.

Parity, carefully. The mirror $P$ maps a solution to a solution if the law is mirror-symmetric. Under $P$, $\partial_i\mapsto-\partial_i$ and $\partial_t\mapsto\partial_t$. So $\nabla\times\vec B$ becomes $(-\nabla)\times\vec B = -\nabla\times\vec B$, $\vec J\mapsto-\vec J$ (charges move the other way), and $\partial_t\vec E\mapsto-\partial_t\vec E$: all three flip, consistently. But $\partial_t\vec B\mapsto+\partial_t\vec B$, which would make the equation fail in the mirror. Biot–Savart confirms the transformation of $\vec B$: $\vec J\mapsto-\vec J$ and $\vec x-\vec y\mapsto-(\vec x-\vec y)$ leave the integrand unchanged.

Is mirror symmetry true? For electromagnetism, yes: no electromagnetic experiment has ever distinguished a process from its mirror image. (The weak nuclear force does distinguish them — Wu's cobalt-60 experiment, 1957 — but it does not enter Maxwell's equations.) Hypothesis (iv) is therefore empirical, and it is the one that can be questioned; the others are the definition of "the simplest local field theory".

What "unique" does not say. It does not exclude terms with more derivatives, or nonlinear ones. Those would be further corrections; experiment, not logic, says they are absent at ordinary field strengths.

24 · the repair

The field between the plates

Symmetry about the axis gives $\vec B = B(s)\hat\phi$ (unit 23's argument for a wire), with $s$ the distance from the axis. Ampère–Maxwell on a circle of radius $s$ between the plates, through the flat disc it bounds:

$$2\pi s\,B = \mu_0\varepsilon_0\,\partial_tE\cdot\pi\min(s,a)^2 .$$

Inside, the loop encloses the fraction $s^2/a^2$ of the displacement current $I$; outside, all of it.

the answer $$B = \frac{\mu_0Is}{2\pi a^2}\ (s\le a),\quad B = \frac{\mu_0I}{2\pi s}\ (s\ge a).$$

At the rim: $\mu_0I/2\pi a = 4.00\ \mu\mathrm T$, the same as beside the wire at the same distance.

Looking along the axis into the gap. The dots are $\varepsilon_0\partial_t\vec E$, uniform over the plate. A loop of radius $a/2$ encloses a quarter of it, so $B$ there is half its rim value; at $2a$ it encloses all of it at twice the distance — half again.

The symmetry argument. Rotation about the axis and reflection in any plane containing the axis map the configuration to itself. Under the reflection, $\vec B$ transforms as a mirror-fixed vector (previous slide): its components in the plane of the mirror flip and its component normal to the mirror is kept. A field that is invariant must therefore be normal to every plane through the axis at points in that plane, i.e. azimuthal. Rotation invariance makes the magnitude depend on $s$ and $z$ only; the $z$-dependence drops out between plates far from the edges.

Reading the figure. The arrows are drawn with length proportional to $|\vec B|$: $\frac12$, $1$, $\frac12$ of the rim value on the three loops. Inside the plate, $2\pi sB$ is proportional to the enclosed flux $\pi s^2$, so $B\propto s$; outside, the enclosed flux stops growing and $B\propto1/s$. The direction is the right-hand rule applied to the displacement current coming out of the page, exactly as for a current.

Dimensions. $[\mu_0I/s] = \mathrm{N\,A^{-2}}\cdot\mathrm A\cdot\mathrm m^{-1} = \mathrm{N\,A^{-1}m^{-1}} = \mathrm T$ ✓. Limit $s\to0$: $B\to0$ on the axis, as symmetry demands ✓.

24 · the repair

$B$ against distance from the axis

why this is exact

With $I$ constant, $\partial_t\vec E$ is constant, so $\vec B$ is static; Faraday then gives $\nabla\times\vec E = \vec 0$, so the uniform $\vec E$ assumed is consistent. All four equations hold between the plates, fringing apart.

what is new, and what is not

Outside the gap the answer is the wire's field: the $S_1$ computation. Inside the gap, where there is no current, there is still a magnetic field, growing linearly with $s$ — the displacement current, visible. It was measured directly only in 1985: $\mu\mathrm T$ fields need care.

$B(s)$ for the running capacitor, computed from the boxed formulas. The shaded strip is the gap's radius, where $\vec J = \vec 0$: the rising line there is made by $\varepsilon_0\partial_t\vec E$ alone. The two pieces meet at $4.00\ \mu$T.

What the graph says. Beyond the rim, a magnetometer could not tell whether it sat beside a wire or beside a capacitor gap: both give $\mu_0I/2\pi s$. Inside the rim it can — beside the wire the field would keep rising as $1/s$, while in the gap it falls linearly to zero on the axis. That difference is the measurable signature of the displacement current.

Continuity at the rim. Both formulas give $\mu_0I/2\pi a = (1.2566\times10^{-6})(1.00)/(2\pi\times0.0500) = 4.000\times10^{-6}$ T at $s = a$ ✓. This is the check that the displacement current through the whole plate area equals $I$.

History. For a century the displacement current was confirmed only through its consequences — above all electromagnetic waves (Hertz, 1887–88). A direct measurement of the magnetic field inside a capacitor was reported by Bartlett and Corle in 1985, at audio frequencies.

section 3

The complete system: energy and momentum

Four equations, and the conservation laws they imply.

24 · the system

Maxwell's equations

Maxwell's equations · in vacuum with sources $$\begin{aligned}\nabla\cdot\vec E &= \rho/\varepsilon_0, & \nabla\times\vec E &= -\partial_t\vec B,\\ \nabla\cdot\vec B &= 0, & \nabla\times\vec B &= \mu_0\vec J + \mu_0\varepsilon_0\partial_t\vec E .\end{aligned}$$

With the Lorentz force $\vec F = q(\vec E + \vec v\times\vec B)$ they are all of classical electromagnetism.

charge conservation is now a theorem

Divergence of the fourth, then the first: $0 = \mu_0\nabla\cdot\vec J + \mu_0\partial_t\rho$. $\square$

Counting, as unit 23's Helmholtz slide predicted: a divergence and a curl for each field. But they split differently in time:

the two curl equations contain $\partial_t$: they are evolution equations, stepping $(\vec E, \vec B)$ forward;
the two divergence equations do not: they are constraints on each instant's data.

the constraints propagate

If the evolution equations hold and $\partial_t\rho + \nabla\cdot\vec J = 0$, and the constraints hold at $t = 0$, they hold for all $t$.

Proof. $\partial_t(\nabla\cdot\vec B) = -\nabla\cdot\nabla\times\vec E = 0$. And $\partial_t(\varepsilon_0\nabla\cdot\vec E - \rho) = \nabla\cdot(\mu_0^{-1}\nabla\times\vec B - \vec J) - \partial_t\rho = -\nabla\cdot\vec J - \partial_t\rho = 0$. $\square$

Integral forms, by Stokes and the divergence theorem, for fixed surfaces and regions: $\oint_{\partial V}\vec E\cdot\mathrm d\vec A = Q_V/\varepsilon_0$; $\oint_{\partial V}\vec B\cdot\mathrm d\vec A = 0$; $\oint_{\partial S}\vec E\cdot\mathrm d\vec r = -\frac{\mathrm d}{\mathrm dt}\int_S\vec B\cdot\mathrm d\vec A$; $\oint_{\partial S}\vec B\cdot\mathrm d\vec r = \mu_0I_S + \mu_0\varepsilon_0\frac{\mathrm d}{\mathrm dt}\int_S\vec E\cdot\mathrm d\vec A$.

Why the count matters. Six unknown functions ($\vec E$, $\vec B$) and eight equations looks over-determined. It is not: the two constraint equations are preserved by the six evolution equations, so they restrict the initial data only. That is the precise sense in which Maxwell's system is "neither over- nor under-determined", which unit 23 promised. The evolution equations are a first-order hyperbolic system — unit 18's classification extends to systems, and the next section shows each component obeys the wave equation — so they take Cauchy data: $\vec E$ and $\vec B$ at $t = 0$, subject to the constraints.

The same structure, later. General relativity has exactly this split: of Einstein's ten equations, four are constraints on the initial data and are preserved by the other six, and the preservation is guaranteed by an identity (the contracted Bianchi identity, unit 33) playing the role that $\nabla\cdot\nabla\times = 0$ plays here. Unit 35 makes the comparison.

History. Maxwell's 1865 paper stated twenty equations in twenty unknowns, in components and with the potentials among the unknowns. The four vector equations are Heaviside's (1885) and, independently, Hertz's; "Maxwell's equations" in this form is their name for them.

24 · the system

Poynting's theorem: energy, locally

field energy density and Poynting vector $$u = \frac{\varepsilon_0}{2}|\vec E|^2 + \frac{1}{2\mu_0}|\vec B|^2,\qquad \vec S = \frac1{\mu_0}\vec E\times\vec B .$$

Poynting's theorem $$\partial_tu + \nabla\cdot\vec S = -\vec J\cdot\vec E .$$

Integrated over a fixed region $V$: the field energy in $V$ falls by what flows out through $\partial V$ plus the work done on the charges inside.

$\vec J\cdot\vec E$ is the power per unit volume the field gives the charges: $q\vec v\cdot\vec E$ per charge, and the magnetic force does no work (unit 23).

proof

Dot Faraday with $\vec B/\mu_0$ and Ampère–Maxwell with $\vec E/\mu_0$ and subtract:

$$\begin{aligned}&\tfrac1{\mu_0}\big(\vec B\cdot\nabla\times\vec E - \vec E\cdot\nabla\times\vec B\big)\\ &\quad= -\tfrac1{\mu_0}\vec B\cdot\partial_t\vec B - \vec J\cdot\vec E - \varepsilon_0\vec E\cdot\partial_t\vec E .\end{aligned}$$

The left side is $\nabla\cdot(\vec E\times\vec B)/\mu_0$ (notes); the right is $-\partial_tu - \vec J\cdot\vec E$. $\square$

what is and is not determined

Only $\nabla\cdot\vec S$ is fixed: $\vec S + \nabla\times\vec W$ satisfies the same law. The choice $\vec E\times\vec B/\mu_0$ is the one relativity confirms (unit 29).

Dimensions: $[\varepsilon_0E^2] = \mathrm{C\,V^{-1}m^{-1}}\cdot\mathrm{V^2m^{-2}} = \mathrm{J\,m^{-3}}$; $[EB/\mu_0] = \mathrm{W\,m^{-2}}$ ✓.

Full proof of the identity

$\nabla\cdot(\vec E\times\vec B) = \vec B\cdot(\nabla\times\vec E) - \vec E\cdot(\nabla\times\vec B)$. Write $\vec E\times\vec B = (E_yB_z - E_zB_y,\ E_zB_x - E_xB_z,\ E_xB_y - E_yB_x)$ and differentiate by the product rule: the six terms in which the derivative falls on $\vec E$ collect to $B_x(\partial_yE_z - \partial_zE_y) + B_y(\partial_zE_x - \partial_xE_z) + B_z(\partial_xE_y - \partial_yE_x) = \vec B\cdot\nabla\times\vec E$, and the six in which it falls on $\vec B$ collect, with the opposite sign, to $-\vec E\cdot\nabla\times\vec B$. (Six on each side: each of the three components has two products.)

The magnetic energy. $\frac1{\mu_0}\vec B\cdot\partial_t\vec B = \partial_t\frac{|\vec B|^2}{2\mu_0}$, and likewise for $\vec E$. Unit 23's electrostatic energy $\frac{\varepsilon_0}{2}\int|\vec E|^2$ is the first half of $u$; the second half is new, though for steady currents it can be checked against the work done against induced EMFs in building the current up (Faraday's law supplies the EMF).

The ambiguity. Any divergence-free addition to $\vec S$ satisfies the local law, and the energy flux through a closed surface is unchanged. Through an open surface it can differ. This is a real ambiguity in the non-relativistic theory; in unit 29, $u$, $\vec S/c$ and the momentum density become components of one symmetric tensor $T^{\mu\nu}$, and symmetry fixes the choice.

History. J. H. Poynting, 1884, and independently Heaviside the same year. Poynting's own first example was the one on the next slide's pattern: a wire carrying current, where the energy dissipated as heat enters through the wire's side (problem P4).

24 · the system · example

Worked: energy enters the capacitor through its rim

worked example 2 — follow the Poynting vector into the charging capacitor

Step 1 — the fields. Between the plates: $\vec E = E(t)\hat z$ uniform, pointing from the positive plate across the gap; $\vec B = \frac{\mu_0Is}{2\pi a^2}\hat\phi$, from the previous section.

Step 2 — the direction. $\hat z\times\hat\phi = -\hat s$. So $\vec S = \frac1{\mu_0}EB\,(-\hat s)$: radially inward, from the space round the capacitor into the gap. Nothing flows along the wire into the plates.

Step 3 — the flux through the rim. The rim is a cylinder of radius $a$ and height $d$, area $2\pi ad$. At $s = a$, $B = \mu_0I/2\pi a$:

$$P_\mathrm{in} = \frac{E}{\mu_0}\cdot\frac{\mu_0I}{2\pi a}\cdot2\pi ad = EId = VI .$$

Side view, gap exaggerated (schematic). $\vec E$ across the gap, $\vec B$ circling the axis — out of the page at the top rim, into it at the bottom — so $\vec S = \vec E\times\vec B/\mu_0$ points inward through the rim, and nothing enters along the wire.

Step 2, the orientation. In the right-handed cylindrical frame, $\hat s\times\hat\phi = \hat z$, $\hat\phi\times\hat z = \hat s$, $\hat z\times\hat s = \hat\phi$. So $\hat z\times\hat\phi = -\hat s$. Here the current flows along $+\hat z$ into the left plate, which charges positively, so $\vec E$ points along $+\hat z$ across the gap, and $\vec B$ circulates as $+\hat\phi$ by the right-hand rule, as round the wire. Inward ✓.

Reading the figure. Only the rim, the cylinder $s = a$ between the plates, is crossed by $\vec S$: on the plates' outer faces $\vec E\approx\vec 0$, and on the faces inside the gap $\vec S$ is radial, parallel to the face. The arrows at top and bottom are the same inward flow seen at two points of the rim.

Where does the energy come from, then? From the fields round the wire and the battery: the wire's surface charges (which keep the current flowing round bends) make an $\vec E$ outside the wire, the current makes a $\vec B$ round it, and $\vec E\times\vec B$ carries energy through the empty space from the battery to the capacitor. The wire guides the energy; it does not carry it. This is strange the first time and it is right: problem P4 shows the same thing for the heat produced in a resistor.

24 · the system · example

Worked: the capacitor's energy budget

worked example 2, continued — does what flows in equal what is stored?

Step 4 — the energy stored. $U = u\cdot\pi a^2d = \frac{\varepsilon_0}{2}E^2\pi a^2d$ plus a magnetic part, which is constant since $B$ is. So

$$\frac{\mathrm dU}{\mathrm dt} = \varepsilon_0E\dot E\pi a^2d = EI d = VI ,$$

using $\varepsilon_0\dot E\pi a^2 = I$ from worked example 1. The two agree, as Poynting's theorem with $\vec J = \vec 0$ in the gap requires ✓.

Step 5 — numbers. When $V = 100$ V: $E = 1.00\times10^5\ \mathrm{V\,m^{-1}}$, $|\vec S| = 3.18\times10^5\ \mathrm{W\,m^{-2}}$ at the rim, rim area $3.14\ \mathrm{cm^2}$, power $100$ W $= VI$ ✓.

Checks. $[EId] = \mathrm{V\,m^{-1}}\cdot\mathrm A\cdot\mathrm m = \mathrm W$ ✓. $I\to0$: $\vec B = \vec 0$, no flux, no change ✓.

The stored energy $\frac12CV^2$ with $V = It/C$, computed. At $V = 100$ V, $6.95$ ns after the start, its slope is $100$ W — the Poynting flux through the rim, $VI$.

Reading the figure. With steady $I$, $Q = It$ and $V = Q/C = It/C$, so $U = \frac12CV^2 = I^2t^2/2C$: a parabola. Its slope $\mathrm dU/\mathrm dt = I^2t/C = VI$ grows with $V$, which is why the inflow through the rim must grow too — it does, because $E$ (and so $|\vec S|$) grows while $B$ at the rim stays at $4.00\ \mu$T. $V = 100$ V is reached at $t = CV/I = 6.954$ ns.

Step 5 in full. $E = V/d = 100/10^{-3} = 1.00\times10^5$; $|\vec S| = EB/\mu_0 = (10^5)(4.000\times10^{-6})/(1.2566\times10^{-6}) = 3.183\times10^5\ \mathrm{W\,m^{-2}}$; $2\pi ad = 2\pi(0.05)(10^{-3}) = 3.1416\times10^{-4}\ \mathrm{m^2}$; product $100.0$ W. The stored energy at that moment is $\frac12CV^2 = \frac12(69.54\ \mathrm{pF})(100)^2 = 0.348\ \mu\mathrm J$, rising at 100 W.

The magnetic energy. With steady $I$ the magnetic energy between the plates, $\int\frac{B^2}{2\mu_0} = \frac{\mu_0I^2d}{16\pi}$, is constant, so it drops out of $\mathrm dU/\mathrm dt$.

24 · the system

Momentum, and the Maxwell stress tensor

the Maxwell stress tensor · the mathematics

At each point, the symmetric bilinear form on $\mathbb R^3$ $$\begin{aligned}T(\vec u,\vec w) &= \varepsilon_0\Big[(\vec E\cdot\vec u)(\vec E\cdot\vec w) - \tfrac12|\vec E|^2\,\vec u\cdot\vec w\Big]\\ &\ + \tfrac1{\mu_0}\Big[(\vec B\cdot\vec u)(\vec B\cdot\vec w) - \tfrac12|\vec B|^2\,\vec u\cdot\vec w\Big].\end{aligned}$$ A field of bilinear forms: a tensor field of type $(0,2)$. Its components are $T_{ij} = T(\hat e_i,\hat e_j)$.

components change as a theorem, not a definition

If $\hat e_i' = \sum_jR_{ji}\hat e_j$ with $R$ orthogonal, then $T'_{ij} = \sum_{k,l}R_{ki}R_{lj}T_{kl}$, i.e. $T' = R^\top TR$. Proof: bilinearity. $\square$

the physics · momentum balance

With field momentum density $\vec g = \varepsilon_0\vec E\times\vec B$, and $\vec f = \rho\vec E + \vec J\times\vec B$ the force per unit volume on the charges, $$\vec f + \partial_t\vec g = \Big(\sum_j\partial_jT_{ij}\Big)_{i} .$$

Full proof of the momentum balance

Eliminate the sources with Gauss and Ampère–Maxwell: $\rho = \varepsilon_0\nabla\cdot\vec E$ and $\vec J = \mu_0^{-1}\nabla\times\vec B - \varepsilon_0\partial_t\vec E$. So $\vec f = \varepsilon_0(\nabla\cdot\vec E)\vec E + \mu_0^{-1}(\nabla\times\vec B)\times\vec B - \varepsilon_0\partial_t\vec E\times\vec B$. Now $\partial_t\vec g = \varepsilon_0\partial_t\vec E\times\vec B + \varepsilon_0\vec E\times\partial_t\vec B = \varepsilon_0\partial_t\vec E\times\vec B - \varepsilon_0\vec E\times(\nabla\times\vec E)$ by Faraday. Adding, and inserting $\mu_0^{-1}(\nabla\cdot\vec B)\vec B = \vec 0$ for symmetry: $$\vec f + \partial_t\vec g = \varepsilon_0\big[(\nabla\cdot\vec E)\vec E - \vec E\times(\nabla\times\vec E)\big] + \mu_0^{-1}\big[(\nabla\cdot\vec B)\vec B - \vec B\times(\nabla\times\vec B)\big].$$ It remains to show that for any $C^1$ field $\vec V$, the $i$-th component of $(\nabla\cdot\vec V)\vec V - \vec V\times(\nabla\times\vec V)$ is $\sum_j\partial_j(V_iV_j - \frac12\delta_{ij}|\vec V|^2)$. First, $\vec V\times(\nabla\times\vec V) = \nabla\frac{|\vec V|^2}2 - (\vec V\cdot\nabla)\vec V$: the $x$-component of the left is $V_y(\partial_xV_y - \partial_yV_x) - V_z(\partial_zV_x - \partial_xV_z) = \partial_x\frac{|\vec V|^2}{2} - V_x\partial_xV_x - V_y\partial_yV_x - V_z\partial_zV_x$. Then $(\nabla\cdot\vec V)V_i + (\vec V\cdot\nabla)V_i = \sum_j\partial_j(V_jV_i)$ by the product rule, and $\partial_i\frac{|\vec V|^2}2 = \sum_j\partial_j(\frac12\delta_{ij}|\vec V|^2)$. $\square$

The transformation law. $T'_{ij} = T(\hat e_i',\hat e_j') = T(\sum_kR_{ki}\hat e_k, \sum_lR_{lj}\hat e_l) = \sum_{k,l}R_{ki}R_{lj}T_{kl}$. This is the course's habit: the tensor is the bilinear map; that its components change by $R^\top TR$ is a consequence. The matrix of $T$ is also the matrix of a linear map $\vec t:\hat n\mapsto\vec t(\hat n)$, using the dot product to identify $\mathbb R^3$ with its dual; in orthonormal bases the two agree, which is why orthonormal bases are used here. Unit 16 separates the two ideas properly and unit 22 meets the same object in an elastic solid, where the linearity in $\hat n$ has to be proved rather than read off a formula.

Dimensions. $[T] = [u] = \mathrm{J\,m^{-3}} = \mathrm{N\,m^{-2}}$ ✓; $[\vec g] = [S]/c^2 = \mathrm{kg\,m^{-2}s^{-1}}$ ✓.

24 · the system

Tension along field lines, pressure across them

Integrated over $V$: the force on the charges in $V$ is $\oint_{\partial V}T(\hat n,\cdot)\,\mathrm dA - \frac{\mathrm d}{\mathrm dt}\int_V\vec g$. The vector $\vec t(\hat n)$ with $\vec t\cdot\vec w = T(\hat n,\vec w)$ is the force per unit area transmitted across a surface with normal $\hat n$ — linear in $\hat n$ because $T$ is bilinear.

Read $T$ on a surface element with unit normal $\hat n$:

$\vec E\parallel\hat n$: $T(\hat n,\hat n) = \varepsilon_0\big[E^2 - \tfrac12E^2\big] = +\tfrac{\varepsilon_0}{2}E^2$ — the field outside pulls the region outward;
$\vec E\perp\hat n$: $T(\hat n,\hat n) = -\tfrac{\varepsilon_0}{2}E^2$ — it pushes the region inward.

The same for $\vec B$, with $\varepsilon_0$ replaced by $1/\mu_0$.

Faraday's picture, made exact by $T$: field lines behave like stretched strings that also repel their neighbours. Both magnitudes are $\tfrac{\varepsilon_0}{2}E^2$, the energy density of the field.

Sign convention. $T(\hat n,\cdot)$ is the force per unit area exerted on the region by the field outside it, across the boundary with outward normal $\hat n$. With only $\vec E$ along $\hat n$, $T(\hat n,\hat n) = +\frac{\varepsilon_0}{2}E^2$: a tension along field lines. With $\vec E\perp\hat n$, $T(\hat n,\hat n) = -\frac{\varepsilon_0}{2}E^2$: a pressure across them. Faraday pictured exactly this — field lines that pull along their length and push sideways — decades before the formula.

Reading the figure. In both panels the force acts on the shaded region, across the boundary drawn in black, from the field on the other side. The bilinear form's two terms do the work: $(\vec E\cdot\hat n)^2$ is there only when the lines cross the surface, and $-\frac12|\vec E|^2$ is there always. For a tilted field the force $\vec t(\hat n)$ is not along $\hat n$ at all — it has a shear part $\varepsilon_0(\vec E\cdot\hat n)\,\vec E_\parallel$, which is why a matrix and not a single number is needed.

24 · the system

The stress tensor at work, and the first energy–momentum tensor

The pull between the plates. Enclose the right plate in a box whose left face lies in the gap, normal $\hat n = -\hat z$; elsewhere $\vec E\approx\vec 0$. There $\vec E = E\hat z$ is parallel to $\hat n$, so

$$\vec t(\hat n)\cdot\hat z = T(-\hat z,\hat z) = -\tfrac{\varepsilon_0}{2}E^2 .$$

Over area $\pi a^2$: a force $\frac{\varepsilon_0}2E^2\pi a^2$ towards the other plate. At $V = 1$ kV: $E = 10^6\ \mathrm{V\,m^{-1}}$, pressure $4.43\ \mathrm{N\,m^{-2}}$, force $0.0348$ N.

check

Unit 23's way: charge $Q = \varepsilon_0E\pi a^2$ in the field of the other plate alone, $E/2$: force $QE/2$ — the same ✓.

The box's only face with a field on it is the one in the gap. There $\vec E\parallel\hat n$, so by the previous slide the field pulls the box — and the plate in it — towards the other plate.

The box, carefully. The force on the charges inside a region is $\oint T(\hat n,\cdot)\,\mathrm dA - \frac{\mathrm d}{\mathrm dt}\int\vec g$, and $\vec g = \varepsilon_0\vec E\times\vec B$ is tiny here and in any case time-independent for steady $I$. Only the face in the gap sees a field. On it $\hat n = -\hat z$ and $T(-\hat z,\hat z) = \varepsilon_0[(\vec E\cdot(-\hat z))(\vec E\cdot\hat z) - \frac12E^2(-\hat z\cdot\hat z)] = \varepsilon_0[-E^2 + \frac12E^2] = -\frac{\varepsilon_0}2E^2$. The magnetic part, $B^2/2\mu_0$ with $B\le4\ \mu$T, is $6.4\ \mu\mathrm{Pa}$ at most — a millionth of the electric pressure — and pushes the other way ($\vec B\perp\hat n$ gives a pressure); it is ignored.

The numbers. $\frac{\varepsilon_0}2E^2 = \frac12(8.854\times10^{-12})(10^{12}) = 4.427\ \mathrm{N\,m^{-2}}$, times $\pi a^2 = 7.854\times10^{-3}\ \mathrm{m^2}$: $0.03477$ N, the weight of about 3.5 g.

The check, in full. The field of one infinite charged sheet is $\sigma/2\varepsilon_0$ on each side; between the plates the two sheets' fields add to $\sigma/\varepsilon_0 = E$. The right plate feels only the left plate's field, $E/2$, so the force is $QE/2 = \varepsilon_0E\pi a^2\cdot E/2$ ✓. The stress tensor got this without ever separating "the field of the other plate" from "the plate's own field", which is its advantage: it needs the total field on a surface and nothing else.

24 · the system

Three densities and their fluxes

three densities and their fluxes

density	flux	law
charge $\rho$	$\vec J$	$\partial_t\rho + \nabla\cdot\vec J = 0$
energy $u$	$\vec S$	$\partial_tu + \nabla\cdot\vec S = -\vec J\cdot\vec E$
momentum $\vec g$	$-T$	$\partial_t\vec g - \operatorname{div}T = -\vec f$

Each row says the same thing: what is in a region changes by what flows through its boundary, plus what is handed to the charges inside. Charge is never handed anywhere; energy and momentum are passed between field and matter.

Note $\vec g = \vec S/c^2$ once $c$ is defined in section 4: the momentum density is the energy flux over $c^2$. The ten quantities $u$, $\vec S$, $\vec g$ and the six $T_{ij}$ become one symmetric tensor on spacetime, $T^{\mu\nu}$, in unit 29 — this is the first energy–momentum tensor of the course.

section 4

Light

The displacement current makes the fields propagate, at a speed computed from two constants.

24 · light

The vacuum wave equation

theorem · the fields obey the wave equation

Where $\rho = 0$ and $\vec J = \vec 0$, every Cartesian component of $\vec E$ and $\vec B$ ($C^2$) satisfies $$\mu_0\varepsilon_0\,\partial_t^2f = \nabla^2f ,$$ the three-dimensional wave equation with speed $1/\sqrt{\mu_0\varepsilon_0}$.

proof

Curl of Faraday, then Ampère–Maxwell: $$\nabla\times(\nabla\times\vec E) = -\partial_t\nabla\times\vec B = -\mu_0\varepsilon_0\partial_t^2\vec E .$$ Unit 23's identity: the left is $\nabla(\nabla\cdot\vec E) - \nabla^2\vec E = -\nabla^2\vec E$, by Gauss with $\rho = 0$. For $\vec B$, curl Ampère–Maxwell and use Faraday and $\nabla\cdot\vec B = 0$. $\square$

both new terms are needed

Without the displacement current the right side is $0$ and $\nabla^2\vec E = \vec 0$: no waves, only quasi-static fields. Without Faraday's $\partial_t\vec B$ the same. The wave is the two induction laws feeding each other: a changing $\vec E$ curls $\vec B$, a changing $\vec B$ curls $\vec E$.

the wave operator

Writing $c = 1/\sqrt{\mu_0\varepsilon_0}$, the d'Alembertian is $$\Box = \frac1{c^2}\partial_t^2 - \nabla^2 ,$$ and the theorem says $\Box\vec E = \Box\vec B = \vec 0$ in vacuum. Unit 18's operator, in three space dimensions.

Sign conventions for $\Box$ differ between books; this course uses the one above until unit 26 fixes a metric.

The $\vec B$ half in full. $\nabla\times(\nabla\times\vec B) = \mu_0\varepsilon_0\partial_t(\nabla\times\vec E) = -\mu_0\varepsilon_0\partial_t^2\vec B$ (with $\vec J = \vec 0$), and $\nabla\times\nabla\times\vec B = \nabla(\nabla\cdot\vec B) - \nabla^2\vec B = -\nabla^2\vec B$. Commuting $\partial_t$ with the curl is Clairaut, which is why $C^2$ is assumed.

The converse fails. Each component solving the wave equation is necessary, not sufficient: $\vec E = (f(z - ct), 0, 0)$ and $\vec B = \vec 0$ gives six solutions of the wave equation but violates Faraday. Maxwell's equations are the wave equation plus the constraints and the coupling between $\vec E$ and $\vec B$; the plane-wave slide shows what the coupling forces.

What unit 18 transfers. Finite propagation speed and uniqueness by energy, both proved there in one dimension, hold in three with the same proofs, the divergence theorem replacing the fundamental theorem of calculus. The energy is Poynting's: $\int u$. Section 6 proves the three-dimensional versions that are needed.

History. Maxwell's 1865 paper derived the wave equation for the potentials and remarked that "light is an electromagnetic disturbance propagated through the field according to electromagnetic laws". Hertz produced and detected the waves with spark gaps in 1887–88, measured their wavelength by standing waves, and found the speed agreed with light's.

24 · light · example

Worked: the speed of light from two bench-top constants

worked example 3 — compute $c$, and compare with Fizeau

Step 1 — dimensions first. $[\varepsilon_0] = \mathrm{C^2N^{-1}m^{-2}}$ (Coulomb's law), $[\mu_0] = \mathrm{N\,A^{-2}} = \mathrm{N\,s^2C^{-2}}$ (Biot–Savart). So

$$[\varepsilon_0\mu_0] = \mathrm{C^2N^{-1}m^{-2}}\cdot\mathrm{N\,s^2C^{-2}} = \mathrm{s^2\,m^{-2}} ,$$

the inverse square of a speed. Electrostatics and magnetostatics, measured separately, together contain a velocity.

Step 2 — the number. With the two constants,

$$\begin{aligned}\varepsilon_0\mu_0 &= 1.11265\times10^{-17}\ \mathrm{s^2m^{-2}},\\ c = \frac1{\sqrt{\varepsilon_0\mu_0}} &= 2.99792\times10^{8}\ \mathrm{m\,s^{-1}} .\end{aligned}$$

Step 3 — compare. Fizeau, 1849, timed light over $2\times8.6$ km with a toothed wheel: about $3.13\times10^8\ \mathrm{m\,s^{-1}}$.

$$\frac{3.13\times10^8}{2.998\times10^8} = 1.044 .$$

Fizeau was 4.4 per cent high. The electrical figure available to Maxwell in 1862 — Weber and Kohlrausch's 1856 ratio of units — was $3.1074\times10^8$, 3.65 per cent high. The two 1860s numbers agreed with each other to about 1 per cent.

Checks. $[1/\sqrt{\varepsilon_0\mu_0}] = \mathrm{m\,s^{-1}}$ ✓. Limit: if $\mu_0\to0$ (no magnetism), $c\to\infty$ — instantaneous action, the static theory ✓.

Step 2 in full. $\varepsilon_0\mu_0 = (8.8541878128\times10^{-12})(1.25663706212\times10^{-6}) = 1.112650\times10^{-17}$, whose square root is $3.335641\times10^{-9}$, and the reciprocal is $2.99792458\times10^8$. Since 2019 this is not a coincidence of measurement but of definition: the SI fixes $c$, and $\varepsilon_0$ and $\mu_0$ are measured with $\varepsilon_0\mu_0c^2 = 1$ exactly. In 1862 they were independent measurements, and that is the whole point of the comparison.

Where the electrical number came from. Weber and Kohlrausch (1856) discharged a Leyden jar whose charge had been measured electrostatically (by force between charges) through a ballistic galvanometer that measured the same charge magnetically (by the kick given to a magnet — unit 06's instrument). The ratio of the two measures of one charge is a velocity. Their "Weber constant" was $4.3945\times10^8\ \mathrm{m\,s^{-1}}$, which is $\sqrt2$ times the modern quantity because of how Weber defined his units; Maxwell used $4.3945\times10^8/\sqrt2 = 3.1074\times10^8$, quoting it in miles, as 193,088 miles per second. He quoted Fizeau as 195,647 miles per second, i.e. $3.149\times10^8$; the figure $3.13\times10^8$ above is the one usually given for Fizeau's own result, the difference being the conversion of his leagues. Either way the electrical and optical numbers agree to between 0.7 and 1.3 per cent. In Foucault's rotating-mirror measurement, also 1862, light did better: $2.98\times10^8$.

Maxwell's sentence (1862, italics his): "we can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena."

24 · light

How good did the measurements have to be?

In 1862 nobody had a reason to expect $1/\sqrt{\varepsilon_0\mu_0}$ to be anywhere near light's speed. The question for a sceptic: could the agreement be chance?

an estimate, with its assumption stated

Prior. Before the theory, the ratio of units could have been any speed; take it spread uniformly in $\log v$ over twelve decades, $1$ to $10^{12}\ \mathrm{m\,s^{-1}}$.

Chance of agreement within a fraction $\delta$. The window $[v_\mathrm{light}(1-\delta), v_\mathrm{light}(1+\delta)]$ has log-width $\approx2\delta$, so

$$\Pr\approx\frac{2\delta}{\ln10^{12}} = \frac{2\delta}{27.6} .$$

With Maxwell's two numbers, $\delta = 0.013$: $\Pr\approx1/1000$. For $\Pr$ to fall below $1/100$ they needed only $\delta\lesssim0.14$.

what the numbers say

Measurements good to ten or fifteen per cent would already have made chance an unlikely explanation. Both 1860s numbers were in fact about 4 per cent high; their mutual 1 per cent was partly luck. The identification did not need precision — it needed two numbers from unrelated experiments to land in the same place on a scale of twelve decades.

what convinced people was not the number

The coincidence made the hypothesis worth taking seriously; what established it was Hertz (1888): waves made by a circuit, reflected, refracted and polarised like light, at the same speed.

The computation. A log-uniform prior on $[1, 10^{12}]$ has density $1/(v\ln10^{12})$. The probability of $[v_0(1-\delta), v_0(1+\delta)]$ is $\ln\frac{1+\delta}{1-\delta}/\ln10^{12}\approx2\delta/27.63$. With $\delta = 0.0133$ (Weber–Kohlrausch against Maxwell's figure for Fizeau): $9.6\times10^{-4}$. Solving $2\delta/27.63 = 0.01$: $\delta = 0.138$.

The assumption is the weak point, and it should be. Twelve decades is generous at the top (no material speed near $10^{12}$ was known) and at the bottom; with a ten-decade range the chance at $\delta = 0.013$ is $1.2\times10^{-3}$ — the conclusion is insensitive to the choice because it enters only through a logarithm. What is not insensitive: whether the two experiments were really independent. They were — one used a Leyden jar and a galvanometer, the other a toothed wheel and a telescope.

A comparison worth making. Laplace's correction to the speed of sound (unit 22) was established by an 18 per cent discrepancy; here, the identification rests on an agreement. The logic is different — a falsification versus a coincidence too improbable to be accidental — and the second kind needs an independent confirmation, which Hertz supplied.

24 · light

Plane waves

plane wave

$\vec E = \operatorname{Re}\big(\vec E_0e^{i(\vec k\cdot\vec x - \omega t)}\big)$, $\vec B = \operatorname{Re}\big(\vec B_0e^{i(\vec k\cdot\vec x - \omega t)}\big)$, with constant $\vec E_0, \vec B_0\in\mathbb C^3$ and wave vector $\vec k\in\mathbb R^3$. On such fields $\nabla\mapsto i\vec k$, $\partial_t\mapsto-i\omega$.

theorem · what Maxwell forces

A plane wave solves the vacuum equations iff $\omega = c|\vec k|$ and

$$\vec k\cdot\vec E_0 = 0,\qquad \vec B_0 = \frac1\omega\,\vec k\times\vec E_0 .$$

So light is transverse, $\vec E\perp\vec B\perp\vec k$, and $|\vec B| = |\vec E|/c$.

A linearly polarised plane wave along $\hat z$, at one instant, computed from $\vec E_0 = E_0\hat x$. $\vec B = \hat k\times\vec E/c$ oscillates in phase with $\vec E$, at right angles to it and to $\vec k$; the whole pattern slides along $\vec k$ at $c$.

Reading the figure. The axes are drawn in oblique projection: $x$ up, $y$ receding to the lower left, $z$ along $\vec k$ to the right. The thin lines are the field vectors at points of the $z$-axis — $\vec E$ in the $xz$-plane, $\vec B$ in the $yz$-plane. With $\vec E_0 = E_0\hat x$ and $\vec k = k\hat z$, the theorem gives $\vec B_0 = \frac{k}{\omega}\hat z\times E_0\hat x = \frac{E_0}{c}\hat y$: real, so $\vec B$ and $\vec E$ peak together, and $\vec E\times\vec B$ points along $+\hat z$ everywhere, where the wave is going.

Taking real parts is allowed because Maxwell's equations are linear with real coefficients: if a complex field solves them, so do its real and imaginary parts (unit 06's device). Products — $u$, $\vec S$ — must be formed from the real fields, not from the complex ones.

24 · light

What a plane wave carries

Proof. Gauss: $i\vec k\cdot\vec E_0 = 0$. Faraday: $i\vec k\times\vec E_0 = i\omega\vec B_0$. Then $\nabla\cdot\vec B = 0$ is automatic, and Ampère–Maxwell, $i\vec k\times\vec B_0 = -i\omega\mu_0\varepsilon_0\vec E_0$, becomes $\vec k\times(\vec k\times\vec E_0) = -|\vec k|^2\vec E_0 = -\omega^2\vec E_0/c^2$. $\square$

Of the four equations, only the two curls involve both fields; they force $\vec B$ given $\vec E$, and back again — consistent only if $\omega = c|\vec k|$.

energy, momentum, pressure

Time-averaged over a period, for real linear $\vec E_0$: $\langle u\rangle = \frac{\varepsilon_0}{2}|\vec E_0|^2$ (half electric, half magnetic), $\langle\vec S\rangle = c\langle u\rangle\hat k$. On an absorber the momentum flux gives a pressure $\langle S\rangle/c$.

Sunlight at the Earth, $1361\ \mathrm{W\,m^{-2}}$: $E_0 = 1013\ \mathrm{V\,m^{-1}}$, $B_0 = 3.38\ \mu\mathrm T$, pressure $4.54\ \mu\mathrm{Pa}$.

The averages. For $\vec E_0$ real, $\vec E = \vec E_0\cos(\vec k\cdot\vec x - \omega t)$ and $\langle\cos^2\rangle = \frac12$, so $\langle\frac{\varepsilon_0}2|\vec E|^2\rangle = \frac{\varepsilon_0}4|\vec E_0|^2$, and the magnetic energy is the same because $|\vec B|^2/\mu_0 = |\vec E|^2/(c^2\mu_0) = \varepsilon_0|\vec E|^2$. Total $\frac{\varepsilon_0}2|\vec E_0|^2$. And $\vec S = \vec E\times\vec B/\mu_0 = |\vec E|^2\hat k/(\mu_0c) = c\,\varepsilon_0|\vec E|^2\hat k = cu\,\hat k$: energy moves at $c$ along $\vec k$. For a general complex $\vec E_0$ replace $|\vec E_0|^2$ by $\vec E_0\cdot\overline{\vec E_0}$.

Sunlight in full. $E_0 = \sqrt{2S/\varepsilon_0c} = \sqrt{2\times1361/(8.854\times10^{-12}\times2.998\times10^8)} = 1012.6\ \mathrm{V\,m^{-1}}$; $B_0 = E_0/c = 3.378\times10^{-6}$ T, a fifteenth of the Earth's $50\ \mu$T; pressure $1361/c = 4.540\times10^{-6}$ Pa on a black surface, twice that on a mirror. Real sunlight is not one plane wave but an incoherent mixture; the numbers are those of a single plane wave carrying the same power.

Radiation pressure follows from the stress tensor: for a plane wave along $\hat z$, $T(\hat z,\hat z) = -u$, a pure pressure equal to the energy density, and $\langle u\rangle = \langle S\rangle/c$. It was measured by Lebedev (1900) and by Nichols and Hull (1901).

24 · light

Polarisation

polarisation · the mathematics

For a wave along $\hat z$, $\vec E_0$ lies in the complex plane $\{\vec v\in\mathbb C^3:\hat z\cdot\vec v = 0\}\cong\mathbb C^2$. Its polarisation is the complex line $[\vec E_0]$, i.e. $\vec E_0$ up to a nonzero complex factor: the factor only rescales and shifts the phase.

theorem · every polarisation is an ellipse

Write $\vec E_0 = \vec a + i\vec b$ with $\vec a, \vec b$ real. At fixed $\vec x = \vec 0$, $\vec E(t) = \vec a\cos\omega t + \vec b\sin\omega t$, which traces an ellipse. It is a segment (linear) iff $\vec a\parallel\vec b$, and a circle iff $\vec a\cdot\vec b = 0$ and $|\vec a| = |\vec b|$.

Proof. $\operatorname{Re}\big((\vec a + i\vec b)(\cos\omega t - i\sin\omega t)\big) = \vec a\cos\omega t + \vec b\sin\omega t$, the image of the unit circle under the linear map $(\cos,\sin)\mapsto$ that combination: an ellipse, degenerate iff the map has rank 1. It is a circle iff the map is a multiple of an orthogonal one. $\square$

Two real parameters — the complex ratio $E_{0y}/E_{0x}$, or $\infty$ — describe every pure polarisation. Superposing two circular ones with equal amplitude gives a linear one: $(1, i) + (1, -i) = (2, 0)$.

The circle case in full. The map $L(\cos\theta, \sin\theta) = \vec a\cos\theta + \vec b\sin\theta$ has matrix with columns $\vec a$, $\vec b$. Its image of the unit circle is a circle iff $L^\top L$ is a multiple of the identity, i.e. iff $|\vec a| = |\vec b|$ and $\vec a\cdot\vec b = 0$. In general the ellipse's semi-axes are the singular values of $L$ (unit 07's spectral theorem applied to $L^\top L$), which is how problem P6 finds the axes of a tilted ellipse.

Why a complex line. Multiplying $\vec E_0$ by $re^{i\alpha}$ scales the wave by $r$ and shifts its time origin by $\alpha/\omega$; neither changes the shape of the ellipse. So the space of polarisations is the set of complex lines in $\mathbb C^2$, the complex projective line — a sphere, the Poincaré sphere of optics. Linear polarisations form its equator and the two circular ones its poles.

History. Malus found polarisation by reflection in 1808; Fresnel and Arago (1819) showed that light polarised at right angles does not interfere, from which Young and Fresnel concluded that light is a transverse wave — a fact with no explanation in an elastic aether and an immediate one here.

24 · light

Polarisations, drawn

The tip of $\vec E(t) = \vec a\cos\omega t + \vec b\sin\omega t$ at a fixed point, for four amplitudes $\vec E_0 = \vec a + i\vec b$ (wave along $\hat z$, out of the page):

$(1,-1,0)$: $\vec b = \vec 0$, a segment. $(1,i,0)$: $\vec a = \hat x$, $\vec b = \hat y$, equal and perpendicular, a circle. $(2,i,0)$: $\vec a = 2\hat x$, $\vec b = \hat y$, an ellipse with axes $2:1$. Last: the two opposite circles (dashed) add to $(2\cos\omega t, 0)$, a segment.

Reading the figure. Each panel is the plane perpendicular to $\hat k = \hat z$, seen with the wave coming towards you, and each curve was computed from $\vec a\cos\omega t + \vec b\sin\omega t$ over one period. The dots mark the two ends of a segment, where the field reverses; the arrowheads mark the sense of rotation. In the last panel $(1,i,0)$ turns anticlockwise and $(1,-i,0)$ clockwise, and at every instant their $y$-components cancel while their $x$-components add.

The sense of rotation. For $(1, i, 0)$ the field at the origin is $(\cos\omega t, \sin\omega t)$: it turns from $\hat x$ towards $\hat y$, anticlockwise about $+\hat z = \hat k$. Such a wave carries angular momentum along $+\hat k$ (positive helicity). The names "left" and "right" circular are attached to this in opposite ways by optics and by particle physics, so this deck avoids them.

section 5

Potentials and gauge

Solve the two homogeneous equations once and for all; what is left over is redundancy.

24 · gauge

Potentials, and what they fail to determine

the homogeneous pair, solved

On $\mathbb R^3\times\mathbb R$, $C^2$ fields satisfy $\nabla\cdot\vec B = 0$ and $\nabla\times\vec E = -\partial_t\vec B$ iff there are potentials $(\varphi, \vec A)$ with

$$\vec B = \nabla\times\vec A,\qquad \vec E = -\nabla\varphi - \partial_t\vec A .$$

Proof. ($\Leftarrow$) div curl $= 0$, curl grad $= 0$. ($\Rightarrow$) Unit 23's Poincaré lemma gives $\vec A$ with $\nabla\times\vec A = \vec B$ (its explicit formula is smooth in $t$). Then $\nabla\times(\vec E + \partial_t\vec A) = \vec 0$, so $\vec E + \partial_t\vec A = -\nabla\varphi$ by unit 04's Poincaré lemma. $\square$

gauge transformation

For any $C^2$ function $\chi(\vec x, t)$: $$\varphi\mapsto\varphi - \partial_t\chi,\qquad \vec A\mapsto\vec A + \nabla\chi .$$

theorem · the fields are exactly the gauge-invariant content

Two pairs of potentials give the same $(\vec E, \vec B)$ iff they differ by a gauge transformation.

Proof. ($\Leftarrow$) $\nabla\times\nabla\chi = \vec 0$ and $-\nabla(-\partial_t\chi) - \partial_t\nabla\chi = \vec 0$. ($\Rightarrow$) Let the difference be $(\varphi', \vec A')$ with zero fields. $\nabla\times\vec A' = \vec 0$ gives $\vec A' = \nabla\chi_0$; then $\nabla(\varphi' + \partial_t\chi_0) = \vec 0$, so $\varphi' + \partial_t\chi_0 = f(t)$. Take $\chi = \chi_0 - \int_0^tf$. $\square$

Mathematics first. The map $\Phi:(\varphi, \vec A)\mapsto(-\nabla\varphi - \partial_t\vec A,\ \nabla\times\vec A)$ is linear. The theorem says its kernel is exactly the image of $\chi\mapsto(-\partial_t\chi, \nabla\chi)$, and the first theorem says its image is exactly the solution space of the homogeneous pair. So the fields are the potentials modulo gauge: $(\vec E,\vec B)$ is the class of $(\varphi,\vec A)$ in the quotient space. A gauge is a choice of representative of each class, usually by imposing a linear condition.

The physics. Classically only $\vec E$ and $\vec B$ enter the Lorentz force, so a gauge transformation changes nothing measurable; it is a redundancy of description, as unit 23 said for statics. Unit 12's Lagrangian $\frac12m|\dot{\vec x}|^2 - q\varphi + q\dot{\vec x}\cdot\vec A$ changes by $q(\partial_t\chi + \dot{\vec x}\cdot\nabla\chi) = q\,\mathrm d\chi/\mathrm dt$ — a total derivative, which leaves the Euler–Lagrange equations alone (unit 12). This is now the time-dependent version.

The smoothness in $t$. Unit 23's Poincaré-lemma formula on a star-shaped domain, $\vec A(\vec x) = \int_0^1\vec B(\lambda\vec x)\times\lambda\vec x\,\mathrm d\lambda$, is an integral of $\vec B$ over a compact parameter range, so it is as smooth in $t$ as $\vec B$ is (differentiation under the integral, unit 18). Then $\vec E + \partial_t\vec A$ is $C^1$ and curl-free, and unit 04's lemma applies at each $t$; the line-integral formula for $\varphi$ is again smooth in $t$.

On a domain with holes the theorem can fail, as the solenoid showed in unit 23: a curl-free $\vec A'$ need not be a gradient, and the loop integral $\oint\vec A'\cdot\mathrm d\vec r$ is then gauge-invariant information that the fields in the domain do not see. On $\mathbb R^3$ that cannot happen.

24 · gauge

The Lorenz gauge: two wave equations

Put the potentials into the two remaining equations (unit 23's curl-curl identity):

$$\begin{aligned}-\nabla^2\varphi - \partial_t\nabla\cdot\vec A &= \rho/\varepsilon_0,\\ \Box\vec A + \nabla\Big(\nabla\cdot\vec A + \tfrac1{c^2}\partial_t\varphi\Big) &= \mu_0\vec J .\end{aligned}$$

the Lorenz gauge

The condition $\nabla\cdot\vec A + \frac1{c^2}\partial_t\varphi = 0$. In it the equations decouple:

$$\Box\varphi = \rho/\varepsilon_0,\qquad \Box\vec A = \mu_0\vec J .$$

it can always be reached

Given any $(\varphi, \vec A)$, let $L = \nabla\cdot\vec A + c^{-2}\partial_t\varphi$. After a gauge transformation $L\mapsto L + \nabla^2\chi - c^{-2}\partial_t^2\chi = L - \Box\chi$. So solve $\Box\chi = L$ — section 6 shows how. $\square$

residual gauge freedom

Lorenz gauge does not fix the potentials: any $\chi$ with $\Box\chi = 0$ preserves it. E.g. $\chi = xy$, or $\chi = \cos(kz - \omega t)$ with $\omega = ck$. The fields are unchanged; the potentials are not.

The first equation needs no gauge: $-\nabla^2\varphi - \partial_t\nabla\cdot\vec A = -\nabla^2\varphi + c^{-2}\partial_t^2\varphi = \Box\varphi$ once the condition holds.

The derivation in full. Gauss: $\nabla\cdot(-\nabla\varphi - \partial_t\vec A) = \rho/\varepsilon_0$, the first line. Ampère–Maxwell: $\nabla\times\nabla\times\vec A = \mu_0\vec J + \mu_0\varepsilon_0\partial_t(-\nabla\varphi - \partial_t\vec A)$, and $\nabla\times\nabla\times\vec A = \nabla(\nabla\cdot\vec A) - \nabla^2\vec A$; collect terms with $c^{-2} = \mu_0\varepsilon_0$: $-\nabla^2\vec A + c^{-2}\partial_t^2\vec A + \nabla(\nabla\cdot\vec A + c^{-2}\partial_t\varphi) = \mu_0\vec J$.

The name. Ludvig Lorenz, Danish, 1867 — not Hendrik Lorentz of the Lorentz force and the Lorentz transformation, although books have spelled it "Lorentz gauge" for a century. The condition is also the one that looks the same in every inertial frame (unit 28: $\partial_\mu A^\mu = 0$), which is why it is the natural gauge for radiation.

The residual freedom, checked. $\Box(xy) = -\nabla^2(xy) = 0$. For $\chi = \cos(kz - \omega t)$, $c^{-2}\partial_t^2\chi = -\frac{\omega^2}{c^2}\chi$ and $\nabla^2\chi = -k^2\chi$, so $\Box\chi = (-\omega^2/c^2 + k^2)\chi = 0$ when $\omega = ck$. The residual $\chi$ are themselves solutions of the wave equation: the freedom that remains propagates, like the fields. Unit 38 meets the same structure in gravitational waves, where the residual freedom of the analogous gauge is used up to reach the transverse-traceless form.

Where the condition is consistent. Taking $\partial_t$ of the Lorenz condition and using the two wave equations: $\Box(\nabla\cdot\vec A + c^{-2}\partial_t\varphi) = \mu_0(\nabla\cdot\vec J + \partial_t\rho) = 0$. So the gauge condition, if it and its time derivative hold initially, is preserved — the same mechanism that preserved the constraints in section 3, and it runs on charge conservation again.

24 · gauge

The Coulomb gauge: an instantaneous potential

the Coulomb gauge, time-dependent

$\nabla\cdot\vec A = 0$, as in unit 23. Then Gauss reads $-\nabla^2\varphi = \rho/\varepsilon_0$ at each instant:

$$\varphi_\mathrm C(\vec x, t) = \frac1{4\pi\varepsilon_0}\int\frac{\rho(\vec y, t)}{|\vec x - \vec y|}\,\mathrm dV_y ,$$

and $\Box\vec A = \mu_0\vec J - \frac1{c^2}\nabla\partial_t\varphi_\mathrm C$.

Move a charge here, and $\varphi_\mathrm C$ on the Moon changes now.

no signal travels faster than light

$\varphi_\mathrm C$ is not measurable; only $\vec E$ is. Since Coulomb and Lorenz potentials differ by a gauge transformation, $\vec E$ is the same in both, and in Lorenz gauge it is built from retarded potentials (section 6), which see only the past light cone. So in Coulomb gauge the instantaneous part of $-\nabla\varphi_\mathrm C$ is cancelled exactly by an instantaneous part of $-\partial_t\vec A$.

why use it, then

With no charge anywhere, $\varphi_\mathrm C = 0$ and $\vec A$ alone describes the radiation, with $\vec E = -\partial_t\vec A$ and $\nabla\cdot\vec A = 0$ making it transverse. Quantum optics lives in this gauge.

The equation for $\vec A$. From the general Ampère–Maxwell line of the previous slide with $\nabla\cdot\vec A = 0$: $\Box\vec A + c^{-2}\nabla\partial_t\varphi_\mathrm C = \mu_0\vec J$. The source on the right, $\mu_0\vec J - c^{-2}\nabla\partial_t\varphi_\mathrm C$, is divergence-free — its divergence is $\mu_0\nabla\cdot\vec J + c^{-2}\partial_t\rho/\varepsilon_0 = \mu_0(\nabla\cdot\vec J + \partial_t\rho) = 0$ — and it is the transverse part of $\mu_0\vec J$ in unit 23's Helmholtz decomposition. Computing the transverse part of a localised current requires an integral over all space, so it is non-local; that non-locality is what cancels the instantaneous $\varphi_\mathrm C$.

It can always be reached. Given $\vec A$, solve $\nabla^2\chi = -\nabla\cdot\vec A$ at each instant (unit 23's Poisson equation, under decay hypotheses) and replace $\vec A$ by $\vec A + \nabla\chi$. The residual freedom is $\nabla^2\chi = 0$. Each component of $\nabla\chi$ is then harmonic, and if it tends to zero at infinity it vanishes identically by the maximum principle (unit 05), leaving only $\chi = \chi(t)$, which does not change $\vec A$ at all and shifts $\varphi$ by a function of time, killed by requiring $\varphi\to0$ at infinity. So Coulomb gauge with decay is complete — the price for that is the instantaneous $\varphi_\mathrm C$.

The argument that $\vec E$ is causal is carried out explicitly for a uniformly moving point charge in problem P8: both gauges' potentials are computed, $\varphi_\mathrm C$ is the instantaneous Coulomb potential of the charge's present position, and the fields agree.

24 · gauge · example

Worked: the capacitor in two gauges

worked example 4 — find potentials for the capacitor's fields in Coulomb gauge, move to Lorenz gauge, and check nothing physical changed

Step 0 — the fields between the plates: $\vec E = E(t)\hat z$ with $\dot E = I/\varepsilon_0\pi a^2$ constant; $\vec B = \beta s\,\hat\phi$ with $\beta = \mu_0I/2\pi a^2$.

Step 1 — $\vec A$. Try $\vec A = A_z(s)\hat z$. Its curl is $-A_z'(s)\hat\phi$, so $A_z = -\frac12\beta s^2$. Divergence: $\partial_zA_z = 0$ — already Coulomb gauge.

Step 2 — $\varphi$. $\partial_t\vec A = \vec 0$, so $-\nabla\varphi = E(t)\hat z$: $\varphi_\mathrm C = -E(t)z$. It changes with $t$ everywhere at once — instantaneous, as the previous slide said.

Step 3 — the Lorenz condition fails. $\nabla\cdot\vec A + c^{-2}\partial_t\varphi_\mathrm C = -\dot E z/c^2\ne0$.

Step 4 — repair it. Need $\Box\chi = L = -\dot Ez/c^2$. As $\dot E$ is constant, try $\chi = \kappa z^3$: $\Box\chi = -6\kappa z$, so $\kappa = \dot E/6c^2$ and

$$\vec A_\mathrm L = \Big(-\tfrac12\beta s^2 + \tfrac{\dot E}{2c^2}z^2\Big)\hat z,\quad \varphi_\mathrm L = -Ez .$$

Step 5 — check the gauge. $\nabla\cdot\vec A_\mathrm L + c^{-2}\partial_t\varphi_\mathrm L = \dot Ez/c^2 - \dot Ez/c^2 = 0$ ✓.

Step 6 — same fields. $\nabla\times\vec A_\mathrm L$: the new term depends on $z$ only and points along $\hat z$, so its curl is $\vec 0$; $\vec B$ unchanged ✓. $-\nabla\varphi_\mathrm L - \partial_t\vec A_\mathrm L = E\hat z - \frac{\ddot E}{2c^2}z^2\hat z = E\hat z$ ✓, as $\ddot E = 0$.

Checks. $[\dot Ez^2/c^2] = \mathrm{V\,m^{-1}s^{-1}}\cdot\mathrm m^2\cdot\mathrm{s^2m^{-2}} = \mathrm{V\,s\,m^{-1}} = \mathrm{T\,m} = [A]$ ✓. $c\to\infty$: the two gauges coincide ✓.

Step 1, the curl. For $\vec A = A_z(s)\hat z$ in cylindrical coordinates, $\nabla\times\vec A = -\partial_sA_z\,\hat\phi$ (the $\hat\phi$-component of the curl is $\partial_zA_s - \partial_sA_z$). So $-A_z' = \beta s$ and $A_z = -\frac12\beta s^2$, up to a constant. In Cartesian coordinates, $A_z = -\frac12\beta(x^2 + y^2)$, and $\nabla\times\vec A = (\partial_yA_z, -\partial_xA_z, 0) = \beta(-y, x, 0) = \beta s\hat\phi$ ✓.

Step 4, the sign. The gauge transformation changes $L$ by $\nabla^2\chi - c^{-2}\partial_t^2\chi = -\Box\chi$, so to cancel $L$ we need $-\Box\chi = -L$, i.e. $\Box\chi = L$. With $\chi = \kappa z^3$, $\partial_t^2\chi = 0$ and $\nabla^2\chi = 6\kappa z$, so $\Box\chi = -6\kappa z = -\dot Ez/c^2$ gives $\kappa = \dot E/6c^2$. Then $\nabla\chi = \frac{\dot E}{2c^2}z^2\hat z$ and $\partial_t\chi = 0$, so $\varphi$ is unchanged.

Size. At the edge of the gap, $z = d = 1$ mm, the added term is $\dot Ed^2/2c^2 = (1.438\times10^{13})(10^{-6})/(2\times8.988\times10^{16}) = 8.0\times10^{-11}\ \mathrm{T\,m}$, against $\frac12\beta a^2 = \frac12\mu_0I/2\pi = 1.0\times10^{-7}\ \mathrm{T\,m}$ at the rim. Gauge terms can be any size; this one happens to be small.

Residual freedom, used. Adding $\chi' = \lambda xy$ keeps Lorenz gauge ($\Box(xy) = 0$) and changes $\vec A_\mathrm L$ by $\lambda(y, x, 0)$ — a different, equally valid Lorenz-gauge potential for the same fields. The checkpoint asks for exactly this kind of verification: two gauges, one pair of fields.

section 6

The retarded Green's function

Every Fourier mode of the field is unit 20's oscillator, and the $i0$ prescription is where causality enters.

24 · Green's function

Fourier transforms in several variables

the transform on $\mathbb R^n$

For $f$ in the Schwartz class $\mathcal S(\mathbb R^n)$ (smooth; $f$ and its partial derivatives fall faster than any power of $|\vec x|$): $$\begin{aligned}\hat f(\vec k) &= \int_{\mathbb R^n}f(\vec x)\,e^{-i\vec k\cdot\vec x}\,\mathrm d^n x,\\ f(\vec x) &= \frac1{(2\pi)^n}\int\hat f(\vec k)\,e^{i\vec k\cdot\vec x}\,\mathrm d^nk .\end{aligned}$$

Proof of inversion. $e^{-i\vec k\cdot\vec x} = \prod_je^{-ik_jx_j}$, so by Fubini (unit 05) the transform is $n$ one-variable transforms in turn, and unit 20's inversion theorem undoes each (with $k_j\mapsto-k_j$ for the sign). $\square$ Likewise $\widehat{\partial_jf} = ik_j\hat f$.

distributions on $\mathbb R^n$

Unit 20's definitions with $\mathbb R$ replaced by $\mathbb R^n$: test functions $C_c^\infty(\mathbb R^n)$, $\langle\partial_jT,\varphi\rangle = -\langle T,\partial_j\varphi\rangle$, $\langle\hat T,\varphi\rangle = \langle T,\hat\varphi\rangle$. Then $\hat\delta = 1$ and $\hat1 = (2\pi)^n\delta$.

on spacetime · this course's convention

In $(\vec x, t)$, combine the space convention above with unit 20's time convention:

$$\hat f(\vec k,\omega) = \iint f(\vec x,t)\,e^{-i\vec k\cdot\vec x + i\omega t}\,\mathrm d^3x\,\mathrm dt .$$

The elementary wave is $e^{i(\vec k\cdot\vec x - \omega t)}$; $\nabla\mapsto i\vec k$, $\partial_t\mapsto-i\omega$, and

$$\widehat{\Box f} = \Big(|\vec k|^2 - \frac{\omega^2}{c^2}\Big)\hat f .$$

the angular integral

For $r = |\vec x|$ and $k = |\vec k|$: $\displaystyle\int_{S^2}e^{i\vec k\cdot\vec x}\,\mathrm d\Omega_{\hat k} = 4\pi\frac{\sin kr}{kr}$.

Proof. Polar axis along $\vec x$: $2\pi\int_0^\pi e^{ikr\cos\theta}\sin\theta\,\mathrm d\theta = 2\pi\frac{e^{ikr} - e^{-ikr}}{ikr}$. $\square$

Why this is taught here. Unit 20 built the Fourier transform and distributions on $\mathbb R$ only. Nothing new is needed for $\mathbb R^n$: every proof there goes through with $\mathbb R$ replaced by $\mathbb R^n$, because the exponential factorises and Fubini lets the integrals be done one variable at a time. The Schwartz class is closed under the transform for the same reason. Plancherel becomes $\int|f|^2 = (2\pi)^{-n}\int|\hat f|^2$.

The table, in $n$ dimensions. $\langle\hat\delta,\varphi\rangle = \langle\delta,\hat\varphi\rangle = \hat\varphi(\vec 0) = \int\varphi$, so $\hat\delta = 1$. And $\langle\hat1,\varphi\rangle = \int\hat\varphi = (2\pi)^n\varphi(\vec 0)$ by inversion at $\vec x = \vec 0$. On spacetime, with the mixed signs, the same computation gives $\widehat{\delta^{(4)}} = 1$, where $\delta^{(4)} = \delta(t)\,\delta^{(3)}(\vec x)$ is the distribution $\varphi\mapsto\varphi(\vec 0, 0)$.

The spacetime signs. Space and time carry opposite signs in the exponent so that a plane wave $e^{i(\vec k\cdot\vec x - \omega t)}$ with $\omega > 0$ moves along $+\vec k$. It is the convention of unit 20 in time, of unit 19's complex series in space, and of every later unit. The inverse is $f = (2\pi)^{-4}\iint\hat f\,e^{i\vec k\cdot\vec x - i\omega t}\,\mathrm d^3k\,\mathrm d\omega$.

The angular integral in full. With $\vec x$ along the polar axis of $\vec k$-space, $\vec k\cdot\vec x = kr\cos\theta$ and $\mathrm d\Omega = \sin\theta\,\mathrm d\theta\,\mathrm d\phi$ (unit 05). Substituting $\mu = \cos\theta$, $\int_{-1}^1e^{ikr\mu}\mathrm d\mu = (e^{ikr} - e^{-ikr})/ikr = 2\sin(kr)/kr$; times $2\pi$. Limit $r\to0$: $4\pi$, the total solid angle ✓.

24 · Green's function

Every mode is an oscillator

a fundamental solution of the wave operator

A distribution $G$ on $\mathbb R^4$ with $\Box G = \delta^{(4)} = \delta(t)\,\delta^{(3)}(\vec x)$. Then $u = G * f$ solves $\Box u = f$ — unit 23's Green's-function idea, with time added.

Transform: $\big(k^2 - \omega^2/c^2\big)\hat G = 1$, so

$$\hat G(\vec k,\omega) = \frac{c^2}{c^2k^2 - \omega^2} .$$

For each fixed $\vec k$ this is $c^2$ times unit 20's undamped oscillator, $1/(\omega_0^2 - \omega^2)$, with $\omega_0 = ck$. Each spatial Fourier mode of the field is a harmonic oscillator of frequency $ck$.

the poles are on the axis

$\omega = \pm ck$ are real. The inversion integral over $\omega$ does not exist, and the equation $\Box G = \delta$ alone does not say which way round them to go. A choice has to be made, and it is a physical choice.

the retarded prescription · $i0$ prescription

As in unit 20, replace $\omega$ by $\omega + i0$:

$$\hat G_\mathrm{ret}(\vec k,\omega) = \frac{c^2}{c^2k^2 - (\omega + i0)^2} ,$$

poles at $\omega = \pm ck - i0$, just below the axis; the contour passes above them.

What "$\omega + i0$" means precisely. $\hat G_\mathrm{ret}$ is the limit, as tempered distributions in $(\vec k,\omega)$, of $c^2/(c^2k^2 - (\omega + i\varepsilon)^2)$ as $\varepsilon\to0^+$. For $\varepsilon > 0$ this is a bounded smooth function (its poles $\pm ck - i\varepsilon$ are off the axis), whose inverse transform in $\omega$ unit 20's contour argument computes exactly; the limit is then taken in the result, which is continuous in $\varepsilon$. That is the order of operations on the next slide.

Why "each mode is an oscillator" is literally true. Transform the vacuum wave equation in space only: $\partial_t^2\tilde u(\vec k, t) = -c^2k^2\tilde u(\vec k, t)$. For each $\vec k$ this is unit 06's harmonic oscillator with $\omega_0 = ck$. The field is a continuum of independent oscillators, one per wave vector — unit 07's normal modes, with the finite chain replaced by all of space. Unit 18 did the same for the string with a discrete set of modes; unit 19 proved they were a basis.

Dimensions. $[\delta^{(4)}] = \mathrm T^{-1}\mathrm L^{-3}$ and $[\Box] = \mathrm L^{-2}$, so $[G] = \mathrm T^{-1}\mathrm L^{-1}$. Then $[\hat G] = [G]\,\mathrm L^3\mathrm T = \mathrm L^2$, and $c^2/(c^2k^2 - \omega^2)$ has dimension $\mathrm L^2\mathrm T^{-2}/\mathrm T^{-2} = \mathrm L^2$ ✓.

24 · Green's function · example

Worked: the retarded Green's function by residues

worked example 5 — invert $\hat G_\mathrm{ret}$: first $\omega$ by contours, then $\vec k$ in spherical coordinates

Step 1 — the $\omega$-integral, $t < 0$. $\tilde G(\vec k,t) = \frac1{2\pi}\int\hat G_\mathrm{ret}e^{-i\omega t}\mathrm d\omega$. For $t < 0$, $|e^{-i\omega t}|\le1$ on the upper half-plane; close upwards, $|\hat G|\le2c^2/R^2$ on the arc: no poles inside, so $\tilde G = 0$ — by contour closing, not inspection.

Step 2 — $t > 0$. Close downwards, enclosing $\pm ck$. This is unit 20's undamped computation with $\omega_0 = ck$, times $c^2$:

$$\tilde G(\vec k,t) = c^2H(t)\,\frac{\sin(ckt)}{ck} .$$

The poles $\pm ck - i0$ sit just below the real axis. $e^{-i\omega t}$ decays in the upper half-plane when $t < 0$ and in the lower when $t > 0$, so that is where the arc goes; only the second encloses anything.

Step 1, the arc bound. On $|\omega| = R$ with $R\ge2ck$, $|c^2k^2 - \omega^2|\ge R^2 - c^2k^2\ge\frac34R^2$, so $|\hat G|\le\frac43c^2/R^2$, below the $2c^2/R^2$ quoted. The arc has length $\pi R$, so its contribution is at most $\frac{4\pi c^2}{3R}\to0$. With $\varepsilon > 0$ the poles sit at $\pm ck - i\varepsilon$, outside the upper half-disc, so Cauchy's theorem gives zero for every $R$; hence $\tilde G_\varepsilon(\vec k, t) = 0$ for $t < 0$ and every $\varepsilon$, and the limit is zero.

Step 2 in full. With $\varepsilon > 0$, the poles of $c^2/(c^2k^2 - (\omega + i\varepsilon)^2)$ are $\omega_\pm = \pm ck - i\varepsilon$, with residues of $\hat G e^{-i\omega t}$ equal to $\mp\frac{c^2e^{-i\omega_\pm t}}{2ck}$. The clockwise contour carries $-2\pi i$: $\tilde G_\varepsilon = \frac{1}{2\pi}(-2\pi i)\frac{c^2}{2ck}\big(-e^{-i\omega_+t} + e^{-i\omega_-t}\big) = \frac{-ic^2}{2ck}e^{-\varepsilon t}\cdot2i\sin ckt = c^2e^{-\varepsilon t}\frac{\sin ckt}{ck}$. Let $\varepsilon\to0$. This is unit 20's "step function, explained" slide with $\omega_0 = ck$, to the letter.

Reading the figure. $|e^{-i\omega t}| = e^{t\operatorname{Im}\omega}$, so for $t < 0$ it is at most $1$ where $\operatorname{Im}\omega \ge 0$ and the upper arc can be added at no cost; for $t > 0$ the lower arc is the cheap one. Because the prescription put both poles below the axis, the answer is zero for every $t < 0$ — causality is the position of two points relative to a line.

24 · Green's function · example

Worked: the wave-vector integral, and the sphere $r = ct$

worked example 5, continued — from one oscillator per $\vec k$ to a function of $(\vec x, t)$

Step 3 — the $\vec k$-integral. $G = (2\pi)^{-3}\int\tilde Ge^{i\vec k\cdot\vec x}\mathrm d^3k$; do the angles by the angular integral:

$$\begin{aligned}G &= \frac{cH(t)}{8\pi^3}\int_0^\infty4\pi\tfrac{\sin kr}{kr}\tfrac{\sin ckt}{k}k^2\,\mathrm dk\\ &= \frac{H(t)\,c}{2\pi^2r}\int_0^\infty\sin kr\,\sin ckt\,\mathrm dk .\end{aligned}$$

Step 4 — the last integral. $\sin kr\sin ckt = \frac12[\cos k(r - ct) - \cos k(r + ct)]$, and $\int_0^\infty\cos kx\,\mathrm dk = \frac12\int_{-\infty}^\infty e^{ikx}\mathrm dk = \pi\delta(x)$ (unit 20's $\hat1 = 2\pi\delta$). So

$$\begin{aligned}&\int_0^\infty\sin kr\,\sin ckt\,\mathrm dk\\ &\quad= \tfrac\pi2\big[\delta(r - ct) - \delta(r + ct)\big].\end{aligned}$$

Step 5 — assemble. For $t > 0$ and $r > 0$, $\delta(r + ct) = 0$, and $\delta(r - ct) = \delta(t - r/c)/c$. The retarded Green's function:

$$G_\mathrm{ret}(\vec x,t) = \frac{\delta(t - r/c)}{4\pi r} .$$

Checks. $[\delta(t)/r] = \mathrm T^{-1}\mathrm L^{-1}$ ✓. Integrate over $t$: $\frac1{4\pi r}$, unit 23's Coulomb Green's function — a source constant in time sees the static answer ✓. Support: the sphere $r = ct$ only.

Step 3. $\mathrm d^3k = k^2\,\mathrm dk\,\mathrm d\Omega$. The factor $c^2/(ck) = c/k$, then $\frac{c}{(2\pi)^3}\cdot4\pi\cdot\frac{1}{r}\int\sin kr\sin ckt\,\mathrm dk$ since $\frac{\sin kr}{kr}\cdot\frac1k\cdot k^2 = \frac{\sin kr}{r}$; and $\frac{4\pi}{8\pi^3} = \frac1{2\pi^2}$.

Step 4 is an identity of distributions, in the variable $r$ with $t$ fixed (or in $t$ with $r$ fixed): the integral does not converge for any single $r$. Justifying every exchange of limits in Steps 3–5 is possible but long. The course's route instead: this slide finds $G_\mathrm{ret}$, and the retarded-potentials slide proves that convolving with it solves $\Box u = f$, by a direct computation with no Fourier transforms at all. A found answer, then verified, is a complete argument.

Step 5, the delta. $\delta(r - ct) = \delta(-c(t - r/c)) = \frac1c\delta(t - r/c)$, by the scaling rule $\delta(\lambda s) = \delta(s)/|\lambda|$ (substitute in the pairing). So $\frac{c}{2\pi^2r}\cdot\frac\pi2\cdot\frac1c\delta(t - r/c) = \frac{\delta(t - r/c)}{4\pi r}$.

The limit check. A source $f(\vec x)$ constant in time gives $u = \int G(\vec x - \vec y, t - s)f(\vec y)\,\mathrm d^3y\,\mathrm ds = \int\frac{f(\vec y)}{4\pi|\vec x - \vec y|}\mathrm d^3y$, the solution of $-\nabla^2u = f$: unit 23's representation formula ✓.

24 · Green's function

Why this prescription is the causal one

theorem · the causal fundamental solution is unique

$G_\mathrm{ret}$ is the only fundamental solution of $\Box$ that vanishes for $t < 0$.

Proof. A second one differs by $D$ with $\Box D = 0$ and $D = 0$ for $t < 0$. Smooth it (convolve with a test function): a smooth solution with zero data in the past, hence zero by the energy argument (notes). $\square$

three accounts that agree

Contours. Poles below $\Leftrightarrow$ $\hat G$ holomorphic above $\Leftrightarrow$ $G = 0$ for $t < 0$: Step 1, and unit 20's theorem.
Dissipation. Any real medium absorbs a little: $\Box + \gamma c^{-2}\partial_t$ has poles at $\pm\sqrt{c^2k^2 - \gamma^2/4} - i\gamma/2$, below the axis. $\gamma\to0^+$ is $+i0$.
Switching on. A source $\propto e^{\varepsilon t}$ grows from nothing at $t = -\infty$; its frequency is $\omega + i\varepsilon$.

Full proof of uniqueness

Energy in a cone. Let $v$ be $C^2$ with $\Box v = 0$. Fix $(\vec x_0, t_0)$ and, for $t\le t_0$, let $B_t$ be the ball $|\vec x - \vec x_0|\le c(t_0 - t)$ and $\mathcal E(t) = \frac12\int_{B_t}\big(c^{-2}v_t^2 + |\nabla v|^2\big)\mathrm d^3x$. Differentiating (unit 18's rule, with a moving boundary contributing the integrand on $\partial B_t$ times the boundary's speed $-c$): $\mathcal E' = \int_{B_t}\big(c^{-2}v_tv_{tt} + \nabla v\cdot\nabla v_t\big) - \frac c2\oint_{\partial B_t}\big(c^{-2}v_t^2 + |\nabla v|^2\big)$. The volume integral is $\int v_t(c^{-2}v_{tt} - \nabla^2v) + \oint v_t\,\partial_nv = \oint v_t\,\partial_nv$ by the divergence theorem and the equation. And $|v_t\,\partial_nv|\le\frac12(c^{-1}v_t^2 + c|\nabla v|^2)$ (the inequality $2xy\le x^2 + y^2$). So $\mathcal E'\le0$. If $v = 0$ for all $t < 0$, then $\mathcal E = 0$ at some $t < 0$, hence for all $t\le t_0$, so $v_t = 0$ and $\nabla v = \vec 0$ in the cone, and $v(\vec x_0, t_0) = 0$. The point was arbitrary: $v\equiv0$. This is unit 18's domain-of-dependence argument, with the divergence theorem for the fundamental theorem of calculus; it needs no decay at infinity.

Smoothing. If $\Box D = 0$ with $D = 0$ for $t < 0$, then for a test function $\psi$ on $\mathbb R^4$, $v = D * \psi$, $v(\vec x, t) = \langle D, \psi(\vec x - \cdot, t - \cdot)\rangle$, is smooth, satisfies $\Box v = (\Box D) * \psi = 0$ (derivatives pass onto either factor), and vanishes for $t$ below the support of $\psi$. So $v\equiv0$ by the cone argument. As $\psi$ runs through an approximate identity, $D * \psi\to D$ in distributions, so $D = 0$. $\square$

The dissipation account in full. In a medium with conductivity $\sigma$, Ohm's law $\vec J = \sigma\vec E$ in Ampère–Maxwell adds $\mu_0\sigma\partial_t$ to the wave operator (the telegraph equation); $\gamma = \mu_0\sigma c^2$. The transform of $\Box + \gamma c^{-2}\partial_t$ is $k^2 - (\omega^2 + i\gamma\omega)/c^2$, zero at $\omega = -i\gamma/2\pm\sqrt{c^2k^2 - \gamma^2/4}$: unit 20's damped oscillator, with both poles in the lower half-plane. The real-axis integral is then retarded automatically, and its $\gamma\to0$ limit is $G_\mathrm{ret}$.

24 · Green's function

The other way round the poles

the other choices solve the equation too

Passing below both poles gives the advanced Green's function $$G_\mathrm{adv}(\vec x,t) = \frac{\delta(t + r/c)}{4\pi r},$$ a wave converging on the origin and arriving at the moment of the kick. Principal values give $\frac12(G_\mathrm{ret} + G_\mathrm{adv})$. $\Box G = \delta$ admits them all; the half-plane is the physics.

What selects $G_\mathrm{ret}$ is a boundary condition in time: no radiation comes in from the infinite past. The PDE is time-symmetric; the world's initial conditions are not.

Left: the real-axis integral passed just above both poles, or just below. Right: what each gives, drawn in spacetime with the kick at the origin — a shell expanding from it on the future cone, or one converging on it along the past cone.

Reading the figure. Passing above both poles is the same as pushing them below a contour on the axis: the $+i0$ of the retarded prescription. Closing as in worked example 5, the answer vanishes for $t < 0$ and lives on $r = ct$, $t > 0$ — the solid future cone, a sphere expanding at $c$. Passing below gives the mirror image under $t\mapsto-t$: nonzero only for $t < 0$, on $r = -ct$, a sphere shrinking onto the origin and arriving at $t = 0$. The diagram shows one line through the origin; at each $t$ the two points of a cone are a whole sphere in space.

The advanced function, checked. Close the $\omega$-contour with poles at $\pm ck + i0$: for $t > 0$ close downwards and get $0$; for $t < 0$ close upwards, anticlockwise, giving $\tilde G = -c^2H(-t)\sin(ckt)/ck$, as in unit 20. The $\vec k$-integral then gives $\frac{c}{2\pi^2r}\cdot\frac\pi2\,\delta(r + ct)$ for $t < 0$, i.e. $\delta(t + r/c)/4\pi r$.

History and a road not taken. Wheeler and Feynman (1945) built a classical electrodynamics on $\frac12(G_\mathrm{ret} + G_\mathrm{adv})$, recovering retarded radiation from the response of a perfectly absorbing universe. It is consistent, and not needed; the course uses $G_\mathrm{ret}$ and the boundary condition above.

24 · Green's function

Retarded potentials

theorem · the retarded solution

Let $f$ be $C^2$ on $\mathbb R^3\times\mathbb R$, zero outside a bounded set in $\vec x$ and for $t < T_0$. Then

$$u(\vec x,t) = \int\frac{f\big(\vec y,\ t - |\vec x - \vec y|/c\big)}{4\pi|\vec x - \vec y|}\,\mathrm d^3y$$

is $C^2$, vanishes for $t < T_0$, and solves $\Box u = f$. It is $G_\mathrm{ret} * f$.

proof: Kirchhoff's formula (two slides on) plus Duhamel's principle — full proof in the notes (N)

$t_\mathrm r = t - |\vec x - \vec y|/c$ is the retarded time: the field at $\vec x$ now reports the source at $\vec y$ as it was when light left it.

Time up, one line of space across. Light from $(\vec y, t_\mathrm r)$ reaches $(\vec x, t)$ along the past light cone of the field point; what the source does after $t_\mathrm r$ has not arrived yet.

Reading the figure. The 45° lines are light rays (time measured in units that make $c = 1$ on the drawing). The integral defining $u(\vec x, t)$ visits every source point $\vec y$ once, each at its own retarded time — the moment its world line, the vertical dashed line, crosses the cone. Points of the source nearer to $\vec x$ are seen as they were more recently.

Full proof of the retarded solution

Write $\Box u = f$ as $u_{tt} - c^2\nabla^2u = c^2f$. For each $s$ let $w(\cdot;s)$ solve the homogeneous equation for $t\ge s$ with $w = 0$ and $w_t = c^2f(\cdot,s)$ at $t = s$. Kirchhoff's formula (two slides on) gives $w(\vec x,t;s) = (t - s)\,M_{c(t-s)}[c^2f(\cdot,s)](\vec x)$, $C^2$ because $f$ is. Put $u(t) = \int_{-\infty}^tw(t;s)\,\mathrm ds$ — a finite integral, since $f = 0$ for $s < T_0$. By differentiation under the integral with a variable limit (unit 18), $u_t = w(t;t) + \int^tw_t\,\mathrm ds = \int^tw_t\,\mathrm ds$ and $u_{tt} = w_t(t;t) + \int^tw_{tt}\,\mathrm ds = c^2f(t) + c^2\nabla^2u$. This is Duhamel's principle, unit 06's convolution idea with an initial-value problem in place of a kick.

To the displayed formula. With $\tau = t - s$: $u = c^2\int_0^\infty\tau\,\frac1{4\pi}\int_{S^2}f(\vec x + c\tau\hat\omega, t - \tau)\,\mathrm d\Omega\,\mathrm d\tau$. Substitute $\varrho = c\tau$: $\tau\,\mathrm d\tau = \varrho\,\mathrm d\varrho/c^2$, so $u = \int_0^\infty\!\!\int_{S^2}\frac{f(\vec x + \varrho\hat\omega, t - \varrho/c)}{4\pi\varrho}\varrho^2\,\mathrm d\Omega\,\mathrm d\varrho$, which is the stated integral in spherical coordinates about $\vec x$. It is $G_\mathrm{ret} * f$: pairing $\delta(t - r/c)/4\pi r$ with $f(\vec x - \cdot, t - \cdot)$ does the time integral at $s = |\vec x - \vec y|/c$.

24 · Green's function

The retarded potentials, and a switch-on

the retarded potentials $$\varphi = \frac1{4\pi\varepsilon_0}\!\int\!\frac{\rho(\vec y,t_\mathrm r)}{|\vec x - \vec y|}\mathrm d^3y,$$ $$\vec A = \frac{\mu_0}{4\pi}\!\int\!\frac{\vec J(\vec y,t_\mathrm r)}{|\vec x - \vec y|}\mathrm d^3y .$$

They satisfy the Lorenz condition because $\partial_t\rho + \nabla\cdot\vec J = 0$ (notes), so they give a solution of all of Maxwell's equations.

the capacitor is switched on

If the charging current starts at $t = 0$, then at a point $1$ m from every part of the circuit nothing changes before $t = 1/c = 3.34$ ns. Static theory said: at once.

The Lorenz condition is exactly charge conservation seen through the retarded integral; it is what makes two separate wave equations a solution of all four Maxwell equations.

Full proof of the Lorenz condition

Let $R = |\vec x - \vec y|$ and $t_\mathrm r = t - R/c$. For a vector field $\vec J(\vec y, t_\mathrm r)$, the chain rule gives $\nabla_x\cdot\vec J(\vec y,t_\mathrm r) = -\frac1c\partial_t\vec J\cdot\nabla_xR$ and $\nabla_y\cdot[\vec J(\vec y,t_\mathrm r)] = (\nabla\cdot\vec J)(\vec y,t_\mathrm r) + \frac1c\partial_t\vec J\cdot\nabla_xR$, since $\nabla_yR = -\nabla_xR$. With $\nabla_x\frac1R = -\nabla_y\frac1R$, these combine to $\nabla_x\cdot\frac{\vec J(\vec y,t_\mathrm r)}{R} = -\nabla_y\cdot\frac{\vec J(\vec y,t_\mathrm r)}{R} + \frac{(\nabla\cdot\vec J)(\vec y,t_\mathrm r)}{R}$. Integrate over $\vec y$: the divergence term vanishes (divergence theorem; $\vec J$ has bounded support, and a small ball about $\vec y = \vec x$ contributes $O(\varepsilon)$). So $\nabla\cdot\vec A = \frac{\mu_0}{4\pi}\int\frac{(\nabla\cdot\vec J)(\vec y,t_\mathrm r)}{R} = -\frac{\mu_0}{4\pi}\int\frac{\partial_t\rho(\vec y,t_\mathrm r)}R = -\mu_0\varepsilon_0\partial_t\varphi$. $\square$ This is unit 23's proof that $\nabla\cdot\vec A_0 = 0$, with retardation.

The switch-on. $1/c = 1/2.99792458\times10^8 = 3.336$ ns. Strictly, $G_\mathrm{ret}$ is supported on the sphere $r = ct$, so the potentials at the point begin to change exactly when the sphere from the nearest source point arrives.

24 · Green's function

Kirchhoff's formula, and sharp wavefronts

spherical mean

$M_r[h](\vec x) = \dfrac1{4\pi}\displaystyle\int_{S^2}h(\vec x + r\hat\omega)\,\mathrm d\Omega$: the average of $h$ over the sphere of radius $r$ about $\vec x$.

Kirchhoff's formula

For $g\in C^3$, $h\in C^2$, the solution of $\Box u = 0$ with $u = g$, $u_t = h$ at $t = 0$ is $$u(\vec x,t) = \partial_t\big(t\,M_{ct}[g](\vec x)\big) + t\,M_{ct}[h](\vec x).$$

Proof (the $h$ term). Let $m(r) = M_r[h](\vec x)$. Darboux's identity (notes): $\nabla^2_xM_r = m'' + \frac2rm'$. For $v = t\,m(ct)$: $v_{tt} = 2cm' + c^2tm''$ and $c^2\nabla^2v = c^2t(m'' + \frac{2m'}{ct})$, equal. $v(0) = 0$, $v_t(0) = m(0) = h$. The $g$ term: $\partial_t$ of a solution is a solution. $\square$

Both terms read the data only through spherical means of radius $ct$: what the next slide turns into sharp fronts.

Full proof of Darboux's identity

$m(r) = \frac1{4\pi}\int_{S^2}h(\vec x + r\hat\omega)\,\mathrm d\Omega$. Differentiating under the integral, $m'(r) = \frac1{4\pi}\int_{S^2}\hat\omega\cdot\nabla h(\vec x + r\hat\omega)\,\mathrm d\Omega = \frac1{4\pi r^2}\oint_{|\vec y - \vec x| = r}\nabla h\cdot\hat n\,\mathrm dA = \frac1{4\pi r^2}\int_{|\vec y - \vec x|\le r}\nabla^2h\,\mathrm dV$ by the divergence theorem (unit 05). In spherical coordinates about $\vec x$ the last integral is $\int_0^r\varrho^2\int_{S^2}\nabla^2h(\vec x + \varrho\hat\omega)\,\mathrm d\Omega\,\mathrm d\varrho$, so $r^2m'(r) = \int_0^r\varrho^2M_\varrho[\nabla^2h](\vec x)\,\mathrm d\varrho$. Differentiate in $r$: $(r^2m')' = r^2M_r[\nabla^2h](\vec x) = r^2\nabla^2_xM_r[h](\vec x)$, the last step because $\nabla^2_x$ passes through the average. Divide by $r^2$: $m'' + \frac2rm' = \nabla^2_xM_r$. $\square$ (It is the radial part of the Laplacian in spherical coordinates, unit 23, which is no accident: $M_r[h](\vec x)$ is $h$ averaged over rotations about $\vec x$.)

The $g$ term. If $v$ solves $\Box v = 0$ with data $(0, g)$, then $\partial_tv$ solves it (the operator has constant coefficients) with data $(\partial_tv(0), \partial_t^2v(0)) = (g, c^2\nabla^2v(0)) = (g, 0)$. That needs $g\in C^3$, so that $v$ is $C^3$. Uniqueness of the solution is the energy argument of Why this prescription is the causal one. The formula's $M_{ct}$ is $G_\mathrm{ret}$ in disguise: $t\,M_{ct}[h] = c^{-2}\,G_\mathrm{ret}(\cdot,t) * h$ — pairing $c\,\delta(r - ct)/4\pi r$ with $h$ gives $c\cdot ct\,M_{ct}[h] = c^2t\,M_{ct}[h]$.

Limits. $t\to0$: $u\to M_0[g] = g$ ✓. Dimensions: $[tM[h]] = \mathrm T\cdot[u]\mathrm T^{-1}$ ✓.

24 · Green's function

A flash is seen as a flash

the strong Huygens principle

$u(\vec x,t)$ depends on the data only on the sphere $|\vec y - \vec x| = ct$ — not inside it.

So data supported in a ball $|\vec y|\le R$ reach a point at distance $D > R$ only for $(D - R)/c\le t\le(D + R)/c$. A flash is seen as a flash: a sharp front and a sharp back, and silence after.

unit 18 had only half of this

In one dimension d'Alembert's $\frac1{2c}\int_{x-ct}^{x+ct}h$ uses the data on the whole interval: a plucked point leaves a permanent step. Finite speed held there; sharp backs do not. Three dimensions is special — and two, next, is not.

Kirchhoff's sphere of radius $ct$ about $\vec x$ grows with $t$. It first touches the data at $ct = D - R$ and leaves them at $ct = D + R$; before and after, $u(\vec x, t)$ averages zeros.

Reading the figure. Each circle is the sphere $|\vec y - \vec x| = ct$ at one moment, and Kirchhoff's formula averages the data over it. While the sphere cuts the ball (solid, second colour) the average can be nonzero; while it is inside or outside the ball's reach (dashed), it sees only zeros. In 1D, d'Alembert's formula instead integrates $h$ over the whole interval $[x - ct, x + ct]$, which keeps containing the data once it has reached them — that is the permanent step of the warning box.

History. Poisson had the formula for spherical means in 1819; Kirchhoff (1882) gave the general representation that bears his name, as a mathematical form of Huygens' 1678 principle of secondary wavelets. The strong principle holds in every odd dimension $n\ge3$ and fails in every even one — cited: Courant and Hilbert, Methods of Mathematical Physics II, ch. VI, on the wave equation in higher dimensions. Problem P10 watches a spherical pulse focus and then vanish exactly.

24 · Green's function

Descent: in two dimensions a flash has a tail

Hadamard's method of descent: a solution in 3D whose data do not depend on $z$ stays independent of $z$, and is a 2D solution. So integrate $G_\mathrm{ret}$ along the $z$-axis. With $\varrho = \sqrt{x^2 + y^2}$:

$$\begin{aligned}G_2(\varrho,t) &= \int_{-\infty}^\infty\frac{\delta\big(t - \sqrt{\varrho^2 + z^2}/c\big)}{4\pi\sqrt{\varrho^2 + z^2}}\,\mathrm dz\\ &= \frac{c\,H(ct - \varrho)}{2\pi\sqrt{c^2t^2 - \varrho^2}} .\end{aligned}$$

supported inside the cone, not on it

Nonzero for every $t > \varrho/c$. After the front passes, the field falls off like $\frac{1}{2\pi t}$ and never stops. At $t = 10\varrho/c$ it is still $0.174$ of its value at $t = 2\varrho/c$.

At a fixed distance $\varrho$, computed: in 3D, $G_\mathrm{ret}$ is a single delta at $t = \varrho/c$ (the spike). $G_2$ jumps to infinity there and then decays like $1/t$; from $t = 2\varrho/c$ to $10\varrho/c$ it falls only from $0.577$ to $0.101$.

If the capacitor's feed wire were long and straight, it would be a 2D source: switched on, its field near the wire would keep arriving after the first front — from ever more distant parts of the wire.

Full computation of $G_2$

Let $s = \sqrt{\varrho^2 + z^2}$. The integrand is even in $z$; on $z > 0$, $\mathrm dz = s\,\mathrm ds/\sqrt{s^2 - \varrho^2}$. So $G_2 = 2\int_\varrho^\infty\frac{\delta(t - s/c)}{4\pi s}\frac{s\,\mathrm ds}{\sqrt{s^2 - \varrho^2}} = \frac1{2\pi}\int_\varrho^\infty\frac{\delta(t - s/c)}{\sqrt{s^2 - \varrho^2}}\,\mathrm ds$. With $\delta(t - s/c) = c\,\delta(s - ct)$ the integral picks out $s = ct$ if $ct > \varrho$ and nothing otherwise: $G_2 = \frac{c}{2\pi\sqrt{c^2t^2 - \varrho^2}}H(ct - \varrho)$. $\square$

Why descent is legitimate. If $f$ does not depend on $z$, neither does $G_\mathrm{ret} * f$ (translate in $z$), and $\Box$ acting on a $z$-independent function is the two-dimensional wave operator. So $(G_\mathrm{ret} * f)(x, y, t) = \int G_2(\cdot) f$ over the plane, with $G_2$ the $z$-integral of $G_\mathrm{ret}$ — Fubini on the pairing. The two-dimensional Kirchhoff formula (Poisson's formula) follows the same way: the sphere of radius $ct$ projects onto the disc $\varrho\le ct$ with area factor $ct/\sqrt{c^2t^2 - \varrho^2}$ from each hemisphere.

The numbers. $G_2(\varrho, t)\propto(c^2t^2 - \varrho^2)^{-1/2}$. At $ct = 2\varrho$: $(3\varrho^2)^{-1/2}$; at $ct = 10\varrho$: $(99\varrho^2)^{-1/2}$; ratio $\sqrt{3/99} = 0.1741$. For $ct\gg\varrho$, $G_2\approx c/(2\pi ct) = 1/(2\pi t)$. The figure plots the dimensionless $2\pi\varrho G_2/c = 1/\sqrt{\tau^2 - 1}$ against $\tau = ct/\varrho$, which takes the values $1/\sqrt3 = 0.577$ and $1/\sqrt{99} = 0.101$ at the two marked times.

Dimensions. In two dimensions $[\delta^{(3)}] = \mathrm T^{-1}\mathrm L^{-2}$, so $[G_2] = \mathrm T^{-1}$, and $c/\sqrt{c^2t^2 - \varrho^2}$ is $\mathrm T^{-1}$ ✓. And the 2D $G_2$ is not a delta on the cone, so the 2D Kirchhoff formula integrates over the whole disc — the strong principle fails in the plane. Problem P11 does this with its hints.

Physically. "Two-dimensional" means a source uniform along a line: a long wire, a cylindrical wave. The tail at a point is the signal from parts of the line further and further away, arriving later and later. A literally two-dimensional world would never hear a clean click.

section 7

Radiation

What an accelerated charge sends to infinity, and a failure this course records.

24 · radiation

Far from a small source

the dipole approximation

Sources within distance $d$ of the origin, varying on a time scale $T$ with $d\ll cT$. At distance $r\gg d$, to leading order in $1/r$,

$$\vec A(\vec x,t) = \frac{\mu_0}{4\pi r}\,\dot{\vec p}(t - r/c),$$

where $\vec p(t) = \int\vec y\,\rho(\vec y,t)\,\mathrm d^3y$ is the electric dipole moment.

Proof. In the retarded $\vec A$, replace $|\vec x - \vec y|$ by $r$ (error $\sim d/r$ relative) and $t_\mathrm r$ by $t - r/c$ (error $\sim d/cT$). Then $\int J_i = -\int y_i\nabla\cdot\vec J = \int y_i\partial_t\rho = \dot p_i$, integrating $\vec J\cdot\nabla y_i$ by parts. $\square$

three zones

$d\ll r\ll cT$: the near zone, quasi-static fields of unit 23, falling like $1/r^3$. $r\gg cT$: the wave zone, fields falling like $1/r$. Only the second carries energy to infinity.

The integral of $\vec J$. $\vec J\cdot\nabla y_i = J_i$, and $\nabla\cdot(y_i\vec J) = J_i + y_i\nabla\cdot\vec J$. Integrate over a ball containing the sources: the divergence term vanishes, so $\int J_i = -\int y_i\nabla\cdot\vec J = \int y_i\partial_t\rho = \dot p_i$. For a single charge $q$ at $\vec x_q(t)$, $\vec p = q\vec x_q$, $\dot{\vec p} = q\vec v$, $\ddot{\vec p} = q\vec a$.

The errors. $|\vec x - \vec y| = r - \hat r\cdot\vec y + O(d^2/r)$. In the denominator the correction is relative $O(d/r)$. In the retarded time it shifts the time by $\hat r\cdot\vec y/c\le d/c$, which changes the source by a relative $O(d/cT)$. Both are dropped. For an atom radiating visible light, $d\sim10^{-10}$ m and $cT\sim5\times10^{-7}$ m: an excellent approximation.

The three zones. $d\ll r\ll cT$: the near zone, where the fields are the quasi-static dipole fields of unit 23, falling as $1/r^3$. $r\gg cT$: the wave zone, $1/r$. The $1/r$ fields carry energy to infinity, because $|\vec S|r^2$ tends to a nonzero limit; the $1/r^2$ and $1/r^3$ fields do not.

24 · radiation

The radiation field

the radiation field

In the wave zone $r\gg cT$, keeping only the $1/r$ terms: $$\vec B = \frac{\mu_0}{4\pi cr}\,\ddot{\vec p}\times\hat r,\qquad \vec E = c\,\vec B\times\hat r ,$$ with $\ddot{\vec p}$ at $t - r/c$. Transverse, $|\vec E| = c|\vec B|$: locally a plane wave moving outward.

Why. On a function of $t - r/c$, $\nabla$ acts as $-\frac{\hat r}{c}\partial_t$ plus terms one power of $r$ smaller. So $\vec B = \nabla\times\vec A\approx-\frac{\hat r}c\times\partial_t\vec A$. For $\vec E$, the Lorenz condition gives $\varphi\approx c\,\hat r\cdot\vec A$ (notes).

Dimensions: $[\mu_0\ddot p/cr] = \mathrm{N\,A^{-2}}\cdot\mathrm{C\,m\,s^{-2}}/(\mathrm{m\,s^{-1}}\cdot\mathrm m) = \mathrm T$ ✓.

At a far point at angle $\theta$ from $\ddot{\vec p}$: $\vec B\propto\ddot{\vec p}\times\hat r$ points into the page, $\vec E = c\vec B\times\hat r$ lies along $\hat\theta$, and $\vec S\propto\vec E\times\vec B$ points out along $\hat r$. Both fields carry the factor $\sin\theta$.

Reading the figure. Put $\ddot{\vec p}$ along $\hat z$ and the field point in the $xz$-plane. Then $\hat z = \cos\theta\,\hat r - \sin\theta\,\hat\theta$, so $\ddot{\vec p}\times\hat r = -|\ddot{\vec p}|\sin\theta\,\hat\theta\times\hat r = |\ddot{\vec p}|\sin\theta\,\hat\phi$, and $\hat\phi = \hat y$ there, into the page with $x$ to the right and $z$ up. $\vec E = c\vec B\times\hat r\propto\hat\phi\times\hat r = \hat\theta$: minus the part of $\ddot{\vec p}$ transverse to $\hat r$. The arc is the wavefront through the point, emitted at the retarded time.

The electric field in full. The Lorenz condition $\nabla\cdot\vec A = -c^{-2}\partial_t\varphi$, with $\nabla\cdot\vec A\approx-\frac{\hat r}c\cdot\partial_t\vec A$, gives $\partial_t\varphi\approx c\,\hat r\cdot\partial_t\vec A$, so $\varphi\approx c\,\hat r\cdot\vec A$ up to a static (Coulomb) part, which falls off as $1/r^2$ and is dropped. Then $-\nabla\varphi\approx\frac{\hat r}c\partial_t\varphi = \hat r(\hat r\cdot\partial_t\vec A)$, and $\vec E = -\nabla\varphi - \partial_t\vec A\approx-\big(\partial_t\vec A - \hat r(\hat r\cdot\partial_t\vec A)\big)$: minus the part of $\partial_t\vec A$ transverse to $\hat r$. Meanwhile $c\vec B\times\hat r = -(\hat r\times\partial_t\vec A)\times\hat r = -\big(\partial_t\vec A - \hat r(\hat r\cdot\partial_t\vec A)\big)$ ✓.

24 · radiation

The Larmor formula

the Larmor formula

A charge $q$ with acceleration $\vec a$, at speed $v\ll c$, radiates power $$P = \frac{q^2|\vec a|^2}{6\pi\varepsilon_0c^3} = \frac{\mu_0|\ddot{\vec p}|^2}{6\pi c},$$ with angular distribution $\propto\sin^2\theta$ about $\vec a$: none along the line of acceleration.

Proof. $\vec S = \frac1{\mu_0}\vec E\times\vec B = \frac c{\mu_0}|\vec B|^2\hat r$ and $|\vec B| = \frac{\mu_0|\ddot{\vec p}|\sin\theta}{4\pi cr}$. So $P = \int|\vec S|r^2\mathrm d\Omega = \frac{\mu_0|\ddot{\vec p}|^2}{16\pi^2c}\int\sin^2\theta\,\mathrm d\Omega$, and $\int\sin^2\theta\,\mathrm d\Omega = \frac{8\pi}3$. $\square$

Dimensions: $[q^2a^2/\varepsilon_0c^3] = \mathrm{C^2m^2s^{-4}}/(\mathrm{C^2N^{-1}m^{-2}}\cdot\mathrm{m^3s^{-3}}) = \mathrm{N\,m\,s^{-1}} = \mathrm W$ ✓.

Power per unit solid angle, $\propto\sin^2\theta$, plotted as distance from the charge in a plane containing $\vec a$. Rotated about $\vec a$ it is a doughnut with no hole: nothing goes along the line of acceleration.

The angular integral. $\int_{S^2}\sin^2\theta\,\mathrm d\Omega = 2\pi\int_0^\pi\sin^3\theta\,\mathrm d\theta = 2\pi\cdot\frac43 = \frac{8\pi}3$. Then $\frac{\mu_0}{16\pi^2c}\cdot\frac{8\pi}3 = \frac{\mu_0}{6\pi c}$, and $\mu_0 = 1/\varepsilon_0c^2$ gives the first form. The average over a period of $\cos^2$ is $\frac12$, so the oscillating dipole radiates $\frac{\mu_0p_0^2\omega^4}{12\pi c}$.

Reading the figure. The curve is $r = \sin^2\theta$ in polar coordinates about the charge, computed; the marked point at $\theta = 40°$ has $r = \sin^240° = 0.413$ of the maximum. The power within an angle $\alpha$ of the plane perpendicular to $\vec a$ is the fraction $\frac32\sin\alpha - \frac12\sin^3\alpha$ of the total (integrate $\sin^3\theta$), which reaches one half at $\alpha = 20.3°$: a fairly fat doughnut.

24 · radiation

A blue sky, and an atom that should not exist

dipole radiation, and the blue sky

$\vec p = \vec p_0\cos\omega t$: $\langle P\rangle = \frac{\mu_0|\vec p_0|^2\omega^4}{12\pi c}$. Molecules driven far below resonance (unit 06) have $|\vec p_0|$ independent of $\omega$, so scattering $\propto\omega^4$: blue at 450 nm is scattered $(700/450)^4 = 5.86$ times as strongly as red at 700 nm.

$\omega^4$ as $(700\ \mathrm{nm}/\lambda)^4$ across the visible, computed: violet is scattered about nine times as strongly as red, blue at 450 nm $5.86$ times.

a failure, recorded and not repaired

A classical electron orbiting a proton is accelerated, so it radiates, loses energy, and spirals in. From the Bohr radius $a_0$, with $r_e = e^2/4\pi\varepsilon_0m_ec^2$:

$$t_\mathrm{collapse} = \frac{a_0^3}{4r_e^2c} = 1.56\times10^{-11}\ \mathrm s .$$

Atoms have existed for thirteen billion years. Classical electrodynamics cannot explain the stability of matter. The repair is quantum mechanics, which is outside this course; the course records the failure and goes on, because every result it needs from electromagnetism concerns fields, not atoms.

Reading the figure. The curve is $(700/\lambda)^4$ with $\lambda$ in nm, the ratio of $\omega^4$ at wavelength $\lambda$ to $\omega^4$ at 700 nm, since $\omega = 2\pi c/\lambda$. It is steep: the short-wavelength end of the visible dominates scattered light, and the sky is blue rather than violet only because sunlight has less violet in it and the eye is less sensitive there.

The collapse, sketched (problem P13 does it with every step). $E = -e^2/8\pi\varepsilon_0r$ for a circular orbit, $a = e^2/4\pi\varepsilon_0m_er^2$, and $\dot E = -P$ gives $\dot r = -\frac43r_e^2c/r^2$; integrate from $a_0$ to $0$. With $a_0 = 5.29177\times10^{-11}$ m and $r_e = 2.81794\times10^{-15}$ m: $t = 1.556\times10^{-11}$ s. The initial power is $4.67\times10^{-8}$ W against a binding energy of $13.6$ eV $= 2.18\times10^{-18}$ J. The orbit shrinks slowly on the scale of one period ($1.52\times10^{-16}$ s at $a_0$), so the circular-orbit assumption is self-consistent until the very end, and $v/c = 1/137$ at the start justifies the non-relativistic Larmor formula.

Why the course does not repair it. The failure is of classical mechanics applied to atoms, not of Maxwell's field equations: quantum electrodynamics keeps Maxwell's equations and changes what matter is. General relativity, where this course is going, is a classical field theory, and nothing in units 25–40 depends on how atoms are held together. It is recorded here because a course that only shows successes teaches the wrong lesson about how physics is built.

History. Joseph Larmor, 1897. Rayleigh explained the blue sky by $\omega^4$ scattering in 1871, from an elastic-aether argument and dimensional analysis — unit 00's method — before the electromagnetic derivation existed.

24 · summary

The capacitor, closed

one capacitor, every section

1 Steady $I$: $\vec B$ static, so Faraday adds nothing and the answers are exact. 2 $\varepsilon_0\partial_tE = 127.3\ \mathrm{A\,m^{-2}}$ carries the current across the gap; $B = 4.00\ \mu$T at the rim. 3 Energy enters through the rim at $VI$; the plates pull with $\frac{\varepsilon_0}2E^2$ per unit area. 4 The same two constants give $c$. 5 $\varphi_\mathrm C = -Ez$ is instantaneous; a $\chi\propto z^3$ reaches Lorenz gauge; $\vec E$, $\vec B$ unchanged. 6 Switched on, it is silent at 1 m for 3.34 ns.

7 · does it radiate?

At steady $I$, $\vec p = Qd\,\hat z$ with $\ddot{\vec p} = \dot Id\,\hat z = \vec 0$: no. Driven at 50 Hz with amplitude $1$ A: $p_0 = Id/\omega = 3.18\times10^{-6}$ C m, and $\langle P\rangle = \mu_0p_0^2\omega^4/12\pi c = 1.1\times10^{-17}$ W.

why the quasi-static picture works

Driven at angular frequency $\omega$, the field between the plates acquires a Faraday correction of relative size $(\omega a/c)^2/4$ at the rim (problem P2):

frequency	$\omega a/c$	correction
50 Hz	$5.2\times10^{-8}$	$6.9\times10^{-16}$
1 GHz	$1.05$	$0.27$: the picture fails

A circuit is quasi-static when it is small compared with $c/\omega$, the reduced wavelength. At 50 Hz that is $950$ km.

The radiated power in full. $Q(t) = (I_0/\omega)\sin\omega t$ for $I = I_0\cos\omega t$, so $p_0 = I_0d/\omega = 10^{-3}/(2\pi\cdot50) = 3.183\times10^{-6}$ C m; $\omega^4 = 9.74\times10^{9}\ \mathrm{s^{-4}}$; $\langle P\rangle = (1.2566\times10^{-6})(1.013\times10^{-11})(9.74\times10^9)/(12\pi\times2.998\times10^8) = 1.10\times10^{-17}$ W. Only the capacitor's own dipole has been counted; the feed wires form a loop whose magnetic-dipole radiation is of the same order of smallness. Either way, the 50 Hz circuit radiates nothing worth measuring, and circuit theory — which ignores radiation entirely — is the $c\to\infty$ limit of Maxwell's equations, justified by these numbers.

The reduced wavelength. $c/\omega = 2.998\times10^8/314.16 = 9.54\times10^5$ m. At 1 GHz it is $4.8$ cm, the size of the plates; there the capacitor is a cavity, and the field between the plates is the Bessel function $J_0(\omega s/c)$ of unit 09 — whose series $1 - (\omega s/c)^2/4 + \dots$ begins with problem P2's correction.

What the running example did not need. Fringing, the resistance of the wires, and the source that drives the current. Each would change numbers in the third figure and none would change a conclusion.

24 · summary

What this unit hands on

unit	what it takes from here
25	The wave operator $\Box$, whose behaviour under a Galilean boost is the crisis; Faraday's flux rule for moving loops.
28	$(\varphi/c, \vec A)$ as one covector field, the Lorenz condition as $\partial_\mu A^\mu = 0$, and $\vec E$, $\vec B$ as the six components of one antisymmetric tensor.
29	$u$, $\vec S$, $\vec g$ and the Maxwell stress tensor as one symmetric $T^{\mu\nu}$, from Noether's theorem.
34	The homogeneous pair as $\mathrm dF = 0$, and the potentials as its solution by the Poincaré lemma.
38	$G_\mathrm{ret}$ radiating gravitational waves; a gauge with residual freedom, spent the same way.

three things to carry forward

1. Consistency can force a law: continuity plus Helmholtz fixed the displacement current's divergence, and symmetry fixed the rest.

2. A wave equation's modes are oscillators, and choosing which way round their poles to go is choosing the arrow of time. In three dimensions the causal choice has sharp fronts.

3. Potentials are fields modulo a redundancy. A gauge is a choice; anything measurable is gauge-invariant; an instantaneous potential is not an instantaneous signal.

Reading. Griffiths, Introduction to Electrodynamics (4th ed.): §7.2–7.3 (induction, the displacement current, Maxwell's equations), §8.1–8.2 (Poynting, the stress tensor, field momentum), §9.1–9.2 (waves, polarisation, energy in plane waves), §10.1–10.2 (potentials, gauges, retarded potentials, Jefimenko), §11.1–11.2 (dipole radiation, Larmor). He treats $\delta$ as a function throughout (his §1.5); read each use as the distribution of unit 20. His §10.3, on point charges in arbitrary motion, is optional here and reads better after unit 28; §11.2.2 onwards (radiation reaction) is optional. Jackson, Classical Electrodynamics (3rd ed.), ch. 6: the displacement current, potentials and gauges, the Green function for the wave equation (done there by transforming in time only, the route of problem P9), and Poynting's theorem; skip the derivation of the macroscopic equations. For Kirchhoff, descent and Huygens: Evans, Partial Differential Equations, §2.4.1, which proves everything this deck does with spherical means and nothing it has not built.

What was deliberately left out. Fields in matter ($\vec D$, $\vec H$, dielectrics, magnetic materials); waveguides and cavities; the Liénard–Wiechert fields of an arbitrarily moving charge and radiation reaction, which are better done after unit 28; and the quantum theory of radiation.

24 · checkpoint

Done when

checkpoint

On paper and without notes, you can:

derive the displacement current from charge conservation alone, by taking the divergence of Ampère's law, and say what hypotheses make it the only repair;
obtain the vacuum wave equation from Maxwell's equations and compute $c$ from $\varepsilon_0$ and $\mu_0$, with its dimensions;
compute the retarded Green's function of $\Box$ by contour integration, proving it vanishes for $t < 0$ by closing the contour, and give a defensible account of why that prescription is the causal one;
show that two different gauges produce identical $\vec E$ and $\vec B$ for a given configuration.

Failing it is unambiguous

A displacement term with the wrong coefficient or a divergence that does not cancel $\mu_0\partial_t\rho$; a $c$ off by a power of ten; a $G_\mathrm{ret}$ nonzero before $t = 0$ or off the sphere $r = ct$; or two sets of potentials whose fields differ.

25 the crisis: apply a Galilean boost to $\Box$ and watch it fail, then find the transformation it survives.

Try the checkpoint on material in neither the deck nor the sheet. Displacement current: a sphere of charge $Q(t)$ leaking radially outward with spherical symmetry has $\vec J = \frac{-\dot Q}{4\pi r^2}\hat r$ outside the charge. Show that the total current $\vec J + \varepsilon_0\partial_t\vec E$ vanishes there, so $\nabla\times\vec B = \vec 0$ — as symmetry demands, since there is no azimuthal direction to point in.

$c$: redo the dimensional analysis with $\mathrm A$, $\mathrm V$ and $\Omega$, and compute the impedance of free space $\sqrt{\mu_0/\varepsilon_0} = 376.73\ \Omega$ as a by-product.

$G_\mathrm{ret}$: redo the $\omega$-integral for the operator $\Box + m^2$ (the Klein–Gordon operator, with $m$ an inverse length). The poles move to $\pm c\sqrt{k^2 + m^2}$ but stay real, so the same prescription and the same closing argument show the retarded function vanishes for $t < 0$ — without computing the $\vec k$-integral, which is harder and has a tail inside the cone.

Gauge: take a plane wave in Coulomb gauge, $\varphi = 0$, $\vec A = A_0\hat x\cos(kz - \omega t)$, and move it by $\chi = \lambda\sin(kz - \omega t)$; show the new potentials are still in Lorenz gauge and give the same fields.

Maxwell: the displacement current, and light

One capacitor, one missing term, seven sections

Faraday: a changing magnetic field curls the electric one

Faraday's law

Faraday's law, pointwise

Four equations, one of them suspect

Maxwell's repair

Ampère's law meets a charging capacitor

Where the inconsistency lives

Worked: the displacement current, from continuity alone

Exactly one term

The field between the plates

$B$ against distance from the axis

The complete system: energy and momentum

Maxwell's equations

Poynting's theorem: energy, locally

Worked: energy enters the capacitor through its rim

Worked: the capacitor's energy budget

Momentum, and the Maxwell stress tensor

Tension along field lines, pressure across them

The stress tensor at work, and the first energy–momentum tensor

Three densities and their fluxes

Light

The vacuum wave equation

Worked: the speed of light from two bench-top constants

How good did the measurements have to be?

Plane waves

What a plane wave carries

Polarisation

Polarisations, drawn

Potentials and gauge

Potentials, and what they fail to determine

The Lorenz gauge: two wave equations

The Coulomb gauge: an instantaneous potential

Worked: the capacitor in two gauges

The retarded Green's function

Fourier transforms in several variables

Every mode is an oscillator

Worked: the retarded Green's function by residues

Worked: the wave-vector integral, and the sphere $r = ct$

Why this prescription is the causal one

The other way round the poles

Retarded potentials

The retarded potentials, and a switch-on

Kirchhoff's formula, and sharp wavefronts

A flash is seen as a flash

Descent: in two dimensions a flash has a tail

Radiation

Far from a small source

The radiation field

The Larmor formula

A blue sky, and an atom that should not exist

The capacitor, closed

What this unit hands on

Done when

Failing it is unambiguous

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