Part C · the analytic toolkit · spine

Static fields: electrostatics and magnetostatics, the same vector calculus twice

Coulomb's law is Newton's with a sign changed, so unit 05 transfers whole; what is new is the boundary-value problem, the vector potential, and a theorem that says how many field equations there must be before you have seen any of them.

needs 05, 10, 19~2½ h slides + 5 h problemsproblems: ./mr problems 23

This is the unit in which vector calculus finally does the job it was invented for. Units 04 and 05 built line integrals, flux, Green, Stokes and Gauss for gravity, where only half of the machinery was ever needed: gravity is a gradient, so curls were always zero and Stokes' theorem had nothing to do. Electromagnetism uses all of it. The electric field is a gradient with sources; the magnetic field is a curl with no sources; and the two together exhaust what a vector field can be, which is the content of the Helmholtz theorem at the end.

Three things are genuinely new. First, boundaries: gravity has no conductors, so unit 05 never had to solve Laplace's equation with a condition on a surface. Here it is the main problem, and it brings uniqueness theorems, Green's functions, separation of variables with unit 19's completeness behind it, and the method of images. Second, the vector potential, whose existence is the Poincaré lemma one degree up and whose failure to be globally a gradient outside a solenoid is unit 04's vortex again. Third, counting: Helmholtz says a decaying field is fixed by its divergence and curl, so a theory of two vector fields needs exactly four equations.

How to read this deck

One problem runs through it: a grounded conducting sphere in a uniform field. Section 1 sets up electrostatics and conductors; section 2 proves that the sphere's problem has exactly one answer and builds the tools that find it; section 3 finds it twice, by Legendre expansion and by images, and reconciles the two term by term; section 4 does magnetostatics and puts the sphere in a magnetic field; section 5 proves Helmholtz and derives the sphere's field a third way.

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.

23 · overview

One sphere, two derivations, five sections

A metal ball of radius $a = 5$ cm, earthed, in a uniform field $E_0 = 10^5\ \mathrm{V\,m^{-1}}$ along $\hat z$. Charge flows onto it until the field inside is zero. Where does it sit, and what field does it make?

the route, in four steps

1. Let $\varphi$ be the potential (volts, $\vec E = -\nabla\varphi$) and $(r,\theta)$ spherical coordinates, $\theta$ from $\hat z$. The metal forces $\varphi = 0$ on $r = a$; far away $\varphi \to -E_0r\cos\theta$, the applied potential; between, $\nabla^2\varphi = 0$.

2. That problem has at most one solution, so any method that finds one has found the one.

3. Every axisymmetric harmonic function is $\sum_\ell(A_\ell r^\ell + B_\ell r^{-\ell-1})P_\ell(\cos\theta)$. Infinity leaves only $A_1 = -E_0$; the surface forces $B_1 = E_0a^3$: $\ \varphi = -E_0\big(r - a^3/r^2\big)\cos\theta$.

4. Or make $E_0$ with two far charges and mirror each in the sphere: the mirror charges form a dipole at the centre with that same potential $E_0a^3\cos\theta/r^2$.

The answer, computed: field lines end on the induced charge at right angles and crowd onto the poles, where the field is $3E_0$.

Sections. 1 electrostatics, conductors · 2 Laplace and Poisson · 3 the sphere twice · 4 magnetostatics · 5 Helmholtz: why four equations. Hover the sphere for its numbers.

The route, step by step. Step 1 is the slide Conductors, which proves that an equilibrium conductor is an equipotential and writes the sphere's problem down. Step 2 is Uniqueness, three ways. Step 3 is Separation in spherical coordinates (the expansion theorem, with unit 19's completeness behind it) and then Worked: the sphere by Legendre expansion, which fixes the coefficients. Step 4 is The Kelvin image in a sphere and Worked: the sphere by images; Two derivations, term by term shows the two routes agree coefficient by coefficient, not just at the end. Section 5 adds a third route, from the induced charge itself.

The problem sheet follows the same order. P1–P2 are the Green's function and the grounded plane; P3–P6 are boundary-value problems — the grounded sphere, a geometry that admits no finite set of images, hemispheres at opposite potentials, and a cylinder; P7 is a uniqueness proof; P8–P12 are magnetostatics, including the solenoid and the monopole; P13 is Helmholtz, stated precisely and used to count Maxwell's equations. Nine of them end in a number.

Why this example. It is the simplest problem in which a conductor's boundary condition does real work — the external field alone does not vanish on the sphere, and something has to be added that does — and it is small enough to be solved exactly in two genuinely different ways. The Legendre route is systematic: it would work for any boundary data. The image route is a trick that happens to work for spheres and planes. Seeing that the trick reproduces the systematic answer coefficient by coefficient is the best way to understand what the trick is.

Units. SI throughout: $\varepsilon_0 = 8.8541878128\times10^{-12}\ \mathrm{F\,m^{-1}}$, $\mu_0 = 1.25663706212\times10^{-6}\ \mathrm{N\,A^{-2}}$, $e = 1.602176634\times10^{-19}\ \mathrm C$. The deck writes $k = 1/4\pi\varepsilon_0 = 8.98755\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}$ where it saves ink.

section 1

Electrostatics is gravity with two signs

Everything unit 05 proved transfers by a dictionary. The differences are the interesting part, and conductors are the biggest of them.

23 · electrostatics

Coulomb's law, and everything unit 05 already proved

the electric field · definition, then law

A static charge density $\rho$ (C m⁻³, $C^1$, compact support) defines $$\vec E(\vec x) = \frac{1}{4\pi\varepsilon_0}\!\int\!\rho(\vec y)\,\frac{\vec x-\vec y}{\lVert\vec x-\vec y\rVert^3}\,\mathrm dV,$$ $$\varphi(\vec x) = \frac{1}{4\pi\varepsilon_0}\!\int\!\frac{\rho(\vec y)}{\lVert\vec x-\vec y\rVert}\,\mathrm dV,$$ and a test charge $q$ at rest feels $q\vec E$. That last clause is Coulomb's law, and it is measured, not proved.

These are unit 05's integrals with $-G\rho_{\mathrm m}$ replaced by $\rho/4\pi\varepsilon_0$. Every theorem there was proved from the integrals, so every theorem transfers with the same substitution.

gravity (unit 05)	electrostatics
$\vec g = -\nabla\Phi$	$\vec E = -\nabla\varphi$
$\oint\vec g\cdot\mathrm d\vec A = -4\pi GM_{\text{enc}}$	$\oint\vec E\cdot\mathrm d\vec A = Q_{\text{enc}}/\varepsilon_0$
$\nabla^2\Phi = 4\pi G\rho_{\mathrm m}$	$\nabla^2\varphi = -\rho/\varepsilon_0$
$\nabla\times\vec g = \vec 0$	$\nabla\times\vec E = \vec 0$
shell theorem, Earnshaw	the same, verbatim

checks

Dimensions: $[\rho/\varepsilon_0] = \mathrm{C\,m^{-3}}/\mathrm{C^2N^{-1}m^{-2}} = \mathrm{N\,C^{-1}m^{-1}} = [\nabla\cdot\vec E]$ ✓. Strength: two protons repel $1.236\times10^{36}$ times more strongly than they attract gravitationally, at any distance.

What transfers, precisely. Unit 05 proved, for any $C^1$ density of compact support: $\vec g = -\nabla\Phi$; Gauss's law; the local form $\nabla^2\Phi = 4\pi G\rho$; the shell theorem; the mean value property and the maximum principle for harmonic functions; Earnshaw's theorem. None of those proofs used the sign of $G$ or the sign of $\rho$ — they used linearity of the integral and the geometry of $1/r^2$. Replacing $-G\rho_{\mathrm m}$ by $\rho/4\pi\varepsilon_0$ throughout therefore gives proofs of the right-hand column. The curl is zero because $\vec E$ is a gradient (unit 04: the curl of a gradient vanishes by Clairaut).

The ratio. $e^2/(4\pi\varepsilon_0Gm_p^2) = 1.2356\times10^{36}$ with $m_p = 1.67262\times10^{-27}$ kg. Both forces go as $1/r^2$, so the ratio has no distance in it — it is a pure number, which is why dimensional analysis (unit 00) cannot predict it. For an electron and a proton it is $2.27\times10^{39}$.

History. Coulomb's torsion-balance measurements are from 1785. Cavendish had the inverse-square law earlier, by the null method that is the modern test — a charged conducting shell has no field inside if and only if the law is exactly inverse-square, which is unit 05's shell theorem read backwards — and did not publish; Maxwell edited his papers in 1879. Modern versions of the null experiment bound a deviation $r^{-2-\epsilon}$ at the level $|\epsilon| \lesssim 10^{-16}$ (cited: Williams, Faller and Hill, Phys. Rev. Lett. 26 (1971) 721).

A convention. Physics writes $\varphi$ for the potential per unit charge (volts) and $q\varphi$ for the potential energy. Unit 04's convention $\vec F = -\nabla V$ with $V$ an energy is $V = q\varphi$.

23 · electrostatics

Two signs: gravity accumulates, electricity screens

difference 1 · no cancellation in gravity

Mass has one sign, so every bit of matter adds to the monopole $M$, and a body's potential depth $GM/R \propto \rho R^2$ grows without bound with its size. Charge has two signs, and opposite charges attract: matter pulls in the opposite charge until it is neutral, and the monopole cancels.

how neutral is neutral?

Two equal bodies, each with a fraction $f$ of its protons uncompensated. Electric repulsion equals gravitational attraction when $(fNe)^2/4\pi\varepsilon_0 = G(Nm_p)^2$, that is $$f = m_p\sqrt{4\pi\varepsilon_0G}/e = 9.0\times10^{-19}.$$ An imbalance of one charge in $10^{18}$ would switch gravity off. The planets orbit by gravity because matter is neutral to better than that.

difference 2 · no shielding of gravity

There is no negative mass, so there is no gravitational conductor and no gravitational Faraday cage. The next slide's conductor — a region where the field is forced to zero by charges rearranging — has no gravitational analogue at all.

what does not change

Superposition, the $1/r^2$ geometry, and hence every differential equation. The two theories differ in their sources and boundary conditions, never in their field equations.

A third difference arrives in section 3: a body's dipole moment can always be removed by choosing the origin at its centre of mass; a neutral body's electric dipole cannot be removed by any choice.

The computation. $m_p\sqrt{4\pi\varepsilon_0G}/e = 1.67262\times10^{-27}\times\sqrt{4\pi\times8.8542\times10^{-12}\times6.6743\times10^{-11}}/1.60218\times10^{-19} = 8.996\times10^{-19}$, which is also $1/\sqrt{1.2356\times10^{36}}$, as it must be: the fraction is the square root of the inverse of the previous slide's ratio. The number of protons cancels, so the answer holds for two dust grains or two planets.

Why gravity wins at large scales anyway. Electric forces are overwhelmingly stronger, but they cancel; gravitational forces are feeble, but they add. On the scale of atoms and people, electricity wins; above roughly the scale of asteroids, the accumulated monopole of gravity wins. Where the crossover sits is a question about material strength, not about the laws.

Screening in general. In a conductor or a plasma, mobile charges rearrange around any inserted charge and cancel its field beyond a short distance. A conductor is the limit in which the cancellation is perfect in the interior — the next slide. The absence of anything like this for gravity is the reason unit 30's equivalence principle can hold: if gravity could be screened, a screened box would be a gravity-free laboratory distinguishable from free fall.

23 · electrostatics

Conductors

conductor

A closed region $K$ containing charges free to move under any force. In electrostatic equilibrium nothing moves, so the force $q\vec E$ on each free charge is zero: $\vec E = \vec 0$ in the interior of $K$.

theorem · what equilibrium forces

In equilibrium, on each connected conductor:

$\rho = 0$ in the interior — all charge sits on the surface, with surface charge density $\sigma$;
$\varphi$ is constant on $K$;
just outside, $\vec E$ is normal to the surface, and $\vec E = (\sigma/\varepsilon_0)\,\hat n$.

Proof. (1) $\rho = \varepsilon_0\nabla\cdot\vec E = 0$. (2) $\nabla\varphi = -\vec E = \vec 0$ on a connected open set. (3) is the next slide's jump conditions with $\vec E = \vec 0$ on the inside. $\square$

what this says about the sphere

"Grounded" means joined by a wire to a reservoir fixing its potential; take that value as $0$. So the problem is: find $\varphi$ outside the ball with

$$\nabla^2\varphi = 0\ \ (r > a),\qquad \varphi = 0\ \ (r = a),$$ $$\varphi + E_0r\cos\theta \to 0\ \ (r\to\infty).$$

The surface charge is then read off as $\sigma = -\varepsilon_0\,\partial_r\varphi\,|_{r=a}$. Nothing about the charge is known in advance; it is an output.

the unknown moved

Unit 05 was given $\rho$ and computed the field. Here part of $\rho$ — the surface charge — is unknown, and what is given is a value of $\varphi$ on a surface. That turns an integral into a boundary-value problem.

The definition is physics in mathematical clothing. A metal has about one conduction electron per atom free to move; in equilibrium they have stopped moving, which is possible only if the force on each vanishes. Real conductors reach equilibrium in a time of order $\varepsilon_0/\sigma_{\text{cond}}$ — about $10^{-19}$ s for copper by that formula, though the true relaxation is limited by the electrons' collision time, near $10^{-14}$ s — so for statics the idealisation is excellent.

Why "the potential at infinity" is the wrong reference here. The applied field's potential $-E_0z$ does not tend to a limit at infinity, so "grounded" cannot mean "at the potential of infinity". It means the ball is held at a fixed potential, and the physically natural choice is the potential the applied field had at the ball's centre, which is $0$. Section 3 shows the induced charge is then zero in total, so the grounded ball and an isolated neutral ball are the same problem.

The condition at infinity is a hypothesis. It says the field made by the induced charge decays. Without it the problem has many solutions — add $c\,(r - a^3/r^2)\cos\theta$ for any $c$. Section 2 proves that with it there is exactly one.

23 · electrostatics

Across a charged surface

jump conditions

Let a surface $S$ carry charge density $\sigma$, with unit normal $\hat n$ pointing from side $-$ to side $+$, and let $\vec E$ be continuous up to $S$ from each side. Then $$(\vec E_+ - \vec E_-)\cdot\hat n = \frac{\sigma}{\varepsilon_0},\qquad (\vec E_+ - \vec E_-)\times\hat n = \vec 0 .$$ The normal component jumps by $\sigma/\varepsilon_0$; the tangential component is continuous.

proof

Normal: Gauss's law on a pillbox of face area $A$ and height $h$ straddling $S$: flux $(\vec E_+ - \vec E_-)\cdot\hat n\,A + O(h) = \sigma A/\varepsilon_0 + o(A)$. Let $h\to0$, divide by $A$, let $A\to0$.
Tangential: $\nabla\times\vec E = \vec 0$, so by Stokes the circulation round a thin rectangle straddling $S$ is zero: $(\vec E_+ - \vec E_-)\cdot\hat t\,\ell + O(h) = 0$ for every tangent $\hat t$. $\square$

Gauss on the pillbox gives the normal jump; the circulation round the thin loop gives tangential continuity. The two arrows, just above and just below, share their sideways part; only the part along $\hat n$ jumps.

checks

A plane sheet: by symmetry $\vec E = \pm\frac{\sigma}{2\varepsilon_0}\hat n$ on the two sides, jump $\sigma/\varepsilon_0$ ✓. A conductor: $\vec E_- = \vec 0$ gives $\vec E_+ = (\sigma/\varepsilon_0)\hat n$ ✓. Dimensions: $[\sigma/\varepsilon_0] = \mathrm{C\,m^{-2}}/\mathrm{C^2N^{-1}m^{-2}} = \mathrm{N\,C^{-1}}$ ✓.

Full proof, with the estimates

Take a point $p \in S$, a small flat disc $\Delta$ of area $A$ in the tangent plane at $p$, and the cylinder $\Delta\times[-h/2, h/2]$ along $\hat n$ (for a curved $S$, bend the faces to follow it; the error is of smaller order). Gauss's law gives $\oint\vec E\cdot\mathrm d\vec A = Q_{\text{enc}}/\varepsilon_0$. The side wall has area $O(\sqrt A\,h)$ and $\vec E$ is bounded near $p$, so its contribution vanishes as $h\to0$. The faces contribute $\int_\Delta\vec E_\pm\cdot\hat n$ in the limit, by continuity from each side. The enclosed charge tends to $\int_\Delta\sigma$ (the volume charge in the box is $O(Ah)$). Dividing by $A$ and shrinking $\Delta$ uses continuity of $\vec E_\pm$ and $\sigma$ at $p$.

Tangential: the rectangle has sides $\ell$ along $\hat t$ and $h$ along $\hat n$. The cleanest justification avoids Stokes across the singular surface altogether: $\vec E = -\nabla\varphi$ with $\varphi$ continuous everywhere, including across $S$ (the potential integral of a bounded surface density converges uniformly, by the same estimate unit 05 used for a volume density), so the circulation of $\vec E$ round any closed loop is zero by the fundamental theorem of line integrals (unit 04). The two long sides give $(\vec E_+ - \vec E_-)\cdot\hat t\,\ell$ in the limit $h\to0$, and the short sides, of length $h$ with $\vec E$ bounded, give $O(h)$.

This is the promise from unit 05, where the jump in $\vec g$ across a massive shell was $-4\pi G\sigma_{\mathrm m}$: the same pillbox, with the dictionary. It is also the first example of a pattern that returns in unit 24: the divergence equation controls the normal component, the curl equation the tangential one.

section 2

Laplace and Poisson: uniqueness, and the Green's function

Before looking for the answer, prove there is only one. Then find the potential of a point charge in every dimension, and see why three is special.

23 · Laplace

Green's identities

For a bounded region $D$ with piecewise-$C^1$ boundary and outward normal $\hat n$, and $u, v \in C^2(\bar D)$: $$\int_D\big(u\nabla^2v + \nabla u\cdot\nabla v\big)\,\mathrm dV = \oint_{\partial D}u\,\partial_nv\,\mathrm dA \tag{I}$$ $$\int_D\big(u\nabla^2v - v\nabla^2u\big)\,\mathrm dV = \oint_{\partial D}\big(u\,\partial_nv - v\,\partial_nu\big)\,\mathrm dA \tag{II}$$ where $\partial_nv = \nabla v\cdot\hat n$.

proof

$\nabla\cdot(u\nabla v) = \nabla u\cdot\nabla v + u\nabla^2v$ by the product rule. Integrate and apply the divergence theorem (unit 05): (I). Swap $u$ and $v$ and subtract: (II). $\square$

what (II) says

If $u$ and $v$ both vanish on $\partial D$, then $\int u\nabla^2v = \int v\nabla^2u$: the Laplacian is a symmetric operator for the inner product $\int_D uv$ under Dirichlet conditions. It is Lagrange's identity of unit 19 with three variables, and every consequence there — real eigenvalues, orthogonal modes — has an analogue here.

what (I) says

With $u = v$ and $\nabla^2u = 0$: $\int_D|\nabla u|^2 = \oint u\,\partial_nu$. The left side is, up to $\varepsilon_0/2$, the field energy in $D$; the identity says a harmonic function's energy is decided on the boundary. The next slide spends this.

Dimensions: each side of (I) is $[u][v]\,\mathrm L$ ✓.

The product rule. In components, $\partial_i(u\,\partial_iv) = \partial_iu\,\partial_iv + u\,\partial_i^2v$; sum over $i$. The divergence theorem needs $u\nabla v \in C^1(\bar D)$, which is why $u, v \in C^2$.

Green's 1828 essay — the one unit 06's notes described, self-published in Nottingham — contains both identities and the representation formula of this section. He used them for exactly this unit's problems: the potential of conductors, and the distribution of charge on them.

A remark on the energy. That $\tfrac{\varepsilon_0}{2}\int|\vec E|^2$ is the energy stored in an electrostatic field is cited here; unit 24 derives it from Poynting's theorem. Nothing in this unit depends on the interpretation — the uniqueness proof uses identity (I) as mathematics.

23 · Laplace

Uniqueness, three ways

uniqueness theorem

Let $D$ be bounded, or the exterior of a bounded set. Poisson's equation $\nabla^2\varphi = -\rho/\varepsilon_0$ in $D$ has at most one solution continuous on $\bar D$ and $C^2$ inside with

Dirichlet data: $\varphi$ given on $\partial D$; or
conductor data: $\partial D$ a union of conductors, each with its total charge given;

and, if $D$ is unbounded, $\varphi \to 0$ at infinity (with $\nabla\varphi = O(r^{-2})$ in case 2).

proof of 1

The difference $w$ of two solutions is harmonic with $w = 0$ on $\partial D$. By the maximum principle (unit 05), $\max w$ and $\min w$ over $\bar D$ are attained on $\partial D$, so $w \equiv 0$. Unbounded $D$: apply this to $D\cap B_R$; $|w| \le \max_{|x| = R}|w| \to 0$. $\square$

proof of 2 · by energy

$w$ is harmonic, constant ($=W_i$) on the $i$-th conductor, and carries zero net charge there, $\oint_{S_i}\partial_nw = 0$ (Gauss). Identity (I) with $u = v = w$: $$\int_D|\nabla w|^2 = \sum_i W_i\oint_{S_i}\partial_nw = 0 .$$ So $\nabla w = 0$: the field is unique, and $\varphi$ up to a constant — fixed by $\varphi\to0$ at infinity, or by one conductor's potential. $\square$

for the sphere

$D$ is the exterior of the ball, the data are Dirichlet ($\varphi = 0$ on $r = a$), and the hypothesis at infinity is imposed on $w$, the difference of two solutions: both contain $-E_0r\cos\theta$, which cancels. So there is at most one answer, and any method that produces one has produced the one.

The third way is the title's third: the representation formula (The representation formula, and the Green's function of a region) writes any solution explicitly in terms of its data, which is uniqueness again. So the maximum principle, the energy identity and the Green's function each prove it.

The detail in case 2 for unbounded $D$. Apply identity (I) on $D\cap B_R$. The extra boundary term on $|x| = R$ is $\oint w\,\partial_nw$, bounded by $\max|w|\cdot O(R^{-2})\cdot4\pi R^2$, which tends to $0$ because $w\to0$. Hence the decay hypotheses.

Why case 1 needs $w$ continuous on $\bar D$. The maximum principle compares interior values with boundary values, so the function has to reach the boundary. A classical counterexample shows what goes wrong without it in two dimensions: on the punctured disc, $\log r$ vanishes on $r = 1$, is harmonic, is not zero — and is not continuous at the puncture, which is part of the boundary.

Mixed data. Some conductors at given potentials and others with given charges also give uniqueness: the boundary sum in proof 2 vanishes term by term either way. Problem P7 asks for the proof with the details.

23 · Green's function

Radial solutions, and where the dimension enters

In $\mathbb R^n$ let $f$ depend on $r = \lVert\vec x\rVert$ only. Since $\partial_ir = x_i/r$, the chain rule (unit 03) gives $\partial_if = f'x_i/r$ and $$\partial_i^2f = f''\frac{x_i^2}{r^2} + f'\Big(\frac1r - \frac{x_i^2}{r^3}\Big).$$ Summing over $i$:

the radial Laplacian $$\nabla^2f = f'' + \frac{n-1}{r}f' = \frac{1}{r^{n-1}}\big(r^{n-1}f'\big)' .$$

So $f$ is harmonic on $\mathbb R^n\setminus\{0\}$ iff $r^{n-1}f' = $ const, iff $$f = \begin{cases}\alpha + \beta\,r^{2-n}, & n \ne 2,\\ \alpha + \beta\log r, & n = 2.\end{cases}$$

the geometry behind it

$r^{n-1}f'$ constant is flux conservation: the field $-f'$ crosses spheres whose area grows as $r^{n-1}$, so it must fall as $r^{1-n}$. The exponent is the dimension of the sphere at infinity, not a fact about charge.

The same field lines cross the circle of radius $r$ and the one of radius $2r$. In $\mathbb R^n$ the outer sphere has $2^{n-1}$ times the area, so the lines — the field — are $2^{n-1}$ times thinner there.

Reading the figure. It is drawn in the plane, so literally it is the case $n = 2$: twelve lines, spread over a circumference $2\pi r$, then over $4\pi r$ — half as dense, a field falling as $1/r$. In three dimensions the circles are spheres, the area quadruples from $r$ to $2r$, and the field falls as $1/r^2$. The integral statement behind the picture is that the flux $\omega_{n-1}r^{n-1}\cdot(-f'(r))$ through the sphere of radius $r$ ($\omega_{n-1}$ the area of the unit sphere) is the same for every $r$ — which is exactly $(r^{n-1}f')' = 0$.

The computation, once more. $\partial_i(x_i/r) = 1/r - x_i^2/r^3$, and $\sum_ix_i^2 = r^2$, so $\sum_i\partial_i(f'x_i/r) = f''\cdot1 + f'(n/r - 1/r)$. The form $r^{1-n}(r^{n-1}f')'$ expands to the same thing. Radial harmonic functions then solve the first-order linear equation $(r^{n-1}f')' = 0$, solved by two quadratures (unit 00).

General $n$. The pattern extends: potential $r^{2-n}$ for $n \ge 3$. The normalisation on the next slide needs the area of the unit sphere $S^{n-1}$, $\omega_{n-1} = 2\pi^{n/2}/\Gamma(n/2)$, and the divergence theorem in $\mathbb R^n$; both are cited for $n \ge 4$ (unit 34 proves Stokes' theorem in every dimension). The cases $n = 1, 2, 3$ need only the fundamental theorem of calculus, Green's theorem (unit 04) and the divergence theorem (unit 05).

23 · Green's function · example

Worked: the constant, in one, two and three dimensions

worked example 1 · the fundamental solution

Want: radial $\Gamma_n$, harmonic off $0$, with unit outward flux of $-\nabla\Gamma_n$ through every sphere about $0$ — the potential of a unit charge with $\varepsilon_0 = 1$.

Step 1. From the last slide, $-\Gamma_n'(r) = c\,r^{1-n}$. With $\omega_{n-1}$ the area of the unit sphere in $\mathbb R^n$, the flux through radius $r$ is $c\,r^{1-n}\cdot\omega_{n-1}r^{n-1} = c\,\omega_{n-1}$, independent of $r$. Set it to 1: $c = 1/\omega_{n-1}$.

Step 2. The "sphere" areas: $\omega_0 = 2$ (two points, $\pm r$), $\omega_1 = 2\pi$, $\omega_2 = 4\pi$.

Step 3. Integrate $-\Gamma_n' = r^{1-n}/\omega_{n-1}$, dropping the constant: $$\Gamma_1 = -\frac{|x|}{2},\qquad \Gamma_2 = -\frac{1}{2\pi}\log r,\qquad \boxed{\Gamma_3 = \frac{1}{4\pi r}}$$

Step 4 · checks. $n=3$: $\varphi = q\Gamma_3/\varepsilon_0 = q/4\pi\varepsilon_0r$, Coulomb ✓. $n=2$: a line charge $\lambda$ has $E = \lambda/2\pi\varepsilon_0s$, as Gauss on a cylinder gives directly ✓. $n=1$: a sheet $\sigma$ has $E = \sigma/2\varepsilon_0$ on each side, as the jump conditions gave ✓.

a number

A line charge of $1\ \mathrm{nC\,m^{-1}}$ at $1$ cm: $E = \lambda/2\pi\varepsilon_0s = 1798\ \mathrm{V\,m^{-1}}$.

the sign convention

Physics wants $\Gamma > 0$ near a positive source, so $\Gamma$ satisfies $-\nabla^2\Gamma = $ "unit source". Some books use $G = -\Gamma$ and $\nabla^2G = $ "unit source". Same object, opposite sign.

dimensions

$[\Gamma_n] = \mathrm L^{2-n}$: charge $\times\,\Gamma_n/\varepsilon_0$ is a potential only when the "charge" is per $\mathrm L^{3-n}$ — per metre for a line ($n=2$), per square metre for a sheet ($n=1$) ✓.

Step 1 in each dimension, honestly. For $n = 3$ the flux integral is unit 05's surface integral over a sphere. For $n = 2$ it is the outward flux through a circle, $\oint\vec F\cdot\hat n\,\mathrm ds$, which Green's theorem (unit 04) relates to the divergence. For $n = 1$ the "sphere" is the pair of points $\pm r$, the outward normals are $\pm1$, and the flux of $-\Gamma_1' = \operatorname{sgn}(x)/2$ is $\tfrac12 + \tfrac12 = 1$.

The checks in full. Line charge: a cylinder of radius $s$ and length $L$ round the line encloses $\lambda L$; by symmetry $\vec E = E(s)\hat s$, so $2\pi sLE = \lambda L/\varepsilon_0$ and $E = \lambda/2\pi\varepsilon_0s$. From $\Gamma_2$: $\varphi = -(\lambda/2\pi\varepsilon_0)\log s$, $E = -\varphi' = \lambda/2\pi\varepsilon_0s$ ✓. The number: $10^{-9}/(2\pi\times8.8542\times10^{-12}\times0.01) = 1797.5\ \mathrm{V\,m^{-1}}$.

What the fundamental solution is not yet. It is tempting to write $-\nabla^2\Gamma = \delta$ and move on. $\delta$ is not a function, and the equation has no classical meaning at $0$; unit 20 builds distributions and makes it an exact statement. This unit needs only what the representation-formula slide proves: $\Gamma$ reproduces a function from its Laplacian and boundary values. That property is what "$-\nabla^2\Gamma = \delta$" is shorthand for.

23 · Green's function

Why three is special

$n$	field $\propto$	potential $\propto$	at $\infty$
1	$r^0$	$r$	grows
2	$r^{-1}$	$\log r$	grows, slowly
3	$r^{-2}$	$r^{-1}$	$\to 0$

what the plot shows

All three potentials have the same value and the same field at $r = 1$; they are $\omega_{n-1}\Gamma_n$ shifted by a constant. Only the $n = 3$ curve has a limit as $r\to\infty$, and only a potential with a limit can have that limit called zero.

Three is the smallest dimension in which a point charge's potential can be normalised to zero at infinity. Unit 00 promised this: dimensional analysis cannot see it; geometry can.

The radial harmonic functions with $f(1) = 0$ and $f'(1) = -1$: in one dimension $1 - r$ falls for ever, in two $-\log r$ falls for ever but slowly, in three $1/r - 1$ levels off at $-1$.

Why this matters physically. In $n = 2$ — which is what a uniform line charge in three dimensions looks like, since nothing depends on $z$ — the potential grows like $\log r$ and cannot be set to zero at infinity. Line charges therefore have potentials defined only relative to a chosen radius, and "grounded at infinity" is meaningless for them. Problem P6's cylinder feels this.

The normalisation in the plot. $\omega_{n-1}\Gamma_n$ has $-(\omega_{n-1}\Gamma_n)' = r^{1-n}$, which is $1$ at $r = 1$ in every dimension; subtracting its value at $r = 1$ gives the plotted $1 - r$ ($n = 1$, from $2\Gamma_1 = -r$ for $r > 0$), $-\log r$ ($n = 2$) and $1/r - 1$ ($n = 3$). The $n = 1$ line is the potential of a charged sheet seen from one side; the $n = 2$ curve, that of a line charge; the $n = 3$ curve, that of a point charge, which in the usual normalisation $\Gamma_3 = 1/4\pi r$ is the curve shifted up by $1$ and scaled by $1/4\pi$.

23 · Green's function

The representation formula, and the Green's function of a region

Green's representation formula

For $u \in C^2(\bar D)$, $D \subset \mathbb R^3$ bounded, and $\vec x \in D$: $$\begin{aligned}u(\vec x) = &\int_D\Gamma(\vec x-\vec y)\big(-\nabla^2u\big)\,\mathrm dV_y\\ &+ \oint_{\partial D}\big(\Gamma\,\partial_nu - u\,\partial_n\Gamma\big)\,\mathrm dA_y,\end{aligned}$$ with $\Gamma = \Gamma_3$ and derivatives in $\vec y$.

proof: identity (II) with a small ball cut out — full proof in the notes (N)

corollary · Poisson in all space

If $\nabla^2\varphi = -\rho/\varepsilon_0$ with $\rho$ of compact support, $\varphi\to0$ and $\nabla\varphi = O(r^{-2})$, then $\varphi = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\vec y)}{\lVert\vec x-\vec y\rVert}\mathrm dV$ — unit 05's integral is the solution.

the Green's function of $D$

$G(\vec x,\vec y) = \Gamma(\vec x-\vec y) - h_{\vec x}(\vec y)$, where $h_{\vec x}$ is harmonic in $D$ and equals $\Gamma(\vec x - \cdot)$ on $\partial D$. So $G = 0$ on $\partial D$, and $$u(\vec x) = \int_DG\,(-\nabla^2u)\,\mathrm dV - \oint_{\partial D}u\,\partial_nG\,\mathrm dA .$$

what $G$ is, physically

$G(\vec x,\cdot)/\varepsilon_0$ is the potential of a unit charge at $\vec x$ in the presence of the grounded surface $\partial D$; $-h_{\vec x}/\varepsilon_0$ is the potential of the charge it induces. Finding $h$ for a sphere is the method of images (section 3).

Full proof of the representation formula

Fix $\vec x \in D$ and $\varepsilon$ with $\bar B_\varepsilon(\vec x) \subset D$. Put $v(\vec y) = \Gamma(\vec x - \vec y) = 1/(4\pi\lVert\vec x-\vec y\rVert)$, harmonic on $D_\varepsilon = D\setminus\bar B_\varepsilon$. Identity (II) on $D_\varepsilon$ with $\nabla^2v = 0$: $$-\int_{D_\varepsilon}v\nabla^2u = \oint_{\partial D}(u\partial_nv - v\partial_nu) + \oint_{S_\varepsilon}(u\partial_nv - v\partial_nu),$$ where on the small sphere $S_\varepsilon$ the normal out of $D_\varepsilon$ points towards $\vec x$. There $v = 1/4\pi\varepsilon$ and $\partial_nv = +1/4\pi\varepsilon^2$ ($v$ increases towards $\vec x$). So $\oint_{S_\varepsilon}u\,\partial_nv = $ the average of $u$ over $S_\varepsilon$, which tends to $u(\vec x)$ by continuity, and $|\oint_{S_\varepsilon}v\,\partial_nu| \le \frac{1}{4\pi\varepsilon}\cdot4\pi\varepsilon^2\max|\nabla u| \to 0$. The volume integral converges as $\varepsilon\to0$ because $v$ is integrable (unit 05's convergence argument). Letting $\varepsilon\to0$ and rearranging gives the formula. $\square$

The corollary. Apply the formula on $B_R(\vec 0)$ with $R$ large. On $|\vec y| = R$, $\Gamma = O(1/R)$ and $\partial_nu = O(R^{-2})$, and $u = o(1)$, $\partial_n\Gamma = O(R^{-2})$; the area is $4\pi R^2$; so the boundary term is $O(1/R) + o(1) \to 0$. What remains is $\varphi = \int\Gamma\,\rho/\varepsilon_0$. This is also a uniqueness proof: any solution with those decay properties equals the integral.

The Green's function formula. Apply (II) to $u$ and the harmonic $h_{\vec x}$ on all of $D$: $\int_Dh\nabla^2u = \oint(h\partial_nu - u\partial_nh)$. Since $h = \Gamma$ on $\partial D$, subtracting this from the representation formula cancels the $\partial_nu$ term and leaves the boxed formula. Whether $h_{\vec x}$ exists for a general $D$ is the Dirichlet problem itself, and is cited (Evans, Partial Differential Equations, §2.2.4, for the construction on balls and half-spaces; Gilbarg and Trudinger, Elliptic Partial Differential Equations of Second Order, §2.8, Perron's method, for general bounded regions with smooth boundary). For the half-space and the ball it is written down explicitly by images, which is why those two are the regions everybody computes.

Symmetry. $G(\vec x,\vec y) = G(\vec y,\vec x)$: identity (II) applied to $G(\vec x,\cdot)$ and $G(\vec y,\cdot)$ with both points cut out. Physically, the potential at $\vec y$ due to a unit charge at $\vec x$ near a grounded conductor equals the potential at $\vec x$ due to a unit charge at $\vec y$ — Green's reciprocity.

23 · Laplace

Earnshaw, transferred

Earnshaw's theorem · electrostatic form

A point charge $q$ in a fixed electrostatic field, in a region free of other charge, has no point of stable equilibrium.

proof

Its potential energy $q\varphi$ is harmonic there. A stable equilibrium would be a strict local minimum of $q\varphi$ at an interior point $\vec x_0$: $q\varphi(\vec x_0) < q\varphi$ on a small sphere round $\vec x_0$. But by the mean value property (unit 05), $q\varphi(\vec x_0)$ is the average of $q\varphi$ over that sphere. Contradiction. $\square$

Unit 05 proved this for gravity. The proof is the same proof; only the potential is renamed.

what it does not forbid

Charges held by non-electric forces (atoms: quantum mechanics), fields that change in time (Paul's ion traps), and magnetic levitation of induced moments. It forbids only static, purely electric trapping of a charge by fixed sources.

What the proof is about. Picture a small sphere round the candidate point. A strict minimum would put the centre below every point of that sphere, hence below their average; the mean value property says the centre is the average. The next slide shows what a harmonic function does instead, on the running example: wherever it curves up in one direction it curves down in another.

History. Samuel Earnshaw, 1842, Trans. Camb. Phil. Soc. 7, 97, in a paper about the luminiferous ether: he showed that a lattice of particles held by inverse-square forces could not be stable, and so could not be the elastic medium light was supposed to need. The theorem outlived the ether by a century. It is the reason matter needs quantum mechanics to be stable, and Paul traps (Nobel 1989) evade it by making the field oscillate so the averaged force is restoring — a time-dependent loophole, not a contradiction.

Why "fixed" matters. A charge near a conductor moves the conductor's induced charge with it, so the field it feels is not fixed. The attraction to a grounded plane (P2) is unstable in the normal direction for a simpler reason — the attraction grows as the charge approaches — but the proof above does not apply to it, and this deck makes no general claim about systems of charges and movable conductors.

23 · Laplace

Earnshaw on the sphere: a saddle, never a well

on the sphere

Along the axis above the north pole, $\varphi = -E_0(z - a^3/z^2)$ decreases steadily, so a positive test charge is pushed away and a negative one pulled in; sideways the curvature has the opposite sign. $\partial_z^2\varphi = +6E_0a^3/z^4 > 0$ on the axis, so $\partial_x^2\varphi + \partial_y^2\varphi = -6E_0a^3/z^4$: every "valley" in one direction is a "ridge" in another.

At $P$, $z = 1.5a$: $\partial_z^2\varphi = 6E_0a^3/z^4 = 1.19\,E_0/a$, and $\partial_x^2\varphi = \partial_y^2\varphi = -0.59\,E_0/a$ — summing to zero, as Laplace requires.

Computed from $\varphi = -E_0(r - a^3/r^2)\cos\theta$: the potential near $P$ minus its linear part. Stepping along the axis it curves up; stepping across, down — no direction is a well for both signs at once.

The axis computation. On the axis $\theta = 0$, $r = z$, $\varphi = -E_0(z - a^3z^{-2})$, so $\partial_z\varphi = -E_0(1 + 2a^3z^{-3})$ and $\partial_z^2\varphi = +6E_0a^3z^{-4}$ (differentiate $-2a^3z^{-3}$: the two minus signs cancel). Laplace's equation forces the transverse second derivatives to sum to $-6E_0a^3/z^4$; by the axial symmetry they are equal, each $-3E_0a^3/z^4$. That is the saddle, explicitly: $\varphi$ curves up along the axis and down across it.

What is plotted. Along the axis, $\varphi(0, z_0 + u) - \varphi(P) - u\,\partial_z\varphi(P)$; across it, $\varphi(u, z_0) - \varphi(P)$ (the first derivative across vanishes at $P$ by symmetry). Both start as $\tfrac12u^2$ times the second derivatives above, $1.19$ and $-0.59$ in units of $E_0/a$ at $z_0 = 1.5a$; the axis curve is lopsided because stepping towards the sphere ($u < 0$) the $a^3/z^2$ term grows fast. Removing the linear part is what makes the curvature visible: $P$ is not an equilibrium at all — the field there is $E_0(1 + 2a^3/z_0^3) = 1.59\,E_0$ — and Earnshaw says that no point outside the ball could be a stable one, for either sign of test charge.

section 3

The grounded sphere, twice

Once systematically, by separating variables with unit 19's completeness behind it; once by a trick, with two fictitious charges; and then the proof that the trick is the system in disguise.

23 · the sphere · Legendre

Separation in spherical coordinates

theorem · axisymmetric harmonic expansion

Let $u$ be $C^2$, harmonic and independent of $\phi$ on a shell $r_1 < r < r_2$. Then there are constants $A_\ell, B_\ell$ with $$u(r,\theta) = \sum_{\ell\ge0}\Big(A_\ell r^\ell + \frac{B_\ell}{r^{\ell+1}}\Big)P_\ell(\cos\theta),$$ the series converging in mean square on each sphere $r = $ const.

proof

Set $u_\ell(r) = \frac{2\ell+1}{2}\int_0^\pi u\,P_\ell(\cos\theta)\sin\theta\,\mathrm d\theta$. Laplace's equation (unit 19's spherical form) says $\partial_r(r^2\partial_ru) = -\Delta_{S^2}u$. Differentiate under the integral (Leibniz, unit 18) and move $\Delta_{S^2}$ onto $P_\ell$ by two integrations by parts in $\theta$, the boundary terms dying with $\sin\theta$: $$\big(r^2u_\ell'\big)' = \ell(\ell+1)\,u_\ell .$$ Euler's equation: $r^s$ solves it iff $s(s+1) = \ell(\ell+1)$, so $s = \ell$ or $-\ell-1$, and these span the two-dimensional solution space. Completeness of the $P_\ell$ in $L^2[-1,1]$ (unit 19) gives the series. $\square$

reading it

$r^\ell P_\ell$ is regular at the origin and grows; $r^{-\ell-1}P_\ell$ decays and is singular at the origin. A problem inside a sphere keeps the first kind; outside, conditions at infinity decide the $A_\ell$.

why completeness is essential

Without it the $u_\ell$ would be numbers attached to $u$ with no guarantee that they rebuild it. Unit 19 proved the $P_\ell$ complete, and that is what turns "try $r^\ell P_\ell$" into "every solution is a sum of these".

Without axial symmetry, the same proof with $Y_\ell^m$ in place of $P_\ell$ gives $\sum_{\ell,m}(A_{\ell m}r^\ell + B_{\ell m}r^{-\ell-1})Y_\ell^m$ (unit 19, completeness cited there).

Full proof of the key step

Write $L u = \frac1{\sin\theta}\partial_\theta(\sin\theta\,\partial_\theta u)$, the axisymmetric $\Delta_{S^2}$. For fixed $r$, $\theta\mapsto u(r,\theta)$ is $C^2$ on $[0,\pi]$ (spherical coordinates are a smooth map, and $u$ is $C^2$). Then, integrating by parts twice, $$\int_0^\pi(Lu)P_\ell\sin\theta\,\mathrm d\theta = \big[\sin\theta\,(\partial_\theta u\,P_\ell - u\,\partial_\theta P_\ell)\big]_0^\pi + \int_0^\pi u\,(LP_\ell)\sin\theta\,\mathrm d\theta .$$ The bracket vanishes at both ends because $\sin 0 = \sin\pi = 0$ and everything multiplying it is bounded. And $LP_\ell(\cos\theta) = -\ell(\ell+1)P_\ell(\cos\theta)$ — Legendre's equation under $x = \cos\theta$, the substitution checked in unit 19's notes. Hence $\int(Lu)P_\ell\sin\theta = -\ell(\ell+1)\int uP_\ell\sin\theta$.

Laplace's equation in spherical coordinates for axisymmetric $u$ reads $\partial_r(r^2\partial_ru) + Lu = 0$. Multiply by $\frac{2\ell+1}2P_\ell\sin\theta$ and integrate over $[0,\pi]$. The first term is $(r^2u_\ell')'$ because $u$, $u_r$, $u_{rr}$ are continuous on $[r_1', r_2']\times[0,\pi]$ for any closed sub-shell, so Leibniz's rule applies twice. The second is $-\ell(\ell+1)u_\ell$. That is the ODE.

The ODE. $r^2u'' + 2ru' - \ell(\ell+1)u = 0$ is linear, second order, with continuous coefficients and leading coefficient $r^2 \ne 0$ on $r > 0$, so its solution space is two-dimensional (units 01 and 06). $r^\ell$ and $r^{-\ell-1}$ are two solutions and are independent (their Wronskian is $-(2\ell+1)r^{-2} \ne 0$).

The series. For fixed $r$, $x\mapsto u(r,\arccos x)$ is continuous on $[-1,1]$ and $u_\ell(r)$ is its Legendre coefficient ($\sin\theta\,\mathrm d\theta = -\mathrm dx$). Unit 19 proved the $P_\ell$ complete, so the Legendre series converges to it in $L^2[-1,1]$. Stronger convergence is true but is not claimed here, because the method never needs it: it only ever identifies coefficients, and mean-square convergence makes coefficients unique.

23 · the sphere · example

Worked: the sphere by Legendre expansion

worked example 2 · the grounded sphere

Step 1 · the form. The data are axisymmetric, the answer is unique, and rotating it about $z$ gives another answer — so it is axisymmetric. On $r > a$: $\varphi = \sum_\ell(A_\ell r^\ell + B_\ell r^{-\ell-1})P_\ell(\cos\theta)$.

Step 2 · infinity. $\varphi + E_0rP_1 \to 0$, and $P_1(\cos\theta) = \cos\theta$. Its $\ell$-th coefficient $A_\ell r^\ell + B_\ell r^{-\ell-1} + E_0r\,\delta_{\ell1}$ must tend to $0$, so $$A_1 = -E_0,\qquad A_\ell = 0\ (\ell \ne 1).$$

Step 3 · the surface. $\varphi = 0$ at $r = a$ for every $\theta$, so every coefficient vanishes there: $A_\ell a^\ell + B_\ell a^{-\ell-1} = 0$, i.e. $$B_\ell = -A_\ell\,a^{2\ell+1}:\qquad B_1 = E_0a^3,\quad B_\ell = 0\ (\ell\ne1).$$

Step 4 · assemble. $$\boxed{\ \varphi = -E_0\Big(r - \frac{a^3}{r^2}\Big)\cos\theta\ }\qquad (r \ge a).$$

checks

Laplace: $r\cos\theta = z$ is harmonic; $r^{-2}\cos\theta = -\partial_z(1/r)$ is a derivative of a harmonic function, so harmonic ✓.
Boundary: $r = a$ gives $0$ ✓. Infinity: the correction $E_0a^3\cos\theta/r^2 \to 0$ ✓.
Dimensions: $[E_0r] = \mathrm{V\,m^{-1}}\cdot\mathrm m = \mathrm V$ ✓.
Limit $a\to0$: the ball disappears and $\varphi = -E_0z$, the bare field ✓.
Special case $\theta = \pi/2$: $\varphi = 0$ on the whole equatorial plane, which is at the ball's potential by the up–down antisymmetry of the problem ✓.

one hypothesis spent silently

Step 3 uses that $\varphi$ is continuous up to $r = a$, so that $u_\ell(r)\to u_\ell(a) = 0$. That is part of what "the potential is $0$ on the sphere" means, and it is exactly the uniqueness theorem's hypothesis.

Step 1 in detail. The expansion theorem needs $\varphi$ axisymmetric. Let $R_\alpha$ be rotation by $\alpha$ about the $z$-axis. $\varphi\circ R_\alpha$ is harmonic, vanishes on the sphere, and satisfies the condition at infinity because $-E_0z$ is invariant. By uniqueness $\varphi\circ R_\alpha = \varphi$. This is unit 05's "symmetric data, unique solution, so symmetric solution", which that unit's notes said would be paid for here.

Step 2 in detail. The coefficient functions of $v = \varphi + E_0r\cos\theta$ are $v_\ell(r) = A_\ell r^\ell + B_\ell r^{-\ell-1} + E_0r\,\delta_{\ell1}$, and $|v_\ell(r)| \le \sqrt{2\ell+1}\,\max_\theta|v(r,\theta)| \to 0$ — Cauchy–Schwarz in $L^2[-1,1]$ with $\int_{-1}^1P_\ell^2 = 2/(2\ell+1)$ (unit 10). A combination $\alpha r^\ell + \beta r^{-\ell-1}$ tends to $0$ as $r\to\infty$ only if $\alpha = 0$, for every $\ell \ge 0$. Apply this to $(A_\ell + E_0\delta_{\ell1})r^\ell + B_\ell r^{-\ell-1}$.

The shells. The theorem was proved on a shell $a < r < r_2$. For larger $r_2$ the coefficient functions are the same functions (they are defined by the same integral), so the constants $A_\ell, B_\ell$ do not depend on $r_2$, and the expansion holds on all of $r > a$.

Inside. The ball is a conductor, so $\vec E = 0$ and $\varphi \equiv 0$ in $r < a$. The potential is continuous across $r = a$; its radial derivative is not, and the jump is the surface charge — next slide.

23 · the sphere

Reading the answer: charge, dipole, and a factor of three

From $\varphi = -E_0(r - a^3/r^2)\cos\theta$: $$E_r = -\partial_r\varphi = E_0\Big(1 + \frac{2a^3}{r^3}\Big)\cos\theta,$$ $$E_\theta = -\frac1r\partial_\theta\varphi = -E_0\Big(1 - \frac{a^3}{r^3}\Big)\sin\theta .$$

three results

Surface charge (jump condition, $\vec E = 0$ inside): $$\boxed{\sigma = \varepsilon_0E_r|_{r=a} = 3\varepsilon_0E_0\cos\theta}$$ Induced dipole: the $r^{-2}$ term is $\dfrac{p\cos\theta}{4\pi\varepsilon_0r^2}$ with $$\boxed{p = 4\pi\varepsilon_0a^3E_0}$$ Field at the poles: $|\vec E| = 3E_0$. At $r = a$, $E_\theta = 0$: the field meets the conductor normally ✓.

Equipotentials of the answer, in steps of $0.4\,E_0a$, computed; field lines faint behind. The level $\varphi = 0$ is the sphere together with the equatorial plane. Above and below the poles the levels crowd together, towards a third of their far spacing at the surface: that is the $3E_0$.

Reading the figure. Far away the equipotentials are the planes $z = $ const of the applied field, evenly spaced. Near the ball they bend to wrap round it, because the ball itself is a level surface. On the axis just above the north pole the first level, $\varphi = -0.4E_0a$, sits at $r - a^3/r^2 = 0.4a$, i.e. $r = 1.15a$: a gap of $0.15a$ where far away the gap is $0.4a$. The field is the potential step divided by the gap, so it averages $2.6E_0$ over that first gap, rising to $3E_0$ at the surface itself. At the equator the levels open out and the field at the surface vanishes.

The dipole, independently. $p = \int z\,\sigma\,\mathrm dA = \int_0^\pi(a\cos\theta)(3\varepsilon_0E_0\cos\theta)2\pi a^2\sin\theta\,\mathrm d\theta = 6\pi\varepsilon_0E_0a^3\cdot\tfrac23 = 4\pi\varepsilon_0a^3E_0$ ✓ — the moment of the charge distribution equals the coefficient read off the potential, as unit 10's multipole expansion says it must.

The polarisability. $p = \alpha E_0$ with $\alpha = 4\pi\varepsilon_0a^3$: a conducting sphere's polarisability is $4\pi\varepsilon_0$ times its volume over $4\pi/3$. Atomic polarisabilities, divided by $4\pi\varepsilon_0$, are volumes of about the atom's size cubed for the same reason — to a first approximation an atom is a small conductor.

Why three. Section 3's cylinder slides show the factor is $2$ for a cylinder and, in $\mathbb R^n$, $n$: it is the dimension again. The field lines crowd onto the poles from all around; the more directions there are to crowd in from, the more they crowd.

23 · the sphere

The sphere, in numbers

with numbers · $a = 5.00$ cm, $E_0 = 1.00\times10^5\ \mathrm{V\,m^{-1}}$

peak $\sigma = 3\varepsilon_0E_0$	$2.656\ \mu\mathrm{C\,m^{-2}}$
charge on the upper hemisphere, $3\pi\varepsilon_0E_0a^2$	$+20.86\ \mathrm{nC}$
total induced charge	$0$
dipole $p$	$1.391\times10^{-9}\ \mathrm{C\,m}$
field at the poles	$3.00\times10^5\ \mathrm{V\,m^{-1}}$

The pole field is a tenth of the roughly $3\times10^6\ \mathrm{V\,m^{-1}}$ at which dry air breaks down; raise $E_0$ tenfold and the ball sparks at its poles first.

checks

$[\varepsilon_0E_0] = \mathrm{C\,m^{-2}}$ ✓; $[\varepsilon_0a^3E_0] = \mathrm{C\,m}$ ✓. Total charge $\propto\int_0^\pi\cos\theta\sin\theta\,\mathrm d\theta = 0$ ✓. Limit $a\to0$: $p\to0$ ✓.

$\sigma = 3\varepsilon_0E_0\,z/a$ is linear in height. A band of the sphere of height $\mathrm dz$ has area $2\pi a\,\mathrm dz$ wherever it is (right), so each triangle's area times $2\pi a$ is a hemisphere's charge: equal and opposite.

The hemisphere charge. $\int_0^{\pi/2}3\varepsilon_0E_0\cos\theta\cdot2\pi a^2\sin\theta\,\mathrm d\theta = 3\varepsilon_0E_0\cdot2\pi a^2\cdot\tfrac12 = 3\pi\varepsilon_0E_0a^2 = 3\pi\times8.8542\times10^{-12}\times10^5\times0.0025 = 2.086\times10^{-8}$ C, about $1.3\times10^{11}$ electrons' worth. The dipole: $4\pi\times8.8542\times10^{-12}\times1.25\times10^{-4}\times10^5 = 1.391\times10^{-9}$ C m.

The figure's shortcut. On the sphere, $z = a\cos\theta$ and $\mathrm dA = 2\pi a^2\sin\theta\,\mathrm d\theta = 2\pi a\,|\mathrm dz|$: zones of equal height have equal area (Archimedes' hat-box theorem, here a one-line change of variable). So the charge above height $z$ is $2\pi a\int_z^a\sigma\,\mathrm dz'$, and the upper hemisphere carries $2\pi a\cdot\tfrac12\cdot a\cdot3\varepsilon_0E_0 = 3\pi\varepsilon_0E_0a^2$ — the same number.

The breakdown field of dry air at sea level is about $3\ \mathrm{MV\,m^{-1}}$ (cited: this is a measured property, varying with pressure and humidity; the engineering literature quotes $3.0$ for uniform fields at one atmosphere).

23 · the sphere · images

The method of images, and why it is allowed

the method of images

To solve a problem in a region $D$ bounded by grounded conductors, place fictitious charges outside $D$ so that the total potential vanishes on $\partial D$. The result is harmonic in $D$ (away from the real charges) and has the right boundary values, so by uniqueness it is the solution in $D$. Outside $D$ it means nothing.

the plane

Charge $q$ at height $d$ above the grounded plane $z = 0$: image $-q$ at depth $d$. Every point of the plane is equidistant from both, so $\varphi = 0$ there. Problem P2 finishes it.

Field lines of $q$ and its image, computed. Above the plane they are the real field: every line from $q$ ends on the plane, at right angles. Dashed, below, is the fictitious continuation to the image — the real field there is zero.

Reading the figure. The region $D$ is the half-space $z > 0$. In it the field of $q$ and $-q$ together is the field of $q$ in the presence of the grounded plane: same Laplace equation away from $q$, same boundary value $0$ on the plane, same decay — so by uniqueness the same function. The lines bend down to meet the plane normally because the plane is an equipotential. Below the plane the two pictures part company: the image picture continues the lines to $-q$, while the real metal has $\vec E = \vec 0$ throughout. The charge that actually ends the lines is the surface charge on the plane, $-q$ in total (problem P2).

What images cannot do. The images must lie outside the region, and there must be finitely many for the method to be a method. Two planes at $90°$ need three images; at $60°$, five; at $120°$, no finite set works — problem P4 proves it.

23 · the sphere · images

The Kelvin image in a sphere

Kelvin image in a sphere

A charge $q$ at distance $R > a$ from the centre of a grounded sphere of radius $a$ has image $$q' = -\frac{a}{R}\,q\quad\text{at distance}\quad b = \frac{a^2}{R}$$ along the same ray. Together they give $\varphi = 0$ on the sphere.

proof · Apollonius

Let $\hat e$ be the unit vector along the ray. For $\vec y$ on the sphere at angle $\gamma$ from it, $|\vec y - b\hat e|^2 = a^2 + b^2 - 2ab\cos\gamma = \frac{a^2}{R^2}\big(R^2 + a^2 - 2aR\cos\gamma\big) = \frac{a^2}{R^2}|\vec y - R\hat e|^2 .$ So $q/|\vec y - R\hat e| + q'/|\vec y - b\hat e| = 0$. $\square$

Limit $a,R\to\infty$ with $R - a = d$ fixed: $q'\to-q$ and $a - b = ad/R \to d$, the plane ✓.

Drawn to scale with $R = 2.5a$, so that $b = 0.4a$ and $q' = -0.4q$; wherever $\vec y$ is on the sphere, its distances to the two charges are in the ratio $a/R$ — the ratio of the charges — and their potentials cancel.

The step $b = a^2/R$. $a^2 + a^4/R^2 - 2a^3\cos\gamma/R = (a^2/R^2)(R^2 + a^2 - 2aR\cos\gamma)$: multiply out the right side and compare term by term. The sphere is the locus of points whose distances to two fixed points are in a fixed ratio $a/R$ — Apollonius' circle, rotated. The two points are inverse to each other in the sphere: $bR = a^2$.

The Green's function of the ball. Kelvin's image supplies the $h_{\vec x}$ of the previous section: for $\vec x$ outside the ball, $G(\vec x,\vec y) = \Gamma(\vec x-\vec y) - \frac{a}{|\vec x|}\Gamma(\vec x^*-\vec y)$ with $\vec x^* = a^2\vec x/|\vec x|^2$, and the same formula works for $\vec x$ inside (the image is then outside). Computing $\partial_nG$ on the sphere gives, for the interior problem, Poisson's integral formula $$u(\vec x) = \frac{a^2 - |\vec x|^2}{4\pi a}\oint_{|\vec y| = a}\frac{u(\vec y)}{|\vec x - \vec y|^3}\,\mathrm dA_y .$$ Full proof. On $|\vec y| = a$, $|\vec x^* - \vec y| = (a/|\vec x|)|\vec x - \vec y|$, and $\nabla_y G\cdot\vec y/a$ reduces, using $(\vec y-\vec x)\cdot\vec y = a^2 - \vec x\cdot\vec y$ and $\frac{|\vec x|^2}{a^2}(\vec y - \vec x^*)\cdot\vec y = |\vec x|^2 - \vec x\cdot\vec y$, to $\partial_nG = (|\vec x|^2 - a^2)/(4\pi a|\vec x-\vec y|^3)$. Put this into the Green's function formula with $\nabla^2u = 0$. This is the solution of the Dirichlet problem on a ball that unit 05's notes said was needed for the converse of the mean value property: at $\vec x = 0$ it is the mean value property.

23 · the sphere · example

Worked: the sphere by images

worked example 3 · two charges and their images

Step 1 · make the uniform field. Put $+Q$ at $z = -R$ and $-Q$ at $z = +R$. Near the origin each contributes $Q/4\pi\varepsilon_0R^2$ along $+\hat z$, so choose $$Q = 2\pi\varepsilon_0R^2E_0 .$$ As $R\to\infty$ the field near the ball tends to $E_0\hat z$ (next slide).

Step 2 · image each. By Kelvin: $+Q$ at $-R$ has image $-Qa/R$ at $-a^2/R$; $-Q$ at $+R$ has image $+Qa/R$ at $+a^2/R$. The four charges give $\varphi = 0$ on the sphere, by linearity.

Step 3 · the images form a dipole. Total image charge $0$. Moment $$p = \frac{Qa}{R}\cdot\frac{a^2}{R} - \frac{Qa}{R}\cdot\Big(-\frac{a^2}{R}\Big) = \frac{2Qa^3}{R^2} = 4\pi\varepsilon_0a^3E_0,$$ independent of $R$, with separation $2a^2/R \to 0$: an ideal dipole at the centre.

Step 4 · the limit. Sources: $-E_0r\cos\theta$. Images: $p\cos\theta/4\pi\varepsilon_0r^2 = E_0a^3\cos\theta/r^2$. Sum: $-E_0(r - a^3/r^2)\cos\theta$ — worked example 2's answer ✓.

Schematic, not to scale: with $R = 1$ m the source charges are $20$ ball radii away and the images only $a^2/R = 2.5$ mm from the centre.

numbers, $R = 1$ m

$Q = 5.563\ \mu\mathrm C$; images $\pm0.2782\ \mu\mathrm C$ at $\pm2.5$ mm; $p = 1.391\times10^{-9}\ \mathrm{C\,m}$ — the same as before, at every $R$.

Step 1. The field of $+Q$ at $(0,0,-R)$, evaluated at the origin, points away from the charge: $+\hat z$, magnitude $Q/4\pi\varepsilon_0R^2$. The field of $-Q$ at $(0,0,R)$ points towards it: also $+\hat z$. Total $2Q/4\pi\varepsilon_0R^2 = E_0$. With $R = 1$ m and $E_0 = 10^5\ \mathrm{V\,m^{-1}}$: $Q = 2\pi\times8.8542\times10^{-12}\times10^5 = 5.563\times10^{-6}$ C.

Step 3's sign. The upper image is $+Qa/R$: the image of the negative charge above is positive, so positive charge collects near the top — the $\sigma \propto +\cos\theta$ of the Legendre answer ✓. Physically: the field points up, pushes positive charge up.

Why the total image charge matters. It is the total induced charge on the sphere (Gauss's law on a large surface enclosing the sphere but not the far charges, in the image picture and the real picture alike — the fields agree outside the ball). It is zero, confirming the claim on the conductors slide that the grounded ball and the isolated neutral ball are the same problem here.

History. William Thomson (later Lord Kelvin) introduced electrical images in 1845–1848, the same years in which he rediscovered Green's essay. The inversion $\vec x\mapsto a^2\vec x/|\vec x|^2$ behind the construction is the Kelvin transform: if $u$ is harmonic, so is $\frac{a}{|\vec x|}u(\vec x^*)$ (cited: Axler, Bourdon and Ramey, Harmonic Function Theory, ch. 4). The image of a point charge is that transform applied to its potential.

23 · the sphere · reconciliation

Two derivations, term by term

one charge and its image, in Legendre form

Charge $q$ at $R\hat z$; image $q' = -qa/R$ at $b\hat z$. By unit 10's generating function, for $b < r < R$: $$4\pi\varepsilon_0\varphi = \sum_\ell\Big(\underbrace{\frac{q}{R^{\ell+1}}}_{A_\ell}r^\ell + \underbrace{\frac{q'b^\ell}{1}}_{B_\ell}\frac{1}{r^{\ell+1}}\Big)P_\ell(\cos\theta).$$ And $q'b^\ell = -\dfrac{qa}{R}\dfrac{a^{2\ell}}{R^\ell} = -\dfrac{q}{R^{\ell+1}}a^{2\ell+1}$, so $$\boxed{B_\ell = -A_\ell\,a^{2\ell+1}\ \text{ for every }\ell}$$ — exactly the Legendre boundary condition of worked example 2, step 3.

what the image is

The source contributes one growing term $A_\ell r^\ell$ per $\ell$. The sphere demands a decaying partner $-A_\ell a^{2\ell+1}r^{-\ell-1}$ for each. A single point charge at $a^2/R$ happens to supply all of them at once, because its coefficients $q'b^\ell$ are geometric in $\ell$ — as the $A_\ell$ are. That is the whole trick.

the limit, term by term

For the pair, $A_\ell = -\frac{2Q}{4\pi\varepsilon_0R^{\ell+1}}$ for odd $\ell$, $0$ for even. With $Q = 2\pi\varepsilon_0R^2E_0$: $A_1 = -E_0$, and $A_3 = -E_0/R^2$, $A_5 = -E_0/R^4$, …, all $\to0$. At $r = a$, $R = 1$ m the $\ell = 3$ term is $(a/R)^2 = 0.25\,\%$ of the $\ell = 1$ term.

The expansions. Unit 10's generating function gives, for $r < R$, $\frac{1}{|\vec x - R\hat z|} = \sum_\ell\frac{r^\ell}{R^{\ell+1}}P_\ell(\cos\theta)$, and for $r > b$, $\frac{1}{|\vec x - b\hat z|} = \sum_\ell\frac{b^\ell}{r^{\ell+1}}P_\ell(\cos\theta)$. Both hold in the annulus $b < r < R$, which contains the sphere $r = a$ because $b = a^2/R < a < R$.

The pair. $+Q$ at $-R\hat z$: $\frac1{|\vec x + R\hat z|} = \sum(-1)^\ell\frac{r^\ell}{R^{\ell+1}}P_\ell$ (since $P_\ell(-x) = (-1)^\ell P_\ell(x)$). $-Q$ at $+R\hat z$: $-\sum\frac{r^\ell}{R^{\ell+1}}P_\ell$. Sum: $\frac{Q}{4\pi\varepsilon_0}\sum_\ell\big((-1)^\ell - 1\big)\frac{r^\ell}{R^{\ell+1}}P_\ell$, which is $-\frac{2Q}{4\pi\varepsilon_0}\frac{r^\ell}{R^{\ell+1}}$ for odd $\ell$ and $0$ for even $\ell$. At $\ell = 1$, with $2Q/4\pi\varepsilon_0 = E_0R^2$: $-E_0r\cos\theta$ ✓.

So what does the limit $R\to\infty$ mean? For each finite $R$ the image construction solves the problem exactly — with a slightly non-uniform applied field, whose $\ell = 3, 5, \dots$ parts are small. The Legendre solution with those same $A_\ell$ is the same function, coefficient by coefficient. As $R\to\infty$ every coefficient converges to the uniform-field one, and so does the solution itself — seen directly by expanding the four Coulomb potentials to first order in $r/R$ and $a^2/Rr$ with $Q = 2\pi\varepsilon_0R^2E_0$, which gives $-E_0z$ from the sources and $E_0a^3\cos\theta/r^2$ from the images, with errors $O(r^2/R^2)$ and $O(a^2/Rr)$. The two derivations are not merely equal at the end; they are equal at every stage.

Why images fail for other shapes. The method needs one charge whose coefficients match the boundary's demand for every $\ell$ at once. For the sphere the demand $B_\ell = -A_\ell a^{2\ell+1}$ is geometric in $\ell$ when $A_\ell$ is, which a point charge supplies. For an ellipsoid the demand mixes different $\ell$ and no finite set of point charges supplies it.

23 · separation

Cylindrical coordinates: the same method, other special functions

In cylindrical coordinates $(s, \phi, z)$ (unit 19's polar Laplacian plus $\partial_z^2$): $$\nabla^2\varphi = \frac1s\partial_s(s\,\partial_s\varphi) + \frac1{s^2}\partial_\phi^2\varphi + \partial_z^2\varphi .$$ Separate $\varphi = S(s)\,\mathrm e^{\mathrm im\phi}Z(z)$ with $m\in\mathbb Z$ by periodicity and $Z'' = k^2Z$:

$$s^2S'' + sS' + (k^2s^2 - m^2)S = 0 .$$

case	radial solutions
$k \ne 0$	$J_m(ks),\ Y_m(ks)$ — Bessel, unit 09
$k = 0$, $m \ne 0$	$s^{m},\ s^{-m}$
$k = 0$, $m = 0$	$1,\ \log s$

A problem with no $z$-dependence has $k = 0$, and then the radial solutions are powers — exactly as on the sphere, where they were $r^\ell$ and $r^{-\ell-1}$.

The separation. Substituting and dividing by $\varphi$: $\frac{1}{sS}(sS')' - \frac{m^2}{s^2} + \frac{Z''}{Z} = 0$. The last term depends on $z$ alone and the rest on $s$ alone, so $Z''/Z = k^2$ is constant (unit 19's argument). Multiply by $s^2$: Bessel's equation in the variable $ks$, whose solutions unit 09 built by Frobenius. The sign of $k^2$ decides the geometry: $Z = \mathrm e^{\pm kz}$ with oscillating $J_m(ks)$ in $s$ — a cylinder with its lid at a potential — or $Z = \cos(kz)$ with $k^2 < 0$, which turns $J_m$ into the modified Bessel functions $I_m$, $K_m$ (cited: Watson, A Treatise on the Theory of Bessel Functions, §3.7). Laplace's equation cannot oscillate in all three directions at once, because $\nabla^2 = 0$ forces the separation constants to sum to zero.

The $k = 0$ equation $s^2S'' + sS' - m^2S = 0$ is Euler's again: $s^\alpha$ works iff $\alpha^2 = m^2$, and for $m = 0$ the double root gives $1$ and $\log s$ — the two-dimensional fundamental solution appearing as a separated solution.

23 · separation

The cylinder beside the sphere

the cylinder beside the sphere

A grounded conducting cylinder, radius $a$, in a transverse field $E_0\hat x$: $k = 0$, $m = 1$, and the same two conditions give $$\varphi = -E_0\Big(s - \frac{a^2}{s}\Big)\cos\phi,\qquad \sigma = 2\varepsilon_0E_0\cos\phi .$$ Peak field $2E_0$ where the sphere had $3E_0$.

the factor is the dimension

In $\mathbb R^n$, $r^{1-n}\cos\theta = \frac{1}{2-n}\partial_z(r^{2-n})$ is harmonic ($n = 2$: $s^{-1}\cos\phi = \partial_x\log s$), and $\varphi = -E_0(r - a^n r^{1-n})\cos\theta$ solves the ball problem, with $E_r(a) = E_0\big(1 + (n-1)\big) = nE_0$. $n = 2$: cylinder ✓. $n = 3$: sphere ✓.

The field on the line through the centre along $\vec E_0$, from the surface outwards: $E_0(1 + (n-1)a^n/r^n)$. It starts at $nE_0$ and falls to $E_0$, faster in higher dimension — the induced term is a dipole, $\propto r^{-n}$.

The cylinder, in two lines. Outside: $\varphi = (A s + B/s)\cos\phi$ with $A = -E_0$ from infinity and $Aa + B/a = 0$, so $B = E_0a^2$. $E_s(a) = -\partial_s\varphi = E_0(1 + a^2/s^2)\cos\phi|_{s=a} = 2E_0\cos\phi$. P6 does it in full, and by images with two line charges.

The general-$n$ claim. $\partial_z$ commutes with $\nabla^2$ (Clairaut), so a $z$-derivative of a harmonic function is harmonic; $\partial_z(r^{2-n}) = (2-n)r^{1-n}\cdot z/r = (2-n)r^{1-n}\cos\theta$ for $n \ne 2$; for $n = 2$ the factor $2 - n$ vanishes and the same role is played by $\partial_x\log s = x/s^2 = s^{-1}\cos\phi$. At $r = a$, $\varphi = 0$; and $E_r = E_0(1 + (n-1)a^nr^{-n})\cos\theta$. Only the uniqueness theorem in $\mathbb R^n$ is cited for $n \ge 4$, via the maximum principle there.

Reading the figure. All three curves are $E_r$ on the ray $\theta = 0$ from the general-$n$ formula above. At the surface they sit at $2$, $3$ and $4$ times the applied field. The sphere's excess $2a^3/r^3$ and the cylinder's $a^2/r^2$ are equal at exactly $r = 2a$, both $25\,\%$: the cylinder starts lower but decays more slowly, and beyond $2a$ it is the stronger.

23 · multipoles

Multipoles, as a boundary-value tool

the exterior expansion

If all charge lies in $r < r_0$, then on $r > r_0$, with $\varphi\to0$, $$\varphi = \frac{1}{4\pi\varepsilon_0}\sum_\ell\frac{q_\ell}{r^{\ell+1}}P_\ell(\cos\theta),\quad q_\ell = \int\rho\,r'^\ell P_\ell(\cos\theta')\,\mathrm dV'$$ for axisymmetric $\rho$: unit 10's multipole moments, now as the $B_\ell$ of the expansion theorem.

proof, without the addition theorem

$\varphi$ is harmonic and axisymmetric on $r > r_0$; decay kills every $A_\ell$. On the axis $P_\ell(1) = 1$, so $4\pi\varepsilon_0\varphi(z\hat z) = \sum B_\ell z^{-\ell-1}$. But unit 10's expansion on the axis, where $\cos\gamma = \cos\theta'$ exactly, gives $\sum q_\ell z^{-\ell-1}$. A power series in $1/z$ has unique coefficients: $B_\ell = q_\ell$. $\square$

difference 3 · the dipole cannot be moved away

Shifting the origin by $\vec c$ changes the dipole to $\vec p - Q\vec c$. Gravity has $Q = M > 0$, so $\vec c = \vec p/M$ — the centre of mass — kills it: unit 10's vanishing $\ell = 1$ term. A neutral charge distribution has $Q = 0$, and its dipole is the same about every origin.

on the sphere

$\sigma = 3\varepsilon_0E_0\cos\theta = 3\varepsilon_0E_0P_1(\cos\theta)$, so $q_\ell = 3\varepsilon_0E_0a^{\ell+2}\cdot2\pi\int_{-1}^1P_1P_\ell\,\mathrm dx$, which is $0$ unless $\ell = 1$ (orthogonality, unit 10) and $4\pi\varepsilon_0E_0a^3$ when it is. The induced field is a pure dipole, exactly ✓.

The axis. A point $z\hat z$ on the axis and a source point at angle $\theta'$ subtend the angle $\gamma = \theta'$ at the origin, so unit 10's $\frac{1}{|\vec x-\vec y|} = \sum\frac{r'^\ell}{r^{\ell+1}}P_\ell(\cos\gamma)$ becomes $\sum\frac{r'^\ell}{z^{\ell+1}}P_\ell(\cos\theta')$ and integrates against $\rho$ to give $\sum q_\ell z^{-\ell-1}$, the series converging for $z > r_0$. Unit 10 needed the addition theorem to get $P_\ell(\cos\theta)$ off the axis; here the expansion theorem supplies the off-axis dependence for free, because a harmonic axisymmetric function is fixed by its axis values. That was the cited step in unit 10, now proved for axisymmetric bodies.

The shift. $\int\rho(\vec y)(\vec y - \vec c)\,\mathrm dV = \vec p - Q\vec c$. Linear algebra: for $Q \ne 0$ exactly one $\vec c$ kills it.

The moments of the sphere. For a surface density on $r' = a$, $q_\ell = \oint\sigma a^\ell P_\ell(\cos\theta')\,a^2\,\mathrm d\Omega'$. With $\mathrm d\Omega = 2\pi\sin\theta\,\mathrm d\theta$ and $x = \cos\theta$: $q_1 = 3\varepsilon_0E_0a^3\cdot2\pi\cdot\tfrac23 = 4\pi\varepsilon_0E_0a^3 = p$ ✓, using $\int_{-1}^1P_1^2 = \tfrac23$.

Why "boundary-value tool". Unit 10 used the expansion to compute a field from a known body. Here it is used the other way: outside any region, whatever is inside, the potential must have this form, and the problem is only to find the coefficients — from boundary data, from the axis, or from the moments.

section 4

Magnetostatics: the field with no sources

A field made by moving charge, with zero divergence everywhere, a curl fixed by the current, a potential that is a vector, and unit 04's vortex hiding in it.

23 · magnetostatics

The Lorentz force, currents, and unit 12's Lagrangian

Lorentz force · the definition of $\vec B$

A charge $q$ moving with velocity $\vec v$ feels $\vec F = q(\vec E + \vec v\times\vec B)$. The vector field $\vec B$ is defined by the velocity-dependent part; that the force has this form is measured.

current density

$\vec J$ (A m⁻²): charge crossing unit area per unit time, so the current through a surface is $I = \int_S\vec J\cdot\mathrm d\vec A$. Charge conservation plus unit 05's localisation lemma give the continuity equation $\partial_t\rho + \nabla\cdot\vec J = 0$. Magnetostatics is the case $\partial_t\rho = 0$, so $\nabla\cdot\vec J = 0$.

the magnetic force does no work

$\vec v\cdot(\vec v\times\vec B) = 0$, so $\frac{\mathrm d}{\mathrm dt}\big(\tfrac12m|\vec v|^2 + q\varphi\big) = 0$ in static fields. This is why unit 04 put magnetism "outside the frame": it is not a gradient, yet it cannot change the energy.

unit 12, paid

Unit 12 showed $L = \tfrac12m|\dot{\vec x}|^2 - q\varphi + q\dot{\vec x}\cdot\vec A$ gives $m\ddot{\vec x} = q(\vec E + \dot{\vec x}\times\vec B)$ if $\vec B = \nabla\times\vec A$ — and took the existence of $\vec A$ on trust. This section proves it: every static $\vec B$ is a curl, because it has no sources.

The continuity equation. For a fixed region $D$, the charge inside changes only by charge crossing $\partial D$: $\frac{\mathrm d}{\mathrm dt}\int_D\rho = -\oint_{\partial D}\vec J\cdot\mathrm d\vec A = -\int_D\nabla\cdot\vec J$. Differentiate under the integral (Leibniz, unit 18) and use unit 05's lemma — a continuous function whose integral over every ball vanishes is zero — to get $\partial_t\rho + \nabla\cdot\vec J = 0$ pointwise. Unit 22 does this for mass and momentum.

No work. $\frac{\mathrm d}{\mathrm dt}\tfrac12m|\vec v|^2 = \vec v\cdot m\dot{\vec v} = q\vec v\cdot\vec E + q\vec v\cdot(\vec v\times\vec B) = -q\vec v\cdot\nabla\varphi = -q\frac{\mathrm d}{\mathrm dt}\varphi(\vec x(t))$ for static $\varphi$. The triple product $\vec v\cdot(\vec v\times\vec B)$ vanishes because it is a determinant with two equal rows (unit 08).

Units. $[\vec B] = \mathrm{N}/(\mathrm C\cdot\mathrm{m\,s^{-1}}) = \mathrm{N\,A^{-1}m^{-1}} = \mathrm T$ (tesla). The Earth's field is about $5\times10^{-5}$ T; a hospital MRI magnet, $1.5$–$3$ T.

A warning about what is being assumed. The Lorentz force and Biot–Savart (next slide) are the magnetic analogues of Coulomb's law: empirical inputs. Everything else in this section is derived from them. In unit 28 both turn out to be consequences of Coulomb's law and special relativity — the magnetic force is what the electric force looks like from a moving frame — but that needs units 25–27.

23 · magnetostatics

Biot–Savart, and the two equations it hides

Biot–Savart law · measured

A steady current density $\vec J$ ($C^1$, compact support, $\nabla\cdot\vec J = 0$) produces $$\vec B(\vec x) = \frac{\mu_0}{4\pi}\int\frac{\vec J(\vec y)\times(\vec x-\vec y)}{\lVert\vec x-\vec y\rVert^3}\,\mathrm dV_y .$$

theorem · the field equations of magnetostatics

With $\vec A_0(\vec x) = \frac{\mu_0}{4\pi}\int\frac{\vec J(\vec y)}{\lVert\vec x-\vec y\rVert}\mathrm dV_y$: $$\vec B = \nabla\times\vec A_0,\qquad \nabla\cdot\vec B = 0,\qquad \nabla\times\vec B = \mu_0\vec J .$$

proof

1. $\nabla_x\times\big(\vec J(\vec y)/|\vec x-\vec y|\big) = \nabla_x\big(1/|\vec x-\vec y|\big)\times\vec J = \vec J\times\frac{\vec x-\vec y}{|\vec x-\vec y|^3}$; differentiate under the integral as unit 05 did for $\vec g$.
2. $\nabla\cdot(\nabla\times\vec A) = 0$ for $C^2$ fields, by Clairaut.
3. $\nabla\times\nabla\times\vec A_0 = \nabla(\nabla\cdot\vec A_0) - \nabla^2\vec A_0$. Poisson, componentwise (unit 05): $\nabla^2\vec A_0 = -\mu_0\vec J$. And $\nabla\cdot\vec A_0 = 0$ because $\nabla\cdot\vec J = 0$ (notes). $\square$

no electrostatic analogue

$\nabla\cdot\vec E = \rho/\varepsilon_0$ has a source; $\nabla\cdot\vec B = 0$ has none. No magnetic charge has ever been found. It is an empirical fact, not a theorem — Biot–Savart builds it in.

Dimensions: $[\mu_0J\cdot\mathrm L] = \mathrm{N\,A^{-2}}\cdot\mathrm{A\,m^{-2}}\cdot\mathrm m = \mathrm T$ ✓.

Full proof of the two identities and of $\nabla\cdot\vec A_0 = 0$

$\nabla\cdot(\nabla\times\vec A) = 0$: $\partial_x(\partial_yA_z - \partial_zA_y) + \partial_y(\partial_zA_x - \partial_xA_z) + \partial_z(\partial_xA_y - \partial_yA_x)$, and the six terms cancel in pairs by Clairaut (unit 03).

$\nabla\times(\nabla\times\vec A) = \nabla(\nabla\cdot\vec A) - \nabla^2\vec A$: the $x$-component is $\partial_y(\partial_xA_y - \partial_yA_x) - \partial_z(\partial_zA_x - \partial_xA_z) = \partial_x(\partial_yA_y + \partial_zA_z) - (\partial_y^2 + \partial_z^2)A_x$; add and subtract $\partial_x^2A_x$ to get $\partial_x(\nabla\cdot\vec A) - \nabla^2A_x$. The other components by cyclic permutation. Clairaut again.

$\nabla\cdot\vec A_0 = 0$. Since $\nabla_x\frac{1}{|\vec x-\vec y|} = -\nabla_y\frac{1}{|\vec x-\vec y|}$, $$\nabla\cdot\vec A_0 = -\frac{\mu_0}{4\pi}\int\vec J(\vec y)\cdot\nabla_y\frac{1}{|\vec x-\vec y|}\mathrm dV_y = \frac{\mu_0}{4\pi}\int\frac{\nabla\cdot\vec J(\vec y)}{|\vec x-\vec y|}\mathrm dV_y = 0,$$ integrating by parts (the divergence theorem on a large ball containing the support, with $\vec J = 0$ on its boundary; the singularity at $\vec y = \vec x$ is handled by cutting out a small ball, whose boundary term is $O(\varepsilon)$). This is the only place the steady-current hypothesis enters, and it is the thread unit 24 pulls: when $\nabla\cdot\vec J \ne 0$, Ampère's law in this form fails.

History. Biot and Savart measured the field of a long wire in 1820, weeks after Ørsted found that a current deflects a compass needle; the law in its differential form is Laplace's reworking of their result. Ampère's force law between current elements followed within months.

23 · magnetostatics

Ampère's law is Stokes' theorem applied to $\vec B$

Ampère's law

For any oriented surface $S$ with boundary loop $\partial S$: $$\oint_{\partial S}\vec B\cdot\mathrm d\vec r = \mu_0\int_S\vec J\cdot\mathrm d\vec A = \mu_0I_{\text{enc}} .$$

Proof. Stokes' theorem (unit 04, for general surfaces cited there with proof in unit 34) and $\nabla\times\vec B = \mu_0\vec J$. $\square$

the surface does not matter — because $\nabla\cdot\vec J = 0$

Two surfaces with the same boundary together bound a region $V$. The difference of their fluxes is $\oint_{\partial V}\vec J\cdot\mathrm d\vec A = \int_V\nabla\cdot\vec J = 0$. A charging capacitor breaks this — unit 24's displacement current is the repair.

One loop $\partial S$, two surfaces on it: the flat disc $S_1$ and the dome $S_2$. The wire pierces both, so both see current $I$. Together they enclose $V$; whatever current enters $V$ through one must leave through the other.

Reading the figure. Orient both surfaces by the same loop (the right-hand rule with the loop's arrow gives "upward" for both). Then $\partial V$ is $S_2$ with its orientation and $S_1$ with the opposite one, so $\oint_{\partial V}\vec J\cdot\mathrm d\vec A = \int_{S_2} - \int_{S_1}$, which the divergence theorem turns into $\int_V\nabla\cdot\vec J$. For steady currents that is zero and the two fluxes agree. A capacitor plate inside $V$ — current arriving through $S_1$ and stopping on the plate, none crossing $S_2$ — is exactly a place where $\nabla\cdot\vec J \ne 0$.

Unit 04's promise, kept. Unit 04 said Stokes' theorem on arbitrary surfaces would "start doing work, as Ampère's law". The work it does is the surface-independence above: the law is a statement about a loop, but its proof passes through a surface, and the answer cannot depend on which one — which forces a condition on the source.

23 · magnetostatics

A long straight wire

a long straight wire

Current $I$ along the $z$-axis; $s$ is the distance from the axis. In Biot–Savart each element contributes along $\hat z\times(\vec x - \vec y)$, which is azimuthal, so $\vec B = B(s)\hat\phi$. Ampère on a circle of radius $s$: $2\pi sB = \mu_0I$, so $$\vec B = \frac{\mu_0I}{2\pi s}\hat\phi = \frac{\mu_0I}{2\pi}\,\vec F_{\mathrm V} .$$

This is unit 04's vortex, scaled: curl-free off the wire, circulation $\mu_0I$ round it. The "hole" in unit 04's domain was the wire. $1$ A at $1$ cm: $2.000\times10^{-5}$ T.

Dimensions: $[\mu_0I/s] = \mathrm{N\,A^{-2}}\cdot\mathrm A/\mathrm m = \mathrm T$ ✓. Limit $s\to\infty$: $B\to0$ ✓.

Looking down the wire, current towards you. $\vec B$ circles anticlockwise with length $\propto 1/s$; along the dashed circle $B$ is constant and parallel to the path, so the circulation is just $2\pi s\,B$.

The symmetry argument, and a trap. The direction came from the integrand, not from unit 05's equivariance theorem, on purpose. $\vec B$ is a pseudovector: under a reflection it transforms with an extra sign (unit 04, "Curl is a vector only under rotations"). Arguing "$\vec B$ must be radial by reflection symmetry in a plane through the wire" gives the wrong answer; with the extra sign the same argument gives azimuthal. The safe route is to look at the integrand.

That $B$ depends only on $s$ follows from invariance of the wire under rotations about $z$ and translations along it, which are rotations and translations and therefore safe.

The number. $\mu_0/2\pi = 2.000\times10^{-7}\ \mathrm{N\,A^{-2}}$ to the precision that matters, so $1$ A at $0.01$ m gives $2.000\times10^{-5}$ T — about 40 % of the Earth's field, which is why Ørsted's compass moved. Until 2019 the ampere was defined through the force between two such wires, which made $\mu_0/2\pi$ exactly $2\times10^{-7}\ \mathrm{N\,A^{-2}}$.

23 · vector potential

The vector potential: Poincaré's lemma one degree up

Poincaré lemma for divergence-free fields

Let $U$ be open and star-shaped about $\vec 0$, and $\vec B \in C^1(U)$ with $\nabla\cdot\vec B = 0$. Then $\vec B$ has a vector potential on $U$, a field $\vec A$ with $\vec B = \nabla\times\vec A$: $$\vec A(\vec x) = \int_0^1t\,\vec B(t\vec x)\times\vec x\,\mathrm dt .$$

proof

By the product rule for curls (notes), with $\nabla\cdot\vec x = 3$, $(\vec B\cdot\nabla)\vec x = \vec B$ and $\nabla_x\cdot[\vec B(t\vec x)] = t(\nabla\cdot\vec B)(t\vec x) = 0$: $$\nabla\times\big(\vec B(t\vec x)\times\vec x\big) = 2\vec B(t\vec x) + t\,\tfrac{\mathrm d}{\mathrm dt}\vec B(t\vec x).$$ Times $t$, this is $\tfrac{\mathrm d}{\mathrm dt}\big(t^2\vec B(t\vec x)\big)$. Integrate over $[0,1]$, moving the curl inside (Leibniz, unit 18): $\nabla\times\vec A = \vec B(\vec x)$. $\square$

the ladder

closed	then locally	unit
$\nabla\times\vec F = \vec 0$	$\vec F = \nabla W$	04
$\nabla\cdot\vec B = 0$	$\vec B = \nabla\times\vec A$	23

Since $\nabla\cdot\vec B = 0$ on all of $\mathbb R^3$ — star-shaped — every static $\vec B$ has a global vector potential. The next slide shows the hypothesis is needed.

The product rule for curls

$\nabla\times(\vec F\times\vec G) = \vec F(\nabla\cdot\vec G) - \vec G(\nabla\cdot\vec F) + (\vec G\cdot\nabla)\vec F - (\vec F\cdot\nabla)\vec G$. Proof, $x$-component: $\partial_y(F_xG_y - F_yG_x) - \partial_z(F_zG_x - F_xG_z) = F_x(\partial_yG_y + \partial_zG_z) + (G_y\partial_y + G_z\partial_z)F_x - G_x(\partial_yF_y + \partial_zF_z) - (F_y\partial_y + F_z\partial_z)G_x$. Add and subtract $F_x\partial_xG_x$ and $G_x\partial_xF_x$: this is $F_x\nabla\cdot\vec G - G_x\nabla\cdot\vec F + (\vec G\cdot\nabla)F_x - (\vec F\cdot\nabla)G_x$. Other components cyclically. $\square$

The four terms in the lemma. $\vec F(\vec x) = \vec B(t\vec x)$, $\vec G = \vec x$. $\nabla\cdot\vec x = 3$. $\nabla\cdot\vec F = t(\nabla\cdot\vec B)(t\vec x) = 0$ by the chain rule. $(\vec x\cdot\nabla_x)\vec B(t\vec x) = t\,(\vec x\cdot\nabla\vec B)(t\vec x) = t\frac{\mathrm d}{\mathrm dt}\vec B(t\vec x)$. $(\vec F\cdot\nabla)\vec x = \vec F$. Total $3\vec B + t\dot{\vec B} - \vec B$. Then $t(2\vec B + t\dot{\vec B}) = \frac{\mathrm d}{\mathrm dt}(t^2\vec B)$, whose integral over $[0,1]$ is $\vec B(\vec x) - 0$. Differentiating under the integral needs the integrand and its $\vec x$-derivatives continuous on $[0,1]\times K$ for compact $K\subset U$: true for $\vec B \in C^1$, and star-shapedness keeps $t\vec x$ in $U$.

23 · vector potential

The monopole: a hole the lemma cannot cross

the hypothesis bites · the monopole

$\vec B = g\,\hat r/r^2$ on $\mathbb R^3\setminus\{0\}$ has $\nabla\cdot\vec B = 0$, but flux $4\pi g$ through the unit sphere. If $\vec B = \nabla\times\vec A$, Stokes on the two hemispheres, whose boundary circles run oppositely, would give flux $0$. So no $\vec A$ exists: one degree above the vortex, a sphere-shaped hole instead of a line.

Here $g$ is a constant, the monopole's strength. The domain $\mathbb R^3\setminus\{0\}$ is not star-shaped about any point: the segment from a point to its mirror image through the origin leaves it.

The unit sphere about the monopole, cut at the equator. Each hemisphere has the equator as its boundary, traversed in opposite senses (the two arrows). If $\vec B$ were a curl, Stokes would give each hemisphere's flux as that boundary circulation — two numbers summing to $0$, not $4\pi g$.

The monopole in detail. $\nabla\cdot(\hat r/r^2) = 0$ off the origin is unit 05's "why $1/r^2$". Flux through the unit sphere: $4\pi g$. If $\vec B = \nabla\times\vec A$ with $\vec A \in C^1$ on $\mathbb R^3\setminus\{0\}$, Stokes on the northern hemisphere gives $\oint_{\text{equator}}\vec A\cdot\mathrm d\vec r$ traversed one way, and on the southern the same integral traversed the other way. Sum: $0 \ne 4\pi g$.

The cohomological reading, for later: unit 04's vortex is a closed 1-form that is not exact on a domain with a loop that cannot be shrunk; the monopole is a closed 2-form that is not exact on a domain with a sphere that cannot be shrunk. Unit 34 names the groups these detect — $H^1$ and $H^2$ of de Rham cohomology — and turns both lemmas into one statement about $\mathrm d^2 = 0$. Dirac showed in 1931 that quantum mechanics permits monopoles if the charge is quantised; none has been seen.

23 · vector potential

Gauge freedom, and the Coulomb gauge

gauge transformation

$\vec A\mapsto\vec A + \nabla\chi$, for any $C^2$ function $\chi$. It leaves $\vec B = \nabla\times\vec A$ unchanged, since the curl of a gradient is zero (unit 04). A choice of $\vec A$ within its class is a gauge.

the Coulomb gauge is always available

Given any $\vec A$ with $f = \nabla\cdot\vec A$ of compact support and $C^1$, let $\chi$ solve $\nabla^2\chi = -f$ (unit 05's integral). Then $\nabla\cdot(\vec A + \nabla\chi) = 0$. In this gauge $$\nabla^2\vec A = -\mu_0\vec J,$$ three copies of Poisson's equation.

magnetostatics is electrostatics three times

In Coulomb gauge each Cartesian component $A_i$ obeys the equation $\varphi$ obeys, with $\mu_0J_i$ for $\rho/\varepsilon_0$. So $\vec A_0$ of the Biot–Savart slide is the Coulomb-gauge potential, and every tool of section 2 applies to it componentwise.

only in Cartesian components

$\nabla^2\vec A$ means $(\nabla^2A_x, \nabla^2A_y, \nabla^2A_z)$. In spherical or cylindrical components the vector Laplacian is not the scalar Laplacian of each component — the unit vectors vary. Use $\nabla(\nabla\cdot\vec A) - \nabla\times\nabla\times\vec A$ there.

The gauge function is itself unique only up to adding a harmonic function: the freedom never quite goes away.

Why the freedom exists. $\vec B$ determines $\vec A$ only up to a field with zero curl, and on a star-shaped domain such a field is a gradient (unit 04's Poincaré lemma). So on $\mathbb R^3$ the gauge transformations are exactly the ambiguity. On a domain with holes there is more ambiguity than gradients — The solenoid: closed but not exact, again.

Is the vector potential physical? Classically, only $\vec B$ enters the force, so $\vec A$ is bookkeeping, and gauge is a redundancy of description. Unit 12 already showed the matching fact in mechanics: a gauge change adds $q\,\mathrm d\chi/\mathrm dt$ to the Lagrangian and leaves the equations of motion alone. In quantum mechanics the circulation $\oint\vec A\cdot\mathrm d\vec r$ round a solenoid is observable even where $\vec B = 0$ — the Aharonov–Bohm effect (1959) — but no gauge-variant quantity ever is.

Unit 24 needs a different gauge for time-dependent problems (Lorenz), in which $\varphi$ and $\vec A$ obey wave equations with the same structure, and which becomes manifestly relativistic in unit 28.

23 · vector potential · example

Worked: a uniform field, in two gauges

worked example 4 · $\vec B = B\hat z$, uniform

Step 1 · the lemma's formula. $\vec B(t\vec x) = \vec B$ is constant, so $$\vec A_{\mathrm S} = \int_0^1t\,\mathrm dt\;\vec B\times\vec x = \tfrac12\vec B\times\vec x = \tfrac B2(-y,\ x,\ 0),$$ the symmetric gauge.

Step 2 · check. $\nabla\times\vec A_{\mathrm S} = \big(0,\ 0,\ \partial_x(\tfrac B2x) - \partial_y(-\tfrac B2y)\big) = (0, 0, B)$ ✓. And $\nabla\cdot\vec A_{\mathrm S} = 0$: already Coulomb gauge.

Step 3 · another. $\vec A_{\mathrm L} = (0,\ Bx,\ 0)$, the Landau gauge: curl $= (0,0,\partial_x(Bx)) = (0,0,B)$ ✓, divergence $0$ ✓.

Step 4 · the gauge function. $\vec A_{\mathrm L} - \vec A_{\mathrm S} = \tfrac B2(y,\ x,\ 0) = \nabla\chi$ with $$\chi = \tfrac B2\,xy,$$ harmonic ($\nabla^2(xy) = 0$) — as it must be, since both potentials are in Coulomb gauge.

checks

Dimensions: $[A] = \mathrm{T\,m}$, and $[q\vec A] = \mathrm{C\,T\,m} = \mathrm{kg\,m\,s^{-1}}$, a momentum, as unit 12's $\vec p = m\dot{\vec x} + q\vec A$ requires ✓.

Circulation of $\vec A_{\mathrm S}$ round the circle of radius $s_0$ in the plane $z = 0$: $\tfrac B2s_0\cdot2\pi s_0 = B\pi s_0^2$, the enclosed flux ✓ (Stokes).

23 · vector potential · example

Two gauges, one field, drawn

$\vec A_{\mathrm S} = \tfrac B2(-y, x, 0)$: a swirl about the origin, growing with distance from it — unchanged by rotations about $z$.

$\vec A_{\mathrm L} = (0, Bx, 0)$: a shear, the same on every horizontal line — unchanged by moving along $y$.

why two

$\vec A_{\mathrm S}$ is symmetric under rotations about $z$: the canonical angular momentum about $z$ is conserved (notes). $\vec A_{\mathrm L}$ is independent of $y$: $p_y$ is cyclic (unit 12). Same physics; a gauge makes different conservation laws visible.

Reading the two pictures. They look nothing alike, and they have the same curl: round the dashed circle of radius $2$ both circulations equal $B\pi\cdot2^2$, the flux through it (Stokes, worked example 4's check). The difference between them is the gradient of $\chi = \tfrac B2xy$, whose field $\tfrac B2(y, x)$ is a saddle-shaped flow — curl-free, so invisible to $\vec B$. Both are drawn at the same scale.

The mechanics. Unit 12's P7 did this field in the symmetric gauge and in the other Landau gauge, $(-By, 0, 0)$, and watched the cyclic coordinate move. In the Landau gauge of this slide $L$ does not depend on $y$, so $p_y = m\dot y + qBx$ is conserved; in the symmetric gauge the conserved quantity is the canonical angular momentum $m(x\dot y - y\dot x) + \tfrac{qB}{2}(x^2 + y^2)$. Proof: in cylindrical coordinates $q\dot{\vec x}\cdot\vec A_{\mathrm S} = \tfrac{qB}2(x\dot y - y\dot x) = \tfrac{qB}2s^2\dot\phi$, so $L = \tfrac12m(\dot s^2 + s^2\dot\phi^2 + \dot z^2) + \tfrac{qB}2s^2\dot\phi$ has $\phi$ cyclic (unit 12), and $p_\phi = ms^2\dot\phi + \tfrac{qB}2s^2$ is that quantity. The motion is the same circle either way — problem P11 finds its frequency.

23 · vector potential

The solenoid: closed but not exact, again

A long solenoid: radius $R$, $n$ turns per metre, current $I$. Ampère on rectangles gives (P9) $$\vec B = \mu_0nI\,\hat z\ \ (s < R),\qquad \vec B = \vec 0\ \ (s > R).$$ Flux $\Phi = \mu_0nI\pi R^2$. In the symmetric gauge, $\vec A = A(s)\hat\phi$ and Stokes on a disc of radius $s$ gives $2\pi sA(s) = $ enclosed flux:

$$\vec A = \frac{\mu_0nI\,s}{2}\hat\phi\ \ (s<R),$$ $$\vec A = \frac{\Phi}{2\pi s}\hat\phi = \frac{\Phi}{2\pi}\vec F_{\mathrm V}\ \ (s > R).$$

The running numbers ($n = 2000\ \mathrm{m^{-1}}$, $I = 1.5$ A, $R = 2$ cm), computed. $\vec B$ stops dead at the wall; $\vec A$ does not — it peaks there and decays as $\Phi/2\pi s$, still $15.08\ \mu\mathrm{T\,m}$ at $5$ cm, where $\vec B = \vec 0$.

Reading the plot. The two panels share the horizontal axis. Inside, $A = Bs/2$ grows linearly, the symmetric gauge of worked example 4; at the wall both formulas give $37.70\ \mu\mathrm{T\,m}$, so $A$ is continuous while its slope jumps — the jump in slope is the jump in $B$, since for $\vec A = A(s)\hat\phi$ the curl is $\frac1s(sA)'\hat z$.

The field, in outline (P9 has the details). Each loop's Biot–Savart contribution is symmetric enough that $\vec B = B(s)\hat z$ for an infinitely long solenoid. Ampère on a rectangle with both long sides outside, at radii $s_1, s_2$, encloses no current: $B(s_1) = B(s_2)$, so $B$ is constant outside; it tends to $0$ far away (the solenoid is a finite-length idealisation whose far field decays), so it is $0$ outside. A rectangle with one side inside encloses $nI\ell$: $B_{\text{in}}\ell = \mu_0nI\ell$.

The potential. By symmetry take $\vec A = A(s)\hat\phi$ (this is a gauge choice, and it is Coulomb: $\nabla\cdot(A(s)\hat\phi) = 0$). Stokes on a disc of radius $s$ perpendicular to the axis: $\oint\vec A\cdot\mathrm d\vec r = 2\pi sA(s) = \int\vec B\cdot\mathrm d\vec A$, which is $\mu_0nI\pi s^2$ inside and $\Phi$ outside. Continuous at $s = R$ ✓.

The numbers. $B = 4\pi\times10^{-7}\times2000\times1.5 = 3.770\times10^{-3}$ T; $\Phi = B\pi(0.02)^2 = 4.737\times10^{-6}$ Wb; at $s = 0.05$ m, $A = \Phi/(2\pi\times0.05) = 1.508\times10^{-5}$ T m. At the wall both formulas give $3.770\times10^{-5}$ T m.

23 · vector potential

Outside the solenoid: closed, not a gradient

the vortex, one rung up the ladder

Outside, $\nabla\times\vec A = \vec B = \vec 0$: $\vec A$ is closed. Yet $\oint\vec A\cdot\mathrm d\vec r = \Phi \ne 0$ round the solenoid, so $\vec A$ is not a gradient on the outside region. It is unit 04's vortex times $\Phi/2\pi$: locally $\nabla(\Phi\phi/2\pi)$, globally not.

what this means

In unit 04 the closed-not-exact object was a force, whose potential was a scalar. Here it is a potential, whose own potential would be a gauge function. On the outside region $\vec A$ is "pure gauge" near every point and not globally: no gauge transformation there removes it, because its circulation is the flux of a field it cannot see.

$n = 2000\ \mathrm{m^{-1}}$, $I = 1.5$ A, $R = 2$ cm: $B = 3.770$ mT, $\Phi = 4.737\ \mu\mathrm{Wb}$, and $|\vec A| = 15.08\ \mu\mathrm{T\,m}$ at $s = 5$ cm, where $\vec B = \vec 0$.

Looking down the solenoid's axis: $\vec B$ out of the page inside the circle, zero outside. $\vec A$ circles the axis everywhere — the same picture as the wire's $\vec B$ — and its circulation round any loop enclosing the solenoid is the whole flux $\Phi$.

Reading the figure. Outside the circle the arrows are $\frac{\Phi}{2\pi}\vec F_{\mathrm V}$, unit 04's vortex: zero curl at every point, yet circulation $\Phi$ round the dashed loop. Inside, where $\vec B \ne \vec 0$, they grow linearly with $s$. Compare the long straight wire, where the same arrangement of arrows was $\vec B$ itself and the "hole" was the wire: here it is $\vec A$, and the hole is the solenoid's interior, the only place the flux lives.

Why "one rung up". The ladder scalar $\to$ vector potential $\to$ field has two steps: $\vec F = \nabla W$ and $\vec B = \nabla\times\vec A$. Unit 04's vortex was a failure at the first step for a force. The solenoid shows the same failure at the first step for the potential of the second: $\vec A$ is determined by $\vec B$ only up to a closed field, and on a domain with a hole "closed" is bigger than "gradient". The monopole (The monopole: a hole the lemma cannot cross) is a failure at the second step itself. Unit 34 says both at once.

23 · magnetostatics · the sphere

The sphere in a magnetic field

Where $\vec J = \vec 0$, $\nabla\times\vec B = \vec 0$, so on a simply connected current-free region $\vec B = -\nabla\psi$ (unit 04), and $\nabla\cdot\vec B = 0$ makes the magnetic scalar potential $\psi$ harmonic. Section 3's machinery applies unchanged; only the boundary condition differs.

the same ball, now a perfect diamagnet

An idealised superconductor excludes $\vec B$, so $\vec B\cdot\hat n = 0$ on its surface: $\partial_r\psi = 0$ at $r = a$ — Neumann data, not Dirichlet. The ball sits in a uniform applied field $\vec B_0 = B_0\hat z$, whose scalar potential is $-B_0r\cos\theta$; so $\psi\to-B_0r\cos\theta$ far away, and $\psi = (-B_0r + C/r^2)\cos\theta$, and $\partial_r\psi(a) = -B_0 - 2C/a^3 = 0$, so $$\psi = -B_0\Big(r + \frac{a^3}{2r^2}\Big)\cos\theta .$$

	grounded conductor in $\vec E_0$	diamagnet in $\vec B_0$
condition	$\varphi = 0$	$\partial_r\psi = 0$
$B_1$	$+E_0a^3$	$-\tfrac12B_0a^3$
field at poles	$3E_0$	$0$
field at equator	$0$	$\tfrac32B_0$

checks

$B_\theta(a) = -\frac1a\partial_\theta\psi = -\tfrac32B_0\sin\theta$ — tangent, as required. Limit $a\to0$: uniform ✓. The induced dipole is opposed to $\vec B_0$: diamagnetic ✓.

The region. The exterior of a ball is simply connected (every loop in it can be shrunk — around the ball if necessary), so unit 04's cited general Poincaré lemma applies. Compare the solenoid, whose exterior is not simply connected, and where $\vec B = \vec 0$ outside but $\vec A$ is not a gradient: the difference between the two examples is entirely the topology of the current-free region.

The expansion. $\psi$ is harmonic and axisymmetric outside the ball, so by the expansion theorem it is $\sum(A_\ell r^\ell + B_\ell r^{-\ell-1})P_\ell$; the condition at infinity leaves $A_1 = -B_0$ only; Neumann at $r = a$ says $\ell A_\ell a^{\ell-1} - (\ell+1)B_\ell a^{-\ell-2} = 0$, so $B_1 = A_1a^3/2 = -B_0a^3/2$ and all other $B_\ell$ vanish. Uniqueness for Neumann data holds up to a constant, which does not affect $\vec B$ (identity (I), as in proof 2 of the uniqueness slide).

The physics, labelled. That a superconductor expels magnetic flux is the Meissner effect (1933), cited as an empirical fact; its explanation is quantum mechanical. The idealisation ignores the thin surface layer in which the field actually decays. The induced dipole opposes the applied field, so the ball is pushed towards weaker field — which is how a magnet can float above a superconductor, the levitation Earnshaw does not forbid because the moment is induced, not fixed.

23 · magnetostatics · the sphere

Conductor and diamagnet, side by side

Both computed: the conductor from worked example 2's $\varphi$, the diamagnet from the $\psi$ of the slide before. Left: $\varphi = 0$ on the surface, so the surface is a level set and field lines must meet it at right angles — they crowd onto the poles ($3E_0$) and leave the equator bare. Right: $\partial_r\psi = 0$, so no line may cross the surface; they part round it, leaving the poles empty and crowding at the equator ($\tfrac32B_0$). Same equation, same expansion, opposite boundary condition.

Reading the pair. Far from the ball the two pictures are the same uniform field. The difference is the sign and size of the induced dipole: $+E_0a^3$ for the conductor, $-\tfrac12B_0a^3$ for the diamagnet. A positive dipole adds to the applied field on the axis and subtracts at the equator; a negative one does the opposite, which is why the crowding moves from the poles to the equator. The factor $\tfrac12$ is what makes the normal component vanish exactly: on the surface the dipole's radial field $2\cdot(-\tfrac12)B_0\cos\theta$ cancels the applied field's $B_0\cos\theta$.

Where the field vanishes. The conductor's surface field is $3E_0\cos\theta$, zero on the equator; the diamagnet's is $\tfrac32B_0\sin\theta$, zero at the poles. Each picture has two places where the field lines split, and they are at right angles to each other.

section 5

Helmholtz: why four equations

A field that decays is determined by its divergence and its curl. So a theory of two vector fields needs four equations, no more and no fewer, and its sources must satisfy a condition.

23 · Helmholtz

The Helmholtz theorem

Helmholtz theorem

Let $\vec F \in C^2(\mathbb R^3)$ with $\vec F\to\vec 0$ at infinity, and suppose $D = \nabla\cdot\vec F$ and $\vec C = \nabla\times\vec F$ are $C^1$ with compact support. Then $$\vec F = -\nabla\Phi + \nabla\times\vec A,$$ $$\Phi = \frac1{4\pi}\!\int\!\frac{D(\vec y)\,\mathrm dV}{\lVert\vec x-\vec y\rVert},\qquad \vec A = \frac1{4\pi}\!\int\!\frac{\vec C(\vec y)\,\mathrm dV}{\lVert\vec x-\vec y\rVert},$$ and $\vec F$ is the only vector field tending to $\vec 0$ with divergence $D$ and curl $\vec C$.

proof of uniqueness

Let $\vec W$ be the difference of two such fields: $\nabla\cdot\vec W = 0$, $\nabla\times\vec W = \vec 0$. Then $\nabla^2\vec W = \nabla(\nabla\cdot\vec W) - \nabla\times\nabla\times\vec W = \vec 0$: each component is harmonic and $\to0$. By the maximum principle on $B_R$, $|W_i| \le \max_{|\vec x| = R}|W_i| \to 0$. So $\vec W = \vec 0$. $\square$

existence: full proof in the notes (N)

the hypotheses are not decoration

Decay. $\vec F = E_0\hat z$ has zero divergence and zero curl and is not zero: without decay, any $\nabla h$ with $h$ harmonic can be added. The running example's applied field is exactly such an $\nabla h$, and Helmholtz applies only to the induced field.

A constraint on the data. $\vec C$ is a curl, so $\nabla\cdot\vec C = 0$ necessarily. Divergence and curl are not independent data: one scalar function, and one vector field that is itself divergence-free.

Here $D$ is a function and $\Phi$ its potential — not the region $D$ of section 2 or the solenoid's flux $\Phi$.

Dimensions: $[D] = [F]/\mathrm L$, so $[\Phi] = [F]\mathrm L^{-1}\cdot\mathrm L^2 = [F]\,\mathrm L$ and $[\nabla\Phi] = [F]$ ✓.

Full proof of existence

Define $\Phi$ and $\vec A$ as on the slide and put $\vec F' = -\nabla\Phi + \nabla\times\vec A$.

Divergence. $\nabla\cdot\vec F' = -\nabla^2\Phi + 0 = D$. The first step uses $\nabla\cdot(\nabla\times\vec A) = 0$; the second is Poisson's equation for the $C^1$, compactly supported density $D$. Unit 05 proved $\nabla^2\big(-G\int\rho/|\vec x-\vec y|\big) = 4\pi G\rho$; dividing by $-4\pi G$ gives $\nabla^2\big(\frac{1}{4\pi}\int\rho/|\vec x-\vec y|\big) = -\rho$, which with $\rho = D$ is $-\nabla^2\Phi = D$.

Curl. $\nabla\times\vec F' = \nabla\times\nabla\times\vec A = \nabla(\nabla\cdot\vec A) - \nabla^2\vec A$. Componentwise Poisson: $-\nabla^2\vec A = \vec C$. And $\nabla\cdot\vec A = 0$ by the integration by parts of the Biot–Savart slide's notes, which used only $\nabla\cdot\vec J = 0$ and compact support — here $\nabla\cdot\vec C = \nabla\cdot(\nabla\times\vec F) = 0$. So $\nabla\times\vec F' = \vec C$.

Decay. $\Phi$ and $\vec A$ are potentials of compactly supported densities, so their gradients are $O(r^{-2})$ (the field of a bounded source) and $\vec F'\to\vec 0$.

Conclusion. $\vec F'$ and $\vec F$ both tend to zero and have the same divergence and curl, so they are equal by uniqueness. $\square$

How much the hypotheses can be weakened. Compact support of $D$ and $\vec C$ can be replaced by decay like $r^{-3-\epsilon}$, and the decay of $\vec F$ is then automatic for the constructed field. Cited: Blumenthal, Math. Ann. 61 (1905) 235, who gave the first careful statement; Helmholtz's own 1858 paper was on vortex motion in fluids and assumed what it needed.

The names. $-\nabla\Phi$ is the longitudinal or curl-free part, and $\nabla\times\vec A$ the transverse or divergence-free part. In unit 20's Fourier language they are the components of $\hat{\vec F}(\vec k)$ parallel and perpendicular to $\vec k$, and the decomposition is orthogonal projection.

23 · Helmholtz

Counting Maxwell's equations before seeing them

the count

Electromagnetism has two vector fields, $\vec E$ and $\vec B$, decaying away from their sources. By Helmholtz each is fixed by its divergence and its curl. So the field equations must specify exactly four things: $$\nabla\cdot\vec E,\quad \nabla\times\vec E,\quad \nabla\cdot\vec B,\quad \nabla\times\vec B .$$ Fewer leaves a field undetermined; a fifth independent equation over-determines it.

statics	source
$\nabla\cdot\vec E = \rho/\varepsilon_0$	Coulomb, via Gauss
$\nabla\times\vec E = \vec 0$	$\vec E$ is a gradient
$\nabla\cdot\vec B = 0$	no magnetic charge
$\nabla\times\vec B = \mu_0\vec J$	Biot–Savart, via Ampère

and why 8 equations for 6 unknowns is not too many

A curl is not three free functions: it is divergence-free, so it carries two functions' worth of data. Divergence (1) + curl (2) = 3 per field — the three components. The count is exact.

the prediction

The curl data must be divergence-free: $\nabla\cdot(\mu_0\vec J) = \nabla\cdot(\nabla\times\vec B) = 0$. So magnetostatics requires $\nabla\cdot\vec J = 0$ — steady currents. When charge piles up somewhere, $\partial_t\rho \ne 0$, $\nabla\cdot\vec J \ne 0$, and the curl equation for $\vec B$ must acquire a term whose divergence cancels $\mu_0\nabla\cdot\vec J = -\mu_0\partial_t\rho$. That term is Maxwell's displacement current, unit 24.

"Fewer leaves a field undetermined", concretely. Drop $\nabla\times\vec E = \vec 0$ and keep $\nabla\cdot\vec E = \rho/\varepsilon_0$: then $\vec E + \nabla\times\vec G$ satisfies the remaining equation for every decaying $C^2$ field $\vec G$ — infinitely many answers. Drop $\nabla\cdot\vec B = 0$: $\vec B + \nabla h$ for any $h$ with $\nabla h$ decaying, e.g. $h$ the potential of any compact density.

"A fifth over-determines it." An extra equation independent of these four either follows from them — and adds nothing — or constrains $\vec E$ or $\vec B$ beyond what their divergence and curl already fixed, in which case, for some choice of sources, it contradicts them. This is a heuristic statement about linear first-order field equations, made precise by Helmholtz for statics; the time-dependent version, where the divergence equations become constraints preserved by the evolution, is unit 24's.

The functional count. "A divergence-free field carries two functions' worth of data" can be made exact: by Helmholtz applied to $\vec C$ itself, a decaying divergence-free field is $\nabla\times\vec A$ with $\nabla\cdot\vec A = 0$, and in unit 20's Fourier variables that is a vector perpendicular to $\vec k$ at each $\vec k$ — a two-dimensional space. Light will have exactly two polarisations for exactly this reason.

Why this point is almost never made. Textbooks present the four equations as found, one experiment each, which is historically true. The logical fact — that four is forced once one knows there are two vector fields that decay — is visible only after Helmholtz, and it is what lets unit 24 derive the displacement current from consistency instead of from an experiment.

23 · the sphere · closed

The sphere, closed: a third derivation

worked example 5 · the sphere from its sources

Step 1 · what Helmholtz sees. The induced field $\vec E_{\mathrm{ind}} = \vec E - E_0\hat z$ decays, is curl-free, and its divergence is the surface layer $\sigma = \sigma_0\cos\theta$. So it is the field of that layer: $\varphi_{\mathrm{ind}}(\vec x) = \frac{1}{4\pi\varepsilon_0}\oint\frac{\sigma(\vec y)}{|\vec x-\vec y|}\mathrm dA_y .$

Step 2 · on the axis. Expand $1/|\vec x-\vec y|$ by unit 10's series, with $r_<, r_>$ the smaller and larger of $z$ and $a$; the $\theta'$-integral against $\cos\theta' = P_1$ keeps only $\ell = 1$, with $\int_{-1}^1P_1^2 = \tfrac23$: $$\varphi_{\mathrm{ind}}(z\hat z) = \frac{\sigma_0a^2}{2\varepsilon_0}\cdot\frac23\cdot\frac{r_<}{r_>^2} .$$

Step 3 · off the axis, by the multipole slide's argument (harmonic, axisymmetric, fixed by the axis): $\varphi_{\mathrm{ind}} = \frac{\sigma_0}{3\varepsilon_0}r\cos\theta$ inside, $\frac{\sigma_0a^3}{3\varepsilon_0}\frac{\cos\theta}{r^2}$ outside.

Step 4 · put in $\sigma_0 = 3\varepsilon_0E_0$. Inside: $\varphi_{\mathrm{ind}} = E_0z$, cancelling the applied $-E_0z$ — total $0$, a conductor at potential $0$ ✓. Outside: $E_0a^3\cos\theta/r^2$ — the dipole of both earlier derivations ✓.

Step 4 on the axis, computed: the layer's own potential (induced) is $E_0z$ inside and a dipole tail outside; added to the applied $-E_0z$ it gives exactly $0$ in the ball.

Dimensions: $[\sigma_0a/\varepsilon_0] = \mathrm V$ ✓. Limit $z\to\infty$: $\varphi_{\mathrm{ind}}\sim p/4\pi\varepsilon_0z^2$ with $p = \tfrac{4\pi}3a^3\sigma_0$ ✓.

Step 1, honestly. Helmholtz as stated wants $C^1$ divergence of compact support, and a surface layer is not a function. The step actually used is the representation formula applied separately inside and outside the sphere, with the jump conditions on $r = a$: the boundary terms from the two sides combine into $\oint\Gamma\,\sigma/\varepsilon_0$ because $\varphi$ is continuous and $\partial_n\varphi$ jumps by $-\sigma/\varepsilon_0$. So the statement "a decaying curl-free field is the field of its divergence" holds for surface divergences too, with the same proof in two pieces.

Step 2 in full. For $\vec x = z\hat z$ and $\vec y$ on the sphere at polar angle $\theta'$, the angle between them is $\theta'$, so $\frac1{|\vec x-\vec y|} = \sum_\ell\frac{r_<^\ell}{r_>^{\ell+1}}P_\ell(\cos\theta')$. With $\mathrm dA = 2\pi a^2\,\mathrm dx$, $x = \cos\theta'$: $\varphi_{\mathrm{ind}} = \frac{\sigma_0\cdot2\pi a^2}{4\pi\varepsilon_0}\sum_\ell\frac{r_<^\ell}{r_>^{\ell+1}}\int_{-1}^1xP_\ell(x)\,\mathrm dx = \frac{\sigma_0a^2}{2\varepsilon_0}\cdot\frac23\cdot\frac{r_<}{r_>^2}$. For $z < a$: $\frac{\sigma_0a^2}{3\varepsilon_0}\frac{z}{a^2} = \frac{\sigma_0z}{3\varepsilon_0}$; for $z > a$: $\frac{\sigma_0a^3}{3\varepsilon_0z^2}$.

Step 3. Inside, $\varphi_{\mathrm{ind}}$ is harmonic, axisymmetric and bounded at the centre, so $\sum A_\ell r^\ell P_\ell$, and on the axis $\sum A_\ell z^\ell = \frac{\sigma_0}{3\varepsilon_0}z$ forces $A_1 = \sigma_0/3\varepsilon_0$ and nothing else. Outside, the same with $B_\ell$.

The uniform interior field of a $\cos\theta$ layer — $-\sigma_0/3\varepsilon_0$ along $\hat z$ — is worth remembering: it is why a uniformly polarised ball has a uniform field inside, the basis of every textbook treatment of dielectric spheres.

23 · the sphere · closed

Three routes, one answer

The running example three ways. Legendre reads the answer off the boundary value; images replace the metal by charges outside the region; Helmholtz integrates the charge the metal actually carries. All three give $\varphi = -E_0(r - a^3/r^2)\cos\theta$.

the division of labour

The third needs $\sigma$ in advance, so it verifies rather than finds. That is the division of labour: boundary-value methods find the charge; the source integral confirms it makes the field.

23 · what next

What this unit hands on

20 distributions	$-\nabla^2\Gamma = \delta$ becomes a theorem; the representation formula is its unpacking.
24 Maxwell	Helmholtz's constraint on the curl data forces the displacement current; the retarded Green's function is $\Gamma$ with a delay; the jump conditions acquire time dependence.
25 the crisis	the Lorentz force's $\vec v\times\vec B$ depends on the frame, which Galilean relativity cannot accommodate.
28 the field tensor	$\vec E$ and $\vec B$ become one antisymmetric tensor field, and $(\varphi, \vec A)$ one covector field; gauge freedom is adding a gradient in four dimensions.
34 forms	the vortex and the monopole are $H^1$ and $H^2$; both Poincaré lemmas are $\mathrm d^2 = 0$ and its converse on a star-shaped set.
35 field equations	must reduce to Poisson's equation — this unit's and unit 05's — in the weak-field limit.

the three sentences to keep

A harmonic function is fixed by its boundary values, and every method of solution is a way of producing the one answer. A field with no sources is a curl, globally on $\mathbb R^3$ and not on a domain with holes. A decaying field is fixed by its divergence and its curl, so field theories count their equations.

What was deliberately left out. Dielectrics and magnetic materials ($\vec D$, $\vec H$, polarisation and magnetisation as bound sources): the $\cos\theta$ layer of the last worked example is the whole of the dielectric sphere, but the constitutive theory is a separate subject and nothing later needs it. Capacitance coefficients and the energy of a system of conductors, beyond the uniqueness proof. Conformal mapping for two-dimensional electrostatics, which needs unit 20's complex analysis — it would solve P4's wedge exactly. The existence theory for the Dirichlet problem on general regions (Perron's method), cited where used.

Reading. Jackson, Classical Electrodynamics (3rd ed.), ch. 1 (§1.1–1.10: Green's theorem, uniqueness, Green's function), ch. 2 §2.1–2.9 (images, and the sphere in a uniform field is §2.5), ch. 3 §3.1–3.3 (Legendre), and ch. 5 §5.1–5.4 (Biot–Savart, Ampère, the vector potential). Skip his §1.11–1.13 (energy, a variational principle, a numerical method) — the energy is unit 24's — and anything in ch. 3 that uses complex contour integrals until unit 20. Jackson writes $\nabla^2(1/r) = -4\pi\delta(\vec x)$ from §1.2 on; read every $\delta$ as the shorthand the worked example on the fundamental solution explains, which the representation formula makes exact and unit 20 makes literal. Griffiths, Introduction to Electrodynamics, ch. 2–3 and 5, is gentler and uses the same shorthand (his §1.5); its Appendix B proves Helmholtz with it, where this deck's notes use Poisson's equation from unit 05 instead.

23 · checkpoint

Done when

checkpoint

On paper and without notes, you can:

solve a spherical boundary-value problem by Legendre expansion and independently by images, and show the two agree coefficient by coefficient;
produce a vector potential for a given divergence-free $\vec B$ in a stated gauge, and verify both its curl and the gauge condition;
state the Helmholtz theorem with its hypotheses, prove its uniqueness half, and use it to say how many field equations electromagnetism must have, and what condition its sources must satisfy, before seeing them.

Failing it is unambiguous

An image placed inside the region; two derivations whose $\ell = 1$ coefficients differ; a claimed $\vec A$ whose curl is not $\vec B$, or no gauge named; or Helmholtz stated without decay — refuted by a uniform field.

24 Maxwell, which needs 20 as well · 22 continuum mechanics, a sibling in this part · 25 after 24, where the Lorentz force meets a moving observer.

Try the checkpoint on a problem that is in neither the deck nor the sheet: the running example's sphere held at potential $V_0$ instead of grounded. By Legendre: the condition at infinity is unchanged, and at $r = a$ the $\ell = 0$ coefficient must now equal $V_0$, so $B_0 = V_0a$ and $\varphi = V_0a/r - E_0(r - a^3/r^2)\cos\theta$. By images: the same two image charges as before, plus a charge $4\pi\varepsilon_0aV_0$ at the centre, which raises the sphere to $V_0$ without changing its shape as an equipotential. The two agree term by term: only $\ell = 0$ is new. With $V_0 = 100$ V and the running example's numbers the central charge is $0.556$ nC, the surface charge is $\sigma = \varepsilon_0V_0/a + 3\varepsilon_0E_0\cos\theta$, and it vanishes not on the equator but at $\cos\theta = -V_0/3E_0a$, i.e. $\theta = 90.38°$.

Then a vector potential: for $\vec B = B\hat z$ in the Landau gauge $\vec A = (-By, 0, 0)$, check the curl is $(0, 0, B)$ and the divergence $0$. Then Helmholtz: write it down, and check that you remembered "tends to zero at infinity" by testing it on $\vec F = E_0\hat z$.

Static fields: electrostatics and magnetostatics, the same vector calculus twice

One sphere, two derivations, five sections

Electrostatics is gravity with two signs

Coulomb's law, and everything unit 05 already proved

Two signs: gravity accumulates, electricity screens

Conductors

Across a charged surface

Laplace and Poisson: uniqueness, and the Green's function

Green's identities

Uniqueness, three ways

Radial solutions, and where the dimension enters

Worked: the constant, in one, two and three dimensions

Why three is special

The representation formula, and the Green's function of a region

Earnshaw, transferred

Earnshaw on the sphere: a saddle, never a well

The grounded sphere, twice

Separation in spherical coordinates

Worked: the sphere by Legendre expansion

Reading the answer: charge, dipole, and a factor of three

The sphere, in numbers

The method of images, and why it is allowed

The Kelvin image in a sphere

Worked: the sphere by images

Two derivations, term by term

Cylindrical coordinates: the same method, other special functions

The cylinder beside the sphere

Multipoles, as a boundary-value tool

Magnetostatics: the field with no sources

The Lorentz force, currents, and unit 12's Lagrangian

Biot–Savart, and the two equations it hides

Ampère's law is Stokes' theorem applied to $\vec B$

A long straight wire

The vector potential: Poincaré's lemma one degree up

The monopole: a hole the lemma cannot cross

Gauge freedom, and the Coulomb gauge

Worked: a uniform field, in two gauges

Two gauges, one field, drawn

The solenoid: closed but not exact, again

Outside the solenoid: closed, not a gradient

The sphere in a magnetic field

Conductor and diamagnet, side by side

Helmholtz: why four equations

The Helmholtz theorem

Counting Maxwell's equations before seeing them

The sphere, closed: a third derivation

Three routes, one answer

What this unit hands on

Done when

Failing it is unambiguous

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