Part 0 · The Toolkit — Chapter 0.7

Fields, Flux, and the Big Theorems

Green, Stokes, Gauss. Three names, three theorems, and one idea, which you already met in Chapter 0.2.

Tools you'll need  — Chapter 0.2: the integral as a limit of tagged sums, both halves of the Fundamental Theorem, and above all the telescoping argument that produced it. Chapter 0.4: the determinant as the factor by which a linear map scales volume, the trace, and the identity $\det(I+\epsilon A)=1+\epsilon\,\mathrm{tr}A+O(\epsilon^{2})$. Chapter 0.5: the inner product, and the spectral theorem for symmetric matrices (used once, in §4). Chapter 0.6: the gradient, the Jacobian, the chain rule along a curve, Clairaut's theorem on mixed partials, and the change-of-variables formula. §3 and §4 each cash a specific theorem from Chapter 0.4. §7 is Clairaut twice.

1 · Scalar and vector fields

This first section is short, and its job is to fix vocabulary. We name the two kinds of object the rest of the chapter operates on, agree on how to draw them, and then flag one thing the notation hides.

A scalar field on a region $U\subseteq\R^{n}$ is a function $\phi:U\to\R$: one number per point. Temperature through a room, pressure through the atmosphere, electrostatic potential, the density of a fluid, the concentration of a drug in tissue. A vector field is a function $\vv F:U\to\R^{n}$: one vector per point. Fluid velocity, the electric field, the gravitational field, the current density of a conserved substance.

We write the components of a vector field as functions of position,

with the index upstairs, following the convention of Chapter 0.6 §4: vector components carry upper indices, derivatives produce lower ones. In three dimensions we will freely write $\vv F = (F^{x},F^{y},F^{z})$ when that is clearer.

1.1 · Field lines

The standard picture of a vector field is a set of field lines: curves that are everywhere tangent to $\vv F$. A field line through $\vv x_{0}$ is the solution of

That is a system of ordinary differential equations, which is Chapter 0.8's subject. Two facts about those solutions are worth stating now, because the pictures depend on them.

The first is that if $\vv F$ is continuously differentiable then through every point there passes exactly one field line. So field lines never cross, except where $\vv F=\vv 0$. (⚑ This is the existence-and-uniqueness theorem for ODEs, which we quote here and Chapter 0.8 states properly.)

The second is about what a field line is when the field is a velocity. In that case the field lines are the paths that particles actually follow. So (0.7.2) is not a visualisation device. It is the equation of motion.

Field-line density is the usual proxy for field strength, and §5 will tell you exactly when that proxy is honest: field lines can only begin or end where the divergence is nonzero, which is why "lines of $\vv E$ start on positive charges and end on negative ones" is a theorem rather than a picture.

1.2 · A vector field is not a vector

This deserves saying once, plainly, because the notation actively hides it and Chapter 3.2 will charge you for the misunderstanding.

A single vector is one arrow. A vector field assigns an arrow to each point, and the arrow attached to $\vv p$ lives in its own private copy of $\R^{n}$. That copy is the space of displacements from $\vv p$. Nothing in the definition lets you compare the arrow at $\vv p$ with the arrow at $\vv q$, because they are elements of two different vector spaces that merely happen to look identical.

In flat space with Cartesian coordinates they are identical, which is why you have never had to think about it. You slide arrows around freely, subtract $\vv F(\vv q)-\vv F(\vv p)$, and get the derivative. That move is the one thing that stops working on a curved space. Take the sphere. A tangent vector at the north pole lies in a plane, a tangent vector at the equator lies in a different plane, and there is no basis-free way to say they point "the same way".

The collection of all these private copies, one vector space glued to every point, is the tangent bundle, and it is the central object of Chapter 3.2. Repairing the subtraction, so that "the derivative of a vector field" means something on a curved space, is the covariant derivative of Chapter 3.3. The leftover from that repair is the gravitational field.

For this chapter we are in flat space with Cartesian coordinates and we will slide arrows around without apology. Just know which cheque is being written.

2 · Line integrals, conservative fields, and a topological surprise

Here is the plan for this section. We ask the first of the chapter's two questions: how much does a field push you along as you walk a path? The answer is the line integral, and §2.1 builds it. Then we ask when that integral depends only on where the path starts and ends, which is §2.2 and §2.3. The answer looks routine until §2.4, where it stops being routine.

2.1 · The definition

Start with the physics that forces the definition. A constant force $\vv F$ acting through a straight displacement $\vv d$ does work $W=\vv F\cdot\vv d$. The dot product is there because only the component of the force along the motion does anything, which is Chapter 0.5's orthogonal projection wearing a physical hat.

Now let the force vary and the path curve. Chop the path $C$ into $N$ short segments with endpoints $\vv x_{0},\ldots,\vv x_{N}$, let $\Delta\vv x_{i}=\vv x_{i}-\vv x_{i-1}$, and pick a sample point on each segment. Over a short enough segment the force is nearly constant and the segment is nearly straight, so

That is a Riemann sum of exactly the kind Chapter 0.2 built the integral from, and we define the line integral as its limit as the segments shrink:

To compute one, parametrise. Let $\vv x(t)$, $t\in[a,b]$, trace out $C$. Then $\Delta\vv x_{i}\approx\dot{\vv x}(t_{i})\,\Delta t_{i}$ by the linearisation of Chapter 0.1, and (0.7.4) becomes an ordinary one-dimensional integral:

Everything from Chapter 0.2 now applies. Before we use the line integral, let's be clear about what it depends on. There are two things it does not depend on and one thing it does.

• It does not depend on which parametrisation you chose.
• It does not depend on how you chopped the curve up.
• It does depend on the direction of travel. Reversing the direction flips the sign.

Both claims are proved in the grind box.

Grind box — the line integral is parametrisation-independent, and orientation is the only thing that matters

Let $\vv x(t)$, $t\in[a,b]$, and $\vv y(u)$, $u\in[c,d]$, be two parametrisations of the same curve, related by a reparametrisation $t=\tau(u)$ with $\tau$ differentiable, $\tau(c)=a$, $\tau(d)=b$ and $\tau'\gt0$ (an increasing change of clock). Then $\vv y(u) = \vv x(\tau(u))$, and by the chain rule along a curve (Chapter 0.6 §5.1),

$$\dot{\vv y}(u) = \dot{\vv x}\big(\tau(u)\big)\,\tau'(u).$$

Substituting into (0.7.5) and using the substitution rule of Chapter 0.2 with $t=\tau(u)$, $\dd t = \tau'(u)\dd u$:

$$\int_{c}^{d}\vv F\big(\vv y(u)\big)\cdot\dot{\vv y}(u)\,\dd u = \int_{c}^{d}\vv F\big(\vv x(\tau)\big)\cdot\dot{\vv x}(\tau)\,\tau'(u)\,\dd u = \int_{a}^{b}\vv F\big(\vv x(t)\big)\cdot\dot{\vv x}(t)\,\dd t.$$

Identical. The line integral is a property of the curve and the field, not of the clock you used. This is the same mechanism that made the ordinary integral independent of the partition: the $\tau'$ produced by the chain rule is exactly cancelled by the $\tau'$ produced by the substitution rule.

Orientation. Now take $\tau'\lt0$, which means running the curve backwards, so $\tau(c)=b$ and $\tau(d)=a$. The same computation goes through but the limits are swapped, and Chapter 0.2's convention $\int_b^a = -\int_a^b$ gives

$$\int_{-C}\vv F\cdot\dd\vv r \;=\; -\int_{C}\vv F\cdot\dd\vv r.$$

So a line integral carries a sign that encodes a direction. That is not a nuisance: it is the entire mechanism of §5. When two adjacent cells share an edge, they traverse it in opposite directions, the two contributions cancel, and only the outer boundary survives. Orientation is what makes the cancellation happen, which is why every theorem in this chapter comes with a right-hand rule attached.

A cheap sanity check on (0.7.4): if $\vv F$ is everywhere perpendicular to the path, every term in the sum vanishes and the integral is zero. A magnetic force does no work for exactly this reason, since $q\vv v\times\vv B$ is perpendicular to $\vv v$ by construction.

2.2 · Gradient fields, and the Fundamental Theorem hiding in plain sight

Suppose $\vv F$ is the gradient of some scalar field: $\vv F = \nabla\phi$. Put that into (0.7.5) and watch what happens.

Two steps there deserve naming. The middle step is Chapter 0.6's chain rule along a curve, $\dv{}{t}\phi(\vv x(t)) = \nabla\phi\cdot\dot{\vv x}$. The last step is the Fundamental Theorem of Calculus from Chapter 0.2. Two tools, one line, and the result is worth stopping on:

That is the Fundamental Theorem again, with a one-dimensional domain sitting inside a three-dimensional space instead of on the real line. The structure is identical: derivative, integrated over a region, equals the undifferentiated thing on the boundary. Hold that sentence. §5 will repeat it twice more with the dimension raised.

2.3 · Four conditions that want to be equivalent

A field that is a gradient is called conservative, because if $\vv F$ is a force then $-\phi$ is a potential energy and the total energy is conserved. Four conditions on a field $\vv F$ defined on a region $U$ compete for that name:

	Condition
(A)	$\vv F = \nabla\phi$ for some single-valued $\phi$ on $U$
(B)	$\int_{C}\vv F\cdot\dd\vv r$ depends only on the endpoints of $C$
(C)	$\oint_{C}\vv F\cdot\dd\vv r = 0$ for every closed loop $C$ in $U$
(D)	$\nabla\times\vv F = \vv 0$ everywhere in $U$

We prove (A) $\Rightarrow$ (B) $\Leftrightarrow$ (C) $\Rightarrow$ (A) now, and (A) $\Rightarrow$ (D) in §7. The remaining arrow, (D) $\Rightarrow$ (C), is the interesting one, and it is false without an extra hypothesis about the shape of $U$.

(A) $\Rightarrow$ (B). Done: (0.7.6) gives $\phi(B)-\phi(A)$, which mentions the path nowhere.

(B) $\Leftrightarrow$ (C). Suppose (B). Any closed loop can be cut at two points into two paths $C_{1}$ and $C_{2}$ running from $A$ to $B$. The loop is then $C_{1}$ followed by $C_{2}$ reversed, so by the orientation rule of the grind box its integral is $\int_{C_{1}} - \int_{C_{2}}$, which is $0$ by (B). Conversely, suppose (C) and let $C_{1},C_{2}$ be any two paths from $A$ to $B$. Then $C_{1}$ followed by reversed $C_{2}$ is a closed loop, so $\int_{C_{1}}-\int_{C_{2}} = 0$. That is (B).

(C) $\Rightarrow$ (A). This is the one that builds something. Assume (C), fix a basepoint $\vv x_{0}\in U$, and define

the integral taken along any path in $U$ from $\vv x_{0}$ to $\vv x$. By (C), and hence (B), the answer does not depend on which path, so $\phi$ is a genuine single-valued function. Now compute $\partial_{i}\phi$ at $\vv x$. Choose the path that reaches $\vv x + t\vv e_{i}$ by first going to $\vv x$ any way you like and then travelling along the straight segment $\vv x + s\vv e_{i}$, $s\in[0,t]$. On that segment $\dot{\vv x}=\vv e_{i}$, so $\vv F\cdot\dot{\vv x} = F^{i}$, and

Divide by $t$ and let $t\to0$. The right-hand side is exactly the object the first half of the Fundamental Theorem (Chapter 0.2) differentiates, and it gives $F^{i}(\vv x)$ provided $F^{i}$ is continuous. Hence $\partial_{i}\phi = F^{i}$ for every $i$, i.e. $\nabla\phi=\vv F$. $\blacksquare$

So (A), (B) and (C) are the same condition wearing three costumes. None of the three is checkable in practice, because you cannot test infinitely many loops. Condition (D) is different. It is a local test, three derivatives evaluated at a point, and you can check it in ten seconds. Which is exactly why the failure of (D) $\Rightarrow$ (C) matters so much.

2.4 · The counterexample, in full

Take the plane with the origin deleted, $U = \R^{2}\setminus\{\vv 0\}$, and on it the field

It is perfectly smooth on $U$, since the only bad point has been removed. Our first job is to compute its curl. In the plane that means the single surviving component $\partial_{x}F^{y}-\partial_{y}F^{x}$, which §4 derives. By the quotient rule,

They are equal, so $\nabla\times\vv F = \vv 0$ at every point of $U$. Condition (D) holds everywhere, with no exceptions and no fine print.

Now integrate around the circle of radius $R$ centred at the origin, parametrised by $\vv x(t)=(R\cos t,\,R\sin t)$ for $t\in[0,2\pi]$. On that circle $x^{2}+y^{2}=R^{2}$, so

and their dot product is $\sin^{2}t+\cos^{2}t = 1$, because the factors of $R$ cancel exactly. That constant is all we need to finish the integral. Therefore

for every radius $R$. Condition (C) fails, spectacularly and by a fixed amount. A curl-free field with a nonzero circulation.

Where did the potential go? It exists locally. On the right half-plane $x\gt0$, the function $\phi=\arctan(y/x)$ has $\nabla\phi = \vv F$ (differentiate it and see). Geometrically $\phi$ is the polar angle $\theta$, and Chapter 0.6's polar grind box already told us $\nabla\theta = \hat{\boldsymbol\theta}/r$, which in Cartesian components is $(-y,x)/r^{2}$. That is precisely (0.7.9).

So the "potential" of this field is the angle. And the angle is not a function on $U$: go once round the origin and it has increased by $2\pi$. It is multivalued, and (0.7.12) is measuring exactly that increment.

The repair to the theorem is a condition on the shape of $U$, not on $\vv F$. A region is simply connected if every closed loop in it can be shrunk continuously to a point without leaving the region. A disc is simply connected. A disc with a puncture is not, because a loop around the puncture has nothing to shrink through.

With that hypothesis the missing arrow works, and §5 supplies the proof in one line. If the loop $C$ bounds a surface $S$ lying inside $U$, Stokes' theorem gives

For the punctured plane there is no such $S$. Any surface spanning a loop around the origin must contain the origin, where $\vv F$ does not exist. So the theorem does not fail. Its hypothesis does.

The circle above was centred on the hole, and the factors of $R$ cancelling looks like the sort of luck a symmetrical choice buys you. The figure below hands you the loop.

radius R

R = 1.00

lumpiness

λ = 0.30

field the vortex ∇(xy)

two turns

reset

∮ F·dr = 6.2831853 2π = 6.2831853 winding ∮/2π = 1.000000

curl along the loop (square of side 1e-5): max 1.796e-9 mean 1.235e-10 — no local test detects anything

loop centre (0.300, -0.200) closest approach to the origin 0.4063

2.5 · Why a physicist should care: the potential knows more than the field

It is tempting to file §2.4 as a curiosity about a contrived function. It is not. It is the reason one of the central objects of modern physics, the gauge potential, is physically real.

Consider an infinite solenoid along the $z$-axis carrying a current. Inside, there is a uniform magnetic field $\vv B = B\hat{\vv z}$. Outside, $\vv B = \vv 0$ exactly: the field is confined. Now §7 will show that $\vv B$ can be written as $\nabla\times\vv A$ for a vector potential $\vv A$, and §5's Stokes theorem gives, for any loop $C$ encircling the solenoid,

That last quantity is the total magnetic flux threading the solenoid. Look at what we have. Outside the solenoid $\vv A$ is curl-free, since its curl is $\vv B$ and $\vv B$ vanishes there. Yet its circulation around the solenoid is $\Phi\neq0$. That is (0.7.9) again, with the solenoid playing the part of the deleted origin. The region outside a solenoid is not simply connected.

Classically this is harmless: a charged particle outside feels force $q\vv v\times\vv B = \vv 0$ and nothing happens. Quantum mechanically it is not harmless. Chapter 6.3 will derive that a charged particle's wavefunction, carried along a path, acquires a phase

So a beam split around either side of the solenoid and recombined shows an interference shift of $q\Phi/\hbar$. That shift is determined entirely by the circulation of $\vv A$, in a region where $\vv B$ is identically zero. This is the Aharonov–Bohm effect, predicted in 1959 and measured (Chambers 1960, then conclusively by Tonomura and co-workers in 1986 with a superconductor-shielded toroidal magnet, which closed the last loopholes about leakage fields). Those are experimental facts, quoted. This book will derive the phase in Chapter 6.3 but cannot derive the measurement.

The moral is exactly §2.4's:

3 · Divergence, derived — and its identity with trace

Now the chapter's second question. Instead of walking along a curve, we surround a point and ask how much of the field leaves. The plan is three steps. First a definition that mentions no coordinates at all, in §3.2. Then the familiar formula, derived from that definition with a box, in §3.3. Then §3.4, where the formula turns out to be something Chapter 0.4 already handed us.

3.1 · Flux

Fix a surface $S$ with a chosen side, and let $\hat{\vv n}$ be the unit normal pointing out of that side. Write $\dd\vv A = \hat{\vv n}\,\dd A$ for the vector area element. The flux of $\vv F$ through $S$ is

The dot product is not decoration. Suppose $\vv F = \rho\vv v$, the mass per unit volume times the velocity of a fluid. In a short time $\dd t$, the material that crosses a small patch of area $\dd A$ is the material that was sitting in an oblique cylinder with base $\dd A$ and slant $\vv v\,\dd t$. The volume of such a cylinder is base times perpendicular height, which is $(\vv v\cdot\hat{\vv n})\,\dd t\,\dd A$. Anything moving parallel to the surface never crosses it, and the dot product is what discards exactly that part of the motion. So the mass crossing per unit time is $\rho\,\vv v\cdot\hat{\vv n}\,\dd A = \vv F\cdot\dd\vv A$, and

3.2 · The definition, before any coordinates

Now shrink the region. If stuff is leaving a small region faster than it enters, something inside is producing it. Define the divergence of $\vv F$ at a point $\vv p$ as the outflow per unit volume in the limit of a vanishing region containing $\vv p$:

Look at what that definition does not contain: no coordinates, no basis, no components. It is built from a flux and a volume, both of which are geometric. That matters, because the formula we are about to derive looks entirely coordinate-dependent, and it will be reassuring to know it cannot be.

3.3 · The box, done properly

Take the region to be a rectangular box centred at $\vv p=(x,y,z)$ with sides $\Delta x$, $\Delta y$, $\Delta z$, so $\abs{V}=\Delta x\,\Delta y\,\Delta z$. It has six faces, which pair up.

The pair perpendicular to $x$. On the face at $x+\tfrac{\Delta x}{2}$ the outward normal is $+\hat{\vv x}$, so the integrand $\vv F\cdot\hat{\vv n}$ is $F^{x}$. On the face at $x-\tfrac{\Delta x}{2}$ the outward normal is $-\hat{\vv x}$ and the integrand is $-F^{x}$. Each face has area $\Delta y\,\Delta z$, and over such a small face $F^{x}$ equals its value at the face's centre plus a correction that integrates away (the grind box does this bookkeeping exactly). So the two faces together contribute

The bracket is a difference of the same function at two nearby points, which is precisely what Chapter 0.1 taught us to expand. Writing $F^{x}(x\pm\tfrac{\Delta x}{2}) = F^{x}(x) \pm \tfrac{\Delta x}{2}\partial_{x}F^{x} + O(\Delta x^{2})$ and subtracting, the values at $x$ cancel and the two first-order terms add:

where the $O(\Delta x^{2})$ terms also cancel by the symmetry of the two-sided difference, leaving $O(\Delta x^{3})$. So the $x$-pair contributes $\partial_{x}F^{x}\;\Delta x\,\Delta y\,\Delta z = \partial_{x}F^{x}\,\abs{V}$, plus something smaller.

The other two pairs. Identical arguments, with $x$ replaced by $y$ and by $z$. Adding all three pairs and dividing by $\abs{V}$:

We want the divergence itself rather than the flux out of a box of finite size, so take the limit as the box shrinks. The correction term $O(\Delta)$ dies with it, and what is left is the formula:

The last form uses the summation convention of Chapter 0.6 §5.2, and note that the index structure is correct: one up, one down, summed. That is the formula every textbook writes down first. Here it is a consequence instead. The definition it came from, (0.7.17), is the one that survives a change of coordinates, a curved space, and Chapter 3.3.

Grind box — the error terms in the box argument, honestly

Two steps above were waved at. Both are repairable, and it is worth seeing how, because the same two gaps reappear in §5.

(i) The field varies across each face. We replaced $\int_{\text{face}}F^{x}\,\dd A$ by $F^{x}(\text{face centre})\times\text{area}$. Expand $F^{x}$ about the face centre in the two in-face directions, writing the offset as $(\eta,\zeta)$:

$$F^{x} = F^{x}_{\text{centre}} + \eta\,\partial_{y}F^{x} + \zeta\,\partial_{z}F^{x} + O(\Delta^{2}).$$

Integrate over the face, which is symmetric about its centre: $\int\eta\,\dd A = 0$ and $\int\zeta\,\dd A=0$, so the two linear terms vanish exactly, not approximately. What remains is

$$\int_{\text{face}}F^{x}\,\dd A = F^{x}_{\text{centre}}\,\Delta y\,\Delta z + O(\Delta^{4}),$$

since the quadratic remainder is $O(\Delta^{2})$ times an area $O(\Delta^{2})$. Divided by $\abs{V}=O(\Delta^{3})$ this is $O(\Delta)$ and vanishes in the limit. The symmetry of the box did the work. Had we used a lopsided region we would have had to keep the linear terms.

(ii) The limit must not depend on the shape of the shrinking region. (0.7.17) says "$V\to0$" without specifying how, but we computed with boxes. For a continuously differentiable $\vv F$ the answer is the same for any reasonable family of shrinking regions, and here is the mechanism. Near $\vv p$ the field is $\vv F(\vv p) + J(\vv p)\,\vv h + o(\abs{\vv h})$ by Chapter 0.6's total derivative. The constant part $\vv F(\vv p)$ contributes zero flux through any closed surface, since as much goes in one side as out the other (or formally, apply §5 to a constant field). The linear part contributes exactly $\mathrm{tr}\,J\cdot\abs{V}$ for any region, which is §3.4. And the $o(\abs{\vv h})$ part contributes $o(\abs{\vv h})\times\text{area} = o(\abs{V})$ for any region whose surface area does not blow up relative to its volume. So the limit is $\mathrm{tr}\,J$ regardless of shape.

That last observation is worth extracting: the divergence only depends on the linear part of the field, i.e. on the Jacobian. Which is the next subsection's entire content.

3.4 · Divergence is the trace of the Jacobian

Compare (0.7.21) with the Jacobian matrix of Chapter 0.6 §2.2, whose entries are $J^{i}{}_{j} = \partial_{j}F^{i}$:

The divergence is the trace of the Jacobian. That is not a curiosity. Combined with Chapter 0.4 it converts the definition into a statement about volume, and it does so in four lines. Here are those four lines.

Let $\vv F$ be a velocity field, so that a particle at $\vv x$ moves to $\vv x + \epsilon\vv F(\vv x)$ in a short time $\epsilon$. Call that map $\Phi_{\epsilon}$. Its derivative is

differentiating $\vv x\mapsto\vv x+\epsilon\vv F(\vv x)$ term by term. Chapter 0.6 §8 proved that a differentiable map multiplies the volume of a small region by $\abs{\det}$ of its derivative. And Chapter 0.4 §6 proved the identity we now need:

Put the three together. A blob of volume $V(0)$ carried by the flow for a time $\epsilon$ has volume

What we actually want is a rate rather than a ratio, so subtract $V(0)$ from both sides, divide by $V(0)$ and by $\epsilon$, and let $\epsilon\to0$. The $O(\epsilon^{2})$ term drops out and we are left with

Integrating (0.7.26) along a trajectory (it is $\dv{}{t}\ln V = \nabla\cdot\vv F$, an equation of the type Chapter 0.1 solved) gives the finite version,

with the integral taken along the path of the blob. The figure below measures both sides of this.

One immediate dividend, banked now and spent in Chapter 1.3. A mechanical system's state is a point $(q,p)$ in phase space, moving with velocity $\dot q = \pdv{H}{p}$, $\dot p = -\pdv{H}{q}$. That is a vector field on phase space, and its divergence is

by Clairaut's theorem (Chapter 0.6 §6.1). Zero divergence, so by (0.7.26) phase-space volume is exactly conserved by Hamiltonian flow. That is Liouville's theorem, one of the deepest facts in classical mechanics and the foundation of statistical mechanics, and it has just cost us two lines.

Grind box — divergence in spherical coordinates, from the flux definition

Because (0.7.17) mentions no coordinates, we can evaluate it in any we like. Take a "box" bounded by the coordinate surfaces $r$ and $r+\dd r$, $\theta$ and $\theta+\dd\theta$, $\phi$ and $\phi+\dd\phi$. From Chapter 0.6 §8.2 its volume is

$$\dd V = r^{2}\sin\theta\;\dd r\,\dd\theta\,\dd\phi,$$

and the three pairs of faces have areas $r^{2}\sin\theta\,\dd\theta\,\dd\phi$ (the $r$-faces), $r\sin\theta\,\dd r\,\dd\phi$ (the $\theta$-faces) and $r\,\dd r\,\dd\theta$ (the $\phi$-faces). Write the field in the orthonormal basis, $\vv F = F^{r}\hat{\vv r} + F^{\theta}\hat{\boldsymbol\theta} + F^{\phi}\hat{\boldsymbol\phi}$.

The $r$-pair. The outward normals are $\pm\hat{\vv r}$, so the net contribution is the difference of $F^{r}\times(\text{area})$ at $r+\dd r$ and at $r$. Here the area itself depends on $r$, which is the entire novelty. So the difference is of the product:

$$\partial_{r}\Big(F^{r}\,r^{2}\sin\theta\Big)\,\dd r\,\dd\theta\,\dd\phi.$$

Divide by $\dd V$: the $\sin\theta$ cancels and we get $\dfrac{1}{r^{2}}\partial_{r}\big(r^{2}F^{r}\big)$.

The $\theta$-pair. Same move. The area $r\sin\theta\,\dd r\,\dd\phi$ depends on $\theta$ through $\sin\theta$, so the net contribution is $\partial_{\theta}\big(F^{\theta}r\sin\theta\big)\dd\theta\,\dd r\,\dd\phi$, and dividing by $\dd V$ gives $\dfrac{1}{r\sin\theta}\partial_{\theta}\big(\sin\theta\,F^{\theta}\big)$.

The $\phi$-pair. The area $r\,\dd r\,\dd\theta$ does not depend on $\phi$, so we get $\partial_{\phi}(F^{\phi}r)\,\dd\phi\,\dd r\,\dd\theta$, and dividing by $\dd V$ gives $\dfrac{1}{r\sin\theta}\partial_{\phi}F^{\phi}$. Adding:

$$\nabla\cdot\vv F = \frac{1}{r^{2}}\pdv{}{r}\big(r^{2}F^{r}\big) + \frac{1}{r\sin\theta}\pdv{}{\theta}\big(\sin\theta\,F^{\theta}\big) + \frac{1}{r\sin\theta}\pdv{F^{\phi}}{\phi}.$$

Every factor in that intimidating formula has now been accounted for: each one is an area or a volume from Chapter 0.6's Jacobian, and the derivatives act on the products because the areas themselves vary. Nothing was memorised.

The one that matters. For a purely radial field $\vv F = f(r)\hat{\vv r}$ only the first term survives, so $\nabla\cdot\vv F = r^{-2}\big(r^{2}f\big)'$. Take $f = 1/r^{2}$:

$$\nabla\cdot\frac{\hat{\vv r}}{r^{2}} \;=\; \frac{1}{r^{2}}\dv{}{r}\big(r^{2}\cdot r^{-2}\big) \;=\; \frac{1}{r^{2}}\dv{}{r}(1) \;=\; 0 \qquad (r\neq0).$$

The inverse-square field is divergence-free everywhere it is defined. Remember that. The warning callout below is about the four words in brackets.

A cross-check worth making. Chapter 3.5 §6.4 will prove the general formula $\nabla\cdot\vv F = \tfrac{1}{\sqrt g}\,\partial_{i}\big(\sqrt g\,F^{i}\big)$, where $\sqrt g$ is the volume factor of Chapter 0.6 §8.3 and the $F^{i}$ are coordinate-basis components. We quote it here only to check it against what we just derived. In spherical coordinates $\sqrt g = r^{2}\sin\theta$, and the coordinate components are related to the orthonormal ones by $F^{\theta}_{\text{coord}} = F^{\theta}/r$ and $F^{\phi}_{\text{coord}} = F^{\phi}/(r\sin\theta)$, because $\partial_{\theta}$ has length $r$ and $\partial_{\phi}$ has length $r\sin\theta$. Substituting reproduces the three terms above exactly. The flux box and the metric formula are the same statement.

4 · Curl, derived — and what a Jacobian is made of

4.1 · The definition

Divergence measured outflow per unit volume. The other thing a field can do near a point is go round, so measure circulation per unit area. Fix a direction $\hat{\vv n}$, take a small loop lying in the plane perpendicular to $\hat{\vv n}$ and traversed counterclockwise as seen from the tip of $\hat{\vv n}$ (the right-hand rule, which is a choice we are making and must then keep), and define

Again: no coordinates. But writing the left side that way asserts something we have not earned yet. It says that the resulting numbers, one for each direction $\hat{\vv n}$, are the components of a single vector. Equivalently, it says that the answer depends linearly on $\hat{\vv n}$. That is not obvious from (0.7.29). The computation below establishes it, because the answer it produces is manifestly a dot product of $\hat{\vv n}$ with a fixed vector.

4.2 · The rectangle

Take $\hat{\vv n}=\hat{\vv z}$ and a rectangle in the plane $z=\text{const}$, centred at $\vv p = (x,y,z)$, with sides $\Delta x$ and $\Delta y$, traversed counterclockwise. Walk the four sides, keeping track of direction:

Side	Direction	Contribution
bottom, at $y-\tfrac{\Delta y}{2}$	$+\hat{\vv x}$	$+F^{x}\big(x,\,y-\tfrac{\Delta y}{2}\big)\,\Delta x$
right, at $x+\tfrac{\Delta x}{2}$	$+\hat{\vv y}$	$+F^{y}\big(x+\tfrac{\Delta x}{2},\,y\big)\,\Delta y$
top, at $y+\tfrac{\Delta y}{2}$	$-\hat{\vv x}$	$-F^{x}\big(x,\,y+\tfrac{\Delta y}{2}\big)\,\Delta x$
left, at $x-\tfrac{\Delta x}{2}$	$-\hat{\vv y}$	$-F^{y}\big(x-\tfrac{\Delta x}{2},\,y\big)\,\Delta y$

Group the two $y$-directed sides together and the two $x$-directed sides together:

Each bracket is the same two-sided difference that appeared in (0.7.19), so each is a derivative times the spacing, up to higher order:

We want circulation per unit area rather than circulation, so divide by the area $\Delta x\,\Delta y$ and let the rectangle shrink. The $O(\Delta^{3})$ term goes with it:

The other two components come from the same computation with the labels rotated $x\to y\to z\to x$, since nothing in the argument singled out $z$:

Each component is manifestly linear in $\vv F$ and the three of them assemble into a vector, which retroactively justifies the notation of (0.7.29). In two dimensions only the last component exists, and it is a scalar. That is the object §2 was computing.

4.3 · Every Jacobian splits in exactly one way

Now the structural fact, which explains what the curl is rather than how to compute it. Any square matrix $J$ can be written as a symmetric part plus an antisymmetric part:

The split exists (just add the two pieces and the transposes cancel) and it is unique: if $J = S'+A'$ with $S'$ symmetric and $A'$ antisymmetric, transposing gives $J^{\mathsf T}=S'-A'$, and adding and subtracting the two equations forces $S'=\tfrac12(J+J^{\mathsf T})$ and $A'=\tfrac12(J-J^{\mathsf T})$. So the decomposition is not a choice.

Apply it to the Jacobian of a velocity field. An antisymmetric matrix has zeros on its diagonal ($A_{ii}=-A_{ii}$ forces $A_{ii}=0$), so $\mathrm{tr}A=0$ and

All of the divergence lives in the symmetric part. And the antisymmetric part, written out with $J^{i}{}_{j}=\partial_{j}F^{i}$, is

Every entry is one of the three components of (0.7.33), up to a sign. Writing $\vv\omega\equiv\nabla\times\vv F$, the matrix is exactly

Check one entry to be sure: $A_{21} = \tfrac12(\partial_{x}F^{y}-\partial_{y}F^{x}) = \tfrac12\omega_{z}$ by (0.7.32), and that is what sits in row 2, column 1 of (0.7.37). The other five follow the same way.

A matrix of that shape does something recognisable. Apply it to a displacement $\vv h$:

And $\vv h\mapsto\vv\Omega\times\vv h$ is precisely the velocity field of a rigid rotation with angular velocity $\vv\Omega$. Put the pieces together with Chapter 0.6's linearisation $\vv F(\vv p+\vv h) = \vv F(\vv p) + J\vv h + o(\abs{\vv h})$:

This is the cleanest possible statement of what divergence and curl are. They are the two invariant pieces of the Jacobian: the trace, and the antisymmetric part. Note also what they are not: they are not everything. The traceless symmetric part of $S$ is left over. It describes pure shear, and it is invisible to both operators. A field can therefore deform a blob dramatically while having zero divergence and zero curl. The figure below is built to show you exactly that, because no static picture can.

4.4 · Why the curl is a vector only in three dimensions

Count. An antisymmetric $n\times n$ matrix is determined by its entries strictly above the diagonal, of which there are $\tfrac12 n(n-1)$. A vector in $n$ dimensions has $n$ components. Those two counts agree when

Only in three dimensions. The table is short and worth memorising:

$n$	antisymmetric components	vector components	curl is…
2	1	2	a scalar
3	3	3	a vector, the lucky coincidence
4	6	4	an antisymmetric matrix, irreducibly

In two dimensions there is one number and it does not point anywhere, which is why §2 wrote $(\nabla\times\vv F)_{z}$ and treated it as a scalar. In four dimensions there are six components and no vector to pack them into, so the antisymmetric object has to stay a matrix.

Grind box — the index form, $\epsilon_{ijk}$, and the dimension count again

Define the Levi-Civita symbol $\epsilon_{ijk}$ to be $+1$ if $(ijk)$ is a cyclic permutation of $(123)$, $-1$ if it is an odd permutation, and $0$ if any index repeats. Then (0.7.33) is the single line below, with one note about heights first. Everything in this chapter is Cartesian and Euclidean, where $\delta_{ij}=\delta^{ij}$ and raising an index changes nothing, so we suppress the distinction and write all of $\epsilon$'s indices down. Chapter 2.4 reinstates it, and from there onward Chapter 0.6 §5.2's one-up-one-down test applies literally.

$$\big(\nabla\times\vv F\big)_{i} \;=\; \epsilon_{ijk}\,\partial_{j}F^{k},$$

summed over $j$ and $k$. Check the case $i=3$: the nonzero terms are $\epsilon_{312}=+1$ and $\epsilon_{321}=-1$, giving $\partial_{1}F^{2}-\partial_{2}F^{1}$, which is (0.7.32). ✓

The contraction identity. We will use

$$\epsilon_{ijk}\,\epsilon_{klm} \;=\; \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl},$$

and here is the argument rather than an assertion. Both sides are antisymmetric under $i\leftrightarrow j$ and under $l\leftrightarrow m$, and both vanish unless $\{i,j\}=\{l,m\}$ as sets with $i\ne j$ (on the left, because the sum over $k$ needs $k\notin\{i,j\}$ and $k\notin\{l,m\}$, and in three dimensions that leaves at most one $k$). So it is enough to check one representative case, $i=l=1$, $j=m=2$: the left side is $\epsilon_{12k}\epsilon_{k12} = \epsilon_{123}\epsilon_{312} = 1$, and the right side is $\delta_{11}\delta_{22}-\delta_{12}\delta_{21} = 1$. ✓ Antisymmetry propagates the check to every other case.

The antisymmetric part in index form. With $\vv\omega=\nabla\times\vv F$,

$$\epsilon_{ijk}\omega_{k} = \epsilon_{ijk}\epsilon_{klm}\partial_{l}F^{m} = \partial_{i}F^{j} - \partial_{j}F^{i} = -2A_{ij},$$

using $A_{ij}=\tfrac12(\partial_{j}F^{i}-\partial_{i}F^{j})$ from (0.7.34). So $A_{ij} = -\tfrac12\epsilon_{ijk}\omega_{k}$, which is (0.7.37) written compactly, and $\omega_{k} = -\epsilon_{kij}A_{ij}$ inverts it.

And now the dimension count, structurally. The repackaging works because $\epsilon_{ijk}$ has three indices and can convert a two-index antisymmetric object into a one-index object. In $n$ dimensions the Levi-Civita symbol has $n$ indices, so it converts an antisymmetric $2$-index object into an antisymmetric $(n-2)$-index one. For $n=3$ that is a vector. For $n=4$ it is another antisymmetric matrix, which is why Chapter 2.6's $\tilde F^{\mu\nu} = \tfrac12\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$ (the "dual" field, which swaps $\vv E$ and $\vv B$) is a matrix and not a vector. The cross product, the curl-as-vector, and the fact that a rotation in three dimensions has an axis are all the same coincidence: $\tfrac12 n(n-1)=n$ at $n=3$. In four dimensions a rotation has no axis. It has an invariant plane instead, and Chapter 2.2's Lorentz transformations are exactly such rotations.

source (x, y) rotation (−y, x) shear (y, x) mixed (xy, x+y)

point x

point y

elapsed t

t = 0.35

∇·F = +2.000 · measured ln(A/A₀)/t = +2.000

(∇×F)_z = +0.000 · measured 2ϑ/t = -0.000

blob: area ×2.014, stretched ×1.42 / ×1.42, turned -0.0°

5 · The three theorems are one theorem

5.1 · The statements

Here are the three results, so you know where we are going.

The divergence theorem (Gauss, Ostrogradsky). For a region $V$ with closed boundary surface $\partial V$, outward normal:

Stokes' theorem. For a surface $S$ with boundary curve $\partial S$, oriented by the right-hand rule relative to the surface normal:

Green's theorem. For a region $R$ in the plane with boundary curve $\partial R$ traversed counterclockwise, and any two functions $P,Q$:

Green's theorem is not independent: put $\vv F = (P,Q,0)$ and take $S$ to be the flat region $R$ with $\hat{\vv n}=\hat{\vv z}$. Then (0.7.32) makes the left side of (0.7.41) into the left side of (0.7.42), and $\vv F\cdot\dd\vv r = P\,\dd x + Q\,\dd y$ makes the right sides agree. Green's theorem is the special case of Stokes' theorem for a flat surface. Two to go.

5.2 · The divergence theorem, derived by cancellation

Chop the region $V$ into $N$ small cells $V_{1},\ldots,V_{N}$. Picture a fine grid of boxes. For each cell, the definition of divergence (0.7.17) says that its outward flux is the divergence at an interior point times its volume, up to something smaller:

Now sum over all cells. Look at the two sides separately.

The right side is a Riemann sum. $\sum_{k}(\nabla\cdot\vv F)(\vv p_{k})\abs{V_{k}}$ is exactly the tagged sum that Chapter 0.2 defined the integral to be the limit of, so as the cells shrink it converges to $\int_{V}(\nabla\cdot\vv F)\,\dd V$.

The left side collapses. Every face of every cell is either an interior face, shared with a neighbouring cell, or a piece of the outer boundary $\partial V$. Consider an interior face shared by cells $k$ and $\ell$. The same field $\vv F$ is integrated over the same patch of surface twice, once with the outward normal of cell $k$ and once with the outward normal of cell $\ell$. Those two normals are opposite, because outward from $k$ is inward to $\ell$. So the two contributions are $+\int\vv F\cdot\hat{\vv n}\,\dd A$ and $-\int\vv F\cdot\hat{\vv n}\,\dd A$, and they cancel exactly. Not approximately: they are the same integral with opposite signs.

Every interior face is shared by exactly two cells, so every interior contribution cancels. What survives is the sum over faces lying on $\partial V$, which is precisely $\oint_{\partial V}\vv F\cdot\dd\vv A$. Equating the two limits gives the theorem:

5.3 · Stokes' theorem, by the identical move

Tile the surface $S$ with small patches $S_{1},\ldots,S_{N}$, each with its own boundary loop oriented consistently with the surface normal. By the definition of curl (0.7.29), each patch satisfies

Sum. The right side is a Riemann sum for $\int_{S}(\nabla\times\vv F)\cdot\dd\vv A$. On the left, every interior edge is shared by exactly two patches. Because the two patches are traversed with a consistent orientation, they walk that shared edge in opposite directions. By the orientation rule of §2.1's grind box, the two line integrals over that edge are negatives of each other and cancel exactly. Only edges on the outer boundary $\partial S$ survive, giving (0.7.41). $\blacksquare$

The consistency of the orientations is where the right-hand rule earns its keep: it is the rule that guarantees adjacent patches disagree about the direction of their shared edge. If you have ever wondered why Stokes' theorem comes with a hand attached, that is why.

5.4 · One theorem

Line up everything this book has proved that has this shape.

Theorem	Region $M$	Boundary $\partial M$	The thing $\omega$	Its derivative $\dd\omega$
FTC (0.2)	interval	two endpoints	$f$	$f'\,\dd x$
Gradient (§2.2)	curve	two endpoints	$\phi$	$\nabla\phi\cdot\dd\vv r$
Stokes / Green	surface	closed curve	$\vv F\cdot\dd\vv r$	$(\nabla\times\vv F)\cdot\dd\vv A$
Divergence	volume	closed surface	$\vv F\cdot\dd\vv A$	$(\nabla\cdot\vv F)\,\dd V$

Every row says the same thing, and the operators $\nabla$, $\nabla\times$, $\nabla\cdot$ are three appearances of one operation applied to objects of three different types. Chapter 3.5 will build the objects properly. They are called differential forms, the operation is the exterior derivative $\dd$, and $\omega$ ranges over forms of degree $0,1,2$. Once those definitions are in place, all four rows become one equation:

⚑ We are quoting the general form, not deriving it: the machinery needed to define $\omega$ and $\dd\omega$ on a manifold of arbitrary dimension is Chapter 3.5's job, and doing it here would be building a cathedral to hang a door. But nothing in the table above is being taken on faith. All four rows have been derived, in Chapter 0.2 and in this chapter, by the same cancellation argument. What Chapter 3.5 adds is notation good enough to say it once.

One structural remark to file. A boundary has no boundary. The boundary of a disc is a circle, and a circle has no endpoints. The boundary of a ball is a sphere, and a sphere has no edge. Symbolically $\partial(\partial M) = \emptyset$. Under the correspondence of the table, that fact about regions must correspond to a fact about derivatives, and it does. It is $\dd^{2}=0$, and §7 proves the two instances of it you already know.

Grind box — what the cancellation argument does and does not establish

The subdivision argument is the right picture, it is how every working physicist thinks about these theorems, and it is not a proof. Three gaps, in increasing order of seriousness.

(i) Uniformity of the remainder. We wrote $o(\abs{V_{k}})$ for each cell and then summed $N$ of them, with $N\to\infty$. A sum of many small errors need not be small. What is needed is that the error is $\varepsilon\abs{V_{k}}$ with a single $\varepsilon$ that works for all cells at once and tends to zero as the grid refines. Then the total error is $\varepsilon\abs{V}\to0$. Uniformity of that kind is supplied by continuity of the partial derivatives on a closed bounded region, which is the same hypothesis, and the same reason, as in Chapter 0.6 §8's grind box.

(ii) The boundary is a staircase. Cells that straddle $\partial V$ are not fully inside, and a grid of boxes approximates a curved boundary by a staircase whose area does not in general converge to the true area even when the enclosed volume does. Repairing this honestly means either using cells adapted to the boundary or a limiting argument that controls the boundary layer's contribution. That control is available for piecewise-smooth boundaries, and genuinely false for sufficiently ugly ones.

(iii) We assumed the definitions converge. (0.7.43) is the definition of divergence read backwards, which presumes the limit in (0.7.17) exists and is approached uniformly over the region. For $C^{1}$ fields it does, by the argument in §3.3's grind box. For fields that are merely differentiable it can fail.

The honest statement of the theorem is therefore: for a compact region with piecewise-smooth boundary and a continuously differentiable field, (0.7.44) holds. That covers everything in this book. What you should take from the argument is the mechanism, which is that interior faces cancel in pairs. The mechanism is what generalises to Chapter 3.5's manifolds, where there is no grid of boxes to draw and the cancellation is enforced algebraically by the antisymmetry of forms instead.

6 · The continuity equation

Now the single most reused equation in physics, and it is three lines from §5.

Let $\rho(\vv x,t)$ be the density of some substance. It could be charge per unit volume, mass per unit volume, or probability per unit volume. Let $\vv J(\vv x,t)$ be its current density, meaning the vector field whose flux through a surface is the rate at which the substance crosses it (§3.1). Fix a region $V$, fixed in space and not moving. The amount inside is

Now impose the physical assumption, which is the entire content of the derivation: the substance is neither created nor destroyed, so the only way the amount inside $V$ can change is by flowing through the boundary. Outflow decreases it, hence the minus sign:

Handle each side. On the left, $V$ is fixed, so the time derivative passes through the integral and lands on the integrand as a partial derivative (differentiating under the integral sign, legitimate here for the same reason as in Chapter 0.2 §4.4: the integrand and its $t$-derivative are continuous and the domain is fixed). On the right, apply the divergence theorem (0.7.44):

And now the step that does the real work: $V$ was arbitrary. We never said which region. So the integral of that bracket vanishes over every region whatsoever, and a continuous function with that property is zero.

The bracket in (0.7.49) is continuous and its integral vanishes over every region, so the lemma applies and the bracket itself is zero everywhere:

6.1 · What it says, and why "local" is the important word

Read it at a point. The density here goes down exactly as fast as the current here diverges, which is to say exactly as fast as stuff flows away from here. That is a much stronger statement than "the total amount in the universe is constant".

To see how much stronger, ask what the weaker version would tolerate. Global conservation would permit a charge to vanish in London and simultaneously appear in Sydney, since the books would balance. (0.7.50) forbids it. To leave a region the substance must cross the boundary, and to get anywhere it must travel through the intervening space.

The global version follows by integrating over all space. Take $V$ to be a ball of radius $R$ and integrate (0.7.50) over it. The divergence theorem turns the second term back into a surface flux:

The surface area grows like $R^{2}$, so if $\vv J$ falls off faster than $1/R^{2}$ the right side tends to zero as $R\to\infty$, and the total amount in the universe is constant. So global conservation is a consequence of local conservation plus a boundary condition at infinity. And when the boundary condition fails, as it does for a radiating system, it fails for good physical reasons.

6.2 · Where it turns up

Constantly. A partial list, all of which this book reaches:

• Charge, in electromagnetism (Chapter 2.6). $\rho$ is charge density and $\vv J$ the electric current. Problem 4 shows that Maxwell's equations do not merely permit this equation. They force it, and the mechanism is §7's identity $\nabla\cdot(\nabla\times\vv A)=0$.
• Probability, in quantum mechanics (Chapter 4.6). With $\rho = \abs{\psi}^{2}$ and a current built from $\psi$ and $\nabla\psi$, the Schrödinger equation implies exactly (0.7.50). That is what "the wavefunction stays normalised" means locally: probability does not teleport.
• Noether currents (Chapters 1.4 and 5.2). Noether's theorem does not produce a conserved number. It produces a conserved current satisfying $\partial_{\mu}J^{\mu}=0$, of which (0.7.50) is the component-by-component reading. Every continuous symmetry gives one.
• Energy and momentum, in general relativity (Chapter 3.6). $\nabla_{\mu}T^{\mu\nu}=0$ is this equation for the energy–momentum tensor, with an index left over because energy and momentum travel together. In general relativity it is not imposed. It is forced by a geometric identity, which is $\dd^{2}=0$ yet again.
• Diffusion (the callout at the end of §7). Combine it with Fick's law and out falls the diffusion equation.

One notational preview, because it explains why the equation looks lopsided. In relativity $\rho$ and $\vv J$ are not separate objects: they are the four components of a single four-vector $J^{\mu}=(\rho c,\,\vv J)$, and (0.7.50) is

a four-dimensional divergence set to zero. The time derivative and the spatial divergence are the same operation, split up by an observer's choice of what counts as time. Chapter 2.4 makes this precise. Note in passing that it is the same "one object, split by an observer" story as $\vv E$ and $\vv B$ in §4.4.

7 · Second-derivative identities, and the Laplacian

Two identities, both one line of Clairaut, both structurally important out of all proportion to their difficulty.

7.1 · The curl of a gradient vanishes

Take $\vv F = \nabla\phi$, so $F^{i}=\partial_{i}\phi$, and put it into (0.7.33). The $z$-component is

by Clairaut's theorem (Chapter 0.6 §6.1), which says mixed partials commute when they are continuous. The other two components are the same statement with the labels rotated. Hence

That completes the chain (A) $\Rightarrow$ (D) of §2.3: a gradient field is curl-free, always, everywhere, with no hypothesis on the domain. It is the converse that needed the domain to be simply connected, and §2.4 is why.

7.2 · The divergence of a curl vanishes

Write it out. With $\vv F = \nabla\times\vv A$, (0.7.21) gives

Six terms. Pair them off: $\partial_{x}\partial_{y}A^{z}$ against $-\partial_{y}\partial_{x}A^{z}$, $-\partial_{x}\partial_{z}A^{y}$ against $+\partial_{z}\partial_{x}A^{y}$, and $\partial_{y}\partial_{z}A^{x}$ against $-\partial_{z}\partial_{y}A^{x}$. Each pair cancels by Clairaut. So

7.3 · Why the vector potential exists

Maxwell's equations contain $\nabla\cdot\vv B = 0$: there are no magnetic charges. Compare that with (0.7.56). Any field written as a curl automatically satisfies that condition, so the ansatz

is consistent with Maxwell for any $\vv A$ whatsoever. The converse says that every divergence-free field can be written this way. That converse is true on regions without holes, and it is the ⚑ Poincaré lemma, which we quote here and Chapter 3.5 proves.

The parallel with §2 is exact and not accidental. There, curl-free implied "is a gradient" only on a simply connected domain. Here, divergence-free implies "is a curl" only on a domain without cavities. Both are instances of one statement about the topology of the region, which Chapter 3.5 calls de Rham cohomology. In each case it is measuring the difference between the fields that are closed and those that are exact.

Now put the two identities side by side, because together they are the skeleton of gauge theory:

• (0.7.56) is why the vector potential exists: it makes $\vv B=\nabla\times\vv A$ compatible with $\nabla\cdot\vv B=0$.
• (0.7.54) is why the vector potential is not unique: replacing $\vv A\to\vv A+\nabla\chi$ for any function $\chi$ leaves $\vv B=\nabla\times\vv A$ unchanged, since $\nabla\times\nabla\chi = \vv 0$.

That second bullet is gauge freedom, and it is the seed of Chapter 6.3, where demanding that $\chi$ be allowed to vary from point to point generates the electromagnetic interaction. Two one-line consequences of Clairaut, and between them they set up the entire structure of the Standard Model.

7.4 · Both identities are $\dd^{2}=0$

Line up the operators by what they act on:

Composing two consecutive arrows gives zero, both times: $\nabla\times\nabla=0$ and $\nabla\cdot\nabla\times=0$. In Chapter 3.5's language the three objects are differential forms of degree $0,1,2$, all three arrows are the same operator $\dd$, and the two identities are the single statement

Which is, by the correspondence of §5.4, the algebraic shadow of $\partial(\partial M)=\emptyset$: a boundary has no boundary. We are flagging that as a promise rather than a derivation, and Chapter 3.5 makes it precise. But you can already see that the two facts you just proved are not two facts.

7.5 · The Laplacian

The remaining composition of two first-order operators is the one that is not zero: divergence of a gradient. Define the Laplacian

Structurally it is the trace of the Hessian of Chapter 0.6 §6. And since the trace is basis-independent (Chapter 0.4 §6), so is the Laplacian, even though the formula above mentions a particular set of axes.

One reading is immediate from §3. $\nabla^{2}\phi$ is the flux of $\nabla\phi$ per unit volume, so it is positive where the gradient field converges on the point, which is where the surroundings are pushing inward. The sharper reading is the following, and it is the one to carry.

The mean-value reading

Expand $\phi$ around $\vv p$ to second order, using Chapter 0.6 §6.2's multivariable Taylor:

Average this over the sphere $\abs{\vv h}=a$. Three averages are needed and all three are pure symmetry.

First order. $\avg{h^{i}}=0$, because $\vv h$ and $-\vv h$ both lie on the sphere and cancel.

Second order. $\avg{h^{i}h^{j}}$ is a symmetric array built from a sphere, and a sphere has no preferred direction, so the only array it can be is a multiple of $\delta^{ij}$: $\avg{h^{i}h^{j}} = c\,\delta^{ij}$. Fix $c$ by taking the trace of both sides, meaning set $j=i$ and sum. The left becomes $\avg{\abs{\vv h}^{2}}=a^{2}$ and the right becomes $3c$ in three dimensions. Hence $c=a^{2}/3$ and

Third order. Cubic in $\vv h$, so it changes sign under $\vv h\to-\vv h$ and averages to zero. The first surviving correction is therefore quartic.

Putting them together, and noting $\delta^{ij}\partial_{i}\partial_{j}\phi = \nabla^{2}\phi$:

An immediate consequence, free of charge. Suppose $\nabla^{2}\phi=0$ everywhere. Such a $\phi$ is called harmonic. Then the value at every point equals its average over every small sphere around it.

So $\phi$ can have no strict interior maximum. At a strict maximum, every nearby value is smaller, so the spherical average would be strictly less than the central value, contradicting equality. Applying the identical argument to $-\phi$, which is harmonic whenever $\phi$ is, rules out strict interior minima too. Hence a harmonic function attains its extremes on the boundary.

That is the maximum principle. It is why you cannot trap a charged particle in a static electric field, which is Earnshaw's theorem: the potential energy is harmonic in charge-free space, so it has no minimum to sit in. And it fell out of a symmetry argument about $\avg{h^{i}h^{j}}$.

Poisson's equation

Combine §3 with §2. Gauss's law in differential form says $\nabla\cdot\vv E = \rho/\epsilon_{0}$ (we take this as the physical input, and Chapter 2.6 derives it). And the electric field is minus the gradient of a potential, $\vv E=-\nabla\phi$, which is possible because the electrostatic field is curl-free and §2.3 then supplies $\phi$. Substituting one into the other:

Poisson's equation, and with $\rho=0$ it is Laplace's equation $\nabla^{2}\phi=0$. Read it through (0.7.63): charge density is exactly the amount by which the potential fails to equal its own local average. Where there is no charge, the potential is the smoothest interpolation of its boundary values that exists.

The same equation governs Newtonian gravity, $\nabla^{2}\Phi = 4\pi G\rho$, and that is not a coincidence of two inverse-square forces. Chapter 3.6 will show that the Einstein field equations reduce to precisely this in the weak-field, slow-motion limit, with $\Phi$ the $00$-component of the metric perturbation. Poisson's equation is what general relativity looks like when you squint.

Grind box — $\nabla^{2}(1/r)=0$, and the $4\pi$ that has gone missing

For a function of $r$ alone, the spherical divergence formula from §3's grind box gives

$$\nabla^{2}f(r) = \nabla\cdot\big(f'(r)\hat{\vv r}\big) = \frac{1}{r^{2}}\dv{}{r}\Big(r^{2}f'(r)\Big).$$

Take $f=1/r$, so $f' = -1/r^{2}$ and $r^{2}f' = -1$, a constant:

$$\nabla^{2}\frac{1}{r} \;=\; \frac{1}{r^{2}}\dv{}{r}(-1) \;=\; 0 \qquad (r\neq0).$$

So the Coulomb potential is harmonic everywhere except at the source. That is consistent with (0.7.64), since $\rho=0$ away from the point charge.

Now the trouble. Apply the divergence theorem to $\vv E = -\nabla(1/r) = \hat{\vv r}/r^{2}$ on a ball of radius $R$ centred at the origin. The left side is $\int_{V}\nabla\cdot\vv E\,\dd V = 0$, since the integrand vanishes at every point where it is defined. The right side is

$$\oint_{\abs{\vv r}=R}\frac{\hat{\vv r}}{r^{2}}\cdot\hat{\vv r}\;\dd A = \frac{1}{R^{2}}\cdot 4\pi R^{2} = 4\pi,$$

independent of $R$. Zero on one side, $4\pi$ on the other. Something is wrong, and it is not the theorem: the theorem requires $\vv E$ to be continuously differentiable everywhere inside $V$, and this one is not even defined at the origin. The hypothesis fails, so the conclusion is not owed to us.

The repair is to stop pretending the origin is not there and write

$$\nabla\cdot\frac{\hat{\vv r}}{r^{2}} \;=\; 4\pi\,\delta^{3}(\vv r),$$

where $\delta^{3}$ is an object that is zero everywhere except the origin and integrates to one. No function does that, which is why $\delta$ is not a function but a distribution, and why Chapter 0.9 exists. The $4\pi$ is not a fudge factor: it is the surface area of the unit sphere, and it is there because the flux computation above returned $4\pi$ for every $R$. Notice the pattern with §2.4, where a deleted point produced a circulation of $2\pi$, which is the circumference of the unit circle. In both cases the removed point is doing all the work, and the number the integral returns is counting it.

8 · Worked examples

9 · Your turn