Part A · the equation of motion · spine

Work, and the arrival of vector calculus

Energy in three dimensions is an integral along a path, and asking when that integral forgets the path is what builds the gradient, the curl and the first theorem of this course that is really about topology.

needs 03~2½ h slides + 5 h problemsproblems: ./mr problems 04

Unit 00 got energy for free. In one dimension, $m\ddot x = F(x)$ multiplied by $\dot x$ integrates on sight, and the antiderivative of $-F$ is the potential energy. In three dimensions the same multiplication gives $\tfrac{\mathrm d}{\mathrm dt}\big(\tfrac12 m\lVert\vec v\rVert^2\big) = \vec F\cdot\vec v$, and integrating that produces an object unit 00 never had to name: an integral along the trajectory. Whether it depends on the trajectory is now a genuine question, and the whole of this unit is the machinery for answering it.

Four things get built, in the order the question forces them. The line integral, and the gradient — which arrives with a distinction, made once here and used for the rest of the course, between the derivative of a scalar field and the vector an inner product turns it into. Then the curl, as a local test, with Green's and Stokes' theorems as the bridge between local and global. Then Taylor and the Hessian, so that a potential can be read as a landscape. Then the inverse and implicit function theorems, from which Lagrange multipliers are derived rather than asserted.

How to read this deck

Two fields run through it, and they are designed to look alike: the Newtonian inverse-square field on $\mathbb R^3$ minus a point, and a plane vortex on $\mathbb R^2$ minus a point. Both pass every local test. One is conservative and the other is not, and the thing that separates them is not the field at all.

Proofs are on the slides where they are one line, and in these notes under Full proof where they are not. Results this course does not prove are marked cited with the unit that proves them. Terms with a dotted underline are glossary entries — click one, or press G. N opens these notes; R stacks the deck into one readable page.

04 · overview

Two punctured spaces, five sections

Two fields, each defined on everything but one point, each passing the same local test at every point. One is the force the Earth exerts; the other turns out to be measuring the hole. Every section tests them both.

A loop round a missing point of space lifts over it and shrinks away; round a missing point of the plane it cannot. Section 3 shows that this, and not the field, decides whether a potential exists.

five sections

1 · work. Line integrals, energy, and the gradient.

2 · conservative fields. Three conditions that are one, and two exceptions.

3 · the curl. Green, Stokes, Poincaré: the local test, and what it cannot see.

4 · landscapes. Taylor, the Hessian, stability from its signature.

5 · constraints. Two function theorems, and multipliers from them.

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

Reading the figure. On the left, the loop around the puncture is moved upwards, out of its plane, until it no longer surrounds the missing point, and then shrunk to a point; it never touches the puncture on the way. That move needs a third dimension. On the right there is nowhere to go: any shrinking of the loop must pass through the hole. The word for "every loop can be shrunk to a point inside the set" is simply connected, defined on the Poincaré-lemma slide; the claim that it settles the question is the content of section 3, and the $2\pi$ is computed on the slide Worked: the vortex, in full.

The problem sheet follows the same order: P1–P2 line integrals and the work–energy theorem, P3–P7 conservativeness and the curl, P8–P9 landscapes and stability, P10–P12 constraints and the two function theorems, P13 the two fields that fall outside the frame. Three of them are proofs and four end in a number.

A word on what is not here. The divergence, the Laplacian, flux through a surface and the change-of-variables formula for multiple integrals are unit 05. This unit needs only integrals along curves, plus — twice, and only over a rectangle — an integral done one variable at a time. That is enough for Green's theorem, and it keeps the whole of section 3 within reach of single-variable integration.

section 1

Work: the integral along a path

Where energy comes from when the force depends on where you are.

04 · work

Why an integral along a path appears at all

A particle obeys $m\ddot{\vec r} = \vec F(\vec r)$. Unit 00's move in one dimension was to multiply by $\dot x$. Do the same with $\vec v = \dot{\vec r}$, using the product rule for the dot product (unit 03):

$$\frac{\mathrm d}{\mathrm dt}\Big(\tfrac12 m \lVert\vec v\rVert^2\Big) = m\,\vec v\cdot\dot{\vec v} = \vec F(\vec r)\cdot\vec v .$$

Integrate from $t_0$ to $t_1$:

$$\Delta\Big(\tfrac12 m\lVert\vec v\rVert^2\Big) = \int_{t_0}^{t_1}\vec F(\vec\gamma(t))\cdot\dot{\vec\gamma}(t)\,\mathrm dt .$$

what unit 00 never had to face

In one dimension the right-hand side is $\int_{x_0}^{x_1}F(x)\,\mathrm dx$: an integral over an interval, and the path between the endpoints is the interval. In three dimensions there are infinitely many paths from $\vec r_0$ to $\vec r_1$, and the integral is taken along one of them.

So the object on the right is new, it deserves a name and a theory, and the first question about it is whether it depends on the path.

It sometimes does. Rub your hands together and return them to where they started: the work is not zero.

The product rule used. Unit 03 proved that $t\mapsto \vec u(t)\cdot\vec w(t)$ is differentiable with derivative $\dot{\vec u}\cdot\vec w + \vec u\cdot\dot{\vec w}$ whenever both factors are. With $\vec u = \vec w = \vec v$ that gives $\tfrac{\mathrm d}{\mathrm dt}\lVert\vec v\rVert^2 = 2\,\vec v\cdot\dot{\vec v}$, which is the only line of calculus on this slide.

Why $\vec F(\vec r)$ and not $\vec F(\vec r, \vec v, t)$. The argument above works for any force, and gives the work–energy theorem in full generality. It is the path-independence question that needs $\vec F$ to be a function of position alone, because only then is there an integral along a path to ask the question about. Friction and magnetism are the two standard forces that are not, and they get a slide of their own in section 2.

Historically the quantity $\tfrac12 mv^2$ is Coriolis' 1829 rescaling of Leibniz's vis viva, $mv^2$; the factor $\tfrac12$ exists precisely so that this identity has no constant in it. The word "work" in its technical sense is Coriolis' too, from the same book, and it was coined for machines: force times distance moved along the line of the force.

04 · work

The line integral, defined

definition

Let $\vec F : U \to \mathbb R^n$ be a continuous vector field on an open $U \subseteq \mathbb R^n$, and let $\vec\gamma : [a,b] \to U$ be piecewise $C^1$. The line integral of $\vec F$ along $\vec\gamma$ is

$$\int_{\vec\gamma}\vec F\cdot\mathrm d\vec r \;:=\; \int_a^b \vec F(\vec\gamma(t))\cdot\dot{\vec\gamma}(t)\,\mathrm dt .$$

When $\vec F$ is a force, this number is the work it does.

parametrisation invariance

If $\varphi : [c,d]\to[a,b]$ is $C^1$ with $\varphi' > 0$, $\varphi(c)=a$, $\varphi(d)=b$, then $\int_{\vec\gamma\circ\varphi} = \int_{\vec\gamma}$. Reversing the orientation multiplies the integral by $-1$.

Why: the chain rule gives $(\vec\gamma\circ\varphi)^{\textstyle\cdot}(s) = \dot{\vec\gamma}(\varphi(s))\varphi'(s)$, and substituting $t = \varphi(s)$ in the one-variable integral is exactly the change of variable $\mathrm dt = \varphi'(s)\,\mathrm ds$.

checks

Dimensions. $[\vec F]\,[\dot{\vec\gamma}]\,[t] = \mathrm{N}\cdot\mathrm{m\,s^{-1}}\cdot\mathrm s = \mathrm{N\,m} = \mathrm J$. Work is an energy.

Special case. Constant $\vec F$, straight $\vec\gamma$ from $\vec r_0$ to $\vec r_1$: the integrand is $\vec F\cdot(\vec r_1 - \vec r_0)$, constant, so the integral is $\vec F\cdot(\vec r_1 - \vec r_0)$ — school physics, force times distance times the cosine.

The definition looks like it depends on the parametrisation. The theorem says it depends only on the image and the direction of travel, which is why one may write $\int_C$ for an oriented curve $C$.

Piecewise $C^1$. $\vec\gamma$ is continuous, and $[a,b]$ splits into finitely many subintervals on each of which $\vec\gamma$ is $C^1$ up to the endpoints. The integral is the sum over the pieces. This is exactly the generality needed for the rectangular loops of section 3, which have four corners.

Full proof of parametrisation invariance. Write $\vec\sigma = \vec\gamma\circ\varphi$. Unit 03's chain rule for a composite $[c,d]\to[a,b]\to\mathbb R^n$ gives $\dot{\vec\sigma}(s) = \varphi'(s)\,\dot{\vec\gamma}(\varphi(s))$. Hence $$\int_c^d \vec F(\vec\sigma(s))\cdot\dot{\vec\sigma}(s)\,\mathrm ds = \int_c^d \big[\vec F(\vec\gamma(\varphi(s)))\cdot\dot{\vec\gamma}(\varphi(s))\big]\,\varphi'(s)\,\mathrm ds .$$ The bracket is a continuous function $h$ of $\varphi(s)$, so the one-variable substitution rule turns the right-hand side into $\int_a^b h(t)\,\mathrm dt$, which is the integral along $\vec\gamma$. If instead $\varphi' < 0$ with $\varphi(c) = b$ and $\varphi(d) = a$, the same substitution runs the limits the other way and produces $-\int_a^b h$.

A warning about the other line integral. $\int_{\vec\gamma} f\,\mathrm ds = \int_a^b f(\vec\gamma(t))\lVert\dot{\vec\gamma}(t)\rVert\,\mathrm dt$ integrates a scalar against arc length (unit 03), and it does not change sign when the curve is reversed, because $\lVert\dot{\vec\gamma}\rVert \ge 0$. The two are different objects and only the vector one is the subject of this unit. The sign behaviour is the quickest way to tell which one a book means: unit 34 will explain that the vector one integrates a $1$-form, and that orientation-sensitivity is what integrating a form always looks like.

04 · work · example

Path matters: one field, three answers

$\vec F(x,y) = (y^2,\ x)$, from $(0,0)$ to $(1,1)$

The straight line $\vec\gamma(t) = (t,t)$, $\dot{\vec\gamma} = (1,1)$: $$\int_0^1\big(t^2\cdot 1 + t\cdot 1\big)\mathrm dt = \tfrac13+\tfrac12 = \tfrac56 \approx 0.8333 .$$

The parabola $\vec\gamma(t) = (t,t^2)$, $\dot{\vec\gamma} = (1,2t)$: $$\int_0^1\big(t^4\cdot 1 + t\cdot 2t\big)\mathrm dt = \tfrac15+\tfrac23 = \tfrac{13}{15} \approx 0.8667 .$$

Right, then up: on $y=0$ the integrand is $y^2\dot x = 0$; on $x=1$ it is $x\dot y = 1$. Total $0 + 1 = \mathbf 1$.

Three paths, three numbers. So $\int_{\vec\gamma}\vec F\cdot\mathrm d\vec r$ genuinely is a functional of the path, not of its endpoints.

The field's arrows grow to the right with height ($F_1 = y^2$) and upwards with $x$ ($F_2 = x$). The corner path runs its whole vertical leg at $x = 1$, where $F_2$ is largest, and collects the most.

Reading the figure. Along each path the integrand is the component of the arrow along the direction of travel, times the speed. The corner path gets nothing on its first leg — on the $x$-axis the field $(0, x)$ points straight up, across the direction of travel — and then the full $F_2 = 1$ for a unit distance. The straight line and the parabola sit between those extremes, and the parabola loses a little on the way across (it is lower, so $y^2$ is smaller) and gains more on the way up (it climbs later, where $x$ is larger), so it collects slightly more than the line.

A coincidence worth noticing. The parabola $x = y^2$ — the reflection of the one used — also gives $\tfrac56$, the same as the straight line, even though the field is not conservative. Agreement on two paths proves nothing; disagreement on two paths proves everything. This is why the test in section 3 is a differential condition rather than a search over paths.

04 · work · example

Path matters: the loop between two paths

The loop out along the parabola and back along the line runs anticlockwise round the thin region between them. Section 3's integrand $1 - 2y$ is positive on the lower part and negative on the upper; the lower part is larger, and the net is $+\tfrac1{30}$.

check, and a forward glance

Going out along the parabola and back along the line is a closed loop with $\oint = \tfrac{13}{15}-\tfrac56 = \tfrac{1}{30}$, not zero.

Section 3 will predict that number from $\partial_x F_2 - \partial_y F_1 = 1 - 2y$ without computing a single line integral: integrate it over the region the loop encloses.

Contrast the inverse-square field on the awkward-path slide, where a path $13$ times longer than necessary gives the same answer in every digit computed.

The loop value, checked against section 3. The region between the line $y = x$ and the parabola $y = x^2$ over $0 \le x \le 1$ has $\int_0^1\int_{x^2}^{x}(1-2y)\,\mathrm dy\,\mathrm dx = \int_0^1\big[(x - x^2) - (x^2 - x^4)\big]\mathrm dx = \tfrac12 - \tfrac13 - \tfrac13 + \tfrac15 = \tfrac{1}{30}$. The loop "out along the parabola, back along the line" runs anticlockwise around that region, and Green's theorem says its circulation is $+\tfrac1{30}$, which is what the two line integrals gave.

Reading the figure. The region is thin — at its widest, at $x = \tfrac12$, it is $x - x^2 = \tfrac14$ across — and the dashed line $y = \tfrac12$ cuts off only a small sliver at the top right, where the integrand is negative. That is why the answer is small and positive. Green's theorem, which makes "integrate over the enclosed region" precise, is proved for rectangles in section 3; this region is not a rectangle, and the general case is cited there.

04 · work

The work–energy theorem

theorem

Let $\vec\gamma$ be a $C^2$ solution of $m\ddot{\vec r} = \vec F$ on $[t_0,t_1]$, with $\vec F$ any force whatsoever — position-dependent, velocity-dependent, time-dependent. Writing $T = \tfrac12 m\lVert\vec v\rVert^2$ for the kinetic energy,

$$T(t_1) - T(t_0) \;=\; \int_{\vec\gamma}\vec F\cdot\mathrm d\vec r .$$

proof

$\dot T = m\,\vec v\cdot\dot{\vec v} = \vec v\cdot\vec F$ by the product rule and the equation of motion. Integrate over $[t_0,t_1]$ and apply the fundamental theorem of calculus on the left; the right-hand side is by definition $\int_{\vec\gamma}\vec F\cdot\mathrm d\vec r$ evaluated along the actual trajectory. $\square$

checks

Dimensions. $\mathrm{kg}\,(\mathrm{m\,s^{-1}})^2 = \mathrm{kg\,m^2\,s^{-2}} = \mathrm J$, and the right-hand side was checked to be joules.

Special case. $\vec F$ constant and vertical, $\vec F = -mg\hat z$: $T_1 - T_0 = -mg\,\Delta z$, the schoolroom exchange of kinetic for height.

Special case. $\vec F \perp \vec v$ throughout: $\dot T = 0$, speed constant. Circular motion at constant speed does no work, and neither does a magnetic field.

The theorem's strength and its weakness. It holds for every force, which is why it can be proved in two lines and why it is almost never enough on its own: the right-hand side is an integral along the trajectory, and the trajectory is what one is trying to find. It becomes a genuine tool exactly when the integral can be evaluated without knowing the path — which is the definition of a conservative field, on the fundamental-theorem slide.

Why the trajectory must be $C^2$. $\dot T$ requires $\vec v$ to be differentiable. The hypothesis is free in practice: a twice-differentiable $\vec r$ that satisfies $m\ddot{\vec r} = \vec F(\vec r,\dot{\vec r},t)$ with $\vec F$ continuous has $\ddot{\vec r}$ equal to a continuous function of $t$, so $\vec r\in C^2$ automatically.

On the phrase "work done by the net force". $\vec F$ here is the total force. Splitting it, $\Delta T = W_{\text{gravity}} + W_{\text{normal}} + W_{\text{friction}} + \dots$, and each piece is its own line integral along the same trajectory. That decomposition is the whole of elementary energy bookkeeping, and it is valid because the line integral is linear in $\vec F$.

04 · work

The derivative is a covector; the gradient is what the inner product makes of it

Unit 03 defined the derivative of $f : U \to \mathbb R$ at $a$ as a linear map $Df(a) : \mathbb R^n \to \mathbb R$ — an element of the dual space $(\mathbb R^n)^*$, that is, a covector. Nothing so far makes it a vector.

Riesz isomorphism

On a finite-dimensional real inner product space, $v \mapsto \langle v, \cdot\rangle$ is a linear isomorphism $V \to V^*$.

Why: injective, since $\langle v,v\rangle = 0$ forces $v=0$; and $\dim V = \dim V^*$.

definition

The gradient $\nabla f(a)$ is the preimage of $Df(a)$: the unique vector with $\langle\nabla f(a),\vec h\rangle = Df(a)\vec h$ for all $\vec h$.

A covector drawn as what it is: the parallel level lines of $\vec h \mapsto Df(a)\vec h$, its value on $\vec h$ the number of lines crossed. "Perpendicular to the lines" needs an inner product; the arrow it picks out, of length one over the spacing, is $\nabla f(a)$.

Physically the covector is primary: work per unit displacement is a linear function of displacement. Force is what the Euclidean inner product makes of it.

Reading the figure. The lines are the sets $\{a + \vec h : Df(a)\vec h = k\}$ for $k = 0, 1, 2, \dots$ — the level sets of the linear approximation to $f$ at $a$. A covector needs nothing more to be drawn: its value on a displacement is read off by counting. The gradient is the one displacement that is perpendicular to the lines and whose length is fixed by $\langle\nabla f,\nabla f\rangle = Df(a)\nabla f$; with the lines a unit apart that length is one unit, so the arrow reaches exactly the next line. Change the inner product and the stack stays the same while the arrow moves.

Why bother, at this point in the course. Because in Minkowski space and on a curved manifold the inner product is no longer the identity matrix, and "the gradient" silently means "the derivative, with indices raised by the metric". Every sign error in units 26 to 33 that is not a sign convention is this confusion. Planting it here costs one slide; it is cashed in unit 16, where a tensor is defined as a multilinear map and the transformation law becomes a theorem, and again in unit 28, where raising an index with $\eta$ flips a sign and the distinction stops being optional.

Two consequences worth keeping. $Df(a)\vec h = \langle\nabla f(a),\vec h\rangle$ is largest, over unit $\vec h$, when $\vec h$ points along $\nabla f(a)$ — Cauchy–Schwarz with equality only for parallel vectors — so the gradient is the direction of steepest increase, with $\lVert\nabla f(a)\rVert$ the rate. And $Df(a)\vec h = 0$ for $\vec h$ tangent to the level set $\{f = f(a)\}$, so the gradient is normal to level sets. Both are statements about the inner product, not about the derivative.

04 · work

The gradient in polar coordinates

components

In an orthonormal basis, $\nabla f(a) = \big(\partial_1 f(a),\dots,\partial_n f(a)\big)$.

Why: $\langle\nabla f, e_i\rangle = Df(a)e_i = \partial_i f(a)$, and there the $i$-th coordinate is the inner product with $e_i$.

the distinction is not pedantry

The components of $\nabla f$ are the partial derivatives only in an orthonormal basis. In polar coordinates $(r,\theta)$, with the orthonormal pair $\hat r = (\cos\theta,\sin\theta)$ along the ray and $\hat\theta = (-\sin\theta,\cos\theta)$ across it, $$\nabla f = \partial_r f\,\hat r + \tfrac1r\partial_\theta f\,\hat\theta ,$$ and that $1/r$ is the inner product showing through. The distinction becomes a definition in unit 16 and an operation with a name — raising an index — in unit 26.

Changing $r$ by $\mathrm dr$ moves the point $\mathrm dr$ metres; changing $\theta$ by $\mathrm d\theta$ moves it $r\,\mathrm d\theta$. So $\partial_\theta f$ is a rate per radian, and per metre along $\hat\theta$ it is $\tfrac1r\partial_\theta f$.

The polar formula, derived. With $x = r\cos\theta$, $y = r\sin\theta$, the chain rule gives $\partial_r f = \partial_x f\cos\theta + \partial_y f\sin\theta = \nabla f\cdot\hat r$ and $\partial_\theta f = r(-\partial_x f\sin\theta + \partial_y f\cos\theta) = r\,\nabla f\cdot\hat\theta$, since $\hat r = (\cos\theta,\sin\theta)$ and $\hat\theta = (-\sin\theta,\cos\theta)$. The pair $(\hat r, \hat\theta)$ is orthonormal, so $\nabla f = (\nabla f\cdot\hat r)\hat r + (\nabla f\cdot\hat\theta)\hat\theta = \partial_r f\,\hat r + \tfrac1r\partial_\theta f\,\hat\theta$. The $1/r$ is there because $\theta$ is an angle and $\partial_\theta$ measures change per radian, not per metre.

The figure is the derivation. The two partial derivatives $\partial_r f$ and $\partial_\theta f$ are the covector $Df$ evaluated on the two coordinate displacements $\partial\vec r/\partial r = \hat r$ and $\partial\vec r/\partial\theta = r\,\hat\theta$. The first has unit length and the second has length $r$ — the side $r\,\mathrm d\theta$ of the shaded cell. The gradient's components in the orthonormal pair are $Df$ on unit vectors, so the second must be divided by $r$. A basis that is orthogonal but not normalised is the simplest case of the general rule: in the coordinate basis $(\partial\vec r/\partial r, \partial\vec r/\partial\theta)$, the components of the gradient are the partials multiplied by the inverse of the matrix of inner products of those basis vectors, here $\mathrm{diag}(1, r^2)^{-1}$ — giving $\partial_\theta f/r^2$ on a basis vector of length $r$, which is $\tfrac1r\partial_\theta f$ on $\hat\theta$.

04 · work

The fundamental theorem of line integrals

theorem

For $V : U \to \mathbb R$ of class $C^1$ on an open $U \subseteq \mathbb R^n$ and $\vec\gamma : [a,b] \to U$ piecewise $C^1$,

$$\int_{\vec\gamma}\nabla V\cdot\mathrm d\vec r \;=\; V(\vec\gamma(b)) - V(\vec\gamma(a)) .$$

proof

On each $C^1$ piece the chain rule of unit 03 gives $\frac{\mathrm d}{\mathrm dt}V(\vec\gamma(t)) = \nabla V(\vec\gamma(t))\cdot\dot{\vec\gamma}(t)$, which is the integrand. The fundamental theorem of calculus turns each piece into a difference of endpoint values, and the sum telescopes because $\vec\gamma$ is continuous at the joins. $\square$

Checks. $\nabla V$ is $\mathrm{J/m} = \mathrm N$, so both sides are joules. With $V = -\vec F_0\cdot\vec r$ the theorem returns $\vec F_0\cdot(\vec r_1-\vec r_0)$, the constant-force answer.

definition

$\vec F$ is conservative on $U$ if $\vec F = -\nabla V$ for some $C^1$ function $V : U \to \mathbb R$, the potential energy. Then $\int_{\vec\gamma}\vec F\cdot\mathrm d\vec r = V(\vec\gamma(a)) - V(\vec\gamma(b))$.

the minus sign

Mathematics writes $\vec F = \nabla\phi$; physics writes $\vec F = -\nabla V$, so that a ball rolls downhill in $V$ and $T+V$ is conserved. This course uses the physics sign, and $V$ always means potential energy, in joules.

What the theorem is. It is the fundamental theorem of calculus, with the interval replaced by a curve and the antiderivative replaced by $V$. Unit 34 makes that literal: both are $\int_{\partial\Omega}\omega = \int_\Omega \mathrm d\omega$ for a $0$-form on a $1$-chain, and Green's and Stokes' theorems are the next two cases. The pattern is worth noticing now, because it is the shape of every integral theorem in this course.

Where $V$ is not unique. On a connected $U$, two potentials for the same field differ by a constant: their difference has zero derivative, hence zero directional derivative along every path, hence is constant along every path, and any two points of a connected open set are joined by a path. If $U$ is disconnected, the constant may differ on each component. Physics fixes the constant by a convention — for the inverse-square field, $V \to 0$ at infinity — and only differences of $V$ ever appear in an answer.

Piecewise $C^1$ and the telescoping. With $a = t_0 < t_1 < \dots < t_k = b$ splitting $[a,b]$ into $C^1$ pieces, the sum is $\sum_{i}\big[V(\vec\gamma(t_i)) - V(\vec\gamma(t_{i-1}))\big]$, which collapses to $V(\vec\gamma(b)) - V(\vec\gamma(a))$ because the same value $V(\vec\gamma(t_i))$ appears twice with opposite signs. Continuity of $\vec\gamma$ at the corners is what makes it the same value.

04 · work · worked example

Worked: the inverse-square field has a potential

$\vec F_{\mathrm N}(\vec r) = -GMm\,\vec r/r^3$ on $U = \mathbb R^3\setminus\{0\}$, where $r = \lVert\vec r\rVert$. Claim: $\vec F_{\mathrm N} = -\nabla V$ with $V(\vec r) = -GMm/r$.

worked: $\vec F_{\mathrm N} = -\nabla V$

Step 1 — differentiate $r$. $r = (x^2+y^2+z^2)^{1/2}$ is $C^\infty$ on $U$, and $\partial_x r = \tfrac12(x^2+y^2+z^2)^{-1/2}\cdot 2x = x/r$, likewise for $y, z$. So $\nabla r = \vec r/r = \hat r$, a unit vector, as it must be: $r$ increases at one metre per metre radially.

Step 2 — chain rule. $V = -GMm\,r^{-1}$, so by the one-variable chain rule applied to $\rho\mapsto -GMm/\rho$ composed with $r$, $$\nabla V = GMm\,r^{-2}\,\nabla r = \frac{GMm}{r^2}\hat r .$$

Step 3 — read off. $-\nabla V = -\dfrac{GMm}{r^2}\hat r = \vec F_{\mathrm N}$. $\square$

checks

Dimensions. $[GMm/r] = \mathrm{m^3\,s^{-2}}\cdot\mathrm{kg}/\mathrm m = \mathrm{kg\,m^2\,s^{-2}} = \mathrm J$ ✓, and $GMm/r^2$ is newtons ✓.

Sign. $V < 0$ everywhere and increases towards $0$ as $r\to\infty$, so the force $-\nabla V$ points towards smaller $r$: gravity attracts ✓.

Limit. Near the surface, $V(R+h) - V(R) \approx \tfrac{GMm}{R^2}h = mgh$ for $h \ll R$ ✓.

so, by the previous theorem

$$\int_{\vec\gamma}\vec F_{\mathrm N}\cdot\mathrm d\vec r = GMm\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$ for every path in $U$ from radius $r_1$ to radius $r_2$. The answer knows nothing but the two radii.

The proof without the potential, for contrast. One can see path-independence directly. Along any $\vec\gamma$, $$\vec F_{\mathrm N}\cdot\dot{\vec\gamma} = -GMm\frac{\vec\gamma\cdot\dot{\vec\gamma}}{\lVert\vec\gamma\rVert^{3}} = GMm\,\frac{\mathrm d}{\mathrm dt}\frac{1}{\lVert\vec\gamma\rVert},$$ because $\frac{\mathrm d}{\mathrm dt}\lVert\vec\gamma\rVert^{-1} = -\lVert\vec\gamma\rVert^{-2}\frac{\mathrm d}{\mathrm dt}\lVert\vec\gamma\rVert = -\lVert\vec\gamma\rVert^{-3}(\vec\gamma\cdot\dot{\vec\gamma})$, using unit 03's product rule for the dot product. The integrand is a total derivative, so the integral depends only on the endpoints. Finding the potential is the systematic version of spotting that.

Why $U$ must exclude the origin. $V$ and $\vec F_{\mathrm N}$ are both unbounded there, and no continuous extension exists. Everything in this unit is a statement about fields on an open set, and which open set is the whole point of section 3. Note that $\mathbb R^3$ minus a point is still simply connected — a loop can be slid off the puncture through the third dimension — which is why this field behaves and the plane vortex does not.

The $mgh$ limit, to one more order. $GMm(\tfrac1R - \tfrac1{R+h}) = \tfrac{GMm}{R}\cdot\tfrac{h/R}{1+h/R} = mgh\big(1 - h/R + O(h^2/R^2)\big)$ with $g = GM/R^2$. At $h = 10$ km the correction is $h/R = 1.6\times10^{-3}$, which is the level at which "uniform gravity" starts to cost you accuracy.

04 · work · worked example

Worked: the same answer on a deliberately awkward path

A $1200$ kg spacecraft goes from $r_1 = 6.771\times10^{6}$ m (a $400$ km orbit) to $r_2 = 4.2164\times10^{7}$ m (geostationary). The work done by gravity, from the theorem:

$$W = GMm\Big(\tfrac{1}{r_2}-\tfrac{1}{r_1}\Big) = -5.9298\times10^{10}\ \mathrm J .$$

Now take a path designed to be as unhelpful as possible: three full turns about the axis, a radial wobble of $\pm 3000$ km, and an out-of-plane wobble of $\pm 5000$ km, ending at the same radius.

straight radial distance	$35\,393$ km
length of the awkward path	$465\,385$ km
ratio	$13.15$
$\int_{\vec\gamma}\vec F_{\mathrm N}\cdot\mathrm d\vec r$, quadrature	$-5.929823671\times10^{10}$ J
$GMm(1/r_2-1/r_1)$	$-5.929823671\times10^{10}$ J

They agree in every digit shown; the difference is a few parts in $10^{16}$, the rounding of double-precision arithmetic.

The path seen from above the pole (its $\pm5000$ km out-of-plane wobble does not show). Both routes run from the same point on a $400$ km orbit to the same point at geostationary radius.

The path, exactly. $\vec\gamma(t) = (\rho\cos 6\pi t,\ \rho\sin 6\pi t,\ 5\times10^6\sin 3\pi t)$ metres for $t\in[0,1]$, with $\rho(t) = r_1 + (r_2-r_1)t + 3\times10^6\sin 4\pi t$. At $t = 0$ and $t = 1$ both wobbles vanish, so the endpoints sit at exactly $r_1$ and $r_2$; $\lVert\vec\gamma\rVert$ stays between those two values throughout, so the path never leaves $U$ and never approaches the singularity. The integral was evaluated by adaptive quadrature on $\int_0^1 \vec F_{\mathrm N}(\vec\gamma)\cdot\dot{\vec\gamma}\,\mathrm dt$ and came out as $-5.9298236712\times10^{10}$ J; the closed form gives the same to a relative difference of about $10^{-16}$, which is the rounding floor of the arithmetic rather than anything physical. The path length, $\int_0^1\lVert\dot{\vec\gamma}\rVert\,\mathrm dt$, was computed the same way.

The numbers. $GM = 3.986004418\times10^{14}\ \mathrm{m^3\,s^{-2}}$ is the Earth's geocentric gravitational constant, known far more precisely than $G$ and $M$ separately — a fact that recurs whenever a spacecraft trajectory is computed. $r_1 = 6.771\times10^6$ m is the Earth's mean radius, $6371$ km, plus $400$ km; $r_2 = 42\,164$ km is the radius at which the orbital period is one sidereal day.

04 · work · worked example

The awkward path: the work along the way

checks

Sign. Negative: the craft climbs, so gravity takes energy from it ✓.

Dimensions. $\mathrm{m^3s^{-2}}\cdot\mathrm{kg}\cdot\mathrm{m^{-1}} = \mathrm J$ ✓.

Limit $r_2\to\infty$. $W \to -GMm/r_1 = -7.0643\times10^{10}$ J, the energy needed to escape from $r_1$. Finite — which is why escape is possible at all ✓.

what the agreement is not

It is not evidence. One path agreeing with the formula, or a thousand, proves nothing: the fundamental theorem's proof is what makes it certain. The computation is here to show what a theorem buys — a $465\,000$ km integral replaced by two divisions.

Gravity's work so far, $\int_0^t \vec F_{\mathrm N}\cdot\dot{\vec\gamma}\,\mathrm dt'$, on both routes. In between they disagree, because at the same $t$ the two craft are at different radii; the running total is always $GMm(1/r - 1/r_1)$ at the current radius $r$, and at $t = 1$ both are at $r_2$.

Reading the figure. Both curves are computed from the parametrisations, the straight one as $\vec r(t) = \big(r_1 + (r_2-r_1)t\big)\hat x$. The awkward curve overshoots early — its radial wobble $3\times10^{6}\sin 4\pi t$ carries it outwards faster at first — and lags later, but at every instant it sits at the value the potential assigns to the craft's current distance from the centre. That is what having a potential means: the work done so far is a function of where you are, not of how you got there.

Where the energy actually comes from. $-5.9298\times10^{10}$ J is the work done by gravity alone. The craft also slows down: the circular speed falls from $7672.6$ to $3074.7\ \mathrm{m\,s^{-1}}$, so $\Delta T = \tfrac12(1200)(3074.7^2 - 7672.6^2) = -2.965\times10^{10}$ J. By the work–energy theorem the engines supply $\Delta T - W_{\text{gravity}} = -2.965 + 5.930 = 2.965\times10^{10}$ J — exactly half the work done against gravity, which is the virial relation $T = -U/2$ for circular orbits. Unit 08 turns this into the Hohmann transfer and its two burns.

section 2

Conservative fields, and the first integral they give

Three conditions that are one condition, and two forces that escape it.

04 · conservative

Three conditions, one property

theorem

For a continuous $\vec F$ on a connected open $U \subseteq \mathbb R^n$, these are equivalent:

(i) $\vec F = -\nabla V$ for some $C^1$ function $V$ on $U$;

(ii) $\int_{\vec\gamma}\vec F\cdot\mathrm d\vec r$ depends only on the endpoints of $\vec\gamma$ — path-independence;

(iii) $\oint_{\vec\gamma}\vec F\cdot\mathrm d\vec r = 0$ for every closed $\vec\gamma$ in $U$ — zero circulation.

proof

(i)$\Rightarrow$(ii) is the fundamental theorem of line integrals. (ii)$\Leftrightarrow$(iii): a closed path is one whose endpoints coincide, so its integral equals that of the constant path, namely $0$; conversely, two paths with common endpoints differ by a closed path, run one forwards and one backwards, and reversal flips the sign.

(ii)$\Rightarrow$(i) is the one with content: fix $\vec p \in U$ and set $V(\vec x) = -\int_{\vec p}^{\vec x}\vec F\cdot\mathrm d\vec r$ along any path. Well defined by (ii); $\nabla V = -\vec F$ by taking the path to end with a short straight segment. full proof in the notes (N)

none of the three is checkable

(ii) and (iii) quantify over infinitely many paths. (i) asks for an antiderivative to be produced out of nowhere. What is wanted is a local, differential test — and section 3 builds one, at the cost of a hypothesis on $U$.

The vortex will satisfy the local test and fail (iii), which is exactly how much the hypothesis matters.

Full proof of (ii) $\Rightarrow$ (i)

$U$ is open and connected, hence path-connected by polygonal paths: the set of points reachable from $\vec p$ by a polygonal path in $U$ is open (around each reachable point sits a ball, and every point of a ball is reachable from its centre by a segment) and its complement in $U$ is open by the same argument, so by connectedness it is all of $U$. Define $V(\vec x) = -\int_{\vec p}^{\vec x}\vec F\cdot\mathrm d\vec r$ along any such path; by (ii) the value does not depend on which.

Fix $\vec x \in U$ and $\varepsilon > 0$ with $B(\vec x,\varepsilon) \subseteq U$. For $\lvert s\rvert < \varepsilon$, reach $\vec x + s\vec e_i$ by any path to $\vec x$ followed by the straight segment $u \mapsto \vec x + u\vec e_i$, $u \in [0,s]$. Then $$V(\vec x + s\vec e_i) - V(\vec x) = -\int_0^s \vec F(\vec x + u\vec e_i)\cdot\vec e_i\,\mathrm du = -\int_0^s F_i(\vec x + u\vec e_i)\,\mathrm du .$$ The integrand is continuous in $u$, so by the fundamental theorem of calculus the right-hand side is differentiable in $s$ with derivative $-F_i(\vec x + s\vec e_i)$; at $s = 0$ this gives $\partial_i V(\vec x) = -F_i(\vec x)$. All partials exist and equal the continuous functions $-F_i$, so $V$ is $C^1$ (unit 03) and $\nabla V = -\vec F$. $\square$

Connectedness is needed only for (ii) $\Rightarrow$ (i), and only to have a single base point. On a disconnected $U$, choose a base point per component; the resulting $V$ is still $C^1$ with $\nabla V = -\vec F$, and the arbitrary constants are independent on each piece.

Continuity of $\vec F$ is needed. Without it the fundamental theorem of calculus step fails, and a field can be path-independent along every piecewise-$C^1$ path while having no differentiable potential.

04 · conservative

Energy is a first integral in three dimensions

theorem

If $m\ddot{\vec r} = \vec F(\vec r) = -\nabla V(\vec r)$ on an open $U$, then $E := \tfrac12 m\lVert\vec v\rVert^2 + V(\vec r)$ is constant along every solution.

Proof. $\dot E = m\vec v\cdot\dot{\vec v} + DV(\vec r)\dot{\vec r} = \vec F\cdot\vec v + \nabla V\cdot\vec v = (-\nabla V + \nabla V)\cdot\vec v = 0$, using the chain rule of unit 03. $\square$

This is unit 00's trick, and no longer a trick: in one dimension every continuous $F(x)$ has an antiderivative, so every one-dimensional autonomous force is conservative. In three dimensions that is false, and the theorem has a hypothesis.

Six unknowns $(\vec r,\vec v)$; one conserved quantity confines the motion to a level set of $E$ in the six-dimensional space of states $(\vec r,\vec v)$, generically five-dimensional. Unit 08 finds three more and solves the orbit completely.

escape speed, from one equation

With $V = -GMm/r$, a body just escaping has $E = 0$ (it arrives at infinity with nothing left). At the surface,

$$\tfrac12 mv_{\mathrm{esc}}^2 = \frac{GMm}{R} \;\Longrightarrow\; v_{\mathrm{esc}} = \sqrt{\frac{2GM}{R}} .$$

$= \sqrt{2(3.986004418\times10^{14})/6.378137\times10^{6}} = 11\,180\ \mathrm{m\,s^{-1}}$.

checks

Dimensions. $\sqrt{\mathrm{m^3s^{-2}}/\mathrm m} = \mathrm{m\,s^{-1}}$ ✓. Independent of $m$, as it must be: $m$ cancels from the equation of motion.

Against the world. $11.18\ \mathrm{km\,s^{-1}}$ is the quoted figure. A real launch needs more, because drag and the climb through the atmosphere are not in this model.

Why $E = 0$ is the escape condition. $E$ is conserved, and $V\to 0$ at infinity. If $E < 0$ then $\tfrac12 mv^2 = E - V \to E < 0$ as $r\to\infty$, which is impossible, so $r$ is bounded: the body is trapped. If $E > 0$ the body arrives at infinity with speed $\sqrt{2E/m}$ to spare. $E = 0$ is the boundary, and the minimum. Unit 08 turns the sign of $E$ into the conic section the orbit is: ellipse, parabola, hyperbola.

The spin of the Earth helps. The equatorial surface moves east at $\omega_{\mathrm E}R_{\mathrm E} = (7.292115\times10^{-5})(6.378137\times10^6) = 465.1\ \mathrm{m\,s^{-1}}$. Launching east, the rocket needs to supply only $11\,179.9 - 465.1 = 10\,714.8\ \mathrm{m\,s^{-1}}$, a saving of $4.16\%$ in speed but $8.15\%$ in energy, since energy goes as $v^2$. That is why Kourou, at $5.2^\circ$ N, is worth an ocean crossing.

The name "first integral". Unit 00 defined it: a function constant along solutions, which lowers the order of the system by one. $E$ is the first of this course's four for the central-force problem; unit 08 adds the three components of angular momentum and finds that two conservation laws suffice. Unit 13 proves that every one of them is a symmetry in disguise.

04 · conservative

Two forces outside the frame, for opposite reasons

friction: a function of the path, not of position

Sliding friction is $\vec F = -\mu N\,\hat v$, with $N$ the magnitude of the normal force pressing the surfaces together, $\mu$ the dimensionless coefficient of friction and $\hat v = \vec v/\lVert\vec v\rVert$. It depends on the direction of motion, not on $\vec r$. So there is no vector field to integrate, and no line integral to be path-independent.

It is also dissipative: on any closed path, $\vec F\cdot\dot{\vec\gamma} = -\mu N\lVert\dot{\vec\gamma}\rVert \le 0$, so $$\oint \vec F\cdot\mathrm d\vec r = -\mu N \times (\text{path length}) < 0 .$$ The loop integral is not merely nonzero — it has a sign, always the same one.

magnetism: velocity-dependent, and does no work

The magnetic part of the Lorentz force is $\vec F = q\,\vec v\times\vec B(\vec r)$ — cited; unit 24 derives it. It depends on $\vec v$, so again there is no field of the kind this unit integrates.

But $\vec F\cdot\vec v = q(\vec v\times\vec B)\cdot\vec v = 0$, since $\vec v\times\vec B \perp \vec v$. By the work–energy theorem the speed is constant, and no energy is exchanged at all.

the moral

Friction is not conservative because it takes energy out. Magnetism is not conservative because it is not that kind of object — and it takes nothing out. Both are velocity-dependent; only one is dissipative.

The same pair, seen again. In unit 03's rotating frame the two fictitious forces behave exactly like this pair. The centrifugal term $-m\vec\omega\times(\vec\omega\times\vec r)$ depends on position only, and the hoop example of section 5 shows it is the gradient of $-\tfrac12 m\omega^2\rho^2$ with $\rho$ the distance from the axis. The Coriolis term $-2m\vec\omega\times\vec v$ depends on velocity and is perpendicular to it, so it does no work — structurally the magnetic force: $-2m\vec\omega\times\vec v = \vec v\times(2m\vec\omega)$, which is $q\vec v\times\vec B$ with $q\vec B$ replaced by $2m\vec\omega$ (problem P13). Unit 30 makes something of the resemblance and unit 38 makes more.

Why "dissipative" is the right word. With friction present, $\tfrac{\mathrm d}{\mathrm dt}(T + V) = \vec F_{\text{fric}}\cdot\vec v \le 0$: the mechanical energy is non-increasing. A function that cannot increase along any motion is what stability theory calls a Lyapunov function, and it is why a damped system settles into a minimum of $V$. Turning that sentence into a theorem — asymptotic stability, with LaSalle's invariance principle for the case where the derivative vanishes on a set rather than at a point — is unit 02's business; this unit needs only the inequality.

Where the energy goes. Nowhere mysterious, but nowhere this course follows: into the internal degrees of freedom of the two surfaces. The total energy of every microscopic constituent is still conserved, because every fundamental force in the Standard Model is conservative or magnetic; "dissipation" is a statement about which variables one has chosen to keep track of. This course never opens that box.

section 3

The curl, and how much a local test can promise

A condition you can check at a point, a theorem converting it into a statement about loops, and the exact size of the gap between them.

04 · curl

A necessary condition, free from Clairaut

theorem

If $\vec F = -\nabla V$ with $V \in C^2$ on an open $U$, then

$$\partial_j F_i = \partial_i F_j \qquad\text{for all } i,j \text{ on } U .$$

proof

$\partial_j F_i = -\partial_j\partial_i V = -\partial_i\partial_j V = \partial_i F_j$, the middle step being Clairaut's theorem of unit 03, whose hypothesis — continuous second partials — is $V \in C^2$. $\square$

A field satisfying it is called closed, or irrotational. Every conservative $C^1$ field is closed.

the test, run twice

$\vec F = (y^2, x)$: $\partial_y F_1 = 2y$, $\partial_x F_2 = 1$. Unequal wherever $y \ne \tfrac12$, so not conservative on any open set meeting $\{y\ne\tfrac12\}$. One differentiation, no paths.

The vortex, $\vec F_{\mathrm V} = (-y, x)/(x^2+y^2)$: with $u = x^2+y^2$, $$\partial_x F_2 = \frac{u - 2x^2}{u^2} = \frac{y^2-x^2}{u^2} = \partial_y F_1 .$$ Equal at every point of $U$. The test passes, with no exceptions.

And yet the vortex is not conservative. The converse of the theorem is false, and the rest of this section is about exactly how false.

The vortex derivatives, in full. With $u = x^2 + y^2$, $F_1 = -y\,u^{-1}$ and $F_2 = x\,u^{-1}$. $$\partial_x F_2 = u^{-1} + x\cdot(-u^{-2})(2x) = \frac{u - 2x^2}{u^2} = \frac{y^2-x^2}{u^2},$$ $$\partial_y F_1 = -u^{-1} - y\cdot(-u^{-2})(2y) = \frac{-u + 2y^2}{u^2} = \frac{y^2-x^2}{u^2}.$$ They are equal at every point of $U = \mathbb R^2\setminus\{0\}$, with no exceptions and no limiting cases. (The display on the slide is compressed; these two lines are the computation.)

Why the converse fails, in one sentence. Closedness is a statement about derivatives at each point; conservativeness is a statement about loops. Getting from one to the other requires filling in the inside of a loop, and whether a loop has an inside that lies in $U$ is a property of $U$.

The $C^2$ hypothesis is not free. Clairaut needs continuous mixed second partials, and unit 03's Peano counterexample $xy(x^2-y^2)/(x^2+y^2)$ shows what happens without them. A $C^1$ field with a merely differentiable potential can, in principle, fail the symmetry. Every field in this course is $C^\infty$ where it is defined, so the hypothesis never bites — but it is the reason the theorem says $C^2$.

04 · curl

What the curl is

$D\vec F(\vec a)$ is a linear map $\mathbb R^3 \to \mathbb R^3$, and every linear map splits uniquely into symmetric and antisymmetric parts, $$D\vec F = \underbrace{\tfrac12\big(D\vec F + D\vec F^{\!\top}\big)}_{\text{symmetric}} + \underbrace{\tfrac12\big(D\vec F - D\vec F^{\!\top}\big)}_{\in\ \mathfrak{so}(3)},$$ while unit 03's hat map identifies $\mathfrak{so}(3)$ with $\mathbb R^3$.

definition

The curl of a $C^1$ field $\vec F$ on $U\subseteq\mathbb R^3$ is twice the vector the hat map assigns to the antisymmetric part of $D\vec F$:

$$\widehat{\tfrac12\,\mathrm{curl}\,\vec F} \;=\; \tfrac12\big(D\vec F - D\vec F^{\!\top}\big).$$

components

$\mathrm{curl}\,\vec F = (\partial_2 F_3 - \partial_3 F_2,\ \partial_3 F_1 - \partial_1 F_3,\ \partial_1 F_2 - \partial_2 F_1)$, and it vanishes exactly when $\vec F$ is closed.

why not the determinant

The familiar determinant with $\hat\imath,\hat\jmath,\hat k$ in its first row is a mnemonic: no matrix has a mixture of vectors and operators for entries. The split into symmetric and antisymmetric parts is what is actually going on, and the next slide draws the two parts.

Full proof of the components formula. Unit 03 fixed the hat map by $\hat\omega\,\vec u = \vec\omega\times\vec u$, whose matrix is $\hat\omega = \begin{pmatrix}0&-\omega_3&\omega_2\\ \omega_3&0&-\omega_1\\ -\omega_2&\omega_1&0\end{pmatrix}$. The antisymmetric part $A$ of $D\vec F$ has entries $A_{ij} = \tfrac12(\partial_j F_i - \partial_i F_j)$. Set $A = \hat\omega$ and read off the three independent entries: $A_{32} = \omega_1$ gives $\omega_1 = \tfrac12(\partial_2 F_3 - \partial_3 F_2)$, $A_{13} = \omega_2$ gives $\omega_2 = \tfrac12(\partial_3 F_1 - \partial_1 F_3)$, and $A_{21} = \omega_3$ gives $\omega_3 = \tfrac12(\partial_1 F_2 - \partial_2 F_1)$. Multiplying by $2$ gives the stated formula. The "if and only if" is immediate: $A = 0$ exactly when all three differences vanish, which is the closedness condition for $n = 3$.

Why the definition is the honest one. The familiar determinant mnemonic with $\hat\imath,\hat\jmath,\hat k$ in the first row is a memory aid and nothing more — there is no determinant of a matrix whose entries are a mixture of vectors and operators. What is actually going on is the decomposition above, and it explains three otherwise unrelated facts at once: why the curl exists only in three dimensions (because $\dim\mathfrak{so}(n) = n$ only for $n = 3$, unit 03), why it is twice an angular velocity, and why it changes sign under a reflection.

Two dimensions. For $\vec F = (F_1, F_2)$ on a planar domain, the single number $\partial_1 F_2 - \partial_2 F_1$ is the third component of the curl of $(F_1, F_2, 0)$, and Green's theorem three slides on is about exactly that number.

04 · curl

Turning and straining: the two parts of the derivative

the motivating case

A body rotating rigidly at $\vec\omega$ has velocity field $\vec F(\vec r) = \vec\omega\times\vec r = \hat\omega\,\vec r$. That map is its own derivative, and it is antisymmetric, so $$\mathrm{curl}(\vec\omega\times\vec r) = 2\vec\omega .$$ The curl is twice the local angular velocity, and the factor $2$ is in the definition to make it so.

checks

Dimensions. $[\vec F]/[\mathrm L]$: for a velocity field, $\mathrm{s^{-1}}$, a rate of turn ✓. Limit. Constant $\vec F$ gives $D\vec F = 0$ and curl $\vec 0$: a uniform flow does not spin a paddle wheel ✓.

Physics: a paddle wheel dropped into a fluid at $\vec a$ spins at $\tfrac12\,\mathrm{curl}\,\vec F(\vec a)$. The symmetric part strains; the antisymmetric part turns.

The two parts of $D\vec F$, each drawn as a field on its own. A rigid rotation is all antisymmetric part, and a paddle wheel in it turns at $\omega$ wherever it is put. A symmetric field turns nothing: it stretches a square into a rectangle along its eigenvectors.

Reading the figure. On the left, $\vec F = \vec\omega\times\vec r$ with $\vec\omega$ out of the page: $D\vec F = \hat\omega$ is antisymmetric, the curl is $2\vec\omega$, and the paddle wheel, which follows the local average turning of the fluid, spins at $\omega = \tfrac12\lVert\mathrm{curl}\,\vec F\rVert$. On the right, $\vec F = (x,-y)$ has $D\vec F = \mathrm{diag}(1,-1)$, symmetric, with curl zero: the flow carries a small square into a rectangle stretched along $x$ and squeezed along $y$ (the dashed outline, schematic), and does not rotate it. A general linear field is a sum of one of each.

The symmetric part is not wasted. It is the rate-of-strain tensor of fluid mechanics, and unit 22 uses it: the stress in a Newtonian fluid is proportional to it, and it must be the symmetric part, because a fluid rotating rigidly is not being strained and must feel no viscous stress.

04 · curl

Curl is a vector only under rotations

equivariance

Let $R \in SO(3)$ and let $\vec G(\vec x) = R\,\vec F(R^{\!\top}\vec x)$ be the field $\vec F$ carried along by the rotation. Then

$$\mathrm{curl}\,\vec G(\vec x) = R\,\big(\mathrm{curl}\,\vec F\big)(R^{\!\top}\vec x).$$

proof

The chain rule gives $D\vec G(\vec x) = R\,D\vec F(R^{\!\top}\vec x)\,R^{\!\top}$, and conjugation by an orthogonal matrix preserves the symmetric–antisymmetric split, since $(RAR^{\!\top})^{\!\top} = RA^{\!\top}R^{\!\top}$. It remains to note $R\,\hat\omega\,R^{\!\top} = \widehat{R\vec\omega}$, which holds because $R(\vec\omega\times\vec u) = (R\vec\omega)\times(R\vec u)$ for $\det R = +1$. $\square$

but not under reflections

Take $R = -I$, which has $\det R = -1$. Then $\vec F \mapsto \vec G$ reverses every vector, $D\vec G(\vec x) = D\vec F(-\vec x)$ — two sign changes cancelling — and so $\mathrm{curl}\,\vec G(\vec x) = +(\mathrm{curl}\,\vec F)(-\vec x)$, whereas a genuine vector field would pick up a minus sign.

The curl is a pseudovector: it transforms correctly under $SO(3)$ and wrongly under $O(3)$.

It is not really a vector at all. Unit 34 identifies what it really is — a $2$-form — at which point the anomaly disappears, because a $2$-form has no reason to transform like a vector. The same footnote applies to the cross product, angular velocity and angular momentum.

Why $R(\vec\omega\times\vec u) = (R\vec\omega)\times(R\vec u)$ needs $\det R = +1$. The triple product $\vec a\cdot(\vec b\times\vec c)$ is the determinant of the matrix with those columns, so it is multiplied by $\det R$ when all three are rotated. Hence $(R\vec a)\cdot\big((R\vec b)\times(R\vec c)\big) = \det R\;\vec a\cdot(\vec b\times\vec c) = \det R\;(R\vec a)\cdot R(\vec b\times\vec c)$ for every $\vec a$, which forces $(R\vec b)\times(R\vec c) = \det R\;R(\vec b\times\vec c)$.

The physical content. A mirror-image experiment has a mirror-image outcome, and any quantity built from a cross product will appear in the mirror with an extra sign. This is harmless as long as pseudovectors are only ever compared with pseudovectors — which is why, in every correct formula of classical physics, the two sides have the same parity. It stops being harmless in 1956, when the weak interaction turned out not to respect it; but that is outside this course.

Equivariance is why Stokes' theorem, two slides on, can be proved for one rectangle. The $xy$-plane case is Green's theorem; equivariance then gives it for a rectangle in any plane, because both sides of the identity transform the same way under $SO(3)$.

04 · curl

Green's theorem in the plane

Green's theorem, for a rectangle

Let $\vec F = (P,Q)$ be $C^1$ on an open set containing the closed rectangle $R = [a,b]\times[c,d]$, and let $\partial R$ be its boundary traversed anticlockwise. Then

$$\oint_{\partial R} P\,\mathrm dx + Q\,\mathrm dy \;=\; \int_a^b\!\!\int_c^d\Big(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Big)\mathrm dy\,\mathrm dx .$$

proof

The two horizontal sides contribute $\int_a^b P(x,c)\,\mathrm dx - \int_a^b P(x,d)\,\mathrm dx = -\int_a^b\!\big[P(x,d)-P(x,c)\big]\mathrm dx = -\int_a^b\!\int_c^d \partial_y P\,\mathrm dy\,\mathrm dx$, by the fundamental theorem of calculus in $y$ at each fixed $x$; the vertical sides contribute $+\int_c^d\!\int_a^b\partial_x Q\,\mathrm dx\,\mathrm dy$ the same way. Add. $\square$

Patching: cut a rectangle into four and run each piece anticlockwise. Every interior edge is run twice, in opposite directions, and cancels; what is left is the outer boundary, once.

So the theorem holds for any finite union of rectangles, hence — by a limiting argument — for regions with piecewise-$C^1$ boundaries. Cited at that level of generality; unit 34 proves it once, for all dimensions at once.

About the iterated integral. The right-hand side is written as an integral in $y$ followed by an integral in $x$, and the proof produces the $Q$ term in the opposite order. That the two orders agree is Fubini's theorem, which unit 05 proves in general; the case needed here — a continuous integrand on a rectangle — is short enough to prove now, and it is the only place in this unit where more than single-variable integration is used. Everything else in section 3 is fundamental-theorem-of-calculus in one variable at a time.

Full proof: the two orders agree for continuous $h$ on a rectangle

$h$ is uniformly continuous on the compact rectangle $[a,b]\times[c,d]$. So the inner integral $x\mapsto\int_c^d h(x,y)\,\mathrm dy$ is continuous (it changes by at most $(d-c)\varepsilon$ when $x$ moves by less than the $\delta$ belonging to $\varepsilon$), and likewise with the roles swapped; both iterated integrals exist. Given $\varepsilon > 0$, cut the rectangle into a grid of cells $R_{kl}$ of diameter less than $\delta$. On each cell $h$ stays within $\varepsilon$ of its value $h_{kl}$ at the cell's corner, so each iterated integral over $R_{kl}$, in either order, lies within $\varepsilon\lvert R_{kl}\rvert$ of $h_{kl}\lvert R_{kl}\rvert$. Iterated integrals over the whole rectangle are the sums of those over the cells (additivity of one-variable integrals over adjacent intervals). Hence both orders lie within $\varepsilon(b-a)(d-c)$ of the same sum $\sum h_{kl}\lvert R_{kl}\rvert$, so they differ by at most $2\varepsilon(b-a)(d-c)$ for every $\varepsilon$, and are equal. $\square$

The signs, once. Anticlockwise on $[a,b]\times[c,d]$ means: bottom left to right ($+\int_a^b P(x,c)\mathrm dx$), right side upwards ($+\int_c^d Q(b,y)\mathrm dy$), top right to left ($-\int_a^b P(x,d)\mathrm dx$), left side downwards ($-\int_c^d Q(a,y)\mathrm dy$). The asymmetry between $P$ and $Q$ in the final formula — one minus sign — comes from the top edge running backwards while the right edge runs forwards.

What the limiting argument needs. To pass from unions of rectangles to a region bounded by a piecewise-$C^1$ Jordan curve, one approximates the region from inside by a grid of squares, checks that the boundary integrals converge (which uses uniform continuity of $\vec F$ on a compact set and rectifiability of the curve), and that the area integrals converge. It is genuinely fiddly, it is done properly in any analysis text under "Green's theorem for a regular region", and nothing later in this course depends on the details. Unit 34's proof via partitions of unity is both shorter and more general, which is the usual sign that the right definitions have been found.

An aside worth having. Taking $P = -y/2$, $Q = x/2$ gives $\oint_{\partial\Omega}\tfrac12(x\,\mathrm dy - y\,\mathrm dx) = \text{area}(\Omega)$: the planimeter formula, which is how mechanical area-measuring instruments work, and which problem P4 uses to get $\pi ab$ for an ellipse from a boundary integral.

04 · curl

Stokes' theorem in $\mathbb R^3$

Stokes' theorem, for a flat rectangle

Let $S$ be a closed rectangle lying in a plane in $\mathbb R^3$ with unit normal $\hat n$, and $\partial S$ its boundary oriented so that $\hat n$ and the traversal obey the right-hand rule. For $\vec F$ of class $C^1$ near $S$,

$$\oint_{\partial S}\vec F\cdot\mathrm d\vec r \;=\; \int_S \big(\mathrm{curl}\,\vec F\big)\cdot\hat n \,\mathrm dA .$$

proof

If $S$ lies in the $xy$-plane then $\hat n = \hat z$ and $(\mathrm{curl}\,\vec F)\cdot\hat z = \partial_x F_2 - \partial_y F_1$, while $\vec F\cdot\mathrm d\vec r = F_1\mathrm dx + F_2\mathrm dy$ on $\partial S$: the statement is Green's theorem. For a rectangle in any other plane, rotate it into the $xy$-plane by some $R\in SO(3)$; both sides transform in the same way, by the equivariance theorem two slides back and by $\hat n \mapsto R\hat n$. $\square$

Schematic. Seen from the tip of $\hat n$, the boundary runs anticlockwise. Rotating the rectangle flat into the $xy$-plane turns the statement into Green's theorem, and both sides rotate alike.

The right-hand rule, stated precisely. Walk along $\partial S$ with $\hat n$ pointing from your feet to your head; the surface should be on your left. Equivalently: if $(\vec t_1, \vec t_2)$ is a positively oriented basis of the plane of $S$ with $\vec t_1\times\vec t_2 = \hat n$, traverse the boundary anticlockwise in the $(\vec t_1,\vec t_2)$ coordinates. Reversing $\hat n$ reverses the boundary orientation and both sides change sign together, so the theorem is insensitive to the choice — as it must be, since a plane has two normals and neither is preferred.

Reading the figure. The tilted rectangle is spanned by $\vec t_1 = \hat y$ and $\vec t_2 = (-\tfrac12, 0, \tfrac{\sqrt3}{2})$, so $\hat n = \vec t_1\times\vec t_2 = (\tfrac{\sqrt3}2, 0, \tfrac12)$, leaning towards the viewer. The dashed rectangle on the floor is its image under a rotation about the $y$-axis that takes $\hat n$ to $\hat z$; on the floor the theorem is Green's, and equivariance carries it back up. The projection is oblique and not to scale.

04 · curl

Stokes' theorem: checks, and the general surface

checks

Dimensions. Left: $[\vec F]\,\mathrm m$. Right: $([\vec F]/\mathrm m)\cdot\mathrm m^2$. Equal ✓.

Special case. $\vec F = -\nabla V$: the curl vanishes, so every loop bounding a rectangle in $U$ has zero circulation — consistent with the fundamental theorem of line integrals ✓.

Special case (of the general version). $\vec F = \vec\omega\times\vec r$ on a disc of radius $\rho$ normal to $\vec\omega$: right-hand side $2\omega\pi\rho^2$; left-hand side $(\omega\rho)(2\pi\rho)$, the field being tangent with magnitude $\omega\rho$. Equal ✓.

General surfaces — any oriented piecewise-$C^1$ surface with boundary — need the surface integral of unit 05 even to state. Cited; unit 34 proves it, and unit 23 is where it starts doing work, as Ampère's law.

The rigid-rotation check, drawn. Round the rim the field is tangent with size $\omega\rho$, so the circulation is $\omega\rho\cdot 2\pi\rho$; the curl is $2\vec\omega$, along the normal, so the right-hand side is $2\omega\cdot\pi\rho^2$. They agree.

Why the flat-rectangle case is not a cheat. Everything this unit needs from Stokes is local: the circulation-density corollary on the next slide, and the Poincaré lemma, which uses only coordinate rectangles. The general theorem is genuinely wanted later — unit 23's Ampère law is a statement about an arbitrary surface spanning a wire — and by then unit 05 has supplied surface integrals and unit 34 supplies the proof.

History. The theorem is Kelvin's: he wrote it to Stokes in a letter of July 1850. Stokes set it as question 8 of the 1854 Smith's Prize examination at Cambridge, which is how it acquired his name and how Maxwell — who sat that paper — first met it. Maxwell's Treatise then made it the standard tool of electromagnetism, and unit 24 uses it in exactly that role.

04 · curl

The curl is circulation per unit area

circulation density

For $\vec F$ of class $C^1$ near $\vec a$, and $S_\varepsilon$ the square of side $\varepsilon$ centred at $\vec a$ with normal $\hat n$,

$$\big(\mathrm{curl}\,\vec F(\vec a)\big)\cdot\hat n \;=\; \lim_{\varepsilon\to0}\frac{1}{\varepsilon^2}\oint_{\partial S_\varepsilon}\vec F\cdot\mathrm d\vec r .$$

proof

By Stokes, the loop integral is $\int_{S_\varepsilon}(\mathrm{curl}\,\vec F)\cdot\hat n\,\mathrm dA$. The integrand is continuous, so it equals $\varepsilon^2$ times its value at some point of $S_\varepsilon$ — the mean value theorem for integrals on a rectangle, one variable at a time. Divide by $\varepsilon^2$ and let $\varepsilon\to0$; that point tends to $\vec a$ and continuity finishes it. $\square$

This is the property a reader should carry: the curl measures how much the field goes round an infinitesimal loop, per unit area enclosed, and $\hat n$ selects the plane of the loop.

checks

Dimensions. $[\vec F]\,\mathrm m / \mathrm m^2 = [\vec F]/\mathrm m$ ✓.

Special case. $\vec F = \vec\omega\times\vec r$ with $\hat n = \hat\omega$: circulation round a circle of radius $\rho$ is $2\pi\rho\cdot\omega\rho$, area is $\pi\rho^2$, ratio $2\omega$ — independent of $\rho$, and equal to $\mathrm{curl}\,\vec F\cdot\hat\omega$ ✓.

the trap this sets

"Zero curl means no circulation" is true for infinitesimal loops, and for every loop that bounds a surface inside $U$. It says nothing about a loop that does not. The rest of this section is about that sentence.

Some books define the curl this way — by the limit, with the component along each of three independent normals — and then prove the component formula. That route makes the physics immediate and the linear algebra invisible; it also has to assume, rather than prove, that the three numbers so obtained are the components of a vector, which is exactly the content that the antisymmetric-part definition supplies. This course takes the mathematical definition and derives the physical characterisation, as it will again for the divergence in unit 05 and for the Riemann tensor in unit 33.

The mean value theorem used. For $h$ continuous on $[\alpha,\beta]\times[\gamma,\delta]$, the iterated integral lies between $(\beta-\alpha)(\delta-\gamma)\min h$ and $(\beta-\alpha)(\delta-\gamma)\max h$, and the intermediate value theorem on the connected rectangle supplies a point where $h$ takes the intermediate value. Both min and max are attained because the rectangle is compact.

04 · curl · worked example

Worked: the vortex, in full

$\vec F_{\mathrm V} = \big(\tfrac{-y}{x^2+y^2},\ \tfrac{x}{x^2+y^2}\big)$ on $U = \mathbb R^2\setminus\{0\}$.

worked: five loops

Step 1 — the curl. Computed on the necessary-condition slide: $\partial_x F_2 - \partial_y F_1 = 0$ at every point of $U$.

Step 2 — the unit circle. $\vec\gamma(t) = (\cos t,\sin t)$. On it $x^2+y^2 = 1$, so $\vec F_{\mathrm V}(\vec\gamma) = (-\sin t,\cos t) = \dot{\vec\gamma}(t)$, and $$\oint \vec F_{\mathrm V}\cdot\mathrm d\vec r = \int_0^{2\pi}\lVert\dot{\vec\gamma}\rVert^2\,\mathrm dt = \mathbf{2\pi} .$$

Step 3 — radius $a$. $\vec F_{\mathrm V}$ is tangent, of magnitude $1/a$, and the circumference is $2\pi a$: again $2\pi$, for every $a > 0$.

Step 4 — a square. The boundary of $[-1,1]^2$, anticlockwise, by quadrature: $6.283185307$, which is $2\pi$ to ten digits.

Step 5 — a loop missing the origin. The circle of radius $1$ about $(3,0)$: $0$ to machine precision, and it must be: the circle lies in the half-plane $\{x > 0\}$, a box on which the field has a potential (the Poincaré lemma, two slides on), so its circulation is zero.

The field's arrows shrink like $1/r$ away from the puncture. Every loop that goes once round the origin collects $2\pi$, the same for circle and square; the loop beside it collects nothing.

04 · curl · worked example

The vortex: closed everywhere, and still no potential

the apparent contradiction

Closed everywhere on $U$, yet a loop in $U$ with circulation $2\pi \ne 0$. By the three-conditions theorem, $\vec F_{\mathrm V}$ has no potential on $U$.

Nothing is wrong. Green's theorem needs the loop to bound a region inside $U$, and the unit disc does not: it contains the puncture, where the curl is not zero but nothing at all.

checks

Dimensionless on both sides ✓. And independent of the loop's shape and size, so long as it encircles the origin once — circle, square, or anything else. That invariance is the signal that something discrete is being counted ✓.

Where a potential would come from. The angle $\theta$ is one: its level sets are the rays, and $\nabla\theta = \vec F_{\mathrm V}$. But going once round, $\theta$ rises by $2\pi$, so it must jump somewhere — here across the cut — and no single-valued choice exists on all of $U$.

A potential does exist locally. On the right half-plane $\{x > 0\}$, the function $\theta = \arctan(y/x)$ satisfies $\partial_x\theta = \frac{-y/x^2}{1+y^2/x^2} = \frac{-y}{x^2+y^2}$ and $\partial_y\theta = \frac{1/x}{1+y^2/x^2} = \frac{x}{x^2+y^2}$, so $\vec F_{\mathrm V} = \nabla\theta$ there. The same works on any half-plane, or on the plane cut along any ray from the origin. What cannot be done is to choose the branch consistently all the way round: $\theta$ increases by $2\pi$ on a circuit, which is the $2\pi$ of step 2. The field is a gradient locally and not globally, and "locally but not globally" is what the slide What the vortex measures is the domain names.

Reading the figure. The faint circles are the field lines of $\vec F_{\mathrm V}$, and they cross every ray at right angles, as the field lines of a gradient must cross its level sets. Remove the cut — the negative $x$-axis — and what is left is a set on which $\theta$, taking values in $(-\pi,\pi)$, is a genuine $C^\infty$ potential (up to the physics sign). Put the cut back and $\theta$ would have to equal both $\pi$ and $-\pi$ there. Any other ray would serve as the cut; there has to be one.

04 · curl

The Poincaré lemma

Poincaré lemma, on a box

Let $\vec F = (P,Q)$ be $C^1$ on an open box $U = I_1\times I_2$ — a product of open intervals, possibly unbounded, so $\mathbb R^2$ and a half-plane both qualify — with $\partial_x Q = \partial_y P$. Then $\vec F = \nabla W$ for some $C^1$ function $W$ on $U$, so $-W$ is a potential energy and $\vec F$ is conservative.

proof

Fix $(x_0,y_0)\in U$. Let $A$ be the path across-then-up and $B$ the path up-then-across, both from $(x_0,y_0)$ to $(x,y)$; both lie in $U$, since $U$ is a box. $A$ followed by $B$ reversed traverses the boundary of a rectangle inside $U$, so by Green's theorem the two integrals are equal. Call the common value $W(x,y)$.

Reading $W$ off $A$, its last leg is $\int_{y_0}^{y}Q(x,t)\,\mathrm dt$ and the first leg does not involve $y$, so $\partial_y W = Q$ by the fundamental theorem of calculus. Reading $W$ off $B$, its last leg is $\int_{x_0}^{x}P(s,y)\,\mathrm ds$, so $\partial_x W = P$. Both partials are continuous, so $W\in C^1$ and $\nabla W = \vec F$. $\square$

$A$, then $B$ backwards, bounds the shaded rectangle — inside $U$, because $U$ is a box — so Green gives the two paths equal integrals.

general version

Closed $\Rightarrow$ conservative on any simply connected open $U$ — every loop in $U$ contracts continuously to a point within $U$. Cited: proved in unit 34 by homotopy invariance. The box case above is the engine of that proof.

The three-dimensional version. Same proof with three legs. On an open box, $W(x,y,z) = \int_{x_0}^{x}F_1(s,y_0,z_0)\mathrm ds + \int_{y_0}^{y}F_2(x,t,z_0)\mathrm dt + \int_{z_0}^{z}F_3(x,y,u)\mathrm du$, and the three permutations of the legs are shown equal by applying Green's theorem in each of the three coordinate planes — once for each of the three equations $\partial_j F_i = \partial_i F_j$. Then each partial is read off the permutation whose last leg runs in that direction.

Why a box, and what it really needs. The proof used only that the two staircase paths, and the rectangle between them, stay in $U$ — which is exactly what a product of intervals guarantees, and what a general convex set does not (a thin diagonal strip contains both corners of a segment and neither corner of the rectangle they span). The conclusion does hold on any convex, indeed any star-shaped, open set, by a different construction: put $W(\vec x) = \int_0^1\vec F(t\vec x)\cdot\vec x\,\mathrm dt$ (star-shaped about $\vec 0$) and differentiate under the integral sign. That is the usual textbook proof; it is cited here, because differentiating under an integral sign is a theorem this course has not proved — it is Theorem 9.42 of Rudin's Principles of Mathematical Analysis (3rd edition). The staircase version needs nothing beyond Green's theorem, and a box is all that anything in this unit or its problem sheet ever uses — $\mathbb R^n$, a half-space and an open rectangle are all boxes.

Orientation, pedantically. The circuit "$A$ then $B$ reversed" runs anticlockwise when $x > x_0$ and $y > y_0$, and clockwise in two of the other three quadrants; Green's theorem gives zero circulation either way, since the integrand $\partial_xQ - \partial_yP$ vanishes. If $x = x_0$ or $y = y_0$ the rectangle is degenerate and the two paths coincide.

Sketch of the simply connected case

Let $\vec\gamma$ be a loop in $U$ and $H : [0,1]^2 \to U$ a continuous homotopy contracting it to a point. $H([0,1]^2)$ is compact, so it has a positive distance to $\partial U$; subdivide the square finely enough that each small square's image lies in a ball inside $U$. On each ball the rectangle case gives a local potential, and the circulation round each small image loop is zero. Summing over the grid, the contributions on shared edges cancel — the same cancellation as in Green's patching picture — leaving the circulation of $\vec\gamma$ equal to that of the constant loop, namely zero. Making "the image of a small square lies in a ball" precise for a merely continuous $H$ takes care, which is why the honest proof waits for unit 34 and the language of forms.

Poincaré's name on it. He stated the general result — every closed differential form is locally exact — in 1887, in a paper on residues of double integrals; Volterra published a proof in 1889. It is the first theorem of what became de Rham cohomology. The one-dimensional case is the existence of antiderivatives, which is why the lemma feels like it should be obvious.

04 · curl

What the vortex measures is the domain

The obstruction is not in $\vec F_{\mathrm V}$: the field is closed, smooth and perfectly well behaved at every point where it exists. The obstruction is that $\mathbb R^2\setminus\{0\}$ has a hole.

what the number is

For a loop $\vec\gamma$ in $\mathbb R^2\setminus\{0\}$, $$n(\vec\gamma) := \frac{1}{2\pi}\oint_{\vec\gamma}\vec F_{\mathrm V}\cdot\mathrm d\vec r$$ is an integer, the winding number of $\vec\gamma$ about the origin. It is unchanged by any continuous deformation of $\vec\gamma$ within the domain.

A circle gives $1$; the same circle traversed backwards gives $-1$; traversed twice, $2$; the loop about $(3,0)$ gives $0$.

Four loops and their winding numbers. Deform any of them without crossing the puncture and $n$ cannot change, since it is an integer that varies continuously.

Why $n(\vec\gamma)$ is an integer. On any sub-interval where the loop stays in a half-plane, $\vec F_{\mathrm V}$ is $\nabla\theta$ for a continuous branch of the angle, so the integral is the total change in $\theta$. Patching the branches along the loop, the integral is the net change of a continuous angle function $\theta(t)$ with $\vec\gamma(t) = r(t)(\cos\theta(t),\sin\theta(t))$. Closing the loop forces $\theta(1) - \theta(0)$ to be a multiple of $2\pi$, since the endpoint is the same point.

Why it is a homotopy invariant. It is an integer and, by an estimate of the same kind as the sketch on the previous slide, it varies continuously as the loop is deformed. A continuous integer-valued function on a connected parameter space is constant.

04 · curl

Closed, conservative, and the holes between them

the summary to keep

closed is a property of the field.
conservative is a property of the field and the domain.
On a simply connected domain they coincide; in general their difference measures the domain's holes.

left open on purpose

How many independent obstructions has a domain? The closed fields modulo the conservative ones form a vector space whose dimension counts the holes: the first de Rham cohomology of $U$, built in unit 34. For $\mathbb R^2\setminus\{0\}$ it is one-dimensional, spanned by $\vec F_{\mathrm V}$.

A calculus question has turned into a topology question. It happens again in unit 36, where deciding whether Schwarzschild's horizon is a real singularity turns out to be a question about which part of the manifold a chart covers.

The hole need not be a point. In $\mathbb R^3$ with a line removed, a loop round the line cannot be lifted off without cutting it, and the field $(-y, x, 0)/(x^2+y^2)$ — the vortex at every height — has circulation $2\pi$ round it and no potential.

Physically. $\vec F_{\mathrm V}$ is the velocity field of an ideal fluid vortex, and the $2\pi$ is its circulation, which Kelvin's theorem says is conserved. It is also the magnetic field outside a long straight wire, up to a constant: unit 23 computes it from Ampère's law and gets $\vec B = \frac{\mu_0 I}{2\pi}\vec F_{\mathrm V}$, so the loop integral is $\mu_0 I$ and the "hole" is the wire. A field with a source inside the hole is the usual way these domains arise in physics.

The three-dimensional cousin. $\mathbb R^3\setminus\{0\}$ is simply connected, so the inverse-square field has no such obstruction — which is the answer to the puzzle the deck opened with. But $\mathbb R^3$ minus a line is not simply connected, and the field $(-y,x,0)/(x^2+y^2)$ on it repeats the vortex's behaviour exactly. Problem P3 asks for that one.

section 4

Potential landscapes

Once a field has a potential, the mechanics can be read off the shape of a surface.

04 · landscapes

Taylor's theorem in $\mathbb R^n$

second-order Taylor

Let $f$ be $C^2$ on an open $U$ and let the segment from $\vec a$ to $\vec a + \vec h$ lie in $U$. Then there is $\theta\in(0,1)$ with

$$f(\vec a+\vec h) = f(\vec a) + Df(\vec a)\vec h + \tfrac12\vec h^{\!\top}H(\vec a+\theta\vec h)\vec h,$$

where $H$ is the matrix of second partials. Consequently $f(\vec a+\vec h) = f(\vec a) + \nabla f(\vec a)\cdot\vec h + \tfrac12\vec h^{\!\top}H(\vec a)\vec h + o(\lVert\vec h\rVert^2)$.

proof

Put $g(t) = f(\vec a + t\vec h)$ on $[0,1]$. The chain rule of unit 03 gives $g'(t) = \sum_i\partial_i f(\vec a+t\vec h)h_i$ and, applied again to each $\partial_i f$, $g''(t) = \sum_{i,j}\partial_j\partial_i f(\vec a+t\vec h)h_ih_j = \vec h^{\!\top}H(\vec a+t\vec h)\vec h$. One-variable Taylor with Lagrange remainder gives $g(1) = g(0)+g'(0)+\tfrac12 g''(\theta)$. $\square$

checks

Dimensions. If $f$ is an energy and $\vec h$ a length, $H$ is an energy per length squared — a spring constant, $\mathrm{N\,m^{-1}}$, and $\tfrac12 k h^2$ is the familiar spring energy ✓.

Special case $n=1$. The statement collapses to $f(a+h) = f(a)+f'(a)h+\tfrac12 f''(a+\theta h)h^2$, the one-variable theorem it was built from ✓.

Special case: $f$ quadratic. $H$ is constant, the remainder is exact, and there is no $o$ term ✓.

The last form is what gets used: near a point where $\nabla f$ vanishes, $f$ looks like a quadratic form, and the rest of this section is about reading quadratic forms.

Where the $o(\lVert\vec h\rVert^2)$ comes from. Write the exact remainder as $\tfrac12\vec h^{\!\top}H(\vec a)\vec h + \tfrac12\vec h^{\!\top}[H(\vec a+\theta\vec h)-H(\vec a)]\vec h$. The bracket has operator norm $\to 0$ as $\vec h\to0$, because $H$ is continuous ($f\in C^2$), so the second piece is bounded by $\tfrac12\lVert H(\vec a+\theta\vec h)-H(\vec a)\rVert\,\lVert\vec h\rVert^2$, which is $o(\lVert\vec h\rVert^2)$.

The segment hypothesis is real. The proof restricts $f$ to the straight line from $\vec a$ to $\vec a+\vec h$, so that line must lie in $U$. On a convex $U$ this is automatic. On a general $U$ it holds for $\lVert\vec h\rVert$ small enough, since $U$ is open — which is all the local statements of this section need.

Higher orders. The same reduction gives the order-$k$ formula, with the $k$-th term $\tfrac1{k!}D^kf(\vec a)(\vec h,\dots,\vec h)$ where $D^kf(\vec a)$ is a symmetric $k$-linear map — symmetric by Clairaut. This is the first place in the course where a multilinear map appears as the natural home of a derivative, and unit 16 will say that such a thing is exactly a tensor. Problem P8 needs the third-order term.

04 · landscapes

The Hessian is a symmetric bilinear form

definition

For $f$ of class $C^2$ on $U$ and $\vec a \in U$, the Hessian of $f$ at $\vec a$ is the map

$$H_{\vec a}(\vec u,\vec w) := \frac{\partial^2}{\partial s\,\partial t}\bigg|_{0}f(\vec a + t\vec u + s\vec w),$$

a bilinear form $\mathbb R^n\times\mathbb R^n\to\mathbb R$. It is defined without reference to any basis.

two theorems, not definitions

(a) $H_{\vec a}$ is symmetric: $H_{\vec a}(\vec u,\vec w) = H_{\vec a}(\vec w,\vec u)$.
(b) Its matrix in the standard basis has entries $\partial_i\partial_j f(\vec a)$.

Why (b): take $\vec u = \vec e_i$, $\vec w = \vec e_j$ in the definition. Why (a): that matrix is symmetric, by Clairaut.

the habit being formed

Define the object as a multilinear map; prove that its components are what everyone writes down. The transformation law under a change of basis is then a theorem: under $\vec x = S\vec x'$ the matrix becomes $S^{\!\top}HS$, not $S^{-1}HS$ — a bilinear form is not an operator, and they transform differently.

The same habit, applied to the inertia tensor (unit 16), the stress tensor (unit 22), the field tensor (unit 28) and the metric (unit 32).

Check. $f(\vec x) = \tfrac12\vec x^{\!\top}A\vec x$ with $A$ symmetric has $H_{\vec a} = A$ at every $\vec a$ ✓ — the second derivative of a quadratic form is the form ✓.

Why $S^{\!\top}HS$ and not $S^{-1}HS$. A bilinear form eats two vectors: $H'(\vec u',\vec w') = H(S\vec u', S\vec w') = (S\vec u')^{\!\top}H(S\vec w') = \vec u'^{\!\top}(S^{\!\top}HS)\vec w'$. A linear operator eats one vector and returns one, so its matrix transforms by $S^{-1}AS$. The two coincide exactly when $S^{\!\top} = S^{-1}$ — that is, when $S$ is orthogonal — which is why the distinction is invisible in Euclidean geometry with orthonormal bases, and unavoidable the moment the inner product is not the identity. Unit 16 is where that becomes the definition of a $(0,2)$-tensor versus a $(1,1)$-tensor; unit 26 is where the inner product stops being positive definite and the distinction starts changing signs.

The definition, unpacked. $\frac{\partial}{\partial t}f(\vec a+t\vec u+s\vec w)$ at general $(t,s)$ is $\sum_i\partial_i f(\vec a+t\vec u+s\vec w)u_i$ by the chain rule; differentiating in $s$ and setting $t=s=0$ gives $\sum_{i,j}\partial_j\partial_i f(\vec a)u_iw_j$. So the definition does produce a bilinear form, and (b) is immediate. The order of the two differentiations can be swapped because the function $(t,s)\mapsto f(\vec a+t\vec u+s\vec w)$ is $C^2$, which is Clairaut again.

Notation. This course writes $H$ for the matrix and $H_{\vec a}(\cdot,\cdot)$ for the form when the difference matters, and $\vec h^{\!\top}H\vec h$ for the associated quadratic form $H_{\vec a}(\vec h,\vec h)$. In unit 32 the same object is written $\nabla_i\nabla_j f$ and is no longer the matrix of second partials, because partial derivatives stop being tensorial — the first casualty of leaving flat space.

04 · landscapes

Classifying a nondegenerate critical point

theorem

Let $f\in C^2$ near $\vec a$ with $\nabla f(\vec a) = \vec 0$, and let $H = H_{\vec a}$.

$H$ positive definite $\Rightarrow$ strict local minimum;
$H$ negative definite $\Rightarrow$ strict local maximum;
$H$ indefinite $\Rightarrow$ saddle: no local extremum;
$H$ singular and semidefinite: degenerate, and the second order decides nothing.

$H$ is symmetric, so its eigenvalues are real (spectral theorem), and the four cases exhaust their sign patterns.

proof of the first case

$m := \min_{\lVert\vec u\rVert = 1}\vec u^{\!\top}H\vec u > 0$, attained because the unit sphere is compact and the form is continuous. By continuity of the second partials there is $\delta > 0$ with $\vec u^{\!\top}H(\vec x)\vec u \ge m/2$ for all unit $\vec u$ and all $\lVert\vec x - \vec a\rVert < \delta$. Taylor then gives, for $0 < \lVert\vec h\rVert < \delta$, $$f(\vec a+\vec h)-f(\vec a) = \tfrac12\vec h^{\!\top}H(\vec a+\theta\vec h)\vec h \ge \tfrac{m}{4}\lVert\vec h\rVert^2 > 0. \;\square$$

degenerate does not mean rare

$f = x^4+y^4$ (minimum), $f = -x^4-y^4$ (maximum) and $f = x^4-y^4$ (saddle) all have $H = 0$ at the origin. No second-order test can separate them, and none exists in general.

Problem P8's $V = x^3-3xy^2$ is degenerate at its only critical point, and the answer needs the third-order term.

checks

$n = 1$. "Positive definite" is $f''(a) > 0$, "indefinite" is impossible, "degenerate" is $f''(a) = 0$ — the school second-derivative test, recovered exactly ✓.

Dimensions. $m$ has the units of $f$ per length squared; $m\lVert\vec h\rVert^2/4$ has the units of $f$ ✓.

The remaining cases. Negative definite: apply the first case to $-f$. Indefinite: pick unit vectors $\vec u$ with $\vec u^{\!\top}H\vec u = \lambda > 0$ and $\vec w$ with $\vec w^{\!\top}H\vec w = -\mu < 0$ — eigenvectors for a positive and a negative eigenvalue will do. Along $t\mapsto \vec a+t\vec u$ the same Taylor estimate gives $f(\vec a+t\vec u) - f(\vec a) \ge \tfrac{\lambda}{4}t^2 > 0$ for small $t\ne0$, and along $t\mapsto\vec a+t\vec w$ it gives $\le -\tfrac{\mu}{4}t^2 < 0$. So every neighbourhood contains points where $f$ is larger and points where it is smaller: no extremum. $\square$

Why compactness is doing the work. Positive definiteness says $\vec u^{\!\top}H\vec u > 0$ for each $\vec u \ne 0$, which on its own gives no uniform lower bound. The minimum over the compact unit sphere converts pointwise positivity into the uniform constant $m$, and it is $m$ that survives the perturbation to nearby $\vec x$. Without compactness — in infinite dimensions — positive definiteness genuinely does not imply a local minimum, and the correct condition is a uniform one. Unit 11 meets this when it asks whether a stationary curve of an action is a minimum.

The determinant test in two dimensions. For $H = \begin{pmatrix}A&B\\B&C\end{pmatrix}$: $\det H = AC - B^2 > 0$ with $A > 0$ is positive definite, $\det H > 0$ with $A < 0$ is negative definite, $\det H < 0$ is indefinite, $\det H = 0$ is degenerate. It is the two-dimensional case of Sylvester's criterion and follows from $\det H = \lambda_1\lambda_2$, $\operatorname{tr}H = \lambda_1+\lambda_2$.

04 · landscapes

Equilibrium, stability, and Lagrange–Dirichlet

definition

$\vec a$ is an equilibrium of $m\ddot{\vec r} = -\nabla V(\vec r)$ if $\nabla V(\vec a) = \vec 0$: then $\vec r(t)\equiv\vec a$ solves the equation. Equilibria of the motion are exactly the critical points of the landscape.

Lagrange–Dirichlet

If $V$ has a strict local minimum at $\vec a$, then $\vec a$ is Lyapunov stable: for every $\varepsilon > 0$ there is $\delta > 0$ such that a motion starting with $\lVert(\vec r-\vec a,\vec v)\rVert < \delta$ keeps $\lVert(\vec r-\vec a,\vec v)\rVert < \varepsilon$ for all later times.

Why: $E(\vec r,\vec v) = \tfrac12 m\lVert\vec v\rVert^2 + V(\vec r) - V(\vec a)$ is zero at $(\vec a,\vec 0)$, positive near it, and constant along motions; a motion that starts with small $E$ cannot climb to the positive minimum of $E$ on a small sphere. full proof in the notes (N)

Check. One dimension, $V = \tfrac12 kx^2$: minimum at $0$, and the motion is the bounded oscillation of unit 00. $V = -\tfrac12 kx^2$: maximum, and the motion is $x_0\cosh(\sqrt{k/m}\,t)$, which leaves every neighbourhood ✓.

Unit 07 turns a nondegenerate minimum into normal modes: diagonalise $H$ against the mass matrix and read off $n$ frequencies $\sqrt{\lambda_i/m}$.

The next slide draws the proof in the space of states.

Full proof of Lagrange–Dirichlet

Choose $\rho > 0$ with $V(\vec r) > V(\vec a)$ for $0 < \lVert\vec r-\vec a\rVert \le \rho$ (strict minimum). Given $\varepsilon$, shrink it if necessary so that $\varepsilon \le \rho$, and let $S$ be the sphere $\lVert(\vec r-\vec a,\vec v)\rVert = \varepsilon$ in the six-dimensional space of states. On $S$, $E > 0$: if $\vec v \ne \vec 0$ the kinetic term is positive and $V(\vec r) \ge V(\vec a)$ because $\lVert\vec r-\vec a\rVert\le\rho$; if $\vec v = \vec 0$ then $\lVert\vec r-\vec a\rVert = \varepsilon$ and $V(\vec r) > V(\vec a)$. $S$ is compact and $E$ continuous, so $c := \min_S E > 0$. $E$ is continuous with $E(\vec a,\vec 0) = 0$, so there is $\delta < \varepsilon$ with $E < c$ on the $\delta$-ball. A motion starting there has $E < c$ for ever, by conservation. If it ever reached distance $\varepsilon$ from $(\vec a,\vec 0)$, the distance — a continuous function of $t$ — would equal $\varepsilon$ at a first time, where the state lies on $S$ and so $E \ge c$: a contradiction. $\square$

Stable, but not asymptotically stable. $\dot E = 0$ exactly, so the motion stays close and nothing more: a marble in a frictionless bowl oscillates for ever and does not settle. Add any dissipation and $\dot E < 0$ off the equilibrium, and LaSalle's invariance principle (unit 02) upgrades the conclusion to asymptotic stability. This is the precise sense in which "systems seek minima of potential energy" is true: they seek them only if something takes energy away.

Lagrange asserted this in the Mécanique analytique (1788) with an argument that quietly assumed $V$ was a quadratic form; Dirichlet gave the proof above in 1846, and it is one of the earliest genuine uses of what is now called a Lyapunov function — nearly half a century before Lyapunov's thesis of 1892.

04 · landscapes

Lagrange–Dirichlet: the proof in a picture

One dimension, $m = 1$, $V = 1 - \cos x$ near its minimum (schematic of the proof, curves computed). $E$ is smallest on $S$ at its two ends, where it equals $c$; inside the $\delta$-ball $E < c$, and since $E$ is conserved, a motion that starts there stays on a level below $c$ and can never reach $S$.

the converse is delicate

A maximum or saddle of $V$ gives an unstable equilibrium whenever the Hessian is nondegenerate: each negative eigenvalue $\lambda$ of $H$ puts a real $+\sqrt{\lvert\lambda\rvert/m}$ in the linearisation's spectrum (notes), and one eigenvalue with positive real part suffices for instability — Lyapunov's first method, cited: Hirsch, Smale and Devaney, Differential Equations, Dynamical Systems, and an Introduction to Chaos, chapter 8. For degenerate $H$ the converse can fail outright (notes).

Reading the figure. The plane is the space of states, position against velocity — for a particle in three dimensions it would be six-dimensional, and $S$ a five-sphere. The closed curves are level sets of $E = \tfrac12 v^2 + V(x)$: near a strict minimum they are closed ovals around the equilibrium. The proof's three moves are the three marked objects: choose the sphere $S$ of radius $\varepsilon$; let $c$ be the least energy on it (attained, since $S$ is compact — here at $(\pm\varepsilon, 0)$, where all the energy is potential); choose $\delta$ so that $E < c$ on the $\delta$-ball. A motion starting there runs round its own level, which lies strictly inside the level $E = c$, and so inside $S$.

The converse, failing. In one dimension take $V(x) = \mathrm e^{-1/x^2}\cos(1/x)$, $V(0) = 0$. It is $C^\infty$, and $0$ is not a minimum: $V < 0$ wherever $\cos(1/x) < 0$, arbitrarily close to $0$. Yet $0$ is stable: on each side there are points arbitrarily close to $0$ where $\cos(1/x) = 1$ and $V > 0$, and a motion that starts between two such barriers with energy below both of their heights can never cross either. For analytic $V$ whether the converse holds is a long-standing open question.

Why the linearisation is what it is. Near $\vec a$, Taylor gives $-\nabla V(\vec a+\vec h) = -H\vec h + o(\lVert\vec h\rVert)$, so the linearised system is $m\ddot{\vec h} = -H\vec h$. Written as a first-order system for $(\vec h,\dot{\vec h})$, its matrix is $\begin{pmatrix}0&I\\-H/m&0\end{pmatrix}$, whose eigenvalues are $\pm\sqrt{-\lambda_i/m}$ over the eigenvalues $\lambda_i$ of $H$: purely imaginary for $\lambda_i > 0$ and a real pair $\pm\sqrt{\lvert\lambda_i\rvert/m}$ for $\lambda_i < 0$. One negative eigenvalue of $H$ therefore supplies one positive real eigenvalue, and instability. (For an eigenvector $\vec w$ of $H$ with eigenvalue $\lambda_i$, try $(\vec h,\dot{\vec h}) = (\vec w, s\vec w)$: the matrix sends it to $(s\vec w, -\lambda_i\vec w/m)$, which is $s$ times it exactly when $s^2 = -\lambda_i/m$.) It also shows why Lagrange–Dirichlet needs a nonlinear argument: at a minimum every eigenvalue is purely imaginary, and purely imaginary eigenvalues are the case in which linearisation cannot decide stability — the nonlinear terms can tip it either way — which is exactly why the energy is used instead.

04 · landscapes · worked example

Worked: two masses, and why the middle is a saddle

Two masses $M$ held at $(\pm d, 0, 0)$; a test mass $m$ at $\vec r$. Its potential energy is the running example's $V$, twice: $$V(\vec r) = -GMm\Big(\frac{1}{r_+}+\frac{1}{r_-}\Big),\quad r_\pm = \lVert\vec r \mp (d,0,0)\rVert .$$

worked: the Hessian at the midpoint

Step 1 — the origin is critical. Swapping the sign of $x$ swaps $r_\pm$, so $V$ is even in each coordinate and every first partial vanishes at $\vec 0$.

Step 2 — the Hessian of one term. For $\phi(\vec s) = 1/\lVert\vec s\rVert$ (differentiated in full in the notes), $$\partial_j\partial_i\phi = \frac{3s_is_j - \lVert\vec s\rVert^2\delta_{ij}}{\lVert\vec s\rVert^5} = \frac{\mathrm{diag}(2,-1,-1)}{d^3}\ \text{ at } \vec s = (\mp d,0,0),$$ the same for both signs, so the two contributions add.

Step 3 — read off and classify. $H_{\vec 0}(V) = \dfrac{GMm}{d^3}\,\mathrm{diag}(-4,\,2,\,2)$: both signs present, indefinite. The midpoint is a saddle — unstable along the line, oscillatory across it.

Level curves of $V$ in the plane of the masses, in units $d = 1$, $GMm = 1$. The level through the midpoint crosses itself there: downhill along the axis, uphill across it — a saddle, as the Hessian said.

Check. $d\to\infty$ sends every eigenvalue to zero ✓, and $GMm/d^3$ is $\mathrm{N\,m^{-1}}$, a spring constant ✓.

Step 2 in full. $\partial_i\lVert\vec s\rVert = s_i/\lVert\vec s\rVert$, so $\partial_i\phi = \partial_i\lVert\vec s\rVert^{-1} = -\lVert\vec s\rVert^{-2}\cdot s_i/\lVert\vec s\rVert = -s_i\lVert\vec s\rVert^{-3}$. Differentiating again with the product rule, $\partial_j\partial_i\phi = -\delta_{ij}\lVert\vec s\rVert^{-3} + 3s_i\lVert\vec s\rVert^{-4}\cdot s_j/\lVert\vec s\rVert = (3s_is_j - \lVert\vec s\rVert^2\delta_{ij})\lVert\vec s\rVert^{-5}$. At $\vec s = (-d,0,0)$: the $(1,1)$ entry is $(3d^2-d^2)/d^5 = 2/d^3$; the $(2,2)$ and $(3,3)$ entries are $(0-d^2)/d^5 = -1/d^3$; every off-diagonal entry has a factor $s_2 = s_3 = 0$. At $\vec s = (+d,0,0)$ the $s_is_j$ terms are unchanged because they are quadratic in $\vec s$. Hence the two contributions are identical and add.

Reading the figure. The contours are computed from $V(x,y,0) = -(1/r_+ + 1/r_-)$ with $d = GMm = 1$, so the midpoint value is $-2$; the figure-of-eight through it is the level $V = -2$. Inside each lobe $V < -2$ and the levels close round one mass; outside, $V > -2$ and they enclose both. Walking along the axis from the midpoint you enter a lobe and go downhill; walking across, you leave both and go uphill. The curvatures there are $-4$ and $+2$ in these units (and $+2$ again out of the page, in $z$, not drawn).

04 · landscapes · worked example

The saddle between two masses, in numbers

numbers: $M = 1\,M_\odot$, $d = 1$ AU

$GM/d^3 = 3.9640\times10^{-14}\ \mathrm{s^{-2}}$. Transverse, $\omega = \sqrt{2GM/d^3} = 2.8157\times10^{-7}\ \mathrm{s^{-1}}$: a period of $\mathbf{258.3}$ days. Along the axis, $\sqrt{4GM/d^3} = 3.9820\times10^{-7}\ \mathrm{s^{-1}}$: growth by $e$ in $\mathbf{29.07}$ days.

the zero trace is not luck

$-4+2+2 = 0$: Step 2's matrix has trace $(3\lVert\vec s\rVert^2 - 3\lVert\vec s\rVert^2)/\lVert\vec s\rVert^5 = 0$ at every point. So for fixed masses in any arrangement, no critical point is a nondegenerate minimum — the Hessian half of Earnshaw's theorem. Unit 05 names the trace $\nabla^2 V$ and removes the word "nondegenerate".

Two slices through the midpoint, exact (solid) against the Hessian's quadratic form (dashed): $\tfrac12(-4)x^2$ along the axis and $\tfrac12(2)y^2$ across it. Close to the saddle the quadratic form is the landscape.

The transverse frequency, derived. Along $y$, $V(0,y,0) - V(\vec 0) \approx \tfrac12(2GMm/d^3)y^2$, so the equation of motion is $m\ddot y = -(2GMm/d^3)y$, a simple harmonic oscillator with $\omega^2 = 2GM/d^3$. With $GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3\,s^{-2}}$ and $d = 1.495978707\times10^{11}$ m: $GM/d^3 = 3.964016\times10^{-14}\ \mathrm{s^{-2}}$, $\omega = 2.815676\times10^{-7}\ \mathrm{s^{-1}}$, and the period is $2.2315\times10^{7}$ s, which is $258.28$ days.

What is being idealised away. Two masses a fixed distance apart are not a solution of anything — they would fall together. The honest version lets them orbit, works in the rotating frame of unit 03, and adds the centrifugal potential $-\tfrac12 m\omega^2\rho^2$ to $V$; the critical points of the sum are the five Lagrange points, and $L_1$, $L_2$, $L_3$ are saddles of exactly the kind computed here. The surprise is $L_4$ and $L_5$, which are maxima of the effective potential and are nevertheless stable — provided the smaller mass is below about $1/25$ of the total, as it is for the Earth–Moon and Sun–Jupiter pairs — because the Coriolis term — velocity-dependent, doing no work, and therefore invisible to this section's energy argument — stabilises them. That is the sharpest available warning that a landscape picture is not the whole of mechanics.

Earnshaw, 1842. Earnshaw proved it for electrostatics, to rule out a class of theories of the luminiferous aether. It is the reason magnetic levitation always needs something extra — diamagnetism, superconductivity, feedback, or spin — and the reason the Paul trap uses a saddle whose orientation oscillates at radio frequency rather than a well.

section 5

Constraints, and where the multiplier comes from

Two theorems proved by the contraction mapping theorem, and Lagrange's rule as a corollary of them.

04 · constraints

The inverse function theorem

inverse function theorem

Let $\vec f : U \to \mathbb R^n$ be $C^1$ on an open $U\subseteq\mathbb R^n$ and let $D\vec f(\vec a)$ be invertible. Then there are open sets $V\ni\vec a$ and $W\ni\vec f(\vec a)$ such that $\vec f : V \to W$ is a bijection, and the inverse $\vec g$ is $C^1$ with

$$D\vec g(\vec f(\vec x)) = \big(D\vec f(\vec x)\big)^{-1}\quad\text{for }\vec x\in V .$$

proof, in outline

Normalise so that $\vec a = \vec 0$, $\vec f(\vec 0)=\vec 0$, $D\vec f(\vec 0)=I$. Solving $\vec f(\vec x) = \vec y$ is finding a fixed point of $\varphi_{\vec y}(\vec x) := \vec x + \vec y - \vec f(\vec x)$. Since $D\varphi_{\vec y} = I - D\vec f$ is small near $\vec 0$, $\varphi_{\vec y}$ is a contraction of a closed ball into itself, and the Banach fixed-point theorem (proved in the notes) supplies exactly one solution. full proof in the notes (N)

The same theorem, with the same completeness hypothesis, that proves Picard–Lindelöf in unit 01 — there on a space of functions, here on a ball in $\mathbb R^n$.

local, and only local

$\vec f(x,y) = (e^x\cos y,\ e^x\sin y)$ has $\det D\vec f = e^{2x}\ne0$ everywhere, so it is invertible near every point; yet $y\mapsto y+2\pi$ changes nothing, so it is not injective on $\mathbb R^2$. The failure is the vortex's again — $y$ is an angle — and an everywhere-nonsingular derivative does not give a global inverse.

check

Differentiate $\vec g(\vec f(\vec x)) = \vec x$ by the chain rule: $D\vec g\,D\vec f = I$. So if $\vec g$ is differentiable at all, its derivative must be $(D\vec f)^{-1}$ — the formula is forced, and the theorem's work is proving $\vec g$ exists and is differentiable ✓.

Full proof

The Banach fixed-point theorem. Let $(X,d)$ be a nonempty complete metric space and $\varphi : X\to X$ with $d(\varphi(x),\varphi(x')) \le k\,d(x,x')$ for some $k < 1$. Then $\varphi$ has exactly one fixed point. Proof. Uniqueness: two fixed points satisfy $d(x,x') \le k\,d(x,x')$, so $d(x,x') = 0$. Existence: from any $x_0$ put $x_{j+1} = \varphi(x_j)$; then $d(x_{j+1},x_j) \le k^jd(x_1,x_0)$, so for $i < j$ the triangle inequality gives $d(x_j,x_i) \le (k^i + \dots + k^{j-1})\,d(x_1,x_0) \le \tfrac{k^i}{1-k}\,d(x_1,x_0)$, and the sequence is Cauchy. By completeness it converges to some $x$, and $\varphi(x) = \lim\varphi(x_j) = \lim x_{j+1} = x$, since a contraction is continuous. $\square$

Normalisation. Replace $\vec f$ by $\vec x \mapsto D\vec f(\vec a)^{-1}\big(\vec f(\vec x+\vec a) - \vec f(\vec a)\big)$. This is $C^1$, sends $\vec 0$ to $\vec 0$, has derivative $I$ at $\vec 0$, and is invertible near $\vec 0$ if and only if the original is invertible near $\vec a$.

A mean value inequality. If $\varphi$ is $C^1$ on a convex set with $\lVert D\varphi\rVert \le k$ throughout, then $\lVert\varphi(\vec x)-\varphi(\vec x')\rVert \le k\lVert\vec x-\vec x'\rVert$: put $h(t) = \varphi(\vec x'+t(\vec x-\vec x'))$, note $h'(t) = D\varphi(\cdot)(\vec x-\vec x')$ by the chain rule, and integrate componentwise, $\lVert\int_0^1 h'\rVert \le \int_0^1\lVert h'\rVert \le k\lVert\vec x-\vec x'\rVert$.

The contraction. $D\vec f$ is continuous and $D\vec f(\vec 0) = I$, so choose $r > 0$ with $\lVert D\vec f(\vec x) - I\rVert \le \tfrac12$ on the closed ball $\bar B_r$. For $\lVert\vec y\rVert \le r/2$ put $\varphi_{\vec y}(\vec x) = \vec x + \vec y - \vec f(\vec x)$; then $D\varphi_{\vec y} = I - D\vec f$ has norm $\le\tfrac12$, so $\varphi_{\vec y}$ is a $\tfrac12$-contraction, and $\lVert\varphi_{\vec y}(\vec x)\rVert \le \lVert\varphi_{\vec y}(\vec x)-\varphi_{\vec y}(\vec 0)\rVert + \lVert\vec y\rVert \le \tfrac12 r + \tfrac12 r = r$, so it maps $\bar B_r$ into itself. $\bar B_r$ is a complete metric space, so Banach gives a unique fixed point, that is, a unique $\vec x\in\bar B_r$ with $\vec f(\vec x) = \vec y$.

Injectivity and a Lipschitz inverse. Applying the mean value inequality to $\varphi_{\vec 0} = \mathrm{id} - \vec f$ gives $\lVert\vec x-\vec x'\rVert - \lVert\vec f(\vec x)-\vec f(\vec x')\rVert \le \tfrac12\lVert\vec x-\vec x'\rVert$, hence $\lVert\vec f(\vec x)-\vec f(\vec x')\rVert \ge \tfrac12\lVert\vec x-\vec x'\rVert$ on $\bar B_r$. So $\vec f$ is injective there and $\vec g := \vec f^{-1}$ is Lipschitz with constant $2$, in particular continuous. Set $W = B_{r/2}$ and $V = \vec f^{-1}(W)\cap B_r$, both open, with $\vec f : V\to W$ a bijection.

Differentiability of $\vec g$. Fix $\vec x_0\in V$, $\vec y_0 = \vec f(\vec x_0)$, $A = D\vec f(\vec x_0)$; $A$ is invertible since $\lVert A - I\rVert\le\tfrac12$. For small $\vec k$, write $\vec x = \vec g(\vec y_0+\vec k)$, so $\vec k = \vec f(\vec x)-\vec f(\vec x_0) = A(\vec x-\vec x_0) + o(\lVert\vec x-\vec x_0\rVert)$. Hence $\vec g(\vec y_0+\vec k)-\vec g(\vec y_0) = \vec x-\vec x_0 = A^{-1}\vec k - A^{-1}o(\lVert\vec x-\vec x_0\rVert)$, and $\lVert\vec x-\vec x_0\rVert \le 2\lVert\vec k\rVert$ by the Lipschitz bound, so the error term is $o(\lVert\vec k\rVert)$. Therefore $D\vec g(\vec y_0) = A^{-1}$. Finally $\vec y\mapsto D\vec g(\vec y) = (D\vec f(\vec g(\vec y)))^{-1}$ is continuous, being a composition of continuous maps with matrix inversion, so $\vec g\in C^1$. $\square$

04 · constraints

The implicit function theorem

implicit function theorem

Let $G : \mathbb R^{n}\times\mathbb R \to \mathbb R$ be $C^1$ near $(\vec a, b)$ with $G(\vec a,b) = 0$ and $\partial_z G(\vec a, b) \ne 0$. Then near $\vec a$ there is a $C^1$ function $\zeta$ with $\zeta(\vec a) = b$ and $$G(\vec x, z) = 0 \iff z = \zeta(\vec x),$$ together with $\nabla\zeta = -\nabla_{\vec x}G\,/\,\partial_z G$.

proof

Apply the inverse function theorem to $\Phi(\vec x,z) := (\vec x,\,G(\vec x,z))$, whose derivative is block triangular with determinant $\partial_zG \ne 0$. The local inverse has the form $(\vec x,u)\mapsto(\vec x, h(\vec x,u))$, and $\zeta(\vec x) := h(\vec x,0)$ does it. Differentiating $G(\vec x,\zeta(\vec x)) = 0$ by the chain rule gives the formula. $\square$

The theorem in one picture, with $n = 1$. Near a point where $\partial_z G \ne 0$ the level set is a graph over $x$; at the folds, where $\partial_z G = 0$, the tangent is vertical and it is not.

The physics reading: a constraint surface $\{G = 0\}$ is, near any point where $\nabla G \ne \vec 0$, a graph — so the system has $n$ genuine degrees of freedom and $\vec x$ can be used as coordinates on it.

Reading the figure. $G(x,z) = x - z^3 + z$ has $\partial_z G = 1 - 3z^2$, zero at $z = \pm 1/\sqrt3$, the two folds of the S. Inside the dashed window round $(a,b) = (0.231, 1.1)$ the curve meets each vertical line once, so it is the graph of a $C^1$ function $\zeta$, with slope $\zeta' = -\partial_xG/\partial_zG = 1/(3z^2 - 1)$. The theorem promises only some window; how large depends on how far $(a,b)$ is from a fold. Over the middle range of $x$ the whole curve is three-valued, and only the window singles out one branch.

The general form. $G : \mathbb R^n\times\mathbb R^m\to\mathbb R^m$ with $D_{\vec z}G(\vec a,\vec b)$ invertible gives $\vec z = \vec\zeta(\vec x)$ with $D\vec\zeta = -(D_{\vec z}G)^{-1}D_{\vec x}G$; the proof is identical, with the same $\Phi$. The version on the slide is $m = 1$, which is all this unit needs — one constraint, one multiplier. Unit 12 uses the general form to say that a mechanical system with $k$ independent holonomic constraints has $3N-k$ degrees of freedom and that any local coordinates on the constraint surface will do, which is the whole point of Lagrangian mechanics.

The block triangular derivative. $D\Phi = \begin{pmatrix}I_n & 0\\ \nabla_{\vec x}G^{\!\top} & \partial_zG\end{pmatrix}$, whose determinant is $\partial_zG$ by expansion along the last column. That the inverse preserves the first $n$ coordinates is immediate from $\Phi$ doing so.

04 · constraints

The implicit function theorem on a sphere

on the running example's level sets

Apply the theorem to $\Psi(\vec r) = V(\vec r) - V_0$ with $V = -GMm/r$ and $V_0 < 0$ — the theorem's $G$, renamed $\Psi$ here because $G$ in $V$ is Newton's constant. The level set $\{\Psi = 0\}$ is the sphere of radius $R := GMm/(-V_0)$, and $\partial_z \Psi = GMm\,z/r^3$, nonzero exactly when $z \ne 0$.

So the sphere is a graph $z = \zeta(x,y)$ near every point except the equator $z = 0$ — precisely where its tangent plane is vertical and where $z$ is two-valued over $(x,y)$. The hypothesis is not a technicality; it is the exact boundary of where the conclusion holds.

Check. On the upper hemisphere $\zeta = \sqrt{R^2-x^2-y^2}$, and $-\nabla_{\vec x}\Psi/\partial_z\Psi = -(x/z, y/z)$, which is $\nabla\zeta$ ✓.

The level sphere cut by the plane $y = 0$. Away from the equator each half is a graph over $(x,y)$. At the equator $\nabla\Psi$ is horizontal, $\partial_z\Psi = 0$, and every vertical line nearby meets the sphere twice.

Reading the figure. $\nabla\Psi = \nabla V = GMm\,\vec r/r^3$ points radially outwards, so it is normal to the sphere everywhere — never zero, which is why the sphere is a smooth surface at all. What changes at the equator is only its direction: it lies in the $xy$-plane, its $z$-component vanishes, and the theorem's hypothesis fails for the choice of $z$ as the dependent variable. Choosing $x$ instead works there, since $\partial_x\Psi \ne 0$ at $(\pm R, 0, 0)$; the sphere is a graph near every point over some coordinate plane, which is what being a level set of a regular value means.

A regular value. $0$ is a regular value of $G$ if $\nabla G \ne \vec 0$ at every point of $\{G = 0\}$. The theorem then says the level set is, near each of its points, the graph of a $C^1$ function of some $n$ of the coordinates — which is the definition of an $n$-dimensional $C^1$ submanifold of $\mathbb R^{n+1}$. Unit 31 takes that as the definition of a manifold, and this is the theorem that supplies the examples.

04 · constraints

Lagrange multipliers, derived

Lagrange multiplier rule

Let $f, g$ be $C^1$ near $\vec a$, with $g(\vec a) = 0$ and $\nabla g(\vec a)\ne\vec 0$. If $f$ restricted to $\{g = 0\}$ has a local extremum at $\vec a$, then there is $\lambda\in\mathbb R$ with

$$\nabla f(\vec a) = \lambda\,\nabla g(\vec a).$$

proof

Let $\vec v \in \ker Dg(\vec a)$. Since $\nabla g(\vec a)\ne\vec 0$, the implicit function theorem makes $\{g=0\}$ a $C^1$ graph near $\vec a$, and that graph supplies a $C^1$ curve $\vec\gamma$ in $\{g=0\}$ with $\vec\gamma(0)=\vec a$, $\dot{\vec\gamma}(0)=\vec v$ (constructed in the notes). Then $t\mapsto f(\vec\gamma(t))$ has a local extremum at $0$, so $\nabla f(\vec a)\cdot\vec v = 0$.

Hence $\nabla f(\vec a) \perp \ker Dg(\vec a) = \big(\mathrm{span}\,\nabla g(\vec a)\big)^{\perp}$, and taking orthogonal complements gives $\nabla f(\vec a)\in\mathrm{span}\,\nabla g(\vec a)$. $\square$

$f = x + y$ on the ellipse $g = x^2/4 + y^2 - 1 = 0$. Where $\nabla f$ has a component along the curve, moving that way raises $f$. At the maximum that component is zero: a level line of $f$ touches the ellipse and $\nabla f = \lambda\nabla g$.

The orthogonal-complement step. $Dg(\vec a)$ is a nonzero linear functional, so its kernel is a hyperplane, of dimension $n-1$. Writing $Dg(\vec a)\vec v = \nabla g(\vec a)\cdot\vec v$, the kernel is exactly $\{\vec v : \vec v \perp \nabla g(\vec a)\} = (\mathrm{span}\,\nabla g(\vec a))^\perp$. In finite dimensions $(W^\perp)^\perp = W$, so $(\ker Dg(\vec a))^\perp = \mathrm{span}\,\nabla g(\vec a)$, and $\nabla f(\vec a)$, being orthogonal to the kernel, lies in it.

Constructing the curve. Say $\partial_n g(\vec a)\ne0$. The implicit function theorem gives $\zeta$ with $g(\vec x', \zeta(\vec x')) = 0$ near $\vec a' $, where $\vec x' $ is the first $n-1$ coordinates. For $\vec v\in\ker Dg(\vec a)$ put $\vec\gamma(t) = \big(\vec a' + t\vec v',\ \zeta(\vec a'+t\vec v')\big)$. Then $\vec\gamma$ lies in $\{g=0\}$, is $C^1$, starts at $\vec a$, and its last component has derivative $\nabla\zeta\cdot\vec v' = -(\nabla_{\vec x'}g\cdot\vec v')/\partial_ng = v_n$, the last step because $\nabla g\cdot\vec v = 0$. So $\dot{\vec\gamma}(0) = \vec v$ as required.

Several constraints. With $g_1,\dots,g_k$ whose gradients at $\vec a$ are linearly independent, the same argument gives $\nabla f(\vec a) = \sum_i\lambda_i\nabla g_i(\vec a)$: the tangent space is the intersection of the $k$ kernels, whose orthogonal complement is the span of the $k$ gradients. Independence of the gradients is the constraint qualification, and it is what the implicit function theorem needs to apply.

04 · constraints

What the multiplier is: the force of the constraint

what $\lambda$ is, physically

A particle rests on a smooth constraint surface $\{g=0\}$ under $\vec F = -\nabla V$. "Smooth" means it can only push normally, so the constraint force is $\vec N = \nu\,\nabla g$ for some number $\nu$. Equilibrium, $-\nabla V + \vec N = \vec 0$, then reads $\nabla V(\vec a) = -\nu\,\nabla g(\vec a)$ — exactly Lagrange's condition with $f = V$, $\lambda = -\nu$, and $\lVert\vec N\rVert = \lvert\lambda\rvert\,\lVert\nabla g\rVert$.

Checks. $\lambda$ carries $[f]/[g]$, so $\lambda\nabla g$ has the units of $\nabla f$ ✓. Special case $\nabla g(\vec a) = \vec 0$: the rule fails and must — $g = x^2+y^2$ constrains to the origin, where $f$ is trivially extremal with no condition on $\nabla f$ ✓.

The multiplier is not bookkeeping. It is the force the constraint exerts, and in unit 12 it is what one computes when the normal force is the answer wanted.

The next slide's bead, at rest in the rotating frame, forces drawn to scale. Being smooth, the hoop can only push normally to itself, here towards the centre; its push exactly cancels the other two, and its size is what the multiplier measures.

"Smooth surface" is a physical idealisation, not a theorem. It says the constraint does no work, which is d'Alembert's principle: the constraint force is orthogonal to every virtual displacement, and the virtual displacements are exactly the tangent vectors to $\{g=0\}$. Unit 12 takes that as the starting axiom and derives the Lagrange equations from it, at which point multipliers reappear as the systematic way to keep the constraint forces when you want them.

Reading the figure. The bead of the worked example, $m = 0.050$ kg on a hoop of radius $0.20$ m spun at $12\ \mathrm{s^{-1}}$, at its tilted equilibrium $70.09^\circ$ from the bottom. In the rotating frame it feels its weight $mg = 0.4905$ N and the centrifugal force $m\Omega^2 x_* = 1.354$ N; their sum is normal to the hoop only at an equilibrium — that is what equilibrium on a smooth constraint means — and there a normal force can cancel it. The multiplier is the size of that normal force divided by the length of the constraint function's gradient, which for $x^2 + z^2 - R^2$ is $2R$ on the hoop.

04 · constraints · worked example

Worked: a bead on a rotating hoop

worked: equilibria and the hoop's force

Step 1 — the effective potential. At equilibrium the bead is at rest, so Coriolis vanishes; the centrifugal term $m\Omega^2x\,\hat x$ is $-\nabla(-\tfrac12 m\Omega^2x^2)$ (notes). So $$V_{\mathrm{eff}} = mgz - \tfrac12 m\Omega^2x^2 ,\qquad c(x,z) = x^2+z^2-R^2 ,$$ writing $c$ for the rule's constraint function, since $g$ here is gravity.

Step 2 — Lagrange's condition. $\nabla V_{\mathrm{eff}} = (-m\Omega^2x,\ mg)$, $\nabla c = (2x,\,2z)$: so $-m\Omega^2 x = 2\lambda x$ and $mg = 2\lambda z$. Step 3 — two branches. Either $x = 0$ (bottom and top), or $\lambda = -\tfrac12 m\Omega^2$, whence $z_* = -g/\Omega^2$ — on the hoop only if $\Omega^2 \ge g/R$.

Step 4 — the numbers. $\Omega^2R/g = 2.936 > 1$, so the tilted pair exists: $z_* = -0.068125$ m, $x_* = 0.18804$ m, angle from the bottom $\arccos(0.340625) = \mathbf{70.09^\circ}$. Step 5 — the force. $\lVert\vec N\rVert = \lvert\lambda\rvert\lVert\nabla c\rVert = \tfrac12 m\Omega^2\cdot 2R = m\Omega^2R = \mathbf{1.440}\ \mathrm N$.

A bead of $m = 0.050$ kg on a smooth hoop of radius $R = 0.20$ m, spun about a vertical diameter at $\Omega = 12\ \mathrm{s^{-1}}$. Work in unit 03's rotating frame: $(x,z)$ in the plane of the hoop, $z$ up, axis at $x = 0$.

Equilibrium angle against spin rate, from $\cos\theta_* = g/(\Omega^2R)$. Above $\Omega_c = \sqrt{g/R} = 7.004\ \mathrm{s^{-1}}$ the bottom turns unstable and a tilted pair appears: one equilibrium becomes three, a pitchfork bifurcation. (The top, always unstable, is not drawn.)

Why the centrifugal term is a gradient. Unit 03's centrifugal force is $-m\vec\Omega\times(\vec\Omega\times\vec r)$. Writing $\vec\rho$ for the component of $\vec r$ perpendicular to $\vec\Omega$, the double cross product is $-\Omega^2\vec\rho$, so the force is $m\Omega^2\vec\rho$, of magnitude $m\Omega^2\rho$ and directed away from the axis. And $\nabla\big(\tfrac12 m\Omega^2\rho^2\big) = m\Omega^2\vec\rho$, so the force is $-\nabla\big(-\tfrac12 m\Omega^2\rho^2\big)$ as claimed. Here the axis is the $z$-axis and $\rho = \lvert x\rvert$ in the plane of the hoop.

Stability. Parametrise the hoop by the angle $\theta$ from the bottom: $x = R\sin\theta$, $z = -R\cos\theta$, and $V_{\mathrm{eff}}(\theta)/m = -gR\cos\theta - \tfrac12\Omega^2R^2\sin^2\theta$. Then $V'_{\mathrm{eff}}/m = R\sin\theta\,(g - \Omega^2R\cos\theta)$, which vanishes at $\theta = 0$, $\pi$ and $\cos\theta = g/(\Omega^2R)$ — the same three branches. $V''_{\mathrm{eff}}(0)/m = R(g - \Omega^2R)$, negative once $\Omega^2 > g/R$, so the bottom turns from a minimum into a maximum exactly where the tilted pair appears; at the tilted point $V''_{\mathrm{eff}}/m = \Omega^2R^2\sin^2\theta_* > 0$, a minimum. That exchange of stability — one stable equilibrium splitting into a stable symmetric pair as a parameter crosses a threshold, with the original turning unstable — is what is called a supercritical pitchfork bifurcation (unit 02 classifies such events), with $\Omega$ as the parameter, and it is the reason a centrifugal governor works.

04 · constraints · worked example

The hoop: checks, and the view from the lab

checks

Dimensions. $m\Omega^2R$ is $\mathrm N$, and so is $\lambda\lVert\nabla c\rVert$ ✓. Limits. At $\Omega^2 = g/R$, $z_* = -R$ and $\lVert\vec N\rVert = mg$: the hoop merely carries the weight ✓. As $\Omega\to\infty$, $z_*\to0$ and $\lVert\vec N\rVert\to\infty$ ✓.

two frames, one number

In the lab the bead runs round a horizontal circle of radius $x_* = 0.188$ m at $12\ \mathrm{s^{-1}}$. That needs a force $m\Omega^2x_* = 1.354$ N towards the axis, and the weight $mg = 0.4905$ N must be held up. The hoop's force does both: $\sqrt{1.354^2 + 0.4905^2} = 1.440$ N ✓, the multiplier's answer.

The same equilibrium seen from the lab, forces to scale. No centrifugal force here: the horizontal part of the hoop's push is what keeps the bead on its circle.

What the $1.440$ N is, from the inertial frame. The bead moves in a horizontal circle of radius $x_* = 0.188$ m at $12\ \mathrm{s^{-1}}$, so it needs a centripetal force $m\Omega^2x_* = 0.050\cdot144\cdot0.188 = 1.354$ N horizontally, and the hoop must also support the weight $mg = 0.4905$ N. The hoop's force is normal to the hoop, so it also has the components to do both: its magnitude is $\sqrt{1.354^2 + 0.4905^2} = 1.440$ N ✓. Two frames, one number.

A small honesty. The hoop's force here is normal to the circle, in the plane of the hoop. A real hoop must additionally push the bead sideways, out of that plane, to keep it rotating — that is the Coriolis force in the rotating frame, which is zero at equilibrium because the bead is at rest there, and is what the hoop supplies as soon as the bead slides.

Reading the figure. Compare it with the figure on What the multiplier is: the force of the constraint. There, in the rotating frame, three forces cancel; here, in the lab, two forces — weight and the hoop's push — do not cancel, and their sum, $1.354$ N horizontally inwards, is mass times the centripetal acceleration $\Omega^2x_*$. The hoop's arrow is the same arrow in both pictures, because the force a body exerts does not depend on the frame it is described in.

04 · what to keep

What to keep

the five lines worth memorising

$\int_{\vec\gamma}\nabla V\cdot\mathrm d\vec r = V(\text{end})-V(\text{start})$

$\langle\nabla f,\vec h\rangle = Df\,\vec h$ — the gradient is the derivative with the inner product applied

$\mathrm{curl}\,\vec F$ is twice the antisymmetric part of $D\vec F$, read through the hat map

$\oint_{\partial S}\vec F\cdot\mathrm d\vec r = \int_S(\mathrm{curl}\,\vec F)\cdot\hat n\,\mathrm dA$

$\nabla f = \lambda\nabla g$ at a constrained extremum, and $\lambda$ is the constraint force

And one sentence: closed is about the field, conservative is about the field and the domain.

What this unlocks.

05: the same programme for flux — divergence, Gauss, and gravity as a field equation rather than a force law.
08: energy as one of the two first integrals that solve the central-force problem completely.
11–12: the potential $V$ as half of a Lagrangian, and constraint forces as multipliers.
16: the Hessian's transformation law, generalised — a tensor is a multilinear map.
23–24: Stokes' theorem doing real work, as Ampère's law and as the curl equations of Maxwell.
34: all three integral theorems as one, and the punctured plane's $2\pi$ explained.

If one thing survives from this unit, make it the last sentence on the left. It is the first time in this course that the answer to a question about a formula depends on the shape of the set the formula lives on, and that pattern — local data, global obstruction — is the whole of units 33 to 40. Curvature is the local data; whether a spacetime can be globally flattened is the obstruction.

The second thing to keep is the habit rather than a formula: define the object as a map, prove the components. It was used three times here — the gradient through Riesz, the curl through the hat map, the Hessian as a bilinear form — and it is used for every object in units 16 onwards. When a later unit's index gymnastics stops making sense, the fix is almost always to go back and ask what map the indices are the components of.

04 · checkpoint

Done when

checkpoint

Given a vector field $\vec F$ on an open set $U \subseteq \mathbb R^n$, you decide whether $\vec F$ is conservative on $U$ and defend the answer, by

producing $V$ explicitly, with $-\nabla V = \vec F$ verified by differentiation; or
exhibiting a specific closed loop in $U$ and computing its nonzero circulation;
including at least one case in which the verdict changes when $U$ changes and $\vec F$ does not.

Falsifiable: "the curl is zero so it is conservative" fails it whenever $U$ is not simply connected, and a potential that does not differentiate back to $\vec F$ fails it always. Record it with ./mr done 04.

Problems

./mr problems 04 — 13 problems, about 5 hours. Hints and worked solutions fold under each problem on the same page.

05 flux, Gauss, and gravity as a field · 08 the central-force problem · 11 the brachistochrone, where the variable is a curve · 34 where the vortex's $2\pi$ is finally explained.

Try the checkpoint cold, on paper, on this trio before looking at the problem sheet. $\vec F_1 = (2xy, x^2)$ on $\mathbb R^2$; $\vec F_2 = (y, -x)$ on $\mathbb R^2$; $\vec F_3 = (-y, x)/(x^2+y^2)$ on the right half-plane $\{x > 0\}$, and then on $\mathbb R^2\setminus\{0\}$. The answers are: conservative with $V = -x^2y$; not conservative, and the unit circle gives $-2\pi$; conservative on the half-plane with $V = -\arctan(y/x)$, and not conservative on the punctured plane, where the unit circle gives $2\pi$. The third field is the one that tests whether the lesson landed: the field did not change.

If the derivation of any of the section-3 results stalls, the usual place is Green's theorem for a rectangle. Redo it from the fundamental theorem of calculus in one variable at a time; every other result in that section is built on it, including the Poincaré lemma's two staircase paths.

Work, and the arrival of vector calculus

Two punctured spaces, five sections

Work: the integral along a path

Why an integral along a path appears at all

The line integral, defined

Path matters: one field, three answers

Path matters: the loop between two paths

The work–energy theorem

The derivative is a covector; the gradient is what the inner product makes of it

The gradient in polar coordinates

The fundamental theorem of line integrals

Worked: the inverse-square field has a potential

Worked: the same answer on a deliberately awkward path

The awkward path: the work along the way

Conservative fields, and the first integral they give

Three conditions, one property

Energy is a first integral in three dimensions

Two forces outside the frame, for opposite reasons

The curl, and how much a local test can promise

A necessary condition, free from Clairaut

What the curl is

Turning and straining: the two parts of the derivative

Curl is a vector only under rotations

Green's theorem in the plane

Stokes' theorem in $\mathbb R^3$

Stokes' theorem: checks, and the general surface

The curl is circulation per unit area

Worked: the vortex, in full

The vortex: closed everywhere, and still no potential

The Poincaré lemma

What the vortex measures is the domain

Closed, conservative, and the holes between them

Potential landscapes

Taylor's theorem in $\mathbb R^n$

The Hessian is a symmetric bilinear form

Classifying a nondegenerate critical point

Equilibrium, stability, and Lagrange–Dirichlet

Lagrange–Dirichlet: the proof in a picture

Worked: two masses, and why the middle is a saddle

The saddle between two masses, in numbers

Constraints, and where the multiplier comes from

The inverse function theorem

The implicit function theorem

The implicit function theorem on a sphere

Lagrange multipliers, derived

What the multiplier is: the force of the constraint

Worked: a bead on a rotating hoop

The hoop: checks, and the view from the lab

What to keep

Done when

Problems

Next