Part B · the action · branch

Hamilton–Jacobi: mechanics was optics all along, and free motion is already geodesic motion

One nonlinear PDE replaces the equations of motion; its solutions are wavefronts, its trajectories are rays, and a particle in a potential is a free particle in a rescaled geometry.

needs 14, 11~2½ h slides + 6 h problemsproblems: ./mr problems 15

Part B's thesis is that nothing in it says anything Newton did not, and that the change of language is worth having for what it makes obvious. This unit is the sharpest case. It contains no new physics at all. What it contains is three things that are invisible in $m\ddot{\vec x} = -\nabla V$ and obvious here.

Frequencies without motion. The frequency of a bounded motion is the derivative of the energy with respect to an area in phase space. You can compute a period without ever finding $x(t)$, and you can compute how it shifts when the potential is perturbed.
Invariance under slow change. The same area is what survives when a parameter of the system is changed slowly — the fact Einstein used in 1911 and Bohr and Sommerfeld built the old quantum theory on.
Geometry. Trajectories at fixed energy are the rays of an optical medium, and the geodesics of a metric. Force-driven motion is free motion in the right geometry — in Newtonian mechanics, a century before general relativity made that the definition of gravity.

It is a branch: no later unit's proofs depend on it except unit 40's separation of Kerr geodesics. But unit 32 will define a geodesic, and unit 30 will argue that gravity is geometry, and both are easier to believe once you have seen, here, that a Kepler ellipse is already a geodesic.

How to read this deck

One problem runs through it: the Kepler problem, instantiated on unit 08's Halley's comet. Section 1 turns the action into a function and writes its PDE; section 2 solves that PDE for Kepler by separation and reads off the orbit with no equation of motion integrated; section 3 extracts the action variables and gets the period by differentiation; section 4 asks what survives slow change, and lets the Sun lose mass; section 5 turns the whole thing into optics and then into geometry, and shows Halley's ellipse as a geodesic.

Proofs are on the slides where they are one line and in these notes under Full proof where they are not; results imported from elsewhere are marked cited. Terms with a dotted underline are in the glossary (G); N opens these notes.

What you are assumed to have. Unit 14's Hamilton's equations, Poisson brackets and type-2 generating functions; unit 11's Euler–Lagrange equation in several dependent variables, the Beltrami identity and conjugate points; unit 12's form-invariance under change of coordinates; unit 08's Binet equation; unit 04's implicit function theorem and line integrals; unit 03's chain rule and Clairaut's theorem; unit 00's improper integrals at a turning point.

15 · overview

One problem, solved without its equation of motion

the move

Make the action a function of its endpoint, $S(q, t)$. It obeys one first-order PDE, $\ \partial_t S + H(q, \partial_q S, t) = 0$. Find enough solutions and the trajectories come out by differentiation — no ODE is integrated.

getting around

A dotted underline is a glossary term: click it, or press G. N opens the notes, where the longer proofs live. A solid underline, like the running example, shows a piece of it on hover.

The plan. 1 the action as a function, and its PDE · 2 separation of variables · 3 action–angle variables: frequencies from areas · 4 adiabatic invariants · 5 wavefronts, rays, and a metric whose geodesics are the trajectories.

the running example · Halley's comet, in four moves

Unit 08's comet, $V = -k/r$ with $k/m = GM_\odot$, in spherical coordinates. No orbit equation is integrated:

1. Guess $S = -Et + L_z\phi + \Theta(\theta) + R(r)$: the PDE splits into three ODEs with constants $\alpha = (E, L, L_z)$. Setting each $\partial S/\partial\alpha$ equal to a constant $\beta$ gives the conic, its tilted plane and its timing.

2. The action $I = \frac1{2\pi}\oint p\,\mathrm dq$ of each coordinate is one integral; their sum makes $H = -mk^2/2I^2$, and the period $2\pi/(\partial H/\partial I) = 75.305$ yr. All three frequencies are equal, so the orbit closes.

3. A Sun losing mass slowly keeps $I$ (adiabatically) and $L$ (exactly), so $M_\odot a$ is fixed: the orbit grows $24$ cm a year.

4. At fixed $E$ the ellipse is a geodesic of $2m(E + k/r)\times$ the flat metric.

The order of the deck is the order of the logic, and the logic has one hinge: Jacobi's theorem, in section 1. Everything afterwards — separation, action variables, adiabatic invariants, the optical picture — is a way of finding solutions of the Hamilton–Jacobi equation or a way of reading them. If Jacobi's theorem is not solid, nothing after it is either, so spend time there.

The four moves, slide by slide. Move 1 is "Kepler, separated" (the guess, and the lemma that splits the equation) followed by "Worked: the orbit, read off" (the $\beta$-equations give the conic and its plane) and "Six constants, six orbital elements". Move 2 is "Two integrals, done once" and "Worked: Kepler's actions, and Kepler's third law", with the closing of the orbit on "Three equal frequencies: why Kepler orbits close". Move 3 is "Kepler, slowly: a Sun that loses mass": $I = m\sqrt{GM_\odot a}$ and $L = I\sqrt{1 - e^2}$ both stay put, so $e$ and $M_\odot a$ do too. Move 4 is "The Jacobi metric" and "Worked: Kepler orbits are geodesics". The symbols $\alpha$, $\beta$ and $I$ are defined properly on "Complete integrals, and the solutions they generate", "Jacobi's theorem" and "The action variable".

Every number on this slide is computed, not quoted, from unit 08's data for the comet: $GM_\odot = 1.32712440\times10^{20}\ \mathrm{m^3\,s^{-2}}$, $\varepsilon = E/m = -2.48740\times10^{7}\ \mathrm{J\,kg^{-1}}$ and $h = L/m = 4.78403\times10^{15}\ \mathrm{m^2\,s^{-1}}$. The period comes out as $75.305$ yr, agreeing with unit 08's $75.31$ yr; the orbital growth uses a solar mass-loss rate of $5.8\times10^{9}\ \mathrm{kg\,s^{-1}}$, discussed on its slide.

Notation. $m$ throughout is the orbiting mass — for the two-body problem, unit 08's reduced mass $\mu$. It is written $m$ here to keep formulas short, and because nothing in this unit is specific to two bodies.

section 1

The action as a function of where you end

Hamilton's principal function, its PDE, and the theorem that turns solutions into trajectories.

15 · the action function

The action, as a function of its endpoint

definition

Fix a start $(q_0, t_0)$. Suppose that for every $(q, t)$ in an open set $U$ there is exactly one solution of the Euler–Lagrange equations running from $q_0$ at $t_0$ to $q$ at $t$, depending $C^2$ on $(q,t)$. Hamilton's principal function is the action along it,

$$S(q, t) = \int_{t_0}^{t} L\big(q(s), \dot q(s), s\big)\,\mathrm{d}s .$$

theorem $$\frac{\partial S}{\partial q_i} = p_i, \qquad \frac{\partial S}{\partial t} = -H\big(q, p, t\big), \qquad\text{i.e.}\quad \mathrm dS = p\cdot\mathrm dq - H\,\mathrm dt ,$$

with $p$ the momentum with which the true path arrives at $(q, t)$.

unit 11's first variation with a free endpoint, then the chain rule along the path — full proof in the notes (N)

Why a physicist wants this: $S$ knows every trajectory from $q_0$ at once, and its gradient is the momentum field of that whole family.

Full proof. The $q$-derivative. Move the endpoint from $q$ to $q + \varepsilon e_i$ at fixed $t$. The true paths $q_\varepsilon(s)$ form a $C^2$ family with $q_\varepsilon(t_0) = q_0$ for all $\varepsilon$, so $\eta(s) = \partial_\varepsilon q_\varepsilon(s)|_{\varepsilon=0}$ has $\eta(t_0) = 0$ and $\eta(t) = e_i$. Differentiating under the integral (legitimate: everything is $C^1$ on a compact interval), $$\frac{\partial S}{\partial q_i} = \int_{t_0}^{t}\big(L_q\cdot\eta + L_{\dot q}\cdot\dot\eta\big)\,\mathrm ds = \Big[L_{\dot q}\cdot\eta\Big]_{t_0}^{t} + \int_{t_0}^{t}\Big(L_q - \frac{\mathrm d}{\mathrm ds}L_{\dot q}\Big)\cdot\eta\,\mathrm ds .$$ The integral vanishes because the path satisfies Euler–Lagrange, and the boundary term is $p(t)\cdot e_i - p(t_0)\cdot 0 = p_i(t)$. This is unit 11's natural-boundary computation with the endpoint moved rather than freed.

The $t$-derivative. Take one true path $q(\cdot)$ and a later time $t'$ on it. By uniqueness, the true path from $(q_0,t_0)$ to $(q(t'), t')$ is the same path restricted, so $S(q(t'), t') = \int_{t_0}^{t'} L\,\mathrm ds$ and its $t'$-derivative is $L$. By the chain rule (unit 03) the same derivative is $\partial_t S + \sum_i \partial_{q_i}S\,\dot q_i = \partial_t S + p\cdot\dot q$. Hence $\partial_t S = L - p\cdot\dot q = -H$, by the definition of $H$ as the Legendre transform (unit 14). $\square$

The next slide draws the hypothesis — one path to each endpoint — for two systems, one where it always holds and one where it fails.

15 · the action function

Fans of paths: where S is a function, and where it stops

Free particle: the path to $(q,t)$ is a straight line, and

$$S = \frac{m(q-q_0)^2}{2(t-t_0)},\quad \partial_q S = m\frac{q - q_0}{t - t_0} = p .$$

One line to each endpoint, at any $t > t_0$. At the marked one the slope is $p/m$, the $p$ that $\partial_q S$ gives. Dashed: levels of $S$; $m = 1$, $q_0 = t_0 = 0$.

Oscillator, frequency $\omega$, with $T = t - t_0$: infinite at $\omega T = \pi$,

$$S = \frac{m\omega\big[(q^2 + q_0^2)\cos\omega T - 2qq_0\big]}{2\sin\omega T} .$$

Paths leaving $q_0$ with every initial velocity. Before $\omega T = \pi$ they fan out and each endpoint has one path; at $\omega T = \pi$ all of them pass through $-q_0$ — a conjugate point (unit 11) — and $S$ is not a function there.

What to look for. The theorem on the previous slide needs "exactly one true path from $(q_0, t_0)$ to $(q,t)$, depending smoothly on the endpoint". On the left that is visibly true: the lines from the start never cross again, so the endpoint names the line, and the line's slope is the arriving velocity — which is why $\partial_q S$ can be a momentum. On the right it holds only for $\omega T < \pi$. The fan of oscillator paths re-converges at half a period, the endpoint $-q_0$ no longer determines the path, and every other endpoint at that instant is reached by none.

Where the hypothesis bites. This is the general picture: the hypothesis holds for $t$ close to $t_0$ and fails at a conjugate point (unit 11). For the oscillator, $S$ above blows up at $\omega T = \pi$ because $\sin\omega T = 0$ in the denominator; that is the conjugate point in the formula. Problem P1 computes the free particle and the uniform field, where no conjugate point ever occurs.

Check. Free particle, $L = \tfrac12 m\dot q^2$: $\partial_q S = m(q-q_0)/(t-t_0)$ is $m$ times the line's slope, i.e. $p$, and $\partial_t S = -m(q-q_0)^2/2(t-t_0)^2 = -p^2/2m = -H$. Dimensions: $[S]$ is energy × time, so $[\partial_q S] = \mathrm{kg\,m\,s^{-1}}$ and $[\partial_t S] = \mathrm J$, as they must be. Oscillator, limit $\omega\to0$: $\cos\omega T\to1$ and $\sin\omega T\approx\omega T$, so $S\to m(q - q_0)^2/2T$, the free particle ✓.

The figures. Both are computed. Left: lines $q = vt$ for $v = -2, -1.5, \dots, 2$, and the curves $S = q^2/2t = \tfrac14, \tfrac34, \tfrac74$; the marked endpoint $(t, q) = (1.5, 1.5)$ sits on the line of slope $1$, and $\partial_q S = q/t = 1$ there. Right: $q(t) = q_0\cos\omega t + (v_0/\omega)\sin\omega t$ with $\omega = q_0 = 1$ and $v_0$ from $-2$ to $2$.

15 · the action function

Kepler's Hamiltonian in spherical coordinates

spherical coordinates

$(r, \theta, \phi)\mapsto \vec x = (r\sin\theta\cos\phi,\ r\sin\theta\sin\phi,\ r\cos\theta)$ on $r > 0$, $0 < \theta < \pi$, $\phi$ mod $2\pi$. The columns of its derivative are

$$\partial_r\vec x = \hat e_r,\quad \partial_\theta\vec x = r\,\hat e_\theta,\quad \partial_\phi\vec x = r\sin\theta\,\hat e_\phi$$

with $\hat e_r, \hat e_\theta, \hat e_\phi$ orthonormal. So by the chain rule

$$\lVert\dot{\vec x}\rVert^2 = \dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\,\dot\phi^2 .$$

the running example

$L = \tfrac12 m\lVert\dot{\vec x}\rVert^2 + k/r$ gives $p_r = m\dot r$, $p_\theta = mr^2\dot\theta$, $p_\phi = mr^2\sin^2\theta\,\dot\phi$, and

$$H = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} + \frac{p_\phi^2}{2mr^2\sin^2\theta} - \frac kr ,$$

with $p_\phi = L_z$ and $p_\theta^2 + p_\phi^2/\sin^2\theta = \lVert\vec L\rVert^2$.

Moving $\theta$ alone sweeps a meridian of radius $r$; moving $\phi$ alone sweeps a circle of latitude of radius $r\sin\theta$. Those radii are the factors in $r\,\hat e_\theta$ and $r\sin\theta\,\hat e_\phi$.

Spherical coordinates are needed here only as generalised coordinates in unit 12's sense: a chart on $\mathbb R^3$ minus the $z$-axis. Unit 05 uses them for integration, with Jacobians; nothing of that is needed, only the chain rule.

Full proof of the speed formula. Differentiate $\vec x = r(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$: $\partial_r\vec x = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta) =: \hat e_r$, $\partial_\theta\vec x = r(\cos\theta\cos\phi, \cos\theta\sin\phi, -\sin\theta) =: r\hat e_\theta$, $\partial_\phi\vec x = r\sin\theta(-\sin\phi, \cos\phi, 0) =: r\sin\theta\,\hat e_\phi$. Each hatted vector has length 1 (sum of squares, using $\sin^2 + \cos^2 = 1$ twice), and the three pairwise dot products are $\sin\theta\cos\theta\cos^2\phi + \sin\theta\cos\theta\sin^2\phi - \sin\theta\cos\theta = 0$, $-\sin\theta\sin\phi\cos\phi + \sin\theta\sin\phi\cos\phi = 0$ and $-\cos\theta\sin\phi\cos\phi + \cos\theta\sin\phi\cos\phi = 0$. By the chain rule $\dot{\vec x} = \dot r\,\hat e_r + r\dot\theta\,\hat e_\theta + r\sin\theta\,\dot\phi\,\hat e_\phi$, and Pythagoras in an orthonormal basis gives the formula. $\square$

Where the chart fails. At $r = 0$ all three columns degenerate; on the $z$-axis $\partial_\phi\vec x = 0$. The derivative has rank less than 3 there, and $(r,\theta,\phi)$ is not a chart. This is why $\phi$'s momentum $p_\phi = mr^2\sin^2\theta\,\dot\phi$ has the factor that vanishes on the axis.

Reading the figure. The three coordinate curves through a point — the ray ($r$ varies), the meridian ($\theta$ varies) and the circle of latitude ($\phi$ varies) — meet at right angles there, which is the orthonormality of the hatted vectors. The speed along each is the radius of the circle it runs on times the angular rate: $r\dot\theta$ on the meridian, $r\sin\theta\,\dot\phi$ on the latitude circle, whose radius is the distance $\rho = r\sin\theta$ to the $z$-axis. That circle shrinks to a point on the axis, which is the chart's failure drawn.

The momentum identities. $p_\phi = m(x\dot y - y\dot x) = L_z$: with $\rho = r\sin\theta$, $x = \rho\cos\phi$, $y = \rho\sin\phi$, one gets $x\dot y - y\dot x = \rho^2\dot\phi$. For $\lVert\vec L\rVert^2$, unit 08's identity $\lVert\vec r\times\vec p\rVert^2 = r^2\lVert\vec p\rVert^2 - (\vec r\cdot\vec p)^2$ with $\vec p = p_r\hat e_r + (p_\theta/r)\hat e_\theta + (p_\phi/r\sin\theta)\hat e_\phi$ and $\vec r\cdot\vec p = rp_r$ gives $\lVert\vec L\rVert^2 = p_\theta^2 + p_\phi^2/\sin^2\theta$.

The Legendre transform. $L$ is a positive-definite quadratic form in the velocities plus a function of position, so unit 14's transform is $H = T + V$ with each $\dot q_i$ replaced by $p_i$ divided by its coefficient; that is the displayed $H$.

15 · the action function

The Hamilton–Jacobi equation

definition

Given $H(q, p, t)$ on an open set of $\mathbb R^n\times\mathbb R^n\times\mathbb R$, the Hamilton–Jacobi equation is the PDE for an unknown $C^2$ function $S(q, t)$:

$$\frac{\partial S}{\partial t} + H\Big(q,\ \frac{\partial S}{\partial q},\ t\Big) = 0 .$$

A solution on an open $U\subset\mathbb R^{n+1}$ is a $C^2$ function making the left side vanish identically on $U$.

corollary of the action-function theorem

Hamilton's principal function solves it: substitute $\partial_q S = p$ into $\partial_t S = -H(q,p,t)$.

what kind of equation

First order, and nonlinear: $H$ is quadratic in $p$, so $S$ enters squared. Unit 05's Poisson equation was linear; this one is not — no superposition.

For the running example, with $H$ from the previous slide: $\ \partial_t S + \dfrac{1}{2m}\Big[(\partial_r S)^2 + \dfrac{(\partial_\theta S)^2}{r^2} + \dfrac{(\partial_\phi S)^2}{r^2\sin^2\theta}\Big] - \dfrac{k}{r} = 0 .$

Why trade an ODE for a PDE? It looks like a bad bargain: Hamilton's equations are $2n$ ordinary differential equations, and this is a partial differential equation, a harder kind of object. The trade pays because of what a single solution buys. One solution $S$ carries a whole $n$-parameter family of trajectories (all those with $p = \partial_q S$), and one complete solution, defined on the complete-integrals slide, carries all of them. And the PDE can often be solved by a trick — separation of variables — that has no analogue for the ODEs. Kepler in spherical coordinates is the showcase.

History. Hamilton found the principal function in 1834–35, in two papers "On a general method in dynamics", having found the characteristic function first in optics in the 1820s — the optical origin is section 5's subject. He had two PDEs, one for each endpoint. Jacobi (1837) saw that one suffices, that any sufficiently rich solution will do rather than the action itself, and that this is what makes the method computational. The equation carries both names for that reason.

On the word "nonlinear". If $S_1$ and $S_2$ solve it, $S_1 + S_2$ generally does not: $(\partial_q(S_1+S_2))^2 \ne (\partial_q S_1)^2 + (\partial_q S_2)^2$. What does survive is adding a constant, since only derivatives of $S$ appear — which is why every complete integral below carries an inessential additive constant, never counted among its parameters.

Check, dimensions. Each term has the dimension of energy: $\partial_t S$ is action per time, and $(\partial_r S)^2/2m$ is momentum squared over mass.

15 · the action function

Complete integrals, and the solutions they generate

definition

A complete integral is a family $S(q, \alpha, t)$, $\alpha\in\mathbb R^n$, of solutions, $C^2$ in all variables, with $\ \det\big(\partial^2 S/\partial q_i\partial\alpha_j\big) \ne 0$. (An additive constant is not counted.)

lemma · envelopes are solutions

If each $S(\cdot,\cdot,a)$ solves the equation and $a(q,t)$ is $C^1$ with $\partial_a S(q,t,a(q,t)) = 0$, then $\Sigma(q,t) = S(q,t,a(q,t))$ solves it too.

Proof. $\partial_q\Sigma = \partial_q S + \partial_a S\,\partial_q a = \partial_q S$, and likewise $\partial_t\Sigma = \partial_t S$. $\square$

The general solution depends on an arbitrary function; envelopes of complete integrals over sub-families produce it.

What the determinant condition says. $\partial^2 S/\partial q\,\partial\alpha$ nonsingular means the map $\alpha\mapsto p = \partial_q S(q,\alpha,t)$ is a local diffeomorphism at each $(q,t)$ (inverse function theorem, unit 04): the family of momentum fields reaches every nearby momentum at every point. That is exactly what makes it rich enough to contain every trajectory, which is the content of Jacobi's theorem next. A family like $S = \alpha_1(q_1 + q_2) - \dots$ that depends on its parameters only through one combination fails it.

The general solution — cited. For a first-order PDE in $n$ variables, a solution depending on an arbitrary function of $n - 1$ variables is produced from any complete integral by restricting $\alpha$ to a sub-family and taking the envelope, as in the lemma; Lagrange showed that locally every solution which is not "singular" arises this way. Proof: Courant and Hilbert, Methods of Mathematical Physics, vol. II, chapter II, §4. This unit never needs the general solution: Jacobi's theorem needs only a complete integral, and that is the whole reason complete integrals are the object of interest.

15 · the action function

The free particle: plane waves and their envelope

a complete integral

$S_a = a(q - q_0) - \dfrac{a^2(t - t_0)}{2m}$ solves $\partial_t S + (\partial_q S)^2/2m = 0$ for every $a$, and $\partial^2_{qa}S = 1 \ne 0$. Each $S_a$ has the single momentum $p = a$ everywhere: a plane wave.

its envelope

$\partial_a S = 0$ gives $a = m(q-q_0)/(t-t_0)$, and the lemma's $\Sigma$ is

$$\Sigma = \frac{m(q - q_0)^2}{2(t - t_0)} ,$$

the free particle's principal function from the fans-of-paths slide — a point source.

At one instant, each plane wave $S_a$ is a straight line in $q$ of slope $a$. Their envelope is the parabola $\Sigma$, touching $S_a$ where $q - q_0 = a(t - t_0)/m$ — the one point a particle of momentum $a$ from $q_0$ has reached. Units $m = t - t_0 = 1$, $q_0 = 0$.

Check of the free-particle example. $\partial_t S + (\partial_q S)^2/2m = -a^2/2m + a^2/2m = 0$ for every $a$. At the envelope, $\Sigma = a(q-q_0) - a^2(t-t_0)/2m$ with $a(t - t_0) = m(q-q_0)$, so $\Sigma = m(q-q_0)^2/(t-t_0) - m(q-q_0)^2/2(t-t_0) = m(q-q_0)^2/2(t-t_0)$. The complete integral is the family of plane waves; the principal function, a "point source", is their envelope. Section 5 makes the optical words literal.

What to look for in the figure. The lines are $S_a(q) = aq - a^2/2$ for $a = -2, -1.5, \dots, 2$, computed; the thick curve is $q^2/2$. At its point of contact with the highlighted line ($a = 1$, at $q = 1$) the two have the same value and the same slope, so the same $\partial_q S$: the envelope inherits, point by point, the momentum of the plane wave that touches it there. That is the lemma's proof seen as geometry, and it is why $\Sigma$ solves the same equation.

15 · the action function

Jacobi's theorem

Let $S(q,\alpha,t)$ be a complete integral and $\beta\in\mathbb R^n$ constant. Solve $\ \dfrac{\partial S}{\partial\alpha_j}(q,\alpha,t) = \beta_j\ $ for $q(t)$ (implicit function theorem: the $q$-Jacobian is the nonsingular matrix $M_{ij} = \partial^2 S/\partial q_i\partial\alpha_j$), and set $\ p(t) = \dfrac{\partial S}{\partial q}\big(q(t),\alpha,t\big)$. Then $(q(t), p(t))$ solves Hamilton's equations.

proof · the $q$ half

Differentiate $\partial_{\alpha_j}S(q(t),\alpha,t) = \beta_j$ in $t$, and the HJ equation in $\alpha_j$:

$$\textstyle\sum_i M_{ij}\,\dot q_i + \partial^2_{t\alpha_j}S = 0, \qquad \sum_i M_{ij}\,\dfrac{\partial H}{\partial p_i} + \partial^2_{t\alpha_j}S = 0 .$$

Subtract (Clairaut makes the mixed partials agree): $\sum_i M_{ij}\big(\dot q_i - \partial H/\partial p_i\big) = 0$, and $M$ is invertible. $\square$

the $p$ half is the same move with a $q$-derivative — in the notes (N)

This is the verification direction: given a complete integral, the curves it defines are trajectories. It needs nothing but the chain rule, Clairaut and the implicit function theorem. The converse — that every trajectory arises this way, from any complete integral — follows because the $2n$ constants $(\alpha,\beta)$ can be matched to any initial data $(q(t_0), p(t_0))$ near a point where the construction works: choose $\alpha$ with $\partial_q S(q(t_0),\alpha,t_0) = p(t_0)$ (possible locally, since $M$ is invertible) and then $\beta = \partial_\alpha S(q(t_0),\alpha,t_0)$; by uniqueness (unit 01) the constructed solution is the trajectory.

Full proof, the $p$ half. Along the curve, by the chain rule, $\dot p_i = \partial^2_{q_i t}S + \sum_k \partial^2_{q_iq_k}S\,\dot q_k$. Differentiate the HJ equation $\partial_t S + H(q, \partial_q S, t) = 0$ with respect to $q_i$: $$\partial^2_{tq_i}S + \frac{\partial H}{\partial q_i} + \sum_k\frac{\partial H}{\partial p_k}\,\partial^2_{q_kq_i}S = 0 .$$ Using $\dot q_k = \partial H/\partial p_k$ (the $q$ half) and Clairaut, the first expression becomes $\dot p_i = \partial^2_{tq_i}S + \sum_k \partial^2_{q_iq_k}S\,\partial H/\partial p_k = -\partial H/\partial q_i$. $\square$

Why it works — unit 14's view. Take $S(q,\alpha,t)$ as a type-2 generating function $F_2(q, P, t)$ with new momenta $P = \alpha$. Unit 14's rules give $p = \partial_q S$, new coordinates $Q = \partial_\alpha S$, and new Hamiltonian $K = H + \partial_t S$, which the HJ equation makes zero. In the new variables every coordinate and every momentum is constant: $Q = \beta$, $P = \alpha$. Solving the HJ equation is finding the canonical transformation to a system that does not move. The determinant condition is what unit 14 requires of a type-2 generating function.

Check. Free particle, $S = aq - a^2t/2m$: $\partial_a S = q - at/m = \beta$, so $q = \beta + at/m$, and $p = \partial_q S = a$ — uniform motion at momentum $a$ with $q(0) = \beta$. Dimensions: $\beta = \partial S/\partial\alpha$ has dimension action over the dimension of $\alpha$; here $\alpha$ is a momentum and $\beta$ a length.

15 · the action function · example

Worked: the oscillator, without its equation

$H = p^2/2m + \tfrac12 m\omega^2q^2$, by Jacobi's theorem

Step 1 — try $S = -Et + W(q)$. The HJ equation becomes an ODE for $W$: $\ \tfrac{1}{2m}W'^2 + \tfrac12 m\omega^2q^2 = E$, so

$$W(q, E) = \int^q\sqrt{2mE - m^2\omega^2 q'^2}\,\mathrm dq' .$$

Step 2 — complete? One parameter $\alpha = E$, and $\partial^2_{qE}S = m/\sqrt{2mE - m^2\omega^2q^2} \ne 0$. Yes.

Step 3 — Jacobi. $\ \beta = \partial_E S = -t + \displaystyle\int^q\frac{m\,\mathrm dq'}{\sqrt{2mE - m^2\omega^2q'^2}} = -t + \frac1\omega\arcsin\Big(q\,\omega\sqrt{\tfrac{m}{2E}}\Big).$

Step 4 — solve for $q$:

$$q(t) = \sqrt{\frac{2E}{m\omega^2}}\,\sin\omega(t + \beta) .$$

Step 5 — the momentum, $p = \partial_q S = \sqrt{2mE - m^2\omega^2q^2} = \sqrt{2mE}\,\cos\omega(t+\beta)$, which is $m\dot q$ ✓.

Check. Amplitude $\sqrt{2E/m\omega^2}$ has dimension $\sqrt{\mathrm{J/(kg\,s^{-2})}} = \mathrm m$ ✓. $\beta$ has dimension time ✓: $-\beta$ is when $q$ last crossed zero upward.

No second-order equation was solved. One integral was done, and one differentiation.

The branch of the square root. $W$ is not single-valued: $W' = \pm\sqrt{2mE - m^2\omega^2q^2}$, positive when moving right and negative when moving left, and the sign flips at the turning points $q = \pm\sqrt{2E/m\omega^2}$. Step 3 took the upper branch, which covers the quarter-cycle from $q = 0$ to the right turning point; continuing through the turning point on the other branch gives the same sine, since $\arcsin$ reflected is again the sine continued. This sign bookkeeping is the one real trap of the method and it recurs for Kepler.

The integral in step 3. With $q = \sqrt{2E/m\omega^2}\,\sin u$, $\mathrm dq = \sqrt{2E/m\omega^2}\cos u\,\mathrm du$ and $\sqrt{2mE - m^2\omega^2q^2} = \sqrt{2mE}\cos u$, so the integrand is $m\sqrt{2E/m\omega^2}/\sqrt{2mE}\,\mathrm du = \mathrm du/\omega$. Hence the integral is $u/\omega = \omega^{-1}\arcsin(q\omega\sqrt{m/2E})$.

Limit. $E\to0$: the amplitude goes to zero and the particle sits at the bottom, as it should. And the action variable of section 3, the area of the ellipse $p^2/2mE + q^2/(2E/m\omega^2) = 1$ divided by $2\pi$, is $\tfrac{1}{2\pi}\,\pi\sqrt{2mE}\sqrt{2E/m\omega^2} = E/\omega$ — which we will meet again.

Problem P2 does the same for a uniform gravitational field, where the answer is a parabola in $t$.

15 · the action function

When H does not depend on time

time separates

If $\partial_t H = 0$, then $S = -Et + W(q)$ solves the HJ equation iff $W$ solves the time-independent equation

$$H\Big(q, \frac{\partial W}{\partial q}\Big) = E .$$

$W$ is Hamilton's characteristic function. Proof: substitute; $\partial_t S = -E$. $\square$

what $W$ measures

Along a trajectory with $p = \partial_q W$, $\ \dot W = \partial_q W\cdot\dot q = p\cdot\dot q$. So

$$W = \int p\cdot\mathrm dq ,$$

the abbreviated action: a line integral (unit 04) with no time in it.

where time went

Taking $E$ as one of the $\alpha$'s, Jacobi gives $\ \partial_E W = t - t_0$: the orbit is found from $W$ alone, and time is recovered afterwards by one derivative. Section 5 builds on this: $W$ sees only the path.

Proof of the box on the right. $\partial_E S = -t + \partial_E W$, and Jacobi's theorem says this is a constant $\beta_E$; write $\beta_E = -t_0$. So $\partial_E W = t - t_0$. The remaining $n-1$ equations $\partial_{\alpha_j}W = \beta_j$ ($j\ge2$) do not contain $t$ at all: they are the equations of the path in configuration space.

Why $W$ is not the action. The action is $\int L\,\mathrm dt = \int(p\cdot\dot q - H)\,\mathrm dt = W - E(t - t_0)$ at fixed energy — the two differ by a term that knows only the elapsed time. That difference is exactly the $-Et$ in $S = -Et + W$. In the physics literature $W$ is sometimes called the "reduced action", and $\int p\,\mathrm dq$ was Maupertuis's "action" long before Hamilton's.

Check. The worked oscillator: $\partial_E W = \int m\,\mathrm dq/\sqrt{2mE - m^2\omega^2q^2}$ is the time taken to travel from $0$ to $q$, since $m/p = 1/\dot q$ and $\int\mathrm dq/\dot q = \int\mathrm dt$. The formula is just "time is distance over speed", assembled by a derivative.

The running example. Kepler's $H$ has no $t$ in it, so the next section looks for $S = -Et + W(r, \theta, \phi)$ and has to solve the time-independent equation in three variables.

section 2

Separation of variables

One PDE in three variables becomes three ODEs in one — when the coordinates are right.

15 · separation

Separation: one equation becomes n

definition

The time-independent HJ equation is separable in coordinates $q$ if it has a complete integral of the form

$$W(q, \alpha) = W_1(q_1, \alpha) + W_2(q_2, \alpha) + \dots + W_n(q_n, \alpha).$$

lemma · the mechanism

If $f(x) = g(y)$ for all $(x,y)$ in a product of intervals, then $f$ and $g$ are the same constant.

Proof. Fix $y_0$: $f(x) = g(y_0)$ for every $x$. $\square$

cyclic coordinates separate

If $q_n$ does not appear in $H$, then $p_n$ is conserved (unit 14) and $W = \alpha_n q_n + W'(q_1,\dots,q_{n-1})$: the equation loses a variable.

In Kepler's $H$, $\phi$ is cyclic; then the $\theta$-dependence occurs only in one bracket; then only $r$ is left. Three uses of the lemma.

How separation is used, in general. Substitute the sum into $H(q, \partial_q W) = E$. If the equation can be rearranged — by multiplying through by some function — into the form $F_1(q_1, W_1') = G(q_2,\dots,q_n, W_2',\dots, W_n')$, the lemma forces both sides to be a constant $\alpha_2$, and $F_1(q_1, W_1') = \alpha_2$ is an ODE in one variable, solved by a quadrature. Repeat on the right-hand side. Each step introduces one separation constant, and together with $E$ there are $n$ of them: exactly the count a complete integral needs.

Proof of the cyclic case. If $\partial H/\partial q_n = 0$, try $W = \alpha_nq_n + W'$. Then $\partial W/\partial q_n = \alpha_n$ is constant and the equation $H(q_1,\dots,q_{n-1}, \partial_{q'} W', \alpha_n) = E$ no longer involves $q_n$. Any complete integral of that $(n-1)$-variable equation, with $\alpha_n$ added as a parameter, is a complete integral of the original: the determinant is block triangular with a $1$ in the $(q_n, \alpha_n)$ slot.

The lemma's hypothesis matters. On a set that is not a product the conclusion can fail, which is why separation is a local statement, on a box in the coordinates.

15 · separation

Stäckel's conditions

theorem · Stäckel conditions

Let $H = \sum_i \psi_i(q)\big[\tfrac12 p_i^2 + w_i(q_i)\big]$, where $\psi$ is the first row of $\Phi^{-1}$ for a nonsingular $n\times n$ matrix $\Phi$ whose $i$-th row depends on $q_i$ only. Then the equation is separable: the $n$ ODEs

$$\tfrac12 W_i'(q_i)^2 + w_i(q_i) = \textstyle\sum_j \Phi_{ij}(q_i)\,\alpha_j$$

give a complete integral with $\alpha_1 = E$. Conversely, for orthogonal coordinates ($H = \tfrac12\sum g^{ii}p_i^2 + V$), separability forces this form.

sufficiency: two lines, in the notes (N) · necessity: cited

Kepler is Stäckel

Take $m = 1$. Rows $\Phi_r = (1, -r^{-2}, 0)$, $\Phi_\theta = (0, 1, -\sin^{-2}\theta)$, $\Phi_\phi = (0,0,1)$, $w_r = -k/r$, $w_\theta = w_\phi = 0$; then $\psi = (1,\ r^{-2},\ r^{-2}\sin^{-2}\theta)$ and $\alpha = (E,\ L^2/2,\ L_z^2/2)$.

Full proof of sufficiency. Let each $W_i$ solve its ODE and put $W = \sum_i W_i$, so $p_i = W_i'$. Then $$H = \sum_i\psi_i\Big[\tfrac12 W_i'^2 + w_i\Big] = \sum_i\psi_i\sum_j\Phi_{ij}\alpha_j = \sum_j(\psi\Phi)_j\,\alpha_j = \alpha_1 ,$$ because $\psi\Phi$ is the first row of $\Phi^{-1}\Phi = I$. So $W$ solves $H = E$ with $E = \alpha_1$. Completeness: $\partial^2 W/\partial q_i\partial\alpha_j = \partial_{\alpha_j}W_i' = \Phi_{ij}/W_i'$, by differentiating the $i$-th ODE in $\alpha_j$. So the matrix is $\mathrm{diag}(1/W_i')\,\Phi$, nonsingular wherever no $W_i'$ vanishes (away from turning points). $\square$

Check on Kepler. With $m$ restored (divide each $\tfrac12 W_i'^2$ by $m$, and $\alpha_2, \alpha_3$ become $L^2/2m$, $L_z^2/2m$), the three ODEs are $\tfrac1{2m}W_r'^2 - k/r = E - L^2/2mr^2$, $\ \tfrac1{2m}W_\theta'^2 = L^2/2m - L_z^2/2m\sin^2\theta$, $\ \tfrac1{2m}W_\phi'^2 = L_z^2/2m$ — exactly the separated equations of the slide "Kepler, separated". And $\psi\Phi = (1,\ -r^{-2} + r^{-2},\ -r^{-2}\sin^{-2}\theta + r^{-2}\sin^{-2}\theta) = (1,0,0)$ ✓.

Necessity — cited. Stäckel proved in 1891 that for orthogonal coordinates this form is also necessary for additive separability. The proof is a longer computation with the separated equations and is given in Goldstein, Classical Mechanics, 2nd edition, appendix D. You do not need it: in practice the theorem is used in the sufficient direction, to recognise coordinates that work.

Physics. The coordinates in which the equation separates are a property of the potential, not of the kinetic energy: Kepler separates in spherical and in parabolic coordinates, which is why a uniform field added to Kepler (the Stark problem) is still solvable — in parabolic coordinates but not spherical ones. Unit 40 meets the most famous separable case of all, the Kerr black hole, where Carter found in 1968 a separation constant that no isometry of the spacetime accounts for.

15 · separation

Kepler, separated

Try $S = -Et + L_z\phi + \Theta(\theta) + R(r)$ in Kepler's HJ equation, and multiply by $2mr^2$:

$$\begin{aligned}&\underbrace{r^2R'^2 - 2mr^2\Big(E + \frac kr\Big)}_{\text{only } r}\\ &\qquad = -\underbrace{\Big(\Theta'^2 + \frac{L_z^2}{\sin^2\theta}\Big)}_{\text{only }\theta}\end{aligned}$$

By the lemma, each bracket is constant. Call the second $L^2$:

three ODEs, three constants $$\Theta'^2 = L^2 - \frac{L_z^2}{\sin^2\theta},$$ $$R'^2 = 2m\Big(E + \frac kr\Big) - \frac{L^2}{r^2},$$

and $\phi$ enters as $L_z\phi$. Meaning: $p_\phi = L_z$, and $p_\theta^2 + p_\phi^2/\sin^2\theta = L^2$ is $\lVert\vec L\rVert^2$ — the separation constants are the energy and two angular momenta.

the complete integral

$S = -Et + L_z\phi \pm\displaystyle\int^\theta\!\sqrt{L^2 - L_z^2/\sin^2\theta'}\,\mathrm d\theta' \pm\int^r\!\sqrt{2m(E + k/r') - L^2/r'^2}\,\mathrm dr'$, with $\alpha = (E, L, L_z)$ and $|L_z| \le L$.

The substitution, line by line. $\partial_t S = -E$, $\partial_\phi S = L_z$, $\partial_\theta S = \Theta'$, $\partial_r S = R'$. The HJ equation becomes $\frac1{2m}\big[R'^2 + \Theta'^2/r^2 + L_z^2/r^2\sin^2\theta\big] - k/r = E$. Multiplying by $2mr^2$ and moving everything to one side gives the displayed equation. The first bracket depends on $r$ only, the second on $\theta$ only; the lemma (applied on a box in $(r,\theta)$) makes the second a constant $C$, and $C \ge L_z^2 > 0$ unless $L_z = 0$, since $\Theta'^2 \ge 0$ and $\sin^2\theta \le 1$. So writing $C = L^2$ with $L \ge |L_z|$ loses nothing.

Is it complete? The matrix $\partial^2 S/\partial q\,\partial\alpha$ in the order $q = (r,\theta,\phi)$, $\alpha = (E, L, L_z)$ is triangular after reordering: $\partial_\phi$ sees only $L_z$ (entry $1$); $\partial_\theta$ sees $L$ and $L_z$ (entry $\partial_L\Theta' = L/\Theta'$); $\partial_r$ sees $E$ (entry $m/R'$). Its determinant is $mL/(R'\,\Theta')\ne0$ away from turning points.

The signs. Each $\pm$ is the sign of the corresponding momentum, $p_\theta = \Theta'$ and $p_r = R'$; both flip at turning points. So $S$ is multivalued on the orbit, one branch per sign pattern. This does not hurt Jacobi's theorem, which is local; section 3 turns the multivaluedness into the action variables.

Dimensions. $L^2$ and $L_z^2/\sin^2\theta$ are angular momentum squared, $(\mathrm{J\,s})^2$; $2mE$ and $L^2/r^2$ are momentum squared; so $\Theta'$ is an angular momentum and $R'$ a momentum, as $p_\theta$ and $p_r$ should be.

15 · separation · example

Worked: the orbit, read off

Kepler's complete integral → the orbit, by $\partial S/\partial\alpha = \beta$

Step 1 — the $L$ equation. $\ \dfrac{\partial S}{\partial L} = \displaystyle\int\frac{L\,\mathrm d\theta}{\sqrt{L^2 - L_z^2/\sin^2\theta}} - \int\frac{L\,\mathrm dr}{r^2\sqrt{2m(E + k/r) - L^2/r^2}} = \beta_L .$

Step 2 — the $\theta$ integral is $\psi$, the angle travelled in the orbital plane from the node line (notes).

Step 3 — the $r$ integral. Put $u = 1/r$, $u_0 = mk/L^2$, $e^2 = 1 + 2EL^2/mk^2$. Then $2mE + 2mku - L^2u^2 = L^2\big[e^2u_0^2 - (u - u_0)^2\big]$ and $\mathrm dr/r^2 = -\mathrm du$, so

$$\begin{aligned}-\!\int\!\frac{L\,\mathrm dr}{r^2\sqrt{\cdots}} &= \int\frac{\mathrm du}{\sqrt{e^2u_0^2 - (u-u_0)^2}}\\ &= -\arccos\frac{u - u_0}{eu_0}.\end{aligned}$$

Step 4 — solve. $\psi - \arccos\big((u-u_0)/eu_0\big) = \beta_L$, so

$$\frac1r = \frac{mk}{L^2}\Big(1 + e\cos(\psi - \beta_L)\Big),$$

a conic of eccentricity $e$ with perihelion at $\psi = \beta_L$ — unit 08's orbit.

Step 5 — the $E$ equation. $\partial S/\partial E = -t + \displaystyle\int\frac{m\,\mathrm dr}{\sqrt{2m(E + k/r) - L^2/r^2}} = \beta_E$: time since perihelion, unit 08's quadrature.

Check. $u_0$ has dimension $\mathrm{kg\cdot J\,m/(J\,s)^2} = \mathrm m^{-1}$ ✓. $E = -mk^2/2L^2$ gives $e = 0$: the circle ✓.

Step 2 in full. Write $L_z = L\cos i$ with $0\le i\le\pi$. Then $\dfrac{L}{\sqrt{L^2 - L_z^2/\sin^2\theta}} = \dfrac{\sin\theta}{\sqrt{\sin^2\theta - \cos^2 i}} = \dfrac{\sin\theta}{\sqrt{\sin^2 i - \cos^2\theta}}$, and $\dfrac{\mathrm d}{\mathrm d\theta}\arcsin\dfrac{\cos\theta}{\sin i} = -\dfrac{\sin\theta}{\sqrt{\sin^2i - \cos^2\theta}}$. So the $\theta$ integral is $-\arcsin(\cos\theta/\sin i)$ on one branch, $+$ on the other; choose the branch of motion towards the north so that it is $+\arcsin(\cos\theta/\sin i) =: \psi$. On a plane through the origin inclined at $i$ to the equator, a point at angle $\psi$ from the node line has height $z = r\sin i\sin\psi$, i.e. $\cos\theta = \sin i\sin\psi$: so this $\psi$ is exactly the in-plane angle from the node.

The $L_z$ equation, the third one. $\partial S/\partial L_z = \phi - \int\dfrac{L_z\,\mathrm d\theta}{\sin^2\theta\sqrt{L^2 - L_z^2/\sin^2\theta}} = \beta_z$. With $x = \cot\theta$, $\mathrm dx = -\mathrm d\theta/\sin^2\theta$ and $1/\sin^2\theta = 1 + x^2$, the integral is $-\int\cos i\,\mathrm dx/\sqrt{\sin^2 i - \cos^2 i\,x^2} = -\arcsin(x\cot i)$. So $\sin(\phi - \beta_z) = \pm\cot\theta\cot i$, which is $\sin\theta\sin i\sin(\phi - \beta_z) = \pm\cos\theta\cos i$ — the equation of a plane through the origin whose normal makes angle $i$ with the $z$-axis and whose node line is at longitude $\beta_z$. The orbit is planar, and we did not assume it.

Step 3's algebra. $L^2[e^2u_0^2 - (u-u_0)^2] = L^2(e^2-1)u_0^2 + 2L^2u_0u - L^2u^2$; with $L^2u_0 = mk$ and $L^2(e^2-1)u_0^2 = 2EL^2 m^2k^2/(mk^2L^4)\cdot L^2 = 2mE$ this is $2mE + 2mku - L^2u^2$ ✓. And $\frac{\mathrm d}{\mathrm du}\arccos(x/c) = -1/\sqrt{c^2 - x^2}$.

Why step 5 is not pursued. The time integral is exactly unit 08's $t(r)$ quadrature; turning it into Kepler's equation is unit 09's subject. The point here is that the method delivers it as a derivative of something already written down.

15 · separation

Six constants, six orbital elements

constant	from	meaning	Halley
$\alpha_1 = E$	separation of $t$	energy; fixes $a = -k/2E$	$E/m = -2.4874\times10^{7}\ \mathrm{J\,kg^{-1}}$
$\alpha_2 = L$	separation of $\theta$	$\lVert\vec L\rVert$; fixes $e$	$L/m = 4.7840\times10^{15}\ \mathrm{m^2\,s^{-1}}$
$\alpha_3 = L_z$	cyclic $\phi$	$L\cos i$; fixes the inclination	$i = 162.26^\circ$, $\ L_z/L = -0.95245$
$\beta_E$	$\partial S/\partial E$	minus the time of perihelion	9 February 1986
$\beta_L$	$\partial S/\partial L$	argument of perihelion	$111.33^\circ$
$\beta_z$	$\partial S/\partial L_z$	longitude of the node	$58.42^\circ$

The six numbers astronomers use to specify an orbit — semi-major axis, eccentricity, inclination, time of perihelion, argument of perihelion, longitude of the ascending node — are, up to reparametrisation, exactly the $2n = 6$ constants $(\alpha, \beta)$ of Jacobi's theorem for Kepler in spherical coordinates. In unit 14's language they are the canonical coordinates $(Q, P) = (\beta, \alpha)$ in which the motion stands still. Celestial mechanics was built on this: perturbation theory for the planets is the study of how these "constants" drift when a small extra force acts, and the equations for their drift (Lagrange's planetary equations) are Hamilton's equations for the perturbation in these variables.

The numbers. $E/m$ and $L/m$ are unit 08's $\varepsilon$ and $h$. $a = -k/2E = GM_\odot/(2\cdot2.4874\times10^7) = 17.832$ AU and $e = \sqrt{1 + 2\varepsilon h^2/(GM_\odot)^2} = 0.96714$, reproducing unit 08. The angles $i$, $\beta_L$, $\beta_z$ and the perihelion date are the observed elements of 1P/Halley (JPL Small-Body Database), measured from the ecliptic and the vernal equinox; they are data, not computed here. $\cos 162.26^\circ = -0.95245$.

Halley goes round backwards. $i > 90^\circ$ means $L_z < 0$ with the $z$-axis at the ecliptic north pole: the comet orbits opposite to the planets, in a plane tilted only $17.74^\circ$ from the ecliptic. Nothing in the method cares; the sign of $L_z$ is just the sign of $\alpha_3$.

15 · separation

The six elements, drawn

$E$ fixes the size, $a = -k/2E$; $L$ then fixes the shape, $e$.
$\vec L$ is normal to the orbital plane, so the tilt $i$ of the plane is the angle from $\hat z$ to $\vec L$: $\ L_z = L\cos i$.
$\beta_z$: the longitude of the node, where the orbit crosses the reference plane going north.
$\beta_L$: the angle from the node to perihelion, in the orbital plane — the $\psi$ of the worked example.
$\beta_E$: when the comet is at perihelion.

branches

Every square root in $S$ changes sign at a turning point, so each $\beta$-equation holds on one arc between turning points and is continued by switching branch. Get the sign wrong and the orbit runs backwards.

Schematic, drawn prograde with $i = 35^\circ$. Two angles place the plane ($i$ and $\beta_z$), one places the ellipse within it ($\beta_L$). Halley's $i = 162.26^\circ$ puts $\vec L$ almost straight down: a plane $17.74^\circ$ from the ecliptic, travelled backwards.

What to look for. The reference plane (for Halley, the ecliptic) and the orbital plane are two planes through the Sun; they meet in the node line. The separation constants say which plane: $L$ and $L_z$ fix the angle $i$ between their normals, and the $L_z$-equation of the worked example — $\sin\theta\sin i\sin(\phi - \beta_z) = \pm\cos\theta\cos i$ — is the equation of exactly this tilted plane, with its node at longitude $\beta_z$. The $L$-equation then measures the angle $\psi$ from the node within that plane, and puts perihelion at $\psi = \beta_L$.

The branches, in the picture. Going round the drawn ellipse, $\theta$ (the angle down from $\hat z$) reaches its minimum at the orbit's most northern point and its maximum at the most southern, and $r$ turns at perihelion and aphelion. Those four points are where $p_\theta$ and $p_r$ change sign, and where the square roots in $S$ change branch.

The figure is a projection of an exact ellipse ($e = 0.3$, chosen for legibility — Halley's $0.967$ would be a needle) with $\beta_z = 50^\circ$, $\beta_L = 110^\circ$ and $i = 35^\circ$, not Halley's elements. The half of the orbit below the reference plane is dashed; the unmarked dot on the node line is the ascending node.

section 3

Action–angle variables: frequencies without motion

An area in phase space whose derivative is a frequency.

15 · action–angle

The action variable

definition · one degree of freedom

Let the level set $\{H(q,p) = E\}$ contain a closed orbit $\gamma_E$. The action variable is

$$I(E) = \frac{1}{2\pi}\oint_{\gamma_E} p\,\mathrm dq ,$$

a line integral (unit 04) once round, in the direction of motion. Two cases:

libration: $\gamma_E$ bounds a region, and $2\pi I$ is its area;
rotation: $q$ is an angle and $\gamma_E$ goes once round the cylinder; $2\pi I$ is the area under it over one period of $q$.

The pendulum, $H = p^2/2 + 1 - \cos q$. Shaded: $2\pi I$ for a libration ($E = 1$) and for a rotation ($E = 2.9$). Dashed: the separatrix, where $I$ is defined but $T = \infty$.

Why the area. For a libration traversed clockwise in the $(q,p)$ plane — $p > 0$ on top moving right, $p < 0$ underneath moving left — the line integral $\oint p\,\mathrm dq$ is $\int(p_{\text{top}} - p_{\text{bottom}})\,\mathrm dq$, which is the enclosed area. Green's theorem (unit 04) says the same: $\oint p\,\mathrm dq$ round a curve is $\pm$ the area it bounds. For a rotation the curve does not bound anything on the cylinder; the integral of $p\,\mathrm dq$ over one period of $q$ is the area between the curve and $p = 0$.

Why $2\pi$. So that the conjugate angle has period $2\pi$ (the angle-variable slide), and so that for the oscillator $I = E/\omega$ exactly. With this normalisation $I$ has the dimension of $p\cdot q$: action, $\mathrm{J\,s}$, which is where the name comes from.

The figure. Both curves are exact level sets of the pendulum's $H$, computed and plotted, in units where $m\ell^2 = 1$ and $g/\ell = 1$. Unit 02 met this picture as a phase portrait; unit 14 read it as level sets of $H$. The new thing is to attach a number, the area, to each closed curve.

Which loop? In one degree of freedom there is no choice. With several, a torus has several independent loops and each gives an action; that is the many-degrees-of-freedom slide.

15 · action–angle

The frequency is a derivative

theorem · frequency theorem

For $H = p^2/2m + V(q)$ and a libration between simple turning points ($V' \ne 0$ there),

$$\frac{\mathrm d}{\mathrm dE}\oint p\,\mathrm dq = T(E), \qquad\text{so}\qquad \omega(E) = \frac{2\pi}{T} = \frac{\mathrm dE}{\mathrm dI} .$$

differentiate $2\int_{x_1}^{x_2}\sqrt{2m(E - V)}\,\mathrm dx$; the integrable singularity needs a line of care — full proof in the notes (N)

check · the oscillator

$I = E/\omega$ (the ellipse's area over $2\pi$), so $\mathrm dE/\mathrm dI = \omega$ ✓ — independent of $E$, which is isochronism.

dimensions

$[E]/[I] = \mathrm J/(\mathrm{J\,s}) = \mathrm s^{-1}$ ✓.

The period is the rate at which phase-space area grows with energy. Find $I(E)$ — one integral — and the frequency follows by differentiation. No $q(t)$.

Full proof. Let $x_1(E) < x_2(E)$ be the turning points, $V(x_i) = E$, with $V' \ne 0$ there, so by the implicit function theorem they move $C^1$ with $E$, and near each, $E - V(x) \approx |V'(x_i)|\,|x - x_i|$, making $1/p$ integrable (unit 00). Write $A(E) = \oint p\,\mathrm dq = 2\int_{x_1(E)}^{x_2(E)} p(E,x)\,\mathrm dx$ with $p = \sqrt{2m(E - V)}$. Take $E' > E$ close to $E$. On $[x_1(E), x_2(E)]$, $$\frac{p(E',x) - p(E,x)}{E' - E} = \frac{2m}{p(E',x) + p(E,x)},$$ which increases to $m/p(E,x)$ as $E'\downarrow E$; by monotone convergence its integral tends to $\int m\,\mathrm dx/p = T/2$. The extra strips $[x_1(E'), x_1(E)]$ and $[x_2(E), x_2(E')]$ have width $O(E' - E)$, and on them $p(E',x) \le \sqrt{2m(E' - E)}$, so they contribute $O\big((E'-E)^{3/2}\big)$, which is $o(E' - E)$. The same argument for $E' \uparrow E$ with the roles swapped gives the left derivative. So $A'(E) = 2\cdot T/2 = T$. $\square$

Since $A = 2\pi I$: $2\pi\,\mathrm dI/\mathrm dE = T$, i.e. $\mathrm dE/\mathrm dI = 2\pi/T = \omega$. And $I(E)$ is strictly increasing (its derivative $T/2\pi$ is positive), so it can be inverted to give $E(I)$: energy as a function of action, which is the Hamiltonian in the new variables.

The general statement — for any $H(q,p)$, not only $p^2/2m + V$ — is the same argument with $\partial p/\partial E = 1/(\partial H/\partial p) = 1/\dot q$ from the implicit function theorem; the bookkeeping at turning points is identical wherever $\partial H/\partial q \ne 0$ there.

At the separatrix the turning points become equilibria, $V' = 0$, the singularity of $1/p$ is no longer integrable, and $T\to\infty$ while $I$ stays finite: $\mathrm dI/\mathrm dE$ blows up logarithmically. Section 4 needs this.

15 · action–angle

The frequency theorem, as an area

Take two neighbouring level curves, $H = E$ and $H = E + \delta E$. At fixed $q$ their momenta differ by

$$\delta p = \frac{\partial p}{\partial E}\,\delta E = \frac{m}{p}\,\delta E = \frac{\delta E}{\dot q} .$$

So the strip between them is wide where the particle is slow, and its area is

$$\begin{aligned}\oint\delta p\,\mathrm dq &= \delta E\oint\frac{\mathrm dq}{\dot q}\\ &= \delta E\oint\mathrm dt = T\,\delta E .\end{aligned}$$

That is $\mathrm d(2\pi I) = T\,\mathrm dE$: the theorem. The proof in the notes of the previous slide only makes the turning points honest.

The pendulum's loops at $E = 1.5$ and $1.7$, and the strip between them, computed. It is thinnest at $q = 0$, where the bob is fastest, and fattest at the ends, where it slows to turn. Its area, $T\,\delta E$ to first order, measures the time spent.

What the picture proves, and what it does not. The strip argument is the frequency theorem's proof with the limits left out. Its weak point is the ends: at a turning point $\dot q = 0$ and $\delta p = \delta E/\dot q$ is infinite, so the first-order formula for the strip's width fails exactly where the strip is fattest. The full proof on the previous slide handles this — the extra slivers near the turning points have area $O(\delta E^{3/2})$ — and that is the only real work in it.

The figure. Units $m\ell^2 = 1$, $g/\ell = 1$, so $H = p^2/2 + 1 - \cos q$, as on the action-variable slide. The loops are the exact level sets at $E = 1.5$ and $E = 1.7$ (both librations, below the separatrix at $E = 2$). The shaded area is $2\pi\big(I(1.7) - I(1.5)\big) = 1.8099$, against $T(1.6)\,\delta E = 1.8058$ from the period at the midpoint energy — agreeing to $0.2$ per cent, the midpoint rule's error, which is second order in $\delta E$ relative to the area.

Dimensions. $\partial p/\partial E = m/p$ is (kg)/(kg m s⁻¹) = s m⁻¹, a time per length; times $\mathrm dq$, a time. The strip's area is momentum × length = J s, and so is $T\,\delta E$ ✓.

15 · action–angle

The angle conjugate to it

definition

Let $W(q, I) = \int^q p(q', E(I))\,\mathrm dq'$ — the characteristic function with $I$ as its constant. The angle variable is

$$w = \frac{\partial W}{\partial I}(q, I) .$$

theorem · action–angle variables

$(q,p)\mapsto(w, I)$ is canonical (type-2 generating function $W(q,I)$, unit 14), the new Hamiltonian is $E(I)$, and

$$\dot I = 0, \qquad \dot w = \frac{\mathrm dE}{\mathrm dI} = \omega(I), \qquad w(t) = \omega t + w_0 .$$

Once round the orbit, $\ \Delta w = \oint\dfrac{\partial p}{\partial I}\,\mathrm dq = \dfrac{\mathrm d}{\mathrm dI}\oint p\,\mathrm dq = \dfrac{\mathrm d}{\mathrm dI}(2\pi I) = 2\pi$.

So $w$ is an honest angle, the motion is a uniform rotation of it, and all the dynamics is in $E(I)$.

Full proof. $W(q,I)$ is a type-2 generating function: $p = \partial_q W$ holds by construction, the new coordinate is $w = \partial_I W$, and since $W$ has no explicit time dependence the new Hamiltonian is $K(w,I) = H(q, p) = E(I)$. This needs $\partial^2 W/\partial q\,\partial I = \partial p/\partial I = (\partial p/\partial E)(\mathrm dE/\mathrm dI) = \omega/\dot q \ne 0$, true away from turning points. Hamilton's equations in the new variables are $\dot I = -\partial K/\partial w = 0$ and $\dot w = \partial K/\partial I = E'(I) = \omega$. For $\Delta w$: interchanging $\mathrm d/\mathrm dI$ with the loop integral is the frequency theorem's computation over again (the loop moves with $I$, but its endpoints are where $p = 0$, so the moving-endpoint terms vanish, as in the proof there). $\square$

Check, the oscillator. $E = \omega I$, $p = \sqrt{2m\omega I - m^2\omega^2q^2}$, and $w = \partial_I W = \int m\omega\,\mathrm dq/\sqrt{2m\omega I - m^2\omega^2 q^2} = \arcsin\big(q\sqrt{m\omega/2I}\big)$. So $q = \sqrt{2I/m\omega}\,\sin w$ and $p = \sqrt{2m\omega I}\cos w$: the ellipse, parametrised by a uniformly rotating angle. Dimensions: $\sqrt{I/m\omega}$ is $\sqrt{\mathrm{J\,s/(kg\,s^{-1})}} = \mathrm m$ ✓.

Why this matters beyond one degree of freedom. $w$ is multivalued as a function of $(q,p)$ — it increases by $2\pi$ each time round — and that is the point: it is a coordinate on a circle. Unit 17 shows that near a torus of an integrable system this is the whole structure (the Liouville–Arnold theorem), and that its breakdown under perturbation is what chaos is.

15 · action–angle

Many degrees of freedom: one loop per coordinate

definition · separable case

If $W = \sum_i W_i(q_i, \alpha)$ and each $q_i$ oscillates (or rotates) between limits fixed by $\alpha$, the orbit fills a subset of the invariant torus $\{p_i = W_i'(q_i),\ \text{each } q_i \text{ in its range}\}$, and

$$I_i(\alpha) = \frac1{2\pi}\oint p_i\,\mathrm dq_i \quad\text{(the loop in the } (q_i, p_i)\text{ plane)}.$$

frequencies

Invert to $\alpha(I)$, so $H = E(I_1,\dots,I_n)$. Then as before $\dot I_i = 0$ and

$$\dot w_i = \omega_i = \frac{\partial H}{\partial I_i} .$$

the general case — cited

Without separation the loops are the $n$ independent topological cycles on the torus. That these tori exist, and the loop integrals depend only on the cycle, is the Liouville–Arnold theorem: unit 17.

Full proof, separable case. $W(q, I) := \sum_i W_i(q_i, \alpha(I))$ is a type-2 generating function to $(w, I)$ with $w_j = \partial W/\partial I_j$ and new Hamiltonian $E(I)$, so $\dot I = 0$ and $\dot w_j = \partial E/\partial I_j$. The periodicity: carrying $q_i$ once round its own loop, with the others fixed, changes $w_j$ by $\oint\partial_{I_j}p_i\,\mathrm dq_i = \partial_{I_j}\oint p_i\,\mathrm dq_i = \partial_{I_j}(2\pi I_i) = 2\pi\delta_{ij}$. So $(w_1,\dots,w_n)$ are angles on an $n$-torus, each advanced by $2\pi$ by exactly one of the coordinate loops. The Jacobian condition is that $\alpha\mapsto I$ be invertible, which holds for Kepler (below) and generically.

Topological cycles. The torus $\mathbb T^n$ is $\mathbb R^n/2\pi\mathbb Z^n$, and its closed loops up to continuous deformation are labelled by $n$ integers — how many times the loop winds in each direction. (That classification is the fundamental group $\mathbb Z^n$; you need only the picture.) Why the loop integral depends only on the class: on the torus, $p = \partial_q W$ locally, so $p\cdot\mathrm dq$ restricted to it is locally exact, and unit 04's argument makes the integral over a loop invariant under deformation. Unit 17 proves this properly.

The torus picture. A motion with $n$ frequencies $\omega_i$ winds round the torus. If the $\omega_i$ are rationally independent it never closes and fills the torus densely (Kronecker's theorem, also in unit 17); if they are commensurate it closes. That dichotomy is the "three equal frequencies" slide, for Kepler.

15 · action–angle

The torus, and its two loops

Two separated coordinates, each on its own closed loop in its own phase plane: a state is a point on each loop, so the motion lives on the product of two circles — a torus. Each action is one loop's area over $2\pi$.

Schematic. The two loops $\gamma_1$ (round the tube) and $\gamma_2$ (round the hole), and an orbit winding with $\omega_2/\omega_1 = \sqrt2$: it never closes, and in time covers the torus.

The same motion seen in each phase plane: two ellipses, areas $2\pi I_1$ and $2\pi I_2$.

What to look for. The torus is not a picture of space; it is the set of states $(q_1, p_1, q_2, p_2)$ that the motion can reach once the two actions are fixed — a two-dimensional surface in a four-dimensional phase space, drawn here as a doughnut because a product of two circles can be drawn that way in three dimensions. Going once round $\gamma_1$ is going once round the first phase-plane loop with the second coordinate frozen; that changes $w_1$ by $2\pi$ and $w_2$ by nothing, which is the $2\pi\delta_{ij}$ of the previous slide's proof.

The example. Two uncoupled oscillators, $H = \tfrac12(p_1^2 + \omega_1^2q_1^2) + \tfrac12(p_2^2 + \omega_2^2q_2^2)$, separate trivially, with $I_i = E_i/\omega_i$ and $w_i = \omega_it + $ const. The orbit on the torus is the straight line $w_2 = \sqrt2\,w_1 + $ const in the angles, wrapped round. With an irrational ratio it never returns to its start and comes arbitrarily close to every point of the torus (Kronecker's theorem, unit 17); with a rational ratio $r/s$ it closes after $s$ turns one way and $r$ the other. Kepler is the second case with $r = s = 1$: the subject of "Three equal frequencies: why Kepler orbits close".

15 · action–angle

Two integrals, done once

lemma · the angular integral

For $0 \le \beta < \alpha$, with the integral between the two zeros of the root,

$$\frac1\pi\int\sqrt{\alpha^2 - \frac{\beta^2}{\sin^2\theta}}\;\mathrm d\theta = \alpha - \beta .$$

lemma · the radial integral

For $A, B, C > 0$ with $B^2 > AC$, between the two zeros $r_1 < r_2$ of the root,

$$\frac1\pi\int_{r_1}^{r_2}\sqrt{-A + \frac{2B}{r} - \frac{C}{r^2}}\;\mathrm dr = \frac{B}{\sqrt A} - \sqrt C .$$

both reduce, by algebra and one substitution, to $\int_a^b\frac{\mathrm du}{\sqrt{(b-u)(u-a)}} = \pi$ — full proofs in the notes (N)

Checks. Angular: $\beta = 0$ gives $\frac1\pi\int_0^\pi\alpha\,\mathrm d\theta = \alpha$ ✓; $\beta\to\alpha$ shrinks the interval to $\theta = \pi/2$, and $0$ ✓. Radial: $B^2 = AC$ makes $r_1 = r_2$, and $B/\sqrt A - \sqrt C = 0$ ✓.

The base integral. For $a < b$ put $u = \tfrac{a+b}2 - \tfrac{b-a}2\cos s$, $s\in[0,\pi]$. Then $(b - u)(u - a) = \big(\tfrac{b-a}2\big)^2\sin^2 s$ and $\mathrm du = \tfrac{b-a}2\sin s\,\mathrm ds$, so $\int_a^b\mathrm du/\sqrt{(b-u)(u-a)} = \int_0^\pi\mathrm ds = \pi$, and $\int_a^b u\,\mathrm du/\sqrt{\cdots} = \int_0^\pi\big(\tfrac{a+b}2 - \tfrac{b-a}2\cos s\big)\mathrm ds = \pi\tfrac{a+b}2$.

Full proof, radial. $-A + 2B/r - C/r^2 = A(r_2 - r)(r - r_1)/r^2$ with $r_1 + r_2 = 2B/A$ and $r_1r_2 = C/A$. So the integrand is $\sqrt A\,(r_2 - r)(r - r_1)/\big(r\sqrt{(r_2-r)(r-r_1)}\big)$, and $(r_2 - r)(r - r_1)/r = (r_1 + r_2) - r - r_1r_2/r$. Three integrals against $\mathrm dr/\sqrt{(r_2-r)(r-r_1)}$: the constant gives $\pi(r_1 + r_2)$; the $r$ gives $\pi(r_1+r_2)/2$; and $\int\mathrm dr/(r\sqrt{\cdots}) = \pi/\sqrt{r_1r_2}$ by $u = 1/r$, which turns it into $\int\mathrm du/\sqrt{r_1r_2(u - 1/r_2)(1/r_1 - u)}$. Total: $\sqrt A\,\pi\big[(r_1+r_2)/2 - \sqrt{r_1r_2}\big] = \pi\big[B/\sqrt A - \sqrt C\big]$. $\square$

Full proof, angular. Write the integrand as $\sqrt{\alpha^2\sin^2\theta - \beta^2}/\sin\theta$ and put $c = \cos\theta$, $\mathrm dc = -\sin\theta\,\mathrm d\theta$: the integral becomes $\alpha\int_{-c_0}^{c_0}\sqrt{c_0^2 - c^2}\,\mathrm dc/(1 - c^2)$ with $c_0^2 = 1 - \beta^2/\alpha^2$. Split $\dfrac{c_0^2 - c^2}{1 - c^2} = 1 - \dfrac{1 - c_0^2}{1 - c^2}$. The first piece gives $\int\mathrm dc/\sqrt{c_0^2 - c^2} = \pi$ (base integral). The second needs $\int_{-c_0}^{c_0}\dfrac{\mathrm dc}{(1-c^2)\sqrt{c_0^2-c^2}}$: with $c = c_0\sin s$ it is $\int_{-\pi/2}^{\pi/2}\dfrac{\mathrm ds}{1 - c_0^2\sin^2 s}$, and with $t = \tan s$ that is $\int_{\mathbb R}\dfrac{\mathrm dt}{1 + (1 - c_0^2)t^2} = \dfrac{\pi}{\sqrt{1 - c_0^2}}$. So the integral is $\alpha\big[\pi - (1 - c_0^2)\pi/\sqrt{1-c_0^2}\big] = \alpha\pi\big(1 - \sqrt{1-c_0^2}\big) = \pi(\alpha - \beta)$. $\square$

Both are sometimes done by contour integration; unit 20 has that tool, and it is not needed.

15 · action–angle · example

Worked: Kepler's actions, and Kepler's third law

the running example's actions

Step 1 — $\phi$ rotates, $p_\phi = L_z$: $\ I_\phi = \frac1{2\pi}\int_0^{2\pi}L_z\,\mathrm d\phi = L_z$.

Step 2 — $\theta$ librates; the angular lemma with $\alpha = L$, $\beta = |L_z|$:

$$I_\theta = \frac1\pi\int\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\,\mathrm d\theta = L - |L_z| .$$

Step 3 — $r$ librates, with $p_r = \sqrt{2mE + 2mk/r - L^2/r^2}$; the radial lemma with $A = -2mE$, $B = mk$, $C = L^2$ ($E < 0$):

$$I_r = \frac1\pi\int_{r_1}^{r_2}\!p_r\,\mathrm dr = k\sqrt{\frac{m}{-2E}} - L .$$

Step 4 — invert. $L$ and $|L_z|$ cancel in the sum $I := I_r + I_\theta + |I_\phi| = k\sqrt{m/(-2E)}$:

$$H = -\frac{mk^2}{2\,(I_r + I_\theta + |I_\phi|)^2} .$$

Step 5 — differentiate. $\omega_r = \omega_\theta = |\omega_\phi| = mk^2/I^3$. With $E = -k/2a$, $I = \sqrt{mka}$ and

$$\omega^2 = \frac{k}{ma^3} = \frac{GM_\odot}{a^3}\quad\text{— Kepler III.}$$

Halley, per unit mass: $I = 1.8816\times10^{16}\ \mathrm{m^2\,s^{-1}}$; $\ T = 2\pi I^3/(GM_\odot)^2 = 75.305$ yr ✓ unit 08.

Checks. Dimensions of $mk^2/I^3$: $\mathrm{kg\,(J\,m)^2/(J\,s)^3} = \mathrm{kg\,m^2\,J^{-1}\,s^{-3}} = \mathrm s^{-1}$ ✓. Circular orbit: $I_r = 0$ iff $L = k\sqrt{m/(-2E)}$ iff $E = -mk^2/2L^2$, the bottom of unit 08's effective potential ✓. Equatorial orbit: $I_\theta = 0$ iff $|L_z| = L$ ✓ — no oscillation in latitude.

Kepler III. $E = -k/2a$ (unit 08) gives $-2E/m = k/(ma)$, so $I = k\sqrt{ma/k} = \sqrt{mka}$ and $\omega = mk^2/(mka)^{3/2} = \sqrt{k/(ma^3)}$; with $k = GM_\odot m$ this is $\sqrt{GM_\odot/a^3}$.

Halley's numbers. Per unit mass, $I/m = GM_\odot/\sqrt{-2\varepsilon} = 1.32712\times10^{20}/7053.2 = 1.88158\times10^{16}$; $I_r/m = 1.88158\times10^{16} - h = 1.40318\times10^{16}$; $|L_z|/m = 0.95245\,h = 4.5565\times10^{15}$ and $I_\theta/m = h - |L_z|/m = 2.275\times10^{14}$. Most of the comet's action is radial, as its eccentricity $0.967$ says, and very little is in latitude, as its $17.7^\circ$ tilt from the ecliptic says. The period $2\pi(I/m)^3/(GM_\odot)^2 = 2.3764\times10^{9}$ s $= 75.305$ Julian years.

The sign of $\omega_\phi$. Step 1 runs the $\phi$ loop in the direction of increasing $\phi$ rather than the direction of motion, so $I_\phi = L_z$ keeps the sign of $L_z$; that is the convention behind every $|I_\phi|$ above, and for a prograde orbit the two directions agree and the modulus does nothing. Then $\partial H/\partial I_\phi = \mathrm{sign}(L_z)\,mk^2/I^3$. For Halley $L_z < 0$, and $\phi$ decreases: the retrograde motion again.

Why the $\theta$ step is honest. $\theta$ oscillates between $\theta_\pm$ with $\sin\theta_\pm = |L_z|/L$ — the orbit's highest and lowest latitudes: $\pm i$ for a prograde orbit, $\pm(180^\circ - i)$ for a retrograde one. That is a libration of $\theta$ and the lemma applies.

15 · action–angle

Three equal frequencies: why Kepler orbits close

degenerate frequencies

For Kepler, $H$ depends on the actions only through their sum, so $\omega_r = \omega_\theta = |\omega_\phi|$. Every coordinate completes its cycle in the same time: after one period all return together, and the orbit closes.

Bertrand, seen from above

All bounded orbits close iff $\omega_r/\omega_\theta$ is a fixed rational for every orbit. Two potentials achieve it:

$-k/r$: ratio $1$ — this slide;
$\tfrac12 m\Omega^2r^2$: $H = \Omega(2I_r + I_\theta + |I_\phi|)$, ratio $2$ — problem P4.

The two closed families. Kepler: the centre of force at a focus, one closest approach per turn ($\omega_r = \omega_\theta$). Oscillator: the centre at the middle, two per turn ($\omega_r = 2\omega_\theta$). Dots: where $r$ is least.

Reading the figure. Each closed orbit is drawn with the centre of force marked. Along Kepler's ellipse $r$ runs from perihelion to aphelion and back once while the angle goes once round: $\omega_r/\omega_\theta = 1$. Along the oscillator's ellipse, centred on the force, $r$ is least at both ends of the minor axis and greatest at both ends of the major axis, so it goes through two radial cycles per turn: $\omega_r/\omega_\theta = 2$, which is $\partial H/\partial I_r = 2\Omega$ against $\partial H/\partial I_\theta = \Omega$. Both ellipses are exact, $e = 0.6$ for Kepler and axis ratio $0.6$ for the oscillator.

Why closing needs a fixed rational ratio. In angle variables the motion is $w_r = \omega_r t$, $w_\theta = \omega_\theta t$ (up to phases), and the orbit in the plane closes iff both return to their starting values mod $2\pi$ at a common time, iff $\omega_r/\omega_\theta \in\mathbb Q$. If the ratio varies continuously with the orbit, some orbits have irrational ratio and do not close; so "all bounded orbits close" forces the ratio to be a constant rational.

Bertrand's theorem — that only $-k/r$ and $r^2$ achieve this among power laws and indeed among all central potentials with bounded orbits — was cited in unit 08, and it stays cited: Bertrand (1873); a proof via the small-oscillation frequency ratio is in Goldstein, Classical Mechanics, 3rd edition, §3.6 (Appendix A of the 2nd edition). What the action-angle view adds is the reason: closure is a statement about $H(I)$, and Kepler's $H$ is the special function of $I_r + L$ alone. The hidden symmetry unit 13 found (the Runge–Lenz vector, the $SO(4)$) is what forces that.

15 · action–angle

Any other central force: the ellipse turns

any central force

The perihelion moves on by $\ 2\pi\big(\omega_\theta/\omega_r - 1\big)\ $ each radial period.

unit 37

General relativity breaks the degeneracy, by $8.0\times10^{-8}$ for Mercury, and the perihelion advances: the $43''$ per century.

Same $e$, two potentials, computed. Kepler's $-k/r$ (thick): $\omega_\theta = \omega_r$, one closed ellipse. Add $-c/r^2$ with $\omega_\theta/\omega_r = 1/0.85$ (thin): perihelion advances $63.5^\circ$ per radial period, shown for six.

Proof that isotropy gives $H(I_r, L)$. For any $V(r)$ the separation of the $\theta$ and $\phi$ parts is word for word the Kepler computation, since only the radial equation contains $V$. So $I_\phi = L_z$ and $I_\theta = L - |L_z|$ always, and the radial action $I_r = \frac1\pi\int\sqrt{2m(E - V) - L^2/r^2}\,\mathrm dr$ is a function of $(E, L)$. Inverting gives $E(I_r, L)$ with $L = I_\theta + |I_\phi|$, and then $\partial E/\partial I_\theta = \pm\partial E/\partial I_\phi = \partial E/\partial L$.

The precession, quantitatively. If $\omega_r \ne \omega_\theta$, the perihelion angle advances by $2\pi(\omega_\theta/\omega_r - 1)$ per radial period. Unit 37 computes that for general relativity's correction and gets $43''$ per century for Mercury.

The perihelion advance. In angle variables, $r$ returns to perihelion each time $w_r$ advances by $2\pi$, which takes $2\pi/\omega_r$; in that time the angle in the plane advances by $2\pi\,\omega_\theta/\omega_r$. The excess over a full turn, $2\pi(\omega_\theta/\omega_r - 1)$, is how far the perihelion has moved on.

The figure. $V = -k/r - c/r^2$ is a central potential whose orbits are known exactly: the extra term only changes $L^2$ to $L^2 - 2mc$ in the radial equation, so Binet's equation becomes $u'' + \gamma^2u = mk/L^2$ with $\gamma^2 = 1 - 2mc/L^2$, and the orbit is $u = (mk/\gamma^2L^2)\big(1 + e\cos\gamma\phi\big)$. The radial period spans $\Delta\phi = 2\pi/\gamma$, so $\omega_\theta/\omega_r = 1/\gamma$. The figure takes $\gamma = 0.85$ and $e = 0.5$, giving $360^\circ(1/0.85 - 1) = 63.5^\circ$ per radial period. Unit 37's general-relativistic orbit equation has the same form near a circular orbit, with $1 - \gamma$ of order $3GM/(c^2r)$ — tiny for Mercury, which is why its rosette turns by only $43''$ a century.

15 · action–angle

How the action responds to a parameter

lemma

Let $H = p^2/2m + V(q, \lambda)$. At fixed energy,

$$\frac{\partial I}{\partial\lambda}\Big|_E = -\frac{1}{\omega}\Big\langle\frac{\partial V}{\partial\lambda}\Big\rangle ,$$

with $\langle\cdot\rangle$ the time average over one period of the unperturbed orbit. Proof. $\partial_\lambda p|_{E,q} = -m\,\partial_\lambda V/p = -\partial_\lambda V/\dot q$, so $\oint\partial_\lambda p\,\mathrm dq = -\oint\partial_\lambda V\,\mathrm dt$; divide by $2\pi$. $\square$

frequency without motion

For $V = \tfrac12 m\omega_0^2q^2 + \lambda U(q)$, to first order in $\lambda$:

$$\begin{aligned}E(I) &= \omega_0 I + \lambda\,\langle U\rangle_I,\\ \omega(I) &= \omega_0 + \lambda\frac{\mathrm d\langle U\rangle_I}{\mathrm dI},\end{aligned}$$

averaging over the unperturbed ellipse of action $I$.

check · a cubic

$U = q^3$ is odd; on $q = \sqrt{2I/m\omega_0}\sin w$ its average is $0$. No first-order shift — true: a cubic term shifts the frequency only at order $\lambda^2$. The quartic is P6; the pendulum is P7.

Full proof of the first-order formula. At $\lambda = 0$, $I = E/\omega_0$. The lemma gives $\partial_\lambda I|_{E,\lambda=0} = -\langle U\rangle/\omega_0$, the average taken over the unperturbed orbit of energy $E$, i.e. of action $E/\omega_0$. So $I(E,\lambda) = E/\omega_0 - \lambda\langle U\rangle_{E/\omega_0}/\omega_0 + O(\lambda^2)$. Solve for $E$: $E = \omega_0 I + \lambda\langle U\rangle_{E/\omega_0} + O(\lambda^2)$, and replacing $\langle U\rangle_{E/\omega_0}$ by $\langle U\rangle_I$ costs another $O(\lambda^2)$ since the two arguments differ by $O(\lambda)$. Differentiate in $I$ for the frequency (the $O(\lambda^2)$ remainder is smooth in $I$, being built from smooth integrals). $\square$ This is first-order perturbation theory in one line; unit 21 builds the systematic version.

The lemma's integral. The interchange of $\partial_\lambda$ with the loop integral is justified exactly as in the frequency theorem: the turning points move with $\lambda$ but $p = 0$ there, and the difference quotient of $p$ is dominated near them. The unperturbed average appears because the orbit being integrated over is the one at the given $\lambda$, and at first order that is the unperturbed one.

Why this is remarkable. The frequency of a nonlinear oscillator is found from an average of the perturbation over the linear motion. No nonlinear ODE is solved, and none of the secular terms that plague naive perturbation theory (unit 21's opening problem) can appear, because nothing here depends on $t$.

Dimensions. $\lambda\langle U\rangle$ is an energy, $I$ an action, so $\lambda\,\mathrm d\langle U\rangle/\mathrm dI$ is a frequency ✓.

section 4

Adiabatic invariants

Change a system slowly, and the area of its orbit is what survives.

15 · adiabatic

Slow change, and what survives it

definition

Let $H = H(q, p, \lambda)$ with $\lambda = \lambda(\varepsilon t)$ for a fixed smooth function $\lambda(\cdot)$, and $\varepsilon > 0$ small. A quantity $F(q,p,\lambda)$ is an adiabatic invariant if there is $C$ independent of $\varepsilon$ with $|F(t) - F(0)| \le C\varepsilon$ for $0 \le t \le 1/\varepsilon$.

theorem · one degree of freedom

If every frozen system $H(\cdot,\cdot,\lambda)$ has closed orbits with $\omega$ bounded away from $0$, the action $I$ is an adiabatic invariant.

Why, at first order: $\dot I = \dot\lambda\,\partial_\lambda I|_{q,p}$, and by the lemma on how the action responds to a parameter

$$\dot I = \frac{\dot\lambda}{\omega}\Big(\frac{\partial H}{\partial\lambda} - \Big\langle\frac{\partial H}{\partial\lambda}\Big\rangle\Big):$$

$O(\varepsilon)$ and of zero mean over each period. It wobbles; it does not drift.

the averaging argument in the notes (N) · the estimate over time $1/\varepsilon$: cited

The first-order computation in full. $I$ as a function on phase space is $I(H(q,p,\lambda), \lambda)$, where $I(E,\lambda)$ is the action of the frozen level set. So at fixed $(q,p)$, $\partial_\lambda I = \partial_E I\cdot\partial_\lambda H + \partial_\lambda I|_E = \omega^{-1}\partial_\lambda H - \omega^{-1}\langle\partial_\lambda H\rangle$, using $\partial_E I = 1/\omega$ (frequency theorem) and the lemma. Along the motion $I$ changes only through $\lambda$ — the frozen flow preserves it, since it is a function of $H$ — so $\dot I = \dot\lambda\,\partial_\lambda I$, and $\dot\lambda = \varepsilon\lambda'(\varepsilon t) = O(\varepsilon)$.

The averaging argument. Over one period $T$ of the motion, $\lambda$ changes by $O(\varepsilon T)$, so the orbit differs from a frozen orbit by $O(\varepsilon)$, and $\int_0^T\dot I\,\mathrm dt = \dot\lambda\,\omega^{-1}\int_0^T(\partial_\lambda H - \langle\partial_\lambda H\rangle)\,\mathrm dt + O(\varepsilon^2) = O(\varepsilon^2)$. Over time $1/\varepsilon$ there are $O(1/\varepsilon)$ periods, so the accumulated drift is $O(\varepsilon)$, on top of a within-period oscillation that is also $O(\varepsilon)$. That is the whole idea.

Cited: the theorem as stated. Making "differs from a frozen orbit by $O(\varepsilon)$" rigorous uniformly over $1/\varepsilon$ periods is done by changing to action-angle variables of the frozen system and averaging over the angle: Arnold, Mathematical Methods of Classical Mechanics, chapter 10, the section on adiabatic invariants; or Landau and Lifshitz, Mechanics, §49, with §51 showing the stronger fact that for analytic $\lambda(\cdot)$ the total change $I(+\infty) - I(-\infty)$ is exponentially small in $1/\varepsilon$.

The hypothesis on $\omega$ matters. "Slow" means slow compared with the motion itself: $\varepsilon$ is really the ratio of the period to the timescale of $\lambda$. If $\omega\to0$ — at a separatrix — the motion is not fast compared with anything, and the theorem fails. That is the subject of "Where adiabatic invariance fails".

15 · adiabatic · example

Worked: the pendulum shortened slowly

the string is drawn up through a hole, from $\ell = 1.00$ m to $0.25$ m, over a minute

Step 1 — the frozen system. Small swings: $H = \dfrac{p_\vartheta^2}{2m\ell^2} + \tfrac12 mg\ell\,\vartheta^2$, an oscillator with $\omega = \sqrt{g/\ell}$ and energy $E = \tfrac12 mg\ell\,\vartheta_0^2$ at amplitude $\vartheta_0$.

Step 2 — the action. $I = E/\omega = \tfrac12 m\sqrt g\,\ell^{3/2}\vartheta_0^2$.

Step 3 — hold it fixed.

$$\vartheta_0 \propto \ell^{-3/4},\quad E \propto \omega \propto \ell^{-1/2},\quad \ell\vartheta_0 \propto \ell^{1/4}.$$

Step 4 — numbers, starting at $\vartheta_0 = 4.00^\circ$, $m = 0.50$ kg:

	$\ell = 1.00$ m	$\ell = 0.25$ m
period	$2.006$ s	$1.003$ s
amplitude	$4.00^\circ$	$11.31^\circ$
arc	$6.98$ cm	$4.94$ cm
swing energy	$0.01195$ J	$0.02391$ J

Check. Slowness at the end: $T\,|\dot\ell|/\ell = 1.003\times0.0125/0.25 = 0.050 \ll 1$ ✓. $I$ has dimension $\mathrm{kg\,m^{1/2}s^{-1}\,m^{3/2}} = \mathrm{J\,s}$ ✓.

Setting it up honestly. With the string length $\ell(t)$ prescribed, the bob's position is $\ell(t)(\sin\vartheta, -\cos\vartheta)$, so $T = \tfrac12 m(\dot\ell^2 + \ell^2\dot\vartheta^2)$ and $V = -mg\ell\cos\vartheta$. The term $\tfrac12 m\dot\ell^2$ is a function of $t$ alone; it is the time derivative of its own integral, so by unit 12's gauge freedom it drops out of the equations. What remains is $L = \tfrac12 m\ell^2\dot\vartheta^2 + mg\ell\cos\vartheta$, with $p_\vartheta = m\ell^2\dot\vartheta$ and $H = p_\vartheta^2/2m\ell^2 - mg\ell\cos\vartheta$. The constant $-mg\ell$ in the small-angle expansion depends on $\lambda = \ell$ but not on $(q,p)$, so it shifts $E$ without changing orbits or $I$; the swing energy in the table is the energy above that floor.

The numbers. $T = 2\pi\sqrt{\ell/g}$ with $g = 9.81$: $2.0061$ s and $1.0030$ s. $\vartheta_1 = 4^\circ\times4^{3/4} = 11.314^\circ$. Arc lengths $\ell\vartheta_0$: $1.00\times0.069813 = 6.981$ cm and $0.25\times0.19746 = 4.937$ cm, ratio $4^{-1/4} = 0.7071$. Swing energies $\tfrac12 mg\ell\vartheta_0^2$: $0.011953$ J, then twice that, $0.023906$ J, since $E\propto\ell^{-1/2}$ and $\ell$ fell by 4. Shortening by $0.75$ m in $60$ s at a steady rate gives $|\dot\ell| = 0.0125\ \mathrm{m\,s^{-1}}$; the slowness parameter is worst at the end. At $11.3^\circ$ the small-angle frequency is still good to $\vartheta_0^2/16 = 0.24\%$ (problem P7's correction).

Where the energy came from. The swing energy doubled. Whoever pulls the string does work against the tension, which is slightly more than $mg$ on average because of the swinging. Problem P10 computes that work and finds exactly the increase predicted here — an independent check that $E/\omega$, and not $E$, is what is conserved.

History. At the first Solvay conference, in 1911, Lorentz asked how a quantum oscillator could keep its energy an integer multiple of $h\nu$ when its frequency is changed; Einstein answered with this pendulum: if the length changes slowly, $E/\nu$ is invariant. Ehrenfest made it a principle — adiabatic invariants are what may be quantised — and Bohr and Sommerfeld quantised $\oint p\,\mathrm dq = nh$. That old quantum theory was superseded, but the action variables were exactly the right objects.

15 · adiabatic

The shortened pendulum, integrated

The full equation $\ell^2\ddot\vartheta + 2\ell\dot\ell\,\dot\vartheta = -g\ell\sin\vartheta$, integrated numerically for the worked example's minute. Dashed: $\pm4^\circ(\ell/1\,\mathrm m)^{-3/4}$, the prediction from $I$ = const. The last swing, at $59.65$ s, peaks at $11.18^\circ$; the dashed curve there is at $11.17^\circ$.

The phase loop at the start and at the end, in $(\vartheta, p_\vartheta)$. Wider and flatter, but the same area, $2\pi I$.

What to look for. Nothing in the simulation knows about action variables: it is Newton's second law for a bob on a string being pulled in, $\frac{\mathrm d}{\mathrm dt}(m\ell^2\dot\vartheta) = -mg\ell\sin\vartheta$, solved step by step. The swing grows smoothly, cycle by cycle, and its envelope lies on the curve that $I = $ const predicts, without drifting off it. On the right the same fact is an area: the loop the motion traces in one period changes shape completely — the amplitude nearly triples, the peak momentum $m\ell^2\dot\vartheta$ falls — and its area does not move.

The numbers. $g = 9.81\ \mathrm{m\,s^{-2}}$, $m = 0.50$ kg, $\ell(t) = 1 - 0.0125\,t$ metres for $0 \le t \le 60$ s, released from rest at $4^\circ$; fourth-order Runge–Kutta with step $10^{-3}$ s. The last peak comes at $t = 59.65$ s, where $\ell = 0.2544$ m and the adiabatic prediction $4^\circ\ell^{-3/4}$ is $11.17^\circ$; the integration gives $11.18^\circ$. (The worked example's $11.31^\circ$ is the value at $\ell = 0.25$ m exactly, which the string reaches between swings.) The two loops on the right are the first and the last complete swings, from one positive peak to the next; their areas, by the shoelace formula, are $0.02396$ and $0.02391\ \mathrm{J\,s}$, against $2\pi I = 2\pi E/\omega = 0.02398\ \mathrm{J\,s}$ from the worked example's numbers. The last loop is drawn while the string is still moving, so it does not quite close; the gap is the $O(\varepsilon)$ wobble of the theorem.

15 · adiabatic

Kepler, slowly: a Sun that loses mass

The Sun loses mass slowly — sunlight carries $L_\odot/c^2$, the solar wind the rest — so $k = GM_\odot m$ in Kepler's $H$ drifts.

two conservation laws

Exact: the force stays central, so $L = m\sqrt{GM_\odot\,a(1-e^2)}$ is conserved.

Adiabatic: $I = I_r + L = m\sqrt{GM_\odot\,a}$ is conserved, because the drift is slow on the orbital timescale.

Divide: $e$ is constant; and

$$M_\odot\,a = \text{const}, \qquad \frac{\dot a}{a} = -\frac{\dot M_\odot}{M_\odot}.$$

numbers

$\dot M_\odot \approx -5.8\times10^{9}\ \mathrm{kg\,s^{-1}}$ ($4.26\times10^{9}$ from sunlight, about $1.5\times10^{9}$ from the wind), so

$$\frac{|\dot M_\odot|}{M_\odot} = 9.1\times10^{-14}\ \text{per year}.$$

Earth's orbit grows by $1.4$ cm per year; Halley's, with $a = 17.83$ AU, by $24$ cm per year.

Check. The slowness parameter is (period)/(timescale of $M$) $\approx 9\times10^{-14}$ for Earth: the adiabatic theorem applies with a vengeance.

Why both $L$ and $I$. $L$ is conserved exactly because the Sun's mass loss is (taken to be) spherically symmetric, so the force on the comet stays central whatever $M_\odot(t)$ does — unit 08. $I_r$ is only adiabatically conserved. Kepler is a system with three degrees of freedom and three equal frequencies, so the one-degree-of-freedom theorem does not apply off the shelf; but after using exact conservation of $\vec L$ to reduce to the radial motion, the radial problem is one degree of freedom with $\omega_r$ bounded away from zero, and the theorem applies to $I_r$. So $I_r + L$ is adiabatically invariant.

Algebra. From the actions slide $I = \sqrt{mka} = m\sqrt{GM_\odot a}$. From unit 08, $L^2 = mk\,a(1 - e^2)$, i.e. $L = m\sqrt{GM_\odot a(1-e^2)}$. Their ratio is $\sqrt{1 - e^2}$, so $e$ is constant; and $I$ constant gives $M_\odot a$ constant. Taking logarithmic derivatives: $\dot a/a = -\dot M_\odot/M_\odot$.

The numbers. $L_\odot = 3.828\times10^{26}$ W, $c^2 = 8.988\times10^{16}\ \mathrm{m^2\,s^{-2}}$, so sunlight carries $4.259\times10^{9}\ \mathrm{kg\,s^{-1}}$. The solar wind's mass flux varies with the solar cycle and is less well known, around $1.5\times10^{9}\ \mathrm{kg\,s^{-1}}$; that figure is an estimate, and the total $5.76\times10^{9}$ inherits its uncertainty. With $M_\odot = 1.988\times10^{30}$ kg and a Julian year of $3.156\times10^{7}$ s, $|\dot M_\odot|/M_\odot = 9.14\times10^{-14}$ per year. Earth: $1.496\times10^{11}\ \mathrm m \times 9.14\times10^{-14} = 1.37$ cm per year (sunlight alone: $1.01$ cm). Halley: $17.83$ AU gives $0.244$ m per year.

Limit. $\dot M_\odot = 0$: nothing changes ✓. And the direction is right: a lighter Sun holds its planets less tightly. Measuring this for Earth is at the edge of what planetary ranging can do; it is one of several centimetre-per-year effects in the astronomical unit.

15 · adiabatic

Where adiabatic invariance fails

at a separatrix

Near the separatrix $\omega\to0$: the period is not short compared with anything, and no change is slow. A pendulum whose amplitude is driven slowly up to the separatrix can cross it, and $I$ jumps by an amount that depends sensitively on the phase at crossing.

at a resonance

With several degrees of freedom there are several frequencies. When a slowly varying system passes through $n_1\omega_1 + n_2\omega_2 = 0$ for integers $n_1, n_2$ not both zero, energy can be exchanged between the modes and the individual $I_i$ need not be invariant.

what does hold

One degree of freedom, away from separatrices: the theorem. Several: averaging holds for most initial conditions, with resonances the exceptions (cited). The comet is the first case — exact conservation of $\vec L$ leaves one radial degree of freedom, never near a separatrix.

Why at a resonance. The averaging argument averaged $\partial_\lambda H$ over one angle. With two angles the average is over the torus, and that average is the time average only if the orbit fills the torus — i.e. if the frequencies are rationally independent. At a resonance the orbit lies on a closed curve in the torus, the time average differs from the torus average, and the zero-mean property is lost. This is the same phenomenon that unit 17 turns into chaos.

Physics. Adiabatic invariance is how magnetic bottles confine plasmas (the magnetic moment of a gyrating charge is an adiabatic invariant), why the Earth's radiation belts exist, and why a slowly compressed gas warms up in the way it does (problem P9 is the one-dimensional version). The name comes from thermodynamics: an "adiabatic" process is slow and exchanges no heat, and for a gas of oscillators the invariance of each $I$ is what makes such a process reversible.

15 · adiabatic

The separatrix: a period without bound

The theorem needs the motion to be fast compared with the change: its small parameter is really

$$\varepsilon_{\text{eff}} = T\,\frac{|\dot\lambda|}{|\lambda|} ,$$

the fraction by which $\lambda$ changes in one period. Hold $\dot\lambda$ fixed and let the energy approach the separatrix: $T\to\infty$, so $\varepsilon_{\text{eff}}\to\infty$. However slowly $\lambda$ moves, near enough to the separatrix it is not slow.

$\omega$ bounded away from $0$ is not a technicality. Crossing a separatrix, $I$ jumps.

The pendulum's period against its energy, computed. Finite on both sides, it grows without bound at the separatrix energy, like $\log\frac{1}{|E - 2|}$: there the motion is slower than any change, however slow.

The figure. $H = p^2/2 + 1 - \cos q$ as on the action-variable slide, separatrix at $E = 2$. Below it, $T = 4K(k)$ with $k = \sqrt{E/2}$; above it, the time for one turn is $T = 4K(k)/\sqrt{2E}$ with $k = \sqrt{2/E}$, where $K(k) = \int_0^{\pi/2}\mathrm d\varphi/\sqrt{1 - k^2\sin^2\varphi}$ is unit 00's complete elliptic integral. At $E = 0$ the period is $2\pi$, the small-oscillation value; as $E\to\infty$ it falls like $2\pi/\sqrt{2E}$, one turn at the speed $\sqrt{2E}$. Near $E = 2$, $K(k)\approx\ln(4/\sqrt{1-k^2})$, which is the logarithm in the caption.

Why at a separatrix. The frequency theorem's proof showed $T\to\infty$ logarithmically as $E$ approaches the separatrix energy. The adiabatic parameter is (period) × (rate of change of $\lambda$), so it grows without bound however small $\varepsilon$ is: the hypothesis "$\omega$ bounded away from zero" is violated, and the averaging argument collapses. The size and statistics of the jump are cited, not proved: the theory of separatrix crossing is surveyed in Arnold, Kozlov and Neishtadt, Mathematical Aspects of Classical and Celestial Mechanics, 3rd ed., chapter 6 (§6.4.7; it is chapter 5, §4 in the 2nd edition).

section 5

Mechanics was optics all along

Trajectories are rays, surfaces of constant action are wavefronts, and the rays are geodesics.

15 · optics

Fermat and Maupertuis

Fermat, 1662 · light

Rays in a medium of refractive index $n(\vec x)$ are the extremals of optical length

$$\int n(\vec x)\,\mathrm ds .$$

Unit 11's Euler–Lagrange equation, parametrised by arc length, is the ray equation

$$\frac{\mathrm d}{\mathrm ds}\Big(n\,\frac{\mathrm d\vec x}{\mathrm ds}\Big) = \nabla n .$$

Maupertuis, 1744 · particles

Trajectories of energy $E$ in a potential $V$ are the extremals of the abbreviated action

$$\int\vec p\cdot\mathrm d\vec x = \int\sqrt{2m\big(E - V(\vec x)\big)}\,\mathrm ds$$

among paths of energy $E$ between the same points — Maupertuis' principle, proved on the slide after this.

the dictionary

Same functional, with $\ n(\vec x) \longleftrightarrow \sqrt{2m(E - V(\vec x))} = \lVert\vec p\rVert$. A particle is a ray in a medium whose index is its own momentum.

Full proof of the ray equation. Parametrise a path as $\vec x(\sigma)$ with any parameter; the functional is $\int F\,\mathrm d\sigma$ with $F(\vec x, \vec x') = n(\vec x)\lVert\vec x'\rVert$. Unit 11's Euler–Lagrange equations in three dependent variables are $\frac{\mathrm d}{\mathrm d\sigma}\partial_{\vec x'}F = \partial_{\vec x}F$, and $\partial_{\vec x'}F = n\,\vec x'/\lVert\vec x'\rVert$, $\partial_{\vec x}F = \lVert\vec x'\rVert\nabla n$. So $\frac{\mathrm d}{\mathrm d\sigma}\big(n\vec x'/\lVert\vec x'\rVert\big) = \lVert\vec x'\rVert\nabla n$. Now choose $\sigma = s$, arc length, so $\lVert\vec x'\rVert = 1$. $\square$ The functional does not care about the parametrisation — the three equations are not independent, which is why one may fix $\sigma$ freely.

Check. Constant $n$: $\mathrm d^2\vec x/\mathrm ds^2 = 0$, straight lines ✓. Unit 11's Snell's law is the ray equation across a sharp interface: the component of $n\,\mathrm d\vec x/\mathrm ds$ along the interface is conserved because $\nabla n$ is normal to it.

History and a warning about names. Maupertuis stated his principle in 1744 as a metaphysical claim about economy in nature, in a form too vague to be wrong; Euler, in the same year, stated it correctly for a particle in a central field; Lagrange extended it; and Jacobi, in his Königsberg lectures of 1842–43, gave the form without time used here. It is often called the "principle of least action", which is doubly misleading: it is stationary, not least (unit 11's conjugate points), and the action in it is $W$, not $\int L\,\mathrm dt$.

The corpuscle puzzle. Rays bend towards larger $n$. In mechanics $n = \lVert\vec p\rVert$, so particles bend towards where they move faster — as Newton's corpuscular theory of light said light does, with light faster in glass. Wave theory says the opposite: in glass light is slower, $n = c/v$. Both "work" because the index is not the speed in both pictures; the wavefronts slide resolves it.

15 · optics

Trajectories are rays

theorem · Maupertuis–Jacobi

Let $n = \sqrt{2m(E - V)}$ on the region $\{V < E\}$. Every trajectory of energy $E$ satisfies the ray equation $\frac{\mathrm d}{\mathrm ds}(n\,\vec T) = \nabla n$, $\vec T = \mathrm d\vec x/\mathrm ds$. Conversely, a solution of it traversed at speed $\mathrm ds/\mathrm dt = n/m$ is a trajectory of energy $E$.

proof

On a trajectory $\lVert\dot{\vec x}\rVert = n/m$ (energy), so $\mathrm ds = (n/m)\,\mathrm dt$ and $n\vec T = m\dot{\vec x} = \vec p$. Hence

$$\frac{\mathrm d}{\mathrm ds}\big(n\vec T\big) = \frac{m}{n}\,\dot{\vec p} = -\frac{m\nabla V}{n}, \qquad \nabla n = \nabla\sqrt{2m(E - V)} = -\frac{m\nabla V}{n} .$$

Equal. Conversely, with $\mathrm dt = m\,\mathrm ds/n$: $\ m\ddot{\vec x} = \dfrac{n}{m}\dfrac{\mathrm d}{\mathrm ds}\big(n\vec T\big) = \dfrac{n}{m}\nabla n = -\nabla V$. $\square$

Check. $V = 0$: $n$ constant, straight lines at constant speed ✓. Dimensions: $n$ is a momentum, $\int n\,\mathrm ds$ an action ✓.

Precisely what is proved. The ray equation is the Euler–Lagrange equation of the abbreviated action $\int n\,\mathrm ds$ (previous slide), so this theorem is Maupertuis' principle: trajectories of energy $E$ are exactly its extremals, as unparametrised curves, with time recovered from $\mathrm dt = m\,\mathrm ds/n$. The mass has not been set to 1; several particles are handled by the same proof in the mass-weighted coordinates $\sqrt{m_i}\,\vec x_i$, in which the kinetic energy is $\tfrac12\lVert\dot y\rVert^2$.

What happens at $V = E$. On the boundary of the allowed region $n = 0$ and the proof breaks: $\mathrm ds/\mathrm dt = 0$ there, the particle stops and turns back. A ray reaching the boundary is a trajectory with a turning point, like a stone thrown straight up.

Why it is not circular. The proof uses energy conservation to link $s$ and $t$. That is legitimate on a trajectory, where energy is conserved (unit 04), and for the converse the parametrisation $\mathrm dt = m\,\mathrm ds/n$ is a choice that builds $\tfrac12 m\lVert\dot{\vec x}\rVert^2 + V = E$ in: $\tfrac12 m(n/m)^2 = n^2/2m = E - V$.

For the running example. Kepler's $n = \sqrt{2m(E + k/r)}$ is spherically symmetric: a spherically stratified medium, like the Earth's atmosphere for light. The worked example "Kepler orbits are geodesics" solves its rays.

15 · optics

A thrown stone, as a ray

Uniform gravity, $V = mgy$. The index $n = \sqrt{2m(E - mgy)}$ depends on height alone: a medium in horizontal layers, densest at the bottom and falling to $n = 0$ at the height $E/mg$ where the stone would stop.

$\nabla n$ is vertical, so the horizontal component of the ray equation says $\frac{\mathrm d}{\mathrm ds}\big(n\,\frac{\mathrm dx}{\mathrm ds}\big) = 0$:

$$n\sin\alpha = \text{const},$$

$\alpha$ the angle to the vertical — Snell's law across the layers. And $n\sin\alpha = \lVert\vec p\rVert\sin\alpha = p_x$: it is conservation of horizontal momentum.

checks

At the top of the arc $\alpha = 90^\circ$ and $n = p_x$: all the momentum is horizontal ✓. Rays bend towards larger $n$ — downwards ✓.

Shading: the index $n$, darkest where the stone is fastest. The arc is a parabola, computed; at each marked point $n\sin\alpha$ takes the same value. Units $m = g = E = 1$, launched at $60^\circ$.

What to look for. Lower down the index is larger and the ray is steeper (small $\alpha$); higher up the index is smaller and the ray leans over, until at the top it runs horizontally. That is exactly what light does in a layered medium when it travels from a dense layer into thinner ones — it bends away from the normal, and if the index falls far enough it turns back (total internal reflection, done smoothly). A mirage is this picture upside down: on a hot road the air near the ground is thinner, $n$ grows upwards, and rays from the sky curve up into the eye.

The numbers. $E = 1$ and launch speed $\sqrt{2E} = \sqrt2$ at $60^\circ$ give $p_x = \sqrt2\cos60^\circ = 0.7071$ and a top at $y = \tfrac12(\sqrt2\sin60^\circ)^2 = 0.75$, below the ceiling $y = 1$. At the marked points the index $n = \sqrt{2(1 - y)}$ and the angle to the vertical were computed from the parabola, and $n\sin\alpha = 0.7071$ at each.

The general statement. Whenever $n$ depends on one Cartesian coordinate only, the momentum along the others is conserved, and the ray equation gives Snell's law in that direction. Unit 14 would say: $V$ is independent of $x$, so $x$ is cyclic and $p_x$ is conserved. The two languages are saying the same sentence.

15 · optics

Wavefronts are surfaces of constant W

the eikonal equation

The time-independent HJ equation for $H = \lVert\vec p\rVert^2/2m + V$ is

$$\lVert\nabla W\rVert^2 = 2m(E - V) = n^2,$$

the equation geometrical optics writes for its phase. The level sets of $W$ are wavefronts.

rays are orthogonal to wavefronts

The trajectories with $\vec p = \nabla W$ move along $\nabla W$, which is normal to the level sets of $W$ (unit 04). So trajectories cross wavefronts at right angles — Huygens' construction.

Uniform gravity: trajectories of one energy from one point (thin), levels of $W$ (thick), and the caustic (dashed), where $W$ folds.

Proof of the orthogonality. If $W$ solves the eikonal equation and $\vec x(t)$ solves $m\dot{\vec x} = \nabla W(\vec x)$, then $\vec x(t)$ is a trajectory: this is problem P3's theorem, that a single solution of the HJ equation already defines a family of trajectories. Along it $\dot{\vec x}$ is parallel to $\nabla W$, and the gradient is orthogonal to every tangent vector of a level set, because the derivative of $W$ along a curve lying in a level set is zero and equals $\nabla W$ dotted with the curve's tangent.

The figure. Units $v_0 = g = 1$, $m = 1$. Rays leave the origin in directions every $15^\circ$ at speed $1$; along each, $W = \int\vec p\cdot\mathrm d\vec x = \int v^2\,\mathrm dt$ was computed exactly and the curves $W = 0.2, 0.4, \dots, 1.0$ drawn. The dashed parabola $y = \tfrac12 - \tfrac12x^2$ is the envelope of the trajectories (the "safety parabola" of artillery), and every ray is drawn only up to where it touches it. Beyond the envelope several rays reach each point and $W$ is multivalued; the wavefronts cannot be continued as level sets of one function. That is exactly what a caustic is in optics — the bright curve on the bottom of a cup of coffee — and the conjugate points of unit 11 are where a ray touches it.

Moving wavefronts. The surfaces $S = -Et + W = $ const move: at time $t$ it is $W = $ const $+ Et$. The normal speed of a level set of $W$ moving so that $W$ grows at rate $E$ is $u = E/\lVert\nabla W\rVert = E/p$. This is the phase velocity of Hamilton's wave. It is inversely proportional to the particle's speed $p/m$. So the particle is fast exactly where the wave is slow — the wave picture's index $\propto 1/u \propto p$ agrees with the particle's index $n = p$. The corpuscle puzzle of the Fermat–Maupertuis slide resolves: mechanics' "light" slows down where the particle speeds up.

15 · optics

The eikonal limit

a wave, and its short-wavelength form

Let $\Delta = \sum_i\partial^2/\partial x_i^2$. A monochromatic wave of wavenumber $k_0$ in a medium obeys Helmholtz's equation $\ \Delta\psi + k_0^2n^2\psi = 0$ (cited: unit 24). Substitute $\psi = A\,e^{ik_0S}$ with real $A, S$. The product and chain rules give, exactly,

$$e^{-ik_0S}\big(\Delta\psi + k_0^2n^2\psi\big) = k_0^2A\big(n^2 - \lVert\nabla S\rVert^2\big) + ik_0\big(2\nabla A\cdot\nabla S + A\,\Delta S\big) + \Delta A .$$

the eikonal limit

For fixed $A \ne 0$ and $S$, the residual is $O(k_0)$ rather than $O(k_0^2)$ as $k_0\to\infty$ iff $\ \lVert\nabla S\rVert^2 = n^2$: the eikonal equation. Short waves travel along rays.

Run it backwards with $n = \sqrt{2m(E-V)}$ and $k_0 = 1/\hbar$, and a wave equation for particles is waiting. Schrödinger, 1926, did exactly that.

Full proof of the identity. $\nabla\psi = (\nabla A + ik_0A\nabla S)e^{ik_0S}$. Differentiating again and summing, $\Delta\psi = \big(\Delta A + 2ik_0\nabla A\cdot\nabla S + ik_0A\Delta S - k_0^2A\lVert\nabla S\rVert^2\big)e^{ik_0S}$. Add $k_0^2n^2\psi$. Complex exponentials are unit 06's algebra; nothing analytic about $\mathbb C$ is used. The "iff" is immediate: the right-hand side is a polynomial in $k_0$ whose $k_0^2$ coefficient is $A(n^2 - \lVert\nabla S\rVert^2)$. $\square$

What is and is not claimed. This shows that the eikonal equation is the condition for $Ae^{ik_0S}$ to solve Helmholtz's equation to leading order. It does not show that true solutions are close to such waves — that is a theorem about asymptotic expansions (the WKB method), which needs unit 21's language and is not proved here. Helmholtz's equation itself, for each component of monochromatic light in a slowly varying medium, is derived in unit 24. The next order, $2\nabla A\cdot\nabla S + A\Delta S = 0$, is the transport equation: it says how the amplitude changes along a ray, and it blows up at caustics, which is why caustics are bright.

History. Hamilton built the whole theory first for optics, in "Theory of systems of rays" (1828–37), and only then noticed that the same mathematics described particles. For ninety years the analogy was an elegance. Schrödinger's second paper of 1926, "Quantisierung als Eigenwertproblem", argued explicitly: classical mechanics is to some wave mechanics as geometrical optics is to wave optics, so find the wave equation whose eikonal limit is the Hamilton–Jacobi equation. That argument, not a derivation, is how quantum mechanics got its wave equation. It is outside this course, but the classical half of it is now yours.

15 · geometry

A metric, defined properly

definition

A Riemannian metric on an open $U\subset\mathbb R^n$ assigns to each $x\in U$ a symmetric, positive-definite bilinear form $g_x$ on $\mathbb R^n$ (the velocities at $x$), with components $g_{ij}(x) = g_x(e_i, e_j)$ smooth in $x$. The length of a curve is $\ \mathcal L[\gamma] = \int\sqrt{g_{\gamma}(\gamma', \gamma')}\,\mathrm d\sigma$, and a geodesic is a critical point of $\mathcal L$ among curves with the same ends.

theorem · components

Let $G = (g_{ij})$ be the matrix of components. If $f_b = \sum_a A_{ab}e_a$ is another basis, then $g(f_a, f_b) = (A^{\mathsf T}GA)_{ab}$.

Proof. Bilinearity: $g(\sum_i A_{ia}e_i, \sum_jA_{jb}e_j) = \sum_{ij}A_{ia}g_{ij}A_{jb}$. $\square$

conformal rescaling

$\tilde g = \Omega^2g$ with $\Omega > 0$ smooth. Lengths change; angles do not: in $\cos\angle(u,v) = g(u,v)/\sqrt{g(u,u)g(v,v)}$ the $\Omega^2$ cancels. $\square$

Physics: kinetic energy is a metric, $T = \tfrac12 g(\dot x, \dot x)$ with $g = m\,\delta$. In spherical coordinates the theorem, with $A$ the derivative of the chart, gives $g = m\,\mathrm{diag}(1, r^2, r^2\sin^2\theta)$.

Mathematics first. The metric is a bilinear form at each point: that is the definition. The array $g_{ij}$ is its representation in one basis, and the rule $G\mapsto A^{\mathsf T}GA$ is a theorem about how that representation changes. Unit 16 builds all tensors this way — a multilinear map first, its components' transformation law as a consequence — and unit 31 replaces "velocities at $x$" with the tangent space of a manifold. Here nothing more than linear algebra is needed.

Notation. $g = m(\mathrm dr^2 + r^2\mathrm d\theta^2 + r^2\sin^2\theta\,\mathrm d\phi^2)$ is shorthand for $g(v,v) = m(v_r^2 + r^2v_\theta^2 + r^2\sin^2\theta\,v_\phi^2)$ for a velocity with components $(v_r, v_\theta, v_\phi)$ in the coordinate basis. The computation is the spherical-coordinates slide: $A$'s columns are $\hat e_r$, $r\hat e_\theta$, $r\sin\theta\,\hat e_\phi$, and $A^{\mathsf T}A = \mathrm{diag}(1, r^2, r^2\sin^2\theta)$ by orthonormality.

Length is independent of parametrisation. If $\sigma = \varphi(\tau)$ with $\varphi' > 0$, then $\gamma'$ scales by $\varphi'$, the square root by $\varphi'$, and $\mathrm d\sigma = \varphi'\,\mathrm d\tau$ absorbs it (unit 04's argument for line integrals). So "critical point of $\mathcal L$" is a property of the unparametrised curve.

On the definition of geodesic. Unit 32 defines geodesics through a connection, as curves whose velocity is parallel-transported, and proves that for the metric's own (Levi-Civita) connection these are exactly the constant-speed critical points of length. The variational definition here is the one unit 11 already used on the sphere, and all that this unit needs.

Conformal. The word means "angle-preserving". Maps between two surfaces that preserve angles are conformal; a metric multiplied by a positive function is the simplest example, and the Jacobi metric is one.

15 · geometry

The Jacobi metric

theorem · the Jacobi metric

Let $g_{\mathrm{kin}}$ be the kinetic metric ($T = \tfrac12 g_{\mathrm{kin}}(\dot x, \dot x)$) and $V < E$ on $U$. On $U$ put the conformal rescaling

$$g_J = 2\big(E - V\big)\,g_{\mathrm{kin}} .$$

The trajectories of energy $E$ are exactly the geodesics of $g_J$, reparametrised so that $g_{\mathrm{kin}}(\dot x,\dot x) = 2(E - V)$.

proof

With $g_{\mathrm{kin}} = m\delta$, $\ \sqrt{g_J(\gamma',\gamma')} = \sqrt{2m(E - V)}\,\lVert\gamma'\rVert = n\lVert\gamma'\rVert$: the $g_J$-length is the abbreviated action $\int n\,\mathrm ds$, whose extremals are the trajectories (Maupertuis–Jacobi). $\square$

what it is not

A different metric for each $E$, and time is gone. It degenerates on $V = E$, where the particle turns.

Check. $V = 0$: $g_J = 2E\,g_{\mathrm{kin}}$, a constant multiple of flat; geodesics are straight lines ✓. With $m = 1$, $g_J = 2(E - V)\,\delta_{ij}$.

Coordinates do not matter. The $g_J$-length of a curve is a number defined without coordinates; its critical points are therefore the same curves in every chart — unit 12's form-invariance argument, applied to this functional. So the theorem, proved in Cartesian coordinates, holds in spherical ones with $g_{\mathrm{kin}} = m(\mathrm dr^2 + r^2\mathrm d\theta^2 + r^2\sin^2\theta\,\mathrm d\phi^2)$. For a system with constraints the kinetic metric is the one induced on the configuration space, and the theorem still holds; the proof without Cartesian coordinates needs unit 32's geodesic equation, and is cited: Landau and Lifshitz, Mechanics, §44, which derives "Jacobi's form of the principle of least action" for any $T = \tfrac12\sum a_{ik}(q)\dot q_i\dot q_k$.

Dimensions. $g_J(v,v)$ is energy × mass × speed², and its square root per unit parameter integrates to an action: a $g_J$-length is measured in $\mathrm{J\,s}$, the unit of $\int p\,\mathrm dq$.

The reparametrisation. A geodesic is a curve; to be a trajectory it must be traversed at the right speed. The condition $g_{\mathrm{kin}}(\dot x, \dot x) = 2(E - V)$ is $2T = 2(E - V)$, energy conservation, which is $\mathrm ds/\mathrm dt = n/m$ in the Cartesian case.

Why "free motion is geodesic motion". With $V = 0$ Newton's first law says free particles move in straight lines at constant speed — geodesics of the flat kinetic metric. The theorem says that with a force, the particle is still free, in the geometry $g_J$. Force has been traded for geometry, and the price is that the geometry depends on $E$ and forgets $t$. Einstein's gravity, in units 30–32, makes the same trade without paying that price.

15 · geometry · example

Worked: Kepler orbits are geodesics

the geodesics of Kepler's Jacobi metric

Step 1 — the metric. In the orbital plane ($\theta = \pi/2$, a reflection-symmetric plane), with $n^2 = 2m(E + k/r)$:

$$g_J = n(r)^2\big(\mathrm dr^2 + r^2\,\mathrm d\phi^2\big),$$ $$\mathcal L = \int n\sqrt{r'^2 + r^2}\,\mathrm d\phi .$$

Step 2 — Beltrami (unit 11): no explicit $\phi$, so $F - r'F_{r'}$ is constant:

$$\frac{n\,r^2}{\sqrt{r'^2 + r^2}} = \ell .$$

This is $n\,r\sin\alpha = \ell$, $\alpha$ the angle to the radius: Bouguer's formula of atmospheric optics. Mechanically, $\lVert\vec p\rVert\,r\sin\alpha = L$: so $\ell = L$.

Step 3 — $u = 1/r$. $r' = -u'/u^2$, and step 2 squared becomes

$$u'^2 = \frac{n^2}{L^2} - u^2 = \frac{2mE}{L^2} + \frac{2mk}{L^2}u - u^2 .$$

Step 4 — differentiate and divide by $2u'$:

$$u'' + u = \frac{mk}{L^2},$$

unit 08's Binet equation. So $u = \frac{mk}{L^2}\big(1 + e\cos(\phi - \phi_0)\big)$, and step 3 at the perihelion fixes $e^2 = 1 + 2EL^2/mk^2$.

Check. Section 2's orbit, exactly ✓. $\ell$ has dimension $n\cdot r$ = momentum × length ✓.

Step 1: why the plane. The reflection $\theta\mapsto\pi - \theta$ preserves $g_J$ and fixes the equatorial plane pointwise. Let $\gamma$ be a curve in the plane that is critical for $\mathcal L$ among curves in the plane. For an out-of-plane variation $\eta$, the reflected variation $\bar\eta$ gives the same first variation (the length is reflection-invariant) but the opposite sign (the out-of-plane component is reversed), so that part of the first variation is zero. Hence $\gamma$ is critical among all curves. And every orbit lies in some plane through the Sun (section 2); rotate it to the equator, which is an isometry of $g_J$.

Step 2 in detail. $F = n(r)\sqrt{r'^2 + r^2}$, $F_{r'} = nr'/\sqrt{r'^2 + r^2}$, so $F - r'F_{r'} = n(r'^2 + r^2 - r'^2)/\sqrt{r'^2 + r^2} = nr^2/\sqrt{r'^2 + r^2}$. The angle $\alpha$ between the curve and the radial direction has $\sin\alpha = r/\sqrt{r'^2 + r^2}$ (the tangent has radial component $r'$ and transverse component $r$ per unit $\phi$), so the constant is $nr\sin\alpha$. Pierre Bouguer used exactly this invariant in the eighteenth century for light in a spherically layered atmosphere. For the particle, $n = \lVert\vec p\rVert$ and $r\lVert\vec p\rVert\sin\alpha = \lVert\vec r\times\vec p\rVert = L$: Bouguer's invariant is the angular momentum.

Step 3. $r'^2 + r^2 = n^2r^4/L^2$ from step 2 squared. With $r = 1/u$, $r' = -u'/u^2$: $u'^2/u^4 + 1/u^2 = n^2/(L^2u^4)$, so $u'^2 = n^2/L^2 - u^2$, and $n^2 = 2mE + 2mku$.

Step 4. $2u'u'' = (2mk/L^2)u' - 2uu'$. Dividing by $2u'$ is allowed where $u' \ne 0$; the resulting equation is linear with smooth coefficients, so it holds also at the isolated apsides by continuity. The circular orbit, $u' \equiv 0$, satisfies step 3 with $u^2 - (2mk/L^2)u - 2mE/L^2 = 0$ having a double root, which is $E = -mk^2/2L^2$ ✓. At the perihelion $u' = 0$ and $u_{\max} = (mk/L^2)(1 + e)$ must be a root of the step-3 quadratic: substituting gives $e^2 = 1 + 2EL^2/mk^2$.

Halley. Per unit mass the "index" $n/m$ is the speed: $54.570\ \mathrm{km\,s^{-1}}$ at perihelion and $0.91164\ \mathrm{km\,s^{-1}}$ at aphelion, ratio $59.86$, so the metric's conformal factor varies by $59.86^2 = 3583$ along the orbit. The total $g_J$-length of one orbit is $\oint\vec p\cdot\mathrm d\vec x = 2\pi(I_r + L) = 2\pi I$ — for Halley, per unit mass, $1.1822\times10^{17}\ \mathrm{m^2\,s^{-1}}$.

15 · geometry

Bouguer's invariant, on an orbit

Step 2 of the worked example found the conserved quantity of the geodesics, $n\,r\sin\alpha = L$. In the medium of index $n = \lVert\vec p\rVert$, a ray near the centre (large $n$, small $r$) and a ray far out (small $n$, large $r$) balance the product.

Halley, at its two apsides

There $\alpha = 90^\circ$, so $n r/m = L/m$:

perihelion: $54.570\ \mathrm{km\,s^{-1}}\times 0.58602$ AU $= 4.7840\times10^{15}\ \mathrm{m^2\,s^{-1}}$;

aphelion: $0.91164\ \mathrm{km\,s^{-1}}\times 35.079$ AU $= 4.7840\times10^{15}\ \mathrm{m^2\,s^{-1}}$ ✓.

Schematic orbit ($e = 0.6$; Halley's would be a needle). The ray's angle $\alpha$ to the radius is $90^\circ$ at the two apsides and not between; $n\,r\sin\alpha$ is the same everywhere — the angular momentum.

What to look for. In a spherically layered medium the "layers" are spheres round the Sun, and the normal to them is the radius. Bouguer's formula is the spherical form of Snell's law from the thrown-stone slide: there the conserved product was $n\sin\alpha$ with $\alpha$ measured from the vertical; here it is $n\,r\sin\alpha$ with $\alpha$ measured from the radius, and the extra $r$ is what the curvature of the layers costs. At the apsides the orbit crosses the radius at right angles and $\sin\alpha = 1$; on the way out $\alpha$ is acute, on the way in obtuse, with the same sine.

The numbers. Perihelion $a(1 - e) = 17.8324\times0.032863 = 0.58602$ AU and aphelion $a(1 + e) = 35.079$ AU, with $1\ \mathrm{AU} = 1.495979\times10^{11}$ m; the speeds are from the Kepler-geodesics notes. Both products reproduce unit 08's $h = 4.7840\times10^{15}\ \mathrm{m^2\,s^{-1}}$, which is the check that the index really is the momentum per unit mass.

15 · geometry

Free motion is geodesic motion

what has been shown

A particle of energy $E$ under a force $-\nabla V$ is a free particle — it follows geodesics — in the geometry $g_J = 2(E - V)g_{\mathrm{kin}}$. Halley's ellipse is a straightest line. This is Newtonian mechanics, 1842–43.

for gravity, the mass drops out

With $V = m\Phi$, $\ g_J/m^2 = 2(\varepsilon - \Phi)\,\delta$ with $\varepsilon = E/m$. Every body with the same $\varepsilon$ sees the same geometry, whatever its mass. For an electric force, $V = q\Phi_e$, the geometry depends on $q/m$.

why this is not yet general relativity

The geometry depends on $\varepsilon$: a fast and a slow body through the same point see different metrics.
Time has been thrown away; the geometry is of space, not of spacetime.

Unit 30 puts time back and finds one metric for every body: possible only because all bodies fall alike, which is the box on the left.

The proof of the left-hand box is one line: substitute $V = m\Phi$ and $E = m\varepsilon$ into $g_J = 2(E - V)\,m\delta$ to get $2m^2(\varepsilon - \Phi)\delta$. Scaling a metric by a constant does not change its geodesics, so the geodesics depend only on $\varepsilon$ and $\Phi$. For a charge, $g_J = 2m(E - q\Phi_e)\delta = 2m^2(\varepsilon - (q/m)\Phi_e)\delta$, and bodies with different $q/m$ at the same $\varepsilon$ follow different geodesics.

The thread to the end of the course. Unit 30 argues that the equality of inertial and gravitational mass — the $m$ cancelling here — is what allows gravity to be geometry at all. Unit 32 defines geodesics of a metric on a manifold and shows free fall is one; unit 37 finds, from the Schwarzschild geodesics, the orbit equation $u'' + u = GM/h^2 + 3GMu^2/c^2$ — step 4 of the Kepler-geodesics example with one extra term. The extra term breaks the degeneracy of the actions slide, which is the $43''$.

What the Jacobi picture is still good for. In unit 17, the curvature of the Jacobi metric is one way to see why some systems are chaotic: geodesics of negatively curved metrics diverge exponentially, and Hadamard and later Krylov made this an argument for instability. That needs unit 33's curvature and is only named here.

15 · close

Halley, closed

the running example, start to finish

1. Kepler's $H$ in spherical coordinates; its HJ equation, one nonlinear PDE.

2. Separated into three ODEs with constants $E, L, L_z$; the conic, its plane and its timing read off from $\partial S/\partial\alpha = \beta$ — the six orbital elements.

3. Actions $I_r = k\sqrt{m/(-2E)} - L$, $I_\theta = L - |L_z|$, $I_\phi = L_z$; $H = -mk^2/2I^2$; period $75.305$ yr by differentiation; three equal frequencies, so the orbit closes.

4. A Sun losing mass: $e$ fixed, $a\propto1/M_\odot$, $24$ cm per year for Halley.

5. Its ellipse is a geodesic of $2m(E + k/r)(\mathrm dr^2 + r^2\mathrm d\phi^2)$, via Bouguer and Binet.

What this unit hands on

16 — the metric as a bilinear form, components as a theorem: now for all tensors.

17 — invariant tori and action–angle variables in general (Liouville–Arnold), and what breaks them.

21 — the anharmonic frequency shift again, by Poincaré–Lindstedt.

32, 37 — geodesics defined by a connection; the degeneracy broken by $3GMu^2/c^2$.

40 — Kerr geodesics separate in the Hamilton–Jacobi equation, with Carter's constant.

Every number above was computed from unit 08's $\varepsilon$ and $h$ for the comet and $GM_\odot$: $I/m = GM_\odot/\sqrt{-2\varepsilon}$, the period $2\pi(I/m)^3/(GM_\odot)^2 = 2.3764\times10^9$ s, the growth rate from $\dot a/a = |\dot M_\odot|/M_\odot = 9.14\times10^{-14}$ per year with $a = 17.832$ AU. The one input that is an estimate rather than a measurement is the solar wind's share of the mass loss.

Reading. Landau and Lifshitz, Mechanics, §§43–44 (the action as a function of coordinates; Maupertuis and Jacobi's form), §§47–49 (the Hamilton–Jacobi equation, separation, adiabatic invariants) and §52 (conditionally periodic motion): terse, complete, and at this level; §§50–51 use canonical transformations from unit 14 and are fine to read now. Goldstein, Classical Mechanics, 3rd edition, chapter 10, is the standard long treatment, with the Kepler problem in action-angle variables in §10.8. Arnold, Mathematical Methods of Classical Mechanics, treats Huygens' principle and the Hamilton–Jacobi equation in chapter 9 in the language of differential forms, which is unit 34: read it after that unit, not now. For the optical side, Born and Wolf, Principles of Optics, chapter 3, derives the eikonal equation from Maxwell's equations — after unit 24.

15 · checkpoint

Done when

checkpoint

You can, on paper and without the deck,

solve a separable Hamilton–Jacobi problem — write the equation, separate it, and read the trajectory off $\partial S/\partial\alpha = \beta$ without integrating an equation of motion;
extract action–angle variables and produce the oscillation frequency of an anharmonic oscillator, such as $V = \tfrac12 kx^2 + \lambda x^4$ to first order in $\lambda$, as $\mathrm dE/\mathrm dI$; and
write the Jacobi metric for a given $V$ and $E$, and check that a known trajectory is one of its geodesics.

Falsifiable: a frequency obtained by solving $\ddot x = \dots$ fails the second; a metric that still contains $t$, or loses the factor $E - V$, fails the third. Record it with ./mr done 15.

Problems

./mr problems 15 — 12 problems, about 6 hours. Hints and worked solutions fold under each problem on the same page.

16 rigid bodies and tensors · 17 integrability and chaos, which builds on section 3 directly · 40, far ahead, which separates Kerr by this unit's method.

Try the checkpoint cold. For the first part, the isotropic oscillator in spherical coordinates (problem P4) is a good test: the separation is Kepler's word for word, and the answer $H = \Omega(2I_r + I_\theta + |I_\phi|)$ can be checked against Cartesian coordinates. For the second, the quartic oscillator (problem P6): $\omega = \omega_0 + 3\lambda A^2/(2m\omega_0)$ to first order, with $\omega_0 = \sqrt{k/m}$. For the third, uniform gravity (problem P11): $g_J = 2m(E - mgy)(\mathrm dx^2 + \mathrm dy^2)$, whose geodesics are parabolas.

If the first part stalls, it is usually at the branch of a square root, or at forgetting that the separation constant must be named before it can be differentiated with respect to. If the second stalls, it is usually at the average $\langle x^4\rangle = 3A^4/8$ over the unperturbed motion. If the third does, recheck that the metric's factor is $E - V$, not $V$ or $E + V$.

Hamilton–Jacobi: mechanics was optics all along, and free motion is already geodesic motion

One problem, solved without its equation of motion

The action as a function of where you end

The action, as a function of its endpoint

Fans of paths: where S is a function, and where it stops

Kepler's Hamiltonian in spherical coordinates

The Hamilton–Jacobi equation

Complete integrals, and the solutions they generate

The free particle: plane waves and their envelope

Jacobi's theorem

Worked: the oscillator, without its equation

When H does not depend on time

Separation of variables

Separation: one equation becomes n

Stäckel's conditions

Kepler, separated

Worked: the orbit, read off

Six constants, six orbital elements

The six elements, drawn

Action–angle variables: frequencies without motion

The action variable

The frequency is a derivative

The frequency theorem, as an area

The angle conjugate to it

Many degrees of freedom: one loop per coordinate

The torus, and its two loops

Two integrals, done once

Worked: Kepler's actions, and Kepler's third law

Three equal frequencies: why Kepler orbits close

Any other central force: the ellipse turns

How the action responds to a parameter

Adiabatic invariants

Slow change, and what survives it

Worked: the pendulum shortened slowly

The shortened pendulum, integrated

Kepler, slowly: a Sun that loses mass

Where adiabatic invariance fails

The separatrix: a period without bound

Mechanics was optics all along

Fermat and Maupertuis

Trajectories are rays

A thrown stone, as a ray

Wavefronts are surfaces of constant W

The eikonal limit

A metric, defined properly

The Jacobi metric

Worked: Kepler orbits are geodesics

Bouguer's invariant, on an orbit

Free motion is geodesic motion

Halley, closed

What this unit hands on

Done when

Problems

Next