Part B · the action · spine

Hamilton: phase space, and flows that preserve volume

Trade n second-order equations for 2n first-order ones, and the new space turns out to carry a bracket, a class of allowed coordinate changes and a preserved volume — none of which configuration space ever showed you.

needs 13, 02~2½ h slides + 5 h problemsproblems: ./mr problems 14

Unit 12 wrote mechanics as the Euler–Lagrange equations of $\int L\,\mathrm dt$, and unit 13 read conservation laws off its symmetries. This unit changes the variables once more: instead of positions and velocities, positions and momenta. Nothing Newton did not know follows from the change, and that is Part B's thesis stated in miniature: the value of a reformulation is entirely in what it makes obvious. Here it makes four things obvious that were hidden before — a phase portrait whose curves are level sets of one function, a bracket under which conserved quantities form an algebra, a large class of coordinate changes that mix $q$ with $p$ and still preserve the equations, and a volume that the motion never changes.

The last of those, Liouville's theorem, is why statistical mechanics can exist at all, and why a good numerical method for planetary orbits is not the same thing as an accurate one. The third is the door to unit 15 (Hamilton–Jacobi) and unit 17 (integrability and chaos). The first piece of mathematics, the Legendre transform, returns in unit 27 for relativistic particles, where it nearly fails.

How to read this deck

Five sections. Section 1 builds the Legendre transform properly, as a supremum, with its degenerate case flagged in advance. Section 2 derives Hamilton's equations and draws what they draw. Section 3 introduces the Poisson bracket and uses it to compute the hidden symmetry of the Kepler problem. Section 4 asks which changes of phase-space coordinates are allowed. Section 5 proves Liouville's theorem and says what it is for.

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

14 · overview

Twice the variables, half the order

the move

Replace the velocity $\dot q$ by the momentum $p = \partial L/\partial\dot q$, and $L$ by its Legendre transform, the Hamiltonian $H(q,p,t)$. The Euler–Lagrange equations become $2n$ first-order equations on phase space: $$\dot q = \frac{\partial H}{\partial p},\qquad \dot p = -\frac{\partial H}{\partial q}.$$

what it buys

1 Orbits are level sets of $H$, and never cross. 2 A bracket $\{f,H\}$ gives every time derivative; brackets of conserved quantities are conserved. 3 Coordinate changes mixing $q$ and $p$ are allowed if their Jacobian preserves one matrix. 4 The flow has zero divergence, so it preserves volume.

the running example

Kepler, with Halley's comet as unit 08 left it. Its orbit becomes a closed curve in the $(r,p_r)$ plane, fenced off by the separatrix $E = 0$; $\vec L$ and the Runge–Lenz vector close under the bracket on $\mathfrak{so}(4)$, the algebra of rotations of $\mathbb R^4$; and the curve's phase area $A(E)$ has derivative the $75.31$-year period, because:

1. The orbits $E$ and $E + \mathrm dE$ bound a thin ring, of area $\mathrm dA$. 2. Across it $H$ rises by $\mathrm dE$: its width is $\mathrm dE/\lVert\nabla H\rVert$. 3. Along it the motion has speed $\lVert\nabla H\rVert$. 4. So time $\mathrm dt$ sweeps area $\mathrm dE\,\mathrm dt$, and $\mathrm dA = T\,\mathrm dE$.

A dotted underline is a glossary term (click, or G); a solid one, like the running example, shows on hover; N opens the notes.

Why the momentum and not the velocity. Unit 01 already turned $m\ddot x = F$ into a first-order system on $\mathbb R^{2n}$, with the pair $(x,\dot x)$. That was bookkeeping: it uses no structure. The pair $(q,p)$ carries structure, and the first sign of it was already in unit 12: when $q_i$ is cyclic, it is $p_i$, not $\dot q_i$, that is conserved. The second sign is in unit 06's Abel identity, which showed that the undamped oscillator's flow preserves area in the $(x,\dot x)$ plane; section 5 shows that in $(q,p)$ coordinates every frictionless flow preserves volume, whatever the forces and whatever the coordinates, while in $(q,\dot q)$ coordinates this fails as soon as the kinetic energy depends on $q$.

The running example, and why this one. The Kepler problem has more conserved quantities than it has any right to — energy, three components of $\vec L$, three of $\vec A$ — and unit 13 explained $\vec A$ only as "the charge of a hidden symmetry". The Poisson bracket is the tool that makes the phrase precise: section 3 computes the brackets and finds that they close on the Lie algebra of the rotation group of $\mathbb R^4$ for bound orbits and of the Lorentz group for unbound ones. The numbers are Halley's, carried over from unit 08 without change: $h = q v_q = 4.78403\times10^{15}\ \mathrm{m^2\,s^{-1}}$ and $\varepsilon = \tfrac12 v_q^2 - GM_\odot/q = -2.48740\times10^{7}\ \mathrm{J\,kg^{-1}}$ from the perihelion distance $q = 0.586025$ AU and speed $v_q = 54.5698\ \mathrm{km\,s^{-1}}$.

The four steps, and where each is done. For Halley the area is $A/\mu = 8.81646\times10^{16}\ \mathrm{m^2\,s^{-1}}$ per unit mass, computed in section 5. They are the picture on Why the area's derivative is the period, in section 5, which draws the ring for two Kepler orbits; the full proof, which also handles the turning points moving with $E$, is in the notes of Worked: Halley's phase area, and its period. Step 2 is the definition of the gradient: moving a distance $\delta$ straight across a level curve changes $H$ by $\lVert\nabla H\rVert\,\delta$. Step 3 is $\lVert X_H\rVert = \lVert J\nabla H\rVert = \lVert\nabla H\rVert$, from One first-order system on $\mathbb R^{2n}$, where $J$ is a quarter-turn. The bracket algebra is computed on The algebra closes, and the separatrix on Worked: Halley's curve, and the separatrix.

A notation warning. $q$ is a generalised coordinate everywhere in this unit except in Halley's perihelion distance, where unit 08's letter is kept; the context always decides. $\mu$ is the reduced mass and $k = GM_\odot\mu$ for gravity, as in unit 08.

section 1

The Legendre transform, as convex duality

Trading a variable for the slope of a function, done so that it cannot go wrong silently.

14 · Legendre

Why a transform, and why the naive one is not enough

We want $p = \partial L/\partial\dot q$ as a new variable in place of $\dot q$. The textbook recipe:

the recipe

Solve $p = \partial L/\partial\dot q$ for $\dot q = v(q,p)$, then put $$H(q,p) := p\cdot v - L(q, v).$$

Unit 11 met this already: Beltrami's first integral $y'F_{y'} - F$ is exactly this expression, and unit 13's energy function $h = \sum\dot q_i\,\partial L/\partial\dot q_i - L$ is too.

where the recipe breaks

"Solve for $\dot q$" needs $\dot q\mapsto\partial L/\partial\dot q$ to be a bijection. For $L = \tfrac12 m\dot x^2$ it is. For $L = c\,\dot x$ it is constant; for $L = \lvert\dot x\rvert$ it is not injective; for $L = -\sqrt{1-\dot x^2}$ its image is all of $\mathbb R$ while its domain is bounded.

the fix

Define the transform by a supremum, which always makes sense, and let convexity decide when the recipe is right. The recipe then becomes a theorem, and its failures become visible instead of silent.

Why these four examples. The first is every Newtonian kinetic energy. The second, linear in the velocity, is the shape of the magnetic term $q\,\dot{\vec x}\cdot\vec A$ in unit 12's Lagrangian for a charge, and of any Lagrangian that is already first order. The third is the arc-length integrand of unit 11 read as a Lagrangian; it is homogeneous of degree one in the velocity, and so is the reparametrisation-invariant form of the relativistic free particle in unit 27. The fourth is the other form of that particle, $-mc^2\sqrt{1-v^2/c^2}$, with $m = c = 1$. All four are handled by the one definition on the next slide.

The name. Legendre used the transform in 1787 for minimal surfaces; Hamilton (1834–35) used it to pass from Lagrange's equations to his own; Fenchel (1949) gave the supremum definition for arbitrary convex functions, which is why the general object is often called the Legendre–Fenchel transform or the convex conjugate. The recipe above is what every mechanics text calls the Legendre transform; this unit's position is that it is a special case of the supremum, valid exactly under the hypotheses of the smooth-duality slide.

14 · Legendre

The definition, by supremum

the Legendre transform

For $f:\mathbb R^n\to\mathbb R$, its transform $f^*:\mathbb R^n\to(-\infty,\infty]$ is $$f^*(p) := \sup_{x\in\mathbb R^n}\big(p\cdot x - f(x)\big).$$ For $f$ defined only on a convex set $C$, take the supremum over $C$.

two facts, no hypotheses

(a) $f^*$ is convex, whatever $f$ is. (b) The Fenchel–Young inequality: $$f(x) + f^*(p) \;\ge\; p\cdot x\quad\text{for all } x, p.$$

proof

(a) Each $p\mapsto p\cdot x - f(x)$ is affine, and a pointwise supremum of affine functions is convex. (b) $f^*(p)$ is at least the value of the bracket at the particular $x$. ∎

$f(x) = x^2/2$, $p = 1.5$. The vertical gap between the line $px$ and the graph is largest where the slope of $f$ is $p$; the tangent there (dashed) meets the axis at $-f^*(p)$.

the first one

$f(v) = \tfrac12 mv^2$: $pv - \tfrac12 mv^2$ is largest at $v = p/m$, so $f^*(p) = p^2/2m$. Dimensions: $[pv] = \mathrm{kg\,m^2\,s^{-2}}$, an energy ✓.

Reading the picture. For each slope $p$, $f^*(p)$ answers: how far must the line $y = px$ be lowered to become a supporting line of the graph of $f$ — a line touching the graph and lying below it? The lowered line $y = px - f^*(p)$ has intercept $-f^*(p)$. So $f^*$ records the graph of $f$ not by its points but by its supporting lines, one per slope. That is the whole idea of duality here: a convex set can be described by the points in it or by the half-planes containing it, and for a convex function the two descriptions carry the same information — which is the involutivity theorem, later in this section.

The value $+\infty$. It is allowed and it carries information. For $f(x) = e^x$ and $p < 0$, $px - e^x\to+\infty$ as $x\to-\infty$, so $f^*(p) = +\infty$: no line of negative slope lies below the exponential. The set where $f^*$ is finite is itself a statement about $f$, namely which slopes $f$ has "at infinity", and this is exactly what goes wrong in the degenerate cases.

Full proof of the example. $g(v) = pv - \tfrac12 mv^2$ is a downward parabola, with $g'(v) = p - mv$ vanishing only at $v = p/m$, where $g = p^2/m - p^2/2m = p^2/2m$. Being a downward parabola, $g$ has its global maximum there. Dimensionally $[p^2/m] = \mathrm{kg^2\,m^2\,s^{-2}}/\mathrm{kg}$, an energy, as $H$ must be.

14 · Legendre

Young's inequality is a Legendre transform

compute it

Let $a > 1$ and $f(x) = \lvert x\rvert^a/a$ on $\mathbb R$. For $p \ge 0$ the supremum of $px - x^a/a$ is over $x \ge 0$ (negative $x$ only lowers it), and the derivative $p - x^{a-1}$ vanishes at $x = p^{1/(a-1)}$. With $b$ defined by $\tfrac1a + \tfrac1b = 1$, so that $\tfrac{1}{a-1} = b - 1$: $$f^*(p) = p\cdot p^{b-1} - \frac{p^{(b-1)a}}{a} = p^b\Big(1 - \frac1a\Big) = \frac{p^b}{b},$$ using $(b-1)a = b$. By symmetry $f^*(p) = \lvert p\rvert^b/b$ for all $p$.

Young's inequality

For $x, p \ge 0$ and $\tfrac1a + \tfrac1b = 1$, $$px \;\le\; \frac{x^a}{a} + \frac{p^b}{b},$$ with equality exactly when $p = x^{a-1}$.

proof

Fenchel–Young with this $f$ and $f^*$. Equality holds exactly when $x$ attains the supremum defining $f^*(p)$, which for this strictly concave bracket is the single point $x = p^{1/(a-1)}$. ∎

Full proof of the exponent algebra. From $\tfrac1a + \tfrac1b = 1$: $b = a/(a-1)$, so $b - 1 = 1/(a-1)$ and $(b-1)a = a/(a-1) = b$. At the critical point $x = p^{b-1}$ the bracket is $p\cdot p^{b-1} - p^{(b-1)a}/a = p^b - p^b/a = p^b/b$. That the critical point is the maximum: $x\mapsto px - x^a/a$ has second derivative $-(a-1)x^{a-2} < 0$ for $x > 0$, so it is strictly concave on $(0,\infty)$, and a critical point of a strictly concave function is its unique global maximum. For $p < 0$ replace $x$ by $-x$. For $p = 0$ the supremum is $0$, attained at $x = 0$, and $0^b/b = 0$ ✓.

Why this matters beyond the inequality. Young's inequality is the step from which Hölder's inequality follows (put $x = |u|/\lVert u\rVert_a$, $p = |w|/\lVert w\rVert_b$ and integrate), and Hölder is the backbone of the $L^p$ spaces unit 19 touches. It is instructive that the inequality is nothing but the Fenchel–Young inequality for one particular pair: the general inequality says every convex function has a partner for which it holds, and the equality case is always "$p$ is the slope of $f$ at $x$". In mechanics that equality case reads $p = \partial L/\partial\dot q$, the definition of the momentum.

14 · Legendre

Young's inequality, as two areas

$a = 3$, $b = \tfrac32$, $x = p = 1.5$. The shaded areas $x^3/3 = 1.125$ and $\tfrac23p^{3/2} = 1.225$ together cover the dashed rectangle $px = 2.25$, and overshoot it by $0.100$. Slide the corner $(x,p)$ onto the curve and the overshoot vanishes: equality.

the same inequality, read off the picture

Under the curve $p = x^{a-1}$ from $0$ to $x$ lies area $\int_0^x t^{a-1}\,\mathrm dt = x^a/a$. To its left, from $0$ to $p$, lies area $\int_0^p s^{1/(a-1)}\,\mathrm ds = p^b/b$, since $\tfrac{1}{a-1} + 1 = b$. Every point of the rectangle $[0,x]\times[0,p]$ is under the curve or to its left. ∎

special case

$a = b = 2$: $px \le \tfrac12(x^2 + p^2)$, the AM–GM inequality, with equality at $p = x$ ✓. And $a = 2$ with $f = \tfrac12 mv^2$ is the kinetic energy of The definition, by supremum.

What to look for. The curve is the graph of $f'$, the slope of $f = x^a/a$; read sideways it is the graph of $(f^*)'$, because the two gradients are inverse functions (The smooth, strictly convex case, later in this section). The area under a derivative is the function, so the two shaded regions are $f(x)$ and $f^*(p)$, and the inequality $f(x) + f^*(p)\ge px$ is the statement that they cover the rectangle. The overshoot is the gap that the Fenchel–Young inequality measures. It is zero exactly when the rectangle's corner sits on the curve, $p = x^{a-1} = f'(x)$ — the equality case, and in mechanics the definition of the momentum.

Why every point is covered. Take $(u, w)$ with $0\le u\le x$, $0\le w\le p$. If $w\le u^{a-1}$ it is under the curve, in the first region. Otherwise $w > u^{a-1}$, i.e. $u < w^{1/(a-1)}$, and it is to the left of the curve at height $w\le p$, in the second. The regions overlap only along the curve, which has no area.

Dimensions. Only $a = 2$ reads as mechanics: $|v|^a$ is an energy per unit mass for that exponent alone, and only then is $\partial f/\partial v$ a momentum. What survives for general $a$ is the scaling — $f = m^{a-1}|v|^a/a$ has conjugate $|p|^b/(b\,m)$, try it — which at $a = b = 2$ is $\tfrac12mv^2$ and $p^2/2m$ ✓.

14 · Legendre

Involutivity: transforming twice gives $f$ back

theorem (involutivity)

If $f:\mathbb R^n\to\mathbb R$ is convex and differentiable, then $f^{**} = f$. Moreover, for each $x$ the supremum defining $f^*(p)$ at $p = \nabla f(x)$ is attained at that $x$.

lemma (supporting line)

$f$ convex and differentiable $\Rightarrow$ $f(y) \ge f(x) + \nabla f(x)\cdot(y-x)$ for all $x,y$.

proof in the notes (N)

proof of the theorem

$f^{**}\le f$. By Fenchel–Young, $p\cdot x - f^*(p)\le f(x)$ for every $p$; take the supremum over $p$.

$f^{**}\ge f$. Fix $x$ and put $p = \nabla f(x)$. The lemma says $p\cdot y - f(y) \le p\cdot x - f(x)$ for every $y$, so the supremum is attained at $y = x$: $$f^*(p) = p\cdot x - f(x).$$ Hence $f^{**}(x) \ge p\cdot x - f^*(p) = f(x)$. ∎

The general case — $f$ convex and lower semicontinuous with values in $(-\infty,\infty]$, no differentiability — is the Fenchel–Moreau theorem, cited: Rockafellar, Convex Analysis, Theorem 12.2. Only the differentiable case is used in this course.

Full proof of the lemma

Fix $x, y$ and put $g(t) = f(x + t(y-x))$ for $t\in\mathbb R$. Then $g$ is convex (the restriction of a convex function to a line) and differentiable, with $g'(0) = \nabla f(x)\cdot(y-x)$ by the chain rule of unit 03. For a convex $g$ the difference quotient $t\mapsto(g(t)-g(0))/t$ is nondecreasing on $t > 0$: for $0 < s < t$, write $s = (1-\lambda)\cdot0 + \lambda t$ with $\lambda = s/t$; convexity gives $g(s)\le(1-\lambda)g(0) + \lambda g(t)$, which rearranges to $(g(s)-g(0))/s \le (g(t)-g(0))/t$. Letting $s\to0^+$ gives $g'(0)\le (g(t)-g(0))/t$ for every $t > 0$; at $t = 1$ this is $f(y) - f(x) \ge \nabla f(x)\cdot(y-x)$. $\square$

Why it matters for mechanics. Involutivity is what makes the passage from Lagrange to Hamilton reversible: $L$ is recovered from $H$ by the same transform, in $p$ instead of $\dot q$. That symmetry is used in unit 15, where the Hamilton–Jacobi function is built from $H$, and in unit 27, where one goes from $H = \sqrt{p^2c^2 + m^2c^4}$ back to a Lagrangian.

14 · Legendre

What the transform cannot see

$f(x) = (x^2-1)^2$. Each slope $p$ has its supporting line; the two of slope $\pm4$ are drawn. Between the minima no supporting line touches the graph, so $f^{**}$ — dashed — replaces the hump by the segment $y = 0$. At $x = 0$ the gap is $f(0) - f^{**}(0) = 1$.

without convexity

$f^{**}$ is the convex envelope of $f$ — the largest convex function below it — and the transform forgets everything non-convex (problem P3).

why, in one line

$f^*(p)$ is fixed by the supporting line of slope $p$ alone. The hump touches none of them, so $f$ and its envelope have the same $f^*$ — and hence the same $f^{**}$.

The picture. The lemma says the tangent plane at $x$ lies below the graph everywhere: a supporting hyperplane. The theorem's proof is then two sentences: every supporting line lies below $f$ (so $f^{**}\le f$), and at each point some supporting line touches (so equality). For a non-convex $f$ the second step fails at every point where no line through $(x, f(x))$ stays below the graph, and there $f^{**}(x) < f(x)$.

The numbers in the figure. $f'(x) = 4x(x^2-1)$, so the supporting line of slope $4$ touches where $x^3 - x - 1 = 0$, at $x = 1.3247$, $f = 0.5698$; by symmetry the slope $-4$ line touches at $-1.3247$. The line of slope $0$ touches at both minima $x = \pm1$ at once, and that is the line the envelope follows between them. Every other supporting line touches one of the outer walls, at some $|x| > 1$ — $f'$ maps $(1,\infty)$ onto $(0,\infty)$ and $f$ is convex there — and never the hump: the inner stretch $|x| < 1$ contributes nothing to any $f^*(p)$.

Why mechanics never meets this. A Lagrangian whose kinetic part is not convex in the velocity would have a Hamiltonian that cannot tell it from its envelope, and transforming back would return the wrong Lagrangian. Every Lagrangian in this course is convex in $\dot q$ — the Legendre condition of unit 11 — so the case is flagged and set aside.

14 · Legendre

The smooth, strictly convex case: gradients are inverse

theorem (smooth duality)

Let $f\in C^2(\mathbb R^n)$ with Hessian $\nabla^2 f$ positive definite everywhere, and let $U = \nabla f(\mathbb R^n)$. Then $\nabla f:\mathbb R^n\to U$ is a bijection onto an open set, and on $U$:

(i) the supremum defining $f^*(p)$ is attained at the unique $x(p)$ with $\nabla f(x) = p$;
(ii) $f^*\in C^2(U)$, with $\nabla f^* = (\nabla f)^{-1}$ and $\nabla^2 f^*(p) = \big(\nabla^2 f(x(p))\big)^{-1}$;
(iii) if $f = f(x; y)$ depends on a parameter $y$, then $\dfrac{\partial f^*}{\partial y} = -\dfrac{\partial f}{\partial y}$, evaluated at $x = x(p)$.

proof of (ii) and (iii), given (i)

$f^*(p) = p\cdot x(p) - f(x(p);y)$, and $x(p)$ is $C^1$ by the inverse function theorem (unit 04). Differentiate: $$\nabla_p f^* = x + (D_px)^{\!\top}\big(p - \nabla_x f\big) = x,$$ since the bracket vanishes at $x(p)$. The $y$-derivative is $(D_yx)^{\!\top}(p - \nabla_xf) - \partial_yf = -\partial_y f$. Differentiating $\nabla f^* = x(p)$ once more gives $D_px$, the inverse of $\nabla^2 f$. ∎

check

$f = \tfrac12 mv^2$: $\nabla f^*(p) = p/m = v$ ✓, and $(f^*)'' = 1/m = (f'')^{-1}$ ✓. The dimensions of (ii): $[\nabla^2 f] = [f]/[x]^2$ and $[\nabla^2 f^*] = [f]/[p]^2 = [x]^2/[f]$ ✓.

Full proof of the first claim and of (i)

$\nabla f$ is injective. For $x\ne y$ put $g(t) = f(x+t(y-x))$; then $g''(t) = (y-x)^{\!\top}\nabla^2f\,(y-x) > 0$, so $g'$ is strictly increasing and $g'(1) - g'(0) = (\nabla f(y) - \nabla f(x))\cdot(y-x) > 0$. Hence $\nabla f(y)\ne\nabla f(x)$.

$U$ is open and the inverse is $C^1$. $D(\nabla f) = \nabla^2f$ is invertible everywhere, so by the inverse function theorem of unit 04 every point of $\mathbb R^n$ has a neighbourhood mapped diffeomorphically onto an open neighbourhood of its image. So $U$ is open, and the global inverse (which exists by injectivity) agrees locally with these local $C^1$ inverses.

(i). For $p\in U$ let $x(p)$ be the preimage. The function $\phi(x) = p\cdot x - f(x)$ is concave, and $\nabla\phi(x(p)) = 0$. By the supporting-line lemma applied to the convex $-\phi$, $-\phi(y)\ge-\phi(x(p))$ for every $y$: a critical point of a concave function is a global maximum. Uniqueness of the maximiser: any other maximiser would be another critical point, contradicting injectivity.

What happens off $U$. $f^*$ may be finite there or not. For $f = e^x$, $U = (0,\infty)$, $f^*(0) = 0$ (a supremum approached as $x\to-\infty$ and never attained), and $f^* = +\infty$ on $p < 0$. For $f = \tfrac12 mv^2$, $U = \mathbb R$ and nothing is lost. For the fourth example of the recipe slide, $f(v) = -\sqrt{1-v^2}$ on $(-1,1)$, $U = \mathbb R$ although the domain is bounded — which is why that case is only nearly degenerate.

(iii) in mechanics. With $x = \dot q$ and the parameters $y = (q, t)$, (iii) says $\partial H/\partial q = -\partial L/\partial q$ and $\partial H/\partial t = -\partial L/\partial t$. Those two identities, with (ii)'s $\partial H/\partial p = \dot q$, are all of Hamilton's equations except the Euler–Lagrange equation itself.

14 · Legendre

The degenerate case, flagged now

linear: $f(v) = c\,v$

$pv - cv$ is unbounded unless $p = c$: $$f^*(p) = \begin{cases}0 & p = c\\ +\infty & \text{else.}\end{cases}$$ The momentum is not a variable at all but pinned: a constraint.

homogeneous: $f(v) = \lvert v\rvert$

$pv - |v|$ is $\le 0$ for all $v$ iff $|p|\le 1$: $$f^*(p) = \begin{cases}0 & |p|\le1\\ +\infty & |p| > 1.\end{cases}$$ On the allowed set the Hamiltonian is identically zero.

nearly: $f(v) = -\sqrt{1-v^2}$

On $|v| < 1$. Here $f'' = (1-v^2)^{-3/2} > 0$ and $f'(v) = v/\sqrt{1-v^2}$ maps $(-1,1)$ onto $\mathbb R$. Solving $p = f'(v)$: $v = p/\sqrt{1+p^2}$, so $$f^*(p) = \frac{p^2+1}{\sqrt{1+p^2}} = \sqrt{1+p^2}.$$

the degenerate Lagrangian

$L$ is degenerate where $\det\big(\partial^2L/\partial\dot q_i\partial\dot q_j\big) = 0$ — unit 11's Legendre condition failing. Then $\dot q\mapsto p$ is not invertible, some momenta are functions of $q$ alone, and $H$ is defined only on a constraint set. Unit 27: the relativistic free particle is the third column when time is the parameter and the second when any parameter is allowed.

Full proof of the three transforms. Linear. $pv - cv = (p-c)v$; if $p\ne c$, take $v\to\pm\infty$ with the sign of $p - c$. Absolute value. If $|p|\le1$ then $pv\le|p||v|\le|v|$, so the bracket is $\le 0$, with $0$ at $v = 0$. If $|p| > 1$ take $v = tp$, $t\to\infty$: $pv - |v| = t|p|(|p|-1)\to\infty$. Square root. $f$ is strictly convex on $(-1,1)$ with $f'$ a bijection onto $\mathbb R$ (it is continuous, strictly increasing, and tends to $\pm\infty$ at $\pm1$), so the smooth-duality theorem applies with $U = \mathbb R$. From $p = v/\sqrt{1-v^2}$: $p^2(1-v^2) = v^2$, $v^2 = p^2/(1+p^2)$, and $v$ has the sign of $p$. Then $\sqrt{1-v^2} = 1/\sqrt{1+p^2}$ and $f^*(p) = pv + \sqrt{1-v^2} = (p^2+1)/\sqrt{1+p^2}$. Check with the theorem: $(f^*)'(p) = p/\sqrt{1+p^2} = v$ ✓. Limit: for small $v$, $f \approx -1 + v^2/2$ and $f^* \approx 1 + p^2/2$ — a unit mass, with the constant shifted ✓.

What "nearly degenerate" means. In the third example $f''\to\infty$ as $|v|\to1$: the velocity saturates while the momentum grows without bound. Transform twice and you get $f$ back (involutivity, with $f^{**}$ restricted to $|v|<1$), so nothing is lost; but a physical theory in which momentum can grow without bound while velocity cannot is a very different theory, and in unit 27 that is special relativity with $m = c = 1$. In the reparametrisation-invariant form of the same particle the Lagrangian is homogeneous of degree one in the velocity, like $|v|$, and its Hamiltonian vanishes identically on a constraint set — the mass shell. Unit 27 handles it.

Constraints. The systematic treatment of degenerate Lagrangians — primary and secondary constraints, first and second class — is Dirac's, and is cited, not used here: P. A. M. Dirac, Lectures on Quantum Mechanics (1964), lecture 1. Problem P2 shows that a Lagrangian linear in velocities can still have perfectly good Hamiltonian dynamics, in fewer variables than expected.

14 · Legendre

The degenerate case, drawn

Left: at the corner of $|v|$ every line of slope $|p|\le1$ through the origin stays below, so nothing needs lowering and $f^*(p) = 0$; a steeper line crosses the graph however far it is lowered, so $f^* = +\infty$. Right: the tangent slopes of $-\sqrt{1-v^2}$ run through all of $\mathbb R$, so every $p$ has its $v = p/\sqrt{1+p^2}$.

The point. Read the transform off the supporting lines, as on The definition, by supremum. A corner supports a whole interval of slopes at one point; a graph with vertical asymptotes supports every slope. The first collapses many momenta onto one velocity, the second stretches a bounded set of velocities over all momenta. Either way $\dot q\mapsto p$ fails to be the bijection the recipe needed.

What to look for. On the left, the supporting line of slope $p$ is the line $y = pv - f^*(p)$ that touches the graph from below. For $|p|\le1$ the line $y = pv$ itself already touches at the corner, so $f^*(p) = 0$ — the Hamiltonian is identically zero on the allowed set, the middle column of the previous slide. For $|p| > 1$ the line eventually overtakes $|v|$, by $(|p|-1)|v|$, and no amount of lowering fixes that: $f^*(p) = +\infty$. The dashed line of slope $1.6$, lowered by $0.6$, still crosses at $v = 1$.

On the right the slopes $f'(v) = v/\sqrt{1-v^2}$ at $v = 0, 0.6, 0.9, 0.97$ are $0, 0.75, 2.06, 3.99$: each tangent is a supporting line, one per slope, and the slopes grow without bound as $v\to1$ while $v$ never gets there. This is the third column, and in unit 27 it is a particle whose speed is bounded by $c$ while its momentum is not. The linear case $f = cv$ is the left picture with the corner flattened out: its graph is a single line, supported only by the slope $c$.

14 · Legendre

Worked: the Hamiltonians of units 08 and 12

two Lagrangians, transformed

General quadratic. $L = \tfrac12\dot q^{\!\top}M(q)\dot q - V(q)$ with $M(q)$ symmetric positive definite. The Hessian in $\dot q$ is $M$: non-degenerate. $p = \partial L/\partial\dot q = M\dot q$, so $\dot q = M^{-1}p$ and $$H = p^{\!\top}M^{-1}p - \tfrac12 p^{\!\top}M^{-1}p + V = \tfrac12 p^{\!\top}M(q)^{-1}p + V(q).$$

Kepler, unit 08. $L = \tfrac12\mu(\dot r^2 + r^2\dot\theta^2) + k/r$: $M = \operatorname{diag}(\mu, \mu r^2)$, $p_r = \mu\dot r$, $p_\theta = \mu r^2\dot\theta$, and $$H = \frac{p_r^2}{2\mu} + \frac{p_\theta^2}{2\mu r^2} - \frac kr .$$

The hoop, unit 12. $L = \tfrac12mR^2\dot\theta^2 + \tfrac12mR^2\Omega^2\sin^2\theta + mgR\cos\theta$. Here the middle term has no $\dot\theta$, so it counts as part of $-V$: $M = mR^2$, $p = mR^2\dot\theta$, and $$H = \frac{p^2}{2mR^2} - \tfrac12 mR^2\Omega^2\sin^2\theta - mgR\cos\theta .$$

Read it. $T + V = \frac{p^2}{2mR^2} + \tfrac12 mR^2\Omega^2\sin^2\theta - mgR\cos\theta$: the $\Omega^2$ term has the opposite sign. $H$ is unit 13's energy function $h$, conserved because $L$ has no $t$; $T + V$ is not, because the motor turning the hoop does work.

Checks. Every term of both $H$s is an energy: $[p_\theta^2/\mu r^2] = (\mathrm{kg\,m^2\,s^{-1}})^2/(\mathrm{kg\,m^2})$ ✓. $\Omega = 0$: the hoop's $H$ is $T + V$, a pendulum ✓.

Full working of the general case. $\partial L/\partial\dot q = M\dot q$ because $M$ is symmetric: the derivative of $\tfrac12\dot q^{\!\top}M\dot q$ in the direction $w$ is $\tfrac12(w^{\!\top}M\dot q + \dot q^{\!\top}Mw) = w^{\!\top}M\dot q$. $M$ positive definite makes the map $\dot q\mapsto M\dot q$ a linear bijection of $\mathbb R^n$, so the smooth-duality theorem applies with $U = \mathbb R^n$ and $\dot q = M^{-1}p$. Then $p\cdot\dot q = p^{\!\top}M^{-1}p$ and $\tfrac12\dot q^{\!\top}M\dot q = \tfrac12 p^{\!\top}M^{-1}MM^{-1}p = \tfrac12p^{\!\top}M^{-1}p$, which gives the boxed $H$. In the polar case $M^{-1} = \operatorname{diag}(1/\mu, 1/\mu r^2)$. So whenever $T$ is quadratic in the velocities and $V$ does not depend on them, $H = T + V$ written in terms of momenta — the familiar case.

Why the hoop is different. The formula $H = T+V$ needs $T$ to be homogeneous quadratic in $\dot q$. The hoop's kinetic energy $\tfrac12 mR^2(\dot\theta^2 + \Omega^2\sin^2\theta)$ has a piece of degree zero in $\dot\theta$, which appears because the constraint (the hoop's position) depends on time. Unit 13 proved that the conserved quantity is then $h = \dot\theta\,\partial L/\partial\dot\theta - L$, and here that is $H$ exactly. Its effective potential $V_{\mathrm{eff}}(\theta) = -\tfrac12 mR^2\Omega^2\sin^2\theta - mgR\cos\theta$ has $V_{\mathrm{eff}}' = mR\sin\theta\,(g - R\Omega^2\cos\theta)$, whose zeros are $\theta = 0$ and, for $\Omega^2 > g/R$, $\cos\theta = g/R\Omega^2$ — unit 12's pitchfork, recovered from $H$ alone.

The $\theta$ coordinate for Kepler is cyclic. $H$ does not contain $\theta$, so $\dot p_\theta = -\partial H/\partial\theta = 0$: $p_\theta = L$ is conserved, and for the rest of the unit it is frozen at Halley's value. That reduces the problem to the $(r,p_r)$ half-plane, which is where section 2 draws it.

section 2

Hamilton's equations, and the picture they draw

One first-order system on $\mathbb R^{2n}$, whose orbits are level sets and never cross.

14 · Hamilton

Hamilton's equations

theorem (Hamilton's equations)

Let $L(q,\dot q,t)$ be $C^2$ with $\partial^2L/\partial\dot q^2$ positive definite and $\dot q\mapsto\partial L/\partial\dot q$ a bijection for each $(q,t)$, and let $H$ be its transform in $\dot q$. Then a $C^2$ curve $q(t)$ solves the Euler–Lagrange equations iff $(q(t), p(t))$ with $p = \partial L/\partial\dot q$ solves $$\dot q = \frac{\partial H}{\partial p},\qquad \dot p = -\frac{\partial H}{\partial q}.$$ Along solutions, $\dfrac{\mathrm dH}{\mathrm dt} = \dfrac{\partial H}{\partial t} = -\dfrac{\partial L}{\partial t}$.

proof

Smooth duality (ii) gives $\partial H/\partial p = \dot q$ — the first equation holds by construction. Part (iii) gives $\partial H/\partial q = -\partial L/\partial q$. So $$\dot p = \frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q}\overset{\text{E–L}}{=}\frac{\partial L}{\partial q} = -\frac{\partial H}{\partial q},$$ and each step reverses. Finally $\frac{\mathrm dH}{\mathrm dt} = H_q\dot q + H_p\dot p + H_t = H_qH_p - H_pH_q + H_t$. ∎

checks

Dimensions: $[H/p] = \mathrm{J}/(\mathrm{kg\,m\,s^{-1}}) = \mathrm{m\,s^{-1}}$ ✓ and $[H/q] = \mathrm N$ ✓. Special case: $H = p^2/2m + V(x)$ gives $\dot x = p/m$, $\dot p = -V'(x)$ — Newton ✓.

The logic, once more slowly. The first equation is not dynamics; it is the inverse of the definition $p = \partial L/\partial\dot q$, delivered by the gradient-inversion half of smooth duality. All the dynamics is in the second equation, which is the Euler–Lagrange equation with its right-hand side rewritten by the passive-variable identity. The converse direction: if $(q,p)$ solves Hamilton's equations, then $\dot q = \partial H/\partial p$ means $p = \partial L/\partial\dot q$ (inverting again), and the second equation read backwards is Euler–Lagrange. Regularity: $q$ is $C^2$ if $(q,p)$ is $C^1$ and $H$ is $C^2$, since $\dot q = \partial H/\partial p(q,p,t)$ is then $C^1$.

The energy statement. $\partial H/\partial t = -\partial L/\partial t$ is part (iii) with the parameter $t$. So $H$ is conserved exactly when $L$ has no explicit $t$, which is unit 13's time-translation theorem in one line. For the hoop, $L$ has no explicit $t$ in the rotating coordinate $\theta$, which is why its $H$ is conserved although energy $T+V$ is not.

History. Hamilton published the equations in 1834–35, in "On a General Method in Dynamics", as a by-product of his optics; the canonical form was anticipated by Lagrange (1809) and Poisson (1809) for perturbation theory, and the name "canonical equations" is Jacobi's.

14 · Hamilton

One first-order system on $\mathbb R^{2n}$

the standard symplectic matrix

Write $z = (q,p)\in\mathbb R^{2n}$ and $$J = \begin{pmatrix}0 & I_n\\ -I_n & 0\end{pmatrix},\qquad J^{\!\top} = -J = J^{-1},\quad J^2 = -I.$$ Hamilton's equations are $\dot z = J\,\nabla H(z)$. The vector field $X_H := J\nabla H$ is the Hamiltonian vector field of $H$.

$H$ is conserved, in one line

For $H$ with no explicit $t$: $\dfrac{\mathrm dH}{\mathrm dt} = \nabla H^{\!\top}J\nabla H = 0$, because $u^{\!\top}Ju = 0$ for every $u$ when $J$ is antisymmetric.

The pendulum, $H = p^2/2 - \cos q$. At each marked point $\nabla H$ points straight across the level curve; $J$ turns it a quarter-turn clockwise into $X_H$, which points along it. Following $X_H$ can therefore never change $H$.

Checking the matrix. $J\nabla H = \begin{pmatrix}0&I\\-I&0\end{pmatrix} \begin{pmatrix}\partial_qH\\\partial_pH\end{pmatrix} = \begin{pmatrix}\partial_pH\\-\partial_qH\end{pmatrix}$ ✓. $J^{\!\top} = \begin{pmatrix}0&-I\\I&0\end{pmatrix} = -J$, and $J^2 = \begin{pmatrix}-I&0\\0&-I\end{pmatrix}$, so $J^{-1} = -J$.

Why $u^{\!\top}Ju = 0$. It is a number, so it equals its transpose $u^{\!\top}J^{\!\top}u = -u^{\!\top}Ju$. This is the entire reason Hamiltonian flows conserve $H$: the gradient is rotated a quarter-turn (in each $(q_i,p_i)$ plane) before it is followed, so the motion is always along the level set. Unit 02 called conservative systems "flows preserving a function" and said the structure behind them was symplectic; $J$ is that structure.

Reading the figure. In one degree of freedom $\nabla H = (\partial_qH, \partial_pH)$ and $X_H = (\partial_pH, -\partial_qH)$: the same two numbers, swapped with one sign changed, which is the clockwise quarter-turn. At the top of each curve ($q = 0$, $p > 0$) the gradient points up and $X_H$ points right, so the motion runs clockwise round the origin. The arrows are drawn at $0.62$ times their true length; they are longest where the curves are closest together, because there $H$ changes fastest across them and the motion is fastest along them.

14 · Hamilton

Orbits never cross

trajectories never cross

If $H\in C^2$ and has no explicit $t$, two distinct orbits of $X_H$ never meet, and no orbit crosses itself except by closing up into a periodic one.

proof

$\nabla H$ is $C^1$, hence locally Lipschitz, so Picard–Lindelöf (unit 01) gives one solution through each point. Two orbits sharing a point are both the solution through it: the same orbit. ∎

what this is not

Unit 01 also wrote a first-order system on $\mathbb R^{2n}$, with $y = (x,\dot x)$. Same count, no structure: the matrix $J$ and the one-line conservation of $H$ are special to $(q,p)$.

Schematic. Two orbits crossing at $z$ would leave it in two directions, but $X_H(z)$ is one vector. The only way back to $z$ is along the same orbit, which then closes up.

What the picture leaves out. Tangency is excluded too: two orbits touching at $z$ with the same direction there are both the solution through $z$, so they coincide. And an orbit may approach a point without reaching it — the separatrix runs into the saddle only as $t\to\infty$ — which is how the pendulum's separatrix and the orbits on either side of it meet "at" the saddle without crossing.

Time-dependent $H$. Uniqueness still holds, but for the non-autonomous system: two solutions cannot pass through the same point at the same time. Projected to phase space their paths can cross at different times — the picture of non-crossing curves is a picture of the autonomous case, or of the extended phase space with $t$ adjoined.

Periodic orbits. If $z(t_1) = z(t_2)$ with $t_1 < t_2$, then by uniqueness $t\mapsto z(t + t_2 - t_1)$ and $t\mapsto z(t)$ are the same solution, so $z$ is periodic with period dividing $t_2 - t_1$. That is the only way an orbit can meet itself.

14 · Hamilton

The phase-space action principle

theorem (phase-space action)

On curves $(q(t),p(t))$, $t\in[t_0,t_1]$, with $q(t_0)$, $q(t_1)$ fixed and $p$ free at both ends, the functional $$S[q,p] = \int_{t_0}^{t_1}\big(p\cdot\dot q - H(q,p,t)\big)\,\mathrm dt$$ is stationary exactly on solutions of Hamilton's equations.

proof

Unit 11's Euler–Lagrange equations for $2n$ dependent variables, with integrand $F = p\cdot\dot q - H$. In $p$: $F_{\dot p} = 0$ and $F_p = \dot q - H_p$, so $0 = \dot q - H_p$. In $q$: $\frac{\mathrm d}{\mathrm dt}F_{\dot q} = \dot p$ and $F_q = -H_q$. The free-end condition for $p$ is $F_{\dot p} = 0$, satisfied identically. ∎

two things it says

1. Put $p = \partial L/\partial\dot q$ back in: $p\cdot\dot q - H = L$, so this is Hamilton's principle of unit 12 with $p$ promoted to an independent variable.

2. The integrand is linear in the velocities $(\dot q,\dot p)$ — degenerate in the sense of the previous section — and still gives good first-order equations. Degeneracy is fatal to the transform, not to the variational principle.

why it is here

Section 4 changes coordinates on phase space. An action whose integrand is $p\cdot\dot q - H$ is what tells you which changes are allowed: those altering $p\cdot\dot q$ by a total derivative.

The free-end condition. Unit 11's natural boundary condition at a free end is that $F_{\dot y}$ vanish there, for each free dependent variable $y$. Here the free variable is $p$ and $F$ does not contain $\dot p$ at all, so the condition is empty. This is why one may fix only $q$ at the ends: fixing $p$ as well would over-determine the first-order system (it has $2n$ equations and admits only $2n$ boundary conditions).

A subtlety in the proof. Unit 11's derivation assumed $F$ of class $C^2$ and admissible variations vanishing at the ends. Here the variations $\delta q$ vanish at the ends and $\delta p$ is free; the boundary term produced by integrating by parts is $[F_{\dot q}\cdot\delta q + F_{\dot p}\cdot\delta p]_{t_0}^{t_1} = [p\cdot\delta q]_{t_0}^{t_1} + 0 = 0$. Nothing else changes.

The same theorem, as unit 15 will use it. Along an actual solution, $S$ as a function of the endpoint $(q_1, t_1)$ has $\partial S/\partial q_1 = p(t_1)$ and $\partial S/\partial t_1 = -H$; that is the Hamilton–Jacobi equation, and it is where the phase-space action goes next.

14 · Hamilton

The pendulum's separatrix, now a level set

$H = p^2/2ml^2 - mgl\cos\theta$, unit 02's pendulum without damping, $p$ in units of $m\sqrt{gl^3}$. Closed curves: libration. Open curves: rotation. Dashed: $H = mgl$, $p = \pm 2m\sqrt{gl^3}\cos(\theta/2)$.

For one degree of freedom every orbit lies in a level set of $H$, and a level-set component containing no equilibrium is exactly one orbit. So the portrait is a contour map and needs no integration.

Linearising at an equilibrium gives $A = JS$ with $S = \nabla^2H$ symmetric, so $\operatorname{tr}A = 0$ and the eigenvalues are $\pm\sqrt{-\det S}$: a centre when $H$ has a minimum or maximum, a saddle when $\det S < 0$.

The separatrix computed. At the up position $(\pi, 0)$, $H = mgl$. The level set $p^2/2ml^2 = mgl(1 + \cos\theta) = 2mgl\cos^2(\theta/2)$ gives $p = \pm2ml^2\sqrt{g/l}\,\cos(\theta/2) = \pm2m\sqrt{gl^3}\cos(\theta/2)$. Dimensions: $m\sqrt{gl^3}$ is $\mathrm{kg}\sqrt{\mathrm{m\,s^{-2}\,m^3}} = \mathrm{kg\,m^2\,s^{-1}}$, an angular momentum ✓. In unit 02 this curve was found by integrating $\ddot\theta = -(g/l)\sin\theta$ once; here it is simply the level set through the saddle. Unit 02 called it the separatrix: the orbits on it take infinite time to reach the saddle, and it divides libration from rotation.

Full proof of the level-set claim. Every orbit lies in a level set because $H$ is conserved. Conversely, let $C$ be a connected component of $\{H = E\}$ containing no equilibrium, in the plane. Then $\nabla H\ne0$ on $C$, so by the implicit function theorem (unit 04) $C$ is locally a $C^1$ curve, and $X_H = J\nabla H$ is a nonvanishing tangent field along it. An orbit starting on $C$ moves along $C$ with speed $\lVert\nabla H\rVert$ bounded below on compact pieces, so it cannot stop, cannot leave, and (if $C$ is a closed curve) traverses all of it in finite time; if $C$ is not closed, the orbit runs along it in both time directions. Either way the orbit is all of $C$.

Full proof of the linearisation claim. At an equilibrium $\nabla H = 0$, and $D(J\nabla H) = J\nabla^2H = JS$. For $n = 1$, $\operatorname{tr}(JS) = S_{21}\cdot1 + S_{12}\cdot(-1) = 0$ by symmetry, and $\det(JS) = \det J\det S = \det S$. A $2\times2$ matrix with trace $0$ has eigenvalues $\pm\sqrt{-\det}$. In unit 02's trace–determinant plane Hamiltonian systems live on the vertical axis $\tau = 0$ — exactly the line where Hartman–Grobman says nothing. Here the conserved $H$ settles it: a nondegenerate minimum of $H$ is Lyapunov stable by unit 02's direct method with $H - H(z_0)$ as Lyapunov function, and nearby orbits are the closed level curves around it. In $2n$ dimensions the same algebra gives: $JS$ is similar to $SJ = -(JS)^{\!\top}$, so eigenvalues come in pairs $\pm\lambda$, and no equilibrium of a Hamiltonian system is asymptotically stable.

14 · Hamilton

Kepler in the $(r, p_r)$ half-plane

$\tfrac12 v_r^2 + h^2/2r^2 - GM_\odot/r = \varepsilon$ at Halley's $h$, with $r$ in units of $r_0 = h^2/GM_\odot$ and $v_r = p_r/\mu$ in units of $GM_\odot/h$, out to $r = 10r_0$. Solid thick: Halley. Dashed: $E = 0$. Faint: two hyperbolic orbits.

Every closed curve is a bound orbit and every open one is a flyby, and none of them touches another: the whole of unit 08's classification of orbits by energy, read off one contour map.

How the figure is made. $H$ at fixed $L$ is $H(r,p_r) = p_r^2/2\mu + V_{\mathrm{eff}}(r)$ with $V_{\mathrm{eff}} = L^2/2\mu r^2 - k/r$, unit 08's effective potential. Dividing by $\mu$ and writing $x = r/r_0$, $y = v_r/(GM_\odot/h)$, the level set $\varepsilon$ becomes $\tfrac12y^2 + \tfrac1{2x^2} - \tfrac1x = \varepsilon/(GM_\odot/h)^2$, and the curves drawn are the values $-0.45, -0.38, -0.30, -0.22, -0.14$, Halley's $-0.0323228$, $0$, $0.18$ and $0.5$ of that dimensionless energy. The minimum value is $-\tfrac12$, at $(x,y) = (1,0)$: the circular orbit.

Why the curves are lopsided. Near $r = 0$ the centrifugal term $L^2/2\mu r^2$ is a wall; at large $r$ the potential flattens to zero, so a bound orbit with energy just below zero spends most of its phase-space length far out at small $|p_r|$. Kepler's second law says it also spends most of its time there.

Where unit 08 was. Unit 08 read all this from the graph of $V_{\mathrm{eff}}$, with the energy as a horizontal line. The phase portrait is that graph unfolded: each horizontal line $E$ becomes the curve $p_r = \pm\sqrt{2\mu(E - V_{\mathrm{eff}})}$.

14 · Hamilton

Worked: Halley's curve, and the separatrix

Halley's level curve, in numbers

Given (unit 08): $h = 4.78403\times10^{15}\ \mathrm{m^2\,s^{-1}}$, $\varepsilon = -2.48740\times10^{7}\ \mathrm{J\,kg^{-1}}$, $GM_\odot = 1.32712440\times10^{20}\ \mathrm{m^3\,s^{-2}}$. Per unit mass, $\varepsilon = \tfrac12 v_r^2 + V(r)$ with $V(r) = h^2/2r^2 - GM_\odot/r$.

Step 1 — the equilibrium. $V'(r) = -h^2/r^3 + GM_\odot/r^2 = 0$ at $r_0 = h^2/GM_\odot = 1.72455\times10^{11}$ m $= 1.15279$ AU, where $V(r_0) = -(GM_\odot)^2/2h^2 = -3.84774\times10^{8}\ \mathrm{J\,kg^{-1}}$. It is a minimum of $H$: a centre, the circular orbit.

Step 2 — where the curve meets $v_r = 0$. $\varepsilon = V(r)$, i.e. $2\varepsilon r^2 + 2GM_\odot r - h^2 = 0$: $r = 0.586025$ and $35.0789$ AU, unit 08's perihelion and aphelion.

Step 3 — its widest point. $v_r^2 = 2(\varepsilon - V)$ is largest where $V$ is least, at $r_0$: $|v_r| = \sqrt{2(\varepsilon - V(r_0))} = 26.829\ \mathrm{km\,s^{-1}}$.

Step 4 — the separatrix. At $\varepsilon = 0$: $v_r^2 = 2GM_\odot/r - h^2/r^2$, zero at $r = h^2/2GM_\odot = r_0/2 = 0.576396$ AU and again only as $r\to\infty$.

checks

Dimensions: $[h^2/GM] = \mathrm{m^4s^{-2}}/\mathrm{m^3s^{-2}} = \mathrm m$ ✓.

Unit 08: $\varepsilon/V(r_0) = 0.0646457 = 1 - e^2$ with $e = 0.96714$ ✓, and $r_0$ is the semi-latus rectum $p = 1.72455\times10^{11}$ m ✓.

Limit: as $\varepsilon\to V(r_0)$ the curve shrinks to the point $(r_0, 0)$ ✓.

a separatrix without a saddle

The pendulum's separatrix runs into a saddle. This one runs into nothing in the finite plane: its "saddle" is at $r = \infty$, $p_r = 0$. What it separates is still the same — bounded from unbounded.

The arithmetic. $h^2 = 2.28869\times10^{31}$, so $r_0 = 2.28869\times10^{31}/1.32712\times10^{20} = 1.72455\times10^{11}$ m, and in AU ($1.495978707\times10^{11}$ m) $1.15279$. $V(r_0) = -(GM)^2/2h^2 = -1.76126\times10^{40}/4.57738\times10^{31} = -3.84774\times10^{8}$. Then $\varepsilon - V(r_0) = 3.59900\times10^{8}$, $2\times$ that is $7.19800\times10^{8}$, and its root is $2.68291\times10^{4}\ \mathrm{m\,s^{-1}}$. The quadratic of step 2 is unit 08's turning-point equation, solved there with the same roots.

Why $\varepsilon/V(r_0) = 1 - e^2$. Unit 08 has $e^2 = 1 + 2EL^2/\mu k^2$; per unit mass $e^2 = 1 + 2\varepsilon h^2/(GM)^2 = 1 - \varepsilon/V(r_0)$. Halley's orbit sits at $6.46\%$ of the depth of the well: very nearly on the separatrix, which is why its level curve in the figure hugs the dashed one near the Sun and why the comet's aphelion is so far out.

The separatrix at infinity. Substitute $u = 1/r$: then near $u = 0$ the reduced system has an equilibrium at $(u, p_r) = (0, 0)$ which is degenerate (not a hyperbolic saddle), and the $E = 0$ curve is the orbit asymptotic to it. That is the precise sense in which "the saddle is at infinity". The parabolic orbit takes infinite time to get there, like a pendulum on its separatrix.

section 3

Poisson brackets: the algebra of observables

One bilinear operation that gives every time derivative, and turns conserved quantities into a Lie algebra.

14 · brackets

The Poisson bracket

the Poisson bracket

For $f, g\in C^\infty(\mathbb R^{2n})$, $$\{f,g\} := \sum_{i=1}^n\Big(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\Big) = \nabla f^{\!\top}J\,\nabla g .$$

properties

1 bilinear; 2 antisymmetric, $\{g,f\} = -\{f,g\}$; 3 Leibniz, $\{f, gh\} = \{f,g\}h + g\{f,h\}$; 4 the Jacobi identity, $$\{f,\{g,h\}\} + \{g,\{h,f\}\} + \{h,\{f,g\}\} = 0.$$

proof of 1–3

1: $\nabla$ is linear. 2: $J^{\!\top} = -J$. 3: $\nabla(gh) = h\nabla g + g\nabla h$. ∎ (4: next slide.)

the fundamental Poisson brackets

$$\{q_i, q_j\} = 0,\quad \{p_i, p_j\} = 0,\quad \{q_i, p_j\} = \delta_{ij}.$$ Bilinearity and Leibniz reduce every bracket of polynomials to these.

dimensions

$[\{f,g\}] = [f][g]/[q][p]$, and $[q][p]$ is an action, $\mathrm{J\,s}$. So $[\{f,H\}] = [f]/\mathrm s$ — a rate, which is what it will turn out to be.

A sign convention to watch. Landau–Lifshitz define the bracket with the opposite sign, $\{f,g\}_{\mathrm{LL}} = \sum(\partial_pf\,\partial_qg - \partial_qf\,\partial_pg)$; Goldstein, Arnold and this course use the one on the slide, for which $\{q,p\} = 1$ and $\dot f = \{f,H\}$. Every bracket identity in a text using the other convention changes sign. Unit 17 and the quantum commutator $[\hat f,\hat g] = i\hbar\{f,g\}$ (Dirac's correspondence, mentioned only) use this one.

The fundamental brackets computed. $\{q_i,p_j\} = \sum_k(\partial q_i/\partial q_k)(\partial p_j/\partial p_k) - 0 = \sum_k\delta_{ik}\delta_{jk} = \delta_{ij}$; $\{q_i,q_j\}$ and $\{p_i,p_j\}$ have a vanishing factor in every term. As matrices: the brackets of the coordinate functions $z_a$ are $\{z_a, z_b\} = J_{ab}$.

History. Poisson introduced the bracket in 1809, in a memoir on the variation of the elements of planetary orbits — the same perturbation problem that later gave unit 10 its secular theory. Jacobi proved the identity that bears his name in lectures of 1842–43, published posthumously in 1866.

14 · brackets

The Jacobi identity, and what it makes of observables

proof of the Jacobi identity

Let $D_f := \{f,\cdot\,\}$. It is a first-order differential operator: $D_f = \sum_i\big(\partial_{q_i}f\,\partial_{p_i} - \partial_{p_i}f\,\partial_{q_i}\big)$. Rewrite the first two terms of the sum: $$\begin{aligned}\{f,\{g,h\}\} + \{g,\{h,f\}\} &= D_fD_gh - D_gD_fh\\ &= [D_f, D_g]\,h.\end{aligned}$$ The commutator of two first-order operators is first order — the second derivatives of $h$ cancel by Clairaut. The third term, $-D_{\{f,g\}}h$, is first order in $h$ too. So the Jacobi sum contains no second derivative of $h$, and by its cyclic symmetry none of $f$ or $g$. But every term of it contains exactly one second derivative. So it is $0$. ∎

a Lie algebra

A real vector space with a bilinear, antisymmetric bracket satisfying Jacobi. So $\big(C^\infty(\mathbb R^{2n}),\{\,,\}\big)$ is one — infinite-dimensional.

A three-dimensional example: $\mathfrak{so}(3)$, unit 03's antisymmetric $3\times3$ matrices, under the commutator $[X,Y] = XY - YX$. Unit 03's hat map is then an isomorphism of Lie algebras, $\widehat{\vec a\times\vec b} = [\hat a,\hat b]$ — one line, in the notes.

Jacobi, reread

It says $[D_f, D_g] = D_{\{f,g\}}$: the map $f\mapsto D_f$ turns brackets of functions into commutators of operators. That is the sense in which functions on phase space generate motions.

Full proof of the two claims used

The commutator is first order. Write $D_f = \sum_a u_a\partial_a$ and $D_g = \sum_b w_b\partial_b$, where $u_a$, $w_b$ are functions (the components of $-J\nabla f$ and $-J\nabla g$, though only their being functions matters) and $\partial_a = \partial/\partial z_a$. Then $D_fD_gh = \sum_{a,b}u_a(\partial_aw_b)\partial_bh + \sum_{a,b}u_aw_b\,\partial_a\partial_bh$, and the second sum is symmetric under exchanging the roles of $u$ and $w$ once $\partial_a\partial_bh = \partial_b\partial_ah$ (Clairaut, unit 03, for $h\in C^2$). So in $D_fD_gh - D_gD_fh$ the second-derivative sums cancel and $[D_f,D_g]h = \sum_b\big(\sum_a(u_a\partial_aw_b - w_a\partial_au_b)\big)\partial_bh$.

Every term has exactly one second derivative. Expand $\{f,\{g,h\}\}$: it is a sum of terms $(\partial f)\,\partial(\partial g\,\partial h)$, and the outer derivative falls on exactly one of the two inner factors, giving (first derivative of $f$)(second derivative of $g$)(first derivative of $h$) or (first derivative of $f$)(first derivative of $g$)(second derivative of $h$). The same holds for the other two terms. Now group the Jacobi sum by which function is differentiated twice. The group in which $h$ is differentiated twice is zero, by the first claim; by cyclic symmetry so are the groups for $f$ and for $g$. There is no other group. $\square$

Jacobi, reread, derived. From the identity, $D_fD_gh - D_gD_fh = -\{h,\{f,g\}\} = \{\{f,g\},h\} = D_{\{f,g\}}h$ for every $h$.

Full proof that the hat map is an isomorphism of Lie algebras

Unit 03 gave the hat map as a linear isomorphism $\mathbb R^3\to\mathfrak{so}(3)$, $\hat a\vec v = \vec a\times\vec v$; that it also carries $\times$ to the commutator is this line. For every $\vec v$, by BAC–CAB (unit 08), $$[\hat a,\hat b]\vec v = \vec a\times(\vec b\times\vec v) - \vec b\times(\vec a\times\vec v) = \big(\vec b(\vec a\cdot\vec v) - \vec v(\vec a\cdot\vec b)\big) - \big(\vec a(\vec b\cdot\vec v) - \vec v(\vec a\cdot\vec b)\big) = \vec b(\vec a\cdot\vec v) - \vec a(\vec b\cdot\vec v),$$ and BAC–CAB again gives $(\vec a\times\vec b)\times\vec v = -\vec v\times(\vec a\times\vec b) = \vec b(\vec v\cdot\vec a) - \vec a(\vec v\cdot\vec b)$, the same vector. So $[\hat a,\hat b] = \widehat{\vec a\times\vec b}$. $\square$

Why a Lie algebra is the right word. The antisymmetric matrices with the commutator are the Lie algebra of the rotation group (unit 03 proved $R^{\!\top}\dot R$ antisymmetric for a moving rotation). In general a Lie algebra is what a group of symmetries looks like infinitesimally, and Jacobi is the one axiom that encodes associativity of the group. So when conserved quantities are found to close under the bracket, they are identifying the symmetry group — which is how the rest of this section uses them.

14 · brackets

Time evolution is a bracket with $H$

theorem

Along any solution of Hamilton's equations, for $f(q,p,t)$, $$\frac{\mathrm df}{\mathrm dt} = \{f, H\} + \frac{\partial f}{\partial t}.$$

proof

Chain rule, then Hamilton: $\dot f = \sum(f_{q_i}\dot q_i + f_{p_i}\dot p_i) + f_t = \sum(f_{q_i}H_{p_i} - f_{p_i}H_{q_i}) + f_t$. ∎

special cases

$f = q_i$: $\dot q_i = \{q_i,H\} = H_{p_i}$ ✓. $f = p_i$: $\dot p_i = -H_{q_i}$ ✓. $f = H$: $\dot H = H_t$ ✓.

corollaries

1. A time-independent $f$ is conserved iff $\{f,H\} = 0$.

2. Poisson's theorem: if $f$ and $g$ are conserved, so is $\{f,g\}$.

proof of 2

Jacobi, rearranged: $\{\{f,g\},H\} = \{f,\{g,H\}\} + \{g,\{H,f\}\} = \{f,0\} + \{g,0\} = 0$. ∎

what Poisson's theorem does not promise

$\{f,g\}$ may be zero, a constant, or a function of $f$, $g$ and $H$. It is a new conserved quantity only sometimes; for Kepler it is never new, and that is the point — the set closes.

The proof of Poisson's theorem with every sign. Jacobi with $(f,g,H)$ cycled so that $H$ sits innermost in each term: $\{f,\{g,H\}\} + \{g,\{H,f\}\} + \{H,\{f,g\}\} = 0$. The first two vanish because $\{g,H\} = 0$ and $\{H,f\} = -\{f,H\} = 0$, and the bracket of anything with the zero function is zero. So $\{H,\{f,g\}\} = 0$, i.e. $\{\{f,g\},H\} = 0$. For time-dependent conserved $f$, $g$ the same argument with $\partial_t$ included works: $\partial_t\{f,g\} = \{\partial_tf,g\} + \{f,\partial_tg\}$.

The history of that caveat. Poisson (1809) presented the theorem as a way to generate new integrals from old ones; Jacobi saw that it usually produces nothing new, and the real content, understood only with Lie's work in the 1870s–80s, is that the conserved quantities of a system form a Lie algebra under the bracket. The Kepler computation that follows is the standard illustration.

Where this goes. In unit 17, $n$ conserved quantities with pairwise vanishing brackets ("in involution") make a system integrable — the Liouville–Arnold theorem — and unit 08 already noticed that Kepler has more than that.

14 · brackets

Noether's converse, now cheap

the flow a function generates

Any $f$ is a Hamiltonian: its flow solves $z' = J\nabla f(z)$ in a parameter $s$, and along it $\mathrm dg/\mathrm ds = \{g, f\}$ for every $g$.

theorem (Hamiltonian Noether, both ways)

For time-independent $f$ and $H$: $f$ is conserved by the flow of $H$ $\iff$ $H$ is invariant under the flow of $f$.

proof

The first is $\{f,H\} = 0$, the second $\{H,f\} = 0$. Antisymmetry. ∎

two generators

Translation. $f = \vec a\cdot\vec p$: $\{\vec r, f\} = \vec a$, $\{\vec p, f\} = \vec 0$. Its flow is $\vec r\mapsto\vec r + s\vec a$.

Rotation. $f = \vec a\cdot\vec L$ with $\vec L = \vec r\times\vec p$: since $\vec a\cdot\vec L = \vec p\cdot(\vec a\times\vec r) = \vec r\cdot(\vec p\times\vec a)$, $$\{\vec r, \vec a\cdot\vec L\} = \vec a\times\vec r,\qquad \{\vec p, \vec a\cdot\vec L\} = -\,\vec p\times\vec a = \vec a\times\vec p .$$ Both rotate about $\vec a$ at unit rate (unit 03's $\dot{\vec x} = \vec\omega\times\vec x$).

why unit 13's converse failed and this one does not

Unit 13 allowed only symmetries moving points of configuration space. The flow of $f$ may mix $q$ and $p$. Allow that, and every conserved quantity has its symmetry.

The rotation brackets in full. $\{r_i, f\} = \partial f/\partial p_i$ and $\{p_i, f\} = -\partial f/\partial r_i$ (the definition with one argument a coordinate). With $f = \vec p\cdot(\vec a\times\vec r)$, $\partial f/\partial\vec p = \vec a\times\vec r$. With $f = \vec r\cdot(\vec p\times\vec a)$ — the same number, by the cyclic property of the scalar triple product from unit 08 — $\partial f/\partial\vec r = \vec p\times\vec a$, so $\{\vec p, f\} = -\vec p\times\vec a = \vec a\times\vec p$. The flow $\vec r' = \vec a\times\vec r$ is rotation about $\vec a$ with angular velocity $\vec a$: unit 03's hat map, $\hat a\vec r$, so $\vec r(s)$ is $\vec r(0)$ turned about $\vec a$ through the angle $s\lVert\vec a\rVert$ — unit 03's rotating frame with $\vec\omega = \vec a$, read as an equation of motion.

What "Noether's converse" means here. Unit 13 proved: a one-parameter group of transformations of configuration space leaving the action (quasi-)invariant gives a conserved quantity. The converse there is false as stated: the Runge–Lenz vector is conserved, and no transformation of configuration space alone produces it. In phase space every conserved $f$ generates a flow preserving $H$; section 4 will show that the flow is also canonical, so it maps solutions to solutions. For $\vec a\cdot\vec A$, that flow changes $\vec r$ at a rate depending on $\vec p$: it is not induced by any map of configuration space, which is the precise meaning of "hidden". It moves a Kepler ellipse to another of the same energy with a different eccentricity.

Completeness. "The flow of $f$" presumes the solutions of $z' = J\nabla f$ exist for all $s$; for $\vec a\cdot\vec p$ and $\vec a\cdot\vec L$ they obviously do. When they do not, the statement holds for the local flow, which is all the infinitesimal statement $\{f,H\} = 0$ needs.

14 · brackets

Every vector rotates the same way under $\vec L$

the Levi-Civita symbol

$\varepsilon_{ijk} := \vec e_i\cdot(\vec e_j\times\vec e_k) = \det[\vec e_i\,\vec e_j\,\vec e_k]$: the sign of the permutation $(i,j,k)$ of $(1,2,3)$, and $0$ if two indices agree. Then $\vec e_i\times\vec e_j = \sum_k\varepsilon_{ijk}\vec e_k$.

lemma

Let $\vec V(\vec r,\vec p)$ be built from $\vec r$ and $\vec p$ by sums, cross products and multiplication by functions of $\vec r\cdot\vec r$, $\vec r\cdot\vec p$, $\vec p\cdot\vec p$. Then $\{\vec V, \vec a\cdot\vec L\} = \vec a\times\vec V$, and so $$\{\vec a\cdot\vec L,\ \vec b\cdot\vec V\} = (\vec a\times\vec b)\cdot\vec V .$$

proof in the notes (N)

the first two brackets of Kepler

Both $\vec L = \vec r\times\vec p$ and $\vec A = \vec p\times\vec L - \mu k\,\vec r/\lvert\vec r\rvert$ qualify. Put $\vec a = \vec e_i$, $\vec b = \vec e_j$: $$\{L_i, L_j\} = \sum_k\varepsilon_{ijk}L_k,\qquad \{L_i, A_j\} = \sum_k\varepsilon_{ijk}A_k .$$

So the three $L_i$ close on unit 03's $\mathfrak{so}(3)$: $\{L_1,L_2\} = L_3$ and cyclically. And $\vec A$ transforms as a vector — which is all the second bracket says.

checks

Dimensions: $[L]^2/[\text{action}] = [L]$ ✓. Special case: from the definition, with $L_3 = xp_y - yp_x$, $\{x, L_3\} = \partial L_3/\partial p_x = -y$; the lemma gives $(\vec e_3\times\vec r)_1 = -y$ ✓.

Full proof of the lemma

Let $D := \{\,\cdot\,, \vec a\cdot\vec L\}$, acting componentwise on vectors. It is a derivation (Leibniz, in the first slot by antisymmetry): $D(fg) = (Df)g + f(Dg)$. The previous slide gives $D\vec r = \vec a\times\vec r$ and $D\vec p = \vec a\times\vec p$. Three closure steps:

Scalars. For vectors $\vec u,\vec w$ with $D\vec u = \vec a\times\vec u$, $D\vec w = \vec a\times\vec w$: $D(\vec u\cdot\vec w) = (\vec a\times\vec u)\cdot\vec w + \vec u\cdot(\vec a\times\vec w) = \vec a\cdot(\vec u\times\vec w) + \vec a\cdot(\vec w\times\vec u) = 0$, by the triple product (unit 08). So $D$ kills $\vec r\cdot\vec r$, $\vec r\cdot\vec p$, $\vec p\cdot\vec p$, and by the chain rule any function $\phi$ of them.

Scalar multiples. $D(\phi\vec u) = (D\phi)\vec u + \phi\,\vec a\times\vec u = \vec a\times(\phi\vec u)$.

Cross products. $D(\vec u\times\vec w) = (\vec a\times\vec u)\times\vec w + \vec u\times(\vec a\times\vec w)$. By BAC–CAB (unit 08), $(\vec a\times\vec u)\times\vec w = -\vec w\times(\vec a\times\vec u) = -\vec a(\vec w\cdot\vec u) + \vec u(\vec w\cdot\vec a)$ and $\vec u\times(\vec a\times\vec w) = \vec a(\vec u\cdot\vec w) - \vec w(\vec u\cdot\vec a)$. The sum is $\vec u(\vec a\cdot\vec w) - \vec w(\vec a\cdot\vec u) = \vec a\times(\vec u\times\vec w)$.

Sums are immediate. By induction on how $\vec V$ is built, $D\vec V = \vec a\times\vec V$. Then $\{\vec b\cdot\vec V, \vec a\cdot\vec L\} = \vec b\cdot(\vec a\times\vec V) = (\vec b\times\vec a)\cdot\vec V$, and antisymmetry gives the boxed form. With $\vec a = \vec e_i$, $\vec b = \vec e_j$: $(\vec e_i\times\vec e_j)\cdot\vec V = \sum_k\varepsilon_{ijk}V_k$. $\square$

The special-case check, written plainly. $\{x, L_3\} = \partial L_3/\partial p_x = -y$, and $(\vec e_3\times\vec r)_1 = (-y, x, 0)_1 = -y$ ✓.

The symbol. $\varepsilon_{ijk}$ is only notation here for the components of the cross product; unit 16 explains in what sense it is and is not a tensor, and unit 28 distinguishes the Levi-Civita symbol from the Levi-Civita tensor.

14 · brackets

Worked: the bracket of Runge–Lenz with itself

$\{A_i, A_j\} = -2\mu H\sum_k\varepsilon_{ijk}L_k$

Step 1 — a better form. BAC–CAB: $\vec p\times(\vec r\times\vec p) = p^2\vec r - (\vec r\cdot\vec p)\vec p$, so $$\vec A = \alpha\,\vec r - \beta\,\vec p,\qquad \alpha = p^2 - \frac{\mu k}{r},\quad \beta = \vec r\cdot\vec p .$$

Step 2 — the brackets needed, each from the definition: $\{r_i,\alpha\} = 2p_i$, $\ \{p_i,\beta\} = -p_i$, $\ \{r_i,\beta\} = r_i$, $\ \{\alpha, p_j\} = \mu k\,r_j/r^3$, and $$\{\alpha,\beta\} = \frac{\mu k}{r} - 2p^2 .$$

Step 3 — expand $\{\alpha r_i - \beta p_i,\ \alpha r_j - \beta p_j\}$ by Leibniz. Only the part antisymmetric in $i,j$ survives; write $\Lambda_{ij} = r_ip_j - r_jp_i = \sum_k\varepsilon_{ijk}L_k$.

$\{\alpha r_i,\alpha r_j\} = \alpha(\{r_i,\alpha\}r_j - \{r_j,\alpha\}r_i) = -2\alpha\Lambda_{ij}$.
$\{\beta p_i,\beta p_j\} = \beta(\{p_i,\beta\}p_j - \{p_j,\beta\}p_i) = 0$.
The cross terms: $\{\alpha r_i,\beta p_j\} = \alpha\beta\delta_{ij} + \alpha r_ip_j + \beta\mu k\,r_ir_j/r^3 + \{\alpha,\beta\}r_ip_j$, whose antisymmetric part is $(\alpha + \{\alpha,\beta\})\Lambda_{ij} = -p^2\Lambda_{ij}$, entering with a minus sign.

Step 4 — add. $$\{A_i,A_j\} = (-2\alpha + p^2)\Lambda_{ij} = \Big(\frac{2\mu k}{r} - p^2\Big)\Lambda_{ij} = -2\mu H\,\Lambda_{ij}.$$

Checks. $[\{A,A\}] = (\mathrm{M^2L^3T^{-2}})^2/\mathrm{ML^2T^{-1}} = \mathrm{M^3L^4T^{-3}} = [\mu HL]$ ✓. Special case $k = 0$: then $\alpha = p^2$ and the same steps give $-p^2\Lambda_{ij}$, which is $-2\mu H\Lambda_{ij}$ for $H = p^2/2\mu$ ✓.

Step 2 in full. $\{r_i, g\} = \partial g/\partial p_i$ and $\{p_i, g\} = -\partial g/\partial r_i$. $\partial\alpha/\partial p_i = 2p_i$. $\partial\beta/\partial r_i = p_i$, so $\{p_i,\beta\} = -p_i$. $\partial\beta/\partial p_i = r_i$. $\{\alpha, p_j\} = \partial\alpha/\partial r_j = \mu k\,r_j/r^3$ (the derivative of $-\mu k/r$). For $\{\alpha,\beta\} = \sum_i(\partial_{r_i}\alpha\,\partial_{p_i}\beta - \partial_{p_i}\alpha\,\partial_{r_i}\beta) = \sum_i(\mu k\,r_i/r^3\cdot r_i - 2p_i\cdot p_i) = \mu k/r - 2p^2$.

Step 3 in full. Leibniz in both slots, with $\{r_i,r_j\} = 0$ and $\{\alpha,\alpha\} = 0$: $\{\alpha r_i,\alpha r_j\} = \alpha\{r_i,\alpha\}r_j + \alpha\{\alpha,r_j\}r_i = 2\alpha(p_ir_j - p_jr_i) = -2\alpha\Lambda_{ij}$. Similarly with $\{p_i,p_j\} = 0$: $\{\beta p_i,\beta p_j\} = \beta(-p_ip_j + p_jp_i) = 0$. The cross terms are $-\{\alpha r_i,\beta p_j\} - \{\beta p_i,\alpha r_j\} = -\{\alpha r_i,\beta p_j\} + \{\alpha r_j,\beta p_i\}$, i.e. minus the antisymmetric part $X_{ij} - X_{ji}$ of $X_{ij} = \{\alpha r_i,\beta p_j\}$. Expanding $X_{ij}$: $\alpha\beta\{r_i,p_j\} + \alpha p_j\{r_i,\beta\} + \beta r_i\{\alpha,p_j\} + r_ip_j\{\alpha,\beta\}$ $= \alpha\beta\delta_{ij} + \alpha r_ip_j + \beta\mu k\,r_ir_j/r^3 + \{\alpha,\beta\}r_ip_j$. The first and third terms are symmetric in $i,j$ and drop out of $X_{ij} - X_{ji} = (\alpha + \{\alpha,\beta\})\Lambda_{ij}$, and $\alpha + \{\alpha,\beta\} = p^2 - \mu k/r + \mu k/r - 2p^2 = -p^2$. So the cross terms contribute $+p^2\Lambda_{ij}$.

Step 4. Total $(-2\alpha + p^2)\Lambda_{ij} = (-p^2 + 2\mu k/r)\Lambda_{ij} = -2\mu(p^2/2\mu - k/r)\Lambda_{ij}$, and $p^2/2\mu - k/r = H$. Finally $\Lambda_{ij} = r_ip_j - r_jp_i = \sum_k\varepsilon_{ijk}L_k$: for $(i,j) = (1,2)$ it is $xp_y - yp_x = L_3$, and the rest follow by cyclic relabelling and antisymmetry.

Nothing here is deeper than Leibniz applied twice and the two brackets of step 2; the length is bookkeeping, and the only place to go wrong is a sign, which the special case $k = 0$ catches.

14 · brackets

The algebra closes: $\mathfrak{so}(4)$, $\mathfrak e(3)$, $\mathfrak{so}(3,1)$

rescale by the energy

Put $\vec M = \vec A/\sqrt{\mp2\mu H}$ on $H \lessgtr 0$; the factor passes through the bracket, as $\{\vec A, H\} = \vec 0$: $$\{L_i,L_j\} = \varepsilon_{ijk}L_k,\qquad \{L_i,M_j\} = \varepsilon_{ijk}M_k,$$ $$\{M_i,M_j\} = \pm\,\varepsilon_{ijk}L_k,$$ with $+$ for bound orbits and $-$ for unbound (sum over $k$). At $H = 0$: $\{A_i,A_j\} = 0$.

the matrix algebras

In $4\times4$ matrices, with $E_{ab}$ the matrix whose only nonzero entry is a $1$ in row $a$, column $b$, let $\ell_i = \hat e_i$ (unit 03's hat, in the top-left block), $m_i = E_{i4} - E_{4i}$, $k_i = E_{i4} + E_{4i}$. Then $[\ell_i,\ell_j] = \varepsilon_{ijk}\ell_k$, $[\ell_i,m_j] = \varepsilon_{ijk}m_k$, $[m_i,m_j] = \varepsilon_{ijk}\ell_k$: $\mathfrak{so}(4)$, antisymmetric matrices. And $[\ell_i,k_j] = \varepsilon_{ijk}k_k$, $[k_i,k_j] = -\varepsilon_{ijk}\ell_k$: $\mathfrak{so}(3,1)$, the $X$ with $X^{\!\top}\eta + \eta X = 0$, $\eta = \operatorname{diag}(1,1,1,-1)$.

what hidden symmetry meant

Bound Kepler orbits have the symmetry algebra of rotations of $\mathbb R^4$ — a space the comet never moves in. Unbound ones have the Lorentz algebra of unit 26. At the separatrix, $E = 0$, the algebra is $\mathfrak e(3)$, rotations and translations of $\mathbb R^3$: the one between.

Unit 13's "SO(4) generator" is this: the flow of $\vec a\cdot\vec A$, preserving $H$, carrying an ellipse to others of the same energy and different eccentricity.

two different algebras

$\mathfrak{so}(4)\not\cong\mathfrak{so}(3,1)$: in $\mathfrak{so}(3,1)$, $\operatorname{ad}k_1 = [k_1,\cdot\,]$ has real eigenvalues $\pm1$; in $\mathfrak{so}(4)$ every $\operatorname{ad}X$ has imaginary ones. Proof in the notes.

Passing the factor through. For $g = g(H)$, Leibniz gives $\{A_ig, A_jg\} = g^2\{A_i,A_j\} + gA_j\{A_i,g\} + gA_i\{g,A_j\} + A_iA_j\{g,g\}$, and $\{A_i, g(H)\} = g'(H)\{A_i,H\} = 0$ because $\vec A$ is conserved (unit 08 proved $\dot{\vec A} = 0$, which by the time-evolution theorem is $\{\vec A,H\} = 0$). So $\{M_i,M_j\} = \{A_i,A_j\}/(-2\mu H) = \varepsilon_{ijk}L_k$ for $H<0$, and $= -\varepsilon_{ijk}L_k$ for $H > 0$. Likewise $\{L_i,M_j\} = g\{L_i,A_j\} = \varepsilon_{ijk}M_k$, using $\{L_i,H\} = 0$. Dimensions: $[A/\sqrt{\mu H}] = \mathrm{M^2L^3T^{-2}}/\mathrm{MLT^{-1}} = \mathrm{ML^2T^{-1}}$, an angular momentum, as it must be to sit beside $\vec L$ ✓.

The matrix relations. $E_{ab}$ is the matrix with a $1$ in row $a$, column $b$, and $E_{ab}E_{cd} = \delta_{bc}E_{ad}$. The $\ell_i$ satisfy $[\ell_i,\ell_j] = \varepsilon_{ijk}\ell_k$ because the hat map is a Lie algebra isomorphism $(\mathbb R^3,\times)\to\mathfrak{so}(3)$ — $\widehat{\vec a\times\vec b} = [\hat a,\hat b]$, proved in the notes of the Jacobi slide — so $[\hat e_i,\hat e_j] = \widehat{\vec e_i\times\vec e_j} = \sum_k\varepsilon_{ijk}\hat e_k$. For one mixed relation: $\ell_1 = E_{32} - E_{23}$, $m_2 = E_{24} - E_{42}$, and $[\ell_1, m_2] = (E_{32}-E_{23})(E_{24}-E_{42}) - (E_{24}-E_{42})(E_{32}-E_{23}) = E_{34} - E_{43} = m_3$ ✓. For one $m$–$m$ relation: $[m_1,m_2] = (E_{14}-E_{41})(E_{24}-E_{42}) - (E_{24}-E_{42})(E_{14}-E_{41}) = -E_{12} + E_{21} = \ell_3$ ✓ (since $\ell_3 = \hat e_3 = E_{21} - E_{12}$). The $k$ relations are the same computation with plus signs, and $[k_1,k_2] = E_{12} - E_{21} = -\ell_3$. The rest follow by cyclic relabelling. The signs of $m_i$ and $k_i$ can be flipped freely: the relations are linear in each $m$ or $k$ on the left exactly when they are on the right.

Full proof that the two are not isomorphic

An isomorphism $\phi$ satisfies $\operatorname{ad}\phi(X) = \phi\circ\operatorname{ad}X\circ\phi^{-1}$, so the set of spectra of the maps $\operatorname{ad}X$ is an invariant. In $\mathfrak{so}(4)$, with the inner product $\langle X,Y\rangle = \operatorname{tr}(X^{\!\top}Y)$, each $\operatorname{ad}X$ is antisymmetric: $\langle[X,Y],Z\rangle = -\langle Y,[X,Z]\rangle$ by cyclicity of the trace and $X^{\!\top} = -X$; so its eigenvalues are purely imaginary. In $\mathfrak{so}(3,1)$, on the span of $k_2$ and $\ell_3$: $[k_1,k_2] = -\ell_3$ and $[k_1,\ell_3] = -[\ell_3,k_1] = -k_2$, so $\operatorname{ad}k_1$ has matrix $\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ there, with eigenvalues $\pm1$. $\square$

So what. For bound orbits $\vec J^\pm = (\vec L\pm\vec M)/2$ satisfy $\{J^\pm_i,J^\pm_j\} = \varepsilon_{ijk}J^\pm_k$ and $\{J^+_i,J^-_j\} = 0$: two commuting copies of $\mathfrak{so}(3)$, i.e. $\mathfrak{so}(4)\cong\mathfrak{so}(3)\oplus\mathfrak{so}(3)$. Pauli used exactly this in 1926 to get the hydrogen spectrum, with its $n^2$-fold degeneracy, before Schrödinger's equation was solved — cited: W. Pauli, Z. Physik 36, 336 (1926). Classically, the extra symmetry is why every bound Kepler orbit closes and why the period depends on the energy alone.

section 4

Canonical transformations

Which changes of phase-space coordinates keep Hamilton's equations — and a way to make them from one function.

14 · canonical

The symplectic form, and the maps that keep it

the symplectic group

A matrix $M\in\mathbb R^{2n\times2n}$ is symplectic if $M^{\!\top}JM = J$. They form a group $\mathrm{Sp}(2n,\mathbb R)$: if $M, N$ are, $(MN)^{\!\top}J(MN) = N^{\!\top}JN = J$, and $M^{-1}$ exists since $(\det M)^2 = 1$.

the symplectic form

The antisymmetric, nondegenerate bilinear form on $\mathbb R^{2n}$ $$\omega(u,v) := \sum_i\big(u_{p_i}v_{q_i} - u_{q_i}v_{p_i}\big) = -\,u^{\!\top}Jv .$$ $M$ is symplectic iff $\omega(Mu,Mv) = \omega(u,v)$ for all $u,v$. Unit 34 writes $\omega = \sum_i\mathrm dp_i\wedge\mathrm dq_i$.

a canonical transformation

A $C^1$ diffeomorphism $\Phi:(q,p)\mapsto(Q,P)$ between open subsets of $\mathbb R^{2n}$ whose Jacobian $D\Phi(z)$ is symplectic at every $z$.

$n = 1$: area

For $2\times2$ matrices $M^{\!\top}JM = (\det M)J$. So in the plane symplectic means $\det M = 1$, and $\omega(u,v) = -\det[u\ v]$ is minus the signed area of the parallelogram on $u, v$. Canonical = area- and orientation-preserving.

Mathematics first. A symplectic form on a real vector space is a bilinear form that is antisymmetric ($\omega(u,v) = -\omega(v,u)$) and nondegenerate ($\omega(u,\cdot) = 0$ only for $u = 0$). Antisymmetry of $\omega$ here: $u^{\!\top}Jv$ is a number, equal to its transpose $v^{\!\top}J^{\!\top}u = -v^{\!\top}Ju$. Nondegeneracy: $J$ is invertible. The linear-algebra theorem (cited: every symplectic form on $\mathbb R^{2n}$ has this matrix in a suitable basis — the linear Darboux theorem, e.g. Arnold, Mathematical Methods of Classical Mechanics, §41) says there is only one such form up to change of basis, so nothing is lost by using this one. Unit 34 promotes $\omega$ to a differential 2-form on a manifold, and there $\omega = \mathrm d\big(\sum p_i\,\mathrm dq_i\big)$ up to sign convention, which is the content of the generating-function slide.

Why $\omega(Mu,Mv) = \omega(u,v)$ is the condition. $\omega(Mu,Mv) = -u^{\!\top}M^{\!\top}JMv$, and this equals $-u^{\!\top}Jv$ for all $u,v$ iff $M^{\!\top}JM = J$.

The $2\times2$ identity. For $M = \begin{pmatrix}a&b\\c&d\end{pmatrix}$, $M^{\!\top}JM = \begin{pmatrix}0&ad-bc\\-(ad-bc)&0\end{pmatrix}$, by multiplying out.

14 · canonical

Volume is not enough: one area per plane

The unit square of each $(q_i,p_i)$ plane, shaded, and its images under $S = \operatorname{diag}(2,1,\tfrac12,1)$ (solid) and $V = \operatorname{diag}(2,\tfrac12,1,1)$ (dashed), coordinates ordered $(q_1,q_2,p_1,p_2)$. Both have determinant $1$. Only $S$ keeps the area in each plane.

in $2n\ge4$ it is more

Preserving $\omega$ preserves the area of the projection onto each $(q_i,p_i)$ plane, summed. Much stronger than preserving volume: in $(q_1,q_2,p_1,p_2)$, $\operatorname{diag}(2,1,\tfrac12,1)$ is symplectic; $\operatorname{diag}(2,\tfrac12,1,1)$ has determinant $1$ and is not.

the test, for diagonal maps

$\operatorname{diag}(d_1,d_2,d_3,d_4)$ is symplectic iff $d_1d_3 = 1$ and $d_2d_4 = 1$: stretch a position only by shrinking its own momentum. $V$ stretches $q_1$ and pays by shrinking $q_2$ — the wrong partner.

The two diagonal examples, with coordinates ordered $(q_1,q_2,p_1,p_2)$ as everywhere in this unit. For $D = \operatorname{diag}(d_1,d_2,d_3,d_4)$, $D^{\!\top}JD$ is $J$ with its two $1$s replaced by $d_1d_3$ and $d_2d_4$ (and the $-1$s likewise). So $D$ is symplectic iff $d_1d_3 = d_2d_4 = 1$: each position may be stretched only if its own momentum is shrunk by the same factor. $\operatorname{diag}(2,1,\tfrac12,1)$ passes. $\operatorname{diag}(2,\tfrac12,1,1)$ stretches $q_1$ and shrinks $q_2$: determinant $1$, so volume is preserved, but $d_1d_3 = 2$, so it is not symplectic. The area of the projection onto the $(q_1,p_1)$ plane doubles and that onto $(q_2,p_2)$ halves; volume, a product, does not notice, and $\omega$, a sum of the two, does.

What to look for. $V$ doubles the shaded square in the first plane and halves it in the second: volume, the product $2\times\tfrac12$, is unchanged, so Liouville's theorem alone could not rule $V$ out. The symplectic form sees each plane separately — $\omega(u,v)$ is minus the sum of the signed areas of the two projections — and that is a strictly stronger condition. Its sharpest form is Gromov's non-squeezing theorem, cited and not used: no canonical map takes a ball of radius $R$ into a cylinder $\{q_1^2 + p_1^2 < \rho^2\}$ with $\rho < R$, though a volume-preserving map easily can (M. Gromov, Invent. Math. 82, 307 (1985)).

14 · canonical

Three tests that are one test

theorem

For a $C^1$ diffeomorphism $\Phi$ with Jacobian $M = D\Phi$, these are equivalent:

(i) $M^{\!\top}JM = J$ everywhere (canonical);
(ii) $\Phi$ preserves brackets: $\{f\circ\Phi, g\circ\Phi\} = \{f,g\}\circ\Phi$ for all $f, g$ — equivalently, the new coordinates have the fundamental brackets $\{Q_i,Q_j\} = \{P_i,P_j\} = 0$, $\{Q_i,P_j\} = \delta_{ij}$;
(iii) for every $K$, $\Phi$ carries solutions of Hamilton's equations for $H = K\circ\Phi$ to solutions for $K$.

proof

Lemma: $M^{\!\top}JM = J\iff MJM^{\!\top} = J$ (invert both sides and use $J^{-1} = -J$).

(ii). By the chain rule $\nabla(f\circ\Phi) = M^{\!\top}\nabla f$, so $\{f\circ\Phi,g\circ\Phi\} = \nabla f^{\!\top}\,MJM^{\!\top}\,\nabla g$. This is $\nabla f^{\!\top}J\nabla g$ for all $f, g$ iff $MJM^{\!\top} = J$. The fundamental brackets are the entries of $MJM^{\!\top}$.

(iii). If $\dot z = J\nabla H$ and $w = \Phi(z)$, then $\dot w = M\dot z = MJM^{\!\top}\nabla K(w)$. This is $J\nabla K$ for every $K$ iff $MJM^{\!\top} = J$. ∎

"for every $K$" matters

$Q = \lambda q$, $P = \lambda p$ is not canonical for $\lambda^2\ne1$, yet maps the oscillator's equations to Hamilton's with $K = \lambda^2H$. One Hamiltonian cannot certify a map (problem P6).

The lemma, fully. If $M^{\!\top}JM = J$ then $\det M\ne0$ (take determinants: $(\det M)^2\det J = \det J$ and $\det J = 1$). Invert: $M^{-1}J^{-1}M^{-\top} = J^{-1}$, i.e. $-M^{-1}JM^{-\top} = -J$, so $J = MJM^{\!\top}$ after multiplying by $M$ on the left and $M^{\!\top}$ on the right. The converse is the same argument with $M^{\!\top}$ in place of $M$.

The fundamental brackets as entries. With $f = w_a$ and $g = w_b$ (coordinate functions in the target), $f\circ\Phi$ is the $a$-th new coordinate as a function of $z$, and $\{w_a\circ\Phi, w_b\circ\Phi\} = e_a^{\!\top}MJM^{\!\top}e_b$. Requiring these to equal $J_{ab}$ for all $a,b$ is $MJM^{\!\top} = J$. So checking the $n(2n-1)$ independent fundamental brackets is a complete test — in the plane, a single one: $\{Q,P\} = 1$.

(iii) with "every $K$". If $MJM^{\!\top}\nabla K = J\nabla K$ at a point for every $K$, take $K$ linear with $\nabla K = e_a$ to get $MJM^{\!\top}e_a = Je_a$ for each $a$. Conversely (i) makes the transformed equations Hamiltonian for every $K$ — which is the physical reason for the definition: a canonical transformation is one you may use without knowing in advance which Hamiltonian you will apply it to.

Time-dependent maps. If $\Phi$ depends on $t$, (iii) holds with $K$ replaced by $K + \partial F/\partial t$ for the generating function $F$ of the next slide; cited, Goldstein, Poole and Safko, Classical Mechanics, 3rd ed., §9.1, and used only in unit 15.

14 · canonical

Generating functions: a canonical map from one function

theorem

A $C^2$ diffeomorphism $\Phi$ on a simply connected domain is canonical iff there is a function $F$ with $$\sum_i p_i\,\mathrm dq_i - \sum_i P_i\,\mathrm dQ_i = \mathrm dF,$$ i.e. $p\cdot\dot q - P\cdot\dot Q = \frac{\mathrm d}{\mathrm dt}F$ along every curve.

proof

The left side is a covector field $a(z) = (p, 0) - M^{\!\top}(P,0)$ on $\mathbb R^{2n}$. By the Poincaré lemma (unit 04) it is a gradient iff $Da - Da^{\!\top} = 0$. Compute: for $(p,0)$ this antisymmetric part is $J$; for $M^{\!\top}(P,0)$ it is $M^{\!\top}JM$ (second derivatives of $Q$ cancel by Clairaut). So $Da - Da^{\!\top} = J - M^{\!\top}JM$. ∎

the four generating functions

Write $F$ in whichever $2n$ of $(q,p,Q,P)$ are independent:

type	variables	equations
$F_1$	$q, Q$	$p = \partial_qF_1$, $\ P = -\partial_QF_1$
$F_2 = F_1 + P\cdot Q$	$q, P$	$p = \partial_qF_2$, $\ Q = \partial_PF_2$
$F_3 = F_1 - p\cdot q$	$p, Q$	$q = -\partial_pF_3$, $\ P = -\partial_QF_3$
$F_4 = F_1 + P\cdot Q - p\cdot q$	$p, P$	$q = -\partial_pF_4$, $\ Q = \partial_PF_4$

three to know

$F_2 = q\cdot P$: the identity. $F_1 = q\cdot Q$: $Q = p$, $P = -q$ — position and momentum are interchangeable. $F_2 = f(q)\cdot P$: the point transformation $Q = f(q)$, $p = Df(q)^{\!\top}P$.

Full proof of the curl computation

Covectors on $\mathbb R^{2n}$ are written as $2n$-component columns, ordered $(q\text{-slots}, p\text{-slots})$. $\sum p_i\,\mathrm dq_i$ is the covector field $\theta(z) = (p, 0)$: component $i\le n$ is $p_i = z_{n+i}$, the rest are $0$. So $(D\theta)_{ij} = \partial\theta_i/\partial z_j = 1$ exactly when $i\le n$ and $j = n+i$: $D\theta = \begin{pmatrix}0&I\\0&0\end{pmatrix}$ and $D\theta - D\theta^{\!\top} = J$. Next, $\sum P_k\,\mathrm dQ_k$ in the $z$ coordinates is $b_i(z) = \sum_{k}\Theta_k(\Phi(z))\,\partial\Phi_k/\partial z_i$ with $\Theta(w) = (P,0)$ — the chain rule for $\mathrm dQ_k$. Differentiate: $\partial_jb_i = \sum_{k,l}(\partial_l\Theta_k)(\Phi)\,M_{lj}M_{ki} + \sum_k\Theta_k\,\partial_i\partial_j\Phi_k$. The last sum is symmetric in $i,j$ (Clairaut, $\Phi\in C^2$), so $Db - Db^{\!\top} = M^{\!\top}(D\Theta - D\Theta^{\!\top})M = M^{\!\top}JM$. Unit 04's Poincaré lemma, in $\mathbb R^N$: a $C^1$ covector field on a simply connected (there: star-shaped suffices) domain is a gradient iff its derivative matrix is symmetric. $\square$

Where the table comes from. For $F_1(q,Q)$: $\mathrm dF_1 = \partial_qF_1\cdot\mathrm dq + \partial_QF_1\cdot\mathrm dQ$, and matching with $p\cdot\mathrm dq - P\cdot\mathrm dQ$ gives the first row; this needs $(q,Q)$ to be independent coordinates, i.e. $\det(\partial^2F_1/\partial q\partial Q)\ne0$ so that $p = \partial_qF_1(q,Q)$ can be solved for $Q$ by the implicit function theorem. $F_2 = F_1 + P\cdot Q$ gives $\mathrm dF_2 = p\cdot\mathrm dq - P\cdot\mathrm dQ + P\cdot\mathrm dQ + Q\cdot\mathrm dP = p\cdot\mathrm dq + Q\cdot\mathrm dP$, whence row two; rows three and four are the same bookkeeping. Each passage is a partial Legendre transform — section 1's operation, done in some of the variables — which is why it adds a product like $P\cdot Q$.

The point transformation. $F_2 = \sum_kf_k(q)P_k$: $Q_k = \partial F_2/\partial P_k = f_k(q)$ and $p_i = \partial F_2/\partial q_i = \sum_k(\partial f_k/\partial q_i)P_k$. Every change of configuration coordinates lifts to a canonical transformation this way, which is unit 12's form-invariance in the Hamiltonian language; but the class of canonical maps is far larger.

14 · canonical

Worked: Cartesian to polar is canonical

Kepler's coordinates, justified

Step 1 — the generating function. A point transformation, type 2: $F_2(x,y,P_r,P_\theta) = P_r\sqrt{x^2+y^2} + P_\theta\,\theta(x,y)$, with $\theta$ the polar angle (locally smooth off the origin).

Step 2 — read off. $r = \partial F_2/\partial P_r = \sqrt{x^2+y^2}$, $\theta = \partial F_2/\partial P_\theta$ ✓, and with $\partial_x\theta = -y/r^2$, $\partial_y\theta = x/r^2$: $$p_x = P_r\frac xr - P_\theta\frac{y}{r^2},\qquad p_y = P_r\frac yr + P_\theta\frac{x}{r^2}.$$

Step 3 — invert. Dot with $(x,y)/r$ and with $(-y,x)$: $$P_r = \frac{xp_x + yp_y}{r} = \hat r\cdot\vec p,\qquad P_\theta = xp_y - yp_x = L_z .$$

Step 4 — second route, the brackets. $\{r, P_r\} = \tfrac xr\cdot\tfrac xr + \tfrac yr\cdot\tfrac yr = 1$, $\{\theta, P_\theta\} = (-\tfrac{y}{r^2})(-y) + \tfrac{x}{r^2}x = 1$, $\{r,P_\theta\} = \tfrac xr(-y) + \tfrac yr x = 0$, $\{\theta, P_r\} = 0$.

Step 5 — the Hamiltonian. $p_x^2 + p_y^2 = P_r^2 + P_\theta^2/r^2$ (orthogonal parts along $\hat r$ and $\hat\theta$), so $H = P_r^2/2\mu + P_\theta^2/2\mu r^2 - k/r$ — the Legendre transform's answer from section 1 ✓.

checks

Dimensions: $[F_2] = [P_r][r] = \mathrm{J\,s}$, an action, as every generating function must be ✓.

The last bracket, $\{P_r, P_\theta\} = 0$, needs a line more: notes.

Step 5 in a picture: $\vec p$ splits into $P_r$ along $\hat r$ and $P_\theta/r$ along $\hat\theta$, at right angles, so $p_x^2 + p_y^2 = P_r^2 + P_\theta^2/r^2$ is Pythagoras.

Step 3 in full. $xp_x + yp_y = P_r(x^2+y^2)/r + P_\theta(-xy + xy)/r^2 = P_rr$. And $xp_y - yp_x = P_r(xy - xy)/r + P_\theta(x^2+y^2)/r^2 = P_\theta$.

The brackets, from the definition. On the $(x,y,p_x,p_y)$ phase space, $\{f,g\} = f_xg_{p_x} + f_yg_{p_y} - f_{p_x}g_x - f_{p_y}g_y$. For $f = r$ or $\theta$, $f_{p_x} = f_{p_y} = 0$, so only the first two terms survive; $\partial P_r/\partial p_x = x/r$, $\partial P_r/\partial p_y = y/r$, $\partial P_\theta/\partial p_x = -y$, $\partial P_\theta/\partial p_y = x$. That gives the four values on the slide. $\{r,\theta\} = 0$ because neither depends on momenta.

$\{P_r,P_\theta\} = 0$. Put $s = xp_x + yp_y$, so $P_r = s/r$. Then $\partial_xP_r = p_x/r - xs/r^3$, $\partial_yP_r = p_y/r - ys/r^3$, and $\partial_xP_\theta = p_y$, $\partial_yP_\theta = -p_x$. The bracket is $(p_x/r - xs/r^3)(-y) + (p_y/r - ys/r^3)(x) - (x/r)(p_y) - (y/r)(-p_x) = -yp_x/r + xp_y/r - xp_y/r + yp_x/r + (xy - xy)s/r^3 = 0$ ✓. Together with $\{r,\theta\} = 0$ all six fundamental brackets hold, so by the three-tests theorem the map is canonical — by an argument independent of the generating function.

Two routes agree. Legendre in polar coordinates, or canonical transformation of the Cartesian $H$: the same Hamiltonian. The Hamiltonian formalism is coordinate-free in the same sense as unit 12's.

Step 5, and the figure. With $\hat r = (x,y)/r$ and $\hat\theta = (-y,x)/r$, step 2 reads $\vec p = P_r\,\hat r + (P_\theta/r)\,\hat\theta$: the coefficients of $P_r$ and $P_\theta$ in $p_x$, $p_y$ are exactly the components of $\hat r$ and of $\hat\theta/r$. The two unit vectors are orthogonal, so the squared length is the sum of the squared parts. The radial part is $\mu\dot r$ and the angular part is $\mu r\dot\theta$, the velocity components of unit 08 times the mass.

The general lesson. Any change of configuration coordinates $Q = f(q)$ extends to phase space by $p = Df^{\!\top}P$, and the extension is canonical. That is why, in practice, one may write the Lagrangian in any coordinates and Legendre-transform there: the result is the same Hamiltonian as transforming a Cartesian one.

14 · canonical

The motion itself is a canonical transformation

theorem

Let $H\in C^2$ and let $\varphi_t$ be its flow: $\varphi_t(z)$ is the state at time $t$ of the solution starting at $z$. Then for each $t$, $\varphi_t$ is a canonical transformation.

proof

Let $Y(t) = D\varphi_t(z)$. It solves $\dot Y = JS(t)Y$, $Y(0) = I$, with $S = \nabla^2H(\varphi_t z)$ symmetric (cited, below). Then $$\tfrac{\mathrm d}{\mathrm dt}(Y^{\!\top}JY) = Y^{\!\top}\big(S^{\!\top}J^{\!\top}J + JJS\big)Y = Y^{\!\top}(S - S)Y = 0,$$ using $J^{\!\top}J = I$, $J^2 = -I$. So $Y^{\!\top}JY = J$ for all $t$. ∎

cited: the variational equation

For a $C^1$ vector field $X$, the flow is $C^1$ in the initial point, and $Y = D\varphi_t(z)$ solves $\dot Y = DX(\varphi_t z)\,Y$. Hirsch, Smale and Devaney, Differential Equations, Dynamical Systems, and an Introduction to Chaos, §17.6. Unit 17 makes it central.

what it is for

Unit 15: if motion is a canonical transformation, look for the one that makes it trivial — new coordinates in which $K = 0$. Its generating function is the Hamilton–Jacobi $S$.

Section 5: symplectic maps have determinant $1$, so the flow preserves volume.

The derivative computed. $\dot Y^{\!\top}JY + Y^{\!\top}J\dot Y = (JSY)^{\!\top}JY + Y^{\!\top}JJSY = Y^{\!\top}S^{\!\top}J^{\!\top}JY + Y^{\!\top}J^2SY = Y^{\!\top}SY - Y^{\!\top}SY = 0$, with $S^{\!\top} = S$, $J^{\!\top}J = -J\cdot J = I$ and $J^2 = -I$. At $t = 0$, $Y^{\!\top}JY = J$. A $C^1$ matrix function with zero derivative is constant.

Why the variational equation is plausible (the proof is cited): formally, differentiate $\partial_t\varphi_t(z) = X(\varphi_t(z))$ in $z$ and exchange the order of derivatives — the chain rule gives $\partial_tD\varphi_t = DX(\varphi_t)D\varphi_t$. What needs proof is that $D\varphi_t$ exists and that the exchange is legitimate; the standard argument applies Gronwall's inequality (unit 01) to the difference quotient, and that is the section cited.

With $H = X_H$'s Hamiltonian. $DX_H = D(J\nabla H) = J\nabla^2H = JS$, which is where the symmetric $S$ comes from.

Numerical meaning. A time-stepping method for mechanics replaces $\varphi_h$ by some approximating map $\psi_h$. If $\psi_h$ is itself symplectic it inherits every consequence of this theorem that does not depend on $H$ — in particular the volume preservation of section 5. Such a method is a symplectic integrator; section 5 puts a number on the alternative.

section 5

Liouville: clouds that deform without shrinking

The flow preserves volume — in three lines — and statistical mechanics, and good integrators, rest on it.

14 · Liouville

Divergence, and how determinants evolve

the divergence

For a $C^1$ vector field $X$ on an open subset of $\mathbb R^N$, $$\operatorname{div}X := \operatorname{tr}DX = \sum_{a=1}^N\frac{\partial X_a}{\partial z_a}.$$ Unit 05 gives it a meaning as flux per unit volume; here only this definition is used.

Liouville's formula

If $\dot Y = A(t)Y$ for $N\times N$ matrices, then $$\frac{\mathrm d}{\mathrm dt}\det Y = \operatorname{tr}A(t)\,\det Y .$$

proof

$Y(t+\epsilon) = (I + \epsilon A + o(\epsilon))Y(t)$, and expanding the determinant by permutations, $\det(I + \epsilon A + o(\epsilon)) = 1 + \epsilon\operatorname{tr}A + o(\epsilon)$: the identity permutation gives $\prod(1 + \epsilon A_{aa}) + o(\epsilon)$; every other one has at least two off-diagonal factors, each $O(\epsilon)$. Multiply by $\det Y(t)$ and let $\epsilon\to0$. ∎

special case

$N = 2$, $A = \begin{pmatrix}0&1\\-\omega_0^2&-\gamma\end{pmatrix}$: $\det Y = W$, the Wronskian, and the formula is unit 06's Abel identity $W(t) = W(0)e^{-\gamma t}$ ✓.

for a flow

With the variational equation: $\det D\varphi_t(z) = \exp\displaystyle\int_0^t\operatorname{div}X(\varphi_s z)\,\mathrm ds$.

The determinant expansion, carefully. $\det B = \sum_\sigma\operatorname{sgn}\sigma\prod_aB_{a\sigma(a)}$. For $B = I + \epsilon A + R$ with $R = o(\epsilon)$: the identity permutation contributes $\prod_a(1 + \epsilon A_{aa} + R_{aa}) = 1 + \epsilon\sum_aA_{aa} + o(\epsilon)$. A permutation $\sigma\ne\mathrm{id}$ moves at least two indices, so its product contains at least two off-diagonal entries, each of size $O(\epsilon)$, and the product is $O(\epsilon^2)$. Hence $\det B = 1 + \epsilon\operatorname{tr}A + o(\epsilon)$, and $(\det Y(t+\epsilon) - \det Y(t))/\epsilon = (\operatorname{tr}A + o(1))\det Y(t)$ by multiplicativity of $\det$. The first step, $Y(t+\epsilon) = Y + \epsilon AY + o(\epsilon) = (I + \epsilon A + o(\epsilon)Y^{-1})Y$, uses $Y(t)$ invertible. For the flow $Y(0) = I$, and the set of $t$ where $\det Y(t)\ne0$ is open and contains $0$; on it the formula gives $\det Y = \exp\int\operatorname{tr}A$, which stays away from $0$ on bounded intervals, so the set is also closed and is everything.

Integrating. $\det Y$ solves the scalar linear equation $u' = a(t)u$ with $u(0) = 1$, whose unique solution (unit 00) is $\exp\int_0^ta$. With $A(t) = DX(\varphi_tz)$, $\operatorname{tr}A = \operatorname{div}X(\varphi_tz)$.

The name. Joseph Liouville proved the formula in 1838 in a note on the "variation of arbitrary constants", and applied it to mechanics in the same year; Abel's identity (1827) is its second-order case. The mechanical theorem next is also Liouville's, and was proved by Jacobi in the form used today.

14 · Liouville

Liouville's theorem, in three lines

Liouville's theorem

Let $H\in C^2$ and $D$ a bounded region of phase space. Then the flow of $H$ preserves volume: $$\operatorname{vol}\varphi_t(D) = \operatorname{vol}D\quad\text{for all } t.$$

proof

1. $\operatorname{div}X_H = \sum_i\Big(\dfrac{\partial}{\partial q_i}\dfrac{\partial H}{\partial p_i} - \dfrac{\partial}{\partial p_i}\dfrac{\partial H}{\partial q_i}\Big) = 0$ by Clairaut.
2. By Liouville's formula, $\det D\varphi_t\equiv e^0 = 1$.
3. By change of variables, $\operatorname{vol}\varphi_t(D) = \int_D\lvert\det D\varphi_t\rvert = \operatorname{vol}D$. ∎

Step 3's change-of-variables formula is proved in unit 05, outside this unit's needs: cited here.

a second proof of step 2

$\operatorname{div}X_H = \operatorname{tr}(JS)$ and $\operatorname{tr}(JS) = \operatorname{tr}\big((JS)^{\!\top}\big) = \operatorname{tr}(-SJ) = -\operatorname{tr}(JS)$. Or: $\varphi_t$ is canonical, and in the plane symplectic already means $\det = 1$.

in which coordinates

In $(q,p)$. In $(q,\dot q)$ it fails whenever the kinetic matrix $M(q)$ varies: $\mathrm dp = M\,\mathrm d\dot q + \dots$ has Jacobian $\det M(q)$, which changes along the orbit.

limits

Damping $\dot p = -V' - \gamma p$: $\operatorname{div} = -\gamma$, volume $\times e^{-\gamma t}$ (units 02, 06). $\gamma = 0$ recovers the theorem ✓.

Step 1 in full. $X_H = (\partial H/\partial p, -\partial H/\partial q)$, so $\operatorname{div}X_H = \sum_i\partial_{q_i}(\partial_{p_i}H) + \sum_i\partial_{p_i}(-\partial_{q_i}H)$, and each pair cancels for $H\in C^2$ by Clairaut (unit 03). The trace argument on the right is the same fact without coordinates: the trace of a product of an antisymmetric and a symmetric matrix is zero.

Symplectic implies $\det = 1$ in general. From $M^{\!\top}JM = J$ one gets $\det M = \pm1$ at once. That it is $+1$ follows for flows from continuity in $t$ (it starts at $1$ and cannot jump), which is all that is needed here; for an arbitrary symplectic matrix it is cited (the Pfaffian identity $\operatorname{Pf}(M^{\!\top}JM) = \det M\operatorname{Pf}J$; Arnold, §41).

The converse, in the plane. A divergence-free $C^1$ vector field $(f,g)$ on a simply connected planar domain is Hamiltonian: $\partial_qf + \partial_pg = 0$ says the field $(-g, f)$ has zero curl, so by unit 04's Poincaré lemma $(-g,f) = \nabla H$ for some $H$, i.e. $f = \partial_pH$, $g = -\partial_qH$. In the plane, "area-preserving" and "Hamiltonian" are the same thing. In $2n\ge4$ they are not: volume preservation is one condition, symplecticity is $n(2n-1)$.

The warning about $(q,\dot q)$. The map $(q,\dot q)\mapsto(q, p = M(q)\dot q)$ has Jacobian determinant $\det M(q)$. Volume in $(q,p)$ is conserved; so volume in $(q,\dot q)$ at time $t$ is the original times $\det M(q_0)/\det M(q_t)$ locally, which is not $1$ when $M$ depends on $q$ — for Kepler in polar coordinates, $\det M = \mu^2r^2$. The momentum is the variable in which the statement is true.

14 · Liouville

What a cloud of initial conditions does

Free particle, $q\mapsto q + pt/m$, with $m = 1$. The rectangle $[0,1]\times[1,2]$ becomes a parallelogram of the same area $1$: the top edge, moving faster, has been sheared ahead.

three flows, one area

Free: $\begin{pmatrix}1&t/m\\0&1\end{pmatrix}$, $\det = 1$ — pure shear; the cloud grows ever longer and thinner.

Oscillator: $\begin{pmatrix}\cos\omega t&\frac{\sin\omega t}{m\omega}\\-m\omega\sin\omega t&\cos\omega t\end{pmatrix}$, $\det = 1$ — it turns, and every period it comes back.

Nonlinear (pendulum, Kepler): each orbit has its own period, so the cloud is sheared along the orbits and wound into a filament of constant area.

The figure's numbers. $m = 1$, $t = 3$: the corners $(0,1),(1,1),(1,2),(0,2)$ go to $(3,1),(4,1),(7,2),(6,2)$. The base is $1$, the height $1$: area $1$ ✓. Problem P9 follows such a rectangle with real numbers for a cloud of atoms, and finds that the spread of momentum at a fixed position shrinks as $1/t$ even though the total spread does not — a statement that only makes sense because area is conserved.

The oscillator matrix. The solution with $q(0) = q_0$, $p(0) = p_0$ is $q = q_0\cos\omega t + (p_0/m\omega)\sin\omega t$, $p = -m\omega q_0\sin\omega t + p_0\cos\omega t$; the determinant is $\cos^2 + \sin^2 = 1$. In the coordinates $(\sqrt{m\omega}\,q, p/\sqrt{m\omega})$ it is a rigid rotation, and the rectangle turns without deforming; in $(q,p)$ it deforms but keeps its area (problem P10).

14 · Liouville

A cloud in the pendulum, wound into a filament

$H = p^2/2 - \cos q$. A disc of radius $0.25$ about $(0, 1.5)$, and the same set of states at $t = 15$ and $t = 45$, computed by integrating its boundary. Its area stays $0.1963 = \pi(0.25)^2$ while it is drawn out along the orbits, the outer ones lagging.

Each orbit inside the separatrix has its own period, longer the closer it is to the separatrix. So the cloud's outer edge falls behind its inner edge, turn after turn, and the disc is wound into a spiral of ever thinner arms.

no attractors

A Hamiltonian flow has no asymptotically stable equilibrium and no attracting limit cycle: a ball in the basin would have to shrink.

How the figure was computed. The boundary circle was followed by a fourth-order Runge–Kutta integration with step $0.005$, inserting new boundary points wherever two neighbours drifted more than $0.025$ apart; the $t = 45$ curve needs about $1350$ of them. The enclosed area, by the shoelace formula, is $0.19632$ at $t = 0$, $15$ and $45$, against $\pi/16 = 0.19635$ for the exact disc — the difference is the polygon's, not the flow's. Energies in the disc run from $-0.219$ to $0.531$, amplitudes from $77^\circ$ to $122^\circ$, and periods from $7.08$ to $8.74$ (unit 00's $4K(\sin\tfrac{\theta_0}{2})$ with $l = g = 1$): in $45$ time units the fastest states make $6.4$ turns and the slowest $5.1$, so the cloud's two edges end up more than a full turn apart.

Full proof of "no attractors"

Suppose $z_0$ is asymptotically stable: there is a ball $B$ around it with $\varphi_t(z)\to z_0$ for every $z\in B$. Given $\delta > 0$, the functions $1[\varphi_t(z)\notin B_\delta(z_0)]$ tend to $0$ pointwise on $B$ as $t\to\infty$, so by dominated convergence (cited: any real-analysis text, e.g. Rudin, Real and Complex Analysis, 1.34) $\operatorname{vol}\{z\in B:\varphi_t(z)\notin B_\delta\}\to0$. Hence $\operatorname{vol}\varphi_t(B)\le\operatorname{vol}B_\delta + o(1)$, which is less than $\operatorname{vol}B$ once $\delta$ is small and $t$ large — contradicting Liouville. The same argument, with a thin tube around the cycle, excludes attracting limit cycles. Unit 02's damped pendulum spirals into its down position precisely because friction breaks the theorem.

14 · Liouville

Why statistical mechanics is possible

the Liouville equation

Let a probability density $\rho_0$ on phase space be carried by the flow: the probability of $\varphi_t(D)$ at time $t$ equals that of $D$ at time $0$. Then $\rho(\varphi_tz, t) = \rho_0(z)$, and $$\frac{\partial\rho}{\partial t} + \{\rho, H\} = 0.$$

proof

Equal probabilities in equal volumes (Liouville) force equal densities, $\rho(\varphi_tz,t) = \rho_0(z)$. Differentiate in $t$ with the time-evolution theorem: $\partial_t\rho + \{\rho,H\} = 0$ at $\varphi_tz$, which is any point. ∎

stationary clouds

$\rho$ is time-independent iff $\{\rho, H\} = 0$ — iff $\rho$ is conserved. So any $\rho = g(H)$ is an equilibrium distribution: uniform on a shell of energy (Gibbs's microcanonical), or $\propto e^{-H/k_BT}$ (canonical).

why this needed Liouville

"Equally likely" means "uniform in some measure", and it must be a measure the dynamics does not change. Liouville says phase-space volume is one. In $(q,\dot q)$ it is not, so there "uniform" would not stay uniform.

The proof, with the volumes. For a small region $D$ around $z$, probability $\approx\rho_0(z)\operatorname{vol}D$ at time $0$ and $\approx\rho(\varphi_tz,t)\operatorname{vol}\varphi_t(D)$ at time $t$. Equality of probabilities with $\operatorname{vol}\varphi_t(D) = \operatorname{vol}D$ gives the densities equal, in the limit of shrinking $D$ (for continuous $\rho$). Then $0 = \frac{\mathrm d}{\mathrm dt}\rho(\varphi_tz,t) = \partial_t\rho + \{\rho,H\}$, by the time-evolution theorem applied to $f = \rho$.

The ensembles. Gibbs (1902) built statistical mechanics on exactly this: a stationary density on phase space, of the form $g(H)$, chosen by further argument — uniform on an energy shell for an isolated system, $e^{-H/k_BT}$ for a system in contact with a heat bath. Liouville's theorem does not say which $g$ nature uses, or why a system out of equilibrium approaches one; that is ergodic theory and the second law, and it is not in this course.

A caution about "filaments". The fine-grained density is carried unchanged by the flow for ever, so its information content never changes. What changes is the density averaged over cells of any fixed size: the filaments of the previous slide become finer than the cells, and the averaged density spreads out. That is the old Gibbs picture of how a reversible dynamics can look irreversible.

14 · Liouville

Why a method for orbits should be symplectic

Oscillator with $m = \omega = 1$, time step $h$: $\dot q = p$, $\dot p = -q$.

Euler's method

$q' = q + hp$, $p' = p - hq$: the step is $\begin{pmatrix}1&h\\-h&1\end{pmatrix}$, with determinant $1 + h^2$. Every step multiplies area — and $q^2 + p^2$, i.e. the energy — by $1 + h^2$.

At $h = 0.01$: $628.3$ steps per period, energy up by a factor $1.06484$ per period: $6.5\%$, orbit after orbit.

symplectic Euler

Update $p$ first and use it: $p' = p - hq$, $q' = q + hp'$. The step is $\begin{pmatrix}1-h^2&h\\-h&1\end{pmatrix}$, with $M^{\!\top}JM = J$ and $\det = 1$: a canonical map.

It conserves exactly the nearby quadratic $q^2 - hqp + p^2$, so the energy error stays of size $h$ for ever — no drift.

the moral

Same cost, same order of accuracy per step. One respects Liouville and one does not, and over many periods that is the only difference that matters.

The numbers. $\lVert(q + hp, p - hq)\rVert^2 = (1+h^2)(q^2+p^2)$, so the energy $\tfrac12(q^2+p^2)$ grows by exactly $1 + h^2$ per step. One period is $2\pi/h = 628.32$ steps at $h = 0.01$, and $(1.0001)^{628.32} = 1.06484$. The amplitude grows by the square root, $1.03191$ per period. After a hundred periods the energy is up by a factor $535$; at $h = 0.1$ it is up by $1.869$ in a single period.

The symplectic step. Substituting $p' = p - hq$ into $q' = q + hp'$ gives $q' = (1-h^2)q + hp$. With $M = \begin{pmatrix}1-h^2&h\\-h&1\end{pmatrix}$, $\det M = 1 - h^2 + h^2 = 1$, which in the plane is the whole symplectic condition. The invariant: seek $S$ symmetric with $M^{\!\top}SM = S$; the solution is $S \propto\begin{pmatrix}1&-h/2\\-h/2&1\end{pmatrix}$, i.e. the quadratic form $q^2 - hqp + p^2$ is conserved exactly (problem P12 verifies it). Its level sets are ellipses within $O(h)$ of the true circles, so the numerical orbit stays on a closed curve near the true one.

Honest limits. A symplectic method conserves a modified Hamiltonian close to $H$ — here exactly, in general to all orders of an asymptotic series (cited: Hairer, Lubich and Wanner, Geometric Numerical Integration, 2nd ed., chapter IX). It does not make the phase along the orbit accurate; that error still grows linearly. Methods of this kind are why long integrations of the Solar System, over hundreds of millions of years, are trustworthy at all.

14 · Liouville

Euler and symplectic Euler, drawn

$h = 0.2$, exaggerated so the shapes show. Euler's points spiral out to radius $1.87$ in one period, $32$ steps; symplectic Euler's points over five periods all lie on one ellipse, $q^2 - hqp + p^2 = 1$, tilted off the dashed circle.

$h = 0.01$, the previous slide's step. Euler's energy climbs by $1.06484$ per period, to $1.874$ after ten; symplectic Euler's wobbles inside $[0.995, 1.005]$ and never drifts.

How the figures were made. Both are the two maps of the previous slide iterated from $(q,p) = (1,0)$, every point plotted (left) or every 25th step (right); nothing is fitted. On the left Euler's radius after $32$ steps is $(1 + h^2)^{16} = 1.873$, and its energy grows by $(1.04)^{31.4} = 3.43$ per period; on the same scale the symplectic points stay on the ellipse to rounding error, the invariant $q^2 - hqp + p^2$ never leaving $1$ in sixteen digits. On the right the Euler curve is $(1.0001)^n$, and after ten periods $(1.0001)^{6283} = 1.874$.

Where the symplectic band comes from. On the invariant ellipse $q^2 + p^2 = 1 + hqp$, and $|qp|\le(q^2+p^2)/2$, so $q^2 + p^2$ lies between $1/(1 + h/2)$ and $1/(1 - h/2)$: $0.99502$ and $1.00503$ at $h = 0.01$. The energy error is of size $h/2$, and it is the same after a billion periods as after one — the phase error is what grows, as the honest-limits note on the previous slide says.

14 · Liouville

Worked: Halley's phase area, and its period

the phase area of Halley's curve

Step 1 — a theorem. For $H = p^2/2m + V(x)$ with closed orbits, the area enclosed by the orbit of energy $E$, $A(E) = \oint p\,\mathrm dx$, satisfies $\mathrm dA/\mathrm dE = T(E)$, the period. (Proof in the notes.)

Step 2 — the area. With $p_r^2 = 2\mu E + 2\mu k/r - L^2/r^2 = 2\mu|E|(r - r_-)(r_+ - r)/r^2$ and the substitution $r = a - c\cos\phi$, $a = \tfrac{r_++r_-}{2}$, $c = \tfrac{r_+-r_-}{2}$ (notes): $$A(E) = 2\int_{r_-}^{r_+}\!p_r\,\mathrm dr = 2\pi\Big(k\sqrt{\frac{\mu}{-2E}} - L\Big).$$

Step 3 — differentiate. $$T = \frac{\mathrm dA}{\mathrm dE} = \pi k\sqrt{\frac\mu2}\,(-E)^{-3/2},$$ which is Kepler's third law $T = 2\pi\sqrt{\mu a^3/k}$ with $a = k/2|E|$ ✓.

Step 4 — Halley, per unit mass: $A/\mu = 2\pi\big(GM_\odot/\sqrt{2|\varepsilon|} - h\big) = 8.81646\times10^{16}\ \mathrm{m^2\,s^{-1}}$, and $T = \pi GM_\odot/\big(\sqrt2\,|\varepsilon|^{3/2}\big) = 75.3052$ yr.

checks

Dimensions: $[A] = [p][r] = \mathrm{J\,s}$, an action; $[\mathrm dA/\mathrm dE] = \mathrm s$ ✓.

Limit $E\to E_{\min} = -\mu k^2/2L^2$: $k\sqrt{\mu/2|E_{\min}|} = L$, so $A\to0$: the curve shrinks to the circular orbit ✓.

Limit $E\to0^-$: $A\to\infty$ and $T\to\infty$ — the separatrix encloses no finite area ✓.

Unit 08: $75.31$ yr ✓.

Full proof of $\mathrm dA/\mathrm dE = T$

$A(E) = 2\int_{x_-(E)}^{x_+(E)}p_E(x)\,\mathrm dx$ with $p_E = \sqrt{2m(E - V)}$, assuming simple turning points ($V'(x_\pm)\ne0$). For $E' > E$, exact algebra gives $p_{E'} - p_E = 2m(E'-E)/(p_{E'}+p_E)$. So $$\frac{A(E') - A(E)}{E'-E} = 2\int_{x_-(E)}^{x_+(E)}\frac{2m\,\mathrm dx}{p_{E'} + p_E} + \frac{2}{E'-E}\int_{\text{ends}}p_{E'}\,\mathrm dx,$$ where "ends" are the two intervals between $x_\pm(E)$ and $x_\pm(E')$. On them $p_{E'}\le\sqrt{2m(E'-E)}$ and their length is $O(E'-E)$ (because $V'(x_\pm)\ne0$), so the second term is $O(\sqrt{E'-E})\to0$. The first integrand increases to $m/p_E$ as $E'\downarrow E$ and is dominated by it, and $m/p_E$ is integrable because $p_E$ vanishes like a square root at simple turning points; by monotone convergence (cited: Rudin, Real and Complex Analysis, 1.26) the first term tends to $2\int m\,\mathrm dx/p_E = 2\int\mathrm dx/|\dot x| = T$. $E' < E$ is the same. $\square$

Full computation of $A(E)$

$2\mu Er^2 + 2\mu kr - L^2 = -2\mu|E|(r-r_-)(r_+-r)$ with $r_-+r_+ = k/|E|$ and $r_-r_+ = L^2/2\mu|E|$ (Vieta). So $\int_{r_-}^{r_+}p_r\,\mathrm dr = \sqrt{2\mu|E|}\int_{r_-}^{r_+}\sqrt{(r-r_-)(r_+-r)}\,\mathrm dr/r$. Put $r = a - c\cos\phi$, $\phi\in[0,\pi]$: $(r-r_-)(r_+-r) = c^2\sin^2\phi$ and $\mathrm dr = c\sin\phi\,\mathrm d\phi$, so the integral is $\int_0^\pi c^2\sin^2\phi\,\mathrm d\phi/(a - c\cos\phi)$. Write $c^2\sin^2\phi = (a - c\cos\phi)(a + c\cos\phi) - (a^2 - c^2)$: the integral becomes $\int_0^\pi(a + c\cos\phi)\,\mathrm d\phi - (a^2-c^2)\int_0^\pi\mathrm d\phi/(a - c\cos\phi) = \pi a - \pi\sqrt{a^2-c^2}$, using $\int_0^\pi\mathrm d\phi/(a - c\cos\phi) = \pi/\sqrt{a^2-c^2}$ (substitute $t = \tan(\phi/2)$: the integral becomes $\int_0^\infty 2\,\mathrm dt/\big((a-c) + (a+c)t^2\big) = \pi/\sqrt{(a-c)(a+c)}$). With $a = k/2|E|$ and $a^2 - c^2 = r_-r_+ = L^2/2\mu|E|$: $\int p_r\,\mathrm dr = \pi\big(k\sqrt{\mu/2|E|} - L\big)$. Double it. $\square$

The numbers. $2|\varepsilon| = 4.97479\times10^{7}$, root $7053.22$; $GM_\odot/7053.22 = 1.88159\times10^{16}$; minus $h = 0.478403\times10^{16}$ gives $1.40318\times10^{16}$; times $2\pi$: $8.81646\times10^{16}\ \mathrm{m^2\,s^{-1}}$. The period: $|\varepsilon|^{3/2} = 1.24056\times10^{11}$, so $T = \pi\times1.32712\times10^{20}/(1.41421\times1.24056\times10^{11}) = 2.37645\times10^{9}$ s $= 75.3052$ yr (Julian year, $365.25$ d). A finite-difference derivative of $A$ in $\varepsilon$ agrees to six figures.

Why the area is the natural variable. $A/2\pi$ is the radial action, and unit 17 uses action variables as the coordinates in which integrable motion is uniform rotation. That $T = \mathrm dA/\mathrm dE$ is a first taste: the frequency is the derivative of the energy with respect to the action.

14 · Liouville

Why the area's derivative is the period

Kepler at fixed $L$, in the units of Kepler in the $(r,p_r)$ half-plane: the orbits $E = -0.30$ and $-0.24$, and the ring between. The cuts mark twelve equal times along the inner orbit. Near the Sun the ring is thin and crossed fast; far out it is wide and crossed slowly; each piece has area $\approx\Delta E\cdot T/12$.

step 1, in a picture

Across the ring $H$ rises by $\Delta E$, so its width is $\Delta E/\lVert\nabla H\rVert$. Along it the motion runs at speed $\lVert X_H\rVert = \lVert\nabla H\rVert$. So a time $\mathrm dt$ sweeps area $\frac{\Delta E}{\lVert\nabla H\rVert}\cdot\lVert\nabla H\rVert\,\mathrm dt = \Delta E\,\mathrm dt$, the same everywhere, and the ring is $\Delta A \approx T\,\Delta E$.

and the cloud

$T$ depends on $E$, so a cloud of comets spread in energy shears along the orbit, as the pendulum's cloud did: fixed area, ever longer filament. Halley's meteor streams are such a filament.

The numbers in the figure. In these units ($r$ in $r_0$, $v_r$ in $GM_\odot/h$, time in $h^3/(GM_\odot)^2$) the area formula of the previous slide is $A(E) = 2\pi\big(1/\sqrt{-2E} - 1\big)$ and the period $T(E) = \pi/\big(\sqrt2\,(-E)^{3/2}\big)$. So the ring's area is $A(-0.24) - A(-0.30) = 0.9574$, and $T\,\Delta E$ at the middle energy is $15.83\times0.06 = 0.9500$: equal to $0.8\%$, the error of using one period for a ring whose two edges have periods $13.52$ and $18.89$. The cuts are the inner orbit's positions at multiples of $13.52/12$, each joined to the outer orbit along $\nabla H$.

What the picture does not prove. It treats the ring as thin, and ignores the two ends where the orbit meets $p_r = 0$: there $\nabla H$ is horizontal and the turning points themselves move as $E$ changes. The proof in the notes of the previous slide handles exactly that, and shows the end contributions vanish as $\Delta E\to0$. Halley's own curve is not drawn because it is $30$ times longer than it is wide; the mechanism is the same.

14 · close

Kepler in phase space, closed

the running example, start to finish

1. Legendre-transforming unit 08's Lagrangian gave $H = p_r^2/2\mu + p_\theta^2/2\mu r^2 - k/r$, and $\theta$ cyclic froze $p_\theta = L$.

2. At Halley's $L$ the $(r,p_r)$ portrait is a contour map: a centre at $r_0 = 1.15279$ AU, Halley's curve from $0.586025$ to $35.0789$ AU and $26.829\ \mathrm{km\,s^{-1}}$ wide, the separatrix $E = 0$ from $0.576396$ AU to infinity.

3. $\{L_i,L_j\} = \varepsilon_{ijk}L_k$, $\{L_i,A_j\} = \varepsilon_{ijk}A_k$, $\{A_i,A_j\} = -2\mu H\varepsilon_{ijk}L_k$: $\mathfrak{so}(4)$, $\mathfrak e(3)$, $\mathfrak{so}(3,1)$ as $E$ crosses zero.

4. Polar coordinates came from $F_2 = P_r r + P_\theta\theta$, and gave the same $H$.

5. Halley's curve encloses $8.81646\times10^{16}\ \mathrm{m^2\,s^{-1}}$ per unit mass, and $\mathrm dA/\mathrm dE$ is its $75.3052$-year period.

What this unit hands on

15 — the motion is a canonical transformation; find the one that trivialises it, and its $F_2$ is the Hamilton–Jacobi $S$.

17 — brackets in involution, action variables like $A/2\pi$, and area-preserving maps.

26, 27 — the Legendre transform of a Lagrangian with bounded velocities, and its degenerate twin.

34 — $\omega = \sum\mathrm dp_i\wedge\mathrm dq_i$ as a differential form, and the generating-function theorem as $\mathrm d^2 = 0$.

Every number above was computed from unit 08's two perihelion measurements and $GM_\odot$ and nothing else: $r_0 = h^2/GM_\odot$; the turning points from the quadratic $2\varepsilon r^2 + 2GM_\odot r - h^2 = 0$; the width $\sqrt{2(\varepsilon + (GM_\odot)^2/2h^2)}$; the separatrix's perihelion $h^2/2GM_\odot$; the area $2\pi(GM_\odot/\sqrt{2|\varepsilon|} - h)$; the period $\pi GM_\odot/(\sqrt2\,|\varepsilon|^{3/2})$.

What is proved and what is cited. Proved on the slides or in these notes: the Legendre transform's convexity, Young, involutivity for differentiable convex functions, smooth duality, Hamilton's equations, the phase-space action principle, non-crossing, the bracket's properties including Jacobi, the time-evolution theorem and Poisson's theorem, all the Kepler brackets, the non-isomorphism of $\mathfrak{so}(4)$ and $\mathfrak{so}(3,1)$, the three-tests theorem, the generating-function theorem, Liouville's formula and theorem up to the change of variables, the Liouville equation, $\mathrm dA/\mathrm dE = T$ and the area of a Kepler orbit. Cited: Fenchel–Moreau in general, the variational equation, the change-of-variables formula (unit 05), the linear Darboux theorem, the dominated and monotone convergence theorems, and the modified-Hamiltonian theory of symplectic integrators.

Reading. Arnold, Mathematical Methods of Classical Mechanics, chapter 3 (§§14–16: Legendre transform, Hamilton's equations, Liouville) is at this level and in this spirit; chapter 8 (§§37–40) has the symplectic structure and brackets but uses differential forms from chapter 7, which this course reaches only in unit 34 — read §§37–40 for the statements and skip the proofs that use forms. Goldstein, Poole and Safko, Classical Mechanics, chapters 8–9, for the generating-function tables and many worked canonical transformations; §9.8 on the Runge–Lenz algebra uses this unit's bracket convention but a different normalisation of $\vec A$, so the constants in front of the brackets are not this unit's. Rockafellar, Convex Analysis, §12, for the Legendre–Fenchel transform in full generality.

14 · checkpoint

Done when

checkpoint

You can, on paper and without the deck,

prove a given transformation canonical by two independent routes — two of $M^{\!\top}JM = J$, the fundamental brackets, a generating function;
perform a Legendre transform including its degenerate case, saying for a linear or degree-one $f$ exactly what $f^*$ is; and
prove Liouville's theorem from Hamilton's equations in three lines.

Falsifiable: two routes that are secretly one, a degenerate case answered "$H$ is undefined", or a Liouville proof assuming a linear flow, each fails. Record it with ./mr done 14.

Problems

./mr problems 14 — 13 problems, about 5 hours, hints and solutions folded under each.

15 Hamilton–Jacobi · 16 tensors · 17 integrability and chaos · 26 Minkowski space.

Try the checkpoint cold. For the first part, take $Q = \arctan(m\omega q/p)$, $P = (p^2 + m^2\omega^2q^2)/2m\omega$ and check your routes against problem P7: $\{Q,P\} = 1$ and $F_1 = \tfrac12m\omega q^2\cot Q$. For the second, $f(v) = \tfrac12 mv^2 + cv$ (answer: $(p-c)^2/2m$) and $f(v) = c\,v$ (answer: $0$ at $p = c$, $+\infty$ elsewhere — the momentum is pinned). For the third, the three lines are on the Liouville slide; the second line needs Liouville's formula, and knowing why is part of the checkpoint.

If the first part stalls, it is usually because the brackets were computed as the entries of $MJM^{\!\top}$ from the same Jacobian — correct, but not independent of $M^{\!\top}JM = J$. Independence means the second route starts from something else: a generating function, or the brackets computed from the formulas for $Q$ and $P$ directly. If the third stalls, reread the determinant slide.