Lecture 5 found angle coordinates on one torus, from its period lattice, and a frequency vector (5.9). It said nothing about how the tori fit together, and its angles are paired with nothing. So the question is

The answer is the action–angle variables of Theorem 6.5. The actions are the loop integrals of $p\cdot\dd q$ over the basis cycles, the angles are conjugate to them, and the frequencies are $\partial K/\partial I$ (Corollary 6.4). The worked example is the librating pendulum (Proposition 6.7), whose frequency, unlike the oscillator’s, depends on its action.

We work near one such torus $\Sigma$, at the value $f$. The phase space is $\R^{2n}$ or a cotangent bundle, as Proposition 5.12 requires; all three examples below are. One fact about the neighbourhood is imported.

Standing assumption (cited: Arnold, §50 A, the problem there). For $f'$ near $f$, the component of $\phase_{f'}$ near $\Sigma$ is again a torus satisfying the hypotheses of Theorem 5.7, and the base point $x_0$ and the lattice basis $e_i$ can be chosen to depend smoothly on $f'$. The smooth lattice basis is the implicit function theorem applied to $\Phi_t(x_0)=x_0$; that the nearby components are tori is the content of Arnold’s problem. We write out neither.

This is the integral (2.8) that Lecture 2 promised. Lecture 2 and Lecture 3 settled the first two of the three things it needs, and the next proposition supplies the third.

A different basis of $\Lambda$ gives different cycles and different actions: the new basis is $e'_i=\sum_jU_{ij}e_j$ with $U$ an integer matrix of determinant $\pm1$, and $I'=UI$. There is no preferred choice in general. What never changes is what the actions are for: they are functions of $F$ whose derivatives are the periods.

In practice nobody computes the lattice first. If the Hamilton–Jacobi equation separates in some coordinates $q_1,\dots,q_n$ — if $S=\sum_iS_i(q_i)$, so that on the level set each $p_i$ is a function of its own $q_i$ and of the constants $f$ only — then the torus is a product of closed curves $C_i$ in the separate $(q_i,p_i)$-planes. Going once round $C_i$ with the other coordinates frozen is a cycle on the torus, and the actions are one-dimensional integrals, $$I_i=\frac{1}{2\pi}\oint_{C_i}p_i\,\dd q_i .$$(6.6) The recipe for the rest of the lecture is: choose the cycle; integrate; invert to get $K(I)$; differentiate to get $\freq$.

$H=\tfrac12(p^{2}+\freq^{2}q^{2})$, the oscillator of Lecture 1. The level set at $E>0$ is an ellipse with semi-axes $\sqrt{2E}/\freq$ and $\sqrt{2E}$, traversed clockwise. By Proposition 6.2 and the sign count of §3.4 the action along the motion is the enclosed area over $2\pi$ (Figure 6.2): $$I=\frac{1}{2\pi}\,\pi\,\frac{\sqrt{2E}}{\freq}\sqrt{2E}=\frac{E}{\freq}, \qquad K(I)=\freq I,\qquad \frac{\dd K}{\dd I}=\freq .$$ The angle: on the upper branch $S=\int^{q}\sqrt{2\freq I-\freq^{2}q^{2}}\,\dd q$, so $\varphi=\partial S/\partial I=\arcsin\big(q\sqrt{\freq/2I}\big)$, that is, $$q=\sqrt{2I/\freq}\,\sin\varphi,\qquad p=\sqrt{2\freq I}\,\cos\varphi .$$ The Jacobian of $(\varphi,I)\mapsto(q,p)$ is $\cos^{2}\varphi+\sin^{2}\varphi=1$, so $\dd q\wedge\dd p=\dd\varphi\wedge\dd I$: canonical, as Theorem 6.5 promised. This is the reference case: the frequency does not depend on the action. Lecture 1 observed that the oscillator’s period is the same on every ellipse; now we see that is the statement $K''(I)=0$, and the pendulum breaks it.

$H=\tfrac12p^{2}-\cos\theta$ on the cylinder $S^{1}\times\R$ ((1.2), (1.1)). With one degree of freedom every regular compact level-set component is a circle and the orbit itself is the cycle, so there is no choice of basis to make. The difficulty is different: the cycle is not the same kind of loop at every energy (Figure 6.3).

• $-1<E<1$, libration: one closed curve round $(0,0)$. It bounds a disc on the cylinder, and $2\pi I$ is the area it encloses.

• $E>1$, rotation: two components, $p>0$ and $p<0$, each going once round the cylinder. Neither bounds anything. Deforming the curve $p=\sqrt{2(E+\cos\theta)}$ down to the zero section $p=0$, on which $p\,\dd\theta$ vanishes, sweeps out the strip beneath it, and $2\pi I$ is the area of that strip — the difference of the two integrals, by Stokes on the strip, both loops taken in the direction of motion. The branch $p<0$ is its mirror image and has the same action: it runs the other way round the cylinder with $p<0$, so $p\,\dd\theta>0$ again.

• $E=1$ is the separatrix, where the level set is not a circle, and $E=-1$ the equilibrium.

The two regimes cannot be merged into one formula: they integrate over loops of different kinds.

Both regimes lead to the same two integrals, and they are defined here, not assumed.

What we need about them: $\mathrm K(0)=\mathrm E(0)=\pi/2$; $\mathrm E(1)=1$; and as $k\to1$, with $k'=\sqrt{1-k^{2}}$, $$\mathrm K(k)=\ln\frac{4}{k'}+o(1)$$(6.8) (cited: DLMF 19.12.1). $\mathrm K$ diverges because at $k=1$ the integrand becomes $1/\cos\psi$.

As $E\to\infty$, $k\to0$ and $I\approx\sqrt{2E}$, $T\approx2\pi/\sqrt{2E}$: a free rotor with $p=\sqrt{2E}$, which does not feel gravity. In both regimes the reader can differentiate (6.9) and (6.11) and find $\dd I/\dd E=T/2\pi$, as Corollary 6.4 says it must.¹

At $E=1$ the level set is the separatrix $p=\pm2\cos(\theta/2)$: the unstable equilibrium $(\pi,0)$, where $\dd H=0$, and two orbits that leave it and return to it. It is not a circle, and Theorem 5.7 does not apply: hypothesis (I) fails at the equilibrium. Approaching it from either side the tori still exist, and two things happen to them. (A third, the jump of the action visible in Figure 6.4, is a reading at the end of this section.)

The period diverges, logarithmically (Figure 6.5). With (6.8): below, $k'^{2}=(1-E)/2$ and $T=4\mathrm K\approx4\ln(4/k')$; above, $k'^{2}=(E-1)/(E+1)\approx(E-1)/2$ and $T=2k\mathrm K\approx2\ln(4/k')$. So $$T=2\ln\frac{32}{1-E}+o(1)\quad(E\uparrow1),\qquad T=\ln\frac{32}{E-1}+o(1)\quad(E\downarrow1).$$(6.12) The coefficient on the rotation side is half the libration one. The reason is visible in Figure 6.3: almost all of the time is spent near the saddle, where the motion is slowest, and a libration loop passes near it twice per period — at $\theta\approx\pi$ and at $\theta\approx-\pi$, which on the cylinder are the same point — while a rotation loop passes once.

The frequency goes to zero. $\freq=2\pi/T\to0$: the torus degenerates into a set on which the motion takes infinitely long to go round.

The action jumps (Figure 6.4). As $E\uparrow1$, $k\to1$ and $I\to8/\pi$, which is $1/2\pi$ times the area of the eye, $\int_{-\pi}^{\pi}4\cos(\theta/2)\,\dd\theta=16$. As $E\downarrow1$, $I\to4/\pi$, one strip of area $8$. One libration torus becomes two rotation tori, and the area is shared between them. The function $I$ is not continuous across the separatrix because the cycle is not.

Why a logarithm. Near the saddle put $\theta=\pi+x$. Then $H\approx\tfrac12p^{2}-\tfrac12x^{2}+1$, the inverted oscillator, whose orbits are the hyperbolas $p^{2}-x^{2}=2(E-1)$. The time to cross a fixed box $\abs{x}\le\delta$ along the hyperbola is $\int\dd x/\sqrt{x^{2}+2(E-1)}\approx\ln\big(\delta^{2}/\abs{E-1}\big)$, which diverges like $\ln(1/\abs{E-1})$; the time spent outside the box stays bounded. One crossing per rotation, two per libration — the factor $2$ in (6.12). The constant $32$ needs the exact integral: it is where (6.8) is used.

The theorem says the motion is a straight line on a torus. It does not say the motion is periodic, and the difference is the subject of Lecture 7.

A change of lattice basis replaces $\freq$ by $B\freq$ with $B$ an integer matrix of determinant $\pm1$, and $k\cdot\freq=0$ becomes $(B^{-\mathsf T}k)\cdot(B\freq)=0$: resonance is a property of the torus, not of the basis. For $n=2$ it says exactly that $\freq_1/\freq_2$ is rational (or a frequency vanishes).

For general $n$ the same holds: a non-resonant orbit is dense in $T^{n}$, and with exactly $r$ independent integer relations $k\cdot\freq=0$ the closure of every orbit is a torus of dimension $n-r$ (cited: Arnold, §51 B and §51 D). So a torus is not an orbit: on a non-resonant torus a single trajectory visits every neighbourhood and never returns exactly, and only when $r=n-1$ (and $\freq\ne0$) is the torus foliated by closed orbits (Figure 6.6).

In the oscillator every torus has the same $\freq=(\freq_1,\freq_2)$, so they are all resonant or all not. That is special. Generically $\freq$ depends on $f$, resonant and non-resonant tori are interleaved, and Lecture 7 asks which of them survive a perturbation.

The other side of the coin is the frequency map. Call an integrable system non-degenerate if $I\mapsto\freq(I)$ is a local diffeomorphism, $\det(\partial^{2}K/\partial I^{2})\ne0$. For $n\ge2$ the resonant tori, the preimages of the rational directions, are then dense but form a set of measure zero (Figure 6.7); for $n=1$ resonance means $\freq=0$, and there may be none at all. Lecture 7 needs exactly this picture. Kepler, in the reading at the end of this lecture, is as degenerate as possible: its frequency map has rank one, and every torus is resonant.

• Free motion in the right variables. In $(\varphi,I)$ the motion is $\dot I=0$, $\dot\varphi=\partial K/\partial I$ (Theorem 6.5): the first of the course’s five ideas, P1, made precise near one torus.

• Frequencies from energies. $\freq=\partial K/\partial I$ (Corollary 6.4), and with one degree of freedom $T=2\pi\,\dd I/\dd E$. The pendulum’s period follows from its action, and diverges at the separatrix.

• The picture Lecture 7 perturbs. A non-degenerate frequency map makes the resonant tori dense and of measure zero, and Lecture 7 asks which tori survive a small perturbation.

• Status. The standing assumption about the neighbouring tori is cited (Arnold, §50 A). Proposition 6.2 is proved modulo Stokes. Theorem 6.5 is sketched, with the patching at turning points omitted. The rotating pendulum’s formulas (6.11) are stated, with their derivation an exercise, and the asymptotics (6.8) and the general-$n$ density of non-resonant orbits are cited. Proposition 6.3, Corollary 6.4 and Proposition 6.7 are proved; the proof of Proposition 6.10 is in a reading.

• The choice of cycles. A different basis of the period lattice gives different actions, related by an integer matrix of determinant $\pm1$ (a reading). The motion, and whether a torus is resonant, do not change.

• Local, not global. Action–angle variables are built near one torus. They break down where the tori degenerate, as at the pendulum’s separatrix, and they need not exist globally (optional lecture A).

A particle in $\R^{3}\setminus\{0\}$ with $H=\tfrac12\abs{p}^{2}-\alpha/\abs{q}$, $\alpha>0$. In spherical coordinates $(r,\theta,\phi)$ — $\theta$ the polar angle, not the pendulum’s, and $\phi$ the azimuth, to be told apart from the angle variables $\varphi$ — $$H=\frac12\Big(p_r^{2}+\frac{p_\theta^{2}}{r^{2}}+\frac{p_\phi^{2}}{r^{2}\sin^{2}\theta}\Big)-\frac{\alpha}{r}.$$(6.13) This is a central force, integrable with $F=(H,\abs{\ell}^{2},\ell_3)$ (Example 5.6), and in these coordinates $$\ell_3=p_\phi,\qquad \abs{\ell}^{2}=p_\theta^{2}+\frac{p_\phi^{2}}{\sin^{2}\theta}.$$ Write $\ell=\abs{\ell}$. Fix $E<0$ and $0<\abs{\ell_3}<\ell<\alpha/\sqrt{-2E}$ (the last so that the radial motion is not a circle). On the level set $$p_\phi=\ell_3,\qquad p_\theta=\pm\sqrt{\ell^{2}-\frac{\ell_3^{2}}{\sin^{2}\theta}},\qquad p_r=\pm\sqrt{2E+\frac{2\alpha}{r}-\frac{\ell^{2}}{r^{2}}},$$(6.14) and each momentum depends only on its own coordinate: Hamilton–Jacobi separates, $S=S_r(r)+S_\theta(\theta)+\ell_3\phi$. The torus is a product of three circles — $r$ oscillating between the roots $r_-<r_+$ of $p_r=0$, $\theta$ between the roots of $p_\theta=0$, and $\phi$ going round — and (6.6) gives three actions (Figure 6.8): $$I_\phi=\frac1{2\pi}\oint p_\phi\,\dd\phi=\ell_3,\qquad I_\theta=\frac1{2\pi}\oint p_\theta\,\dd\theta=\ell-\abs{\ell_3},\qquad I_r=\frac1{2\pi}\oint p_r\,\dd r .$$ The first is immediate. The second is a one-variable integral of the same kind as the next one; it is an exercise, and we use it as stated.

Adding the three actions, $I_r+I_\theta+\abs{I_\phi}=\alpha/\sqrt{-2E}$, so $$K(I)=-\frac{\alpha^{2}}{2\big(I_r+I_\theta+\abs{I_\phi}\big)^{2}} .$$(6.15) The energy depends on the actions only through their sum. By (6.3), with $\mathcal J=I_r+I_\theta+\abs{I_\phi}$, $$\freq_r=\freq_\theta=\abs{\freq_\phi}=\frac{\alpha^{2}}{\mathcal J^{3}}$$(6.16) (with $\freq_\phi$ of the sign of $I_\phi$). All three frequencies are equal: every torus is resonant, in the sense of Definition 6.9, as resonant as it can be. The radial oscillation, the swing in $\theta$ and the turn in $\phi$ all close at the same moment, so every bounded Kepler orbit with $\ell\ne0$ is closed: $\varphi(t)=\varphi(0)+\freq(1,1,\pm1)\,t$ returns to its start after $2\pi/\freq$. (The tori excluded above, $\ell_3=0$ or $\abs{\ell_3}=\ell$, differ only in the choice of polar axis, and the same holds there; circular orbits are closed anyway. A bounded orbit with $\ell=0$ is radial and falls into the centre in finite time.) The common period is $T=2\pi\mathcal J^{3}/\alpha^{2}$; with the semi-major axis $a=\tfrac12(r_-+r_+)=\alpha/(-2E)$ this is $T=2\pi a^{3/2}/\sqrt\alpha$, Kepler’s third law.

In celestial mechanics the actions are taken as $(\mathcal J,\ell,\ell_3)$ — the Delaunay variables — whose conjugate angles are the mean anomaly, the argument of pericentre and the longitude of the node. Since $K$ depends on $\mathcal J$ alone, only the first angle moves. We name them and do not use them.

Add a small $\beta/r^{2}$ to the potential. It only replaces $\ell^{2}$ by $\ell^{2}+2\beta$ in $p_r$, so the proof of Proposition 6.11 goes through unchanged and $$I_r=\frac{\alpha}{\sqrt{-2E}}-\sqrt{\ell^{2}+2\beta},\qquad K=-\frac{\alpha^{2}}{2\big(I_r+\sqrt{\ell^{2}+2\beta}\big)^{2}} .$$ The motion is still in a plane, and $\ell$ is the action of the angle in it. Now $\partial K/\partial\ell=\freq_r\,\ell/\sqrt{\ell^{2}+2\beta}$: the angular frequency differs from the radial one. In one radial period the orbit turns by $2\pi\ell/\sqrt{\ell^{2}+2\beta}$, not $2\pi$, so the pericentre moves on every pass, and unless the ratio is rational the orbit never closes; it fills the annulus $r_-\le r\le r_+$ densely (Arnold, §8 C) (Figure 6.9). The degeneracy (6.16) is the whole reason Kepler orbits close, and it is fragile.

Equal frequencies are the footprint of extra integrals. For Kepler the extra one is the Laplace–Runge–Lenz vector $$A=p\times\ell-\alpha\,\frac{q}{\abs{q}} .$$(6.17) It is conserved, $\pb{A_i}{H}=0$ (an exercise); it lies in the orbital plane, $\ell\cdot A=0$; and it points from the centre to the pericentre, with length $\abs{A}=\alpha e$ (Figure 6.10). A conserved vector pointing at the pericentre is exactly the statement that the pericentre does not move.

Count the integrals. $H$, the three $\ell_i$ and the three $A_i$ are seven functions on a six-dimensional phase space, tied by two relations, $$\ell\cdot A=0,\qquad \abs{A}^{2}=\alpha^{2}+2H\,\abs{\ell}^{2},$$ and $5$ of them are independent. The Jacobian has rank $5$ at one sample point,² so some $5\times5$ minor is not identically zero; the minors are real-analytic on the connected set $q\ne0$, and such a function vanishing on an open set vanishes identically, so the minor is nonzero on a dense open set.

Kepler is maximally superintegrable: $5=2\cdot3-1$. The oscillator with equal frequencies is the other standard example.

The extra integrals explain the closed orbits a second way. Fix $E<0$ and $\ell\ne0$. Then $\ell$ fixes the orbital plane and the sense of motion, $A$ the eccentricity $e=\abs{A}/\alpha$ and, if $e>0$, the direction of the pericentre, and $H$ the size: one ellipse in $q$-space. Over each of its points $p$ is fixed as well, by $p=\ell\times(A+\alpha q/\abs{q})/\abs{\ell}^{2}$, which inverts (6.17) because $p\perp\ell$. So the joint level set of $(H,\ell,A)$ is a single closed curve in phase space. $\hvf{H}$ does not vanish on it ($p\ne0$ since $\ell\ne0$), so the orbit is the whole curve and is periodic. The integrals do not all commute: $\pb{\ell_i}{A_j}=\sum_k\epsilon_{ijk}A_k$ and $\pb{A_i}{A_j}=-2H\sum_k\epsilon_{ijk}\ell_k$ (an exercise), so at most $n=3$ of them can be the integrals of Theorem 5.7 at once.

Which central potentials share Kepler’s luck? Bertrand’s theorem: among smooth central potentials that have bounded orbits at all, the only ones all of whose bounded non-collision orbits are closed are, up to an additive constant, $-\alpha/r$ with $\alpha>0$ and $\tfrac12\freq^{2}r^{2}$ (cited: Arnold, §8 D, through a sequence of problems).

The old quantum theory quantised the actions, $$\oint_{\gamma_i}p\cdot\dd q=2\pi\hbar\,n_i,\qquad n_i\in\Z, \qquad\text{that is,}\qquad I_i=\hbar\,n_i .$$(6.18) For the oscillator it gives $E=\hbar\freq n$, which misses the zero-point energy $\tfrac12\hbar\freq$; for Kepler, (6.15) with $\mathcal J=\hbar n$, $n=1,2,\dots$, gives $E=-\alpha^{2}/2\hbar^{2}n^{2}$, the Bohr levels. The correction $n\mapsto n+\tfrac12$ at each turning-point cycle, and the reason (6.18) works as well as it does, belong to Part II. What matters now is its shape: quantisation is a condition on the actions, and Lecture 10 will find the Bethe equations to be this condition for a many-body system.