Exercises — Action-angle variables — integrable systems

WHAT: $\oint pdq$ is literally the geometric area enclosed by the loop — that is the entire content of the symbol (exactly the shaded region in Figure s01). WHY: by Green's theorem $\oint pdq=\iint dqdp=\text{area}$. A circle of radius $R$ has area $\pi R^2$, so $\oint pdq=\pi R^2,J=\frac{1}{2\pi }\oint pdq=\frac{R^2}{2}.$ No calculus needed — recognition of "action = area / $2\pi$" is the whole skill here.

(i) Below the critical energy → libration. The angle $\theta$ swings between $\pm {\theta }_{\max }$ and reverses — the round-trip contour is ${\theta }_{\min }\to {\theta }_{\max }\to {\theta }_{\min }$ (picks up the usual factor of 2). (ii) Above the critical energy → rotation. Here $\theta$ keeps increasing and never reverses, so one period is $\theta :0\to 2\pi$ (the pendulum returns to the same physical state after a full turn, traversed once). (iii) Exactly at the critical energy → the separatrix, an infinite-period edge case. At the knife-edge energy the pendulum does not complete a $2\pi$ turn in finite time: it asymptotically creeps toward the unstable top ($\theta \to \pi$) and takes infinitely long to arrive, so $T\to \infty$ and $\omega \to 0$. This curve in phase space is called the separatrix — it is the boundary between the libration region (closed ovals inside) and the rotation region (open waves outside), and it is not a genuine periodic orbit at all. Action-angle variables break down right on it (the frequency vanishes), which is exactly why perturbation theory is hardest near separatrices.

Step A — get $p(q,E)$. Solve $H=E$ for $p$: $p=\pm \sqrt{2mE-m^2{\omega }_0^2{q}^2}$. Step B — recognise the ellipse. Divide $H=E$ by $E$: $\frac{q^2}{2E/(m{\omega }_0^2)}+\frac{p^2}{2mE}=1,$ an ellipse with semi-axes $a=\sqrt{2E/(m{\omega }_0^2)}$ along $q$ and $b=\sqrt{2mE}$ along $p$. Step C — area / $2\pi$. Ellipse area $=\pi ab=\pi \sqrt{\frac{2E}{m{\omega }_0^2}}\sqrt{2mE}=\frac{2\pi E}{{\omega }_0}$. Hence $J=\frac{1}{2\pi }\cdot \frac{2\pi E}{{\omega }_0}=\frac{E}{{\omega }_0}\Rightarrow E={\omega }_{0}J.$ Step D — differentiate. $\omega =\frac{dE}{dJ}={\omega }_0.$ ✔ The frequency dropped out with no trajectory solved.

First, the definition tie-in: comparing $H=\frac{1}{2}m{\omega }_0^2{q}^2$ (from L2.1) with $H=\frac{1}{2}k{q}^2$ shows the potential coefficient is $k=m{\omega }_0^2$; this $k$ is a fixed property of the spring, while ${\omega }_0=\sqrt{k/m}$ depends on the mass hanging on it. From L2.1, $J=E/{\omega }_0$. Doubling $m$ gives ${\omega }_0^{\prime }=\sqrt{k/(2m)}={\omega }_0/\sqrt{2}$. (a) $J^{\prime }=E/{\omega }_0^{\prime }=\sqrt{2}E/{\omega }_0=\sqrt{2}J$ — the loop encloses $\sqrt{2}$ times more area at the same energy (heavier ⇒ larger momentum at each $q$). (b) $\omega ={\omega }_0^{\prime }={\omega }_0/\sqrt{2}$ — slower, as expected for a heavier bob.

Step A: $p=\sqrt{2mE}$, constant in magnitude (no forces between walls). Step B — the loop. In phase space the orbit is a rectangle (see Figure s02): top edge $p=+\sqrt{2mE}$ (moving right, $q:0\to L$), bottom edge $p=-\sqrt{2mE}$ (moving left, $q:L\to 0$). Both legs contribute $+pL$: $\oint pdq=pL+(-p)(-L)=2pL.$ Step C: $J=\frac{1}{2\pi }2pL=\frac{pL}{\pi }=\frac{L\sqrt{2mE}}{\pi }$. Invert: $E=\frac{{\pi }^2{J}^2}{2m{L}^2}$. Step D: $\omega =\frac{dE}{dJ}=\frac{{\pi }^{2}J}{m{L}^2}=\frac{\pi v}{L}$ with $v=p/m$; period $T=\frac{2\pi }{\omega }=\frac{2L}{v}$. ✔ Exactly the round-trip bounce time.

WHAT/WHY — scaling instead of brute force. We want the $E$-dependence, so use a substitution that pulls all the $E$'s outside the integral. $J=\frac{1}{2\pi }\oint \sqrt{2m(E-\alpha q^4)}dq.$ Fix the contour first. The motion is libration: $q$ runs between the two turning points where $p=0$, i.e. $E-\alpha q^4=0\Rightarrow q=\pm q_{\max }$ with $q_{\max }=(E/\alpha {)}^{1/4}$. The closed loop is $-q_{\max }\to +q_{\max }\to -q_{\max }$. Substitute $q=q_{\max }u={\left(\frac{E}{\alpha }\right)}^{1/4}u$, so $q^4=\frac{E}{\alpha }{u}^4$ and $dq={\left(\frac{E}{\alpha }\right)}^{1/4}du$. As $q$ runs from $-q_{\max }$ to $+q_{\max }$, the new variable $u$ runs from $-1$ to $+1$; the round-trip (there-and-back) doubles it. Then $J=\frac{1}{2\pi }\oint \sqrt{2mE}\sqrt{1-u^4}{\left(\frac{E}{\alpha }\right)}^{1/4}du=\underset{\text{a pure number }C(\text{no }E,\text{finite, }>0)}{\underbrace{\frac{\sqrt{2m}}{2\pi {\alpha }^{1/4}}\cdot 2{\int }_{-1}^1\sqrt{1-u^4}du}}{E}^{1/2+1/4}.$ Here $C$ is just a fixed positive number: the dimensionful constants $m,\alpha$ times the definite integral ${\int }_{-1}^1\sqrt{1-u^4}du\approx 1.748$ (a plain number, the area under one hump of $\sqrt{1-u^4}$), with the extra factor 2 for the return leg. It carries no energy dependence, which is the whole point. So $J=C{E}^{3/4}$, i.e. $E=(J/C{)}^{4/3}\propto J^{4/3}$. Differentiate: $\omega =\frac{dE}{dJ}\propto J^{1/3}\propto (E^{3/4}{)}^{1/3}=E^{1/4}$. Interpretation: unlike the SHO (whose $\omega$ is energy-independent because the well is quadratic), a steeper well squeezes higher-energy orbits faster ⇒ higher frequency. This energy-dependence of $\omega$ is exactly what makes anharmonic systems non-isochronous.

WHY no factor of 2: rotation means $\varphi$ increases monotonically through $2\pi$; the contour is traversed once, not there-and-back. $J=\frac{1}{2\pi }\oint p_{\varphi }d\varphi =\frac{1}{2\pi }{p}_{\varphi }\cdot 2\pi =p_{\varphi }.$ Here the action equals the momentum — a special symmetric case (this is the honest version of the "$J=p$" temptation). Then $E=p_{\varphi }^2/2I=J^2/2I$ and $\omega =\frac{dE}{dJ}=\frac{J}{I}=\frac{p_{\varphi }}{I}=\dot{\varphi }.$ The angle-variable frequency equals the actual angular velocity, as it must for uniform circular motion.

Set up: $\oint pdq=2pL=nh$. With $h=2\pi \hslash$, this is $J=\frac{2pL}{2\pi }=\frac{pL}{\pi }=n\hslash$, so $p=\frac{n\pi \hslash }{L}$. Energy: $E_n=\frac{p^2}{2m}=\frac{n^2{\pi }^2{\hslash }^2}{2m{L}^2}$. Compare: this matches the exact answer identically — the box is one of the rare cases where semiclassical quantization is exact (no zero-point offset, because the walls are hard). Contrast with the SHO, where Bohr–Sommerfeld gives $E_n=\hslash {\omega }_0(n+\frac{1}{2})$: the $+\frac{1}{2}$ comes from the smooth turning points (Maslov correction), absent for hard walls.

Key idea: slow ⇒ $J$ stays fixed. From L2.3, $J=\frac{L\sqrt{2mE}}{\pi }$, so $J^2=\frac{L^2\cdot 2mE}{{\pi }^2}\propto L^{2}E$. Holding $J$ constant: $L_0^2{E}_0=L_1^2{E}_1\Rightarrow \frac{E_1}{E_0}={\left(\frac{L_0}{L_1}\right)}^2.$ Sanity check: squeezing the box ($L_1<L_0$) raises the energy — the particle speeds up, exactly like a ball bouncing between an approaching wall gaining speed. If $L_1=L_0/2$, then $E_1/E_0=4$.

Picture the torus surface unrolled into a unit square whose left/right edges are glued and whose top/bottom edges are glued (Figure s03). A point at $({\theta }_1/2\pi ,{\theta }_2/2\pi )$ moves in a straight line of slope ${\omega }_2/{\omega }_1$; when it leaves one edge it re-enters the opposite edge. (a) Rational ratio ⇒ closed orbit. Write ${\omega }_1=3\Omega$, ${\omega }_2=2\Omega$ with $\Omega$ a common base rate. The state repeats when both angles have advanced by whole multiples of $2\pi$: ${\theta }_1$ needs time $2\pi /{\omega }_1$ per turn, ${\theta }_2$ needs $2\pi /{\omega }_2$. The common period is the least $T$ with ${\omega }_{1}T=2\pi p$ and ${\omega }_{2}T=2\pi q$ for integers $p,q$. Smallest: $p=3,q=2$ gives $T=\frac{2\pi \cdot 3}{{\omega }_1}=\frac{2\pi }{\Omega }$. The curve closes after ${\theta }_1$ has looped 3 times and ${\theta }_2$ has looped 2 times — a (3,2) torus knot (left panel of Figure s03). (b) Irrational ratio ⇒ never closes. No integers $p,q$ satisfy $\sqrt{2}=p/q$, so the orbit never exactly repeats; instead it winds forever and densely fills the whole torus surface (quasi-periodic motion — right panel of Figure s03). (c) KAM link (one line): the KAM theorem says that under small perturbations of an integrable system, tori with sufficiently irrational frequency ratios survive (deform but persist), while those with rational or near-rational ratios are the first to break into chaos.

Invert. Set $S\equiv J_r+L$. Then $k\sqrt{\frac{m}{2|E|}}=S$, so $\frac{m{k}^2}{2|E|}=S^2$, giving $|E|=\frac{m{k}^2}{2S^2}\Rightarrow E=-\frac{m{k}^2}{2(J_r+L{)}^2}.$ Differentiate for the frequency. Since $E$ depends only on the sum $J_r+L$, the two frequencies are equal: ${\omega }_r=\frac{\partial E}{\partial J_r}=\frac{m{k}^2}{(J_r+L{)}^3}=\frac{\partial E}{\partial L}={\omega }_{\varphi }.$ Meaning: ${\omega }_r={\omega }_{\varphi }$ means the radial and angular motions share one frequency — the orbit closes after exactly one radial oscillation per revolution ⇒ a closed ellipse (no precession). This "accidental" degeneracy of the Kepler problem is why only $1/r$ (and $r^2$) potentials give closed orbits — the deep reason planetary ellipses don't slowly rotate. Finish the quantization (Bohr–Sommerfeld). Semiclassically each action is an integer multiple of $\hslash$: write the radial quantum number $n_r=J_r/\hslash$ and the angular one $\ell =L/\hslash$, so the sum $S/\hslash =n_r+\ell \equiv n$ is a single principal quantum number ($n=1,2,3,\dots$). Substitute $J_r+L=n\hslash$: $E_n=-\frac{m{k}^2}{2(n\hslash {)}^2}=-\frac{m{k}^2}{2{\hslash }^2{n}^2}.$ With the Coulomb coupling $k=\frac{e^2}{4\pi {\epsilon }_0}$ this is exactly the hydrogen spectrum $E_n=-\frac{m{e}^4}{2(4\pi {\epsilon }_0{)}^2{\hslash }^2{n}^2}=-\frac{13.6\text{eV}}{n^2}$. The degeneracy ${\omega }_r={\omega }_{\varphi }$ we found above is the classical shadow of the fact that hydrogen levels depend only on $n$ (not on $\ell$ separately) — the celebrated $\ell$-degeneracy.

The action is the phase-space area divided by ==$2\pi$. For libration you pick up a factor of 2 because the loop goes there and back; for rotation the contour is traversed once. At exactly the critical pendulum energy the orbit is the separatrix, whose period is infinite== (frequency $\to 0$). The frequency is obtained by ==differentiating $E(J)$== (i.e. $\omega =dE/dJ$). A quantity conserved under slow parameter changes is called an adiabatic invariant, and the conserved thing is ==$J$== (not the energy). On an $n$-torus, an irrational frequency ratio makes the orbit densely fill the torus (quasi-periodic); a Diophantine ratio is one that is hard to approximate by fractions and so survives perturbation.