Worked examples — Action-angle variables — integrable systems

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The scenario matrix

Before we compute, let us list every cell a 1-degree-of-freedom periodic system can land in. A cell = one qualitatively different phase-space loop or one different question type.

Cell	Name	What is different about it	Example
A	Smooth well (libration)	Loop is a closed oval; $p$ has two signs $\pm$	Ex 1 (SHO)
B	Hard walls (libration)	$\Vert p\Vert$ is constant, loop is a rectangle	Ex 2 (box)
C	Power-law well	Loop is neither oval nor rectangle; $\omega$ depends on $E$	Ex 3 (quartic)
D	Pendulum below the barrier	Swings back and forth — libration, same as A but with $\sin$	Ex 4
E	Pendulum above the barrier	Whirls all the way round — rotation, a different contour	Ex 5
F	Attractive $1/r$ (bound orbit)	Loop with turning points $r_{\min },r_{\max }$; a real astronomy word problem	Ex 6
G	Degenerate / zero input	Zero energy, zero amplitude — does the recipe survive the limit?	Ex 7
H	Exam twist	A relation that looks like it needs the trajectory but does not	Ex 8

We now hit each cell in turn. Read "Forecast" and guess before scrolling — that is where the learning lives.

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Ex 1 — Cell A: the smooth well (harmonic oscillator)

The phase-space picture (why an ellipse?). The equation $\frac{p^2}{2mE}+\frac{q^2}{2E/m{\omega }_0^2}=1$ is the equation of an ellipse — a squashed circle. Its two half-widths (called semi-axes, the distance from centre to edge along each axis) are

$a=\sqrt{\frac{2E}{m{\omega }_0^2}}(\text{along }q),b=\sqrt{2mE}(\text{along }p).$

Step 1 — Solve $H=E$ for $p$. $p=\pm \sqrt{2mE-m^2{\omega }_0^2{q}^2}.$ Why this step? $J$ needs $p$ as a function of $q$; rearranging the energy equation is the only place $p$ appears. The $\pm$ is the top and bottom halves of the oval (blue = $+$, red = $-$ in the figure).

Step 2 — Get the area without integrating. The area of an ellipse is $\pi ab$ (area of a circle $\pi r^2$ stretched by $a/r$ one way, $b/r$ the other). So $\oint pdq=\pi ab=\pi \sqrt{\frac{2E}{m{\omega }_0^2}}\sqrt{2mE}=\frac{2\pi E}{{\omega }_0}.$ Why this step? $\oint pdq$ is the enclosed area — a known formula beats an integral.

Step 3 — Divide by $2\pi$ and invert. $J=\frac{1}{2\pi }\cdot \frac{2\pi E}{{\omega }_0}=\frac{E}{{\omega }_0}\Rightarrow E={\omega }_{0}J.$

Step 4 — Differentiate for the frequency. $\omega =\frac{dE}{dJ}={\omega }_0.$ Why this step? $\omega =\partial H/\partial J$; since $E=H(J)={\omega }_{0}J$ is a straight line, its slope is ${\omega }_0$ — independent of energy. This is the special property that makes the oscillator isochronous (same period at any amplitude).

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Ex 2 — Cell B: hard walls (particle in a box)

The phase-space picture. Between walls there is no force, so $|p|=\sqrt{2mE}$ is constant. The particle moves right at $+p$, hits the wall, instantly reverses to $-p$, moves back. In the $(q,p)$ plane that is a rectangle: top edge at $+p$, bottom edge at $-p$, joined by vertical jumps at the walls.

Step 1 — Momentum. $p=\sqrt{2mE}$, constant. Why? No potential means all energy is kinetic.

Step 2 — Area = the whole loop (there-and-back). $\oint pdq=\underset{\text{going right, }+p}{\underbrace{pL}}+\underset{\text{coming back, }-p\times -dq}{\underbrace{pL}}=2pL.$ Why this step? A full cycle is out and back — traversing the box twice. The bottom edge contributes a second positive $pL$ because both $p$ and $dq$ flip sign there (a negative times a negative). This is the classic factor-of-2 trap.

Step 3 — Action and inversion. $J=\frac{1}{2\pi }2pL=\frac{pL}{\pi }=\frac{L\sqrt{2mE}}{\pi }\Rightarrow E=\frac{{\pi }^2{J}^2}{2m{L}^2}.$

Step 4 — Frequency and period. $\omega =\frac{dE}{dJ}=\frac{{\pi }^{2}J}{m{L}^2}=\frac{\pi v}{L},T=\frac{2\pi }{\omega }=\frac{2L}{v},$ using $v=p/m$. Why this step? Unlike the spring, $E\propto J^2$ is a parabola, so its slope $\omega$ grows with $J$ — a faster particle bounces more often.

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Ex 3 — Cell C: a power-law well (quartic oscillator)

The idea — scaling instead of a hard integral. We do not need the messy integral; we need only how $J$ scales with $E$. Write the turning point $q_{\max }$ where $p=0$: $\alpha q_{\max }^4=E$, so $q_{\max }=(E/\alpha {)}^{1/4}$.

Step 1 — Rescale the integral. $J=\frac{1}{2\pi }\oint \sqrt{2m(E-\alpha q^4)}dq.$ Substitute $q=q_{\max }x$ so $x$ runs over a fixed range independent of $E$: $J=\frac{1}{2\pi }\oint \sqrt{2mE}\sqrt{1-x^4}{q}_{\max }dx=C\sqrt{E}{E}^{1/4}=C{E}^{3/4},$ where $C$ collects all the $E$-independent constants and $q_{\max }\propto E^{1/4}$. Why this step? Pulling every $E$ outside the integral turns an unsolvable integral into a pure power law — all we need for the frequency.

Step 2 — Invert. $J\propto E^{3/4}\Rightarrow E\propto J^{4/3}.$

Step 3 — Frequency. $\omega =\frac{dE}{dJ}\propto J^{1/3}\propto E^{1/4}.$ Why this step? $\omega$ increases with energy: a stiffer-than-spring well ($q^4$ rises faster than $q^2$) makes the particle turn around sooner, so it oscillates faster at higher amplitude. The period shrinks — the opposite of your maybe-guess.

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Ex 4 — Cell D: pendulum below the barrier (libration)

The two regimes of a pendulum. The angle $\varphi$ is measured from straight down. If total energy $E<mg\ell$ (energy at the top), the bob cannot reach the top; it swings back and forth — a libration loop (closed oval), our Cell D. If $E>mg\ell$ it whirls all the way over — Cell E (next example). The dividing energy $E=mg\ell$ is the separatrix.

Step 1 — Small-angle limit. For small $\varphi$, $\cos \varphi \approx 1-{\varphi }^2/2$, so $H\approx \frac{p^2}{2m{\ell }^2}+\frac{1}{2}mg\ell {\varphi }^2-mg\ell .$ Why this step? Near the bottom the $\cos$ well looks parabolic — this is exactly the harmonic form of Ex 1 with mass $M=m{\ell }^2$ and stiffness $k=mg\ell$.

Step 2 — Read off the frequency by analogy. For an oscillator $H=\frac{p^2}{2M}+\frac{1}{2}M{\Omega }^2{\varphi }^2$ we need $M{\Omega }^2=k$: ${\Omega }^2=\frac{k}{M}=\frac{mg\ell }{m{\ell }^2}=\frac{g}{\ell }\Rightarrow {\omega }_0=\sqrt{\frac{g}{\ell }}.$ Why this step? We already solved the harmonic case in Ex 1 ($\omega ={\omega }_0$, energy-independent); mapping onto it saves all the work.

Step 3 — Action near the bottom. Using Ex 1's result $J=E^{\prime }/{\omega }_0$ (with $E^{\prime }=E+mg\ell$ measured from the well bottom): $J\approx \frac{E+mg\ell }{\sqrt{g/\ell }}.$

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Ex 5 — Cell E: pendulum above the barrier (rotation)

Why a different contour (the whole point of this cell). In libration (Ex 4) $\varphi$ reaches a maximum, stops, and returns — the loop closes back on itself, $p$ takes both signs. In rotation, $p$ never changes sign: the bob keeps going the same way, and $\varphi$ increases by a full $2\pi$ each turn. The "one period" is one full revolution, so the contour runs $\varphi :0\to 2\pi$ once, not there-and-back.

Step 1 — Momentum stays one-signed. $p=+\sqrt{2m{\ell }^2(E+mg\ell \cos \varphi )}>0\text{for all }\varphi (\text{since }E>mg\ell ).$ Why this step? $E>mg\ell$ guarantees the bracket is always positive — the bob never runs out of kinetic energy, so it never stops. No $\pm$; just $+$.

Step 2 — The action integral, single sweep. $J=\frac{1}{2\pi }{\int }_0^{2\pi }\sqrt{2m{\ell }^2(E+mg\ell \cos \varphi )}d\varphi .$ Why this step? One full turn $=0\to 2\pi$ is exactly one period, so no factor of 2 here (contrast the box in Ex 2, which needed it). This is the single most common mistake: rotation and libration use different contours.

Step 3 — High-energy limit (fast spinning). When $E\gg mg\ell$, $\cos \varphi$ barely matters: $J\approx \frac{1}{2\pi }{\int }_0^{2\pi }\sqrt{2m{\ell }^{2}E}d\varphi =\sqrt{2m{\ell }^{2}E}=|p|,E\approx \frac{J^2}{2m{\ell }^2},$ so $\omega =\frac{dE}{dJ}=\frac{J}{m{\ell }^2}=\frac{p}{m{\ell }^2}=\dot{\varphi }$ — the frequency is just the spin rate. Why? A fast rotor ignores gravity; it behaves like a free rotor whose "period" is one revolution.

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Ex 6 — Cell F: gravity word problem (radial action of a bound orbit)

Step 1 — The radial turning points. Set $p_r=0$: the bob's radial motion reverses where all radial kinetic energy is spent. The two roots of $E=\frac{L^2}{2m{r}^2}-\frac{k}{r}$ are $r_{\min },r_{\max }$ — perigee and apogee. Between them, $r$ librates like a 1-DOF well.

Step 2 — Radial action (standard contour integral). $J_r=\frac{1}{2\pi }\oint p_{r}dr=\frac{1}{2\pi }\oint \sqrt{2m(E+\frac{k}{r}-\frac{L^2}{2m{r}^2})}dr=-L+k\sqrt{\frac{m}{-2E}}.$ Why this step? This is a known residue integral; the point for us is the result, which isolates $E$ cleanly.

Step 3 — Invert for energy. Solve for $E$: $J_r+L=k\sqrt{\frac{m}{-2E}}\Rightarrow E=-\frac{m{k}^2}{2(J_r+L{)}^2}.$ Why this step? $E$ depends on $J_r$ and $L$ only through the sum $J_r+L$ — a hint that the two frequencies will be equal (a degeneracy special to $1/r$).

Step 4 — Frequency. $\omega =\frac{\partial E}{\partial J_r}=\frac{m{k}^2}{(J_r+L{)}^3}.$ Since $E$ depends only on $J_r+L$, ${\omega }_r={\omega }_{\text{angular}}$ — the radial and angular periods coincide, so the orbit closes into a fixed ellipse (no precession). Why? This equality is exactly why Kepler orbits do not drift — a hallmark of the pure $1/r$ law.

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Ex 7 — Cell G: degenerate / zero input (does the recipe survive?)

Step 1 — Oscillator at $E\to 0$. From Ex 1, $J=E/{\omega }_0\to 0$ and $\omega ={\omega }_0$ (constant). Why this step? The loop area shrinks to zero (a point at the origin), so $J\to 0$; but the shape stays an ellipse of the same aspect ratio, so the frequency is unchanged. A point still has a well-defined frequency — the limiting frequency of infinitesimally small swings. No blow-up.

Step 2 — Box at $E\to 0$. From Ex 2, $J=L\sqrt{2mE}/\pi \to 0$ and $\omega =\pi v/L$ with $v=\sqrt{2E/m}\to 0$, so $\omega \to 0$, $T=2L/v\to \infty$. Why this step? A particle with zero speed never completes a bounce — the period diverges. This is the correct, physical degenerate limit: no motion, no period.

Step 3 — The lesson. The recipe is robust at the boundary: for the spring the degenerate limit is finite (isochronous), for the box it diverges (speed-controlled). Neither produces $0/0$ nonsense because we take limits of $J(E)$ and $\omega (E)$ that were built from areas, which vanish smoothly.

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Ex 8 — Cell H: the exam twist (frequency without solving anything)

Step 1 — Don't look for the trajectory. The whole reason action–angle exists is that $\omega =dE/dJ$ needs only the $J-E$ relation. Why this step? The parent's magic line: frequency comes from the area-vs-energy curve, never from integrating $\dot{q},\dot{p}$.

Step 2 — Invert. $J=a{E}^{2/3}\Rightarrow E=(J/a{)}^{3/2}=a^{-3/2}{J}^{3/2}.$

Step 3 — Differentiate. $\omega =\frac{dE}{dJ}=\frac{3}{2}{a}^{-3/2}{J}^{1/2}.$ Rewrite in terms of $E$ using $J=a{E}^{2/3}$, so $J^{1/2}=a^{1/2}{E}^{1/3}$: $\omega =\frac{3}{2}{a}^{-3/2}\cdot a^{1/2}{E}^{1/3}=\frac{3}{2a}{E}^{1/3},T=\frac{2\pi }{\omega }=\frac{4\pi a}{3}{E}^{-1/3}.$ Why this step? $T\propto E^{-1/3}$ decreases with energy — a stiffening well (like a milder version of the quartic in Ex 3). You solved it with algebra alone; the "impossible-looking" problem was two lines.

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Recall Self-test (cover the answers)

A pendulum with $E$ just above $mg\ell$ — which cell? ::: Cell E (rotation), but barely — near the separatrix the period diverges. Why does the box get a factor of 2 but the whirling pendulum does not? ::: Box librates (out-and-back, $\pm p$); the whirling pendulum rotates (one sweep $0\to 2\pi$, single-signed $p$). The spring's frequency is independent of $E$. Which matrix property causes that? ::: $E={\omega }_{0}J$ is a straight line, so its slope $\omega =dE/dJ$ is constant.

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Connections

• Parent topic — the recipe
• Hamiltonian mechanics — every example is a Hamiltonian $H(q,p)$
• Canonical transformations and Generating functions — the machinery turning $(q,p)$ into $(\theta ,J)$
• Hamilton–Jacobi equation — where $W(q,J)$ comes from
• Poisson brackets and Liouville–Arnold theorem — the multi-DOF (torus) generalization
• Adiabatic invariants — why $J$ is preserved under slow changes (Ex 7's shrinking loop)
• Bohr–Sommerfeld quantization — Ex 1's $E={\omega }_{0}J$ becomes the SHO spectrum
• KAM theorem — what survives when these integrable systems are perturbed