Action-angle variables — integrable systems: Question bank

Before the questions, one line so no symbol is unearned:

Definition The characters, in plain words

$q$ = the position coordinate (where the thing is). $p$ = its momentum (mass × speed for simple cases).
$H$ = the Hamiltonian, the function that gives the system's total energy in terms of $(q, p)$ ; see Hamiltonian mechanics. For time-independent systems its numerical value is conserved, and we call that conserved number the energy $E$ . So " $H = E$ " just means "the energy function currently equals the constant $E$ ."
$J = \frac{1}{2 π} \oint p d q$ = the action: the area of the closed loop the motion traces in the $(q, p)$ picture, divided by $2 π$ . Think "size of the loop." Since $J$ is built from $E$ via the loop, we write $E = H (J)$ once the loop area is expressed as a function of $J$ .
Orientation of $\oint$ : the loop is traversed in the direction the physical motion actually goes (counter-clockwise for a standard oscillator loop), so the enclosed area — and hence $J$ — comes out positive. A rotor going the other way flips the sign; by convention we take $J \geq 0$ by orienting the contour with the motion, or equivalently using $J = \frac{1}{2 π} \oint p d q$ .
$θ$ = the angle: a clock hand that ticks forward at constant rate, telling you where along the loop you are. It goes up by $2 π$ each cycle.
$ω = \partial H / \partial J$ = the frequency, how fast that clock hand ticks.
$n$ = the number of degrees of freedom (independent position coordinates the system needs); a pendulum has $n = 1$ , a planet in a plane $n = 2$ .
$F_{1}, \dots, F_{n}$ = conserved quantities (functions of $(q, p)$ whose values stay fixed along the motion — energy, momenta, angular momenta). ${F_{i}, F_{j}}$ = the Poisson bracket, a "do these two quantities interfere?" test; if it is zero they are in involution (compatible constants). See Poisson brackets.

Intuition Why

$\dot{J} = - \partial H / \partial θ$ even applies In any canonical coordinates, motion obeys Hamilton's equations — for the pair $(θ, J)$ these read $\dot{θ} = \partial H / \partial J$ and $\dot{J} = - \partial H / \partial θ$ . This is exactly why we insisted the transformation to $(θ, J)$ be canonical (via a generating function): canonicity is precisely the property that guarantees Hamilton's equations keep their form in the new variables. So once $H$ has no $θ$ , $\dot{J} = - \partial H / \partial θ = 0$ follows immediately.

True or false — justify

J

the same thing as the instantaneous momentum

p

False.

p

changes moment-to-moment along the loop;

J

is the whole loop's area over

2 π

and stays constant. They only coincide in special symmetric cases like circular orbits where

J = L

Is the action

J

always conserved along the motion?

True — but only after the transformation, because the new Hamiltonian

H (J)

has no

θ

, so Hamilton's equation

\dot{J} = - \partial H / \partial θ = 0

. It is a constant of the motion by construction, not by luck.

For the harmonic oscillator, does the frequency

ω = \partial H / \partial J

depend on the energy?

False. From

E = ω_{0} J

we get

ω = ω_{0}

, a constant. The SHO is special (isochronous); most systems have energy-dependent frequencies.

Is every bounded, periodic 1-DOF system solvable with action-angle variables?

True for one degree of freedom — any bounded 1-DOF motion is automatically integrable, since energy conservation alone traces a closed curve you can integrate.

Is every multi-DOF Hamiltonian system integrable?

False. You need

n

(the number of degrees of freedom) independent conserved quantities in involution; most systems (double pendulum, three-body) lack them and are chaotic. Integrable systems are rare exceptions.

Does

\oint d θ = 2 π

hold over exactly one period?

True — that is precisely why

J

carries the

1/2 π

factor, so the angle advances by a clean

2 π

per cycle (see the recall derivation in the parent).

If the frequencies

ω_{i}

on a torus are irrationally related, does the orbit ever close?

False. Irrational ratios make the motion quasi-periodic — it never repeats and densely fills the whole torus instead.

Are the level sets

{F_{i} = c_{i}}

of an integrable system always spheres?

False. The Liouville–Arnold theorem says the bounded level set (where the conserved

F_{i}

take fixed values

c_{i}

) is an

n

-dimensional torus

T^{n}

(a doughnut), not a sphere.

Spot the error

"To find

J

for a swinging pendulum, integrate

p d q

from

q_{m i n}

q_{m a x}

once."

Wrong. The

\oint

is the full closed loop:

q_{m i n} \to q_{m a x} \to q_{m i n}

. For libration that round trip usually gives a factor of 2 versus the one-way integral.

"Since energy is conserved,

H

already depends only on

J

in the original

(q, p)

Wrong. In raw

(q, p)

H

still depends on

q

. Only after the canonical transformation to

(θ, J)

does the

θ

-dependence vanish, giving

H = H (J)

J

has units of energy, since it comes from the energy

E

Wrong.

J = \frac{1}{2 π} \oint p d q

has units [momentum × length] = [energy × time] = angular momentum. That's exactly why it pairs cleanly with

ℏ

in Bohr–Sommerfeld.

"The angle

θ

speeds up and slows down as the particle moves around the loop."

Wrong. By construction

θ = ω t + θ_{0}

advances at constant rate. The uneven-looking motion in

(q, p)

is repackaged into steady winding — that's the whole payoff.

"For a particle bouncing in a box, one period is a single trip across the box."

Wrong. One full cycle is out and back, so

\oint p d q = 2 p L

. Missing the round-trip factor halves your action and doubles your frequency.

"Two conserved quantities always mean the system is integrable in 2 DOF."

Incomplete. You also need them in involution:

{F_{1}, F_{2}} = 0

. Conserved but non-commuting constants do not give you the torus structure.

"Action-angle variables are found by directly solving Hamilton's equations for the trajectory."

Wrong. The point is you avoid solving the trajectory: you compute an area

J (E)

, invert to

E (J)

, and differentiate — the frequency drops out without any time integration.

"The sign of

\oint p d q

doesn't matter, so

J

can be negative."

Wrong in spirit. We fix the loop's orientation to follow the actual motion (or take the absolute value), so

J \geq 0

measures a genuine area. A blindly signed integral could come out negative for a rotor circulating the "other way."

Why questions

Why divide by

2 π

in the definition of

J

So that the conjugate angle

θ

advances by a clean

2 π

per period, since

\oint d θ = \frac{d}{dJ} \oint p d q = \frac{d}{dJ} (2 π J) = 2 π

. It makes the "clock hand" close neatly.

Why is

J (E)

usually monotonic, so we can invert to

E (J)

Bigger energy

E

encloses a bigger loop, hence larger area, so

J

increases with

E

and the inverse

E = H (J)

is single-valued. Caveat: this fails exactly at the separatrix, where

J \to \infty

and the smooth

J (E)

relation breaks — so monotonic inversion holds only away from that boundary.

Why do we use a type-2 generating function

W (q, J)

rather than another type?

A type-2 generator naturally takes old coordinate

q

and new momentum

J

as its variables, delivering

p = \partial W / \partial q

and

θ = \partial W / \partial J

— exactly the pairing we want, canonical by construction.

Why does making

H

depend on

J

only immediately give the dynamics?

Because Hamilton's equations in

(θ, J)

then read

\dot{J} = - \partial H / \partial θ = 0

(action frozen) and

\dot{θ} = \partial H / \partial J = ω

= const, so the motion is just linear winding — trivially solved.

Why is

ω = d E / dJ

called "magic"?

It hands you the oscillation frequency from an area-versus-energy relation alone, with no trajectory solving. All the dynamics collapse into one derivative.

Why does the harmonic oscillator's phase orbit being an ellipse make

J

so easy?

Because

\oint p d q

is the enclosed area, and an ellipse's area is

π ab

— a known formula. No integration needed, just geometry.

Edge cases

For a rotor (angle

q

increasing monotonically through

2 π

, not swinging back), what changes in

\oint p d q

The contour is different:

q

runs once through

0 \to 2 π

with

p

not reversing sign, so there is no there-and-back factor of 2. Orientation matters here — circulating the opposite way flips the sign, so you take the loop along the actual motion to keep

J \geq 0

At the separatrix (pendulum with just enough energy to reach the top), what happens to the period and

ω

The period diverges to infinity and

ω \to 0

, because the motion asymptotically stalls at the unstable top. Here

J (E)

ceases to be monotonic (

J \to \infty

) and action-angle variables break down right on the separatrix.

What is the action of a genuinely free particle (unbounded, no walls)?

Undefined — the motion is not periodic, so there is no closed loop to enclose an area. Action-angle machinery needs bounded motion; add walls (the box example) to make it periodic.

If two frequencies on a torus satisfy

ω_{1} / ω_{2} = 2/3

exactly, does the orbit close?

Yes. A rational ratio means the winding repeats after a finite number of turns, so the trajectory closes into a knot on the torus rather than filling it.

What happens to a KAM torus as you turn on a small perturbation?

Sufficiently irrational tori survive (slightly deformed), while rational/resonant ones break up first — the near-integrable regime the KAM theorem describes.

Connections

Liouville–Arnold theorem — supplies the torus that every "edge case" here lives on.
Canonical transformations and Generating functions — the machinery behind "why type-2 $W (q, J)$ ."
Poisson brackets — the involution test in the integrability traps.
Adiabatic invariants and Bohr–Sommerfeld quantization — where the unit of $J$ pays off.
KAM theorem — the fate of the last edge-case torus.

Recall One-line self-test

Cover everything above: can you state why the round-trip factor of 2 appears, why $ω = d E / dJ$ needs no trajectory, and why most systems are not integrable? ::: Round-trip = the $\oint$ is the whole closed loop (out and back); $ω = d E / dJ$ because area-vs-energy already encodes the frequency; non-integrable because $n$ commuting constants of motion almost never exist.