Hamilton–Jacobi and action-angle methods

Start with a free particle

Supply a point particle of constant mass m = 2 kg with no force. Use a fixed inertial straight coordinate q, positive to the right. Momentum is p = mv, where v = dq/dt is signed velocity. Choose q(0) = −1 m and p(0) = 4 kg m/s. The resulting velocity is 2 m/s, so on 0 ≤ t ≤ 2 s the position is q(t) = −1 m + (2 m/s)t.

For this model the Hamiltonian is the energy-valued function H(q, p) = p²/(2m), with potential energy zero. Hamilton’s equations read dq/dt = ∂H/∂p and dp/dt = −∂H/∂q. A partial derivative changes its named input while holding the other input fixed. Here the two equations give dq/dt = p/m and dp/dt = 0; the energy is 4 J.

Hamilton–Jacobi theory asks whether one function can encode the motion. Prepare with Algebra, functions, graphs, and logarithms and Geometry, trigonometry, and coordinate systems. First follow the supplied free-particle history, oscillator formulas, diagram and table. For an optional return to the partial-derivative and chain-rule checks, use Derivatives, Taylor series, and local approximation; for the action integral, use Integrals and multivariable calculus. We introduce the Hamiltonian and phase conventions locally before using the method’s derivative conditions.

Check an action function

Let S(q, t; α) = αq − α²t/(2m), where α is a constant parameter with momentum units. Treat q and t as independent inputs; hold α fixed when taking their partial derivatives. S has units J s: both momentum times position and energy times time have these units. This is a family of functions, not yet one particle history.

Write S_q = ∂S/∂q and S_t = ∂S/∂t. Differentiating the supplied family gives S_q = α and S_t = −α²/(2m). Thus H(q, S_q) + S_t = α²/(2m) − α²/(2m) = 0 for every q, t and finite α in this free model. This equation is the one-coordinate Hamilton–Jacobi equation: H(q, ∂S/∂q, t) + ∂S/∂t = 0.

The equation is a partial differential equation because S depends on more than one independent variable. In this example its position derivative supplies momentum, and its time derivative cancels the Hamiltonian. Those derivatives have different units: S_q is momentum, while S_t is energy. Finding a suitable S for a more complicated Hamiltonian can be difficult.

Recover a history from two constants

Differentiate S with respect to α while holding q and t fixed: S_α = q − αt/m. Set this expression equal to a constant β, with length units. Together, p = S_q = α and β = S_α give q(t) = β + αt/m. Choose α = 4 kg m/s and β = −1 m to recover the original history.

Here α selects momentum and β selects initial position. A single energy value would not select both: opposite momenta have the same free-particle energy, and different initial positions can share that energy. Initial conditions choose the two constants.

For a sufficiently smooth Hamiltonian and action family, the method uses p = S_q and β = S_α to recover a solution locally where the relation can be inverted. In one coordinate a nonzero mixed derivative S_qα supplies the needed local condition. Our S_qα = 1, and the inversion is explicit. This does not promise a globally invertible solution for every Hamiltonian or every boundary problem.

Separate two kinds of time change

At α = 4 kg m/s, S_t = −4 J: this partial derivative holds q fixed. Along the moving history, q changes too. The chain rule gives dS/dt = S_q(dq/dt) + S_t = (4 kg m/s)(2 m/s) − 4 J = 4 J. The minus sign in S_t therefore does not say that S decreases along this history.

The Lagrangian for this free particle is L = mv²/2 = 4 J. Integrating it from 0 to 2 s gives an action of 8 J s. The table agrees: S(q(2), 2; α) − S(q(0), 0; α) = 4 J s − (−4 J s) = 8 J s. The selected S along this solution differs from the accumulated action by its initial constant.

An action accumulated along a history, a solution family S(q, t; α), and an action value specified by endpoint data are related descriptions with different inputs. A fixed-endpoint problem requires its own boundary data and a suitable solution branch. Checking our free family alone does not settle every such problem or prove that stationary action is always a minimum.

Choose a periodic oscillator

Now supply a separate ideal undamped oscillator on a fixed inertial straight q axis. Its constant mass is m = 1 kg and its stiffness is k = 9 N/m. Use H = p²/(2m) + kq²/2 and select energy E = 2 J. The positive rate ν = √(k/m) = 3 s⁻¹ will describe numerical radian phase.

Define the phase angle as θ = φ rad: φ is the dimensionless numerical radian value, so it can be the argument of sine and cosine. Choose φ = νt with initial phase zero. Then q = A cos φ and p = −P sin φ, where A = √(2E/k) = 2/3 m and P = √(2mE) = 2 kg m/s. These supplied formulas satisfy dq/dt = p/m and dp/dt = −kq.

One full modeled cycle takes T = 2π/ν = (2π/3) s. We draw only 0 ≤ t ≤ T. The physical angular frequency is 3 rad/s; ν is its numerical radian-phase rate, not the cycle frequency. The phase-space curve records position and momentum together. Its vertical direction is momentum, not another direction in physical space.

Calculate the normalized action

For this positive-energy periodic orbit, choose the action convention I = (1/(2π))∮p dq. The loop sign ∮ means integrate once around the whole closed phase curve, following increasing φ. In our q–p axes this is clockwise. The integral adds momentum times each signed position increment; its unit is kg m²/s = J s.

Use q = A cos φ, so dq = −A sin φ dφ. With p = −P sin φ, their product is p dq = AP sin²φ dφ. Over 0 to 2π, the integral of sin²φ is π: write sin²φ = (1 − cos 2φ)/2, integrate the constant term, and note that the cosine term completes whole cycles with zero integral. Thus ∮p dq = πAP = (4π/3) J s and I = AP/2 = 2/3 J s.

Equivalently, I = E/ν for this oscillator. The loop integral is an oriented phase-space area with action units, not a physical area in square metres or the area measured in screen pixels. We fixed both its orientation and the factor 1/(2π). A convention that instead calls ∮p dq the action uses a different numerical normalization and corresponding phase scale.

Read the same states in two diagrams

The upper panel is the exact position–momentum ellipse at E = 2 J. Four different marker shapes identify the cardinal phases. The circle at φ = 0 and 2π is shared because the initial and final physical states coincide. The short leftward tangent arrow at the bottom shows increasing phase; it is not a force arrow or a speed scale.

The lower panel uses unwrapped φ horizontally and I vertically. I = 2/3 J s stays horizontal while φ progresses from 0 to 2π. Its five markers include both endpoints. Unlike the ellipse, this chosen phase interval has two displayed endpoints; it is a representation of one cycle, not a second spatial motion.

Read the units and each axis scale separately. The upper axes measure different quantity kinds, and their drawn scales need not be equal. Both panels and the table use exact model values, with no measurement uncertainty implied. The lower panel shows only this positive-action orbit; its zero tick is not an assigned phase for the equilibrium state.

Use action-angle coordinates locally

Allow I to vary over positive values while keeping m and ν fixed. Write q = √(2I/(mν)) cos φ and p = −√(2mνI) sin φ. On a local phase branch this is a canonical coordinate change: it preserves the form of Hamilton’s equations. One-coordinate confirmation is q_φp_I − q_Ip_φ = 1, where the subscripts name partial derivatives.

To check that condition, let A = √(2I/(mν)) and P = √(2mνI). Then q_φ = −A sin φ, q_I = A cos φ/(2I), p_φ = −P cos φ and p_I = −P sin φ/(2I). The difference is [AP/(2I)](sin²φ + cos²φ) = 1 because AP = 2I. This local check requires I > 0.

The transformed Hamiltonian is H = νI, independent of φ. In these coordinates Hamilton’s equations become dφ/dt = ∂H/∂I = ν and dI/dt = −∂H/∂φ = 0. That is the simplification: a constant action and uniform phase reconstruct the changing q and p. I and S both have action units, but I labels the periodic orbit while S is the action function used in the Hamilton–Jacobi equation.

Phase is understood modulo 2π, so no single real-valued phase branch covers the whole closed orbit without a cut. At E = I = 0, q = p = 0 and phase is undefined. Other Hamiltonian systems need their own regularity, invertibility and periodic-orbit checks; this oscillator calculation does not establish global action-angle coordinates for arbitrary motion. The next lesson, Rigid bodies and classical field Lagrangians, changes the kinds of systems being described.

Worked examples

Try it yourself

1. Keep the free model’s m = 2 kg and β = −1 m, but choose α = −2 kg m/s. Find velocity, q at t = 1 s, energy and S at that event.

   Show the answer and reasoning

   v = −1 m/s, q(1 s) = −2 m, H = 1 J and S = 3 J s.

   Use v = α/m and q = β + αt/m. Squaring the momentum gives H = α²/(2m) = 1 J; then S = (−2 kg m/s)(−2 m) − (1 J)(1 s) = 3 J s.

2. For the main free history, a learner claims S_t = −4 J means S decreases as the particle moves. What term is missing, and what is the correct along-history rate?

   Show the answer and reasoning

   The missing term is S_q(dq/dt) = +8 J; the along-history rate is +4 J.

   S_t holds position fixed. Along the history the position changes, so the chain rule adds (4 kg m/s)(2 m/s) to −4 J.

3. At φ = π/2 in the main oscillator, give q, p, H and I. Is the particle at rest?

   Show the answer and reasoning

   q = 0 m, p = −2 kg m/s, H = 2 J and I = 2/3 J s. It is not at rest.

   cos(π/2) = 0 and sin(π/2) = 1. The velocity is p/m = −2 m/s, so crossing the origin here does not mean zero velocity.

4. Integrate p dq around the same main oscillator ellipse in decreasing φ instead. What is that oriented integral? May it replace the increasing-phase integral without changing the stated convention?

   Show the answer and reasoning

   The reversed integral is −(4π/3) J s. It cannot replace the increasing-phase integral unchanged.

   Reversing the traversal reverses a line integral’s sign. Our positive normalized I uses the expressly chosen increasing-phase orientation; the physical orbit’s energy has not become negative.

5. Keep m = 1 kg and k = 9 N/m, but choose a different ideal oscillator orbit with E = 4 J. Find its normalized action and period. Does the original diagram show this new orbit?

   Show the answer and reasoning

   I = 4/3 J s and T = (2π/3) s. The original diagram shows only the E = 2 J orbit.

   The unchanged ν = 3 s⁻¹ gives I = E/ν and T = 2π/ν. Doubling the orbit energy changes its action, while this ideal linear model’s period stays the same.

6. For the oscillator equilibrium q = p = 0, what are H and I? Can the state select a unique phase φ?

   Show the answer and reasoning

   H = 0 J and I = 0 J s. No unique phase is selected.

   Both energy terms vanish. In q = √(2I/(mν)) cos φ and p = −√(2mνI) sin φ the amplitudes vanish at I = 0 for every φ; the positive-action phase chart fails there.

What to carry forward

• A Hamilton–Jacobi solution uses partial derivatives with specified independent inputs; its parameter constants can reconstruct motion where the required inversion works.
• A partial time derivative of S is different from the derivative along a moving history.
• State the orientation and normalization when calculating an action loop integral.
• For the selected positive-energy oscillator, H = νI gives constant action and uniform numerical radian phase.
• A phase-space diagram is not a physical-space trajectory, and periodic phase needs local branches and a separate zero-action check.