advanced38 min studyLesson 135 in Physics

The Hamilton–Jacobi equation

Replace trajectory equations with a generating-function equation, recover motion from its constants, and recognize local and global limits.

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What you’ll understand

  • Derive the Hamilton–Jacobi equation using K = 0.
  • Recover a free trajectory from a complete principal function.
  • Distinguish a complete integral from an endpoint action and a global single-valued solution.

Instead of solving directly for a trajectory, imagine finding a generating function that transforms the motion into constant coordinates. Hamilton–Jacobi theory makes that proposal precise. The unknown is a function of configuration and time rather than a separate curve for each initial state. Its derivatives recover momenta and integration constants. This can simplify separable problems and expose the connection between action, wavefronts, and mechanics, but solving its partial differential equation is not automatically easier than solving the original motion.

Ask for a transformation with a vanishing new Hamiltonian

Use a type-2 generating function S whose new momenta are constants denoted α. Setting the transformed Hamiltonian K to zero gives a first-order partial differential equation for S. Old momentum is the position gradient of S, and the new constant coordinates are derivatives with respect to α. A complete integral contains n independent α parameters for n degrees of freedom, with suitable nondegeneracy so the transformation actually provides the needed initial data locally.

The phrase principal function is sometimes used for several closely related action functions. A complete integral parameterized by constants is not identical to the action between a fixed initial endpoint and a variable final endpoint, although both can satisfy Hamilton–Jacobi equations in appropriate domains. Keep track of which endpoint or parameter is fixed before differentiating. Without those distinctions, one can confuse the initial momentum label with a spatial coordinate or mistake one selected family of paths for all solutions.

∂S∂t+H ⁣(q,∂S∂q,t)=0,pi=∂S∂qi,βi=∂S∂αi\frac{\partial S}{\partial t}+H\!\left(q,\frac{\partial S}{\partial q},t\right)=0,\qquad p_i=\frac{\partial S}{\partial q_i},\quad \beta_i=\frac{\partial S}{\partial\alpha_i}
S(q,α,t) is used as a type-2 generator. When the mixed-coordinate transformation is nonsingular, α and β are constants because K = 0.↔ Scroll wide equations horizontally.

Solve a free-particle complete integral

For a free particle, choose S linear in position with slope α. The Hamilton–Jacobi equation then fixes the time-dependent term. Differentiating with respect to position says that α is the constant momentum. Differentiating with respect to α gives a constant β that is the initial position at the chosen time origin. Solving that algebraic relation for q reconstructs the familiar straight trajectory. The method has moved integration into a generating function rather than introduced a new force law.

When the Hamiltonian is autonomous, a common separation is S=W(q,α)−EtS=W(q,\alpha)-Et. The reduced function W satisfies a time-independent equation involving energy E. This does not mean that W alone records the entire physical time dependence; the omitted linear time term and the derivatives with respect to the constants are what recover timing. At turning points, branches of the momentum can meet and a single expression for W may need to be replaced by several local branches.

H=p22m,S(q,α,t)=αq−α2t2m,β=q−αtmH=\frac{p^2}{2m},\qquad S(q,\alpha,t)=\alpha q-\frac{\alpha^2t}{2m},\qquad \beta=q-\frac{\alpha t}{m}
This complete integral has p = α and q = β + αt/m. The independent constant β labels the initial position.↔ Scroll wide equations horizontally.

An endpoint action has a useful local geometry

Fix a free particle at the origin at time zero and ask for the action to reach q at positive time t. The classical velocity is q divided by t, so its action is a quadratic function of q. Its position derivative gives the final momentum. Its time derivative, with the final position held fixed, is minus the final energy. This is an especially clear case where the meaning of held fixed matters: following one trajectory while differentiating would be a different operation.

In more complicated systems, different classical paths can reach the same endpoint. Their actions define separate branches, and caustics can make a single-valued local gradient description fail. Hamilton–Jacobi theory therefore supplies local structure and powerful separability tools without guaranteeing a global smooth solution. Its resemblance to short-wavelength wave phases is meaningful, but this classical equation alone does not include wave interference, tunneling probabilities, or quantum measurement. Those require additional physical structure and controlled approximation.

Send(q,t)=mq22t,∂Send∂q=mqt,∂Send∂t=−mq22t2,t>0S_{\mathrm{end}}(q,t)=\frac{mq^2}{2t},\qquad \frac{\partial S_{\mathrm{end}}}{\partial q}=\frac{mq}{t},\qquad \frac{\partial S_{\mathrm{end}}}{\partial t}=-\frac{mq^2}{2t^2},\quad t>0
This endpoint action fixes the initial event at q = 0, t = 0. It is not the same parameterization as the complete integral S(q,α,t).↔ Scroll wide equations horizontally.

Recover a trajectory from two constants

For a free particle of mass m=2m=2, use the complete integral with α=6\alpha=6 and β=−1\beta=-1. Find q and p at t = 4 and verify the Hamilton–Jacobi equation.

  1. Differentiate S=αq−α2t/4S=\alpha q-\alpha^2t/4 with respect to q. This gives p=α=6p=\alpha=6, so the physical velocity is p/m=3p/m=3.

  2. Differentiate with respect to α: β=q−αt/2\beta=q-\alpha t/2. Therefore q=β+αt/2=−1+6(4)/2=11q=\beta+\alpha t/2=-1+6(4)/2=11. At t = 0 the same relation returns q = −1.

  3. At fixed q and α, St=−α2/4=−9S_t=-\alpha^2/4=-9. The Hamiltonian evaluated at p=Sq=6p=S_q=6 is p2/4=9p^2/4=9. Their sum is zero, as required; the chosen trajectory also has constant energy 9.

The result

At t = 4, the particle has q = 11 and p = 6. The constants specify initial position and conserved momentum independently.

Explore the model

Interactive lab

Interactive lab · function

Inspect a quadratic endpoint action

For normalized m = 1 and the positive endpoint time t = 1, set a = 0.5 in y = ax² and interpret x as q. Predict that moving x from 0.5 to 1 doubles final momentum and quadruples endpoint action, then inspect value and tangent slope. The lab varies a and x, not physical time automatically, and its area readout is not the mechanical action. It illustrates this free endpoint-action slice rather than solving a general Hamilton–Jacobi equation.

Inspect a quadratic endpoint actionThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn action function's gradient supplies momentum when its arguments and fixed endpoints are specified.

Check your understanding

In the free-particle complete integral, what does β = ∂S/∂α represent with the chosen time origin?

Try it yourself

Verify the endpoint action S=q2/(2t)S=q^2/(2t) for unit mass at q = 4 and t = 2. Find its momentum and show that St+H=0S_t+H=0 with the derivative taken at fixed q.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The derivative of 1/t is −1/t². Do not differentiate along the trajectory q(t).

Compare with the solution

The position derivative is Sq=q/t=2S_q=q/t=2, so H=p2/2=2H=p^2/2=2. The explicit time derivative is St=−q2/(2t2)=−16/8=−2S_t=-q^2/(2t^2)=-16/8=-2, and their sum vanishes. The action value itself is 16/4=416/4=4. Momentum, energy, and action are distinct quantities even though this simple quadratic example relates all three by derivatives.

Take it with you

  • The Hamilton–Jacobi equation asks for a generating function with K = 0.
  • A complete integral includes enough independent parameters to reconstruct local trajectories.
  • Endpoint conditions and branch structure determine how an action function can be used.

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