The Hamilton-Jacobi Equation

Statement

For a system with Hamiltonian H(q, p, t), there exists a canonical transformation to new coordinates that are all constants of motion, generated by Hamilton's principal function S(q₁,…,q_n, α₁,…,α_n, t). Requiring the transformed Hamiltonian to vanish yields the first-order partial differential equation H(q, ∂S/∂q, t) + ∂S/∂t = 0, whose complete integral reduces the dynamics to the algebraic inversion of 2n constants.

Why it matters

The Hamilton-Jacobi equation (HJE) is the deepest classical formulation of mechanics: instead of solving 2n coupled first-order ordinary differential equations, one seeks a single scalar field S over configuration space whose gradient is the momentum. When a complete integral can be found, the entire motion follows by differentiation and algebra, with no further integration. This is the natural home of separation of variables, action-angle coordinates, and adiabatic invariants.

It is also the bridge between particle and wave. The surfaces S = const behave exactly as wavefronts, with trajectories running orthogonal to them, and the substitution ψ ~ exp(iS/ℏ) turns the Schrödinger equation into the HJE at leading order in ℏ. Hamilton discovered this optical-mechanical analogy a century before it became the semiclassical (WKB) limit of quantum mechanics.

Assumptions

Derivation

Result

Limiting cases

• Conservative system (∂H/∂t = 0): S separates as S = W(q, α) − E t, giving the time-independent HJE H(q, ∂W/∂q) = E = α₁, with W Hamilton's characteristic function.
• Fully separable system: S = Σ_i S_i(q_i, α) collapses the PDE into n decoupled ODEs — the route to action-angle variables and Liouville-integrable tori.
• Free particle: S is linear in q, S = p·q − (p²/2m)t; the wavefronts are planes moving at the phase speed.
• Semiclassical (ℏ → 0) limit: writing ψ = A exp(iS/ℏ), the leading order of the Schrödinger equation is exactly the HJE; the wave collapses onto classical rays.

Breaks when

Failure modes

• Confusing S the field with the action number. Hamilton's principal function S(q, t) is a field over configuration space; the action functional evaluated on one path is a number equal to S along the actual trajectory. They coincide only on-shell.
• Keeping p independent after substitution. Once p_i = ∂S/∂q_i is inserted, the equation is a PDE for S(q, t) alone; treating p and q as separate variables double-counts degrees of freedom.
• Dropping the ∂S/∂t term. Writing H = 0 instead of H + ∂S/∂t = 0; the whole method rests on K = 0, i.e. on that explicit time derivative.
• Putting explicit t into W. In the conservative case W(q, α) carries no time dependence; all time sits in the −Et piece. Writing H(q, ∂W/∂q, t) is wrong.
• Sign error in S = W − Et. The energy term is subtracted (so that ∂S/∂t = −E cancels H = E). Writing +Et flips the sign of the momentum and the direction of motion.
• Miscounting constants. A complete integral needs n non-trivial constants; the extra additive constant in S is physically irrelevant. Using n+1 "essential" constants over-parametrises the solution.

Discussion

The power of the HJE is that it trades a system of equations for one equation whose complete integral already contains every trajectory. The generating function S pushes all the dynamics into a canonical transformation to coordinates that simply stand still. In this frame "solving the motion" means differentiating a known function and inverting algebraic relations β_i = ∂S/∂α_i. This is why separable problems — the Kepler problem, the symmetric top, a charge in crossed fields — are solved so cleanly here: separability of S is exactly the existence of enough conserved α_i, and it delivers action-angle variables directly.

The geometric content is the optical-mechanical analogy. Because p = ∇S, the momentum is everywhere normal to the surfaces S = const, so those surfaces act as wavefronts and the trajectories as rays orthogonal to them. The propagation of an S = const surface in time is governed by H + ∂S/∂t = 0, structurally identical to the eikonal equation (∇Φ)² = n²ω²/c² of geometrical optics. Hamilton built this correspondence for optics and mechanics alike; the constant-S fronts move at a phase velocity that is generally different from the particle (group) velocity, foreshadowing the two velocities of a quantum wave packet.

The deepest connection is to quantum mechanics. Substituting ψ = A(q, t) exp(iS(q, t)/ℏ) into the Schrödinger equation and grouping by powers of ℏ gives, at order ℏ⁰, exactly the Hamilton-Jacobi equation (1/2m)(∇S)² + V + ∂S/∂t = 0, and at order ℏ¹ the continuity equation ∂(A²)/∂t + ∇·(A²∇S/m) = 0 for the probability density. The classical action S is therefore the phase of the quantum wavefunction in the ℏ → 0 limit, and the neglected −(ℏ²/2m)∇²A/A term is precisely Bohm's quantum potential. WKB quantization, the Bohr-Sommerfeld rule ∮ p dq = (n + ½)2πℏ, and instanton tunneling rates all descend from this expansion.

Common misconceptions. Hamilton's principal function is not "the action of the true path" as a mere number — it is a field whose derivatives generate momenta and orbits; its value at a point equals the accumulated action only along the actual trajectory reaching that point. Nor is K = 0 a physical statement that energy vanishes: it is a bookkeeping choice for the transformed frame, in which the real energy still appears as the constant α₁ = E. And separability is a property of a chosen coordinate system, not of the physics: the same problem may separate in one set of coordinates and not another.

Worked examples

Problems

1. Write the time-dependent HJE for a particle of mass m in uniform gravity, H = p²/2m + mgz, and find z(t) by separation. Evaluate for m = 1 kg, g = 9.8 m/s², E = 49 J, released from rest at the top.
   Solution

   HJE: (1/2m)(∂S/∂z)² + mgz + ∂S/∂t = 0. Separate S = W(z) − Et: (1/2m)(dW/dz)² = E − mgz, so p = dW/dz = √(2m(E − mgz)). Then β = ∂S/∂E = ∫ m dz/√(2m(E − mgz)) − t = −(1/g)√(2(E − mgz)/m) − t. Solving: E − mgz = ½mg²(t + β)², i.e. z(t) = E/mg − ½g(t + β)². Released from rest means the turning point (top) is at t + β = 0, so β = 0 and z(t) = E/mg − ½gt². Numerically E/mg = 49/(1·9.8) = 5 m, so z(t) = 5 − 4.9t² m — free fall from a 5 m peak, reaching z = 0 at t = √(10/9.8) ≈ 1.01 s.
2. For a free particle in three dimensions, H = (p_x² + p_y² + p_z²)/2m, show that S separates additively and find W. Interpret the surfaces S = const.
   Solution

   Try S = W_x(x) + W_y(y) + W_z(z) − Et. The HJE becomes (1/2m)[(dW_x/dx)² + (dW_y/dy)² + (dW_z/dz)²] = E. Each term depends on its own variable, so each derivative is a separation constant: dW_x/dx = α_x, etc., with α_x² + α_y² + α_z² = 2mE. Hence W = α_xx + α_yy + α_zz = p·r, and S = p·r − Et. The surfaces S = const are planes normal to p (the momentum p = ∇S is constant and orthogonal to them); as t advances they translate along p — plane wavefronts, exactly the free-particle de Broglie wave phase.
3. Starting from H(q, p) with no explicit time dependence, prove that S = W(q, α) − Et solves the full HJE and reduces it to H(q, ∂W/∂q) = E.
   Solution

   With S = W(q) − Et we have ∂S/∂q_i = ∂W/∂q_i (independent of t) and ∂S/∂t = −E. Substituting into H(q, ∂S/∂q) + ∂S/∂t = 0 gives H(q, ∂W/∂q) − E = 0, i.e. H(q, ∂W/∂q) = E. This is consistent because the left side is time-independent while the only t-dependence sat in the −Et term, whose derivative supplied exactly the constant −E. The separation constant E is identified as α₁, the conserved energy, confirming that time-translation symmetry is what makes the reduction possible (energy thread).
4. Verify dimensionally that every term of the time-independent HJE for the oscillator, (1/2m)(dW/dq)² + ½mω²q² = E, carries units of energy, and confirm [W] = [S].
   Solution

   [W] = [S] = J·s (action), since W differs from S only by the term Et which has units J·s. Then [dW/dq] = J·s/m = kg·m/s = [p]; squaring and dividing by mass: (kg·m/s)²/kg = kg·m²/s² = J. Second term: [mω²q²] = kg·(1/s)²·m² = kg·m²/s² = J. Right side [E] = J. All three terms are energies, and [W] = J·s matches the action dimension of Hamilton's principal function.
5. A particle experiences a constant force, V(x) = −Fx (so H = p²/2m − Fx). Using the HJE, find x(t). Take m = 2 kg, F = 4 N, E = 0, starting at x = 0 with p = 0.
   Solution

   HJE (conservative): (1/2m)(dW/dx)² − Fx = E, so p = dW/dx = √(2m(E + Fx)). Then β = ∂S/∂E = ∫ m dx/√(2m(E + Fx)) − t = (1/F)√(2m(E + Fx)) − t. Invert: E + Fx = F²(t + β)²/2m, i.e. x(t) = −E/F + (F/2m)(t + β)². With E = 0 and p = 0 at x = 0 the turning point is t + β = 0, so β = 0 and x(t) = (F/2m)t² = (4/4)t² = t² m. This is uniform acceleration a = F/m = 2 m/s² (since x = ½a t² = ½·2·t²), recovering Newton's second law. At t = 3 s, x = 9 m.