Time-Independent Hamilton-Jacobi Equation

The full Hamilton-Jacobi equation is a magnificent machine with one awkward feature: it asks for a function of $n+1$ variables — all the coordinates and the time — and first-order nonlinear PDEs do not forgive extra variables. Yet the systems mechanics cares about most, from planets to pendulums to molecules, share a mercy: their Hamiltonians do not depend explicitly on time. Energy is conserved. And whenever a variable's absence from the Hamiltonian buys a conservation law, Hamilton-Jacobi theory converts that conservation law into a separation: the time dependence of the principal function peels off as the trivial drift $-Et$ , and what remains is an equation over configuration space alone. That remainder — $H(q,\partial W/\partial q) = E$ , the time-independent Hamilton-Jacobi equation — is the working form of the theory. Nearly every problem ever actually solved by Hamilton-Jacobi methods begins here, not with the full equation: the Kepler orbit, the symmetric top, the hydrogen atom of the old quantum theory, the geodesics threading a rotating black hole. It is a single scalar PDE in which the conserved energy $E$ enters as a parameter labeling a family of solutions — a structure that quietly prefigures the eigenvalue problems of quantum mechanics, where the same letter $E$ plays the same role for deeper reasons. This article owns the equation itself: how the time separates (the argument is a template for every separation to come), what a complete integral looks like and where the energy sits among its constants, and the one-dimensional reduction to quadrature that turns any conservative 1D problem — solvable in elementary functions or not — into a definite integral you can always evaluate.

Key takeaways

When $\partial H/\partial t = 0$ , the Hamilton-Jacobi equation separates additively in time: $S(q,\alpha,t) = W(q,\alpha) - Et$ , with the separation constant $E$ equal to the conserved value of $H$ .
The remainder is the time-independent Hamilton-Jacobi equation $H(q,\partial W/\partial q) = E$ — a first-order PDE over configuration space alone, in which $E$ appears as a parameter labeling the solution family.
A complete integral carries $n$ constants $\alpha$ with the energy among them — either literally as $\alpha_1 = E$ , or as a function $E(\alpha_1,\dots,\alpha_n)$ ; both conventions are canonical and the second yields uniform drifts $\dot\beta_i = \partial E/\partial\alpha_i$ , the germ of frequencies.
In one dimension the equation is solved outright: $W = \int\!\sqrt{2m(E-V)}\,\mathrm{d}q$ , and the timetable $t - t_0 = \partial W/\partial E = \int\mathrm{d}q/\sqrt{(2/m)(E-V)}$ reduces any conservative 1D motion to a definite integral.
The function $W$ that solves the equation — its geometry, its role as a generator with $K = E$ — is Hamilton's Characteristic Function; systematic multi-coordinate separation is its own art (Separation of Variables in Hamilton-Jacobi Theory).
Key idea: conservation of energy lets you peel off the time dependence of the principal function as $-Et$ , converting the time-dependent Hamilton-Jacobi PDE into the time-independent equation $H(q,\partial W/\partial q) = E$ for the characteristic function $W$ , whose $\alpha$ -derivatives give the trajectory by quadrature.

Where this lives today

The time-independent Hamilton-Jacobi equation is the classical half of the WKB approximation: writing $\psi = e^{iW/\hbar}$ in the time-independent Schrödinger equation returns equation (1) at leading order in $\hbar$ , which is why semiclassical spectroscopy — computing molecular vibrational levels from classical trajectories and their quadratures — still runs on the machinery of this page. (The full two-way dictionary is a later chapter: Connection to the Schrödinger Equation.) The most spectacular modern employer, though, is general relativity. Geodesics around a rotating black hole are governed by a time-independent HJ equation in the Kerr metric, and Carter's 1968 discovery that it separates — yielding a hidden fourth constant of motion beyond energy, axial angular momentum and rest mass — is the reason black-hole orbits are integrable rather than chaotic. Every ray-traced image of an accretion flow, including the Event Horizon Telescope reconstructions of M87 and Sgr A, and every extreme-mass-ratio inspiral waveform template being prepared in 2026 for the LISA gravitational-wave mission, is built on quadratures of the separated Kerr equation — Jacobi's method, running at the edge of an event horizon.

The physical picture

Picture the principal function $S$ as a surface over the $(q,t)$ plane. For a generic, driven system that surface is genuinely two-dimensional terrain: ridges and valleys that change shape as time runs, with no pattern repeating. Now switch off the explicit time dependence of $H$ . The terrain acquires a rigid symmetry: every slice at fixed $t$ is the same profile $W(q)$ , merely lowered by an amount proportional to how late the slice is taken. The whole surface is one static profile sliding downhill at constant rate $E$ . Nothing about the landscape's shape involves time — time only sets how far the sheet has sunk.

Figure

Peeling time off the principal function — a static profile W sinking at constant rate E

That picture is the separation ansatz $S = W - Et$ , and it explains why the time-independent equation deserves its own life. All questions about the motion's geometry — where the system may go, how fast it passes each point, where it must turn back — are questions about the static profile, answerable without ever mentioning time. The profile's slope at each point is the momentum there, $p = \partial W/\partial q$ , and demanding that kinetic plus potential energy reproduce the same $E$ everywhere gives the equation the slope must satisfy. Time re-enters only at the end, through the gentlest possible door: the rate at which the frozen sheet sinks. The second thing to internalize early is where the energy lives in the resulting equation. In Hamilton's equations, $E$ is something you compute after solving. In the time-independent Hamilton-Jacobi equation, $E$ is promoted to an input: a dial on the equation itself, with each setting selecting a different family of solutions $W(q; E)$ . Some dial settings correspond to bound motion trapped between turning points, others to unbounded escape, and the character of the solution changes qualitatively as the dial crosses special values. A single equation whose solutions are labeled by an energy parameter, with qualitative changes at special energies — any student of quantum mechanics will feel the shape of things to come. Classically every $E$ in the allowed range is admissible; what the quantum theory adds is a single-valuedness condition that selects a discrete subset. The full story of that upgrade belongs to a later chapter (Connection to the Schrödinger Equation); here we only note that the classical equation already has the eigenvalue problem's silhouette.

Separating the time

The argument that peels time off is short, and it is worth doing in the general separation-of-variables style, because the same three-move template — assume additive form, substitute, watch the variables disentangle — reappears every time a cyclic coordinate is separated in the chapters ahead.

Derivation ASeparating time from the full Hamilton-Jacobi equation

We are deriving the time-independent Hamilton-Jacobi equation $H(q,\partial W/\partial q) = E$ , together with the split $S = W - Et$ , from the full equation $H(q,\partial S/\partial q,t) + \partial S/\partial t = 0$ under the single hypothesis $\partial H/\partial t = 0$ . The strategy is additive separation: posit that $S$ splits into a spatial piece plus a purely temporal piece, substitute, and argue that a function of coordinates alone can equal a function of time alone only if both are the same constant. Watch the logic of that last step — it is the engine of every separation argument in this chapter, and it fails instantly if $H$ retains any explicit $t$ .

Step 1. Posit the additive ansatz

S(q,\alpha,t) = W(q,\alpha) + T(t)

Why additive rather than multiplicative: only derivatives of $S$ enter the Hamilton-Jacobi equation, and a sum cleanly routes $\partial S/\partial q_i = \partial W/\partial q_i$ to the spatial slots and $\partial S/\partial t = \mathrm{d}T/\mathrm{d}t$ to the temporal one, with no cross-contamination.

Step 2. Substitute into the full equation. Because $H$ carries no explicit $t$ , every appearance of time is now confined to the $T$ -term:

H\!\left(q,\frac{\partial W}{\partial q}\right) = -\frac{\mathrm{d}T}{\mathrm{d}t}

Step 3. Read the equation as a standoff. The left side is a function of the coordinates only; the right side is a function of time only; and the equation asserts they are equal for all $q$ and all $t$ independently. Vary $t$ holding $q$ fixed: the left side cannot change, so the right side is constant. Call it $E$ :

H\!\left(q,\frac{\partial W}{\partial q}\right) = E, \qquad \frac{\mathrm{d}T}{\mathrm{d}t} = -E

Why this is the crux: separation constants are born exactly here, as the referee of a standoff between groups of variables that cannot talk to each other. If $H$ depended explicitly on $t$ , the left side would change with time at fixed $q$ , no standoff would occur, and no constant would exist — the hypothesis $\partial H/\partial t = 0$ is not decorative.

Step 4. Integrate the temporal piece: $T = -Et$ (the integration constant is additive and can be absorbed into $W$ ). Assembling,

S(q,\alpha,t) = W(q,\alpha) - Et, \qquad H\!\left(q,\frac{\partial W}{\partial q}\right) = E

Why $E$ is the physical energy: along the motion, $\partial S/\partial t = -E$ ; but the full equation says $\partial S/\partial t = -H$ . Hence $E$ equals the conserved value of the Hamiltonian — the mechanical energy whenever the coordinates are time-independent (scleronomic). *Note: the identification $\partial S/\partial t = -E = -H$ is consistent with the endpoint-variation relations of the action, and setting $H = p^2/2m + V$ recovers energy conservation in the form $|\partial W/\partial q|^2 = 2m(E-V)$ .*

The output of the separation is worth displaying on its own, since it is the equation the rest of the chapter orbits:

H\!\left(q_1,\dots,q_n,\;\frac{\partial W}{\partial q_1},\dots,\frac{\partial W}{\partial q_n}\right) = E \qquad (1)

where: $W(q,\alpha)$ — Hamilton's characteristic function, the time-free generator whose geometry (wavefronts, abbreviated action) and dynamics (the $K = E$ canonical transformation) are developed in Hamilton's Characteristic Function; $E$ — the conserved energy, entering as a parameter. Structurally, (1) says: replace every momentum in the conservation-of-energy statement by the gradient of one unknown scalar. For a particle in a potential it is $|\nabla W|^2 = 2m(E - V)$ — pointwise, purely local, and first-order: no second derivatives, no time, no initial conditions. That locality is what makes the equation solvable coordinate by coordinate when the geometry cooperates.

The complete integral, and where the energy sits

What counts as "solving" equation (1)? As with the full equation, not a general solution but a complete integral: a family $W(q;\alpha_1,\dots,\alpha_n)$ with $n$ independent constants, $\det(\partial^2 W/\partial q_i\partial\alpha_j) \neq 0$ . But there is a wrinkle special to the time-independent equation: the energy $E$ is already sitting in the equation as a parameter, so one of the constants is supplied before you start. Two bookkeeping conventions coexist in the literature, and confusing them costs real errors:

\textbf{(a)}\quad \alpha_1 = E \qquad\text{or}\qquad \textbf{(b)}\quad E = E(\alpha_1,\dots,\alpha_n) \qquad (2)

where: in convention (a) the energy is itself the first new momentum, and the remaining $\alpha_2,\dots,\alpha_n$ arise as further separation constants; in convention (b) the $n$ constants are whatever the solution naturally produces (momentum components, separation constants), and the energy is expressed as a function of them. Both are legitimate canonical choices related by a relabeling of the constants. The dynamics distinguishes them only in how the conjugate coordinates drift: with the transformed Hamiltonian $K = E(\alpha)$ , Hamilton's equations give

\dot\beta_i = \frac{\partial E}{\partial \alpha_i} \qquad (3)

where: in convention (a) this is $\dot\beta_1 = 1$ , $\dot\beta_{j\geq 2} = 0$ — one clock, everything else frozen — while in convention (b) each $\beta_i$ drifts uniformly at its own rate $\partial E/\partial\alpha_i$ . Those constant drift rates are not a nuisance; they are the first appearance of the frequencies that action-angle theory will extract wholesale (Frequencies from Action-Angle Variables). The following derivation makes convention (b) concrete in the simplest possible setting.

Go deeper · Derivation B · 1 minA complete integral where the energy is a function of the constants

Derivation BA complete integral where the energy is a function of the constants

The goal is a worked structural example of convention (b): the free particle in two dimensions, whose complete integral carries two momentum constants with the energy a function of them, and whose conjugate coordinates drift at rates $\partial E/\partial\alpha_i$ that are visibly the velocity components. The strategy is to solve equation (1) by the most naive ansatz available — linear in the coordinates — and then run the canonical machinery honestly, using the principal function $S = W - E(\alpha)t$ when differentiating with respect to constants. The step to watch is exactly that: once $E$ depends on the $\alpha_i$ , the $-Et$ term contributes to $\partial S/\partial\alpha_i$ , and forgetting it produces frozen coordinates that should be drifting.

Step 1. The time-independent equation for $H = (p_x^2 + p_y^2)/2m$ reads

\frac{1}{2m}\left[\left(\frac{\partial W}{\partial x}\right)^{\!2} + \left(\frac{\partial W}{\partial y}\right)^{\!2}\right] = E

Try $W = \alpha_x x + \alpha_y y$ with constants $\alpha_x,\alpha_y$ . Substituting: $E = (\alpha_x^2 + \alpha_y^2)/2m$ . Why this is already complete: two coordinates, two independent constants, and $\det(\partial^2 W/\partial q_i\partial\alpha_j) = \det(\mathbb{1}) = 1 \neq 0$ . The energy did not need to be one of the constants — it emerged as a function of them.

Step 2. Identify the physical meaning of the constants: $p_x = \partial W/\partial x = \alpha_x$ and $p_y = \partial W/\partial y = \alpha_y$ — the conserved momentum components. The solution family is a plane wavefront family, one plane per direction and magnitude of momentum.

Step 3. Form the principal function and differentiate with respect to the constants:

S = \alpha_x x + \alpha_y y - \frac{\alpha_x^2 + \alpha_y^2}{2m}\,t, \qquad \beta_x = \frac{\partial S}{\partial \alpha_x} = x - \frac{\alpha_x}{m}t, \quad \beta_y = y - \frac{\alpha_y}{m}t

Why $S$ and not $W$ here: the frozen coordinates $\beta_i$ are conjugates in the $K = 0$ scheme of the principal function; since $E$ depends on both constants, the $-E(\alpha)t$ term feeds each derivative. Using bare $W$ would lose the $-(\alpha_i/m)t$ drift terms.

Step 4. Invert: $x(t) = \beta_x + (\alpha_x/m)t$ , $y(t) = \beta_y + (\alpha_y/m)t$ — uniform straight-line motion, with the drift rates $\partial E/\partial\alpha_i = \alpha_i/m$ equal to the velocity components, exactly as equation (3) promised. Note: the same computation in convention (a) — constants $(E,\theta)$ with $\alpha_x = \sqrt{2mE}\cos\theta$ — gives $\beta_1 = t + \mathrm{const}$ and a frozen $\beta_\theta$ fixing the line's direction, the same trajectories in different bookkeeping.

One dimension: the equation solved outright

For a single degree of freedom the time-independent Hamilton-Jacobi equation is not merely simplified — it is finished. With $H = p^2/2m + V(q)$ , equation (1) is an algebraic condition on the derivative of $W$ :

\frac{1}{2m}\left(\frac{\mathrm{d}W}{\mathrm{d}q}\right)^{\!2} + V(q) = E \quad\Longrightarrow\quad \frac{\mathrm{d}W}{\mathrm{d}q} = \pm\sqrt{2m\,[E - V(q)]} \qquad (4)

where: the sign is the direction of the momentum, flipping at turning points where $E = V$ ; between turning points the momentum's magnitude at every $q$ is fixed by energy conservation alone, before any trajectory is known. One integration delivers the characteristic function, and one differentiation delivers the timetable:

W(q,E) = \int_{q_0}^{q}\!\sqrt{2m\,[E - V(q')]}\;\mathrm{d}q', \qquad t - t_0 = \frac{\partial W}{\partial E} = \int_{q_0}^{q}\frac{\mathrm{d}q'}{\sqrt{\tfrac{2}{m}\,[E - V(q')]}} \qquad (5)

where: the second integral is recognizable as $\int \mathrm{d}q/v(q)$ — elapsed time as distance over local speed — but derived here as a transformation equation, the statement that the coordinate conjugate to $E$ advances as $t$ . Equation (5) is the complete solution of every conservative one-dimensional problem in mechanics: pendulum, quartic well, screened Coulomb, whatever — solvable in elementary functions or not, the motion is this definite integral, ready for exact evaluation when luck allows and numerical quadrature when it does not.

Go deeper · Derivation C · 1 minOne-dimensional motion reduced to a single quadrature

Derivation COne-dimensional motion reduced to a single quadrature

We are deriving the quadrature formulas (5) as canonical-transformation statements — not as a rearrangement of $\tfrac12 m\dot q^2 = E - V$ — because the derivation route is what generalizes to $n$ degrees of freedom, where no naive rearrangement exists. The strategy: solve equation (4) for $\mathrm{d}W/\mathrm{d}q$ , integrate to get $W(q,E)$ with the energy as the single constant $\alpha_1 = E$ (convention (a)), then invoke the transformation equations, in which the coordinate conjugate to the energy advances at unit rate. Watch the branch choice at the square root: it must be made consistently along each leg of the motion, and the two branches join smoothly at turning points, where the integrand of the timetable diverges — integrably, as the last step checks.

Step 1. From equation (4), choose the branch matching the motion's current direction (say $p > 0$ , moving right) and integrate:

W(q,E) = \int_{q_0}^{q}\sqrt{2m\,[E - V(q')]}\;\mathrm{d}q'

Why one constant suffices: $n = 1$ , so a complete integral needs exactly one nonadditive constant, and the energy parameter $E$ already in the equation does the job — nothing else needs inventing.

Step 2. Apply the transformation equations of the $W$ -generated map (Hamilton's Characteristic Function, or directly: $K = E$ , so the conjugate coordinate runs at $\dot\beta = \partial K/\partial E = 1$ ):

\beta = \frac{\partial W}{\partial E} = t - t_0

Step 3. Differentiate under the integral sign:

t - t_0 = \frac{\partial}{\partial E}\int_{q_0}^{q}\sqrt{2m\,[E-V]}\;\mathrm{d}q' = \int_{q_0}^{q}\frac{m\;\mathrm{d}q'}{\sqrt{2m\,[E - V(q')]}} = \int_{q_0}^{q}\frac{\mathrm{d}q'}{\sqrt{\tfrac{2}{m}[E - V(q')]}}

Why differentiation beats integration: $W$ itself is often a hard integral, but its $E$ -derivative — the physically wanted timetable — has a simpler integrand. In practice one differentiates first and integrates after, evaluating only the easier integral.

Step 4. Check the turning-point behavior. Near a simple turning point $q_T$ (where $V(q_T) = E$ , $V'(q_T) \neq 0$ ), the integrand diverges as $[E - V]^{-1/2} \sim |q - q_T|^{-1/2}$ — an integrable singularity, so the arrival time at the turning point is finite, while the momentum branch flips sign there for the return leg. Why this matters: it certifies that bound motion assembled from the two branches is smooth and periodic, with period $T(E) = 2\int_{q_-}^{q_+}\mathrm{d}q/\sqrt{(2/m)(E-V)}$ — the formula the worked examples will exercise. *As a check: for $V = \tfrac12 m\omega^2q^2$ the integral gives $t - t_0 = (1/\omega)\arcsin(q\sqrt{m\omega^2/2E})$ , i.e. simple harmonic motion with period $2\pi/\omega$ , amplitude-independent as required.*

Figure

The one-dimensional quadrature machine — from the potential to momentum to timetable

Which equation answers which question

you want	use	notes
the time-free PDE itself	eq. (1)	replace each $p_i$ by $\partial W/\partial q_i$ in $H = E$
where the energy sits among the constants	eq. (2)	convention (a): $\alpha_1 = E$ ; convention (b): $E(\alpha)$
drift rates of the conjugate coordinates	eq. (3)	$\dot\beta_i = \partial E/\partial\alpha_i$ — the germ of frequencies
momentum magnitude at each point (1D)	eq. (4)	fixed by $E$ and $V$ alone, before any solving
the full 1D solution	eq. (5)	$W$ by one integral; timetable by its $E$ -derivative
period of bound 1D motion	$T = 2\displaystyle\int_{q_-}^{q_+}\!\frac{\mathrm{d}q}{\sqrt{(2/m)(E-V)}}$	turning points $q_\pm$ solve $V = E$

Limiting cases

Case	Result	Meaning
$\partial H/\partial t \neq 0$	no separation	the standoff argument fails; retreat to the full equation (Generating Function to a Trivial Hamiltonian)
one degree of freedom	eq. (5) closes the problem	every conservative 1D system reduced to a definite integral — no exceptions
free particle	$W = \mathbf{p}\cdot\mathbf{q}$ , $E = p^2/2m$	linear landscape; complete integral in convention (b); straight lines recovered
$E \to V_{\min}$ in a well	turning points merge	motion shrinks to the stable equilibrium; period tends to the harmonic value $2\pi/\omega$ with $\omega^2 = V''/m$
$E$ at a potential maximum	timetable integral diverges	approach to unstable equilibrium takes infinite time — the separatrix, where period $\to\infty$
$n \geq 2$ , no further structure	eq. (1) holds but may admit no complete integral	conserved energy alone does not make a system integrable — see the warning below

Common trapConserved energy does not mean solvable — the time separation is only the first step

Because equation (1) looks so much simpler than the full equation, it is easy to conclude that every conservative system is now solved. False — and the failure is not technical but structural. The time separation removes one variable using one conservation law. For $n \geq 2$ the remaining PDE in $n$ coordinates still needs $n-1$ more separations, each requiring its own conserved quantity — and most conservative Hamiltonians simply do not have them. The double pendulum and the three-body problem are perfectly conservative; their time-independent HJ equations exist and are correct; and no complete integral exists globally for either, which is precisely why they exhibit chaos. A second, subtler trap: $E$ equals the mechanical energy $T + V$ only for scleronomic systems (time-independent coordinate relations). With rotating or moving constraints, $H$ is conserved and the separation goes through, but the conserved quantity is the Jacobi integral, not $T + V$ — a bead on a steadily rotating wire conserves $H$ while its kinetic-plus-potential energy oscillates.

Worked examples

1 · A dropped stone, by canonical transformation (confidence builder)

A stone of mass $m = 1\ \mathrm{kg}$ is dropped from rest at height $h = 20\ \mathrm{m}$ in gravity $g = 9.8\ \mathrm{m/s^2}$ , with $V(q) = mgq$ . Solve the time-independent Hamilton-Jacobi equation and find the fall time to the ground and the impact momentum — then check against elementary kinematics.

Step 1. Set the dial. Dropped from rest at $q = h$ : the energy parameter is $E = V(h) = mgh = 1\times 9.8\times 20 = 196\ \mathrm{J}$ , and $q = h$ is the turning point of the motion ( $E = V$ there, momentum zero).

Step 2. The equation and its gradient: $\dfrac{1}{2m}\left(\dfrac{\mathrm{d}W}{\mathrm{d}q}\right)^{2} + mgq = E$ , so $\dfrac{\mathrm{d}W}{\mathrm{d}q} = -\sqrt{2m(E - mgq)}$ , taking the descending branch ( $p < 0$ ).

Step 3. Timetable from the $E$ -derivative, eq. (5), integrating from the start $q_0 = h$ down to height $q$ :

t = \int_{q}^{h}\frac{\mathrm{d}q'}{\sqrt{\tfrac{2}{m}(E - mgq')}} = \int_q^h \frac{\mathrm{d}q'}{\sqrt{2g\,(h - q')}} = \sqrt{\frac{2(h-q)}{g}}

using $E = mgh$ so that $E - mgq' = mg(h - q')$ .

Step 4. Numbers at the ground $q = 0$ : $t = \sqrt{2\times 20/9.8} = \sqrt{4.082} = \boxed{2.02\ \mathrm{s}}$ ; impact momentum $p = -\sqrt{2m(E - 0)} = -\sqrt{2\times 1\times 196} = \boxed{-19.8\ \mathrm{kg\,m/s}}$ (downward, speed $19.8\ \mathrm{m/s}$ ).

Step 5. Check: elementary kinematics gives $t = \sqrt{2h/g} = 2.02\ \mathrm{s}$ ✓ and $v = gt = 9.8\times 2.02 = 19.8\ \mathrm{m/s}$ ✓. The machinery is overkill here by design — the point is watching the dial ( $E$ ), the branch choice, and the $E$ -derivative timetable work on a case where the answer is known cold.

2 · The quartic oscillator — a period no elementary method delivers (exam level)

A particle of mass $m = 1\ \mathrm{kg}$ moves in the quartic well $V(q) = \lambda q^4$ with $\lambda = 1\ \mathrm{J/m^4}$ . Unlike the harmonic oscillator, this system has no closed-form $q(t)$ in elementary functions. Use the quadrature (5) to find (a) the oscillation period at energy $E = 1\ \mathrm{J}$ , (b) the scaling of period with energy, and (c) the period at $E = 16\ \mathrm{J}$ . The dimensionless integral $\int_0^1 \mathrm{d}x/\sqrt{1-x^4} = 1.3110$ may be used.

Step 1. Turning points: $\lambda q_\pm^4 = E$ gives $q_\pm = \pm(E/\lambda)^{1/4}$ . At $E = 1\ \mathrm{J}$ : $q_\pm = \pm 1\ \mathrm{m}$ ; the amplitude is $A = (E/\lambda)^{1/4}$ .

Step 2. Period as four quarter-swings of the timetable integral:

T = 4\int_0^{A}\frac{\mathrm{d}q}{\sqrt{\tfrac{2}{m}\,(E - \lambda q^4)}} = 4\sqrt{\frac{m}{2E}}\int_0^A \frac{\mathrm{d}q}{\sqrt{1 - \lambda q^4/E}}

Step 3. Rescale to the dimensionless variable $x = q/A$ (so $\lambda q^4/E = x^4$ ):

T = 4\sqrt{\frac{m}{2E}}\;A\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^4}} = 4\sqrt{\frac{m}{2E}}\left(\frac{E}{\lambda}\right)^{\!1/4}\times 1.3110

Step 4. (b) Read off the scaling: $T \propto E^{-1/2}\cdot E^{1/4} = E^{-1/4}$ — the quartic oscillator speeds up with amplitude, in sharp contrast to the harmonic oscillator's amplitude-independent period. Why: the walls steepen faster than linearly, so larger swings meet disproportionately stronger restoring force.

Step 5. (a) Numbers at $E = 1\ \mathrm{J}$ : $T = 4\times\sqrt{0.5}\times 1\times 1.3110 = 4\times 0.7071\times 1.3110 = \boxed{3.71\ \mathrm{s}}$ .

Step 6. (c) At $E = 16\ \mathrm{J}$ : the scaling gives $T = 3.71\times 16^{-1/4} = 3.71/2 = \boxed{1.85\ \mathrm{s}}$ — quadrupling the amplitude's fourth power halves the period. Direct evaluation: $4\sqrt{1/32}\times 2\times 1.3110 = 4\times 0.1768\times 2\times 1.3110 = 1.85\ \mathrm{s}$ ✓. No equation of motion was solved at any point; the entire dynamics came out of the static equation (1) and one $E$ -derivative.

Origins · 1 minCarl Gustav Jacob Jacobi and how this was discovered

Carl Gustav Jacob Jacobi

1804–1851 · Königsberg and Berlin

Jacobi read Hamilton's 1834-35 dynamics papers with a mixture of admiration and impatience. Hamilton's own scheme required his function to satisfy two partial differential equations simultaneously — and, as Jacobi pointedly observed, appeared to require knowing the motion before constructing the function that was supposed to deliver it. Jacobi's 1837 repair became the theory taught ever since: a single first-order PDE, any complete integral of which — however obtained, by separation, ansatz or luck — generates the full solution of the dynamics. The shift of emphasis from "the" characteristic function to "any complete integral" turned a beautiful representation theorem into a working method. His Königsberg lectures of 1842-43 (published posthumously as the Vorlesungen über Dynamik) drove the method through the solar system's classic problems, and his separation of the geodesic equation on a triaxial ellipsoid — using the confocal coordinates that now bear his name — remains the standard demonstration that the right coordinates can crack a problem no force of ODE manipulation will open. Analyst first and mechanician second, Jacobi saw the theory as a chapter of PDE theory and elliptic function theory at once; mechanics has been the richer for both allegiances.

History

1834

— Hamilton's dynamics papers introduce the principal and characteristic functions, each satisfying a pair of PDEs — the seed, in an unwieldy form.

1837

— Jacobi reduces the scheme to a single first-order PDE and proves that one complete integral suffices to solve the motion — the time-independent workhorse takes its modern shape.

1843

— Jacobi separates the geodesic problem on the triaxial ellipsoid in confocal coordinates, the virtuoso demonstration of the separated

W

1913

— Bohr quantizes the hydrogen atom; Sommerfeld, Wilson and Ishiwara soon phrase the rule as

\oint p\,\mathrm{d}q = nh

applied to the separated pieces of

W

— old quantum theory runs on this equation.

1968

— Carter separates the Hamilton-Jacobi equation for geodesics in the Kerr black-hole spacetime, exposing a hidden fourth constant; every modern black-hole orbit code descends from that separation.

Going further

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §10.3 · Landau & Lifshitz, Mechanics, 3rd ed., §47-48 · Taylor, Classical Mechanics, Ch. 13 · Morin, Introduction to Classical Mechanics, Ch. 15 (Appendix) · Arnold, Mathematical Methods of Classical Mechanics, §46-47 Key results to keep in mind: $\partial S/\partial t=-E$ yields $H=E$ ; the 1D quadrature gives $t-t_0=\int \mathrm{d}q/\sqrt{(2/m)(E-V)}$ ; the harmonic-oscillator action evaluates to $J=2\pi E/\omega$ with frequency $\omega/2\pi$ . Worked-example numbers on this page (fall time $2.02\ \mathrm{s}$ , impact momentum $19.8\ \mathrm{kg\,m/s}$ ; quartic period $3.71\ \mathrm{s}$ at $E=1\ \mathrm{J}$ via $\int_0^1\mathrm{d}x/\sqrt{1-x^4}=1.3110$ , scaling $T\propto E^{-1/4}$ ) follow directly from the displayed formulas.

Builds on: Hamilton-Jacobi Equation · Hamilton's Principal Function · Generating Function to a Trivial Hamiltonian Leads to: Hamilton's Characteristic Function · Separation of Variables in Hamilton-Jacobi Theory · Action-Angle Variables · Liouville Integrable Systems

In the separation argument, why is ∂S/∂t forced to be constant? Couldn't it vary along the motion?

The forcing happens off the motion, which is the subtlety. The Hamilton-Jacobi equation is an identity over the whole $(q,t)$ domain, not a statement along one trajectory. After substituting the ansatz, the equation says a function of $q$ alone equals a function of $t$ alone for every pair $(q,t)$ independently: pick any fixed $q$ and vary $t$ — the left side cannot respond, so the right side must be flat. That is a much stronger statement than anything a single trajectory could enforce, and it is available only because we are solving for a field $S(q,t)$ rather than an orbit. If $H$ carried explicit time, the left side would vary with $t$ at fixed $q$ , the standoff would collapse, and no separation constant would exist.

Is E always the mechanical energy T plus V?

Only for scleronomic systems — those whose coordinate definitions do not involve time. The separation needs $\partial H/\partial t = 0$ , and the constant it produces is the conserved value of $H$ . When the transformation to generalized coordinates is time-independent, $H = T + V$ and $E$ is the mechanical energy. But with moving constraints — a bead on a uniformly rotating wire, a pendulum with a driven pivot rewritten in the co-moving frame — $H$ can be time-independent while differing from $T + V$ by centrifugal-type terms; the conserved $E$ is then the Jacobi integral. The separation works identically; only the name of $E$ changes. The practical rule: compute $H$ honestly from the Legendre transformation and never assume it equals $T + V$ without checking the constraints.

For one degree of freedom the complete integral has just one constant. Where did the second constant of motion go?

It is there — it is $\beta = \partial W/\partial E - t$ , hiding in the transformation equations rather than in $W$ . A 1D problem needs two constants: the energy $E$ fixes which orbit (the shape), and $\beta$ (equivalently $t_0$ ) fixes where on it the system sits at $t = 0$ (the phase). The complete integral only ever carries the momentum-like half of the constants; the coordinate-like half is generated by differentiation. Counting both halves, 1D delivers $2 = 2n$ constants, exactly right. The same split persists for all $n$ : the $\alpha$ 's label the invariant set, the $\beta$ 's locate the system on it.

The equation looks like an eigenvalue problem with E as the eigenvalue. Is that analogy real or cosmetic?

Structurally real, with one honest difference. Real: $E$ enters as a spectral-style parameter; solutions change character at special values (bound vs unbounded, separatrix energies); and under $\psi = e^{iW/\hbar}$ the equation is literally the $\hbar \to 0$ limit of the time-independent Schrödinger equation, with the same $E$ in the same slot. The difference: classically every $E$ in the allowed range is admissible — the spectrum is continuous — because nothing constrains $W$ globally. Quantization enters only when a global single-valuedness condition is imposed on $e^{iW/\hbar}$ , which selects discrete $E$ via $\oint p\,\mathrm{d}q = (n + \tfrac12)h$ . So the classical equation supplies the entire structure of the eigenvalue problem; the quantum theory adds one boundary condition. The full dictionary is the later-chapter article Connection to the Schrödinger Equation.

How do I choose between the plus and minus branches of the momentum square root?

By the direction of motion on the current leg, and consistently along that whole leg. Moving in the direction of increasing $q$ means $p > 0$ : take the plus branch; on the return leg take minus. The two branches are two sheets of a multivalued $W$ that join at the turning points, where $p = 0$ and the square root vanishes. For a single transit (scattering, falling) one branch suffices — as in worked example 1, which used the descending branch throughout. For bound motion you alternate sheets each half-period; quantities built from full cycles, like the period integral or $\oint p\,\mathrm{d}q$ , automatically count both branches (hence the factor 2 in $T = 2\int_{q_-}^{q_+}$ , or 4 with quarter-swing symmetry). Choosing one global sign for a bound orbit is the classic error — it describes a particle that never comes back.

If energy is conserved, why can a conservative system still be chaotic? Doesn't this equation solve it?

The equation exists for every conservative system; a complete integral of it does not. Solving the PDE in the useful sense means finding $W$ with $n$ independent constants, and each constant beyond $E$ must come from somewhere — an additional independent conserved quantity in involution with the others. Generic conservative systems with $n \geq 2$ have no such extra conservation laws: for the double pendulum or the three-body problem, energy is the only global constant, the PDE in the remaining coordinates does not separate in any coordinate system, and no smooth family $W(q;\alpha)$ covers phase space. Trajectories then wander over the $(2n-1)$ -dimensional energy surface instead of lying on nice $n$ -dimensional sheets — that wandering is chaos. The time-independent HJ equation is thus also a diagnostic: integrability is precisely the situation in which it admits a complete integral.

Why differentiate W with respect to E instead of just using energy conservation to get the speed?

For one degree of freedom the two routes coincide — $\partial W/\partial E$ reproduces $\int\mathrm{d}q/v$ exactly, and if 1D were the whole world the canonical detour would be pedantry. The derivative route is taught because it is the one that generalizes. In $n$ degrees of freedom there is no "the speed along the path" to invert: the orbit geometry and the schedule tangle across coordinates. The transformation equations untangle them mechanically — $\partial W/\partial\alpha_j = \beta_j$ for the non-energy constants yields time-free orbit equations (the path's shape), while $\partial W/\partial E = t - t_0$ yields the schedule — with no ingenuity required at the point of use. The 1D case is the training version of a tool whose value appears at $n = 2$ and never stops growing.

Can the time-separation trick be repeated for other variables?

Yes — that is exactly the program of the rest of the chapter. The argument needed only this: a variable absent from $H$ (here $t$ ), an additive ansatz splitting it off, and a standoff forcing a constant. A cyclic coordinate $q_k$ (absent from $H$ ) separates identically: $W = \alpha_k q_k + W'(\text{rest})$ , with the conserved momentum $\alpha_k$ as the new constant — for the azimuthal angle in a central field this is conservation of $L_z$ appearing as a separation constant. Non-cyclic coordinates can also separate when the Hamiltonian's structure (not a symmetry visible as absence) permits an additive split with its own standoff argument — the systematic theory of when and how, coordinate system by coordinate system, is Separation of Variables in Hamilton-Jacobi Theory, with the sharp algebraic criterion given by the Staeckel Conditions for Separability.