Generating Function to a Trivial Hamiltonian

The trivial-Hamiltonian idea did not retire with classical mechanics — it changed fields. In optimal control theory the exact same move (encode the whole solution family in one scalar function of state and time, then demand a PDE that makes the rest trivial) is the Hamilton-Jacobi-Bellman equation, the backbone of dynamic programming. Every time a quadrotor plans a minimum-energy path or a portfolio model prices a hedging strategy, a descendant of $K \equiv 0$ is being solved numerically. Modern reinforcement learning sits one step further down the same lineage: a value function is a principal function estimated by samples, and the Bellman backup is the discretized statement that a properly chosen scalar makes the remaining decision problem trivial. In 2026 the sharpest working edge is in robotics and autonomous systems, where Hamilton-Jacobi reachability analysis — solving the HJ PDE on a grid, or with neural PDE solvers for systems too high-dimensional for grids — certifies the safe operating envelope of drones and self-driving vehicles: the level sets of the computed $S$ are provable "cannot-crash" boundaries. The physicist's freeze-the-dynamics trick has become a safety certificate you can ship.

Watch a pendulum's phase point trace its oval in the $(q,p)$ plane. An observer using those coordinates has genuine work to do: the point moves, and predicting where it will be requires integrating Hamilton's equations. Now imagine an observer whose coordinate grid is itself carried along by the flow — a chart drawn on the moving fluid of phase space rather than on the container. To that observer the phase point never goes anywhere. It sits at the same grid intersection forever, because the grid deforms in perfect sympathy with the dynamics. All the motion has been absorbed into the dictionary between the two charts.

The Hamiltonian formulation makes this dream precise. In any canonical chart the motion is governed by some Hamiltonian; the question is only how simple that Hamiltonian can be made. There is a natural ladder here, and it is worth climbing consciously. One cyclic coordinate makes one momentum constant. If every coordinate were cyclic — $K = K(P)$ only — every momentum would be constant and every coordinate would drift linearly in time; this is the destination of the action-angle construction. But there is one rung higher: make $K$ vanish identically. Then not even linear drift survives. Coordinates and momenta alike freeze, and the system's entire history is compressed into $2n$ numbers.

Why does the top rung require a time-dependent generator? Because a time-independent canonical transformation can only reshuffle the energy surface, never flatten it: if $\partial F_2/\partial t = 0$, then $K = H$ takes the same values as before, and a nonzero Hamiltonian keeps generating motion. The extra term $\partial F_2/\partial t$ in the transformation law is the only lever that can cancel $H$ rather than merely disguise it. The generator must therefore change in time at exactly the rate the dynamics accumulates — and the quantity that does precisely that is the classical action. That the action would reappear here, as the demolition charge that levels the Hamiltonian, is the most elegant twist in classical mechanics. One more orientation point before the algebra. Freezing the motion does not destroy information — it relocates it. In the new chart the trajectory is a dead point, but the map back from $(Q,P)$ to $(q,p)$ is time-dependent, and all the orbital richness lives in that map. Solving mechanics this way splits cleanly into two stages: a hard analytic stage (find the generator $S$ by solving one PDE) and a purely algebraic stage (differentiate $S$, invert the relations). The Hamilton-Jacobi chapter is the study of when and how the hard stage can actually be carried out.

Everything rests on the type-2 transformation rules, established in Type-2 Generating Function and Generating Functions: the Four Types. A generator $F_2(q,P,t)$ — old coordinates and new momenta as its arguments — defines the transformation implicitly and prescribes how the Hamiltonian changes:

where: $F_2(q,P,t)$ — the type-2 generating function; $(q_i,p_i)$ — the old canonical pair; $(Q_i,P_i)$ — the new pair; $K$ — the Hamiltonian that governs the new variables. The type-2 shape is the natural one for our program: we intend the new momenta to be constants, so a generator that takes those constants directly as arguments — and hands back the old momenta and new coordinates by differentiation — is exactly the right interface. (The near-identity structure of type-2 generators, $F_2 = \sum_i q_iP_i$ giving the identity map, also means the construction deforms smoothly from "no transformation at all", which matters for perturbation theory later in the chapter.) Now impose the design goal. We want the new Hamiltonian to vanish for all values of the new variables and all times. Renaming the sought generator $S$, equation (1) turns the demand $K \equiv 0$ into a closed condition on $S$ alone:

where: $S(q,\alpha,t)$ — Hamilton's principal function, the type-2 generator that achieves $K\equiv 0$; the $n$ constants $\alpha_i$ label the family of solutions and play the role of the new momenta $P_i$. Equation (2) is the Hamilton-Jacobi equation, treated in full as the flagship of this chapter (Hamilton-Jacobi Equation). Notice what happened structurally: the momenta in $H$ have been replaced by gradients of an unknown scalar. One function of $n+1$ variables now carries the entire dynamical problem. The counting deserves a hard look, because it is the honest price tag of the method. Hamilton's equations are $2n$ first-order ODEs — hard work, but of a familiar kind. Equation (2) is a single first-order PDE, generally nonlinear because $H$ is typically quadratic in the momenta and hence quadratic in $\partial S/\partial q$. Trading $2n$ ODEs for one nonlinear PDE is not obviously progress; for a generic Hamiltonian it is a strictly worse deal. The trade pays off in exactly one circumstance: when the PDE separates into $n$ independent one-dimensional problems, each solvable by a single quadrature. When it does — and Separation of Variables in Hamilton-Jacobi Theory maps out when that happens — the method becomes the most powerful integration engine classical mechanics possesses. What must a useful solution of (2) look like? Not the general solution, which involves arbitrary functions, but a complete integral: a solution $S(q_1,\dots,q_n;\,\alpha_1,\dots,\alpha_n;\,t)$ depending on $n$ independent, nonadditive constants $\alpha_i$ — one constant per coordinate, with independence meaning $\det(\partial^2 S/\partial q_i\,\partial\alpha_j) \neq 0$. That determinant condition is what lets the $\alpha_i$ serve as honest new momenta and makes the implicit relations in (1) solvable. (An additive constant $S \to S + c$ changes nothing, since only derivatives of $S$ appear; that is why the count is $n$ and not $n+1$.) With a complete integral in hand, the new variables are constants and the trajectory is recovered by differentiation and inversion:

where: $\beta_i$ — the $n$ frozen new coordinates. The middle set of relations is the heart of the payoff: $n$ algebraic equations connecting the $q_j$, the time $t$, and $2n$ constants. Solve them for $q_j(t;\alpha,\beta)$ — algebra, not integration — then substitute into the last relation to read off $p_j(t)$. Together $(\alpha_i,\beta_i)$ are exactly the $2n$ integration constants that $n$ second-order equations of motion demand: the count closes perfectly, which is the formal sign that no information was lost in the freezing.

Which equation answers which question

There is something suspicious about $K = 0$. The Hamiltonian is (often) the energy — how can a coordinate change abolish the energy? The resolution is that $K$ is not obtained from $H$ by substitution alone; the extra term $\partial S/\partial t$ is an honest reassignment of bookkeeping, and the function whose time derivative can pace the dynamics exactly is the action. Along the actual motion, the generator obeys a remarkable identity. Take the total time derivative of $S(q,\alpha,t)$ along a trajectory (the $\alpha_i$ are constants, so only $q$ and $t$ vary):

where: the first equality is the chain rule, the second uses $p_i = \partial S/\partial q_i$ together with the Hamilton-Jacobi equation $\partial S/\partial t = -H$, and the third is the inverse Legendre transformation. Integrated,

where: the integral runs along the true trajectory. The generator that trivializes the Hamiltonian is Hamilton's principal function — the classical action accumulated along the motion, studied as an object in its own right in Hamilton's Principal Function. This is why the cancellation works: $S$ grows at rate $L = \sum p\dot q - H$, so its explicit time dependence contributes exactly $-H$ after the $\sum p\dot q$ part is accounted to the coordinate dependence. The action is the unique ledger whose columns balance the Hamiltonian to zero.

For conservative systems there is a standard first simplification that the rest of the chapter leans on. When $\partial H/\partial t = 0$, the time dependence of $S$ can be split off additively:

where: $W$ — Hamilton's characteristic function, and $E$ — the conserved energy, which joins (or generates) the constant set $\alpha_i$. The reduced, time-free PDE on the right is the Time-Independent Hamilton-Jacobi Equation, and the object $W$ — which generates its own, subtly different canonical transformation with $K = E$ rather than $K = 0$ — is the subject of Hamilton's Characteristic Function. Here it is enough to note the division of labor: $S$ freezes everything; $W$ freezes the momenta and lets one clock-like coordinate run. Both live inside the same trivial-Hamiltonian program.

1 · A thrown ball, frozen and thawed (confidence builder)

A ball of mass $m = 0.5\ \mathrm{kg}$ moves vertically in uniform gravity, $H = p^2/2m + mgq$ with $g = 9.8\ \mathrm{m/s^2}$. Its energy is $E = 100\ \mathrm{J}$ and its frozen coordinate is $\beta = -2\ \mathrm{s}$ (the constants conjugate to $\alpha = E$ carry units of time here). Find the trajectory, the apex height, and the position and momentum at $t = 0$.

Step 1. Solve the Hamilton-Jacobi equation. Since $H$ is conserved, use $S = W - Et$ with $\tfrac{1}{2m}(W')^2 + mgq = E$, so $W' = \sqrt{2m(E - mgq)}$ (upward branch) and $S = \int\!\sqrt{2m(E-mgq)}\,\mathrm{d}q - Et$, taking $\alpha = E$ as the one constant.

Step 2. Freeze and differentiate. The conjugate constant is $\beta = \partial S/\partial E = \int \frac{m\,\mathrm{d}q}{\sqrt{2m(E-mgq)}} - t = -\frac{\sqrt{2m(E-mgq)}}{mg} - t = -\frac{p}{mg} - t$, using $p = \partial S/\partial q = \sqrt{2m(E-mgq)}$.

Step 3. Thaw by algebra. Solving the Step 2 relation: $p(t) = -mg\,(t + \beta)$, and substituting into $p^2 = 2m(E - mgq)$ gives $q(t) = \dfrac{E}{mg} - \dfrac{g}{2}\,(t+\beta)^2$ — the parabola of projectile motion, recovered without integrating a single equation of motion. The apex sits at $t = -\beta = 2\ \mathrm{s}$, height $q_{\max} = E/mg = 100/(0.5\times 9.8) = \boxed{20.4\ \mathrm{m}}$.

Step 4. Evaluate at $t=0$: $p(0) = -0.5\times 9.8\times(0-2) = \boxed{+9.8\ \mathrm{kg\,m/s}}$ (moving upward, speed $19.6\ \mathrm{m/s}$) and $q(0) = 20.41 - \tfrac{9.8}{2}(2)^2 = 20.41 - 19.6 = \boxed{0.81\ \mathrm{m}}$.

Step 5. Check the freeze. Energy: $p^2/2m + mgq = (9.8)^2/1.0 + 4.9\times 0.81 = 96.04 + 3.96 = 100.0\ \mathrm{J} = E$ ✓. The two constants $(\alpha,\beta) = (100\ \mathrm{J}, -2\ \mathrm{s})$ encode the whole flight: $\alpha$ fixes the shape (apex height), $\beta$ fixes the timetable (when the apex occurs).

2 · A time-dependent force, where energy methods fail (exam level)

A particle of mass $m = 2\ \mathrm{kg}$ feels a linearly growing force $F(t) = \lambda t$ with $\lambda = 3\ \mathrm{N/s}$, so $H = \dfrac{p^2}{2m} - \lambda t\,q$. Energy is not conserved and no characteristic function exists — the full principal function is required. Solve by Hamilton-Jacobi with the ansatz $S = a(t)\,q + b(t)$, then find $q$ and $p$ at $t = 2\ \mathrm{s}$ given $\alpha = 4\ \mathrm{kg\,m/s}$ and $\beta = 1\ \mathrm{m}$.

Step 1. Insert the ansatz into the HJ equation $\tfrac{1}{2m}(\partial S/\partial q)^2 - \lambda t q + \partial S/\partial t = 0$. With $\partial S/\partial q = a(t)$ and $\partial S/\partial t = \dot a\,q + \dot b$:

Step 2. Match coefficients — the equation must hold for all $q$ (the PDE is a pointwise identity, not an on-shell statement). Thus $\dot a = \lambda t$ giving $a(t) = \tfrac{\lambda t^2}{2} + \alpha$, and $\dot b = -a^2/2m$ giving $b(t) = -\tfrac{1}{2m}\displaystyle\int_0^t \left(\tfrac{\lambda \tau^2}{2} + \alpha\right)^{\!2}\mathrm{d}\tau$. The constant $\alpha$ is the single new momentum.

Step 3. Freeze and differentiate: $\beta = \dfrac{\partial S}{\partial \alpha} = q - \dfrac{1}{m}\displaystyle\int_0^t\left(\tfrac{\lambda\tau^2}{2}+\alpha\right)\mathrm{d}\tau = q - \dfrac{1}{m}\left(\dfrac{\lambda t^3}{6} + \alpha t\right)$.

Step 4. Thaw: $q(t) = \beta + \dfrac{\alpha}{m}\,t + \dfrac{\lambda}{6m}\,t^3$ and $p(t) = \partial S/\partial q = a(t) = \alpha + \dfrac{\lambda t^2}{2}$. Sanity check against Newton: $\dot p = \lambda t = F(t)$ ✓, and $\alpha = p(0)$, $\beta = q(0)$ — the frozen constants are the initial conditions.

Step 5. Numbers at $t = 2\ \mathrm{s}$: $p = 4 + \tfrac{3\times 4}{2} = \boxed{10\ \mathrm{kg\,m/s}}$ and $q = 1 + \tfrac{4\times 2}{2} + \tfrac{3\times 8}{12} = 1 + 4 + 2 = \boxed{7\ \mathrm{m}}$.

Step 6. The moral: with $H$ explicitly time-dependent there is no conserved energy, no $W$, no energy-surface picture at all — yet $K\equiv 0$ still froze the system into two constants $(\alpha,\beta) = (4,1)$ and returned the exact cubic trajectory. The trivial-Hamiltonian program is strictly more general than energy conservation.

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 10 · Landau & Lifshitz, Mechanics, 3rd ed., §47 · Morin, Introduction to Classical Mechanics, Ch. 15 (Hamiltonian methods) · Taylor, Classical Mechanics, Ch. 13 · Arnold, Mathematical Methods of Classical Mechanics, §47 All equations are standard results of Hamilton-Jacobi theory. The key relations are: the type-2 transformation rule $K=H+\partial F_2/\partial t$ with $p_i=\partial F_2/\partial q_i$ and $Q_i=\partial F_2/\partial P_i$; the Hamilton-Jacobi PDE obtained by imposing $K=0$; the free-particle result $q(t)=\beta+(\alpha/m)t$; and the identity $\mathrm{d}S/\mathrm{d}t=L$. Worked-example numbers on this page (apex $20.4\ \mathrm{m}$, $q(0)=0.81\ \mathrm{m}$, $p(0)=9.8\ \mathrm{kg\,m/s}$; cubic trajectory $q(2)=7\ \mathrm{m}$, $p(2)=10\ \mathrm{kg\,m/s}$) follow directly from the boxed formulas.

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