Hamilton-Jacobi equation: Visual walkthrough — Physics deep dive

Step 0 — The stage: what "phase space" and a "trajectory" mean

Before any equation, fix the picture in your head.

A mechanical system with one degree of freedom has a position $q$ (where the thing is) and a momentum $p$ (how much motion it carries: mass times velocity for a simple particle).
Plot $q$ horizontally and $p$ vertically. This plane is phase space. One dot $(q, p)$ is the complete state of the system right now.
As time $t$ flows, the dot moves, tracing a curve — the trajectory.

Figure s01 — Phase space and one trajectory. The blue curve is the history; the yellow dot is the state right now.

Look at the blue curve: the dot at each instant is the system. Newton would push this dot step by step. We are going to find a change of coordinates that freezes the dot in a new frame.

Step 1 — The wish: coordinates where nothing moves

WHAT. We introduce brand-new coordinates $(Q, P)$ that describe the same physics, chosen so that in these new axes the trajectory is a single stationary point — it does not move at all.

WHY. If $\dot{Q} = 0$ and $\dot{P} = 0$ (recall the dot means $d / d t$ , the time rate of change), then $Q$ and $P$ are just fixed numbers. The motion is then read off by pure algebra: no differential equation left to solve. This is the entire dream of Hamilton-Jacobi.

PICTURE. Same trajectory, two coordinate grids. In the old $(q, p)$ grid the state sweeps a curve; in the new $(Q, P)$ grid it is one frozen dot.

Figure s02 — Old moving vs new frozen. The white arrow is the canonical map that sends the sweeping curve to a single green dot.

$old, moving (q, p) ⟶ new, frozen (Q, P)$

$q, p$ — old position and momentum, the moving dot.
$Q, P$ — new position and momentum; our goal is that these are constants.

The question of the whole chapter: which change of variables achieves this, and is it allowed?

Step 2 — "Allowed" means canonical: the rulebook we must obey

WHAT. Not every change of variables is legal. A legal one is a canonical transformation — it keeps the shape of the laws of motion intact.

WHY. Hamilton's Equations read $\overset{q}{˙} = \frac{\partial H}{\partial p}, \overset{p}{˙} = - \frac{\partial H}{\partial q} .$

$\overset{q}{˙}, \overset{p}{˙}$ — the time rates of change of position and momentum (dot = $d / d t$ ).
$H$ — the Hamiltonian, the system's energy written as a function of $q$ and $p$ .
$\partial H / \partial p$ — "how fast $H$ changes if I nudge $p$ while holding $q$ fixed"; it turns out to equal $\overset{q}{˙}$ .
The minus sign in the second equation is what makes phase-space flow swirl rather than spread — it is the signature of Hamiltonian dynamics.

A canonical transformation is precisely one after which the new variables obey the same-shaped equations with some new energy $K$ : $\dot{Q} = \frac{\partial K}{\partial P}, \dot{P} = - \frac{\partial K}{\partial Q} .$ $K$ is the new Hamiltonian (playfully, the "Kamiltonian").

PICTURE. Both grids carry the same swirl pattern. A canonical map is a bendy re-labelling of phase space that never breaks the swirl.

Figure s03 — The swirl is preserved. Each blue arrow is the flow $(\overset{q}{˙}, \overset{p}{˙})$ ; a canonical map must not tear this pattern.

Step 3 — The bridge between old and new: a generating function

WHAT. How do we actually specify a canonical map? Through a single generating function. We use the type-2 kind, $S (q, P, t)$ — it depends on the old position $q$ and the new momentum $P$ .

WHY type-2 and not another? We want the new momenta $P$ to be the constants we froze in Step 1. A type-2 function treats $P$ as an independent input, so it is the natural tool to "design constant new momenta."

WHERE the three relations come from. A canonical transformation can be built by demanding that the old and new descriptions give the same action principle. In the action principle the old variables contribute $p \overset{q}{˙} - H$ and the new ones contribute $P \dot{Q} - K$ . These two may differ only by the total time-derivative of some function — otherwise the variational principle would change and the map would not be canonical. Choosing that function to depend on $(q, P, t)$ (type-2), the bookkeeping is cleanest if we add $\frac{d}{d t} (P Q)$ , which gives $p \overset{q}{˙} - H = - Q \dot{P} - K + \frac{d S}{d t}, S = S (q, P, t) .$ Expand $\frac{d S}{d t} = \frac{\partial S}{\partial q} \overset{q}{˙} + \frac{\partial S}{\partial P} \dot{P} + \frac{\partial S}{\partial t}$ and match the coefficients of the independent rates $\overset{q}{˙}$ and $\dot{P}$ (and the leftover constant piece). Matching $\overset{q}{˙}$ gives $p = \partial S / \partial q$ ; matching $\dot{P}$ gives $Q = \partial S / \partial P$ ; the leftover gives $K = H + \partial S / \partial t$ :

$p = \frac{\partial S}{\partial q}, Q = \frac{\partial S}{\partial P}, K = H + \frac{\partial S}{\partial t} .$

Read each one as a sentence:

$p = \frac{\partial S}{\partial q}$ — the old momentum is the slope of $S$ as you slide $q$ (holding $P, t$ fixed). (This is the relation that will let us hunt momenta as derivatives.)
$Q = \frac{\partial S}{\partial P}$ — the new coordinate is the slope of $S$ as you slide the new momentum.
$K = H + \frac{\partial S}{\partial t}$ — the new energy equals the old energy plus how fast $S$ itself changes in time. (The explicit-time term $\partial S / \partial t$ is the knob we will exploit.)

WHEN is this a genuine transformation? For these slope relations to be invertible — so that from $(q, P)$ we can actually recover $(p, Q)$ and vice versa — the second-derivative matrix (the Hessian) of $S$ must not degenerate: $det (\frac{\partial ^{2} S}{\partial q \partial P}) \neq = 0.$ If this determinant were zero, the map would collapse dimensions and fail to be a bona-fide canonical transformation. We assume it holds.

PICTURE. $S$ sits like a landscape above the $(q, P)$ plane; its slopes in the two directions emit the values $p$ and $Q$ , while its slope in time feeds the extra term into $K$ .

Figure s04 — The $S$ -landscape emits $p$ and $Q$ . Slope in the $q$ -direction gives $p$ ; slope in the $P$ -direction gives $Q$ .

Step 4 — The killer demand: force the new energy to vanish

WHAT. We now make the boldest possible choice: demand $K \equiv 0 everywhere.$

WHY. Feed $K = 0$ into the new Hamilton's equations from Step 2: $\dot{Q} = \frac{\partial K}{\partial P} = 0, \dot{P} = - \frac{\partial K}{\partial Q} = 0.$

$\dot{Q} = 0$ — the new coordinate never changes (dot = $d / d t$ ).
$\dot{P} = 0$ — the new momentum never changes.

Both are constant — exactly the frozen dot of Step 1. We are buying trivial dynamics by imposing the strongest condition available on $K$ . This is not a random trick; $K = 0$ is the mathematical spelling of "make the new variables stand still."

PICTURE. A flat energy surface ( $K = 0$ over all of new phase space) means zero slope in every direction, so both flow rates die — the dot is nailed in place.

Figure s05 — Flat new energy freezes the dot. With $K = 0$ everywhere, every slope of $K$ vanishes, so $\dot{Q} = \dot{P} = 0$ .

Step 5 — Assemble the equation: substitute the rulebook into $K = 0$

WHAT. Take the type-2 relation for $K$ and set it to zero: $K = H + \frac{\partial S}{\partial t} = 0 ⟹ H + \frac{\partial S}{\partial t} = 0.$

Now use the other type-2 relation, $p = \partial S / \partial q$ , to eliminate every momentum inside $H$ . Wherever $H$ asked for $p$ , we hand it $\partial S / \partial q$ instead:

$H (q, \frac{\partial S}{\partial q}, t) + \frac{\partial S}{\partial t} = 0$

Term by term:

$H (q, \cdot, t)$ — the energy function, now fed the derivative of $S$ in the momentum slot.
$\frac{\partial S}{\partial q}$ — stands in for the old momentum $p$ .
$\frac{\partial S}{\partial t}$ — the time-slope of $S$ , which used to be the extra piece of $K$ ; setting $K = 0$ moved it here.

WHY this is a win. We started with a system of coupled ordinary differential equations (Hamilton's equations). We traded them for one PDE. That trade is only worthwhile because the PDE often separates (Step 7).

PICTURE. A flow chart: the two type-2 relations pour into the box $K = 0$ and out drops the HJ equation.

Figure s06 — Assembling the HJ equation. Both type-2 relations feed the demand $K = 0$ ; the HJ PDE drops out.

Step 6 — Reading the meaning: $S$ is the action

WHAT. First, name the actor we are about to meet.

Now follow $S (q, P, t)$ along the real trajectory and compute its total time rate of change. The chain rule gives $\frac{d S}{d t} = \frac{\partial S}{\partial q} \overset{q}{˙} + \frac{\partial S}{\partial t} .$

(The $\partial S / \partial P$ term drops out here because $P$ is a constant along the motion, so $\dot{P} = 0$ .) Now substitute the two facts we already proved:

$\frac{\partial S}{\partial q} = p$ (Step 3),
$\frac{\partial S}{\partial t} = - H$ (Step 5, rearranged),

so $\frac{d S}{d t} = p \overset{q}{˙} - H = L .$

$p \overset{q}{˙}$ — momentum times velocity ( $\overset{q}{˙} = d q / d t$ ).
$p \overset{q}{˙} - H$ — this combination is exactly the Lagrangian $L$ just defined.

Integrate: $S = \int L d t + const$ .

WHY this matters. $S$ is not an abstract convenience — it is the action, the very quantity minimized in Hamilton's principle. The HJ equation is the action obeying its own local law.

PICTURE. As the dot advances along the blue trajectory, $S$ accumulates the area $\int L d t$ ; the growth rate of $S$ at every instant is $L$ .

Figure s07 — $S$ accumulates the action. The shaded area under $L$ is the growing value of $S$ ; its slope at each instant equals $L$ .

Step 7 — Degenerate/simplifying case: when $H$ ignores time

WHAT. If $H$ has no explicit $t$ , energy is conserved. We separate variables by guessing the split $S (q, t) = W (q) - E t .$

$W (q)$ — the characteristic function, carrying all the $q$ -dependence.
$- E t$ — a clean straight-line-in-time piece; $E$ is the conserved energy.
Then $\frac{\partial S}{\partial t} = - E$ automatically.

WHY. Substituting into the HJ equation, the two terms peel apart: $H (q, \frac{\partial W}{\partial q}) - E = 0 ⟹ H (q, \frac{\partial W}{\partial q}) = E .$ The time has vanished; we are left with an ordinary differential equation for $W (q)$ . This only works because $H$ was time-blind — that is the degenerate condition.

PICTURE. $S$ as a surface over $(q, t)$ : a fixed shape $W (q)$ sliding downhill at constant rate $E$ as time increases — a rigid ramp translating in $t$ .

Figure s08 — Rigid ramp sliding in time. The shape $W (q)$ is fixed; increasing $t$ just lowers the whole curve at constant rate $E$ .

Step 8 — Sanity check on the simplest system: the free particle

WHAT. Let $H = \frac{p ^{2}}{2 m}$ (a mass $m$ with no forces). Its position is still our one coordinate $q$ — no new letter, the free particle simply lives on the $q$ -line.

Step-through (each move justified):

HJ: $\frac{1}{2 m} (\frac{\partial S}{\partial q})^{2} + \frac{\partial S}{\partial t} = 0$ — replace $p$ by $\partial S / \partial q$ .
Separate $S = W - E t$ : $\frac{1}{2 m} (W^{'})^{2} = E \Rightarrow W^{'} = 2 m E$ — energy conserved, peel off time. Here $W^{'} \equiv d W / d q$ .
Integrate: $W = 2 m E q$ , so $S = 2 m E q - E t$ .
Take $P = E$ as the new momentum. The new coordinate is constant: $Q = \frac{\partial S}{\partial E} = \frac{m}{2 E} q - t \equiv - t_{0} .$ Here $t_{0}$ is simply the integration constant — it is the fixed value the constant $Q$ takes, and it has the units of a time, so we name it $- t_{0}$ ; physically $t_{0}$ is the instant at which the particle passes $q = 0$ (its initial-time offset).
Solve for $q$ : $q = \frac{2 E}{m} (t - t_{0})$ — uniform velocity $v = 2 E / m$ , and at $t = t_{0}$ the particle is at $q = 0$ . ✅

WHY. This is the humblest possible test: a free particle must move in a straight line at constant speed, and HJ reproduces exactly that. The machinery is trustworthy.

PICTURE. In old space, $q$ marches linearly with $t$ ; in new coordinates the pair $(Q, P) = (- t_{0}, E)$ is a single frozen dot — the promise of Step 1 delivered.

Figure s09 — Free particle recovered. The blue line is $q (t)$ at constant speed; the red dot marks $t_{0}$ where $q = 0$ ; the green square is the frozen new state.

The one-picture summary

Everything above collapses into a single arrow diagram: from the two type-2 relations, through the $K = 0$ demand, to the HJ equation, and out to frozen constants $Q, P$ plus the recovered trajectory.

Figure s10 — The whole derivation in one picture. Type-2 relations → demand $K = 0$ → HJ PDE → ( $S$ is the action) → frozen $Q, P$ → trajectory by algebra.

Recall Feynman retelling — the whole walk in plain words

You have a dot wandering through a swirl (phase space). You wish for a magic pair of goggles that makes the dot sit perfectly still. The goggles are a canonical re-labelling — allowed only if it keeps the swirl intact. To build them you invent one landscape function $S$ : its slopes automatically tell you the old momentum ( $p = \partial S / \partial q$ ) and the new position ( $Q = \partial S / \partial P$ ), as long as the landscape is not degenerate (its curvature determinant is non-zero). Then you make the boldest wish possible: let the new energy be exactly zero. Zero energy means zero slope everywhere, means the dot can't move — frozen, just as desired. Writing that wish out, "old energy plus the time-slope of $S$ equals zero," and swapping every momentum for a slope of $S$ , gives you one equation: the Hamilton-Jacobi equation. The new momentum $P$ just rides along as a dial you set. And the punchline: that landscape $S$ turns out to be the action you already knew — its growth rate is the Lagrangian $L = p \overset{q}{˙} - H$ . Solve one PDE, follow the frozen constants back, and the whole motion drops out by algebra. That is the magic map that walks the maze for you.

Recall Rapid self-test

In new coordinates, what makes $\dot{Q} = \dot{P} = 0$ ? ::: The demand $K = 0$ , since $\dot{Q} = \partial K / \partial P$ and $\dot{P} = - \partial K / \partial Q$ .
Which type-2 relation turns $H$ into a PDE? ::: $p = \partial S / \partial q$ .
As the HJ PDE is solved, what role does $P$ play? ::: A held-fixed parameter (a dial); each value gives one solution, and $P$ becomes the conserved new momentum.
What condition lets the generating function define a real canonical map? ::: A non-degenerate Hessian, $det (\partial^{2} S / \partial q \partial P) \neq = 0$ .
What is $S$ physically? ::: The action $\int L d t$ along the real path, with $L = p \overset{q}{˙} - H$ .
When does $W$ exist and what equation does it solve? ::: When $H$ is time-independent; $H (q, \partial W / \partial q) = E$ .
Free-particle answer for $q (t)$ , and what is $t_{0}$ ? ::: $q = 2 E / m (t - t_{0})$ ; $t_{0}$ is the integration constant, the instant the particle passes $q = 0$ .

Where next: Action-Angle Variables apply this machinery to periodic motion, and the Schrodinger Equation is what HJ becomes when you let $S$ live inside an exponential.