The principle of least action

Drag a knot ○ up or down and watch S go up

The action of every path you try

Experiments

Start here

The blue curve is a path a particle could take between two fixed events: where it is, up the plot, against time, across. The table above scores it by its action S, beside the true path's, the path the particle really takes.

Drag a knot ○ up or down and watch S go up; drag it sideways to change when the particle passes there. Try a wrong path bends the whole path in one click, and Back to the true path returns.

Look for this: nothing you draw gets below the true path's S (until the experiments about saddles), and the record beside the path keeps every value you tried. The theory is at the end, under How it works.

Drawing a path. The path runs through knots (○), and each knot is an event: a place and a time. Drag a knot up or down to change where the particle is then; drag it sideways to change when it gets there. A drag moves the knot one way only, whichever way you start (hold Shift to move it both ways at once). The end points (□) keep their times; drag one up or down to change where the trip starts or ends, and the true path follows. The velocity is the slope: point at the path to read it, and while you drag a knot the velocities just before and after it show. From the keyboard: click the plot, choose a knot with [ and ], move it with the arrow keys, and press R to go back to the true path.

Reading it. The table at the top compares your path with the true one: the action $S$, its kinetic part $\int\tfrac12 m\dot x^2\,dt$ and its potential part $-\int V\,dt$, and $\Delta S$, yours less the true path's. Beside the path, every change you make adds a point to the record of $S$, newest at the right, with the true path's $S$ dashed: you see yourself going up, and coming back down. The buttons under Try a wrong path make the obvious wrong ones (Rush, then dawdle passes the same places at the wrong times), and Back to the true path returns. Under the path, choose what else to see: the path in phase space, its energy, or Feynman's arrows.

Every experiment starts on the true path (dashed, under yours until you move it), so whatever you do shows what it costs.

1. Throw and catch: the parabola is a balance

This is the opening set-up. A ball thrown straight up and caught where it was thrown, $T=2$ s later, follows $x=\tfrac12 gt(T-t)$, with $S=-mg^2T^3/24=-33.333$ ($m=1$, $g=10$, $V=mgx$). The table splits it: $\int\tfrac12 m\dot x^2dt=33.333$ and $-\int V\,dt=-66.667$. Drag the top knot up a little: $S$ goes up, and so does its kinetic part. Then press Bend ↑: the ball goes higher, and spends longer high up, so $-\int V\,dt$ falls to $-85.04$, which lowers $S$; but it must move faster, and $\int\tfrac12 m\dot x^2dt$ rises to 54.27. $S=-30.77$: higher than the true path. Go back and press Bend ↓: now the kinetic part falls to 17.52, but the potential part rises to $-48.29$, and again $S=-30.77$. The potential part changed by $\mp18.38$, and the kinetic part by the same $\pm18.38$ plus 2.56 both times: the first-order changes cancel, and what is left is second order, the same either way. That is what stationary means. Straight line keeps the ball on the ground: $S=0$.

2. Try to beat the straight line

A free particle ($V=0$, $m=1$) must get from $x=0$ to $x=4$ in $T=2$ s. The true path is a straight line at constant speed, with $S=m(\Delta x)^2/2T=4$. Try to beat it. Drag the middle knot up from $x=2$ to $x=3$: $S=8.13$. Drag it sideways instead, from $t=1$ to $t=1.3$, so the particle gets there late and must hurry afterwards: $S=8.08$. Rush, then dawdle gives 4.32. Nothing gets below 4, because with $V=0$ the action is $\tfrac12 m\int\dot x^2dt$, and the average of $\dot x^2$ is never less than the square of the average of $\dot x$, which is fixed at $\Delta x/T$: equal only at constant speed. Hold Shift and slide the knot along the line itself: $S$ hardly changes (it is 4.000 while the knot is on the line), as the path is the same line through a different point of it.

3. Which way is downhill?

The throw again, but the path starts bent (Bend ↓, 7 knots, $S=-30.775$). The arrows show which way each knot should go to lower $S$ fastest, up or down and sooner or later. Follow them by hand, or press Roll downhill: every knot slides up or down by its own slope, and the roll stops, after 253 steps, on the true path again, $S=-33.333$. The arrows' up-down part is the Euler-Lagrange equation itself: $$\delta S=\int\left(\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial\dot x}\right)\delta x\,dt,$$ so a knot's slope is the bracket averaged around the knot, and where the bracket is zero everywhere no small change helps. Their sideways part, $\partial S/\partial t_k$, is the next experiment. (Roll downhill leaves the knots' times alone: the path settles on the true path through knots at the times they have.)

4. Only the true path conserves energy

The view under the path plots the energy $E=\dot x\,\partial L/\partial\dot x-L=\tfrac12 m\dot x^2+V$ along your path. On the true path it is flat, $E=50$ (the ball's $\tfrac12 m(gT/2)^2$ as it leaves the hand). Drag a knot and it wanders. Now choose Straight segments: each segment has its own constant speed and its own energy (with $V$ averaged along it), and the best segments through these 6 knots all have $E=48.0$. Drag a knot sideways: the energies of the segments on either side part, and the slope of $S$ with the knot's time is exactly the jump, $$\frac{\partial S}{\partial t_k}=E_{\text{after}}-E_{\text{before}},$$ because a segment of duration $h$ has $S=m\Delta x^2/2h-hV$, and $\partial S/\partial h=-E$. So asking $S$ to be stationary as knots move in time is asking the energy not to jump: energy conservation, which the action principle gives whenever $L$ does not depend on $t$ (Noether's theorem).

5. Phase space: the true path keeps to one energy

A harmonic oscillator, $L=\tfrac12 m\dot x^2-\tfrac12 m\omega^2x^2$, from $x=1$ to $x=0.5$ in $T=2$: $S=\frac{m\omega}{2\sin\omega T}\left[(x_1^2+x_2^2)\cos\omega T-2x_1x_2\right]=-0.8359$. The view draws your path in phase space, $(x,p)$ with $p=\partial L/\partial\dot x=m\dot x$, over the contours of $H=p^2/2m+\tfrac12 m\omega^2x^2$, which are ellipses. The true path runs along one of them, and with its two ends at different places it covers only an arc. Drag a knot and yours crosses the contours: its energy changes, though nothing acts to change it.

6. A closed orbit

To be a loop in phase space a path must end where it began, $x(T)=x(0)$, with the same momentum. Choose Periodic (for the oscillator and the pendulum): the first and last knots become one point (■, drag either), and the path is smooth across its ends. Why that is the true orbit's own condition: with the shared end $a=x(0)=x(T)$ free to move, $\partial S/\partial a=p(T)-p(0)$, so a path that is stationary in $a$ too has $p(T)=p(0)$, and closes up.

Here a pendulum, $L=\tfrac12\dot\theta^2+\cos\theta$ ($m=g=\ell=1$), with $T=8$. A swing takes longer the bigger it is, from $2\pi$ for small ones, so $T=8$ picks the swing that takes 8: amplitude 1.86 rad (107°), energy $E=\tfrac12\dot\theta^2-\cos\theta=0.288$. The true orbit goes once round its contour, with $S=\oint p\,d\theta-ET=6.689$. Two orbits pass through the start, moving up first or down first. Drag the start point to $\theta=0.3$ and go back to the true path: the orbit through it is the same loop, started elsewhere, with the same $S$. Shorter than $2\pi$, no swing fits; longer than $4\pi$, a swing of half the period, made twice, fits too.

A closed orbit is a saddle of $S$. Bend ↑ and $S$ rises, to 8.358; but drag the start point a little and press Roll downhill, and the whole loop climbs to the top: balanced upside down, $\theta=\pi$, the pendulum has the least action of all, $S=-T=-8$, since $L=\tfrac12\dot\theta^2+\cos\theta$ is never below $-1$.

For the harmonic oscillator every swing takes $2\pi/\omega$: there Periodic fixes $T$ at one period, every amplitude is a true orbit, and every one has $S=\tfrac12 m[x\dot x]_0^T=0$. Bend ↑ and the loop grows, with $S$ still 0.000.

7. When the action is not least

The oscillator again, but now the trip takes $T=4$, longer than half a period ($\pi/\omega=\pi$). The true path still makes $S$ stationary, $S=1.2005$, but as a saddle. Drag the third knot up by 0.5: $S=1.659$, up, as at a minimum. Now press Bend ↑, which moves every knot at once along half a sine wave: $S=1.002$, below the true path's; Bend ↓ the same. Roll downhill from there runs away: the bend grows until the path hits the edge of the plot, $S=-9.47$. Why: for a change $\delta x=a\sin(\pi t/T)$ the action changes by $$\delta^2S=\frac m2\int_0^T\left(\delta\dot x^2-\omega^2\delta x^2\right)dt=\frac{ma^2T}{4}\left[\left(\frac\pi T\right)^2-\omega^2\right],$$ negative once $T>\pi/\omega$. Set $T=3$ and bending raises $S$ again (from $-7.923$ to $-7.885$). At $t=\pi/\omega$ every true path leaving $x_1$ arrives at $-x_1$, whatever its speed: a conjugate point. Past the first one the action is no longer least, only stationary; the number of conjugate points, the Morse index, counts the independent directions in which $S$ goes down.

8. More than one true path

A pendulum from $\theta=0$ to $\theta=1$ in 8 seconds. Five motions do it, each a true path, listed under the plots with its action and Morse index:

$S=-1.924$, a minimum: rising slowly almost to the top (3.03 rad, 174°), and back down to 1;
$S=1.899$, also a minimum: the other way first, almost to the top on that side ($-2.95$), and back up through 0 to 1;
$S=5.115$, a saddle (index 1): the other way first to $-2.09$, across to $2.09$, and back down to 1;
$S=8.051$ (index 1) and $S=8.057$ (index 2): up, down to the other side and back up to 1, swinging to $\pm1.35$ and to $\pm1.12$.

The path starts on the one of least action, where Bend ↑ and Bend ↓ both raise $S$ (to 0.07 and $-0.15$). Click the third in the list: on that saddle, Bend ↑ lowers $S$, from 5.116 to 5.081.

9. The twin paradox

Time runs up and space across, in years and light-years ($c=1$). The end points are two events at the same place, 10 years apart, and the world line starts as the twin who stays home, drawn in straight legs. A free particle's action is $$S=-mc^2\int\sqrt{1-\dot x^2/c^2}\,dt=-mc^2\tau,$$ where $\tau$ is the time its own clock shows. Drag the middle event sideways to $x=1.5$: the traveller goes out and back at $0.6c$, each leg taking 2.5 years by the home clock and 2 by the traveller's, so $\tau=9$ and $S=-9$, more than the stay-at-home's $-10$. The straight world line has the least action, and so the most proper time. Drag an event up or down to change when the traveller is there. Past $x=2.5$ the legs would be faster than light: no proper time, no action.

10. Fermat's principle

Light from one point to another takes the path of least optical length $\int n\,ds$, which is $c$ times its travel time. Here air ($n_1=1$) is above and glass ($n_2=1.5$) below, and $x$ plays the part of time: a knot is a point the ray passes through. The true ray has optical length 13.955 (9 knots draw it to 13.964); the straight line, 14.577. It meets the surface at $\theta_1=69.0°$ from the normal and leaves at $\theta_2=38.5°$: $n_1\sin\theta_1=0.934=n_2\sin\theta_2$, Snell's law. Why: $L=n(y)\sqrt{1+y'^2}$ does not depend on $x$, so its "energy" $H=y'\,\partial L/\partial y'-L=-n/\sqrt{1+y'^2}=-n\sin\theta$ is conserved along the ray (see it flat in the Energy view). It is minus the light's momentum along the surface (in units of $\hbar\omega/c$), which a surface that is the same all along $x$ cannot change. Then try Light: mirage, $n=1+0.05y$, the air thinner low down, as over a hot road: the true ray between two points at $y=0$ dips to $y=-0.65$ (optical length 9.893, against 10.000 straight). Light from the sky reaches you from below, and the road looks wet.

11. The brachistochrone

A bead slides without friction from rest at the top left to the bottom right, and the "action" is the time it takes, $\int ds/v$ with $v=\sqrt{2gy}$, $y$ measured down. The fastest curve is a cycloid, $x=a(\phi-\sin\phi)$, $y=a(1-\cos\phi)$: here $a=1$, $g=1$, and it takes $\pi\sqrt{a/g}=3.1416$ (8 knots draw it to 3.156). Press Straight line: the incline is shorter but slower, 3.724. Bend ↑ (shallower): 3.223; Bend ↓ (deeper): 3.209. Diving steeply at first gains speed early. As for light, $x$ plays the part of time, and the start stays put.

12. Straight segments: more knots, less action

Feynman's way of thinking, and E. F. Taylor's way of teaching: a path of straight segments, at constant speed along each. Here the throw again, through 4 knots. The path starts as the best one through them: its knots lie exactly on the parabola, every segment has the same energy (44.44, in the Energy view), and $S=-29.630$, above the true $-33.333$ by $g^2Th^2/24=3.704$, where $h=T/3$ is a segment's duration. Add knots with the Knots slider and it comes down as $h^2$: 5 knots give $-31.250$, 9 give $-32.813$, 16 give $-33.185$.

13. Feynman's arrows: why the classical path

In quantum mechanics every path contributes to the amplitude to get from one event to the other, each with an arrow $e^{iS/\hbar}$: length 1, turned by $S/\hbar$. The view takes one knot (ringed: click a knot to choose it) and lets it be at every height while the others stay put. Each height is a path; the view adds their arrows head to tail. (This is one of the integrals in the path integral: over the position at one time.) Near the height where $S$ is stationary the arrows point the same way and add up: the green part of the spiral, and the green band on the path plot. Away from it $S$ changes quickly, the arrows spin round, curl up into the ends of the spiral, and cancel. How near is near: for the throw's middle knot $S\approx S^*+\tfrac12S''(h-h^*)^2$ with $S''=8.27$, so the arrows within a quarter turn lie within $\sqrt{\pi\hbar/S''}$ of the true height: 0.62 at $\hbar=1$, 0.20 at $\hbar=0.1$, 0.06 at $\hbar=0.01$. Make $\hbar$ smaller and only paths ever nearer the true one count: the classical limit. Drag the knot away and your path's arrow (blue) moves into a curl, where its neighbours cancel it.

14. Friction from a Lagrangian

$L=e^{\gamma t}(\tfrac12\dot x^2-\tfrac12x^2)$ looks like a curiosity, but its Euler-Lagrange equation, $\frac{d}{dt}\left(e^{\gamma t}\dot x\right)=-e^{\gamma t}x$, is $\ddot x+\gamma\dot x+x=0$: a damped oscillator ($\gamma=0.3$ here). The Energy view plots $E=\dot x\,\partial L/\partial\dot x-L=e^{\gamma t}(\tfrac12\dot x^2+\tfrac12x^2)$, which wobbles even on the true path, because $L$ depends on $t$: it is $e^{\gamma t}$ times the mechanical energy $\tfrac12\dot x^2+\tfrac12x^2$, which decays. Over this trip ($T=6$) the true path is a saddle, index 1: the damped oscillator's first conjugate point is at $t=\pi/\omega_d=3.18$, with $\omega_d=\sqrt{1-\gamma^2/4}$.

15. Your own potential

Type any $V(x,t)$: a double well, $\tfrac14x^4-x^2$, to start with, from $x=-1$ to $x=1$ in $T=3$. The true paths are found numerically, by shooting: trying every starting speed and keeping those that arrive. There are six: $S=1.368$ (a minimum, straight over the hump), 4.360 twice (index 1: back out to $x=-2.17$ first, or past the end to $2.17$ and back), 11.02 and 30.40 (index 2) and 67.92 (index 3). The path starts on the one of least action. A potential that depends on $t$ does not conserve energy.

How it works

Between two events, a start $(t_1,x_1)$ and an end $(t_2,x_2)$, a particle could go along any path $x(t)$. Every path has an action, the kinetic energy less the potential energy added up along the way: $$S[x]=\int_{t_1}^{t_2}L(x,\dot x,t)\,dt,\qquad L=\tfrac12 m\dot x^2-V(x).$$ The path the particle really takes makes $S$ stationary: change it a little, any way you like, and $S$ changes only at second order. That is the Euler-Lagrange equation, $\frac{d}{dt}\frac{\partial L}{\partial\dot x}=\frac{\partial L}{\partial x}$, which for this $L$ is Newton's $m\ddot x=-dV/dx$. Usually the true path has the least action of all the paths near it (a minimum); over a long enough trip it can be a saddle, with some nearby paths below it, and the Morse index counts how many independent ways down there are (experiment 7).

What is computed: the path is a cubic spline (or straight segments) through the knots, and $S$ is its integral by Gauss quadrature between the knots; the true paths are found by shooting, trying every starting velocity and keeping those that arrive, and their actions are checked against the closed forms where there are any. A few knots can only approximate a curved true path, so the best path through them may have a little more action than the true one; the table says when that is all the difference is.

New in 2026 for BrainFlux. Source: github.com/gallicchio/brainflux, MIT licence.