CLASSICAL MECHANICS / VISUAL NOTE

The path nature chooses

Least action says the real trajectory is balanced against every tiny detour: nudge the path and, to first order, its action does not change.

That global rule becomes the local Euler–Lagrange equation.
CM–01 / ONE LINKED EXPERIMENT

Nudge the path. Watch the action respond.

The question: which complete trajectory makes the action locally flat?

START HERE

Drag η once. The orange trajectory and orange point move together. Your goal is to make the action slope on the right become zero.

01 / CHANGE THE MOVIE

A neighboring path

The start, finish, and travel time stay fixed. Only the history between them changes.

Building path space…
t = 0t = 1 s
PATH NUDGE
+0.32 m
MIDPOINT
1.55 m
ENDPOINTS
fixed
02 / READ THE SCORE

Its position in action space

The same η becomes a point on the cyan curve. Its tangent measures the first-order response.

Building action space…
TARGET · η = 0− variation+ variation
ACTION CHANGE
0.25 J·s
FIRST VARIATION
+1.58
TARGET
0.00
ONE CONTROL / BOTH VIEWSTest a neighboring pathMove toward the center to recover the physical trajectory.
η +0.32 m
NOT STATIONARY YET
This nudge changes the action at first order.

The local slope is +1.58 J·s/m. Move η toward zero until the tangent becomes horizontal.

δS=0⟹ddt(∂L∂q˙)−∂L∂q=0\delta S=0\quad\Longrightarrow\quad\frac{d}{dt}\left(\frac{\partial L}{\partial\dot q}\right)-\frac{\partial L}{\partial q}=0
THEORY / AFTER THE VISUAL

From a whole path to a local law

The visualization compares complete histories. The derivation explains why the winning history can be identified by a differential equation at every instant.

01

Perturb the path

Start with a candidate path and add a small, otherwise arbitrary deformation. The deformation must vanish at the fixed endpoints.

qε(t)=q(t)+εη(t),η(0)=η(T)=0q_\varepsilon(t)=q(t)+\varepsilon\eta(t),\qquad \eta(0)=\eta(T)=0
02

Differentiate the action

The first variation measures the initial slope of the action as the path moves in the direction η.

δS=ddεS[q+εη]∣ε=0=∫0T ⁣(∂L∂qη+∂L∂q˙η˙)dt\delta S=\left.\frac{d}{d\varepsilon}S[q+\varepsilon\eta]\right|_{\varepsilon=0}=\int_0^T\!\left(\frac{\partial L}{\partial q}\eta+\frac{\partial L}{\partial \dot q}\dot\eta\right)dt
03

Move the derivative

Integration by parts transfers the time derivative from the arbitrary variation η to the momentum-like term. The boundary contribution disappears because η is zero at both ends.

δS=∫0T ⁣[∂L∂q−ddt(∂L∂q˙)]η(t) dt\delta S=\int_0^T\!\left[\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)\right]\eta(t)\,dt
THE FUNDAMENTAL STEP

η is arbitrary between the endpoints.

For the integral to vanish for every allowed deformation, the expression multiplying η must vanish point by point. That is the Euler–Lagrange equation.

Picture η as a tiny bump placed at one chosen instant. If the bracketed expression were nonzero there, that bump would change the action. Because the bump can be placed anywhere, the expression must be zero everywhere.

ddt(∂L∂q˙)−∂L∂q=0\boxed{\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)-\frac{\partial L}{\partial q}=0}
FOR THE BALL IN THE VISUALIZATION

With kinetic energy 12mq˙2\tfrac12m\dot q^2 and gravitational potential mgqmgq, the local law becomes ordinary constant-acceleration motion.

L=12mq˙2−mgq⟹mq¨+mg=0L=\frac12m\dot q^2-mgq\qquad\Longrightarrow\qquad m\ddot q+mg=0
INTUITION / PLAIN LANGUAGE

Think of a path as a whole movie.

A path is not one position of the ball. It is the complete movie: where the ball is at every moment from the fixed start to the fixed finish.

The action gives that entire movie one score. It rewards and penalizes different combinations of motion and height, compressing the whole trajectory into a single number.

Now reshape the movie by an almost invisible amount. For most invented paths, the score immediately moves up or down. At the physical path, those first-order changes cancel. The score is locally flat.

  1. 01Whole motionChoose a complete path.
  2. 02One scoreCompute its action.
  3. 03Tiny nudgeReshape the path slightly.
  4. 04No first-order changeThat path is stationary.
WHY MUST THE FIRST-ORDER CHANGE VANISH?

If a tiny nudge changed the action linearly in one direction, reversing that same nudge would reverse the sign and lower the action. A minimum cannot allow that. Its linear change must be zero, leaving only second-order and smaller effects.

Nature is not previewing every possible future.

Stationary action and Newton’s local force law are mathematically equivalent descriptions of the same motion. Euler–Lagrange is the frame-by-frame rule hidden inside the whole-movie statement.

Further reading: Feynman Lectures, Vol. II, Chapter 19 — The Principle of Least Action.