Part 3 of 13. Noether's theorem is about systems described by a Lagrangian, a function whose integral along the path actually taken is stationary. This part sets up that framework and measures it.
Noether's theorem is about a group acting on a physical system, and the system is specified by a Lagrangian . Its integral along a path is the action, and the path actually followed makes the action stationary.
In June 1696 Johann Bernoulli asked the readers of the Acta Eruditorum which curve between two points at different heights carries a bead, released from rest, from one to the other in the least time. Leibniz, Jakob Bernoulli, Tschirnhaus, l'Hôpital and (anonymously) Newton all found the answer, a cycloid. What is minimized is a number attached to a whole curve,
with the bead's speed fixed by energy conservation.
At t = 0.583 s the cycloid bead has arrived; the others are still sliding. Arrival times: straight 0.639 s, sagging arc 0.600 s, cycloid 0.583 s.
Figure 1. Three beads released together from rest, g = 9.81 m/s², end 1 m across and 1 m down, each integrated along its curve from energy conservation. The default frame is the cycloid bead's arrival. The shortest path is the slowest.
Restricting to a one-parameter family of curves turns that functional into an ordinary function you can plot.
a = 0.350: T = 0.6000 s.
Figure 2. Travel time over the family (amber), integrated as in Figure 1. The family's best member still misses the cycloid's time (dashed), because the cycloid is not in the family. Drag on the plot or use the slider.
In 1834 Hamilton made this a rule for all of mechanics: the path between fixed endpoints is a stationary point of the action
For a ball of unit mass thrown straight up, is kinetic minus potential energy:
Nothing in that formula says "parabola". Figure 3 lets the action find it by lowering one knot at a time.
S = 0.0000 for the flat path.
Figure 3. A trial path (blue) of 8 straight segments with , its action computed exactly. Each sweep moves every knot to the height minimizing with its neighbours fixed; the second panel plots per sweep. The knots settle onto the true motion (dashed violet), whose smooth action the segmented path approaches from above. Drag a knot or use arrow keys (left and right pick one), then relax.
The flat path has action zero. The parabola beats it by spending the middle of the second high up, where is very negative, and paying in kinetic energy near the ends.
Perturb a path by , vanishing at the endpoints, and integrate by parts. Noether's theorem reuses exactly this computation.
Call the bracket the residual . If every bump must leave the action unchanged to first order, the residual must vanish:
For the thrown ball this is . Figure 4 checks step (4) by actually perturbing paths.
Figure 4. Top: a thrown-ball path and the bumped path ( exaggerated). Bottom: residual (amber), bump (blue), their product (shaded). The readout compares a finite difference of the action at with . On the true parabola both are zero wherever the bump sits.
The derivation only uses stationarity, and over long enough intervals the true path is not a minimum. Take an oscillator pinned at at both ends; staying at rest is a solution, and the action of is exactly times that of .
Figure 5. Oscillator with (period 2 s). Top: variations , , on . Bottom: action of , integrated numerically. Mode turns downward once s, and rest becomes a saddle. The readout checks the curvature against .
For a pendulum of length , with angle from the downward vertical,
with Euler-Lagrange equation . The difference swings through both signs; the sum stays put, and Hamilton's principle does not say why. Part 6 will: does not depend on time.
T = 0.00 · V = 4.91 · L = −4.91 · E = 4.91
Figure 6. A pendulum released at with , integrated with RK4 at a fixed 1/240 s step, over a 4-second window. Energy (green) holds; its largest relative departure in the readout is the integrator's own error.
Noether's theorem adds one assumption to the derivation above, a continuous symmetry of the Lagrangian, and gets a conservation law out. The next part asks what such a symmetry is.