Part 5 of 13. Every continuous symmetry of a Lagrangian gives a quantity that stays constant along the motion. The proof is five lines; the figures measure it.
Part 4 sorted transformations into strict symmetries, which leave unchanged, and quasi-symmetries, which change it by a total derivative . Either kind gives a quantity
that is constant along every solution of the equations of motion. This is the first theorem of Noether's 1918 paper Invariante Variationsprobleme (Nachr. Ges. Wiss. Göttingen, 235–257), stated there for fields as well as particles.
Steps (2) to (4) never use the symmetry. On any solution, for any generator, . The symmetry is needed only at the last step, to make the left side a total derivative as well. Figure 1 checks the identity on a pendulum with two generators that are not symmetries.
Figure 1. A pendulum, , released from and sampled every 0.1 s for 6 s. Each dot plots against , both by finite differences of the sampled path, never from the equation of motion. Dots from the true motion (green) lie on the diagonal. Dots from a trial path bent by (red) miss it by the Euler-Lagrange residual times .
Two blocks on a frictionless track, joined by a spring: . Shifting both blocks together, , leaves the stretch alone, so and Noether gives , the total momentum. A second spring from block 1 to the wall adds , which a shift does change.
Figure 2. Top: the blocks (), block 1 launched at speed 0.6, integrated with RK4 at and replayed on a 10 s loop. Bottom: the state in the plane of the two momenta. With each momentum swings, but the point never leaves the dashed line . Turn on the wall spring and the path spreads off it.
How far off depends on how badly the symmetry is broken. Figure 3 runs the same experiment for every on the slider, at two step sizes.
Figure 3. Largest change in total momentum over 10 s against , on log axes, at (circles) and 0.0025 (crosses). Halving the step does not move the points, so the drift is physics. At what remains is rounding: the spring pushes the blocks equally and oppositely at every RK4 stage, so the sum cannot change. Click a point to load it into Figure 2.
Run the derivation without step (1) and it still says something. Steps (2) to (4) give on every solution, so
For the pendulum under , and Noether's formula gives , which gravity changes at exactly that rate. Figure 4 pairs three Lagrangians with three generators and checks the formula for each.
Figure 4. Each cell integrates its Lagrangian for 6 s (RK4, ) and plots , with the range of Q below it. Only two of the nine pairs are symmetries, and both are on the free particle. The large plot shows the selected cell: solid, and dashed, computed separately by trapezoid rule from along the path. The two agree wherever Q drifts.
A ball in uniform gravity has . A sideways shift is a strict symmetry, so is conserved. An upward shift changes by the constant , which is a total derivative: . With Noether gives . Leave out F and you get , which falls at rate .
Figure 5. Left: the flight (), integrated by velocity Verlet at , a dot every 0.1 s. Right: each dot's velocity drawn from one origin. The tips stack on a vertical line, so holds, and step down evenly, so does not. The green segments add to each tip and bring every one back to the launch velocity. Drag the ringed tip, or focus it and use the arrow keys, to change the launch.
Energy needs a generator that moves time rather than , which is Part 6, along with angular momentum from rotation. Part 7 finds a fourth conserved quantity in the Kepler problem, and Part 8 gets electric charge from a phase rotation. Noether's second theorem, for symmetries that vary from point to point, is Part 9.