Part 6 of 13. Applied to shifts in time and space and to rotations, Noether's theorem gives energy, momentum and angular momentum, and it says which quantity counts as the conserved energy.
Energy is older than Noether's theorem: Leibniz argued for vis viva, and Lagrange and Hamilton built mechanics around it. What the theorem adds is a test and a name. If the Lagrangian does not mention explicitly, a quantity called the Hamiltonian is conserved. It is often , but not always.
Time translation moves the parameter of the path, not its values, so the formula of Part 5 does not apply as it stands. The direct route is to differentiate along a solution.
The bracket is the Hamiltonian,
which Hamilton made the central function of his mechanics in the 1830s and which appears here as the Noether charge of time translation. When with quadratic in , Euler's theorem on homogeneous functions gives , so .
With the pendulum has , and a solution stays on one level curve of in the plane. A kick is a sudden change of that the Lagrangian does not describe, so the theorem says nothing about that instant, only about the motion between kicks.
Figure 1. Top: the phase plane shaded by (lighter is higher), with the separatrix dashed; wraps at . Bottom: along the RK4 trajectory (step 0.01). The default run kicks twice as the bob passes the bottom moving right, and the second kick lifts it over the separatrix into rotation. Green stretches hold; red steps are the kicks. Click the plane to seed a new trajectory.
Every curve on this page comes from a numerical integrator, so even a conserved wanders a little. The two kinds of drift separate cleanly: the integrator's error shrinks as the step shrinks, and a physical change does not.
Figure 2. Largest over the run against the step size, both on log scales, for the pendulum of Figure 1 () under RK4 (circles) and under the symplectic leapfrog scheme (squares), and for the driven oscillator of Figure 3 under RK4 (triangles). The pendulum's drift falls along a straight line as the step shrinks, and the driven oscillator's stays put. Longer runs raise the RK4 line; the leapfrog line does not move, because leapfrog stays close to a slightly perturbed for very long times instead of accumulating error.
Step (4) says more than "conserved or not". It gives the rate, . For an oscillator pushed by an external force, , that rate is , which has either sign: the drive can feed energy in or take it out.
Figure 3. The oscillator with , from at rest. Top: the measured change with (solid), and the theorem's prediction accumulated separately (dashed). Bottom: the integrand, shaded blue where the drive adds energy and grey where it removes it. At the force is constant, , and holds; near the drive is resonant.
The reduction needs to be quadratic in the velocities. A bead sliding on a hoop that a motor spins at a fixed rate about its vertical diameter breaks that. With ,
The term carries no , so it enters with the opposite sign from . The Lagrangian has no explicit , so holds, while changes by exactly the work the motor does to keep the rate fixed.
Figure 4. The spinning hoop and bead, released at from rest relative to the hoop, and (green) and (red) over 20 time units, with the motor's accumulated work, , added to the starting (dashed). Above the bottom of the hoop is no longer stable and the bead swings about . At the two curves coincide.
The same bookkeeping sorts the textbook cases.
| System | conserved? | ? | |
|---|---|---|---|
| Oscillator, pendulum | yes | yes | |
| Driven oscillator | no | no | |
| Spinning hoop | yes | no | |
| Charge in static | yes | yes |
The charged particle, with , has a potential linear in velocity. That term cancels out of , leaving the kinetic energy plus , as it must, since the magnetic force does no work. The vector potential appears instead in the canonical momentum .
Translations and rotations act on the configuration itself, so they go straight into from Part 5. Take two particles joined by a spring:
Shifting both by leaves velocities and separation unchanged, so , , and
Add a uniform downward field, a term in . A horizontal shift still leaves alone. A vertical shift changes it by the constant , the time derivative of , so it is a quasi-symmetry and the formula returns
Figure 5. Two unit masses on a spring (, ), RK4: the masses with the dashed path of their center of mass, and , , and the quasi-symmetry charge (dashed). Green lines hold and red ones drift: falls at while the charge built from it stays flat. Turn gravity off and all three hold.
For one particle in a central potential, , a small rotation with keeps every length, so and
The line from the origin sweeps area at rate , so conservation of is Kepler's second law, for every central force. Stiffen the potential along one axis, , and the continuous rotations are gone. The reflections and the half-turn that survive are discrete, with no one-parameter family to differentiate, so they give no charge.
Figure 6. One unit mass from the same start, , , in the central potential and in the stiffened one, over equipotential contours, with the area swept in the last time unit shaded, and for both. The readout measures the shaded area against .
The theorem is a claim about pairs: this transformation leaves unchanged, so that charge is constant. Figure 7 tests both halves of each pair separately for the systems above.
Figure 7. Rows are Lagrangians; columns are the four transformations and their charges. The lower line of each cell tests the symmetry: under a small shift, rotation or time step, at 40 seeded random states. The upper line tests the charge: its largest drift over 20 time units of RK4 from a fixed start. Green cells hold, red cells drift. The spring with loses only energy. Gravity's vertical shift changes by a constant, the quasi-symmetry of Figure 5, so itself drifts. Select a cell (click, or arrow keys) to plot its charge.
Part 7 finds a conserved vector in the Kepler problem that none of these symmetries explains.