Part 4 of 13. Noether's theorem needs a symmetry of the Lagrangian, a stronger condition than a symmetry of the equations of motion. This part defines it and measures it.
Part 3 described a system by its Lagrangian. Noether's theorem starts from a transformation of the coordinates that leaves the Lagrangian unchanged, or changes it only by a total time derivative. This part makes that precise.
An infinitesimal transformation nudges the coordinates along a generator :
A constant shift has , a rescaling , a Galilean boost . To first order the Lagrangian changes by
If at every , the transformation is a strict symmetry. If for some function , it is a quasi-symmetry: the action changes only by , a boundary term that fixed-endpoint variations never see.
In the plane, rotation about a point has generator . The kinetic energy is unchanged by any rotation, so : the rotation is a symmetry exactly when the generator runs along the level curves of everywhere.
Figure 1. Colour is , computed by finite differences on a 61 × 61 grid (blank where it vanishes); thin curves are level sets of , arrows the rotation generator about the ring-marked centre. Drag the centre, or focus it and use arrow keys. Round bowl ; tilted, added; stiffened, added. The search scans centres on a grid and refines the best, reporting the smallest largest-value it can reach.
A linear tilt only slides the bowl, so rotation about the new bottom is still a symmetry and the search finds it. Stiffening one axis makes the level curves ellipses, and no centre works: angular momentum is lost about every point.
A symmetry of is a stronger condition than a symmetry of the equations of motion. Rescaling maps free-particle solutions to solutions, since is linear, but it multiplies the action by . That is not a total derivative, and Noether's theorem gets nothing from it. The test that matters is whether the transformation moves every path's action by the same amount.
Figure 2. First panel: eight paths from to : the true motion (violet) and seven seeded detours (grey), with their images under the transformation (dashed blue). Second panel: the change in action for each path, by Simpson's rule. Shift , scale , boost ; the free particle has , the oscillator . The readout also checks whether the image of the true motion still solves the Euler-Lagrange equation.
Equal down the column means the transformation cannot change which path is stationary: that is a strict or quasi-symmetry. Scaling the free particle still sends the true motion to a true motion, yet the dots scatter.
Whether is a total derivative can be tested without guessing . Since , the change must be affine in velocity, , with so that and have a common source.
Figure 3. Four Lagrangians, minus , (), and , against five generators. Every cell is computed in the page: by finite differences at 36 sample points , then a second difference in for the affine test and a mixed difference for . Solid green: strict. Dashed green: quasi. Grey: neither. Pick a cell for its numbers; the closed forms shown are checked against the numerical values.
Two cells deserve a look. A uniform shift under gravity is not a strict symmetry, because it changes the potential energy by a constant, but a constant is of something. And the oscillator, which has no translation symmetry at all, has a time-dependent quasi-symmetry : shifting by a solution of its own equation of motion.
The Galilean boost of a free particle of mass is the standard example:
The action changes by , endpoints only, so the stationary path is unchanged. Part 5 shows that each symmetry gives a conserved charge , where the boundary piece enters with a minus sign. Drop it and the charge drifts.
Figure 4. Each system integrated with RK4 at step 0.01 over 8 s, from , (oscillator ). Red: alone. Green: with subtracted. The readout gives each one's largest departure from its starting value; the green one's is integrator error.
Quasi-symmetries are common. A gauge transformation of the electromagnetic potentials changes a charged particle's Lagrangian by , a total derivative, and leaves its motion alone. The next part takes a Lagrangian and a symmetry, strict or quasi, and derives the conserved quantity.