2.1.9 · D5Analytical Mechanics
Question bank — Noether's theorem — symmetry ↔ conservation law
True or false — justify
Reflection (parity) is a symmetry, so it must give a conserved Noether charge.
False. Noether needs a continuous parameter you can differentiate with; reflection is discrete, so there is no to build a charge — it gives selection rules, not conserved quantities.
If the Lagrangian changes under a transformation, the action cannot be invariant.
False. may change by a total time derivative ; then is just endpoint values, so the action is still invariant and the charge becomes .
Energy is conserved in every isolated mechanical system.
False. Energy (the function ) is conserved only when has no explicit time dependence. A time-dependent potential breaks time-translation symmetry and energy drifts.
A cyclic coordinate always conserves the quantity .
False. It conserves the canonical momentum , which equals only for simple free particles; with a magnetic field , and in curvilinear coordinates it looks different entirely.
If does not depend explicitly on time, then always.
False. only when the kinetic energy is a quadratic function of the velocities (via Euler's theorem on homogeneous functions). Time-dependent constraints or velocity-linear terms break this even when .
Every continuous symmetry of the equations of motion gives a Noether charge.
False. Noether's theorem requires invariance of the action (or of up to a boundary term), which is stronger. Some symmetries of the EL equations do not leave the action invariant and yield no conserved charge.
For a free particle, both linear and angular momentum are conserved simultaneously.
True. The free-particle has no dependence (translation symmetry → ) and is rotationally invariant (rotation symmetry → ), so all three charges hold at once.
Spot the error
" means never changes."
The coordinate still evolves; what is constant is its conjugate momentum . A cyclic coordinate says doesn't depend on , not that is frozen.
"Space-translation invariance means doesn't depend on position at all."
It means depends only on relative positions , not on the absolute origin. The center-of-mass coordinate is cyclic; internal separations can still enter .
"For a time shift, because time is external, not a coordinate."
For time translation the trajectory carries forward, so . Plugging this (with the boundary term ) into Noether is exactly what produces the energy function .
" is the Noether charge."
The charge uses the canonical momentum , not the velocity: . Using drops the mass and any field terms.
"Rotation gives , and that's the final answer."
It's correct but should be simplified with the scalar-triple-product identity to — revealing it is angular momentum about .
"We used Euler–Lagrange to prove Noether, so Noether only holds at the end of the motion."
EL holds at every instant along a physical path, so the substitution is valid throughout — the charge is constant for all , not just at a special moment.
"Since came from the chain rule, it needs small but not ."
The relation is essential — it's what lets Step 3 recognize the product rule and collapse everything into a single . Without it the charge never appears.
Why questions
Why does Noether's theorem need the symmetry to be continuous and not just any symmetry?
Because the proof takes an infinitesimal transformation and expands to first order in ; you need a smooth knob to differentiate. Discrete symmetries have no such infinitesimal generator.
Why does "symmetry" translate into a conserved quantity rather than something else?
A symmetry means (or a boundary term), and the algebra forces . Setting that to zero says a specific combination has zero rate of change — i.e. it's conserved.
Why is the energy charge the special one that needs the boundary term ?
Under a time shift the Lagrangian isn't strictly invariant; it changes by its own total derivative . That is a boundary term, so we subtract , giving instead of a naive .
Why can a magnetic (velocity-dependent) potential still respect momentum conservation despite ?
The conserved object is the canonical momentum . Translation invariance conserves this full expression, even though the mechanical momentum alone is not conserved.
Why does rotational symmetry about conserve only the component , not all of ?
The symmetry is invariance under rotation about that one axis . Noether hands you the charge along the specific direction you rotated about; other components are conserved only if those rotations are also symmetries.
Why is a cyclic coordinate the "cheap" version of Noether's theorem?
A cyclic coordinate is precisely translation symmetry in (shifting leaves unchanged), so Euler–Lagrange immediately gives — the full Noether machinery collapses to one line.
Edge cases
What happens to the Noether charge when the parameter appears in the transformation but and is not a total time derivative?
There is no conserved charge — the transformation simply isn't a symmetry of the action. Noether only speaks when or .
A pendulum with a support driven up and down as : is energy conserved?
No. The explicit in the constraint makes , breaking time-translation symmetry; the mechanical energy oscillates as the drive pumps energy in and out.
A particle in a uniform gravitational field moving in the – plane: which momenta survive?
has no , so is conserved (horizontal translation symmetry), but appears in , so is not conserved — gravity picks out a direction and breaks vertical translation.
A perfectly free particle in empty space at rest (): is the rotation charge still "conserved"?
Yes, trivially — it equals zero and stays zero. Conservation means the rate of change is zero; a charge that is constantly is a perfectly valid conserved quantity (a degenerate but legitimate case).
If a system has a symmetry but the trajectory happens to make everywhere (e.g. rotation of a particle sitting exactly on the axis), does Noether fail?
It doesn't fail; the charge is just for that trajectory and stays . The theorem's guarantee (constant in time) still holds — the constant simply happens to be zero.
Does a symmetry that only holds at one instant (not for all ) give a conserved charge?
No. Invariance must hold along the whole path so that for all . A one-instant coincidence gives you nothing conserved.
Recall One-line summary of every trap
Noether needs a continuous symmetry of the action (allowing boundary terms), the charge uses canonical momenta, and each conserved quantity survives only while its specific symmetry is unbroken — energy dies under explicit time dependence, under gravity, etc.