2.1.19 · D5Analytical Mechanics
Question bank — Principle of least action — Hamilton's principle derivation
True or false — justify
The action is always minimized along the true path.
False — only demands a stationary point; over long times (past a conjugate/focal point) the true path can be a saddle. The honest name is "principle of stationary action."
The Lagrangian equals the total energy .
False — the Lagrangian is . Only makes Euler–Lagrange reproduce ; the conserved reappears later as the Hamiltonian.
The action is a number that depends on a single instant of the motion.
False — is a functional: it eats the entire path over and returns one number, so it depends on the whole route, not one instant.
Hamilton's principle and Newton's second law give different predictions for a particle.
False — for a 1D particle the Euler–Lagrange equation yields exactly , so they are equivalent, not competing.
During the variation , the wiggle may be nonzero at the endpoints.
False — endpoints are fixed data, so ; this is precisely what kills the boundary term in the integration by parts.
When we take partial derivatives of , we treat and as independent variables.
True — inside the partial-derivative "slot" they are independent; the link only re-enters during the variation and is what integration by parts exploits.
Units of action are the same as Planck's constant.
True — has units energytime , matching ; this coincidence is the bridge to the path integral where is a phase.
The Euler–Lagrange equation is a statement true at every instant along the path.
True — the Fundamental Lemma forces the bracketed expression to vanish for every , so although is a global quantity, its stationarity condition is local in time.
Spot the error
", and since is a minimum, always."
The first-order condition says nothing about the second-order sign; can be positive (minimum), negative (maximum) or vanish (saddle), so the claim over-asserts.
"Integrating by parts gives and nothing else."
The boundary term is dropped without stating why; it vanishes only because is zero at both fixed endpoints.
"Because is arbitrary, means the integral of the bracket is zero."
Wrong conclusion — arbitrariness of forces the bracket itself to vanish pointwise, not merely its integral (else a localized would expose a nonzero integral).
"For a free particle , so ."
A slot-confusion error: treats as fixed, and has no explicit , so ; the writer differentiated the wrong variable.
"Euler–Lagrange is , so momentum is always conserved."
Only when (the coordinate is cyclic) does ; in general the momentum changes at rate — that's Noether in disguise.
"For the falling body , so ."
Sign slip — the derivative of is , giving , i.e. downward acceleration; the wrong sign would send the body up.
"We can drop after differentiating and set to get the true path."
No — the true path is ; we differentiate and evaluate the derivative at , since the true path is the reference from which we wiggle.
Why questions
Why do we integrate the second term by parts instead of leaving ?
To convert into so a common factor can be pulled out of the whole integrand, letting the Fundamental Lemma act.
Why must the endpoints be held fixed?
They are the given start and end data, and fixing them makes vanish there so the boundary term dies, leaving the clean Euler–Lagrange equation; free endpoints would instead generate extra natural boundary conditions.
Why rather than ?
Because plugging into Euler–Lagrange reproduces , the correct dynamics; would give wrong equations — it is the conserved energy, not the action's integrand.
Why is the Fundamental Lemma of the calculus of variations needed at all?
Because it upgrades "the weighted integral is zero for every wiggle" into "the weight (the bracket) is identically zero," which is the actual equation of motion.
Why does Hamilton's principle need no force vectors or coordinate axes?
Because is a scalar and the action is a scalar integral, so the whole formulation is coordinate-free — you choose any generalized coordinate and the machinery still works.
Why is action measured in the same units as physically meaningful?
In the path-integral view every path contributes a phase ; classical stationary-action paths are where neighbouring phases add coherently, so having units of is what makes that phase dimensionless.
Why does making stationary "look like" Fermat's principle in optics?
Fermat makes optical path time stationary while Hamilton makes action stationary; both are variational "laziness" laws, and the same integration-by-parts machinery yields their governing equations.
Edge cases
What is the Euler–Lagrange equation when does not depend on explicitly (a cyclic coordinate)?
Then , so and the conjugate momentum is conserved — a symmetry directly gives a conservation law.
For a free particle, what does stationary action select among all fixed-time routes from to ?
Constant velocity (straight-line, uniform motion), because any speeding-up or slowing-down raises the time-average of and hence raises .
What happens to the derivation if is nonzero somewhere but the bracket is nonzero only at one isolated point?
You can choose an bumped exactly where the bracket is nonzero, making the integral nonzero — contradicting ; hence the bracket cannot be nonzero even at a single point.
What does reduce to when the potential is zero everywhere?
The equation of motion becomes , i.e. momentum constant — Newton's first law emerges as the degenerate, force-free case.
If the motion happens over a very long time past a conjugate point, is the true path still a minimum of ?
Not necessarily — it remains stationary, but it may become a saddle, which is exactly why "least action" is a misleading name for the general principle.
What role does play, and what happens in the limit ?
measures how far the trial path strays from the true one; stationarity is the statement that the first-order-in- change of vanishes as , so only the linear term matters.
Can two different paths both be stationary between the same endpoints?
Yes — past a focal/conjugate point multiple stationary paths (like multiple ray trajectories) can connect the same endpoints, each satisfying Euler–Lagrange but differing in whether they minimize .
Recall One-line self-test
If you can state why the boundary term dies, why , and why stationary minimum without peeking, you own this topic. ::: Endpoints fixed there; reproduces ; is only a first-order condition so it also admits saddles and maxima.