Question bank — Hamilton's principle — least action

False. $L=T-V$, while total energy is $T+V$. They only coincide when $V=0$; otherwise they differ by $2V$.

False. The exact condition is $\delta S=0$ (stationary). Past a conjugate/focal point the true path becomes a saddle, so it is a stationary point but not a minimum.

False. The Fundamental Lemma needs it to hold for all smooth $\eta$ vanishing at the ends; a single $\eta$ tells you almost nothing.

True. Adding a total time derivative $\frac{dF(q,t)}{dt}$ to $L$ changes $S$ only by a fixed boundary term, so $\delta S$ and hence the E–L equations are unchanged.

False. That is its selling point: writing $T$ and $V$ in well-chosen coordinates makes constraint forces (like a pendulum's tension) drop out automatically — no force decomposition needed.

False. It vanishes because the endpoints are held fixed; that is what kills the boundary term $[\frac{\partial L}{\partial \dot{q}}\eta {]}_{t_1}^{t_2}$ in the integration by parts, not any infinity.

False. $T+V$ is the same constant along the true path, so it carries no information distinguishing paths; the difference $T-V$, integrated, is what gets extremized.

True in spirit. For $L=\frac{1}{2}m{\dot{x}}^2-V$, E–L gives $m\ddot{x}=-V^{\prime }$ directly, and the same principle generalizes cleanly to fields and relativity where "force" is awkward. See Newton's Laws.

False. $S=\int (T-V)dt$ can be negative, zero, or positive depending on how kinetic and potential trade off; its sign is irrelevant — only its stationarity matters.

Missing $\frac{d}{dt}$. The momentum term must be time-differentiated: $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$, because $p=\partial L/\partial \dot{q}$ changes with time.

Wrong problem. Hamilton's principle fixes $q(t_1),q(t_2)$. Freeing the ends leaves a surviving boundary term and imposes natural boundary conditions — a different variational problem entirely.

Misread. Stationarity is only at $\epsilon =0$: ${\frac{dS}{d\epsilon }|}_0=0$. The action still bends for nonzero $\epsilon$; it's flat only at the true path.

No. Inside the partial derivative $q,\dot{q},t$ are treated as independent slots; only after forming the E–L equation do we remember $\dot{q}=dq/dt$ and apply $\frac{d}{dt}$.

Unnecessary. Choosing coordinate $\theta$ and writing $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta$ encodes all the geometry; tension does no work and never appears. See Lagrangian Mechanics.

Wrong argument. $S$ eats an entire path $q(\cdot )$ and returns one number after integrating $t$ out; it depends on the function, not on any single time.

It's essential. By-parts moves the derivative off $\dot{\eta }$ onto $\frac{\partial L}{\partial \dot{q}}$, so that every term multiplies the same $\eta$ — only then can the lemma factor $\eta$ out and expose the E–L bracket.

Because $T+V$ is constant along the true motion and distinguishes nothing; the difference measures the moment-to-moment trade between kinetic and potential energy, and its integral is what nature extremizes.

The true path must beat all nearby competitors. Requiring ${\frac{dS}{d\epsilon }|}_0=0$ for arbitrary $\eta$ is exactly what forces the E–L integrand's bracket to vanish everywhere.

It vanishes because $\eta (t_1)=\eta (t_2)=0$ (fixed ends). This is the pivot of the whole derivation — without it a leftover boundary contribution would spoil the clean E–L equation.

Newton asks "what force acts right now, right here?"; Hamilton asks "which entire path scores best?". One is a local instantaneous rule, the other a statement about the whole trajectory — yet they agree.

A constant shift changes $L$ by a constant, hence $S$ by a fixed amount independent of the path; $\delta S$ is unaffected, so the E–L equations and the motion are identical.

Both extremize an integral over a path (action for matter, optical path length/time for light). Fermat's Principle is the optics cousin: light takes the stationary-time route just as a particle takes the stationary-action route.

If $L$ doesn't change under some shift (in time, position, or angle), E–L makes the matching momentum conserved — the content of Noether's Theorem. E.g. $L$ independent of $q$ gives $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=0$, so $p=\partial L/\partial \dot{q}$ is conserved.

A genuine minimum for the ordinary case: any wiggle adds kinetic energy without any potential to compensate, so $S$ strictly increases.

The coordinate is cyclic/ignorable: $\partial L/\partial q=0$, so $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=0$ and the conjugate momentum $p=\partial L/\partial \dot{q}$ is conserved.

Then a Hamiltonian-like quantity $H=\dot{q}\frac{\partial L}{\partial \dot{q}}-L$ is conserved; for scleronomic systems this equals $T+V$, the total energy. See Hamiltonian Mechanics.

Yes — the wiggle argument runs independently in each direction, giving one E–L equation per coordinate, since ${\eta }_i$ can be varied separately.

Possible past a conjugate point (e.g. many geodesics between antipodes of a sphere). Stationarity is a local condition, so several paths can each be stationary; extra analysis decides which is a minimum.

Only if it satisfies E–L. E.g. a pendulum hanging at $\theta =0$ gives $\ddot{\theta }=-\frac{g}{\ell }\sin 0=0$, a valid stationary (equilibrium) solution; at $\theta =\pi$ it's stationary but unstable.

The action integral shrinks toward zero and endpoints nearly coincide; the true path is essentially forced to be the tiny straight segment, and stationarity becomes trivial — no room to wiggle.

No. It isn't even a competitor: Hamilton's principle only compares paths sharing the given $q(t_1)$ and $q(t_2)$, so an endpoint-violating curve is simply out of the contest.