Question bank — Hamilton-Jacobi equation

False that they're equivalent, true that $K=0$ is not the only option: any constant $K$ still gives $\dot{Q},\dot{P}$ from its derivatives, but $K=0$ is chosen because it makes all new variables constant with zero extra bookkeeping.

False. $S$ is the time integral of $L$ along the true path ($S=\int Ldt$); $L$ is the integrand, $S$ is the accumulated action.

False. $H$ quadratic in $p$ becomes quadratic in $\partial S/\partial q$, so terms like $(\partial S/\partial q{)}^2$ appear — the equation is genuinely nonlinear in the unknown $S$.

True. Those $n$ constants ${\alpha }_i$ are precisely the new constant momenta $P_i$; with fewer you cannot generate the full canonical transformation, and one extra additive constant is trivial because only derivatives of $S$ matter.

False. $W$ requires energy conservation (no explicit $t$ in $H$). If $H=H(q,p,t)$ genuinely depends on time, you cannot split off $-Et$ and $W$ is undefined.

False. The new variables $Q,P$ are constant, but the physical position $q$ is recovered by inverting $Q=\partial S/\partial P$ for $q(t)$ — the particle moves; only the cleverly chosen labels stand still.

False. Trajectories come from derivatives of $S$ (through $p$ and $Q$); adding a constant to $S$ changes no derivative, hence no physics.

True — but the single equation is a first-order PDE, trading many ordinary equations for one partial one; the difficulty is compressed, not deleted.

Error: the type-2 relations are $p=\partial S/\partial q$ and $Q=\partial S/\partial P$. There is no relation $q=\partial S/\partial p$; $S$ depends on $q$ and $P$, not on $p$.

Error: $K=0$ gives $H=-\partial S/\partial t$, not $H=0$. The energy-like $H$ is generally nonzero; it is exactly balanced by the time-slope of $S$.

Error: $S$ is the action along the actual (extremal) trajectory, not an arbitrary path. Off the true path, $\int Ldt$ is not Hamilton's principal function.

Error: $E$ is a constant (the conserved energy), one of the separation constants ${\alpha }_i$. It is the same number at every $q$; the equation determines $\partial W/\partial q$ given that fixed $E$.

Error: choosing $P=E$ is a convenient option, not a rule. Any independent constant of integration ${\alpha }_i$ may be labelled a new momentum; $E$ is just the natural pick when energy is conserved.

Error: the equation hides them, but its complete integral carries $n$ constants, and identifying $P_i={\alpha }_i$ is what lets $Q_i=\partial S/\partial P_i$ produce the solution. Drop them and you have no transformation.

Error: $\partial W/\partial t=0$ (it's time-independent), but $S=W-Et$ has $\partial S/\partial t=-E\ne 0$. That $-E$ is exactly what feeds the full HJ equation.

Type-2 $F_2=S(q,P,t)$ treats the new momenta $P$ as independent variables, which is precisely what we want to hold constant; type-1 depends on $Q$ instead, awkward when the goal is constant $P$. See Canonical Transformations.

Hamilton's equations give $\dot{Q}=\partial K/\partial P$ and $\dot{P}=-\partial K/\partial Q$. If $K$ is identically zero, both partials vanish everywhere, so $\dot{Q}=\dot{P}=0$ — the new variables are frozen constants.

HJ solves for the action $S$, whose first slope in $q$ already encodes momentum ($p=\partial S/\partial q$); the "second order" of Newton is absorbed because momentum is now a derivative of the unknown, not a separate variable.

Separation $S=W(q)-Et$ needs $\partial S/\partial t=-E$ to be a constant. This works only if $H$ (hence $E$) is conserved; an explicitly time-dependent $H$ would not yield a constant $E$ to peel off. See Separation of Variables.

Because $S=\int Ldt$ is the action, and along the true path $dS/dt=\sum p_i{\dot{q}}_i-H=L$; the HJ equation is then just the statement that this same action satisfies $H(q,\partial S/\partial q,t)+\partial S/\partial t=0$.

When motion is periodic, choosing the new constant momenta as action variables (integrals $\oint pdq$) makes $W$ generate angle variables that advance linearly in time — HJ is the machinery that produces them.

In the limit of small wavelength, the phase of a quantum wavefunction obeys the HJ equation with $S$ as that phase; classical mechanics is the "geometric optics" of quantum mechanics, and HJ is its wavefront equation.

A single particular solution lacks the free constants; only a complete integral (with all $n$ independent ${\alpha }_i$) can be differentiated with respect to those constants to yield $Q_i$ and thereby the full family of trajectories.

$S=\sqrt{2mE}x-Et\to 0$: zero energy means zero momentum ($p=\sqrt{2mE}\to 0$) and a particle sitting still, so the action along the (motionless) path vanishes.

No for $E>0$: $v=\sqrt{2E/m}$ is finite. At exactly $E=0$ the particle is at rest ($v=0$), a legitimate degenerate case, not a singularity.

$W^{\prime }=\sqrt{2m(E-\frac{1}{2}m{\omega }^2{x}^2)}\to 0$: momentum vanishes there (the particle instantaneously stops), and the integrand in $W$ has an integrable square-root edge, not a true blow-up.

Yes — the full HJ equation $H(q,\partial S/\partial q,t)+\partial S/\partial t=0$ is unchanged and valid; only the shortcut $S=W-Et$ and the characteristic function $W$ become unavailable.

The transformation still works locally, but the frequencies of motion match, signalling a degeneracy; in action-angle language extra conserved quantities appear and orbits close in fewer independent periods.

Exactly one — commonly $E$ (or $\alpha =P$). It is the new constant momentum; the paired new coordinate $Q=\partial S/\partial E$ is the constant that fixes the phase or start time (e.g. $-t_0$ for the free particle).

$Q$ is a frozen constant equal to (minus) the time the particle passed the origin; inverting it, $x=\sqrt{2E/m}(t-t_0)$, recovers uniform motion — the constant encodes when, not where.