Past exam of the classical mechanics course of Part B of the physics course of the University of Oxford / 2016 / 4

Fix an initial event $(q_i,t_i)$. For each reachable final event $(q_f,t_f)$, evaluate the action on a classical path satisfying those endpoints. On a chosen local branch this produces Hamilton's principal function $S(q_f,t_f)$, rather than a functional of every possible path. Multiple classical paths may make this function multivalued, so the derivation is local to a branch.

Allow the final event to vary. The integrated interior variation vanishes by the Euler-Lagrange equations. The endpoint variation must distinguish the total coordinate change from the change at fixed time: $\delta q(t_f)=\delta q_f-{\dot{q}}_f\delta t_f$. ConsequentlyHence $p_i=\partial S/\partial q_i$ and $\partial S/\partial t=-H$. Eliminating the momenta yields the Hamilton-Jacobi equationA complete integral of the Hamilton-Jacobi equation depends on $n$ suitable independent constants for $n$ coordinates; differentiating it with respect to those constants supplies the conjugate integration constants needed to reconstruct trajectories.

The required Hamilton-Jacobi equation isBecause $x,y$ and time are cyclic in the Hamiltonian, use separation $S=ax+by-Et+W(z)$. The remaining equation isFor $g\ne 0$, each sign branch has the separated complete integralIt is real in the classically allowed region $Q\ge 0$, and $p_z=s\sqrt{Q}$. At a turning point one matches the ascending and descending branches. If $g=0$, the integral is simply $W=p_{z}z$ with constant $p_z$.

A convenient equivalent complete integral, expressed using initial-momentum constants $\mathbf{\alpha }$, is globally polynomial in time:Substitution verifies that its time derivative cancels the kinetic term and $mgz$. It avoids a square-root branch change when extracting the complete trajectory.

Apply Jacobi's reconstruction to the polynomial complete integral in solution. Its parameter derivatives are constant:ThereforeAt $t=0$ the constants $\mathbf{\beta }$ are the initial positions and $\mathbf{\alpha }$ are the initial momenta. Direct differentiation of $S$ gives $p_x={\alpha }_x$, $p_y={\alpha }_y$, $p_z={\alpha }_z-mgt$, confirming the same motion. The six constants provide arbitrary initial data; the solution covers both ascent and descent, not merely one local square-root branch of the separated action.

An autonomous Hamiltonian system with $n$ degrees of freedom is Liouville integrable if it has $n$ conserved functions $I_1,\dots ,I_n$, including the Hamiltonian, whose differentials are independent on a regular open set and which are in involution:The equations can then be reduced locally to quadratures. The statement about action-angle coordinates on invariant tori additionally requires compact connected regular common level sets; integrable unbounded motion need not be periodic or lie on a torus.

For the uniform-gravity Hamiltonian, take the three integrals $p_x,p_y,H$. Since the potential does not depend on $x,y$,For $g\ne 0$, the $mgdz$ term in $dH$ makes these three differentials independent. Thus the system is Liouville integrable, as the explicit separated solution already demonstrates.

Replacing $mgz$ by a sufficiently smooth arbitrary function $f(z)$ preserves the same commuting integrals. They are independent wherever $p_z$ and $f^{\prime }(z)$ do not vanish simultaneously. The vertical motion reduces toThe horizontal motion remains uniform. Thus an elementary antiderivative is unnecessary: the system is still integrable by a one-dimensional quadrature on regular branches, joined at turning points. Critical equilibria may be singular common level sets without destroying generic integrability. If $f$ is constant, the three momenta themselves give an everywhere independent commuting set, avoiding the rank loss of $p_x,p_y,H$ at $p_z=0$.