Past exam of the classical mechanics course of the University of Oxford / 2023 / 3

Define the canonical momenta byTo obtain an ordinary unconstrained Hamiltonian, assume a regular Lagrangian: the velocity Hessian matrix $({\partial }^{2}L/\partial {\dot{q}}_i\partial {\dot{q}}_j)$ is invertible. The inverse function theorem then lets us solve locally for $\dot{\mathbf{q}}=\mathbf{v}(\mathbf{q},\mathbf{p},t)$. The phase space is the cotangent bundle $T^{\ast }\mathcal{Q}$, with local canonical coordinates $(\mathbf{q},\mathbf{p})$ and dimension $2n$. The Legendre transform in mechanics definesThe velocities in this formula have been eliminated in favour of phase space variables. A hyperregular Lagrangian makes this construction global; regularity alone guarantees it locally. Hamilton's equations are

In canonical coordinates on phase space, the Poisson bracket isIt treats $t$ as a parameter. Along a solution of Hamilton's equations, the chain rule givesTherefore the Hamiltonian conservation law isThe total derivative follows the trajectory; the explicit partial derivative holds $(\mathbf{q},\mathbf{p})$ fixed.

The Poisson bracket is antisymmetric, so $\{H,H\}=0$. The Hamiltonian energy balance consequently givesFor a regular Lagrangian the Legendre transform in mechanics gives $H_t=-L_t$, evaluated at the corresponding velocity. In particular, a Lagrangian with no explicit time dependence has time-translation symmetry. Noether theorem associates this continuous symmetry of the action with the conserved quantity ${\sum }_i{\dot{q}}_i\partial L/\partial {\dot{q}}_i-L=H$. For an ordinary natural mechanical Lagrangian $L=T-V$, with $T$ quadratic in the velocities, this quantity is the total energy $T+V$.

Set $D={\partial }_t+\{\cdot ,H\}$. The hypotheses mean $Df=Dg=0$ as identities on phase space. By the Jacobi identity for the Poisson bracket,Also, the product rule applied to the defining coordinate formula givesAdding these identities provesConsequentlyThis proves the Poisson bracket of time-dependent conserved quantities theorem, including explicit time dependence of $f$ and $g$.

Since $H$ is a conserved quantity, the Hamiltonian energy balance gives $H_t=0$. Since $f$ is a conserved quantity, $f_t+\{f,H\}=0$. Take an explicit partial derivative with respect to time, holding the phase space coordinates fixed:The last Poisson bracket vanishes because $H_t=0$. ThusThis is the time derivative of a conserved quantity for an autonomous Hamiltonian. The conservation hypotheses are understood to hold for every trajectory, as identities on phase space.

The canonical momentum of the harmonic oscillator is $p=\partial L_0/\partial \dot{q}=m\dot{q}$. Substituting $\dot{q}=p/m$ into the Legendre transform in mechanics gives

Work on $I>0$, with a local choice of the periodic angle $\theta$. The inverse formulas areIn particular, ${\theta }_q=p/(2I)$, ${\theta }_p=-q/(2I)$, $I_q=\alpha q$ and $I_p=p/\alpha$. Their Poisson bracket isThe other fundamental Poisson brackets vanish by antisymmetry. Thus this is a canonical transformation for every $\alpha >0$, on local charts away from the origin. At $I=0$ the angle is undefined.

In the new coordinates the harmonic oscillator Hamiltonian isFor $I$ to be a conserved quantity throughout phase space, Hamilton's equations require $\dot{I}=-{\partial }_{\theta }{H}_0=0$ identically. The two coefficients must coincide. For the usual $m>0$ and $\omega >0$, positivity of $\alpha$ selectsThese are harmonic oscillator action-angle variables; the action variable is $I=H_0/\omega$.

A type-two generating function for a canonical transformation uses the mixed variables $(q,P)$ and definesFor a local invertible canonical transformation, require $F_{2,qP}\ne 0$ so that $p=F_{2,q}(q,P)$ can be solved for $P$. The differential identity $d{F}_2=pdq+QdP$ gives $d(F_2-PQ)=pdq-PdQ$, expressing preservation of the canonical symplectic form.

On the interior of the principal inverse-cosine branch, $0<q<\pi$, the given canonical transformation hasIntegrating $Q={\partial }_P{F}_2$ suggests $F_2=\cos q\sin P+C(q)$. Its other derivative is ${\partial }_q{F}_2=-\sin q\sin P+C^{\prime }(q)$; comparison with $p$ requires $C^{\prime }(q)=0$. Hence a suitable type-two generating function for a canonical transformation isBoth derivative equations reproduce the specified transformation exactly. Its mixed derivative is $F_{2,qP}=-\sin q\cos P$, so it defines a local canonical transformation where $0<q<\pi$ and $\cos P\ne 0$, equivalently where $|Q/\cos P|<1$. At $q=0,\pi$ or $\cos P=0$, the stated formulas fail to give a regular coordinate chart. These restrictions explain the square-root branch and the local meaning of canonicity. This is a trigonometric type-two canonical transformation.

For a time-dependent type-two generating function for a canonical transformation, the differential identitygives $K=H+{\partial }_t{F}_2$, with the partial derivative taken at fixed $(q,P)$. Here $F_2=\cos q\sin P$ has no explicit time dependence, so the Hamiltonian correction isAs always, adding a function of time alone to the new Hamiltonian leaves Hamilton's equations unchanged. Choosing a time-independent type-two generating function for a canonical transformation fixes that irrelevant freedom and gives the displayed $\Lambda =0$.