Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S7 / 2

The canonical momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If the velocity-momentum relation can be locally inverted, the Legendre transform defines $H(q,p,t)={\sum }_i{p}_i{\dot{q}}_i-L$, with the velocities expressed in terms of $q,p,t$. Its differential isThe Euler-Lagrange equations then yield Hamilton's equations,A singular velocity Hessian instead requires a constrained-Hamiltonian treatment.

The canonical momentum is $p=\gamma m\dot{r}$, where $\gamma =(1-{\dot{r}}^2/c^2{)}^{-1/2}$. Eliminating velocity givesIts Hamilton equations are $\dot{r}=c^{2}p/H$ and $\dot{p}=0$. In the limit $|p|\ll mc$, the Taylor expansion is $H=m{c}^2+p^2/(2m)+O(p^4/(m^3{c}^2))$, and $\dot{r}\simeq p/m$. The additive rest energy does not change the classical free-particle equations.

Minimal electromagnetic coupling changes the canonical momentum to $p=\gamma m\dot{r}+qA$. The mechanical momentum is $\pi =p-qA$. The Legendre transform yieldsThe cancellation of $q\dot{r}\cdot A$ in the Legendre transform does not remove magnetic effects: $A$ remains in the relation between canonical and mechanical momenta. The expression is the relativistic kinetic-plus-rest energy with an electrostatic potential-energy term, expressed in canonical variables.

Let $\mathcal{E}=\sqrt{m^2{c}^4+c^2{\pi }^2}$. Hamilton's equations giveSubtracting $qd{A}_i/dt$ from the second equation gives the familiar Lorentz force, $\dot{\pi }=q(E+\dot{r}\times B)$, with $E=-\nabla \varphi -{\partial }_{t}A$ and $B=\nabla \times A$. This also checks the signs of both potential terms.