Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 8D (Classical Dynamics)

Under the stated gauge transformation, the charged-particle electromagnetic Lagrangian becomesThusThe action changes only by $q[f(t_2)-f(t_1)]$, whose variation vanishes for fixed endpoints. By total-time-derivative invariance of a Lagrangian, the Euler-Lagrange equation is unchanged.

In Cartesian components,andThe Euler-Lagrange equation therefore givesUsingand $[\dot{r}\times B{]}_i={\dot{r}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, this becomes the Lorentz force

Let $\hat{b}=B/|B|$. For a constant uniform magnetic field, choose the symmetric gaugewhich has $\nabla \times A=B$ and $A\cdot \hat{b}=0$. With $E=0$, one may take $\varphi =0$. The coordinate $r_{\parallel }=r\cdot \hat{b}$ is then absent from the Lagrangian, so it is an ignorable coordinate. Its generalized momentumis conserved. Hencewhich is the canonical momentum parallel to a uniform magnetic field in this gauge.