Past exam of the theoretical physics I course of the University of Cambridge / 2014 / Paper 1 / 2

Choose electric scalar potential $\Phi =-Ex$ and the symmetric gauge $\mathbf{A}=B(-y,x,0)/2$, whose curl is $B\hat{\mathbf{z}}$. Minimal coupling gives $L=m{\dot{\mathbf{r}}}^2/2-k{\mathbf{r}}^2/2+q\mathbf{A}\cdot \dot{\mathbf{r}}-q\Phi$. Since $x=\rho \cos \theta$ and $x\dot{y}-y\dot{x}={\rho }^2\dot{\theta }$,The magnetic term is velocity dependent; it is not an ordinary position-only potential.

The canonical momenta areTaking the Legendre transform givesIts four Hamilton equations, in a polar chart with $\rho >0$, areCanonical angular momentum includes the vector-potential contribution and is not simply mechanical angular momentum.

Direct variation of the Lagrangian givesFor a static configuration, $k\rho =qE\cos \theta$ and $qE\rho \sin \theta =0$. Since $qE>0$ and the radius is positive,The magnetic force vanishes at rest and does not shift this equilibrium. In canonical variables it has $p_{\rho }=0$ but $p_{\theta }=qB{\rho }_0^2/2$, rather than zero canonical angular momentum.

Substitution of the proposed sine and cosine displacements into either linearized equation gives $m{\omega }^2+qB\omega -k=0$. ThusThese are signed angular frequencies, not two positive frequency magnitudes. To first order, with $\epsilon \ll {\rho }_0$,The orbit is therefore a circle of radius $\epsilon$ about $({\rho }_0,0)$ at this order. The positive root runs counterclockwise; the negative root runs clockwise and has the larger magnitude. A positive charge in $B\hat{\mathbf{z}}$ alone circles clockwise. Here the spring permits an additional counterclockwise mode by providing inward force against its magnetic force.