Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 36B (Electrodynamics)

Use units $c=1$, consistent with comparing $B$ directly to $E$, and metric signature $(+---)$. For an arbitrary worldline parameter $s$, an action for a charged relativistic particle isThe first term has variation $-m{\dot{x}}_{\mu }/\sqrt{{\dot{x}}^2}$ in its momentum, and the coupling gives $-q{A}_{\mu }$. The Euler-Lagrange equations therefore readWhen $s$ is proper time, $u^{\mu }={\dot{x}}^{\mu }$ has $u^{\mu }{u}_{\mu }=1$, giving the covariant Lorentz forceWith $F^{i0}=E_i$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, put $e=qE/m$, $b=qB/m$, $\omega =\sqrt{b^2-e^2}$ for nonzero charge. The velocity equations becomeThus ${\ddot{u}}^z=-{\omega }^2{u}^z$, with $u^z(0)=0$, ${\dot{u}}^z(0)=e$. Integrating with $u^0(0)=1$, $u^x(0)=0$ gives $u^z=(e/\omega )\sin \omega s$, $u^0=1+(e^2/{\omega }^2)(1-\cos \omega s)$ and $u^x=-(be/{\omega }^2)(1-\cos \omega s)$. Integrate once more from the origin:For zero charge the trajectory is simply rest, $x^0=s$, with all spatial coordinates zero. The charged trajectory oscillates in $z$ while drifting in the negative $x$ direction, the direction of $\mathbf{E}\times \mathbf{B}$. Its displacement-to-time ratio over a proper-time period is $-E/B$. A boost at that drift velocity removes the electric field and leaves circular magnetic motion; in the original frame the orbit has repeating cycloidal cusps where the particle is momentarily at rest. Energy oscillates rather than growing indefinitely because $B>E$. In SI units the corresponding magnetic-dominated condition is $cB>E$.