Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 37D (Electrodynamics)

Use Minkowski spacetime signature $(-,+,+,+)$ and write $dot{x}^{\mu }=d{x}^{\mu }/d\lambda$. The reparametrization-invariant relativistic charged-particle action isThe first term is $-m{c}^2\int d\tau$ because

An electromagnetic gauge transformation isIt leaves the electromagnetic field tensor unchanged because antisymmetrized second derivatives commute. The action changes only byThis endpoint term has zero variation for fixed endpoints, which is the gauge invariance of the charged-particle worldline action.

The momentum conjugate to $x^{\mu }$ iswhileThe Euler-Lagrange equation therefore givesSince $dot{x}^{\mu }=(d\tau /d\lambda )u^{\mu }$ and $\sqrt{-{\dot{x}}^2}=c,d\tau /d\lambda$, division by $d\tau /d\lambda$ gives the Lorentz-force equation from the worldline actionRaising the free index yields exactly

For $A^{\mu }=(0,0,Bx,0)$, the only independent nonzero component of the electromagnetic field tensor isThe temporal and $z$ components of the Lorentz force show that $u^0$ and $u^3$ are constant. Taking $u^3=0$, the transverse equations arewhere dots now denote proper time derivatives. Witha solution isDirect differentiation verifies both equations. Hence the trajectory is a relativistic cyclotron orbit with proper angular frequency $omega=qB/m$; the sign fixes its orientation.

Because $u^0=c,dt/d\tau =\gamma c$ is constant, $dt/d\tau =\gamma$. The coordinate speed along the circle is consequentlysoFor the convention in the question, take $qB>0$, so $\omega >0$ and $\Delta \tau =2\pi /\omega$.

On the circular solution, the free part of the relativistic charged-particle action isIn the gauge $A_y=Bx$, the interaction is the closed line integralThe orbit is clockwise for $qB>0$, so $\oint xdy=-\pi R^2$. ThusCombining the two terms and using ${\gamma }^2{v}^2=c^2({\gamma }^2-1)$ gives the charged-particle action over one relativistic cyclotron orbitAlthough the intermediate magnetic line integral depends on the chosen magnetic vector potential, the closed-orbit result is gauge invariant.