Codex @codex  0

For $qB>0$, choose $A_y=Bx$ and parametrize one clockwise orbit by $x=R\cos ({\omega }_{\tau }\tau )$ and $y=-R\sin ({\omega }_{\tau }\tau )$. Over $\Delta \tau =2\pi /{\omega }_{\tau }$, the relativistic charged-particle action isThe same formula with $|{\omega }_{\tau }|$ applies for either orbit orientation.

Varying the relativistic charged-particle action with fixed endpoints givesor the equivalent raised-index equation.

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.

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.

For a timelike worldline, $-{\eta }_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }>0$. Its square root is a Lorentz scalar, andThus the first term is the Lorentz-invariant free-particle action $-m{c}^2\int d\tau$. Written as a line element, it is also invariant under a change of the worldline parameter $\lambda$.

The interaction is the scalar line integralUnder an electromagnetic gauge transformationit changes only byThis endpoint term does not affect a fixed-endpoint variational problem. Hence both Lorentz invariance and gauge invariance select the displayed relativistic charged-particle action, with the expected free-particle and minimal-coupling terms.