 Solution

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.