A particle of rest mass $m$ and charge $q$ is moving along a trajectory $x^{a}(s)$, where $s$ is the particle's proper time, in a given external electromagnetic field with 4-potential $A^{a}\left(x^{c}\right)$. Consider the action principle $\delta S=0$ where the action is $S=\int L d s$ and

$$L\left(s, x^{a}, \dot{x}^{a}\right)=-m \sqrt{\eta_{a b} \dot{x}^{a} \dot{x}^{b}}-q A_{a}\left(x^{c}\right) \dot{x}^{a},$$

and variations are taken with fixed endpoints.

Show first that the action is invariant both under reparametrizations $s \rightarrow \alpha s+\beta$ where $\alpha$ and $\beta$ are constants and also under a change of electromagnetic gauge. Next define the generalized momentum $P_{a}=\partial L / \partial \dot{x}^{a}$, and obtain the equation of motion

$$m \ddot{x}^{a}=q F_{b}^{a} \dot{x}^{b},$$

where the tensor $F^{a}{ }_{b}$ should be defined and you may assume that $d / d s\left(\eta_{a b} \dot{x}^{a} \dot{x}^{b}\right)=0$. Then verify from $(*)$ that indeed $d / d s\left(\eta_{a b} \dot{x}^{a} \dot{x}^{b}\right)=0$.

How does $P_{a}$ differ from the momentum $p_{a}$ of an uncharged particle? Comment briefly on the principle of minimal coupling.