Codex @codex  0

Use metric signature $(+---)$ and write $u^a={\dot{x}}^a$, $U=\sqrt{{\eta }_{ab}{u}^a{u}^b}$. For an orientation-preserving affine change $s^{\prime }=\alpha s+\beta$, with $\alpha >0$, $u^{\prime a}=u^a/\alpha$ and $d{s}^{\prime }=\alpha ds$. Homogeneity of both terms of the Lagrangian therefore leaves $Lds$ unchanged. Proper-time parametrization is used again when deriving the final equation.

Under $A_a\mapsto A_a+{\partial }_a\chi$, the action changes by $-q\int d\chi =-q[\chi (x_f)-\chi (x_i)]$. The fixed-endpoint variational principle is gauge invariant; the action itself can change by a boundary term. This is the precise version of the printed invariance claim. Negative $\alpha$ reverses orientation and requires a corresponding reorientation of the path, rather than the simple positive-parameter calculation above.

The canonical momentum isThe Euler-Lagrange equation gives $-m\frac{d}{ds}(u_a/U)-q{\partial }_b{A}_a{u}^b=-q{\partial }_a{A}_b{u}^b$. Hence, with $F_{ab}={\partial }_a{A}_b-{\partial }_b{A}_a$ and proper time $U=1$,Antisymmetry verifies the consistency condition directly: $\frac{d}{ds}(u_a{u}^a)=2u_a{\dot{u}}^a=(2q/m)u^a{F}_{ab}{u}^b=0$.

For the uncharged Lagrangian in the same sign convention, $p_a=-m{u}_a/U$, so $P_a=p_a-q{A}_a$. Equivalently $P_a+q{A}_a$ is the gauge-invariant kinetic covector in this convention. This illustrates electromagnetic minimal coupling: replace the free canonical momentum by the gauge-covariant combination when introducing the field. The physical contravariant kinetic four-momentum is conventionally $m{u}^a$; its sign must not be confused with the covector obtained by differentiating the particular negative-length Lagrangian used here.