Gauge-compensated charged Killing energy

In any coordinate chart, $({\iota }_{\xi }F{)}_a={\xi }^b({\mathrm{\partial }}_b{A}_a-{\mathrm{\partial }}_a{A}_b)$ and $d({\iota }_{\xi }A{)}_a={\mathrm{\partial }}_a({\xi }^b{A}_b)$. Adding these identities gives $({\iota }_{\xi }F{)}_a+d({\iota }_{\xi }A{)}_a={\xi }^b{\mathrm{\partial }}_b{A}_a+A_b{\mathrm{\partial }}_a{\xi }^b=({\mathcal{L}}_{\xi }A{)}_a$. Using ${\mathcal{L}}_{\xi }A=d{\mathrm{\Lambda }}_{\xi }$ and [F1], this yields ${\iota }_{\xi }F=d{\mathrm{\Lambda }}_{\xi }-d({\iota }_{\xi }A)=-d{\mathrm{\Psi }}_{\xi }$.

Metric compatibility and $P=Mw$ yield $\frac{d}{d{s}_m}g(P,\xi )=cqg(F^{\mathrm{♯}}(w),\xi )+M{w}^a{w}^b{\mathrm{\nabla }}_b{\xi }_a=cq({\iota }_{\xi }F)(w)=-cqw({\mathrm{\Psi }}_{\xi })\mathrm{.}$ The middle Killing term vanishes because $w^a{w}^b$ is symmetric and ${\mathrm{\nabla }}_{(a}{\xi }_{b)}=0$; the force contraction has the displayed sign since $g(F^{\mathrm{♯}}(w),\xi )={\xi }^c{F}_{cb}{w}^b$.

Since $d{\mathrm{\Psi }}_{\xi }\mathrm{/}d{s}_m=w({\mathrm{\Psi }}_{\xi })$ and $q{\mathrm{\Phi }}_{\xi }=-cq{\mathrm{\Psi }}_{\xi }$, $\frac{d{H}_{\xi }}{d{s}_m}=-\frac{d}{d{s}_m}g(P,\xi )-cqw({\mathrm{\Psi }}_{\xi })=0.$ The curve is $C^2$ and the fields are smooth, so $H_{\xi }$ is $C^1$ and hence continuous on $J$. Its derivative is zero by step 2.1; since the open interval $J$ is order-convex, [F3] makes $H_{\xi }$ constant there.

Under $A^{\mathrm{\prime }}=A+d\chi$, $d{A}^{\mathrm{\prime }}=F$ because the coordinate mixed partials of the smooth scalar $\chi$ commute. Also $({\mathcal{L}}_{\xi }d\chi {)}_a={\xi }^b{\mathrm{\partial }}_b{\mathrm{\partial }}_a\chi +({\mathrm{\partial }}_a{\xi }^b){\mathrm{\partial }}_b\chi ={\mathrm{\partial }}_a({\xi }^b{\mathrm{\partial }}_b\chi )$, so ${\mathrm{\Lambda }}_{\xi }^{\mathrm{\prime }}:={\mathrm{\Lambda }}_{\xi }+\xi (\chi )$ satisfies ${\mathcal{L}}_{\xi }{A}^{\mathrm{\prime }}=d{\mathrm{\Lambda }}_{\xi }^{\mathrm{\prime }}$. Since $({\iota }_{\xi }d\chi )=\xi (\chi )$, ${\mathrm{\Psi }}_{\xi }^{\mathrm{\prime }}={\mathrm{\Psi }}_{\xi }$ and $H_{\xi }^{\mathrm{\prime }}=H_{\xi }$. Replacing ${\mathrm{\Lambda }}_{\xi }$ by ${\mathrm{\Lambda }}_{\xi }+C_0\mathrm{/}c$ leaves $d{\mathrm{\Lambda }}_{\xi }$ fixed and sends ${\mathrm{\Phi }}_{\xi }$ to ${\mathrm{\Phi }}_{\xi }+C_0$, so $H_{\xi }$ changes by $q{C}_0$.