Conserved quantities from Killing vectors in the presence of electric charge

Question

I know that for particles carrying no electric charge, given a Killing vector $K_{\mu}$, we have a conserved quantity $K_{\mu}p^{\mu}$ along geodesics, where $p^{\mu}$ is a tangeant vector. However, if $e \neq 0$, the equation of motion isn't simply $p^{\nu}\nabla_{\nu}p^{\mu} = 0$ anymore and so the above doesn't hold. I thought that the quantity $K_{\mu}(p^{\mu} - e A^\mu$), with $A^\mu$ the vector potential, would be conserved instead, but I'm having trouble showing this. Is this quantity really conserved and, if not, what is the correct quantity?

Answer 1

Hints:

1. Write down the action $S[x]$ for a relativistic point charge in a background metric $g=g_{\mu\nu}\mathrm{d}x^{\mu}\odot\mathrm{d}x^{\nu}$ and a background gauge field $A=A_{\mu}\mathrm{d}x^{\mu}$.

2. Assume that the vertical infinitesimal transformation $\delta x^{\mu}=\epsilon K^{\mu}$ is a symmetry, i.e. that the Lie derivatives $$ {\cal L}_Kg~=~0\quad\text{and}\quad{\cal L}_KA~=~0 $$ vanish.

3. Deduce that the conserved Noether charge is $$ Q ~=~K^{\mu}p_{\mu}, $$ where $$ p_{\mu}~:=~\frac{\partial L}{\partial \dot{x}^{\mu}}~=~ p^{\rm kin}_{\mu} \pm q A_{\mu} $$ is the canonical/conjugate 4-momentum. Here we are using the signature convention $(\mp,\pm,\pm,\pm)$, respectively.