# Conserved quantities from Killing vectors in the presence of electric charge

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?

• A general procedure would be to start with lagrangian of the system :(free particle+electromagnetic field+interaction $J^{\mu}A_{\mu}$), then evaluate $\delta S=0$ by considering variation of coordinates $\delta x^{\mu}=\alpha K^{\mu}$ and then solve for E-L equations of motion
– KP99
Commented Jul 12, 2021 at 16:48

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.
• Right. That's the main point. Btw $\odot$ stands for the symmetric tensor product. Commented Jul 14, 2021 at 18:35