What is the four-current $J_{\mu}$ of a single charged particle in curved spacetime?

Question

I read this paper Phys.Rev.D 88.084059 which gives the formula (Eq.51) of four-potential using Kirchhoff representation with Green function and parallel propagator in curved spacetime.

Then I want to find out the four-potential of a single charged particle but I don’t know how to write down the four-current $J_{\mu}$ of it in curved spacetime. So could anyone help me?

Is it correct if I write down:

$J_{\mu}(x) = e g_{\mu\nu}(x) \int d\tau g^{\nu}_{\ \nu’}(x,X(\tau))U^{\nu’}(\tau)\delta_{4}(x,X(\tau))$

where the first $g_{\mu\nu}$ is the spacetime metric at point $x$ and the second $g^{\nu}_{\nu’}$ is the parallel propagator from $X(\tau)$ to $x$ (Eq.25 in the paper) . $X(\tau),U(\tau)$ are the trajectory and four-velocity of the charged particle and $\delta_{4}(x,x’)$ is the invariant Dirac distribution.

This review

• Poisson, E., Pound, A., & Vega, I. (2011). The motion of point particles in curved spacetime. Living Reviews in Relativity, 14(1), 7, https://doi.org/10.12942/lrr-2011-7

claimed that in page 113 the propagator term should be added, so, should I?

Thank you.

Answer 1

First, parameterize the trajectory of the particle by proper time: $\mathbf{X}(s)\equiv {{X}^{\alpha }}(s)$ and write ${{u}^{\alpha }}=d{{X}^{\alpha }}/ds$. The 4-current is ${{J}^{\alpha }}(\mathbf{x})=\int{ds}\ {{\delta }^{4}}(\mathbf{x}-\mathbf{X}(s))\ {{u}^{\alpha }}$ as a function of $\mathbf{x}=(t,x,y,z)$.