Equation of motion for a charged particle

Question

To compute the equations of motion of a neutral testparticle in the graviational field, one needs the metric tensor $g_{\mu \nu}$ and $g^{\mu \nu}$ to compute the Christoffel-symbols

$${\Gamma^{\rm i}_{\rm j k} = \sum _{\rm s=1}^4 \ \frac{{{g}}^{\rm i s}}{2} \left(\frac{\partial {g}_{\rm s j}}{\partial {\rm x^k}}+\frac{\partial {g}_{\rm s k}}{\partial {\rm x^j}}-\frac{\partial {g}_{\rm j k}}{\partial {\rm x^s}}\right)}$$

and get the coordinate acceleration of the test particle:

$${{\rm \ddot x^{i} = -\sum _{j=1}^4 \sum _{k=1}^4 \ \dot x^j \ \dot x^k \ \Gamma^{i}_{j k}}}$$

But in the vicinity of a charge, there is not only the metric tensor, which might look like this

$$g_{\mu \nu}=\left( \begin{array}{cccc} g_{\rm t t} & 0 & 0 & g_{\rm t \phi} \\ 0 & g_{\rm r r} & 0 & 0 \\ 0 & 0 & g_{\rm \theta \theta} & 0 \\ g_{\rm t \phi} & 0 & 0 & g_{\rm \phi \phi} \\ \end{array} \right)$$

where the $g$-components are functions of the coordinates and mass, charge and spin, but also a Coulomb-potential, which looks like

$$\rm A_{\alpha}=(Q/r,0,0,0)$$

How is the Coulomb-potential plugged into the geodesic equation? The metric tensor is a $4 \times 4$ matrix, but the Coulomb potential seems to be $1 \times 4$, how is this added together to find the geodesics $\rm \ddot x^i$ of a charged particle of charge $\rm q$ in the vicinity of a dominant mass of charge $\rm Q$?

On Wikipedia I found the equation

$${{\rm \ddot x^i = - \Gamma^i_{j k}{\dot x^j}{\dot x^k}\ +\frac{q}{m} \ {F^{i k}} \ {\dot x^j}} \ {g_{\rm j k}}}$$

but there is no definition of ${\rm F^{i k}}$, which is, as Chrisoph and Eddy mentioned, the electromagnetic tensor. But what is the electromagnetic tensor for the Kerr-Newman metric in spherical (Boyer Lindquist) coordinates?

Edit: thanks for the answears, that's what I got so far: Screenshot

Answer 1

You stopped reading too early. Quoting the Wikipedia page you linked to:

The resulting equation of motion is as follows:

$$ {d^2 x^\mu \over ds^2} =- \Gamma^\mu {}_{\alpha \beta}{d x^\alpha \over ds}{d x^\beta \over ds}\ +{q \over m} {F^{\mu \beta}} {d x^\alpha \over ds}{g_{\alpha \beta}}. $$

with

$$ {g_{\alpha \beta}}{d x^\alpha \over ds}{d x^\beta \over ds}=-1. $$

Here, $F^{\mu\nu}$ denotes the electromagnetic field tensor given by

$$ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu $$

in it's incarnation with lowered indices.

Regarding your edit, in Boyer-Lindquist coordinates, $A$ is given by

$$ A = \frac{Qr}{\rho^2}(dt-a\sin^2\theta d\phi) + \frac{1}{\rho^2}P\cos\theta \left(a dt -(r^2+a^2)d\phi\right) $$

where

$$ \rho^2(r,\theta) = r^2+a^2 \cos^2 \theta $$

according to this paper I just googled.

Answer 2

$F$ is the Electromagnetic tensor

Edit: this answer was posted in answer to the original question which can be paraphrased as 'I found this equation on Wikipedia, what does this term mean,' and it is clear from the OPs subsequent edit that this helped narrow down what they actually wanted to know.