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