+
\epsilon^{\mu\nu\alpha\beta} B_{\alpha} n_{\beta}\,, \\
^{*\!}F^{\mu\nu} &=& n^{\mu} B^{\nu} - n^{\nu} B^{\mu} -
\epsilon^{\mu\nu\alpha\beta} E_{\alpha} n_{\beta}\,,\end{aligned}$$ with $\epsilon^{\mu\nu\alpha\beta} := \eta^{\mu\nu\alpha\beta} /
{\sqrt{-g}}$, $g$ the determinant of the 4-metric and $\eta^{\mu\nu\alpha\beta}$ the Levi-Civitá symbol. The same can be done for the electric four-current $$I^{\mu} := q n^{\mu} + J^{\mu}\,,
\label{current1}$$ where $q$ and $J^{\mu}$ are the charge density and the electric current density for an Eulerian observer, respectively. Using these definitions and performing a 3+1 decomposition of Eqs.  and with respect to the normal vector $n^{\mu}$, we arrive at the following evolution equations, $$\begin{aligned}
&(\partial_t - {\cal L}_{\boldsymbol{\beta}}) E^{i} - \epsilon^{ijk} \nabla_j (\alpha
B_k) + \alpha \gamma^{ij} \nabla_j \psi = \nonumber \\
& \hskip 5.0cm \alpha K E^i - \alpha J^i\,, 
\label{maxwellext_3+1_eq1a}\\
&(\partial_t - {\cal L}_{\boldsymbol{\beta}}) \psi + \alpha \nabla_i E^i   = \alpha
q -\alpha \kappa \psi\,,
\label{maxwellext_3+1_eq1b}\\
&(\partial_t - {\cal L}_{\boldsymbol{\beta}}) B^{i} + \epsilon^{ijk} \nabla_j
(\alpha E_k) + \alpha \gamma^{ij} \nabla_j \phi = \alpha K B^i\,,
\label{maxwellext_3+1_eq1c} \\ 
& (\partial_t - {\cal L}_{\boldsymbol{\beta}}) \phi
+ \alpha \nabla_i B^i = -\alpha \kappa \phi\,,
\label{maxwellext_3+1_eq1d}\end{aligned}$$ where $\gamma_{ij}$ is the spatial 3-metric, $K := K^i_{\ i}$ is the trace of the extrinsic curvature $K_{ij}$, $\alpha$ is the lapse and $\boldsymbol{\beta}$ is the shift 