The Question
Given Maxwell's equations of the form \begin{align} \bar{\nabla}\times \bar{B} = \dfrac{4\pi}{c} \bar{J} + \partial_0 \bar{E} \\ \bar{\nabla}\times \bar{E} = -\partial_0 \bar{B} \\ \bar{\nabla} \cdot \bar{B}=0 \\ \bar{\nabla} \cdot \bar{E} = 4 \pi \rho, \end{align} what is the physical significance of the following transformation of Maxwell's equations: \begin{align} -i \bar{\nabla} \times \bar{G} + \bar{\nabla} G_0 = \dfrac{4\pi}{c} \bar{R} + \partial_{0}\bar{G} \tag{Amp-Far Dipole} \\ \bar{\nabla} \cdot \bar{G} - \partial_{0}G_0 = 4\pi R_0 , \tag{Gauss Dipole} \end{align} where $\bar{G} =(i\bar{r} \times\bar{F}-x_0 \bar{F}) $, $G_0 = (\bar{r} \cdot \bar{F})$, $\bar{R} = (\rho c \bar{r}-x_0\bar{J}+i \bar{r} \times\bar{J})$, $R_0 = (\bar{r}\cdot\bar{J} - x_0\rho c)/c$ and $\bar{F} = \bar{E} - i \bar{B}$.
The Definitions
In this post, I refer to (Amp-Far Dipole) and (Gauss Dipole) as the dipole equations, because I do not know what there actual names are or who first published these equations. I just stumbled upon them by accident on pencil and paper.
$x_0 = ct$ is the time variable multiplied by the speed of light for condensed notation purposes.
$\bar{R}$ is the complex combination of the electric dipole field density $\rho c \bar{r}-x_0\bar{J}$ and the magnetic dipole field density $\bar{r} \times\bar{J}$.
$\bar{R}$ is interpreted as a fictitious current in the dipole equations (Amp-Far Dipole). The corresponding fictitious charge density of $\bar{R}$ is $R_0$, which is equal to the Minkowski inner product of the four-position and the four-current.
Physical Consequences
Though $R_0$ and $\bar{R}$ are fictitious charge and current, they are conserved as a current when $G_0 = 0$. This implies that $G_0$ breaks charge conservation of the fictitious charge and current $R_0$ and $\bar{R}$.
An interesting consequence of the dipole equations is they are identical to Maxwell's equations when $G_0 = 0$.
Complex Formulation of Maxwell's Equations
I first write Ampere's law, Faraday's law and Gauss' law in complex form \begin{align} -i\bar{\nabla} \times \bar{F} = \dfrac{4\pi}{c} \bar{J} + \partial_{0} \bar{F} \tag{Amp-Far} \\ \bar{\nabla} \cdot \bar{F} = 4\pi \rho \tag{Gauss} , \end{align} where $\bar{F} = \bar{E} + i \bar{B}$.
Formulation of Ampere-Faraday Law in Dipole form
I use the following differential vector calculus identity \begin{align} \bar{r} \times (\bar{\nabla} \times) + \bar{r} (\bar{\nabla} \cdot) + x_0(\partial_{0}) = \bar{\nabla} \times (\bar{r} \times) + \bar{\nabla} (\bar{r} \cdot) + \partial_{0}(x_0) \end{align} to transform (Amp-Far) into the following: \begin{align} \bar{r} \times (\bar{\nabla} \times \bar{F}) + \bar{r} (\bar{\nabla} \cdot \bar{F}) + x_0(\partial_{0} \bar{F}) =\\ \bar{r} \times \left( i\dfrac{4\pi}{c} \bar{J} + i\partial_{0} \bar{F} \right) + \bar{r} \left(4\pi \rho\right) + x_0\left(-i\bar{\nabla} \times \bar{F} - \dfrac{4\pi}{c} \bar{J}\right) =\\ \dfrac{4\pi}{c} (i\bar{r} \times\bar{J}) + \partial_{0} (i\bar{r} \times\bar{F}) + 4\pi \left( \rho \bar{r} \right) - i\bar{\nabla} \times (x_0\bar{F}) - \dfrac{4\pi}{c} (x_0\bar{J}) =\\ \bar{\nabla} \times (\bar{r} \times \bar{F}) + \bar{\nabla} (\bar{r} \cdot \bar{F}) + \partial_{0}(x_0 \bar{F}) , \end{align} which reduces to the following expression \begin{align} -i \bar{\nabla} \times \left( i\bar{r} \times \bar{F} - x_0\bar{F} \right) + \bar{\nabla} (\bar{r} \cdot \bar{F}) =\\ \dfrac{4\pi}{c} \left( \rho c \bar{r} - x_0\bar{J} + i \bar{r} \times\bar{J} \right) + \partial_{0} \left( i\bar{r} \times\bar{F} - x_0 \bar{F} \right) . \end{align} One can perform the following substitutions $\bar{G} =(i\bar{r} \times\bar{F}-x_0 \bar{F}) $, $G_0 = (\bar{r} \cdot \bar{F})$, and $\bar{R} = (\rho c \bar{r}-x_0\bar{J}+i \bar{r} \times\bar{J})$ to obtain \begin{align} -i \bar{\nabla} \times \bar{G} + \bar{\nabla} G_0 = \dfrac{4\pi}{c} \bar{R} + \partial_{0}\bar{G} . \tag{Amp-Far Dipole} \end{align}
Formulation of Gauss' Law in Dipole form
I use the following differential vector calculus identity \begin{align} -x_0 (\nabla\cdot) + \bar{r}\cdot (-i\bar{\nabla}\times) = \bar{\nabla} \cdot (i(\bar{r}\times)- x_0) \end{align} to transform (Gauss) into the following: \begin{align} -x_0 (\nabla\cdot\bar{F}) + \bar{r}\cdot (-i\bar{\nabla}\times\bar{F}) =\\ -x_0 (4\pi \rho) + \bar{r}\cdot \left(\dfrac{4\pi}{c} \bar{J} + \partial_{0} \bar{F}\right) =\\ \dfrac{4\pi}{c} (\bar{r}\cdot\bar{J} - x_0\rho c) + \partial_{0} (\bar{r}\cdot\bar{F}) =\\ \bar{\nabla} \cdot (i\bar{r}\times\bar{F} - x_0\bar{F}) , \end{align} which reduces to the following expression \begin{align} \bar{\nabla} \cdot (i\bar{r}\times\bar{F} - x_0\bar{F}) - \partial_{0} (\bar{r}\cdot\bar{F}) = \dfrac{4\pi}{c} (\bar{r}\cdot\bar{J} - x_0\rho c) . \end{align} One can perform the following substitutions $R_0 = (\bar{r}\cdot\bar{J} - x_0\rho c)/c$ to obtain \begin{align} \bar{\nabla} \cdot \bar{G} - \partial_{0}G_0 = 4\pi R_0 . \tag{Gauss Dipole} \end{align}