\textbf{E} / v_0 ) + \emph{i} ( \nabla^* \cdot \textbf{E} / v_0 )
+ ( \nabla^* \times \textbf{B} - \partial_0 \textbf{E} / v_0 ) + \nabla^* \cdot \textbf{B}  ~.\end{aligned}$$

Comparing the variables on both sides of the equal sign in the above will reason out the Maxwell’s equations as follows, $$\begin{aligned}
&& \nabla^* \cdot \textbf{B} = 0 ~,
\\
&& \partial_0 \textbf{B} + \nabla^* \times \textbf{E} / v_0 = 0 ~,
\\
&& \nabla^* \cdot \textbf{E} / v_0 = - \mu_e \textbf{S}_0   ~,
\\
&& \nabla^* \times \textbf{B} - \partial_0 \textbf{E} / v_0 = - \mu_e \textbf{S} ~,\end{aligned}$$ where $\textbf{S}_0 = S_0 \emph{\textbf{I}}_0$, $S_0 = q V_0$. $\textbf{E} = \Sigma E_k \emph{\textbf{I}}_k$, $\textbf{B} = \Sigma B_k \emph{\textbf{I}}_k$. The density of electric current is $\textbf{S} = \Sigma S_k \emph{\textbf{I}}_k$. For the charged particles, there may be, $\mathbb{V}_e = \mathbb{V}_g \circ \textbf{I} ( \emph{\textbf{I}}_j )$. And the unit $\textbf{I} ( \emph{\textbf{I}}_j )$ is one function of $\emph{\textbf{I}}_j$, with $\textbf{I} ( \emph{\textbf{I}}_j )^* \circ \textbf{I} ( \emph{\textbf{I}}_j ) = 1$.

In a similar way the electromagnetic potential $\mathbb{A}_e$, electromagnetic source $\mathbb{S}_e$, radius vector $\mathbb{R}_e$ , and velocity $\mathbb{V}_e$ etc are restricted by the component selection of the definitions. That is, the coordinates with 