field is $\textbf{A} = \Sigma A_k \emph{\textbf{I}}_k$. $\textbf{A}_0 = A_0 \emph{\textbf{I}}_0$; $A_0 = \phi / v_0$, with $\phi$ being the scalar potential of electromagnetic field.

For the sake of convenience the paper adopts the gauge condition, $\textbf{F}_0 = - \partial_0 \textbf{A}_0 + \nabla \cdot \textbf{A} = 0$. Therefore the above can be written as $\textbf{F} = \emph{i} \textbf{E} / v_0 + \textbf{B}$. Herein the electric field intensity is $\textbf{E} / v_0 = \partial_0 \textbf{A} + \nabla \circ \textbf{A}_0$, and the magnetic flux density is $\textbf{B} = \nabla \times \textbf{A}$. $\textbf{F}_0 = F_0 \emph{\textbf{I}}_0$. $\textbf{F} = \Sigma F_k \emph{\textbf{I}}_k$.

In the definition of electromagnetic source Eq.(4), the electromagnetic source is, $\mathbb{S}_e = \emph{i} \textbf{S}_0 + \textbf{S}$. Substituting the electromagnetic source $\mathbb{S}_e$ and electromagnetic strength $\mathbb{F}_e$ into the definition of electromagnetic source can separate the definition as, $$- \mu_e ( \emph{i} \textbf{S}_0 + \textbf{S} ) = ( \emph{i} \partial_0 + \nabla )^* \circ ( \emph{i} \textbf{E} / v_0 + \textbf{B} ) ~,$$ where the electromagnetic coefficient is $\mu_e = 1 / ( \varepsilon_e v_0^2)$, with $\varepsilon_e$ being the permittivity.

The above can be rewritten as, $$\begin{aligned}
- \mu_e ( \emph{i} \textbf{S}_0 + \textbf{S} ) = \emph{i} ( \partial_0 \textbf{B} + \nabla^* \times 