1 / ( \varepsilon_g v_0^2)$, and $\varepsilon_g = - 1 / (4 \pi G)$, with $G$ being the gravitational constant. The density of linear momentum is, $\textbf{s} = \Sigma s_k \emph{\textbf{i}}_k$.

The above can be rewritten as, $$\begin{aligned}
- \mu_g ( \emph{i} s_0 + \textbf{s} ) =  \emph{i} ( \partial_0 \textbf{b} + \nabla^* \times \textbf{g} / v_0 ) + \emph{i} ( \nabla^* \cdot \textbf{g} / v_0 )
+ ( \nabla^* \times \textbf{b} - \partial_0 \textbf{g} / v_0 ) + \nabla^* \cdot \textbf{b}  ~.\end{aligned}$$

Comparing both sides of the equal sign in the above will yield, $$\begin{aligned}
&& \nabla^* \cdot \textbf{b} = 0 ~,
\\
&& \partial_0 \textbf{b} + \nabla^* \times \textbf{g} / v_0 = 0 ~,
\\
&& \nabla^* \cdot \textbf{g} / v_0 = - \mu_g s_0   ~,
\\
&& \nabla^* \times \textbf{b} - \partial_0 \textbf{g} / v_0 = - \mu_g \textbf{s} ~,\end{aligned}$$ where $s_0 = m v_0$. $\textbf{g} = \Sigma g_k \emph{\textbf{i}}_k$, $\textbf{b} = \Sigma b_k \emph{\textbf{i}}_k$.

Eqs.(9)-(12) are the gravitational field equations. Because the gravitational constant $G$ is weak and the velocity ratio $\textbf{v} / v_0$ is tiny, the gravity produced by the linear momentum $\textbf{s}$ can be ignored in general. When $\textbf{b} = 0$ and $\textbf{a} = 