$\Sigma^{R}_{\mu}$, where: $$\Sigma^{(0)}_\mu = \Gamma_{\omega} V_\mu
+\frac{\Gamma_{\rho}}{2} \boldsymbol {\tau}\cdot {\mathbf b}_{\mu}
+ \frac{e}{2} (1+\tau_3 ) A_{\mu},$$ $$\Sigma^{R}_{\mu}=\left( \frac{\partial \Gamma_v}{\partial
{\rho}} V^\nu j_\nu  + \frac{1}{2}
\frac{\partial \Gamma_\rho}{\partial \rho}  {\mathbf b}_\nu  \cdot \mathbf j_3^\nu  - \frac{\partial
\Gamma_\phi}{\partial {\rho}} \bar \psi \psi \phi\right)
u_\mu ~, \label{REAR}$$ where $\Sigma^{(0)}_\mu$ is the usual vector self-energy, $\hat\rho u_\mu=j_\mu$ with $u^2=1$ $j_\nu=\bar \psi \gamma_\nu \psi$, $\mathbf j_3^\nu= \bar \psi {\boldsymbol \tau} \gamma^\nu \psi$ and, as a result of the derivative of the Lagrangian with respect to $\rho$ a new term appears, $\Sigma^{R}_{\mu}$, which is called rearrangement self-energy and has been shown to play an essential rôle in the applications of the theory. This term guarantees the thermodynamical consistency and the energy-momentum conservation. For more detailed calculations, at zero and finite temperatures, please refer to [@previous].

In the static case there are no currents in the nucleus and the spatial vector components are zero. Therefore, the mesonic equations of motion become: $$\nabla^2 \phi = m_s^2\phi- \Gamma_s \rho_s,
\label{elphi}$$ $$\nabla^2 V_0 = m_v^2 V_0 - \Gamma_v \rho,
\label{elV0}$$ $$\nabla^2 b_0 =m_\rho^2 b_0 -\frac{\Gamma_\rho}{2} \rho_3,
\label{elb0}$$ $$\nabla^2 A_0 =-e \rho_p,\label{elA0}$$ where $\rho_s=<\bar \psi \psi>$ is the scalar density, $\rho=\rho_p+\rho_n$, $\rho_3=\rho_p-\rho_n$ and $\rho_p$ and $\rho_n$ are the proton and neutron densities.

