## 2 Devising the Modeling Equations  
We begin with observing that the following continuity equation holds:  
\[\frac{\partial\rho}{\partial t}=-\text{div}(\rho\vec{V})=-\frac{1}{r^{2}} \frac{\partial(\rho vr^{2})}{\partial r} \tag{3}\]  
which is obtained by applying the divergence operator to both terms of Equation (2). The next step is to express the model equations in tensor form, by the use of four vectors. Among many others, we suggest [18; 19; 20], as possible referring books in general relativity. We proceed by introducing the system of coordinates in space-time: \(x_{\alpha}=(ct,r,\theta,\phi)\). We then need to consider the electromagnetic tensor \(F_{\alpha\beta}\). In the special case we are examining, its non-vanishing entries are \(F_{10}=E\), \(F_{01}=-E\), \(F^{10}=-E\), \(F^{01}=E\). We also recall that spherical coordinates are characterized by the metric tensor \(g_{\alpha\beta}=\text{diag}(1,-1,-r^{2},-r^{2}\sin^{2}\theta)\). In other terms, we have:  
\[(ds)^{2}=c^{2}(dt)^{2}-(dr)^{2}-r^{2}(d\theta)^{2}-r^{2}\sin^{2}\theta(d\phi) ^{2} \tag{4}\]  
Moreover, we get: \(\sqrt{-g}=r^{2}\sin\theta\), where \(g\) denotes the determinant of the metric tensor. In this space-time, it is straightforward to compute [PERSON]'s symbols  
\[\Gamma^{1}_{22}=-r\qquad\qquad\Gamma^{1}_{33}=-r\sin^{2}\theta\qquad\qquad \Gamma^{3}_{31}=\Gamma^{3}_{13}=\Gamma^{2}_{12}=\Gamma^{2}_{21}=1/r\]\[\Gamma_{33}^{2}=-\sin\theta\cos\theta\qquad\qquad\Gamma_{23}^{3}=\Gamma_{32}^{3}= \frac{\cos\theta}{\sin\theta} \tag{5}\]  
In addition, the Ricci's tensor \(R_{\alpha\beta}\) is identically zero, as well as the scalar curvature \(R\). These calculations and the following ones have been also checked with software for symbolic manipulation. It is also worthwhile to define the electromagnetic stress-energy tensor \(U_{\alpha\beta}\), that, in the simplified case we are examining, is diagonal and can be written as a function of \(E\) as follows:  
\[U_{00}=\tfrac{1}{2}E^{2},\quad U_{11}=-\tfrac{1}{2}E^{2},\quad U_{22}=\tfrac{1 }{2}E^{2}r^{2},\quad U_{33}=\tfrac{1}{2}E^{2}r^{2}\sin^{2}\theta \tag{6}\]  
\[U^{00}=\tfrac{1}{2}E^{2},\quad U^{11}=-\tfrac{1}{2}E^{2},\quad U^{22}=\frac{E ^{2}}{2r^{2}},\quad U^{33}=\frac{E^{2}}{2r^{2}\sin^{2}\theta} \tag{7}\]  
It will be useful to introduce the velocity 4-vector \(V_{\alpha}=(c,-v,0,0)\), with its contravariant version \(V^{\alpha}=(c,v,0,0)\). We also need a kind of mass tensor \(M_{\alpha\beta}\), akin to that of a perfect fluid:  
\[M_{\alpha\beta}=\mu^{-1}\rho\ V_{\alpha}V_{\beta}+\epsilon_{0}^{-1}\ g_{\alpha \beta}\Pi_{\alpha\beta} \tag{8}\]  
with \(\Pi_{\alpha\beta}=\text{diag}(\mathcal{E},-p,-p,-p)\). Here, \(\mathcal{E}\) denotes an energy density per unit of volume and \(p\) a pressure density per unit of surface. We are assuming to be in a vacuum so that \(\epsilon_{0}\) is the dielectric constant. If we want to study the case of a different isotropic medium, it will be enough to change the dielectric constant accordingly. Finally, there is a constant \(\mu\) which is dimensionally equivalent to Coulomb/Kg. The magnitude of this constant depends on the type of application one has in mind. An estimate of \(\mu\) was provided for example in [5], Appendix H, and, in that circumstance, it turned out to be approximately of the order of the ratio between the elementary charge and the electron mass. The multiplicative term \(\mu\) is necessary because \(\rho\) is not a density of mass, but a density of charge per unit of volume. This setting is justified by the fact that we want to remain within a pure electromagnetic context. The information is not transported through the dust of massive particles as it may happen in some plasma but travels by means of compression and rarefaction waves altering the divergence of the electric field. By the way, if we also have matter at the exterior of our ball, appropriate corrections can be easily taken into account.