## 3 Solving [PERSON]'s Equations
\begin{array}{cccc} \mathcal{E}\tau^{2}&0&0&0\\ 0&p\sigma^{2}&0&0\\ 0&0&pr^{2}&0\\ 0&0&0&pr^{2}\sin^{2}\theta\end{array} \right) \tag{23}\]  
In the end, we arrive at a set of five differential equations to be solved. To go on, it is worthwhile to set \(\tau=1/\sigma\). In this way we are left with only one unknown, for which we must have:  
\[G_{00}=\frac{1}{r^{2}\sigma^{5}}\bigg{(}2r\frac{\partial\sigma}{\partial r} -\sigma+\sigma^{3}\bigg{)}=\chi\bigg{(}\frac{\epsilon_{0}E^{2}}{2\sigma^{2}} -\frac{\mathcal{E}}{\sigma^{2}}-\frac{\epsilon_{0}\varrho c^{2}}{\mu\sigma^{4 }}\bigg{)} G_{01}=G_{10}=\frac{2}{r\sigma}\frac{\partial\sigma}{\partial t}=\chi \frac{\epsilon_{0}\varrho cv}{\mu}\]  
\[G_{11}=-\frac{1}{r^{2}\sigma}\bigg{(}2r\frac{\partial\sigma}{\partial r}- \sigma+\sigma^{3}\bigg{)}=\chi\bigg{(}-\frac{\epsilon_{0}E^{2}}{2}\sigma^{2} -pv\sigma^{2}-\frac{\epsilon_{0}\varrho v^{2}}{\mu}\sigma^{4}\bigg{)}\]  
\[G_{22}=-\frac{r}{\sigma^{4}}\bigg{(}2\sigma\frac{\partial\sigma}{\partial r}- 3r\bigg{(}\frac{\partial\sigma}{\partial r}\bigg{)}^{2}+r\sigma^{4}\bigg{(} \frac{\partial\sigma}{\partial t}\bigg{)}^{2}+r\sigma^{5}\ \frac{\partial^{2}\sigma}{\partial t^{2}}+r\ \sigma\ \frac{\partial^{2}\sigma}{\partial r^{2}}\bigg{)}=\chi \bigg{(}\frac{\epsilon_{0}E^{2}}{2}r^{2}-pr^{2}\bigg{)} \tag{24}\]The equation relative to \(G_{33}\) is trivially obtained from the one corresponding to \(G_{22}\) after multiplication by \(\sin^{2}\theta\). We successively evaluate \(\mathcal{E}\) from the first equation:  
\[\mathcal{E}=\frac{\epsilon_{0}E^{2}}{2}-\frac{1}{\chi}\bigg{[}\frac{c}{v}\frac{ 2}{r\sigma^{3}}\frac{\partial\sigma}{\partial t}+\frac{1}{r^{2}\sigma^{3}} \bigg{(}2r\frac{\partial\sigma}{\partial r}-\sigma+\sigma^{3}\bigg{)}\bigg{]} \tag{25}\]  
By suitably combining \(G_{00}\) and \(G_{11}\) we may recover \(p\):  
\[p=-\mathcal{E}+\frac{\epsilon_{0}\rho}{\mu}\bigg{(}\frac{c^{2}}{\sigma^{2}}+v^ {2}\sigma^{2}\bigg{)}=-\frac{\epsilon_{0}E^{2}}{2}+\frac{1}{\chi}\bigg{[} \frac{1}{r^{2}\sigma^{3}}\bigg{(}2r\frac{\partial\sigma}{\partial r}-\sigma+ \sigma^{3}\bigg{)}-\frac{v}{c}\frac{2\sigma}{r}\frac{\partial\sigma}{ \partial t}\bigg{]} \tag{26}\]  
In such a way, \(\rho\), \(\mathcal{E}\), and \(p\) are evaluated as functions of the given \(E\) and the unknown \(\sigma\). These will be substituted into the equation relative to \(G_{22}\). The equation for \(G_{33}\) is equivalent. After implementing the substitution: \(\sigma(t,r)=1/\sqrt{1+\omega(t,r)}\), and using the expression of \(p\) in (26), we get from the last equation in (24):  
\[\frac{1}{r}\bigg{(}-\frac{\omega}{r}-r\sigma^{6}\bigg{(}\frac{\partial\omega} {\partial t}\bigg{)}^{2}+\frac{r\sigma^{4}}{2}\frac{\partial^{2}\omega}{ \partial t^{2}}+\frac{r}{2}\frac{\partial^{2}\omega}{\partial r^{2}}+\frac{v} {c}\sigma^{4}\frac{\partial\omega}{\partial t}\bigg{)}=\chi\epsilon_{0}E^{2} \tag{27}\]  
where \(\sigma^{2n}=1/(1+\omega)^{n}\). This is the final equation to be solved. Unfortunately, we do not have a closed expression for \(\omega\) as a function of \(E\). We can, however, make some considerations. If \(\omega\) does not depend on \(t\) and \(E=Q/r^{2}\), equation (27) reduces to:  
\[-\frac{\omega}{r^{2}}+\frac{1}{2}\frac{\partial^{2}\omega}{\partial r^{2}}= \chi\epsilon_{0}\frac{Q^{2}}{r^{4}} \tag{28}\]  
The general solution in this case is \(\omega(r)=-K/r+\chi\epsilon_{0}Q^{2}/2r^{2}\), where \(K\) is an arbitrary real number. We have \(\rho=0\), \(\mathcal{E}=0\) and \(p=0\). Therefore \(M_{\alpha\beta}=0\). The final expression of \(\sigma\) exactly corresponds to that of the Reissner-Nordstrom metric, where the parameter \(K=2M>0\) is usually related to an effective mass \(M\). This stationary metric simulates a charged black hole. In the phenomenon we are currently studying there are no physical masses, and \(K\) is just a constant to be fixed according to some boundary conditions.