kre10.txt \documentclass[12pt]{article} \renewcommand{\thefootnote}{\fnsymbol{footnote}} \setcounter{footnote}{0} \renewcommand{\baselinestretch}{1} \setlength{\oddsidemargin}{-0.30in} \setlength{\textwidth}{7.00in} \setlength{\topmargin}{-0.60in} \setlength{\textheight}{9.10in} \begin{document} \title{Essay on the Unified Theory of the Classical Fields of Gravitation and Electromagnetism} \author{Kurt Reichling\footnote{kr@e-kr.org}} \date{version 1.0, March 2002} \maketitle \begin{center} \bf Extract \rm \end{center} \noindent We would like to introduce a new attempt to establish a unified theory of the classical fields of gravitation and electromagnetism which complies with the geometric paradigm of the theory of relativity of Albert Einstein. The theory we shall describe is geometrically unified in the sense that the fields equations, and the Hamiltonian function from which they derive, are formally unified entities (i.e. they are not the sum of several independent parts) which depend only on the metric of the particular spacetime being considered. \bigskip \bigskip \bigskip \begin{center} \bf Remarks \end{center} \medskip \noindent In this essay, we make use of well known results from the theory of relativity of Albert Einstein without giving explicit references. We implicitly refer to [1] for a detailed explanation of these results. \bigskip \noindent Copyright \copyright\ Kurt Reichling. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.1 or any later version published by the Free Software Foundation; with no Invariant Sections, with no Front-Cover Texts, and with no Back-Cover Texts. A copy of the license is included in the section entitled ``GNU Free Documentation License''. \bigskip \noindent The source code and other machine readable formats of the last version of this essay can be found on the internet at the {\it http://www.e-kr.org} address. Available formats are at least ``TeX'', ``DVI'' and ``pdf'' formats. \clearpage \section{Introduction} \noindent The attempts to geometrically unify the classical fields of gravitation and electromagnetism can be classified in two categories (see [1]): \begin{itemize} \item {those consisting in abandoning the Riemannian geometry in the 4-dimensional spacetime by considering geometries with affine connections (symmetric or asymmetric). This is the case with the attempts of H. Weyl, A.S. Eddington, E. Schr\"{o}dinger and A. Einstein; and} \item {those consisting in keeping the Riemannian geometry by increasing the dimension of the spacetime being considered. This is the case with the attempts of T. Kaluza, O. B. Klein and A. Einstein.} \end{itemize} \noindent Our approach belongs to the second category by considering a ``double'' 4-dimensional spacetime composed of the usual 4-dimensional spacetime and the corresponding dual 4-dimensional spacetime. \bigskip \noindent The linearization of the fields equations by the so-called ``Einstein-Infeld-Hoffmann'' (EIH) method, will allow us to: \begin{itemize} \item {deduct the ``Newton-Maxwell-Lorentz'' results of classical physics. That is, the gravitational field and the electromagnetic field of charged mass points and their equations of motion;} \item {explain why the Lorentz transformation must be linear in classical physics. This property is imposed in classical physics by the sake of simplicity, without being formally justified as it is the case in the theory we shall describe; and} \item {show that the electromagnetic field could not exist without the gravitational field and by consequence, that we are not allowed to describe the electromagnetic field without considering the gravitational field.} \end{itemize} \noindent It is important to say in advance that these deductions will hold separately in each of the two 4-dimensional spacetime being considered. During the linearization process, we will see that by imposing a formal restriction to the equations and to their solutions, the things happen as if there was only one 4-dimensional spacetime to consider. \clearpage \section{The structure of the spacetime and the metric} \noindent The structure of the spacetime on which the theory is based is double, for it is composed of the usual 4-dimensional spacetime $\left(z^{1}, z^{2}, z^{3}, z^{4}\right)$ and the corresponding dual 4-dimensional spacetime $\left(z^{\bar{1}}, z^{\bar{2}}, z^{\bar{3}}, z^{\bar{4}}\right)$. \bigskip \noindent $z^{1}, z^{2}, z^{3}$ represent the usual spacial coordinates $x^{1}, x^{2}, x^{3}$ and $z^{\bar{1}}, z^{\bar{2}}, z^{\bar{3}}$ represent the corresponding dual spacial coordinates $x^{\bar{1}}, x^{\bar{2}}, x^{\bar{3}}$. In order to facilitate the calculations, $z^{4}$ which represents the usual time coordinate will be chosen imaginary: $z^{4}= ict$. The same for $z^{\bar{4}}$ which represents the corresponding dual time coordinate: $z^{\bar{4}}=ic\bar{t}$. \bigskip \noindent This particular structure of the spacetime allows the introduction of the following metric: \begin{equation} ds^{2}=dz_{A}dz^{A}=g_{A B}dz^{A}dz^{B}=g_{\alpha \beta}dz^{\alpha}dz^{\beta}+ g_{\alpha\bar{\beta}}dz^{\alpha}dz^{\bar{\beta}}+g_{\bar{\alpha}\beta}dz^{\bar{\alpha}}dz^{\beta}+ g_{\bar{\alpha}\bar{\beta}}dz^{\bar{\alpha}}dz^{\bar{\beta}} \end{equation} \[ \mbox{with }g_{AB}=g_{BA}\mbox{ symmetric} \mbox{ , }g_{AB}g^{BC}=\delta_{A}^{C} \mbox{ , }dz_{A}=g_{AB}dz^{B} \mbox{ and }z^{A}=\left(z^{\alpha} , z^{\bar{\alpha}}\right) \] \bigskip \noindent \bf Remark: \rm \medskip \noindent The following notations are used in this essay: \begin{itemize} \item{Lower Latin indices run over the values from 1 to 3 (spatial coordinates), Greek indices run over the values from 1 to 4 (spatial and time coordinates) and upper Latin indices run over the values from the usual and the corresponding dual Greek indices. Repeated dummy indices indicate an implicit summation:} \medskip \begin{tabular}{lcl} . $g_{ab}dz^{a}dz^{b}= \sum_{a=1}^{3}\sum_{b=1}^{3}g_{ab}dz^{a}dz^{b}$ & , & $ g_{a\bar{b}}dz^{a}dz^{\bar{b}}= \sum_{a=1}^{3}\sum_{\bar{b}=\bar{1}}^{\bar{3}}g_{a\bar{b}}dz^{a}dz^{\bar{b}}$ \\ [.10in] . $g_{\alpha\beta}dz^{\alpha}dz^{\beta}= \sum_{\alpha=1}^{4}\sum_{\beta=1}^{4}g_{\alpha\beta}dz^{\alpha}dz^{\beta}$ & , & $ g_{\alpha\bar{\beta}}dz^{\alpha}dz^{\bar{\beta}}= \sum_{\alpha=1}^{4}\sum_{\bar{\beta}=\bar{1}}^{\bar{4}}g_{\alpha\bar{\beta}}dz^{\alpha}dz^{\bar{\beta}}$ \\ [.10in] \multicolumn{3}{l}{. $g_{AB}dz^{A}dz^{B}= g_{\alpha\beta}dz^{\alpha}dz^{\beta}+g_{\alpha\bar{\beta}}dz^{\alpha}dz^{\bar{\beta}}+ g_{\bar{\alpha}\beta}dz^{\bar{\alpha}}dz^{\beta}+g_{\bar{\alpha}\bar{\beta}}dz^{\bar{\alpha}}dz^{\bar{\beta}}$} \end{tabular} \item {$\partial_{A}f=\displaystyle{\frac{\partial f}{\partial z^{A}}}$ , $\partial_{\alpha}f= \displaystyle{\frac{\partial f}{\partial z^{\alpha}}}$ , $\partial_{\bar{\alpha}}f= \displaystyle{\frac{\partial f}{\partial z^{\bar{\alpha}}}}$ are partial derivatives} \item {$D_{A}f$ , $D_{\alpha}f$ , $D_{\bar{\alpha}}f$ are covariant derivatives} \item {$z^{\bar{\alpha}}$ is the corresponding dual vector of the usual vector $z^{\alpha}$} \item {$z^{A}=\left(z^{\alpha} , z^{\bar{\alpha}}\right)$ with $z^{\alpha} = \left(z^{1}, z^{2}, z^{3}, z^{4}\right)$ and $z^{\bar{\alpha}} = \left(z^{\bar{1}}, z^{\bar{2}}, z^{\bar{3}}, z^{\bar{4}}\right)$} \item {$\acute{z}^{A}=\acute{z}^{A}\left(z^{B}\right)$ with $\acute{z}^{\alpha}= \acute{z}^{\alpha}\left(z^{\beta}\right)$ and $\acute{z}^{\bar{\alpha}}= \acute{z}^{\bar{\alpha}}\left(z^{\bar{\beta}}\right)$ are coordinate transformations} \end{itemize} \clearpage \section{The coordinate transformations and the tensors} \noindent The coordinate transformations that we shall consider are of the form $\acute{z}^{A}= \acute{z}^{A}\left(z^{B}\right)$ with \\ $\acute{z}^{\alpha}= \acute{z}^{\alpha}(z^{\beta})$ and $\acute{z}^{\bar{\alpha}}= \acute{z}^{\bar{\alpha}}(z^{\bar{\beta}})$ . \bigskip \noindent Under these particular coordinate transformations, a contravariant vector $V^{A}$ will transform according to: $\acute{V}^{A}=V^{K}\displaystyle{\frac{d\acute{z}^{A}}{dz^{K}}} \Rightarrow \acute{V}^{\alpha}=V^{K}\displaystyle{{\frac{d\acute{z}^{\alpha}}{dz^{K}}}}= V^{\epsilon}\displaystyle{{\frac{d\acute{z}^{\alpha}}{dz^{\epsilon}}}}+ V^{\bar{\epsilon}}\overbrace{\displaystyle{{\frac{d\acute{z}^{\alpha}}{dz^{\bar{\epsilon}}}}}}^{=0} \Rightarrow \acute{V}^{\alpha}= V^{\epsilon}\displaystyle{{\frac{d\acute{z}^{\alpha}}{dz^{\epsilon}}}} \Rightarrow\acute{V}^{\bar{\alpha}}= V^{\bar{\epsilon}}\displaystyle{{\frac{d\acute{z}^{\bar{\alpha}}}{dz^{\bar{\epsilon}}}}}$ \bigskip \noindent A covariant vector $V_{A}$ will transform according to: $\acute{V}_{A}=V_{K}\displaystyle{{\frac{dz^{K}}{d\acute{z}^{A}}}} \Rightarrow \acute{V}_{\alpha}=V_{K}\displaystyle{{\frac{dz^{K}}{d\acute{z}^{\alpha}}}}= V_{\epsilon}\displaystyle{{\frac{dz^{\epsilon}}{d\acute{z}^{\alpha}}}}+ V_{\bar{\epsilon}}\overbrace{\displaystyle{{\frac{dz^{\bar{\epsilon}}}{d\acute{z}^{\alpha}}}}}^{=0} \Rightarrow \acute{V}_{\alpha}= V_{\epsilon}\displaystyle{{\frac{dz^{\epsilon}}{d\acute{z}^{\alpha}}}} \Rightarrow \acute{V}_{\bar{\alpha}}= V_{\bar{\epsilon}}\displaystyle{{\frac{dz^{\bar{\epsilon}}}{d\acute{z}^{\bar{\alpha}}}}}$ \bigskip\noindent The generalization to tensors of any rank is immediate: \bigskip \begin{tabular}{lcl} $\acute{T}_{..C}^{AB.D}= T_{..M}^{KL.N}\displaystyle{{\frac{d\acute{z}^{A}}{dz^{K}}}} \displaystyle{{\frac{d\acute{z}^{B}}{dz^{L}}}}\displaystyle{{\frac{dz^{M}}{d\acute{z}^{C}}}} \displaystyle{{\frac{d\acute{z}^{D}}{dz^{N}}}}$ & , & $\acute{T}_{..\gamma}^{\alpha\bar{\beta}.\delta}= T_{..\tau}^{\epsilon\bar{\theta}.\sigma}\displaystyle{{\frac{d\acute{z}^{\alpha}}{dz^{\epsilon}}}} \displaystyle{{\frac{d\acute{z}^{\bar{\beta}}}{dz^{\bar{\theta}}}}} \displaystyle{{\frac{dz^{\tau}}{d\acute{z}^{\gamma}}}} \displaystyle{{\frac{d\acute{z}^{\delta}}{dz^{\sigma}}}}$ \end{tabular} \bigskip \noindent We may also mix upper Latin indices and Greek indices: \medskip $\acute{T}_{..B}^{A\bar{\alpha}.\beta}= T_{..L}^{K\bar{\epsilon}.\theta}\displaystyle{{\frac{d\acute{z}^{A}}{dz^{K}}}} \displaystyle{{\frac{d\acute{z}^{\bar{\alpha}}}{dz^{\bar{\epsilon}}}}} \displaystyle{{\frac{dz^{L}}{d\acute{z}^{B}}}} \displaystyle{{\frac{d\acute{z}^{\beta}}{dz^{\theta}}}}$ \bigskip \noindent The inverse transformation is similar: \medskip $T_{..B}^{A\bar{\alpha}.\beta}= \acute{T}_{..L}^{K\bar{\epsilon}.\theta}\displaystyle{{\frac{dz^{A}}{d\acute{z}^{K}}}} \displaystyle{{\frac{dz^{\bar{\alpha}}}{d\acute{z}^{\bar{\epsilon}}}}} \displaystyle{{\frac{d\acute{z}^{L}}{dz^{B}}}} \displaystyle{{\frac{dz^{\beta}}{d\acute{z}^{\theta}}}}$ \bigskip \bigskip \noindent \bf The particular case of the metric tensor: \rm \medskip \noindent When we consider the metric tensor $g_{AB}$, we obtain: \begin{equation} \acute{g}_{AB}=g_{KL} \displaystyle{{\frac{dz^{K}}{d\acute{z}^{A}}}} \displaystyle{{\frac{dz^{L}}{d\acute{z}^{B}}}} \Rightarrow \acute{g}_{\alpha\beta}=g_{\epsilon\theta} \displaystyle{{\frac{dz^{\epsilon}}{d\acute{z}^{\alpha}}}} \displaystyle{{\frac{dz^{\theta}}{d\acute{z}^{\beta}}}} \Rightarrow \acute{g}_{\alpha\bar{\beta}}=g_{\epsilon\bar{\theta}} \displaystyle{{\frac{dz^{\epsilon}}{d\acute{z}^{\alpha}}}} \displaystyle{{\frac{dz^{\bar{\theta}}}{d\acute{z}^{\bar{\beta}}}}} \end{equation} \medskip \noindent These relations mean that $\acute{g}_{\alpha\beta}$ depend only on $g_{\epsilon\theta}$ and that $\acute{g}_{\alpha\bar{\beta}}$ depend only on $g_{\epsilon\bar{\theta}}$ . In particular, if $g_{\epsilon\bar{\theta}}=0$ , we obtain $\acute{g}_{\alpha\bar{\beta}}=0$ whatever $g_{\epsilon\theta}$ . It is important to notice that these results are due to the particular coordinate transformations being considered. \clearpage \section{The Christoffel symbols} \noindent The calculation $\int \delta ds=0$ (the variation is operated on the $z^{A}$), with $ds^{2}= g_{AB}dz^{A}dz^{B}$ leads to the following equation: \begin{equation} \displaystyle{{\frac{d^{2}z^{C}}{d^{2}s}}}+ \Gamma_{AB}^{C}\displaystyle{{\frac{dz^{A}}{ds}}}\displaystyle{{\frac{dz^{B}}{ds}}}=0 \mbox{ , }\Gamma_{AB}^{C}= \textstyle{\frac{1}{2}}g^{CK}\left(\partial_{A}g_{BK}+\partial_{B}g_{AK}-\partial_{K}g_{AB}\right) \mbox{ , }\Gamma_{AB}^{C}=\Gamma_{BA}^{C}\mbox{ symmetric} \end{equation} \bigskip \noindent The Christoffel symbols $\Gamma_{AB}^{C}$ can be developed into Greek indices: \medskip \noindent \begin{tabbing} $\Gamma_{\alpha\beta}^{\gamma}$\= $=\textstyle{\frac{1}{2}}g^{\gamma K}\left(\partial_{\alpha}g_{\beta K}+ \partial_{\beta}g_{\alpha K}- \partial_{K}g_{\alpha\beta}\right)$ \\ [.10in] \>$\Rightarrow\Gamma_{\alpha\beta}^{\gamma}= \textstyle{\frac{1}{2}}g^{\gamma\epsilon}\left(\partial_{\alpha}g_{\beta\epsilon}+ \partial_{\beta}g_{\alpha\epsilon}- \partial_{\epsilon}g_{\alpha\beta}\right)+ \textstyle{\frac{1}{2}}g^{\gamma\bar{\epsilon}}\left(\partial_{\alpha}g_{\beta\bar{\epsilon}}+ \partial_{\beta}g_{\alpha\bar{\epsilon}}- \partial_{\bar{\epsilon}}g_{\alpha\beta}\right)$ \end{tabbing} \noindent \begin{tabbing} $\Gamma_{\alpha\bar{\beta}}^{\gamma}$\= $=\textstyle{\frac{1}{2}}g^{\gamma K}\left(\partial_{\alpha}g_{\bar{\beta} K}+ \partial_{\bar{\beta}}g_{\alpha K}- \partial_{K}g_{\alpha\bar{\beta}}\right)$ \\ [.10in] \>$\Rightarrow\Gamma_{\alpha\bar{\beta}}^{\gamma}= \textstyle{\frac{1}{2}}g^{\gamma\epsilon}\left(\partial_{\alpha}g_{\bar{\beta}\epsilon}+ \partial_{\bar{\beta}}g_{\alpha\epsilon}- \partial_{\epsilon}g_{\alpha\bar{\beta}}\right)+ \textstyle{\frac{1}{2}}g^{\gamma\bar{\epsilon}}\left(\partial_{\alpha}g_{\bar{\beta}\bar{\epsilon}}+ \partial_{\bar{\beta}}g_{\alpha\bar{\epsilon}}- \partial_{\bar{\epsilon}}g_{\alpha\bar{\beta}}\right)$ \end{tabbing} \noindent \begin{tabbing} $\Gamma_{\bar{\alpha}\beta}^{\gamma}$\= $=\textstyle{\frac{1}{2}}g^{\gamma K}\left(\partial_{\bar{\alpha}}g_{\beta K}+ \partial_{\beta}g_{\bar{\alpha}K}- \partial_{K}g_{\bar{\alpha}\beta}\right)$ \\ [.10in] \>$\Rightarrow\Gamma_{\bar{\alpha}\beta}^{\gamma}= \textstyle{\frac{1}{2}}g^{\gamma\epsilon}\left(\partial_{\bar{\alpha}}g_{\beta\epsilon}+ \partial_{\beta}g_{\bar{\alpha}\epsilon}- \partial_{\epsilon}g_{\bar{\alpha}\beta}\right)+ \textstyle{\frac{1}{2}}g^{\gamma\bar{\epsilon}}\left(\partial_{\bar{\alpha}}g_{\beta\bar{\epsilon}}+ \partial_{\beta}g_{\bar{\alpha}\bar{\epsilon}}- \partial_{\bar{\epsilon}}g_{\bar{\alpha}\beta}\right)$ \end{tabbing} \noindent \begin{tabbing} $\Gamma_{\bar{\alpha}\bar{\beta}}^{\gamma}$\= $=\textstyle{\frac{1}{2}}g^{\gamma K}\left(\partial_{\bar{\alpha}}g_{\bar{\beta} K}+ \partial_{\bar{\beta}}g_{\bar{\alpha} K}- \partial_{K}g_{\bar{\alpha}\bar{\beta}}\right)$ \\ [.10in] \>$\Rightarrow\Gamma_{\bar{\alpha}\bar{\beta}}^{\gamma}= \textstyle{\frac{1}{2}}g^{\gamma\epsilon}\left(\partial_{\bar{\alpha}}g_{\bar{\beta}\epsilon}+ \partial_{\bar{\beta}}g_{\bar{\alpha}\epsilon}- \partial_{\epsilon}g_{\bar{\alpha}\bar{\beta}}\right)+ \textstyle{\frac{1}{2}}g^{\gamma\bar{\epsilon}}\left(\partial_{\bar{\alpha}}g_{\bar{\beta}\bar{\epsilon}}+ \partial_{\bar{\beta}}g_{\bar{\alpha}\bar{\epsilon}}- \partial_{\bar{\epsilon}}g_{\bar{\alpha}\bar{\beta}}\right)$ \end{tabbing} \bigskip \noindent And so forth for $\Gamma_{\bar{\alpha}\bar{\beta}}^{\bar{\gamma}}$ , $\Gamma_{\bar{\alpha}\beta}^{\bar{\gamma}}$ , $\Gamma_{\alpha\bar{\beta}}^{\bar{\gamma}}$ , $\Gamma_{\alpha\beta}^{\bar{\gamma}}$ . \section{The covariant derivatives} The covariant derivatives are expressed by: \bigskip \begin{tabular}{lcl} \medskip $D_{C}V^{A}=\partial_{C}V^{A}+ \Gamma_{KC}^{A}V^{K}$ & , & $D_{C}V_{A}=\partial_{C}V_{A}- \Gamma_{AC}^{K}V_{K}$ \\ \medskip $D_{C}T^{AB}=\partial_{C}T^{AB}+ \Gamma_{KC}^{A}T^{KB}+ \Gamma_{KC}^{B}T^{AK}$ & , & $D_{C}T_{AB}=\partial_{C}T_{AB}- \Gamma_{AC}^{K}T_{KB}- \Gamma_{BC}^{K}T_{AK}$ \\ \medskip $D_{C}T_{A}^{B}=\partial_{C}A_{A}^{B}- \Gamma_{AC}^{K}T_{K}^{B}+ \Gamma_{KC}^{B}T_{A}^{K}$ & , & $D_{C}\left(V^{A}V^{B}\right)=D_{C}V^{A}V^{B}+ V^{A}D_{C}V^{B}$ \\ \medskip $D_{C}\left(V^{A}V_{B}\right)=D_{C}V^{A}V_{B}+ V^{A}D_{C}V_{B}$ & , & $D_{C}\left(V_{A}V_{B}\right)=D_{C}V_{A}V_{B}+ V_{A}D_{C}V_{B}$ \end{tabular} \bigskip \noindent These expressions can be developed into Greek indices: \bigskip $D_{\gamma}T^{\alpha\beta}=\partial_{\gamma}T^{\alpha\beta}+\Gamma_{K\gamma}^{\alpha}T^{K\beta}+ \Gamma_{K\gamma}^{\beta}T^{\alpha K}=\partial_{\gamma}T^{\alpha\beta}+ \Gamma_{\epsilon\gamma}^{\alpha}T^{\epsilon\beta}+ \Gamma_{\bar{\epsilon}\gamma}^{\alpha}T^{\bar{\epsilon}\beta}+ \Gamma_{\epsilon\gamma}^{\beta}T^{\alpha\epsilon}+ \Gamma_{\bar{\epsilon}\gamma}^{\beta}T^{\alpha \bar{\epsilon}}$ \medskip $D_{\bar{\gamma}}T^{\alpha\beta}=\partial_{\bar{\gamma}}T^{\alpha\beta}+\Gamma_{K \bar{\gamma}}^{\alpha}T^{K \beta}+ \Gamma_{K \bar{\gamma}}^{\beta}T^{\alpha K}=\partial_{\bar{\gamma}}T^{\alpha\beta}+ \Gamma_{\epsilon\bar{\gamma}}^{\alpha}T^{\epsilon\beta}+ \Gamma_{\bar{\epsilon}\bar{\gamma}}^{\alpha}T^{\bar{\epsilon}\beta}+ \Gamma_{\epsilon\bar{\gamma}}^{\beta}T^{\alpha\epsilon}+ \Gamma_{\bar{\epsilon}\bar{\gamma}}^{\beta}T^{\alpha\bar{\epsilon}}$ \medskip $D_{\gamma}\left(V^{\alpha}V^{\beta}\right) =D_{\gamma}V^{\alpha}V^{\beta}+V^{\alpha}D_{\gamma}V^{\beta}$ \clearpage \section{The covariant derivatives of the metric tensor} \noindent When the Christoffel symbols are expressed by (3), we have: \begin{equation} D_{C}g_{AB}=D_{C}g^{AB}=0 \end{equation} \bigskip \noindent This relation can be developed into Greek indices: \[ D_{\gamma}g_{\alpha\beta}=D_{\bar{\gamma}}g_{\alpha\beta}= D_{\gamma}g_{\alpha\bar{\beta}}=D_{\bar{\gamma}}g_{\alpha\bar{\beta}}= D_{\gamma}g^{\alpha\beta}=D_{\bar{\gamma}}g^{\alpha\beta}= D_{\gamma}g^{\alpha\bar{\beta}}=D_{\bar{\gamma}}g^{\alpha\bar{\beta}}=0 \] \section{The Riemann and the Ricci curvature tensors} \noindent The calculation $T_{CAB}=D_{C}D_{A}V_{B}-D_{A}D_{C}V_{B}$ leads to the result $T_{CAB}= R_{CAB}^{D}V_{D}$ \medskip \noindent where $R_{CAB}^{D}= \partial_{A}\Gamma_{BC}^{D}- \partial_{C}\Gamma_{AB}^{D}+ \Gamma_{AL}^{D}\Gamma_{CB}^{L}- \Gamma_{AB}^{L}\Gamma_{LC}^{D}$ is the Riemann curvature tensor. \bigskip \noindent By contracting the Riemann curvature tensor, we obtain the Ricci curvature tensor: \begin{equation} R_{AB}= R_{KAB}^{K}= \partial_{A}\Gamma_{BK}^{K}- \partial_{K}\Gamma_{AB}^{K}+ \Gamma_{AL}^{K}\Gamma_{KB}^{L}- \Gamma_{AB}^{L}\Gamma_{LK}^{K} \mbox{ , symmetric} \end{equation} \medskip \noindent The symmetry of $R_{AB}$ follows from the fact that $\Gamma_{AK}^{K}= \textstyle{\frac{1}{2}}\partial_{A}\ln\left|g\right|$ , where $g=$ determinant of $g_{AB}=\left|g_{AB}\right|$ \bigskip \bigskip \noindent $R_{AB}$ can be developed into Greek indices: \bigskip \noindent \begin{tabbing} $R_{\alpha\beta}$\= $=\partial_{\alpha}\Gamma_{\beta K}^{K}- \partial_{K}\Gamma_{\alpha\beta}^{K}+ \Gamma_{\alpha L}^{K}\Gamma_{K\beta}^{L}- \Gamma_{\alpha\beta}^{L}\Gamma_{LK}^{K}$ \\ [.05in] \>$\Rightarrow R_{\alpha\beta}= \partial_{\alpha}\Gamma_{\beta\epsilon}^{\epsilon}+ \partial_{\alpha}\Gamma_{\beta\bar{\epsilon}}^{\bar{\epsilon}}- \partial_{\epsilon}\Gamma_{\alpha\beta}^{\epsilon}- \partial_{\bar{\epsilon}}\Gamma_{\alpha\beta}^{\bar{\epsilon}}+ \Gamma_{\alpha L}^{\epsilon}\Gamma_{\epsilon\beta}^{L}+ \Gamma_{\alpha L}^{\bar{\epsilon}}\Gamma_{\bar{\epsilon}\beta}^{L}- \Gamma_{\alpha\beta}^{L}\Gamma_{L \epsilon}^{\epsilon}- \Gamma_{\alpha\beta}^{L}\Gamma_{L\bar{\epsilon}}^{\bar{\epsilon}}$ \\ [.05in] \>$\Rightarrow R_{\alpha\beta}=$\= $\partial_{\alpha}\Gamma_{\beta\epsilon}^{\epsilon}+ \partial_{\alpha}\Gamma_{\beta\bar{\epsilon}}^{\bar{\epsilon}}- \partial_{\epsilon}\Gamma_{\alpha\beta}^{\epsilon}- \partial_{\bar{\epsilon}}\Gamma_{\alpha\beta}^{\bar{\epsilon}}+ \Gamma_{\alpha\theta}^{\epsilon}\Gamma_{\epsilon\beta}^{\theta}+ \Gamma_{\alpha\bar{\theta}}^{\epsilon}\Gamma_{\epsilon\beta}^{\bar{\theta}}+ \Gamma_{\alpha\theta}^{\bar{\epsilon}}\Gamma_{\bar{\epsilon}\beta}^{\theta}+ \Gamma_{\alpha\bar{\theta}}^{\bar{\epsilon}}\Gamma_{\bar{\epsilon}\beta}^{\bar{\theta}}$ \\ [.05in] \>\>$-\Gamma_{\alpha\beta}^{\theta}\Gamma_{\theta\epsilon}^{\epsilon}- \Gamma_{\alpha\beta}^{\bar{\theta}}\Gamma_{\bar{\theta}\epsilon}^{\epsilon}- \Gamma_{\alpha\beta}^{\theta}\Gamma_{\theta\bar{\epsilon}}^{\bar{\epsilon}}- \Gamma_{\alpha\beta}^{\bar{\theta}}\Gamma_{\bar{\theta}\bar{\epsilon}}^{\bar{\epsilon}}$ \end{tabbing} \bigskip \noindent \begin{tabbing} $R_{\alpha\bar{\beta}}$\= $=\partial_{\alpha}\Gamma_{\bar{\beta}K}^{K}- \partial_{K}\Gamma_{\alpha\bar{\beta}}^{K}+ \Gamma_{\alpha L}^{K}\Gamma_{K\bar{\beta}}^{L}- \Gamma_{\alpha\bar{\beta}}^{L}\Gamma_{LK}^{K}$ \\ [.05in] \>$\Rightarrow R_{\alpha\bar{\beta}}= \partial_{\alpha}\Gamma_{\bar{\beta}\epsilon}^{\epsilon}+ \partial_{\alpha}\Gamma_{\bar{\beta}\bar{\epsilon}}^{\bar{\epsilon}}- \partial_{\epsilon}\Gamma_{\alpha\bar{\beta}}^{\epsilon}- \partial_{\bar{\epsilon}}\Gamma_{\alpha\bar{\beta}}^{\bar{\epsilon}}+ \Gamma_{\alpha L}^{\epsilon}\Gamma_{\epsilon\bar{\beta}}^{L}+ \Gamma_{\alpha L}^{\bar{\epsilon}}\Gamma_{\bar{\epsilon}\bar{\beta}}^{L}- \Gamma_{\alpha\bar{\beta}}^{L}\Gamma_{L \epsilon}^{\epsilon}- \Gamma_{\alpha\bar{\beta}}^{L}\Gamma_{L\bar{\epsilon}}^{\bar{\epsilon}}$ \\ [.05in] \>$\Rightarrow R_{\alpha\bar{\beta}}=$\= $\partial_{\alpha}\Gamma_{\bar{\beta}\epsilon}^{\epsilon}+ \partial_{\alpha}\Gamma_{\bar{\beta}\bar{\epsilon}}^{\bar{\epsilon}}- \partial_{\epsilon}\Gamma_{\alpha\bar{\beta}}^{\epsilon}- \partial_{\bar{\epsilon}}\Gamma_{\alpha\bar{\beta}}^{\bar{\epsilon}}+ \Gamma_{\alpha\theta}^{\epsilon}\Gamma_{\epsilon\bar{\beta}}^{\theta}+ \Gamma_{\alpha\bar{\theta}}^{\epsilon}\Gamma_{\epsilon\bar{\beta}}^{\bar{\theta}}+ \Gamma_{\alpha\theta}^{\bar{\epsilon}}\Gamma_{\bar{\epsilon}\bar{\beta}}^{\theta}+ \Gamma_{\alpha\bar{\theta}}^{\bar{\epsilon}}\Gamma_{\bar{\epsilon}\bar{\beta}}^{\bar{\theta}}$ \\ [.05in] \>\>$-\Gamma_{\alpha\bar{\beta}}^{\theta}\Gamma_{\theta\epsilon}^{\epsilon}- \Gamma_{\alpha\bar{\beta}}^{\bar{\theta}}\Gamma_{\bar{\theta}\epsilon}^{\epsilon}- \Gamma_{\alpha\bar{\beta}}^{\theta}\Gamma_{\theta\bar{\epsilon}}^{\bar{\epsilon}}- \Gamma_{\alpha\bar{\beta}}^{\bar{\theta}}\Gamma_{\bar{\theta}\bar{\epsilon}}^{\bar{\epsilon}}$ \end{tabbing} \clearpage \section{The principle of equivalence} \noindent One consequence of the principle of equivalence is that we can make ``appear'' a gravitational field from a flat spacetime by applying a coordinate transformation. Another consequence of the principle of equivalence is that we can locally make the gravitational field ``disappear'' by applying an appropriate coordinate transformation. \medskip \noindent A similar result does not exist for the electromagnetic field. We can not make appear an electromagnetic field from a flat spacetime by applying a coordinate transformation. Also, we can not locally make the electromagnetic field disappear by applying an appropriate coordinate transformation. \bigskip \noindent The metric being considered in this essay is particularly adapted to this situation. A flat spacetime is characterized by: \[ g_{AB}=\delta_{AB}\Rightarrow g_{\alpha\beta}=\delta_{\alpha\beta} \mbox{ and } g_{\alpha\bar{\beta}}=0 \mbox{ where } \delta_{\alpha\beta}=1 \mbox{ if } \alpha=\beta \mbox{ and } \delta_{\alpha\beta}=0 \mbox{ if } \alpha\not=\beta \] \bigskip \noindent From the equations (2), we see that when $g_{\alpha\beta}=\delta_{\alpha\beta}$ , we obtain $\acute{g}_{\alpha\beta}\not=\delta_{\alpha\beta}$ when we apply a coordinate transformation. Also, we see that when $g_{\alpha\bar{\beta}}=0$ , we obtain $\acute{g}_{\alpha\bar{\beta}}=0$ when we apply a coordinate transformation. It is therefore natural to consider $g_{\alpha\beta}$ as the gravitational field potentials and to consider $g_{\alpha\bar{\beta}}$ as the electromagnetic field potentials. \bigskip \noindent From the equations (2) and the previous paragraph, we also see that the only way to locally obtain a flat spacetime ($g_{AB}=\delta_{AB}$) by applying an appropriate coordinate transformation, is to have no electromagnetic field ($g_{\alpha\bar{\beta}}=0$). \section{The fields equations} \noindent As the Hamiltonien function, we shall consider $H=R\sqrt{\left|g\right|}d\Omega=0$ where $R=g^{KL}R_{KL}$ , \medskip \noindent $g=$ determinant of $g_{AB}=\left|g_{AB}\right|$ and $d\Omega= dz^{1}dz^{2}dz^{3}dz^{4}dz^{\bar{1}}dz^{\bar{2}}dz^{\bar{3}}dz^{\bar{4}}$ .\bigskip \noindent The calculation $\delta H=0$ (the variation is operated on the $g_{AB}$) leads to the following fields equations: \begin{equation} G_{AB}=R_{AB}-\textstyle{\frac{1}{2}}g_{AB}R=0 \mbox{ , with } D_{K}G^{AK}=0 \mbox{ (from the Bianchi identities)} \end{equation} \bigskip \noindent For $G_{AB}=0$ , we conclude that $R=0$ and by consequence, that $R_{AB}=0$ . \clearpage \section{The linearization of the fields equations: the classical method} \noindent We shall assume that the components of the metric tensor can be expanded into a series: \[ g_{AB}=\delta_{AB}+h_{AB}=\delta_{AB}+\lambda\mathop{h_{AB}}_{1}+\lambda^2\mathop{h_{AB}}_{2}+\cdots \] \noindent By making use of this series, the fields equations can be expanded into successive approximations. If we limit ourselves to the first approximation, we obtain: \medskip \noindent \begin{tabbing} $\displaystyle{\mathop{R_{AB}}_{1}}$\= $=\partial_{A}\displaystyle{\mathop{\Gamma_{BL}^{L}}_{1}}- \partial_{L}\displaystyle{\mathop{\Gamma_{AB}^{L}}_{1}} \mbox{ with } \displaystyle{\mathop{\Gamma_{BL}^{L}}_{1}}= \textstyle{\frac{1}{2}}\left(\partial_{B}\displaystyle{\mathop{h_{LL}}_{1}}\right) \mbox{ and } \displaystyle{\mathop{\Gamma_{AB}^{L}}_{1}}= \textstyle{\frac{1}{2}}\left(\partial_{A}\displaystyle{\mathop{h_{BL}}_{1}}+ \partial_{B}\displaystyle{\mathop{h_{AL}}_{1}}- \partial_{L}\displaystyle{\mathop{h_{AB}}_{1}}\right)$ \\ \>$\Rightarrow\displaystyle{\mathop{R_{AB}}_{1}}= \textstyle{\frac{1}{2}}\left(\partial_{L}\partial_{L}\displaystyle{\mathop{h_{AB}}_{1}}+ \partial_{A}\partial_{B}\displaystyle{\mathop{h_{LL}}_{1}}- \partial_{L}\partial_{A}\displaystyle{\mathop{h_{BL}}_{1}}- \partial_{L}\partial_{B}\displaystyle{\mathop{h_{AL}}_{1}}\right)$ \end{tabbing} \medskip \noindent The calculation is simplified by the introduction of the following quantities: \medskip \noindent \begin{tabbing} $\gamma_{AB}$\= $=h_{AB}-\textstyle{\frac{1}{2}}\delta_{AB}h\Leftrightarrow h_{AB}=\gamma_{AB}-\textstyle{\frac{1}{6}}\delta_{AB}\gamma\Rightarrow \displaystyle{\mathop{h_{AB}}_{1}}=\displaystyle{\mathop{\gamma_{AB}}_{1}}- \textstyle{\frac{1}{6}}\delta_{AB}\displaystyle{\mathop{\gamma}_{1}}\mbox{ (for }\delta_{KK}=8\mbox{)}$ \\ \>$\Rightarrow\displaystyle{\mathop{R_{AB}}_{1}}= \textstyle{\frac{1}{2}}\left(\partial_{L}\partial_{L}\displaystyle{\mathop{\gamma_{AB}}_{1}}- \textstyle{\frac{1}{6}}\delta_{AB}\partial_{L}\partial_{L}\displaystyle{\mathop{\gamma}_{1}}- \partial_{L}\partial_{A}\displaystyle{\mathop{\gamma_{BL}}_{1}}- \partial_{L}\partial_{B}\displaystyle{\mathop{\gamma_{AL}}_{1}}\right)$ \\ \>$\Rightarrow\displaystyle{\mathop{R}_{1}}= \textstyle{\frac{1}{2}}\left(-\textstyle{\frac{1}{3}}\partial_{L}\partial_{L}\displaystyle{\mathop{\gamma}_{1}}- 2\partial_{L}\partial_{T}\displaystyle{\mathop{\gamma_{LT}}_{1}}\right)$ \\ \>$\Rightarrow\displaystyle{\mathop{G_{AB}}_{1}}= \displaystyle{\mathop{R_{AB}}_{1}}- \textstyle{\frac{1}{2}}\delta_{AB}\displaystyle{\mathop{R}_{1}}= \textstyle{\frac{1}{2}}\left(\partial_{L}\partial_{L}\displaystyle{\mathop{\gamma_{AB}}_{1}}+ \delta_{AB}\partial_{L}\partial_{T}\displaystyle{\mathop{\gamma_{LT}}_{1}}- \partial_{L}\partial_{A}\displaystyle{\mathop{\gamma_{BL}}_{1}}- \partial_{L}\partial_{B}\displaystyle{\mathop{\gamma_{AL}}_{1}}\right)$ \end{tabbing} \medskip \noindent If we impose $\partial_{L}\gamma_{LA}=0$ for the choice of the coordinate system, we obtain the following linearization of the fields equations $\displaystyle{\mathop{G_{AB}}_{1}}=0$: \[ \partial_{L}\partial_{L}\mathop{\gamma_{AB}}_{1}=0 \] \medskip \noindent These equations can be developed into Greek indices: \medskip \noindent $\partial_{\epsilon}\partial_{\epsilon}\displaystyle{\mathop{\gamma_{\alpha\beta}}_{1}}+ \partial_{\bar{\epsilon}}\partial_{\bar{\epsilon}}\displaystyle{\mathop{\gamma_{\alpha\beta}}_{1}}=0$ \noindent $\partial_{\epsilon}\partial_{\epsilon}\displaystyle{\mathop{\gamma_{\alpha\bar{\beta}}}_{1}}+ \partial_{\bar{\epsilon}}\partial_{\bar{\epsilon}}\displaystyle{\mathop{\gamma_{\alpha\bar{\beta}}}_{1}}=0$ \medskip \noindent At this point, we could say that the first equation is formally equivalent to the classical Poisson equation $\partial_{l}\partial_{l}\phi=0$ of the gravitational field in the empty space. We could also say that the second equation is formally equivalent to the classical d'Alembert equation $\partial_{\epsilon}\partial_{\epsilon}A_{l}=0$ of the electromagnetic field in the empty space. However, it is not clear which part of $\gamma_{\alpha\beta}$ plays the role of the gravitational potential $\phi$ and which part of $\gamma_{\alpha\bar{\beta}}$ plays the role of the electromagnetic potentials $A_{a}$. For this, we need to establish the classical equations of motions for charged mass points, which will be the subject of the next section. \clearpage \section{The linearization of the fields equations: the ``Einstein-Infeld-Hoffmann'' (EIH) method} \noindent The so-called ``Einstein-Infeld-Hoffmann'' (EIH) method of linearization is more complicated than the classical one, but it leads to more physical results. For a detailed description of this method, we refer to [1, 2, 3]. \medskip \noindent The basic idea behind the EIH method is to take into account the fact that the time derivative of a quantity is small relatively to the quantity itself and to its spatial derivatives. If $\phi$ is such a quantity, it is assumed that $\partial_{a}\phi$ and $\partial_{\bar{a}}\phi$ are of the same order of magnitude than $\phi$, and that $\partial_{4}\phi$ and $\partial_{\bar{4}}\phi$ are of the order $c^{-1}\phi$ (c being the speed of light). \medskip \noindent As for the classical linearization method, the fields equations will be expanded by making use of the following quantities: \[ g_{AB}=\delta_{AB}+h_{AB} \mbox{ , } \gamma_{AB}=h_{AB}-\textstyle{\frac{1}{2}}\delta_{AB}h\Leftrightarrow h_{AB}=\gamma_{AB}-\textstyle{\frac{1}{6}}\delta_{AB}\gamma \] \noindent Following the EIH method, we shall assume that theses quantities can be expanded into the following series: \bigskip \noindent \begin{tabular}{lcl} $\gamma_{ab}=c^{-4}\displaystyle{\mathop{\gamma_{ab}}_{4}}+ c^{-6}\displaystyle{\mathop{\gamma_{ab}}_{6}}+\cdots$ & , & $\gamma_{a\bar{b}}=c^{-4}\displaystyle{\mathop{\gamma_{a\bar{b}}}_{4}}+ c^{-6}\displaystyle{\mathop{\gamma_{a\bar{b}}}_{6}}+\cdots$ \\ $\gamma_{a4}=c^{-3}\displaystyle{\mathop{\gamma_{a4}}_{3}}+ c^{-5}\displaystyle{\mathop{\gamma_{a4}}_{5}}+\cdots$ & , & $\gamma_{a\bar{4}}=c^{-3}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}+ c^{-5}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{5}}+\cdots$ \\ $\gamma_{44}=c^{-2}\displaystyle{\mathop{\gamma_{44}}_{2}}+ c^{-4}\displaystyle{\mathop{\gamma_{44}}_{4}}+\cdots$ & , & $\gamma_{4\bar{4}}=c^{-2}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ c^{-4}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{4}}+\cdots$ \\ $h_{ab}=c^{-2}\displaystyle{\mathop{h_{ab}}_{2}}+ c^{-4}\displaystyle{\mathop{h_{ab}}_{4}}+\cdots$ & , & $h_{a\bar{b}}=c^{-4}\displaystyle{\mathop{h_{a\bar{b}}}_{4}}+ c^{-6}\displaystyle{\mathop{h_{a\bar{b}}}_{6}}+\cdots$ \\ $h_{a4}=c^{-3}\displaystyle{\mathop{h_{a4}}_{3}}+ c^{-5}\displaystyle{\mathop{h_{a4}}_{5}}+\cdots$ & , & $h_{a\bar{4}}=c^{-3}\displaystyle{\mathop{h_{a\bar{4}}}_{3}}+ c^{-5}\displaystyle{\mathop{h_{a\bar{4}}}_{5}}+\cdots$ \\ $h_{44}=c^{-2}\displaystyle{\mathop{h_{44}}_{2}}+ c^{-4}\displaystyle{\mathop{h_{44}}_{4}}+\cdots$ & , & $h_{4\bar{4}}=c^{-2}\displaystyle{\mathop{h_{4\bar{4}}}_{2}}+ c^{-4}\displaystyle{\mathop{h_{4\bar{4}}}_{4}}+\cdots$ \end{tabular} \bigskip \noindent The different power expansion of the $\gamma$-series is justified by the fact that we shall impose \\ $\partial_{L}\gamma_{LA}=0\Rightarrow \partial_{l}\gamma_{la}+\partial_{4}\gamma_{4a}+ \partial_{\bar{l}}\gamma_{\bar{l}a}+\partial_{\bar{4}}\gamma_{\bar{4}a}=0 \mbox{ and } \partial_{l}\gamma_{l4}+\partial_{4}\gamma_{44}+ \partial_{\bar{l}}\gamma_{\bar{l}4}+\partial_{\bar{4}}\gamma_{\bar{4}4}=0$ for the choice of the coordinate system. \bigskip \noindent The power expansion of the $h$-series follows from the $\gamma$-series. The following components are needed: \medskip \noindent \begin{tabular}{lcl} $\displaystyle{\mathop{h_{ab}}_{2}}=-\textstyle{\frac{1}{3}}\delta_{ab}\displaystyle{\mathop{\gamma_{44}}_{2}} \Rightarrow \displaystyle{\mathop{h_{ll}}_{2}}=-\displaystyle{\mathop{\gamma_{44}}_{2}}$ & , & $\displaystyle{\mathop{h_{a\bar{b}}}_{2}}=0$ \\ $\displaystyle{\mathop{h_{ab}}_{4}}=\displaystyle{\mathop{\gamma_{ab}}_{4}}- \textstyle{\frac{1}{3}}\delta_{ab}\displaystyle{\mathop{\gamma_{ll}}_{4}}- \textstyle{\frac{1}{3}}\delta_{ab}\displaystyle{\mathop{\gamma_{44}}_{4}}\Rightarrow \displaystyle{\mathop{h_{ll}}_{4}}=-\displaystyle{\mathop{\gamma_{44}}_{4}}$ & , & $\displaystyle{\mathop{h_{a\bar{b}}}_{4}}=\displaystyle{\mathop{\gamma_{a\bar{b}}}_{4}}$ \\ $\displaystyle{\mathop{h_{a4}}_{3}}=\displaystyle{\mathop{\gamma_{a4}}_{3}}$ & , & $\displaystyle{\mathop{h_{a\bar{4}}}_{3}}=\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}$ \\ $\displaystyle{\mathop{h_{44}}_{2}}=\textstyle{\frac{2}{3}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ & , & $\displaystyle{\mathop{h_{4\bar{4}}}_{2}}=\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ $\displaystyle{\mathop{h_{44}}_{4}}=\textstyle{\frac{2}{3}}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{3}}\displaystyle{\mathop{\gamma_{ll}}_{4}}$ & , & $\displaystyle{\mathop{h_{4\bar{4}}}_{4}}=\displaystyle{\mathop{\gamma_{4\bar{4}}}_{4}}$ \end{tabular} \bigskip \noindent We shall further impose that $\displaystyle{\mathop{\gamma_{44}}_{2}}= \displaystyle{\mathop{\gamma_{\bar{4}\bar{4}}}_{2}}$ , $\displaystyle{\mathop{\gamma_{44}}_{4}}= \displaystyle{\mathop{\gamma_{\bar{4}\bar{4}}}_{4}}$ and $\displaystyle{\mathop{\gamma_{ll}}_{4}}= \displaystyle{\mathop{\gamma_{\bar{l}\bar{l}}}_{4}}$ . \clearpage \noindent For $g_{AL}g^{LB}=\delta_{A}^{B}$ , we have: \medskip \noindent \begin{tabular}{lcl} $\displaystyle{\mathop{h^{ab}}_{2}}=-\displaystyle{\mathop{h_{ab}}_{2}}= \textstyle{\frac{1}{3}}\delta_{ab}\displaystyle{\mathop{\gamma_{44}}_{2}}$ & , & $\displaystyle{\mathop{h^{a\bar{b}}}_{2}}=-\displaystyle{\mathop{h_{a\bar{b}}}_{2}}=0$ \\ $\displaystyle{\mathop{h^{a4}}_{3}}=-\displaystyle{\mathop{h_{a4}}_{3}}= -\displaystyle{\mathop{\gamma_{a4}}_{3}}$ & , & $\displaystyle{\mathop{h^{a\bar{4}}}_{3}}=-\displaystyle{\mathop{h_{a\bar{4}}}_{3}}= -\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}$ \\ $\displaystyle{\mathop{h^{44}}_{2}}=-\displaystyle{\mathop{h_{44}}_{2}}= -\textstyle{\frac{2}{3}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ & , & $\displaystyle{\mathop{h^{4\bar{4}}}_{2}}=-\displaystyle{\mathop{h_{4\bar{4}}}_{2}}= -\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \end{tabular} \bigskip \noindent By making use of these series, the fields equations can be expanded into successive approximations. As we shall see below, the first one leads to the mass and the charge conservation laws, and the second one leads to the equations of motion. \subsection{The first approximation: the mass and the charge conservation laws} \bigskip \noindent For the components of the Christoffel symbols, we obtain the following expressions: \medskip \noindent $\displaystyle{\mathop{\Gamma_{ab}^{c}}_{2}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{bc}}_{2}}+ \textstyle{\frac{1}{2}}\partial_{b}\displaystyle{\mathop{h_{ac}}_{2}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{h_{ab}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{ab}^{c}}_{2}}= -\textstyle{\frac{1}{6}}\delta_{bc}\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{6}}\delta_{ac}\partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{6}}\delta_{ab}\partial_{c}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{ab}^{4}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{b4}}_{3}}+ \textstyle{\frac{1}{2}}\partial_{b}\displaystyle{\mathop{h_{a4}}_{3}}- \textstyle{\frac{1}{2}}\partial_{4}\displaystyle{\mathop{h_{ab}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{ab}^{4}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{b4}}_{3}}+ \textstyle{\frac{1}{2}}\partial_{b}\displaystyle{\mathop{\gamma_{a4}}_{3}}+ \textstyle{\frac{1}{6}}\delta_{ab}\partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{a4}^{c}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{4c}}_{3}}+ \textstyle{\frac{1}{2}}\partial_{4}\displaystyle{\mathop{h_{ac}}_{2}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{h_{a4}}_{3}}\Rightarrow \displaystyle{\mathop{\Gamma_{a4}^{c}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{4c}}_{3}}- \textstyle{\frac{1}{6}}\delta_{ac}\partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{\gamma_{a4}}_{3}}$ \\ $\displaystyle{\mathop{\Gamma_{a4}^{4}}_{2}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{44}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{a4}^{4}}_{2}}= \textstyle{\frac{1}{3}}\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{44}^{c}}_{2}}= -\textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{h_{44}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{44}^{c}}_{2}}= -\textstyle{\frac{1}{3}}\partial_{c}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{44}^{4}}_{3}}= \textstyle{\frac{1}{2}}\partial_{4}\displaystyle{\mathop{h_{44}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{44}^{4}}_{3}}= \textstyle{\frac{1}{3}}\partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{a\bar{b}}^{c}}_{2}}= \textstyle{\frac{1}{2}}\partial_{\bar{b}}\displaystyle{\mathop{h_{ac}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{a\bar{b}}^{c}}_{2}}= -\textstyle{\frac{1}{6}}\delta_{ac}\partial_{\bar{b}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{a\bar{b}}^{4}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{\bar{b}4}}_{3}}+ \textstyle{\frac{1}{2}}\partial_{\bar{b}}\displaystyle{\mathop{h_{a4}}_{3}}\Rightarrow \displaystyle{\mathop{\Gamma_{a\bar{b}}^{4}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{\bar{b}4}}_{3}}+ \textstyle{\frac{1}{2}}\partial_{\bar{b}}\displaystyle{\mathop{\gamma_{a4}}_{3}}$ \\ $\displaystyle{\mathop{\Gamma_{a\bar{4}}^{c}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{\bar{4}c}}_{3}}+ \textstyle{\frac{1}{2}}\partial_{\bar{4}}\displaystyle{\mathop{h_{ac}}_{2}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{h_{a\bar{4}}}_{3}}\Rightarrow \displaystyle{\mathop{\Gamma_{a\bar{4}}^{c}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{\bar{4}c}}_{3}}- \textstyle{\frac{1}{6}}\delta_{ac}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}$ \\ $\displaystyle{\mathop{\Gamma_{a\bar{4}}^{4}}_{2}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{\bar{4}4}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{a\bar{4}}^{4}}_{2}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{\bar{4}4}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{4\bar{4}}^{c}}_{2}}= -\textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{h_{4\bar{4}}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{4\bar{4}}^{c}}_{2}}= -\textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{4\bar{4}}^{4}}_{3}}= \textstyle{\frac{1}{2}}\partial_{\bar{4}}\displaystyle{\mathop{h_{44}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{4\bar{4}}^{4}}_{3}}= \textstyle{\frac{1}{3}}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{4\bar{b}}^{c}}_{3}}= \textstyle{\frac{1}{2}}\partial_{\bar{b}}\displaystyle{\mathop{h_{4c}}_{3}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{h_{4\bar{b}}}_{3}}\Rightarrow \displaystyle{\mathop{\Gamma_{4\bar{b}}^{c}}_{3}}= \textstyle{\frac{1}{2}}\partial_{\bar{b}}\displaystyle{\mathop{\gamma_{4c}}_{3}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{\gamma_{4\bar{b}}}_{3}}$ \\ $\displaystyle{\mathop{\Gamma_{4\bar{b}}^{4}}_{2}}= \textstyle{\frac{1}{2}}\partial_{\bar{b}}\displaystyle{\mathop{h_{44}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{4\bar{b}}^{4}}_{2}}= \textstyle{\frac{1}{3}}\partial_{\bar{b}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{ab}^{\bar{c}}}_{2}}= -\textstyle{\frac{1}{2}}\partial_{\bar{c}}\displaystyle{\mathop{h_{ab}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{ab}^{\bar{c}}}_{2}}= \textstyle{\frac{1}{6}}\delta_{ab}\partial_{\bar{c}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{ab}^{\bar{4}}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{b\bar{4}}}_{3}}+ \textstyle{\frac{1}{2}}\partial_{b}\displaystyle{\mathop{h_{a\bar{4}}}_{3}}- \textstyle{\frac{1}{2}}\partial_{\bar{4}}\displaystyle{\mathop{h_{ab}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{ab}^{\bar{4}}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{b\bar{4}}}_{3}}+ \textstyle{\frac{1}{2}}\partial_{b}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}+ \textstyle{\frac{1}{6}}\delta_{ab}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{a4}^{\bar{c}}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{4\bar{c}}}_{3}}- \textstyle{\frac{1}{2}}\partial_{\bar{c}}\displaystyle{\mathop{h_{a4}}_{3}}\Rightarrow \displaystyle{\mathop{\Gamma_{a4}^{\bar{c}}}_{3}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{4\bar{c}}}_{3}}- \textstyle{\frac{1}{2}}\partial_{\bar{c}}\displaystyle{\mathop{\gamma_{a4}}_{3}}$ \\ $\displaystyle{\mathop{\Gamma_{a4}^{\bar{4}}}_{2}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{4\bar{4}}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{a4}^{\bar{4}}}_{2}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{44}^{\bar{c}}}_{2}}= -\textstyle{\frac{1}{2}}\partial_{\bar{c}}\displaystyle{\mathop{h_{44}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{44}^{\bar{c}}}_{2}}= -\textstyle{\frac{1}{3}}\partial_{\bar{c}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{44}^{\bar{4}}}_{3}}= \partial_{4}\displaystyle{\mathop{h_{4\bar{4}}}_{2}}- \textstyle{\frac{1}{2}}\partial_{\bar{4}}\displaystyle{\mathop{h_{44}}_{2}}\Rightarrow \displaystyle{\mathop{\Gamma_{44}^{\bar{4}}}_{3}}= \partial_{4}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}- \textstyle{\frac{1}{3}}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \clearpage \noindent For the components of the Ricci curvature tensor, we obtain the following expressions: \medskip \noindent \begin{tabbing} $\displaystyle{\mathop{R_{ab}}_{2}}= \partial_{a}\displaystyle{\mathop{\Gamma_{bl}^{l}}_{2}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{b4}^{4}}_{2}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{b\bar{l}}^{\bar{l}}}_{2}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{b\bar{4}}^{\bar{4}}}_{2}}- \partial_{l}\displaystyle{\mathop{\Gamma_{ab}^{l}}_{2}}- \partial_{\bar{l}}\displaystyle{\mathop{\Gamma_{ab}^{\bar{l}}}_{2}} \Rightarrow\displaystyle{\mathop{R_{ab}}_{2}}= -\textstyle{\frac{1}{6}}\delta_{ab}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}\right)$ \\ [.05in] $\displaystyle{\mathop{R_{a4}}_{3}}$\= $=\partial_{a}\displaystyle{\mathop{\Gamma_{4l}^{l}}_{3}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{44}^{4}}_{3}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{4\bar{l}}^{\bar{l}}}_{3}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{4\bar{4}}^{\bar{4}}}_{3}}- \partial_{l}\displaystyle{\mathop{\Gamma_{a4}^{l}}_{3}}- \partial_{4}\displaystyle{\mathop{\Gamma_{a4}^{4}}_{2}}- \partial_{\bar{l}}\displaystyle{\mathop{\Gamma_{a4}^{\bar{l}}}_{3}}- \partial_{\bar{4}}\displaystyle{\mathop{\Gamma_{a4}^{\bar{4}}}_{2}}$ \\ \>$\Rightarrow\displaystyle{\mathop{R_{a4}}_{3}}= -\textstyle{\frac{1}{2}}\partial_{a}\partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\partial_{a}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{1}{2}}\partial_{l}\left(\partial_{l}\displaystyle{\mathop{\gamma_{a4}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{l4}}_{3}}\right)+ \textstyle{\frac{1}{2}}\partial_{\bar{l}}\left(\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{a4}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{\bar{l}4}}_{3}}\right)$ \\ $\displaystyle{\mathop{R_{44}}_{2}}= -\partial_{l}\displaystyle{\mathop{\Gamma_{44}^{\bar{l}}}_{2}}- \partial_{\bar{l}}\displaystyle{\mathop{\Gamma_{44}^{\bar{l}}}_{2}} \Rightarrow\displaystyle{\mathop{R_{44}}_{2}}= \textstyle{\frac{1}{3}}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}\right)$ \\ [.05in] $\displaystyle{\mathop{R_{a\bar{b}}}_{2}}= \partial_{a}\displaystyle{\mathop{\Gamma_{\bar{b}l}^{l}}_{2}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{\bar{b}4}^{4}}_{2}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{\bar{b}\bar{l}}^{\bar{l}}}_{2}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{\bar{b}\bar{4}}^{\bar{4}}}_{2}}- \partial_{l}\displaystyle{\mathop{\Gamma_{a\bar{b}}^{l}}_{2}}- \partial_{\bar{l}}\displaystyle{\mathop{\Gamma_{a\bar{b}}^{\bar{l}}}_{2}} \Rightarrow\displaystyle{\mathop{R_{a\bar{b}}}_{2}}=0$ \\ [.05in] $\displaystyle{\mathop{R_{a\bar{4}}}_{3}}$\= $=\partial_{a}\displaystyle{\mathop{\Gamma_{\bar{4}l}^{l}}_{3}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{\bar{4}4}^{4}}_{3}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{\bar{4}\bar{l}}^{\bar{l}}}_{3}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{\bar{4}\bar{4}}^{\bar{4}}}_{3}}- \partial_{l}\displaystyle{\mathop{\Gamma_{a\bar{4}}^{l}}_{3}}- \partial_{4}\displaystyle{\mathop{\Gamma_{a\bar{4}}^{4}}_{2}}- \partial_{\bar{l}}\displaystyle{\mathop{\Gamma_{a\bar{4}}^{\bar{l}}}_{3}}- \partial_{\bar{4}}\displaystyle{\mathop{\Gamma_{a\bar{4}}^{\bar{4}}}_{2}}$ \\ \>$\Rightarrow\displaystyle{\mathop{R_{a\bar{4}}}_{3}}= -\textstyle{\frac{1}{2}}\partial_{a}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\partial_{a}\partial_{4}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{1}{2}}\partial_{l}\left(\partial_{l}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{l\bar{4}}}_{3}}\right)+ \textstyle{\frac{1}{2}}\partial_{\bar{l}}\left(\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{\bar{l}\bar{4}}}_{3}}\right)$ \\ $\displaystyle{\mathop{R_{4\bar{4}}}_{2}}= -\partial_{l}\displaystyle{\mathop{\Gamma_{4\bar{4}}^{l}}_{2}}- \partial_{\bar{l}}\displaystyle{\mathop{\Gamma_{4\bar{4}}^{\bar{l}}}_{2}} \Rightarrow\displaystyle{\mathop{R_{4\bar{4}}}_{2}}= \textstyle{\frac{1}{2}}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\right)$ \\ $\displaystyle{\mathop{R}_{2}}= \displaystyle{\mathop{R_{ll}}_{2}}+ \displaystyle{\mathop{R_{44}}_{2}}+ \displaystyle{\mathop{R_{\bar{l}\bar{l}}}_{2}}+ \displaystyle{\mathop{R_{\bar{4}\bar{4}}}_{2}} \Rightarrow\displaystyle{\mathop{R}_{2}}= -\textstyle{\frac{1}{3}}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}\right)$ \end{tabbing} \bigskip \noindent The fields equations of the first approximation become: \bigskip \noindent \begin{equation} \left. \begin{tabular}{l} $\displaystyle{\mathop{G_{ab}}_{2}}=0$ \\ $\displaystyle{\mathop{G_{a4}}_{3}}= -\textstyle{\frac{1}{2}}\partial_{a}\partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\partial_{a}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{1}{2}}\partial_{l}\left(\partial_{l}\displaystyle{\mathop{\gamma_{a4}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{l4}}_{3}}\right)+ \textstyle{\frac{1}{2}}\partial_{\bar{l}}\left(\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{a4}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{\bar{l}4}}_{3}}\right)=0$ \\ [.15in] $\displaystyle{\mathop{G_{44}}_{2}}= \textstyle{\frac{1}{2}}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}\right)=0$ \\ [.20in] $\displaystyle{\mathop{G_{a\bar{b}}}_{2}}=0$ \\ [.10in] $\displaystyle{\mathop{G_{a\bar{4}}}_{3}}= -\textstyle{\frac{1}{2}}\partial_{a}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\partial_{a}\partial_{4}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{1}{2}}\partial_{l}\left(\partial_{l}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{l\bar{4}}}_{3}}\right)+ \textstyle{\frac{1}{2}}\partial_{\bar{l}}\left(\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{\bar{l}\bar{4}}}_{3}}\right)=0$ \\ [.15in] $\displaystyle{\mathop{G_{4\bar{4}}}_{2}}= \textstyle{\frac{1}{2}}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\right)=0$ \end{tabular} \right| \end{equation} \bigskip \noindent We shall impose the following conditions for the choice of the coordinate system: \begin{equation} \partial_{L}\gamma_{LA}=0\Rightarrow \partial_{l}\gamma_{la}+\partial_{4}\gamma_{4a}+ \partial_{\bar{l}}\gamma_{\bar{l}a}+\partial_{\bar{4}}\gamma_{\bar{4}a}=0 \mbox{ and } \partial_{l}\gamma_{l4}+\partial_{4}\gamma_{44}+ \partial_{\bar{l}}\gamma_{\bar{l}4}+\partial_{\bar{4}}\gamma_{\bar{4}4}=0 \end{equation} \bigskip \noindent The fields equations of the first approximation become: \bigskip \noindent \begin{equation} \left. \begin{tabular}{lcl} $\displaystyle{\mathop{G_{ab}}_{2}}=0$ & , & $\displaystyle{\mathop{G_{a\bar{b}}}_{2}}=0$ \\ [.10in] $\displaystyle{\mathop{G_{a4}}_{3}}= \textstyle{\frac{1}{2}}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{a4}}_{3}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{a4}}_{3}}\right)=0$ & , & $\displaystyle{\mathop{G_{a\bar{4}}}_{3}}= \textstyle{\frac{1}{2}}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}\right)=0$ \\ [.10in] $\displaystyle{\mathop{G_{44}}_{2}}= \textstyle{\frac{1}{2}}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}\right)=0$ & , & $\displaystyle{\mathop{G_{4\bar{4}}}_{2}}= \textstyle{\frac{1}{2}}\left(\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\right)=0$ \\ [.10in] $\partial_{l}\displaystyle{\mathop{\gamma_{la}}_{4}}+ \partial_{4}\displaystyle{\mathop{\gamma_{4a}}_{3}}+ \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{l}a}}_{4}}+ \partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{4}a}}_{3}}=0$ & , & $\partial_{l}\displaystyle{\mathop{\gamma_{l4}}_{3}}+ \partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{l}4}}_{3}}+ \partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{4}4}}_{2}}=0$ \end{tabular} \right| \end{equation} \clearpage \noindent The mathematical world of classical physics is based on a 4-dimensional spacetime. In order to take this fact into account, we shall further restrict the equations (9) so that they hold separately in each of the two 4-dimensional spacetime being considered: \bigskip \noindent \begin{equation} \left. \begin{tabular}{lcl} $\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}} \left(z^{\alpha}\right)= \partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{\bar{4}\bar{4}}}_{2}} \left(z^{\alpha}\right)=0$ & , & $\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}} \left(z^{\bar{\alpha}}\right)= \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{4}\bar{4}}}_{2}} \left(z^{\bar{\alpha}}\right)=0$ \\ $\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{a4}}_{3}} \left(z^{\alpha}\right)= \partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{\bar{a}\bar{4}}}_{3}} \left(z^{\alpha}\right)=0$ & , & $\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{a4}}_{3}} \left(z^{\bar{\alpha}}\right)= \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{a}\bar{4}}}_{3}} \left(z^{\bar{\alpha}}\right)=0$ \\ $\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \left(z^{\alpha}\right)= \partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{\bar{4}4}}_{2}} \left(z^{\alpha}\right)=0$ & , & $\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \left(z^{\bar{\alpha}}\right)= \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{4}4}}_{2}} \left(z^{\bar{\alpha}}\right)=0$ \\ $\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}} \left(z^{\alpha}\right)= \partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{\bar{a}4}}_{3}} \left(z^{\alpha}\right)=0$ & , & $\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}} \left(z^{\bar{\alpha}}\right)=\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{a}4}}_{3}} \left(z^{\bar{\alpha}}\right)=0$ \\ $\partial_{l}\displaystyle{\mathop{\gamma_{la}}_{4}}\left(z^{\alpha}\right)+ \partial_{4}\displaystyle{\mathop{\gamma_{4a}}_{3}}\left(z^{\alpha}\right)=0$ & , & $\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{l}\bar{a}}}_{4}}\left(z^{\bar{\alpha}}\right)+ \partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{4}\bar{a}}}_{3}}\left(z^{\bar{\alpha}}\right)=0$ \\ $\partial_{l}\displaystyle{\mathop{\gamma_{l4}}_{3}}\left(z^{\alpha}\right)+ \partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}\left(z^{\alpha}\right)=0$ & , & $\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{l}\bar{4}}}_{3}}\left(z^{\bar{\alpha}}\right)+ \partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{4}\bar{4}}}_{2}}\left(z^{\bar{\alpha}}\right)=0$ \\ $\partial_{l}\displaystyle{\mathop{\gamma_{l\bar{a}}}_{4}}\left(z^{\alpha}\right)+ \partial_{4}\displaystyle{\mathop{\gamma_{4\bar{a}}}_{3}}\left(z^{\alpha}\right)=0$ & , & $\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{l}a}}_{4}}\left(z^{\bar{\alpha}}\right)+ \partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{4}a}}_{3}}\left(z^{\bar{\alpha}}\right)=0$ \\ $\partial_{l}\displaystyle{\mathop{\gamma_{l\bar{4}}}_{3}}\left(z^{\alpha}\right)+ \partial_{4}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\left(z^{\alpha}\right)=0$ & , & $\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{l}4}}_{3}}\left(z^{\bar{\alpha}}\right)+ \partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{4}4}}_{2}}\left(z^{\bar{\alpha}}\right)=0$ \end{tabular} \right| \end{equation} \bigskip \noindent It is clear that the solutions to the equations (10) are also solutions to the equations (9) if they have the following form: \begin{equation} \gamma_{AB}=\gamma_{AB}\left(z^{\alpha}, z^{\bar{\alpha}}\right)= \gamma_{AB}\left(z^{\alpha}\right)+\gamma_{AB}\left(z^{\bar{\alpha}}\right)\Rightarrow \gamma_{\epsilon\theta}=\gamma_{\bar{\epsilon}\bar{\theta}} \mbox{ , } \gamma_{\epsilon\bar{\theta}}=\gamma_{\bar{\epsilon}\theta} \mbox{ and } \partial_{\epsilon}\partial_{\bar{\theta}}\gamma_{AB}=0 \end{equation} \bigskip\noindent At this point, we can understand why the Lorentz transformation must be linear in classical physics. For the metric transforms according to (2), we conclude that the only coordinate transformations that preserve the form of (11) are the linear coordinate transformations. From the theory of relativity, we know that the Lorentz transformation is the only linear coordinate transformation which preserves the form of the Maxwell equations. The conclusion that the Lorentz transformation must be linear provides the explanation to the fact that certain coordinate systems (the {\it inertial systems}) are privileged in classical physics. \clearpage \noindent As the solutions of equations (10), we shall choose the following expressions: \bigskip \noindent \begin{equation} \left. \begin{tabular}{lcl} $\displaystyle{\mathop{\gamma_{44}}_{2}}= -\sum_{k=1}^{N}\stackrel{k}{\mu}\left(\frac{1}{\stackrel{k}{r}}+ \frac{1}{\stackrel{k}{\bar{r}}}\right)$ & , & $\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}= +\sum_{k=1}^{N}\stackrel{k}{\rho}\left(\frac{1}{\stackrel{k}{r}}+ \frac{1}{\stackrel{k}{\bar{r}}}\right)$ \\ [.20in] $\displaystyle{\mathop{\gamma_{a4}}_{3}}= -\sum_{k=1}^{N}\stackrel{k}{\mu}\left(\frac{\partial_{4}\stackrel{k}{y^{a}}}{\stackrel{k}{r}}+ \frac{\partial_{\bar{4}}\stackrel{k}{\bar{y}^{a}}}{\stackrel{k}{\bar{r}}}\right)$ & , & $\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}= +\sum_{k=1}^{N}\stackrel{k}{\rho}\left(\frac{\partial_{4}\stackrel{k}{y^{a}}}{\stackrel{k}{r}}+ \frac{\partial_{\bar{4}}\stackrel{k}{\bar{y}^{a}}}{\stackrel{k}{\bar{r}}}\right)$ \\ [.20in] $\displaystyle{\mathop{\gamma_{ab}}_{4}}= -\sum_{k=1}^{N}\stackrel{k}{\mu}\left(\frac{\partial_{4}\stackrel{k}{y^{a}} \partial_{4}\stackrel{k}{y^{b}}}{\stackrel{k}{r}}+ \frac{\partial_{\bar{4}}\stackrel{k}{\bar{y}^{a}} \partial_{\bar{4}}\stackrel{k}{\bar{y}^{b}}}{\stackrel{k}{\bar{r}}}\right)$ & , & $\displaystyle{\mathop{\gamma_{a\bar{b}}}_{4}}= +\sum_{k=1}^{N}\stackrel{k}{\rho}\left(\frac{\partial_{4}\stackrel{k}{y^{a}} \partial_{4}\stackrel{k}{y^{b}}}{\stackrel{k}{r}}+ \frac{\partial_{\bar{4}}\stackrel{k}{\bar{y}^{a}} \partial_{\bar{4}}\stackrel{k}{\bar{y}^{b}}}{\stackrel{k}{\bar{r}}}\right)$ \\ [.20in] \multicolumn{3}{l}{$\mbox{ with } \stackrel{k}{r}= \sqrt{\left(z^{l}-\stackrel{k}{y^{l}}\right)\left(z^{l}-\stackrel{k}{y^{l}}\right)} \Rightarrow \partial_{a}\stackrel{k}{r}= \displaystyle{\frac{\left(z^{a}-\stackrel{k}{y^{a}}\right)}{\stackrel{k}{r}}} \mbox{ and } \partial_{4}\stackrel{k}{r}= -\displaystyle{\frac{\partial_{4}\stackrel{k}{y^{l}}\left(z^{l}- \stackrel{k}{y^{l}}\right)}{\stackrel{k}{r}}}$} \\ [.20in] \multicolumn{3}{l}{$\mbox{ and }\stackrel{k}{\bar{r}}= \sqrt{\left(z^{\bar{l}}-\stackrel{k}{y^{\bar{l}}}\right)\left(z^{\bar{l}}-\stackrel{k}{y^{\bar{l}}}\right)} \Rightarrow \partial_{\bar{a}}\stackrel{k}{\bar{r}}= \displaystyle{\frac{\left(z^{\bar{a}}-\stackrel{k}{y^{\bar{a}}}\right)}{\stackrel{k}{\bar{r}}}} \mbox{ and }\partial_{\bar{4}}\stackrel{k}{\bar{r}}= -\displaystyle{\frac{\partial_{\bar{4}}\stackrel{k}{y^{\bar{l}}}\left(z^{\bar{l}}- \stackrel{k}{y^{\bar{l}}}\right)}{\stackrel{k}{\bar{r}}}}$} \end{tabular} \right| \end{equation} \bigskip \noindent These solutions represents $N$ charged mass points. The $3N$ functions $\stackrel{k}{y^{l}}$ of $z^{4}$ determine the locations of the $N$ charged mass points at every time $z^{4}$ . \bigskip \noindent We shall now prove that the masses $\stackrel{k}{\mu}$ and the charges $\stackrel{k}{\rho}$ are constant. If we take into account the results of (10) and (11) to rewrite the fields equations (7), we obtain: \medskip \noindent $\displaystyle{\mathop{G_{a4}}_{3}}= -\textstyle{\frac{1}{2}}\partial_{a}\partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{2}}\partial_{l}\left(\partial_{l}\displaystyle{\mathop{\gamma_{a4}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{l4}}_{3}}\right)=0$ \noindent $\displaystyle{\mathop{G_{a\bar{4}}}_{3}}= -\textstyle{\frac{1}{2}}\partial_{a}\partial_{4}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{1}{2}}\partial_{l}\left(\partial_{l}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{l\bar{4}}}_{3}}\right)=0$ \bigskip \noindent Each of these equations contain the divergence of a skewsymmetric expression which is equivalent to a curl. According to the Stockes's theorem, the integral of a curl taken over a closed surface vanishes. If we apply this result to a closed surface which encloses the $pth$ charged mass point, we obtain (with $\eta_{a}$ being the cosine of the angle between the $z^{a}$ coordinate and the normal of the surface element $dS$): \noindent \begin{tabbing} $\displaystyle{\oint_{S}}\displaystyle{\mathop{G_{a4}}_{3}}\eta_{a}dS=0$\= $\Rightarrow-\textstyle{\frac{1}{2}}\displaystyle{\oint_{S}}\partial_{a}\partial_{4} \displaystyle{\mathop{\gamma_{44}}_{2}}\eta_{a}dS+ \overbrace{\textstyle{\frac{1}{2}}\displaystyle{\oint_{S}}\partial_{l}\left(\partial_{l} \displaystyle{\mathop{\gamma_{a4}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{l4}}_{3}}\right)\eta_{a}dS}^{=0}=0\Rightarrow \partial_{4}\displaystyle{\oint_{S}}\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}}\eta_{a}dS=0$ \\ [.05in] \>$\Rightarrow\displaystyle{\frac{d\stackrel{p}{\mu}}{{dz}^{4}}}=0 \Rightarrow \stackrel{p}{\mu}$ is constant. \end{tabbing} \noindent \begin{tabbing} $\displaystyle{\oint_{S}}\displaystyle{\mathop{G_{a\bar{4}}}_{3}}\eta_{a}dS=0$\= $\Rightarrow-\textstyle{\frac{1}{2}}\displaystyle{\oint_{S}}\partial_{a}\partial_{4} \displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\eta_{a}dS+ \overbrace{\textstyle{\frac{1}{2}}\displaystyle{\oint_{S}}\partial_{l}\left(\partial_{l} \displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}- \partial_{a}\displaystyle{\mathop{\gamma_{l\bar{4}}}_{3}}\right)\eta_{a}dS}^{=0}=0\Rightarrow \partial_{4}\displaystyle{\oint_{S}}\partial_{a}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\eta_{a}dS=0$ \\ [.05in] \>$\Rightarrow\displaystyle{\frac{d\stackrel{p}{\rho}}{{dz}^{4}}}=0 \Rightarrow \stackrel{p}{\rho}$ is constant. \end{tabbing} \clearpage \subsection{The second approximation: the equations of motion} \bigskip \noindent For the components of the Christoffel symbols, the following expressions are needed: \noindent \begin{tabbing} $\displaystyle{\mathop{\Gamma_{ab}^{c}}_{4}}$\= $=\textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{bc}}_{4}}+ \textstyle{\frac{1}{2}}\partial_{b}\displaystyle{\mathop{h_{ac}}_{4}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{h_{ab}}_{4}}- \textstyle{\frac{1}{2}}\displaystyle{\mathop{h_{cl}}_{2}}\partial_{a}\displaystyle{\mathop{h_{bl}}_{2}}- \textstyle{\frac{1}{2}}\displaystyle{\mathop{h_{cl}}_{2}}\partial_{b}\displaystyle{\mathop{h_{al}}_{2}}+ \textstyle{\frac{1}{2}}\displaystyle{\mathop{h_{cl}}_{2}}\partial_{l} \displaystyle{\mathop{h_{ab}}_{2}}$ \\ \>$\Rightarrow \displaystyle{\mathop{\Gamma_{ab}^{c}}_{4}}=$\= $\textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{bc}}_{4}}- \textstyle{\frac{1}{6}}\delta_{bc}\partial_{a}\displaystyle{\mathop{\gamma_{ll}}_{4}}- \textstyle{\frac{1}{6}}\delta_{bc}\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{4}}+ \textstyle{\frac{1}{2}}\partial_{b}\displaystyle{\mathop{\gamma_{ac}}_{4}}- \textstyle{\frac{1}{6}}\delta_{ac}\partial_{b}\displaystyle{\mathop{\gamma_{ll}}_{4}}- \textstyle{\frac{1}{6}}\delta_{ac}\partial_{b}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{\gamma_{ab}}_{4}}$ \\ \>\>$-\textstyle{\frac{1}{6}}\delta_{ab}\partial_{c}\displaystyle{\mathop{\gamma_{ll}}_{4}}- \textstyle{\frac{1}{6}}\delta_{ab}\partial_{c}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{18}}\delta_{bc}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{18}}\delta_{ac}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{18}}\delta_{ab}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{c}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{a4}^{4}}_{4}}$\= $=\textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{44}}_{4}}- \textstyle{\frac{1}{2}}\displaystyle{\mathop{h_{44}}_{2}}\partial_{a}\displaystyle{\mathop{h_{44}}_{2}}- \textstyle{\frac{1}{2}}\displaystyle{\mathop{h_{4\bar{4}}}_{2}}\partial_{a} \displaystyle{\mathop{h_{4\bar{4}}}_{2}}$ \\ \>$\Rightarrow \displaystyle{\mathop{\Gamma_{a4}^{4}}_{4}}= \textstyle{\frac{1}{3}}\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{6}}\partial_{a}\displaystyle{\mathop{\gamma_{ll}}_{4}}- \textstyle{\frac{2}{9}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{a}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{a\bar{b}}^{c}}_{4}}$\= $=\textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{\bar{b}c}}_{4}}+ \textstyle{\frac{1}{2}}\partial_{\bar{b}}\displaystyle{\mathop{h_{ac}}_{4}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{h_{a\bar{b}}}_{4}}- \textstyle{\frac{1}{2}}\displaystyle{\mathop{h_{cl}}_{2}}\partial_{\bar{b}} \displaystyle{\mathop{h_{al}}_{2}}$ \\ \>$\Rightarrow \displaystyle{\mathop{\Gamma_{a\bar{b}}^{c}}_{4}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{\bar{b}c}}_{4}}+ \textstyle{\frac{1}{2}}\partial_{\bar{b}}\displaystyle{\mathop{\gamma_{ac}}_{4}}- \textstyle{\frac{1}{6}}\delta_{ac}\partial_{\bar{b}}\displaystyle{\mathop{\gamma_{ll}}_{4}}- \textstyle{\frac{1}{6}}\delta_{ac}\partial_{\bar{b}}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{2}}\partial_{c}\displaystyle{\mathop{\gamma_{a\bar{b}}}_{4}}- \textstyle{\frac{1}{18}}\delta_{ac}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{b}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{4\bar{b}}^{4}}_{4}}$\= $=\textstyle{\frac{1}{2}}\partial_{\bar{b}}\displaystyle{\mathop{h_{44}}_{4}}- \textstyle{\frac{1}{2}}\displaystyle{\mathop{h_{44}}_{2}}\partial_{\bar{b}} \displaystyle{\mathop{h_{44}}_{2}}- \textstyle{\frac{1}{2}}\displaystyle{\mathop{h_{4\bar{4}}}_{2}}\partial_{\bar{b}} \displaystyle{\mathop{h_{4\bar{4}}}_{2}}$ \\ \>$\Rightarrow \displaystyle{\mathop{\Gamma_{4\bar{b}}^{4}}_{4}}= \textstyle{\frac{1}{3}}\partial_{\bar{b}}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{6}}\partial_{\bar{b}}\displaystyle{\mathop{\gamma_{ll}}_{4}}- \textstyle{\frac{2}{9}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{b}}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{\bar{b}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ $\displaystyle{\mathop{\Gamma_{ab}^{\bar{c}}}_{4}}$\= $=\textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{h_{b\bar{c}}}_{4}}+ \textstyle{\frac{1}{2}}\partial_{b}\displaystyle{\mathop{h_{a\bar{c}}}_{4}}- \textstyle{\frac{1}{2}}\partial_{\bar{c}}\displaystyle{\mathop{h_{ab}}_{4}}+ \textstyle{\frac{1}{2}}\displaystyle{\mathop{h_{cl}}_{2}}\partial_{l} \displaystyle{\mathop{h_{ab}}_{2}}$ \\ \>$\Rightarrow \displaystyle{\mathop{\Gamma_{ab}^{\bar{c}}}_{4}}= \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{b\bar{c}}}_{4}}+ \textstyle{\frac{1}{2}}\partial_{b}\displaystyle{\mathop{\gamma_{a\bar{c}}}_{4}}- \textstyle{\frac{1}{2}}\partial_{\bar{c}}\displaystyle{\mathop{\gamma_{ab}}_{4}}+ \textstyle{\frac{1}{6}}\delta_{ab}\partial_{\bar{c}}\displaystyle{\mathop{\gamma_{ll}}_{4}}+ \textstyle{\frac{1}{6}}\delta_{ab}\partial_{\bar{c}}\displaystyle{\mathop{\gamma_{44}}_{4}}+ \textstyle{\frac{1}{18}}\delta_{ab}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{c}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \end{tabbing} \medskip\noindent For the components of the Ricci curvature tensor, the following expressions are needed: \medskip \noindent \begin{tabbing} $\displaystyle{\mathop{R_{ab}}_{4}}$\= $=$\= $\partial_{a}\displaystyle{\mathop{\Gamma_{bl}^{l}}_{4}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{b4}^{4}}_{4}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{b\bar{l}}^{\bar{l}}}_{4}}+ \partial_{a}\displaystyle{\mathop{\Gamma_{b\bar{4}}^{\bar{4}}}_{4}}- \partial_{l}\displaystyle{\mathop{\Gamma_{ab}^{l}}_{4}}- \partial_{4}\displaystyle{\mathop{\Gamma_{ab}^{4}}_{3}}- \partial_{\bar{l}}\displaystyle{\mathop{\Gamma_{ab}^{\bar{l}}}_{4}}- \partial_{\bar{4}}\displaystyle{\mathop{\Gamma_{ab}^{\bar{4}}}_{3}}+ \displaystyle{\mathop{\Gamma_{at}^{l}}_{2}}\displaystyle{\mathop{\Gamma_{lb}^{t}}_{2}}+ \displaystyle{\mathop{\Gamma_{a\bar{t}}^{l}}_{2}}\displaystyle{\mathop{\Gamma_{lb}^{\bar{t}}}_{2}}$ \\ \>\>$+\displaystyle{\mathop{\Gamma_{a4}^{4}}_{2}}\displaystyle{\mathop{\Gamma_{4b}^{4}}_{2}}+ \displaystyle{\mathop{\Gamma_{a\bar{4}}^{4}}_{2}}\displaystyle{\mathop{\Gamma_{4b}^{\bar{4}}}_{2}}+ \displaystyle{\mathop{\Gamma_{at}^{\bar{l}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{l}b}^{t}}_{2}}+ \displaystyle{\mathop{\Gamma_{a\bar{t}}^{\bar{l}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{l}b}^{\bar{t}}}_{2}}+ \displaystyle{\mathop{\Gamma_{a4}^{\bar{4}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{4}b}^{4}}_{2}}+ \displaystyle{\mathop{\Gamma_{a\bar{4}}^{\bar{4}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{4}b}^{\bar{4}}}_{2}}- \displaystyle{\mathop{\Gamma_{ab}^{t}}_{2}}\displaystyle{\mathop{\Gamma_{tl}^{l}}_{2}}- \displaystyle{\mathop{\Gamma_{ab}^{\bar{t}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{t}l}^{l}}_{2}}- \displaystyle{\mathop{\Gamma_{ab}^{t}}_{2}}\displaystyle{\mathop{\Gamma_{t4}^{4}}_{2}}- \displaystyle{\mathop{\Gamma_{ab}^{\bar{t}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{t}4}^{4}}_{2}}$ \\ \>\>$-\displaystyle{\mathop{\Gamma_{ab}^{t}}_{2}}\displaystyle{\mathop{\Gamma_{t\bar{l}}^{\bar{l}}}_{2}}- \displaystyle{\mathop{\Gamma_{ab}^{\bar{t}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{t}\bar{l}}^{\bar{l}}}_{2}}- \displaystyle{\mathop{\Gamma_{ab}^{t}}_{2}}\displaystyle{\mathop{\Gamma_{t\bar{4}}^{\bar{4}}}_{2}}- \displaystyle{\mathop{\Gamma_{ab}^{\bar{t}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{t}\bar{4}}^{\bar{4}}}_{2}}$ \\ \>$\Rightarrow\displaystyle{\mathop{R_{ab}}_{4}}=$\= $-\textstyle{\frac{1}{2}}\partial_{a}\partial_{l}\displaystyle{\mathop{\gamma_{bl}}_{4}}- \textstyle{\frac{1}{2}}\partial_{b}\partial_{l}\displaystyle{\mathop{\gamma_{al}}_{4}}+ \textstyle{\frac{1}{2}}\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{ab}}_{4}}+ \textstyle{\frac{1}{2}}\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{ab}}_{4}}$ \\ \>\>$-\textstyle{\frac{1}{6}}\delta{ab}\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{tt}}_{4}}- \textstyle{\frac{1}{6}}\delta{ab}\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{tt}}_{4}}- \textstyle{\frac{1}{6}}\delta{ab}\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{6}}\delta{ab}\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{4}}$ \\ \>\>$-\textstyle{\frac{1}{2}}\partial_{a}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{b\bar{l}}}_{4}}- \textstyle{\frac{1}{2}}\partial_{b}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{a\bar{l}}}_{4}}$ \\ \>\>$-\textstyle{\frac{1}{6}}\delta_{ab}\partial_{4}\partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{6}}\delta_{ab}\partial_{\bar{4}}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{3}}\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{2}{3}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}- \displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ \>\>$-\textstyle{\frac{1}{18}}\delta_{ab}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{18}}\delta_{ab}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{18}}\delta_{ab}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{18}}\delta_{ab}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ \>\>$-\textstyle{\frac{1}{2}}\partial_{a}\partial_{4}\displaystyle{\mathop{\gamma_{b4}}_{3}}- \textstyle{\frac{1}{2}}\partial_{b}\partial_{4}\displaystyle{\mathop{\gamma_{a4}}_{3}}$ \\ \>\>$-\textstyle{\frac{1}{2}}\partial_{a}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{b\bar{4}}}_{3}}- \textstyle{\frac{1}{2}}\partial_{b}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}$ \end{tabbing} \noindent \begin{tabbing} $\displaystyle{\mathop{R_{44}}_{4}}$\= $=$\= $\partial_{4}\displaystyle{\mathop{\Gamma_{4l}^{l}}_{3}}+ \partial_{4}\displaystyle{\mathop{\Gamma_{4\bar{l}}^{\bar{l}}}_{3}}+ \partial_{4}\displaystyle{\mathop{\Gamma_{4\bar{4}}^{\bar{4}}}_{3}}- \partial_{l}\displaystyle{\mathop{\Gamma_{44}^{l}}_{4}}- \partial_{\bar{l}}\displaystyle{\mathop{\Gamma_{44}^{\bar{l}}}_{4}}- \partial_{\bar{4}}\displaystyle{\mathop{\Gamma_{44}^{\bar{4}}}_{3}}+ \displaystyle{\mathop{\Gamma_{4\bar{4}}^{l}}_{2}}\displaystyle{\mathop{\Gamma_{l4}^{\bar{4}}}_{2}}+ \displaystyle{\mathop{\Gamma_{4t}^{4}}_{2}}\displaystyle{\mathop{\Gamma_{44}^{t}}_{2}}+ \displaystyle{\mathop{\Gamma_{4\bar{t}}^{4}}_{2}}\displaystyle{\mathop{\Gamma_{44}^{\bar{t}}}_{2}}+ \displaystyle{\mathop{\Gamma_{4\bar{4}}^{\bar{l}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{l}4}^{\bar{4}}}_{2}}$ \\ \>\>$+\displaystyle{\mathop{\Gamma_{4t}^{\bar{4}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{4}4}^{t}}_{2}}+ \displaystyle{\mathop{\Gamma_{4\bar{t}}^{\bar{4}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{4}4}^{\bar{t}}}_{2}}- \displaystyle{\mathop{\Gamma_{44}^{t}}_{2}}\displaystyle{\mathop{\Gamma_{tl}^{l}}_{2}}- \displaystyle{\mathop{\Gamma_{44}^{\bar{t}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{t}l}^{l}}_{2}}- \displaystyle{\mathop{\Gamma_{44}^{t}}_{2}}\displaystyle{\mathop{\Gamma_{t\bar{l}}^{\bar{l}}}_{2}}- \displaystyle{\mathop{\Gamma_{44}^{\bar{t}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{t}\bar{l}}^{\bar{l}}}_{2}}- \displaystyle{\mathop{\Gamma_{44}^{t}}_{2}}\displaystyle{\mathop{\Gamma_{t\bar{4}}^{\bar{4}}}_{2}}- \displaystyle{\mathop{\Gamma_{44}^{\bar{t}}}_{2}}\displaystyle{\mathop{\Gamma_{\bar{t}\bar{4}}^{\bar{4}}}_{2}}$ \\ \>$\Rightarrow\displaystyle{\mathop{R_{44}}_{2}}=$\= $\textstyle{\frac{1}{3}}\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{4}}+ \textstyle{\frac{1}{3}}\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{6}}\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{tt}}_{4}}- \textstyle{\frac{1}{6}}\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{tt}}_{4}}$ \\ \>\>$-\textstyle{\frac{2}{3}}\partial_{4}\partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{3}}\partial_{\bar{4}}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{44}}_{2}}- \partial_{4}\partial_{l}\displaystyle{\mathop{\gamma_{4l}}_{3}}- \partial_{4}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{l}}}_{3}}- \partial_{4}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ \>\>$-\textstyle{\frac{2}{9}}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{2}{9}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{1}{2}}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}- \textstyle{\frac{1}{2}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \end{tabbing} \clearpage \noindent \begin{tabbing} $\displaystyle{\mathop{R}_{4}}$\= $=$\= $\delta_{lt}\displaystyle{\mathop{R_{lt}}_{4}}+ \displaystyle{\mathop{R_{44}}_{4}}+ \delta_{\bar{l}\bar{t}}\displaystyle{\mathop{R_{\bar{l}\bar{t}}}_{4}}+ \displaystyle{\mathop{R_{\bar{4}\bar{4}}}_{4}}$ \\ \>$\Rightarrow\displaystyle{\mathop{R}_{4}}=$\= $-\partial_{l}\partial_{t}\displaystyle{\mathop{\gamma_{lt}}_{4}}- \partial_{\bar{l}}\partial_{\bar{t}}\displaystyle{\mathop{\gamma_{\bar{l}\bar{t}}}_{4}}- \textstyle{2}\partial_{l}\partial_{\bar{t}}\displaystyle{\mathop{\gamma_{l\bar{t}}}_{4}}- \textstyle{\frac{1}{3}}\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{3}}\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{4}}- \textstyle{\frac{1}{3}}\partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{tt}}_{4}}- \textstyle{\frac{1}{3}}\partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{tt}}_{4}}$ \\ \>\>$-\textstyle{\frac{4}{3}}\partial_{4}\partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{4}{3}}\partial_{\bar{4}}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{10}{9}}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{10}{9}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ \>\>$-\textstyle{\frac{3}{2}}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}- \textstyle{\frac{3}{2}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}- \displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{l}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}- \displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{\bar{l}}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ \>\>$-\textstyle{2}\partial_{l}\partial_{4}\displaystyle{\mathop{\gamma_{l4}}_{3}}- \textstyle{2}\partial_{\bar{l}}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{l}\bar{4}}}_{3}}- \textstyle{2}\partial_{l}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{l\bar{4}}}_{3}}- \textstyle{2}\partial_{\bar{l}}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{l}\bar{4}}}_{3}}- \textstyle{2}\partial_{4}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \end{tabbing} \noindent For the components of the fields equations, the following expressions are needed: \medskip \noindent \begin{tabbing} $\displaystyle{\mathop{G_{ab}}_{4}}$\= $=$\= $\displaystyle{\mathop{R_{ab}}_{4}}-\textstyle{\frac{1}{2}}\delta_{ab}\displaystyle{\mathop{R}_{4}}$ \\ \>$\Rightarrow\displaystyle{\mathop{G_{ab}}_{4}}=$\= $\textstyle{\frac{1}{2}}\partial_{l}\left(\partial_{l}\displaystyle{\mathop{\gamma_{ab}}_{4}}- \partial_{b}\displaystyle{\mathop{\gamma_{al}}_{4}}\right)+ \textstyle{\frac{1}{2}}\partial_{l}\left(\delta_{ab}\partial_{t}\displaystyle{\mathop{\gamma_{lt}}_{4}}- \delta_{al}\partial_{t}\displaystyle{\mathop{\gamma_{bt}}_{4}}\right)+ \textstyle{\frac{1}{2}}\partial_{\bar{l}}\left(\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{ab}}_{4}}- \partial_{b}\displaystyle{\mathop{\gamma_{a\bar{l}}}_{4}}\right)+ \textstyle{\frac{1}{2}}\partial_{l}\left(\delta_{ab}\partial_{\bar{t}}\displaystyle{\mathop{\gamma_{l\bar{t}}}_{4}}- \delta_{al}\partial_{\bar{t}}\displaystyle{\mathop{\gamma_{b\bar{t}}}_{4}}\right)$ \\ \>\>$+\textstyle{\frac{1}{2}}\partial_{l}\left(\delta_{ab}\partial_{4}\displaystyle{\mathop{\gamma_{l4}}_{3}}- \delta_{al}\partial_{4}\displaystyle{\mathop{\gamma_{b4}}_{3}}\right)+ \textstyle{\frac{1}{2}}\partial_{l}\left(\delta_{ab}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{l\bar{4}}}_{3}}- \delta_{al}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{b\bar{4}}}_{3}}\right)+ \textstyle{\frac{1}{2}}\delta_{ab}\partial_{\bar{l}}\left(\partial_{t}\displaystyle{\mathop{\gamma_{t\bar{l}}}_{4}}+ \partial_{4}\displaystyle{\mathop{\gamma_{4\bar{l}}}_{3}}+ \partial_{\bar{t}}\displaystyle{\mathop{\gamma_{\bar{t}\bar{l}}}_{4}}+ \partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{4}\bar{l}}}_{3}}\right)$ \\ [.10in] \>\>$-\textstyle{\frac{1}{2}}\partial_{b}\partial_{4}\displaystyle{\mathop{\gamma_{a4}}_{3}}- \textstyle{\frac{1}{3}}\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{2}{3}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{2}}\delta_{ab}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{2}}\delta_{ab}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ [.10in] \>\>$-\textstyle{\frac{1}{2}}\partial_{b}\partial_{\bar{4}}\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}- \textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}- \displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{3}{4}}\delta_{ab}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{3}{4}}\delta_{ab}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \\ \>\>$+\textstyle{\frac{1}{2}}\delta_{ab}\partial_{4}\left(\partial_{l}\displaystyle{\mathop{\gamma_{l4}}_{3}}+ \partial_{4}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{l}4}}_{3}}+ \partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{4}4}}_{2}}\right)+ \textstyle{\frac{1}{2}}\delta_{ab}\partial_{\bar{4}}\left(\partial_{l}\displaystyle{\mathop{\gamma_{l\bar{4}}}_{3}}+ \partial_{4}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{\bar{l}\bar{4}}}_{3}}+ \partial_{\bar{4}}\displaystyle{\mathop{\gamma_{\bar{4}\bar{4}}}_{2}}\right)$ \end{tabbing} \noindent If we impose the conditions (8) and (11), we obtain: \medskip \noindent \begin{tabbing} $\displaystyle{\mathop{G_{ab}}_{4}}=$\= $\textstyle{\frac{1}{2}}\partial_{l}\left(\partial_{l}\displaystyle{\mathop{\gamma_{ab}}_{4}}- \partial_{b}\displaystyle{\mathop{\gamma_{al}}_{4}}\right)+ \textstyle{\frac{1}{2}}\partial_{l}\left(\delta_{ab}\partial_{t}\displaystyle{\mathop{\gamma_{lt}}_{4}}- \delta_{al}\partial_{t}\displaystyle{\mathop{\gamma_{bt}}_{4}}\right)+ \textstyle{\frac{1}{2}}\partial_{\bar{l}}\left(\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{ab}}_{4}}- \partial_{b}\displaystyle{\mathop{\gamma_{a\bar{l}}}_{4}}\right)+ \textstyle{\frac{1}{2}}\partial_{l}\left(\delta_{ab}\partial_{\bar{t}}\displaystyle{\mathop{\gamma_{l\bar{t}}}_{4}}- \delta_{al}\partial_{\bar{t}}\displaystyle{\mathop{\gamma_{b\bar{t}}}_{4}}\right)$ \\ \>$+\textstyle{\frac{1}{2}}\partial_{l}\left(\delta_{ab}\partial_{4}\displaystyle{\mathop{\gamma_{l4}}_{3}}- \delta_{al}\partial_{4}\displaystyle{\mathop{\gamma_{b4}}_{3}}\right)$ \\ [.10in] \>$-\textstyle{\frac{1}{2}}\partial_{b}\partial_{4}\displaystyle{\mathop{\gamma_{a4}}_{3}}$\= $-\textstyle{\frac{1}{3}}\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{2}{3}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{2}}\delta_{ab}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{2}}\delta_{ab}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}$ \\ [.10in] \>\>$-\textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}- \displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{3}{4}}\delta_{ab}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{3}{4}}\delta_{ab}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ \end{tabbing} \medskip \noindent From now on, the argument proceeds as in the case of the mass and charge conservation laws. If we form the integrals of $\displaystyle{\mathop{G_{ab}}_{4}}\eta_{b}$ over a closed surface S, then these integrals must vanish. We have: \begin{equation} \left. \begin{tabular}{l} $\displaystyle{\oint_{S}}\displaystyle{\mathop{G_{ab}}_{4}}\eta_{b}dS=0\Rightarrow$ \\ $\textstyle{\frac{1}{2}}\overbrace{\displaystyle{\oint_{S}} \partial_{l}\left(\partial_{l}\displaystyle{\mathop{\gamma_{ab}}_{4}}- \partial_{b}\displaystyle{\mathop{\gamma_{al}}_{4}}\right)\eta_{b}dS}^{=0}+ \textstyle{\frac{1}{2}}\overbrace{\displaystyle{\oint_{S}}\partial_{l}\left(\delta_{ab}\partial_{t} \displaystyle{\mathop{\gamma_{lt}}_{4}}- \delta_{al}\partial_{t}\displaystyle{\mathop{\gamma_{bt}}_{4}}\right)\eta_{b}dS}^{=0}+ \textstyle{\frac{1}{2}}\overbrace{\displaystyle{\oint_{S}}\partial_{\bar{l}}\left(\partial_{\bar{l}} \displaystyle{\mathop{\gamma_{ab}}_{4}}- \partial_{b}\displaystyle{\mathop{\gamma_{a\bar{l}}}_{4}}\right)\eta_{b}dS}^{=0}$ \\ $+\textstyle{\frac{1}{2}}\overbrace{\partial_{l}\left(\delta_{ab}\partial_{\bar{t}} \displaystyle{\mathop{\gamma_{l\bar{t}}}_{4}}- \delta_{al}\partial_{\bar{t}}\displaystyle{\mathop{\gamma_{b\bar{t}}}_{4}}\right)\eta_{b}dS}^{=0}+ \textstyle{\frac{1}{2}}\overbrace{\displaystyle{\oint_{S}}\partial_{l}\left(\delta_{ab}\partial_{4} \displaystyle{\mathop{\gamma_{l4}}_{3}}- \delta_{al}\partial_{4}\displaystyle{\mathop{\gamma_{b4}}_{3}}\right)\eta_{b}dS}^{=0}$ \\ [.10in] $+\displaystyle{\oint_{S}}\left(-\textstyle{\frac{1}{2}}\partial_{b}\partial_{4} \displaystyle{\mathop{\gamma_{a4}}_{3}}\right. \left.-\textstyle{\frac{1}{3}}\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}- \textstyle{\frac{2}{3}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{2}}\delta_{ab}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}+ \textstyle{\frac{1}{2}}\delta_{ab}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}\right.$ \\ [.10in] $\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ } \mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ } \left.-\textstyle{\frac{1}{2}}\partial_{a}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}- \displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{3}{4}}\delta_{ab}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}+ \textstyle{\frac{3}{4}}\delta_{ab}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\right)\eta_{b}dS=0$ \end{tabular} \right| \end{equation} \clearpage \noindent We shall choose for $S$ a small spherical surface with the radius $R$, the center of which is the $pth$ charged mass point. We have: \medskip \begin{tabular}{lcl} $\eta_{a}=\displaystyle{\frac{1}{R}}\left(z^{a}-\stackrel{p}{y^{a}}\right)$ & , & $\eta_{l}\eta_{l}=1$ \\ $\partial_{a}R=\eta_{a}$ & , & $\partial_{4}R=-\partial_{4}\stackrel{p}{y^{l}}\eta_{l}$ \\ $\partial_{b}\eta_{a}=\displaystyle{\frac{1}{R}}\left(\delta_{ab}-\eta_{a}\eta_{b}\right)$ & , & $\partial_{4}\eta_{a}=\displaystyle{\frac{1}{R}}\left(\eta_{a}\eta_{l}- \delta_{al}\right) \partial_{4}\stackrel{p}{y^{l}}$ \end{tabular} \bigskip \noindent We may rewrite the solutions (12): \medskip $\displaystyle{\mathop{\gamma_{44}}_{2}}= -\stackrel{p}{\mu}\left(\frac{1}{R}+\frac{1}{\bar{R}}\right)- \sum_{k=1}^{N, k\not{=}p}\stackrel{k}{\mu}\left(\frac{1}{\stackrel{k}{r}}+ \frac{1}{\stackrel{k}{\bar{r}}}\right)$ $\displaystyle{\mathop{\gamma_{a4}}_{3}}= -\stackrel{p}{\mu}\left(\frac{\partial_{4}\stackrel{p}{y^{a}}}{R}+ \frac{\partial_{\bar{4}}\stackrel{p}{\bar{y}^{a}}}{\bar{R}}\right)- \sum_{k=1}^{N, k\not{=}p}\stackrel{k}{\mu}\left(\frac{\partial_{4}\stackrel{k}{y^{a}}}{\stackrel{k}{r}}+ \frac{\partial_{\bar{4}}\stackrel{k}{\bar{y}^{a}}}{\stackrel{k}{\bar{r}}}\right)$ $\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}= +\stackrel{p}{\rho}\left(\frac{1}{R}+\frac{1}{\bar{R}}\right)+ \sum_{k=1}^{N, k\not{=}p}\stackrel{k}{\rho}\left(\frac{1}{\stackrel{k}{r}}+ \frac{1}{\stackrel{k}{\bar{r}}}\right)$ $\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}= +\stackrel{p}{\rho}\left(\frac{\partial_{4}\stackrel{p}{y^{a}}}{R}+ \frac{\partial_{\bar{4}}\stackrel{p}{\bar{y}^{a}}}{\bar{R}}\right)+ \sum_{k=1}^{N, k\not{=}p}\stackrel{k}{\rho}\left(\frac{\partial_{4}\stackrel{k}{y^{a}}}{\stackrel{k}{r}}+ \frac{\partial_{\bar{4}}\stackrel{k}{\bar{y}^{a}}}{\stackrel{k}{\bar{r}}}\right)$ \bigskip \noindent We have: \medskip \begin{tabbing} \indent $\stackrel{k}{r}$\= $=\sqrt{\left(z^{l}-\stackrel{k}{y^{l}}\right)\left(z^{l}-\stackrel{k}{y^{l}}\right)}= \stackrel{k}{r}=\sqrt{\left(\stackrel{p,k}{y^{l}}+R\eta_{l}\right)\left(\stackrel{p,k}{y^{l}}+R\eta_{l}\right)} \mbox { with } \stackrel{p,k}{y^{l}}=\stackrel{p}{y^{l}}-\stackrel{k}{y^{l}}$ \\ [.10in] \>$\Rightarrow\stackrel{k}{r}=\stackrel{p,k}{r}\sqrt{\left(1+\displaystyle{\frac{2\stackrel{p,k}{y^{l}}\eta_{l}R} {(\stackrel{p,k}{r})^{2}}}+\displaystyle{\frac{R^{2}}{(\stackrel{p,k}{r})^{2}}}\right)}\mbox{ with } \stackrel{p,k}{r}=\sqrt{\left(\stackrel{p,k}{y^{l}}\right)\left(\stackrel{p,k}{y^{l}}\right)}\Rightarrow \stackrel{k}{r}=\stackrel{p,k}{r}\left(1+\displaystyle{\frac{\stackrel{p,k}{y^{l}}\eta_{l}} {(\stackrel{p,k}{r})^{2}}}R+\cdots\right)$ \end{tabbing} \bigskip \noindent We therefore obtain: \medskip $\displaystyle{\mathop{\gamma_{44}}_{2}}= -\stackrel{p}{\mu}\left(\frac{1}{R}+\frac{1}{\bar{R}}\right)- \sum_{k=1}^{N, k\not{=}p}\stackrel{k}{\mu}\left[\frac{1}{\stackrel{p,k}{r}}\left(1- \displaystyle{\frac{\stackrel{p,k}{y^{l}}\eta_{l}}{(\stackrel{p,k}{r})^{2}}}R+\cdots\right)+ \frac{1}{\stackrel{p,k}{\bar{r}}}\left(1- \displaystyle{\frac{\stackrel{p,k}{\bar{y}^{l}}\bar{\eta}_{l}}{(\stackrel{p,k}{\bar{r}})^{2}}}\bar{R}+ \cdots\right)\right]$ $\displaystyle{\mathop{\gamma_{a4}}_{3}}= -\stackrel{p}{\mu}\left(\frac{\partial_{4}\stackrel{p}{y^{a}}}{R}+ \frac{\partial_{\bar{4}}\stackrel{p}{\bar{y}^{a}}}{\bar{R}}\right)- \sum_{k=1}^{N, k\not{=}p}\stackrel{k}{\mu}\left[\frac{\partial_{4}\stackrel{k}{y^{a}}}{\stackrel{p,k}{r}}\left(1- \displaystyle{\frac{\stackrel{p,k}{y^{l}}\eta_{l}}{(\stackrel{p,k}{r})^{2}}}R+\cdots\right)+ \frac{\partial_{\bar{4}}\stackrel{k}{\bar{y}^{a}}}{\stackrel{p,k}{\bar{r}}}\left(1- \displaystyle{\frac{\stackrel{p,k}{\bar{y}^{l}}\bar{\eta}_{l}}{(\stackrel{p,k}{\bar{r}})^{2}}}\bar{R}+ \cdots\right)\right]$ $\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}= +\stackrel{p}{\rho}\left(\frac{1}{R}+\frac{1}{\bar{R}}\right)+ \sum_{k=1}^{N, k\not{=}p}\stackrel{k}{\rho}\left[\frac{1}{\stackrel{p,k}{r}}\left(1- \displaystyle{\frac{\stackrel{p,k}{y^{l}}\eta_{l}}{(\stackrel{p,k}{r})^{2}}}R+\cdots\right)+ \frac{1}{\stackrel{p,k}{\bar{r}}}\left(1- \displaystyle{\frac{\stackrel{p,k}{\bar{y}^{l}}\bar{\eta}_{l}}{(\stackrel{p,k}{\bar{r}})^{2}}}\bar{R}+ \cdots\right)\right]$ $\displaystyle{\mathop{\gamma_{a\bar{4}}}_{3}}= +\stackrel{p}{\rho}\left(\frac{\partial_{4}\stackrel{p}{y^{a}}}{R}+ \frac{\partial_{\bar{4}}\stackrel{p}{\bar{y}^{a}}}{\bar{R}}\right)+ \sum_{k=1}^{N, k\not{=}p}\stackrel{k}{\rho}\left[\frac{\partial_{4}\stackrel{k}{y^{a}}}{\stackrel{p,k}{r}}\left(1- \displaystyle{\frac{\stackrel{p,k}{y^{l}}\eta_{l}}{(\stackrel{p,k}{r})^{2}}}R+\cdots\right)+ \frac{\partial_{\bar{4}}\stackrel{k}{\bar{y}^{a}}}{\stackrel{p,k}{\bar{r}}}\left(1- \displaystyle{\frac{\stackrel{p,k}{\bar{y}^{l}}\bar{\eta}_{l}}{(\stackrel{p,k}{\bar{r}})^{2}}}\bar{R}+ \cdots\right)\right]$ \clearpage \noindent For the values of the integrals (13) do not depend on the size of S, they must be independent of $R$. If the integrands are expanded into power series with respect to $R$, then all those terms which contain $R$ in any power other than $R^{-2}$ cannot contribute to the integrals. We shall therefore reduce the necessary computations by expanding all terms in the integrands into power series and by neglecting all expressions which are not multiplied by $R^{-2}$ . We have: \medskip $\partial_{b}\partial_{4}\displaystyle{\mathop{\gamma_{a4}}_{3}}= +\displaystyle{\frac{\stackrel{p}{\mu}}{R^{2}}}\eta_{b}\partial_{4}\partial_{4}\stackrel{p}{y^{a}}$ + other terms which are not multiplied by $R^{-2}$ . \medskip $\partial_{a}\displaystyle{\mathop{\gamma_{44}}_{2}}\partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}= +\sum_{k=1}^{N, k\not{=}p}\displaystyle{\frac{\stackrel{p}{\mu}\stackrel{k}{\mu}}{(\stackrel{p,k}{r})^{3}}} \displaystyle{\frac{1}{R^{2}}}\left(\eta_{a}\stackrel{p,k}{y^{b}}+\eta_{b}\stackrel{p,k}{y^{a}}\right)$ + other terms which are not multiplied by $R^{-2}$ . \medskip $\displaystyle{\mathop{\gamma_{44}}_{2}}\partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{44}}_{2}}= +\sum_{k=1}^{N, k\not{=}p}\displaystyle{\frac{\stackrel{p}{\mu}\stackrel{k}{\mu}}{(\stackrel{p,k}{r})^{3}}} \displaystyle{\frac{1}{R^{2}}}\eta_{l}\stackrel{p,k}{y^{l}}\left(\delta_{ab}-3\eta_{a}\eta_{b}\right)$ + other terms which are not multiplied by $R^{-2}$ . \medskip $\delta_{ab}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}\partial_{l}\displaystyle{\mathop{\gamma_{44}}_{2}}= +2\delta_{ab}\sum_{k=1}^{N, k\not{=}p}\displaystyle{\frac{\stackrel{p}{\mu}\stackrel{k}{\mu}}{(\stackrel{p,k}{r})^{3}}} \displaystyle{\frac{1}{R^{2}}}\eta_{l}\stackrel{p,k}{y^{l}}$ + other terms which are not multiplied by $R^{-2}$ . \medskip $\delta_{ab}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{44}}_{2}}=0$ + other terms which are not multiplied by $R^{-2}$ . \medskip $\partial_{a}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}= -\sum_{k=1}^{N, k\not{=}p}\displaystyle{\frac{\stackrel{p}{\rho}\stackrel{k}{\rho}}{(\stackrel{p,k}{r})^{3}}} \displaystyle{\frac{1}{R^{2}}}\left(\eta_{a}\stackrel{p,k}{y^{b}}+\eta_{b}\stackrel{p,k}{y^{a}}\right)$ + other terms which are not multiplied by $R^{-2}$ . \medskip $\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\partial_{a}\partial_{b}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}= -\sum_{k=1}^{N, k\not{=}p}\displaystyle{\frac{\stackrel{p}{\rho}\stackrel{k}{\rho}}{(\stackrel{p,k}{r})^{3}}} \displaystyle{\frac{1}{R^{2}}}\eta_{l}\stackrel{p,k}{y^{l}}\left(\delta_{ab}-3\eta_{a}\eta_{b}\right)$ + other terms which are not multiplied by $R^{-2}$ . \medskip $\delta_{ab}\partial_{l}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}\partial_{l} \displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}= -2\delta_{ab}\sum_{k=1}^{N, k\not{=}p}\displaystyle{\frac{\stackrel{p}{\rho}\stackrel{k}{\rho}}{(\stackrel{p,k}{r})^{3}}} \displaystyle{\frac{1}{R^{2}}}\eta_{l}\stackrel{p,k}{y^{l}}$ + other terms which are not multiplied by $R^{-2}$ . \medskip $\delta_{ab}\partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}} \partial_{\bar{l}}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}=0$ + other terms which are not multiplied by $R^{-2}$ . \bigskip \noindent We therefore obtain: \begin{tabbing} $\displaystyle{\mathop{G_{ab}}_{4}}=$ \= $-\textstyle{\frac{1}{2}}\displaystyle{\frac{\stackrel{p}{\mu}}{R^{2}}}\eta_{b} \partial_{4}\partial_{4}\stackrel{p}{y^{a}}$ \\ \>$-\textstyle{\frac{1}{3}}\displaystyle{\sum_{k=1}^{N, k\not{=}p}} \displaystyle{\frac{\stackrel{p}{\mu}\stackrel{k}{\mu}}{(\stackrel{p,k}{r})^{3}}}\displaystyle{\frac{1}{R^{2}}} \left(\eta_{a}\stackrel{p,k}{y^{b}}+\eta_{b}\stackrel{p,k}{y^{a}}\right)- \textstyle{\frac{2}{3}}\displaystyle{\sum_{k=1}^{N, k\not{=}p}} \displaystyle{\frac{\stackrel{p}{\mu}\stackrel{k}{\mu}}{(\stackrel{p,k}{r})^{3}}}\displaystyle{\frac{1}{R^{2}}} \eta_{l}\stackrel{p,k}{y^{l}}\left(\delta_{ab}-3\eta_{a}\eta_{b}\right)+ \delta_{ab}\sum_{k=1}^{N, k\not{=}p}\displaystyle{\frac{\stackrel{p}{\mu}\stackrel{k}{\mu}}{(\stackrel{p,k}{r})^{3}}} \displaystyle{\frac{1}{R^{2}}}\eta_{l}\stackrel{p,k}{y^{l}}$ \\ \>$+\textstyle{\frac{1}{2}}\displaystyle{\sum_{k=1}^{N, k\not{=}p}} \displaystyle{\frac{\stackrel{p}{\rho}\stackrel{k}{\rho}}{(\stackrel{p,k}{r})^{3}}}\displaystyle{\frac{1}{R^{2}}} \left(\eta_{a}\stackrel{p,k}{y^{b}}+\eta_{b}\stackrel{p,k}{y^{a}}\right)+ \displaystyle{\sum_{k=1}^{N, k\not{=}p}} \displaystyle{\frac{\stackrel{p}{\rho}\stackrel{k}{\rho}}{(\stackrel{p,k}{r})^{3}}}\displaystyle{\frac{1}{R^{2}}} \eta_{l}\stackrel{p,k}{y^{l}}\left(\delta_{ab}-3\eta_{a}\eta_{b}\right)- \textstyle{\frac{3}{2}}\delta_{ab}\displaystyle{\sum_{k=1}^{N, k\not{=}p}}\displaystyle{\frac{\stackrel{p} {\mu}\stackrel{k}{\mu}}{(\stackrel{p,k}{r})^{3}}}\displaystyle{\frac{1}{R^{2}}}\eta_{l}\stackrel{p,k}{y^{l}}$ \\ [.05in] \> + other terms which do not contribute to the integrals (13) \end{tabbing} \begin{tabbing} $\Rightarrow\displaystyle{\mathop{G_{ab}}_{4}}\eta_{b}=$ \= $-\textstyle{\frac{1}{2}}\displaystyle{\frac{\stackrel{p}{\mu}}{R^{2}}} \partial_{4}\partial_{4}\stackrel{p}{y^{a}}+ \sum_{k=1}^{N, k\not{=}p}\displaystyle{\frac{\stackrel{p}{\mu}\stackrel{k}{\mu}} {(\stackrel{p,k}{r})^{3}}}\displaystyle{\frac{1}{R^{2}}} \left(-\textstyle{\frac{1}{3}}\stackrel{p,k}{y^{a}}+2\eta_{a}\eta_{l}\stackrel{p,k}{y^{l}}\right)+ \sum_{k=1}^{N, k\not{=}p}\displaystyle{\frac{\stackrel{p}{\rho}\stackrel{k}{\rho}} {(\stackrel{p,k}{r})^{3}}}\displaystyle{\frac{1}{R^{2}}} \left(+\textstyle{\frac{1}{2}}\stackrel{p,k}{y^{a}}-3\eta_{a}\eta_{l}\stackrel{p,k}{y^{l}}\right)$ \\ [.05in] \> + other terms which do not contribute to the integrals (13) \end{tabbing} \clearpage \noindent We have: \medskip \begin{tabular}{lcl} $\displaystyle{\oint_{S}}\eta_{l}\stackrel{p,k}{y^{l}}\eta_{a}dS= \textstyle{\frac{4}{3}}\pi{R^{2}}\stackrel{p,k}{y^{a}}$ & and & $\displaystyle{\oint_{S}}dS=4\pi{R^{2}}$ \end{tabular} \bigskip \noindent We therefore obtain: \begin{tabbing} $\displaystyle{\oint_{S}}\displaystyle{\mathop{G_{ab}}_{4}}\eta_{b}dS$\= $=-2\pi\stackrel{p}{\mu}\partial_{4}\partial_{4}\stackrel{p}{y^{a}}+ \textstyle{\frac{4}{3}}\pi\displaystyle{\sum_{k=1}^{N, k\not{=}p}} \displaystyle{\frac{\stackrel{p}{\mu}\stackrel{k}{\mu}}{(\stackrel{p,k}{r})^{3}}}\stackrel{p,k}{y^{a}}- 2\pi\displaystyle{\sum_{k=1}^{N, k\not{=}p}} \displaystyle{\frac{\stackrel{p}{\rho}\stackrel{k}{\rho}}{(\stackrel{p,k}{r})^{3}}}\stackrel{p,k}{y^{a}}=0$ \\ [.10in] \>$\Rightarrow -\stackrel{p}{\mu}\partial_{4}\partial_{4}\stackrel{p}{y^{a}}+ \textstyle{\frac{2}{3}}\displaystyle{\sum_{k=1}^{N, k\not{=}p}} \displaystyle{\frac{\stackrel{p}{\mu}\stackrel{k}{\mu}}{(\stackrel{p,k}{r})^{3}}}\stackrel{p,k}{y^{a}}- \displaystyle{\sum_{k=1}^{N, k\not{=}p}} \displaystyle{\frac{\stackrel{p}{\rho}\stackrel{k}{\rho}}{(\stackrel{p,k}{r})^{3}}}\stackrel{p,k}{y^{a}}=0$ \\ [.10in] \end{tabbing} \noindent If we pose: \begin{equation} \stackrel{k}{\mu}=\textstyle{\frac{3}{2}}G\stackrel{k}{m}\mbox{ and } \stackrel{k}{\rho}=\sqrt{\textstyle{\frac{3}{2}}GK}\stackrel{k}{q} \end{equation} \bigskip \noindent We obtain, for ${dz}^{4}=icdt$ , the classical equations of motion for charged mass points: \begin{equation} \stackrel{p}{m}\stackrel{p}{{\ddot{y}}^{a}}+ \displaystyle{\sum_{k=1}^{N, k\not{=}p}}G \displaystyle{\frac{\stackrel{p}{m}\stackrel{k}{m}}{(\stackrel{p,k}{r})^{3}}}\stackrel{p,k}{y^{a}}- \displaystyle{\sum_{k=1}^{N, k\not{=}p}}K \displaystyle{\frac{\stackrel{p}{q}\stackrel{k}{q}}{(\stackrel{p,k}{r})^{3}}}\stackrel{p,k}{y^{a}}=0 \end{equation} \bigskip \noindent We recognize in (15) the Newton inertial force, the Newton gravitational force and the \\ [.05in] Coulomb electrostatic force. We conclude from (12) and (15) that $\displaystyle{\mathop{\gamma_{44}}_{2}}$ represents the gravitational potential and that $\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}$ represents the electrostatic potential. Therefore, we conclude from the equation $\partial_{l}\displaystyle{\mathop{\gamma_{l\bar{4}}}_{3}}+ \partial_{4}\displaystyle{\mathop{\gamma_{4\bar{4}}}_{2}}=0$ of (10) that $\displaystyle{\mathop{\gamma_{\alpha\bar{4}}}_{2}}$ represent the electromagnetic potentials $A_{a}$. Finally, we conclude from (14) that the electromagnetic field could not exist without the gravitational field ($\stackrel{k}{\rho}=0$ if $G=0$). \section{conclusion} \noindent We have shown that the linearization of the equations (6) leads to the ``Newton-Maxwell-Lorentz'' results of classical physics. That is, the gravitational field and the electromagnetic field of charged mass points and their equations of motion. We have also seen that the formal conditions (11) must be imposed to the solutions of the linearized equations (10) to achieve these results. \bigskip \noindent The formal conditions (11) cannot be directly imposed to the solutions of the equations (6). This mean that in the most general case, a physical interpretation must be given to the double spacetime being considered. We believe that this can only be achieved by finding exact solutions (generalized Schwarzschild solutions) to the equations (6). We did not try to achieve this result in this essay. \clearpage \section{References} \noindent [1] P. G. Bergmann, ``Introduction to the theory of relativity'', Dover Publications INC, New York, 1976. \medskip \noindent [2] A. Einstein, L. Infeld and B. Hoffmann, ``The gravitational equations and the problem of motion'', Annals of Mathematics, vol. 39, p. 65-100, 1938. \medskip \noindent [3] A. Einstein and L. Infeld, ``The gravitational equations and the problem of motion. 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