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\title{Motion of a Charged Particle in a Non-symmetrical Gravitational Theory}\author{ S. M. Saydur Rahman\\	 \\         Department of Physics\\          University of Cincinnati \\         Cincinnati, Ohio 45221\\[1em]} \date{\today}

\maketitle

\begin{abstract}
A new theory of gravity in which the geometry of space-time has a non-symmetric field structure is described here. The combined electromagnetic and gravitational field in this theory maintain the validity of the weak equivalance principle. The field equations of the non-symmetric gravitational theory are derived by considering the presence  of electromagnetic magnetic field by adopting a non-minimal coupling, which ensures the validity of equivalance principle. The static, spherically symmetric solution of a charged point particle is also obtained.
\end{abstract}
\pagebreak
\section{Introduction}\label{S:intro}
In 1979 Moffat [1] studied the nonsymmetrical gravitational theory (NGT).
Soon after establishing the NGT, Moffat studied the field produced by an electric charge [2] adopting a minimal coupling of the electromagnetic field to gravitation. In 1989 Will [3] pointed that such a coupling violates the weak equivalance principle (WEP) because it predicts that the test body gravitational acceleration depends upon its internal structure. Then by writing a more general coupling, Mann et al. [4] were to able to maintain the validity of the WEP. In this paper we have studied the field of a charge particle in a non-symmetric gravitational theory, by using the Lagrangian given by Mann et al[4] and Ragusa [5]. In sec-2 we first study the consequences of this more general coupling on the electromagnatic field equations and corresponding energy-momentum tensor. In sec-3 we study the field produced by an electric charge in the new scheme. 

\section{The electromagnatic action}\label{S:The}
The electromagnetic action given by Mann et al. is,
\begin{equation}\label{E:cong1}
I_{em} = -\frac{1}{16}\int d^4x\sqrt{-g}f g^{\mu\alpha}
g^{\nu\beta}[ZF_{\mu\nu} F_{\alpha\beta}+(1-Z)F_{\alpha\nu} F_{\mu\beta} + YF_{\mu\alpha} F_{\nu\beta}]
\end{equation}
where, $F_{\mu\nu} = \partial_\mu A_\nu-\partial_\nu A_\mu$, is the electromagnetic field tensor (with vector potential $A_\mu$). 

In the action (1), $Z$ and $Y$ are constants while $f$ is a scalar function which must be unity in the limit when the antisymmetric part of $g_{\mu\nu}$ vanishes, $g_{[\mu\nu]}\to 0$, implying that $f$ is a function of $\frac{\sqrt{-g}}{\sqrt{-\gamma}}$, where $g = detg_{\mu\nu}$ and $\gamma$ is the determinant of symmetric part, $\gamma = det \gamma_{(\mu\nu)}$. According to Mann et. al.[4] we have, 
\begin{equation}\label{E:cong3}
f = \frac{\sqrt{-g}}{\sqrt{-\gamma}}
\end{equation}
Following Ragusa[5] we take the choice (3) and get the resulting action in the form,
\begin{equation}\label{E:cong4}
I_{em} = \int d^4x\sqrt{-g} L_{em}
\end{equation}
where,

\begin{eqnarray}\label{E:cong6}
L_{em} = -\frac{1}{16}f F_{\mu\nu}F_{\alpha\beta} [g^{(\mu\alpha)}g^{(\nu\beta)} + (2Z-1) g^{[\mu\alpha]}g^{[\nu\beta]}\\ \nonumber
+ (1-Z) (g^{(\mu\nu)}g^{(\alpha\beta)} + g^{[\mu\nu]}g^{[\alpha\beta]})]
\end{eqnarray}
where $f$ is given by Eq. (2). We have then one parameter left, Z, but we will find later that the field of a spherically symmetric charge turns out to be independent of $Z$. Now varying the Lagrangian with respect to $g^{\mu\nu}$ we obtain,

\begin{equation}\label{E:cong7}
\delta(\sqrt{(-g)} L_{em}) = \frac{\sqrt{-g}}{2}\delta g^{\mu\alpha}E_{\mu\alpha}
\end{equation}\\
where,

\begin{align}
E_{\mu\alpha} & = & \frac{1}{4\pi}[
\frac{1}{4}(fg_{\mu\alpha}-2\frac{\partial f} {\partial g^{\mu\alpha}})g^{\lambda\sigma}g^{\nu\beta}
{ZF_{\lambda\nu}F_{\sigma\beta} + (1-Z) (F_{\sigma\nu}F_{\lambda\beta} + F_{\lambda\sigma}F_{\nu\beta})}\\
\nonumber
- fg^{\nu\beta} (ZF_{\mu\nu}F_{\alpha\beta} +(1-Z)(F_{\mu\beta}F_{\alpha\nu} + F_{\mu\alpha}F_{\nu\beta}))]
\end{align}\\
\\
is the energy-momentum tensor of the electromagnetic field. We can easily show that, $E_{\mu\alpha}$ is a traceless tensor. 
We now consider the variation with respect to $A_{\mu}$ of the fields plus interaction Lagrangian, $L_{em} = -\sqrt{-g}J^{\mu}A_{\mu}$. The total Lagrangian is,
\begin{equation}
L = \sqrt{-g}L_{em} - \sqrt{-g}J^{\mu}A_{\mu} 
\end{equation}
Therefore, we get the Euler-Lagrange equation of motion,
\begin{displaymath}
\frac{\partial{L}}{\partial{A_{\mu}}} - \partial_{\nu}[\frac{\partial L}{\partial (\partial_{\mu}A_{\nu})}] = 0
\end{displaymath}
which gives,
\begin{eqnarray}\label{E:cong}
\partial_{\nu}[\sqrt{-g}f[g^{(\mu\alpha)}g^{(\nu\beta)} + (2Z-1) g^{[\mu\alpha]}g^{[\nu\beta]} + (1-Z)\\ 
\nonumber (g^{(\mu\nu)}g^{(\nu\beta)} + g^{[\mu\nu]}g^{[\alpha\beta]})]]  = -4\pi J^{\mu}
\end{eqnarray}\\
This is the inhomogenous Maxwell equation in the presence of the non-symmetric field. 

\section{The Field of a Charged Particle}\label{S:field}
We can write the gravitational and electromagnetic Lagrangian as,
\begin{equation}
L = \sqrt{-g} [g^{\mu\nu} R(W) + L_{em}]
\end{equation}\\
By using the variational principle we get the gravitational vacuum field equation of the NGT are [2],
\begin{equation}
g^{\mu+\nu-} = 0
\end{equation}
\begin{equation}
(\sqrt{-g} g^{[\mu\nu]})_{,\nu} = 0
\end{equation}
\begin{equation}
R_{\mu\nu}(\Gamma) = 8\pi E_{(\mu\nu)}
\end{equation} 
\begin{equation}
R_{[\mu\nu,\sigma]}(\Gamma) = 8\pi E_{[\mu\nu,\sigma]}
\end{equation}\\
where,
\begin{equation}
R_{\mu\nu}(\Gamma) = \Gamma^{\beta}_{\mu\nu,\beta} - \frac{1}{2}(\Gamma^{\beta}_{(\mu\beta),\nu} + \Gamma^{\beta}_{(\nu\beta),\mu}) - \Gamma^{\beta}_{\alpha\nu} \Gamma^{\alpha}_{\mu\beta} + \Gamma^{\alpha}_{\mu\nu}\Gamma^{\beta}_{(\alpha\beta)}
\end{equation}
For a static spherically symmetric field, $g_{\mu\nu}$ in spherical polar coordinates has the form:

$\qquad\qquad\qquad\qquad\quad g_{00} = \gamma(r)\qquad g_{11} = -\alpha(r)\\$

$\qquad\qquad\qquad\qquad\quad g_{22} = -r^{2}\qquad g_{33} = -r^{2}\sin^{2}\theta\\$
\begin{equation}
g_{01} = -w(r) = -g_{10}
\end{equation}\\
and all other components are equal to zero.
Then from Eq.(10) we get,
\begin{equation}
w^{2} = \alpha\gamma\frac{l^{4}}{l^{4}+r^{4}}
\end{equation}
where, $l^{2}$ is a constant, the conserved fermionic charge.
As, $g = (w^{2}-\alpha\gamma)r^{4}\sin^{2}\theta$ and $\gamma = det g_{\mu\nu} -\alpha\gamma r^{4}\sin^{2}\theta$ we get from Eq.(2),
\begin{equation}
f = (1-\frac{w^{2}}{\alpha\gamma})^\frac{1}{2} = \sqrt{1-\frac{g^{01}g^{10}}{g^{00}g^{11}}}
\end{equation}
The electric field is $E(r) = F_{01}$. Then, outside the source, Eq.(7) yields, for $\mu = 0$,
\begin{equation}
\partial_{r}(\frac{r^{2}E}{\sqrt{\alpha\gamma}}) = 0
\end{equation}
independent of Z. By integrating the above equation we get,
\begin{equation}
E = \frac{Q}{r^{2}}\sqrt{\alpha\gamma}
\end{equation}
where $Q$ is a constant of integration whic can be identified with the charge of the particle. Now from the energy-momentum tensor  and Eq.(18) we obtain the following non-zero components of $E_{\mu\nu}$:
\begin{equation}
4\pi E_{00} = \frac{1}{2}\frac{Q^{2}}{r^{4}}\frac{\gamma}{\sqrt{1-\frac{w^{2}}{\alpha\gamma}}} (1+ \frac{w^{2}}{\alpha\gamma})
\end{equation}
\begin{equation}
E_{11} = -\frac{\alpha}{\gamma}E_{00}
\end{equation}
\begin{equation}
4\pi E_{22} = \frac{Q^{2}}{2r^{4}}\frac{1}{\sqrt{1-\frac{w^{2}}{\alpha\gamma}}}
\end{equation}
\begin{equation}
E_{33} = \sin^{2}\theta E_{22}
\end{equation}
\begin{equation}
4\pi E_{[01]} = \frac{Q^{2}\sqrt{1-\frac{w^{2}}{\alpha\gamma}}}{r^{4}w}
\end{equation}
The energy-momentum tensor is independent of Z and therefore, the same is true for the gravitational field equations. Followinf [2] and [6], $R_{\mu\nu}$ is calculated. Since the only non-zero component of $R_{[\mu\nu]}$ is $R_{[10]}$, Eq.(12) is identically satisfied. As in [2] we obtain, from $\alpha^{-1}R_{00} + \gamma^{-1}R_{11} = 0$ by taking $\alpha = e^{\lambda}$ and $\gamma = e^{\nu}$,
\begin{equation}
\frac{\lambda^{\prime} + \nu^{\prime}}{r} + \frac{4}{r^{2}}(\frac{l^{4}}{l^{4}+r^{4}}) = 0
\end{equation}
which integrates to,
\begin{equation}
\alpha\gamma = \frac{l^{4}}{l^{4}+r^{4}}
\end{equation}
where the constant of integration has been chosen in such a way that $\alpha\gamma$ becomes equal to Reissner-N\"{o}rdstr\"{o}m value when $l^{2} = 0$.\\
Now from Eq.(25) and Eq.(15) we get,
\begin{equation}
w = \frac{l^{2}}{r^{2}}
\end{equation}
Using $\alpha = e^{\lambda}$ and $\gamma = e^{\nu}$, the $R_{22}$ equations give,
\begin{equation}
\frac{\nu^{\prime}}{r} e^{-\lambda} + \frac{\lambda^{\prime}}{r}  e^{-\lambda} - - \frac{2}{r^{2}} (e^{-\lambda}-1) - 4\frac{l^{4}}{l^{4}+r^{4}} e^{-\lambda} = \frac{Q^{2}}{r^{2}} (1-\frac{l^{4}}{l^{4}+r^{4}})^{\frac{1}{2}}
\end{equation}
This differs from the corresponding equation of Ref [2] by the presence of the square root  factor in the right hand side. Using Eq.(15), Eq.(24) and Eq.(27) we get a equation for $\lambda$, which by integrating we obtained,
\begin{equation}
\alpha = (1 - \frac{2m}{r} - \frac{Q^{2}}{r} f(r))^{-1}
\end{equation}\
where
\begin{equation}
f(r) = \int \frac{l^{4}+r^{4}}{r^{4}} dr
\end{equation}\\
Then from Eq.(25) we get,
\begin{equation}
\gamma = (1 + \frac{l^{4}}{r^{4}}) (1 - \frac{2m}{r} + \frac{Q^{2}}{r} f(r))
\end{equation}\\
For large values of $r\gg l$, we have $f(r)\to-r^{-l}$ and the solutions (28) and (30) becomes identical with those obtained in Ref [2]. For small values of $r$, $f(r)\to -l^{2}r^{-3}$ and therefore
\begin{equation}
\alpha = (1 - \frac{2m}{r} - \frac{Q^{2}l^{2}}{3r^{4}})^{-1}
\end{equation}
and,
\begin{equation}
\gamma = (1 + \frac{l^{4}}{r^{4}}) (1 - \frac{2m}{r} + \frac{Q^{2}l^{2}}{3r^{4}}) \qquad\qquad (r\ll l)
\end{equation}\\
The electric field is, from (17) and (25),
\begin{equation}
E = \frac{Q}{r^{2}} (1 + \frac{l^{4}}{r^{4}})^{\frac{1}{2}}
\end{equation}\\
For large values of $r$ the electric field becomes the Coulomb field but for small values of $r$ it varies as $r^{-4}$.
\section{Conclusion}\label{S:conclu}
By adopting a non-minimal coupling that ensures the validity of the equivalance principle [4] we have derived the field equations of Moffat's [1,6] non-symmetric gravitational field when electromagnetic fields are present. We have worked out the energy-momentum tensor, $E_{\mu\nu}$, by varying the Lagrangian.We have shown that $E_{\mu\nu}$ is a traceless tensor. We give a static, spherically symmetric solution of the field equations. Using this solution, we derived the non-zero components of the energy-momentum tensor. We show that the electromagnetic action is independent of the free parameter $Z$. It is also found that the energy-momentum tensor is independent of $Z$ and therefore the same will be true for gravitational field. The solution of the field  equations for a charged particle is also obtained. Apart from the usual Reissner-N\"{o}rdstr\"{o}m solution, the electric field departs strongly from the Coulomb field value obtained in the Reissner-N\"{o}rdstr\"{o}m case. For large values of $r$, the electric field becomes the usual Coulomb field. At small distances, it varies as $r^{-4}$.

\begin{thebibliography}{99}
\bibitem{}  J. W. Moffat (1979): \textquotedblleft New theory of gravitation\textquotedblright  {\itshape Phys. Rev.} {\bf D-19, No-12}, 3554-3558.
\bibitem{}
 J. W. Moffat (1979): \textquotedblleft Static spherically solution for the field of a charged particle in a  theory of gravity\textquotedblright {\itshape Phys. Rev.} {\bf {D-19, No-12}}, 3562-3564.
\bibitem{}
 C. M. Will (1989): \textquotedblleft Violation of Weak Equivalance Principle in Theories of Gravity with a Nonsymmetric Metric\textquotedblright {\itshape Phys. Rev. Lett.} {\bf Vol-62, No-4}, 369-372.
\bibitem{}
 R. B. Mann, J. H. Palmer and J. W. Moffat (1989): Comment on \textquotedblleft Violation of Weak Equivalance Principle in Theories of Gravity with a Nonsymmetric Metric\textquotedblright {\itshape Phys. Rev. Lett.} {\bf Vol-62, No-23}, 2765.
\bibitem{}
 S. Ragusa (1997): \textquotedblleft Field of a Charged Particle in a Nonsymmetrical Gravitational Theory\textquotedblright {\itshape Brazilian Journal of Physics} {\bf Vol-27, No-2}, 76-79.
\bibitem{}
 J. W. Moffat (1987): \textquotedblleft Test -particle motion in the nonsymmetric gravitational theory\textquotedblright {\itshape Phys. Rev.} {\bf D-35, No-12}, 3733-3747.

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