% Words:   637
\chapter{\color{blue}The space-like geodesics in the Reissner-Nordstr\"om
space-time}
The equations governing space-like geodesics are
\begin{equation}
    \textcolor{blue}{\left(\frac{{\mathrm d}r}{{\mathrm d}\tau}\right)^2+
    \frac\varDelta{r^2} \left(\frac{L^2}{r^2}- 1\right) = E^2} \,; \quad
    \frac{{\mathrm d}t}{{\mathrm d}\tau} = E\frac{r^2}\varDelta \,; \quad
    \mbox{and} \quad \frac{{\mathrm d}\varphi}{{\mathrm d}\tau} =
    \frac{L}{r^2} \,.
    \label{r65n}
\end{equation}
\begin{figure}
    \epsfxsize = 13cm
    \epsfbox{Reis-Nor/slpotent.eps}
    \caption{\textcolor{blue}{The effective potential appropriate for
    space-like
    trajectory
    (comp. equation~\ref{RNslpotent}).} $Q = 0.8$ and \textcolor{blue}{$L$ is
    a)~3, b)~5, c)~10 and d)~15.}}
    \label{RNslpotentpic}
\end{figure}
\textcolor{blue}{As in part~\ref{sch}, we interpret
\begin{equation} \label{RNslpotent}
    {\cal V} = \frac\varDelta{r^2} \left(\frac{L^2}{r^2}-1\right)
    \,
\end{equation}
as `potential energy'.  In the third graph of
fig.~\ref{RNslpotentpic}, the unstable orbits correspond to
maximum potential and the stable orbits correspond to minimum
potential. Points of inflexion in the second graph indicate
unstable orbits.}

By considering $r$ in equation~(\ref{r65n}) as a function of $\varphi$ and
replacing $r$ by $u =
1/r$ as the independent variable we obtain the equation (cf.~equation
(\ref{s94}))
\begin{equation}\color{blue}
    \left(\frac{{\mathrm d}u}{{\mathrm d}\varphi}\right)^2 = -Q_*^2u^4+
    2Mu^3- \left(1- \frac{Q_*^2}{L^2}\right)u^2- \frac{2M}{L^2}u +\frac{E^2+ 1}{L^2} \,.
    \label{r67n}
\end{equation}


\section{The radial geodesics}
The equations governing the radial space-like geodesics can be obtained by
setting $L = 0$ in equations (\ref{r65n}); thus,
\begin{equation}
    \textcolor{blue}{\left(\frac{{\mathrm d}r}{{\mathrm d}\tau}\right)^2 =
    E^2+
    \frac\varDelta{r^2}} \,, \quad \frac{{\mathrm d}t}{{\mathrm d}\tau} =
    E\frac{r^2}\varDelta \,, \quad \mbox{and} \quad \frac{{\mathrm
    d}\theta}{{\mathrm d}\tau} = \frac{{\mathrm d}\varphi}{{\mathrm
    d}\tau} = 0 \,.
    \label{r70n}
\end{equation}
\textcolor{blue}{Accordingly,
\begin{equation}
    \frac{{\mathrm d}r}{{\mathrm d}t} =
    \pm\left(\frac\varDelta{Er^2}\right) \sqrt{E^2+ \frac\varDelta{r^2}} \,.
    \label{r71n}
\end{equation}
The solution for the arc lenght, $\tau$, is
\begin{multline}
    \tau = \frac{r_\mathrm{i}\sqrt{E^2+\frac{\varDelta_\mathrm{i}}
    {r_\mathrm{i}^2}}-r\sqrt{E^2
    +\frac{\varDelta}{r^2}}}{E^2+1} +
    \frac{M}{\left(E^2+1\right)^{3/2}}\cdot\\
    \cdot\left(\arctanh\frac{E^2+1-\frac{M}{r_\mathrm{i}}}
    {\sqrt{E^2+1}
    \sqrt{E^2+\frac{\varDelta_\mathrm{i}}{r_\mathrm{i}^2}}}
    -\arctanh\frac{E^2+1-\frac{M}{r}}{\sqrt{E^2+1}
    \sqrt{E^2+\frac{\varDelta}{r^2}}}\right) %\nonumber
    \label{r72a}
\end{multline}
and for coordinate t,
%\newpage
\begin{multline}
    t = \frac{M(4M+1)}{\left(E^2+1\right)^{3/2}}\cdot \\*
    \cdot\left(\arctanh\frac{E^2+1-\frac{M}{r_\mathrm{i}}}
    {\sqrt{E^2+1}
    \sqrt{E^2+\frac{\varDelta_\mathrm{i}}{r_\mathrm{i}^2}}}
    -\arctanh\frac{E^2+1-\frac{M}{r}}{\sqrt{E^2+1} \sqrt{E^2
    +\frac{\varDelta}{r}}}\right)+
    \\*
    +\left(Q_*^2 -\frac{4M^2}{E^2 +1}\right) \ln\frac{r\sqrt{E^2
    +\frac{\varDelta}{r^2}}}{r_\mathrm{i}\sqrt{E^2
    +\frac{\varDelta_i}{r_\mathrm{i}^2}}}+\\*
    +2M\sqrt{E^2 +1} \ln\frac{r(E^2 +1) -M +r\sqrt{E^2 +1} \sqrt{E^2
    +\frac{\varDelta}{r^2}}}{r_\mathrm{i}(E^2 +1) -M
    +r_\mathrm{i}\sqrt{E^2+1} \sqrt{E^2
    +\frac{\varDelta_\mathrm{i}}{r_\mathrm{i}^2}}}+ \\*
    +\frac{Q_*^2}{2} \left(\frac{\sqrt{E^2
    +\frac{\varDelta_\mathrm{i}}{r_\mathrm{i}^2}}}{r_\mathrm{i}}
    -\frac{\sqrt{E^2
    +\frac{\varDelta}{r^2}}}{r}\right)
    -2M\left(\sqrt{E^2+\frac{\varDelta_\mathrm{i}}{r_\mathrm{i}^2}}
    -\sqrt{E^2
    +\frac{\varDelta}{r^2}}\right)+\\*
    +\frac{r_\mathrm{i}\sqrt{E^2+\frac{\varDelta_\mathrm{i}}
    {r_\mathrm{i}^2}} -r\sqrt{E^2
    +\frac{\varDelta}{r^2}}}{E^2 +1}\,,
    \label{r72b}
\end{multline}
where the initial condition for $r$ is given by
\begin{equation}
    E^2 +1 = \frac{Q_*^2 +2Mr_\mathrm{i}}{r_\mathrm{i}^2}
    \label{r72c}
\end{equation}}
for $r(0) = r_\mathrm{i}$ and $\dot{r}(0) = 0$.



\section{\color{blue}Unbound orbits}
These orbits are governed by equation (\ref{s109}), where
\begin{equation}\label{r89}\color{blue}
    {\mathrm f}(u) = -Q_*^2u^4 +2Mu^3 -\left(1
    -\frac{Q_*^2}{L^2}\right)u^2 -\frac{2M}{L^2}u +\frac{E^2+1}{L^2}\,.
\end{equation}

\vspace{-7mm} Letting $u_1, u_2, u_3$, and $u_4$ denote the roots
of the quartic equation f$(u) = 0$, we have again several
possibilities of their location (cf.~figures~\ref{picture},
\ref{picture2}, \ref{picture3}, \ref{picture4}, \ref{picture5} and
\ref{picture6} in Chapter~\ref{RNnull}).

\textcolor{blue}{First we find the condition for parameters $M$,
$L$ and $Q_*$ for the limiting case in fig.~\ref{picture5}, which
happens when
\begin{equation}
    {\mathrm f}'(u) = -4Q_*^2u^3 +6Mu^2 -2\left(1
    -\frac{Q_*^2}{L^2}\right)u -\frac{2M}{L^2} = 0 %\,.
    \label{r90}
\end{equation}
and
\begin{equation}\label{rns90}
  {\mathrm f}''(u) = -12Q_*^2u^2 +12Mu -2\left(1-\frac{Q_*^2}{L^2}\right) =
  0 \,.
\end{equation}
%has one real and one double root. That leads us to equation
%\begin{equation}\label{rnscond}
%  \left(-2Q_*^2L^2+2Q_*^4+3M^2L^2\right)^3 =
%  27L^2M^2\left(-M^2L^2+Q_*^2L^2+3Q_*^4\right)^2 \,.
%\end{equation}
Since we require $M$ to be positive we have
\begin{equation}\label{rnscondM}
    M =
    %\sqrt{3\left(23Q_*^4+26Q_*^2L^2-L^4\right)
    %+\sqrt{3(3L^2+Q_*^2)\left(L^2+11Q_*^2\right)^3}}
    \frac{\sqrt{6\left[16Q_*^2L^2-\left(L^2+Q_*^2\right)^2\right]
    \pm2\sqrt{3(3L^2-Q_*^2)\left(5Q_*^2+L^2\right)^3}}}{12L}
\end{equation}
or
\begin{multline}\label{rnscondL}
  L^2 =
  \Biggl|\left\{3M^{2/3}\left[2Q_*^4\left(8Q_*^2-9M^2\right)
  \sqrt{(Q_*^2-M^2)\left(4Q_*^2-5M^2\right)^3}+ \right.\right.\Biggr.\\
  +8154Q_*^4M^6 -5177Q_*^6M^4 +1472Q_*^8M^2 -128Q_*^{10}+
  \\
  \biggl.+1728M^{10} -6040Q_*^2M^8
  \biggr]^{1/3} \\
  -3M^{4/3}\left(64Q_*^6 -257Q_*^4M^2 -144M^6 +336Q_*^2M^4\right)\\
  /\left[2Q_*^4\left(8Q_*^2-9M^2\right)
  \sqrt{(Q_*^2-M^2)\left(4Q_*^2-5M^2\right)^3}\right.\\
  +8154Q_*^4M^6 -5177Q_*^6M^4 +1472Q_*^8M^2 -128Q_*^{10} +1728M^{10}
  \\
  \left.\biggl.\biggl. -6040Q_*^2M^8
  \biggr]^{1/3} +36M^4 +8Q_*^4 -42M^2Q_*^2\biggr\}
  /\left(8Q_*^2 -9M^2\right)\right| \,.
\end{multline}}
The conditions for the occurrence of circular orbits are equations
\begin{equation}\label{r89'}
  \mbox{f}(u) = 0
\end{equation}
and (\ref{r90}). From these equations it follows that energy $E$
and the angular momentum $L$ of a circular orbit of radius
$r_\mathrm{c} = 1/u_\mathrm{c}$ is given by
\bal%egin{equation}
    E^2 &= -\frac{(1 -2Mu_\mathrm{c} +Q_*^2u_\mathrm{c}^2)^2}{1 -3Mu_\mathrm{c}
    +2Q_*^2u_\mathrm{c}^2}
    \label{r91}
    %\end{equation}
    \intertext{and}
    %\begin{equation}
    \color{blue}
    L^2 &\textcolor{blue}{= -\frac{M -u_\mathrm{c}Q_*^2}{u_\mathrm{c}(1
    -3Mu_\mathrm{c}
    +2Q_*^2u_\mathrm{c}^2)}}\,.
    \label{r92}
\eal%nd{equation}
These equations require, in particular, that
\begin{equation}\color{blue}
    1 -3Mu_\mathrm{c} +2Q_*^2u_\mathrm{c}^2 < 0\,.
    \label{r93}
\end{equation}and
\begin{equation}\label{rn93l}\color{blue}
  \frac M{Q_*^2} \geq u_\mathrm{c}
\end{equation}
\textcolor{blue}{Inequality (\ref{r93}) shows that the maximum
radius for space-like circular orbits, in the case \( M^2 \geq
      Q_*^2 \),
 is the radius of the circular photon-orbit.}
Since \textcolor{blue}{for \( M^2 < Q_*^2 < 9M^2/8 \)} we have two
distinct circular photon-orbits (one stable and one unstable),
\textcolor{blue}{space-like circular
orbits exist in the region \( r_\mathrm{cs} < r < r_\mathrm{cu} \). %(all
%stable)
When \(Q_*^2 \geq 9M^2/8 \), there is no space-like circular orbit
at all. Inequality \eqref{rn93l} shows minimum radius for
space-like circular orbit. So in figure~\ref{RNslpotentpic} the
stable orbits in the last picture are not allowed.} Analytical
solutions for general and circular orbits are formally the same as
in Chapter~\ref{RNnull}.
