% 807 words
\chapter{The motion of a charged particle}\label{rncharge}
A charged test particle does not follow a geodesic path, but we will look
briefly at them to compare their motion with particles without
charge. Their motion is determined by Lagrangian (from
\cite[chapter 5]{chandrasekhar})
\begin{equation}\label{r103}
  2{\cal L} = \left[\frac\varDelta{r^2}\left(\frac{\text dt}{\text
  d\tau}\right)^2 -\frac{r^2}\varDelta\left(\frac{\text dr}{\text
  d\tau}\right)^2 -r^2\left(\frac{\text d\theta}{\text
  d\tau}\right)^2 -r^2\sin^2\theta\left(\frac{\text
  d\varphi}{\text d\tau}\right)^2
  \right] +2\frac{qQ_*}r\frac{\text dt}{\text d\tau}\,,
\end{equation}
where $q$ denotes the charge per unit mass of the test particle. It is the last term in which the Lagrangian of the charged particle differs from Lagrangian of particle without charge.
The equations of motion which follow from Lagrangian \eqref{r103} are
(cf.~equations \eqref{r65})
\begin{equation}\label{r104}
  p_t = \frac\varDelta{r^2}\frac{\text dt}{\text d\tau}
  +\frac{qQ_*}r = E = \text{constant} \,,\qquad r^2\frac{\text d\varphi}
  {\text d\tau} = L = \text{constant} \,,
\end{equation}
and
\begin{equation}\label{r105}
  \left(\frac{\text dr}{\text d\tau}\right)^2
  +\frac\varDelta{r^2}\left(1 +\frac{L^2}{r^2}\right) =
  \left(\frac\varDelta{r^2}\frac{\text dt}{\text d\tau}\right)^2 =
  \left(E -\frac{qQ_*}r\right)^2 \,;
\end{equation}
and in place of equation \eqref{r67} we now have
\begin{equation}\label{r106}
    \left(\frac{\text du}{\text d\varphi}\right)^2 = -Q_*^2u^4
    +2Mu^3 -\left[1 +Q_*^2\frac{(1-q^2)}{L^2}\right]u^2 +
    \frac2{L^2}(M-qQ_*E)u -\frac{1-E^2}{L^2} \,.
\end{equation}

If a particle should have a turning point as it arrives at the
event horizon, then its energy, according to equation
\eqref{r105}, will be
\begin{equation}\label{r107}
  E = qQ_*/r_+ \,;
\end{equation}
and this will be negative if $qQ_* < 0$. This fact gives rise to
the possibility of extracting energy from the black hole, energy
that is associated with its charge. (Processes of this kind are
considered in \cite[chapter 7]{chandrasekhar}.)


\section{\color{blue}Radial motion}
Let us briefly examine radial motion of a charged particle. The governing equations can be obtained by setting $L = 0$ in equation \eqref{r105}; \textcolor{blue}{thus
\begin{equation}\label{rc105}
  \left(\frac{\text dr}{\text d\tau}\right)^2 = \left(E
  -\frac{qQ_*}r\right)^2 -\frac\varDelta{r^2}\,, \qquad \text{and} \qquad
  \frac{\text dt}{\text d\tau} = \left(E
  -\frac{qQ_*}r\right)\frac{r^2}\varDelta\,.
\end{equation}
Accordingly,
\begin{equation}\label{rc7t}
  \frac{\text dr}{\text dt} = \pm
  \frac\varDelta{r^2\Biggl(E-\cfrac{qQ_*}r\Biggr)^2}
  \sqrt{\left(E-\frac{qQ_*}r\right)^2-\frac\varDelta{r^2}}\,.
\end{equation}
Since $\varDelta > 0$ in the interval $0 \leq r < r_-$ , then for $q^2 < 1$ , $(Er -qQ_*)^2 -\varDelta$ will vanish for some finite value of $0 < r < r_-$ and the trajectory will have a~turning point inside the Cauchy horizon (like in the case of a particle without charge). If $qQ_* > 0$ the particle can have a turning point anywhere depending on relations between parameters $E\,,\ q\,,\ M$ and $Q_*$. And finally if $qQ_* < 0$ there is turning point only if $E^2 < 1$ (if $q^2 > 1$ holds).}

The formal solutions of equations \eqref{rc105} and \eqref{rc7t} can be readily written down in terms of elementary integrals. We shall not write them since they are too complicated to exhibit any more features that what can be deduced from a simple inspection of the equations.


\section{\color{blue}Circular orbits}
\textcolor{blue}{The conditions for the occurrence of circular orbits are
\bal
    \begin{split}\label{r89c}
        \text f(u) &= -Q_*^2u^4
        +2Mu^3 -\left[1 +Q_*^2\frac{(1-q^2)}{L^2}\right]u^2 +\\+
        &\frac2{L^2}(M-qQ_*E)u -\frac{1-E^2}{L^2} =\\ &= 0\,,
    \end{split}
    %\raisetag{25pt}
    \\
    \intertext{and}
    \text f\,'(u) &=
    -4Q_*^2u^3 +6Mu^2-2\left(1
    +Q_*^2\frac{1-q^2}{L^2}\right)u +2\frac{M-qQ_*E}{L^2} = 0\,.
    \label{r90c}
\eal}
From these equations, it follows that the energy $E$ and the
angular momentum $L$ of a circular orbit of radius \(r_\mathrm{c}
= 1/u_\mathrm{c}\) is given by (cf. equations \eqref{rn91} and \eqref{rn92}) \textcolor{blue}{
\bal
    \begin{split}\label{r91c}
        E^2 &= \left\{1 -(4M-qQ_*E)u_c+\right.
        \\&+2\left[Q_*^2
        +2M(M
        -qQ_*E)\right]u_c^2-Q_*^2\left[3(M-qQ_*E)+\right.\\
        &\left.\left.+M\left(1-q^2\right)\right]u_c^3
        +Q_*^4(1-q^2)u_c^4
        \right\}/
        \left(1-3Mu_c+2Q_*^2u_c^2\right)
    \end{split}
    \\
    \intertext{and}
    L^2 &= \frac{M-qQ_*E-Q_*^2(1-q^2)u_c}{u_c(1-3Mu_c+2Q_*^2u_c^2)}\,.
    \label{r92c}
\eal}
It is possible to give explicit solutions of these equations, but for easier comparison with time-like geodesic and a clearer understanding of the roles of signs of $qQ_*$ and $(1 -q^2)$, we have left them in this implicit form.

The minimum radius for a stable circular orbit will occur at a point of inflexion of the function f$(u)$ (see figure \ref{picture5t}), i.e., we must supplement equation \eqref{r89c} and \eqref{r90c} with the further equation
\begin{equation}\label{r98c}\color{blue}
  \text f\,''(u) =
  -12Q_*^2u^2+12Mu-2\left(1+Q_*^2\frac{1-q^2}{L^2}\right) =
  0\,.
\end{equation}
Eliminating \(L^2\) in this equation with the aid of
equation (\ref{r92c}), we obtain the equation (cf. equation \eqref{r99})\textcolor{blue}{
\begin{multline}\label{r99c}
    4Q_*^4\left(q^2-1\right)u^3+\\
    +3Q_*^2\left[2(M-qQ_*E)
    +M\left(1-q^2\right)\right]u^2-6M(M-qQ_*E)u+M-qQ_*E = 0\,,
    \raisetag{51pt}
\end{multline}
or, alternatively
\begin{equation}\label{r100c}
  r_c^3-6Mr_c^2
  +3Q_*^2\frac{\left[2(M-qQ_*E)-M\left(1-q^2\right)\right]r_c}{M-qQ_*E}
  -4Q_*^4\frac{1-q^2}{M-qQ_*E} = 0\,.
\end{equation}
(It may be noted here that equation \eqref{r100c} gives $r_c = 6M$ when $Q_*^2 = 0$, in agreement with the value for the Schwarzschild geometry; and when $q = 0$ is identical to the equation for time-like geodesic (cf. \cite[\S 40(b)]{chandrasekhar}).)}


\section{\color{blue}The bound orbits $\left(E^2 <1\right)$}
As in chapter \ref{rt}, the geometry of the orbits described in
the invariant plane is separated to two parts distinguished by
$E^2 <1$ and $E^2 \geq 1$ (see equation \eqref{r106}). This will
determine which orbits are bound and which ones are unbound. The
geometry of the orbits will be determined by the disposition of
the roots of the equation f$(u) = 0$ (see figure~\ref{tlpicture}).
Orbits look essentially as time-like geodesics.
\textcolor{blue}{Analytical solutions are formally the same as in
Chapter~\ref{rt}.}


\section{\color{blue}The unbound orbits ($E^2 > 1$)}
When \(E^2 > 1\), the constant term in f\((u)\) is positive. The
equation f\((u) = 0\) must, therefore, allow a negative root. The
cases we must distinguish look essentially as space-like, unbound time-like  and null
geodesics. \textcolor{blue}{Analytical solutions are formally the same as
in Chapter~\ref{RNnull}. An example of unbound orbits of a~charged particle is in figure \ref{RNcharpl}.}
\begin{figure}
    \epsfxsize=13cm
    \epsfbox{Reis-Nor/charplot.eps}
    \caption{\textcolor{blue}{Orbits of the first and the     second
    kind of a charged particle} with $Q_* =
    0.8,\ \textcolor{blue}{M = 1.2,\ L = 6,\ E =
    1.1}$     \textcolor{blue}{and $q = 2$.}  The
    thin circle
    represents the unstable circular-orbit $r_{\mathrm c}$ ($u_{\mathrm c} =
    1/r_{\mathrm c} =
    0.2225$). The orbit of the first kind starts from infinity and ends in infinity,
    the orbit of the second kind starts and ends certain distance from the black hole.}
    \label{RNcharpl}
\end{figure}
