\color{black}
\chapter{The time-like geodesics}
\label{kerrtime}
Finally we look at the time-like geodesics in Kerr space-time. We expand discussion of \cite{chandrasekhar}. 

For time-like geodesics $\delta_1 = 1$. Writing $R$ and $\varTheta$ instead of $R/E^2$ and $\varTheta/E^2$ we have
\begin{equation}\label{k267}
  R = r^4 +(a^2-\xi^2-\eta^2)r^2 +2M[\eta+\left(\xi-a\right)^2] -a^2\eta -\frac{r^2\varDelta}{E^2}
\end{equation}
and
\begin{equation}\label{k269}
  \varTheta = \eta +a^2\left(1-\frac1{E^2}\right)\arccos\theta -\xi^2\cot^2\theta \,.
\end{equation}


\section{The $\theta$-motion}
As in sections \ref{nulteta} and \ref{spaceteta} we use $\mu = \cos\theta$ as a variable of integration. Then the right-hand side of equation \eqref{k193} that describes the orbit on $(r, \theta)$-plane becomes integral \eqref{k196} where
\begin{equation}\label{k271}
  \varTheta_\mu = \eta -\left[\xi^2+\eta-a^2\left(1-\frac1{E^2}\right)\right]\mu^2 -a^2\left(1-\frac1{E^2}\right)\mu^4 \,.
\end{equation}

As in the cases of time-like geodesics in Schwarzschild and Reissner-Nordstr\"om space-times we must distinguish orbits with $E^2 < 1$ (bound orbits), $E^2 = 1$ (marginally bound orbits), and $E^2 > 1$ (unbound orbits).

\subsection{The bound orbits}
It is useful to replace $a^2$ by
\begin{equation}\label{k272}
  \alpha^2 = a^2\left(\frac1{E^2} -1\right) > 0\,.
\end{equation}
Then $\varTheta_\mu$ will be written in the form
\begin{equation}\label{k273}
  \varTheta_\mu = \eta -\left(\xi^2+\eta+\alpha^2\right)\mu^2 +\alpha^2\mu^4\,.
\end{equation}

Following \cite[page 362]{chandrasekhar}, the solution of integral \eqref{k196} is
\begin{equation}\label{k275}
  \int\limits_0^\mu \frac{\text d\mu}{\sqrt{\varTheta_\mu}} = \frac{\F(\psi, k)}{\alpha\mu_+} \,,
\end{equation}
where
\bal
    \mu_\pm^2 &= \frac{(\xi^2+\eta+\alpha^2) \pm\sqrt{\left(\xi^2+\eta+\alpha^2\right)^2-4\alpha^2\eta}} {2\alpha^2}\,,
    \label{k276}\\
    k &=\frac{\mu_-}{\mu_+} \label{k277k}\\
    \intertext{and}
    \sin\psi &= \frac\mu{\mu_-} \label{k277psi}\,,
\eal
with $\mu^2$ between zero and $\mu_-^2$ and $\eta$ non-negative.


\subsection{The marginally bound orbits}
For these orbits $E^2 = 1$ and so equation \eqref{k273} becomes
\begin{equation}\label{k278}
  \varTheta_\mu = \eta -(\xi^2+\eta)\mu^2 \,.
\end{equation}
Negative values of $\eta$ are not allowed and the solution is \cite[page 362]{chandrasekhar}
\begin{equation}\label{k279}
  \int\limits^\mu \frac{\text d\mu}{\sqrt{\varTheta_\mu}} = -\frac{\arccos\left(\mu\sqrt{\dfrac{\xi^2+\eta}\eta}\right)} {\sqrt{\xi^2+\eta}}\,.
\end{equation}

Both bound and marginally bound orbits cross the equatorial plane and oscillate about it.


\subsection{The unbound orbits}\label{kttetaunbd}
For orbits with $E^2 > 1$, after replacing $a^2$ by
\begin{equation}\label{k280}
  \alpha^2 = a^2\left(1-\frac1{E^2}\right) > 0 \,,
\end{equation}
the discussion in section \ref{spaceteta} applies.


\section{The $r$-motion}
We consider two simple integrable cases.

\subsection{The circular orbits}
In this section the orbits with constant radial coordinate will be considered. They occur when equations \eqref{k281space} hold. The constraints on the constants of motion, $\xi,\ \eta$ and $E^2$, can be obtained by the same procedure as in section \ref{knullcirc}, with an additional term, $-r^2\varDelta/E^2$, in the definition for $R$. Instead of equations \eqref{k216} and \eqref{k217} there are now equations
\begin{equation}\label{k282}
  3r^4 +r^2a^2 -\eta\left(r^2-a^2\right) -\frac{r^2}{E^2}\left(3r^2-4Mr+a^2\right) = r^2\xi^2
\end{equation}
and
\begin{equation}\label{k283}
  r^4 -a^2Mr +\eta(a^2-Mr) -\frac{r^3}{E^2}(r-M) = rM\left(\xi^2-2a\xi\right)\,;
\end{equation}
and, as in sections \ref{knullcirc} and \ref{kspacecirc}, these equations can be solved for $\xi$ and $\eta$.

Eliminating $\eta$ from above equations gives (cf. equations \eqref{k218} and \eqref{k284space})
\begin{multline}\label{k284}
  a^2(r-M)\xi^2 -2aM\left(r^2-a^2\right)\xi \\ -\left\{\left(r^2+a^2\right)\left[r\left(r^2+a^2\right) -M\left(3r^2-a^2\right)\right] -\frac{r\varDelta}{E^2}\right\} 
= 0 \,.
\end{multline}
The solution of this equation is (cf. equations \eqref{k220} and \eqref{k285space})
\begin{equation}\label{k285}
  \xi = \frac{M(r^2-a^2) \pm r\varDelta\sqrt{1-\dfrac1{E^2}\left(1-\dfrac Mr\right)}} {a(r-M)} \,,
\end{equation}
and the corresponding solution for $\eta$ is (cf. equations \eqref{k225} and \eqref{k286space})
\begin{multline}\label{k286}
  \eta a^2(r-M) = 
   \frac{r^3}{r-M}[4a^2M-r\left(r-3M\right)^2]\;+ \\+ \frac{r^2}{E^2} \left[r\left(r-2M\right)^2-a^2M\right]
-\frac{2r^3M\varDelta}{r-M} \left[1\mp\sqrt{1+\frac1{E^2}\left(1-\frac Mr\right)}\right] \,.
\end{multline}

The particle in an orbit with constant radius, $r_s$, has values $\xi_s$ and $\eta_s$ given by above equations. Because $r_s$ is a double root of the equation $R = 0$, $R$ can be reduced to
\begin{equation}\label{k287}
  \text R = \left(r-r_s\right)^2 \left\{r^2\left(1-\frac1{E^2}\right) +2rr_s\left[1-\frac1{E^2}\left(1-\frac M{r_s}\right)\right] -\frac{a^2\eta_s}{r_s^2}\right\} \,.
\end{equation}
Substituting \eqref{k233}, the integral on the left-hand side of equation \eqref{k193} becomes (cf. equation \eqref{k290space})
\begin{multline}\label{k290}
  \int\frac{\text dr}{\sqrt\text R} = - \int\text dx \left\{\left(1-\frac1{E^2}\right) +\left[4r_s\left(1-\frac1{E^2}\right) +\frac{2M}{E^2}\right]x\: + \right.\\
\left. +\left[3r_s^2\left(1-\frac1{E^2}\right) -\frac{a^2\eta_s}{r_s^2} +\frac{2Mr_s}{E^2}\right]x^2\right\}^{-\frac12} \,.
\end{multline} With the definition \eqref{k291space}, the integral is
\bal
    \int\frac{\text dr}{\sqrt R} &= \pm\frac{\ln(B +2Cx+2\sqrt{CX})} {\sqrt C} &(C > 0)\,, \label{k292a}\\
    &= \pm \frac{\arcsinh\frac{2Cx+B}{\sqrt D}} {\sqrt C} &(C > 0,\ D > 0)\,, \label{k292b}\\
    &= \mp\frac{\arcsin\frac{2CX+B}{\sqrt{-D}}} {\sqrt{-C}} &(C < 0,\ D < 0), \label{k292c}\,
\eal
where
\begin{equation}\label{k293}
  D = -4\left\{\left[r_s\left(1-\frac1{E^2}\right)+\frac M{E^2}\right]^2 +\frac{\alpha^2\eta_s}{{r_s}^2}\left(1-\frac1{E^2}\right)\right\}\,.
\end{equation}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/tlmgcirc.eps}
         \caption{An example of a marginally bound time-like geodesic with constant radius $r_s = 2.154$. Parameters are $M = 1$, and $a=0.8$.}
         \label{tlmgcirc}
\end{figure} 
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/tlbdcirc.eps}
         \caption{An example of a stable spherical time-like geodesic. Parameters are $M = 1$, $e = -0.08 $ and $a^2=0.84$.}
         \label{tlbdcirc}
\end{figure} 
Examples of orbits with constant radius are in figures \ref{tlmgcirc} and \ref{tlbdcirc}.


{\color{blue}
\subsection{Orbits with negative $\eta$}
Because of the conditions on parameters $\eta$, $\alpha$ and $xi$ from section \ref{kttetaunbd}, integration of the left-hand side of equation \eqref{k193} gives
\begin{equation}\label{ktimebd}
  \int\frac{\text dr}{\sqrt\text R} = g[\F(\phi),k)+\K(k)] \,,
\end{equation}
where
\bal
    g &= \frac1{\sqrt{A+B}}\,, \\
    \phi &= \arccos\frac{(A-B)r+u_2B-u_1A}{(A+B)r-u_2B-u_1A}\,, \\
    k &= \frac{\sqrt{\dfrac{(A+B)^2-(u_2-u_1)^2}{AB}}}{2} \,,\\
    A &= \sqrt{\left(u_2-b_1\right)^2+a_1^2}\,, \\
    B &= \sqrt{\left(u_1-b_1\right)^2+a_1^2}\,, \\
    a_1 &= \frac{\sqrt{-(u_3-\bar{u_3})^2}}2\,, \\
    \intertext{and}
    b_1 &= \frac{u_3+\bar{u_3}}2\,,
\eal
where $u_1,\ u_2,\ u_3$ and $\bar{u_3}$ are the solutions of the equation $R = 0$. Example of this kind of orbit is in figure \ref{tim-Cart}.
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/tim_Cart.eps}
         \caption{An example of a time-like orbit with negative Carter constant. Parameters are $M = 1$, $e = -0.08 $, $\xi = -0.5$, $\eta = -1/9$, $E = 3$ and $a^2=0.84$.}
         \label{tim-Cart}
\end{figure} 


\section{The $\varphi$-motion}
\subsection{Marginally bound orbits} 
We substitute $E^2 = 1$ in the equations \eqref{k287} and \eqref{k269}, then
\bal\label{ktmgR}
  R &= \left(r-r_s\right)^2 \left(2rM-\frac{a^2\eta_s}{{r_s}^2}\right) \\
  \varTheta &= \eta_s -{\xi_s}^2\cot^2\theta \,.
\eal
The $\varphi$-motion is determined by integral
\begin{equation}\label{ktmgfi}
  \varphi = a\int[(r^2+a^2)-aL]\frac{\text dr}{\varDelta\sqrt R} +\int(L\csc^2\theta-a)\frac{\text d\theta}{\sqrt\varTheta} \,.
\end{equation}

Using method of partial fractions, the integral over $r$ will become
\begin{equation}\label{ktIr}
 I_R = -\frac1{\varDelta_s}\int \frac{Ar+B}{\varDelta\sqrt{2Mr-\dfrac{a^2\eta_s}{{r_s}^2}}}{\text dr} +\frac{a^2-aL+{r_s}^2}{\varDelta_s} \int\frac{\text dr}{(r-r_s)\sqrt{2Mr-\dfrac{a^2\eta_s}{{r_s}^2}}} \,,
\end{equation}
where
\bal
    \varDelta_s &= r_s^2 -2Mr_s +a^2 \,,\\
    A &= 2r_sM +2a^2 -aL \\
    \intertext{and}
    B &= 2r_sa^2-r_saL-2Ma^2+2MaL \,.
\eal
The first integral can be further reduced to
\begin{multline}\label{ktmgI1}
  I_1 = 4M\int\frac{At^2+\gamma}{t^4+2t^2\alpha+\beta}{\text dt} =\\ = 4M \int\frac{At^2+\gamma}{(t^2+\alpha^2-\sqrt{\alpha^4-\beta}) (t^2-\sqrt\beta+\sqrt{\alpha^4-\beta})}\text dt\,,
\end{multline}
where
\bal
    \alpha &= \frac{a^2\eta_s}{{r_s}^2}-2M^2 \,,\\
    \beta &= \left(\frac{a^2\eta_s}{{r_s}^2}\right)^2 -4M^2\frac{a^2\eta_s}{{r_s}^2} +4M^2a^2 \\
    \intertext{and}
    \gamma &= A\frac{a^2\eta_s}{{r_s}^2}+2MB \,.
\eal
Using the partial fractions method again we find
\begin{equation}\label{ktmgI1fin}
  I_1 = 4M\arctan(2t) \left(\frac{\text{const}_1}{\sqrt{\alpha^2-\sqrt{\alpha^4-\beta}}} + \frac{\text{const}_2} {\sqrt{\sqrt{\alpha^4-\beta}-\sqrt\beta}}\right) \,,
\end{equation}
where
\begin{multline}
    \text{const}_1 = -5\gamma\beta^{3/2} -3\gamma\alpha^2\beta +3\gamma\alpha^6 +4\gamma\alpha^2*\sqrt{\beta}\sqrt{\alpha^4-\beta} +5\gamma\sqrt{\beta}\alpha^4\\ +6\gamma\alpha^4\sqrt{\alpha^4-\beta} -10\gamma\beta\sqrt{\alpha^4-\beta} 
+A\beta^{3/2}\alpha^2 -5A\beta^{3/2}\sqrt{\alpha^4-\beta} -13A\alpha^4\beta \\
    +7A\alpha^2\beta\sqrt{\alpha^4-\beta} +3A\alpha^8 -3A\alpha^6\sqrt{\alpha^4-\beta} \\ +A\alpha^4\sqrt\beta\sqrt{\alpha^4-\beta} -A\sqrt\beta\alpha^6 +10A\beta^2
\end{multline}
and
\begin{multline}
    \text{const}_2 = 5\gamma\beta^{3/2}+\\ +3\gamma\alpha^2\beta -3\gamma\alpha^6 -4\gamma\alpha^2\sqrt\beta\sqrt{\alpha^4-\beta} -5\gamma\sqrt\beta\alpha^4 -6\gamma\alpha^4\sqrt{\alpha^4-\beta}+\\ +10\gamma\beta\sqrt{\alpha^4-\beta} +15A\beta^2+\\+ 5A\beta^{3/2}\sqrt{\alpha^4-\beta} -A\beta^{3/2}\alpha^2 -7A\alpha^2\beta\sqrt{\alpha^4-\beta} +A\sqrt\beta\alpha^6 +\\ +3A\alpha^6\sqrt{\alpha^4-\beta} -21A\alpha^4\beta -A\alpha^4\sqrt\beta\sqrt{\alpha^4-\beta} +6A\alpha^8\,.
\end{multline}
The second integral in \eqref{ktIr} is
\begin{equation}\label{I2}
  I_2 =
    \begin{cases}
        \dfrac{2\arctan\sqrt{\dfrac{2Mr-\dfrac{a^2\eta_s} {{r_s}^2}}{-D}}} {\sqrt{-D}}& \text{for $D < 0$}\\
        \dfrac{\ln\left|\dfrac{\sqrt{2Mr-\dfrac{a^2\eta_s} {{r_s}^2}}-\sqrt D} {\sqrt{2Mr-\dfrac{a^2\eta_s} {{r_s}^2}}+\sqrt D}\right|} {\sqrt{D}}& \text{for $D > 0$}.,
    \end{cases}
\end{equation}
where
\begin{equation}\label{D}
  D = 2Mr_s -\frac{a^2\eta_s}{{r_s}^2} \,.
\end{equation}

We substitute $x = \sin\theta$ in the integral over $\theta$ in \eqref{ktmgfi}. Then
\begin{equation}\label{Ix}
\begin{split}
  I_x &= \int\frac{L-a^2x^2} {x\sqrt{x^2(\eta_s+{\xi_s}^2)-{\xi_s}^2}}\text dx = \\
  &= \frac{\xi_s\arccos\dfrac{\xi_s}{\abs x\sqrt{\eta_s+{\xi_s}^2}} -a^2\sqrt{x^2(\eta_s+{\xi_s}^2)-{\xi_s}^2}}{\eta_s+{\xi_s}^2}\,.
\end{split}
\end{equation}
} 