% 73 slov
\part{Geodesics in the Kerr space-time}\label{kerr}
\chapter{Introduction to geodesics in the Kerr space-time}
\label{kerprefa}
In this part we will deal with geodesics in Kerr space-time (black
hole characterized by mass $M$ and angular momentum per unit mass
$a$, see Appendix \ref{ik}). We shall use the following terminology:
\begin{description}
  \item[slowly rotating Kerr space-time] occurs when $0 < a^2 < M^2$
  \item[extreme Kerr space-time] occurs when $a^2 = M^2$
  \item[rapidly rotating Kerr space-time] occurs when $M^2 < a^2$
\end{description}
with Schwarzschild space time occurring when $a = 0$. In this chapter there is a~brief derivation of equations of motion, and discussion of different types of orbits in relation to relationship among parameters (energy, angular momentum, Carter constant). This will make later chapters easier to follow. Lemmas from \cite{onil} are in this thesis without proofs (only with relevant references). Lemmas modified or additional ones are followed by their proofs. The pictures are in ``spherical" coordinates; the accurate definitions can be found in Appendix \ref{programy} in the files that generate the figures.

A first indication of the differences between these \emph{rotation
types} is given by the horizon function \eqref{k45}.
\begin{itemize}
    \item For Schwarzschild space-time, $\varDelta$ has roots $0,\
    2M$ (see Part \ref{sch}).
    \item For slow Kerr, $\varDelta$ has two roots: $0 < r_\pm = M
    \pm\sqrt{M^2 -a^2} < 2M$.
    \item For extreme Kerr, $\varDelta  = (r -M)^2$, so $r = M$ is
    a double root.
    \item For fast Kerr (rapidly rotating), $\varDelta$ has no
    real roots.
\end{itemize}



%The following function will pervade the study of Kerr
%space-time:
%\begin{equation}
%    \rho^2 = r^2 +a^2\cos\vartheta \label{onil58-1}\,.
%\end{equation}
%Its Schwarzschild meaning can be seen by setting $a = 0$. Thus,
%$\rho^2$ obviously generalizes $r^2$.

There are four independent first-integrals on the Kerr space-time:
$\delta_1 = \langle \dot\gamma, \dot\gamma\rangle$, energy $E$,
angular momentum about the axis $L$ and the fourth, called the
\emph{Carter constant}.


\section{The nature of the Kerr space-time}
Unlike the Schwarzschild and the
Reissner-Nordstr\"om space-times, in the Kerr metric  there are  three subsets on
which the  metric fails (see \cite[Section 2.2]{onil}).
\begin{enumerate}
    \item On the \emph{horizon} $H:\ \varDelta = 0$. For the fast
    Kerr
    space-time there is no horizon; for the extreme Kerr space-time
    there is a~single horizon $H:\ r = M$; and for the slow Kerr
    space-time there are \emph{two} horizons $H_+:\ r = r_+$ and
    $H_-:\ r = r_-$. As in the Reissner-Nordsrt\"om geometry the
    surface $r = r_+$ represents an event horizon, and surface $r
    = r_-$ represents a~Cauchy horizon.
    \item On the \emph{ring singularity} $\Sigma:\ \rho^2 = 0$,
   because $\rho^2 = 0$ iff both
    $r = 0$ and $\cos\theta = 0$. Thus $\Sigma$ is  the equatorial circle $\theta = \pi/2$ in $S^2$ at
    radius $r = 0$. Sometimes the circle itself is
    called the ring singularity, with $\Sigma =
    S^1\times\mathbf R^1$ its history through time.
    \emph{Singularity along this ring is the only singularity of
    the Kerr space-time.}
    \item On the \emph{axis} $A:\ \sin\theta = 0$. For spherical
    coordinates, $\sin\theta = 0$ gives the
    $z$-axis, but  now
     there are \emph{two} such axes (enduring through time).
    %In the sphere $S^2$, $\sin\theta = 0$ pick out the north %pole
    %$p:\ \theta = 0$ and the south pole $-p:\ \theta = \pi$, and
    t%hen in the $\mathbf R^2\times S^2$ each pole gives a %complete
    %``$z$-axis" enduring through time.
\end{enumerate}

As in the case of the Reissner-Nordstr\"om space-time, we must
distinguish three regions $A,\ B,\ C$:
\begin{itemize}
    \item For slow Kerr, where there are two horizons $r = r_\pm$,
    \begin{description}
        \item[$A$]: $r < r_-$,
        \item[$B$]: $r_- < r < r_+$,
        \item[$C$]: $r > r_+$.
    \end{description}
    \item For extreme Kerr, where there is a single horizon $r =
    M$,
    \begin{description}
        \item[$A$]: $r < M$,
        \item[$C$]: $ r > M$.
    \end{description}
    \item For fast Kerr , since there is no horizon, $\mathbf
    R^2\times S^2 -\Sigma$ itself can be regarded as a single
    block $A = C$.
\end{itemize}


\section{The general equations of motion}
In this section, we shall consider the reduction of the equations of
geodesic motion, which follows from discoveries of Carter, and
Walker and Penrose (\cite{chandrasekhar}). Alternative derivation of the basic equations follows from separability of the Hamilton-Jacobi equation and can be found in \cite[\S62(a)]{chandrasekhar}.

From the general Lagrangian
%\begin{multline}%\label{kch147}
%  2{\cal L} = \left(1-\frac{2Mr}{\rho^2}\right)\dot t^2+\\
%  +\frac{4aMr\sin^2\theta}{\rho^2}\dot t\dot\varphi
%  -\frac{\rho^2}\varDelta\dot r^2 -\rho^2\dot\theta^2 -\left(r^2
%  +a^2 +\frac{2a^2Mr}{\rho^2}\right) \left(\sin^2\theta\right)\dot\varphi^2
%  \,,
%\end{multline}
we derive the energy and the angular momentum integrals:
\bal
    +p_t &= \left(1 -\frac{2Mr}{\rho^2}\right)\dot t
    +\frac{2aMr\sin^2\theta}{\rho^2}\dot\varphi = E =
    \text{constant}\label{kch148}
    \\
    \intertext{and}
    -p_\varphi &= -\frac{2aMr\sin^2\theta}{\rho^2}\dot t +\left(r^2
    +a^2
    +\frac{2a^2Mr}{\rho^2}\sin^2\theta\right)
    \left(\sin^2\theta\right)\dot\varphi = L = \text{constant}.
    \label{kch149}
\eal
Further, there are the integrals (\cite[chapter 7]{chandrasekhar}) \begin{equation}\label{kch150}
   2\rho^2(\vek k\cdot\vek l)(\vek k\cdot\vek n) -r^2\abs{\vek k}^2 = K
\end{equation}
and
\begin{equation}\label{kch151}
  2\rho^2(\vek k\cdot\vek m)(\vek k\cdot\bar{\vek m}) +   \left(a^2\cos^2\theta\right)\abs{\vek k} = K \,,
\end{equation}
where $\vek l,\ \vek n,\ \vek m$, and $\bar{\vek m}$ are the basis
null vectors, and $\vek k$ is an affinely parametrized geodesic.
We set
\begin{equation}\label{kch152}
    \abs{\vek k}^2 = \delta_1 =
    \begin{cases}
        1& \text{for time-like geodesics},\\
        0& \text{for null geodesics},\\
        -1& \text{for space-like geodesics}.
    \end{cases}
\end{equation}
In the Kerr geometry the integrals \eqref{kch150} and \eqref{kch151} have the explicit forms
\begin{equation}\label{kch154}
  \frac{\left[\varDelta\dot t
  -\left(a\varDelta\sin^2\theta\right)\dot\varphi\right]^2} {\varDelta}
  -\frac{\rho^4}\varDelta\dot r^2 -\delta_1r^2 = K
\end{equation}
and
\begin{equation}\label{kch155}
  \left[(a\sin\theta)\dot t -\left(r^2
  +a^2\right)(\sin\theta)\dot\varphi\right]^2 +\rho^4\dot\theta^2
  +\delta_1a^2\cos^2\theta = K\,.
\end{equation}
Also, by equations \eqref{kch148} and \eqref{kch149}, we can write
\begin{equation}\label{kch156}
  (a\sin\theta)\dot t-\left[\left(r^2
  +a^2\right)\sin\theta\right]\dot\varphi = aE\sin\theta -L\csc\theta
\end{equation}
and
\begin{equation}\label{kch157}
  \varDelta\dot t -\left(a\varDelta\sin^2\theta\right)\dot\varphi =
  \left(r^2 +a^2\right)E -aL\,.
\end{equation}
After inserting these expressions in equations \eqref{kch154} and \eqref{kch155}, equations
\bal\label{kch160}
    \rho^4\dot r^2 &= \text R(r) = \left[\left(r^2 +a^2\right)E
    -aL\right]^2
    -\varDelta\left(\delta_1r^2 +K\right)\\
    \intertext{and}
    \rho^4\dot\theta^2 &= \Theta(\theta) = -\left(aE\sin\theta -L\csc\theta\right)^2 +(-\delta_1a^2\cos^2\theta +K) \,
    \label{kch161}
\eal
are obtained.
Equations \eqref{kch156} and \eqref{kch157} also provide the complementary pair of equations
\begin{equation}\label{kch162}
  \rho^2\dot t = \frac{\varSigma^2E-2aMrL}\varDelta
\end{equation}
and
\begin{equation}\label{kch163}
  \rho^2\dot\varphi = \frac{2aMrE+(\rho^2-2Mr)L\csc^2\theta}\varDelta \,.
\end{equation}


Any function of first-integrals is again a first-integral. A particularly useful case is the first-integral $Q$ defined by the relation
\begin{equation}\label{kch182}
  K = Q +(L -aE)^2 \,.
\end{equation}
Following \cite{onil}, we call $Q$ or $K$, \emph{the Carter constant}. Each form has its advantages. Often $K$ gives simpler formulas, and $K \geq 0$ \emph{holds for nonspace-like geodesics} (by the $\theta$-equation \eqref{kch161}), but the geometric meaning of $Q$ is usually clearer. For example, $Q$ is crucial for the behavior of geodesics near $r = 0$, and $Q$ decisively influences the $\theta$-motion of geodesics.

With the above identification, the equations of motions \eqref{kch160}--\eqref{kch163} are equivalent to
\begin{gather}
    \int^r\frac{\text dr}{\sqrt\text R} = \int^\theta\frac{\text d\theta}{\sqrt\Theta} \,,\label{kch178}\\
    \tau = \int^r\frac{r^2}{\sqrt\text R}\text dr +a^2\int^\theta\frac{\cos^2\theta}{\sqrt\Theta}\text d\theta \,,\label{kch179}\\
    t = \tau E +2M\int^rr\left[r^2E-a(L-aE)\right]\frac{\text dr}{\varDelta\sqrt\text R} \,,\label{kch180}\\
    \intertext{and}
    \varphi= a\int^r\left[\left(r^2+a^2\right)E-aL\right]\frac{\text dr}{\varDelta\sqrt\text R} +\int^\theta\left(L\csc^2\theta-aE\right)\frac{\text d\theta}{\sqrt\Theta} \,.\label{kch181}
\end{gather}


\section{Control of the $r$ coordinate}
In this section, effect of the inequality $R \geq 0$ on the $r$ coordinate of a Kerr geodesic $\gamma$ will be studied. The radial coordinate $r$ determines such critical issues as escape to infinity and passage through a horizon. Substituting\footnote{comp. with \cite[page 208]{onil}} $\varDelta = r^2 -2Mr +a^2$ and $\mathbb P = (r^2 +a^2)E -aL$ into the definition R$(r) = \mathbb P^2 -\varDelta(\delta_1r^2 +K)$ gives fundamental result:
\begin{proposition}\label{4.6.1}
    For a geodesic $\gamma$ with $\delta_1 = \langle \dot\gamma,             \dot\gamma\rangle$, energy $E$, angular momentum $L$, and Carter
    constant $Q$,
    \[\emph R(r) = \left(E^2 -\delta_1\right)r^4 +2M\delta_1r^3     +\mathfrak Xr^2
    +2MKr -a^2Q\,,\]
    where $\mathfrak X = a^2(E^2 -\delta_1) -L^2 -Q$ and $K =     Q
    +(L -aE)^2$\,.
\end{proposition}

%In view of the $r$-equation $\rho^4\dot r^2 =$ R, it is clear that the dependence of the radial motion of Kerr geodesics on the basic first-integrals $\delta_1,\ L,\ E,\ Q$ is expressed by the way they appear in the coefficients of the polynomial R$(r)$.

R$(r)$ is unchanged if the pair $E,\ L$ is replaced by $-E,\ -L$. On the other hand, R$(r)$ \emph{is} changed (provided $E \neq L \neq 0$) if only \emph{one} of the two parameters $E,\ L$ changes sign, because $K = Q +L^2 -2aLE +a^2E^2$. Therefore, in Kerr space-time the direct and retrograde orbits with the same $\delta_1,\ E,\ K$ and $\abs L$ generally vary in their radial motion (except in the special cases $E = 0$ or $L = aE$).

  The polynomial R$(r)$ of a Kerr geodesic $\gamma$ is determined by the first-integrals $\delta_1,\ E,\ L,\ Q$.%, and the radial coordinate $r$


\section{$r$-$L$ plots}\label{section 4.7}
In this section we follow \cite{onil}, who concentrates on time-like geodesics. We expand his discussion to null geodesics (which are analogous and are usually easier to deal with), and space-like geodesics (which have more varied trajectories).

If the Kerr parameters $M,\ a$ and the causal parameter $\delta_1$ are fixed then the function R$(r)$ of Proposition \ref{4.6.1} depends on $E,\ L,\ Q$. Since the resulting $r$-line cannot enter the region R $<$ 0, the set $Z$: R = 0 in $\textbf R^4$ is the key as it provides the boundary points of R $<$ 0. However, it is possible to show the qualitative character of orbits by considering R only as a~function of the variables $r$ and $L$--for a few selected constant values of $E$ and $Q$.
The region of $Z$ in the $r$-$L$ plane is said to be \emph{allowed} if R $>$ 0, and \emph{forbidden} if R $<$ 0. A~diagram of this situation is called the $r$-$L$ plot.

The set $Z$: R = 0 is usually a smooth curve.

The abbreviation $e = E^2 -\delta_1$ is often convenient. This
first-integral is called \emph{effective energy}. It could also be
called the \emph{escape parameter} (\cite{onil}). (In the
literature the notation $\Gamma$ has also been used.)

Initially some properties of $r$-$L$ plots that are independent of causal character will be examined.

In some contexts it is useful to regard the function in Proposition\ref{4.6.1}  as a polynomial R$(L)$ in $L$, with coefficients determined by the remaining first-integrals and $r$. Specifically,
\begin{equation}\label{ko215}
  \text R(L) = r(2M-r)L^2-4MaErL+\text R(r,0)\,,
\end{equation}
where R$(r,0)$ is the function R with $L = 0$ ($r$-axis).

Using \eqref{ko215} explicit formulas for the implicitly defined set R = 0 can be derived.
\begin{lemma}\label{4.7.1}
    In any
    $r$-$L$ plot the set \emph R = 0 is the union of the graphs of (the real     values
    of) the two functions
    \[ L_\pm(r) = \frac{2MaEr\pm\sqrt{\emph D(r)}}{r(2M-r)}\, ,\]
    for $r \neq 0,\ \neq 2M$, where \emph D$(r) = r\varDelta(r)\phi(r)$ is one     fourth of
    the discriminant of \emph R$(L)$, and the discriminant polynomial $\phi(r)$     is
    \[ \phi(r) = \left(E^2-\delta_1\right)r^3+2M\delta_1r^2-Qr+2MQ\,.\]
\end{lemma}
For a proof of this lemma see \cite[page 215]{onil}.

The polynomial R$(L)$ is quadratic in $L$ (except when $r = 0$ and $r = 2M$), so its graph is parabolic with \emph{vertex} at $L_\text v(r) = 2MaE/(2M-r)$. The number $\mu(r)$ of points on any vertical line  is 0, 1, or 2 as the modified discriminant D$(r)$ is negative, zero or positive, respectively. When $\mu(r) = 2$ the two points are placed symmetrically on both sides of $L_\text v(r)$. The curve $r \rightarrow (r,L_\text v(r))$ in an $r$-$L$ plot is called the \emph{midline}. %This curve is a rectangular hyperbola with asymptotes the $r$-axis and the line $r = 2M$, and (for $E > 0$) is above the $r$-axis for $r < 2M$, below it for $r > 2M$.

When $\mu(r) = 1$, there is just one point of R = 0 on the line $L \rightarrow (r,L)$, the vertex $(r, L_\text v(r))$.

The formula \eqref{ko215} shows that when $r = 0$, R has the constant value $-a^2Q$, and that the behavior of R on and near the line $r = 2M$ is following:
\begin{lemma}
    In an $r$-$L$ plot the function $L \rightarrow$ \emph{R}$(2M,L)$ is linear     with
    negative slope (if $E > 0$). Thus there is a critical momentum $L_0$     such
    that \emph{R}$(2M,L) < 0$ if $L > L_0$ and\:\footnote{comp. \cite[lemma
    4.7.2]{onil}} \emph{R}$(2M,L) > 0$ if $L < L_0$.
    Furthermore, if $0 < \delta \ll M$, then \emph{R}$(2M+\delta) < 0$ and
    \emph{R}$(2M-\delta) > 0$ for all sufficiently large $\abs L$.
\end{lemma}
Proof of this lemma is in \cite[page 216]{onil}.

Investigation of $r$-$L$ plots frequently comes in vertical strips:
The $B$ strip $r_- < r < r_+$ is particularly simple for nonspacelike
geodesics; because they cannot turn in $B$, R is never zero there.
\begin{lemma}
    In the $r$-$L$ plot of a time-like (or null) geodesic $\gamma$ in slow
    Kerr space-time, we have \emph R $> 0$ on the closed vertical strip $r_- \leq r
    \leq r_+$ \emph{except} that \emph R $= 0$ at the two points $(r_-, L_-)$ and
    $(r_+, L_+)$, where
    \begin{equation}\label{13.4}
        L_\pm = L_\text v(r_\pm) = 2MaE/(2M-r_\pm) = 2MaE/r_\mp =
        2Mr_\pm E/a \,.
    \end{equation}
At these points the curve \emph R $= 0$ is smooth and tangent
(from outside the strip) to the lines $r = r_-$ and $r = r_+$.
\end{lemma}
\begin{proof}
    For any geodesic, the formula R $= \mathbb P^2 -\varDelta(\delta_1r^2
    +K)$ yields\footnote{cf. \cite[page 217]{onil}} R$(r_\pm) = \mathbb P(r_\pm)^2$, where $\mathbb P(r_\pm) =
    (r_\pm^2+a^2)E-La$. So R$(r_\pm) > 0$ except when $\mathbb P(r_\pm)
    = 0$, and that is at $L_\pm$. The other formulas follow because $r_-+r_+ =
    2M$  and $r_-r_+ = a^2$.

    Since R is never zero on the $B$ strip, therefore R $>$ 0 there.

     Because $L_\pm$ is a minimum point on $r = r_\pm$, then $\frac{\partial\text R}{\partial L} = 0$  at the points $(r_\pm,
    L_\pm)$.

    To show that $\frac{\partial\text R}{\partial r} \neq 0$ at
    these point we write
    \begin{equation}\label{p4.7.3}
        \frac{\partial\text R}{\partial r} = 2\mathbb
        P\frac{\partial\mathbb P}{\partial r} - 2(r
        -M)\left(\delta_1^2 +K\right) -2\delta_1r\varDelta \,,
    \end{equation}
    which at $(r_\pm,L_\pm)$  becomes $-2(r_\pm -M)(\delta_1r^2
    +K) \neq 0$. This inequality holds for both $\delta_1 = 0$ and
    $\delta_1 = 1$ (null and time-like geodesics respectively).
\end{proof}
\begin{remark}
    A crucial effect of the Carter constant $Q$  on (time-like) orbits
    derives from the
    fact that the function \emph R, which depends quadratically on $L$, depends
    \emph{linearly}  on $Q$. In fact, the formulas \emph R $= \mathbb P^2
    -\varDelta(r^2+K)$ and $K = Q+(L-aE)^2$ give \emph R$(r) = -\varDelta(r)Q
    +\text{\emph R}_0(r)$, where $\text{\emph R}_0(r)$ is obtained by setting $Q = 0$.
    (Thus $\text{\emph R}_0(r)$ governs the $r$-equation for geodesics in the
    equator.) It follows that when the other first-integrals are held constant,
    \emph{as $Q$ increases, forbidden regions expand}. In fact, the strip
    $r_- \leq r  \leq r_+$, where $\varDelta \leq 0$, is always in
    a~allowed\:\footnote{cf. \cite[Remark 4.7.7]{onil}}
    region; so elsewhere, as $Q$ increases, \emph R$(r)$ decreases.
\end{remark}

\begin{lemma}
    For time-like geodesics, $\phi(r) = er^3 +2Mr^2 -Qr +2MQ$,
    where $e = E^2 -1$.
    \begin{enumerate}
        \item If $Q > 0$, then $\phi$ has no roots in $0 \leq r
        \leq 2M$. If also $e > 0$, then $\phi$ has exactly one negative root,
        hence either zero or two positive roots. If $e < 0$, $\phi$ has no
        negative roots, hence either one or three positive
        roots\footnote{\color{blue}\cite[lemma 4.7.5]{onil} has the following statement in
        addition to the above: For $e =
        0$, $\phi$ has no real roots. That is not true, because if $e =
        0$ then we have the polynomial $2Mr^2-Qr+2MQ$, which has discriminant
        $Q(Q-16M^2)$. That is negative only when $Q < 16M^2$.}.
        \item If $Q < 0$, so $e > 0$ then $\phi$ has a unique root
        in $0 < r <2M$, and either zero or two roots in $r < 0$.
    \end{enumerate}
\end{lemma}
Proof of this lemma can be found in \cite[page 218]{onil}.
\begin{lemma}\color{blue}
    For null geodesics, $\phi(r) = er^3 -Qr +2MQ$,
    where $e = E^2$.
    \begin{enumerate}
        \item If $Q > 0$, then $\phi$ has no roots in $0 \leq r \leq
        2M$. If also $e > 0$, then $\phi$ has exactly one negative
        root, hence either zero or two positive roots. For $e =
        0$, $\phi$ has one real positive root $r = 2M$.
        \item As for time-like geodesics.
    \end{enumerate}
\end{lemma}
\begin{proof}
    \begin{enumerate}
        \item For $e > 0$, proof is the same as for time-like
        geodesics. \textcolor{blue}{For $e = 0$ direct solution of the equation $-Qr+2MQ = 0$
        gives $2M$.}
        \item As for time-like geodesics.
    \end{enumerate}
\end{proof}
\begin{lemma}\color{blue}
    For space-like geodesics $\phi(r) = er^3-2Mr^2-Qr+2MQ$, where
    $e = E^2 +1$, $\phi$ has exactly one negative root, hence either zero or two positive roots.
\end{lemma}
\begin{proof}\color{blue}
    For $r < 0$ the signs for Descartes' rule are - - + +, so there is a unique negative root.
\end{proof}

The ambiguities 0/2 and 1/3 in the preceding lemmas can be settled
precisely. The discriminant $d$ of a real cubic polynomial determines
the number of its real roots: one if $d < 0$, three if $d > 0$,
and repeated roots if $d = 0$.

For the time-like geodesics we find the discriminant of the polynomial
$\phi$ to be $d = d(e, Q) = \discrim\phi(r) = -4Q\text f(e, Q)$,
where
\begin{equation}\label{onil218f}
    \text f(e, Q) = -eQ^2+M^2(27e^2+18e-1)Q+16M^4\,.
\end{equation}
The regions in the $e$-$Q$ plane where $d
> 0,\ d = 0,$ or $\ d < 0,$ have to be found. Let $\beta = \beta(e) = 27e^2
+18e -1$. Then f = 0 for
\begin{equation}\label{onil218Q}
    Q_\pm(e) = \frac{M^2\left(\beta \pm\sqrt{\beta^2
    +64e}\right)}{2e}\,.
\end{equation}

The functions \eqref{onil218Q} are real  only when $\beta^2 +64e \geq 0$, which always holds if $e > 0$. The factorization
\begin{equation}\label{onil218fac}
    \beta^2 +64e = \left(27e^2 +18e -1\right)^2 +64e = (e+1)\left(9e+1\right)^3
\end{equation}
demonstrates that $\beta^2 +64e > 0$ iff $e > -1/9$. The zero at $e = -1$ can be neglected since $Q < 0$ there (\cite[chapter 4]{onil}). Thus the explicit formula emerges:
\begin{equation}\label{onil219Q}
  Q_\pm(e) = \frac{M^2[27e^2 +18e -1 \pm(e +1)^{1/2}(9e +1)^{3/2}]}{2e}\,.
\end{equation}

Finally, the functions $Q_-(e),\ Q_+(e)$ on $e \geq -1/9$ will be examined. They have the same value  at $ e = -1/9$ ($12M^2$), but thereafter they differ considerably. For $Q_+$, solving the equation f = 0, where f is introduced by the equation \eqref{onil218f}, we define $Q_+(0) = 16M^2$. For $e \gg 1$, $Q_+(e) \sim \beta(e)/e \sim M^2(27e +18)$, so $Q_+$ becomes almost linear (see figure \ref{4.8}).

However, $Q_-$ approaches $+\infty$ as $e \rightarrow 0$ from the left. This is clear from\footnote{cf. \cite[p.~219]{onil}}
\begin{equation}\label{onil219pul}
  (Q_+ +Q_-)/2 = \frac12\frac{M^2(27e^2+18e-1)}e\,.
\end{equation}
since $Q_+$ is bounded near $e = 0$. For $e < 0$, $Q_-(e) > Q_+(e)$.
%For example, with $M = 1$, at $e = -0.01$ we find $Q_+(-.01) \approx 15.7$  but $Q_-(-.01) > 100$.

$Q_-(e)$ is always negative for $e > 0$, approaching $-\infty$ when $e \rightarrow 0$, and approaching 0 when $e \rightarrow +\infty$.

\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/r-Ltimel2.eps}
         \caption{The          sign of $d(e, Q) =
         \discrim(\phi, r)$ on the relevant portion of the $e$-$Q$ plane for the time-like             geodesics is showed on this $e$-$Q$ \emph{chart}.
         If          $d(e,
         Q) < 0$, then in the $r$-$L$ plot determined by $(e, Q)$, the curve R
         = 0 has only one vertical tangent in addition to those at $(r_\pm,          L_\pm)$.
         When $d > 0$ there are three such points. Repeated roots occur when $d =
         0$. (In this plot $a = \sqrt{19}$.)}
         \label{4.8}
\end{figure}

 There are three restrictions for a point of the $e$-$Q$ plane to correspond to an $r$-$L$ plot:
\begin{enumerate}
  \item $e \geq -1$, because $e = E^2 -1$,
  \item $Q > 0$ if $e < 0$, because vortical geodesics have escape energy,
  \item $Q \geq -a^2e$ if $e > 0$, by Carter's inequality (see \cite[chapter 4]{onil}).
\end{enumerate}
Therefore in figure \ref{4.8} the third quadrant is irrelevant, and so is the area in the fourth quadrant below the line $Q = -a^2e$.

For the null geodesics the discriminant of the polynomial
$\phi$ is $d = d(e, Q) = \discrim\phi(r) = -4eQ^2\text f(e, Q)$, where
\begin{equation}\label{onil218fnull}\color{blue}
  \text f(e, Q) = -Q+27M^2e\,.
\end{equation}
Again, we will find the region in the $e$-$Q$ plane where $d > 0,\ d = 0$ and $d < 0$. The solution of equation f = 0 is
\begin{equation}\label{onil218Qnull}\color{blue}
  Q = 27M^2e \,,
\end{equation}
which is a real-valued linear function (see figure \ref{4.8null}).

\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/r-Lnull2.eps}
         \caption{\textcolor{blue}{This $e$-$Q$ \emph{chart} shows the          sign of $d(e, Q) =
         \discrim(\phi, r)$ on the relevant portion of the $e$-$Q$ plane for the null geodesics.}
         When          $d(e,
         Q) < 0$, then in the $r$-$L$ plot determined by $(e, Q)$ the curve R
         = 0 has only one vertical tangent except those at $(r_\pm,          L_\pm)$.
         When $d > 0$ there are three such points. Repeated roots occur when $d =
         0$. (In this plot $a = \sqrt{19}$.)}
         \label{4.8null}
\end{figure}
There are two restrictions or the point of the $e$-$Q$ plane to correspond to an $r$-$L$ plot:
\begin{enumerate}\color{blue}
  \item $e \geq 0$, since $e = E^2$; and
  \item $Q \geq -a^2e$, by Carter's inequality.
\end{enumerate}
Thus in figure \ref{4.8null} the region in fourth quadrant below the line $Q = -a^2e$ is irrelevant.

\textcolor{blue}{For the space-like geodesics the discriminant of the polynomial $\phi$ is, as for time-like geodesics, $d = d(e, Q) = \discrim(\phi r) = -4Q\text f(e, Q)$, but this time
\begin{equation}\label{onil418fspace}
  \text f(e, Q) = -eQ^2+M^2(27e^2-18e-1)Q-16M^4\,.
\end{equation}
As before, our goal is to find the regions in the $e$-$Q$ plane where $d > 0,\ d = 0,\ d < 0$, respectively. This time, let $\beta = 27e^2-18e-1$. Again we solve the equation f = 0. In this case, the solution is (cf. equation \eqref{onil218Q})
\begin{equation}\label{onil218Qspace}
  Q_\pm(e) = \frac{M^2\left(\beta \pm\sqrt{\beta^2-64e}\right)}{2e}\,.
\end{equation}
These functions are real only when $\beta^2 -64e \geq 0$. The factorization \begin{equation}\label{onil218fakspace}
  \beta^2-64e = \left(27e^2-18e-1\right)^2 -64e = (e-1)\left(9e-1\right)^3
\end{equation}
shows that $\beta^2-64e > 0 \Leftrightarrow e > 1$. Thus we get the explicit formula: \begin{equation}\label{onil219Qspace}
  Q_\pm(e) = \frac{M^2[27e^2 -18e -1 \pm(e -1)^{1/2}(9e -1)^{3/2}]}{2e}\,.
\end{equation}}

\textcolor{blue}{At $e = 1$ the functions $Q_-(e),\ Q_+(e)$ have the same value, $4M^2$, but thereafter are quite different. For $e \gg 1$, $Q_+(e) \sim \beta(e)/e \sim M^2(27e -18)$, so as figure \ref{4.8space} shows, $Q_+$ becomes almost linear.}

\textcolor{blue}{For $e \geq 1,\ Q_-(e)$ is always positive, approaching 0 as $e \rightarrow \infty$. (Note that $Q_-(e) > Q_+(e)$ for $e > 1$.)}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/r-Lspace2.eps}
         \caption{\textcolor{blue}{This $e$-$Q$ \emph{chart} shows the          sign of $d(e, Q) =
         \discrim(\phi, r)$ on the relevant portion of the $e$-$Q$ plane for the space-like geodesics.}
         When          $d(e,
         Q) < 0$, then in the $r$-$L$ plot determined by $(e, Q)$ the curve R
         = 0 has only one vertical tangent apart from those at $(r_\pm,          L_\pm)$.
         There are three such points for $d > 0$ . Repeated roots occur when $d =
         0$. % in region 0 < Q < Q_minus d > 0
         (In this plot $a = \sqrt{19}$.)}
         \label{4.8space}
\end{figure}

For the point of the $e$-$Q$ plane to correspond to an $r$-$L$ plot the following inequalities must hold:
\begin{enumerate}\color{blue}
  \item $e \geq 1$, since $e = E^2 +1$, and
  \item $Q \geq -a^2e$, by Carter's inequality.
\end{enumerate}
Thus, as before, the area in the fourth quadrant below the line $Q = -a^2e$ is irrelevant, as is the region $e < 1$.

\begin{remark}\label{4.7.6}
    Summary: The number of zero the discriminant polynomial $\phi(r) =
    er^3 +2M\delta_1r^2 -Qr +2MQ$ has determines the qualitative character of $r$-$L$. For given
    $e$ and $Q$, the sign of a second discriminant, $d(e,
    Q) = \discrim(\phi, r)$ determines this number. Therefore $d$ becomes a function on the $e$-$Q$ chart in figures
    \ref{4.8}, \ref{4.8null} and \ref{4.8space} (for time-like, null and space-like geodesics
    respectively). The regions $d > 0$ and $d < 0$ in the chart are separated by graphs of the functions
    \bal
        Q_\pm& = \frac{M^2}{2e}\left[27e^2 +18e -1 \pm\left(e +1\right)^{1/2}(9e
        +1)^{3/2}\right]                                && \text{for time-like geodesics,}\\
        \color{blue}Q&\color{blue} = 27M^2e             && \text{for null geodesics,}\\
        \color{blue}Q_\pm&\color{blue} = \frac{M^2}{2e}\left[27e^2 -18e -1 \pm\left(e
        -1\right)^{1/2}\left(9e -1\right)^{3/2}\right]  && \text{for space-like geodesics.}
    \eal

Hence, for the time-like geodesics, if $e > 0$, the negative root of $\phi$ is its only root when $0 < Q < Q_\pm$, but when $Q > Q_\pm$, then $\phi$ has three roots. These roots are all simple, so they represent the vertical tangents in the $r$-$L$ plot determined by $e$ and $Q$. Similarly, if $e < 0$, the negative root of $\phi$ is its only root unless both $-1/9 < e \leq 0$ and $Q_+ < Q < Q_-$ hold\footnote{cf. \cite[Remark 4.7.6]{onil}}, giving three roots. \color{blue} If $Q < 0$ the conditions giving three roots of $\phi$ are $Q > -a^2e$ and $Q > Q_-$.

For the null geodesics, we can see from the chart that when $-a^2e < Q < 27M^2e$ the negative root of $\phi$ is its only root,  but if $Q > 27M^2e$, then $\phi$ has three roots.

For the space-like geodesics, $\phi$ has three roots when $0 < Q < Q_-$ or $Q > Q_+$ otherwise, that is for $-a^2e < Q < 0$ and for $Q_- < Q < Q_+$, the negative root is its only root.
\end{remark}

\begin{lemma}\label{4.7.5}
    For $e \neq 0$ and $Q > 0$, let $L_0 = L_\text{\emph v}(r_0)$, where $r_0$ is a root of
    $\phi(r)$. Then:
    \begin{enumerate}
      \item \emph R = 0 has a vertical tangent at $(r_0, L_0) \Leftrightarrow r_0$ is a simple
      root of $\phi(r)$; and
      \item $(r_0, L_0)$ is a critical point of the function \emph R$(r, L) \Leftrightarrow
      r_0$ is a repeated root of $\phi(r)$. If $\phi''(r_0) > 0$, then $(r_0, L_0)$ is a saddle
      point, with adjacent allowed regions along the midline, and if $\phi''(r_0) < 0$, then
      $(r_0, L_0)$ is an isolated maximum point, surrounded by a forbidden region.
    \end{enumerate}
\end{lemma}
For the proof of this lemma see \cite[section 4.7]{onil}.

\begin{proposition}\label{4.7.8}
    For $Q > 0$, consider the $r$-$L$ plot of $(e, Q)$ where $d(e, Q) = 0$ (so
    $e \geq -1/9$ and $ Q = Q_\pm(e) > 0$). Excluding $e = -1/9$
    and\footnote{\cite[proposition 4.7.8]{onil} does not exclude $e = 0$.} $e =
    0$,
    \emph R $= 0$
    has
    a~unique vertical tangent point $(r_1, L_1)$ with $r_1 \neq \text{\emph
    R}_\pm$ and
    a~unique critical point $(r_2, L_2)$, both necessary on the midline, so
    $L_i = L_\text{\emph v}(r_i)$. There are three cases:
    \begin{enumerate}
      \item If $e > 0$ and $Q = Q_+(e)$, then $r_1 < 0 < 2M < r_2$. Further,
        $(r_2, L_2)$ is a saddle point of the function \emph R$(r, L)$, with
        adjacent allowed regions along the midline.
      \item If $-1/9 < e < 0$ and $Q = Q_+(e)$, then $2M < r_2 < r_1$, in fact,
        \begin{equation}
          2M < r_2 < \frac{2M}{3\abs e} < r_1 < \frac{2M}{\abs e}\,,
        \end{equation}
        and $(r_2, L_2)$ is a saddle point, as above.
      \item If $-1/9 < e < 0$ and $Q = Q_-(e)$, then $2M < r_1 < r_2$, in fact,
        \begin{equation}\label{onil222prop}
          2M < r_1 < \frac{2M}{3\abs e} < r_2 < \frac{2M}{\abs e}\,,
        \end{equation}
        and $(r_2, L_2)$ is an isolated \emph R $= 0$ point in a forbidden
        region.
    \end{enumerate}
\end{proposition}
\begin{proof}
    See \cite[page 222]{onil}.
    %\begin{enumerate}
    %  \item For $e > 0$ see proof in \cite[page 222]{onil}.
    %  \item For $e < 0$, lemma \ref{4.7.5} asserts that there are no negative
    %    roots, but there are positive roots of total multiplicity one or three.
    %    The former can be excluded since $d = 0$. Since $e = -1/9$ is excluded,
    %    for multiplicity three there must be two roots, say, $r_1$ simple and
    %    $r_2$ double. \textcolor{blue}{Because $\phi''(r_2) > 0$, $r_2$ is a
    %    saddle point, and so
    %    $r_2 < \frac{2M}{3\abs e}$ holds, too. Equating coefficients in the
    %    equation $\phi(r) = e (r -r_1)(r -r_2)^2$ gives
    %    \begin{equation}\label{gproof1}
    %        r_1 +2r_2 = \frac{2M}{\abs e}\,, \qquad r_2(2r_1 +r_2) = \frac
    %        Qe\,, \qquad r_1r_2^2 = \frac{2MQ}{\abs e}\,,
    %    \end{equation}
    %    from which follows that $r_1 < \frac{2M}{\abs e}$.}
    %  \item For
    %\end{enumerate}
\end{proof}

    To find the critical points $(r_2, L_2)$ in proposition \ref{4.7.8} we know that $r_2$ is a solution of both quadratic equations
\begin{equation}\label{onil222r2}
  Mr^2 -Qr +3MQ = 0\,, \qquad \text{and} \qquad er^2 -M(3e-1)r -4M^2 = 0 \,,
\end{equation}
which come from eliminating $e$ and $Q$, respectively, from the equations $\phi(r) = 0,\ \phi'(r) = 0$. Then, by the general midline formula \eqref{13.4}, $L_2 = L_\text v(r_2) = 2MaE/(2M-r_2)$.

\textcolor{blue}{For $e = 1$ and $d = 0$ we have a double root at $r = 4M$, which is a saddle point, since $\phi''(4M) > 0$.}


\section{First-integrals and orbits}
The most important question about a time-like geodesic is whether or not it has escape energy (determined by the sign of $e = E^2 -1$). Next (also for a~null and space-like geodesic) is the sign of the Carter constant $Q$. This section deals with the ``standard" case $Q > 0$. Such geodesics do not reach $r = 0$ and have $\theta$ motion oscillating about $\pi/2$. %Both these properties fail in the special cases $Q < 0$ and $Q = 0$.

Depending on shape of allowed and forbidden regions for slow black holes we distinguish four cases. ``Continents" configuration has two large forbidden areas  --  one for direct and one for retrograde orbits. In ``barrier" configuration these two ``continents" are joined and form a barrier preventing geodesics to fall in a black hole. ``Bay" configuration has an allowed region in a shape of bay, and in ``lake" configuration this bay is reduced to a ``lake" -- an  allowed region surrounded by a forbidden area.

\subsection{Continents configuration}\label{continets}
Perhaps the most common is case of escape energy $e > 0$ and moderate Carter constant $0 < Q < Q_+(e)$ (\cite[section 4.8]{onil}), where $Q_+(e)$ are given precisely by the formulas in Remark \ref{4.7.6}. See figures \ref{4.9time} and \ref{4.9timef} for $r$-$L$ plots for the time-like geodesics, figures \ref{4.9null} and \ref{4.9nullf} for the null geodesics, and \ref{4.9space}, \ref{4.9spacef}, \ref{4.9space1} and \ref{4.9space1f} for the space-like geodesics ($Q <Q_+(e)$, $Q < Q_-(e)$). These plots are distinguished by two ``continents" on the $r > 0$ side. There the situation is similar to Newtonian gravitation. Particles falling from $r = +\infty$ have flyby orbit if their angular momentum $\abs L$ is large, those with smaller $\abs L$ continue through the horizons.

In the strip $0 < r < 2M$ there are many bound orbits, both direct and retrograde. No particle, no matter how high is its energy, can reach $r = 0$ in $r > 0$ region.
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/t-orbits.eps}
         \caption{Plot for a time-like geodesic with escape energy $e = 0.4$ and Carter constant $Q = 1$. Forbidden regions are the continents $W,\ N$ and $S$. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that angular momentum. Most particles infalling from $r = +\infty$ have flyby orbit. (Additional parameters: $M = 1,\ a^2 = 0.84$.)}
         \label{4.9time}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/t-orbitsf.eps}
         \caption{Plot for a time-like geodesic with escape energy $e = 0.4$ and Carter constant $Q = 1$. \textcolor{blue}{In a fast black hole forbidden regions $W$ and $N$ are joined (there is only one vertical tangent--in $r<0$ region, see lemma \ref{4.7.5}).} On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that angular momentum. Most particles infalling from $r = +\infty$ have flyby orbit. (Additional parameters: $\color{blue}M = 1,\ a^2 = 19$.)}
         \label{4.9timef}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/n-orbits.eps}
         \caption{\textcolor{blue}{Continents configuration: an $r$-$L$ plot for a null geodesic} with escape energy $e = 0.4$ and Carter constant $Q = 1$. Forbidden regions are the continents $W,\ N$ and $S$. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that given angular momentum. Most particles infalling from $r = +\infty$ have flyby orbit. (Additional parameters: $M = 1,\ a^2 = 0.84$.)}
         \label{4.9null}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/n-orbitsf.eps}
         \caption{\textcolor{blue}{Continents configuration: an $r$-$L$ plot for a null geodesic} with escape energy $e = 0.4$ and Carter constant $Q = 1$. \textcolor{blue}{The continents $W,\ N$ are joined and together with continent $S$ are forbidden for a photon to enter.} On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that given angular momentum. Most particles infalling from $r = +\infty$ have flyby orbit. (Additional parameters: $\color{blue}M = 1,\ a^2 = 19$.)}
         \label{4.9nullf}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/s-orbits.eps}
         \caption{\textcolor{blue}{Continents configuration: an $r$-$L$ plot for a space-like geodesic with escape energy $e = 1.04$ and Carter constant $Q = 5$.} Forbidden regions are the continents $W,\ N$ and $S$. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that angular momentum. Most particles infalling from $r = +\infty$ have flyby orbit. (Additional parameters: $M = 1,\ a^2 = 0.84$.)}
         \label{4.9space}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/s-orbitsf.eps}
         \caption{\textcolor{blue}{Continents configuration: an $r$-$L$ plot for a space-like geodesic with escape energy $e = 1.04$ and Carter constant $Q = 5$. In a fast black hole forbidden regions $W$ and $N$ are joined.} On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a~geodesics with that given angular momentum. Most particles infalling from $r = +\infty$ have flyby orbit. (Additional parameters: $\color{blue}M = 1,\ a^2 = 19$.)}
         \label{4.9spacef}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/s-orbits1.eps}
         \caption{\textcolor{blue}{Continents configuration: an $r$-$L$ plot for a space-like geodesic with escape energy $e = 1.04$ and Carter constant $Q = 1$.} Forbidden regions are the continents $W,\ N$, $S$ and the oval $U$. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that angular momentum. Most particles infalling from $r = +\infty$ have flyby orbit. (Additional parameters: $M = 1,\ a^2 = 0.84$.)}
         \label{4.9space1}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/s-orbits1f.eps}
         \caption{\textcolor{blue}{Continents configuration: an $r$-$L$ plot for a space-like geodesic with escape energy $e = 1.04$ and Carter constant $Q = 1$. In a fast black hole oval $U$ disappears.} On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that given angular momentum. Most particles infalling from $r = +\infty$ have flyby orbit. (Additional parameters: $\color{blue}M = 1,\ a^2 = 19$.)}
         \label{4.9space1f}
\end{figure}


\subsection{Barrier configuration}
As $Q$ increases from small values, the continents $N$ and $S$ become larger. At the critical point of the function R$(r, L)$ they meet (\cite[section 4.8]{onil}) and then they merge into a~single \emph{barrier} as shown in figures \ref{4.11time}, \ref{4.11timef}, \ref{4.11null}, \ref{4.11nullf}, \ref{4.11space} and \ref{4.11spacef}. Hence the large values of the Carter constant $Q$ provide the complete protection against a~crush into a black hole, no matter how small angular momentum $L$ may be.
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/t-orbits2.eps}
         \caption{Barrier configuration: an $r$-$L$ plot for a time-like geodesic with escape energy $e = 0.4$ and Carter constant $Q = 28$. Regions $N$ and $S$ merge, forming a barrier between $r = +\infty$ and the horizons. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that angular momentum. All particles infalling from $r = +\infty$ have flyby orbit. (Additional parameters: $M = 1,\ a^2 = 0.84$.)}
         \label{4.11time}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/t-orbits2f.eps}
         \caption{Barrier configuration: an $r$-$L$ plot for a time-like geodesic with escape energy $e = 0.4$ and Carter constant $Q = 28$. \textcolor{blue}{Regions $W$, $N$ and $S$ merge.} On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that particular angular momentum. All particles infalling from $r = +\infty$ have flyby orbits. (Additional parameters: $\color{blue}M = 1,\ a^2 = 19$.)}
         \label{4.11timef}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/n-orbits2.eps}
         \caption{\textcolor{blue}{Barrier configuration: an $r$-$L$ plot for a null geodesic} with escape energy $e = 0.4$ and Carter constant $Q = 28$. Regions $N$ and $S$ merge, forming a barrier between $r = +\infty$ and the horizons. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that particular angular momentum. All particles falling from $r = +\infty$ have flyby orbits. (Additional parameters: $M = 1,\ a^2 = 0.84$.)}
         \label{4.11null}
\end{figure}\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/n-orbits2f.eps}
         \caption{\textcolor{blue}{Barrier configuration: an $r$-$L$ plot for a null geodesic} with escape energy $e = 0.4$ and Carter constant $Q = 28$. \textcolor{blue}{In a fast black hole regions $N$, $W$ and $S$ merge.} On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that particular angular momentum. All particles falling from $r = +\infty$ have flyby orbit. (Additional parameters: $\color{blue}M = 1,\ a^2 = 0.84$.)}
         \label{4.11nullf}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/s-orbits2.eps}
         \caption{\textcolor{blue}{Barrier configuration: an $r$-$L$ plot for a space-like geodesic with escape energy $e = 1.04$ and Carter constant $Q = 28$.} Regions $N$ and $S$ merge, forming a barrier between $r = +\infty$ and the horizons. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that particular angular momentum. All particles falling from $r = +\infty$ have flyby orbits. (Additional parameters: $M = 1,\ a^2 = 0.84$.)}
         \label{4.11space}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/s-orbits2f.eps}
         \caption{\textcolor{blue}{Barrier configuration: an $r$-$L$ plot for a space-like geodesic with escape energy $e = 1.04$ and Carter constant $Q = 28$. Regions $N$, $W$ and $S$ merge.} On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that particular angular momentum. All particles falling from $r = +\infty$ have flyby orbits. (Additional parameters: $\color{blue}M = 1,\ a^2 = 19$.)}
         \label{4.11spacef}
\end{figure}


\subsection{Bay configuration}
If a particle has less than escape energy, all orbits are bound and none meet $r < 0$ (\cite[section 4.8]{onil}). Figures \ref{4.12} and \ref{4.12f} show $r$-$L$ plots for this case.

Comparison with figures \ref{4.9time} and \ref{4.9timef} shows that for the same value of $Q$, the decrease of energy from $e > 0$ to $e < 0$ (that is, from $E > 1$ to $E < 1$) has enclosed the large allowed region on the $r > 0 $ side.

For fixed $Q$, there are both retrograde and direct orbits in the exterior when $e < 0$ is near zero, but for $e$ nearer $-1$ all bound orbits disappear. (See \cite[section 4.8]{onil}.)
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/t-orbits3.eps}
         \caption{Bay configuration: an $r$-$L$ plot for a time-like geodesic with escape energy $e = -0.2$ and Carter constant $Q = 1$. Forbidden regions are those with $R < 0$. For $r < r_-$ the plot still resembles the $e > 0$ case in figure \ref{4.9time}. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that angular momentum. All orbits are bound. (Additional parameters: $M = 1,\ a^2 = 0.84$.)}
         \label{4.12}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/t-orbits3f.eps}
         \caption{Bay configuration: an $r$-$L$ plot for a time-like geodesic with escape energy $e = -0.2$ and Carter constant $Q = 1$. Forbidden regions are those with $R < 0$. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that particular angular momentum. All orbits are bound. (Additional parameters: $\color{blue}M = 1,\ a^2 = 19$.)}
         \label{4.12f}
\end{figure}


\subsection{Lake configuration}
For $-1/9 < e < 0$ and $Q_+(e) \leq Q \leq Q_-(e)$, forbidden regions enlarge with increasing $Q$ (\cite[section 4.8]{onil}). When $Q = Q_+(e)$ they meet (at a saddle point of R$(r, L)$, see the second assertion of Proposition \ref{4.7.8}) and thereafter they merge, forming a~``lake" as shown in figures \ref{4.13} and \ref{4.13f} .
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/t-orbits4.eps}
         \caption{Lake configuration: an $r$-$L$ plot for a~time-like geodesic with escape energy $e = -0.08$ and Carter constant $Q = 14.2$. Forbidden regions are those with $R < 0$. For $r < r_-$ the plot still resembles the $e > 0$ case in figure \ref{4.9time}. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a~geodesics with that particular angular momentum. All orbits are bound. (Additional parameters: $M = 1,\ a^2 = 0.84$.)}
         \label{4.13}
\end{figure}
\begin{figure}
         \epsfxsize=13cm
         \epsfbox{kerr/t-orbits4f.eps}
         \caption{Lake configuration: an $r$-$L$ plot for a~time-like geodesic with escape energy $e = -0.08$ and Carter constant $Q = 14.2$. Forbidden regions are those with $R < 0$. On each horizontal line $L =$ const, the maximal segments on which $R \geq 0$ give ranges of $r(\tau)$ for a geodesics with that particular angular momentum. All orbits are bound. (Additional parameters: $\color{blue}M = 1,\ a^2 = 19$.)}
         \label{4.13f}
\end{figure}

Every $r$-$L$ plot with $-1/9 < e < 0$ and $Q_+(e) < Q < Q_-(e)$ has this lake. Because $d > 0$, $\phi$ has two simple roots $2M < r_1 < r_2$. % and
The boundary of the lake is the closed curve formed by the two curves $L_-(r)$ and $L_+(r)$, which meet with vertical tangents at $r_1$ and $r_2$. The midline runs along the middle of the lake.

Hence there are many bound orbits in the Kerr exterior, given by horizontal lines within the lake, and they all are stably bound.

If $Q$ continues to increase, the forbidden regions grow and therefore the lake shrinks until at $Q = Q_-(e)$ it vanishes. But, as specified by the third assertion of Proposition \ref{4.7.8}, there remains an isolated maximum point of R$(r, L)$ at which R = 0. This point represents a stable spherical orbit.
