% 1970 words
\chapter{\color{blue}The space-like geodesics in the Schwarzschild space-time}
\label{s}
The constancy of the Lagrangian with equations (\ref{s85}) and (\ref{s88}) gives
\begin{equation}
\frac{E^2}{1-2M/r}-\frac{\dot{r}^2}{1-2M/r}-\frac{L^2}{r^2} =
    2{\cal L} = -1\,.
    \label{s89}
\end{equation}
This can be rewritten in the form
\begin{equation}
    \left(\frac{\mbox{d}r}{\mbox{d}\tau}\right)^2 +
    \left(1-\frac{2M}{r}\right) \left(\frac{L^2}{r^2}-1\right) = E^2\,.
   \label{s90}
\end{equation}
%A geodesic is \emph{a curve whose tangent is paralelly transported
%along it}. The equations governing the geodesics in a space-time
%with the line element,
%\begin{equation}
%    \mbox{d}s^2 = g_{ij}\mbox{d}x^i\mbox{d}x^j\,,
%    \label{s78}
%\end{equation}
%can be derived from the Lagrangian
%\begin{equation}
%    2{\cal L} =
%    g_{ij} \frac{\mbox{d}x^i}{\mbox{d}\tau}
%    \frac{\mbox{d}x^j}{\mbox{d}\tau}\,,
%    \label{s79}
%\end{equation}
%where $\tau$ is some affine parameter along the geodesic.
%For the Schwarzschild space-time, the Lagrangian is
%\begin{equation}
%    {\cal L} = \frac{1}{2}\left[\left(1 -\frac{2M}{r}\right)\dot t^2
%    -\frac{\dot r^2}{1 -2M/r} -r^2\dot\theta^2
%    -(r^2\sin^2\theta)\dot\varphi^2\right]\,,
%    \label{s80}
%\end{equation}
%where dot denotes differentiation with respect to $\tau$. The
%corresponding ca\-no\-ni\-cal momenta are
%\begin{equation}
    %\left.
%    \begin{aligned}%{l}
%        p_t = {\displaystyle \frac{\partial{\cal L}}{\partial\dot t} = %\left(1
%        -\frac{2M}{r}\right)\dot t\,,\quad p_\varphi = %-\frac{\partial{\cal
%        L}}{\partial\dot\varphi}} =
%        (r\sin\theta)^2\dot\varphi\,,\\
%        p_r = -{\displaystyle\frac{\partial{\cal L}}{\partial\dot r} =
%        \left(1 -\frac{2M}{r}\right)^{-1}\dot r\, \quad
%        \mbox{and} \quad p_\theta = -\frac{\partial{\cal
%        L}}{\partial\dot\theta} = r^2\dot\theta}\,.
%    \end{aligned}
%    %\right\}
%    \label{s81}
%\end{equation}
%By rescaling the affine parameter $\tau$, we can arrange that
%$2{\cal L}$ has the value $-1$~for space-like geodesics.
%Further integrals of the motion follow from the equations
%\begin{equation}
%    \frac{\mbox d p_t}{\mbox d\tau} = \frac{\partial{\cal L}}{\partial
%    t} = 0 \quad \mbox{and} \quad \frac{\mbox d p_\varphi}{\mbox d\tau}
%    = -\frac{\partial{\cal L}}{\partial\varphi} = 0\,.
%    \label{s84}
%\end{equation}
%Thus,
%\begin{equation}
%    p_t = \left(1 -\frac{2M}{r}\right)\frac{\mbox d t}{\mbox d\tau} =
%    \mbox{constant} = E \quad \mbox{(say)}
%    \label{s85}
%\end{equation}
%and
%\begin{equation}
%    p_\varphi = r^2\sin^2\theta\frac{\mbox d\varphi}{\mbox d\tau} =
%    \mbox{constant}\,,
%    \label{s86}
%\end{equation}
%where $E$ is the `kinetic energy' as we can see from equation
%(\ref{s90}) bellow. Moreover, from the equation of
%motion,
%\begin{equation}
%    \frac{\mbox d p_\theta}{\mbox d\tau} = \frac{\mbox d}{\mbox
%    d\tau}\bigl(r^2\dot\theta\bigr) = -\frac{\partial{\cal %L}}{\partial\theta}
%    =
%2\left(r^2\sin\theta\cos\theta\right)\left(\frac{\mbox d\varphi}{\mbox
%    d\tau}\right)^2\,,
%    \label{s87}
%\end{equation}
%it follows that geodesic is described in an invariant plane which we may
%distinguish by \(\theta = \pi/2\). Equation (\ref{s86}) then gives
%\begin{equation}
%    p_\varphi = r^2\frac{\mbox d\varphi}{\mbox d\tau} = \mbox{constant}
%    = L \quad \mbox{(say)},
%    \label{s88}
%\end{equation}
%where $L$ denotes angular momentum about an axis normal to the invariant
%plane.
%
%
%
%From the integral of the motion follows
%\begin{equation}
%    \frac{\mbox{d}\varphi}{\mbox{d}\tau} = \frac{L}{r^2}\,.
%    \label{s91}
%\end{equation}
By considering $r$ as a function of $\varphi$, we obtain equation
\begin{equation}
    \left(\frac{\mbox{d}r}{\mbox{d}\varphi}\right)^2 =
\left(E^2+1\right)\frac{r^4}{L^2}-\frac{2Mr^3}{L^2}-r^2+2Mr\,.
    \label{s92}
\end{equation}
Letting
\begin{equation}
    u~= r^{-1}\,,
    \label{s93}
\end{equation}
{\color{blue}
we obtain the basic equation of the problem (cf. equation (\ref{s214null})):
\begin{equation}
    \left(\frac{\mbox{d}u}{\mbox{d}\varphi}\right)^2 =
    2Mu^3-u^2-\frac{2M}{L^2}u+\frac{E^2+1}{L^2}\,.
    \label{s94}
\end{equation}
The solution can be completed by direct quadratures of the equations
\begin{equation}
    \frac{\mbox{d}\tau}{\mbox{d}\varphi} = \frac{1}{Lu^2} \qquad
    \mbox{and} \qquad \frac{\mbox{d}t}{\mbox{d}\varphi} =
    \frac{E}{Lu^2(1-2Mu)}\,.
    \label{s95}
\end{equation}
%
%Again, by considering $r$ as a function of $\varphi$
%and replacing $r$ by \(u = r^{-1}\) as the independent variable,
%we obtain equation (cf. equations (\ref{s94}) and (\ref{s94time}))
%\begin{equation} \label{s214null}
%     \left(\frac{\mbox{d}u}{\mbox{d}\varphi}\right)^2 =
%     2Mu^3-u^2+\frac{1}{D^2} =\mbox{f}(u) \qquad \mbox{(say)} \,,
%\end{equation}
%where
%\begin{equation} \label{ip}
%    D = \frac L E
%\end{equation}
%denotes the impact parameter.
We observe that in contrast to the earlier definition
\eqref{s225null}, f($u$), as now defined, has a linear term in
$u$; and, as in the case of null geodesics, the constant term is
always positive.}




















{\color{blue}\section{\color{blue}The radial geodesics}\label{sr}
The radial geodesics have zero angular momentum. The equations governing
these geodesics are
\begin{equation}
    \left(\frac{\mbox{d}r}{\mbox{d}\tau}\right)^2 = \left(E^2+1\right)
    -\frac{2M}{r}
    \qquad \mbox{and} \qquad \frac{\mbox{d}t}{\mbox{d}\tau} =
    \frac{E}{1-2M/r}\,.
    \label{s96}
\end{equation}
We shall consider trajectories of particles which start from rest
at some finite distance $r_i$ and fall towards the center. The
starting distance is related to the constant $E$ by
\begin{equation}
    r_\mathrm i = \frac{2M}{E^2+1}\,.
    \label{s97}
\end{equation}
As we can see, $r_\mathrm i$ cannot be bigger than $2M$. (The space-like radial geodesics always ``start" under the horizon.)

The equations of motion are most conveniently integrated in terms of
a~variable $\eta$ where
\begin{equation}
    r = r_\mathrm i\cosh^2{\frac{\eta}{2}}\,.
    \label{s98}
\end{equation}
Clearly when $\eta = 0$ then $r = r_\mathrm i$; and the value of
$\eta$ when $r$ crosses the horizon at $r = 2M$ is
\begin{equation}
    \eta_\mathrm{H} = 2\,\mbox{arcsinh}E\,.
    \label{s99}
\end{equation}}
{\color{blue}Since for $r = 0$
\begin{equation}
    \eta_0 = 2\,\mathrm{i}\pi
\end{equation}
is not real, radial space-like geodesics do not reach the singularity.

In terms of $\eta$, the equations to be integrated are
\begin{equation}
    \left(\frac{{\mathrm d}r}{\mathrm{d}\tau}\right)^2 = \left(E^2 +
    1\right)\tanh^2\frac{\eta}{2} \quad \mbox{and} \quad \frac{{\mathrm
    d}t}{\mathrm{d}\tau} =
    \frac{E\cosh^2\frac{\eta}{2}}{\cosh^2\frac{\eta}{2} -
    \cosh^2\frac{\eta_\mathrm{H}}{2}}\,,
    \label{s100}
\end{equation}}
{\color{blue}together with
\begin{equation}
    \frac{{\mathrm d}r}{\mathrm{d}\eta} =
    r_\mathrm i\sinh\frac{\eta}{2}\cosh\frac{\eta}{2}\,.
    \label{s101}
\end{equation}}
%Since we are considering infalling particles,
{\color{blue}To have $\tau$ positive with $\eta$ positive,
\begin{equation}
    \frac{{\mathrm d}r}{\mathrm{d}\tau} = \left(1 +
    E^2\right)^{1/2}\tanh\frac{\eta}{2} =
\left(\frac{2M}{r_i}\right)^{1/2}\tanh\frac{\eta}{2}\,.
    \label{s102}
\end{equation}
From equations (\ref{s101}) and (\ref{s102}), we now obtain
\begin{equation}
    \frac{\mathrm{d}\tau}{\mathrm{d}\eta} =
    \left(\frac{r_\mathrm i^3}{2M}\right)^{1/2}\cosh^2\frac{\eta}{2} =
    (1 + \cosh\eta)\sqrt{\frac{r_\mathrm i^3}{8M}}\,.
    \label{s103}
\end{equation}
Therefore,
\begin{equation}
    \tau = (\eta + \sinh\eta)\sqrt{\frac{r_\mathrm i^3}{8M}}\,,
    \label{s104}
\end{equation}
where we have assumed that $\tau = 0$ at starting point $\eta = 0$.} {\color{blue}From
equation (\ref{s104}) it follows that the particle crosses the horizon at
finite arc lenght,}{\color{blue}
\begin{equation}
    \tau_\mathrm{H} = (\sinh\eta_\mathrm{H} +
    \eta_\mathrm{H})\sqrt{\frac{r_\mathrm i^3}{8M}}\,.
    \label{s105}
\end{equation}
The equation to be integrated to obtain $t$ is
\begin{equation}
    \frac{{\mathrm d}t}{\mathrm{d}\eta} =
    E\frac{\cosh^4\frac{\eta}{2}}{\cosh^2\frac{\eta}{2} -
\cosh^2\frac{\eta_\mathrm{H}}{2}}\sqrt{\frac{r_\mathrm i^3}{2M}}\,.
    \label{s106}
\end{equation}
On integration, this equation gives
\begin{equation}
    t = E\left(\frac{r_\mathrm i^3}{2M}\right)^{1/2}\left[\frac{1}{2}(\eta +
    \sinh\eta) + \left(1 + E^2\right)\eta\right] -
    4M\,\mathrm{arctanh} \frac{\tanh\frac{\eta}{2}}{\tanh\frac{\eta_\mathrm{H}}{2}}\,.
    \label{s107}
\end{equation}
According to this equation,
\begin{equation}
    t\rightarrow -\infty \quad \mbox{as} \quad \eta\rightarrow\eta_\mathrm{H} - 0\,
    .
    \label{s108}
\end{equation}
Solutions (\ref{s104}) and (\ref{s107}) are illustrated in
fig.~\ref{radial}.
\begin{figure}
%    \begin{flushleft}
         \epsfxsize=13cm
     %    \vspace{14cm}
         \epsfbox{schwarzs/radial.eps}
         \caption{\color{blue}Proper time ($\tau$) and coordinate time ($t$)
         along a space-like radial geodesic; initial condition is
         $r_\mathrm i=M$.}
         \label{radial}
%    \end{flushleft}
\end{figure}
Because the space-like geodesics do not relate to any known physical object, the interpretation of this section is not clear. One might speculate that while the arc lenght of a tachyon increases, its coordinate time decreases, thus the particle moves backward in time. Also from the equation \eqref{s97} that $r_i$ is never 0 unless $M = 0$, therefore radially moving tachyon does not reach singularity. }

%Transformation between Schwarzschild coordinates $(r,t)$ and
%Krus\-kal-Sze\-ke\-res
%coordinates $(u,v)$ is given by
%\begin{equation}
%    \left.
%    \begin{array}{l}
%        u = (r/2M-1)^{1/2}\mbox{e}^{r/4M}\cosh(t/4M)\\
%        v = (r/2M-1)^{1/2}\mbox{e}^{r/4M}\sinh(t/4M)
%    \end{array}
%    \right\} \mbox{when}\ r > 2M\,,
%    \label{MTV31.13a}
%\end{equation}
%\begin{equation}
%    \left.
%    \begin{array}{l}
%        u = (1-r/2M)^{1/2}\mbox{e}^{r/4M}\sinh(t/4M)\\
%        v = (1-r/2M)^{1/2}\mbox{e}^{r/4M}\cosh(t/4M)\,,
%    \end{array}
%    \right\} \mbox{when}\ r < 2M\,.
%    \label{MTV31.13b}
%\end{equation}
%when \(r<2M\).
%You can see solution (\ref{s107}) in Kruskal-Szekeres coordinates in
%fig.~\ref{Schkrrad}.
%\begin{figure}
%     \epsfxsize=14cm
     %\epsfysize=15cm
%     \epsfbox{schwarzs/Schkrrad.eps}
%     \caption{The radial geodesic, as seen in Kruskal-Szekeres
%     coordinate system. (Comp. fig.~\ref{radial}.)}
%     \label{Schkrrad}
%\end{figure}



\section{\color{blue}The unbound orbits}\label{su}
These orbits are governed by the equation
\begin{equation}
    \left(\frac{\mbox{d}u}{\mbox{d}\varphi}\right)^2 = \mbox{f}(u)\,,
    \label{s109}
\end{equation}
where{\color{blue}
\begin{equation}
    \mbox{f}(u) = 2Mu^3 - u^2 - \frac{2M}{L^2}u + \frac{E^2 + 1}{L^2}\,.
    \label{s110}
\end{equation}}
Letting $u_1, u_2$, and $u_3$ denote the roots of the cubic equation
f$(u)=0$, we have
\begin{equation}
    u_1u_2u_3 = -\frac{E^2 + 1}{2ML^2}
    \label{s111}
\end{equation}
and
\begin{equation}
    u_1 + u_2 + u_3 = \frac{1}{2M}\,.
    \label{s112}
\end{equation}
Since $-\left(E^2+1\right)<0$, the equation f$(u)=0$ must always
allow a negative real root. From (\ref{s112}) it follows that we
cannot have all roots negative.

 The different cases
that must be distinguished are, therefore, essentially the same as
those for the null geodesics. There are no bound orbits in case of space-like geodesics!


The different cases we have distinguished can also be inferred by
interpreting
\begin{equation}
    {\cal V} = \left(1-\frac{2M}{r}\right)
    \left(\frac{L^2}{r^2}-1\right)
    \label{s113}
\end{equation}
as 'potential energy` (see equation (\ref{s90})). (In
fig.~\ref{potentia} there are displayed some examples of potential
energy.) The unstable orbits correspond to maximum potential.
\begin{figure}
%    \begin{flushleft}
         \epsfxsize=13cm
         \epsfbox{schwarzs/potentia.eps}
         \caption{The effective potential appropriate for space-like
         trajectory (comp. equation (\ref{s113})). $L/M$ is a)~0, b)~3,
         c)\textcolor{blue}{~4.33}, d)~5,
         e)~10, f)~15.} %See also
         %\cite{chandrasekhar}, \cite{mtw} or \cite{hlr}}
         \label{potentia}
%    \end{flushleft}
\end{figure}




\subsection{\color{blue}Orbits of the first kind} \label{su1}
%\addtocontents{toc}{aaa}
Orbits of the first kind occur in cases ($\alpha$), and ($\beta$); and
we shall write the real roots of equation f$(u) = 0$ as
\begin{equation}
    u_{1} = -\frac{1}{l}(e-1)\,, \ u_{2} = \frac{1}{l}(e+1)\,,
    \ \mbox{and}\
    u_{3} = \frac{1}{2M}-\frac{2}{l}\,.
    \label{s114}
\end{equation}

To be in conformity with ordering, $u_{1} < u_{2} \leq u_{3}$, we must
require
\begin{equation}
    l \geq 2M(e+3)\,.
    \label{s116}
\end{equation}
Defining
\begin{equation}
    \mu = \frac{M}{l}
    \label{s117}
\end{equation}
we have the important inequality
\begin{equation}
    \mu \leq \frac{1}{2(e+3)}
    \label{s118}
\end{equation}
or
\begin{equation}
    1-6\mu-2\mu e \geq 0\,.
    \label{s191}
\end{equation}

{\color{blue}The condition, that
\begin{equation}
    \mbox{f}(u) = 2M\left(u+\frac{e-1}{l}\right)
    \left(u-\frac{e+1}{l}\right)
    \left(u-\frac{1}{2M}+\frac{2}{l}\right)
    \label{s119}
\end{equation}
agrees with its definition (\ref{s110}), yields relations,
\begin{equation}
    \frac{M}{L^2} =
\frac{1}{l^2}\left[M\left(3+e^2\right)-l\right]\quad\mbox{and}\quad
    \frac{E^2+1}{L^2} =
    \frac{1}{l^3}(4M-l)\left(1-e^2\right)\,,
    \label{s120}
\end{equation}
or, in terms of $\mu$,
\begin{equation}
    \frac{1}{L^2} =
\frac{1}{lM}\left[\mu\left(3+e^2\right)-1\right]\quad\mbox{and}\quad
    \frac{E^2+1}{L^2}  =
    \frac{1}{l^2}(4\mu-1)\left(1-e^2\right)\,.
    \label{s121}
\end{equation}}
{\color{blue}We write, since $e^2 >1$,
\begin{equation}
    \frac{1}{L^2} =
\frac{1}{lM}\left[\mu\left(3+e^2\right)-1\right]\quad\mbox{and}\quad
    \frac{E^2+1}{L^2} =
    \frac{1}{l^2}(1-4\mu)\left(e^2-1\right)\,.
    \label{s192}
\end{equation}}
{\color{blue}Since $L^2 > 0$ and $E^2+1 > 0$
\begin{equation}
    \mu\left(3+e^2\right)-1 > 0 \,,
    \label{s193}
\end{equation}
and}
{\color{blue}\begin{equation}
    \mu < \frac{1}{4}\,.
    \label{s194}
\end{equation}
\begin{figure}
     \epsfxsize=13cm
     \epsfbox{schwarzs/koef_mue.eps}
     \caption{\textcolor{blue}{Area marked} {\fontfamily{phv}\fontseries{b}\selectfont A}
     {\color{blue}designates where allowed pairs of
     $e$ and $\mu$ lie.}}
     \label{koef_mue}
\end{figure}
The allowed range of $\mu$
%\begin{equation}
%    0 < \mu \leq \frac 1{12}
%    \label{s194}
%\end{equation}
and $e$,}
%\begin{equation}
%    e > 3 \,.
%    \label{s196}
%\end{equation}
{\color{blue}according to (\ref{s191}) and (\ref{s193}), can be seen in
fig.~\ref{koef_mue}.

An alternative expression for $E^2$ may be noted here:
\begin{equation}
    \frac{E^2}{L^2} = -\frac{1}{lM}(2\mu e-2\mu+1)(2\mu e +2\mu-1)\,.
    \label{s123}
\end{equation}}

%We may associate with orbits of the first kind an impact
%parameter, $D$, and velocity at infinity $V$; these are related to
%$L$ and $E$ by
%\begin{equation}
%    D^2 = \frac{L^2}{V^2} = \frac{L^2E^2}{E^2+1} \qquad\left(E^2 =
%    \frac{1}{V^2-1}\right)\,.
%    \label{s198}
%\end{equation}

We shall make a substitution
\begin{equation}
    u~= \frac{1}{l}(1+e\cos{\chi})\,.
    \label{s199}
\end{equation}
Since $e > 1$,
\begin{equation}
    u~= 0\quad\mbox{when}\quad\chi = \arccos{\left(-e^{-1}\right)} =
    \chi_{\infty} \qquad\mbox{(say)}\,.
    \label{s200}
\end{equation}
But the perihelion passage still occurs when $\chi = 0$. The allowed
range of $\chi$~is, therefore,
\begin{equation}
    0 \leq \chi < \chi_{\infty} = \arccos{\left(-e^{-1}\right)}\,.
    \label{s201}
\end{equation}
The substitution (\ref{s199}) reduces equation (\ref{s109}) to the simple
form
\begin{equation}
    \left(\frac{\mbox{d}\chi}{\mbox{d}\varphi}\right)^2 = 1 -
    2\mu(3+e\cos{\chi}) = 1-6\mu+2\mu e-4\mu e\cos^2{\frac{\chi}{2}}\,,
    \label{s126}
\end{equation}
or, alternatively,
\begin{equation}
    \pm\frac{\mbox{d}\chi}{\mbox{d}\varphi} = \left(1-6\mu+2\mu
e\right)^{1/2}\left(1-k^2\cos^2{\frac{\chi}{2}}\right)^{1/2}\,,
    \label{s127}
\end{equation}
where
\begin{equation}
    k^2 = \frac{4\mu e}{1-6\mu+2\mu e}\,.
    \label{s128}
\end{equation}
The inequality (\ref{s118}) guarantees that
\begin{equation}
    k^2 \leq 1\quad\mbox{and}\quad 1-6\mu+2\mu e > 0\,.
    \label{s129}
\end{equation}
It is now apparent that the solution for $\varphi$ can be
expressed in terms of the Jacobian elliptic integral
\begin{equation}
    \F(\psi, k) =
\int\limits_0^\psi\frac{\mbox{d}\gamma}{\sqrt{1-k^2\sin^2{\gamma}}}\,,
    \label{s130}
\end{equation}
where
\begin{equation}
    \psi = \frac{1}{2}(\pi-\chi)\,.
    \label{s131}
\end{equation}
Thus, keeping in mind restriction in the range of $\chi$, we may now
write the solution for $\varphi$ in the form
\begin{equation}
    \varphi = \frac{2}{\sqrt{1-6\mu+2\mu
    e}}\left[\K(k)-\F \left(\frac{\pi}{2}-\frac{\chi}{2},k\right)
    \right]\,,
    \label{s202}
\end{equation}
where $\K(k)$ denotes the complete elliptic integral,
\begin{equation}
    \K(k) =
\int\limits_0^{\pi/2}\frac{\mbox{d}\gamma}{\sqrt{1-k^2\sin^2\gamma}}\,.
    \label{s159}
\end{equation}
In writing the solution for $\varphi$ in the form (\ref{s202}), we
have taken the origin of $\varphi$ at the perihelion passage where
$\chi = 0$. The trajectory goes off to infinity, asymptotically,
along the direction
\begin{equation}
    \varphi = \varphi_\infty = \frac{2}{\sqrt{1-6\mu+2\mu
    e}}\bigl[\K(k)-\F(\psi_\infty, k)\bigr]
    \quad\left(\psi_\infty = \frac{1}{2}\arccos\left(e^{-1}\right)\right)\,.
    \label{s203}
\end{equation}

The solution can be now completed by direct integration of the equation
(\ref{s95}) for $t$ and $\tau$. Thus,
\begin{equation}
    \tau = \frac{1}{L}\int\frac{\mathrm{d}\varphi}{u^2} =
    \frac{1}{L} \int\
\frac{\mathrm{d}\varphi}{\mathrm{d}\chi}\frac{\mathrm{d}\chi}{u^2}\,,
    \label{s133}
\end{equation}
and
\begin{equation}
    t = \frac{E}{L} \int \frac{\mathrm{d}\varphi}{\mathrm{d}\chi}
    \frac{\mathrm{d}\chi}{u^2(1 - 2Mu)}\,.
    \label{s134}
\end{equation}
By making use of equations (\ref{s121}), (\ref{s123}), (\ref{s199}), and
(\ref{s127}), we obtain
{\color{blue}
\begin{multline}
    \tau =-\frac{l}{\left(e^2-1\right)[k^2(e-1)-2e]\sqrt{\mu}}
    \left\{\frac{2e^2\sin\chi\sqrt{1-k^2+k^2
    \sin^2\frac{\chi}{2}}}
    {2e\sin^2\frac{\chi}{2}-e-1}+\right.\\
    +2e\E\left(\frac{\pi}{2}-\frac{\chi}{2},k\right)
    +\F\left(\frac{\pi}{2}-\frac{\chi}{2},k\right)+\\
    \left.+ \Pi\left(\frac{\pi}{2}-\frac{\chi}{2},\frac{2e}{e-1},
    k\right)
    \left[k^2(e+3)-\frac{4e}{e-1}\right]\right\}
    \sqrt\frac{\mu(3+e^2)-1}{1-2\mu(3-e)}
    \label{s135}
\end{multline}
and
\begin{multline}
    t =
    -\frac{2e^2\sin{\chi}}
{(e^2-1)\left(1-e+2e\cos^2\frac{\chi}2\right)[k^2(e-1)-2e^2]}\cdot
    \\*
    \cdot \sqrt\frac{(1-2\mu+2\mu e)
    (1-2\mu-2\mu e)\left(1-k^2\cos^2\frac{\chi}2\right)}{\mu(1-6\mu+2\mu
    e)} + \\
    +\cfrac{-8\Pi\biggl(\cfrac\pi 2-\cfrac\chi 2,\cfrac{4\mu e}{1-2\mu+2\mu
    e},k\biggr)}
    {(e-1)(e^2-1)\left(1-e+2e\cos^2\frac\chi 2\right)[k^2(e-1)-2e^2]}
    \Bigl\{(e-1)
    \bigl[2e(e+\bigr.\Bigr.
    \\
    +
    1)\left(e^2k^2-ek^2-k^2-2e^2+2e\right)\cos^2\frac\chi 2+k^2\left(e^2
    +1\right)^2+ \\
    \left.\left.+2e^2\left(1-k^2e^2\right)\right] -2ek^2(e+1)
    \left[(e-1)\left(1+e\cos^2\frac\chi 2\right)
    -e^2\right]\right\}\cdot \\*
    \cdot\sqrt\frac{\mu^3(1-2\mu-2\mu e)}{(2\mu
    e-2\mu+1)(1-6\mu+2\mu e)}
    -\Pi\left(\frac\pi 2-\frac\chi 2,\frac{2e}{e-1},k\right)\cdot
    \\*
    \cdot\frac{\{(e^2-1)[k^2(1-4\mu
    e+4\mu)+8\mu e]+2k^2(e-1)
-4e\}}{\left(e-1\right)^2\left[k^2(e-1)-2e^2\right]}\cdot \\*
    \cdot\sqrt\frac{(1-2\mu+2\mu e)(1-2\mu-2\mu e)}{\mu(1-6\mu+2\mu e)}+
    \\
    +\frac{-\F\left(\frac\pi 2-\frac\chi 2,k\right)[k^2(e-1)-2e]}
    {(e^2-1)[k^2(e-1)-2e^2]}\sqrt\frac{(1-2\mu+2\mu e)(1-2\mu-2\mu e)}
    {\mu(1-6\mu+2\mu e)}+ \\
    +\frac{-2e\E\left(\frac\pi 2-\frac\chi 2,k\right)}
    {(e^2-1)[k^2(e-1)-2e^2]}\sqrt\frac{(1-2\mu+2\mu e)(1-2\mu-2\mu e)}
    {\mu(1-6\mu+2\mu e)}\,,
    \label{s136}
\end{multline}
where
\begin{gather}\label{110.03}
  \E(\psi, k) = \int\limits_0^\psi
  \sqrt\frac{1-k^2\gamma^2}{1-\gamma^2}{\mbox{d}\gamma}\\
  \intertext{and}
  \Pi(\psi, \alpha^2, k) = \int\limits_0^\psi \frac{\mbox
  d\gamma}{(1-\alpha^2t^2)\sqrt{(1-t^2)(1-k^2t^2)}}
\end{gather}
are elliptic integrals.

%The last two equations are in figure~\ref{plottime} together with $1/u$
%coordinate.


\subsubsection{The case $2\mu(3+e) = 1$}} \label{ss1}
In this case, the perihelion distance,
$r_{\scriptstyle\mathrm p}$,
%and the aphelion $r_{\rm\scriptsize ap}$,
is given by
\begin{equation}
    r_{\scriptstyle\mathrm p} = \frac{l}{1+e} = 2M\frac{3+e}{1+e} %\qquad
    %\mbox{and} \qquad
    %r_{\rm\scriptsize ap} = 2M\frac{3+e}{1-e}
    \label{s146}\,.
\end{equation}
{\color{blue}
For these orbits
\begin{equation}
    \frac{L^2}{M^2} = \frac{4(3+e)^2}{(e-3)(e+1)} \qquad \mbox{and}
    \qquad
    E^2+1 = \frac{e^2-1}{e^2-9}\,.
    \label{s195}
\end{equation}
According to the allowed range of $e$ (see fig.~\ref{koef_mue}), %(equation (\ref{s196})),
%    \begin{equation}
%        1 < e < 3
%        \label{s196}
%    \end{equation}
the corresponding perihelion-distances must be
\begin{equation}
     2M < r_{\scriptstyle\mathrm p} < 3M\,.
     \label{s197}
\end{equation}}

It is convenient to go back to the equation (\ref{s126}); it gives
\begin{equation}
    \left(\frac{\mbox{d}\chi}{\mbox{d}\varphi}\right)^2 = 4\mu
    e\sin^2\frac{\chi}{2} \qquad \mbox{or} \qquad
    \frac{\mbox{d}\chi}{\mbox{d}\varphi} = -2\left(\mu
    e\right)^{1/2}\sin\frac{\chi}{2}\,,
    \label{s149}
\end{equation}
where the negative sign has been chosen so that $\varphi$ may
increase when $\chi$~decreases. %from its aphelion value $\pi$.
The
required solution of equation (\ref{s149}) is
\begin{equation}
    \varphi = -\frac{1}{\sqrt{\mu
    e}}\ln\tan\frac{\chi}{4}  \,.%\qquad
    %(\varphi_\infty = -\\quad\mbox{when}\quad\chi = \pi)
    \label{s150}
\end{equation}
The asymptotic direction is
\begin{equation}
    \varphi_\infty = -\frac 1{\sqrt{\mu e}}\ln{\tan{\frac{\chi_\infty}4}}
    \label{s150'}
\end{equation}
(see $\chi_\infty$ in equation (\ref{s200})).





\subsection{\color{blue}Orbits of the second kind} \label{su2}
To obtain the solution for these orbits, we now make a substitution
\begin{equation}
    u~= \left(\frac{1}{2M} - \frac{2}{l}\right) + \left(\frac{1}{2M} -
    \frac{3 + e}{l}\right)\tan^2\frac{\xi}{2}\,.
    \label{s154}
\end{equation}
By this substitution,
\bal
    %\left.
    \begin{alignedat}{3}
        &u~= u_3 = {\displaystyle\frac1{2M} - \frac2l} \quad
        &\mbox{when}
        &\quad&
        &\xi = 0 \\
        \text{and}\\
        &u~\rightarrow \infty  &\mbox{as} &&&\xi \rightarrow \pi&
    \end{alignedat}
    %\right\}
    \label{s155}
\eal
we further find that equation (\ref{s109}) now reduces to
\begin{equation}
\left(\frac{\mathrm{d}\xi}{\mathrm{d}\varphi}\right)^2 = (1 - 6\mu +
    2\mu e)\left(1 - k^2\sin^2\frac{\xi}{2}\right)
    \label{s156}
\end{equation}
with the same definition of $k^2$ as for the orbits of the first kind.
Thus we may now write
\begin{equation}
    \varphi = \frac{2}{\sqrt{1 - 6\mu +2\mu e}}\F\left(\frac{\xi}{2},
    k\right) \,.
    \label{s157}
\end{equation}
At aphelion, $\xi = 0$ and $\varphi = 0$; and at the singularity $\xi
\rightarrow \pi$ and $\varphi$ takes the finite value
\begin{equation}
    \varphi_0 = \frac{2}{\sqrt{1 - 6\mu + 2\mu e}}\K(k)\,.
    \label{s158}
\end{equation}
\begin{figure}
    \epsfxsize=13cm
    \epsfbox{schwarzs/222_2.eps}
    \caption{{\color{blue}The orbit of the first kind arrives from infinity, and after reaching
    the perihelion,
    goes
    again to infinity. Orbit of the second, starting from a certain
    distance from the black hole (aphelion), falls into the
    singularity.
    Eccentricity is
    $e = 5$} and latus rectum is $l = 4.5$ ($\color{blue}M = 1/7$ in the scale along the
    coordinate axes.)}
    \label{222}
\end{figure}
\textcolor{blue}{In fig.~\ref{222}, an example of unbound orbits is illustrated (as given
by solutions (\ref{s150}) and (\ref{s157})).}

The solution for the arc lenght $\tau$ and the coordinate time $t$ can
be obtained by integrating the equations
\begin{equation}
    \frac{\mathrm{d}\tau}{\mathrm{d}\xi} =
    \frac{1}{Lu^2}\frac{\mathrm{d}\varphi}{\mathrm{d}\xi} \quad
    \mbox{and} \quad \frac{{\mathrm d}t}{\mathrm{d}\xi} =
    \frac{E}{Lu^2(1 - 2Mu)}\frac{\mathrm{d}\varphi}{\mathrm{d}\xi} \,.
    \label{s160}
\end{equation}
\begin{figure}
    \epsfxsize=13cm
    \epsfbox{schwarzs/plottime2.eps}
    \caption{{\color{blue}Proper time ($\tau$) and
        Schwarzschild time ($t$)
    along space-like geodesics of the first and second kind;}
    ${\color{blue}M = 1/7},\ l = 4.5$ and {\color{blue}$e = 6$.}}
    \label{plottime}
\end{figure}
{\color{blue}(Numerical solution for integrals (\ref{s133}), (\ref{s134}) and
(\ref{s160}) are shown in figure~\ref{plottime}.)
By making use of equations (\ref{s121}),
(\ref{s123}),
(\ref{s154}), and
(\ref{s156}), we obtain
\begin{multline}
    \tau = {\displaystyle\frac{kl} {4\mu-1}\sqrt{\frac{\mu
    e^2+3\mu-1}e}}
    \left\{\frac{(2\mu e+6\mu-1)\E\left(\frac\xi 2,k\right)}
    {(e+1)[2\mu(e+1)+k^2(4\mu-1)]}
    +\right.\\
    +\frac{-2(2\mu e-2\mu+1)\F\left(\frac\xi 2,k\right)}
    {\mu\left(e+1\right)^2} +(2\mu
    e+6\mu-1)\Pi\left(\frac\xi 2,\frac{2\mu(e+1)}{1-4\mu},k\right)\cdot \\
    \cdot{\displaystyle\frac{k^2(4\mu-1)(4\mu e+1)+4\mu(e+1)(\mu
    e-3\mu+1)}
{2\mu(4\mu-1)\left(e+1\right)^2\left[4\mu-1+2\mu(e+1)\sin^2\frac\xi
    2\right]}}+ \\
    \left.+\frac{\mu(2\mu
    e+6\mu-1)\left(1-k^2\sin^2\frac\xi
    2\right)\sin\xi\cos\frac\xi 2}
{[2\mu(e+1)+k^2(4\mu-1)]\left[4\mu-1+2\mu(e+1)\sin^2\frac\xi
    2\right]}\right\}
    \label{s161}
\end{multline}
and
\begin{multline}
    t ={\displaystyle\frac{kl} {3\mu e+6\mu-1-4\mu^2e}
    \left\{\frac{\left(2\mu e+6\mu -1\right)^2\E\left(\frac\xi 2, k\right)}
    {(1-4\mu)(e+1)[k^2(1-4\mu) -2\mu(e+1)]}+\right.} \\
    +{\displaystyle\frac{-[4\mu^3(3e^3+20e^2+31e+18) -4\mu^2(4e^2+9e-3)
    +\mu(e-6) +1]\F\left(\frac\xi 2, k\right)}
    {2\mu(1-4\mu)\left(e+1\right)^2(1-2\mu-3\mu e)}}+ \\
    +{\displaystyle\frac{(2\mu
    e+6\mu-1)^2\Pi\biggl(\cfrac\xi
    2, \cfrac{2\mu(e+1)}{1-4\mu}, k\biggr)}
{2\mu\left(1-4\mu\right)^2\left(e+1\right)^2[k^2(1-4\mu)
    -2\mu(e+1)]}}\cdot \\
    \cdot\left[k^2(1-4\mu)\left(64\mu^3e^2+144\mu^3
    e+96\mu^3-40\mu^2-72\mu^2e-24\mu^2e^2-2\mu+ \right.\right.\\
    \left.\bigl.+5\mu e-1\bigr)
    -4\mu^3(e+1)\left(28\mu e^3-9e^2-19e+60\mu e+48\mu-2\right)\right]+ \\
    +\displaystyle{\frac{-(3\mu e+6\mu -1)\Pi\biggl(\cfrac\xi 2, \cfrac{1
    -2\mu-
    3\mu
    e}{4\mu}, k\biggr)}{(1-2\mu-3\mu e)(3\mu e+6\mu-1-4\mu^2 e)}}
    +\\
    \left. +{\displaystyle\frac{-\mu(2\mu
    e+6\mu-1)\sqrt{1-k^2\sin^2\frac\xi 2}\sin^2\xi}
    {(1-4\mu)[k^2(1-4\mu) -2\mu(e+1)]\left[1 -4\mu -2\mu\sin^2\frac\xi
    2(e+1)\right]}}\right\}\cdot \\
    \cdot{\displaystyle\sqrt{\frac{(1-2\mu+2\mu e)(1-2\mu-2\mu
    e)}e}} \,.
    \label{s162}
\end{multline}

\subsubsection{The case $2\mu(3+e) = 1$}} \label{ss2}
When $2\mu(3+e) = 1$, the roots of
$f(u) = 0$ are
\begin{equation}
    u_1 = -\frac{1}{l}(e-1) \quad \mbox{and} \quad u_2 = u_3 =
    \frac{1}{4M}+\frac{e-1}{2l} = \frac{e+1}{l}\,;
    \label{s166}
\end{equation}
and the substitution that is suggested is
\begin{equation}
    u~= \frac{1}{l}\left(1+e+2e\tan^2\frac{\xi}{2}\right)\,.
    \label{s167}
\end{equation}
By this substitution,
\begin{equation}
    u~= u_2 = u_3 = \frac{e+1}{l} \ \mbox{when} \ \xi = 0\,,\ \mbox{and} \ u~\rightarrow \infty \ \mbox{when} \ \xi
    \rightarrow \pi\,.
    \label{s168}
\end{equation}
We further find that equation (\ref{s109}) reduces to
\begin{equation}
    \left(\frac{\mbox{d}\xi}{\mbox{d}\varphi}\right)^2 = 4\mu
    e\sin^2\frac{\xi}{2}\,.
    \label{s169}
\end{equation}
We observe that this is exactly the same equation we obtained when
considering the orbits of the first kind in the same context, $u_2 =
u_3$ (see equation (\ref{s149})). And, as before, we shall write
\begin{equation}
    \varphi = -\frac{1}{\sqrt{\mu
    e}}\ln\tan\frac{\xi}{4}\,.
    \label{s170}
\end{equation}
Along this orbit, $\varphi = 0$ when $\xi = \pi$ and $r \rightarrow
0$; and $\varphi \rightarrow \infty$ as $\xi \rightarrow
\infty$ and we approach the aphelion at $r = \frac{l}{1+e}$.
% ------------------------------------------------------------
% poznamka viz dale ( o uplne variete ):    viz str. 110
\begin{figure}
    \epsfxsize = 13cm
    \epsfbox{schwarzs/pic2_2.eps}
    \caption{{\color{blue}The case $2\mu(3+e) = 1$. The orbit of the first kind comes from
    infinity, and after spiralling infinitely around the black
    hole, reaches the perihelion.
    The orbit of the second kind starts form aphelion, and after
    spiralling around the black hole infinitely, reaches
    singularity.
    Eccentricity is $e = 6$} ($M = 1/7$ in the scale along the
    coordinate axes.)}
    \label{pic2}
\end{figure}
{\color{blue}In fig.~\ref{pic2} there is an example of solutions (\ref{s150}) and
(\ref{s170}).}



\subsection{\color{blue}The orbits with imaginary eccentricities} \label{si}
Finally, we have to consider the case when f$(u) = 0$ allows only one
real root and pair of complex-conjugate roots (the case ($\gamma$) of
fig.~\ref{fig}). It is evident that the corresponding orbits, while
arriving from infinity, will fall into the singularity though they may
circle the origin one or more times before doing so. As in section~\ref{ni}, we shall
characterize these orbits by an imaginary eccentricity i$e$ and
write roots of f$(u) = 0$ as
in equations
(\ref{s171}).
With the present definition, we can obtain the equations which replace
equations (\ref{s121}) and (\ref{s123}) by simply writing $-e^2$ wherever
$+e^2$ occurs and conversely. {\color{blue}We thus obtain
\begin{equation}
    %\left.
    \begin{aligned}%{3}
        &{\displaystyle\frac{1}{L^2} = \frac{1}{lM}\left[\mu\left(3
        -e^2\right) -1\right]\,, \quad
        \frac{E^2+1}{L^2} = \frac{1}{l^2}(4\mu-1)\left(1+e^2\right)}\,,\\
        \mbox{and}\\
        &{\displaystyle\frac{E^2}{L^2} = \frac{1}{lM}}\left(4\mu^2e^2+4\mu^2
        -4\mu+1\right)\,.
    \end{aligned}
    %\right\}
    \label{s172}
\end{equation}
Accordingly, we must now require
\begin{equation}
    \mu > \frac14 \quad \mbox{and} \quad 3\mu -\mu e^2 -1 > 0 \,.
    \label{s205}
\end{equation}
%For $1/3 > \mu > 1/4$, the second inequality imposes no restrictions on
%$e^2$; but for $ \mu > 1/4$,
The second inequality imposes restrictions
\begin{equation}
    e^2 < 3 - \frac{1}{\mu} \qquad \mbox{and} \qquad \mu > 1/3\,.
    \label{s206}
\end{equation}}

The analysis of section~\ref{si} applies equally to the present considerations
with the only proviso that \eqref{s206} holds.
Since these orbits are unbounded, the solution for
$\varphi$ is given by the same equation (\ref{s209}) as in
section~\ref{si}.
The origin of $\varphi$ is at the singularity (where $\xi = \pi$ and
$\psi = +\pi/2$). But the lower limit on $\psi$, say $\psi_\infty$, as the
orbit goes off to infinity, can be obtained from equation (\ref{s184}) by
inserting for sin$\frac\xi2$ and cos$\frac\xi2$ the values appropriate
for $\xi_\infty$ given in equation (\ref{s207}). We find
\begin{equation}
    \sin^2\psi_\infty =
\frac1{{\mathsf\triangle}+6\mu-1}\left({\mathsf\triangle}+
6\mu-1-2e^2\frac{4\mu-1}{e^2+1}\right)\,.
    \label{s210}
\end{equation}
\begin{figure}
    \epsfxsize = 13cm
    \epsfbox{schwarzs/s-l_imag2.eps}
        The orbit of the first kind arrives from infinity, and after reaching
    the perihelion,
    goes
    again to infinity. Orbit of the second, starting from a certain
    distance from the black hole (aphelion), falls into the
    singularity.
    \caption{{\color{blue}Orbits with $l = 0.16$} and with imaginary eccentricity $e =\
    $i \newline ($M = 1/7$ in the scale along the
    coordinate axes). The orbit arrives from infinity, and falls into the
    singularity.}
    \label{s-l imag}
\end{figure}
{\color{blue}In fig.~\ref{s-l imag} an example of solution (\ref{s209}) is illustrated.

\begin{figure}
    \epsfxsize = 13 cm
    \epsfbox{schwarzs/pl_itime2.eps}
    \caption{{\color{blue}Proper time ($\tau$) and
    Schwarzschild time ($t$)
    along a space-like geodesic with an imaginary eccentricity;}
    $M = 1/7,{\color{blue}\ l = 0.16}$ {\color{blue}and $e =
    $ i}.}
    \label{pl_itime}
\end{figure}
The solution for the arc lenght $\tau$ and the coordinate time
$t$ can be obtained by integrating the equations (\ref{s160}).
Numerical solution for integrals (\ref{s133}) and (\ref{s134}) are
shown in figure~\ref{pl_itime}. By making use of equations
(\ref{s172}), (\ref{s175}),  and (\ref{s181}), we obtain
%\newpage
\begin{multline}
    \tau = \frac{\mu\left(3-e^2\right)-1}{\sqrt\mu}
    \left\{\frac{\tan\frac\xi4-1}{2\mu e^3\left[4\mu\left(e^2-1\right)
    -\left(e^2+1\right)\right]}\cdot
    \right.\\*
    \cdot
    \E\left(\sqrt{\frac{\left(\tan\frac\xi4+1\right)
    \left(2\mu e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-6\mu+1\right)}
    {\left(\tan\frac\xi4-1\right)
    \left(2\mu e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}+6\mu-1\right)}},
    \right.\\*
    \left.
    \sqrt{\frac{(6\mu-1)\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}
    -12\mu(3\mu-1) -1}
{(1-6\mu)\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-12\mu(3\mu-1)-1}}\right)\cdot
    \\*
    \cdot
    \sqrt{\frac{\left(\tan\frac\xi4+1\right)
    \left(\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-2\mu e+6\mu-1\right)}
    {\left(\tan\frac\xi4-1\right)
    \left(\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-2\mu e-6\mu+1\right)}}
    \cdot\\*
    \cdot
\sqrt{\frac{(6\mu-1)\left(\tan^2\frac\xi4-1\right)+4\mu e\tan\frac\xi4}
    {(6\mu-1)\left(\tan^2\frac\xi4-1\right)-4\mu e\tan\frac\xi4}}
    +\\
+\frac1{\sqrt{2(1-6\mu)(6\mu-1)+\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}}}
    \cdot
    \\*
    \cdot
    \frac{(6\mu-1)\left(\tan^2\frac\xi4-1\right)-4\mu e\tan\frac\xi4}
    {\mu e\left(e^2+1\right)(4\mu-1)\left[(6\mu-1)\left(\tan^2\frac\xi4
    -1\right)+4\mu e\tan\frac\xi4\right]}\cdot
    \\*
    \cdot
    \F\left(\sqrt{\frac{\left(\tan\frac\xi4+1\right)
    \left(2\mu e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-6\mu+1\right)}
    {\left(\tan\frac\xi4-1\right)
    \left(2\mu e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}+6\mu-1\right)}}, \right.
    \\*
    \left.
    \sqrt{\frac{1-6\mu+\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}}
    {1-6\mu-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}}}\right)
    %\cdot\nonumber\\
    %\right.\cdot \nonumber\\
    %\cdot
    \Biggl[\left(\left(e^2+1\right) %\cdot\nonumber\\
    %&\cdot
    (1-4\mu) \right.\Biggr.\cdot\\*
    \cdot
    (2\mu e+6\mu-1) %\right.\right.+\nonumber\\
+\left.\left(4\mu\left(e^2+1\right)+e^2-1\right)\sqrt{4\mu^2e^2+36\mu^2
    -12\mu+1}\right) \cdot\\*\cdot
    \sqrt{(6\mu-1)\left(1-\tan^2\frac\xi4\right)+4\mu
    e\tan\frac\xi4}
    +\\
    +\left(\left(\mu e-4\mu e^2-3\mu+e^2+1\right)\right. %\cdot\nonumber\\
    %&\cdot
    \sqrt{4\mu^2e^2+36\mu^2-12\mu+1} -3\mu e +\\
    \left.
    +e^2 %+\nonumber\\\left.
    +1 -2\mu e^3 %+\nonumber\\
    %\left.
    +12\mu %+\nonumber\\&
    +\mu^2e^3 -9\mu -10e^2\mu +22\mu^2e^2\right)
    \cdot\\*\cdot\left.
    \sqrt{(6\mu-1)\Bigl(1-\tan^2\frac\xi4\Bigr)+4\mu
    e\tan\frac\xi4}\right] +\\
    +\frac1{4e^2\left(\tan\frac\xi4-1\right)^2
    \left(\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}-2\mu e\right)\sqrt{\mu^5\left(e^2
    +1\right)}} \cdot\\*
    \cdot\frac{\sqrt{\left(\tan^2\frac\xi4-1\right)
    (6\mu-1)+4\mu e\tan\frac\xi4}}{\left(\tan^2\frac\xi4-1\right)(6\mu-1)
    -4\mu e\tan\frac\xi4} \cdot %\nonumber
    %\end{eqnarray}
    \label{s211}
    \\*
\cdot\Pi\left(\sqrt\frac{\left(\tan\frac\xi4+1\right)\left(2\mu e
    -\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}-6\mu+1\right)}
    {\left(\tan\frac\xi4-1\right)\left(2\mu
    e-\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}+6\mu-1\right)},
    \right.\\*
    \frac{\left(2\mu e-\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}+6\mu-1\right)\left(e
    -1+\sqrt{e^2+1}\right)}
    {\left(2\mu
e-\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}-6\mu+1\right)\left(e+1+\sqrt{e^2
    +1}\right)},
    \\*
    \left.
\frac{1-6\mu+\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}+6\mu-1}{1-6\mu
    -\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}+6\mu-1}\right)
    \cdot\\*
    \cdot
    \left[2(\mu
    e^2-3\mu+1) %\cdot\nonumber\\\cdot
    \left(2\mu
    e-\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}+6\mu-1\right)
    \right.\cdot\\*
    \cdot
\left(1+2(e-1)\tan\frac\xi4-4e\tan^2\frac\xi4+2(e+1)\tan^3\frac\xi4
    -\tan^4\frac\xi4\right)
    +\\
    +\left.2 +\sqrt2\right] %+\nonumber\\
+\frac{\mu\left(3-e^2\right)-1}{(4\mu-1)\left(e^2+1\right)^{3/2}}
    \cdot\\*\cdot
\Pi\left(\sqrt\frac{\left(\tan^2\frac\xi4+1\right)\left(2\mu e
    -\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}-6\mu+1\right)}
    {\left(\tan^2\frac\xi4-1\right)\left(2\mu
    e-\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}+6\mu-1\right)},
    \right.\\* %\left.
    \frac{\left(2\mu e-\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}+6\mu-1\right)\left(e
    -1-\sqrt{e^2+1}\right)}
    {\left(2\mu
e-\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}-6\mu+1\right)\left(e+1-\sqrt{e^2
    +1}\right)},
    \\* \left.
\sqrt\frac{(6\mu-1)\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}+12\mu(1-3\mu)-1}
{(1-6\mu)\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}+12\mu(1-3\mu)-1}\right)
    \cdot\\*\cdot
    \sqrt\frac{1-6\mu}{2e} %\cdot\nonumber\\
%    \nonumber
    %\cdot
    \sqrt\frac{6\mu-1-2\mu e+\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}}
    {1-6\mu-2\mu e+\sqrt{4\mu^2e^2-36\mu^2-12\mu+1}}
    \cdot\nonumber\\*\cdot
\sqrt\frac{(1-6\mu)\left(1-\tan^2\frac\xi4\right)+4\mu e\tan\frac\xi4}
    {(6\mu-1)\left(1-\tan^2\frac\xi4\right)+4\mu e\tan\frac\xi4}+
    \\
    +\frac{(e-1)\tan\frac\xi4+e+1}{2(e^2+1)(1 -4\mu)\left(1+2e\tan\frac\xi4
    -\tan^2\frac\xi4\right)}
    \sqrt\frac{1+\tan\frac\xi4}{1-\tan\frac\xi4} \cdot\\*
\cdot\left.\sqrt\frac{(6\mu-1)\left(1-\tan^2\frac\xi4\right)+4\mu
    e\tan\frac\xi4}\mu\right\}
\end{multline}
and\enlargethispage{1cm}
\begin{multline}%\label{S212}
  t = l \sqrt\frac{2\mu e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}+6\mu-1}{2\mu
  e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-6\mu+1}
  \cdot\\*\cdot
  \sqrt\frac{(6\mu-1)\left(\tan^2\frac\xi4-1\right)+4\mu
e\tan\frac\xi4}{(6\mu-1)\left(\tan^2\frac\xi4-1\right)-4\mu
  e\tan\frac\xi4} \cdot\\*\cdot
\left[\Pi\left(\sqrt\frac{\left(\tan\frac\xi4+1\right)\left(2\mu e
  -\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-6\mu+1\right)}
  {\left(\tan\frac\xi4-1\right)\left(2\mu
e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}+6\mu-1\right)},\right.\right.
  \\*
\frac{\left(\sqrt{e^2+1}+1-e\right)\left(\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}
  -2\mu e-6\mu+1\right)}
{\left(\sqrt{e^2+1}-1-e\right)\left(\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-2\mu
  e+6\mu-1\right)}, \\*
\left.\sqrt\frac{6\mu-1-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}}{6\mu-1
  +\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}}\right)
  \left(\frac1{\mu\sqrt
  e}+\frac{\sqrt{e^2+1}+e}{\sqrt{e^2+1}-e}\right) \cdot\\*\cdot
  \sqrt\frac{4\mu^2e^2+4\mu^2-4\mu+1}{e^2+1} +\\+\frac2{\sqrt e}
\Pi\left(\sqrt\frac{\left(\tan\frac\xi4+1\right)\left(2\mu
  e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-6\mu+1\right)}
  {\left(\tan\frac\xi4-1\right)\left(2\mu
  e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}+6\mu-1\right)},
  \right. \\*
  \frac{\left(1-2\mu+2\mu
e+\sqrt{4\mu^2e^2+4\mu^2-4\mu+1}\right)}{\left(2\mu-1+2\mu
  e+\sqrt{4\mu^2e^2+4\mu^2-4\mu+1}\right)} \cdot\\*\cdot
  \frac{\left(\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-2\mu
e-6\mu+1\right)}{\left(\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-2\mu
  e+6\mu-1\right)}, \\*
\left.\sqrt\frac{6\mu-1-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}}{6\mu-1
  +\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}}\right)
  +\\
  +\frac2{\sqrt e} \Pi\left(\sqrt\frac{\left(\tan\frac\xi4+1\right)\left(2\mu
  e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-6\mu+1\right)}
  {\left(\tan\frac\xi4-1\right)\left(2\mu
  e-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}+6\mu-1\right)},
  \right. \\*
  \frac{\left(2\mu-1-2\mu
e+\sqrt{4\mu^2e^2+4\mu^2-4\mu+1}\right)}{\left(1-2\mu-2\mu
  e+\sqrt{4\mu^2e^2+4\mu^2-4\mu+1}\right)} \cdot\\*\cdot
  \frac{\left(\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-2\mu
e-6\mu+1\right)}{\left(\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}-2\mu
  e+6\mu-1\right)}, \\*
\left.\left.\sqrt\frac{6\mu-1-\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}}{6\mu-1
  +\sqrt{4\mu^2e^2+36\mu^2-12\mu+1}}\right)
  \sqrt\frac{\tan\frac\xi4-1}{\tan\frac\xi4+1}\right] \,.
\end{multline}}
