% Words:   1648
\chapter{The null geodesics in the Schwarzschild
space-time}\label{n}
% A geodesic is \emph{a curve whose tangent is parallely 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 the
 Lagrangian has the value $0$~for null geodesics which describe the motion of
 massless freely falling particles.

 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}
are % 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{s90null}). Moreover, since %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
a geodesic is described in an invariant plane
(\(\theta = \pi/2\)), equation (\ref{s86}) 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 $L$ represents angular momentum about an axis normal to the invariant
 plane.

 The constancy of the Lagrangian for null geodesics 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} = 0\,.
%     \label{s89null}
% \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) \frac{L^2}{r^2} = E^2\,.
     \label{s90null}
 \end{equation}
 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 regarding $r$ as a function of $\varphi$
and substituting $r$ with \(u = r^{-1}\) as the independent variable,
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}
is obtained, where
\begin{equation} \label{ip}
 D = L/E
\end{equation}
represents the impact parameter.  This equation must be considered together with
 (\ref{s85}) and
                               (\ref{s91}).



\section{The radial geodesics}
The relevant equations are
\begin{equation}
     \frac{\mbox{d}r}{\mbox{d}\tau} = \pm E
\end{equation}
because \(\mbox d\varphi/\mbox d\tau = 0\), and so \(L = 0\).  Then from
(\ref{s85}),
%Accordingly,
%\begin{equation}\label{s217null}
%    \frac{\mbox{d}r}{\mbox{d}\tau} = \pm \left(1
%    -\frac{2M}r\right) \,,
%\end{equation}
after integration,
\begin{equation}\label{s218null}
    t = \pm r_* +\mbox{constant}_\pm \,,
\end{equation}
where
\begin{equation}\label{s219null}
    r_* = r +2M\ln\left(\frac r{2M} -1\right) \,.
\end{equation}

%The variable $r_*$ is customarily defined by the equation
%(cf.~equation (\ref{s74}))
%\begin{equation}\label{s220null}
%    \frac{\mbox d}{\mbox dr_*} = \frac\varDelta{r^2}\frac{\mbox
%    d}{\mbox dr} \,,
%\end{equation}
%where
%\begin{equation}\label{s221null}
%    \varDelta = r^2 -2Mr = r^2\mbox e^{2\nu}
%\end{equation}
%is the \emph{horizon function}. %With $

Equation (\ref{s218null}) must be compared with the equation
\begin{equation}\label{s223null}
    r = \pm E\tau +\mbox{constant}_\pm
\end{equation}
which relates $r$ to the proper time $\tau$. From equations
(\ref{s218null}) and (\ref{s223null}) it is evident that while the radial
geodesic crosses the horizon in its own proper time, it takes an infinite
coordinate-time to reach the horizon.


\section{The unbound orbits} \label{nu}
%Returning to the general equation (\ref{s214null}), we first
%consider
The different cases that must be distinguished
are related to the location of the roots of the cubic
equation
\begin{equation}\label{s225null}
    \mbox f(u) = 2Mu^3 -u^2 + \frac1{D^2} = 0 \,.
\end{equation}
The sum and the product of the roots $u_1$, $u_2$, and $u_3$ of
this equation are given by
\begin{equation}\label{s226null}
    u_1 +u_2 +u_3 = \frac1{2M} \qquad \mbox{and} \qquad u_1u_2u_3
    = -\frac1{2MD^2} \,.
\end{equation}
The equation f\((u) = 0\) must allow a negative real root;
and the two remaining roots may either be real (distinct or
coincident) or be a complex-conjugate pair.
The three cases are
distinguished in fig.~\ref{fig}.
%\setlength{\unitlength}{1mm}
\begin{figure}
    \begin{picture}(360,142)
        \multiput(0,62)(124,0){3}{\line(1,0){105}}
        \multiput(26,24)(124,0){3}{\line(0,1){110}}
        \multiput(5,127)(124,0){3}{f($u$)}
\multiput(0,0)(124,10){3}{\qbezier(9,52)(15,90)(35,90)}
\multiput(0,0)(124,10){3}{\qbezier(35,90)(45,90)(54,68)}
\multiput(0,0)(124,10){3}{\qbezier(54,68)(60,52)(78,52)}
\multiput(0,0)(124,10){3}{\qbezier(78,52)(88,52)(109,87)}
        \put(355,59){$u$}
        \put(133,62){\line(-1,-6){2}}
        \put(257,72){\line(-1,-6){4}}
        \multiput(12,54)(122,0){3}{$u_1$}
        \put(48,54){$u_2$}
        \put(92,54){$u_3$}
        \put(200,62){\circle*{4}}
        \put(182,54){$u_2=u_3$}
        \put(50,0){($\alpha$)}
        \put(179,0){($\beta$)}
        \put(303,0){($\gamma$)}
        \linethickness{2pt}
        \put(26,62){\line(1,0){31}}
        \put(92,62){\line(1,0){15}}
        \multiput(150,62)(124,0){2}{\line(1,0){79}}
    \end{picture}
    \caption{\color{blue}The disposition of the roots of a cubic f$(u)=0$.}
    \label{fig}
\end{figure}
\begin{itemize}
    \item Case ($\alpha$): For every pair of values $E$ and $L$, which
    permits three real roots $u_1<0<u_2<u_3$, there exists two
    distinct orbits limited, respectively, to the intervals
    $0<u\leq u_2$ and $u\geq u_3$, i.e. orbits of the first kind and
    orbits of the second kind. The orbits of the first kind are the
    analogues of Newtonian hyperbolae, while the orbits of the second kind are unique to Einsteinian gravity.     \item Case ($\beta$): The two kinds of orbits coalesce as they
    asymptotically approach a common circle by
    spiralling round it from opposite sides an infinite number of times.
    \item Case ($\gamma$): These orbits are most conveniently parameterized by imaginary eccentricity. They are similar to the radial geodesics in view of the fact that they start from infinity and
end in the singularity.
\end{itemize}









\subsection{The geodesics of the first kind}\label{sn1} First the
case when all the roots of the cubic equation f\((u) = 0\) are
real and the two positive roots are distinct will be looked at. Let the
roots be
\begin{equation}\label{s246null}
    u_1 = \frac{P-2M-Q}{4MP}\ (<0)\,, \quad u_2 = \frac1P \,,
    \quad \mbox{and} \quad u_3 = \frac{P-2M+Q}{4MP} \,,
\end{equation}
where $P$ denotes the perihelion distance and $Q$ is a constant to
be soon specified. The sum of the roots has been arranged to
be equal to $\frac1{2M}$ as required by equation
(\ref{s226null}). The arrangement of
the roots, \(u_1 < u_2 < u_3\), demands that
\begin{equation}\label{s247null}
    Q +P -6M > 0 \,.
\end{equation}

Next, % evaluating
%\begin{equation}\label{s248null}
%    \mbox f(u) = 2M(u-u_1)(u-u_2)(u-u_3)
%\end{equation}
with $u_1$, $u_2$, and $u_3$ as specified in equation
(\ref{s246null}), and %comparing the result with
the expression
(\ref{s225null}), we obtain the relations
\begin{equation}\label{s249null}
    Q^2 = (P-2M)(P+6M)
\end{equation}
and
%\begin{equation}\label{s250null}
%    \frac1{D^2} = \frac{Q^2-\left(P-2M\right)^2}{8MP^3} \,.
%\end{equation}
%With the aid of the first of these relations, the second
%simplifies to give
\begin{equation}\label{s251null}
    D^2 = \frac{P^3}{P-2M} \,.
\end{equation}

The relation (\ref{s249null}) and the inequality
(\ref{s247null}) give
\begin{equation}\label{s253null}
    P > 3M \,.
\end{equation}
Additionally, from equation (\ref{s251null}), the
inequality
\begin{equation}\label{s254null}
    D > 3M\sqrt 3 = D_{\mbox{\scriptsize c}} \qquad (\mbox{say})
   % \,.
\end{equation}
can be obtained.
Hence, the orbits in question lay entirely outside the circle \(r = 3M\),
and have impact
parameters bigger than the critical value.


By a substitution
\begin{equation}\label{s255null}
    u -\frac1P = -\frac{Q-P+6M}{8MP}(1+\cos\chi) \,,
\end{equation}
or, equivalently,
\begin{equation}\label{s256null}
    u +\frac{Q-P+2M}{4MP} = \frac{Q-P+6M}{8MP}(1-\cos\chi) \,,
\end{equation}
it can be determined that \begin{equation}\label{s257null}
    %\left.
    \begin{aligned}%{clr}
        u = \frac{\displaystyle 1}{\displaystyle P} \qquad& \mbox{when}
        \qquad
        \chi = \pi \,, & \mbox{and}\\
        u = 0 \ \mbox{and} \ r \rightarrow \infty \ & \mbox{when}
        \quad \sin^2{\frac\chi2} = \frac{\displaystyle Q-P+2M}{\displaystyle
        Q-P+6M} =
        \sin^2{\frac{\chi_\infty}2} & \mbox{(say)} \,.
    \end{aligned}
    %\right\}
\end{equation}
The substitutions (\ref{s255null}) and
(\ref{s256null}) also reduce equation (\ref{s214null}) to
\begin{equation}\label{s258null}
    \left(\frac{\mbox d\chi}{\mbox d\varphi}\right)^2 = \frac
    QP \left(1-k^2\sin^2{\frac\chi 2}\right) \,,
\end{equation}
where
\begin{equation}\label{s259null}
    k^2 = \frac{Q-P+6M}{2Q} \,,
\end{equation}
and $Q, P$ are defined in \eqref{s249null} and \eqref{s251null}
respectively. The solution for $\varphi$ than can be represented
by the Jacobian elliptic integral in the form
\begin{equation}\label{s260null}
    \varphi = 2[\K(k)-\F(\chi/2, k)]\sqrt{\frac PQ} \,,
\end{equation}
where the origin of $\varphi$ has been chosen at the perihelion
passage when \(\chi = \pi\) (see equations (\ref{s257null})).% The
%asymptotic value of $\varphi$, as \(r \rightarrow \infty\) is
%given by
%\begin{equation}\label{s261null}
%    \varphi_\infty = 2[\K(k)-\F(\chi_\infty/2, k)]\sqrt{\frac PQ} \,,
%\end{equation}
%where $\chi_\infty$ is specified in (\ref{s257null}).


\subsubsection{The critical orbits}
The conditions for occurrence of two coincident roots can be
described as follows. The derivative of equation (\ref{s225null})
allows \(u = (3M)^{-1}\) as a root; and if
\begin{equation}\label{s228null}
   % D^2 = 27M^2 \qquad \mbox{or}\qquad
    D = 3M\sqrt 3 \,, \end{equation}
\(u = (3M)^{-1}\) will be a double root of equation (\ref{s225null}).
From the equation
(\ref{s226null}), f\((u) = 0\) can be found.
%\begin{equation}\label{s229null}
%    u_1 = -\frac1{6M} \qquad \mbox{and} \qquad u_2 = u_3 =
%    \frac1{3M} \,.
%\end{equation}
The full equation, suitable for this case,
\begin{equation}\label{s230null}
    \left(\frac{\mbox du}{\mbox d\varphi}\right)^2 =
    2M \left(u+\frac1{6M}\right) \left(u-\frac1{3M}\right)^2
\end{equation}
is satisfied by the substitution
\begin{equation}\label{s231null}
    u = -\frac1{6M} +\frac1{2M}\tanh^2\frac{\varphi-\varphi_0}2
    \,,
\end{equation}
where $\varphi_0$ is a constant of integration. For $\varphi_0$ such
that
\begin{equation}\label{s232null}
    \tanh^2\frac{\varphi_0}2 = \frac13 \,,
\end{equation}
it can be inferred that
\begin{equation}\label{s233null}
    u = 0 \qquad \mbox{and} \qquad r \rightarrow \infty \qquad
    \mbox{when} \qquad \varphi = 0 \,.
\end{equation}
In addition it can be observed that
\begin{equation}\label{s234null}
    u = \frac1{3M} \qquad \mbox{when} \qquad \varphi \rightarrow
    \infty \,.
\end{equation}
So, a null geodesic, which comes from infinity with an impact
parameter \( D = 3M\sqrt3 \), asymptotically approaches the circle of
radius
\(3M\) (unstable circular orbit), by spiralling around it.


\subsection{The geodesics of the second kind}\label{sn2}
In this section, in place of (\ref{s255null}), the substitution
\begin{equation}\label{s271null}
    u = \frac1P +\frac{Q+P-6M}{4MP}\sec^2{\frac\chi2} %\,,
\end{equation}
is made
to obtain the null geodesics, whose ranges are \( u_3 \leq u <
\infty \). %By this substitution,
%\begin{equation}\label{s272null}
    %\left.
%    \begin{aligned}%{c}
%        u \ \mbox{is at aphelion when} \ &u = u_3 =
%        {\displaystyle\frac{Q+P-2M}{4MP}} \ \mbox{and} \ \chi = 0\\
%        \mbox{and} \qquad u \rightarrow \infty \ \mbox{and} \ r \rightarrow
%        0 \ \mbox{when} \ &\chi = \pi \,.
%    \end{aligned}
    %\right\}
%\end{equation}
This reduces equation
(\ref{s214null}) to the same form (\ref{s258null}) with the same
value of $k^2$. We may now write (cf. equation (\ref{s260null}))
\begin{equation}\label{s273null}
    \varphi = 2\F(\chi/2, k)\sqrt{\frac PQ} \,,
\end{equation}
where the origin of $\varphi$ is now at aphelion passage.
\begin{figure}
    \epsfxsize=13cm
    \epsfbox{schwarzs/null2.eps}
    \caption{Orbits of the first and
    the second
    kind with $P = 1$. 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.
    ($\color{blue}M = 1/7$ in the scale along the
    coordinate axes).}
    \label{Schwnull}
\end{figure}
An example of general null geodesics of the first and second kind is in figure~\ref{Schwnull}.


\subsubsection{The critical orbits}
With the orbit (\ref{s231null}), there must be an associated orbit
of the second kind, which starts at the singularity and,
from the opposite side, asymptotically approaches the same circle at \( r =
3M
\), by spiralling around it. Such an orbit can be
obtained by the substitution
\begin{equation}\label{s235null}
    u = \frac1{3M} + \frac1{2M}\tan^2\frac\xi2
\end{equation}
in the same equation (\ref{s230null}). %We find that the equation
%then reduces to
%\begin{equation}\label{s236null}
%    \left(\frac{\mbox d\xi}{\mbox d\varphi}\right)^2 =
%    \sin^2\frac\xi2 \,;
%\end{equation}
%and we may take
%\begin{equation}\label{s237null}
%    \varphi = 2\ln\tan\frac\xi4 \qquad \mbox{or} \qquad \tan\frac\xi4 =
%    \mbox{e}^{\varphi/2}
%\end{equation}
Inserting the appropriate solution in equation
(\ref{s235null}), we obtain
\begin{equation}\label{s238null}
    u = \frac1{3M} +\frac{2\mbox e^\varphi}{M\left(\mbox e^\varphi
    -1\right)^2}\,.
\end{equation}
Along this orbit
\begin{equation}\label{s239null}
    %\left.
    \begin{aligned}%{ll}
        & u \rightarrow \infty \qquad \mbox{and} \qquad r \rightarrow 0
        \qquad \mbox{when} \qquad  \varphi \rightarrow 0\\
        \mbox{and} &\\
        & u \rightarrow (3M)^{-1} \qquad \mbox{as} \qquad
        \varphi \rightarrow \infty \,.     \end{aligned}
    %\right\}
\end{equation}
%it has accordingly attributes that were stated.
The orbit approaches the circle at $\frac{1}{3M}$,
asymptotically, by spiraling around it an infinite number of times (cf.
section \ref{sn1}).
There is one important difference from the behavior of the orbits of the
first kind: the circle at $\frac{1}{3M}$ is
the perihelion for the orbits of the first kind, whereas it is the
aphelion for the orbits of the second kind (see figure~\ref{Schwnullcirc}).
\begin{figure}
    \epsfxsize=13cm
    \epsfbox{schwarzs/nullcirc.eps}
    \caption{An example of the orbits of the orbits of the first
    kind, that starts from infinity,
    and the second kind, that starts from a certain
    distance from the black hole (radius of the unstable circular orbit) and falls into the
    singularity. The two orbits coalesce ($D = 3M\sqrt3$).
    ($\color{blue}M = 1/7$ in the scale along the coordinate
    axes.)}
    \label{Schwnullcirc}
\end{figure}



\subsection{The orbits with imaginary eccentricities}\label{ni}
To conclude this chapter, the orbits when the equation f\( (u) = 0 \)
has a pair of complex-conjugate roots, besides a negative real
root, will be regarded. The
roots shall be expressed in terms of an imaginary eccentricity i$e\ (e>0)$
\begin{equation}
    u_1 = \frac{1}{2M}-\frac{2}{l}\,, \quad u_2 =
    \frac{1+\mbox{i}e}{l}\,, \quad \mbox{and} \quad u_3 =
    \frac{1-\mbox{i}e}{l}\,,
    \label{s171}
\end{equation}
where latus rectum, $l$, is some positive constant and the eccentricity
$e$ is more than 1. %On evaluating
%\begin{equation}\label{s274null}
%    \mbox f(u) = 2M \left(u-\frac1{2M}+\frac2l\right)
%    \left[\left(u-\frac1l\right)^2 +\frac{\mbox e^2}{l^2}\right]
%    \,,
%\end{equation}
%and comparing it with f\((u)\) given in equation (\ref{s214null}),
%we obtain the relations
%\begin{equation}\label{s275null}
%    l -M\left(3-\mbox e^2\right) = 0 \qquad \mbox{and} \qquad \frac1{2MD^2} =
%    \left(\frac2l-\frac1{2M}\right)\frac{1+\mbox e^2}{l^2}
%\end{equation}
%or, alternatively, in terms of \(\mu = M/l \), we have
%\begin{equation}\label{s276null}
%    \mbox e^2 = \frac{3\mu-1}\mu \qquad \mbox{and} \qquad
%    \frac{D^2}{M^2} = \frac1{\mu\left(4\mu-1\right)^2} \,.
%\end{equation}
%These relations require
%\begin{equation}\label{s277null}
%    \mu > 1/3 \qquad \mbox{and} \qquad D < 3M\sqrt3 \,.
%\end{equation}
These orbits have an impact parameter less than the orbits
studied in sections \ref{sn1} and \ref{sn2}.



Now, parameters $\mbox e^2$ and $\mu$ are no longer
 independent but related: \( \mbox e^2 = 3
-\mu^{-1} \).
With the assumption (\ref{s171}) regarding the roots of f$(u) = 0$, the
equation to be looked at is
\begin{equation}
    \left(\frac{\mbox{d}u}{\mbox{d}\varphi}\right)^2 =
    2M\left(u-\frac1{2M}+\frac2l\right)
\left[\left(u-\frac1l\right)^2+\frac{e^2}{l^2}\right]%\,.
    \label{s174}
\end{equation}
and the substitution is
\begin{equation}
    u~= \frac{1}{l}\left(1+e\tan\frac{\xi}{2}\right)\,.
    \label{s175}
\end{equation}
%In view of the requirement, $\mu > 1/4$,
The range of $\xi$ must be
terminated at $\xi_\infty$ where
\begin{equation}
   %\left.
    \begin{aligned}%{l}
        &{\displaystyle\tan\frac{\xi_\infty}2 = -e^{-1}} \,,\\
        \mbox{or},\\
        &{\displaystyle\sin\frac{\xi_\infty}2 = -\frac1{\sqrt{1+e^2}} \quad
        \mbox{and} \quad
        \cos\frac{\xi_\infty}2 = \frac{e}{\sqrt{1+e^2}}} \,.
    \end{aligned}
    %\right\}
    \label{s207}
\end{equation}
For, when $\xi = \xi_\infty$, $u$ becomes zero and when $\xi <
\xi_\infty$,
$u$ becomes negative. %Therefore, as $\xi \rightarrow \xi_\infty+0$, $u
%\rightarrow 0$ and $r \rightarrow \infty$. The allowed range of $\xi$
%is, therefore,
%\begin{equation}
%    \xi_\infty < \xi \leq \pi\,.
%    \label{s208}
%\end{equation}

With the substitution \eqref{s175}, equation (\ref{s174})
reduces to
\begin{equation}
    \frac{\mbox{d}\xi}{\mbox{d}\varphi} = \pm\sqrt{2[6\mu-1+2\mu
    e\sin\xi+(6\mu-1)\cos\xi]}\,.
    \label{s181}
\end{equation}
The solution
for $\varphi$ can be expressed in terms of Jacobian elliptic integral
%Thus,
\begin{equation}
    \pm\varphi =
    \frac1{\sqrt{{\mathsf\triangle}}}
\int\limits^\psi\frac{\mbox{d}\gamma}{\sqrt{1-k^2\sin^2\gamma}}\,,
    \label{s182}
\end{equation}
where
\begin{eqnarray}
    {\mathsf\triangle} &=& \sqrt{(6\mu-1)^2+4\mu^2e^2}\,,
    \label{s180}\\
    k^2 &=& \frac{{\mathsf\triangle}+6\mu-1}{2{\mathsf\triangle}}\,,
    \label{s183}
\end{eqnarray}
and
\begin{equation}
    \sin^2\psi =%& \frac{{\mathsf\triangle}-2\mu
    %e\sin\xi-(6\mu-1)\cos\xi}{{\mathsf\triangle}+6\mu-1}
    %\nonumber\\
    %&=&
    \frac{{\mathsf\triangle}+6\mu-1-2\left[2\mu
e\sin\frac\xi2+(6\mu-1)\cos\frac\xi2\right]\cos\frac\xi2}
{{\mathsf\triangle}
+6\mu-1}\,.
    \label{s184}
\end{equation}
The solution for $\varphi$ can be written as
\begin{equation}
    \varphi = \frac{\K(k)-\F(\psi, k)}{\sqrt{{\mathsf\triangle}}}\,.
    \label{s209}
\end{equation}
In particular, as the orbit goes off to infinity,
\begin{equation}\label{s278null}
    \varphi_\infty = \frac{\mbox K(k) - \mbox F(\psi_\infty, k)}
    {\sqrt\triangle}
\end{equation}
where $\psi_\infty$ %(on substituting for $\mbox e^2$ its present
is given by
\begin{equation}\label{s279null}
    \sin^2\psi_\infty = \frac{\triangle+1}{\triangle+6\mu-1} \,.
\end{equation}
%where it should be remembered that \( \mu > 1/3 \); and further
%that
%\begin{equation}\label{s280null}
%    \triangle = \sqrt{48\mu^2-16\mu+1} \,.
%\end{equation}
