% 3872 words
%documentclass{book}
%\begin{document}
\sloppy
\chapter{Derivation of the metrics}
\label{metriky}
The general Theory of Relativity (GTR) was developed by Einstein in
order to discuss gravitation. He postulated the {\em principle of
covariance\/}, which asserts that the laws of physics must be
independent of the space-time coordinates. Einstein also
introduced the {\em principle of equivalence\/}, which states that
in an arbitrary gravitational field no local experiment can
distinguish a freely falling nonrotating system (local inertial
system) from a uniformly moving system in the absence of a
gravitational field.


The metric field depends on the gravitating masses and energies
which are present. This is expressed quantitatively in {\em
Einstein's field equations\/},
\begin{equation}
    R_{ij} = \frac{8\pi G}{c^4}\left(T_{ij} -\frac{1}{2}g_{ij}T\right)
    \qquad (T =
    g^{ij}T_{ij})
    \label{chandr1-236}\,
\end{equation}
where $T_{ij}$ denotes the {\em energy-momentum tensor\/} of the matter
and of the field (other than gravitational) which may be present and $G$
is the constant of gravitation (which, like $c$, we shall set equal to
1).
They are the core of GTR. Einstein discovered these after a long and
difficult search, and announced them in November 1915. He was also able
to show that this theory reduces to the Newtonian one, in lowest order,
for the case of weak field and slow motions of the masses which serve as
sources of the gravitational field.

Today we know that objects having extremely strong gravitational
field exist in the Universe. Not even the most repulsive nuclear
forces can always prevent the final collapse to a black hole. For
such dramatic events GTR must be used.



\section{Derivation of the spherically symmetric Schwarzschild
solution}
\label{is} The Schwarzschild metric is a spherically
symmetric solution of Einstein's equation for the vacuum. A~metric
of the form
\begin{equation}
    \mbox{d}s^2 = \mbox{e}^{2\nu}\left(\mbox{d}t\right)^2
    -\mbox{e}^{2\mu_2}\left(\mbox{d}r\right)^2
    -r^2[\left(\mbox{d}\theta\right)^2
    +\left(\mbox{d}\varphi\right)^2\sin^2\theta]\,,
    \label{straumann311}
\end{equation}
where $\nu$ and $\mu_2$ are functions of $t$ ( \(= x^0 =
-\mbox{i}x^4\)) and $r$, is spherically
symmetrical.

We must now insert the ansatz (\ref{straumann311}) into the field
equations. For this, it is necessary to compute the Ricci tensor (or
the Einstein tensor) corresponding to the metric
(\ref{straumann311}), which can be accomplished with the help of
the Cartan calculus \cite{chandrasekhar, straumann} (the
traditional computation using the Christoffel symbols can be found
in \cite{spain}). The only non-vanishing components
of the Riemann tensor are
\begin{equation}
    \begin{aligned}
        -R_{1212} &= -R_{2323} = +\frac{\mbox{e}^{-2\mu_2}}{r}\mu_{2,r}\,
        ,\\
        -R_{1414} &= -R_{3434} = -\frac{\mbox{e}^{-2\mu_2}}{r}\nu_{,r}\,
        ,\\
        -R_{2334} &= +R_{1214} =
        -\frac{\mbox{e}^{-\nu-\mu_2}}{r}\mu_{2,4}\,
        ,\label{s64}\\
        -R_{1313} &= r^{-2}\left(1-\mbox{e}^{-2\mu_2}\right)\, ,\\
        \text{and}\qquad \\
        -R_{2424}
        &=-\mbox{e}^{-\mu_2-\nu}
        \left[\left(\mbox{e}^{\mu_2-\nu}\mu_{2,4}\right)_{,4}
        +\left(\mbox{e}^{\nu-\mu_2}\nu_{,r}\right)_{,r}\right]\, .
    \end{aligned}
\end{equation}
%where these are the tetrad components in a frame with the
%basis vector's
%\begin{equation}
%    e_{(a)i} =
%    \begin{vmatrix}
%        \mbox{e}^\nu & 0             & 0               & 0  \\
%        0            & -r\sin\theta  & 0               & 0  \\
%        0            & 0             & -\mbox{e}^\mu_2 & 0  \\
%        0            & 0             & 0               & -r \\
%    \end{vmatrix}
%    \, .
%  %  \label{s65}
%\end{equation}

The field equations now follow from setting the various components
of the Ricci tensor equal to zero. %Thus,
%\begin{equation}
%    -R_{42} = R_{1412} +R_{3432} =
%    -2\frac{\mbox{e}^{-\nu-\mu_2}}{r}\mu_{2,4} = 0 \,.
%   % \label{s66}
%\end{equation}
Therefore
\begin{equation}
    \mu_2 \equiv \mu_2(r) 
    \label{s67}
\end{equation}
%Considering next the equation,
%\begin{equation}
%    -R_{44} +R_{22} = R_{1414} +R_{3434} -(R_{1212} +R_{2323}) = 0 \,,
%   % \label{s68}
%\end{equation}
%we find
%\begin{equation}
%    \frac{2}{r}\,\mbox{e}^{-2\mu_2}\left(\nu +\mu_2\right)_{,r} = 0 \,.
%   % \label{s69}
%\end{equation}
and
\begin{equation}
    \nu = -\mu_2 \,.
    \label{s71}
\end{equation}
The solution can now be completed by considering the equation
\begin{equation}
    R_{33} = R_{1313} +R_{2323} +R_{4343} = 0 \,.
    \label{s72}
\end{equation}
Hence
\begin{equation}
    -2\mu_{2,r}\frac{\mbox{e}^{-2\mu_2}}{r} = \frac{1}{r}\left(1
    -\mbox{e}^{-2\mu_2}\right) \,.
    \label{s73}
\end{equation}
The solution of this equation is
\begin{equation}
    \mbox{e}^{-2\mu_2} = 1 -2M/r \,,
    \label{s74}
\end{equation}
where M is a constant, and it can be identified physically with the rest
mass of the gravitating particle by comparison with Newtonian limit at
large distances. We thus obtain the \emph{Schwarzschild
solution} in the most familiar form
(and also in the form in which Schwarzschild
originally wrote it)
\begin{equation}
    \mbox{d}s^2 = \left(1-\frac{2M}{r} \right)\left(\mbox{d}t\right)^2
    -\frac{\left(\mbox{d}r\right)^2}{1-2M/r}
    -r^2\left[\left(\mbox{d}\theta\right)^2
    +\left(\mbox{d}\varphi\right)^2\sin^2\theta\right]\,.
    \label{s60}
\end{equation}
It consists of an event horizon (at $r=2M$) and a singularity at
the center. Since the metric coefficients are explicitly
independent of time and there is no dragging of the inertial frame
the space-time is \emph{static} as experienced by an observer
external to the horizon. The radial coordinate $r$ is a
`luminosity distance' in the sense that the surface area of a
sphere of `radius' $r$ is $4\pi r^2$, $t$~is a
measure  of proper time for an observer at rest at infinity. %The metric
%represents the space-time external to a spherical distribution of
%inertial mass $M$.

If a signal is emitted from inside the horizon, it will not reach a
distant observer. The stellar matter is cut off from the outside world.
The horizon is the boundary of the region which is causally connected to
a distant observer. Thus it acts like a one way membrane through which
energy and information can pass to the interior, but not to the exterior. The
singularity has no causal connection to an external observer; he
cannot ``see" it.

An external observer far away from the collapsing star will see it
reach the horizon only after an infinitely long time as a result of
the gravitational time dilatation. However, in practice, the star
will become invisible, since the red shift will start to increase,
and luminosity decreases. Afterwards, we are dealing with a
``black hole".


\section{Reissner--Nordstr\"om solution}
\label{irn}
A~spherically symmetric solution of the coupled
equations of Einstein and of Maxwell represents
a~black hole with mass and charge. % Since it represents a
%spherically symmetric space-time,
We can, as in section~\ref{is}, write the line
element
\begin{equation}
    \mbox{d}s^2 = \mbox{e}^{2\mu_2}\left[\left(\mbox{d}x^0\right)^2
    -\left(\mbox{d}x^2\right)^2\right]
    -\mbox{e}^{2\mu_3}\left[\left(\mbox{d}\theta\right)^2
    +\left(\mbox{d}\varphi\right)^2\sin^2\theta\right]\,,
    \label{r2}
\end{equation}
where $\mu_2$ and $\mu_3$ are functions only of $x^0$ and $x^2$.
As we are now considering a space-time, in which an 
electromagnetic field is present, we cannot set the Ricci tensor to
zero, as we did in \ref{is}. Instead we must set
$
    R_{ab} = -2\left(\eta^{cd}F_{ac}F_{bd}
    -\frac{1}{4}\eta_{ab}F_{ef}F^{ef}\right)
  %  \label{r3}
$,
where $F_{ab}$ denotes the tetrad components of the Maxwell tensor and
$\eta_{ab}$ denotes the Minkowskian metric of the chosen ortho-normal
tetrad-frame.

\subsection{The solution of Maxwell equations}
The axial components of the Maxwell tensor %($F_{ab}$)
must vanish. %Its polar components can be written down as
%\cite{chandrasekhar}
%\begin{equation}
%    \begin{aligned}
%       &\left(\mbox{e}^{2\mu_3}F_{02}\right)_{,2} +\mbox{e}^{\mu_2
%       +\mu_3}F_{03}\cot\theta &= 0\, ,\\
%       &\left(\mbox{e}^{2\mu_3}F_{02}\right)_{,0} +\mbox{e}^{\mu_2
%        +\mu_3}F_{23}\cot\theta &= 0\, ,\\
%       & \left(\mbox{e}^{\mu2 +\mu_3}F_{23}\right)_{,2} -\left(\mbox{e}^{\mu_2
%        +\mu_3}F_{03}\right)_{,0} &= 0\label{r4}\\
%        \text{and} \qquad\\
%       & \left(\mbox{e}^{\mu2 +\mu_3}F_{23}\right)_{,0} -\left(\mbox{e}^{\mu_2
%        +\mu_3}F_{03}\right)_{,2} &= 0\, .
%    \end{aligned}
%\end{equation}
As none of the components of the Maxwell tensor can depend
on $\theta$, %it follows that
%\begin{equation}
%    F_{03} = F_{23} =0 \, .
%  %  \label{r5}
%\end{equation}
%and
%\begin{equation}
%    \left(\mbox{e}^{2\mu_3}F_{02}\right)_{,2} =
%    \left(\mbox{e}^{2\mu_3}F_{02}\right)_{,0} = 0
%    \, .
%  %  \label{r6}
%\end{equation}
the only non-vanishing component of the Maxwell tensor is
$F_{02}$ which must have the form
\begin{equation}
    F_{02} = -Q_*\mbox{e}^{-2\mu_3}\, ,
    \label{r7}
\end{equation}
where $Q_*$ is a constant.

With \(F_{03} = F_{23} = 0\) and $F_{02}$ given by equation (\ref{r7}),
we find that
\begin{equation}
    -R_{44} = R_{00} = R_{11} = -R_{22} = R_{33} =
    Q_*^2\mbox{e}^{-4\mu_3}\, ,\ \mbox{and}\ R_{42} = 0\, .
    \label{r8}
\end{equation}


\subsection{Derivation of the Reissner--Nordstr\"om metric}
A similar derivation to one for the Schwarzschild metric in section~\ref{is},
can be given. Hence, with a metric of the form
(\ref{straumann311}), we have the same expressions (\ref{s64}) for the
non-vanishing components of the Riemann tensor. And with (\ref{r8}), the
conclusions expressed in (\ref{s67}) and (\ref{s72}) continue to hold.
But in place of (\ref{s73}) there is now (see
(\ref{r8}))
\begin{equation}
    2\mu_{2,r}\frac{\mbox{e}_{-2\mu_2}}{r} +\frac{1}{r^2}\left(1
    -\mbox{e}^{-2\mu_2}\right) = R_{33} = \frac{Q_*}{r^4}\, .
  %  \label{r61}
\end{equation}
The solution of this equation is
\begin{equation}
    \mbox{e}^{-2\mu_2} = 1 -\frac{2M}{r} +\frac{Q_*^2}{r^2} \quad
    \left(= \mbox{e}^{2\nu} = \frac{\varDelta}{r^2}\right)\, ;
    \label{r62}
\end{equation}
and we have the solution in its standard form
\begin{equation}
    \mbox{d}s^2 = \frac{\varDelta}{r^2}\left(\mbox{d}t\right)^2
    -\frac{r^2}{\varDelta}\left(\mbox{d}r\right)^2
    -r^2\left[\left(\mbox{d}\theta\right)^2
    +\left(\mbox{d}\varphi\right)^2\sin^2\theta\right]\, .
    \label{r48}
\end{equation}

Asymptotically the electric field is radial and the corresponding
Gaussian flux integral is \(4\pi Q_*\). This shows that $Q_*$ is
the \emph{charge} of the black hole.
%Now we can see (comparing this with (\ref{r2}) and (\ref{r7}), and
%classical value for electric field around a spherical body) that
%$Q_*$ is in fact $\rho\over4\epsilon_0\pi$, but we shall refer to it as
%``charge".




\section{Axisymmetric space-time -- The Kerr Solution}
\label{ik}
A~metric appropriate to stationary axisymmetric space-times can be
written as
\begin{equation}
    \mbox{d}s^2 = \mbox{e}^{2\nu}\left(\mbox{d}t\right)^2
    -\mbox{e}^{2\psi}\left(\mbox{d}\varphi -\omega\mbox{d}t\right)^2
    -\mbox{e}^{2\mu_2}\left(\mbox{d}x^2\right)^2
    -\mbox{e}^{2\mu_3}\left(\mbox{d}x^3\right)^2\,
    ,
    \label{k1}
\end{equation}
where \(\nu, \psi, \omega\), and $\mu_3$ are functions of $x^2$ and
$x^3$ with the freedom to impose coordinate conditions on $\mu_2$ and
$\mu_3$.


 The vanishing and the non-vanishing components
of the Riemann tensor can be found (see \cite{chandrasekhar}).
%are \bal
%    R_{1212} &=
%    +\mbox{e}^{-\psi-\mu_2}\left(\mbox{e}^{\psi-\mu_2}\psi_{,2}\right)_{,2}
%    +\mbox{e}^{-2\mu_3}\psi_{,3}\mu_{2,3}
%    +\frac{1}{4}\,\mbox{e}^{2\psi-2\mu_2-2\nu}{\omega_{,2}}^2\,\nonumber
%    ,\\
%    R_{1313} &=
%    +\mbox{e}^{-\psi-\mu_3}\left(\mbox{e}^{\psi-\mu_3}\psi_{,3}\right)_{,3}
%    +\mbox{e}^{-2\mu_2}\psi_{,2}\mu_{3,2}
%    +\frac{1}{4}\,\mbox{e}^{2\psi-2\mu_3-2\nu}{\omega_{,3}}^2\,\nonumber
%    ,\\
%    R_{1010} &=
%    -\mbox{e}^{-2\mu_2}\psi_{,2}\nu_{,2}
%    -\mbox{e}^{-2\mu_3}\psi_{,3}\nu_{,3}
%    -\frac{1}{4}\,\mbox{e}^{2\psi-2\nu}
%    \left(\mbox{e}^{-2\mu_2}{\omega_{,2}}^2
%    +\mbox{e}^{-2\mu_3}{\omega_{,3}}^2\right)\,\nonumber ,\\
%    R_{2323} &=
%    +\mbox{e}^{-\mu_2-\mu_3}
%    \left[\left(\mbox{e}^{\mu_2-\mu_3}\mu_{2,3}\right)_{,3}
%    +\left(\mbox{e}^{\mu_3-\mu_2}\mu_{3,2}\right)_{,2}\right]\, ,\nonumber\\
%    R_{2020} &=
%    -\mbox{e}^{-\mu_2-\nu}\left(\mbox{e}^{\nu-\mu_2}\nu_{,2}\right)_{,2}
%    -\mbox{e}^{-2\mu_3}\nu_{,3}\mu_{2,3}
%    +\frac{3}{4}\,\mbox{e}^{2\psi-2\mu_2-2\nu}{\omega_{,2}}^2\,\nonumber
%    ,\\
%    R_{3030} &=
%    -\mbox{e}^{-\mu_3-\nu}\left(\mbox{e}^{\nu-\mu_3}\nu_{,3}\right)_{,3}
%    -\mbox{e}^{-2\mu_2}\nu_{,2}\mu_{3,2}
%    +\frac{3}{4}\,\mbox{e}^{2\psi-2\mu_3-2\nu}{\omega_{,3}}^2\,\nonumber
%    ,\\
%    R_{1213} &=
%    +\mbox{e}^{-\psi-\mu_3}\left(\mbox{e}^{\psi-\mu_2}\psi_{,2}\right)_{,3}
%    -\mbox{e}^{-\mu_2-\mu_3}\psi_{,3}\mu_{3,2}
%    +\frac{1}{4}\,\mbox{e}^{2\psi-\mu_2-\mu_3-2\nu}\omega_{,2}\omega_{,3}\,
%    \nonumber
%    ,\\
%    \begin{split}
%        R_{1220} &= +\mbox{e}^{\psi-2\mu_2-\nu}\omega_{,2}\left(\psi_{,2}
%        -\frac{1}{2}\nu_{,2}\right)
%        +\frac{1}{2}\,\mbox{e}^{-\mu_2-\nu}
%        \left(\mbox{e}^{\psi-\mu_2}\omega_{,2}\right)_{,2}
%        \\
%        & \quad
%        +\frac{1}{2}\,\mbox{e}^{\psi-2\mu_3-\nu}\omega_{,3}\mu_{2,3}\,
%        ,\\
%        R_{1330} &= +\mbox{e}^{\psi-2\mu_3-\nu}\omega_{,3}\left(\psi_{,3}
%        -\frac{1}{2}\nu_{,3}\right)
%        +\frac{1}{2}\,\mbox{e}^{-\mu_3-\nu}
%        \left(\mbox{e}^{\psi-\mu_3}\omega_{,3}\right)_{,3}
%        \\
%        & \quad +\frac{1}{2}\,\mbox{e}^{\psi-2\mu_2-\nu}\omega_{,2}\mu_{3,2}\,
%        ,
%    \end{split}\\
%    R_{3002} &=
%    +\mbox{e}^{-\nu-\mu_3}\left(\mbox{e}^{\nu-\mu_2}\nu_{,2}\right)_{,3}
%    -\mbox{e}^{-\mu_2-\mu_3\nu{,_3}\mu_{3,2}}
%    -\frac{3}{4}\,\mbox{e}^{2\psi-\mu_2-\mu_3-2\nu}\omega_{,2}\omega_{,3}\,
%    ,\nonumber\\
%    R_{1230} &=
%    +\frac{1}{2}\,
%    \mbox{e}^{-\mu_3-\nu}\left(\mbox{e}^{\psi-\mu_2}\omega_{,2}\right)_{,3}
%    +\frac{1}{2}\,\mbox{e}^{\psi-\mu_2-\mu_3-\nu}\omega_{,3}(2\psi_{,2}
%    -\mu_{3,2} -\nu_{,2})\, ,\nonumber\\
%    R_{1302} &=
%    -\frac{1}{2}\,
%    \mbox{e}^{-\mu_2-\nu}\left(\mbox{e}^{\psi-\mu_3}\omega_{,3}\right)_{,2}
%    -\frac{1}{2}\,\mbox{e}^{\psi-\mu_2-\mu_3-\nu}\omega_{,2}\left(2\psi_{,3}
%    -\mu_{2,3} -\nu_{,3}\right)\, ,\nonumber\\
%    R_{1023} &=
%    -\frac{1}{2}\,
%    \mbox{e}^{-\mu_2-\mu_3}
%    \left[\left(\mbox{e}^{\psi-\nu}\omega_{,3}\right)_{,2}
%    -\left(\mbox{e}^{\psi-\nu}\omega_{,2}\right)_{,3}\right]\,.\nonumber
%    \label{k3}
%\eal
%The components which vanish are
%%\begin{equation}
%%    \left.
%%    \begin{array}{c}
%%        R_{1210} = R_{1310} = R_{1223} = R_{1332} = R_{1002} = R_{1003}\\
%%        = R_{2330} = R_{3220} = 0 \,.
%%    \end{array}
%%    \right\}
%%\end{equation}
%\begin{equation}
%     R_{1210} = R_{1310} = R_{1223} = R_{1332} = R_{1002} = R_{1003}
%     = R_{2330} = R_{3220} = 0 \,.
%  %   \label{k4}
%\end{equation}

The components of the Ricci ($R_{ab}$) and the Einstein ($G_{ab}$)
tensors are given by linear combinations of the components
of the Riemann tensor \cite{chandrasekhar}. By setting them
equal to zero, the equations governing stationary
axisymmetric vacuum space-times can be obtained. It will be
sufficient to consider only the following equations:
\begin{gather}
    \begin{split}
        &\mbox{e}^{-2\mu_2}\left[\nu_{,2,2} +\nu_{2}\left(\psi +\nu -\mu_2
        +\mu_3\right)_{,2}\right] +\\
        &+\mbox{e}^{-2\mu_3}\left[\nu_{,3,3} +\nu_{,3}\left(\psi +\nu
        +\mu_2 -\mu_3\right)_{,3}\right] =\\
        &=+\frac{1}{2}\,
        \mbox{e}^{2\psi-2\nu}\left(\mbox{e}^{-2\mu_2}{\omega_{,2}}^2
        +\mbox{e}^{-2\mu_3}{\omega_{,3}}^2\right) \qquad \left(R_{00} =
        0\right) \,,
    \end{split}\label{k5}\\
    \begin{split}
        &\mbox{e}^{-2\mu_2}\left[\psi_{,2,2} +\psi_{,2}\left(\psi +\nu -\mu_2
        +\mu_3\right)_{,2}\right] \\
        &+\mbox{e}^{-2\mu_3}\left[\psi_{,3,3} +\psi_{,3}\left(\psi +\nu
        +\mu_2 -\mu_3\right)_{,3}\right] =\\
        &=-\frac{1}{2}\,
        \mbox{e}^{2\psi-2\nu}\left(\mbox{e}^{-2\mu_2}{\omega_{,2}}^2
        +\mbox{e}^{-2\mu_3}{\omega_{,3}}^2\right) \qquad \left(R_{11} =
        0\right) \,,
    \end{split}\label{k6}\\
    %\begin{equation}
    \left(\mbox{e}^{3\psi-\nu-\mu_2+\mu_3}\omega_{,2}\right)_{,2}
    +\left(\mbox{e}^{3\psi-\nu+\mu_2-\mu_3}\omega_{,3}\right)_{,3} = 0
    \qquad\left(R_{00}
    = 0\right) \,,
    \label{k7}\\
    %\end{equation}
    \begin{split}
        &\left(\psi +\nu\right)_{,2,3} -\left(\psi +\nu\right)_{,2}\mu_{2,3}
        - \left(\psi
        +\nu\right)_{,3}\mu_{3,2} +\psi_{,2}\psi_{,3} +\nu_{,2}\nu_{,3} =
        \\& =
        \frac{1}{2}\,\mbox{e}^{2\psi-2\nu}\omega_{,2}\omega_{,3} \qquad
        \left(R_{23}
        =
        0\right),
    \end{split}\label{k8}\\
    \begin{split}
        &\mbox{e}^{-2\mu_3}\left[\left(\psi +\nu\right)_{,3,3} +\left(\psi
        +\nu\right)_{,3}\left(\nu
        -\mu_3\right)_{,3} +\psi_{,3}\psi_{,3}\right] \\
        &+\mbox{e}^{-2\mu_2}\left[\nu_{,2}\left(\psi +\mu_3\right)_{,2}
        +\psi_{,2}\mu_{3,2}\right]
        =
        \\
        &=-\frac{1}{4}\,
        \mbox{e}^{2\psi-2\nu}\left(\mbox{e}^{-2\mu_2}{\omega_{,2}}^2
        -\mbox{e}^{-2\mu_3}{\omega_{,3}}^2\right) \qquad \left(G_{22} =
        0\right) \,,
    \end{split}\label{k9}\\
    \begin{split}
        &\mbox{e}^{-2\mu_2}\left[\left(\psi +\nu\right)_{,2,2} +\left(\psi
        +\nu\right)_{,2}\left(\nu
        -\mu_2\right)_{,2} +\psi_{,2}\psi_{,2}\right] \\
        &+\mbox{e}^{-2\mu_3}\left[\nu_{,3}\left(\psi +\mu_2\right)_{,3}
        +\psi_{,3}\mu_{2,3}\right]
        =
        \\
        &= +\frac{1}{4}\,
        \mbox{e}^{2\psi-2\nu}\left(\mbox{e}^{-2\mu_2}{\omega_{,2}}^2
        -\mbox{e}^{-2\mu_3}{\omega_{,3}}^2\right) \qquad \left(G_{33} =
        0\right) \,.
    \end{split}\label{k10}
\end{gather}

%Letting
%\begin{equation}
%    \beta = \psi +\nu
%   % \label{k11}
%\end{equation}
%we can rewrite equations (\ref{k5}) and (\ref{k6}) in the forms
%\bal
%    \left(\mbox{e}^{\beta+\mu_3-\mu_2}\nu_{,2}\right)_{,2}
%    +\left(\mbox{e}^{\beta+\mu_2-\mu_3}\nu_{,3}\right)_{,3} &=
%    +\frac{1}{2}\,
%    \mbox{e}^{3\psi-\nu}\left(\mbox{e}^{\mu_3-\mu_2}{\omega_{,2}}^2
%    +\mbox{e}^{\mu_2-\mu_3}{\omega_{,3}}^2\right)\, ,
%   % \label{k12}\raisetag{28pt}\\
%    \left(\mbox{e}^{\beta+\mu_3-\mu_2}\psi_{,2}\right)_{,2}
%    +\left(\mbox{e}^{\beta+\mu_2-\mu_3}\psi_{,3}\right)_{,3} &=
%    -\frac{1}{2}\,
%    \mbox{e}^{3\psi-\nu}\left(\mbox{e}^{\mu_3-\mu_2}{\omega_{,2}}^2
%    +\mbox{e}^{\mu_2-\mu_3}{\omega_{,3}}^2\right)\, .
%   % \label{k13}
%\eal
The sum and the difference of equations (\ref{k5}) and (\ref{k6}) give
\begin{gather}
    \left(\mbox{e}^{\mu_3-\mu_2}{\mbox{e}^\beta}_{,2}\right)_{,2}
    +\left(\mbox{e}^{\mu_2-\mu_3}{\mbox{e}^\beta}_{,3}\right)_{,3} = 0 \,,
    \label{k14}\\
    %\end{equation}
    \begin{split}
        &\left[\mbox{e}^{\beta+\mu_3-\mu_2}
        \left(\psi -\nu\right)_{,2}\right]_{,2}
        +\left[\mbox{e}^{\beta+\mu_2-\mu_3}
        \left(\psi -\nu\right)_{,3}\right]_{,3} = \\
        &=
        -\mbox{e}^{3\psi-\nu}\left(\mbox{e}^{\mu_3-\mu_2}{\omega_{,2}}^2
        +\mbox{e}^{\mu_2-\mu_3}{\omega_{,3}}^2\right) \,,
        \label{k15}
    \end{split}
\end{gather}
where $\beta = \psi +\nu$.

The addition of equations (\ref{k9}) and (\ref{k10}) gives the same
equation (\ref{k14}), while subtracting them gives
\begin{multline}
    4\mbox{e}^{\mu_3-\mu_2}\left(\beta_{,2}\mu_{3,2}
    +\psi_{,2}\nu_{,2}\right)
    -4\mbox{e}^{\mu_2-\mu_3}\left(\beta_{,3}\mu_{2,3}
    +\psi_{,3}\nu_{,3}\right) =
    \\
    =2\mbox{e}^{-\beta}
    \left[\left(\mbox{e}^{\mu_3-\mu_2}{\mbox{e}^\beta}_{,2}\right)_{,2}
    -\left(\mbox{e}^{\mu_2-\mu_3}{\mbox{e}^\beta}_{,3}\right)_{,3}\right]
    -\mbox{e}^{2\psi-2\nu}\left(\mbox{e}^{\mu_3-\mu_2}{\omega_{,2}}^2
    -\mbox{e}^{\mu_2-\mu_3}{\omega_{,3}}^2\right) \,.
    \label{k16}
\end{multline}

Exercising the gauge-freedom, we have to impose  coordinate condition on
$\mu_2$ and $\mu_3$ %, we may specify
%\begin{equation}
%    \mbox{e}^{2(\mu_3-\mu_2)} = \varDelta\left(x^2,x^3\right)
%   % \label{k17}
%\end{equation}
in some convenient manner. % Equation (\ref{k14}) then becomes an equation
%for \(\beta = \psi +\nu\). Equations (\ref{k7}) and (\ref{k15}) then
%become a pair of coupled equations for \(\psi -\nu\) and $\omega$.
The metric (\ref{k1}) we now write in the form
\begin{equation}
    \mbox{d}s^2 = \mbox{e}^\beta\left[\chi\left(\mbox{d}t\right)^2
    -\frac{1}{\chi}\left(\mbox{d}\varphi -\omega\mbox{d}t\right)^2 \right]
    -\frac{\mbox{e}^{\mu_2+\mu_3}}{\sqrt\varDelta}
    \left[\left(\mbox{d}x^2\right)^2
    +\varDelta\left(\mbox{d}x^3\right)^2\right]\,,
    \label{k18}
\end{equation}
where
\begin{equation}
    \chi = \mbox{e}^{-\psi+\nu}\,.
    \label{k19}
\end{equation}


\subsection{The choice of gauge}
\label{kgauge}
It is useful to choose the polar angle $\theta$ (with respect
to the axis of symmetry) as the spacial coordinate $x^3$. We shall
suppose that the metric allows an \emph{event horizon}, which we
shall define as a smooth
two--dimensional, null surface, which is spanned by the Killing 
vectors $\partial/\partial t$ and $\partial/\partial\varphi$.

With the stationarity and axisymmetry of the space-time,
the equation of the event horizon is
$
    N\left(x^2,x^3\right) = 0
   % \label{k35}
$.
The condition to be null is
$
    g^{ij}N_{,i}N_{,j} = 0 
   % \label{k36}
$.
For the chosen form of metric this equation gives
\begin{equation}
    \mbox{e}^{2(\mu_3-\mu_2)}{N_{,r}}^2 +{N_{,\theta}}^2 = 0 \,,
    \label{k37}
\end{equation}
where \(x^2 = r\). Because of the gauge freedom we can suppose that
\begin{equation}
    \mbox{e}^{2(\mu_3-\mu_2)} = \varDelta(r) \,,
    \label{k38}
\end{equation}
where $\varDelta(r)$ is some function of $r$. % From equation (\ref{k37}) it
%follows that the equation of the null surface is given by
%\begin{equation}
%    \varDelta(r) = 0 \,.
%   % \label{k39}
%\end{equation}
%The second condition that the null-surface be spanned by
%$\partial/\partial t$ and $\partial/\partial\varphi$ requires that the
%determinant of the metric of the subspace, ($t,\varphi$), vanish on
%\(\varDelta(r) =0 \):
%\begin{equation}
%    \mbox{e}^{2\beta} = 0 \quad \mbox{on} \quad \varDelta(r) = 0 \,.
%   % \label{k40}
%\end{equation}
We shall assume that $\mbox{e}^\beta$ is separable in the variables $r$
and $\theta$,
\begin{equation}
    \mbox{e}^\beta = \varDelta^{1/2}f(\theta) \,,
    \label{k42}
\end{equation}
where $f(\theta)$ is some function of $\theta$ which is regular on the axis
\(\theta = 0\).

With $\mbox{e}^{\mu_3-\mu_2}$ and $\mbox{e}^\beta$ given by equations
(\ref{k38}) and (\ref{k42}),  %equation (\ref{k14}) for $\mbox{e}^\beta$
%becomes
%\begin{equation}
%    \left(\varDelta^{1/2}{\varDelta^{1/2}}_{,r}\right)_{,r}
%    +\frac{1}{f}f_{,\theta,\theta}
%    = 0 \,.
%   %\label{k43}
%\end{equation}
a~solution of equation \eqref{k14}, compatible with the requirements of
regularity on the axis and convexity of the horizon, is determined by
$
    \varDelta_{,r,r} =2$ and  $f = \sin\theta$;
   % \label{k44}
and the appropriate solution for $\varDelta$ is
\begin{equation}
    \varDelta = r^2 -2Mr +a^2 \,,
    \label{k45}
\end{equation}
where $M$ and $a$ are constants.

Choice of gauge consistent with the existence of
an event horizon gives the solution
$
    \mbox{e}^{\mu_3-\mu_2} = \varDelta^{1/2}$ and
    $\mbox{e}^\beta = \varDelta^{1/2}\sin\theta
$,
where $\varDelta$ is given by equation (\ref{k45}).

With the solutions for $\mbox{e}^{\mu_3-\mu_2}$ and $\mbox{e}^\beta$
%given in equations (\ref{k46})
and by using
$
    \mu = \cos\theta
    %\label{k48}
$
instead of $\theta$ as the variable indicated by index `3', the equations (\ref{k15}) and
(\ref{k7}), respectively become
\begin{equation}
    \left[\varDelta\left(\psi -\nu\right)_{,2}\right]_{,2}
    +\left[\delta\left(\psi -\nu\right)_{,3}\right]_{,3} =
    -\mbox{e}^{2\left(\psi-\nu\right)}\left(\varDelta{\omega_{,2}}^2
    +\delta{\omega_{,3}}^2\right) \,,
    \label{k49}
\end{equation}
and
\begin{equation}
    \left[\varDelta\mbox{e}^{2(\psi-\nu)}\omega_{,2}\right]_{,2}
    +\left[\delta\mbox{e}^{2(\psi-\nu)}\omega_{,3}\right]_{,3} = 0 \,,
    \label{k50}
\end{equation}
where $
    \delta = 1 -\mu^2 = \sin^2\theta $.
Using the equation (\ref{k19}), the
equations (\ref{k49}) and (\ref{k50}) will have the forms
\begin{gather}
    \left(\frac\varDelta\chi\chi_{,2}\right)_{,2}
    +\left(\frac\delta\chi\chi_{,3}\right)_{,3} = \frac
    1{\chi^2}\left(\varDelta{\omega_{,2}}^2 +\delta{\omega_{,3}}^2\right) \,,
    \label{k52}\\
    %\end{equation}
    %\begin{equation}
    \left(\frac\varDelta{\chi^2}\omega_{,2}\right)_{,2}
    +\left(\frac\delta{\chi^2}\omega_{,3}\right)_{,3} = 0 \,.
    \label{k53}
\end{gather}
Letting
$X = \chi +\omega$ and $Y = \chi -\omega$,
   % \label{k56}
we obtain the pair of symmetric equations.
%\bal
%    \frac 1 2(X +Y)\left[\left(\varDelta X_{,2}\right)_{,2}
%    +\left(\delta X_{,3}\right)_{,3}\right] &=
%    \varDelta{X_{,2}}^2 +\delta{X_{,3}}^2 \,,\label{k57}\\
%    \frac 1 2(X +Y)\left[\left(\varDelta Y_{,2}\right)_{,2}
%    +\left(\delta Y_{,3}\right)_{,3}\right] &=
%    \varDelta{Y_{,2}}^2 +\delta{Y_{,3}}^2 \,.
%   % \label{k58}
%\eal

Equations (\ref{k8}) and (\ref{k16}) become
\begin{equation}
    -\frac\mu\delta\left(\mu_3 +\mu_2\right)_{,2} +\frac{r
    -M}\varDelta\left(\mu_3
    +\mu_2\right)_{,3} = \frac 2{\left(X +Y\right)^2}\left(X_{,2}Y_{,3}
    +Y_{,2}X_{,3}\right)
    \label{k59}
\end{equation}
and
\begin{multline}
    2(r -M)\left(\mu_3 +\mu_2\right)_{,3}
    +\frac{r-M}\varDelta\left(\mu_3 +\mu_2\right)_{,3} =
    \\
    = \frac 4{\left(X+Y\right)^2}\left(\varDelta X_{,2}Y_{,2}
    -\delta X_{,3}Y_{,3}\right)
    -3\frac{\left(r-M\right)^2-\varDelta}\varDelta +\frac{\mu^2-\delta}\delta
    \,.
    \label{k60}
\end{multline}

%Equations (\ref{k57})--(\ref{k60}) can be written in more symmetrical
%forms by changing $r$ to the new variable
%\begin{equation}
%    \left.
%    \begin{array}{lc}
%        &\eta = (r -M)/\sqrt{M^2 -a^2}\\
%        \mbox{when}&\\
%        &\varDelta = (M^2 -a^2)(\eta^2 -1) \,.
%    \end{array}
%    \right\}
%    \label{k61}
%\end{equation}
%Thus, we find
%\begin{equation}
%    \frac 1 2(X +Y)\{[(\eta^2 -1)X_{,\eta}]_{,\eta} +[(1
%    -\mu^2)X_{,\mu}]_{,\mu}\} = (\eta^2 -1){X_{,\eta}}^2 +(1
%    -\mu^2){X_{,\mu}}^2 \,,
%    \label{k62}
%\end{equation}
%\begin{equation}
%    \frac 1 2(X +Y)\{[(\eta^2 -1)Y_{,\eta}]_{,\eta} +[(1
%    -\mu^2)Y_{,\mu}]_{,\mu}\} = (\eta^2 -1){Y_{,\eta}}^2 +(1
%    -\mu^2){Y_{,\mu}}^2 \,,
%    \label{k63}
%\end{equation}
%\begin{equation}
%    -\frac\mu{1-\mu^2}(\mu_3 +\mu_2)_{,\eta} +\frac\eta{\eta^2 -1}(\mu_3
%    +\mu_2)_{,\mu} = \frac 2{(X+Y)^2}(X_{,\eta}Y_{,\mu}
%    +Y_{,\eta}X_{,\mu}) \,,
%    \label{k64}
%\end{equation}
%\begin{eqnarray}
%    2\eta(\mu_3 +\mu_2)_{,\eta} +2\mu(\mu_3 +\mu_2)_{,\mu} = \nonumber\\
%    = \frac
%    4{(X+Y)^2}[(\eta^2 -1)X_{,\eta}Y_{,\eta} -(1
%    -\mu^2)X_{,\mu}Y_{,\mu}] -\frac 3{\eta^2-1} +\frac 1{1-\mu^2} \,.
%    \label{k65}
%\end{eqnarray}
%
%A convenient form of equations (\ref{k62}) and (\ref{k63}) is obtained
%by the transformation
%\begin{equation}
%    X = \frac{1+F}{1-F} \qquad \mbox{and} \qquad Y = \frac{1+G}{1-G} \,.
%    \label{k66}
%\end{equation}
%We find
%\begin{equation}
%    (1 -FG)\{[(\eta^2 -1)F_{,\eta}]_{,\eta} +[(1 -\mu^2)F_{,\mu}]_{,\mu}
%    = -2G[(\eta^2 -1){F_{,\eta}}^2 + (1 -\mu^2){F_{,\mu}}^2 \,,
%    \label{k68}
%\end{equation}
%and
%\begin{equation}
%    (1 -FG)\{[(\eta^2 -1)G_{,\eta}]_{,\eta} +[(1 -\mu^2)G_{,\mu}]_{,\mu}
%    = -2F[(\eta^2 -1){G_{,\eta}}^2 + (1 -\mu^2){G_{,\mu}}^2 \,.
%    \label{k698}
%\end{equation}
%The metric functions $\chi$ and $\omega$ are related to $F$ and $G$ by
%\begin{equation}
%    \chi = \frac{1-FG}{(1-F)(1-G)} \qquad \mbox{and} \qquad \omega =
%    \frac{F-G}{(1-F)(1-G)} \,.
%    \label{k69}
%\end{equation}



\subsubsection{The Ernst equation}
Additional form of the basic equations that is related to the finding of all
stationary axisymmetric solutions of Einstein's equations for the
vacuum is due to Ernst.

%First, we observe that equation (\ref{k53}) allows $\omega$ to be
%derived from a potential $\varPhi$ in the manner:
%\begin{equation}
%    \varPhi_{,2} = \frac\delta{\chi^2}\omega_{,3} \qquad \mbox{and} \qquad
%    \varPhi_{,3} = -\frac\varDelta{\chi^2}\omega_{,2} \,.
%    \label{k87}
%\end{equation}
The potential $\varPhi$ is governed by the equation
\begin{equation}
    \left(\frac{\chi^2}\delta\varPsi_{,2}\right)_{,2} +
    \left(\frac{\chi^2}\varDelta\varPsi_{,3}\right)_{,3} \,;
    \label{k88}
\end{equation}
and equation (\ref{k52}) in terms of $\varPhi$ becomes
\begin{equation}
    \left[\varDelta\left(\ln\chi\right)_{,2}\right]_{,2}
    +\left[\delta\left(\ln\chi\right)_{,3}\right]_{,3} =
    \frac{\chi^2}\varDelta{\varPhi_{,3}}^2 +\frac{\chi^2}\delta{\varPhi_{,2}}^2
    \,.
    \label{k89}
\end{equation}
Letting
$
    \varPsi = \sqrt{\varDelta\delta}/\chi \, ,
   % \label{k90}
$
equations (\ref{k88}) and (\ref{k89})
can be reduced to (\cite{chandrasekhar})
\begin{gather}
    \varPsi\left[\left(\varDelta\varPsi_{,2}\right)_{,2}
    +\left(\delta\varPsi_{,3}\right)_{,3}\right] =
    \varDelta\left({\varPsi_{,2}}^2 -{\varPsi_{2}}^2\right)
    +\delta\left({\varPsi_{,3}}^2
    -{\varPsi_{,3}}^2\right)
    \label{k91}\,,\\
    \varPsi[(\varDelta\varPhi_{,2})_{,2} +(\delta\varPhi_{,3})_{,3}] =
    2\varDelta\varPsi_{,2}\varPhi_{,2} +2\delta\varPsi_{,3}\varPhi_{,3} \,.
    \label{k92}
\end{gather}
Now expressing $\varPsi$ and $\varPhi$ as the real and imaginary parts of
a~complex function $Z$,
%\begin{equation}
%    Z~= \varPsi +\mbox{i}\varPhi
%   % \label{k93}
%\end{equation}
equations (\ref{k91}) and (\ref{k92}) can be combined into the single
equation
\begin{equation}
    \left[\left(\varDelta Z_{,2}\right)_{,2}
    +\left(\delta Z_{,3}\right)_{,3}\right]
    \operatorname{Re}Z =
    \varDelta{Z_{,2}}^2 +\delta{Z_{,3}}^2 \,.
    \label{k94}
\end{equation}

Using the transformation,
\begin{equation}
    Z~= -\frac{1+E}{1-E} \,,
    \label{k95}
\end{equation}
(see \cite{chandrasekhar}), equation (\ref{k94}) becomes
\begin{equation}
    \left(1 -EE^*\right)\left[\left(\varDelta E_{,2}\right)_{,2}
    +\left(\delta E_{,3}\right)_{,3}\right] =
    -2E^*\left(\varDelta{E_{,2}}^2 +\delta{E_{,3}}^2\right) \,.
    \label{k96}
\end{equation}
This is Ernst's equation.

Associated with a solution \((\chi,\omega)\), we have the conjugate
solution $(\tilde\chi, \tilde\omega)$.
%\begin{equation}
%    \tilde\chi = \frac\chi{\chi^2-\omega^2} \qquad \mbox{and} \qquad
%    \tilde\omega = \frac\omega{\chi^2-\omega^2} \,;
%   % \label{k97}
%\end{equation}
(analysis in \cite{chandrasekhar}). Accordingly, a
conjugate Ernst-equation may be derived. Thus by defining
\begin{gather}
    \tilde\varPsi = \frac{\sqrt{\varDelta\delta}}{\tilde\chi} =
    \mbox{e}^{\psi+\nu}\frac{\chi^2-\omega^2}\chi = \mbox{e}^{2\nu}
    -\omega \,\mbox{e}^{2\psi} \,;
    \label{k98}\\
    %\end{equation}
    %\begin{equation}
    \tilde\varPhi_{,2} = \frac\delta{\tilde\chi^2}\tilde\omega_{,3}
    \,;\:\tilde\varPhi_{,3} = -\frac\varDelta{\tilde\chi^2}\tilde\omega_{,2}
    = -\frac{\tilde\varPsi^2}\delta\tilde\omega_{,2} \,;
    \label{k99}
\end{gather}
%and
%\begin{equation}
%    \tilde Z~= \tilde\varPsi +\mbox{i}\tilde\varPhi = -\frac{1+\tilde
%    E}{1-\tilde E} \,;
%   % \label{k100}
%\end{equation}
and chaning back to the variables $\eta$ and $\mu$, we have
\begin{multline}
    \left(1 -\tilde E\tilde E^*\right)\left\{\left[\left(\eta^2
    -1\right)\tilde E_{,\eta}\right]_{,\eta} +\left[\left(1
    -\mu^2\right)\tilde E_{,\mu}\right]_{,\mu}\right\} = \\
    = -2\tilde E^*\left[\left(\eta^2 -1\right)\bigl(\tilde
    E_{,\eta}\bigr)^2 +\left(1 -\mu^2\right)\bigl(\tilde
    E_{,\mu}\bigr)^2\right] \,.
    \label{k102}
\end{multline}



\subsection{The derivation of the Kerr metric}
\label{k-der}
Ernst's equation (\ref{k102}) allows an elementary solution.
%\begin{equation}
%    \tilde E = -p\eta -\mbox{i}q\mu \,,
%    \label{k104}
%\end{equation}
%where
%\begin{equation}
%    p^2 +q^2 = 1
%    \label{k105}
%\end{equation}
%and $p$ and $q$ are real constants.
The solutions for $\tilde\varPsi$
and $\tilde\varPhi$ ca be found by separating the real and
imaginary parts of the corresponding solution for $\tilde Z$, and
reverting to the variable $r$ (see \cite[p.~286]{chandrasekhar}).
%\begin{equation}
%    \tilde
%    \varPsi = \frac 1{\rho^2}\left(\varDelta -a^2\delta\right) \qquad
%    \mbox{and}
%    \qquad \tilde\varPhi = \frac{2aM\mu}{\rho^2} \,,
%    \label{k111}
%\end{equation}
%where
%\begin{equation}
%    \rho^2 = r^2 +a^2\mu^2 = r^2 +a^2\cos^2\theta
%    \label{112}
%\end{equation}
Then with equations (\ref{k99}) and (\ref{k98}) and with the
knowledge of $\mbox{e}^{2\beta} (= \nolinebreak \varDelta\delta)$,
the solutions for $\omega$ and $\mbox{e}^{2\psi}$ take simple
forms because of the identities
\begin{equation}
    \left[\left(r^2 +a^2\right) \mp a\sqrt{\varDelta\delta}\right]
    \bigl(\sqrt\varDelta \pm a\sqrt\delta\bigr)
    = \rho^2\sqrt\varDelta \pm 2aMr\sqrt\delta \,,
    \label{k119}
\end{equation}
and
\begin{equation}
    \varSigma^2\left(\varDelta -a^2\delta\right) = \rho^4\varDelta -4a^2M^2r^2\delta \,,
    \label{k120}
\end{equation}
where
\begin{equation}
    \varSigma^2 = \left(r^2 +a^2\right)^2 -a^2\varDelta\delta \,.
    \label{k121}
\end{equation}
Therefore, using the identity (\ref{k120}), %we find from equation
%that
\begin{equation}
    \mbox{e}^{2\psi} = \frac{\delta\varSigma^2}{\rho^2} \,,
    \label{k122}
\end{equation}
and
\begin{equation}
    \omega = \frac{2aMr}{\varSigma^2}\,.
    \label{k123}
\end{equation}
Further,
\begin{equation}
    \mbox{e}^{2\nu} = \mbox{e}^{2\beta-2\psi}
    =\frac{\rho^2\varDelta}{\varSigma^2} \,.
    \label{k124}
\end{equation}
and
\begin{equation}
    \chi = \mbox{e}^{-\psi+\nu} =
    \frac{\rho^2}{\varSigma^2}\sqrt{\frac\varDelta\delta} \,.
    \label{k125}
\end{equation}

To complete the solution, using the formulae that give the derivatives of $X$
and $Y$,equations (\ref{k59}) and
(\ref{k60}) can be reduced. Their solution is given by (see
\cite[p.~288]{chandrasekhar})
\begin{equation}
    \mbox{e}^{\mu_3+\mu_2} = \frac{\rho^2}{\varDelta} \,;
    \label{k131}
\end{equation}
and because \(\mbox{e}^{\mu_3-\mu_2} = \sqrt\varDelta\), the solutions
for $\mbox{e}^{\mu_3}$ and $\mbox{e}^{\mu_2}$, separately, are
\begin{equation}
    \mbox{e}^{2\mu_2} = \rho^2/\varDelta \qquad \mbox{and} \qquad
    \mbox{e}^{2\mu_2} = \rho^2 \,.
    \label{k132}
\end{equation}
The solution for all metric coefficients are completed; and the final
metric is
\begin{equation}
    \mbox{d}s = \rho^2\frac\varDelta{\varSigma^2}\left(\mbox{d}t\right)^2
    -\frac{\varSigma^2}{\rho^2}\left(\mbox{d}\varphi
    -\frac{2aMr}{\varSigma^2}\mbox{d}t\right)^2\sin^2\theta
    -\frac{\rho^2}\varDelta\left(\mbox{d}r\right)^2
    -\rho^2\left(\mbox{d}\theta\right)^2 \,.
    \label{k133}
\end{equation}
From the asymptotical behavior we conclude that the parameter $M$
is to be identified with the mass of the black hole, and $a$ is to
be interpreted as the angular momentum per unit of black hole.





%\section{Analytic Form of the The Kerr-Newman family}
%%By choice of gauge
%%\begin{equation}
%%    \varDelta = r^2 -2Mr +a^2 +Q^2
%%    \label{straumann711}
%%\end{equation}
%%(compare with \(\varDelta = r^2 -2Mr +a^2\) in Kerr solution)
%%we have the Kerr--Newman solution (which for \( Q = 0\) becomes Kerr
%%solution) given by (\ref{k133}) .
%As in section \ref{irn} we start with a metric of the form
%\begin{equation}
%    \mbox{d}s^2 = \mbox{e}^{2\nu}\left(\mbox{d}t\right)^2
%    -\mbox{e}^{2\psi}\left(\mbox{d}\varphi -\omega\mbox{d}t\right)^2
%    -\mbox{e}^{2\mu_2}\left(\mbox{d}x^2\right)^2
%    -\mbox{e}^{2\mu_3}\left(\mbox{d}x^3\right)^2
%    \,,
%    \label{ikn1}
%\end{equation}
%where \(\nu, \psi, \omega, \mu_2\), and $\mu_3$ are functions only of
%$x^2$ and $x^3$, with the freedom to impose a coordinate condition on
%$\mu_2$ and $\mu_3$.
%
%Considering Maxwell's equations in a space-time with a metric of
%a~chosen form, we can find the components of the Maxwell tensor. (Detailed
%process is in \cite[chapters 2 and 11]{chandrasekhar}.) Thus in
%stationary axisymmetric space-time
%\begin{equation}
%    F_{01} = F_{23} = 0 \,;
%    \label{ikn2}
%\end{equation}
%and
%\begin{gather}
%    \left(\mbox{e}^{\psi+\mu_2}F_{12}\right)_{,3}
%    -\left(\mbox{e}^{\psi+\mu_3}F_{13}\right)_{,2} = 0 \,,
%    \label{ikn3}\\
%    %\end{equation}
%    %\begin{equation}
%    \left(\mbox{e}^{\psi+\mu_3}F_{02}\right)_{,2}
%    +\left(\mbox{e}^{\psi+\mu_2}F_{03}\right)_{,3} = 0 \,,
%    \label{ikn4}\\
%    %\end{equation}
%    %\begin{equation}
%    \left(\mbox{e}^{\nu+\mu_3}F_{12}\right)_{,2}
%    +\left(\mbox{e}^{\nu+\mu_2}F_{13}\right)_{,3}
%    = \mbox{e}^{\psi+\mu_3}F_{02}\omega_{,2}
%    +\mbox{e}^{\psi+\mu_2}F_{03}\omega_{,3} \,,
%    \label{ikn5}\\
%    %\end{equation}
%    %\begin{equation}
%    \left(\mbox{e}^{\nu+\mu_2}F_{02}\right)_{,3}
%    -\left(\mbox{e}^{\nu+\mu_3}F_{03}\right)_{,2}
%    = \mbox{e}^{\psi+\mu_2}F_{12}\omega_{,3}
%    -\mbox{e}^{\psi+\mu_3}F_{03}\omega_{,3} \,.
%    \label{ikn6}
%\end{gather}
%
%
%Two pairs of components, \((F_{12},F_{13})\) and \((F_{02},
%F_{03})\), of the Maxwell tensor, can be derived from two
%potentials, $A$ and $B$, in the manner
%\begin{equation}
%    %\left.
%    \begin{array}{cc}
%        \mbox{e}^{\psi+\mu_2}F_{12} = A_{,2} \,;&
%        \mbox{e}^{\psi+\mu_2}F_{03} = -B_{,2} \,,\\
%        \mbox{e}^{\psi+\mu_3}F_{13} = A_{,3} \,;&
%        \mbox{e}^{\psi+\mu_3}F_{02} = +B_{,2} \,.
%    \end{array}
%    %\right\}
%    \label{ikn7}
%\end{equation}
%
%With the
%energy-momentum tensor of the Maxwell field given by
%\begin{equation}
%    T_{ab} = \eta_{cd}F_{ac}F_{bd} -\frac 1 4\eta_{ab}F_{ef}F^{ef}  \,,
%    \label{ikn10}
%\end{equation}
%the Einstein field-equations are given by
%\begin{equation}
%    R_{ab} = -2T_{ab} = -2\eta^{cd}F_{ac}F_{bd} -\eta_{ab}\left({F_{02}}^2
%    +{F_{03}}^2 -{F_{12}}^2 -{F_{13}}^2\right) \,,
%    \label{ikn11}
%\end{equation}
%where the various components of $R_{ab}$ are calculated with the aid of equation
%(\ref{ikn7}), and then equated with the corresponding expressions
%for the Ricci (or the Einstein) tensors in terms of metric. The
%expressions for the latter can be directly read off from Chapter \ref{ik},
%equations (\ref{k5})--(\ref{k13}). We find that equations (\ref{k12}),
%(\ref{k13}), (\ref{k7}), (\ref{k8}), (\ref{k9}), and (\ref{k10}) are now
%replaced, respectively, by
%\begin{gather}
%    \begin{split}
%        &\left(\mbox{e}^{\beta+\mu_3-\mu_2}\nu_{,2}\right)_{,2}
%        +\left(\mbox{e}^{\beta+\mu_2-\mu_3}\nu_{,3}\right)_{,3}
%        =\frac 1 2
%        \,\mbox{e}^{3\psi-\nu}\left(\mbox{e}^{\mu_3-\mu_2}{\omega_{,2}}^2
%        +\mbox{e}^{\mu_2-\mu_3}{\omega_{,3}}^2\right) + \\
%        &+\mbox{e}^{-\psi+\nu}
%        \left[\mbox{e}^{\mu_3-\mu_2}\left({A_{,2}}^2 +{B_{,2}}^2\right)
%        +\mbox{e}^{\mu_2-\mu_3}\left({A_{,3}}^2 +{B_{,3}}^2\right)\right] \,,
%    \end{split}\label{ikn16}\raisetag{38pt}\\
%    \begin{split}
%        &\left(\mbox{e}^{\beta+\mu_3-\mu_2}\psi_{,2}\right)_{,2}
%        +\left(\mbox{e}^{\beta+\mu_2-\mu_3}\psi_{,3}\right)_{,3}
%        =-\frac 1 2\,
%        \mbox{e}^{3\psi-\nu}\left(\mbox{e}^{\mu_3-\mu_2}{\omega_{,2}}^2
%        \right.+
%        \\
%        &\left.+\mbox{e}^{\mu_2-\mu_3}{\omega_{,3}}^2\right)
%        -\mbox{e}^{-\psi+\nu}\left[\mbox{e}^{\mu_3-\mu_2}\left({A_{,2}}^2
%        +{B_{,2}}^2\right)
%        +\mbox{e}^{\mu_2-\mu_3}\left({A_{,3}}^2 +{B_{,3}}^2\right)\right],
%    \end{split}\label{ikn17}\\
%    %\begin{equation}
%    \left(\mbox{e}^{3\psi-\nu-\mu_2+\mu_3}\omega_{,2}\right)_{,2}
%    +\left(\mbox{e}^{3\psi-\nu+\mu_2-\mu_3}\omega_{,3}\right)_{,3} =
%    4\left(A_{,2}B_{,3}
%    -A_{,3}B_{,2}\right) \,,
%    \label{ikn18}\\
%    %\end{equation}
%    \begin{split}
%        &\left(\psi +\nu\right)_{,2,3} -\left(\psi +\nu\right)_{,2}\mu_{2,3}
%        -\left(\psi
%        +\nu\right)_{,3}\mu_{3,2} +\psi_{,2}\psi_{,3} +\nu_{,2}\nu_{,3} =
%        \\
%        &=\frac 1
%        2\,\mbox{e}^{2\psi-2\nu}\omega_{,2}\omega_{,3}
%        -2\mbox{e}^{-2\psi}\left(A_{,2}A_{,3} +B_{,2}B_{,3}\right) \,,
%    \end{split}\label{ikn19}\\
%    \begin{split}
%        &\mbox{e}^{-2\mu_3}\left[\left(\psi +\nu\right)_{,3,3}
%        +\left(\psi +\nu\right)_{,3}\left(\nu
%        -\mu_3\right)_{,3} +{\psi_{,3}}^2 \right] +\\
%        &+\mbox{e}^{-2\mu_2}\left[\nu_{,2}\left(\psi
%        +\mu_3\right)_{,2} +\psi_{,2}\mu_{3,2}\right] = %\nonumber\\
%        -\frac 1 4\,
%        \mbox{e}^{2\psi-2\nu}\left(\mbox{e}^{-2\mu_2}{\omega_{,2}}^2
%        -\mbox{e}^{-2\mu_3}{\omega_{,3}}^2\right) +\\
%        &+\mbox{e}^{-2\psi}\left[\mbox{e}^{-2\mu_2}\left({A_{,2}}^2
%        +{B_{,2}}^2\right)
%        -\mbox{e}^{-2\mu_3}\left({A_{,3}}^2 +{B_{,3}}^2\right)\right] \,,
%    \end{split}\label{ikn20}\raisetag{38pt}
%\end{gather}
%and
%\begin{multline}
%    \mbox{e}^{-2\mu_2}\left[\left(\psi +\nu\right)_{,2,2}
%    +\left(\psi +\nu\right)_{,2}\left(\nu
%    -\mu_2\right)_{,2} +{\psi_{,2}}^2 \right]
%    +\mbox{e}^{-2\mu_3}\left[\nu_{,3}\left(\psi
%    +\mu_2\right)_{,3} +\right.\\
%    \left.+\psi_{,3}\mu_{2,3}\right]  %\nonumber\\
%    =+\frac 1 4\,
%    \mbox{e}^{2\psi-2\nu}\left(\mbox{e}^{-2\mu_2}{\omega_{,2}}^2
%    -\mbox{e}^{-2\mu_3}{\omega_{,3}}^2\right) %-\nonumber\\
%    -\mbox{e}^{-2\psi}\left[\mbox{e}^{-2\mu_2}\left({A_{,2}}^2
%    +{B_{,2}}^2\right)
%    +\right.\\
%    \left.+\mbox{e}^{-2\mu_3}\left({A_{,3}}^2 +{B_{,3}}^2\right)\right] \,,
%    \label{ikn21}
%\end{multline}
%where $\beta$ is written in equation~(\ref{k11}).
%
%The sum of equations (\ref{ikn16}) and (\ref{ikn17}) is
%(\ref{k14}) (same as for pure vacuum), and their difference
%\begin{multline}
%    \left[\mbox{e}^{\beta+\mu_3-\mu_2}\left(\psi -\nu\right)_{,2}\right]_{,2}
%    +\left[\mbox{e}^{\beta+\mu_2-\mu_3}\left(\psi -\nu\right)_{,3}\right]_{,3}
%    %=\nonumber\\
%    =-\mbox{e}^{3\psi-\nu}\left(\mbox{e}^{\mu_3-\mu_2}{\omega_{,2}}^2
%    +\right.\\
%    \left.+\mbox{e}^{\mu_2-\mu_3}{\omega_{,3}}^2\right)
%    -2\mbox{e}^{-\psi+\nu}\left[\mbox{e}^{\mu_3-\mu_2}\left({A_{,2}}^2
%    +{B_{,2}}^2\right)
%    +\mbox{e}^{\mu_2-\mu_3}\left({A_{,3}}^2 +{B_{,3}}^2\right)\right] \,.
%    \label{ikn24}
%\end{multline}
%
%The addition of equations (\ref{ikn20}) and (\ref{ikn21}) yields the
%same equation (\ref{k14}), while substraction gives
%\begin{multline}
%    4\mbox{e}^{\mu_3 -\mu_2}\left(\beta_{,2}\mu_{3,2}
%    +\psi_{,2}\nu_{,2}\right)
%    -4\mbox{e}^{\mu_2 -\mu_3}\left(\beta_{,3}\mu_{2,3}
%    +\psi_{,3}\nu_{,3}\right)
%    =\\
%    = 2\mbox{e}^{-\beta}\left[\left(\mbox{e}^{\mu_3
%    -\mu_2}{\mbox{e}^\beta}_{,2}\right)_{,2}
%    -\left(\mbox{e}^{\mu_2 -\mu_3}{\mbox{e}^\beta}_{,3}\right)_{,3}\right]
%    -\mbox{e}^{2\psi -2\nu}\left(\mbox{e}^{\mu_3 -\mu_2}{\omega_{,2}}^2
%    -\mbox{e}^{\mu_2 -\mu_3}{\omega_{,3}}^2\right) +\\
%    +4\mbox{e}^{-2\psi}\left[\mbox{e}^{\mu_3 -\mu_2}\left({A_{,2}}^2
%    +{B_{,2}}^2\right)
%    -\mbox{e}^{\mu_2-\mu_3}\left({A_{,3}}^2 +{B_{,3}}^2\right)\right] \,.
%    \label{ikn25}
%\end{multline}
%
%
%\subsection{The choice of gauge}
%As in section \ref{kgauge} we exercise the gauge freedom and impose a
%coordinate condition on $\mu_2$ and $\mu_3$. And by making further use
%of equation (\ref{k14}), we can arrange that the space-time we are
%considering has a smooth event horizon when a certain quadratic function
%in $r$, namely (cf. equations (\ref{k45}) and (\ref{k46})),
%\begin{equation}
%    \varDelta(r) = r^2 -2Mr + M_0^2 \,,
%    \label{ikn26}
%\end{equation}
%vanishes and
%\begin{equation}
%    \mbox{e}^{2(\mu_3-\mu_2)} = \varDelta(r) \qquad \mbox{and} \qquad
%    \mbox{e}^\beta = \varDelta^{1/2}\sin\theta \,.
%    \label{ikn27}
%\end{equation}
%In equation (\ref{ikn26}), $M$ and $M_0$ are constants.
%
%We write equation (\ref{ikn18}) in the form
%\begin{multline}
%    \left(\mbox{e}^{3\psi-\nu-\mu_2+\mu_3}\omega_{,2} -2AB_{,3}
%    +2BA_{,3}\right)_{,2}+\\
%     +\left(\mbox{e}^{3\psi-\nu+\mu_2-\mu_3}\omega_{,3}
%    +2AB_{,2}
%    +2BA_{,2}\right)_{,3} %=\\
%    = 0 \,.
%    \label{ikn30}
%\end{multline}
%
%After writing equations (\ref{ikn5}) and (\ref{ikn6}) in terms of
%potentials $A$ and $B$, we can combine them into a single equation, for
%the {\em complex potential},
%\begin{equation}
%    H = A~+\mbox{i}B \,.
%    \label{ikn28}
%\end{equation}
%Also equation (\ref{ikn30}) can be written in terms of $H$.
%
%Finally with the solution for $\mbox{e}^{\mu_3-\mu_2}$ and $\mbox{e}^\beta$
%given in equation (\ref{ikn27}) and by using (\ref{k48}) instead of
%$\theta$, as the variable indicated by the index `3', we can bring
%equations (\ref{ikn5}) with (\ref{ikn6}), and equation (\ref{ikn30}) to
%the forms
%\begin{equation}
%    \left(\frac\chi{\sqrt{\varDelta\delta}}\varDelta H_{,2}\right)_{,2}
%    +\left(\frac\chi{\sqrt{\varDelta\delta}}\delta H_{,3}\right)_{,3} =
%    \mbox{i}\left(\omega_{,2}H_{,3} -\omega_{,3}H_{,2}\right)
%    \label{ikn33}
%\end{equation}
%and
%\begin{equation}
%    \left(\frac\varDelta{\chi^2}\omega_{,2}
%    -2\operatorname{Im}{HH^*_{,3}}\right) +\left(\frac\delta{\chi^2}\omega_{,3}
%    +2\operatorname{Im}{HH^*_{,2}}\right) = 0 \,,
%    \label{ikn34}
%\end{equation}
%where $\chi$ is as in (\ref{k19}) and $\delta$ in (\ref{k51}). In
%similar fashion, equation (\ref{ikn24}), expressed in terms of
%$\chi$ and $H$, becomes
%\begin{equation}
%    \left[\varDelta\left(\ln\chi\right)_{,2}\right]_{,2}
%    +\left[\delta\left(\ln\chi\right)_{,3}\right]_{,3} = \frac
%    1{\chi^2}\left(\varDelta{\omega_{,2}}^2 +\delta{\omega_{,3}}^2\right)
%    +\frac{2\chi}{\sqrt{\varDelta\delta}}\left(\varDelta\abs{H_{,2}}^2
%    +\delta\abs{H_{,3}}^2\right) \,.
%    \label{ikn37}
%\end{equation}
%
%We observe that equation (\ref{ikn34}) allows us to define a potential
%$\varPhi$ by the equations (cf.~equation (\ref{k87})).
%\begin{equation}
%    -\varPsi_{,2} = \frac\delta{\chi^2}\omega_{,3}
%    +2\operatorname{Im}HH^*_{,2} \,, \qquad +\varPsi_{,3} =
%    \frac\varDelta{\chi^2}\omega_{,2} -2\operatorname{Im}HH^*_{,3} \,,
%    \label{ikn38}
%\end{equation}
%for equation (\ref{ikn34}) is no
%more that the integrability condition for these equations. Solving these
%equations for $\omega_{,2}$, and $\omega_{,3}$, we have
%\begin{equation}
%    \omega_{,2} = \frac{\chi^2}\varDelta\left(\varPhi_{,3}
%    +2\operatorname{Im}HH^*_{,3}\right) \, \quad \omega_{,3} =
%    -\frac{\chi^2}\delta\left(\varPhi_{,2}
%    +2\operatorname{Im}HH^*_{,2}\right) \,;
%    \label{ikn39}
%\end{equation}
%
%It is now convenient to define function $\varPsi$ as in equation
%(\ref{k90}). In terms of $\varPsi$, the integrability condition for
%equations (\ref{ikn39}), takes the form
%\begin{multline}
%    \varPsi\left\{\left[\varDelta\left(\varPhi_{,2}
%    +2\operatorname{Im}HH^*_{,2}\right)\right]_{,2}
%    +\left[\delta\left(\varPhi_{,3}
%    +2\operatorname{Im}HH^*_{,3}\right)\right]_{,3}\right\} =\\
%    = 2\varDelta\left(\varPsi_{,2}
%    +2\operatorname{Im}HH^*_{,2}\right)\varPsi_{,2}
%    +2\delta\left(\varPhi_{,3} +2\operatorname{Im}HH^*_{,3}\right)\varPsi_{,3}
%    \,.
%    \label{ikn43}
%\end{multline}
%
%Equation (\ref{ikn37}) written in terms of $\varPsi$ and
%$\varPhi$ becomes
%\begin{multline}
%    \varPsi\left[\left(\varDelta\varPsi_{,2}\right)_{,2}
%    +\left(\delta\varPsi_{,3}\right)_{,3}\right]
%    = \varDelta{\varPsi_{,2}}^2 +\delta{\varPsi_{,3}}^2
%    -2\varPsi\left(\varDelta\abs{H_{,2}}^2+\right. \\
%    \left.+\delta\abs{H_{,3}}^2\right)
%    -\varDelta\left(\varPhi_{,2} +2\operatorname{Im}HH^*_{,2}\right)^2
%    -\delta\left(\varPhi_{,3} +\operatorname{Im}HH^*_{,3}\right)^2 \,;
%    \label{ikn44}
%\end{multline}
%while in place of equation (\ref{ikn33}), we have
%\begin{multline}
%    \varPsi\left[\left(\varDelta H_{,2}\right)_{,2}
%    +\left(\delta H_{,3}\right)_{,3}\right] = \\
%    = \varDelta
%    H_{,2}\left[\varPsi_{,2} +\mbox{i}\left(\varPhi_{,2}
%    +2\operatorname{Im}HH^*_{,2}\right)\right]
%    +\delta H_{,3}\left[\varPsi_{,3} +\mbox{i}\left(\varPhi_{,3}
%    +2\operatorname{Im}HH^*_{,3}\right)\right] \,.
%    \label{ikn45}
%\end{multline}
%
%Finally, we shall state as lemmas two direct consequences of the
%foregoing equations.
%\begin{lemma}
%    \begin{multline}
%        \varPsi\left[\left(\varDelta\abs{H}^2_{,2}\right)_{,2}
%        +\left(\delta\abs{H}^2_{,3}\right)_{,3}\right] %= \nonumber\\
%        = 2\varPsi\left[\varDelta\abs{H_{,2}}^2 +\delta\abs{H_{,3}}^2\right]
%        +\\
%        +\varDelta\left[\abs{H}^2_{,2}\varPsi_{,2}
%        +2\left(\varPsi_{,2}
%        +2\operatorname{Im}HH^*_{,2}\right)\operatorname{Im}HH^*_{,2}\right]
%        +\\
%        +\delta\left[\abs{H}^2_{,3}\varPsi_{,3}
%        +2\left(\varPsi_{,3}
%        +2\operatorname{Im}HH^*_{,3}\right)\operatorname{Im}HH^*_{,3}\right]
%        \,.
%        \label{ikn47}
%    \end{multline}
%    \label{l1}
%\end{lemma}
%\begin{lemma}
%    \begin{multline}
%        2\varPsi\left[\left(\varDelta\operatorname{Im}HH^*_{,2}\right)_{,2}
%        +\left(\delta\operatorname{Im}HH^*_{,3}\right)_{,3}\right] = %\nonumber\\
%        2\varDelta\varPsi_{,2}\operatorname{Im}HH^*_{,2} +\\
%        +2\delta\varPsi_{,3}\operatorname{Im}HH^*_{,3}
%        -\left[\varDelta{\abs{H}^2}_{,2}\left(\varPsi_{,2}
%        +2\operatorname{Im}HH^*_{,2}\right) +\right.\\
%        \left.+\delta{\abs{H}^2}_{,3}\left(\varPsi_{,3}
%        +2\operatorname{Im}HH^*_{,3}\right)\right] \,.
%        \label{ikn48}
%    \end{multline}
%    \label{l2}
%\end{lemma}
%
%
%
%\subsubsection{The Ernst equations}
%Equation (\ref{ikn45}) can be written in the form \cite[p.~569]{chandrasekhar}
%\begin{multline}
%    \varPsi\left[\left(\varDelta H_{,2}\right)_{,2}
%    +\left(\delta H_{,3}\right)_{,3}\right] %=\nonumber\\
%    =-2H^*\left(\varDelta{H_{,2}}^2 +\delta{H_{,3}}^2\right) \\
%    +\varDelta H_{,2}\left(\varPsi +\abs{H}^2 +\mbox{i}\varPhi\right)_{,2}
%    +\delta
%    H_{,3}\left(\varPsi +\abs{H}^2 +\mbox{i}\varPhi\right)_{,3} \,.
%    \label{ikn50}
%\end{multline}
%
%Next, eliminating the terms in \(\operatorname{Im}HH^*_{,2}\) and
%\(\operatorname{Im}HH^*_{,3}\) on the left-hand sides of equation
%(\ref{ikn43}) with the aid of lemma~\ref{l2}, we are left with
%\begin{multline}
%    \varPsi\left[\left(\varDelta\varPhi_{,2}\right)_{,2}
%    +\left(\delta\varPhi_{,3}\right)_{,3}\right]
%    %=\nonumber\\
%    = \varDelta\left[\varPhi_{,2}\left(2\varPsi +\abs{H}^2\right)_{,2}
%    +2\left(\varPsi
%    +\abs{H}^2\right)_{,2}\operatorname{Im}HH^*_{,2}\right] +\\
%    +\delta\left[\varPhi_{,3}\left(2\varPsi
%    +\abs{H}^2\right)_{,3} +2\left(\varPsi
%    +\abs{H}^2\right)_{,3}\operatorname{Im}HH^*_{,3}\right] \,.
%    \label{ikn51}
%\end{multline}
%
%Similarly, combining equations (\ref{ikn44}) and (\ref{ikn47}), we
%obtain
%\begin{multline}
%    \varPsi\left\{\left[\varDelta\left(\varPsi
%    +\abs{H}^2\right)_{,2}\right]_{,2}
%    +\left[\delta\left(\varPsi
%    +\abs{H}^2\right)_{,3}\right]_{,3}\right\} = \\
%    = \varDelta\left({\varPsi_{,2}}^2 %+\nonumber\\
%    +\abs{H}^2_{,2}\varPsi_{,2} -{\varPhi_{,2}}^2
%    -2\varPhi_{,2}\operatorname{Im}HH^*_{,2}\right) +\\
%    +\delta\left({\varPsi_{,3}}^2
%    +\abs{H}^2_{,3}\varPsi_{,3} -{\varPhi_{,3}}^2
%    -2\varPhi_{,3}\operatorname{Im}HH^*_{,3}\right) \,.
%    \label{ikn52}
%\end{multline}
%
%It is now clear that the left-hand sides of equations (\ref{ikn51}) and
%(\ref{ikn52}) can be combined into a single expression for the complex
%function,
%\begin{equation}
%    Z = \varPsi +\abs{H}^2 +\mbox{i}\varPhi \,.
%    \label{ikn53}
%\end{equation}
%The terms with the coefficient $\delta$ will combine to give
%\begin{equation}
%    {Z_{,3}}^2 -2H^*H_{,3}Z_{,3} \,.
%    \label{ikn57}
%\end{equation}
%Thus equation (\ref{ikn51}) and (\ref{ikn52}) are equivalent to the
%single complex equation
%\begin{equation}
%    \varPsi\left[\left(\varDelta Z_{,2}\right)_{,2}
%    +\left(\delta Z_{,3}\right)_{,3}\right] =
%    \varDelta{Z_{,2}}^2 +\delta{Z_{,3}}^2 -2H^*\left(\varDelta Z_{,2}H_{,2}
%    +\delta Z_{,3}H_{,3}\right) \,;
%    \label{ikn58}
%\end{equation}
%and equation (\ref{ikn50}) with the terms in the second line
%written in terms of $Z$~becomes
%\begin{equation}
%    \varPsi\left[\left(\varDelta H_{,2}\right)_{,2}
%    +\left(\delta H_{,3}\right)_{,3}\right] =
%    \varDelta H_{,2}Z_{,2} +\delta H_{,3}Z_{,3} -2H^*\left(\varDelta
%    {H_{,2}}^2
%    +\delta {H_{,3}}^2\right) \,.
%    \label{ikn59}
%\end{equation}
%
%These last two equations are the {\em Ernst equations}, they replace the
%single equation (\ref{k94}) applicable for pure vacuum.
%
%
%\subsubsection{The operation of conjugation}
%In \cite[\S52(a)]{chandrasekhar}, it is shown that given a metric
%of the form (\ref{ikn1}), we can derive a {\em conjugate metric}
%of the same form by the transformation
%\begin{equation}
%    t \rightarrow \mbox{i}\varphi \qquad\mbox{and}\qquad \varphi
%    \rightarrow \mbox{i}t \,,
%    \label{ikn67}
%\end{equation}
%when $\chi$ and $\omega$ are replaced by
%\begin{equation}
%     \tilde\chi = -\frac\chi{\chi^2 -\omega^2} \qquad \mbox{and} \qquad
%     \tilde\omega = \frac\omega{\chi^2 -\omega^2} \,.
%     \label{ikn68}
%\end{equation}
%We can define a potential $\tilde H$ by the equations
%\begin{equation}
%    %\left.
%    \begin{aligned}%{l}
%        -\tilde H_{,3} =
%        \chi{\displaystyle\frac{\sqrt\varDelta}{\sqrt\delta}H_{,2}}
%        -\mbox{i}\omega H_{,3} \,,\\
%        +\tilde H_{,2} =
%        \mbox{i}\omega H_{,2}
%        +\chi{\displaystyle\frac{\sqrt\delta}{\sqrt\varDelta}H_{,3}} \,,
%    \end{aligned}
%    %\right\}
%    \label{ikn70}
%\end{equation}
%Solving these equations for $H_{,2}$ and $H_{,3}$, we obtain
%\begin{equation}
%    %\left.
%    \begin{aligned}%{l}
%        H_{,2} &= \tilde\chi{\displaystyle\frac{\sqrt\delta}{\sqrt\varDelta}\tilde
%        H_{,3}}
%        +\mbox{i}\tilde\omega\tilde H_{,2} \,,\\
%        H_{,3} &=
%        -\tilde\chi{\displaystyle\frac{\sqrt\varDelta}
%        {\sqrt\delta}\tilde H_{,2}}
%        +\mbox{i}\tilde\omega\tilde H_{,3} \,.
%    \end{aligned}
%    %\right\}
%    \label{ikn71}
%\end{equation}
%where $\tilde\chi$ and $\tilde\omega$ are defined as in equations
%(\ref{ikn68}). After ascertaining how the potential $H$ transforms
%under conjugation, it appears (\cite[p.~572]{chandrasekhar}) that
%conjugation results in replacing $H$ by $\tilde H$. This inference
%can be confirmed by directly verifying that $\tilde\chi$,
%$\tilde\omega$ and $\tilde H$ do in fact satisfy equations of
%exactly the same form as equations (\ref{ikn30}) and
%(\ref{ikn37}).
%
%

%\subsection{The derivation of the Kerr--Newman solution}
%As in the case of Kerr solution (section~\ref{k-der}), the
%Kerr--Newman solution follows from a consideration of the simplest
%solution of Ernst's equation from the conjugate metric.
%
%Linear relation of the form
%\begin{equation}
%    H = Q(Z +1) \,,
%    \label{ikn76}
%\end{equation}
%where $Q$ is a complex constant, is consistent with the pair of
%equations (\ref{ikn58}) and (\ref{ikn59}), and both these
%equations reduce to the same equation,
%\begin{multline}
%    \frac 1 2\left[\left(1 -2\abs{Q}^2\right)(Z +Z^*)
%    -2\abs{Q}^2\left(\abs{Z}^2
%    +1\right)\right]\left[\left(\varDelta Z_{,2}\right)_{,2}
%    +\left(\delta Z_{,3}\right)_{,3}\right] =\\
%    = \left[1 -2\abs{Q}^2(Z^*
%    +1)\right]\left(\varDelta{Z_{,2}}^2 +\delta{Z_{,3}}^2\right) \,.
%    \label{ikn78}
%\end{multline}
%The problem is thus effectively reduced to solving the
%Ernst equation for the vacuum. Accordingly, in seeking solutions of
%equation (\ref{ikn78}), we make the same transformation as in
%Chapter~\ref{ik}, (see equation
%(\ref{k95})). By changing the variable $r$ to
%\begin{equation}
%    \eta = (r -M)/\sqrt{M^2 -M_0^2} \,,
%    \label{ikn83}
%\end{equation}
%the equation governing $E$ takes form
%\begin{multline}
%    \left(1 -4\abs{Q}^2 -\abs{E}^2\right)\left\{\left[\left(\eta^2
%    -1\right)E_{,\eta}\right]_{,\eta} +\left[\left(1
%    -\mu^2\right)E_{,\mu}\right]_{,\mu}\right\} =\\
%    = -2E^*\left[\left(\eta^2 -1\right){E_{,\eta}}^2 +\left(1
%    -\mu^2\right){E_{,\mu}}^2\right] \,.
%    \label{ikn85}
%\end{multline}
%
%The derivation of the Kerr--Newman solution now proceeds as in
%section~\ref{k-der}. We consider equation (\ref{ikn85}) for the
%conjugate function, $\tilde E$; and we verify that it allows the
%simple solution
%\begin{equation}
%    \tilde E = -p\eta -\mbox{i}q\mu \,,
%    \label{ikn86}
%\end{equation}
%where $p$ and $p$ are two real constants related in the manner
%\begin{equation}
%    p^2 +q^2 = 1 -4\abs{Q}^2 \,.
%    \label{ikn87}
%\end{equation}
%
%For $\tilde E$ given by equation (\ref{ikn86}) and reverting to the
%variable $r$, we find that the solutions are much simplified by the
%choice %compatible
%\begin{equation}
%    p = \left(M^2 -M_0^2\right)^{1/2} \qquad \mbox{and} \qquad q = a/M \,,
%    \label{ikn94}
%\end{equation}
%where $a$ is a constant. However, to be consistent with the relation
%(\ref{ikn87}), we must require that
%\begin{equation}
%    M_0^2 = a^2 + Q_* \quad\mbox{(say),}
%    \label{ikn97}
%\end{equation}
%where
%\begin{equation}
%    Q_*^2 = a^2 +4\abs{Q}^2M^2 \,.
%    \label{ikn99}
%\end{equation}
%The expression for $\varDelta$ now becomes
%\begin{equation}
%    \varDelta = r^2 -2Mr +a^2 +Q_*^2 \,.
%    \label{ikn100}
%\end{equation}
%
%Therefore, with the choices (\ref{ikn94}) and (\ref{ikn99}), the
%solutions take the simple forms
%\begin{gather}
%    \operatorname{Re}\tilde Z = -\frac{\varDelta -a^2\delta -Q_*^2}{\rho^2} \,;
%    \quad \operatorname{Im}\tilde Z = \tilde\varPhi = -\frac{2aM\mu}{\rho^2}
%    \,,
%    \label{ikn101}\\
%    %\end{equation}
%    %\begin{equation}
%    \tilde\varPsi = \frac{\sqrt{\varDelta\delta}}{\tilde\chi} =
%    -\frac{\varDelta -a^2\delta}{\rho^2} \,,\qquad \mbox{and} \qquad
%    \tilde H^2 = \frac{Q_*^2}{\rho^2} \,.
%    \label{ikn102}
%\end{gather}
%Also by equation (\ref{ikn76}),
%\begin{equation}
%    \tilde H = Q\bigl(\tilde Z +1\bigr) = q_*\frac{r -\mbox{i}a\mu}{\rho^2} \,;
%    \label{ikn103}
%\end{equation}
%and, therefore,
%\begin{equation}
%    \tilde A = Q_*\frac r{\rho^2} \qquad\mbox{and}\qquad \tilde B =
%    -aQ_*\frac\mu{\rho^2} \,.
%    \label{ikn104}
%\end{equation}
%
%From solutions (\ref{ikn104}) we deduce:
%\begin{equation}
%    %\left.
%    \begin{array}{ll}
%        \tilde A_{,2} = {\displaystyle\frac{Q_*} {\rho^4}}\left(-r^2
%        +a^2\mu^2\right) \,, &\tilde
%        A_{,3} = {\displaystyle\frac{Q_*} {\rho^4}}\left(-2a^2r\mu\right)
%        \,,\\
%        \tilde B_{,3} = a{\displaystyle\frac{Q_*} {\rho^4}}\left(-r^2
%        +a^2\mu^2\right) \,;
%        &\tilde B_{,2} = {\displaystyle\frac{Q_*}
%        {\rho^4}}\left(+2ar\mu\right) \,.
%    \end{array}
%    %\right\}
%    \label{ikn105}
%\end{equation}
%With the aid of these equations we find:
%\begin{equation}
%    %\left.
%    \begin{aligned}%{l}
%        2\operatorname{Im}{\tilde H\tilde H^*}_{,2} = 2\bigl(\tilde
%        B\tilde A_{,2}
%        -\tilde A\tilde B_{,2}\bigr) = -2{\displaystyle\frac{Q_*^2}
%        {\rho^4}}a\mu \,,\\
%        2\operatorname{Im}{\tilde H\tilde H^*}_{,3} = 2\bigl(\tilde
%        B\tilde A_{,3}
%        -\tilde A\tilde B_{,3}\bigr) = +2{\displaystyle\frac{Q_*^2}
%        {\rho^4}}ar \,.
%    \end{aligned}
%    %\right\}
%    \label{ikn106}
%\end{equation}
%
%From equations (\ref{ikn39}), (\ref{ikn101}) and (\ref{ikn106}) it now
%follows that
%\begin{equation}
%    %\left.
%    \begin{aligned}%{l}
%        +{\displaystyle\frac\varDelta{\tilde\chi^2}}\tilde\omega_{,2} &=
%        \tilde\varPhi_{,3} +2{\operatorname{Im}HH^*}_{,3} =
%        -{\displaystyle\frac{2aM\left(r^2
%        -a^2\mu\right)} {\rho^4}} +2{\displaystyle\frac{Q_*^2} {\rho^4}}ar
%        \,,\\
%        -{\displaystyle\frac{\delta} {\tilde\chi^2}}\tilde\omega_{,3} &=
%        \tilde\varPhi_{,2} +2{\operatorname{Im}HH^*}_{,2}
%        ={\displaystyle\frac{4aMr\mu} {\rho^4}
%        -2\frac{Q_*^2}{\rho^4}a\mu} \,.
%    \end{aligned}
%    %\right\}
%    \label{ikn107}
%\end{equation}
%On the other hand, according to the solution for $\tilde\varPsi$ given
%in equation (\ref{ikn102}),
%\begin{equation}
%    \frac\varDelta{\tilde\chi^2} = \frac{\left(\varDelta
%    -a^2\delta\right)^2}{\rho^4\delta} \qquad\mbox{and}\qquad
%    \frac\delta{\tilde\chi^2} = \frac{\left(\varDelta
%    -a^2\delta\right)^2}{\rho^4\varDelta} \,.
%    \label{ikn108}
%\end{equation}
%The equations which determine $\tilde\omega$ are, therefore,
%\begin{equation}
%    %\left.
%    \begin{aligned}%{l}
%        \tilde\omega_{,2} &= -{\displaystyle\frac{2a\delta}{\left(\varDelta
%        -a^2\delta\right)^2}}\left[M\left(r^2 -a^2\mu^2\right) -Q_*^2r\right]
%        \,,\\
%        \tilde\omega_{,3} &=
%        -{\displaystyle\frac{2a\varDelta\mu}{\left(\varDelta
%        -a^2\delta\right)^2}}\left(2Mr -Q_*^2\right) \,.
%    \end{aligned}
%    %\right\}
%    \label{ikn109}
%\end{equation}
%From these equations we readily find that (cf. equations (\ref{ikn68})
%and (\ref{ikn100}))
%\begin{equation}
%    \tilde\omega = \frac{a\delta}{\varDelta -a^2\delta}\left[\left(r^2
%    +a^2\right)
%    -\varDelta\right]\,;
%    \label{ikn111}
%\end{equation}
%and, similarly, from equations (\ref{ikn68}) and (\ref{ikn102}), we find
%\begin{equation}
%    \tilde\chi = -\frac{\rho^2}{\varDelta
%    -a^2\delta}\sqrt{\varDelta\delta} \,.
%    \label{ikn112}
%\end{equation}
%
%With the aid of the solutions (\ref{ikn111}) and (\ref{ikn112}) for
%$\tilde\omega$ and $\tilde\chi$, we can solve for the metric functions,
%$\mbox{e}^{2\nu}$, $\mbox{e}^{2\psi}$, and $\omega$ exactly as we did in
%section~\ref{k-der}. We find:
%\begin{gather}
%    \mbox{e}^{2\psi} = \frac{\delta\varSigma^2}{\rho^2} \,, \quad
%    \mbox{e}^{2\nu} = \frac{\rho^2\varDelta}{\varSigma^2} \,, \quad
%    \mbox{e}^{2\beta} = \mbox{e}^{2(\psi +\nu)} = \varDelta\delta \,,
%    \label{ikn113}\\
%    %\end{equation}
%    %\begin{equation}
%    \omega = \frac a{\varSigma^2}\left(r^2 +a^2 -\varDelta\right) \qquad
%    \mbox{and}
%    \qquad \chi = \mbox{e}^{-\psi +\nu} =
%    \frac{\rho^2\sqrt\varDelta}{\varSigma^2\sqrt\delta} \,,
%    \label{ikn114}
%\end{gather}
%where $\varSigma$ is defined as in equation (\ref{k121}).
%%\begin{equation}
%%    \varSigma^2 = (r^2 +a^2)^2 -a^2\delta\varDelta \,.
%%    \label{ikn115}
%%\end{equation}
%We observe that these solutions, apart from the definition of
%$\varDelta$, are exactly the same as the solutions
%(\ref{k121})--(\ref{k125}) for the Kerr metric.
%
%Returning to the equations (\ref{ikn105}), we can now obtain, by making
%use of equations (\ref{ikn11}) and (\ref{ikn112}), the solutions for the
%untilded quantities in accordance with equations (\ref{ikn71}); thus,
%\begin{equation}
%    %\left.
%    \begin{aligned}%{l}
%        A_{,r} &= +{\displaystyle\frac{Q_*}
%        {\rho^4}}2a^2r\sin^2\theta\cos\theta \,\\
%        A_{,\theta} &= -{\displaystyle\frac{Q_*} {\rho^4}}\left(r^2
%        +a^2\right)\left(-r^2
%        +a^2\cos^2\theta\right)\sin\theta \,\\
%        B_{,r} &= -{\displaystyle\frac{Q_*} {\rho^4}}a\left(-r^2
%        +a^2\cos^2\theta\right)\sin^2\theta
%        \,,\\
%        B_{,\theta} &= -{\displaystyle\frac{Q_*}{\rho^4}}2ar\left(r^2
%        +a^2\right)\sin\theta\cos\theta \,,
%    \end{aligned}
%    %\right\}
%    \label{ikn116}
%\end{equation}
%where, to avoid any ambiguity, the variables to which differentiations
%refer are explicitly indicated. According to the foregoing solutions,
%\begin{equation}
%    A_{,r}A_{,\theta} +B_{,r}B_{,\theta} = 0 \,.
%    \label{ikn117}
%\end{equation}
%
%Finally, to determine \((\mu_2 +\mu_3)\) and complete the solution for
%the metric, we turn to equations (\ref{ikn19}) and (\ref{ikn25}). By
%comparison with the corresponding equations (\ref{k8}) and (\ref{k16})
%for the vacuum, we observe that, by virtue of the relation
%(\ref{ikn117}), equation (\ref{ikn19}) is the same as equation
%(\ref{k8}) while equation (\ref{ikn25}) has the additional term.
%Therefore, while equation (\ref{k59}) stays, in place of equation
%(\ref{k60}), we now have
%\begin{multline}
%    2(r -M)\left(\mu_2 +\mu_3\right)_{,2} +2\mu\left(\mu_2 +\mu_3\right)_{,3}
%    =\\
%    = {\displaystyle\frac 4{\left(X
%    +Y\right)^2}\left(\varDelta X_{,2}Y_{,2} -\delta X_{,3}Y_{,3}\right)
%    -3\frac{\left(r -M\right)^2
%    -\varDelta}\varDelta +\frac{\mu^2 +\delta}\delta
%    -4\frac{Q_*^2}{\rho^2}} ,
%    \label{ikn121}
%\end{multline}
%where $X$ and $Y$ are determined by equations (\ref{k56}).
%
%%In view of the formal identity of the solution for $\chi$ and $\omega$
%%(as written in equations (\ref{ikn114})) with the corresponding
%%solutions for the Kerr metric, we are led to the same equation
%%(\ref{k129}).
%With the occurrence of the additional term \( -4Q_*^2/\rho^2\) on
%the right-hand side of equation (\ref{ikn121}) we find that
%\begin{equation}
%    \frac\varDelta{\chi^2}X_{,2}Y_{,2} -\frac\delta{\chi^2}X_{,3}Y_{,3}
%    = \frac{\left(r -M\right)^2}\varDelta -\frac{4Mr}{\rho^2}
%    -\frac{\mu^2}\delta
%    +4\frac{Q_*^2}{\rho^2} \,.
%    \label{ikn124}
%\end{equation}
%By combining equation (\ref{ikn121}) and (\ref{ikn124}), we are again
%led to the same equation as in section~\ref{k-der}. The solution for
%\((\mu_2 +\mu_3)\) is, therefore, the same as for the Kerr metric,
%\begin{equation}
%    \mbox{d}s = \rho^2\frac\varDelta{\varSigma^2}\left(\mbox{d}t\right)^2
%    -\frac{\varSigma^2}{\rho^2}\left(\mbox{d}\varphi
%    -\frac{2aMr}{\varSigma^2}\mbox{d}t\right)^2\sin^2\theta
%    -\frac{\rho^2}\varDelta\left(\mbox{d}r\right)^2
%    -\rho^2\left(\mbox{d}\theta\right)^2 \,,
%\end{equation}
%except for definition of $\varDelta$.
%
%The parameters $M$, $a$ and $Q$ are most easily interpreted from the
%asymptotic form of the fields. In \cite{straumann}, it is shown, by
%performing the flux integrals for the energy and angular momentum, that
%$M$ is the total mass and that
%\begin{equation}
%    S = aM
%    \label{straumann722}
%\end{equation}
%is the total angular momentum. There is also shown that $Q_*$ is
%the {\em charge} of the black hole.
