# Analytical calculation of Kerr and Kerr-Ads black holes in $f(R)$ theory

Ping Li<EMAIL_ADDRESS>College of Mathematics and Physics, Hunan
University of Arts and Sciences, 3150 Dongting Dadao, Changde City, Hunan
Province 415000, China  Hunan Province Key Laboratory Integration and Optical
Manufacturing Technology, 3150 Dongting Dadao, Changde City, Hunan Province
415000, China  Yong-qiang Liu<EMAIL_ADDRESS>Division of
Mathematica and Theoretical Physics, Shanghai Normal University, 100 Guilin
Road, Shanghai 200234, China  Jiang-he Yang<EMAIL_ADDRESS>College of
Mathematics and Physics, Hunan University of Arts and Sciences, 3150 Dongting
Dadao, Changde City, Hunan Province 415000, China  Center for Astrophysics,
Guangzhou University, 230 West Ring Road, Guangzhou, Guangdong Province
510006, China  Siwei Xu<EMAIL_ADDRESS>College of Mathematics and
Physics, Hunan University of Arts and Sciences, 3150 Dongting Dadao, Changde
City, Hunan Province 415000, China  Hunan Province Key Laboratory Integration
and Optical Manufacturing Technology, 3150 Dongting Dadao, Changde City, Hunan
Province 415000, China  Xiang-hua Zhai<EMAIL_ADDRESS>Division of
Mathematica and Theoretical Physics, Shanghai Normal University, 100 Guilin
Road, Shanghai 200234, China

###### Abstract

In this paper, we extend Chandrasekhar’s method of calculating rotating black
holes into $f(R)$ theory. We consider the Ricci scalar is a constant and
derive the Kerr and Kerr-Ads metric by using the analytical mathematical
method. Suppose that the spacetime is a 4-dimensional Riemannian manifold with
a general stationary axisymmetric metric, we calculate Cartan’s equation of
structure and derive the Einstein tensor. In order to reduce the solving
difficulty, we fix the gauge freedom to transform the metric into a more
symmetric form. We solve the field equations in the two cases of the Ricci
scalar $R=0$ and $R\neq 0$. In the case of $R=0$, the Ernst’s equations are
derived. We give the elementary solution of Ernst’s equations and show the way
to obtain more solutions including Kerr metric. In the case of $R\neq 0$, we
reasonably assume that the solution to the equations consists of two parts:
the first is Kerr part and the second is introduced by the Ricci scalar.
Giving solution to the second part and combining the two parts, we obtain the
Kerr-Ads metric. The calculations are carried out in a general $f(R)$ theory,
indicating the Kerr and Kerr-Ads black holes exist widely in general $f(R)$
models. Furthermore, the whole solving process can be treated as a standard
calculation procedure to obtain rotating black holes, which can be applied to
other modified gravities.

##  I Introduction

The black hole solutions in $f(R)$ gravity have been widely discussed. Due to
the extension of gravitational action, there may be more possible black hole
solutions in $f(R)$ theory. But this also brings more mathematical
difficulties to search the black hole solutions. In order to obtain the metric
with analytical functions, some special calculation methods are often
necessary. In the static spherically symmetric (SSS) case, the Birkhoff
theorem is no longer valid for general $f(R)$ theories. Multamäki and Vilja
showed how to construct SSS solutions in some special $f(R)$ models
Multamaki:2006 . Capozziello _et. al._ constructed the Noether symmetry method
to calculate the solutions Capozziello2007 ; Capozziello2010 . Later, the
Lagrange multiplier method was introduced by Sebastiani Sebastiani:2011  and
Nashed Nashed:2019 . The Lagrangian multiplier method treats the Ricci scalar
$R$ as a new variable and uses the Lagrangian multiplier to constrain it. In
Refs. Nashed:2019 ; Elizalde:2020 ; Nashed:2021-1 , the authors obtained a new
class of analytic SSS black hole solutions with a Ricci scalar
$R\propto\frac{1}{r^{2}}$ in $f(R)=R-2\alpha\sqrt{R}$ model. This new black
hole was also obtained in Ref. Li:2022  and may be considered as a spacetime
corresponding to a fixed deficit angle $\delta\theta=\pi$ Li:2022-1 .

It is more difficult to derive axially symmetric solutions than SSS solutions
in $f(R)$ gravity. An interesting consideration may be the generation of
axially symmetric solutions from SSS solutions through a set of complex
coordinate transformations. This method was first introduced by Newman and
Janis in general relativity (GR) Newman:1965 . Since then, this algorithm has
been widely used as a powerful tool to obtain the Kerr-like rotating metrics
from the corresponding SSS metrics. For recent studies, please refer to
Shao:2021 ; Kubiznak:2021 ; Kamenshchik:2023 ; Fernandes:2023 . Capozziello
_et. al._ have shown that the Newman-Janis algorithm also works in $f(R)$
gravity Capozziello2010 . Later, the radiating Kerr-Newman black hole is
obtained by using the Newman-Janis algorithm in $f(R)$ gravity Ghosh:2013 .
Recently, using a modified Newman-Janis algorithm, Chaturvedi obtained the
rotating axisymmetric solutions in $f(R)$ gravity theories with a constant
Ricci scalar Chaturvedi:2023 . However, the Newman-Janis algorithm does not
depend on the Einstein field equations but directly works on the solutions.
All solutions obtained by this method should be checked whether they fulfill
the field equations. In quadratic gravity models, the Newman-Janis algorithm
is not suitable for generating axisymmetric metrics Cadoni:2011 ; Hansen:2013
; Ayzenberg:2015 . Thus, this method can not be used as a universal method for
deriving axially symmetric solutions.

There are few papers aiming to derive the axisymmetric solutions through
strict mathematical procedures. As far as we know, there are two different
roads to obtain the Kerr metric using the strict analytical calculations. One
of them originally came from Carter Carter , who used an additional physical
condition to simplify the line element. Suppose a test particle traveling in
the black hole spacetime, the correspondence of quantum mechanics requires
that the Klein-Gordon equation is separable. Therefore, the spacetime of black
hole should have a special structure and the rotating metric could be written
in a separable canonical form. Using the separable canonical metric form,
Carter obtained the axisymmetric solutions including Kerr and Kerr-Ads
solutions in Ref. Carter .

The other road is given by Chandrasekhar Chandrasekhar1983 . In GR,
Chandrasekhar used a general metric of stationary axisymmetric spacetime to
derive the Kerr and Kerr-Newman solutions. After lots of variable
substitutions, the Ernst’s equations are obtained. As is shown in
Ref.Chandrasekhar1983 , the Kerr solution is a special case of the elementary
solutions of the Ernst’s equations. The biggest benefit of this process is the
strict analytical calculation, which also makes the method universal.

In this paper, we derive the axisymmetric solutions using the method mainly
adopted from Chandrasekhar. Unlike in GR, the field equations of $f(R)$
gravity would be highly coupled and not easy to solve. Fortunately, after
adopting some strategies, we do solve the field equations analytically. In the
cases of $R=0$ and $R\neq 0$, we obtain Kerr and Kerr-Ads solutions,
respectively. What counts is that the whole process is highly universal and
can be easily generalized to derive the axisymmetric solutions in other
modified gravitational theories.

This paper is organized as follows. In Sec.II, we review the vacuum field
equations with a constant Ricci scalar in a general $f(R)$ theory. In Sec.III,
based on a general stationary axisymmetric metric, we calculate the Cartan’s
structure equation and obtain the Einstein tensor. In Sec.IV, we choose a
highly symmetrical gauge to fix the metric form and derive the field
equations. In Sec.V, variables are transformed to the conjugate metric and
some preprocess is done for solving. In Sec.VI, the Kerr metric is derived by
solving the Ernst’s equations. In Sec.VII, the Kerr-Ads metric is derived. The
last section is a detailed conclusion.

In this paper, the Greek letters $\alpha,\beta,\gamma,...=0,1,2,3$ represent
indices of the natural coordinate base, The Latin letters $a,b,c,d=0,1,2,3$
represent indices of moving frame. And the Latin letter $i,j,k=1,2,3$
represent the spatial indices.

##  II The Action

The action of $f(R)$ theory is

$S=\int d^{4}x\sqrt{-g}(f(R)+\mathcal{L}_{m}),$ (1)

where $\mathcal{L}_{m}$ is the Lagrangian density of matter. We set $8\pi
G=1$. The field equations can be written as

$f_{R}R_{\alpha\beta}-\frac{1}{2}fg_{\alpha\beta}+(g_{\alpha\beta}\square+\nabla_{\alpha}\nabla_{\beta})f_{R}=T_{\alpha\beta},$
(2)

where $f_{R}$ stands for the derivative of $f$ with respect to $R$. The
Einstein tensor is defined by

$G_{\alpha\beta}=R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R.$ (3)

Using Einstein tensor, the field equation (2) is expressed as

$f_{R}G_{\alpha\beta}+\frac{g_{\alpha\beta}}{2}(Rf_{R}-f)+(g_{\alpha\beta}\square+\nabla_{\alpha}\nabla_{\beta})f_{R}=T_{\alpha\beta}.$
(4)

In this paper, we consider $f_{R}\neq 0$ and in vacuum case
$T_{\alpha\beta}=0$. In order to compare with GR, we express the vacuum field
equation as

$G_{\alpha\beta}+\mathcal{T}_{\alpha\beta}=0,$ (5)

where

$\displaystyle\mathcal{T}_{\alpha\beta}$
$\displaystyle=\frac{g_{\alpha\beta}}{2}(R-\frac{f}{f_{R}})+\mathcal{K}_{\alpha\beta},$
(6) $\displaystyle\mathcal{K}_{\alpha\beta}$
$\displaystyle=\frac{1}{f_{R}}(g_{\alpha\beta}\square+\nabla_{\alpha}\nabla_{\beta})f_{R}.$
(7)

The Bianchi identity $\nabla^{\alpha}G_{\alpha\beta}=0$ indicates
$\nabla^{\alpha}\mathcal{T}_{\alpha\beta}=0$. Through some mathematical
process, one obtains

$R_{,\alpha}=-\frac{2f_{R}^{2}}{ff_{RR}}\nabla^{\beta}\mathcal{K}_{\beta\alpha},$
(8)

where $R_{,\alpha}$ stands for the derivative of $R$ with respect to
$x^{\alpha}$. The above equation (8) is an extra equation constraining $R$ in
$f(R)$ theory.

Furthermore, if the Ricci scalar $R$ is a constant, one obtains
$\mathcal{K}^{\alpha}{}_{\beta}=0$ immediately. Then, the constraint equation
(8) is satisfied automatically. Only the diagonal terms of
$\mathcal{T}_{\alpha\beta}$ are non-zero and have the same value, i.e.

$\mathcal{T}^{0}{}_{0}=\mathcal{T}^{1}{}_{1}=\mathcal{T}^{2}{}_{2}=\mathcal{T}^{3}{}_{3}=\frac{1}{2}(R-\frac{f}{f_{R}}).$
(9)

Because the trace of Einstein tensor $G^{\alpha}{}_{\alpha}$ is $-R$, one
obtains $\frac{f}{f_{R}}=\frac{R}{2}$ if tracing the vacuum field equation.
Then, we have $\frac{1}{2}(R-\frac{f}{f_{R}})=\frac{R}{4}$. We emphasize that
the constraint $\frac{f}{f_{R}}=\frac{R}{2}$ is always true for constant Ricci
scalar case in $f(R)$ theory. And one can’t use it to rebuild the $f(R)$
model.

##  III The Einstein tensor in a stationary axisymmetric metric

The ansätz of stationary axisymmetric metric is chosen to be

$ds^{2}=-e^{2\nu(x^{2},x^{3})}dt^{2}+e^{2\psi(x^{2},x^{3})}(d\varphi-q(x^{2},x^{3})dt)^{2}+e^{2\mu_{2}(x^{2},x^{3})}(dx^{2})^{2}+e^{2\mu_{3}(x^{2},x^{3})}(dx^{3})^{2},$
(10)

where $(x^{0},x^{1},x^{2},x^{3})=(t,\varphi,x^{2},x^{3})$ are the coordinate
axes. And all unknown variables $\nu,\psi,\mu_{2},\mu_{3},q$ are the functions
of $x^{2}$ and $x^{3}$. As is shown by Chandrasekhar Chandrasekhar1983 , the
metric (10) is the most general form of stationary axisymmetric metric.

We use the method of moving frame to calculate the Einstein tensor
$G_{\alpha\beta}$. We consider the spacetime is a Riemannian manifold that is
torsion-free and metric compatible. Then, the geometric properties of the
manifold are fully described by the frame 1-forms $\omega^{a}$ and the Riemann
Levi-Civita (RLC) connection forms $\omega_{ab}$. They satisfy Cartan’s
equation of structure

$\displaystyle\mathrm{d}\omega^{a}+\omega^{a}{}_{b}\wedge\omega^{b}$
$\displaystyle=0,$ (11)
$\displaystyle\mathrm{d}\omega^{a}{}_{b}+\omega^{a}{}_{c}\wedge\omega^{c}{}_{b}$
$\displaystyle=\Omega^{a}{}_{b},$ (12)

where $\mathrm{d}$ is the exterior differential operator and $\wedge$ stands
for Cartan wedge. $\Omega^{a}{}_{b}$ are curvature 2-forms. In the coordinate
frame $\omega_{\alpha}=\partial_{\alpha}$, they can be written in terms of
Christoffel symbols $\Gamma^{\gamma}_{\alpha\beta}$ as

$\displaystyle\omega_{\alpha}{}_{\beta}$
$\displaystyle=\Gamma^{\alpha}_{\beta\gamma}dx^{\gamma},$ (13)
$\displaystyle\Omega^{\alpha}{}_{\beta}$
$\displaystyle=\frac{1}{2}(\Gamma^{\alpha}_{\beta\sigma,\tau}+\Gamma^{\alpha}_{\gamma\tau}\Gamma^{\gamma}_{\beta\sigma})dx^{\tau}\wedge
dx^{\sigma}.$ (14)

Based on the metric (10), the frame 1-forms are chosen to be

$\omega^{0}=e^{\nu}dt,\quad\omega^{1}=e^{\psi}(d\varphi-
qdt),\quad\omega^{2}=e^{\mu_{2}}dx^{2},\quad\omega^{3}=e^{\mu_{3}}dx^{3}.$
(15)

These forms are orthogonal in the Minkowski metric $\eta_{ab}=diag(-1,1,1,1)$.
One may also reversely express Eq. (15) to

$dt=e^{-\nu}\omega^{0},\quad
d\varphi=qe^{-\nu}\omega^{0}+e^{-\psi}\omega^{1},\quad
dx^{2}=e^{-\mu_{2}}\omega^{2},\quad dx^{3}=e^{-\mu_{3}}\omega^{3}.$ (16)

On the other hand, the RLC connection forms in Riemannian manifold satisfy

$\omega^{0}{}_{i}=\omega^{i}{}_{0},\quad\omega^{i}{}_{j}=-\omega^{j}{}_{i}.$
(17)

The above equations (16) and (17) are used to simplify the calculation
results.

Exterior differentiating the frame (15), one obtains

$\displaystyle\mathrm{d}\omega^{0}$
$\displaystyle=-e^{-\mu_{2}}\nu_{,2}\omega^{0}\wedge\omega^{2}-e^{-\mu_{3}}\nu_{,3}\omega^{0}\wedge\omega^{3},$
(18) $\displaystyle\mathrm{d}\omega^{1}$
$\displaystyle=-(e^{-\mu_{2}}\psi_{,2}\omega^{1}-e^{\psi-\nu-\mu_{2}}q_{,2}\omega^{0})\wedge\omega^{2}-(e^{-\mu_{3}}\psi_{,3}\omega^{1}-e^{\psi-\nu-\mu_{3}}q_{,3}\omega^{0})\wedge\omega^{3},$
(19) $\displaystyle\mathrm{d}\omega^{2}$
$\displaystyle=-e^{-\mu_{3}}\mu_{2,3}\omega^{2}\wedge\omega^{3},$ (20)
$\displaystyle\mathrm{d}\omega^{3}$
$\displaystyle=-e^{-\mu_{2}}\mu_{3,2}\omega^{3}\wedge\omega^{2}.$ (21)

Comparing the results to the first Cartan’s equation of structure (11), we can
extract the nonzero RLC connection forms

$\displaystyle\omega^{0}{}_{1}$
$\displaystyle=\omega^{1}{}_{0}=\frac{1}{2}e^{\psi-\nu-\mu_{2}}q_{,2}\omega^{2}+\frac{1}{2}e^{\psi-\nu-\mu_{3}}q_{,3}\omega^{3},$
(22) $\displaystyle\omega^{0}{}_{2}$
$\displaystyle=\omega^{2}{}_{0}=e^{-\mu_{2}}\nu_{,2}\omega^{0}+\frac{1}{2}e^{\psi-\nu-\mu_{2}}q_{,2}\omega^{1},$
(23) $\displaystyle\omega^{0}{}_{3}$
$\displaystyle=\omega^{3}{}_{0}=e^{-\mu_{3}}\nu_{,3}\omega^{0}+\frac{1}{2}e^{\psi-\nu-\mu_{3}}q_{,3}\omega^{1},$
(24) $\displaystyle\omega^{2}{}_{1}$
$\displaystyle=-\omega^{1}{}_{2}=\frac{1}{2}e^{\psi-\nu-\mu_{2}}q_{,2}\omega^{0}-e^{-\mu_{2}}\psi_{,2}\omega^{1},$
(25) $\displaystyle\omega^{3}{}_{1}$
$\displaystyle=-\omega^{1}{}_{3}=\frac{1}{2}e^{\psi-\nu-\mu_{3}}q_{,3}\omega^{0}-e^{-\mu_{3}}\psi_{,3}\omega^{1},$
(26) $\displaystyle\omega^{2}{}_{3}$
$\displaystyle=-\omega^{3}{}_{2}=e^{-\mu_{3}}\mu_{2,3}\omega^{2}-e^{-\mu_{2}}\mu_{3,2}\omega^{3}.$
(27)

Inserting the frame forms and the RLC connection forms into the second
Cartan’s equation of structure (12), one can obtain the curvature 2-forms
$\Omega^{a}{}_{b}$. The direct calculation shows the non-zero components of
curvature forms are

$\displaystyle\Omega^{0}{}_{1}$
$\displaystyle=-(e^{-2\mu_{2}}\psi_{,2}\nu_{,2}+e^{-2\mu_{3}}\psi_{,3}\nu_{,3}+\frac{1}{4}e^{2(\psi-\nu-\mu_{2})}q^{2}_{,2}+\frac{1}{4}e^{2(\psi-\nu-\mu_{3})}q^{2}_{,3})\omega^{0}\wedge\omega^{1}$
$\displaystyle+\frac{1}{2}e^{-\mu_{2}-\mu_{3}}((e^{\psi-\nu}q_{,3})_{,2}-e^{\psi-\nu}q_{,2})_{,3})\omega^{2}\wedge\omega^{3},$
(28) $\displaystyle\Omega^{0}{}_{2}$
$\displaystyle=(e^{-\nu-\mu_{2}}(e^{\nu-\mu_{2}}\nu_{,2})_{,2}-\frac{3}{4}e^{2(\psi-\nu-\mu_{2})}q_{,2}^{2}+e^{-2\mu_{3}}\nu_{,3}\mu_{2,3})\omega^{2}\wedge\omega^{0}$
$\displaystyle+(e^{-\nu-\mu_{3}}(e^{\nu-\mu_{2}}\nu_{,2})_{,3}-\frac{3}{4}e^{2(\psi-\nu)-\mu_{2}-\mu_{3}}q_{,2}q_{,3}-e^{-\mu_{2}-\mu_{3}}\nu_{,3}\mu_{3,2})\omega^{3}\wedge\omega^{0}$
$\displaystyle-\frac{1}{2}(e^{-\psi-\mu_{2}}(e^{2\psi-\nu-\mu_{2}}q_{,2})_{,2}+e^{\psi-\nu-2\mu_{2}}q_{,2}\psi_{,2}+e^{\psi-\nu-2\mu_{3}}q_{,3}\mu_{2,3})\omega^{1}\wedge\omega^{2}$
$\displaystyle-\frac{1}{2}(e^{-\psi-\mu_{3}}(e^{2\psi-\nu-\mu_{2}}q_{,2})_{,3}+e^{\psi-\nu-\mu_{2}-\mu_{3}}q_{,3}\psi_{,2}-e^{\psi-\nu-\mu_{2}-\mu_{3}}q_{,3}\mu_{3,2})\omega^{1}\wedge\omega^{3},$
(29) $\displaystyle\Omega^{0}{}_{3}$
$\displaystyle=(e^{-\nu-\mu_{3}}(e^{\nu-\mu_{3}}\nu_{,3})_{,3}-\frac{3}{4}e^{2(\psi-\nu-\mu_{3})}q_{,3}^{2}+e^{-2\mu_{2}}\nu_{,2}\mu_{3,2})\omega^{3}\wedge\omega^{0}$
$\displaystyle+(e^{-\nu-\mu_{2}}(e^{\nu-\mu_{3}}\nu_{,3})_{,2}-\frac{3}{4}e^{2(\psi-\nu)-\mu_{2}-\mu_{3}}q_{,2}q_{,3}-e^{-\mu_{2}-\mu_{3}}\nu_{,2}\mu_{2,3})\omega^{2}\wedge\omega^{0}$
$\displaystyle-\frac{1}{2}(e^{-\psi-\mu_{2}}(e^{2\psi-\nu-\mu_{3}}q_{,3})_{,2}+e^{\psi-\nu-\mu_{2}-\mu_{3}}q_{,2}\psi_{,3}-e^{\psi-\nu-\mu_{2}-\mu_{3}}q_{,2}\mu_{2,3})\omega^{1}\wedge\omega^{2}$
$\displaystyle-\frac{1}{2}(e^{-\psi-\mu_{3}}(e^{2\psi-\nu-\mu_{3}}q_{,3})_{,3}+e^{\psi-\nu-2\mu_{3}}q_{,3}\psi_{,3}+e^{\psi-\nu-2\mu_{2}}q_{,2}\mu_{3,2})\omega^{1}\wedge\omega^{3},$
(30) $\displaystyle\Omega^{1}{}_{2}$
$\displaystyle=-\frac{1}{2}(e^{-\nu-\mu_{2}}(e^{\psi-\mu_{2}}q_{,2})_{,2}+2e^{\psi-\nu-2\mu_{2}}q_{,2}\psi_{,2}-e^{\psi-\nu-2\mu_{2}}\nu_{,2}q_{,2}+e^{\psi-\nu-2\mu_{3}}q_{,3}\mu_{2,3})\omega^{2}\wedge\omega^{0}$
$\displaystyle-\frac{1}{2}(e^{-\nu-\mu_{3}}(e^{\psi-\mu_{2}}q_{,2})_{,3}+2e^{\psi-\nu-\mu_{2}-\mu_{3}}q_{,3}\psi_{,2}-e^{\psi-\nu-\mu_{2}-\mu_{3}}\nu_{,2}q_{,3}+e^{\psi-\nu-\mu_{2}-\mu_{3}}q_{,3}\mu_{3,2})\omega^{3}\wedge\omega^{0}$
$\displaystyle+(e^{-\psi-\mu_{2}}(e^{\psi-\mu_{2}}\psi_{,2})_{,2}+\frac{1}{4}e^{2(\psi-\nu-\mu_{2})}q_{,2}^{2}+e^{-2\mu_{3}}\psi_{,3}\mu_{2,3})\omega^{2}\wedge\omega^{1}$
$\displaystyle+(e^{-\psi-\mu_{3}}(e^{\psi-\mu_{2}}\psi_{,2})_{,3}+\frac{1}{4}e^{2(\psi-\nu)-\mu_{2}-\mu_{3}}q_{,2}q_{,3}-e^{-\mu_{2}-\mu_{3}}\psi_{,3}\mu_{3,2})\omega^{3}\wedge\omega^{1},$
(31) $\displaystyle\Omega^{1}{}_{3}$
$\displaystyle=-\frac{1}{2}(e^{-\nu-\mu_{2}}(e^{\psi-\mu_{3}}q_{,3})_{,2}+2e^{\psi-\nu-\mu_{2}-\mu_{3}}q_{,2}\psi_{,3}-e^{\psi-\nu-\mu_{2}-\mu_{3}}\nu_{,3}q_{,2}+e^{\psi-\nu-\mu_{2}-\mu_{3}}q_{,2}\mu_{2,3})\omega^{2}\wedge\omega^{0}$
$\displaystyle-\frac{1}{2}(e^{-\nu-\mu_{3}}(e^{\psi-\mu_{3}}q_{,3})_{,3}+2e^{\psi-\nu-2\mu_{3}}q_{,3}\psi_{,3}-e^{\psi-\nu-2\mu_{2}}\nu_{,3}q_{,3}+e^{\psi-\nu-2\mu_{2}}q_{,2}\mu_{3,2})\omega^{3}\wedge\omega^{0}$
$\displaystyle+(e^{-\psi-\mu_{2}}(e^{\psi-\mu_{3}}\psi_{,3})_{,2}+\frac{1}{4}e^{2(\psi-\nu)-\mu_{2}-\mu_{3}}q_{,2}q_{,3}-e^{-\mu_{2}-\mu_{3}}\psi_{,2}\mu_{2,3})\omega^{2}\wedge\omega^{1}$
$\displaystyle+(e^{-\psi-\mu_{3}}(e^{\psi-\mu_{3}}\psi_{,3})_{,3}+\frac{1}{4}e^{2(\psi-\nu-\mu_{3})}q_{,3}^{2}-e^{-2\mu_{2}}\psi_{,2}\mu_{3,2})\omega^{3}\wedge\omega^{1},$
(32) $\displaystyle\Omega^{2}{}_{3}$
$\displaystyle=\frac{1}{2}e^{\psi-\nu-\mu_{2}-\mu_{3}}((\psi-\nu)_{,3}q_{,2}-(\psi-\nu)_{,2}q_{,3})\omega^{0}\wedge\omega^{1}$
$\displaystyle-e^{-\mu_{2}-\mu_{3}}((e^{\mu_{2}-\mu_{3}}\mu_{2,3})_{,3}+(e^{\mu_{3}-\mu_{2}}\mu_{3,2})_{,2})\omega^{2}\wedge\omega^{3}.$
(33)

The Ricci tensor $R_{ab}$ is defined by the inner product of the curvature
2-form $\Omega_{ab}$ and the dual vectors $e_{a}$

$R_{ab}=\Omega^{c}{}_{a}(e_{c},e_{b})\equiv\langle\Omega^{c}{}_{a};e_{c},e_{b}\rangle,$
(34)

where $\langle;\rangle$ is the inner product. A normalized orthometric basis
$\omega^{a}$ and its dual basis $e_{a}$ satisfy

$\langle\omega^{a};e_{b}\rangle=\delta^{a}_{b}.$ (35)

Note that the inner product can also operate on the $n$-form and $n$ dual
vectors. For the 2-form basis $\omega^{a}\wedge\omega^{b}$ and two dual basis
vectors $e_{c},e_{d}$, we have

$\langle\omega^{a}\wedge\omega^{b};e_{c},e_{d}\rangle=\delta^{ab}_{cd},$ (36)

where $\delta^{ab}_{cd}$ is the generalized Kronecker-$\delta$ function.

Calculating Eq. (34), one obtains $R_{ab}$. Since the Ricci tensor
$R^{\alpha}{}_{\beta}$ and $R^{a}{}_{b}$ have obvious physical significance,
they have the same values whether they are in the moving frame or in the
natural coordinates. So does the Ricci scalar $R$. Then, inserting the results
into Eq. (3), we obtain $G^{\alpha}{}_{\beta}$. The calculation is
straightforward and tedious. For more information about this calculation, we
recommend the book Chandrasekhar1983 . At last, we obtain nonzero components
of $G^{\alpha}{}_{\beta}$ as

$\displaystyle G^{0}{}_{0}=$ $\displaystyle
e^{-2\mu_{2}}\left(\psi_{,2,2}+\psi^{2}_{,2}+\psi_{,2}(\mu_{3}-\mu_{2})_{,2}\right)+e^{-2\mu_{3}}\left(\psi_{,3,3}+\psi^{2}_{,3}+\psi_{,3}(\mu_{2}-\mu_{3})_{,3}\right)$
$\displaystyle+e^{-\mu_{2}-\mu_{3}}\left((e^{\mu_{2}-\mu_{3}}\mu_{2,3})_{,3}+(e^{\mu_{3}-\mu_{2}}\mu_{3,2})_{,2}\right)+\frac{1}{4}e^{2(\psi-\nu-\mu_{2})}q_{,2}^{2}+\frac{1}{4}e^{2(\psi-\nu-\mu_{3})}q_{,3}^{2},$
(37) $\displaystyle G^{1}{}_{1}=$ $\displaystyle
e^{-2\mu_{2}}\left(\nu_{,2,2}+\nu^{2}_{,2}+\nu_{,2}(\mu_{3}-\mu_{2})_{,2}\right)+e^{-2\mu_{3}}\left(\nu_{,3,3}+\nu^{2}_{,3}+\nu_{,3}(\mu_{2}-\mu_{3})_{,3}\right)$
$\displaystyle+e^{-\mu_{2}-\mu_{3}}\left((e^{\mu_{2}-\mu_{3}}\mu_{2,3})_{,3}+(e^{\mu_{3}-\mu_{2}}\mu_{3,2})_{,2}\right)-\frac{3}{4}e^{2(\psi-\nu-\mu_{2})}q_{,2}^{2}-\frac{3}{4}e^{2(\psi-\nu-\mu_{3})}q_{,3}^{2},$
(38) $\displaystyle G^{2}{}_{2}=$ $\displaystyle
e^{-2\mu_{2}}\left(\psi_{,2}\nu_{,2}+(\psi+\nu)_{,2}\mu_{3,2}\right)+e^{-2\mu_{3}}\big{(}(\psi+\nu)_{,3,3}+\nu^{2}_{,3}+\psi^{2}_{,3}+\psi_{,3}\nu_{,3}$
$\displaystyle-(\psi+\nu)_{,3}\mu_{3,3}\big{)}+\frac{1}{4}e^{2(\psi-\nu-\mu_{2})}q_{,2}^{2}-\frac{1}{4}e^{2(\psi-\nu-\mu_{3})}q_{,3}^{2},$
(39) $\displaystyle G^{3}{}_{3}=$ $\displaystyle
e^{-2\mu_{3}}\left(\psi_{,3}\nu_{,3}+(\psi+\nu)_{,3}\mu_{2,3}\right)+e^{-2\mu_{2}}\big{(}(\psi+\nu)_{,2,2}+\nu^{2}_{,2}+\psi^{2}_{,2}+\psi_{,2}\nu_{,2}$
$\displaystyle-(\psi+\nu)_{,2}\mu_{2,2}\big{)}-\frac{1}{4}e^{2(\psi-\nu-\mu_{2})}q_{,2}^{2}+\frac{1}{4}e^{2(\psi-\nu-\mu_{3})}q_{,3}^{2},$
(40) $\displaystyle G^{0}{}_{1}=$
$\displaystyle-\frac{1}{2}e^{-(2\psi+\mu_{2}+\mu_{3})}\left((e^{3\psi-\nu+\mu_{3}-\mu_{2}}q_{,2})_{,2}+(e^{3\psi-\nu+\mu_{2}-\mu_{3}}q_{,3})_{,3}\right),$
(41) $\displaystyle G^{2}{}_{3}=$
$\displaystyle-e^{-\mu_{2}-\mu_{3}}\left((\psi+\nu)_{,2,3}+\nu_{,2}\nu_{,3}+\psi_{,2}\psi_{,3}-(\psi+\nu)_{,2}\mu_{2,3}-(\psi+\nu)_{,3}\mu_{3,2}\right)$
$\displaystyle+\frac{1}{2}e^{2(\psi-\nu)-\mu_{2}-\mu_{3}}q_{,2}q_{,3}.$ (42)

And the Ricci scalar $R$ is expressed as

$\displaystyle R=$
$\displaystyle-2e^{-2\mu_{2}}\left((\psi+\nu)_{,2,2}+\psi^{2}_{,2}+\nu^{2}_{,2}+\psi_{,2}\nu_{,2}+(\psi+\nu)_{,2}(\mu_{3}-\mu_{2})_{,2}\right)$
$\displaystyle-$ $\displaystyle
2e^{-2\mu_{3}}\left((\psi+\nu)_{,3,3}+\psi^{2}_{,3}+\nu^{2}_{,3}+\psi_{,3}\nu_{,3}+(\psi+\nu)_{,3}(\mu_{2}-\mu_{3})_{,3}\right)$
$\displaystyle-$ $\displaystyle
2e^{-\mu_{2}-\mu_{3}}\left((e^{\mu_{2}-\mu_{3}}\mu_{2,3})_{,3}+(e^{\mu_{3}-\mu_{2}}\mu_{3,2})_{,2}\right)+\frac{1}{2}e^{2(\psi-\nu-\mu_{2})}q_{,2}^{2}+\frac{1}{2}e^{2(\psi-\nu-\mu_{3})}q_{,3}^{2}.$
(43)

##  IV The Field equations

Similar to Ref.Chandrasekhar1983 , we consider the following equations

$\displaystyle G^{2}{}_{2}+G^{3}{}_{3}$ $\displaystyle=-\frac{R}{2},$ (44)
$\displaystyle G^{2}{}_{2}-G^{3}{}_{3}$ $\displaystyle=0,$ (45) $\displaystyle
G^{1}{}_{1}+G^{0}{}_{0}$ $\displaystyle=-\frac{R}{2},$ (46) $\displaystyle
G^{1}{}_{1}-G^{0}{}_{0}$ $\displaystyle=0,$ (47) $\displaystyle G^{0}{}_{1}$
$\displaystyle=0.$ (48)

Since the Bianchi identity $\nabla^{\alpha}G_{\alpha\beta}=0$ is satisfied,
only 5 components of Einstein equations are independent. The direct
calculation of (44)-(48) gives

$\displaystyle(e^{\psi+\nu+\mu_{3}-\mu_{2}}(\psi+\nu)_{,2})_{,2}+(e^{\psi+\nu+\mu_{2}-\mu_{3}}(\psi+\nu)_{,3})_{,3}$
$\displaystyle=-\frac{R}{2}e^{\psi+\nu+\mu_{2}+\mu_{3}},$ (49) $\displaystyle
e^{2\mu_{3}}(e^{\psi+\nu-\mu_{2}-\mu_{3}}(\psi+\nu)_{,2})_{,2}-e^{2\mu_{2}}(e^{\psi+\nu-\mu_{2}-\mu_{3}}(\psi+\nu)_{,3})_{,3}$
$\displaystyle-2e^{\psi+\nu+\mu_{3}-\mu_{2}}\psi_{,2}\nu_{,2}$
$\displaystyle+2e^{\psi+\nu+\mu_{2}-\mu_{3}}\psi_{,3}\nu_{,3}-\frac{1}{2}(e^{3\psi-\nu+\mu_{3}-\mu_{2}}q_{,2}^{2}-e^{3\psi-\nu+\mu_{2}-\mu_{3}}q_{,3}^{2})$
$\displaystyle=0,$ (50)
$\displaystyle(e^{\psi+\nu+\mu_{3}-\mu_{2}}(\psi+\nu)_{,2})_{,2}+(e^{\psi+\nu+\mu_{2}-\mu_{3}}(\psi+\nu)_{,3})_{,3}$
$\displaystyle-2e^{\psi+\nu+\mu_{3}-\mu_{2}}\psi_{,2}\nu_{,2}$
$\displaystyle-2e^{\psi+\nu+\mu_{2}-\mu_{3}}\psi_{,3}\nu_{,3}+2e^{\psi+\nu}((e^{\mu_{3}-\mu_{2}}\mu_{3,2})_{,2}+$
$\displaystyle(e^{\mu_{2}-\mu_{3}}\mu_{2,3})_{,3})$
$\displaystyle-\frac{1}{2}(e^{3\psi-\nu+\mu_{3}-\mu_{2}}q_{,2}^{2}+e^{3\psi-\nu+\mu_{2}-\mu_{3}}q_{,3}^{2})$
$\displaystyle=-\frac{R}{2}e^{\psi+\nu+\mu_{2}+\mu_{3}},$ (51)
$\displaystyle(e^{\psi+\nu+\mu_{3}-\mu_{2}}(\nu-\psi)_{,2})_{,2}+(e^{\psi+\nu+\mu_{2}-\mu_{3}}(\nu-\psi)_{,3})_{,3}$
$\displaystyle=e^{3\psi-\nu+\mu_{3}-\mu_{2}}q_{,2}^{2}+e^{3\psi-\nu+\mu_{2}-\mu_{3}}q_{,3}^{2},$
(52)
$\displaystyle(e^{3\psi-\nu+\mu_{3}-\mu_{2}}q_{,2})_{,2}+(e^{3\psi-\nu+\mu_{2}-\mu_{3}}q_{,3})_{,3}$
$\displaystyle=0.$ (53)

The system of equations (49)-(53) is difficult to solve. In order to obtain
the solution, we need to make substitution of variables and transform the
metric (10) to a more symmetric form.

There is a gauge freedom in the plane of $(x^{2},x^{3})$ that can rotate the
plane. In some special position, the metric can be transformed to

$e^{2\mu_{2}(x^{2},x^{3})}(dx^{2})^{2}+e^{2\mu_{3}(x^{2},x^{3})}(dx^{3})^{2}=\rho^{2}(r,\mu)\left(\frac{dr^{2}}{\Delta_{r}(r)}+\frac{d\mu^{2}}{\Delta_{\mu}(\mu)}\right),$
(54)

where the new coordinates $(r,\mu)$ are related to $(x^{2},x^{3})$ by the
relationships

$\displaystyle e^{2\mu_{2}(x^{2},x^{3})}$
$\displaystyle=\frac{\rho^{2}(r,\mu)}{\Delta_{r}},$ (55) $\displaystyle
e^{2\mu_{3}(x^{2},x^{3})}$
$\displaystyle=\frac{\rho^{2}(r,\mu)}{\Delta_{\mu}}.$ (56)

In the Boyer-Lindquist coordinate system, $r$ is the radial coordinate and
$\mu=\cos\theta$. The function $\Delta_{r}$ depends only on $r$ and
$\Delta_{\mu}$ depends only on $\mu$. In order to solve the field equations, a
highly symmetrical metric form is required. Similar to $(x^{2},x^{3})$, there
is also a gauge freedom in the plane of $(t,\varphi)$. According to
Chandrasekhar1983 , this gauge freedom is chosen as

$-e^{2\nu}dt^{2}+e^{2\psi}(d\varphi-qdt)^{2}=e^{\kappa}\left(-\chi
dt^{2}+\frac{1}{\chi}(d\varphi-qdt)^{2}\right),$ (57)

where $\kappa=\kappa(r,\mu)$. More precisely, we expect $e^{\kappa}$ is a
separable variable, i.e. $e^{\kappa}=h(r)l(\mu)$. As is shown in
Chandrasekhar1983 , $h(r)=\sqrt{\Delta_{r}}$. For symmetry reasons, we choose
$l(\mu)=\sqrt{\Delta_{\mu}}$. Therefore, after the gauge freedoms are fixed,
the metric (10) has a form

$ds^{2}=\sqrt{\Delta_{r}\Delta_{\mu}}\left(-\chi
dt^{2}+\frac{1}{\chi}(d\varphi-
qdt)^{2}\right)+\rho^{2}\left(\frac{dr^{2}}{\Delta_{r}}+\frac{d\mu^{2}}{\Delta_{\mu}}\right),$
(58)

where $\chi=\chi(r,\mu),q=q(r,\mu)$. Now the unknown variables turn into
$\chi,q,\rho^{2},\Delta_{r},\Delta_{\mu}$. Comparing (58) with (10), we have

$e^{2\nu}=\chi\sqrt{\Delta_{r}\Delta_{\mu}},~{}~{}e^{2\psi}=\frac{\sqrt{\Delta_{r}\Delta_{\mu}}}{\chi},~{}~{}e^{2\mu_{2}}=\frac{\rho^{2}}{\Delta_{r}},~{}~{}e^{2\mu_{3}}=\frac{\rho^{2}}{\Delta_{\mu}}.$
(59)

Plugging these expressions into (49)-(53), we obtain

$\displaystyle\Delta_{r,rr}+\Delta_{\mu,\mu\mu}$ $\displaystyle=-R\rho^{2},$
(60)
$\displaystyle\Delta_{r,rr}-\Delta_{\mu,\mu\mu}-\frac{1}{4}\frac{\Delta_{r,r}^{2}}{\Delta_{r}}+\frac{1}{4}\frac{\Delta_{\mu,\mu}^{2}}{\Delta_{\mu}}-\Delta_{r,r}(\ln\rho^{2})_{,r}+$
$\displaystyle\Delta_{\mu,\mu}(\ln\rho^{2})_{,\mu}$
$\displaystyle+\frac{\Delta_{r}}{\chi^{2}}(\chi^{2}_{,r}-q^{2}_{,r})-\frac{\Delta_{\mu}}{\chi^{2}}(\chi^{2}_{,\mu}-q^{2}_{,\mu})$
$\displaystyle=0,$ (61)
$\displaystyle\Delta_{r,rr}+\Delta_{\mu,\mu\mu}-\frac{1}{4}\frac{\Delta_{r,r}^{2}}{\Delta_{r}}-\frac{1}{4}\frac{\Delta_{\mu,\mu}^{2}}{\Delta_{\mu}}+2\Delta_{r}(\ln\rho^{2})_{,rr}+$
$\displaystyle 2\Delta_{\mu}(\ln\rho^{2})_{,\mu\mu}$
$\displaystyle+\Delta_{r,r}(\ln\rho^{2})_{,r}+\Delta_{\mu,\mu}(\ln\rho^{2})_{,\mu}+\frac{\Delta_{r}}{\chi^{2}}(\chi^{2}_{,r}-q^{2}_{,r})+\frac{\Delta_{\mu}}{\chi^{2}}(\chi^{2}_{,\mu}-q^{2}_{,\mu})$
$\displaystyle=-R\rho^{2},$ (62)
$\displaystyle\frac{1}{\chi^{2}}(\Delta_{r}q_{,r}^{2}+\Delta_{\mu}q_{,\mu}^{2})-\left(\frac{\Delta_{r}}{\chi}\chi_{,r}\right)_{,r}-\left(\frac{\Delta_{\mu}}{\chi}\chi_{,\mu}\right)_{,\mu}$
$\displaystyle=0,$ (63)
$\displaystyle\left(\frac{\Delta_{r}}{\chi^{2}}q_{,r}\right)_{,r}+\left(\frac{\Delta_{\mu}}{\chi^{2}}q_{,\mu}\right)_{,\mu}$
$\displaystyle=0.$ (64)

The above Eqs.(60)-(64) determine all five unknown variables
$\chi,q,\rho^{2},\Delta_{r},\Delta_{\mu}$. And we find $r$ and $\mu$ always
appear in pairs. This is because we use the metric (58) to simplify the field
equations and this is also a key point to obtain the solutions.

##  V Conjugate metric

The forms of Eqs. (60), (63) and (64) are relatively simple and two variables
$r$ and $\mu$ always appear symmetrically, which is a breakthrough for solving
the field equations. In the special case $R=0$, Eqs. (60), (63) and (64)
contain four unknowns $\chi,q$ and $\Delta_{r},\Delta_{\mu}$. The paired
appearance of $r$ and $\mu$ indicates that the dependence of $\Delta_{r}$’s
configuration on $r$ is similar to that of $\Delta_{\mu}$ on $\mu$. In this
sense, simultaneously solving Eqs. (60), (63) and (64) is possible. However,
if $R$ is a non-zero constant, Eqs. (60), (63) and (64) contain an extra
unknown function $\rho^{2}$. Strictly speaking, in the non-zero Ricci scalar
case, the closed system of field equations contain all five Eqs. (60)-(64),
which brings great difficulty to do analytical solution. To avoid this
situation, we presuppose the form of $\rho^{2}$ to keep the system of Eqs.
(60), (63) and (64) remain closed. In this section, we do some pre-processing
on Eqs. (63) and (64).

Similar to Ref. Chandrasekhar1983 , we use new variables to re-express Eqs.
(63) and (64). Defining

$\Phi_{,r}=\frac{\Delta_{\mu}}{\chi^{2}}q_{,\mu},~{}~{}\Phi_{,\mu}=-\frac{\Delta_{r}}{\chi^{2}}q_{,r},$
(65)

one can rewrite Eq. (64) as $\Phi_{,\mu,r}=\Phi_{,r,\mu}$. At the same time,
since $q_{,r,\mu}=q_{,\mu,r}$, the variable $\Phi$ satisfies

$\left(\frac{\chi^{2}}{\Delta_{\mu}}\Phi_{,r}\right)_{,r}+\left(\frac{\chi^{2}}{\Delta_{r}}\Phi_{,\mu}\right)_{,\mu}=0.$
(66)

Furthermore, letting

$\Psi=\frac{\sqrt{\Delta_{r}\Delta_{\mu}}}{\chi},$ (67)

we can reexpress Eqs. (63) and (66) as

$\displaystyle\Psi[(\Delta_{r}\Psi_{,r})_{,r}+(\Delta_{\mu}\Psi_{,\mu})_{,\mu}]$
$\displaystyle=\Delta_{r}[\Psi_{,r}^{2}-\Phi_{,r}^{2}]+\Delta_{\mu}[\Psi_{,\mu}^{2}-\Phi_{,\mu}^{2}]+\frac{\Psi^{2}}{2}(\Delta_{r,rr}+\Delta_{\mu,\mu\mu}),$
(68)
$\displaystyle\Psi[(\Delta_{r}\Phi_{,r})_{,r}+(\Delta_{\mu}\Phi_{,\mu})_{,\mu}]$
$\displaystyle=2\Delta_{r}\Phi_{,r}\Psi_{,r}+2\Delta_{\mu}\Phi_{,\mu}\Psi_{,\mu}.$
(69)

In order to give a simple solving process, we need to solve Eqs. (68) and (69)
in a conjugate metric. With the coordinate transformation

$t\rightarrow i\varphi,~{}~{}~{}\varphi\rightarrow-it,$ (70)

one can make the metric transformation and obtain the conjugate metric as

$\displaystyle\chi dt^{2}-\frac{1}{\chi}(d\varphi-qdt)^{2}\rightarrow$
$\displaystyle\chi
dt^{2}+\frac{2q}{\chi}dtd\varphi-\frac{\chi^{2}-q^{2}}{\chi}d\varphi^{2}$
$\displaystyle=$
$\displaystyle\tilde{\chi}dt^{2}-\frac{1}{\tilde{\chi}}(d\varphi-\tilde{q}dt)^{2},$

where

$\tilde{\chi}=\frac{\chi}{\chi^{2}-q^{2}},~{}~{}~{}\tilde{q}=\frac{q}{\chi^{2}-q^{2}}.$

Similarly, one can also define

$\displaystyle\tilde{\Psi}$
$\displaystyle=\frac{\sqrt{\Delta_{r}\Delta_{\mu}}}{\tilde{\chi}},$ (71)
$\displaystyle\tilde{\Phi}_{,r}$
$\displaystyle=\frac{\Delta_{\mu}}{\tilde{\chi}^{2}}\tilde{q}_{,\mu},$ (72)
$\displaystyle\tilde{\Phi}_{,\mu}$
$\displaystyle=-\frac{\Delta_{r}}{\tilde{\chi}^{2}}\tilde{q}_{,r}.$ (73)

The new variables $\tilde{\Psi}$ and $\tilde{\Phi}$ also satisfy Eqs. (68) and
(69).

##  VI The derivation of Kerr metric

We solve Eqs. (60), (63) and (64) simultaneously in the two cases of $R=0$ and
$R\neq 0$. In this section, we give the standardized solving procedure in the
case of $R=0$. The main process is rooted from Chandrasekhar1983 .

In the case of $R=0$, we denote

$\Delta_{r}=\Delta_{r}^{K},~{}\Delta_{\mu}=\Delta_{\mu}^{K},~{}\tilde{\Phi}=\tilde{\Phi}^{K},~{}\tilde{\Psi}=\tilde{\Psi}^{K}.$

Eqs. (63) and (64) reduce to

$\displaystyle\tilde{\Psi}^{K}[(\Delta^{K}_{r}\tilde{\Psi}^{K}_{,r})_{,r}+(\Delta^{K}_{\mu}\tilde{\Psi}^{K}_{,\mu})_{,\mu}]$
$\displaystyle=\Delta^{K}_{r}[(\tilde{\Psi}_{,r}^{K})^{2}-(\tilde{\Phi}_{,r}^{K})^{2}]+\Delta^{2}_{\mu}[(\tilde{\Psi}_{,\mu}^{K})^{2}-(\tilde{\Phi}_{,\mu}^{K})^{2}],$
(74)
$\displaystyle\tilde{\Psi}^{K}[(\Delta^{K}_{r}\tilde{\Phi}^{K}_{,r})_{,r}+(\Delta^{K}_{\mu}\tilde{\Phi}^{K}_{,\mu})_{,\mu}]$
$\displaystyle=2\Delta^{K}_{r}\tilde{\Phi}^{K}_{,r}\tilde{\Psi}^{K}_{,r}+2\Delta^{K}_{\mu}\tilde{\Phi}^{K}_{,\mu}\tilde{\Psi}^{K}_{,\mu}.$
(75)

In the conjugate metric, Eq. (60) remains the same. The shortest and simplest
route to the Kerr metric is to give the solution of Ernst’s equations. In
order to obtain the Ernst’s equations, we further express $\tilde{\Psi}^{K}$
as the real part and $\tilde{\Phi}^{K}$ as the imaginary part of a complex
function $Z$, i.e.

$Z=\tilde{\Psi}^{K}+i\tilde{\Phi}^{K}.$ (76)

Then, Eqs. (74) and (75) can be combined into

$\mathrm{Re}(Z)[(\Delta^{K}_{r}Z_{,r})_{,r}+(\Delta^{K}_{\mu}Z_{,\mu})_{,\mu}]=\Delta^{K}_{r}Z_{,r}^{2}+\Delta^{K}_{\mu}Z_{,\mu}^{2},$
(77)

where $\mathrm{Re}(Z)$ is the real part of $Z$. One can prove that if $Z$ is
the solution of Eq. (77), $\frac{Z}{1+iCZ}$ is also a solution, where $C$ is a
real constant. The transformation $Z\rightarrow\frac{Z}{1+iCZ}$ is called the
Ehler’s transformation. By the transformation

$Z=-\frac{1+E}{1-E},$ (78)

one can rewrite the above Eq. (77) to

$(1-EE^{*})[(\Delta^{K}_{r}E_{,r})_{,r}+(\Delta^{K}_{\mu}E_{,\mu})_{,\mu}]=-2E^{*}[\Delta^{K}_{r}E_{,r}^{2}+\Delta^{K}_{\mu}E_{,\mu}^{2}].$
(79)

Eqs. (60) and (79) are called the Ernst’s equations.

As is said before, the dependence of $\Delta^{K}_{r}$’s configuration on $r$
is similar to that of $\Delta^{K}_{\mu}$ on $\mu$. It is not difficult to show
that $\Delta^{K}_{r}(r)$ is a quadratic function of $r$. Similarly,
$\Delta^{K}_{\mu}(\mu)$ is a quadratic function of $\mu$. Thus, the Ernst’s
equations permit the elementary solution

$\displaystyle\Delta^{K}_{r}$ $\displaystyle=r^{2}-1,$ (80)
$\displaystyle\Delta^{K}_{\mu}$ $\displaystyle=1-\mu^{2},$ (81) $\displaystyle
E$ $\displaystyle=-C_{r}r-iC_{\mu}\mu,$ (82)

where $C_{r}^{2}+C_{\mu}^{2}=1$ and $C_{r},C_{\mu}$ are real constants. Then,
the corresponding solutions for $\tilde{\Psi}^{K},\tilde{\Phi}^{K}$ are

$\displaystyle\tilde{\Psi}^{K}$
$\displaystyle=\frac{C_{r}^{2}\Delta^{K}_{r}-C_{\mu}^{2}\Delta^{K}_{\mu}}{(C_{r}r+1)^{2}+C_{\mu}^{2}\mu^{2}},$
(83) $\displaystyle\tilde{\Phi}^{K}$
$\displaystyle=\frac{2C_{\mu}\mu}{(C_{r}r+1)^{2}+C_{\mu}^{2}\mu^{2}}.$ (84)

However, they are not unique solutions of Ernst’s equations (60) and (79). For
any quadratic functions

$\displaystyle\Delta^{K}_{r}$ $\displaystyle=r^{2}+2Ar+B,$ (85)
$\displaystyle\Delta^{K}_{\mu}$ $\displaystyle=-(\mu^{2}+2C\mu+D),$ (86)

where $A,B,C,D$ are real constants, we can obtain similar solutions. Defining

$\displaystyle u$ $\displaystyle=\frac{r+A}{\sqrt{A^{2}-B}},$ (87)
$\displaystyle v$ $\displaystyle=\frac{\mu+C}{\sqrt{C^{2}-D}},$ (88)

we have

$\displaystyle\Delta^{K}_{r}$
$\displaystyle=(A^{2}-B)(u^{2}-1)\equiv(A^{2}-B)\Delta^{K}_{u},$ (89)
$\displaystyle\Delta^{K}_{\mu}$
$\displaystyle=(C^{2}-D)(1-v^{2})\equiv(C^{2}-D)\Delta^{K}_{v}.$ (90)

Direct mathematical calculations show that $\Delta^{K}_{u},\Delta^{K}_{v},E$
also satisfy the Ernst’s equations, i.e.

$\displaystyle\Delta^{K}_{u,uu}+\Delta^{K}_{v,vv}$ $\displaystyle=0,$ (91)
$\displaystyle(1-EE^{*})[(\Delta^{K}_{u}E_{,u})_{,u}+(\Delta^{K}_{v}E_{,v})_{,v}]$
$\displaystyle=-2E^{*}[\Delta^{K}_{u}E_{,u}^{2}+\Delta^{K}_{v}E_{,v}^{2}].$
(92)

In virtue of these processes, we can obtain many new solutions of Ernst’s
equations.

Kerr metric is one of the above solutions, which is

$\displaystyle\Delta^{K}_{r}$ $\displaystyle=r^{2}-2Mr+a^{2},$ (93)
$\displaystyle\Delta^{K}_{\mu}$ $\displaystyle=1-\mu^{2},$ (94) $\displaystyle
E$ $\displaystyle=-C_{r}\frac{r-M}{\sqrt{M^{2}-a^{2}}}-iC_{\mu}\mu.$ (95)

The constants are chosen to be $C_{r}=\frac{\sqrt{M^{2}-a^{2}}}{M}$ and
$C_{\mu}=\frac{a}{M}$. From
$\tilde{Z}=-\frac{1+E}{1-E}=\tilde{\Psi}+i\tilde{\Phi}$, we have

$\displaystyle\tilde{\Psi}^{K}$
$\displaystyle=\frac{\Delta^{K}_{r}-a^{2}\Delta^{K}_{\mu}}{r^{2}+a^{2}\mu^{2}},$
(96) $\displaystyle\tilde{\Phi}^{K}$
$\displaystyle=\frac{2aM\mu}{r^{2}+a^{2}\mu^{2}}.$ (97)

In the original metric, the unknown functions are

$\displaystyle\rho^{2}$ $\displaystyle=r^{2}+a^{2}\mu^{2},$ (98)
$\displaystyle\chi$
$\displaystyle=\frac{(r^{2}+a^{2}\mu^{2})\sqrt{\Delta_{r}^{K}\Delta_{\mu}^{K}}}{(r^{2}+a^{2})^{2}\Delta^{K}_{\mu}-a^{2}(1-\mu^{2})^{2}\Delta^{K}_{r}},$
(99) $\displaystyle q$
$\displaystyle=a\frac{(r^{2}+a^{2})\Delta^{K}_{\mu}-(1-\mu^{2})\Delta^{K}_{r}}{(r^{2}+a^{2})^{2}\Delta^{K}_{\mu}-a^{2}(1-\mu^{2})^{2}\Delta^{K}_{r}}.$
(100)

If we define $\mu=\cos\theta$, the solutions transform to the standard Kerr
metric form in the Boyer-Lindquist coordinate system, which is

$ds^{2}=-\frac{\rho^{2}\sin^{2}\theta\Delta^{K}_{r}}{\Sigma_{K}^{2}}dt^{2}+\frac{\Sigma_{K}^{2}}{\rho^{2}}\left(d\varphi-\frac{(r^{2}+a^{2})-\Delta^{K}_{r}}{\Sigma_{K}^{2}}a\sin^{2}\theta
dt\right)^{2}+\frac{\rho^{2}}{\Delta^{K}_{r}}dr^{2}+\rho^{2}d\theta^{2},$
(101)

where

$\displaystyle\rho^{2}$ $\displaystyle=r^{2}+a^{2}\cos^{2}\theta,$ (102)
$\displaystyle\Sigma_{K}^{2}$
$\displaystyle=(r^{2}+a^{2})^{2}\sin^{2}\theta-a^{2}\sin^{4}\theta\Delta^{K}_{r},$
(103) $\displaystyle\Delta^{K}_{r}$ $\displaystyle=r^{2}-2Mr+a^{2}.$ (104)

##  VII The derivation of Kerr-Ads metric

In the case of $R\neq 0$, we still expect the system of Eqs. (60), (63) and
(64) to be closed. For this purpose, we try to presuppose the function
$\rho^{2}$. The choice of $\rho^{2}$ is arbitrary. However, since $r$ and
$\mu$ always appear in similar configurations, we can expect that $\rho^{2}$
should also have a similar configuration for $r$ and $\mu$. Notice that the
left-hand side of Eq. (60) does not have terms $\Delta_{r}\Delta_{\mu}$ and
$\Delta_{r,r}\Delta_{\mu,\mu}$, we can expect that term $r\mu$ is not included
in $\rho^{2}$. One may try such terms as $r+\mu$, $r^{2}+\mu^{2}$,
$r^{3}+\mu^{3}$…. Considering the form $\rho^{2}=r^{2}+a^{2}\mu^{2}$ in the
case of $R=0$, we can set the function $\rho^{2}=r^{2}+\mu^{2}$ as the
simplest choice.

The solution in the case of $R\neq 0$ may be obtained by the generalization of
the solution in the case of $R=0$. In the spherical case $a=0$, it is not
difficult to find that the Ricci scalar is a linear term in the metric. In the
axisymmetric case $a\neq 0$, we expect this algebraic property is still true.
Thus, we decompose the solutions into two parts

$\displaystyle\Delta_{r}$ $\displaystyle=\Delta_{r}^{K}+R\Delta_{r}^{R},$
(105) $\displaystyle\Delta_{\mu}$
$\displaystyle=\Delta_{\mu}^{K}+R\Delta_{\mu}^{R},$ (106)
$\displaystyle\tilde{\Phi}$
$\displaystyle=\tilde{\Phi}^{K}+R\tilde{\Phi}^{R},$ (107)
$\displaystyle\tilde{\Psi}$
$\displaystyle=\tilde{\Psi}^{K}+R\tilde{\Psi}^{R},$ (108)

where $\Delta_{r}^{R},\Delta_{\mu}^{R},\tilde{\Phi}^{R},\tilde{\Psi}^{R}$ are
the solutions introduced by $R$. Notice that

$\Delta_{r}^{R}=\frac{\partial\Delta_{r}}{\partial
R},~{}\Delta_{\mu}^{R}=\frac{\partial\Delta_{\mu}}{\partial
R},~{}\tilde{\Phi}^{R}=\frac{\partial\tilde{\Phi}}{\partial
R},~{}\tilde{\Psi}^{R}=\frac{\partial\tilde{\Psi}}{\partial R}.$

Plugging the solutions (105) - (108) into Eqs. (60), (63) and (64), and taking
the partial differential of $R$, we obtain

$\displaystyle\Delta^{R}_{r,rr}+\Delta^{R}_{\mu,\mu\mu}$
$\displaystyle=-(r^{2}+\mu^{2}),$ (109)
$\displaystyle\tilde{\Psi}^{R}[(\Delta^{R}_{r}\tilde{\Psi}^{R}_{,r})_{,r}+(\Delta^{R}_{\mu}\tilde{\Psi}^{R}_{,\mu})_{,\mu}]$
$\displaystyle=\Delta^{R}_{r}[(\tilde{\Psi}_{,r}^{R})^{2}-(\tilde{\Phi}_{,r}^{R})^{2}]+\Delta^{R}_{\mu}[(\tilde{\Psi}_{,\mu}^{R})^{2}-(\tilde{\Phi}_{,\mu}^{R})^{2}]$
$\displaystyle+\frac{(\tilde{\Psi}^{R})^{2}}{2}(\Delta^{R}_{r,rr}+\Delta^{R}_{\mu,\mu\mu}),$
(110)
$\displaystyle\tilde{\Psi}^{R}[(\Delta^{R}_{r}\tilde{\Phi}^{R}_{,r})_{,r}+(\Delta^{R}_{\mu}\tilde{\Phi}^{R}_{,\mu})_{,\mu}]$
$\displaystyle=2\Delta^{R}_{r}\tilde{\Phi}^{R}_{,r}\tilde{\Psi}^{R}_{,r}+2\Delta^{R}_{\mu}\tilde{\Phi}^{R}_{,\mu}\tilde{\Psi}^{R}_{,\mu}.$
(111)

To solve Eq. (109), we let

$\Delta^{R}_{r,rr}=-(r^{2}+k_{1}^{2}),\quad\Delta^{R}_{\mu,\mu\mu}=-(\mu^{2}-k_{1}^{2}),$
(112)

where $k_{1}$ is a constant. Therefore, one can easily obtain the solution

$\Delta^{R}_{r}=-(\frac{r^{4}}{12}+\frac{k_{1}^{2}}{2}r^{2}),\quad\Delta^{R}_{\mu}=-\frac{\mu^{4}}{12}+\frac{k_{1}^{2}}{2}\mu^{2}.$
(113)

We still express $\tilde{\Psi}^{R}$ as the real part and $\tilde{\Phi}^{R}$ as
the imaginary part of a complex function $Z$, i.e.

$Z=\tilde{\Psi}^{R}+i\tilde{\Phi}^{R}.$ (114)

Thus, Eqs. (110) and (111) can be expressed as

$\mathrm{Re}(Z)[(\Delta^{R}_{r}Z_{,r})_{,r}+(\Delta^{R}_{\mu}Z_{,\mu})_{,\mu}]=\Delta^{R}_{r}Z_{,r}^{2}+\Delta^{R}_{\mu}Z_{,\mu}^{2}-\frac{\mathrm{Re}^{2}(Z)}{2}(r^{2}+\mu^{2}),$
(115)

where $\mathrm{Re}^{2}(Z)=\mathrm{Re}(Z)\cdot\mathrm{Re}(Z)$. It’s not hard to
prove that Eq. (115) permits the solution

$Z=-\frac{1}{12}(r+i\mu)^{2}-\frac{k_{1}^{2}}{2}.$ (116)

From (114), we have

$\displaystyle\tilde{\Psi}^{R}$
$\displaystyle=-\frac{1}{12}(r^{2}-\mu^{2})-\frac{k_{1}^{2}}{2},$ (117)
$\displaystyle\tilde{\Phi}^{R}$ $\displaystyle=-\frac{r\mu}{6}.$ (118)

The above solution also can be extended to the choice of
$\rho^{2}=r^{2}+k_{2}^{2}\mu^{2}$. Notice that the solution (117) can also be
written as

$\tilde{\Psi}^{R}=\frac{\Delta^{R}_{r}-\Delta^{R}_{\mu}}{r^{2}+\mu^{2}},$
(119)

coinciding with the solution (96) where $a=1$. We then can introduce a new
constant $k_{2}$ to the solution. Defining $\bar{\mu}=k_{2}\mu$ and replacing
$\mu$ with $\bar{\mu}$ in Eqs. (60), (63) and (64), we know that
$\Delta^{R}_{r},\Delta^{R}_{\bar{\mu}},\tilde{\Psi}^{R}(r,\bar{\mu}),\tilde{\Phi}^{R}(r,\bar{\mu})$
are also solutions. The function $\rho^{2}$ turns into
$\rho^{2}=r^{2}+\bar{\mu}^{2}=r^{2}+k_{2}^{2}\mu^{2}$. Since
$\frac{\partial}{\partial\bar{\mu}}=\frac{\partial}{\partial\mu}\frac{\partial\mu}{\partial\bar{\mu}}=\frac{1}{k_{2}}\frac{\partial}{\partial\mu}$,
Eqs. (60), (63) and (64) can be reproduced when letting
$\Delta^{R}_{\mu}=\frac{1}{k_{2}^{2}}\Delta^{R}_{\bar{\mu}}$. Therefore, for
$\rho^{2}=r^{2}+k_{2}^{2}\mu^{2}$, Eqs. (60), (63) and (64) permit the
solution

$\displaystyle\Delta^{R}_{r}$
$\displaystyle=-(\frac{r^{4}}{12}+\frac{k_{1}^{2}}{2}r^{2}),$ (120)
$\displaystyle\Delta^{R}_{\mu}$
$\displaystyle=-\frac{k_{2}^{2}\mu^{4}}{12}+\frac{k_{1}^{2}}{2}\mu^{2},$ (121)
$\displaystyle\tilde{\Psi}^{R}$
$\displaystyle=-\frac{1}{12}(r^{2}-k_{2}^{2}\mu^{2})-\frac{k_{1}^{2}}{2},$
(122) $\displaystyle\tilde{\Phi}^{R}$ $\displaystyle=-\frac{k_{2}r\mu}{6}.$
(123)

Till now, we already have the solutions both for
$\Delta_{r}^{K},\Delta_{\mu}^{K},\tilde{\Phi}^{K},\tilde{\Psi}^{K}$ and
$\Delta_{r}^{R},\Delta_{\mu}^{R},\tilde{\Phi}^{R},\tilde{\Psi}^{R}$. Combining
them together and letting $k_{1}=k_{2}=a$, we obtain the final solution

$\displaystyle\Delta_{r}$
$\displaystyle=r^{2}-2Mr+a^{2}-\frac{R}{12}r^{2}(r^{2}+a^{2}),$ (124)
$\displaystyle\Delta_{\mu}$
$\displaystyle=1-\mu^{2}-\frac{R}{12}a^{2}\mu^{2}(\mu^{2}-1),$ (125)
$\displaystyle\tilde{\Psi}$
$\displaystyle=\frac{\Delta_{r}-a^{2}\Delta_{\mu}}{r^{2}+a^{2}\mu^{2}},$ (126)
$\displaystyle\tilde{\Phi}$
$\displaystyle=\frac{2Ma\mu}{r^{2}+a^{2}\mu^{2}}-R\frac{ar\mu}{6}.$ (127)

In the original metric, the unknown variables $\rho^{2},\chi,q$ can also be
expressed as

$\displaystyle\rho^{2}$ $\displaystyle=r^{2}+a^{2}\mu^{2},$ (128)
$\displaystyle\chi$
$\displaystyle=\frac{(r^{2}+a^{2}\mu^{2})\Delta_{r}^{\frac{1}{2}}\Delta_{\mu}^{\frac{1}{2}}}{(r^{2}+a^{2})^{2}\Delta_{\mu}-a^{2}(1-\mu^{2})^{2}\Delta_{r}},$
(129) $\displaystyle q$
$\displaystyle=a\frac{(r^{2}+a^{2})\Delta_{\mu}-(1-\mu^{2})\Delta_{r}}{(r^{2}+a^{2})^{2}\Delta_{\mu}-a^{2}(1-\mu^{2})^{2}\Delta_{r}}.$
(130)

Choosing $\mu=\cos\theta$, we can transform the solution into the Boyer-
Lindquist coordinate

$ds^{2}=-\frac{\rho^{2}\Delta_{r}\Delta_{\theta}}{\Sigma^{2}}dt^{2}+\frac{\Sigma^{2}}{\rho^{2}}\left(d\varphi-\frac{(r^{2}+a^{2})\Delta_{\theta}-\sin^{2}\theta\Delta_{r}}{\Sigma^{2}}adt\right)^{2}+\frac{\rho^{2}}{\Delta_{r}}dr^{2}+\frac{\rho^{2}\sin^{2}\theta}{\Delta_{\theta}}d\theta^{2},$
(131)

where

$\displaystyle\rho^{2}$ $\displaystyle=r^{2}+a^{2}\cos^{2}\theta,$ (132)
$\displaystyle\Sigma^{2}$
$\displaystyle=(r^{2}+a^{2})^{2}\Delta_{\theta}-a^{2}\sin^{4}\theta\Delta_{r},$
(133) $\displaystyle\Delta_{r}$
$\displaystyle=-\frac{R}{12}(r^{4}+a^{2}r^{2})+r^{2}-2Mr+a^{2},$ (134)
$\displaystyle\Delta_{\theta}$
$\displaystyle=\sin^{2}\theta\left(1+\frac{R}{12}a^{2}\cos^{2}\theta\right).$
(135)

This is the Kerr-Ads metric in a general general $f(R)$ theory.

##  VIII Conclusion and discussion

As far as we know, at least three roads can reach rotating black holes in
gravitational theories. The first is the Newman-Janis algorithm, which
directly works on the solutions. The second is Carter’s consideration of the
spacetime’s structure. The third is from Chandrasekhar, which is based on a
strict analytical calculation. In this paper, we extend the Chandrasekhar’s
method into $f(R)$ theory. As an analytical calculation, the solving process
is universal and can be established as a standard analytical calculation
procedure to obtain the Kerr and Kerr-Ads metric, which is easy to be applied
in other modified gravities.

We worked in a general $f(R)$ gravity theory and began with a general
stationary axisymmetric metric, which have 5 unknowns and all of them depend
on two variables. We use the moving frame to calculate the Einstein tensor.
Suppose the spacetime is a 4-dimensional Riemannian manifold, which are fully
described by RLC connection and curvature. Through calculating the Cartan’s
equation of structure, we extract the RLC connection forms and derive the
curvature forms. Furthermore, we construct the Einstein tensor by using the
curvature forms.

We chose the gauge freedom to transform the axisymmetric metric into a more
symmetrical form, which can effectively reduce the difficulty of analytical
calculation. Based on the symmetrical metric, we derived the field equations
in $f(R)$ gravity. In order to give a simple solving process, we further
transformed the variables in the field equations into the conjugate
coordinates. At last, we solved them separately according to whether the Ricci
scalar is zero.

In the case of $R=0$, the field equations reduce to the Ernst’s equations. The
Kerr metric can be obtained by generalizing the elementary solution of Ernst’s
equations. In the case of $R\neq 0$, the solution is decomposed into two
parts: one is the Kerr solution and the second is introduced by $R$. Through
some mathematical process, we analytically calculated the solution for the
second part. Combining the two parts, we obtained the Kerr-Ads solution.

Acknowledgement This work is partially supported by the National Natural
Science Foundation of China (NSFC U2031112) and the Scientific Research
Foundation of Hunan University of Arts and Sciences (E07023026). We also
acknowledge the science research grants from the China Manned Space Project
with NO. CMS-CSST-2021-A06.

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