Deriving the Kerr metric
(i.e. paint-by-numbers feat. Chandrasekhar)
David Wagner
dwagner@th.physik.uni-frankfurt.de
September 23, 2021
Deriving the Kerr metric
David Wagner
In this document we will attempt to derive the Kerr metric (in natural units
G = c = 1), describing spacetime
outside a uniformly rotating massive body, e.g. a black hole. This calculation is by no means simple and will
take a while. We will follow Chandrasekhar tightly, who presented a derivation in his paper from 1978 [1].
Contents
1 Preliminaries and goal
1
2 Basic quantities characterising spacetime
2
3 The Einstein eld equations
10
4 Null surface and gauge xing
11
5 Reducing the eld equations
12
6 Deriving and solving Ernst's equation
14
7 Properties of the Kerr metric
18
7.1
Asymptotic properties
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
7.2
Special surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
7.3
Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
1 Preliminaries and goal
In general the line element
ds2
is given as
ds2 = dxµ dxµ = gµν dxµ dxν .
(1)
Since we want a vacuum solution for an axisymmetric mass distribution (this diers from spherical symmetry
in such a way that it incorporates attening due to centrifugal forces, such that azimuthal symmetry is
broken), we can simplify this generic form considerably. Firstly, we will for now work in spherical coordinates
t, r, θ, φ. Now due to the rotational symmetry and the fact that spacetime should be static, we can infer that
ds2 should be invariant under changes t → −t, φ → −φ (this corresponds to reversing the time direction
and thus the rotation). This excludes the (t, θ), (t, r), (θ, φ), (r, φ) components of the metric, leaving only
the (t, t), (r, r), (θ, θ), (φ, φ) and (t, φ) components. Notice that in contrast to the Schwarzschild case, there
is one o-diagonal term.
Furthermore, due to axisymmetry the metric should only depend on
r
and
θ
(just as in electrodynamics).
This dependence in combination with the o-diagonal term is what complicates the eld equations immensely,
but will not deter us from obtaining a solution.
So far we thus have reduced the line element to
ds2 = gtt dt2 + grr dr2 + gθθ dθ2 + gφφ dφ2 + gtφ dtdφ .
After rewriting
µ2
and
µ3
gtt := −e2ν + e2ψ ω 2 , gtφ := −2ωe2ψ , gφφ := e2ψ , grr := e2µ2
and
gθθ := e2µ3
(2)
(where the terms
originate from Chandrasekhar numbering of coordinates), we can write
ds2 = −e2ν dt2 + e2ψ (dφ − ωdt)2 + e2µ2 dr2 + e2µ3 dθ2 ,
where ν, ψ, ω, µ2 and µ3 are functions of r and θ . Notice that we are obviously working to get
(− + ++) convention; this however, can in the end easily reverted by multiplying with −1.
In the end, our goal will be to obtain the following values for the interesting functions:
e2ν =
ρ2 ∆
Σ2
1
(3)
a result in the
Deriving the Kerr metric
David Wagner
Σ2 sin2 (θ)
ρ2
2
ρ
e2µ2 =
∆
e2µ3 =ρ2
2aM r
ω=
Σ2
e2ψ =
with
∆ =r2 + a2 − 2M r
ρ2 =r2 + a2 cos2 (θ)
Σ2 =(r2 + a2 )2 − a2 ∆ sin2 (θ) .
2 Basic quantities characterising spacetime
The Einstein eld equations in vacuum imply that the Ricci-Tensor vanishes:
Rµν = 0 .
(4)
Now one could go on to explicitly compute the dierent components of the Ricci tensor using the general
metric given by Eq. 3, however this is not practical, as the equations become unwieldy (which is one of the
reasons why it has taken nearly 50 years to arrive at an axisymmetric solution).
Instead we use Cartan's calculus of exterior forms and apply its structure equations. Firstly, we decompose
the metric into tetrads as
gµν := eaµ ebν ηab .
(− + ++)
µ
convention for reasons mentioned above). These tetrads are matrices transforming the coordinate basis dx
∗
to an orthonormal basis of the cotangent space Tx M of the spacetime manifold M at point x [3]
Here latin indices are raised and lowered by the Minkowski metric
ηab
(here dened in the
ea := eaµ dxµ .
Furthermore, their inverse
Eµa
transforms to an orthonormal basis of the respective tangent space
Tx M
Ea := Eµa ∂µ .
The tetrads thus give a coordinate system at each spacetime point that is locally at and thus easier to work
with. The price to pay is of course that the basis vectors are now dependent on the spacetime point
x
we
evaluate them at.
We now need to nd the connection
ωab
on our manifold, which is related to the Christoel symbols and
determines the curvature.
Since we assume a torsion-free connection (i.e. the absence of spin), the connection has to full the Cartan
structure equations:
0 =dea + ω ab ∧ eb
R
a
a
b =dω b
+
ω ac
∧ω
(5)
c
b
.
From Eq. 3 we can read o the tetrad one-forms as
e0 =eν dt
dt =e−ν e0
e1 =eψ (dφ − ωdt)
dφ =ωe−ν e0 + e−ψ e1
2
(6)
Deriving the Kerr metric
David Wagner
e2 =eµ2 dr
dr =e−µ2 e2
e3 =eµ3 dθ
dθ =e−µ3 e3 .
The exterior derivative of a one-form reads
d ◦ (fi dxi ) =
∂fi j
dx ∧ dxi .
∂xj
(7)
For the zeroth dierential form, we get
d ◦ e0 =d ◦ (eν dt)
∂ν
ν ∂ν
=e
dr ∧ dt +
dθ ∧ dt
∂r
∂θ
=(∂r ν)e−µ2 e2 ∧ e0 + (∂θ ν)e−µ3 e3 ∧ e0 .
(8)
Likewise, we obtain for the rst form
h
i
d ◦ e1 =d ◦ eψ dφ − eψ ωdt
=eψ [∂r ψdr ∧ dφ + ∂θ ψdθ ∧ dφ − (ω∂r ψ + ∂r ω) dr ∧ dt − (ω∂θ ψ + ∂θ ω) dθ ∧ dt]
h
=eψ ∂r ψe−µ2 e2 ∧ ωe−ν e0 + e−ψ e1 + ∂θ ψe−µ3 e3 ∧ ωe−ν e0 + e−ψ e1
i
− (ω∂r ψ + ∂r ω) e−µ2 e2 ∧ e−ν e0 − (ω∂θ ψ + ∂θ ω) e−µ3 e3 ∧ e−ν e0
=∂r ψe−µ2 e2 ∧ e1 + ∂θ ψe−µ3 e3 ∧ e1 − ∂r ωeψ−ν−µ2 e2 ∧ e0 − ∂θ ωeψ−ν−µ3 e3 ∧ e0 .
(9)
The dierential of the second one-form reads
de2 =eµ2 [∂r µ2 dr ∧ dr + ∂θ µ2 dθ ∧ dr]
=e−µ3 ∂θ µ2 e3 ∧ e2 .
(10)
Here we used that the wedge product of a one-form with itself vanishes, since it is antisymmetric.
The dierential of the last one-form gives
de3 =eµ3 [∂r µ3 dr ∧ dθ + ∂θ µ3 dθ ∧ dθ]
=e−µ2 ∂r µ3 e2 ∧ e3 .
(11)
Combining these dierentials with the rst structure equation, we obtain
−ω 0b ∧ eb =(∂r ν)e−µ2 e2 ∧ e0 + (∂θ ν)e−µ3 e3 ∧ e0
−ω
1
−µ2 2
b
b
∧ e =∂r ψe
1
−µ3 3
e ∧ e + ∂θ ψe
1
(12)
ψ−ν−µ2 2
e ∧ e − ∂r ωe
0
e ∧ e − ∂θ ωe
−ω 2b ∧ eb =e−µ3 ∂θ µ2 e3 ∧ e2
−ω
3
Furthermore,
that
ωab
b
b
−µ2
∧ e =e
ωab = −ωba
2
ψ−ν−µ3 3
0
e ∧e
(13)
(14)
3
∂ r µ3 e ∧ e .
(15)
has to be fullled (this is equivalent to the metric compatibility condition), such
has six independent components. To solve these equations, it is helpful decomposing the connections
into their respective tetrad components:
ω ab = (ω ab )c ec .
(16)
This decomposition gives (due to the fact that the tetrads are orthogonal) 24 equations for 24 variables.
From the rst two equations we get
(ω 01 )0 =0
(ω 10 )1 =0
3
Deriving the Kerr metric
David Wagner
(ω 02 )0 =e−µ2 ∂r ν
(ω 10 )2 − (ω 12 )0 =eψ−ν−µ2 ∂r ω
(ω 03 )0 =e−µ3 ∂θ ν
(ω 10 )3 − (ω 13 )0 =eψ−ν−µ3 ∂θ ω
(ω 12 )1 =e−µ2 ∂r ψ
(ω 01 )2 − (ω 02 )1 =0
(ω 02 )3 − (ω 03 )2 =0
(ω 13 )2 − (ω 12 )3 =0
(ω 13 )1 =e−µ3 ∂θ ψ ,
(ω 03 )1 − (ω 01 )3 =0
while the third and fourth equations give us
(ω 20 )1 − (ω 21 )0 =0
(ω 30 )1 − (ω 31 )0 =0
(ω 20 )2 =0
(ω 30 )2 − (ω 32 )0 =0
(ω 20 )3 − (ω 23 )0 =0
(ω 30 )3 =0
(ω 21 )2 =0
(ω 31 )2 − (ω 32 )1 =0
(ω 23 )2 =e−µ3 ∂θ µ2
(ω 32 )3 =e−µ2 ∂r µ3
(ω 21 )3 − (ω 23 )1 =0
(ω 31 )3 =0 .
The solution of this system of equations reads
1
1
ω 01 = ω 10 = eψ−ν−µ2 ∂r ωe2 + eψ−ν−µ3 ∂θ ωe3
2
2
1
ω 02 = ω 20 =e−µ2 ∂r νe0 + eψ−ν−µ2 ∂r ωe1
2
1
0
3
−µ3
0
ω 3 = ω 0 =e ∂θ νe + eψ−ν−µ3 ∂θ ωe1
2
1
ω 12 = −ω 21 = − eψ−ν−µ2 ∂r ωe0 + e−µ2 ∂r ψe1
2
1
1
3
ω 3 = −ω 1 = − eψ−ν−µ3 ∂θ ωe0 + e−µ3 ∂θ ψe1
2
2
3
−µ3
ω 3 = −ω 2 =e ∂θ µ2 e2 − e−µ2 ∂r µ3 e3 .
(17)
(18)
(19)
(20)
(21)
(22)
Next we employ Eq. 6 to construct the components of the curvature 2-form, which will give us the Ricci
tensor at once.
Explicitly we have to evaluate
R0 1 = R1 0 =dω 01 + ω 0b ∧ ω b 1
R0 2 = R2 0 =dω 02 + ω 0b ∧ ω b 2
R0 3 = R3 0 =dω 03 + ω 0b ∧ ω b 3
R1 2 = −R2 1 =dω 12 + ω 1b ∧ ω b 2
R1 3 = −R3 1 =dω 13 + ω 1b ∧ ω b 3
R2 3 = −R3 2 =dω 23 + ω 2b ∧ ω b 3 .
Since we will spell out everything here in detail to give a thorough calculation, we will start by evaluating
the exterior derivatives:
1
dω 01 = eψ−ν−µ2 −µ3 (∂θ ψ − ∂θ ν) ∂r ω + ∂r2 ω − (∂r ψ − ∂r ν) ∂θ ω − ∂θ2 ω e3 ∧ e2 .
2
The second form is more work:
dω 02 =e−2µ2 ∂r (ν − µ2 )∂r ν + ∂r2 ν e2 ∧ e0 + e−µ2 −µ3 [∂θ (ν − µ2 )∂r ν + ∂r ∂θ ν] e3 ∧ e0
4
(23)
Deriving the Kerr metric
1
+ e2ψ−ν−µ2
2
David Wagner
∂r (2ψ − ν − µ2 )∂r ω + ∂r2 ω dr ∧ dφ
+ [∂θ (2ψ − ν − µ2 )∂r ω + ∂θ ∂r ω] dθ ∧ dφ − ∂r (2ψ − ν − µ2 )ω∂r ω + (∂r ω)2 + ω∂r2 ω dr ∧ dt
+ [∂θ (2ψ − ν − µ2 )ω∂r ω + ∂θ ω∂r ω + ω∂θ ∂r ω] dθ ∧ dt
=e−2µ2 ∂r (ν − µ2 )∂r ν + ∂r2 ν e2 ∧ e0 + e−µ2 −µ3 [∂θ (ν − µ2 )∂r ν + ∂r ∂θ ν] e3 ∧ e0
h
i
1 2ψ−ν−µ2 + e
∂r (2ψ − ν − µ2 )∂r ω + ∂r2 ω e−µ2 ωe−ν e2 ∧ e0 + e−ψ e2 ∧ e1
2
h
i
+ [∂θ (2ψ − ν − µ2 )∂r ω + ∂θ ∂r ω] e−µ3 ωe−ν e3 ∧ e0 + e−ψ e3 ∧ e1
− ∂r (2ψ − ν − µ2 )ω∂r ω + (∂r ω)2 + ω∂r2 ω e−µ2 −ν e2 ∧ e0
−µ3 −ν 3
0
+ [∂θ (2ψ − ν − µ2 )ω∂r ω + ∂θ ω∂r ω + ω∂θ ∂r ω] e
e ∧e
2
2 1 2ψ−2ν
−2µ2
∂r (ν − µ2 )∂r ν + ∂r ν − (∂r ω) e
=e
e2 ∧ e0
2
1 2ψ−2ν 3
−µ2 −µ3
e ∧ e0
+e
∂θ (ν − µ2 )∂r ν + ∂r ∂θ ν − ∂θ ω∂r ω e
2
1
+ eψ−ν−2µ2 ∂r (2ψ − ν − µ2 )∂r ω + ∂r2 ω e2 ∧ e1
2
1
+ eψ−ν−µ2 −µ3 [∂θ (2ψ − ν − µ2 )∂r ω + ∂θ ∂r ω] e3 ∧ e1 .
(24)
2
The third one-form looks similar:
dω 03 =e−µ2 −µ3 [∂r (ν − µ3 )∂θ ν + ∂r ∂θ ν] e2 ∧ e0 + e−2µ3 ∂θ (ν − µ3 )∂θ ν + ∂θ2 ν e3 ∧ e0
1 2ψ−ν−µ3
+ e
[∂r (2ψ − ν − µ3 )∂θ ω + ∂r ∂θ ω] dr ∧ dφ
2
+ ∂θ (2ψ − ν − µ3 )∂θ ω + ∂θ2 ω dθ ∧ dφ − [∂r (2ψ − ν − µ3 )ω∂θ ω + ∂r ω∂θ ω + ω∂r ∂θ ω] dr ∧ dt
2
2
+ ∂θ (2ψ − ν − µ3 )ω∂θ ω + (∂θ ω) + ω∂θ ω dθ ∧ dt
=e−µ2 −µ3 [∂r (ν − µ3 )∂θ ν + ∂r ∂θ ν] e2 ∧ e0 + e−2µ3 ∂θ (ν − µ3 )∂θ ν + ∂θ2 ν e3 ∧ e0
i
h
1 2ψ−ν−µ3
+ e
[∂r (2ψ − ν − µ3 )∂θ ω + ∂r ∂θ ω] e−µ2 ωe−ν e2 ∧ e0 + e−ψ e2 ∧ e1
2
h
i
+ ∂θ (2ψ − ν − µ3 )∂θ ω + ∂θ2 ω e−µ3 ωe−ν e3 ∧ e0 + e−ψ e3 ∧ e1
− [∂r (2ψ − ν − µ3 )ω∂θ ω + ∂r ω∂θ ω + ω∂r ∂θ ω] e−µ2 −ν e2 ∧ e0
+ ∂θ (2ψ − ν − µ3 )ω∂θ ω + (∂θ ω)2 + ω∂θ2 ω e−µ3 −ν e3 ∧ e0
1 2ψ−2ν 2
−µ2 −µ3
=e
∂r (ν − µ3 )∂θ ν + ∂r ∂θ ν − ∂r ω∂θ ω e
e ∧ e0
2
−2µ3
2
2 1 2ψ−2ν
+e
∂θ (ν − µ3 )∂θ ν + ∂θ ν − (∂θ ω) e
e3 ∧ e0
2
1
+ eψ−ν−µ2 −µ3 [∂r (2ψ − ν − µ3 )∂θ ω + ∂r ∂θ ω] e2 ∧ e1
2
1 ψ−ν−2µ3 + e
∂θ (2ψ − ν − µ3 )∂θ ω + ∂θ2 ω e3 ∧ e1 .
2
5
(25)
Deriving the Kerr metric
David Wagner
The fourth one-form's dierential gives
1
1
dω 12 = − eψ−ν−2µ2 ∂r (ψ − µ2 )∂r ω + ∂r2 ω e2 ∧ e0 − eψ−ν−µ2 −µ3 [∂θ (ψ − µ2 )∂r ω + ∂r ∂θ ω] e3 ∧ e0
2
2
i
−µ2 h −ν 2
ψ−µ2
2
+e
∂r (ψ − µ2 )∂r ψ + ∂r ψ e
ωe e ∧ e0 + e−ψ e2 ∧ e1
h
i
+ [∂θ (ψ − µ2 )∂r ψ + ∂r ∂θ ψ] e−µ3 ωe−ν e3 ∧ e0 + e−ψ e3 ∧ e1
− ∂r (ψ − µ2 )∂r ψω + ∂r2 ψω + ∂r ψ∂r ω e−ν−µ2 e2 ∧ e0
−ν−µ3 3
0
− [∂θ (ψ − µ2 )∂r ψω + ∂r ∂θ ψω + ∂r ψ∂θ ω] e
e ∧e
1 −µ2 ψ−ν−µ2
2
−µ2
=−e
∂r (ψ − µ2 )∂r ω + ∂r ω + e ∂r ψ∂r ω e2 ∧ e0
e
2
1 −µ2
− eψ−ν−µ3
e
[∂θ (ψ − µ2 )∂r ω + ∂r ∂θ ω] + e−µ2 ∂r ψ∂θ ω e3 ∧ e0
2
−µ2 −µ3
+e
[∂θ (ψ − µ2 )∂r ψ + ∂r ∂θ ψ] e3 ∧ e1 + e−2µ2 ∂r (ψ − µ2 )∂r ψ + ∂r2 ψ e2 ∧ e1 .
(26)
The fth form is very similar:
1
dω 13 = − eψ−ν−µ2 −µ3 [∂r (ψ − µ3 )∂θ ω + ∂r ∂θ ω + ∂θ ψ∂r ω] e2 ∧ e0
2
1
− eψ−ν−2µ3 ∂θ (ψ − µ3 )∂θ ω + ∂θ2 ω + ∂θ ψ∂θ ω e3 ∧ e0
2
+ e−µ2 −µ3 [∂r (ψ − µ3 )∂θ ψ + ∂r ∂θ ψ] e2 ∧ e1 + e−2µ3 ∂θ (ψ − µ3 )∂θ ψ + ∂θ2 ψ e3 ∧ e1 .
(27)
For the last form's exterior derivative we obtain
dω 23 =e−µ2 −µ3 eµ2 −µ3 ∂θ (µ2 − µ3 )∂θ µ2 + ∂θ2 µ2 + eµ3 −µ2 ∂r (µ2 − µ3 )∂r µ3 + ∂r2 µ3 e3 ∧ e2 .
(28)
Next we evaluate the wedge products appearing in the second structure equations one by one:
ω 0b ∧ ω b 1 =ω 02 ∧ ω 21 + ω 03 ∧ ω 31
"
#
2 2
1 ψ−ν−µ2
1 ψ−ν−µ3
−2µ2
−2µ3
=
e
∂r ω +
e
∂θ ω + e
∂r ν∂r ψ + e
∂θ ν∂θ ψ e1 ∧ e0 ,
2
2
ω 0b ∧ ω b 2 =ω 01 ∧ ω 12 + ω 03 ∧ ω 32
"
2 #
1 ψ−ν−µ2
−2µ3
= e
∂θ ν∂θ µ2 −
e
∂r ω
e2 ∧ e0
2
"
#
2
1
eψ−ν ∂r ω∂θ ω e3 ∧ e0
− e−µ2 −µ3 ∂r µ3 ∂θ ν +
2
1
+ eψ−ν−µ2 −µ3 eµ2 −µ3 ∂θ ω∂θ µ2 + eµ3 −µ2 ∂r ω∂r ψ e2 ∧ e1
2
1
+ eψ−ν−µ2 −µ3 [−∂θ ω∂r µ3 + ∂θ ω∂r ψ] e3 ∧ e1 ,
2
ω 0b ∧ ω b 3 =ω 01 ∧ ω 13 + ω 02 ∧ ω 23
"
#
1 ψ−ν 2
−µ2 −µ3
=−e
e
∂r ω∂θ ω + ∂r ν∂θ µ2 e2 ∧ e0
2
" #
2
1 ψ−ν−µ3
+ −
e
∂θ ω + e−2µ2 ∂r ν∂r µ3 e3 ∧ e0
2
6
(29)
(30)
Deriving the Kerr metric
David Wagner
1
+ eψ−ν−µ2 −µ3 [∂r ω∂θ ψ − ∂r ω∂θ µ2 ] e2 ∧ e1
2
1
+ eψ−ν−µ2 −µ3 eµ3 −µ2 ∂r ω∂r µ3 + eµ2 −µ3 ∂θ ω∂θ ψ e3 ∧ e1 ,
2
ω 1b ∧ ω b 2 =ω 10 ∧ ω 02 + ω 13 ∧ ω 32
1
1
= eψ−ν e−2µ2 ∂r ω∂r ν − e−2µ3 ∂θ ω∂θ µ2 e2 ∧ e0 + eψ−ν−µ2 −µ3 ∂θ ω [∂r ν + ∂r µ3 ] e3 ∧ e0
2 "
2
#
2
1 ψ−ν−µ2
−2µ3
+
e
∂r ω + e
∂θ µ2 ∂θ ψ e2 ∧ e1
2
#
"
1 ψ−ν 2 −µ2 −µ3
−µ2 −µ3
e
e
∂θ ω∂r ω − e
∂r µ3 ∂θ ψ e3 ∧ e1 ,
+
2
ω 1b ∧ ω b 3 =ω 10 ∧ ω 03 + ω 12 ∧ ω 23
1
1
= eψ−ν−µ2 −µ3 ∂r ω [∂θ ν + ∂θ µ2 ] e2 ∧ e0 + eψ−ν e−2µ3 ∂θ ω∂θ ν − e−2µ2 ∂r ω∂r µ3 e3 ∧ e0
2 "
2
#
2
1 ψ−ν
+
e
e−µ2 −µ3 ∂r ω∂θ ω − e−µ2 −µ3 ∂θ µ2 ∂r ψ e2 ∧ e1
2
#
"
2
1 ψ−ν−µ3
e
∂θ ω + e−2µ2 ∂r µ3 ∂r ψ e3 ∧ e1 ,
+
2
ω 2b ∧ ω b 3 =ω 20 ∧ ω 03 + ω 21 ∧ ω 13
1
= eψ−ν−µ2 −µ3 [∂r ω∂θ ν − ∂r ν∂θ ω + ∂r ψ∂θ ω + ∂θ ψ∂r ω] e1 ∧ e0 .
2
(31)
(32)
(33)
(34)
Having computed all of this, we are able to read o the Riemann tensor by considering how it is related to
the curvature two-form:
Rab = Rabcd ec ∧ ed .
ˆ
Note:
(35)
There might be a factor of 1/2 missing here, which will not bother us anyway, however, since
we are looking for vacuum solutions.
Furthermore, we should remember some symmetries of the Riemann tensor:
Rabcd = −Rbacd = −Rabdc = Rcdab .
We obtain thus
R
0
110
R0 132
R0 220
R0 230
R0 221
R0 231
R0 320
1 2ψ−2ν µ3 −µ2
2
µ2 −µ3
2
µ3 −µ2
µ2 −µ3
e
e
(∂r ω) + e
(∂θ ω) + e
=e
∂r ν∂r ψ + e
∂θ ν∂θ ψ
4
1
= eψ−ν−µ2 −µ3 (∂θ ψ − ∂θ ν) ∂r ω + ∂r2 ω − (∂r ψ − ∂r ν) ∂θ ω − ∂θ2 ω
2
−µ2 −µ3
µ3 −µ2
µ2 −µ3
2
2 3 2ψ−2ν+µ3 −µ2
=e
e
∂r (ν − µ2 )∂r ν + ∂r ν − (∂r ω) e
+e
∂θ ν∂θ µ2
4
3
=e−µ2 −µ3 ∂θ (ν − µ2 )∂r ν + ∂r ∂θ ν − ∂θ ω∂r ω e2ψ−2ν − ∂r µ3 ∂θ ν
4
1 ψ−ν−µ2 −µ3 µ3 −µ2
= e
e
∂r (3ψ − ν − µ2 )∂r ω + eµ3 −µ2 ∂r2 ω + eµ2 −µ3 ∂θ ω∂θ µ2
2
1
= eψ−ν−µ2 −µ3 [∂θ (2ψ − ν − µ2 )∂r ω + ∂θ ∂r ω − ∂θ ω∂r µ3 + ∂θ ω∂r ψ]
2
3 2ψ−2ν
−µ2 −µ3
=e
∂r (ν − µ3 )∂θ ν + ∂r ∂θ ν − ∂r ω∂θ ω e
− ∂r ν∂θ µ2
4
−µ2 −µ3
7
(36)
(37)
(38)
(39)
(40)
(41)
(42)
Deriving the Kerr metric
R
0
330
−µ2 −µ3
=e
David Wagner
µ2 −µ3
2
2 3 2ψ−2ν+µ2 −µ3
µ3 −µ2
e
∂θ (ν − µ3 )∂θ ν + ∂θ ν − (∂θ ω) e
+e
∂r ν∂r µ3
4
1
R0 321 = eψ−ν−µ2 −µ3 [∂r (2ψ − ν − µ3 )∂θ ω + ∂r ∂θ ω + ∂r ω∂θ ψ − ∂r ω∂θ µ2 ]
2
1
R0 331 = eψ−ν−µ2 −µ3 eµ2 −µ3 ∂θ (3ψ − ν − µ3 )∂θ ω + eµ2 −µ3 ∂θ2 ω + eµ3 −µ2 ∂r ω∂r µ3
2
1 µ3 −µ2 1 µ2 −µ3
1
ψ−ν−µ2 −µ3
2
R 220 = − e
e
∂r (3ψ − ν − µ2 )∂r ω + ∂r ω + ∂θ ψ∂r ω + e
∂θ ω∂θ µ2
2
2
1
1
µ2 −µ3
1
ψ−ν−µ2 −µ3
[∂θ (ψ − µ2 )∂r ω + ∂r ∂θ ω] + e
∂θ ψ + ∂r (2ψ − ν − µ3 ) ∂θ ω
R 230 = − e
2
2
1 2ψ−2ν
1
−µ2 −µ3
∂θ (ψ − µ2 )∂r ψ + ∂r ∂θ ψ − ∂r µ3 ∂θ ψ + e
R 231 =e
∂θ ω∂r ω
4
1
R1 221 =e−µ2 −µ3 eµ2 −µ3 ∂θ µ2 ∂θ ψ + eµ3 −µ2 ∂r (ψ − µ2 )∂r ψ + ∂r2 ψ + e2ψ−2ν+µ3 −µ2 (∂r ω)2
4
1
R1 320 = − eψ−ν−µ2 −µ3 [∂r (ψ − µ3 )∂θ ω + ∂r ∂θ ω − ∂r ω∂θ ν − ∂r ω∂θ µ2 ]
2
1
R1 330 = − eψ−ν−µ2 −µ3 eµ2 −µ3 ∂θ (ψ − ν − µ3 )∂θ ω + ∂θ2 ω + eµ3 −µ2 ∂r ω∂r µ3
2
1
−µ2 −µ3 1 2ψ−2ν
R 321 =e
e
∂r ω∂θ ω − ∂θ µ2 ∂r ψ + ∂r (ψ − µ3 )∂θ ψ + ∂r ∂θ ψ
4
1 2ψ−2ν+µ2 −µ3
e
(∂θ ω)2 + eµ3 −µ2 ∂r µ3 ∂r ψ + eµ2 −µ3 ∂θ (ψ − µ3 )∂θ ψ + ∂θ2 ψ
R1 331 =e−µ2 −µ3
4
µ2 −µ3 2
−µ2 −µ3
R 332 =e
e
∂θ (µ2 − µ3 )∂θ µ2 + ∂θ2 µ2 + eµ3 −µ2 ∂r (µ2 − µ3 )∂r µ3 + ∂r2 µ3
1
R2 310 = eψ−ν−µ2 −µ3 [∂r ω∂θ ν − ∂r ν∂θ ω + ∂r ψ∂θ ω + ∂θ ψ∂r ω] .
2
(43)
(44)
(45)
(46)
(47)
(48)
(49)
(50)
(51)
(52)
(53)
(54)
(55)
From these quantities we can obtain all other non-vanishing components of the Riemann tensor as well since
the rst pair of indices commutes if one of them is nonzero, while it anticommutes if both are nonzero. The
second pair of indices always anticommutes. The reason for these simple symmetries even for the Riemann
tensor of the rst kind (i.e. with one upper index) lies in the use of tetrads, which lets us pull the rst two
indices of this tensor with the Minkowski metric. This can be seen by considering Ref. [3], section 3.2.
Now the components of the Ricci tensor are
R00 =R1 010 + R2 020 + R3 030
1
−µ2 −µ3
=e
− e2ψ−2ν eµ3 −µ2 (∂r ω)2 + eµ2 −µ3 (∂θ ω)2
2
µ3 −µ2
2
µ2 −µ3
2
+e
∂r (ψ + ν + µ3 − µ2 )∂r ν + ∂r ν + e
∂θ (ψ + ν + µ2 − µ3 )∂θ ν + ∂θ ν
(56)
R11 =R0 101 + R2 121 + R3 131
µ3 −µ2
−µ2 −µ3 1 2ψ−2ν
e
e
(∂r ω)2 + eµ2 −µ3 (∂θ ω)2
=−e
2
+ eµ3 −µ2 ∂r (ψ + ν + µ3 − µ2 )∂r ψ + ∂r2 ψ + eµ2 −µ3 ∂θ (ψ + ν + µ2 − µ3 )∂θ ψ + ∂θ2 ψ
(57)
R01 =R2 021 + R3 031
1 −2ψ+ν−µ2 −µ3
3ψ−ν+µ3 −µ2
3ψ−ν+µ2 −µ3
∂r ω + ∂θ e
∂θ ω
∂r e
= e
2
R22 =R0 202 + R1 212 + R3 232
8
(58)
Deriving the Kerr metric
David Wagner
1
eµ3 −µ2 ∂r (ν − µ2 )∂r ν + ∂r2 ν − (∂r ω)2 e2ψ−2ν+µ3 −µ2
2
µ2 −µ3
µ3 −µ2
+e
∂θ µ2 ∂θ (ψ + ν) + e
∂r (ψ − µ2 )∂r ψ + ∂r2 ψ
µ2 −µ3
2
µ3 −µ2
2
+e
∂θ (µ2 − µ3 )∂θ µ2 + ∂θ µ2 + e
∂r (µ2 − µ3 )∂r µ3 + ∂r µ3
=−e
−µ2 −µ3
(59)
R33 =R0 303 + R1 313 + R2 323
1
−µ2 −µ3
eµ2 −µ3 ∂θ (ν − µ3 )∂θ ν + ∂θ2 ν + ∂θ (ψ − µ3 )∂θ ψ + ∂θ2 ψ − (∂θ ω)2 e2ψ−2ν+µ2 −µ3
=−e
2
+ eµ3 −µ2 (∂r ψ + ∂r ν) ∂r µ3 + eµ2 −µ3 ∂θ (µ2 − µ3 )∂θ µ2 + ∂θ2 µ2 + eµ3 −µ2 ∂r (µ2 − µ3 )∂r µ3 + ∂r2 µ3
(60)
R23 =R0 203 + R1 213
1 2ψ−2ν
−µ2 −µ3
∂θ (ν − µ2 )∂r ν + ∂r ∂θ (ν + ψ) − ∂θ ω∂r ω e
=−e
− ∂θ (ν + ψ) ∂r µ3 + ∂θ (ψ − µ2 )∂r ψ
2
(61)
Lastly we need the Ricci scalar, which can be computed by considering the contraction of the Ricci tensor.
Note that this is quite easy as we are still working in the tetrad basis and thus can pull indices with the
Minkowski metric (still in the
− + ++
convention to be consistent).
R =Raa
=R0 0 + R1 1 + R2 2 + R3 3
= − R00 + R11 + R22 + R33
µ2 −µ3
−µ2 −µ3 1 2ψ−2ν
e
(∂θ ω)2 + eµ3 −µ2 (∂r ω)2
e
=2e
4
µ3 −µ2
−e
∂r (ψ + ν + µ3 − µ2 )∂r ν + ∂r2 ν + ∂r (ψ + µ3 − µ2 )∂r ψ + ∂r2 ψ + ∂r (µ2 − µ3 )∂r µ3 + ∂r2 µ3
2
2
µ2 −µ3
2
.
−e
∂θ (ψ + ν + µ2 − µ3 )∂θ ν + ∂θ ν + ∂θ (ψ + µ2 − µ3 )∂θ ψ + ∂θ ψ + ∂θ (µ2 − µ3 )∂θ µ2 + ∂θ µ2
(62)
This enables us nally to write down the components of the Einstein tensor
Gab = Rab −
R
ηab ,
2
(63)
where the metric is Minkowskian once again due to the employed tetrad frame.
Thus we obtain
1
− e2ψ−2ν eµ3 −µ2 (∂r ω)2 + eµ2 −µ3 (∂θ ω)2
4
µ3 −µ2
−e
∂r (ψ + µ3 − µ2 )∂r ψ + ∂r2 ψ + ∂r (µ2 − µ3 )∂r µ3 + ∂r2 µ3
µ2 −µ3
2
2
−e
∂θ (ψ + µ2 − µ3 )∂θ ψ + ∂θ ψ + ∂θ (µ2 − µ3 )∂θ µ2 + ∂θ µ2
G00 =e−µ2 −µ3
G11 = − e
−µ2 −µ3
(64)
3 2ψ−2ν µ3 −µ2
e
e
(∂r ω)2 + eµ2 −µ3 (∂θ ω)2
4
− eµ3 −µ2 ∂r (ν + µ3 − µ2 )∂r ν + ∂r2 ν + ∂r (µ2 − µ3 )∂r µ3 + ∂r2 µ3
µ2 −µ3
2
2
−e
∂θ (ν + µ2 − µ3 )∂θ ν + ∂θ ν + ∂θ (µ2 − µ3 )∂θ µ2 + ∂θ µ2
9
(65)
Deriving the Kerr metric
David Wagner
1 2ψ−2ν µ2 −µ3
e
e
(∂θ ω)2 − eµ3 −µ2 (∂r ω)2 − eµ3 −µ2 [∂r (ψ + µ3 )∂r ν + ∂r µ3 ∂r ψ]
4
µ2 −µ3
2
−e
∂θ (ψ + ν − µ3 )∂θ ν + ∂θ (ν + ψ) + ∂θ (ψ − µ3 )∂θ ψ
−µ2 −µ3
G22 = − e
1 2ψ−2ν µ3 −µ2
e
e
(∂r ω)2 − eµ2 −µ3 (∂θ ω)2 − eµ2 −µ3 [∂θ (ψ + µ2 )∂θ ν + ∂θ µ2 ∂θ ψ]
4
µ3 −µ2
2
.
−e
∂r (ψ + ν − µ2 )∂r ν + ∂r (ν + ψ) + ∂r (ψ − µ2 )∂r ψ
G33 = − e
−µ2 −µ3
(66)
(67)
3 The Einstein eld equations
Having now all these objects at our disposal, we can nally conrm the starting equations of Ref. [1]; for
this we dene
X :=eµ3 −µ2 (∂r ω)2 + eµ2 −µ3 (∂θ ω)2
β :=ψ + ν
(68)
(69)
and employ that in vacuum
Rµν = Gµν = 0 ,
leading to
1 2ψ−2ν
e
X =eµ3 −µ2 ∂r (ψ + ν + µ3 − µ2 )∂r ν + ∂r2 ν + eµ2 −µ3 ∂θ (ψ + ν + µ2 − µ3 )∂θ ν + ∂θ2 ν
2
1
− e2ψ−2ν X =eµ3 −µ2 ∂r (ψ + ν + µ3 − µ2 )∂r ψ + ∂r2 ψ + eµ2 −µ3 ∂θ (ψ + ν + µ2 − µ3 )∂θ ψ + ∂θ2 ψ
2
1
− e2ψ−2ν X =eµ3 −µ2 ∂r (ψ + µ3 − µ2 )∂r ψ + ∂r2 ψ + ∂r (µ2 − µ3 )∂r µ3 + ∂r2 µ3
4
+ eµ2 −µ3 ∂θ (ψ + µ2 − µ3 )∂θ ψ + ∂θ2 ψ + ∂θ (µ2 − µ3 )∂θ µ2 + ∂θ2 µ2
3 2ψ−2ν
e
X =eµ3 −µ2 ∂r (ν + µ3 − µ2 )∂r ν + ∂r2 ν + ∂r (µ2 − µ3 )∂r µ3 + ∂r2 µ3
4
+ eµ2 −µ3 ∂θ (ν + µ2 − µ3 )∂θ ν + ∂θ2 ν + ∂θ (µ2 − µ3 )∂θ µ2 + ∂θ2 µ2
0 =∂r e3ψ−ν+µ3 −µ2 ∂r ω + ∂θ e3ψ−ν+µ2 −µ3 ∂θ ω .
Also we can take the dierence of
G22
and
G33
(70)
(71)
(72)
(73)
(74)
to obtain
1
0 = e2ψ−2ν eµ2 −µ3 (∂θ ω)2 − eµ3 −µ2 (∂r ω)2
2
− eµ2 −µ3 ∂θ (ν − µ2 − µ3 )∂θ ν + ∂θ2 β + ∂θ (ψ − µ2 − µ3 )∂θ ψ
+ eµ3 −µ2 ∂r (ν − µ2 − µ3 )∂r ν + ∂r2 β + ∂r (ψ − µ2 − µ3 )∂r ψ ,
(75)
which can be rewritten as
h i
2e−β ∂r eβ+µ3 −µ2 ∂r β − ∂θ eβ+µ2 −µ3 ∂θ β =4eµ3 −µ2 (∂r β∂r µ3 + ∂r ψ∂r ν) − 4eµ2 −µ3 (∂θ β∂θ µ2 + ∂θ ψ∂θ ν)
(76)
+ e2ψ−2ν eµ3 −µ2 (∂r ω)2 − eµ2 −µ3 (∂θ ω)2 .
Equations 70 and 71 can be rewritten as
1 3ψ−ν
e
X =∂r eψ+ν+µ3 −µ2 ∂r ν + ∂θ eψ+ν+µ2 −µ3 ∂θ ν
2
1
− e3ψ−ν X =∂r eψ+ν+µ3 −µ2 ∂r ψ + ∂θ eψ+ν+µ2 −µ3 ∂θ ψ .
2
10
(77)
(78)
Deriving the Kerr metric
David Wagner
The sum and dierence of these two gives
0 =∂r eβ+µ3 −µ2 ∂r β + ∂θ eβ+µ2 −µ3 ∂θ β
h
i
h
i
e3ψ−ν X =∂r eβ+µ3 −µ2 ∂r (ν − ψ) + ∂θ eβ+µ2 −µ3 ∂θ (ν − ψ) .
(79)
(80)
Noticing that we can rewrite Eq. 74 as
e3ψ−ν X = ∂r e3ψ−ν+µ3 −µ2 ω∂r ω + ∂θ e3ψ−ν+µ2 −µ3 ω∂θ ω ,
(81)
we can insert this result into Eq. 80 and obtain
3ψ−ν+µ3 −µ2
0 = ∂r e
1
1
2ν−2ψ
2ν−2ψ
3ψ−ν+µ2 −µ3
2
2
e
∂θ (ψ − ν) + ∂θ ω
e
∂r (ψ − ν) + ∂r ω
+ ∂θ e
,
2
2
which, after dening
χ := eν−ψ ,
reads
h
i
h
i
0 = ∂r e3ψ−ν+µ3 −µ2 ∂r (χ2 − ω 2 ) + ∂θ e3ψ−ν+µ2 −µ3 ∂θ (χ2 − ω 2 ) .
Comparing the expression above with Eq. 74, we notice that the quantities
ω
and
(82)
χ2 − ω 2
are determined
by the same equation.
4 Null surface and gauge xing
Our problem, for a vanishing angular velocity, should reduce ot the Schwarzschild metric, which features an
event horizon spanned by the Killing vector
∂t .
This event horizon constitutes a so-called
null surface,
as
the normal vectors to this surface are null-vectors.
Now in our problem (as it should reduce to the Schwarzschild case in the limit of vanishing rotation) it is
reasonable to assume that there is a smooth null surface spanned by the two Killing vectors
∂t
and
∂φ
as
well.
As done in Ref. [1], we denote the equation determining the surface by
N (r, θ) = 0 .
Since the gradient of this equation gives the respective normal vector, the condition that the equation above
describes a null surface is given by
g ij ∂i N ∂j N = 0 ,
(83)
e2µ3 −2µ2 (∂r N )2 + (∂θ N )2 = 0 .
(84)
which (for our Ansatz) reduces to
Now, as we are free to impose any dieomorphism due to the gauge freedom, we choose
e2µ3 −2µ2 = ∆(r) ,
where
∆(r)
is an unspecied function of
is because the term
∂θ N
r
(85)
only. This implies that on the event horizon
should vanish separately, and thus
In order for the null surface to be spanned by
∂t
and
∂φ ,
N
should not depend on
θ
∆(r) = 0
holds (this
at all).
the determinant of the induced
(t, φ)−submetric
has to vanish[2] (intuitively, this is because a null surface should have a vanishing two-volume (i.e. surface),
and in the resulting integral the invariant volume element contains the determinant of this submetric). From
Eq. 3 we can compute this determinant and arrive at
eβ = 0 ,
11
(86)
Deriving the Kerr metric
which must hold when
and
θ
David Wagner
∆ = 0.
Next we suppose that, in line with the conditions above,
is separable in
r
as
eβ =
where
eβ
f (θ)
p
∆(r)f (θ) ,
(87)
is as of now not specied.
Inserting these expressions for
eβ
and
e2µ3 −2µ2
0 = ∂r
|
into Eq. 79, we obtain
√
√ ∂2f
∆∂r ∆ + θ .
f
{z
}
(88)
1 2
∂ ∆
2 r
One solution (where the space of solutions is restricted by e.g. the requirement of convexity of the horizon)
of this equation is given by
∂r2 ∆ = 2 and f (θ) = sin(θ).
Introducing two constants
be tied to the mass and angular momentum of the black hole), we can thus write
M and a (which will later
∆ as
∆(r) = r2 + a2 − 2M r .
(89)
5 Reducing the eld equations
Using our solutions
eµ3 −µ2 =
√
eβ =
∆,
√
∆ sin(θ)
(90)
and substituting
µ := cos(θ) ,
δ := 1 − µ2 = sin2 (θ) ,
we can rewrite Eqs. 74 and 80 as
0 =∂r e2ψ−2ν ∆∂r ω + ∂µ e2ψ−2ν δ∂µ ω
e2ψ−2ν ∆(∂r ω)2 + δ(∂µ ω)2 =∂r [∆∂r (ν − ψ)] + ∂µ [δ∂µ (ν − ψ)] ,
where we used that
√
X=
(91)
(92)
δ
∆(∂r ω)2 + √ (∂µ ω)2 .
∆
These equations can be expressed as
∆
δ
0 =∂r
∂r ω + ∂µ
∂µ ω
χ2
χ2
2 (∆∂r χ∂r ω + δ∂µ χ∂µ ω) =χ [∂r (∆∂r ω) + ∂µ (δ∂µ ω)]
1 ∆
δ
2
2
∆(∂r ω) + δ(∂µ ω) =∂r
∂r χ + ∂µ
∂µ χ
χ2
χ
χ
⇐⇒
⇐⇒
∆(∂r ω)2 + δ(∂µ ω)2 + ∆(∂r χ)2 + δ(∂µ χ)2 =χ [∂r (∆∂r χ) + ∂µ (δ∂µ χ)] ,
where we employed the denition of
χ
(93)
(94)
from earlier.
Next we dene
X := χ + ω ,
Y := χ − ω ,
(95)
allowing us to rewrite Eqs. 93 and 94 as
2 ∆(∂r X )2 − ∆(∂r Y)2 + δ(∂µ X )2 − δ(∂µ Y)2
=(X + Y) [∂r (∆∂r X ) − ∂r (∆∂r Y) + ∂µ (δ∂µ X ) − ∂µ (δ∂µ Y)]
2 ∆(∂r X )2 + ∆(∂r Y)2 + δ(∂µ X )2 + δ(∂µ Y)2
(96)
=(X + Y) [∂r (∆∂r X ) + ∂r (∆∂r Y) + ∂µ (δ∂µ X ) + ∂µ (δ∂µ Y)] .
(97)
12
Deriving the Kerr metric
David Wagner
Adding and subtracting these two equations, we end up with a pair of symmetric expressions:
1
∆(∂r X )2 + δ(∂µ X )2 = (X + Y) [∂r (∆∂r X ) + ∂µ (δ∂µ X )]
2
1
∆(∂r Y)2 + δ(∂µ Y)2 = (X + Y) [∂r (∆∂r Y) + ∂µ (δ∂µ Y)] .
2
(98)
(99)
We should note that, with these denitions, Eq. 76 can be expressed as
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
√
√ i
√ 2 h √
√
∂r
∆∂r ∆ − ∂µ
δ∂µ δ
∆
√
√
√
1
1
δ
√ ∂µ ∆δ∂µ µ2 + ∂µ ψ∂µ ν
=4 ∆ √ ∂r ∆δ∂r µ3 + ∂r ψ∂r ν − 4 √
∆
∆δ
∆δ
1 √
δ
+ 2
∆(∂r ω)2 − √ (∂µ ω)2
χ
∆
h
√
√
√ i
√
√
√
2
√
(∂r ∆)2 + ∆∂r2 ∆ − (∂µ δ)2 − δ∂µ2 δ
∆
√
√
√
1
δ
1
√ ∂µ ∆δ∂µ µ2 + ∂µ ψ∂µ ν
=4 ∆ √ ∂r ∆δ∂r µ3 + ∂r ψ∂r ν − 4 √
∆
∆δ
∆δ
1 − √
∆ ∂r X ∂r Y − χ2 (∂r (ν − ψ))2 + δ ∂µ X ∂µ Y − χ2 (∂µ (ν − ψ))2
χ2 ∆
√
√
√ i
2 h √ 2 √ 2√
√
(∂r ∆) + ∆∂r ∆ − (∂µ δ)2 − δ∂µ2 δ
∆
r
√ i
√
1
δ √
1 h √ 2
(∂r ∆) − (∂µ δ)2 − √ [∆∂r X ∂r Y − δ∂µ X ∂µ Y]
∂µ δ∂µ µ2 + √
=4∂r ∆∂r µ3 − 4
2
∆
∆
χ ∆
h √
√ 2 √ 2√ i
√ 2√
2
2 (∂r ∆) + ∆∂r ∆ − (∂µ δ) − δ∂µ δ
h √
√ i
1
=4(r − M )∂r µ3 + 4µ∂µ µ2 + (∂r ∆)2 − (∂µ δ)2 − 2 [∆∂r X ∂r Y − δ∂µ X ∂µ Y]
χ
h
√ 2 √ 2√ i
√ 2√
2 (r − M )∂r (µ3 − µ2 ) + ∆∂r ∆ − (∂µ δ) − δ∂µ δ
h √
√ i
1
=4(r − M )∂r µ3 + 4µ∂µ µ2 + 2µ∂µ (µ3 − µ2 ) + (∂r ∆)2 − (∂µ δ)2 − 2 [∆∂r X ∂r Y − δ∂µ X ∂µ Y]
χ
1
0 =2(r − M )∂r (µ2 + µ3 ) + 2µ∂µ (µ2 + µ3 ) − 2 [∆∂r X ∂r Y − δ∂µ X ∂µ Y]
χ
√ 2
√ 2√
√ 2
√ 2√
+ (∂r ∆) − 2 ∆∂r ∆ + (∂µ δ) + 2 δ∂µ δ
1
0 =2(r − M )∂r (µ2 + µ3 ) + 2µ∂µ (µ2 + µ3 ) − 2 [∆∂r X ∂r Y − δ∂µ X ∂µ Y]
χ
√
√
√
1
+ (∂r ∆)2 − 2 ∆∂r2 ∆ − 1 −
δ
1
M 2 − a2 1
0 =2 [(r − M )∂r + µ∂µ ] (µ2 + µ3 ) − 2 [∆∂r X ∂r Y − δ∂µ X ∂µ Y] + 3
− .
(100)
χ
∆
δ
In this series of elementary reductions (cf. Ref. [1], p. 410), we employed that
√
√
1
∂r ∆ = ∂r (µ3 − µ2 ) ∆ = √ (r − M )
∆
and
2∂µ (µ3 − µ2 ) =
13
∂µ ∆
=0.
∆
Deriving the Kerr metric
David Wagner
The remaining task is now as follows: Solve Eqs. 98 and 99 to get expressions for
after which Eq. 100 may be solved for
µ2 + µ3 .
X,Y
and thus for
χ, ω ,
Then the metric, which after our calculations so far can
already be described through
√
2
ds =
eµ2 +µ3
1
2
dr2 + ∆dθ2 ,
∆δ −χdt + (dφ − ωdt) + √
χ
∆
2
(101)
will be fully specied.
6 Deriving and solving Ernst's equation
In order to solve Eqs. 98 and 99, we are now going to rewrite them into a single complex expression that is
called Ernst's equation.
For this we dene
f := X Ye2ψ =
Since we know from Eq. 82 that
ω
√
∆δ
χ2 − ω 2
,
χ
W :=
ω
ω
= 2
XY
χ − ω2
χ2 − ω 2 = X Y satisfy the same equation,
2
2
f
f
∂r W + ∂θ
∂θ W = 0 .
∂r
∆
δ
and
it holds that
(102)
This can be seen by expanding the expression:
2
f
f2
∂r W + ∂µ
∂µ W
∂r
δ
∆
∆
δ
2
2
=∂r 2 (X Y) ∂r W + ∂µ 2 (X Y) ∂µ W
χ
χ
∆
δ
=∂r 2 (X Y∂r ω − ω∂r (X Y)) + ∂µ 2 (X Y∂µ ω − ω∂µ (X Y))
χ
χ
∆
∆
δ
δ
=X Y∂r 2 ∂r ω − ω∂r 2 ∂r (X Y) + X Y∂µ 2 ∂µ ω − ω∂µ 2 ∂µ (X Y)
χ
χ
χ
χ
=0 ,
W
where Eq. 91 was used at the end. This implies that
∂r g =
f2
∂µ W ,
∆
may be derived from a potential
∂µ g = −
g,
(103)
where
f2
∂r W ,
δ
such that
∇ × ∇g = 0
in the two-dimensional
r, µ−subspace.
The potential
∂r
g
(104)
itself obviously satises
∆
δ
∂r g + ∂µ
∂µ g = 0 ,
f2
f2
(105)
which can be rewritten as
f [∂r (∆∂r g) + ∂µ (δ∂µ g)] = 2∆∂r g∂r f + 2δ∂µ g∂µ f .
(106)
Along the same lines, we can see that
∂r
∆
∂r f
f
+ ∂µ
δ
∂µ f
f
= −f
14
2
1
1
(∂r W)2 + (∂µ W)2
δ
∆
(107)
Deriving the Kerr metric
David Wagner
holds, which can be expanded to
f [∂r (∆∂r f ) + ∂µ (δ∂µ f )] = ∆(∂r f )2 + δ(∂µ f )2 − ∆(∂r g)2 − δ(∂µ g)2 .
(108)
Now we introduce the complex variable
Z := f + ig ,
through the use of which we can combine Eqs. 106 and 108 into
<(Z) [∂r (∆∂r Z) + ∂µ (δ∂µ Z)] = ∆(∂r Z)2 + δ(∂µ Z)2 .
(109)
Finally introducing the transformation
Z=
E − Ē
1+E
1 − E Ē
+
,
=
1−E
|1 − E|2 |1 − E|2
(110)
we nd
1 − E Ē
2
2
4∆
4δ
2
2
∂r (∆∂r E) +
∂µ (δ∂µ E) +
(∂r E) +
(∂µ E)
|1 − E|2 (1 − E)2
(1 − E)2
(1 − E)3
(1 − E)3
4∆
4δ
=
(∂r E)2 +
(∂µ E)2
(1 − E)4
(1 − E)4
2
2
⇐⇒ (1 − E Ē)
∂r (∆∂r E) +
∂µ (δ∂µ E)
(1 − E)2
(1 − E)2
4∆
4δ
2
2
= − Ē(1 − E)
(∂
E)
+
(∂
E)
r
µ
(1 − E)3
(1 − E)3
⇐⇒ (1 − E Ē) [∂r (∆∂r E) + ∂µ (δ∂µ E)] = −Ē 2∆(∂r E)2 + 2δ(∂µ E)2 .
This expression, when using the new variable
η 2 :=
(r − M )2
∆
= 2
+1,
2
2
M −a
M − a2
can be rewritten as
(1 − E Ē) ∂η [(η 2 − 1)∂η E] + ∂µ [(1 − µ2 )∂µ E] = −Ē 2(η 2 − 1)(∂η E)2 + 2(1 − µ2 )(∂µ E)2 ,
which constitutes Ernst's equation.
This equation is solved by
E = −pη − iqµ , p2 + q 2 = 1 ,
as can be directly veried. In order to do this we rstly evaluate the right-hand side of Eq. 111:
(1 − E Ē) ∂η [(η 2 − 1)∂η E] + ∂µ [(1 − µ2 )∂µ E]
=(1 − p2 η 2 − q 2 µ2 ) ∂η [(η 2 − 1)(−p)] + ∂µ [(1 − µ2 )(−iq)]
=(1 − p2 η 2 − q 2 µ2 ) (−2ηp + 2iqµ)
=2 −ηp + iqµ + η 3 p3 − ip2 qη 2 µ + ηq 2 pµ2 − iq 3 µ3 .
The right-hand side gives
− Ē 2(η 2 − 1)(∂η E)2 + 2(1 − µ2 )(∂µ E)2
=(pη − iqµ) 2(η 2 − 1)p2 + 2(1 − µ2 )(−q 2 )
=2 p3 η 3 − p3 η − iqη 2 p2 µ + iqp2 µ − q 2 pη + iq 3 µ + q 2 pηµ2 − iq 3 µ3
15
(111)
Deriving the Kerr metric
David Wagner


=2 p3 η 3 − pη (p2 + q 2 ) −iqη 2 p2 µ + iq (p2 + q 2 ) µ + q 2 pηµ2 − iq 3 µ3  ,
| {z }
| {z }
=1
=1
such that the proposed solution does indeed solve Ernst's equation.
Thus we can split up the complex variable
Z
into its real and imaginary parts
Z = Z1 + iZ2
and obtain
1 − E Ē
1 − p2 η 2 − q 2 µ 2
p2 (η 2 − 1) − q 2 (1 − µ2 )
=
=
−
|1 − E|2
(1 + pη)2 + q 2 µ2
(1 + pη)2 + q 2 µ2
2qµ
E − Ē
=−
.
Z2 ==(Z) =
2
|1 − E|
(1 + pη)2 + q 2 µ2
Z1 =<(Z) =
Since the equation of motion for
Z,
Eq. 109, is invariant under a change
Z → −Z ,
(112)
(113)
we will use the negative
of the solutions above.
Resubstituting the variable
η,
we arrive at
(114)
2 pq2 (M 2 − a2 )µ
Z2 = h
.
i2
√
2
p−1 M 2 − a2 + r − M + (M 2 − a2 ) pq 2 µ2
(115)
Next we choose
√
p=
which manifestly fulls
q2
(M 2
p2
− a2 )δ
Z1 = h
i2
√
2
p−1 M 2 − a2 + r − M + (M 2 − a2 ) pq 2 µ2
∆−
p2 + q 2 = 1
M 2 − a2
,
M
q=
a
,
M
and leads us to
∆ − a2 δ
r 2 + µ2 a 2
2aM µ
Z2 = 2
.
r + a2 µ2
Z1 =
(116)
(117)
Introducing the variable
ρ2 := r2 + a2 µ2
and reverting back to the functions
f
and
g,
we obtain
f =(χ2 − ω 2 )e2ψ = e2ν − ω 2 e2ψ =
g=
Employing the dening relations of
∆ − a2 δ
ρ2
2aM µ
.
ρ2
g,
(118)
(119)
we nd the constraints
4raM µ
(∆ − a2 δ)2
=
∂µ W
ρ4
∆ρ4
2aM (ρ2 − a2 µ2 )
(∆ − a2 δ)2
∂µ g =
=
−
∂r W ,
ρ4
δρ4
∂r g = −
16
(120)
(121)
Deriving the Kerr metric
David Wagner
which are fullled by
W=
Combining the expressions for
f
W,
and
ω
2aM rδ
=
.
χ2 − ω 2
∆ − a2 δ
(122)
we nd
ωe2ψ =
2aM rδ
.
ρ2
(123)
We can combine the expression above with Eq. 118 to obtain
∆ − a2 δ 2ψ
∆δρ4 − 4a2 M 2 r2 δ 2
β
2 4ψ
e
=
e
−
ω
e
=
,
ρ2
ρ4
where we used our solution for
eβ .
These expressions for
ω
and
e2ψ
(124)
are solved by
δΣ2
ρ2
2aM r
,
ω=
Σ2
e2ψ =
where
Σ2 =
(125)
(126)
ρ4 ∆ − 4a2 M 2 r2 δ
= (r2 + a2 )2 − a2 ∆δ .
∆ − a2 δ
Furthermore, we nd the solution for
e2ν
by considering the identity
e2ν = e2β−2ψ ,
yielding
e2ν =
In the nal step we can nd the solution for
µ2 + µ3 ,
ρ2 ∆
.
Σ2
(127)
for which we rst need to nd the objects
X
√
∆ − a2 δ
∆+a δ
X =√ h √
√ i=√ h
√ i
δ ρ2 ∆ − 2aM r δ
δ r2 + a2 + a ∆δ
√
√
∆ − a2 δ
∆−a δ
Y =√ h √
√ i.
√ i=√ h
δ ρ2 ∆ + 2aM r δ
δ r2 + a2 − a ∆δ
and
Y:
√
(128)
(129)
Here we used the decomposition
√
√ √
√ ρ4 ∆ − 4a2 M 2 r2 δ = ρ2 ∆ − 2aM r δ ρ2 ∆ + 2aM r δ .
The derivatives of these quantities can be computed to
√ i
√ h
µ ∆ r2 + a2 + a2 δ + 2a ∆δ
∂µ X =
,
h
√ i2
3
δ 2 r2 + a2 + a ∆δ
√ i
√ h
µ ∆ r2 + a2 + a2 δ − 2a ∆δ
∂µ Y =
.
h
√ i2
3
δ 2 r2 + a2 − a ∆δ
√
√
√
(r −
− ( ∆ + a δ)2r ∆
∂r X =
,
√ h
√ i2
∆δ r2 + a2 + a ∆δ
M )ρ2
√
√
√
(r − M )ρ2 − ( ∆ − a δ)2r ∆
∂r Y =
,
√ h
√ i2
∆δ r2 + a2 − a ∆δ
These results can now be inserted into Eq. 100 to obtain
[(r − M )∂r + µ∂µ ] (µ2 + µ3 ) =2 −
17
(r − M )2
rM
−2 2 ,
∆
ρ
(130)
Deriving the Kerr metric
David Wagner
which is solved (as is easily checked) by
ρ2
eµ2 +µ3 = √ .
∆
(131)
With this result we are nally done; the line element (reverting back to the (+
− −−)
convention) is now
given by
2
Σ2
2aM r
ρ2
ρ2
2
2
dt
−
dr2 + ∆dθ2 .
ds = 2 ∆dt − 2 sin (θ) dφ −
2
Σ
ρ
Σ
∆
2
(132)
In matrix notation, we have



(gµν ) = 


1−
2M r
ρ2
0
0
0
2aM r
ρ2
0
2
− ρ∆
0
0
−ρ2
0
0


 ,

h
i

2M a2 r sin2 (θ)
2
2
2
− r +a +
sin
(θ)
ρ2
sin2 (θ)
Σ2
ρ2 ∆
0
− ρ∆2
0
0
0
− ρ12
2aM r
ρ2 ∆
0
0
sin (θ)
− ∆−a
ρ2 ∆ sin2 (θ)

 0
(g ) = 
 0

µν
2aM r
ρ2 ∆

0
0
2aM r
ρ2

sin2 (θ)

0
0
2
2
(133)


 .


(134)
7 Properties of the Kerr metric
Having, after a lot of work, nally having established the Kerr metric, we want to take a look at some of its
properties.
7.1
Asymptotic properties
Firstly, when letting
a → 0,
we obtain

1 − 2M
r
 0

(gµν ) = 
 0
0
0
− 1−12M
r
0
0
0
0
0
0
−r2
0
0
−r2 sin2 (θ)
which is the usual Schwarzschild metric. Thus the parameter
black hole and
When letting
r
rS = 2M
M



 ,

(135)
has to be identied with the mass of the
(in natural units) is the Schwarzschild radius.
go to innity, we obtain (up to order
O(r−3 ))
1 − 2M
0
r

0
− 1−12M

r
(gµν ) = 

0
0
2aM
2
0
r sin (θ)

0
0
−r2
0
2aM
r

sin2 (θ)

0

 .

0
2
2
−r sin (θ)
(136)
dφdt cross term, which should correspond
a → 0, this term of course vanishes. Further-
Manifestly, this object constitutes the Schwarzschild metric with a
to the rotation of the body in question. We observe that for
more, if we exclude all orders smaller than
O(1),
we are left with the usual Minkowski metric in spherical
coordinates, thus letting us conclude that the Kerr metric is asymptotically at.
If we calculate the rate of rotation
Ω for
Eq. 136 (cf. [4]), we nd, when identifying
a=
J
M (with
J
the total
angular momentum of the body) that
Ω=
3(J · r̂)r̂ − J
,
r3
18
(137)
Deriving the Kerr metric
David Wagner
in agreement with the known Lense-Thirring eect, which constitutes a weak-eld approximation of the exact
Kerr solution.
Finally, for
M → 0,
we nd

1
0

r2 +a2 cos2 (θ)
0 − r2 +a2
(gµν ) = 
0
0
0
0

0
0


0
0
 ,
2
2
2

−[r + a cos (θ)]
0
0
− r2 + a2 sin2 (θ)
(138)
which is the at Minkowski metric in ellipsoidal coordinates.
7.2
Special surfaces
As was the case with the Schwarzschild metric, we are facing some interesting behaviours in Eq. 135, which
we will now take a closer look at.
The rst one occurs when
∆ = 0,
which translates to
rH,±
r r 2
rS
=
±
2
S
2
− a2 .
(139)
This makes it apparent that the one event horizon from the Schwarzschild metric at
rS
up into an inner and an outer event horizon, with the distance between the two increasing
that for
a
suciently large,
leading to a
∆
can
not
a = 0) splits
with a. Notice
(for
become zero at all, implying that there will be no event horizon, thus
naked singularity, with singularityas it is predicted by GR (we will in the next subsection show
the existence of such a singularity in the rst place).
Secondly, the
tt
component of the metric changes sign when
rE,±
rS
=
±
2
r r 2
These two surfaces are called the inner and outer
S
2
ρ2 = 2M r;
this leads to
− a2 cos2 (θ) .
ergospheres.
Since
cos2 (θ) ≤ 1,
(140)
the outer ergosphere lies
outside the outer event horizon, touching it at the poles, while the inner ergosphere lies within the inner
event horizon.
Inside this (outer) ergosphere, every particle on a timelike path has to co-rotate with the
central mass.
Due to the respective particles not yet having crossed the event horizon, they may still be ejected from the
black hole, having gained energy through the forced co-rotation. Thus the rotating black hole emits highly
energetic particles, which is called the
7.3
Penrose process
and is one theory to explain gamma-ray bursts.
Singularities
In order to see where the singularities of the Kerr metric lie, we would have to express the components of the
Riemann tensor in terms of our solution; this however, we will not do right now, because this was enough
work already, but refer to Ref. [2], where it becomes clear that the components of the Riemann tensor only
diverge for
ρ = 0.
This in turn can only occur for
r=0
and
θ=
π
2 (Here we also can see that the inner
ergosphere touches the singularity in the equatorial plane). In order to clarify the meaning of this singularity
we rstly change to new time and angular variables
r 2 + a2
dr
∆
a
dφ̃ =dφ − dr .
∆
du =dt −
Hereafter (and after a few steps in Ref. [2]) we introduce
dx0 =du + dr
19
Deriving the Kerr metric
David Wagner
h
i
x = r cos(φ̃) + a sin(φ̃) cos(θ)
h
i
y = r sin(φ̃) − a cos(φ̃) cos(θ)
z =r cos(θ)
x2 + y 2 =(r2 + a2 ) sin2 (θ) ,
after which the line element can be brought into the following form:
2M r3
ds = (dx ) − dx − dy − dz − 4
r + a2 z 2
2
0 2
2
2
2
1
z
dx − 2
[r (xdx + ydy) + a (xdy − ydx)] − dz
2
r +a
r
0
2
.
The advantage of this form lies in the fact that it reduces to a cartesian coordinate system in the limit of
M → 0,
thus removing the degeneracy of spherical (or ellipsoidal) coordinates at
Here we can clearly see that the point of the singularity
r = 0, θ =
r = 0.
π
2 corresponds to
x2 + y 2 = a2
in the equatorial plane (z
= 0).
(141)
Thus the Kerr metric actually features a
ring singularity
with radius
a,
that
of course shrinks to a point in the limit of vanishing rotation.
As a nal remark, the existence of a ring singularity makes it obvious that we are (somehow) allowed to
continue our solution to negative values of
r.
This entails many interesting consequences, which we will
however not deal with here and refer to e.g. Ref. [5].
References
[1] S. Chandrasekhar
The Kerr Metric and Stationary Axisymmetric Gravitational Fields
Proceedings
of the Royal Society of London. Series A, Mathematical and Physical Sciences, Vol. 358, No. 1695 (Jan.
16, 1978), pp. 405-420
url:
https://www.jstor.org/stable/79479
[2] S. Chandrasekhar
The Mathematical Theory of Black Holes
[3] T. Eguchi, P. B. Gilkey, A. J. Hanson
Oxford University Press, 1983
Graviation, gauge theories and dierential geometry
PHYSICS REPORTS (Review Section of Physics Letters) 66. No. 6 (1980) 213393
[4] C. Chakraborty
Lense-Thirring Precession in Strong Gravitational Fields
https://indico.cern.ch/event/336103/contributions/786924/attachments
/1203051/1813620/chakraborty_proc.pdf
url:
[5] W. Israel
Source of the Kerr Metric
Phys. Rev. D 2, 641 (1970)
20