Black hole lecture notes

7/25/2019 Black hole lecture notes

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Chapter 3

The Kerr solution

As we have seen, the solution of Einsteins equations describing theexterior of an isolated, spherically symmetric object (the Schwarzschildsolution) is quite simple. Indeed, it has been found in 1916, imme-diately after the derivation of Einsteins equation. In the case ofa rotating body, instead, the problem (which is very relevant: as-trophysical bodiesdo rotate) is much more difficult: we dont knowany analytic, exact solution describing the exterior of a rotating star(even if we know approximate solutions).

But we know the exact solution describing a rotating, stationary,axially symmetric black hole. It is the Kerr solution, derived in1963 by R. Kerr. We say that this metric describes a black hole,because it is a solution of Einstein equations in vacuum (T = 0)and it has a curvature singularity covered by an horizon: like in thecase of Schwarzschild spacetime, everything falling inside the holecannot escape.

We stress that while, thanks to Birkhofftheorem, the Schwarzschildmetric forr >2Mdescribes the exterior of any spherically symmet-ric isolated object (a star, a planet, a stone, etc.), the Kerr metricoutside the horizon can only describe the exterior of a black hole. 1

1

Actually, there is no proof that it cannot exist a stellar model matching with Kerr metricat the surface of the star, but such a model has never been found, and it is common beliefthat it is unlikely to exist.

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3.1 The Kerr metric in Boyer-Lindquist coordi-

nates

The explicit form of the metric is the following:

ds2 = dt2 +

dr2

+d2

+ (r2 +a2)sin2 d2

+2Mr

(a sin2 d dt)2 (3.1)

where

(r) r2 2M r+a2(r, ) r2 +a2 cos2 . (3.2)

The coordinates (t,r, ,) in which it is expressed the Kerr metricin (3.1) are called Boyer-Lindquist coordinates.

The Kerr metric depends on two parameters, M anda; compar-ing (3.1) with the far field limit metric of an isolated object (18.3),we see that Mrepresents the mass of the black hole, and Ma itsangular momentum, as measured from infinity.

Some properties of the Kerr metric can be directly seen by lookingat the line element (3.1):

It is stationary: it does not depend explicitly on time.

It is axisymmetric: it does not depend explicitly on .

It is not static: it is not invariant for time reversal t t. It is invariant for simultaneous inversion oft and ,

t t , (3.3)

as can be expected: the time reversal of a rotating object pro-duces an object which rotates in the opposite direction.

In the limitr , the Kerr metric (3.1) reduces to Minkowskimetric in polar coordinates; then, the Kerr spacetime isasymp-totically flat.

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In the limita 0 (withM= 0), it reduces to the Schwarzschildmetric: r(r 2M), r2 thends2 (12M/r)dt2 + (12M/r)1dr2 + r2(d2 +sin2 d2) .

(3.4)

In the limit M 0 (with a = 0), it reduces to

ds2 = dt2+ r2 +a2 cos2

r2 +a2 dr2+(r2+a2 cos2 )d2+(r2+a2)sin2 d2

(3.5)

which is the metric of flat space in spheroidal coordinates:

ds2 = dt2 +dx2 +dy2 +dz2 (3.6)where

x =

r2 +a2 sin cos

y =

r2 +a2 sin sin

z = r cos . (3.7)

Indeed,

dx = r

r2 +a2sin cosdr+

r2 +a2 cos cosd

r2 +a2 sin sin d

dy = r

r2 +a2sin sindr+

r2 +a2 cos sind+

r2 +a2 sin cosd

dz = cos dr r sin d (3.8)thus

dx2 +dy2 +dz2 =

r2

r2 +a2sin2 + cos2

dr2

+

(r2 +a2)cos2 +r2 sin2

d2 + (r2 +a2)sin2 d2

=r2 +a2 cos2

r2 +a2 dr2 + (r2 +a2 cos2 )d2 + (r2 +a2)sin2 d2 .

(3.9)

The metric (3.1) is singular for = 0 and for = 0. By com-puting the curvature invariants (like for instance RR

)one finds that they are regular at = 0, and singular at = 0.Thus = 0 is a true, curvature singularity of the manifold,whereas (as we will show) = 0 is a coordinate singularity.

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Notice that in the Schwarzschild limit (a= 0), =r2 = 0 givesthe curvature singularity, while (for r = 0) =r(r 2M) = 0gives the coordinate singularity at the horizon.

The metric has the form

g=

gtt 0 0 gt0

0 0

0 0 0gt 0 0 g

(3.10)

with

gtt =

1 2Mr

gt = 2Mr

a sin2

g =

r2 +a2 +

2Mra2

sin2

sin2 . (3.11)

The g component can be rewritten in a different way, which willbe useful later:

g = (r2 +a2)sin2 +2Mra

2 sin4

= sin2

(r2 +a2 cos2 )(r2 +a2) + 2Mra2 sin2

=

sin2

(r2 +a2)2 (r2 +a2)a2 sin2 + 2Mra2 sin2

= sin2

(r2 +a2)2 a2 sin2 . (3.12)

Let us compute the inverse metric. To get g, we only have toinvert the t block in (3.10), while the inversion of the r part is

trivial. The metric in the t block is

gab=

gtt gtgt g

(3.13)

and its determinant is

g = gttg g2t

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=

1 2M r

r2 +a2 +

2M ra2

sin2

sin2 4M

2r2a2

2 sin4

=

r2 +a2 +2Mra2

sin2

sin2 + (r2 +a2)

2Mr

sin2

= (r2 +a2)sin2 +2Mr

sin2 a2 sin2 +r2 +a2

= (r2 +a2)sin2 + 2M r sin2 = sin2 (3.14)therefore

gab = 1 sin2

g gtgt gtt

(3.15)

and

g =

gtt 0 0 gt

0

0 00 0 1

0

gt 0 0 g

(3.16)

with

gtt = 1

r2 +a2 +

2Mra2

sin2

gt = 2Mr

a

g = a2 sin2 sin2

(3.17)

where we have used the fact that

2M r sin2

=r2 +a2 cos2 2Mr

sin2 = a2 sin2 sin2

. (3.18)

3.2 Symmetries of the metric

Being stationary and axisymmetric, the Kerr metric admits twoKilling vector fields:

k t

m

(3.19)

or equivalently, in coordinates (t,r, ,),

k (1, 0, 0, 0) m (0, 0, 0, 1) . (3.20)

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In the motion of a particle with four-velocity u, there are then twoconserved quantities:

E uk= ut L= um= u . (3.21)In the case of massive particles, for which the four-momentum isP =mu, they are the energy at infinity per mass unit and the an-gular momentum per mass unit, respectively. In the case of masslessparticle, we can choose properly the affine parameter (as we will al-ways do in the following) so that the four-momentum coincides with

the four-velocity: P

=u

; thus, for massless particles E is the en-ergy at infinity andL the angular momentum.It can be shown that k, m are the only Killing vector fields of

the Kerr metric; thus, any Killing vector field is a linear combinationof them.

3.3 Frame dragging and ZAMO

Let us consider an observer, with timelike four-velocity u, whichfalls into the black hole with zero angular momentum

L= u= 0 . (3.22)

This implies that at r , where the metric becomes flat, alsou = 0, and its angular velocity is zero. Such observer is conven-tionally named ZAMO, which stands for zero angular momentumobserver. The contravariant component of the velocity does notvanish (except in the limit r ):

u =gtut = 0 (3.23)then the trajectory of the ZAMO has a non-zero angular velocity:

ddt

=dddtd

=u

ut= 0 . (3.24)

To compute in terms of the metric (3.1), which is given in covari-antform, we use the fact that

u= 0 =gu +gtu

t (3.25)

thus

=u

ut = gt

g. (3.26)

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We have

gt = 2Mra

sin2 (3.27)

and, due to (3.12),

g=sin2

(r2 +a2)2 a2 sin2 , (3.28)

therefore the angular velocity of a ZAMO is

=

2Mar

(r2 +a2)2 a2 sin2 . (3.29)Notice that

(r2 +a2)2 > a2 sin2 (r2 +a2 2Mr) (3.30)thus we always have /(Ma)> 0: the angular velocity has the samesign as the angular momentum Ma of the black hole, namely, themotion of the ZAMO is corotating with the black hole.

We can conclude that an observer which approaches a Kerr blackhole with a trajectory which has zero angular velocity at infinity(and then zero angular momentum) is draggedby the gravitational

field of the black hole, acquiring an angular velocity corotating withthe black hole.

3.4 Horizon structure of the Kerr metric

3.4.1 Removal of the singularity at = 0

To show that = 0 is a coordinate singularity, we make a coordi-nate transformation that brings the metric into a form which is notsingular at = 0, and then extend the spacetime; such coordinatesare called Kerr coordinates. They are the generalization, to rotat-

ing black holes, of the Eddington-Finkelstein coordinates derived inSchwarzschild spacetime. To begin with, we need to find two familiesof null geodesics, one ingoing and one outgoing, and to determine thecorresponding null coordinates (u, v), i.e. the quantities which areconstant in any of these geodesics. In the case of Kerr geometry, thespacetime cannot be decomposed in a product of two-dimensionalmanifolds, thus the study of null geodesics is more complex than in

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the Schwarzschild case. The Kerr metric admits two special familiesof null geodesics, named principal null geodesics, given by

u =dx

d =

dt

d,dr

d,d

d,d

d

=

r2 +a2

, 1, 0,

a

, (3.31)

where the sign plus (minus) corresponds to outgoing (ingoing) geodesics.In the Schwarzschild limit these are the usual outgoing and ingoinggeodesics u = (1/(1 2M/r), 1, 0, 0), but in the Kerr case theyacquire an angular velocity d/d proportional to a and diverging

when = 0.We will show explicitly that (3.31) are geodesics later, in the

coordinate frame we are going to define; here we check that they arenull:

guu = 0 . (3.32)

We have

gdx

d

dx

d =

dt

d

2+

1

dr

d

2+

d

d

2

+(r2 +a2)sin2 d

d2

+2Mr

a sin2 d

d dt

d2

.

(3.33)

First, we notice that

dt

d a sin2 d

d=

r2 +a2 a2 sin2

=

. (3.34)

Then,

guu = (r

2 +a2)2

2 +

+ (r2 +a2)sin2

a2

2+

2Mr

2

= 1

2 (r2 +a2)(r2 +a2) + (r2 +a2 cos2 )(r2 +a2

2Mr)

+sin2 a2(r2 +a2) + (r2 +a2 cos2 )2Mr

= 0 (3.35)

and the tangent vector (3.31) is null.Let us consider the ingoing geodesics, whose tangent vector we

call

l =

r2 +a2

, 1, 0, a

; (3.36)

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let us parametrize the geodesics in terms ofr :

dt

dr = r

2 +a2

d

dr = a

. (3.37)

We want these geodesics to be coordinate lines of our new system;thus, one of our coordinates is r, while the others are quantitieswhich are constant along a geodesic of the family. One of these is, which is constant along the considered geodesics; the remainingtwo coordinates are given by

v t+T(r) + (r) (3.38)

whereT(r) and (r) are solutions of2

dT

dr =

r2 +a2

d

dr =

a

(3.39)

so that, along a geodesic of the family,

dv

dr =

d

dr 0 (3.40)and the tangent vector of the ingoing principal null geodesics (3.36)is, in the new coordinates, simply

l = (0, 1, 0, 0) . (3.41)We can now compute the metric tensor in the coordinate system(v,r, ,). We recall that, in Boyer-Lindquist coordinates,

ds2 = dt2+

dr2

+d2

+(r2+a2)sin2 d2+

2Mr

(a sin2 ddt)2 .

(3.42)

We have

dv = dt+r 2 +a2

dr ; dt= dv r

2 +a2

dr

d = d+ a

dr ; d= d a

dr , (3.43)

2Notice that (3.39) have an unique solution, with the only arbitrariness of the choice ofthe origins ofv and , because the right-hand sides of (3.39) depend on r only.

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then

dt2 = dv2 (r2 +a2)2

2 dr2 + 2

r2 +a2

dvdr

(r2 +a2)sin2 d2 = (r2 +a2)sin2 d2 + (r2 +a2)a2

2sin2 dr2

2(r2 +a2)a

sin2 drd , (3.44)

dr2 + d2 do not change (r, are also coordinates in the new

frame), the parenthesis in the last term of (3.42) reduces to

dt a sin2 d = dv a sin2 d r2 +a2 a2 sin2

dr

= dv a sin2 d

dr , (3.45)

thus

2Mr

(dt a sin2 d)2 =2Mr

dv2 +

2M r

a2 sin4 d2

4Mr

dvdr+4M r

a sin2 ddr+

2Mr

2 dr2 4M r

a sin2 dvd

(3.46)

and, putting all together, we have that the coefficient ofdvdr is

2r2 +a2

4Mr

= 2 , (3.47)

the coefficient ofddr is

2(r2 +a2)a

sin2 +4Mr

a sin2 = 2a sin2 , (3.48)

and the coefficient ofdr 2 is

(r2 +a2)2

2 +

r 2 +a2

2 a2 sin2 +

2M r2

(r2 +a2 cos2 )

=

(r

2 +a2)(r2 +a2 2Mr)2

+r 2 +a2 2M r

2 a2 sin2

=

r

2 +a2 a2 sin2

= 0 . (3.49)

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Therefore,

ds2 =

1 2Mr

dv2 + 2dvdr+ d2

+(r2 +a2)2 a2 sin2

sin2 d2

2a sin2 drd 4Mra

sin2 dvd . (3.50)

The coordinates (v,r, ,) are called Kerr coordinates. They re-

duce to the Eddington-Finkelstein coordinates for a = 0. In thisframe, it is easy to show that l are tangent vector to geodesics;indeed, since l = (0, 1, 0, 0),ll;=l

l = rr = 0 rr = 0 gr,r =grr,= 0

(3.51)and this is the case, because in (3.50) grr = 0 and gvr , gr do notdepend on r .

In the Kerr coordinates, differently from the the Boyer-Lindquistcoordinates, the metric is not singular at = 0. Thus, after chang-ing coordinates to the Kerr frame, we can extend the manifold, to

include also the submanifold

= 0, and removing the correspond-ing coordinate singularity.We note, for later use, that being

gvr = 1 grr = gr = 0 gr = a sin2 , (3.52)we have

l = (1, 0, 0, a sin2 ) . (3.53)Notice also that, as we have shown above,

g = (r2 +a2)2 a2 sin2

sin2

= (r2 +a2)sin2 +2Mr

a2 sin4 (3.54)

and

2Mr

(dv a sin2 d)2 =2Mr

dv2 +a2 sin4 d2 2a sin2 dvd

(3.55)

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therefore the metric in Kerr coordinates can also be written in thesimpler form

ds2 = dv2 + 2dvdr+ d2 + (r2 +a2)sin2 d2 2a sin2 drd+

2M r

(dv a sin2 d)2 . (3.56)

If we want an explicit time coordinate, we can define

t v r (3.57)so that the metric (3.56) becomes

ds2 = dt2 +dr2 + d2 + (r2 +a2)sin2 d2 2a sin2 drd+

2Mr

(dt+dr a sin2 d)2 . (3.58)

3.4.2 The horizon

Here we study the submanifold

=r2 +a2 2Mr= 0 , (3.59)

which is a coordinate singularity in Boyer-Lindquist coordinates

ds2 = dt2+

dr2

+d2

+(r2+a2)sin2 d2+

2Mr

(a sin2 ddt)2 .

(3.60)Whena2 > M2, the equation = 0 has no real solution. In this casethere is no horizon, and the Kerr solution does not describe a blackhole. In this situation, the singularity = 0 is not covered by anyhorizon (naked singularity), and this would bring to paradoxesin our universe. For this reason, and for the reason that knownastrophysical processes are believed to give rise to black holes with|a|< M, this situation is generally considered unphysical. Here and

in the following, then, we will restrict our analysis to the case

a2 M2 (3.61)(the limiting case a2 =M2 is called extremal black hole).

We have(r) = (r r+)(r r) (3.62)

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with

r+ M+

M2 a2r

M

M2 a2 (3.63)solutions of Eq. (3.60). The surfaces of the coordinate singularity = 0 are then r = r+ andr = r.

Let us consider now the surfaces r constant= 0, whosenormal is

n= ,= (0, 1, 0, 0) . (3.64)

From (3.64), (3.16)

nng =grr =

. (3.65)

Thus, on the surfaces r = r+ and r = r, where = 0, nn = 0,

and these surfaces are null hypersurfaces, i.e. horizons. Being r+>r

, we can say that r =r+ is the outer horizon, and r = r is theinner horizon. Actually, we should have used the Kerr coordinatesto make this computation, but the result would be the same; indeed,the surfacesr = const.are the same in the two frames, and the signofnn

is the same as well. It can be easily checked that the resultit the same by computing nn in Kerr coordinates.

The two horizons separate the spacetime in three regions:

I. The region with r > r+. Here the r = const.hypersurfaces aretimelike. Ther limit, where the metric becomes flat, isin this region; so we can consider this region the exterior of theblack hole.

II. The region with r

< r < r+. Here the r = const. hyper-surfaces are spacelike. An object which falls inside the outerhorizon, can only continue falling to decreasing values ofr, untilit reaches the inner horizon and pass to region III.

III. The region with r < r

. Here the r = const.hypersurfaces aretimelike. This region contains the singularity, which we willstudy in section 3.6.

In the case of extremal black holes, when a2 =M2, the two horizonscoincide, and region I I disappears.

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If we consider the outer horizon r+ as a sort of surface of theblack hole, then we could conventionally consider the angular ve-locity at r = r+ of an observer which falls radially from infinity -i.e., an observer with zero angular momentum, or ZAMO - as a sortof angular velocity of the black hole. The angular velocity of aZAMO is given by (3.29):

=d

dt =

2Mar

(r2 +a2)2 a2 sin2 . (3.66)

Atr = r+, = 0 thus

= 2Mar+(r2++a

2)2 H (3.67)

which is a constant. In this sense, we can say that a black holerotates rigidly.

The quantity H= (r+) can be expressed in a simpler way. Wehave

(r+ M)2 =M2 a2 (3.68)therefore

r2++a2 = 2M r+ (3.69)

andH=

a

2Mr+=

a

r2++a2

. (3.70)

3.5 The infinite redshift surface and the ergo-

sphere

While in Schwarzschild spacetime the horizon is also the surfacewhere gtt changes sign, in Kerr spacetime these surfaces do not co-incide. We have that

gtt = 1 +2Mr

= 1

r2 2Mr+a2 cos2

= 1

(r rS+)(r rS) = 0 (3.71)

when r = rS+ and when r = rS, where

rS M

M2 a2 cos2 . (3.72)

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These surfaces are calledinfinite redshift surfaces, because if a sourcelocated on a point Pem near the black hole emits a light signal withfrequencyem, it will be observed at infinity with frequency

obs=

gtt(Pem)

gtt(Pobs)em (3.73)

thus if at Pem gtt= 0, obs = 0.The coefficient of r2 in (3.71) is negative, so gtt < 0 outside

[rS

, rS+], andgtt > 0 inside that interval. On the other hand, beingM2 a2 cos2 M2 a2, the horizons, located at

r= M

M2 a2 , (3.74)fall inside the interval [rS

, rS+]:

rS r< r+ rS+. (3.75)They coincide at = 0, , i.e. on the symmetry axis, while at the

!"#

#"

$#%&'()$#$

)*+&,

Figure 3.1: The ergosphere and the outer horizon

equatorial planrS+= 2M and rS

= 0.Therefore, there is a regionoutside the outer horizonwheregtt>0 3. This region, i.e.

r+< r < rS+ (3.76)

is called ergoregion, and its outer boundary r = rS+ is called ergo-sphere. Notice that, being the ergosphere outside the outer horizon,

3This does not happen in Schwarzschild spacetime, where gtt > 0 only inside the horizon

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an observer from infinity can go inside the ergoregion and come backto infinity.

In the ergoregion the killing vector k = (1, 0, 0, 0) becomesspacelike:

kkg=gtt> 0 . (3.77)

We define a static observeran observer (i.e. a timelike curve) withtangent vector proportional to k. The coordinates r, , are con-stant along its worldline, therefore this observer is still in the Boyer-Lindquist coordinate system (3.60). Such an observer cannot exist

inside the ergosphere, becausek

is spacelike there; in other words,an observer inside the ergosphere cannot stay still, but is forced tomove.

Astationary observeris an observer which does not see the metricchange in its motion. Then, its tangent vector must be a killingvector, i.e. it must be a combination of the two killing vectors ofthe Kerr metric, k= /t andm = /:

u =k +m

|k+m| = (ut, 0, 0, u) =ut(1, 0, 0,) (3.78)

where we have defined the angular velocity of the observer

ddt

= u

ut . (3.79)

In other words, the worldline has constant r and . The observercan only move along a circle, with angular velocity . Indeed, insuch orbits it does not see the metric change, being the spacetimeaxially symmetric.

A stationary observer can exist provided

uug= (ut)2

gtt+ 2gt+2g

= 1 (3.80)

i.e.2g+ 2gt+gtt< 0 . (3.81)

To solve (3.81), let us consider the equation

2g+ 2gt+gtt = 0 (3.82)

whose solutions are

=gt

g2t gttg

g. (3.83)

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The discriminant is (using eq.(3.14))

g2t gttg= sin2 . (3.84)Thus, a stationary observer cannot exist when 0 , (3.85)

the coefficient of2 in (3.81) is positive, and the inequality (3.81)is satisfied, outside the outer horizon (where r > r+, so > 0 andthen

< +), for

< < +. (3.86)

On the outer horizon r = r+, = 0 and = +, so (3.81) hasno solution, whereas equation (3.82), corresponding to a stationarynull worldline (for instance, a photon), has one solution only; theonly possible stationary null worldline on the horizon has

= gtg

= H (3.87)

i.e. it has the ZAMO angular velocity. This is another reason whythe angular velocity of the ZAMO at the horizon is considered as theblack hole angular velocity: it is the only possible angular velocityof a stationary particle on the outer horizon.

On the infinite redshift surface, gtt= 0 so (being gt< 0)

=gt

g2t

g= 0 . (3.88)

As expected, for r rS+ 0, and = 0 belongs to the interval(3.86), thus the static observer (which has = 0) is allowed, while

for r < rS+ > 0 and the static observer is not allowed.

3.6 The singularity of the Kerr metric

Let us consider the curvature singularity

=r2 +a2 cos2 = 0 . (3.89)

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If we interpret the Boyer-Lindquist coordinates t, r, , as sphericalpolar coordinates, like in Schwarzschild spacetime, it is not clear atall whereis the singularity: = 0 at r = 0, =/2, not at r = 0,= /2, but this has no meaning in polar coordinates; we need acoordinate system which has not the coordinate singularity r = 0,so that we can distinguish and analyze the curvature singularity.

3.6.1 The Kerr-Schild coordinates

In order to understand the singularity structure, we now change

coordinate frame, to the so-called Kerr-Schild coordinates, whichare well defined in r = 0. Let us start with the metric in Kerrcoordinates (t, r, ,), given in eq. (3.58):

ds2 = dt2 +dr2 + d2 + (r2 +a2)sin2 d2 2a sin2 drd+

2Mr

(dt+dr a sin2 d)2 . (3.90)

The Kerr-Schild coordinates (t ,x,y ,z ) are defined by

x =

r2 +a2 sin cos+ arctan

a

r

y = r2

+a2

sin sin+ arctan

a

r

z = r cos . (3.91)

In the next section we will derive the form of the metric in Kerr-Schild coordinates, showing that the coordinate singularity r = 0can be removed in this frame; therefore, in this frame we only havethe curvature singularity; to understand the structure of the curva-ture singularity, then, we must consider it in this frame.

We have

x2 +y2 = (r2 +a2)sin2

z2 = r2 cos2 (3.92)

thusx2 +y2

r2 +a2+

z2

r2 = 1 , (3.93)

then the surfaces with constant r are ellipsoids (Figure 3.2), and

x2 +y2

a2 sin2 z

2

a2 cos2 = 1 , (3.94)

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then the surfaces with constant are half-hyperboloids (Figure 3.3).In Figures 3.2, 3.3 we have represented the r = const, = const

-

. /

#01

Figure 3.2: r = const ellipsoidal surfaces in the Kerr-Schild frame; the thickline represents the r = 0 disk.

-

. / #01

!

!#$

!#%

!

!%

Figure 3.3: = const half-hyperboloidal surfaces in the Kerr-Schild frame; thethick ring represents the r = 0, = /2 singularity.

surfaces in the Kerr-Schild ( t ,x,y ,z ) frame. This means that x, y ,z are represented as Euclidean coordinates, and r, are considered asfunctions ofx, y ,z .

Notice that if we look at Kerr spacetime where r is sufficientlylarge, the r, coordinates behave like ordinary polar coordinates.But closer to the black hole, their nature changes: r = 0 is not a

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single point but a disk,

x2 +y2 a2 , z= 0 (3.95)and this disk is parametrized by the coordinate . In particular,

r= 0 =

2 (3.96)

corresponds to the ring

x2 +y2 =a2 , z= 0 . (3.97)

This is the structure of the singularity of the Kerr metric: it is aring singularity. Inside the ring, the metric is perfectly regular.

3.6.2 The metric in Kerr-Schild coordinates

By calling = arctan a/r, we have

r2 sin2 = a2 cos2 (3.98)

thus

r2 = (r2 +a2)cos2

a2 = (r2 +a2)sin2 (3.99)

and, rewriting (3.91) as

x = sin

r2 +a2(cos cos sin sin)y = sin

r2 +a2(sin cos+ cos sin)

z = r cos (3.100)

and substituting (3.99) we have

x = sin (r cos a sin)y = sin (r sin+a cos)

z = r cos . (3.101)

Differentiating,

dx = cos (r cos a sin)d+ sin cosdr sin (r sin+a cos)ddy = cos (r sin+a cos)d+ sin sindr+ sin (r cos a sin)ddz = r sin d+ cos dr (3.102)

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thus

dx2 +dy2 +dz2 = dr2 +

r2 sin2 + (r2 +a2)cos2

d2

+(r2 +a2)sin2 d2 2sin2 adrd= dr2 + d2 + (r2 +a2)sin2 d2 2a sin2 drd .

(3.103)

Then, the metric (3.90) is the Minkowski metric plus the term

2Mr

(dt+dr

a sin2 d)2 . (3.104)

Being

= r2 +a2 cos2 = r2 +a2z2

r2 , (3.105)

the factor 2Mr/is easily expressed in Kerr-Schild coordinates:

2Mr

=

2M r3

r4 +a2z2. (3.106)

The one-form dt+dr a sin2 dis more complicate to transform.We will prove that

dt+dr

a sin2 d= dt+

r(xdx+ydy) a(xdy ydx)

r2

+a2

+z dz

r

.

(3.107)First of all, let us express the differentials (3.102) as

dx = cos

sin xd+ sin cosdr yd

dy = cos

sin yd+ sin sindr+xd

dz = r sin d+ cos dr . (3.108)We have

xdx+ydy = cos

sin (x2 +y2)d+ sin (x cos+ysin)dr

= sin cos (r2 +a2)d+ sin2 rdr(3.109)

ydx xdy = (x2 +y2)d+ sin (y cos x sin)dr= (r2 +a2)sin2 d+ sin2 adr

(3.110)

zdx = r2 sin cos d+r cos2 dr (3.111)

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then

(xdx+ydy) r

r2 +a2+ (ydx xdy) a

r2 +a2+

zdz

r

=

r sin cos d+

r2

r2 +a2sin2 dr

+

a sin2 d+ a

2

r2 +a2sin2 dr

+ r sin cos d+ cos

2 d= dr a sin

2

d (3.112)which proves (3.107). The metric in Kerr-Schild coordinates is then

ds2 = dt2 +dx2 +dy2 +dz2

+ 2Mr3

r4 +a2z2

dt+

r(xdx+ydy ) a(xdy ydx)r2 +a2

+z dz

r

2.

(3.113)

Notice that the metric has the form

g=+ Hll (3.114)

with

H 2Mr3r4 +a2z2

(3.115)

and, in Kerr-Schild coordinates,

ldx =

dt+

r(xdx+ydy ) a(xdy ydx)r2 +a2

+z dz

r

(3.116)

while in Kerr coordinates

ldx = dt+dr a sin2 d = dv+a sin2 d (3.117)

thus l is exactly the null vector (3.53), i.e. the generator of theprincipal null geodesics which have been used to define the Kerr

coordinates. The form (3.114), called Kerr-Schild form, has beenthe starting point for Kerr to derive his solution.

3.6.3 Some strange features of the inner region of the Kerr

metric

If we took seriously the Kerr metric and its ring singularity, we findsome really weird features. We should keep in mind that what we are

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going to discuss has no direct link with astrophysical observations,since only the region r > r+ is causally connected to us. Further-more, we are considering an ethernal black hole, and it is unlikelythat these properties apply also to astrophysical Kerr black holes,originating (at finite time) from gravitational collapse. It is only forcompleteness in our discussion on the Kerr metric that I briefly dis-cuss such features, which, although fascinating, should not be takentoo seriously.

Maximal extension of the Kerr metric

As we have discussed above, the Kerr metric in Kerr-Schild coordi-nates is

ds2 = dt2 +dx2 +dy2 +dz2

+ 2Mr3

r4 +a2z2

dt+

r(xdx+ydy) a(xdy ydx)r2 +a2

+z dz

r

2(3.118)

wherer is a function of (t ,x,y ,z ), defined implicitly by

r4

(x2 +y2 +z2

a2)r2

a2z2 = 0 . (3.119)

This metric is not singular inside the ring, i.e. at r = 0, = /2,or, equivalently,

r= 0 , x2 +y2 < a2 . (3.120)

It is singular at the ring r = 0, = /2, i.e. r = 0, x2 +y2 = a2;this ring is a true curvature singularity: indeed the scalar invariantRR

diverges there.We can then extend the spacetime manifold to r = 0, = /2,

removing the coordinate singularity at the interior of the ring. Asdiscussed in the case of Schwarzschild spacetime, we have to ex-tend the spacetime manifold so that the geodesics can be extended

across the ring itself. But this extension cannot simply consist in theinclusion of the hypersurface corresponding to the coordinate sin-gularity (in this case, the interior of the ring), as we did to removethe horizon singularity.

To understand this problem, let us consider an observer fallingto the center of the ring through the = 0 axis; along its geodesic,x = y = 0 and r = z. It arrives at z = r = 0 (which is not

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!#%

!%

#01!#$

!'#(

!#(

!

!

!#%

!%

#01!#$

!'#(

!#(

!

!

# 3 1 # 4 1

Figure 3.4: Removal of the disc singularity. The top of the disc in the spacetimewithr 0 (left panel) is identified with the bottom of the spacetime withr 0(right panel), and viceversa. Crossing the ring, an observer can pass from ther >0 region to the r 0region) avoiding discontinuities in dr/d and in .

a singularity of the spacetime) with = 0 and a finite value ofdz/d = dr/d; then, jumps to and dr/d changes sign. Onecould object that we should not worry aboutr, , sincex, y ,z , which

are the coordinates in this frame, behave regularly; on the otherhand, if we compute the curvature scalars, we find that they arediscontinuous as we pass through the ring. Therefore, we have notreally removed the coordinate singularity inside the ring.

Let us consider, now, the equation (3.119) forr: it admits two realsolutions forr(there are other two, but they are complex conjugate),one positive and one negative. Therefore, for each set of Kerr-Schildcoordinates there are two different real values of r, with oppositesigns. We have then two different (asymptotically flat) spacetimesdescribed by the metric (3.118), one withr >0 and one withr 0 with the bottom of the disc x2 +y 2 < a2,z= 0 in the spacetime with r


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Actually, the maximal extension of Kerr spacetime is even larger.A detailed study of geodesic completeness would go far beyond theselectures, we only give the final result. Requiring that all (timelikeor null) geodesics which do not hit the curvature singularity canbe extended, forward and backwards, for an infinite amount of theaffine parameter, one finds that it is necessary to patch togetheran infinity of copies of Kerr spacetime, both with r > 0 and withr


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inner horizon can either cross the ring, escaping to the asymptoti-cally flat region r 0, and so on. Such copies of the region Iare causally connected.

On the other hand, we should remind that this scheme only de-scribes an eternal black hole. In the case of a black hole originatingfrom a gravitational collapse this multiplication of spacetimes dis-appears; indeed, our region I cannot receive signals from a regionII, because they should come from t , when the black hole

was not yet born.

Causality violations

To conclude the discussion of the ring singularity of Kerr spacetime,we show another weird feature of the region close to the singularity.

# 0 1

# 4 1

Figure 3.6: Close timelike curve in Kerr spacetime.

Let us consider a curve consisting in a ring just outside thesingular ring, in the spacetime with r


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= 1

r2(r4 +r2a2 +2Mra2) = r2+a2 +

2Ma2

r


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curvature singularity inside the horizon. Black holes form in thegravitational collapse of stars, if they are sufficiently massive.

When a star has collapsed producing a black hole, we can expectthat, after some time, it settles down to a stationary state as a resultof gravitational waves emission. It is then reasonable to considerstationary black holes (i.e. black holes admitting a killing vectorfield which is timelike at r sufficiently large).

There are some remarkable theorems on stationary black holes,derived by S. Hawking, W. Israel, B. Carter, which prove the fol-

lowing: A stationary black hole is axially symmetric.

Any stationary, axially symmetric black hole, without electriccharge, is described by the Kerr solution.

Any stationary, axially symmetric black hole is described by theso-called Kerr-Newman solution, which is a generalizationof Kerr solution with nonvanishing electric charge and nonvan-ishing electromagnetic fields, characterized by the mass M, theangular momentum J, and the charge Q of the black hole.

Furthermore, we remark that any staticblack hole is sphericallysymmetric, and, if it has no electric charge, it is described by theSchwarzschild solution; in presence of electric charge, it is describedby the Reissner-Nordstrom solution, which is the non-rotatinglimit of the Kerr-Newman solution, and is characterized by the massMand the charge Q. We have not considered in these lectures theReissner-Nordstrom and the Kerr-Newman solutions because it iswidely believed that they are not astrophysically relevant; indeed,if an astrophysical black hole has an electric charge, it would likelylose it in a very short timescale, due to the interactions with thesurrounding matter.

We can conclude that a general stationary black hole is character-

ized by three quantities only: the mass M, the angular momentumJ, and the charge Q. All other features of the star which has col-lapsed to the black hole are not features of the final black hole. Thisresult has been summarized with the sentence: A black hole hasno hair. For this reason, the unicity theorems are also called nohair theorems.

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