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4.2 The Kerr Solution

Back

This is obtained from KN by setting e = 0. Then

_ = r2 􀀀 2Mr + a2 (4.10)

(_ = r2 + a2 cos2 _) (4.11)

The Kerr metric is important astrophysically since it is a good approximation

to the metric of a rotating star at large distances where all multipole

moments except l = 0 and l = 1 are unimportant. The only known solution

of Einstein's equations for which Kerr is exact for r > R is the Kerr solution

itself (for which T__ = 0), i.e. it has not been matched to any known

non-vacuum solution that could represent the interior of a star, in contrast

to the Schwarzschild solution which is guaranteed by Birkho_'s theorem to

be the exact exterior spacetime that matches on to the interior solution for

any spherically symmetric star.

The Kerr metric in BL coordinates has coordinate singularities at

(a) _ = 0 (i.e on axis of symmetry)

(b) _ = 0

Write

_ = (r 􀀀 r+) (r 􀀀 r􀀀) (4.12)

where

r_ = M _

M2 􀀀 a2 (4.13)

There are 3 cases to consider

(i) M2 < a2: r_ are complex, so _ has no real zeroes, and there are

no coordinate singularities there. The metric still has a coordinate

singularity at _ = 0. More signi_cantly, it has a curvature singularity

at _ = 0, i.e.

r = 0; _ = _=2 (4.14)

The nature of this singularity is best seen in Kerr-Schild coordinates

(~t; x; y; z) (which also removes the coordinate singularity at _ = 0).

These are de_ned by

x + iy = (r + ia) sin _ exp

Z _

d_ +

(4.15)

z = r cos _ (4.16)

~t =

Z _

dt +

r2 + a2

􀀀 r (4.17)

which implies that r = r(x; y; z) is given implicitly by

r4 􀀀

􀀀

x2 + y2 + z2 􀀀 a2_

r2 􀀀 a2z2 = 0 (4.18)

In these coordinates the metric is

ds2 = 􀀀d~t2 + dx2 + dy2 + dz2 (4.19)

2Mr3

r4 + a2z2

r(x dx+ y dy) 􀀀 a(x dy 􀀀 y dx)

r2 + a2 +

zdz

+ d~t

which shows that the spacetime is at (Minkowski) when M = 0.

The surfaces of constant ~t; r are confocal ellipsoids which degenerate

at r = 0 to the disc z = 0; x2 + y2 _ a2.

.......................... .... .... ... ..... .. .. .. ..... ... .... .. ....... ....... .. .. ... .. ... .. ... ... .. . .. . .. ... .. ... . .. . .. .. . .. ... . .. . .. . .. . .. ... .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. ... . .. . .... . .. ... . .. ... . .. .. ... . .. . .. . .... ... ..... ..... ...... .. .. ........ .. ...... .. .... .. .. .. .. ..... ... ... ... .... ......................................................................................................................................................................................................................................................................................... .................................................. ...... ..... .... ... ... ... ... ... .. ... .. ..... .. .... .... .... .. .. .. ...... .. ... .. ..... .. .. .. ........ .. .. . .. ..... ... .. ... ... .. . .. . .... ... . .. . ... .. ... .. ... . .. ... . .. . .. .. . .. . ... .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. .. ... . .. . .. ... . .. . .. . .. . .. .. . .. . ... .. ... ... ... .. ... .. . .. . .. ... .. . .. .. . .. .. ...... ....... ....... .. ....... .... .. .... .. .... .... ..... .. .. .. ..... ... ... ... ... ... .... .... ..... ..............................................................................................................................................................................................................................................................

.................................................................. ................................................................... ........ ..... ..... .... .... ... ... ... ... ... .. ... .. ... .. .. ... .... .. ... .... .... .... .. .... ... .. ...... .. ... .. .. ... .. .. ........ .. .. . .. .. ... .. ... .. ... ... .. ... ... .. . ... ... .. .. . .. . ... .. ... .. ... . ... .. . ... .. .. . .. . .. . .. . .. . .. . .. . .. . .. . ... .. . .. . .. . .. . .. .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .... . .. ... . ... .. . .. . .... . .. . ... .... . .. . .. . .. ... ... ..... ..... ..... .. ... .. .. ...... .. .. .. ....... ..... .. .. ... .. .... .. .... .... .. .. .. .. ... .. .. .. ... .. .. ..... ... ... ... .. .... ... ... .... ..... ..... ....... .............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. ................................................................................................................................................................................................................................................................................................................................................................................................... ................. .... .. . .. .... ... .

..................................... .. .... ... .... ................................................................................................... .... .... ....

.. .. .. .. .. ... .. .. ....................

.......................

....................................

........ . . . ... ..

........................................................................................................................................................ ..........

.......... .......... ........ ....... ........ .......... ..........

...............

...

.....................................

...... . . . . . . .. . . . . . . . . .

..................... .... .. ... .. .. .. .

..........................................................................

.....................................

disc

(y axis suppressed)

􀀀a a

r = constant

r decreasing

x2 + y2 _ a2

z = 0

r = 0

_ = _=2 corresponds to the boundary of the disc at x2 + y2 = a2 so

the curvature singularity occurs on the boundary of the disc, i.e. on

the `ring'

x2 + y2 = a2; z = 0 (4.20)

There is no reason to restrict r to be positive. The spacetime can be

analytically continued through the disc to another asymptotically at

region with r < 0.

Causal structure

Because we now have only axial symmetry we really need a 3-dim

spacetime diagram to encode the causal structure, but the _ = 0; _=2

submanifolds are totally-geodesic, i.e. a geodesic that is initially tangent

to the submanifold remains tangent to it, so we can draw 2-dim

CP diagrams for them.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.......... . . . . . .......................................

.. . .. . ... . .. . .. . ....................

. . . . . . . . . . . . . . .. ...............

=􀀀

i􀀀

naked

singularity

at r = 0

(boundary of disk)

i􀀀

=􀀀

_ = 0

r < 0 r > 0

r = 0

_ = _=2

For _ = _=2 each point in the diagram represents a circle (0 _ _ < 2_).

Each ingoing radial geodesic hits the ring singularity at r = 0, which

is clearly naked. For _ = 0 we are considering only geodesics on the

axis of symmetry. Ingoing radial null geodesics pass through the disc

at r = 0 into the other region with r < 0. We can summarize both

diagrams by the single one.

ring-singularity

at r = 0

..........

...... ...... ....................... .... .. . . . . . . . . . . . . .. . . . . . . . .

i􀀀

=􀀀

r < 0 r > 0

The spacetime is unphysical for another reason. Consider the norm of

the Killing vector _eld m = @=@_:

m2 = g__ = a2 sin2 _

1 +

Ma2

2 sin4 _

1 + a2

r2 cos2 _

(4.21)

Let r=a = _ (small) and consider _ = _=2 + _. Then

m2 = a2 +

+ O(_); for _ _ 1 (4.22)

< 0 for su_ciently small negative _

So m becomes timelike near the ring-singularity on the r < 0 branch.

But the orbits of m are closed, so the spacetime admits closed timelike

curves (CTCs). This constitute a global violation of causality.

Moreover because of the absence of a horizon these CTCs may be

deformed to pass through any point of the spacetime (Carter). They

also miss the singularity by a distance _ M, for M _ a, and M can

be arbitrarily large. Since the ring singularity would be naked for

M2 < a2, then even if the white hole region is replaced by a collapsing

star, we can invoke cosmic censorship to rule out M2 < a2.

(ii) M2 > a2. We still have a ring-singularity but now the metric (in BL

coordinates) is singular at r = r+ and r = r􀀀. These are coordinate

singularities. To see this we de_ne new coordinates v and _ by

dv = dt +

􀀀

r2 􀀀 a2

dr (4.23)

d_ = d_ +

dr (4.24)

This yields the Kerr solution in Kerr coordinates (v; r; _; _) which are

analogous to ingoing EF for Schwarzschild:

ds2 = 􀀀

􀀀

_ 􀀀 a2 sin2 _

dv2 + 2dv dr 􀀀

2a sin2 _

􀀀

r2 + a2 􀀀 _

dv d_

􀀀2a sin2 _d_ dr +

h􀀀

r2 + a2_2

􀀀 _a2 sin2 _

sin2 _d_2 + _d_2

(4.25)

This metric is non singular when _ = 0, i.e. when r = r+ or r = r􀀀.

Proposition The hypersurfaces r = r_ are Killing horizons of the Killing vector _elds

__ = k +

r2_ + a2

m (4.26)

with surface gravities

__ =

r_ 􀀀 r_

􀀀

r2_ + a2

_ (4.27)

Proof Let N_ be the hypersurfaces r = r_. The normals are

l_ = f_g_rjN_

@_; for some non-zero functions f_ (4.28)

= 􀀀

r2_ + a2

r2_ + a2 cos2 _

r2_ + a2

| {z }

(Exe(r4c.i2s9e)

First

l2_ /

gvv +

r2 + a2 gv_ +

(r2 + a2)2 g__

_____

_=0

= 0 (4.30)

so N_ are null hypersurfaces. Since __jN_ / l_, they are Killing horizons

of __. It remains to compute __D__

_. This gives the result for

__ (Exercise).

This result can be used to _nd KS type coordinates that cover 4 regions

around a BK axis of each Killing horizon, and the _ = 0 and _ = _=2

CP diagram of the maximal analytic extension of M2 > a2 Kerr can

be found. Note that the diagram can be extended in_nitely in both

time directions.

ring

singularity

r = 0

􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 ��􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀 􀀀􀀀􀀀

...........

..........

...........

..........

.......... ...........

..........

...........

..........

..... .....

.....

.. .. .

.....................................

.. .. .. . . . . . . .. . . .

.................................. . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . .

..... ... .. ...... .. ...................

. .. .. .. .. . . . . . . . . . . ............

..................................... . . . . . . . . . . . . . . . . . .. .. .. .. .. .

..................... .. .. .. . .. .. .. ...

.. .. .. .. . . . . . . . .. . . .

.....................................

.. .. .. .. .. .. . . . . . .. . . .

.... .... .... ....... ... . .... . .. ..... ... ...............................................................

.................... .. .. ... .. .... .. ..

.. .... .... .... .... .... .... .... .

...............................

.. .... .... .... .... .... .... .... .

...............................

r = 􀀀1

=􀀀

II i+

cauchy horizon at r = r􀀀

future event horizon

at r = r+

III

internal 1 H􀀀

r > r+

r < 0

I =r < r + +H+

CTCs

4.2.1 Angular Velocity of the Horizon

The event horizon is a Killing horizon of

_ = k + Hm (4.31)

where

H =

r2+

+ a2 =

M2 + pM4 􀀀 J2

i (4.32)

In coordinates for which k = @=@t and m = @=@_ we have that

__@_ (_ 􀀀 Ht) = 0 (4.33)

i.e. _ = Ht+constant, on orbits of _, whereas _ is constant on orbits

of k. Note that k is unique. Consider

(k + _m)2 = gtt + 2_gt_ + _2g__ (4.34)

As long as gt_ is _nite and g__ _ r2 as r ! 1, we have (k + _m)2 _

_2r2 > 0 (if _ 6= 0) as r ! 1. So there can be only one Killing vector

k that is timelike at 1 and normalized s.t. k2 ! 􀀀1 as r ! 1.

Thus particles on orbits of _ rotate with angular velocity H relative to

static particles, those on orbits of k, and hence relative to a stationary

frame at 1. Since the null geodesic generators of the horizon follow

orbits of _ the black hole is rotating with angular velocity H.

Lemma _ _ k = 0 on a Killing horizon, N, of _.

Proof

_ _ kjN = _2

N 􀀀 H_ _mjN (4.35)

= 􀀀 H_ _mjN (since _2 = 0 on N) (4.36)

Now, N is a _xed point set of m, since m is Killing (Choose coordinates

s.t. m = @=@_. The metric is _ independent, so the position of the

horizon is independent of _). So m must be tangent to N or l _m = 0

where l is normal to N. But _ / l on N, so _ _mjN = 0. Hence result.

Consistency checks (See Question III.3)

_2 = 0 implies that

k2 + 2Hm_ k 􀀀m2H = 0; on N (4.37)

But _ _ k = 0 implies that

k2 + Hm_ k = 0; on N (4.38)

Consistency requires

D _ (k _ m)2 􀀀 k2m2

= 0 (4.39)

For Kerr, D = _sin2 _ = 0 on N 􀀀 .

Also

H = 􀀀

m _ k

= 􀀀

gtt

gt_

____

in BL coordinates (4.40)

= 􀀀a2 sin2 _

􀀀2a sin2 _

􀀀

r2+

+ a2

_ (4.41)

r2+

+ a2

􀀀 : (4.42)

(iii) M2 = a2 Extreme Kerr

In this case we have a degenerate (_ = 0) Killing horizon at r = M of

the Killing vector _eld

_ = k + Hm; H =

(4.43)

The CP diagram is

internal

future event horizon

(and future Cauchy horizon)

at r = M

ring

singularity

at r = 0

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ................................................................................................................................................................................................................................................................

.... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... ...... ...... ..... ...... .............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...... ..... ...... ...

.......... .

......... .

......... .......... .

......... .

􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀 􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀􀀀

................ . . . . . . . . . . . . . . . . ..................... . . .. .. .. .. .. .. .

............... .. ..... .. .. .... ..

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

. . . . . . ........................................ . . . .. .. .. .. .. .. .

....................................... . . . . . . . . . . . . . . . . . . ...............

=􀀀

r = 􀀀1

H+

r > M

r < 0

r > 0

r < M

So there can be only one Killing vector k for which k _ k ! 􀀀1 as

r ! 1.

N.B. If you change the sign of r in the Kerr metric this e_ectively changes

the sign of M.

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