P.K.Townsend - Black Holes
.pdfCausal 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.
= =2
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r. |
i0
= 0
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.
81
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.....
ring-singularity
at r = 0
i
The spacetime is unphysical for another reason. Consider the norm of the Killing vector eld m = @=@ :
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m2 |
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Let r=a = (small) and consider = =2 + . Then
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0 for su ciently small negative |
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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
82
singularities. To see this we de ne new coordinates v and by |
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(4.24) |
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This yields the Kerr solution in Kerr coordinates (v; r; ; ) which are |
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analogous to ingoing EF for Schwarzschild: |
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dv2 + 2dv dr |
2a sin2 r + a |
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2a sin2 d dr + |
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a2 sin2 i |
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This metric is non singular when = 0, i.e. when r = r+ or r = r . |
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Proposition |
The hypersurfaces r = r are Killing horizons of the Killing vector elds |
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with surface gravities |
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(4.27) |
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Proof Let N be the hypersurfaces r = r . The normals are |
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(Exercise)(4.29 |
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/ gvv + r2 + a2 gv + (r2 + a2)2 g =0 = 0 |
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83
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.
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CTCs ..... |
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. ..
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future event horizon at r = r+
4.2.1Angular Velocity of the Horizon
The event horizon is a Killing horizon of
= k + H m |
(4.31) |
84
where |
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H = |
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(4.32) |
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r+2 + a2 |
2M hM2 + p |
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M4 J2 |
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In coordinates for which k = @=@t and m = @=@ we have that |
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@ ( H t) = 0 |
(4.33) |
i.e. = H t+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)22r2 > 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 |
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(since 2 = 0 on |
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(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.
85
Consistency checks (See Question III.3)
2 = 0 implies that
k2 + 2 H m k m2 H = 0; on N
But k = 0 implies that
k2 + H m k = 0; on N
Consistency requires
D (k m)2 |
2 |
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For Kerr, D = sin |
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(4.37)
(4.38)
(4.39)
(4.40)
(4.41)
(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 + H m; H = |
a |
(4.43) |
2M |
The CP diagram is
86
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future event horizon
(and future Cauchy horizon) at r = M
i0
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.
87
4.3The Ergosphere
Although k is timelike at 1 it need not be timelike everywhere outside the horizon. For Kerr,
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so k is timelike provided that |
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r2 + a2 cos2 2Mr > 0 |
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(4.46) |
(or r < M M2 a2 cos2 , but this is not physically relevant).
The boundary of this region, i.e. the hypersurface p
r = M + M2 a2 cos2 |
(4.47) |
is the ergosphere. The ergosphere intersects the event horizon at = 0; , but it lies outside the horizon for other values of . Thus, k can become spacelike in a region outside the event horizon. This is called the ergoregion.
ergosphere .
event . horizon
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= =2
4.4The Penrose Process
Suppose that a particle approaches a Kerr black hole along a geodesic. If p is its 4-momentum we can identify the constant of the motion
E = p k |
(4.48) |
88
as its energy (since E = p0 at 1). Now suppose that the particle decays into two others, one of which falls into the hole while the other escapes to
1.
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By conservation of energy
E2 = E E1 |
(4.49) |
Normally E1 > 0 so E2 < E, but in this case
E1 = p1 k |
(4.50) |
which is not necessarily positive in the ergoregion since k may be spacelike there. Thus, if the decay takes place in the ergoregion we may have E2 > E, so energy has been extracted from the black hole.
4.4.1Limits to Energy Extraction
For particles passing through the horizon at r = r+ we have
p 0 |
(4.51) |
Since is future-directed null on the horizon and p is future-directed timelike or null. Since = k + H m,
E H L 0 |
(4.52) |
89
where L = p m is the component of the particle's angular momentum in the direction de ned by m (only this component is a constant of the motion). Thus
L |
E |
(4.53) |
H |
If E is negative, as it is for particle 1 in the Penrose process then L is also negative, so the hole's angular momentum is reduced. We end up with a hole of mass M + M and angular momentum J + J where M = E andJ = L so
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(4.55) |
(This quantity must increase in the Penrose process).
Lemma A = 8 hM2 + pM4 J2i is the `area of the event horizon', of a
Kerr black hole (i.e. area of intersection of H+ with partial Cauchy surface, e.g. area of v = constant, r = r+ in Kerr coordinates (See Question III.5).
Corollary Energy extraction by Penrose process is limited by the requirement that A 0. This is a special case of the second law of black hole mechanics.
4.4.2Super-radiance
The Penrose process has a close analogue in the scattering of radiation by a Kerr black hole. For simplicity, consider a massless scalar eld . Its stress tensor is
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90