Chapter 13. Rotating Black Holes


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1 Chapter 13 Rotating Black Holes The Schwarzschild solution, which is appropriate outside a spherical, nonspinning mass distribution, was discovered in It was not until 1963 that a solution corresponding to spinning black holes was discovered by New Zealander Roy Kerr. This solution leads to the possible existence of a family of rotating, deformed black holes that are called Kerr black holes. Angular momentum is complicated: In Newtonian gravity rotation produces centrifugal effects but does not influence the gravitational field directly. But angular momentum implies rotational energy, so in general relativity rotation of a gravitational field is itself a source for the field. 307
2 308 CHAPTER 13. ROTATING BLACK HOLES 13.1 The Kerr Solution: Rotating Black Holes The Schwarzschild solution corresponds to the simplest black hole, which can be characterized by a single parameter, the mass M. Other solutions to the field equations of general relativity permit black holes with more degrees of freedom. The most general black hole can possess 1. Mass 2. Charge 3. Angular momentum as distinguishing quantities. Of particular interest are those solutions where we relax the restriction to spherical symmetry and thus permit the black hole to be characterized by angular momentum in addition to mass (Kerr black holes). Rotating black holes are of particular interest in astrophysics because they are thought to power quasars and other active galaxies, Xray binaries, and gammaray bursts. Unlike for Schwarzschild black holes, it is possible to devise mechanisms that permit energy and angular momentum to be extracted from a (classical) rotating black hole
3 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES The Kerr Metric The Kerr metric corresponds to the line element ( ds 2 = 1 2Mr ) dt 2 4Mrasin2 θ ρ 2 ρ 2 ( + r 2 + a 2 + 2Mra2 sin 2 ) θ ρ 2 with the definitions dϕdt + ρ2 dr2 + ρ 2 dθ 2 sin 2 θdϕ 2 a J/M ρ 2 r 2 + a 2 cos 2 θ r 2 2Mr+ a 2. The coordinates (t,r,θ,ϕ) are called Boyer Lindquist coordinates. The parameter a, termed the Kerr parameter, has units of length in geometrized units. The parameter J will be interpreted as angular momentum and the parameter M as the mass for the black hole.
4 310 CHAPTER 13. ROTATING BLACK HOLES The Kerr solution ( ds 2 = 1 2Mr ) dt 2 4Mrasin2 θ ρ 2 ρ 2 ( + r 2 + a 2 + 2Mra2 sin 2 ) θ ρ 2 dϕdt + ρ2 dr2 + ρ 2 dθ 2 sin 2 θdϕ 2 a J/M ρ 2 r 2 + a 2 cos 2 θ r 2 2Mr+ a 2. has the following properties. Vacuum solution: the Kerr metric is a vacuum solution of the Einstein equations, valid in the absence of matter. Reduction to Schwarzschild metric: If the black hole is not rotating (a = J/M = 0), the Kerr line element reduces to the Schwarzschild line element. Asymptotically flat: The Kerr metric becomes asymptotically flat for r >> M and r >> a. Symmetries: The Kerr metric is independent of t and ϕ, implying the existence of Killing vectors ξ t = (1,0,0,0) ξ ϕ = (0,0,0,1) (stationary metric) (axially symmetric metric). Unlike the Schwarzschild metric, the Kerr metric has only axial symmetry. The metric has offdiagonal terms g 03 = g 30 = 2Mrasin2 θ ρ 2 (inertial frame dragging).
5 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES 311 For the Kerr metric ( ds 2 = 1 2Mr ) ρ 2 dt 2 4Mrasin2 θ ρ 2 ( + r 2 + a 2 + 2Mra2 sin 2 ) θ ρ 2 dϕdt + ρ2 dr2 + ρ 2 dθ 2 sin 2 θdϕ 2 a J/M ρ 2 r 2 + a 2 cos 2 θ r 2 2Mr+ a 2. Surfaces of constant Boyer Lindquist coordinates r and t do not have the metric of a 2sphere. Singularity and horizon structure: 0 at r ± = M ± M 2 a 2 and ds, assuming a M. 1. This is a coordinate singularity. 2. As a 0 we find that r + 2M, which coincides with the Schwarzschild coordinate singularity. 3. Thus r + corresponds to the horizon that makes the Kerr solution a black hole. On the other hand, the limit ρ 0 corresponds to a physical singularity with associated infinite components of spacetime curvature, similar to the case for the Schwarzschild solution.
6 312 CHAPTER 13. ROTATING BLACK HOLES Extreme Kerr Black Holes The horizon radius r ± = M ± M 2 a 2 exists only for a M. Thus, there is a maximum angular momentum J max for a Kerr black hole since a = J/M and a max = M, J max = a max M = M 2. Black holes for which J = M 2 are termed extreme Kerr black holes. Nearextreme black holes may develop in many astrophysical situations: 1. Angular momentum transfer through accretion disks in either binary systems or around supermassive black holes in galaxy cores tends to spin up the central object. 2. Massive stars collapsing to black holes may have significant initial angular momentum
7 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES Cosmic Censorship That the horizon exists for a Kerr black hole only under restricted conditions raises the question of whether a singularity could exist in the absence of a horizon ( naked singularity ). Cosmic Censorship Hypothesis: Nature conspires to censor spacetime singularities in that all such singularities come with event horizons that render them invisible to the outside universe. No known violations (observationally or theoretically). It cannot at this point be derived from any more fundamental concept and must be viewed as only an hypothesis. All theoretical attempts to add angular momentum to a Kerr black hole in order to cause it to exceed the maximum angular momentum permissable for existence of the horizon have failed.
8 314 CHAPTER 13. ROTATING BLACK HOLES The Kerr Horizon The area of the horizon for a black hole is of considerable importance because of the area theorem, which states that the horizon area of a classical black hole can never decrease in any physical process. The Kerr horizon corresponds to a constant value of r = r +. Since the metric is stationary, the horizon is also a surface of constant t. Setting dr = dt = 0 in the Kerr line element gives the line element for the 2dimensional horizon, ( ) 2 dσ 2 = ρ+ 2 dθ 2 2Mr+ + sin 2 θdϕ 2, ρ + where ρ 2 + is defined by ρ 2 r 2 ++ a 2 cos 2 θ This is not the line interval of a 2sphere. Thus the Kerr horizon has constant Boyer Lindquist coordinate r = r + but it is not spherical.
9 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES 315 The metric tensor for the Kerr horizon is g = ρ+ 2 0 ( ) 2 2Mr+ 0 sin 2 θ ρ + The area of the Kerr horizon is then A K = 2π 0 dϕ π 0 = 2Mr + 2π 0 = 8πMr + = 8πM detg dθ π dϕ sinθ dθ 0 ( M + M 2 a 2 ). The horizon area for a Schwarzschild black hole is obtained by setting a = 0 in this expression, corresponding to vanishing angular momentum, in which case r + = 2M and A S = 16πM 2, as expected for a spherical horizon of radius 2M.
10 316 CHAPTER 13. ROTATING BLACK HOLES Orbits in the Kerr Metric We may take a similar approach as in the Schwarzschild metric to determine the orbits of particles and photons. In the Kerr metric are not confined to a plane (only axial, not full spherical symmetry). Nevertheless, we shall consider only motion in the equatorial plane (θ = 2 π ) to illustrate concepts. The Kerr line element with that restriction is then ( ds 2 = 1 2M ) dt 2 4Ma dϕdt r r ) + r2 dr2 + (r 2 + a 2 + 2Ma2 dϕ 2. r There are two conserved quantitities, ε = ξ t u l = ξ ϕ u, At large distances from the gravitational source ε may be interpreted as the conserved energy per unit rest mass. l may be interpreted as the angular momentum component per unit mass along the symmetry axis. The corresponding Killing vectors are ξ t = (1,0,0,0) ξ ϕ = (0,0,0,1). and the norm of the 4velocity provides the usual constraint u u = 1.
11 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES 317 From the Kerr metric ε = g 00 u 0 + g 03 u 3 l = g 30 u 0 + g 33 u 3, and solving for u 0 = dt/dτ and u ϕ = dϕ/dτ gives dt dτ = 1 ) [(r 2 + a 2 + 2Ma2 r ε 2Ma ] l r dϕ dτ = 1 [( 1 2M r ) l+ 2Ma ] ε. r By similar steps as for the Schwarzschild metric, this leads to the equations of motion ε = 1 2 ( ) dr 2 +V eff(r,ε,l) dτ V eff (r,ε,l) M r + l2 a 2 (ε 2 1) 2r 2 M(l aε)2 r 3.
12 318 CHAPTER 13. ROTATING BLACK HOLES Frame Dragging A striking feature of the Kerr solution is frame dragging: loosely, the black hole drags spacetime with it as it rotates. This arises ultimately because the Kerr metric contains offdiagonal components g 03 = g 30. One consequence is that a particle dropped radially onto a Kerr black hole will acquire nonradial components of motion as it falls freely in the gravitational field. Let s investigate by calculating dϕ/dr for a particle dropped from rest (ε = 1) with zero angular momentum (l = 0) onto a Kerr black hole. Since ε and l are conserved, from earlier equations, dϕ dτ = 1 [( 1 2M r ε = 1 2 ) l+ 2Ma ] ε r dϕ dτ = 1 ( 2Ma r ), ( ) dr 2 +V eff(r,ε,l) dr ( ) dτ dτ = 2M 1+ a2 r r 2, which may be combined to give dϕ dr = dϕ/dτ [ 2M dr/dτ = 2Ma r r )] 1/2 (1+ a2 r 2. The particle is dragged in angle ϕ as it falls radially inward, even though no forces act on it.
13 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES 319 This effect is often termed dragging of inertial frames and produces a detectable gyroscopic precession termed the Lense Thirring effect. NASA s Gravity Probe B has 4 gyroscopes aboard a satellite in an orbit almost directly over the poles. It is looking for gyroscopic precession giving direct evidence of frame dragging by the rotating gravitational field of the Earth. In the Kerr metric p ϕ p 3 = g 3µ p µ = g 33 p 3 + g 30 p 0, p t p 0 = g 0µ p µ = g 00 p 0 + g 03 p 3, p 0 = m dt p 3 = m dϕ dτ dτ, and combining these relations gives an expression for dϕ/dt, dϕ = p3 dt p 0 = g33 p 3 + g 30 p 0 g 00 p 0 + g 03. p 3 If we consider a zero angular momentum particle, then p 3 = 0 and ω(r,θ) dϕ dt = g30 g 00 = gϕt g tt. ω(r,θ) is termed the angular velocity of a zero angular momentum particle; it measures the frame dragging.
14 320 CHAPTER 13. ROTATING BLACK HOLES The Kerr line element defines the covariant components of the Kerr metric g µν. To evaluate ω(r,θ) in we need g µν, which may be obtained as the matrix inverse of g µν. The metric is diagonal in r and θ so g rr = g 1 rr = ρ 2 g θθ = g 1 θθ = 1 ρ 2, and we need only evaluate g 1 = ( gtt g tϕ g ϕt g ϕϕ ) 1 to obtain the other nonzero entries for g µν. Letting D = detg = g tt g ϕϕ (g tϕ ) 2, the matrix inverse is ( g 1 = 1 gϕϕ g tϕ D g ϕt g tt ). Inserting explicit expressions for the g µν and carrying out some algebra yields ω(r,θ) = gϕt g tt = 2Mra (r 2 + a 2 ) 2 a 2 sin 2 θ.
15 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES The Ergosphere With capable propulsion, an observer could remain stationary at any point outside the event horizon of a Schwarzschild black hole. No amount of propulsion can enable an observer to remain stationary inside a Schwarzschild horizon (causality violation). We now demonstrate that, for a rotating black hole, even outside the horizon it may be impossible for an observer to remain stationary.
16 322 CHAPTER 13. ROTATING BLACK HOLES For a stationary observer ( ) u µ dt obs = (u0 obs,0,0,0) = dτ,0,0,0. Writing the normalization u obs u obs = 1 for the Kerr metric gives u obs u obs = g 00 (u 0 obs )2 = 1 But ( g 00 = 1 2Mr ) ( ) ( 2Mr r 2 + a 2 cos 2 ) θ 2Mr ρ 2 = 1 r 2 + a 2 cos 2 = θ r 2 + a 2 cos 2 θ vanishes if r 2 2Mr+ a 2 cos 2 θ = 0 Generally then, solving the above quadratic for r g 00 = 0 on the surface defined by r e (θ) = M + M 2 a 2 cos 2 θ, g 00 > 0 inside this surface. g 00 < 0 outside this surface. Therefore, since (u 0 obs )2 must be positive, the condition u obs u obs = g 00 (u 0 obs )2 = 1 cannot be satisfied for any observer having r r e (θ).
17 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES 323 Horizon g 00 < 0 g 00 > 0 g 00 > 0 r+ θ re(θ) Black Hole Black Hole Ergosphere Cutaway "Side View" "Top View" Figure 13.1: The ergosphere of a Kerr black hole. Note that the horizon lies at a constant value of r but it is not a spherical surface. Comparing r ± = M ± we see that M 2 a 2 r e (θ) = M + M 2 a 2 cos 2 θ, If a 0 the surface r e (θ) generally lies outside the horizon r +, except at the poles, where the two surfaces are coincident (Figure 13.1). If a = 0, the Kerr black hole reduces to a Schwarzschild black hole, in which case r + and r e (θ) define the same surface. The region lying between r e (θ) and the horizon r + is termed the ergosphere, for reasons that will become more apparent shortly.
18 324 CHAPTER 13. ROTATING BLACK HOLES From preceding considerations, there can be no stationary observers within the ergosphere. Further insight comes from motion of photons within the ergosphere. Assume photons within the ergosphere moving tangent to a circle at constant r in the equatorial plane θ = 2 π, so dr = dθ = 0. Since they are photons, ds 2 = 0 and from the Kerr line element with dr = dθ = 0, g 00 dt 2 + 2g 03 dtdϕ + g 33 dϕ 2 = 0. Divide both sides by dt 2 to give a quadratic equation in dϕ/dt Thus g g 03 dϕ dt + g 33 ( dϕ dt ) 2 = 0 dϕ dt = 2g 03 ± 4g g 33g 00 = g 03 g 33 ± 2g 33 (g03 g 33 ) 2 g 00 g 33, where + is associated with motion opposite black hole rotation. is associated with motion with black hole rotation.
19 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES 325 Now g 00 vanishes at the boundary of the ergosphere and it is positive inside the boundary. Setting g 00 = 0 in gives two solutions, dϕ = g 03 ± dt g 33 (g03 g 33 ) 2 g 00 g 33, dϕ = 0 } dt{{} opposite rotation dϕ = 2g 03, dt g }{{ 33 } with rotation The photon sent backwards against the hole rotation at the surface of the ergosphere is stationary in ϕ! Obviously a particle, which must have a velocity less than a photon, must rotate with the black hole irrespective of the amount of angular momentum that it has. Inside the ergosphere g 00 > 0, so all photons and particles must rotate with the hole. In essence, the frame dragging for r < r e (θ) is so severe that speeds in excess of light would be required for an observer to remain at rest with respect to infinity.
20 326 CHAPTER 13. ROTATING BLACK HOLES Horizon g 00 < 0 g 00 > 0 g 00 > 0 r+ θ re(θ) Black Hole Black Hole Ergosphere Cutaway "Side View" "Top View" In the limit a = J/M 0, where the angular momentum of the Kerr black hole vanishes, r e (θ) = M + M 2 a 2 cos 2 θ 2M and becomes coincident with the Schwarzschild horizon. Rotation has extended the region r < 2M of the spherical black hole where no stationary observers can exist to a larger region r < r e (θ) surrounding the rotating black hole where no observer can remain at rest because of frame dragging effects. This ergospheric region lies outside the horizon, implying that a particle could enter it and still escape from the black hole.
21 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES Extraction of Rotational Energy from Black Holes We have seen that quantum vacuum fluctuations allow mass to be extracted from a Schwarzschild black hole as Hawking radiation, but it is impossible to extract mass from a classical Schwarzschild black hole (it all lies within the event horizon). However, the existence of separate surfaces defining the ergosphere and the horizon for a Kerr black hole implies the possibility of extracting rotational energy from a classical black hole. The simplest way to demonstrate the feasibility of extracting rotational energy from a black hole is through a Penrose process.
22 328 CHAPTER 13. ROTATING BLACK HOLES g 00 < 0 E 2 E 0 E 1 g 00 > 0 Black Hole Ergosphere Figure 13.2: A Penrose process. A Penrose process is illustrated schematically in Fig A particle falls into the ergosphere of a Kerr black hole and decays into two particles. One falls through the horizon; one exits the ergosphere and escapes to infinity. The decay within the ergosphere is a local process. By equivalence principle arguments, it may be analyzed in a freely falling frame according to the usual rules of scattering theory. Therefore, energy and momentum are conserved in the decay, implying that in terms of 4momenta p p 0 = p 2 + p 1 at the point of decay (subscripts label particles, not components). If the particle scattering to infinity has rest mass m 2, its energy is E 2 = p 2 ξ t = m 2 ε ξ t = (1,0,0,0) where ε = ξ t u = ξ t (p/m) has been used.
23 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES 329 g 00 < 0 E 2 E 0 E 1 g 00 > 0 Black Hole Ergosphere Taking the scalar product of p 0 = p 2 + p 1 with ξ t yields the requirement, since E = ξ t p. p 0 ξ t = (p 2 + p 1 ) ξ t E 2 = E 0 E 1. If particle 1 were to reach instead of crossing the event horizon of the black hole, E 1 would necessarily be positive so 1. E 2 < E 0 2. Less energy would be emitted than put in.
24 330 CHAPTER 13. ROTATING BLACK HOLES g 00 < 0 E 2 E 0 E 1 g 00 > 0 Black Hole Ergosphere However... For the Killing vector ξ t = (1,0,0,0), ξ t ξ t = g µν ξ µ t ξ ν t = g 00 ξ 0 t ξ 0 t = g 00 = ( 1 2Mr ) ρ 2, and g 00 vanishes on the surface r e (θ) and is positive inside it. Therefore, within the ergosphere ξ t is a spacelike vector: ξ t ξ t = g 00 > 0 (within the ergosphere). By arguments similar to those concerning Hawking radiation, 1. E 1 is not an energy within the ergosphere but is a component of spatial momentum, which can have either a positive or negative value (!). 2. For those decays where the trajectories are arranged such that E 1 < 0, we obtain from E 2 = E 0 E 1 that E 2 > E 0 and net energy is extracted in the Penrose process.
25 13.1. THE KERR SOLUTION: ROTATING BLACK HOLES 331 The energy extracted in the Penrose process comes at the expense of the rotational energy of the Kerr black hole. For trajectories with E 1 < 0, the captured particle adds a negative angular momentum, thus reducing the angular momentum and total energy of the black hole to just balance the angular momentum and total energy carried away by the escaping particle. A series of Penrose events could extract all the angular momentum of a Kerr black hole, leaving a Schwarzschild black hole. No further energy can then be extracted from the resulting spherical black hole (except by quantum Hawking processes).
26 332 CHAPTER 13. ROTATING BLACK HOLES The Penrose mechanism establishes proof of principle in a simple model that the rotational energy of Kerr black holes is externally accessible in classical processes. Practically, Penrose processes are not likely to be important in astrophysics because the required conditions are not easily realized. Instead, the primary sources of emitted energy from black hole systems likely come from 1. Complex electromagnetic coupling of rotating black holes to external accretion disks and jets. 2. Gravitational energy released by accretion onto black holes. There is very strong observational evidence for such processes in a variety of astrophysical environments.
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