chapter 13 rotating black holes - university of...

Chapter 13

Rotating Black Holes

The Schwarzschild solution, which is appropriate outsidea spherical, non-spinning mass distribution, was discov-ered in 1916. It was not until 1963 that a solution corre-sponding to spinning black holes was discovered by NewZealander Roy Kerr. This solution leads to the possibleexistence of a family of rotating, deformed black holesthat are calledKerr black holes.

Angular momentum is complicated: In Newtonian grav-ity rotation produces centrifugal effects but does not in-fluence the gravitational field directly. But angular mo-mentum implies rotational energy, so in general relativityrotation of a gravitational field is itself a source for thefield.

307

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 massM.

• Other solutions to the field equations of general relativity permitblack 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 re-striction to spherical symmetry and thus permit the black holeto be characterized by angular momentum in addition to mass(Kerr black holes).

• Rotating black holes are of particular interest in astrophysics be-cause they are thought to power quasars and other active galax-ies, X-ray binaries, and gamma-ray bursts.

• Unlike for Schwarzschild black holes, it is possible to devisemechanisms that permit energy and angular momentum to beextracted from a (classical) rotating black hole

13.1. THE KERR SOLUTION: ROTATING BLACK HOLES 309

13.1.1 The Kerr Metric

The Kerr metric corresponds to the line element

ds2 =−

(

1−2Mrρ2

)

dt2−4Mrasin2θ

ρ2 dϕdt+ρ2

∆dr2+ρ2dθ2

+

(

r2+a2+2Mra2sin2θ

ρ2

)

sin2θdϕ2

with the definitions

a≡ J/M ρ2 ≡ r2+a2cos2θ ∆ ≡ r2−2Mr +a2.

• The coordinates(t, r,θ ,ϕ) are calledBoyer–Lindquist coordi-nates.

• The parametera, termed theKerr parameter,has units of lengthin geometrized units.

• The parameterJ will be interpreted as angular momentum andthe parameterM as the mass for the black hole.


The Kerr solution

ds2 =−

(

1−2Mrρ2

)

dt2−4Mrasin2θ

ρ2 dϕdt+ρ2

∆dr2+ρ2dθ2

+

(

r2+a2+2Mra2sin2θ

ρ2

)

sin2θdϕ2

a≡ J/M ρ2 ≡ r2+a2cos2θ ∆ ≡ r2−2Mr +a2.

has the following properties.

• Vacuum solution:the Kerr metric is a vacuum solution of theEinstein equations, valid in the absence of matter.

• Reduction to Schwarzschild metric:If the black hole is not rotat-ing (a= J/M = 0), the Kerr line element reduces to the Schwarzschildline element.

• Asymptotically flat:The Kerr metric becomes asymptotically flatfor r >> M andr >> a.

• Symmetries:The Kerr metric is independent oft andϕ, implyingthe existence of Killing vectors

ξt = (1,0,0,0) (stationary metric)

ξϕ = (0,0,0,1) (axially symmetric metric).

Unlike the Schwarzschild metric, the Kerr metric has only axialsymmetry.

• The metric has off-diagonal terms

g03 = g30 = −2Mrasin2θ

ρ2 (inertial frame dragging).


For the Kerr metric

ds2 =−

(

1−2Mrρ2

)

dt2−4Mrasin2θ

ρ2 dϕdt+ρ2

∆dr2+ρ2dθ2

+

(

r2+a2+2Mra2sin2θ

ρ2

)

sin2θdϕ2

a≡ J/M ρ2 ≡ r2+a2cos2θ ∆ ≡ r2−2Mr +a2.

• Surfaces of constant Boyer–Lindquist coordinatesr andt do nothave the metric of a 2-sphere.

• Singularity and horizon structure:∆ → 0 at

r± = M±√

M2−a2

andds→ ∞, assuminga≤ M.

1. This is a coordinate singularity.

2. As a→ 0 we find thatr+ → 2M, which coincides with theSchwarzschild coordinate singularity.

3. Thusr+ corresponds to the horizon that makes the Kerr so-lution a black hole.

On the other hand, the limitρ → 0 corresponds to a physical sin-gularity with associated infinite components of spacetime curva-ture, similar to the case for the Schwarzschild solution.


13.1.2 Extreme Kerr Black Holes

• The horizon radius

r± = M±√

M2−a2

exists only fora≤ M.

• Thus, there is a maximum angular momentumJmax for a Kerrblack hole sincea = J/M andamax = M,

Jmax = amaxM = M2.

• Black holes for whichJ = M2 are termedextreme Kerr blackholes.

• Near-extreme black holes may develop in many astrophysicalsituations:

1. Angular momentum transfer through accretion disks in ei-ther binary systems or around supermassive black holes ingalaxy cores tends to spin up the central object.

2. Massive stars collapsing to black holes may have significantinitial angular momentum


13.1.3 Cosmic Censorship

That the horizon exists for a Kerr black hole only underrestricted conditions raises the question of whether a sin-gularity could exist in the absence of a horizon (“nakedsingularity”).

Cosmic Censorship Hypothesis:Nature conspires to “cen-sor” spacetime singularities in that all such singularitiescome with event horizons that render them invisible to theoutside universe.

• No known violations (observationally or theoreti-cally).

• It cannot at this point be derived from any more fun-damental concept and must be viewed as only an hy-pothesis.

• All theoretical attempts to add angular momentum toa Kerr black hole in order to cause it to exceed themaximum angular momentum permissable for exis-tence of the horizon have failed.


13.1.4 The Kerr Horizon

The area of the horizon for a black hole is of consider-able importance because of the area theorem, which statesthat the horizon area of a classical black hole can neverdecrease in any physical process.

• The Kerr horizon corresponds to a constant value ofr = r+.

• Since the metric is stationary, the horizon is also asurface of constantt.

• Settingdr = dt = 0 in the Kerr line element gives theline element for the 2-dimensional horizon,

dσ2 = ρ2+dθ2 +

(2Mr+

ρ+

)2

sin2θdϕ2,

whereρ2+ is defined byρ2 ≡ r2

+ +a2cos2θ

• This is not the line interval of a 2-sphere.

Thus the Kerr horizon has constant Boyer–Lindquist co-ordinater = r+ but it is not spherical.


The metric tensor for the Kerr horizon is

g =

ρ2+ 0

0

(2Mr+

ρ+

)2

sin2θ

The area of the Kerr horizon is then

AK =

∫ 2π

0dϕ∫ π

0

√

detg dθ

= 2Mr+

∫ 2π

0dϕ∫ π

0sinθ dθ

= 8πMr+ = 8πM(

M +√

M2−a2)

.

The horizon area for a Schwarzschild black hole is ob-tained by settinga = 0 in this expression, correspondingto vanishing angular momentum, in which caser+ = 2Mand

AS = 16πM2,

as expected for a spherical horizon of radius2M.


13.1.5 Orbits in the Kerr Metric

• We may take a similar approach as in the Schwarzschild metricto determine the orbits of particles and photons.

• In the Kerr metric are not confined to a plane (only axial, notfullspherical symmetry).

• Nevertheless, we shall consider only motion in the equatorialplane (θ = π

2) to illustrate concepts.

The Kerr line element with that restriction is then

ds2 =−

(

1−2Mr

)

dt2−4Ma

rdϕdt

+r2

∆dr2+

(

r2+a2+2Ma2

r

)

dϕ2.

There are two conserved quantitities,

ε = −ξt ·u ℓ = ξϕ ·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 perunit mass along the symmetry axis.

The corresponding Killing vectors are

ξt = (1,0,0,0) ξϕ = (0,0,0,1).

and the norm of the 4-velocity provides the usual constraint

u·u = −1.


From the Kerr metric

−ε = g00u0+g03u

3 ℓ = g30u0+g33u

3,

and solving foru0 = dt/dτ anduϕ = dϕ/dτ gives

dtdτ

=1∆

[(

r2+a2+2Ma2

r

)

ε −2Ma

rℓ

]

dϕdτ

=1∆

[(

1−2Mr

)

ℓ+2Ma

rε]

.

By similar steps as for the Schwarzschild metric, this leadsto theequations of motion

ε2−12

=12

(drdτ

)2

+Veff(r,ε, ℓ)

Veff(r,ε, ℓ) ≡−Mr

+ℓ2−a2(ε2−1)

2r2 −M(ℓ−aε)2

r3 .


13.1.6 Frame Dragging

A striking feature of the Kerr solution isframe dragging:loosely, theblack hole drags spacetime with it as it rotates.

• This arises ultimately because the Kerr metric contains off-diagonalcomponentsg03 = g30.

• One consequence is that a particle dropped radially onto a Kerrblack hole will acquire non-radial components of motion as itfalls freely in the gravitational field.

Let’s investigate by calculatingdϕ/dr for a particle dropped from rest(ε = 1) with zero angular momentum (ℓ = 0) onto a Kerr black hole.Sinceε andℓ are conserved, from earlier equations,

dϕdτ

=1∆

[(

1−2Mr

)

ℓ+2Ma

rε]

−→dϕdτ

=1∆

(2Ma

r

)

,

ε2−12

=12

(drdτ

)2

+Veff(r,ε, ℓ) −→drdτ

=

√

2Mr

(

1+a2

r2

)

,

which may be combined to give

dϕdr

=dϕ/dτdr/dτ

= −2Mar∆

[2Mr

(

1+a2

r2

)]−1/2

.

The particle is dragged in angleϕ as it falls radially in-ward, even though no forces act on it.


This effect is often termed “dragging of inertial frames” and producesa detectable gyroscopic precession termed theLense–Thirring effect.

NASA’s Gravity Probe B has 4 gyroscopes aboard a satel-lite in an orbit almost directly over the poles. It is lookingfor gyroscopic precession giving direct evidence of framedragging by the rotating gravitational field of the Earth.

In the Kerr metric

pϕ ≡ p3 = g3µ pµ = g33p3+g30p0,

pt ≡ p0 = g0µ pµ = g00p0+g03p3,

p0 = mdtdτ

p3 = mdϕdτ

,

and combining these relations gives an expression fordϕ/dt,

dϕdt

=p3

p0 =g33p3+g30p0

g00p0+g03p3.

If we consider a zero angular momentum particle, thenp3 = 0 and

ω(r,θ) ≡dϕdt

=g30

g00 =gϕt

gtt .

ω(r,θ) is termed theangular velocity of a zero angularmomentum particle; it measures the frame dragging.


The Kerr line element defines the covariant components of theKerrmetric gµν . To evaluateω(r,θ) in we needgµν , which may be ob-tained as the matrix inverse ofgµν . The metric is diagonal inr andθso

grr = g−1rr =

∆ρ2 gθθ = g−1

θθ =1

ρ2,

and we need only evaluate

g−1 =

(

gtt gtϕ

gϕt gϕϕ

)−1

to obtain the other non-zero entries forgµν . Letting

D = detg = gttgϕϕ − (gtϕ)2,

the matrix inverse is

g−1 =1D

(

gϕϕ −gtϕ

−gϕt gtt

)

.

Inserting explicit expressions for thegµν and carrying out some alge-bra yields

ω(r,θ) =gϕt

gtt =2Mra

(r2+a2)2−a2∆sin2θ.


13.1.7 The Ergosphere

• With capable propulsion, an observer could remain stationaryat any point outside the event horizon of a Schwarzschild blackhole.

• No amount of propulsion can enable an observer to remain sta-tionary inside a Schwarzschild horizon (causality violation).

• We now demonstrate that, for a rotating black hole, even out-side the horizon it may be impossible for an observer to remainstationary.


For a stationary observer

uµobs= (u0

obs,0,0,0) =

(dtdτ

,0,0,0

)

.

Writing the normalizationuobs·uobs = −1 for the Kerr metric gives

uobs·uobs = g00(u0obs)

2 = −1

But

g00=−

(

1−2Mrρ2

)

=−

(

1−2Mr

r2+a2cos2θ

)

=−

(r2+a2cos2θ −2Mr

r2+a2cos2θ

)

vanishes ifr2−2Mr +a2cos2θ = 0

Generally then, solving the above quadratic forr

• g00 = 0 on the surface defined by

re(θ) = M +√

M2−a2cos2θ ,

• g00 > 0 inside this surface.

• g00 < 0 outside this surface.

Therefore, since(u0obs)

2 must be positive, the condition

uobs·uobs = g00(u0obs)

2 = −1

cannot be satisfied for any observer havingr ≤ re(θ).


r+

Cutaway "Side View"

Ergosphere

Black HoleBlack Hole

re(θ)

"Top View"

θ

Horizon

g00

> 0

g00

> 0g

00 < 0

Figure 13.1:The ergosphere of a Kerr black hole. Note that the horizon lies at aconstant value ofr but it is not a spherical surface.

Comparing

r± = M±√

M2−a2 re(θ) = M +√

M2−a2cos2θ ,

we see that

• If a 6= 0 the surfacere(θ) generally lies outside the horizonr+,except at the poles, where the two surfaces are coincident (Fig-ure 13.1).

• If a = 0, the Kerr black hole reduces to a Schwarzschild blackhole, in which caser+ andre(θ) define the same surface.

The region lying betweenre(θ) and the horizonr+ istermed theergosphere, for reasons that will become moreapparent shortly.


From preceding considerations, there can be no stationaryobservers within the ergosphere. Further insight comesfrom motion of photons within the ergosphere.

• Assume photons within the ergosphere moving tangent to a cir-cle at constantr in the equatorial planeθ = π

2 , sodr = dθ = 0.

• Since they are photons,ds2 = 0 and from the Kerr line elementwith dr = dθ = 0,

g00dt2+2g03dtdϕ +g33dϕ2 = 0.

• Divide both sides bydt2 to give a quadratic equation indϕ/dt

g00+2g03dϕdt

+g33

(dϕdt

)2

= 0

Thus

dϕdt

=−2g03±

√

4g203−4g33g00

2g33

=−g03

g33±

√(

g03

g33

)2

−g00

g33,

where

··· + is associated with motion opposite black hole rotation.

··· − is associated with motion with black hole rotation.


• Now g00 vanishes at the boundary of the ergosphere and it ispositive inside the boundary.

• Settingg00 = 0 in

dϕdt

= −g03

g33±

√(

g03

g33

)2

−g00

g33,

gives two solutions,

dϕdt

= 0︸︷︷︸

opposite rotation

dϕdt

= −2g03

g33︸︷︷︸

with rotation

,

The photon sent backwards against the holerotation at the surface of the ergosphere is sta-tionary inϕ!

• Obviously a particle, which must have a velocity less than apho-ton, must rotate with the black hole irrespective of the amountof angular momentum that it has.

• Inside the ergosphereg00 > 0, so all photons and particles mustrotate with the hole.

• In essence, the frame dragging forr < re(θ) is so severe thatspeeds in excess of light would be required for an observer toremain at rest with respect to infinity.


r+

Cutaway "Side View"

Ergosphere

Black HoleBlack Hole

re(θ)

"Top View"

θ

Horizon

g00

> 0

g00

> 0g

00 < 0

• In the limit a = J/M → 0, where the angular momentum of theKerr black hole vanishes,

re(θ) = M +√

M2−a2cos2θ → 2M

and becomes coincident with the Schwarzschild horizon.

• Rotation has extended the regionr < 2M of the spherical blackhole where no stationary observers can exist to a larger regionr < re(θ) surrounding the rotating black hole where no observercan remain at rest because of frame dragging effects.

• This ergospheric region liesoutsidethe horizon, implying that aparticle could enter it and still escape from the black hole.


13.1.8 Extraction of Rotational Energy from Black Holes

We have seen that quantum vacuum fluctuations allowmass to be extracted from a Schwarzschild black hole asHawking radiation, but it is impossible to extract massfrom a classical Schwarzschild black hole (it all lieswithin the event horizon).

• However, the existence of separate surfaces defining the ergo-sphere and the horizon for a Kerr black hole implies the possi-bility of extracting rotational energy from aclassicalblack hole.

• The simplest way to demonstrate the feasibility of extracting ro-tational energy from a black hole is through aPenrose process.


E0

E1

E2

Black

Hole

Ergosphere

g00

> 0

g00

< 0

Figure 13.2:A Penrose process.

A Penrose process is illustrated schematically in Fig. 13.2.

• A particle falls into the ergosphere of a Kerr black hole andde-cays into two particles. One falls through the horizon; one exitsthe ergosphere and escapes to infinity.

• The decay within the ergosphere is a local process. By equiva-lence principle arguments, it may be analyzed in a freely fallingframe according to the usual rules of scattering theory.

• Therefore, energy and momentum are conserved in the decay,implying that in terms of 4-momentap

p0 = p2 + p1

at the point of decay (subscripts label particles, not components).

• If the particle scattering to infinity has rest massm2, its energy is

E2 = −p2·ξt = m2ε ξt = (1,0,0,0)

whereε = −ξt ·u = −ξt ·(p/m) has been used.


E0

E1

E2

Black

Hole

Ergosphere

g00

> 0

g00

< 0

• Taking the scalar product ofp0 = p2+ p1 with ξt

p0 ·ξt = (p2 + p1) ·ξt

yields the requirement,

E2 = E0−E1.

sinceE = −ξt ·p.

• If particle 1 were to reach∞ instead of crossing the event horizonof the black hole,E1 would necessarily be positive so

1. E2 < E0

2. Less energy would be emitted than put in.


E0

E1

E2

Black

Hole

Ergosphere

g00

> 0

g00

< 0

However . . .

• For the Killing vectorξt = (1,0,0,0),

ξt ·ξt = gµνξ µt ξ ν

t

= g00ξ 0t ξ 0

t = g00 = −

(

1−2Mrρ2

)

,

andg00 vanishes on the surfacere(θ) and is positive inside it.

• Therefore,within the ergosphereξt is a spacelike vector:

ξt ·ξt = g00 > 0 (within the ergosphere).

• By arguments similar to those concerning Hawking radiation,

1. −E1 is not an energy within the ergosphere but is a compo-nent of spatial momentum, which can have either apositiveor negative value (!).

2. For those decays where the trajectories are arranged suchthat E1 < 0, we obtain fromE2 = E0 −E1 that E2 > E0 andnet energy is extracted in the Penrose process.


The energy extracted in the Penrose process comes at theexpense of the rotational energy of the Kerr black hole.

• For trajectories withE1 < 0, the captured particleadds a negative angular momentum, thus reducingthe angular momentum and total energy of the blackhole to just balance the angular momentum and totalenergy carried away by the escaping particle.

• A series of Penrose events could extract all the an-gular momentum of a Kerr black hole, leaving aSchwarzschild black hole.

• No further energy can then be extracted from theresulting spherical black hole (except by quantumHawking processes).


The Penrose mechanism establishes proof of principle ina simple model that the rotational energy of Kerr blackholes is externally accessible in classical processes.

• Practically, Penrose processes are not likely to be im-portant in astrophysics because the required condi-tions are not easily realized.

• Instead, the primary sources of emitted energy fromblack hole systems likely come from

1. Complex electromagnetic coupling of rotatingblack holes to external accretion disks and jets.

2. Gravitational energy released by accretion ontoblack holes.

There is very strong observational evidence for such pro-cesses in a variety of astrophysical environments.

chapter 13 rotating black holes - university of...

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