ñòð. 14 |

Ë™ r h k = 2 c 2 k2 âˆ’ 1

12 1

(13.45)

where we have identified the effective potential per unit mass as

c2 h2 âˆ’ a2 c2 k2 âˆ’ 1 h âˆ’ ack 2

r h k =âˆ’ + âˆ’ (13.46)

Veff

2r 2 r3

r

There are several points to note here. First, Veff has the same r-dependence as

the corresponding expression for the Schwarzschild case, derived in Chapter 9;

it is only that the coefficients of the last two terms are more complicated in the

Kerr case. The graph of Veff therefore has the same general shape as those shown

in Figure 9.4. Indeed, as one would expect, in the limit a â†’ 0 equation (13.46)

reduces to the Schwarzschild result. When a = 0, however, one must be careful

in interpreting (13.46) as an effective potential, since it depends on the energy k

333

13.12 Equatorial motion of massive particles with zero angular momentum

of the particle (as well as the usual dependence on the angular momentum h).

Nevertheless, by differentiating (13.45) with respect to , one finds that the radial

acceleration of a particle is still given by r = âˆ’dVeff /dr. Similarly, the stability of

Â¨

circular orbits, for example, may be deduced by considering the sign of d2 Veff /dr 2

in the usual manner. Also, an incoming particle will fall into the black hole only

if the parameters h and k defining its trajectory are such that the maximum value

of Veff r h k exceeds 2 c2 k2 âˆ’ 1 .

1

In our discussion of the Schwarzschild geometry in Chapter 9, in addition to the

energy equation it was reasonably straightforward also to derive the â€˜shapeâ€™ equa-

tion for a general massive particle orbit and, equivalently, a simple expression for

d /dr. Unfortunately, it is algebraically very complicated (and unilluminating)

to obtain the equivalent expressions for the Kerr geometry, even in the case of

equatorial orbits. It is therefore natural to confine our attention to special cases

in which the symmetry of the orbit makes the algebra more manageable. We are

once again unfortunate, however, since the Kerr solution does not admit radial

geodesics (either null or non-null). In a loose sense, the reason is that the rotating

object â€˜dragsâ€™ the surrounding space and the geodesics with it. Nevertheless, it is

still reasonably straightforward to consider motion with zero angular momentum

and motion in a circle.

13.12 Equatorial motion of massive particles with zero angular momentum

For a particle falling into a Kerr black hole whose angular momentum about

the black hole is zero, we have h = 0. Setting h = 0 reduces the complexity of

the geodesic equations (13.40â€“13.44) somewhat. To simplify the equations still

further, however, we will also consider the limit in which the particle starts at

rest from infinity, in which case k = 1. In this case the particle will initially be

moving radially.

Using these values of h and k, the geodesic equations become

2 a2

1

Ë™

t= r +a +

2 2

r

Ë™ = 2 ac

r

2 c2 a2

r=

Ë™ 1+ 2

2

r r

From these expression, we see that both t and Ë™ are infinite at the horizons (when

Ë™

= 0), which is an illustration of the fact that both t and are â€˜bad coordinatesâ€™

in these regions. Interestingly, the singular behaviours of the t and coordinates

Ë™

â€˜cancelâ€™ in the expression for r 2 .

334 The Kerr geometry

The above equations may in turn be used to obtain expressions relating differ-

entials of the coordinates along the particle trajectory. In particular, we find that

âˆ’1

1/2

Ë™ 2 c2 a2 2 a2

dr r

= =âˆ’ 1+ 2 r +a +

2 2

Ë™

dt t r r r

âˆ’1

Ë™ 2 a2

2 ac 2

d

== r +a +

2

Ë™

dt t r r

âˆ’1/2

Ë™ a2

2a 2

d

= =âˆ’ 1+ 2

Ë™

dr r r r r

We note that both dt/dr and d /dr become infinite at the horizons (when

= 0), but d /dt remains finite there. The above equations may be integrated

numerically in a straightforward way to obtain the trajectory of the massive

particle. In Figure 13.4, we plot such a trajectory âˆš the ct r -plane and in the

in

âˆš

x y -plane (where x = r 2 + a2 cos and y = r 2 + a2 sin ) for a particle

= 8 0 at t = 0 in a Kerr geometry with

that passes through the point r

rotation parameter a = 0 8 . In particular, we note that both plotted curves have

30

4

2

20

ct/Âµ

y/Âµ

0

10

â€“2

â€“4

0

2468 â€“4 â€“2 0 2 4

r/Âµ x/Âµ

Figure 13.4 The trajectory of an initially radially moving massive particle falling

from rest at infinity in a Kerr geometry with rotation parameter a = 0 8 . The

trajectory (solid line) is plotted in the ct âˆš -plane (left) and the x y -plane

r

âˆš

(right), where x = r 2 + a2 cos and y = r 2 + a2 sin . The locations of the

horizons (broken lines) and ring singularity (dotted line) are also indicated. The

points correspond to unit intervals of c / , where is the proper time and we

have taken = t = 0 at r = 8 .

335

13.13 Equatorial circular motion of massive particles

discontinuities at the horizons, which are shown as broken lines. This illustrates

the pathology of the t- and -coordinates in these regions. The points in each plot

correspond to unit intervals of c / , where is the proper time and we have taken

= 0 at r = 8 . The proper time increases steadily, without becoming singular;

the particle reaches the ring singularity in the equatorial plane at c / = 10 2. In

the right-hand plot we also note the effect of frame-dragging on the trajectory of

the initially radially moving particle.

13.13 Equatorial circular motion of massive particles

For circular motion, we require that r = 0 and, for the particle to remain in a

Ë™

Â¨

circular orbit, that the radial acceleration r must also vanish. In terms of the

effective potential defined in (13.46), for a circular orbit at r = rc we thus require

dVeff

Veff rc h k = 2 c2 k2 âˆ’ 1 =0

1

and (13.47)

dr r=rc

These two equations determine the values of the constants k and h that correspond

to a circular orbit at some assigned radius r = rc .

Obtaining analytic expressions for k and h is, algebraically, considerably more

complicated than in the Schwarzschild case. The derivation is simplified some-

what, however, by working in terms of u = 1/r. Making this substitution into

(13.46) and then differentiating the resulting expression with respect to u, we

find that

âˆ’ c2 u + 2 h2 âˆ’ a2 c2 k2 âˆ’ 1 u2 âˆ’ h âˆ’ ack 2 u3 = 2 c2 k2 âˆ’ 1

1 1

âˆ’ c2 + h2 âˆ’ a2 c2 k2 âˆ’ 1 u âˆ’ 3 h âˆ’ ack 2 u2 = 0

where the second equality holds since dVeff /dr = dVeff /du du/dr , and there-

fore dVeff /du = 0 implies that dVeff /dr = 0. The algebra is further eased by

introducing the variable x = h âˆ’ ack, so that the two equations above become

âˆ’ c2 u + 2 x2 + 2ackx + a2 c2 u2 âˆ’ x2 u3 = 2 c2 k2 âˆ’ 1

1 1

(13.48)

âˆ’ c2 + x2 + 2ackx + a2 c2 u âˆ’ 3 x2 u2 = 0 (13.49)

Subtracting (13.48) from u times (13.49) and performing a simple rearrangment

of (13.49), we obtain

c2 k2 = c2 1 âˆ’ u + x2 u3 (13.50)

2xacku = x2 u 3 u âˆ’ 1 âˆ’ c2 a2 u âˆ’ (13.51)

336 The Kerr geometry

These are the basic equations to be used for obtaining analytic expressions for

the constants k and h.

Eliminating k between (13.50) and (13.51), we quickly obtain a quadratic

equation for x2 ,

u2 3 u âˆ’ 1 2 âˆ’ 4a2 u3 x4 âˆ’ 2c2 u 3 u âˆ’ 1 a2 u âˆ’ âˆ’ 2ua2 u âˆ’ 1 x2

+ c4 a2 u âˆ’ =0

2

Using the standard formula for the roots of a quadratic equation, one finds (after

some straightforward but substantial algebra) that

âˆš âˆš2

c2 a u Â±

x=

2

(13.52)

u 1 âˆ’ 3 u âˆ“ 2a u3

As we shall see below, the upper signs corresponds to the counter-rotating circular

orbit and the lower signs to the co-rotating one. Furthermore, in order to obtain

x we must choose either the positive or negative square root of (13.52). As we

might expect from our discussion of the stability of massive particle orbits in the

Schwarzschild case (see Chapter 9), the possibility exists for a circular orbit at

a given coordinate radius to be either stable or unstable. It is straightforward to

show that it is the negative root of (13.52) that corresponds to the stable case, in

which we are most interested. We therefore consider only the solution

âˆš âˆš

c a uÂ±

x=âˆ’ (13.53)

u 1 âˆ’ 3 u âˆ“ 2a u3 1/2

Inserting this solution into (13.50), for a stable circular orbit of inverse coordinate

radius u we find that

1âˆ’2 uâˆ“a u3

k= (13.54)

1 âˆ’ 3 u âˆ“ 2a u3 1/2

the energy of a particle of rest mass m0 being E = km0 c2 . The corresponding

value of the specific angular momentum for the orbit is obtained by calculating

h = x + ack, which gives

âˆš

1 + a2 u2 Â± 2a u3

c

h = âˆ“âˆš (13.55)

u 1 âˆ’ 3 u âˆ“ 2a u3 1/2

We note that, as expected, in the limit a â†’ 0 the expressions (13.54) and (13.55)

reduce to the corresponding results in the Schwarzschild case derived in Chapter 9.

337

13.14 Stability of equatorial massive particle circular orbits

13.14 Stability of equatorial massive particle circular orbits

It is worthwhile considering in some detail the stability of equatorial massive-

particle orbits. Of particular astrophysical interest is the stability of the circular

orbits discussed above and, especially, the coordinate radius of the innermost

stable circular orbit in the co- and counter-rotating cases.

For a circular orbit of coordinate radius r = rc we require the condition (13.47).

In addition, for marginal stability we require that (for r = rc )

2

d2 Veff d2 Veff dVeff d2 u 2

du 3 d Veff dV

+ 2 eff

= + =u =0

dr 2 du2 du dr 2 du2

dr du

where u = 1/r. Since dVeff /du = 0 for a circular orbit, this additional requirement

amounts to d2 Veff /du2 = 0. From (13.49), this reads

x2 + 2ackx + a2 c2 âˆ’ 6 x2 u = 0

which may be more conveniently written as

x2 + 2ackx + a2 c2 h2 âˆ’ a2 c2 k2 âˆ’ 1

u= =

6 x2 6 x2

Inserting the expressions (13.53â€“13.55) for x, k and h respectively into this

equation and simplifying, one finds that

1 âˆ’ 3a2 u2 âˆ’ 6u âˆ“ 8a u3 = 0

Finally, using u = 1/r, one obtains an implicit equation for the coordinate radius

r of the innermost stable circular orbit,

âˆš

r 2 âˆ’ 6 r âˆ’ 3a2 âˆ“ 8a r =0 (13.56)

where, once again, the upper sign corresponds to the counter-rotating orbit and

the lower sign to the co-rotating orbit. In the limit a = 0, we see that we recover

r = 6 for the innermost stable circular orbit in the Schwarzschild case. In the

extreme Kerr limit a = we find, by inspection, that r = 9 for the counter-

rotating orbit and r = for the co-rotating case.

âˆš

The general solution to the above quartic equation in r can be found analyti-

cally by standard methods, but the resulting expressions are algebraically messy.

It is more instructive instead to solve the equation numerically and plot the results

for a range of a/ values, as shown in Figure 13.5 (left-hand panel). Also of

particular interest is the energy E of a particle in the innermost co- and counter-

rotating stable circular orbits. Using the expression (13.54), in the right-hand

panel of Figure 13.5 we plot k = E/ m0 c2 for these orbits as a function of a/ .

The difference between the energy E = km0 c2 of a particle in an orbit and the

energy m0 c2 of the particle at rest at infinity is the gravitational binding energy

338 The Kerr geometry

8 counter-rotating

counter-rotating

1

6

E /m0c2

co-rotating

r /Âµ

co-rotating

4

0.5

2

0

0

0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

a /Âµ a /Âµ

Figure 13.5 The scaled coordinate radii r/ (left) and the constant ks =

E/ m0 c2 (right) for the innermost stable co-rotating and counter-rotating circu-

lar orbits in the equatorial plane of the Kerr geometry, as functions of a/ .

of the orbit. As discussed in Chapter 9, the binding energy of a particle in an

accretion disc around a compact object can be released. As the particle loses

angular momentum, owing to turbulent viscosity, it gradually moves inwards,

releasing gravitational energy mostly as radiation, until it reaches the innermost

stable circular orbit, at which point it spirals rapidly inwards onto the compact

object. The efficiency acc of the accretion disc is the fraction of the rest mass

energy that can be released in making the transition from rest at infinity to

the innermost stable circular orbit and is given by acc = 1 âˆ’ k. We see from

Figure 13.5 (right-hand panel) that for all values of a/ the co-rotating orbit is

the more bound, and the corresponding binding energy is greatest for an extreme

Kerr black hole a/ = 1 . In this case

1

= 1 âˆ’ âˆš â‰ˆ 42%

acc

3

and so an accretion disc around such an object could convert nearly one-half

of the rest mass energy of its constituent particles into radiation. For a realistic

astrophysical Kerr black hole that has been â€˜spun-upâ€™ by the accretion process one

expects that a/ â‰ˆ 0 998, in which case acc â‰ˆ 32%, which is still substantially

larger than the value of 5 7% in the Schwarzschild case.

13.15 Equatorial trajectories of photons

For photons, the null geodesics in the equatorial plane are governed by (13.40)

and the â€˜energyâ€™ equation (13.43) with 2 = 0, which reads

a2 c2 k2 âˆ’ h2 2 h âˆ’ ack 2

r =c k +

Ë™ +

2 22

(13.57)

r2 r3

339

13.16 Equatorial principal photon geodesics

As in our discussion of photon trajectories in the Schwarzschild geometry in

Chapter 9, it is useful to introduce the parameter b = h/ ck . By considering the

limit r â†’ , one again finds that b may be interpreted as an impact parameter

for the trajectories that extend to infinity. We note that for r â†’ the constant k

is positive and so b (or h) has the same sign as Ë™ in this limit.

One may rewrite the energy equation (13.57) in the form

Ë™

r2 1

+ Veff r b = 2 (13.58)

h2 b

where we have identified the effective potential as

1 2

2 2

a a

r b = 2 1âˆ’ âˆ’ 1âˆ’ (13.59)

Veff

r b r b

As was the case for massive particles, Veff has the same r-dependence as the

corresponding expression for the Schwarzschild case, derived in Chapter 9, and

so the graph of Veff has the same general shape. Indeed, as one would expect,

in the limit a â†’ 0 equation (13.59) reduces to the Schwarzschild result. When

a = 0, however, one must again be careful in interpreting (13.59) as an effective

potential, since it depends on the value b (and hence k) of the particle trajectory.

Nevertheless, by differentiating (13.58) with respect to , one finds that the radial

acceleration of a particle is still given by r = âˆ’h2 dVeff /dr. (In fact, by a rescaling

Â¨

â†’ h of the affine parameter , the explicit h-dependence is removed from

this result and (13.58).) Similarly, the stability of a circular orbit, for example,

may be deduced by considering the sign of d2 Veff /dr 2 in the usual manner.

13.16 Equatorial principal photon geodesics

As might be expected, radial photon geodesics do not exist in the equatorial plane

of the Kerr geometry. Nevertheless, we can obtain information about the radial

variation of the light-cone structure by investigating the principal null geodesics.

These are defined by the condition b = a. The system of equations (13.38),

(13.39), (13.57) then reduces to

Ë™

t = r 2 + a2 k/

Ë™ = ack/

r = Â±ck

Ë™

where the plus sign and the minus sign in the last equation correspond respectively

to outgoing and incoming photons. We can see that such geodesics play the same

role in the Kerr geometry as do the radial geodesics in the Schwarzschild case, in

340 The Kerr geometry

that the radial coordinate is described at a uniform rate with respect to the affine

parameter.

Choosing r = +ck for outgoing photons, we find that

Ë™

Ë™

Ë™ r 2 + a2

dt t d a

== ==

Ë™ Ë™

dr r c dr r

Using the fact that > 0 in region I, it follows that dr/dt > 0 in region I, thus

confirming that these equations correspond to outgoing photons. If we restrict our

attention to the case a2 < 2 , the equations can be immediately integrated to give

2 2

r r

ct = r + + âˆ’1 + âˆ’ âˆ’1

ln ln

2 âˆ’ a2 2 âˆ’ a2

r+ râˆ’

+ constant (13.60)

r âˆ’ r+

a

= + constant

ln (13.61)

r âˆ’ râˆ’

2 âˆ’ a2

2

The solution corresponding to incoming photons is obtained by choosing r = âˆ’ck

Ë™

and has the same form as above but with t â†’ âˆ’t and â†’ âˆ’ . In Figure 13.6 we

plot an incoming principal null geodesic in the ct r -plane and in the x y -plane

âˆš âˆš

(with x = r 2 + a2 cos and y = r 2 + a2 sin ) for a photon that passes through

4

30

2

20

y/Âµ

ct/Âµ

0

10

â€“2

â€“4

0

2468 â€“4 â€“2 0 2 4

r/Âµ x/Âµ

Figure 13.6 A principal null geodesic in a Kerr geometry with rotation param-

eter a = 0 8 . The trajectory (solid line) is plotted in theâˆš ct r -plane (left)

âˆš

and in the x y -plane (right); x = r 2 + a2 cos and y = r 2 + a2 sin . The

locations of the horizons (broken lines) and the ring singularity (dotted line) are

also indicated.

341

13.17 Equatorial circular motion of photons

= 8 0 at t = 0 in a Kerr geometry with rotation parameter

the point r

a=08 .

We note that, in the limit a â†’ 0, (13.60) reduces to the equation for a radial

photon trajectory in the Schwarzschild geometry, as presented in Section 11.3.

Indeed, the null geodesics considered above play the same role as the null radial

geodesics in the Schwarzschild case, giving information about the radial variation

of the light-cone structure. We can draw a spacetime diagram of the light-cone

structure using these equations and we find in region I a diagram analogous to that

obtained for the Schwarzschild geometry in t r coordinates in Chapter 11:

the light-cones narrow down as r â†’ r+ . On r = r+ both t and become infinite,

again indicating that this is a coordinate singularity.

13.17 Equatorial circular motion of photons

For circular photon motion we require r = 0 and, for the photon to remain in a

Ë™

Â¨

circular orbit, the radial acceleration r must also vanish. In terms of the effective

potential defined in (13.59), for a circular orbit at r = rc we thus require

1 dVeff

Veff rc b = =0

and (13.62)

b2 dr r=rc

These two equations determine a single value r = rc (different for prograde and

retrograde orbits) for which there exists a circular orbit, and the corresponding

value of the constant b.

Using the expression (13.59), the above conditions yield respectively

bâˆ’a

rc = 3 (13.63)

b+a

b+a = 27 bâˆ’a

3 2

(13.64)

These equations may be solved by setting y = b + a in (13.64), solving the

resulting cubic equation and substituting the resulting value of b into (13.63). One

may easily verify that the result can be written as

2 a

cosâˆ’1 Â±

rc = 2 1 + cos (13.65)

3

âˆš

b=3 rc âˆ’ a (13.66)

where the upper sign in (13.65) corresponds to retrograde orbits and the lower

sign to prograde orbits. In the limit a â†’ 0 we recover the conditions for a circular

photon orbit in the Schwarzschild case, obtained in Chapter 9, namely rc = 3

âˆš

and b = 3 3 . As in the Schwarzschild case, circular photon orbits in the Kerr

geometry are unstable.

342 The Kerr geometry

13.18 Stability of equatorial photon orbits

In our discussion of the stability of photon orbits in the Schwarzschild geometry,

it was useful to consider the effective potential for photon motion. As mentioned

above, however, when a = 0 one must be careful in interpreting (13.59) as an

effective potential since it depends on the value b (and hence k) of the particle

trajectory. Nevertheless, we can still investigate the stability of the photon orbits

by factorising the energy equation (13.57).5 One finds that

r 2 + a2 2 âˆ’ a2 r r âˆ’2

4 ra

r=

Ë™ c 2 k2 âˆ’ ckh âˆ’ 2

2

h2

2 + a2 2 âˆ’ a2 r +a 2 2 âˆ’ a2

r4 r

r 2 + a2 2 âˆ’ a2

= c2 k âˆ’ V+ r c2 k âˆ’ Vâˆ’ r (13.67)

2r 4

c

where VÂ± r do not depend on k and are given by

1/2

2 ra Â± r 2 1/2 g

r= 2 ch = Â± âˆ’ tt

2

(13.68)

VÂ± h

r + a2 2 âˆ’ a2 g

The first property to notice is that if < 0 then the functions VÂ± r are complex

and so there are no (real) solutions to the equation r = 0. This shows that the

Ë™

photon orbit has no turning points. Thus once a photon crosses the surface = 0,

it cannot turn around and return back across the surface. Therefore = 0 defines

an event horizon in the equatorial plane (in fact, as we showed earlier, that = 0

defines the event horizon is true in the general case).

The qualitative features of photon trajectories may be deduced by plotting the

functions VÂ± r . We choose first the case ah > 0 (angular momentum in the

same sense as the rotation of the source) and confine our attention to r > r+ (i.e.

outside the outer horizon). The curves are plotted in Figure 13.7. It is clear from

(13.67) that photon propagation is only possible if c2 k > V+ or c2 k < Vâˆ’ , since

Ë™

we require r 2 > 0. Thus, at any given coordinate radius r, photon propagation

cannot occur if c2 k has a value lying in the region between the curves Vâˆ’ r

and V+ r . However, we must also remember, from (13.38), that c2 k = pt , the

covariant time component of the photonâ€™s 4-momentum. This is the energy of

the photon relative to a fixed observer at infinity. We are used to the idea of

â€˜positive-energyâ€™ photons with pt > 0. They may come in from infinity and either

reach a minimum r or plunge into the black hole, depending on whether they

encounter the hump in V+ r .

What about photons for which pt < Vâˆ’ ? Some of these have pt > 0 but others

have pt < 0. Which photons of these types, if any, can actually exist? Near the

5

This approach is based on that presented in B. Schutz, A First Course in General Relativity, Cambridge

University Press, 1985.

343

13.18 Stability of equatorial photon orbits

V

Ï‰+h

V+(r)

r

Vâ€“(r)

r = r+ r = rs+

Figure 13.7 The factored effective-potential diagram for equatorial photon

orbits with positive angular momentum ah > 0 . The quantity + is the value

of at r = r+ . Photon propagation is forbidden in the shaded region.

horizon in the Kerr metric, the â€˜energyâ€™ pt relative to an observer at infinity has no

obvious physical meaning. The important requirement is that to an observer near

the horizon the photon has a positive energy. A convenient observer, although

any would suffice, is one who resides at fixed r in the equatorial plane, circling

the hole with a fixed angular velocity (this observer is not on a geodesic, so

would need to be in a spaceship). As discussed in Section 13.8, the observerâ€™s

4-velocity u in t r coordinates has components

= ut 1 0 0

u

Thus, he measures a photon energy

E = p Â· u = pt ut + p u = ut pt âˆ’ h

h. Since pt = c2 k, we thus require

The photon must therefore have pt >

c2 k > (13.69)

h

From our discussion in Section 13.8 about observers in the ergoregion, we know

that is restricted to lie in the range âˆ’ < < + , where Â± are given by

(13.28). Comparing (13.28) with (13.68) we see that, remarkably, VÂ± = Â± h.

Thus, any photon with c2 k > V+ also satisfies the condition (13.69) and so is

allowed, while any photon with c2 k < Vâˆ’ violates (13.69) and is forbidden. We

conclude that here there is nothing qualitatively different from our discussion of

photon orbits in the Schwarzschild geometry.

For photons moving in the opposite direction to the holeâ€™s rotation ah < 0 ,

new features do appear. If ah < 0 it is clear from (13.68) that the shapes of the

VÂ± r curves are simply turned upside down (see Figures 13.8 and 13.7). From

(13.67) directly, we again see that in the region between the curves Vâˆ’ r and

344 The Kerr geometry

V

E V+(r)

Penrose

r

process

Vâ€“(r)

â€“E

Ï‰+h

r = r+ r = rs+

Figure 13.8 The factored effective-potential diagram for equatorial photon

orbits with negative angular momentum ah < 0 . The quantity + is the value

of at r = r+ . Photon propagation is forbidden in the shaded region. The Penrose

process is also illustrated (see the text for details).

V+ r there is no photon propagation. Moreover, the condition (13.69) means

that photons must have c2 k > V+ but from Figure 13.8 we see that in the region

r < rS+ (the ergoregion) some of these photons can have c2 k < 0. We can now

understand in an alternative manner an idealised version of the Penrose process,

discussed in Section 13.9. At some point between r+ and rS+ it is allowable to

create two photons, one having pt = E and the other having pt = âˆ’E, so that

their total energy is zero. Then the â€˜positive-energyâ€™ photon could be directed in

such a way as to leave the hole and reach infinity, while the â€˜negative-energyâ€™

photon is necessarily trapped and inevitably crosses the horizon. The net effect is

that the positive-energy photon will leave the hole, carrying its energy to infinity.

Thus energy has been extracted.

13.19 Eddingtonâ€“Finkelstein coordinates

We have seen throughout our discussion of the Kerr geometry that the Boyerâ€“

Lindquist coordinates t and are â€˜badâ€™ in the region near the horizons. By analogy

with our discussion of removing the coordinate singularity in the Schwarzschild

geometry, we may use the equations for the principal photon geodesics, (13.60)

and (13.61), to obtain a coordinate transformation that extends the solution through

r = r+ .

Working with these equations in differential form, we have

r 2 + a2

c dt = âˆ’ (13.70)

dr

a

d =âˆ’ (13.71)

dr

345

13.19 Eddingtonâ€“Finkelstein coordinates

for the ingoing photons. For advanced Eddingtonâ€“Finkelstein coordinates

we want ingoing principal photon trajectories to be straight lines.

t r

Thus, for such a trajectory, we require

c dt = âˆ’dr d =d =0

and

From (13.70) and (13.71), we see immediately that the required transformations are

2r

c dt = c dt + dr

a

=d +

d dr

The Kerr solution in advanced Eddingtonâ€“Finkelstein coordinates then takes the

form

2r 4r

ds2 = 1 âˆ’ 2 c2 dt 2 âˆ’ 2 c dt dr

4 ra sin2

2r

âˆ’ 1+ dr + 2

c dt d

2 2

2 r 2 + a2 a sin2

+ âˆ’ 2 2

dr d d

2

2 ra2 sin4

âˆ’ r + a sin +

2

2 2 2

d

2

If we define the advanced time parameter p = ct + r (such that dp = 0 along the

photon geodesic), the Kerr solution can also be written as

4 ra sin2

2r

ds = 1 âˆ’ dp âˆ’ 2 dp dr + + 2a sin2 dr d âˆ’

2 2 2 2

dp d d

2 2

2 ra2 sin4

âˆ’ r + a sin +

2

2 2 2

d

2

One may alternatively straighten the outgoing photon geodesics by introducing

retarded Eddingtonâ€“Finkelstein coordinates tâˆ— âˆ— r and the retarded time

âˆ— âˆ’ r, in an analogous manner.

parameter q = ct

Figure 13.9 shows a spacetime diagram along the equator of a Kerr black hole

using advanced Eddingtonâ€“Finkelstein coordinates. As in the Schwarzschild solu-

tion, the event horizon at r + marks a surface of â€˜no returnâ€™. Once a particle has

crossed the event horizon, its future is directed towards region III, which contains

the singularity â€“ you can never return back to region I. Unlike the Schwarzschild

solution, the singularity in the Kerr solution is timelike (the singularity in the

Schwarzschild solution is spacelike). In theory, this means that it is possible to

avoid the singularity by moving along a timelike path; in other words, if we

346 The Kerr geometry

Event horizon Event horizon

r = râ€“ r = r+

III II I

r=0

Figure 13.9 spacetime diagram of the Kerr solution in advanced Eddingtonâ€“

Finkelstein coordinates.

were in a spaceship (and ignoring the intense tidal forces which would make this

experiment impractical) we could manoeuvre along a path to avoid the singularity.

Indeed, by performing a maximal extension of the Kerr geometry in an analo-

gous way to the Kruskal extension of the Schwarzschild geometry described in

Chapter 9, one finds that a particle may re-cross the surface r = râˆ’ and eventually

emerge from r = r+ into a different asymptotically flat spacetime (in an analogous

way to that described for the Reissnerâ€“NordstrÃ¶m geometry in Section 12.6).

However, you should not take the internal structure of the Kerr solution too seri-

ously. As mentioned above, region III also contains closed timelike curves (at

r < 0), which are very bad news because they violate causality. Most theorists

would hope that quantum gravity comes to the rescue and prevents causality

violation. At present we do not really know what happens within region III.

Figure 13.10 shows a schematic illustration of the light-cone structure in the

equatorial plane of the Kerr solution, which also illustrates the frame-dragging

effect. As we approach the infinite redshift surface S + , any particle travelling

against the direction of rotation has to travel at the speed of light just to remain

stationary (relative to a fixed observer at infinity). At smaller r, in the ergoregion,

the light-cones are tipped over, so that photons (and massive particles) are forced

to travel in the direction of rotation. At the event horizon r + , the lightcones tip

over so far that the future is directed towards region II.

347

13.20 The slow-rotation limit and gyroscope precession

Event horizon r = r â€“

Ring singularity

Ergosphere

Event horizon r = r + Stationary limit

surface S +

Figure 13.10 Frame dragging in the equatorial plane of the Kerr solution.

13.20 The slow-rotation limit and gyroscope precession

Since the full Kerr solution is rather complicated, it is useful to consider the

simpler approximate form for the common limiting case of a slowly rotating body.

Thus, we will only keep terms in the Kerr metric to first order in a. Writing the

resulting metric in terms of the angular momentum J = Mac of the rotating body,

in Boyerâ€“Lindquist coordinates we obtain

4GJ

ds2 = dsSchwarzschild + sin2 d dt

2

(13.72)

c2 r

where the first term on the right-hand side is the standard Schwarzschild

line element. In the slow-rotation limit, Boyerâ€“Lindquist coordinates tend to

Schwarzschild coordinates. This metric is useful for performing calculations of,

for example, the general-relativistic effects due to the rotation of the Earth. In

fact, for terrestrial applications, and many other astrophysical situations, we may

also assume the gravitational field to be weak, in which case the line element

becomes

2GM 2GM

ds2 = c2 1 âˆ’ dt2 âˆ’ 1 + dr 2 + r 2 d + r 2 sin2 d

2 2

c2 r c2 r

4GJ

+ sin2 d dt (13.73)

2r

c

It is often also convenient to work in Cartesian coordinates defined by

x = r sin cos y = r sin sin y = r cos

348 The Kerr geometry

for which the line element is easily shown to take the form

2GM 2GM

ds2 = c2 1 âˆ’ dt2 âˆ’ 1 + dx2 + dy2 + dz2

c2 r c2 r

4GJ

+ x dy âˆ’ y dx dt (13.74)

c2 r 3

where r is now defined by r = x2 + y2 + z2 .

To illustrate the usefulness of the slow-rotation limit, we now consider the

precession of a gyroscope induced by the frame-dragging effect of a slowly rotat-

ing body, such as the Earth. As discussed in Chapter 10, in general a gyroscope

in orbit around a massive non-rotating body will precess simply as a result of

the spacetime curvature induced by the massive body (geodesic precession). If

the central body is also rotating then there is an additional precessional effect,

which we now discuss. Let us consider the thought experiment shown schemati-

cally in Figure 13.11. A gyroscope falls freely down the rotation axis of a slowly

rotating body. Initially the spin axis is oriented perpendicular to the rotation axis.

By symmetry, if the body were not rotating then the spin axis would remain

fixed with respect to infinity (e.g. pointing constantly to one distant star), thus

for this particular orbit there is no geodesic precession of the gyroscope. By this

measure, the local inertial frames on the axis are not rotating with respect to

infinity. However, if instead the body were rotating, even with a small angular

momentum, then the gyroscope would precess, indicating that the local inertial

frames are rotating with respect to infinity.

z

s

J

Figure 13.11 A gyroscope (solid circle) falling down the rotation axis of a

spinning body.

349

13.20 The slow-rotation limit and gyroscope precession

We can use the metric (13.74) to calculate the precession rate of the gyro-

scope on its downward â€˜polar plungeâ€™ trajectory. As shown in Section 10.5, the

spin 4-vector s is parallel-transported along the geodesic trajectory. Thus, its

components satisfy

ds

+ s u =0 (13.75)

d

For the physical arrangement under consideration, the initial 4-velocity u and the

spin 4-vector s of the gyroscope in Cartesian coordinates have the forms

= ut 0 0 uz = 0 sx sy 0

and

u s

Moreover, these forms remain valid at all later times, since the trajectory is a

polar plunge and s Â· u is conserved along it. Thus, in (13.75), the only equations

we need to consider are

dsx

+ u+ u+ u+ u =0

x xt x xz x yt x yz

(13.76)

xt s xz s yt s yz s

d

dsy

+ u+ u+ u+ u =0

y xt y xz y yt y yz

(13.77)

xt s xz s yt s yz s

d

To continue with our calculation, we must first find the connection coeffi-

cients appearing in the above equations. This is most easily achieved using the

â€˜Lagrangianâ€™ approach, writing down the Eulerâ€“Lagrange equation for x and

remembering that on the polar axis all terms proportional to some positive power

of x or y are zero and r = z. One finds that the only non-zero connection coeffi-

cients of the form x are

2GJ 2GM

= =âˆ’

x x

and

yt xz

c2 z z + 2GM/c2

c2 z3 z-axis

z-axis

By considering the symmetry properties of the metric (13.74), one immediately

deduces that, on the polar axis, the only non-zero connection coefficients of the

form y are

2GJ 2GM

z-axis = âˆ’ =âˆ’

y y

and

xt yz

c2 z z + 2GM/c2

c2 z3 z-axis

These connection coefficients can now be substituted into equations (13.76,

13.77), which can be solved once ut and uz have been determined from the

geodesic equations. Assuming, however, that the speed of the falling gyroscope

is non-relativistic, then to leading order in 1/c we may take ut â‰ˆ 1 and uz â‰ˆ 0.

Thus, in this approximation, equations (13.76, 13.77) reduce to

dsx dsy

2GJ 2GJ

= âˆ’ 2 3 sy = 2 3 sx

and

d cz d cz

350 The Kerr geometry

Hence, as it falls, the gyroscope precesses in the same direction as the body is

rotating, i.e. the local inertial frames are dragged with respect to infinity. This is

called the Lensâ€“Thirring effect. At a height z the rate of precession is

2GJ

LT =

c2 z3

It should be noted that we have calculated this precession rate in a Cartesian

coordinate system in which the centre of the gravitating body is at rest and

the gyroscope is falling. Fortunately, an observer free-falling with the gyroscope

would measure the same precession rate since the Lorentz transformation that

connects the two frames is a boost along the z-axis, which does not affect the

transverse components sx and sy of the spin vector. Of course, the Lensâ€“Thirring

effect also results in the precession of gyroscopes following trajectories other

than the polar plunge considered here, but determining the rate of precession in

general requires a considerably longer calculation (see Exercise 17.26).

Exercises

13.1 Verify that the Boyerâ€“Lindquist form of the Kerr metric satisfies the empty-space

Einstein field equations.

Note: This exercise is only for the truly dedicated reader!

13.2 Show that the Boyerâ€“Lindquist form of the Kerr metric can be written in the forms

(13.12) and (13.13).

13.3 Calculate the contravariant components g of the Kerr metric in Boyerâ€“Lindquist

coordinates.

13.4 Show that, in the limit â†’ 0, the Kerr metric tends to the Minkowksi metric.

13.5 Show that the Kerrâ€“Schild form of the Kerr metric can be transformed into the

Boyerâ€“Lindquist form by the coordinate transformations (13.16 â€“13.19).

13.6 Consider the 2-surfaces defined by t = constant and r = rÂ± in the Kerr geometry.

Show that, for each surface, the circumference around the â€˜polesâ€™ is less than the

circumference around the equator. Show that the same is true for the 2-surfaces

defined by t = constant and r = rSÂ± .

13.7 Show that the proper area of the event horizon r = r+ in the Kerr geometry is

given by

A=4 r + + a2

2

Hence show that, for fixed , the area A is a maximum for a = 0. Conversely, for

fixed A, show that is a minimum for a = 0. Comment on your results.

351

Exercises

13.8 An observer is at fixed (r, ) coordinates in the ergoregion of a Kerr black hole

and has angular velocity = d /dt with respect to a second observer at rest at

infinity. Show that the allowed range for is given by âˆ’ < < + , where

1/2

= Â±c

Â± 2

and , and 2 have their usual meanings in the Kerr metric.

Use your answer to Exercise 13.7 to show that the area of the event horizon r = r+

13.9

in the Kerr geometry may be written as

âŽ¡ âŽ¤

2

8G cJ âŽ¦

A = 4 âŽ£M 2 + M 4 âˆ’

c G

where M and J are the mass and angular momentum of the black hole. Hence

show that if the mass and angular momentum change by M and J respectively

then the corresponding change in the proper area of the horizon is given by

8G a J

A= Mâˆ’ H

2 âˆ’ a2 c2

c H

where H is the â€˜angular velocity of the horizonâ€™, defined in (13.29). Thus show

that the area of the event horizon must increase in the Penrose process.

13.10 Show that, in the equatorial plane = /2 of the Kerr geometry, the contravariant

metric components in Boyerâ€“Lindquist coordinates are

2 a2

1 2a

g tt = r 2 + a2 + gt =

c2 r cr

1 2

g rr = âˆ’ =âˆ’ 1âˆ’

g

r2 r

13.11 Show that the geodesic equations for particle motion in the equatorial plane of the

Kerr geometry may be written in Boyerâ€“Lindquist coordinates as

2 a2

1 2a

Ë™

t= r +a + kâˆ’

2 2

h

r cr

Ë™=1 2 ac 2

k+ 1âˆ’ h

r r

a2 c 2 k 2 âˆ’ âˆ’ h2 h âˆ’ ack

2 2 2

2 2

r = ck âˆ’

Ë™ + + +

2 22 2

r2 r3

r

where 2 = c2 for a massive particle and = 0 for a photon. Verify that these

equations reduce to the Schwarzschild case in the limit a â†’ 0.

13.12 The trajectory of an infalling particle of mass m in the equatorial plane of a

Kerr black hole is characterised by the usual parameters k and h. If the particle

352 The Kerr geometry

eventually falls into the black hole, show that the mass and the angular momentum

of the hole are changed in such a way that

M â†’ M + kmc2 J â†’ J + mh

Show further that the corresponding change a in the rotation parameter of the

black hole is given by

m

a= h âˆ’ ack

cM

If the particle falls into an extreme Kerr black hole, for which a = , show that a

naked singularity would be created if

h > 2ck

However, by determining the maximum value of the effective potential Veff r h k

defined in (13.46) for a = , show that a particle with h > 2ck can never fall

into the black hole.

13.13 For a Kerr black hole, using the Boyerâ€“Lindquist coordinates show that, for a

particle in circular orbit in the = /2 plane, the coordinate angular velocity

= d /dt satisfies

1/2

c

= 1/2 Â± r 3/2

a

This is the Kerr-metric analogue to 2 = GM/r 3 for the Schwarzschild metric.

Here the plus sign corresponds to prograde orbits, the minus to retrograde orbits.

13.14 If a particleâ€™s motion is initially in the = /2 plane in a Kerr metric, show that

the motion will remain in this plane.

13.15 Show that the values of the parameters k and h for a circular orbit of coordinate

radius r = rc , given in (13.54) and (13.55) respectively, satisfy the requirements

dVeff

Veff rc h k = 2 c2 k2 âˆ’ 1 =0

1

and

dr r=rc

Show further that for the orbit to be stable one requires

âˆš

rc âˆ’ 6 rc âˆ’ 3a2 âˆ“ 8a rc = 0

2

13.16 An observer (not necessarily free-falling) orbits a Kerr black hole in the equatorial

plane in a circular orbit. His â€˜angular velocity with respect to a distant observerâ€™

is = d /dt. Find the components ut u ut and u in terms of r and a.

13.17 Suppose that the circular orbit considered in Exercise 13.16 lies outside the horizon

r+ but inside the stationary limit rS+ . Show that under these circumstances must

be non-zero, i.e. the observer cannot remain at rest relative to a distant observer.

If the orbiting observer is in the region râˆ’ < r < r+ , show that the orbit cannot be

circular.

353

Exercises

13.18 Show that the effective potential for photon orbits in the equatorial plane of the

Kerr geometry is given by

1 2

a a

2 2

Veff r b = 1âˆ’ âˆ’ 1âˆ’

r2 b r b

13.19 For a circular photon orbit of coordinate radius r = rc in the Kerr geometry,

show that

2 a

cosâˆ’1 Â±

rc = 2 1 + cos

3

âˆš

b=3 rc âˆ’ a

where the upper sign in the first equation corresponds to retrograde orbits and the

lower sign to prograde orbits. Hence show that, for an extreme Kerr black hole

(a = ), rc = 4 for a retrograde orbit and rc = for a prograde orbit.

13.20 For a photon orbit in the equatorial plane of the Kerr geometry, show that

r 2 + a2 2 âˆ’ a2

r=

Ë™ c 2 k âˆ’ V+ r c 2 k âˆ’ Vâˆ’ r

2

c2 r 4

where

1/2

2 ra Â± r 2 1/2 g

VÂ± r = 2 ch = Â± âˆ’ tt

2

h

r + a2 2 âˆ’ a2 g

13.21 The general axisymmetric stationary metric can be written in the form

ds2 = A dt2 âˆ’ B d âˆ’ dt 2 âˆ’ C dr 2 âˆ’ D d 2

where A, B, C, D and are functions only of the coordinates r and . Alice

is an astronaut in a powered spaceship that maintains fixed r coordinates in

the equatorial plane = /2 (at a position for which gtt > 0). She simultaneously

emits two photons in opposite tangential directions in the equatorial plane and uses

a prearranged system of mirrors to cause each photon to move along a circular

(non-geodesic) path of constant r. Show that the coordinate angular velocities of

the two photons are given by

1/2

d A

= Â±

dt B

Hence show that the two photons do not arrive back with Alice simultaneously

but are separated by a time interval

4B

=

c Aâˆ’B 2 1/2

as measured by Aliceâ€™s on-board clock. Comment on the physical significance of

this result.

354 The Kerr geometry

13.22 Bob is in a powered spaceship following a circular orbit r = constant in the

equatorial plane of the geometry in Exercise 13.21. His angular velocity is such

that the component u of his 4-velocity is zero. Using the same arrangement of

mirrors as in Exercise 13.21, he performs an experiment similar to Aliceâ€™s. Show

that for Bob the two photons arrive back to him simultaneously.

13.23 Which, if any, of the photons considered in Exercises 13.21 and 13.22 is redshifted

from its original frequency on arriving back with Alice or Bob? Explain your

reasoning.

13.24 An isolated thin rigid spherical shell has mass M and radius R. If the shell is set

spinning slowly, with angular momentum J , show that inertial frames within the

shell rotate with angular velocity

2GJ

=

c 2 R3

Comment briefly on how this result is related to Machâ€™s principle.

14

The Friedmannâ€“Robertsonâ€“Walker geometry

We now discuss the application of general relativity to modelling the behaviour

of the universe as a whole. In order to do this, we make some far-reaching

assumptions, but only those consistent with our observations of the universe. As

in our derivations of the Schwarzschild and Kerr geometries, we begin by using

symmetry arguments to restrict the possible forms for the metric describing the

overall spacetime geometry of the universe.1

14.1 The cosmological principle

When we look up at the sky we see that the stars around us are grouped into a

large-density concentration â€“ the Milky Way Galaxy. On a slightly larger scale,

we see that our Galaxy belongs to a small group of galaxies (called the Local

Group). Our Galaxy and our nearest large neighbour, the Andromeda galaxy,

dominate the mass of the Local Group. On still larger scales we see that our

Local Group sits on the outskirts of a giant supercluster of galaxies centred in the

constellation of Virgo. Evidently, on small scales matter is distributed in a highly

irregular way but, as we look on larger and larger scales, the matter distribution

looks more and more uniform. In fact, we have very good evidence (particularly

from the constancy of the temperature of the cosmic microwave background in

different directions on the sky) that the universe is isotropic on the very largest

scales, to high accuracy. If the universe has no preferred centre then isotropy also

implies homogeneity. We therefore have good physical reasons to study simple

cosmological models in which the universe is assumed to be homogeneous and

1

For a detailed discussion, see, for example, J. N. Islam, An Introduction to Mathematical Cosmology,

Cambridge University Press, 1992.

355

356 The Friedmannâ€“Robertsonâ€“Walker geometry

isotropic2 . We thus assume the cosmological principle, which states that at any

particular time the universe looks the same from all positions in space at a

particular time and all directions in space at any point are equivalent.

14.2 Slicing and threading spacetime

The intuitive statement of the cosmological principle given above needs to be

made more precise. In particular, how does one define a â€˜particular timeâ€™ in general

relativity that is valid globally, when there are no global inertial frames? Also,

since observers moving relative to one another will view the universe differently,

according to which observers do we demand the universe to appear isotropic?

In general relativity the concept of a â€˜moment of timeâ€™ is ambiguous and is

replaced by the notion of a three-dimensional spacelike hypersurface. To define

a â€˜timeâ€™ parameter that is valid globally, we â€˜slice upâ€™ spacetime by introducing

a series of non-intersecting spacelike hypersurfaces that are labelled by some

parameter t. This parameter then defines a universal time in that â€˜a particular

timeâ€™ means a given spacelike hypersurface. It should be noted, however, that

we may construct the hypersurfaces t = constant in any number of ways. In a

general spacetime there is no preferred â€˜slicingâ€™ and hence no preferred â€˜timeâ€™

coordinate t.

It is useful at this point to introduce the idealised concept of fundamental

observers, who are assumed to have no motion relative to the overall cosmological

fluid associated with the â€˜smeared-outâ€™ motion of all the galaxies and other matter

in the universe. A fundamental observer would, for example, measure no dipole

moment in his observations of the cosmic microwave background radiation; an

observer with a non-zero peculiar velocity would observe such a dipole as a

result of the Doppler effect arising from his motion relative to the cosmological

fluid. Adopting Weylâ€™s postulate, the timelike worldlines of these observers are

assumed to form a bundle, or congruence, in spacetime that diverges from a point

in the (finite or infinitely distant) past or converges to such a point in the future.

These worldlines are non-intersecting, except possibly at a singular point in the

past or future or both. Thus, there is a unique worldline passing through each

(non-singular) spacetime point. The set of worldlines is sometimes described as

providing threading for the spacetime.

The hypersurfaces t = constant may now be naturally constructed in such a way

that the 4-velocity of any fundamental observer is orthogonal to the hypersurface.

2

It is worth noting that isotropy about every point automatically implies homogeneity. However, homogeneity

does not necessarily imply isotropy. For example, a universe with a large-scale magnetic field that pointed

in one direction everywhere and had the same magnitude at every point would be homogeneous but not

isotropic.

357

14.3 Synchronous coordinates

Spacelike

t = t2

hypersurface

Simultaneity surface

in local Lorentz frame

Worldline

of observer

t = t1

Figure 14.1 Representation (with one spatial dimension suppressed) of spacelike

hypersurfaces on which fundamental observers are assumed to lie. The worldline

of any fundamental observer is orthogonal to any such surface.

Thus, the surface of simultaneity of the local Lorentz frame of any such observer

coincides locally with the hypersurface (see Figure 14.1). Each hypersurface may

therefore be considered as the â€˜meshing togetherâ€™ of all the local Lorentz frames

of the fundamental observers.

14.3 Synchronous coordinates

The spacelike hypersurfaces discussed above are labelled by a parameter t, which

may be taken to be the proper time along the worldline of any fundamen-

tal observer. The parameter t is then called the synchronous time coordinate. In

addition, we may also introduce spatial coordinates x1 x2 x3 that are constant

along any worldline. Thus each fundamental observer has fixed x1 x2 x3

coordinates, and so the latter are called comoving coordinates. Since each hyper-

surface t = constant is orthogonal to the observerâ€™s worldline, the line element

takes the form

ds2 = c2 dt2 âˆ’ gij dxi dxj for i j = 1 2 3 (14.1)

where the gij are functions of the coordinates t x1 x2 x3 .

We may verify that the metric (14.1) does indeed incorporate the properties

described in the previous section, as follows. Let x be the worldline of a

358 The Friedmannâ€“Robertsonâ€“Walker geometry

fundamental observer, where is the proper time along the worldline. Then, by

construction, x is given by

x0 = x1 = constant x2 = constant x3 = constant (14.2)

Since dxi = 0 along the worldline, we obtain ds = c d = c dt and so t = ,

thereby showing that the proper time along the worldline is indeed equal to t.

Thus, from (14.2), it is clear that the 4-velocity of a fundamental observer in

comoving coordinates is

dx

â‰¡ =1000 (14.3)

u

d

Since any vector lying in the hypersurface t = constant has the form a =

0 a1 a2 a3 , we see that

g u a =0

because g0i = 0 for i = 1 2 3. Hence, the observerâ€™s 4-velocity is orthogonal to

the hypersurface, as we required. Finally, we may show that the worldline given

by (14.2) satisfies the geodesic equation

d2 x dx dx

+ =0

d2 dd

= 0. This quantity is given by

Using (14.3), we see that we require only that 00

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