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Московский государственный университет им. М.В. Ломоносова

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Физика

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P.K.Townsend - Black Holes

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as the future directed ( k J > 0) energy ux 4-vector of . Now consider the following region, S, of spacetime with a null hypersurface N H+ as one boundary.

H+. . .

. 2. . . N

. 1

. n

. i0

Note: is `outward directed' normal to N, as determined by continuity


Assume that @ = 0 at i0. Since D j = 0 we have
0 =	ZS d4x p		gD j = Z@S dS j		(4.59)
=	Z 2	dS j Z 1		dS j ZN dS j	(4.60)
=	E2 E1 ZN dS j				(4.61)

where Ei is the energy of the scalar eld on the spacelike hypersurface i.

The energy going through the horizon is therefore

E = E1 E2 =		dS j	(4.62)
	Z	N
=	Z	dA dv j ;	(v is Kerr coordinate) (4.63)

The energy ux lost/unit time (power) is therefore


P =		dA j =			dA ( @ )(k D )			(4.64)
(since k = 0 on horizon by previous Lemma)
		@			@		@
=	Z	dA		+ H				(4.65)
			@v			@	@v

For a wave-mode of angular-frequency !
= 0 cos (!v ) ; 2 Z (angular quantum no.)			(4.66)
The time average power lost across the horizon is
1		2
P =		0A!(! )	(4.67)
	2
where A is the area of the horizon.
P is positive for most values of !, but for ! in the range
0 < ! < H			(4.68)

it is negative, i.e. a wave-mode with !; satisfying the inequality is ampli ed by the black hole.

Remarks

i)Process is positive only for 6= 0 because the ampli ed eld must also take away angular momentum from the hole.

ii)Process is similar to stimulated emission in atomic physics, which suggests the possibility of a spontaneous emission e ect. This can be shown to occur in the quantum theory so any black with an ergoregion cannot be stable quantum mechanically.

iii)We have neglected the back-reaction of on the metric. When corrected for back-reaction the metric can be stationary only if @ =@ = 0, but then j = 0 and the black hole energy doesn't change, i.e. strictly speaking super-radiance is incompatible with stationarity.

Chapter 5

Energy and Angular

Momentum

5.1Covariant Formulation of Charge Integral

In the usual Minkowski space formulation with charge density (~x; t), the charge in a volume V is written as

V dV =

(5.1)

V dV r E by Maxwell's eqs.

I@V dS~ E~

by Gauss' law

(5.2)

where surface integral is over boundary of V . Note that,

r~ E~ =

@iq(3)gEi;

dV = d3x q(3)g

(5.3)

(3)g

(3)

where g is the determinant of the 3-metric, so

Z dV r~ E~ = Z d3x @i q

(3)gEi = Z dSiEi :

(5.4)

The Lorentz covariant formulation uses the similar result

@ q

(4)gF = D F :

(5.5)

(4)g

The volume V is replaced by an arbitrary spacelike hypersurface (partial Cauchy surface) with boundary @ . The volume element on is a non-

spacelike co-vector (1-form) dS . Given the current density 4-vector j (x)

we write	Z dS j
Q =	Z dS j	(5.6)

We can choose (at least locally) to be t = constant, in which case dS =


~				0	= , we recover the previous expression for Q. Now use
(dV; 0). Since j					= , we recover the previous expression for Q. Now use
Maxwell's equations. D F = j to rewrite Q as
Q =	Z dS D F					(5.7)
=	1		I@ dS F by Gauss' law			(5.8)

		2

where dS is the area element of @ . When is t = constant the only non-vanishing components of dS are

dS0i = dSi0 dSi	(5.9)
in which case
Q = I@ dSi F 0i	(5.10)

But F 0i = F i0 = Ei, so we recover the previous formula.

5.2ADM energy

We cannot de ne energy in the same way because this is associated with a conserved symmetric tensor T , rather than a vector. This is not unexpected because a locally conserved energy can exist only in a spacetime admitting a timelike Killing vector eld.

[Unlike photons, which do not carry charge, gravitons do carry energy ) possibility of energy exchange between matter and its gravitational eld.]

We can still de ne a total energy in asymptotically at spacetimes as a surface integral at in nity because @=@t is asymptotically Killing in such

spacetimes. In this case
g ! as r ! 1 ( Minkowski metric)		(5.11)
We shall assume that, in Cartesian coordinates,
1
h = g = O		(5.12)
	r

which will justify a linearization of Einstein's equations near 1.

Exercise Show that G = 8 GT becomes the Pauli-Fierz equation

h + h;

2h(;) = 16 G T

(5.13)

where

= @ @

(5.14)

= h

(5.15)

= h; = h ;

(5.16)

T = T

(5.17)

Take the trace to get

h h ; = 8 GT

(5.18)

We shall rst consider a weak static dust source

zero pressure for `dust'

(5.19)

T = 0

= 0	for	static
4 G 1	for	weak
T0i = 0

Since source is static we may assume static h , i.e. h = 0. Then = = 0 component of (5.13) becomes

r2h00 = 8 GT00									(5.20)
while (5.18) becomes
r2h00 + r2hjj hij;ij								= 8 GT00	(5.21)
		i\|					}
		i\|	i		jj{z	j	}
@ (@				h		@ hij )
Add (5.20) and (5.21) to get
1		@i (@j hij @ihjj )
T00 =								(Cartesian coordinates)	(5.22)
T00 =	16 G							(Cartesian coordinates)	(5.22)

Since the source is weak we can assume that the spacetime is almost Minkowski, i.e. we treat h as a eld on Minkowski spacetime. The total energy is now

found by integrating T00 over all space.

E =	t = constant d3x T00		(5.23)
	all space
Using Gauss' law we can rewrite result as the surface integral
E =	1	I1 dSi (@j hij @ihjj ) (Cartesian coordinates)	(5.24)

	16 G

But this depends only on the asymptotic data, so we may now change the source in any way we wish in the interior without changing E, provided that the asymptotic metric is unchanged. So formula for E is valid in general.

This is the ADM formula for the energy of asymptotically at spacetimes.

5.2.1Alternative Formula for ADM Energy

Subtract (5.21) from (5.20) to get

@i (@j hij @ihjj ) = 2r2h00

This allows us to rewrite ADM formula as

E =

I1 dSi

@ih00

8 G

But (Exercise)

+ O

gij 0j0

@ih00

( = a ne connection)

and hence

I1 dSi gij 0j0

E =

4 G

I1 dS0i Dik0

where k =

; dSi dS0i

4 G

But k is asymptotically Killing, i.e.

D k + D k = O

I1 dS D k

E =

8 G

(5.25)

(5.26)

(5.27)

(5.28)

(5.29)

(5.30)

(5.31)

5.3 Komar Integrals

Let V be a volume of spacetime on a spacelike hypersurface , with boundary @V . To every Killing vector eld we can associate the Komar integral

	c
Q (V ) =			I@V dS D	(5.32)
	16 G
for some constant c. Using Gauss' law
	c
Q (V ) =		ZV dS D D		(5.33)
	8 G
Lemma D D = R for Killing vector eld .
Proof By contraction of previous `Killing vector Lemma.'

Using Lemma,
Q (V ) =		c		ZV	dS R

		8 G
=	c Z		dS		T

=dS J ( )

where

J ( ) = c T	1	T

	2

Proposition @ J ( ) = 0.

		(5.34)
1
	T	(by Einstein's eqs.) (5.35)
2	T	(by Einstein's eqs.) (5.35)
		(5.36)

(5.37)

Proof Using D T = 0 we have

D J =

c T D

T D

(5.38)

}

0 for

Killing vector

(by Einstein's eqs.)

(5.39)

for Killing vector eld

(5.40)

(In this last step, choose coordinates s.t. @ = @=@ , then the metric is-independent (@g =@ = 0), so R is too (@R=@ = 0)).

Since J ( ) is a `conserved current', the charge Q (V ) is time-independent provided J ( ) vanishes on @V , just as for electric charge.

Exercise = k (time-translation Killing vector eld)
	1
E(V ) =	8 G I@V	dS D k	(5.41)

i.e. c = 2, is xed by comparison with previous formula derived for total energy, i.e. by choosing V = 2-sphere at spatial 1.

Exercise Verify that E(V ) = M for Schwarzschild, for any V with @V in exterior (r > 2M) spacetime.

5.3.1Angular Momentum in Axisymmetric Spacetimes

Return to Komar integral. Let = m = @=@ and choose c = 1 to get


	1		I@V	dS D m
	J(V ) =							(5.42)
	J(V ) =	16 G						(5.42)
Note here factor of 1=2 relative to Komar integral for the energy.
			use Gauss' law to write J(V ) =			RV	dS	J (m) where
To check coe cient,				1	T m	RV
	J (m) = T m							(5.43)
	J (m) = T m			2				(5.43)

If we choose V to be on t = constant hypersurface, and m = @=@ , then dS m = 0, so

J(V ) = ZV dV T 0					m = ZV dV T 0			2x1 T 0	1x2	(5.44)
	coordinates						i
in Cartesian						x ; i = 1; 2; 3 where
	@				@
m = x1			x2							(5.45)
		@x2		@x1
For a weak source, g and
J(V ) "3jk ZV d3x xj T k0										(5.46)

which is result for 3rd component of angular momentum of eld in Minkowski spacetime with stress tensor T .

So the total angular momentum of an asymptotically at spacetime is found by taking @V to be a 2-sphere at spatial in nity

	1
J =	16 G I1 dS D m	(5.47)

5.4Energy Conditions

T satis es the dominant energy condition if for all future-directed timelike vector elds v, the vector eld

j(v) v T @

(5.48)

is future-directed non-spacelike, or zero.

All physically reasonable matter satis es this condition, e.g. for massless scalar eld (with T = @ @ 12 g (@ )2):

j (v)		=	v @ @ +								1	v (@ )2				(5.49)

											2
2			1			2	)2		2					2
j	(v)	=			v		(@ 0			0			if v		< 0	(5.50)
j	(v)	=		4	v		(@ 0			0			if v		< 0	(5.50)
							\|	{z	}

so j(v) is timelike or null if v is timelike. Since v is assumed future-directed, j(v) will be too if v j > 0. Allowing for j = 0 means that we have to prove that v j 0. Now


v j =	(v @ )		2	1	2				2
					v			(@ )			(5.51)
				2	v			(@ )	@		(5.51)
	1				1					v(v @ )	2
=		(v @ )2 +					v2				(5.52)
=	2	(v @ )2 +				2	v2			v2	(5.52)

But (v @ )2 0 and v2 > 0 for timelike v, so we have to prove that

@ v(v @ ) 2 0 (5.53) v2

i.e. that


@	v(v @ )	is spacelike or zero. This follows from
	v2
@	v(v @ )		= 0	(5.54)
	v2

since v V < 0 for any non-zero timelike or null vector for timelike v (choose

		~
coordinates s.t. v = (1; 0)). So if v V = 0 then V cannot be timelike or null.
	Since v j = v v T , the dominant energy condition implies that
v v T 0 for all timelike v. By continuity it also implies the
	Weak energy condition
	v v T 0			8 non-spacelike v		(5.55)

	There is also the
	Strong energy condition
		1
	v v T			g T 0	8 non-spacelike v	(5.56)
	v v T		2	g T 0	8 non-spacelike v	(5.56)

Note, Dominant 6,Strong.

The strong energy condition is needed to prove the singularity theorems, but the dominant energy condition is the physically important one. (An in ationary universe violates the strong energy condition). For example it is needed for the

Positive Energy Theorem (Shoen & Yau, Witten)

The ADM energy of an asymptoticallyat spacetime satisfying G = 8 GT is positive semi-de nite, and vanishes only for Minkowski spacetime with T = 0, provided that

i)9 an initially non-singular Cauchy surface (otherwise M < 0 Schwarzschild would be a counter-example).

ii)T satis es the dominant energy condition (clearly, some condition on T is necessary).

iii)Some other technical assumptions which we ignore here.

100

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