Добавил:
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:

P.K.Townsend - Black Holes

.pdf
Скачиваний:
20
Добавлен:
07.05.2013
Размер:
769.18 Кб
Скачать

Now apply Gauss' law to the Komar integral for the total energy (=

mass).

 

 

 

 

 

 

 

 

M =

 

1

Z dS R k

1

IH dS D k

(insert = k + H m)

 

 

4 G

8 G

=

Z dS ( 2T k + T k )

1

IH dS

(D H D m (6).65)

 

8 G

 

 

 

 

 

 

 

 

(6.66)

since H is constant on H. For T = T (F ) (T (F ) = 0)we have

 

M = 2 Z dS T

1

 

IH dS D

 

(F )k + 2 H JH

 

 

(6.67)

8 G

for an isolated black hole. Using (6.64) we have

 

M = 2 Z dS T

1

IH dS D

 

(F ) + 2 H J

 

(6.68)

8 G

For simplicity, we now suppose that T (F ) = 0, i.e. the black hole has zero charge (see Questions III.7&8 for general case). Then

 

1

 

M = 2 H J

8 G IH dS D

(6.69)

Lemma

dS = ( n n ) dA on H

(6.70)

where n is s.t. n = 1.

Proof n and are normals to H, so we have to check coe cients. In coordinates such that

 

t

 

.

n

 

.

..

................. .. x

... H

111

 

 

1

 

 

 

=

p

 

(1; 1; 0; 0)

(6.71)

2

 

 

1

 

 

n

=

p

 

(1; 1; 0; 0)

(6.72)

2

we should have jdS01j = dA. We do if dS is as given. [There is still a sign ambiguity. Fix by requiring sensible results].

Thus

 

1

 

dS

 

D

=

1

 

 

 

D ) n

 

(6.73)

 

 

 

 

 

 

 

 

 

 

 

8 G IH

 

4 G IH dA (

 

 

 

 

 

 

 

 

}

 

 

 

 

 

 

 

 

 

th

 

 

 

 

 

 

 

 

 

|

 

{z

 

 

 

 

 

 

 

 

 

 

I

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

dA n

 

 

( is constant by 0 (6law).74

 

 

 

 

 

 

4 G

 

 

 

 

 

 

 

 

=

A

|{z}

 

 

 

 

(6.75)

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

4 G

where A is the \area of the horizon" (i.e. H). Hence

M =

A

+ 2 H J

(6.76)

4

This is Smarr's formula for the mass of a Kerr black hole. [Exercise: Check, using previous results for , H , and A]. In the Q 6= 0 case, this formula generalizes to

M =

A

+ 2 H J + H Q

(6.77)

4

where H is the co-rotating electric potential on the horizon (see Question III.6&7).

6.2.3First Law

If a stationary black hole of mass M, charge Q and angular momentum J, with future event horizon of surface gravity , electric surface potential H and angular velocity H , is perturbed such that it settles down to another black hole with mass M + M charge Q+ Q and angular momentum J + J, then

 

 

 

dM =

8 dA + H dJ + H dQ

(6.78)

1) De nition of H and proof for Q 6= 0 in Q. III.6&7.

112

2)This statement of the rst law uses the fact that the event horizon of a stationary black hole must be a Killing horizon.

`Proof' for Q = 0 (Gibbons)

Uniqueness theorems imply that

M = M(A; J)

(6.79)

But A and J both have dimensions of M2 (G = c = 1) so the function M(A; J) must be homogeneous of degree 1=2. By Euler's theorem for homogeneous functions

A

@M

+ J

@M

=

 

 

1

 

M

 

 

(6.80)

 

 

@J

 

 

 

2

 

 

 

 

@A

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

A + H J by Smarr's formula

(6.81)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8

Therefore

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A

@M

 

 

 

 

+ J

 

@M

H

= 0

(6.82)

 

 

 

 

 

 

 

@A

 

 

8

@J

But A and J are free parameters so

 

 

 

@M

 

=

 

 

;

 

 

@M

= H

 

 

(6.83)

 

 

 

 

 

 

 

 

 

 

 

 

 

@A

 

8

 

 

 

@J

 

 

6.2.4The Second Law (Hawking's Area Theorem)

If T satis es the weak energy condition, and assuming that the cosmic censorship hypothesis is true then the area of the future event horizon of an asymptotically at spacetime is a non-decreasing function of time.

Technically the cosmic censorship assumption is that the spacetime is `strongly asymptotically predictable' which requires the existence of a globally hyperbolic submanifold of spacetime containing both the exterior spacetime and the horizon. A theorem of Geroch states that in this case there

exists a family of Cauchy hypersurfaces ( ) such that ( 0) D+ ( ( )) if 0 > .

113

H+.

H0

( 0)

( )

We can choose to be the a ne parameter on a null geodesic generator of H+ . The \area of the horizon" A( ) is the area of the intersection of ( ) with H+. The second law states that A( 0) A( ) if 0 > .

Idea of proof To show that A( ) cannot decrease with increasing it is su cient to show that each area element, a, of H has this property. Recalling that

da

 

= a

(6.84)

d

 

 

we see that the second law holds if 0 everywhere on H+. To see that this is true, recall that if < 0 the geodesics must converge to a focus or caustic, i.e. nearby geodesics to a given one passing through a point p must intersect at nite a ne distance along it. The rst point q for which this happens is called the point conjugate to p on .

q

..... .

.. .

.. .

.. ..

... .

.

.

. ..

. . p

.

.

.

. . . ..

.

.

.. . ..................... conjugate

point

Points on beyond q are no longer null separated. They are timelike separated from p. An example illustrating this is light rays in a at 2-dim cylindrical spacetime.

114

.

..

 

 

. .....

...

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

....

....

 

 

 

.

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

....

.......

 

 

 

 

 

 

 

 

 

 

 

.

 

 

........

....

 

 

 

 

.

 

 

 

 

 

 

......... .... .... .... ....... .....

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

.

...

 

 

 

 

 

 

 

 

 

 

 

 

. .

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

.

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

 

 

 

 

..

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

 

 

..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.. . .

 

 

. .. ..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.. ..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

. .. ....

.... .....

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

. .

..... ...

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

........

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

.

 

 

 

 

 

 

 

 

 

 

....

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

.

.

 

 

 

 

 

 

 

 

 

 

...

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

...

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

...

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

. .

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

....

 

 

 

 

. ..

 

 

 

 

 

 

 

 

 

 

...

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

. ..

.

. . .

 

. . . . .

.

 

. .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

. .

. . . . ..

 

 

 

 

 

. .

. . . . .

. . .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

. .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

point on beyond q

can be reached by timelike curve from p

conjugate point to p on (on far side of cylinder)

.

pspace

The existence of a conjugate point to the future of a null geodesic generator in H+ would mean that this generator of H+ has a nite endpoint, in contradiction to Penrose's theorem, so the hypothetical conjugate point cannot exist. Thus it must be that 0 everywhere on H+ and hence the second law.

= 0 only for stationary spacetimes.

115

Example Formation of black hole from pressure-free spherically-symmetric gravitational collapse. Illustrate by a Finkelstein diagram

.

 

 

 

t

.

. . . . . . . . . . . . . . .

..

.

 

 

 

.

 

 

.

H+

H+

A ( 2) 16M2

.( 2 1)

.

.

. . . . . . . . . . . . .

..

.

.

.

.( 1)

 

 

 

 

A ( 1) 6= 0

 

..

.

 

 

........... past endpoint of

 

 

.....

.

 

 

. . . . . . . ..... . . .

........

.

all generators of H

+

 

......... .... ..... ..............

 

 

 

Star

A ( 0) = 0

( 0)

.

A = 0 on ( 0). A 6= 0 on ( 1) and it has increased to its nal value of A = 16M2 for a stationary Schwarzschild black hole on ( 2).

116

Consequences of 2nd Law

(1)Limits to e ciency of mass/energy conversion in black hole collisions. Consider Finkelstein diagram of two coalescing black holes.

.M3

 

H+

 

H+

.

gravitational

new generators enter .. ..

 

 

 

 

.

.. ..... .............

 

 

.

radiation

the horizon H+

 

 

 

at this caustic

 

..

 

 

 

 

 

 

.

 

 

 

 

 

 

.

 

 

 

 

 

 

...

 

 

 

 

 

 

.

 

 

 

 

 

.

..

.

.

 

 

 

 

 

.

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

.

 

..

 

 

 

 

 

 

.

 

 

. . . . . . . .

M1

Then energy radiated is M1 to energy conversion is

. . . . . . . .

M2

+ M2 M3, so the e ciency, , of mass

=

M1 + M2 M3

= 1

 

M3

(6.85)

 

M1 + M2

 

M1 + M2

 

Assuming that the two black holes are initially approximately stationary, so A1 = 16 M12 and A2 = 16 M22 the area theorem says that

A3 216 M12

+ M22

 

 

 

 

(6.86)

But 16 M3

A3 (with equality at late times), so

 

M3 q

 

 

 

 

 

 

 

 

 

M12 + M22

 

 

 

 

 

(6.87)

Thus

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

p

 

 

 

 

 

 

2

 

 

1

 

 

2

 

 

 

 

 

 

M

2

+ M2

 

 

1

 

 

1

 

 

 

 

1

 

2

 

1

p

 

 

(6.88)

 

 

M + M

 

 

 

 

 

 

 

 

 

The radiated energy could be used to do work, so the area theorem limits the useful energy that can be extracted from black holes in the same way that the 2nd law of thermodynamics limits the e ciency of heat engines.

117

(2)Black holes cannot bifurcate. Consider M3 ! M1 + M2 (with M1 > 0 and M2 > 0). The area theorem now says that

q

M3 M12 + M22 M1 + M2 (6.89)

but energy conservation requires M3 M1 + M2 (with M3 M1 M2 being radiated away). We have a contradiction so the process cannot occur.

118

Chapter 7

Hawking Radiation

7.1Quantization of the Free Scalar Field

Let (x) be a real scalar eld satisfying the Klein-Gordon equation.

 

D @ m2 (x) = 0

(7.1)

Let f g span the space S of solutions. We shall assume that the spacetime is globally hyperbolic, i.e. that 9 a Cauchy surface . A point in the space S then corresponds to a choice of initial data on . The space S has a natural symplectic structure.

Z

$

^ =

dS @

 

 

 

;

 

 

 

 

(= ^ )

 

(7.2)

$

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where @ is de ned by

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

$

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f @ g = f@g g@f

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.3)

`Natural' means that ^ does not depend on the choice of .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

$

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

( ^ ) ( ^ ) 0 = ZS d x p gD @

(7.4)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

...

 

 

..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

..

 

 

 

 

 

 

 

 

 

 

 

. ... ... ...

 

 

 

 

 

 

.. ...

.. ..

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

.

..

 

.

 

... ...

...

 

.

..

 

 

.

.

 

 

 

.

.

.

 

.

.

.

 

.

..

.

..

..

.

 

..

.

 

..

.

 

.

.

 

.

.

 

 

. .

 

.

.

.

 

 

 

 

 

 

 

..

..

 

.

 

 

.

 

 

.

..

.S.

 

 

.

..

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

...

...

...

 

 

 

..

...

..

..

...

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

...

 

...

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

..

 

 

 

 

 

 

 

 

 

 

119

$
dS i @

But

 

 

@

 

 

 

2

 

2

 

D

 

$

=

 

 

(D @ )

 

(D @ )

(7.5)

 

 

 

=

m

m = 0 ;

(7.6)

using the Klein-Gordon equation in the last step.

The antisymmetric form ^ can be brought to a canonical block

diagonal form, with 2 2 blocks

0

1

, by a change of basis (Darboux's

1

0

theorem). Thus, real solutions of the Klein-Gordon equation can be grouped

in pairs ( ; 0) with ^ 0

= 1. It then follows that the complex solution

= ( i 0) =p

2

has unit norm if we de ne its norm jj

jj by jj jj2 = ^ 0

or, equivalently,

 

 

 

jj jj

2

= i Z dS

$

 

 

@

:

(7.7)

More generally, we can introduce a complex basis f ig of solutions of the

Klein-Gordon equation with hermitian inner product de ned by

Z

( i; j) = i j ;

and we can choose this basis such that ( i; is not positive de nite, however, because jj choose the basis f ig such that

j) = ij .

jj2 = jj

0 ( i; j) = ij

 

i;

j

=

0

 

1

B

 

 

 

 

 

 

C

@

 

 

 

 

A

B ( i ; j) = 0

 

i ;

j

=

 

ij

C

(7.8)

This inner product jj2. In fact, we can

(7.9)

We could interpret the complex solution = Pi ai i as the wavefunction of a free particle since ( ; ) is positive-de nite when restricted to such solutions, but this cannot work when interactions are present. It is also inapplicable for real scalar elds. A real solution of the K-G equation can be written as

(x) = X [ai

i(x) + ai

i (x)]

(7.10)

To quantize we pass to the quantum eld

 

(x) = X hai

i(x) + aiy

i (x)i

(7.11)

120

Соседние файлы в предмете Физика