Walker.D_Perturbations_Of_Black_Holes

8/7/2019 Walker.D_Perturbations_Of_Black_Holes

1/34

Perturbations of Black Holes

David Walker

Supervisor: Dr. B. Nolan

15th April 2002

This dissertation is submitted in partial fulfilment of the requirements forthe Mathematical Sciences B. Sc degree. This work is entirely the work ofthe author, except where indicated otherwise.

Signature:

1


2/34

Contents

1 Mathematical Background 4

1.1 Manifolds and tensors . . . . . . . . . . . . . . . . . . . . . . 41.2 The metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 The metric connection . . . . . . . . . . . . . . . . . . . . . . 71.4 Covariant differentiation . . . . . . . . . . . . . . . . . . . . . 81.5 The Riemann tensor . . . . . . . . . . . . . . . . . . . . . . . 81.6 The Ricci tensor . . . . . . . . . . . . . . . . . . . . . . . . . 91.7 The Einstein tensor . . . . . . . . . . . . . . . . . . . . . . . . 9

2 The Vacuum Field Equations in a Spherically Symmetric

Space-time 10

2.1 The non-zero Ricci tensor components . . . . . . . . . . . . . 102.2 The non-zero Einstein tensor components . . . . . . . . . . . . 132.3 The non-zero mixed Einstein tensor components . . . . . . . . 142.4 The contracted Bianchi identities . . . . . . . . . . . . . . . . 142.5 The line element . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Waves Propagating in Schwarzschild Space-time 21

3.1 The wave equation . . . . . . . . . . . . . . . . . . . . . . . . 213.2 Separation of variables . . . . . . . . . . . . . . . . . . . . . . 23

3.3 Stability analysis of the perturbations . . . . . . . . . . . . . . 263.3.1 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.3.2 = i . . . . . . . . . . . . . . . . . . . . . . . . . . 293.3.3 Time dependence . . . . . . . . . . . . . . . . . . . . . 31

2


3/34

Summary

Perturbations of stars and black holes has been one of the main topics ofrelativistic astrophysics for the last few decades. They are of particularimportance today, because of their relevance to gravitational wave astronomy.In this project we shall study the perturbations of Schwarzschild black holes.The motivation for us to study these perturbations will be to assess thestability of the spherically symmetric Schwarzschild solution for black holes.The assumption of spherical symmetry on the black hole makes the solutionphysically unrealistic. In order to determine if this solution has physicalsignificance, we need to study its stability. We do this by studying smallperturbations of the solution, that is, gravitational waves which impinge onthe black hole. These are described by the wave equations in the black holespace-time. We shall then solve these equations to obtain answers as towhether the black-hole solutions are accidents that arise as a consequence ofthe assumption of spherical symmetry only, or will they exist if perturbed?These studies will indeed show that the black hole solutions are stable undersmall perturbations, that all asymmetries are eventually radiated away andthat asymptotically in time, the system settles down to a Schwarzschild blackhole.

3


4/34

1 Mathematical Background

General Relativity is Einsteins theory of the gravitational field. It describesgravity as the curvature of the space-time manifold. In this section, we willdefine some objects which are the basics of the theory.

1.1 Manifolds and tensors

A manifold can be considered as a space which is locally similar to n-dimensional Euclidian space Rn. A more detailed description of a manifold isnot required for the scope of this paper. We will just take an n-dimensionalmanifold to be a set of points such that each point possesses a set of ncoordinates(x1, x2, , xn), where each coordinate ranges over a subset ofthe reals, which may, in particular range from to +.

In physics we want to be able to obtain the same physical results for a prob-lem, no matter what co-ordinate system we use. Therefore, the equations ofphysics must not depend on any particular co-ordinate system.If we take an equation T(xi) = 0 in a co-ordinate system xi, then the objectT must transform to T = 0 where T is the representative of T in anotherco-ordinate system xi.In order to achieve this, we consider objects called tensors whose transfor-

mation laws are now defined.

Note: It should be pointed out at this stage, that from now on, wheneveran index is repeated, it will imply a summation over the index from 1 to n,which is the dimension of the manifold. Repeated indexes are called dummyindices and can be replaced by another index that has not already been used.

A contravariant tensor of rank 1 is defined to be a set of quantities, Xa

in the xa-coordinate system which are associated with some point P on themanifold and which transforms under a change of coordinates according to

Xa =xa

xbXb

where the transformation matrix xa

xbis evaluated at the point P.

4


5/34

We can generalize this definition to obtain contravariant tensors of higher

rankXab... =

xa

xcxb

xd. . . X cd...

Similarly a covariant tensor of rank 1 is defined to be a set of quantities,Xa in the x

a-coordinate system which are associated with some point P onthe manifold and which transforms under a change of coordinates accordingto

Xa =xb

xaXb

where the transformation matrixxb

xa is again evaluated at the point P.

We can again generalize this definition to obtain covariant tensors of higherrank

Xab... =xc

xaxd

xb. . . X cd...

We can now define mixed tensors. A mixed tensor of contravariant rank 1and covariant rank 1 satisfies

Xab =xa

xc

xd

xb

Xcd

Mixed tensors of higher rank transform in the same way that of covariantand contravariant tensors of higher rank do. A mixed tensor of contravariantrank q and covariant rank q is said to have valence (p, q).

On some region of the manifold we can define a tensor field to be someassociation of a tensor of the same valence to every point in that region, thatis

P Ta...b... (P),

where Ta...b... (P) is the value of the tensor evaluated at P.

5


6/34

1.2 The metric

The metric measures distances on the manifold. If we take coordinates

xa = (x0, x1, x2, x3) = (t,r,,)

then the four-dimensional spherically symmetric line element is

ds2 = edt2 edr2 r2d2 r2 sin2 d2

where = (t, r) and = (t, r) are arbitrary functions of t and r and ds isthe distance between adjacent space-time points.

Any symmetric covariant tensor field of rank 2, for example gab(x), defines ametric, which implies that the above metric can be written in the form

ds2 = gab(x)dxadxb

From this, we get an expression for gab for our metric

gab = diag(e, e, r2, r2 sin2 )

We define the determinant of the metric by

g = det(gab)

= e+r4 sin2

The metric is non-singular if g = 0, in which case the inverse of gab existsand is denoted gab.

Hence its contravariant form is

gab = diag(e, e, r2, r2 sin2 )

6


7/34

1.3 The metric connection

We define an affine connection abc to be a quantity which transforms ac-cording to

abc =xa

xdxe

xbxf

xcdef +

xa

xd2xd

xbxc

If our manifold contains an affine connection and a metric, we can define ametric connection to be a special connection which is a combination of themetric and its partial derivatives.

abc =1

2

gad (bgdc + cgdb dgcb)

where a =

xais the partial derivative taken with respect to xa.

Note: The connection is symmetric, (abc = acb) .

We need the metric connection because it is an object which will allow us todifferentiate tensors in such a way that our result will be a tensor.

The metric connection is computed as follows:

000 =1

2g0d (0gd0 + 0gd0 dg00) .

Setting d = 0 to keep g0d non-zero, we get

000 =1

2g00 (0g00 + 0g00 0g00)

=1

2e(t(e

))

=1

2

7


8/34

Computing the other non-zero elements of abc, we get

001 =12

, 011 =12

e,

100 =12

e , 101 =12

, 111 =12

, 122 = re , 133 = re

sin2 ,

212 = r1 , 233 = sin cos ,

313 = r1 , 323 = cot

1.4 Covariant differentiation

If we take the partial derivative of a tensor, we find that the result does nottransform as per the usual tensor transformation law. The reason for this isdue to the curvature of space-time. We deal with this problem by defininga covariant derivative to be a partial derivative plus some correction tomake the result covariant. We therefore define a covariant derivative of atensor of rank 1 to be

cXa = cX

a + abcXb

If abc is an affine connection then cXa is a tensor of type (1,1).

The covariant derivative also has the property that is reduces to the par-tial derivative on a scalar field, i.e.

a = a

The expression for a covariant derivative of a general tensor is given by

cXa...b... = cX

a...b... +

adcX

d...b... +

dbcX

a...d... . . .

1.5 The Riemann tensor

For a Xa, contravariant tensor of rank 1, we define its commutator to be

cdXa dcX

a.

Remembering that cXa = cX

a + abcXb and using the definition for the

covariant derivative of a general tensor, we find that if the connection issymmetric

cdXa dcX

a = RabcdXb

8


9/34

where Rabcd, the Riemann tensor is defined by

Rabcd = cabd d

abc +

ebd

aec

ebc

aed

The fact that the derivatives are non-commuting indicates the presence of acurvature in the space-time Note that if the Riemann tensor vanishes, thecommutator of the tensor also vanishes and hence the manifold is called affineflat.

1.6 The Ricci tensor

We define the Ricci tensor to be a contraction of the Riemann tensor, i.e.

Rab = Rcacb

= ccab b

cac +

dab

cdc

dac

cdb

If we impose another contraction, we can define the Ricci Scalar R as

R = gabRab

1.7 The Einstein tensor

A combination of the Ricci tensor and the Ricci scalar defines the Einsteintensor as

Gab = Rab 1

2gabR.

We shall need this tensor to solve the vacuum field equations

Gab = 0

which shall arise in the next section.

9


10/34

2 The Vacuum Field Equations in a Spheri-

cally Symmetric Space-time

We begin by assuming spherical symmetry. The vacuum field equations ofgeneral relativity are given by

Rab = 0.

Now the Einstein tensor Gab = 0 if and only if Rab = 0, so we will look forsolutions to

Gab = gacGcb = 0,

to determine the unknown functions and in the spherically symmetricline element

ds2 = e(r,t)dt2 e(r,t)dr2 r2d2 r2 sin2 d2

The solutions to Gab = 0 are derived in the following way.

2.1 The non-zero Ricci tensor components

We begin with the Riemann Tensor.

Rabcd = cabd d

abc +

ebd

aec

ebc

aed

With this in mind, and the fact that the Ricci Tensor is defined by

Rab = Rcacb

Rbd = ccbd d

cbc +

ebd

cec

ebc

ced

The elements of the Ricci Tensor are derived in the following way

R00 = cc00 0

c0c +

e00

cec

e0c

ce0

10


11/34

R00 = 0000 0

000 +

e00

0e0

e00

0e0

+ 1100 0

101 +

e00

1e1

e01

1e0

+ 2200 0

202 +

e00

2e2

e02

2e0

+ 3300 0

303 +

e00

3e3

e03

3e0

Seeing which of these con be non-zero, we get

R00 = 1100 0101 + e001e1 e011e0 + e002e2 + e003e3

= r

1

2e

t

1

2

+ e00

1e1

e01

1e0 +

e00

2e2 +

e00

3e3

= r

1

2e

t

1

2

+ 000

101

001

100 +

000

202 +

000

303

+ 100111

101

110 +

100

212 +

100

313

+ 200121

201

120 +

200

222 +

200

323

+ 300131

301

130 +

300

232 +

300

333

again taking the non-zero elements, we are left with

R00 = r

1

2e

t

1

2

+ 000

101

001

100 +

100

111

101

110 +

100

212 +

100

313

=1

2(

e

+

e

(

)) 1

2 +1

2 1

2

1

2e

1

2 +

1

2e

1

2

1

2

1

2

+1

2er1 +

1

2er1

11


12/34

Finally, working through this yields

R00 =1

2e +

1

4e

2

1

4e

1

2 +

1

4

1

42 + r1e

Similarly, the other non-zero elements of the Ricci tensor are

R01 = r1

R11 = 1

2 +

1

4e2

1

4e +

1

2e

1

4

2+

1

4 + r1

R22 = 1

2re +

1

2re e + 1

R33 = sin2 R22

Now we use the fact that R = gabRab and letting a = b to get the non-zeroelements of the Ricci scalar R, we find that

R = g00R00 + g11R11 + g

22R22 + g33R33

= eR00 eR11 r

2R22 r2 sin2 sin2 R22

= eR00 eR11 2r

2R22

= e

1

2e +

1

4e

2

1

4e

1

2 +

1

4

1

42 + r1e

e

1

2 +

1

4e2

1

4e +

1

2e

1

4

2+

1

4 + r1

2r2

1

2re +

1

2re e + 1

Simplifying this equation, we see that

R =1

2e

2 + 2

1

2e

2 + 2

+2r1e ( )+2r2

e 1

12


13/34

2.2 The non-zero Einstein tensor components

We can now calculate the Einstein tensor Gab in the following way.

Gab = Rab 1

2gabR

This gives us

G00 = R00 1

2g00R

=1

2e +

1

4e

2

1

4e

1

2 +

1

4

1

42 + r1e

1

2e

1

2e

2 + 2

1

2e

1

2e

2 + 2

+ 2r1e ( ) + 2r2

e 1

Simplifying this equation gives

G00 = e

e

r

1

r2 +1

r2Similarly, the other non-zero elements are

G01 =1

r,

G11 =

r+

1

r2

e

r2,

G22 =r2

2e

r

r+ +

2

2

2

+

r2

2e

2

2

2

,

G33 = sin2 G22.

13


14/34

2.3 The non-zero mixed Einstein tensor components

We can now evaluate the components of the mixed Einstein tensor Gab

Gab = gacGcb

G00 = g0cGc0

= g00G00 + g01G10 + g

02G20 + g03G30

= g00G00

= e

e

e

r

1

r2

+

1

r2

= e

r 1

r2

+ 1

r2

and similarly, the other non-zero components of the mixed Einstein tensorare

G01 =1

re,

G10 = 1

re,

G11 = e

r

+1

r2 + 1

r2

,

G22 =1

2e

2+

r

r

2

2

+

1

2e

+

2

2

2

,

G33 = G22.

2.4 The contracted Bianchi identities

It can be shown that the Riemann tensor satisfies the identity

Rabcd + Radbc + R

acdb = 0

We can raise and lower indices of tensors according to

T... ......a... = gabT...b...

... ...

andT...a...... ... = g

abT... ......b...

14


15/34

Note: The metrics gab and gab also have the property that

gabgbc = ca

where ca is known as the Kronecker delta and has the effect of turning theindex c into a.

Lowering the first index with the metric, we find that

Rabcd + Rabdc + Racdb = gaeRebcd + gaeR

ebdc + gaeR

ecdb

= gae (Rebcd + R

ebdc + R

ecdb)

= 0.

It can be shown that the Riemann tensor satisfies a set of identities calledthe Bianchi identities

aRdebc + cRdeab + bRdeca = 0

This implies that

gdb [aRdebc + cRdeab + bRdeca] = 0.

Since agdb = 0 and agdb = 0, we can take gdb in and out of covariant

derivatives at will. We get

aRbebc + cR

beab + bR

beac = 0

From the definition of the Riemann tensor, it follows directly that the tensoris anti-symmetric on its last pair of indices, i.e.

Rabcd = Rabdc.

Now, using this anti-symmetry on the indices b and a, we get

aRbebc cR

beba + bR

beac = 0

and so by a contraction,

aRec cRea + bRbeac = 0.

15


16/34

These equations are called the contracted Bianchi identities.

Let us now contract a second time on the indices e and c,

gec

aRec cRea + bRbeac

= 0

giving us

aRcc cR

ca + bR

bcca = 0

agcdRdc 2bR

ba = 0

2bRba ag

cdRdc = 0

2bRb

a aR = 0

Since aR = babR, we get

b

Rba

1

2baR

= 0

Now raising the index a with gca we get

b

Rcb

1

2gcbR

= 0

which gives uscG

cb = 0

Now lowering the index b with gba, we get

cGcb = c

gbaGca

= gbacG

ca

= 0

This gives us the contracted Bianchi identities

bGba = 0

16


17/34

2.5 The line element

We shall now determine the functions (t, r) and (t, r) for the sphericallysymmetric line element


using the the field equations Gab = 0 and also the contracted Bianchi identi-ties bG

ba = 0.

If bGba = 0, then

aGab +

aacG

cb

cbaG

ac = 0

In the case of spherical symmetry, this shows that

if G00 = G01 = G

11 = 0,

then G22 = 0

Note: This is proven by setting a = 1 for the contracted Bianchi identitiesbG

ba = 0, and remembering that the only non-zero elements of G

ab are

G00, G01, G

10, G

11, G

22 and G

33. So we are left with 3 independent equations to

solve, which are

G00 = e

r

1

r2

+

1

r2= 0,

G10 = 1

re = 0,

G11 = e

r+

1

r2

+

1

r2= 0.

or more simply

e

r 1r2

+ 1r2 = 0, (1)

e

r+

1

r2

1

r2= 0, (2)

= 0. (3)

17


18/34

Adding (1) and (2) together, we get

+ = 0

Integrating this, we find that

+ = f(t)

where f(t) is a function of integration.

Now, from (3), since

= 0,

= g(r)

where g(r) is another function of integration. This implies that (1) is anordinary differential equation and can be written in the form

re e + 1 = 0

e re = 1

(re) = 1

re

= r + c e = 1 +

c

r

e =

1 +c

r

1

Now looking at


From above we have + = f(t), this implies that = + f(t)

Thus,

edt2 = e+f(t)dt2

= eef(t)dt2

= edT2

18


19/34

where,

dT2 = ef(t)dt2

dT = ef(t)2 dt

T =

e

f(t)2 dt


becomesds2 = edT2 edr2 r2d2 r2 sin2 d2.

Renaming T as t we get

ds2 = edt2 edr2 r2d2 r2 sin2 d2.

Now substituting for e we get

ds2 =

1 +c

r

dt2

1 +

c

r

1dr2 r2d2 r2 sin2 d2.

Now for flat space-time, we know that

e = 1.

For a small perturbation, i.e. slowly moving bodies in a weak gravitationalfield that are a large distance from the source, we can write

e = 1 + 2,

where is small and the constant 2 is added for convenience.

It can be shown that in this approximation, satisfies the equations of the

Newtonian gravitational potential, and so for a spherical body of mass m,

= m

r.

This implies that

e = 1 2m

r

19


20/34

and comparing this with our current line element, we find that c = 2m.

We can now define the Schwarzschild line element as

ds2 =

1

2m

r

dt2

1

2m

r

1dr2 r2d2 r2 sin2 d2

where m may be thought of as the mass of a spherical object which generatesthe gravitational field.

20


21/34

3 Waves Propagating in Schwarzschild Space-

time

In this section we shall derive the wave equation for curved space-time. Weshall then use the process of separation of variables to find that the equationsgoverning the perturbations of a spherically symmetric system are separablein all four of the variables t,r, and . We shall then analyze the solutionsto these equations to determine whether the system is stable.

3.1 The wave equation

The wave equation in four dimensions is

2

t2

2

x2

2

y2

2

z2= 0

or

2

t2+ 2 = 0.

Now in Minkowski Space-time

ab = diag(1, 1, 1, 1)

ab

ab = 0.For Special Relativity we can denote this

= gaba(b) = 0

This form of the wave equation allows us to generalize to curved space-time.

Next we denote the contravariant vector field

b = b = b

This gives us

a(b) = ab

= ab cabc

= ab cabc

21


22/34

and hence

= gab(ab cabc)where

gab = diag

1

2m

r

,

1 2m

r

1, r2, r2 sin2

and

gab = diag

1

2m

r

1,

1

2m

r

, r2, r2 sin2

Therefore computing gives

= g00(20 c00c) + g

11(21 c11c)

+ g22(22 c22c) + g

33(23 c33c).

Computing the relevant connections yields

c00c =m

r2 1 2m

r rc11c =

m

r2

1

2m

r

1

r

c22c = r

1

2m

r

r

c33c = r sin2

1

2m

r

r sin cos

and so our wave equation is

=

1 2mr

12t

1 2m

r

2r

2(r m)

R2

r

r2(csc2 2 + cot + 2 )

= 0

22


23/34

3.2 Separation of variables

We begin by letting = P(t, r)Y(, )

=

1

2m

r

12t P(t, r)Y(, )

1

2m

r

2r P(t, r)Y(, )

2(r m)

r2rP(t, r)Y(, )

1

r2

csc2 P(t, r)2Y(, ) + P(t, r)

2 Y(, ) + cot P(t, r)Y(, )

= 0

Separating these out and assuming that P(t, r)Y(, ) = 0, we see that

r2

1 2m

r

12t P(t, r)

P(t, r) r2

1

2m

r

2r P(t, r)

P(t, r) 2(r m)

rP(t, r)

P(t, r)= c

and

csc2 2Y(, )

Y(, )+ cot

Y(, )

Y(, )+

2 Y(, )

Y(, )= c

We begin by looking for solutions of the form Y(, ) = ()() and sub-stituting into the above equation, we find that

1

sin2

+

+ cot

= c

The solution to this equation is a well known one. The subject is known asSpherical Harmonics. It has been studied extensively and yields a solutionof the form

(Aln cos(n) + Bln sin(n))sinn P(n)l (cos )

wheren = 0, 1, . . . , l , l = 0, 1, 2, . . . ,

Aln, Bln are constants,

23


24/34

and P(n)l (cos ) is the nth derivative of the Legendre polynomial Pl with re-

spect to cos .

So we get

Y(, ) =

l=0

n=l

(Aln cos(n) + Bln sin(n))sin

n P(n)

l (cos )

We now have to deal with P(t, r). Strictly speaking we should be work-ing with Pln(r, t) but for convenience we shall drop the subscripts. So

r2

1 2m

r

12t P(t, r)

P(t, r)r2

1

2m

r

2r P(t, r)

P(t, r)2(rm)

rP(t, r)

P(t, r)= n(n+1)

Using another separation of variables by letting P(t, r) = W(t)Q(r) we findthat

W

(t)

W(t)=

1

2m

r

2Q

(r)

Q(r)+

2(r m)(r 2m)

r3Q

(r)

Q(r)

n(n + 1)(r 2m)

r3= 2

This gives usW(t) = eit

and1

2m

r

2Q

(r)+2(r m)(r 2m)

r3Q

(r)+

2

n(n + 1)(r 2m)

r3

Q(r) = 0

Letting Q(r) = R(r)r

, we find that

Q(r) = R(r)r2

+ R(r)r

,

Q(r) = 2R(r)

r3 2

R(r)

r2+

R(r)

r

24


25/34

Substituting these expressions for Q(r), Q(r), and Q(r) into the above equa-

tion we find that

(r 2m)2

r3R

(r)+2m(r 2m)

r4R

(r)+

2

r

(r 2m)

r5(rn(n + 1) + 2m)

R(r) = 0.

Now we make a substitution using tortoise coordinates R(r) = T(r),where

r = r + 2m ln r

2m 1

we find that

R(r) =r

(r 2m)T(r),

R(r) =r2

(r 2m)2T(r)

2m

(r 2m)2T(r)

Again substituting these expressions for R(r), R(r), and R(r) into the aboveequation, we find that

T

(r) +

2

1

2m

rrn(n + 1) + 2m

r3

T(r) = 0

The importance of the variable r arises from the fact that it ranges from to + exhausts the entire part of space-time that is accessible to observersoutside the horizon.

This equation can be written in the form

T

(r) + V(r)T(r) = 0,

where the potential V(r) is given by

V(r) = 2

1

2m

r

rn(n + 1) + 2m

r3

= 2 V(r)

25


26/34

3.3 Stability analysis of the perturbations

For each value of C, we have a solution P(t, r) satisfying

P(t, r) = W(t)Q(r)

= eitQ(r)

r

= eitT(r)

r

wherer = r + 2m ln

r

2m 1

and T(r) satisfies the ordinary differential equation

T(r) +

2 V(r)

T(r) = 0. (4)

Note: The effective potential V(r) is defined by

V(r) =

1

2m

r

rn(n + 1) + 2m

r3

,

and crucially for what follows, satisfies V(r) > 0.

Also, since

r = r + 2m ln r

2m 1

we get that

r r

r 2m r

Next we define

limrV(r) = V

limr2m

V(r) = V

26


27/34

We can now approximate the equation

T

(r) + V(r)T(r) = 0,

by

T

(r) + VT(r) = 0. (5)

The solutions to these equations describe the asymptotic behavior of thesolution at r = (spacial infinity) and r = 2m (the event horizon).

3.3.1 = 0

We want to show that there are no time dependent solutions, i.e. = 0,which are regular at both r = 2m and r = .

If = 0, then both V = 0 and V = 0.So (5) becomes

T(r) = 0

and we get the solutions

T(r) = Ar + B (at r = )

T(r) = ar + b (at r = +).

Now we can only get regular solutions at r = if A = a = 0.This gives us the solutions

T(r) = B (at r = )

T(r) = b (at r = +).

Now from (4) with = 0, we find that

T

(r) = V(r)T(r).

Suppose that B > 0, thenT(r) > 0

27


28/34

since both V(r) > 0 and T(r) > 0 for sufficiently large negative r.

Now, as r

T(r) B

T(r) 0.

Hence T(r) increases away from B and so stays positive. Thus the conditionthat T(r) > 0 persists.From this we can say that the function does not have an inflection point andtherefore cannot level off again. This implies that the function has no localmaximum and hence that T(r) > 0.

But as r +

T(r) b

T(r) 0.

r*

T

Figure 1: A function that starts with zero slope, increases, and the ends withzero slope must have an inflection point.

We know that the function cannot start at B with zero slope, increase, andthen level off again with zero slope at b without having an inflection point, asin Figure 1 and therefore there is a contradiction and this case can be ruledout.

28


29/34

The where B < 0 can be ruled out in a similar fashion. The case where

B = 0 simply leads to the trivial solution T(r) 0 which we are not inter-ested in.

This proves that no physical static solution exists which is regular at both theevent horizon (r = 2m) and at spatial infinity (r = ), i.e. that the blackhole cannot support an external static field which could perhaps destroy theblack hole nature of the space-time.

3.3.2 = i

We shall now show that equation (4) with = i, where R, cannothave solutions which are finite at both r = and r = +.

Note: This corresponds to P(t, r) = eit T(R)r

increasing exponentially intime.

In this case V = 2 and V = 2, so by approximating

T

(r) + V(r)T(r) = 0

using (5), we find that the solutions to

T

(r) 2T(r) = 0

describe the asymptotic behavior of the solution at r = and r = +.

The solutions to these equations are

T(r) = Aer + Ber (at r = )

T(r) = aer + ber (at r = +)

Now we can only get regular solutions at r = if B = a = 0.This gives us the solutions

T(r) = Aer (at r = )

T(r) = ber (at r = +)

Now from (4) with = i, we find that

T(r) =

2 + V(r)

T(r).

29


30/34

Suppose that A > 0, then

T

(r) > 0

since 2 > 0, V(r) > 0 and T(r) > 0 for sufficiently large negative r.

Now, as r

T(r) 0

T(r) 0.

Hence T(r) increases away from 0 and so stays positive. Thus the conditionthat T(r) > 0 persists.

From this we can say that the function does not have an inflection pointand therefore cannot level off again. This implies that the function has nolocal maximum and hence that T(r) > 0.

But as r +

T(r) 0

T(r) 0.

r*

T

Figure 2: A function that starts with zero slope, increases to a maximum,and then decreases to zero must have inflection points.

30


31/34

We know that the function cannot start at 0 with zero slope, increase to

a maximum, and then decrease to 0 with zero slope at 0 without havinginflection points, as in Figure 2 and therefore there is a contradiction andthis case can be ruled out.

The case where A < 0 can be ruled out in a similar fashion and the casewhere A = 0 is simply the trivial case T(r) 0 which we are not interestedin.

This shows that a perturbation which is initially finite cannot undergo expo-nential growth and hence is a good indicator of the stability of the solutions.

3.3.3 Time dependence

When we include a time dependence rather than factoring it out, the equa-tions governing perturbations of the Schwarzschild black hole can be ex-pressed in the form

2Z

t2=

2Z

r2

V(r)Z, (6)

where the function Z characterizes the perturbation, r is the tortoise coor-dinate and V(r) is a smooth positive potential which goes to zero at both

spacial infinity (r = ) and the event horizon (r = ).

We know that the complex conjugate Z ofZ must satisfy the same equation.A useful bound on the growth of Z is attained as follows.

Multiplying (6) by Zt

and rearranging gives us

Z

t

2Z

t2+

Z

tV Z =

Z

t

2Z

r2

and integrating both sides

Z

t2

Zt2

+ V Z

Zt

dr =

Z

t2

Zr2

dr. (7)

An integration by parts of the right hand side of (7) yields

Z

t

2Z

r2

dr =

Zt Zr

r=

r=

Z

r

2Z

tr

dr

31


32/34

It can be shown that Zt Zr

r=

r

= 0

.This leaves us with

Z

t

2Z

r2

dr =

Z

r

2Z

tr

dr

and substituting this into (7) gives us

Z

t

2Z

t2+ V Z

Z

t dr =

Z

r

2Z

tr dr

or

Z

t

2Z

t2+

Z

r

2Z

tr+ V Z

Z

t

dr = 0.

If we add this equation to its complex conjugate we get

Z

t

2Z

t2+

Z

t

2Z

t2

+

Z

r

2Z

tr+

Z

r

2Z

tr

+ V

Z

Z

t+ Z

Z

t

dr = 0

or

t

Zt

Zt

+

t

Zr

Zr

+ V

t

ZZ

= 0.

Noticing that xx = |x|2, the above equation becomes

t

Zt2

+

Zr2

+ V |Z|2

dr = 0.

Now assuming that the derivatives are continuous, we can take the derivativeoutside of the integral to give

t

Zt 2

+Zr

2

+ V |Z|2

dr

= 0.

and integrating this with respect to t we have our bound on the integral

Zt2

+

Zr2

+ V |Z|2

dr = C

32


33/34

where C is some constant.

Zt2

dr < C, since V > 0

Zt is bounded

The bound on the time derivative of Z implies that Z cannot grow fasterthan a linear function of time. This is a very good indicator that the fieldremains finite for all times t > 0.

33


34/34

Bibliography

[1] DInverno R. (1992). Introducing Einsteins Relativity, Oxford Univer-sity Press.

[2] Chandrasekhar S. (1983). The Mathematical Theory of Black Holes, Ox-ford University Press.

[3] Hawking, S. W. and Ellis, G. F. R. (1973). The large scale structureof space-time, Cambridge University Press.

[4] Boyce, W. E. and DiPrima, R. C. (1986). Elementary Differential Equa-tions and Boundary Value Problems, John Wiley & Sons, Inc.

[5] Wald, R. M. (1978). Note on the stability of the Schwarzschild metric,Enrico Fermi Institute, University of Chicago.

34

Documents

Walker.D_Perturbations_Of_Black_Holes