exam general relativity 1

7/27/2019 Exam General Relativity 1

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Exam to General Relativity

(Winter term 2012/2013)

Prof. V. F. Mukhanov, LMU Munchen 01.02.2012

Solution!

Problem: 1 2 3

Points:


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Solution!

1 Christoffel symbols and Geodesic equation

(12 Points)The Christoffel symbols transform under general coordinate transformationsx x(x) as

(x) =x

xx

xx

x(x) +

2x

xxx

x. (1)

a) Prove that under infinitesimal general coordinate transformations x x = x + x, x = (x), the variation

(x) (x)

(x) (2)

can be written as

=

+

+

+ . (3)

Hint: Expand (1) and keep only terms first order in . Also keep in mindthat in (2) both terms are functions of x, not x

(4 Points)

b) Show that the geodesic equation

d2x

d2+ (x)

dx

d

dx

d= 0 (4)

is invariant under general coordinate transformations of x by verifyingthat (4) varies under x = (x) into itself.

(4 Points)

c) Determine the (infinitesimal) reparametrizations () that leavethe geodesic equation invariant by varying (4) under

x = x

d

d

where is a function only of .

Hint: Check for which () the variation of the geodesic equation is zeroby organizing the variation into terms proportional to , and .

(4 Points)

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Solution!

a) From x x = x (x) we have to first order in

x

x=

,x

x= +

,2x

xx=

.

Insertion into (1) gives (again to first order in )

(x) = (x) +

+

+ . (5)

Next, we need (x) rather than (x), so we compute

(x) = (x ) =

(x)

,

ignoring again higher order terms in . Using this in (5) we recover (3),which is what we wanted to prove.

b) With x = (x) we get by the chain rule

x = x , x =

xx x .

Varying the left-hand side of (4), using (3), then gives

x + (x)x x

= x + x

x + ,xx x + 2x

x

= xx

x ,x x 2

xx

+

+

+

+

x x

=

x + (x)x x

= 0 ,

using the geodesic equation again. Thus, the geodesic equation is invariant.

c) From x = x we get

x = x + x , x = ...x + 2x + x .

Therefore,

x + (x)x x

=

...x + 2x + x + , x

xx + 2

x + x

x

= d

d

x + (x)x

x

+ 2

x + (x)x x

+ x

= x ,

using that first two terms vanish by the geodesic equation. Since x isgenerally non-zero we conclude

= 0 () = a + b ,

with a, b constant. Therefore, only affine parameters are allowed.

Remark: One can write an invariant equation using a covariant derivative

Dx =

d

dx e1e x , e() :=

g x()x()

asDx

+ x x = 0 .

The usual equation is recovered by choosing the gauge e = const. (e = 0).

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Solution!

2 Perturbation theory (12 Points)

Consider small perturbations around the flat Minkowski background,

g = + h ,

where |h| 1. The linearized Ricci tensor is

R =1

2 (h, + h,)

1

2(h + h,) ,

where h = h. Assume only scalar perturbations, i. e. ,

h00 = 2 , h0i = 0 , hij = 2ij .

a) Give the components of the Einstein tensor.Consistency check: the Ricci scalar reads

R = 2 + 6 4.

(6 Points)

b) Solve the Einstein equations for matter of the form

T = 0

0M (r) ,

using perturbation theory. We impose boundary conditions such that theperturbations vanish at spatial infinity.

Hint: You may use the standard solution of Poissons equation.

(4 Points)

c) What is a possible extension of this solution to an exact solution of thefull Einstein equations (i. e. without using perturbation theory)?

(2 Points)

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Solution!

a) The Einstein tensor is

G = R 12

gR

From above we have

R =1

2 (h, + h,)

1

2(h + h,)

R = h h

With only scalar perturbations we have h = 2 6 and the Ricci scalaris

R = 2 + 2 2 + 6 = 2 + 6 4

So the Einstein tensor is

G = 12

(h, + h,) 12

(h + h,) ( + 3 2)

For the different components we get:

G00 =1

2 (h0,0 + h0,0)

1

2(h00 + h,00) 00( + 3 2) =

= h00,00 1

2(h00 + h,00) ( + 3 2) =

= 2 + 3 3 + 2 =

= 2

G0i =1

2 (h0,i + hi,0)

1

2(h0i + h,0i) 0i( + 3 2) =

= 12

(h0,i + hi,0) 12

h,0i =

=1

2h00,i0

1

2hii,0i

1

2h,0i =

= + 3 =

= 2

Gij =1

2 (hi,j + hj,i)

1

2(hij + h,ij) ij( + 3 2) =

= 1

2hii,ji

1

2hjj,ij

1

2(hij + h,ij) + ij( + 3 2) =

= 2,ij (ij + ,ij 3,ij) + ij + 3ij 2ij =

= 2,ij ,ij + 3,ij + ( + + 3 2)ij =

= ( ),ij + ij

2 ( )

b) From the off-diagonal components we get

=

As the source is time independent, so are the metric perturbations, so =0 and the diagonal spatial components are satisfied. The 00 componentfinally reads

= 4GM(r)

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Solution!

The solution to this equation is

= Gm

r

This gives the following metric to first order

g =

1 2GMr 0 0 00 1 2GMr 0 00 0 1 2GMr 00 0 0 1 2GMr

c) The full solution should be the Schwarzschild metric of a black hole, asthe matter source is a point mass at the origin. The above metric is the

Taylor expansion of the metric in isotropic coordinates to first order.

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Solution!

3 Universe filled by radiation (16 Points)

Consider a universe described by a metric g and filled by the electromagneticfield only. The energy-momentum tensor of the field is given by

T = gFF +

1

4gFF

,

where F A A is the electromagnetic field strenght and A is theelectromagnetic potential.

a) By using the electromagnetic field equation, i.e. F = 0, show that

T is conserved: T = 0.

Hint: Use that the following Bianchi identity holds in a curved spacetime:

F

+

F

+

F

= 0.

(4 Points)

b) We specialize now to a flat FRW universe,

ds2 = dt2 a2(t)dr2 .

Find T00, T0i and Tij expressed through Ei F0i and Bi 12ijkFjk ,

where ijk is the Levi-Civita symbol.

Hint: In order to express Fij via Bk, use the following property of theLevi-Civita symbol: ijkimn = jmkn jnkm.

(4 Points)

c) Assuming the electromagnetic field is isotropic, i.e. Ei Bi ni atany point of spacetime, where n = r/r, determine the (averaged) energydensity T00 and pressure p defined by Tij pgij . Here, theaverage is over the solid angle , ..

d4 (..). Give the equation of

state by determining p/.

Hint: Recall that by symmetry reasons ninj ij and determine thefactor of proportionality.

(4 Points)

d) Find a(t) and (t) from the Friedmann equations, where has been definedin the previous subproblem.

Hint: assume that a(t = 0) = 0.

(4 Points)

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Solution!

a) One has

T = FF +

1

2FF ,

where we have used the field equation of the electromagnetic field and thefact that the covariant derivative of the metric tensor is zero.

Exchanging and in the first term of RHS into and , respectively,one has

T = FF +

1

2FF

= F

F 1

2F

= F

1

2

F F

1

2F

=1

2F

F + F + F

F[F] = 0 ,

where we have used in the third and fourth lines the skew-symmetry ofthe field strength F.

b) First, lets calculate FF:

FF = F00F

00 + 2F0iF0i + FijF

ij = 2

a2 |E|

2 |B|2

a2 ,

where |E|2 and |B|2 have been defined as ijEiEj and ijBiBj , respec-

tively.

Hence, one has

T00 = F0iF0i +

1

4FF

=1

2a2

|E|2 +

|B|2

a2

,

T0i = FikF0k =

1

a2E B

i

,

Tij = Fi0Fj0 FikFk

j +1

4gijFF

= EiEj 1

a2BiBj +

1

2ij

|E|2 +

|B|2

a2

.

c) Since

ninj 1

4

d ninj =

1

3ij ,

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Solution!

one immediately obtains

T00 =1

2a2

|E|2 +

|B|2

a2

= ,

T0i = 0 ,

Tij = 1

6a2

|E|2 +

|B|2

a2

gij = pgij .

Consequently, w p/ = 1/3.

d) Friedmann equations read

a

2

=

8 G

3 a

2

a = 4G

3

+ 3p

a ,

where we have put the speed of light c to be unity and dot means a diffe-rentiation w.r.t. time t. Taking into account that p = 13 and eliminating from the second equation by using the first one, one gets

aa + a2 = 0 aa =C12

a =

C1t + C2 ,

where C1,2 are constants of integration. Assuming that a(t = 0) = 0, onehas that C2 = 0, so that one finally finds

a t1/2 , and =3

32Gt2.

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exam general relativity 1

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