spherically symmetric solutions for five dimensional topological gravity

5/25/2018 Spherically Symmetric Solutions for Five Dimensional Topological Gravity

1/58

AMERICAN UNIVERSITY OF BEIRUT

Spherically Symmetric Solutions for Five Dimensional

Topological Gravity

by

SAMAR JEAN BECHARA

A thesissubmitted in partial fulfillment of the requirements

for the degree of Master of Science

to the Department of Physicsof the Faculty of Arts and Sciences

at the American University of Beirut

Beirut, LebanonJanuary 2011


2/58


SPHERICALLY SYMMETRIC SOLUTIONS FOR FIVE DIMENSIONALTOPOLOGICAL GRAVITY

by

SAMAR JEAN BECHARA

Approved by:

Dr.Ali Chamseddine, Professor AdvisorPhysics

Dr.Jihad Touma, Professor Member of CommitteePhysics

Dr.Mounib El Eid, Professor Member of CommitteePhysics

Date of thesis defense: January 17, 2011


3/58


THESIS RELEASE FORM

I, SAMAR JEAN BECHARA

authorize the American University of Beirut to supply copies of mythesis to libraries or individuals upon request.

do not authorize the American University of Beirut to supply copies of mythesis to libraries or individuals for a period of two years starting with the dateof the thesis defense.

Signature

Date


4/58

ACKNOWLEDGEMENTS

I would like to express my gratitude to all those who helped me in completing thisthesis.

Above all, my gratefulness goes to my supervisor, Dr. Ali Chamseddine, forhis continous help and follow up on the content of this thesis and to my committeemembers, Dr. Jihad Touma and Dr. Mounib Eid, for their valuable comments.Moreover, I would like to thank all the professors who taught me at AUB through-out the years.

I also offer my special regards and blessings to all my friends and colleagues,especially Ola Malaeb, for their moral and technical support. Your friendship andcollaboration means a great deal to me.

It would not have been possible to complete this thesis without the supportand encouragement of my family members. Words cannot express the thanks I oweto each one of you for always being there for me. I would never be able to pay backall the love and sacrifice you have provided me with.

v


5/58

AN ABSTRACT OF THE THESIS OF

Samar Bechara for Master of ScienceMajor : Physics

Title : Spherically Symmetric Solutions for Five Dimensional Topological Gravity

Einstein constructed the theory of general relativity in terms of a metricand a connection coefficient. After the attempt for constructing a gauge theoryof gravity by Utiyama and Kibble, another formulation of general relativity wasintroduced. This formulation was in terms of the veirbeins and the spin connections.By gauging the SO(1,5) and the SO(2,4) groups, Chamseddine then constructeda topological action that contained a Gauss-Bonnet term, an Einstein term, anda cosmological constant. Taking a spherically symmetric five-dimensional metricansatz, we obtained a set of 7 solutions for the derived equations of motion. Three ofthese solutions were considered physical and were analyzed. None of them containedsingularities except for coordinate singularities in the de Sitter space, where a set ofcoordinate transformations has been applied to eliminate such a singularity. Finally,

after reducing the metric in five dimensions to a metric in four dimensions with anadditional scalar field equal toe4

4, we realized that the same set of solutions was still

valid.

vi


6/58

Contents

1 Introduction 2

1.1 Historical Background . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Aim and Organization . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Solutions for the Five-Dimensional Equations of Motion 82.1 Constructing the Five-Dimensional Spherically Symmetric Line Ele-

ment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 Equations of Motion in Component Form . . . . . . . . . . . . . . . . 112.3 Calculating the Curvature tensor components . . . . . . . . . . . . . 132.4 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Solutions For Equations of Motion . . . . . . . . . . . . . . . . . . . . 16

3 Analyzing the Physical Solutions 20

3.1 Analyzing the First Physical Solution . . . . . . . . . . . . . . . . . . 203.1.1 Geodesic Equations . . . . . . . . . . . . . . . . . . . . . . . . 233.1.2 Coordinate Transformations . . . . . . . . . . . . . . . . . . . 26

3.2 Analyzing the Second Physical Solution . . . . . . . . . . . . . . . . . 283.2.1 Geodesic Equations . . . . . . . . . . . . . . . . . . . . . . . . 323.2.2 Coordinate Transformations . . . . . . . . . . . . . . . . . . . 34

3.3 Analyzing the Third Physical Solution . . . . . . . . . . . . . . . . . 363.3.1 Geodesic Equations . . . . . . . . . . . . . . . . . . . . . . . . 393.3.2 Coordinate Transformations . . . . . . . . . . . . . . . . . . . 41

4 Dimensionally Reducing the Action 46

5 Conclusion and Future Work 50

vii


7/58

Chapter 1

Introduction

1.1 Historical Background

According to Einsteins theory of general relativity, postulated at the end of 1915,

gravity is considered as a manifestation of the curvature of spacetime.[1]

Three main principles lead to Einsteins formulation. These principles are:

1. The principle of general covariance: laws of physics must take the same form in

any reference frame.

2. The principle of equivalence: All objects fall at the same acceleration

irrespective of their masses.

3. Machs principle: the local inertial properties of an object are determined bythe total mass distribution in the universe.

The first principle lead Einstein to suggest that the equations should be

written in tensorial form. The second and third principles lead respectively to the

ideas that gravity should be considered as curvature of spacetime and that this

curvature is determined by the distribution of matter.

In the theory of relativity, we consider a manifold M, called Reimannian

manifold, in which the notion of distance is given by:

1


8/58

ds2 =gdxdx (1.1)

where g is a symmetric tensor called the metric.

On the manifold M, we define coordinates{xa}. Each vector on themanifold can then be expressed as a linear expansion of a set of coordinate basis

vectors{ea} such that v=vaea. The same applies for tensors.Since we are dealing with a curved space, we have to take into account

that not only the coordinates of a tensor change during a coordinate

transformation but also the basis vectors themselves.

This is why the notion of covariant derivatives has been introduced.The

covariant derivative of a vector v, for instance, with respect to a coordinate is

given by:

v = (ve) = (v)e+ v(e)= (v

+ v)e (1.2)

where

e= e

and

are known as the connection coefficients.

For a torsionless manifold, the connection is symmetric in its last two

indices and can be written in terms of the metric tensor according to:

=1

2g(g+ g g) (1.3)

We can define the curvature of the manifold in terms of the connection

coefficients as:

2


9/58

R = + (1.4)

If the curvature tensor vanishes, the manifold considered is flat.

The Einstein field equations for general relativity are given by

G=R 12

gR = 8T (1.5)

where Tis called the stress-energy tensor.[1]

Another formulation of general relativity is given in terms of the veilbeins

and spin connections.

This formulation arises when we try to build gravity as a gauge theory

that is invariant under a Poincare transformation. The veilbeins are then the

gauge fields associated with translations and spin connections are the gauge fields

associated with local lorentz transformations. [2]

The veilbein can be understood as follows: at any point P on the manifold, we can

define a set of basis vectors{ea} such that this set is orthonormal. Orthonormalityimplies that

g(ea, eb) =ab

where represents the Minkowski metric.1

This set of orthonormal basis vectors is generally known as the veilbein. One way

to define this set is by taking:

ea =x

xa (1.6)

where x correspond to the coordinates of a point P on the manifold and

xa are the coordinates of its mapping in the tangent space of M at P.[4]

Note that Greek indices are usually used to denote coordinates in the

1in this paper, the signature of the metric is taken as (+,-,-,...,-)

3


10/58

curved space, and Latin indices are used to denote coordinates on the flat, tangent

space.

In terms of the veilbein, the notion of distance is defined by:

ds2 = abea eb (1.7)

= abeae

bdx

dx =gdxdx (1.8)

The relation between the metric and the veilbein is then given by:

g=eaea (1.9)

The spin connection is defined such that the covariant derivative of the

veilbein vanishes[2]. The covariant derivative of the veilbein is given by:

Dea=e

a+ w

abe

b

e

a (1.10)

are the Christoffel connections and wab are the spin conections.

For a torsionless manifold, the spin connection can be expressed in terms of the

veilbeins [2]. It is given by:

wab =1

2eae

b((e) (e) + (e)) (1.11)

where

= (ec ec)ec (1.12)

The curvature in terms of the spin connection can then be defined as:

Rab=ab ab + ak bk ak bk (1.13)

4


11/58

The relation between Rab(w) and R() satisfies:

R() =eaebR

ab(w) (1.14)

The veilbein and the spin connection were actually introduced when

Utiyama [5], in 1955, followed by Kibble [6] in 1960 , attempted to construct

gravity as a gauge theory in which the lagrangian is invariant under general

inhomogeneous lorentz transformations. This has been done by introducing the

veirbein and the spin connection as gauge fields. It was then understood [7] that if

we consider the veirbein2 and the spin connection as independent gauge fields,

then the Einstein action would have a constrained symmetry where the torsion is

set to zero.

Based on this understanding, Chamseddine was able to construct [2, 3] a

topological gauge theory of gravity in five dimensions and then to generalize it to

all odd dimensions. The action is based on the Chern-Simons five-form given by

5 =

F F A 12

F A A A + 110

A A A A A

(1.15)

where F =dA + A A.

By letting

Aab =ab, Aa5 =ea, a= 1, ..., 5 (1.16)

the action

I5 = kM5

5 (1.17)

takes the form:

2In four dimensions, the veilbein is called veirbein, and in five dimensions it is referred to asfunfbein.

5


12/58

I5 = 3 abcde(eaRbc

Rde

2

3

ea

eb

ec

Rde +

1

5

2ea

eb

ec

ed

ee) (1.18)

The first term, which is quadratic in R, is known as a Gauss-Bonnet term.

The second term, which is linear in R, is the Einstein term, and the third term is

the cosmological constant term, where is a cosmological constant that could take

the values:1, 0, 1.3

The equations of motion for this action has been derived by

Chamseddine[2, 3]. They are given by:

abcde(Rab ea eb) (Rcd ec ed) = 0 (1.19)

abcde(Rab ea eb) Tc = 0 (1.20)

whereTa =dea + ab eb (1.21)

is the torsion.

1.2 Aim and Organization

Our aim, in the second chapter, is to construct a spherically symmetricfive-dimensional space-time configuration for which we find exact solutions of the

above equations of motion. In the third chapter, we analyze particular solutions of

physical interest. In the fourth chapter, we reduce our five-dimensional action to a

four-dimensional action plus a scalar field, where the scalar field is taken to be e44.

We then check that the obtained solutions are still valid.

3In this paper, we have relabelled in [3] to be.

6


13/58

Chapter 2

Solutions for the Five-Dimensional

Equations of Motion

2.1 Constructing the Five-Dimensional

Spherically Symmetric Line Element

The general expression for the line element in 5 dimensions is given by:

ds2 =gdxdx

where = 0, 1, 2, 3, 4.

We assume that the metric is isotropic in three of its spatial coordinates

x1, x2, and x3.

Denote: xs=(x1, x2, x3) , x4 and x0 t .Because of isotropy, the metric ds2 depends on x1, x2, and x3 only through

their rotational invariants: xs

xs

r2 , dxs

dxs,and xs

dxs. Thus, the most general

form of the metric must be:

7


14/58

ds2 = A(t, r)dt2

B(t, r)dt(xs

dxs)

C(t, r)d(xs

dxs)

D(t, r)(xs

dxs)2

E(t, r)(dxs dxs) F(t, r)d2

Transform xs (x1, x2, x3) into spherical polar coordinates.

x1 = r sin cos

x2 = r sin sin

x3 = r cos (2.1)

The rotational invariants are now given by: xs xs=r2 ,xs dxs=rdr ,anddxsdxs=dr2 + r2d2 + r2 sin2 d2.

ds2 = A(t, r)dt2 B(t, r)rdrdt C(t, r)rdrd D(t, r)r2dr2

E(t, r)(dr2 + r2d2 + r2 sin2 d2) F(t, r)d2 (2.2)

Collecting together terms, absorbing factors of r into our functions, and

redefining A,B,C,D,E, and F, we get:

ds2 = A(t, r)dt2 B(t, r)drdt C(t, r)drd D(t, r)dr2

E(t, r)(d2 + sin2 d2) F(t, r)d2 (2.3)

Define a new radial coordinate by r2 =E(t, r). Again by redefining A, B, C, D,

and F, we get:

8


15/58

ds2 = A(t, r)dt2

B(t, r)drdt

C(t, r)drd

D(t, r)dr2

r2(d2 + sin2 d2) F(t, r)d2 (2.4)

Introduce a new time-like coordinate t defined by:

dt= (t, r)

A(t, r)dt 12

B(t, r)dr

(t, r) is an integrating factor that makes the RHS an exact differential.

Squaring, we get:

dt2 = 2

A2dt2 ABdrdt +14

B2dr2

Thus,

Adt2 Bdrdt = 1A2

dt2 B2

4Adr2

Define new functions:

A= 1

A2

B = B2

4A+ D

The metric now takes the form:

ds2 = A(t, r)dt2 B(t, r)dr2 C(t, r)drd r2(d2 + sin2 d2) F(t, r)d2 (2.5)

Similarly, introduce a new spacelike coordinate:

d= (t, r)[F(t, r)d+12

C(t, r)dr]

9


16/58

(t, r) is an integrating factor that makes the RHS an exact differential.

Squaring, we get:

d2 = 2[F2d2 + CF drd+1

4C2dr2]

Thus,

F d2 + Cddr= 1

F2d2 C

2

4Fdr2

Define new functions:

F = 1

F2

C= B C2

4F

The metric now becomes diagonal and takes the form:

ds2 = A(t, r)dt2 B(t, r)dr2 r2(d2 + sin2 d2) C(t, r)d2 (2.6)

We take as an additional assumption the metric to be static, i.e. A,B, and

C are now independent of t. Redefining the functions A,B, and C and removing

the bars from the coordinates, the metric takes the form:

ds2 =A2(r)dt2 B2(r)dr2 r2(d2 + sin2 d2) C2(r)d2 (2.7)

2.2 Equations of Motion in Component Form

As mentioned before, the equations of motion derived from the five-dimensional

action are given by:

abcde(Rab

ea

eb)

(Rcd

ec

ed) = 0 (2.8)

abcde(Rab ea eb) Tc = 0 (2.9)

10


17/58

where

Ta =dea + ab eb (2.10)

is the torsion, and the curvature R is given by:

Rab =dab + ac bc (2.11)

where is the spin connection and e is the funfbein.

Considering both the funfbein and the spin connection as gauge fields, the

Einstein action becomes invariant only under a constrained symmetry for which the

torsion is set to zero[3]. Therefore, we are only interested in the case where Ta = 0

Writing the first set of equations of motion in component form we obtain:

abcde(Rab eaeb)(Rcd eced) = 0 (2.12)

Of course, this is a set of 25 different equations of motion where

e= 1, 2, 3, 4, 5 and = 1, 2, 3, 4, 5.

Our aim is to find a solution for the equations of motion using the metric

ansatz derived above.

Relabeling dummy indices in the above equation and using the fact that:

abcde= e[ae

be

ce

de

]e (2.13)

The equations of motion take the form:

abcdeRabR

cd 2(2!)e[cede]eRcd+ 4!2eedet(e1) (2.14)

11


18/58

2.3 Calculating the Curvature tensor

components

Remember that our static, spherically symmetric metric is given by:

ds2 =A2(r)dt2 B2(r)dr2 r2(d2 + sin2 d2) C2(r)d2 (2.15)

In terms of the funfbeins, the line element is given by:

ds2 =abeae

bdx

dx (2.16)

where in our case, the funfbein has the form:

A(r) 0 0 0 0

0 B(r) 0 0 00 0 r 0 0

0 0 0 r sin() 0

0 0 0 0 C(r)

(2.17)

ab is the five dimensional flat space-time metric with signature (+,-,-,-,-).

Since we are only interested in the case where torsion is set to zero, we can

derive the spin connections from the funfbeins using equation 1.11.

The non-zero s turn out to be:

010 = 100 = A(r)

B(r)

122 = 212 = 1

B(r)

133 = 313 = sin B(r)

12


19/58

233 = 323 = cos

144 = 414 = C(r)

B(r) (2.18)

(2.19)

where prime denotes derivatives with respect to r, and dotted indices denote

curved indices.

The curvatures are then calculated from the spin connections using

equation 1.13.

Rab=ab ab + ak bk ak bk (2.20)

After calculating the curvature tensor components, we discover that the

only non-zero components are:

R0101 =

A(r)B(r)

B(r)2 A(r)

B(r)

R0202 = A(r)

B(r)2

R0303 = A(r)

B(r)2sin

R0404 = A(r)C(r)

B(r)2

R1212 = B(r)

B(r)2

R1313 = B(r)B(r)2

sin

R1414 = C(r)B(r)

B(r)2 +

C(r)

B(r)

R2323 = sin

1

B(r)2 1

R2424 = C(r)

B(r)2

R3434 = C(r)

B(r)2

sin

(2.21)

13


20/58

where here also prime denotes derivatives with respect to r.

2.4 Equations of Motion

Plugging the curvature tensor components obtained into the 25 equations of

motion, we notice that there are only 5 non-trivial equations, those for which

e= r, from which we obtain 3 independent equations (e=r=0,1,5) for the three

unknowns A(r), B(r), and C(r).

The simplified equations obtained are the following:

242B(r)r2C(r)sin 16rC(r)sin

B(r) 8C(r)sin

B(r) + 8B(r)C(r)sin

+8r2C(r)B(r)sin

B(r)2 8r

2C(r)sin

B(r) +

16rC(r)B(r)sin

B(r)2

24B(r)C(r)sin

B(r)4 +

8B(r)C(r)sin

B(r)2 8C

(r)sin

B(r)3 +

8C(r)sin

B(r) = 0

242r2A(r)C(r)sin 16rA(r)C(r)sin

B(r)2 8A(r)C(r)sin

B(r)2

+8A(r)C(r)sin 8r2A(r)C(r)sin

B(r)2 16rA

(r)C(r)sin

B(r)2

+24A(r)C(r) sin

B(r)4 8A(r)C(r)sin

B(r)2 = 0

242r2A(r)B(r)sin 8A(r)sin B(r)

+ 8A(r)B(r)sin +16rA(r)B(r)sin

B(r)2

16rA(r)sin B(r)

+8r2A(r)B(r)sin B(r)2

8r2A(r)sin B(r)

14


21/58

24A(r)B(r)sin

B(r)4 +

8A(r)B(r)sin

B(r)2 +

8 sin A(r)

B(r)3 8sin A

(r)

B(r) = 0

2.5 Solutions For Equations of Motion

After inserting the above three equations into MAPLE, we obtained 7 different

solutions.

First let us state the solutions obtained:

solution 1:

A(r) = 0

B(r) = 33r2 + 3

C(r) =C1LegendreP

10 1

2, 52

,r

+ C2LegendreQ

10 1

2, 52

,r

(r2 + 1)1/4

(2.22)

solution 2:

A(r) = C1

r2 + 1

B(r) = 1r2 + 1

C(r) = C(r)

(2.23)

solution 3:

A(r) = C3r

B(r) = 3

9r2

+ 3C(r) = C1r+ C2

3r2 + 1

15


22/58

(2.24)

solution 4:

A(r) = 0

B(r) = B(r)

C(r) = Y(r)

(2.25)

such that Y(r) is the solution for

1B(r)(B(r)2r2 + 1 B(r)2) (Y

(r)B(r)3r2 Y(r)B(r) + Y(r)B(r)3

2Y(r)rB(r)3 Y(r)B(r)2r2B(r) + 3Y(r)B(r) + Y(r)B(r)2B(r)

+3B(r)52Y(r)r2 + B(r)3Y(r) B(r)5Y(r) + 2B(r)2Y(r)B(r)r) = 0

(2.26)

solution 5:

A(r) = A(r)

B(r) = 1r2 + 1

C(r) = C1

r2 + 1

(2.27)

solution 6:

B(r) = B(r)

C(r) = 0

A(r) = Y(r)

16


23/58

(2.28)

such that Y(r) is the solution for

1B(r)(B(r)2r2 + 1 B(r)2) (Y

(r)B(r)3r2 Y(r)B(r) + Y(r)B(r)3

2Y(r)rB(r)3 Y(r)B(r)2r2B(r) + 3Y(r)B(r) + Y(r)B(r)2B(r)

+3B(r)52Y(r)r2 + B(r)3Y(r) B(r)5Y(r) + 2B(r)2Y(r)B(r)r) = 0

(2.29)

solution 7:

A(r) = r

C2

exp

2 + B(r)2r2

r dr

dr+ C3

B(r) = 1r2 + C4

C(r) = C1exp

B(r)2(3A(r)r2B(r)2 + 2rA(r) + A(r)

B(r)2A(r))

2B(r)2A(r)r+ B(r)2r2A(r) + 3A(r) A(r)B(r)2 dr(2.30)

Note that in all solutions C1, C2, C3, and C4 are constants of integration.

Notice that in the first solution, we have A(r) = 0, which in other words

means that the time component in our line element vanishes. This makes the

solution non-physical. We will thus discard this solution.

For the same reason, we discard solution 4 that contains A(r) = 0.

Solution 6 which contains C(r) = 0 will also be discarded because it is not

five-dimensional.

We are left out with 4 solutions: solutions 2, 3, 5,and 7.

solution 7 turns out to be mathematically complicated and will not be

discussed in this paper.

The remaining three solutions will be analyzed throughout the following

17


24/58

chapter.

18


25/58

Chapter 3

Analyzing the Physical Solutions

In this chapter, we analyze the three physical solutions that we have obtained.

3.1 Analyzing the First Physical Solution

The first physical solution to be analyzed is solution 2. Recall that it is given by:

A(r) = C1

r2 + 1

B(r) = 1r2 + 1

C(r) = C(r)

(3.1)

where C1 is a constant of integration.

Plugging the solutions for A,B, and C into the metric ansatz, we obtain:

ds2 =C21(r2 + 1)dt2 1

r2 + 1dr2 r2d2 r2 sin2 d2 C(r)2d2 (3.2)

We absorb the constants of integration by redefining our coordinates such that:

t C1t

19


26/58

The metric finally becomes:

ds2 = (r2 + 1)dt2

1

r2

+ 1

dr2

r2d2

r2 sin2 d2

C(r)2d2 (3.3)

for =0 the metric takes the form:

ds2 =dt2 dr2 r2d2 r2 sin2 d2 C(r)2d2 (3.4)


ds2 = (r2 + 1)dt2 1r2 + 1

dr2 r2d2 r2 sin2 d2 C(r)2d2 (3.5)

for =-1 the metric takes the form:

ds2 = (r2 + 1)dt2 1r2 + 1 dr2 r2d2 r2 sin2 d2 C(r)2d2 (3.6)

It is clear that no singularities exist for = 0, 1. However, for = 1, wehave:

B(r) for r= 1

To check whether this singularity is a coordinate singularity or a real

singularity, we need to calculate RR for this metric.

To calculate this term, we start by calculating the Christoffel symbols.

These are given by equation (1.3). The non-zero components turn out to be:

001 = 010=

r

r

2

+ 1100 = (r

2 + 1)r

20


27/58

111 = r

r2 + 1

122 = (r2 + 1)r

133 = (r2 + 1)r sin2

144 = (r2 + 1)C(r)C(r)

212 = 212=

1

r

233 = sin cos

313 = 331=

1

r

3

32

= 3

23

=cos

sin

414 = 441=

C(r)

C(r) (3.7)

From the Christoffel symbols we can calculate the components of the

Reimann curvature tensor, given by equation (1.4).

The non-zero components are then given by:

R0101 =

R0202 = (r2 + 1)r2

R0303 = (r2 + 1)r2 sin2

R0404 = (r2 + 1)rC(r)C(r)

R1212 = r2

r2 + 1

R1313 = r2

r2 + 1sin2

R1414 = C(r)(C(r)r2 + C(r) + rC(r))

r2 + 1

R2323 = r4 sin2

R2424 = (r2 + 1)rC(r)C(r)

R3434 = (r2 + 1)rC(r)C(r)sin2

(3.8)

21


28/58

Taking= 1, We can now calculate RR. The result obtained is:

RR =

16r2C(r)2

C(r)2 16C

(r)2

C(r)2 +

8C(r)2

C(r)2r2+

4r4C(r)2

C(r)2

8r2C(r)2

C(r)2 +

8r3C(r)C(r)

C(r)2 +

4C(r)2

C(r)2 8rC

(r)C(r)

C(r)2 + 24

We notice that the above expression does not blow up for r= 1, as long as

C(r) is a well-behaved function having no roots. Therefore, we conclude that thisis a coordinate singularity.

3.1.1 Geodesic Equations

Now let us derive the geodesic equations for this metric.

We can now derive the geodesic equations for the coordinates: t,r,,, from the

following formula:

x + xx = 0

where dot represents the derivative with respect to an affine parameter .

However,we choose to derive the geodesic equaitons by varying a Lagrangian. The

Lagrangian, L= gxx, for this metric is given by:

L= (r2 + 1)t2 1r2 + 1

r2 r22 r2 sin2 2 C(r)22 (3.9)

The geodesic equations are then obtained from the Euler-Lagrange

equations:

dd

Lx

Lx

= 0 (3.10)

22


29/58

d

d2(r2 + 1)t= 0

r

r2 + 1+ rt2 r

(r2 + 1)2 r(2 + sin2 2) C(r)C(r)2 = 0

+2r

r sin cos 2 = 0d

d

2r2 sin2

= 0

d

d

2C(r)2

= 0

(3.11)

This set of equations can be easily simplified to:

(r2 + 1)t= k

r

r2 + 1+ rt2 r

(r2 + 1)2 r(2 + sin2 2) C(r)C(r)2 = 0

+2r

r sin cos 2 = 0

r2= h

C(r)2=l

(3.12)

where k has the significance of energy and h and l have the significance of angularmomenta along the and axes respectively.

Note that = 2

is a solution for the third equation. Since our metric is

spherically symmetric, we can, without loss of generality, confine our attention to

particles moving in the equatorial plane, = 2

.

The 4 remaining equations now become:

(r2 + 1)t= k

23


30/58

r

r2 + 1+ rt2 r

(r2 + 1)2 r2 C(r)C(r)2 = 0

r2= h

C(r)2=l

(3.13)

This set of equations is valid for both null and non-null geodesics.

We now replace the complicated r-equation by a first integral of the

geodesic equations.

For a non-null geodesic, this is given by:

gxx = 1 (3.14)

where we have taken c2 = 1.

For a null geodesic, it is given by:

gxx = 0 (3.15)

We are interested in trajectories of photons. Therefore, the set of null geodesic

equations becomes:

(r2 + 1)t= k

(r2 + 1)t2 1r2 + 1

r2 r22 C(r)22 = 0

r2= h

C(r)2=l

(3.16)

To perform a coordinate transformation that will get rid of the coordinate

24


31/58

singularity, we use the analogous to Eddington-Finkelstein coordinates. Therefore,

we are interested in the radial motion of photons. For radial motion, we have

= 0 and = 0.

The first integral reduces to:

(r2 + 1)t2 1r2 + 1

r2 = 0 (3.17)

which upon simplification gives:

dr

dt = (r2

+ 1) (3.18)

for = 1 we have:dt= dr

1 r2 (3.19)

upon integration, this gives:

t= 1

2ln1

r

1 + r + constant (outgoing photons)

t= +1

2ln

1 r1 + r + constant (incoming photons) (3.20)

3.1.2 Coordinate Transformations

We will now perform a series of coordinate transformations that will get rid of the

coordinate singularities.

First, denote the above integration constants by u and v and take them as new

coordinates.

We have:

u= t +1

2ln

1 r1 +r

(outgoing photons)

v=t 12

ln1 r1 + r

(incoming photons) (3.21)25


32/58

differentiating we get:

du= dt 11 r2 dr

dv= dt + 1

1 r2 dr (3.22)

Multiplying du and dv together by (1 r2) we get:

(1 r2)dudv= (1 r2)dt2 11 r2 dr2 (3.23)

Substituting into the line element, we get:

ds2 = (1 r2)dudv r2d2 r2 sin2 d2 C(r)2d2 (3.24)

r is now viewed as a function ofu and v defined implicitly by:

ln

1 r1 + r = u v2 (3.25)

From the above equation we can rewrite equation (3.24) as

ds2 = (1 +r)2 exp

v u2

dudv r2d2 r2 sin2 d2 C(r)2d2 (3.26)

Now let

V= expv

2U= exp

u2

(3.27)

26


33/58

Differentiating, we get:

dV =1

2

exp v

2 dv

dU=1

2exp

u2

du (3.28)

We then multiply dU by dV to get:

dUdV =1

4exp

v u

2

dudv (3.29)

Substituting back into the line element, we get:

ds2 = 4(1 + r)2dUdV r2d2 r2 sin2 d2 C(r)2d2 (3.30)

Notice that the line element has become well behaved at r= 1. We have

got rid of the singularity at r= 1 by performing this series of coordinate

transformations.

3.2 Analyzing the Second Physical Solution

The second physical solution to be analyzed is solution 3. Recall that it is given by:

A(r) = C3r

B(r) = 3

9r2 + 3

C(r) = C1r+ C2

3r2 + 1

(3.31)

27


34/58

where C1, C2, and C3 are constants of integration.


ds2 =C23r2dt2 3

3r2 + 1dr2r2d2r2 sin2 d2(C1r+C23r2 + 1)2d2 (3.32)


t C3t and C1.We also denote C2

C1by C, where C is some integration constant that can be fixed by

boundary conditions.


ds2 =r2dt2 33r2 + 1

dr2 r2d2 r2 sin2 d2 (r+ C

3r2 + 1)2d2 (3.33)


ds2

=r2

dt2

3dr2

r2

d2

r2

sin2

d2

(r+ C)2

d2

(3.34)


ds2 =r2dt2 33r2 + 1

dr2 r2d2 r2 sin2 d2 (r+ C

3r2 + 1)2d2 (3.35)


ds2 =r2dt2 33r2 + 1 dr2 r2d2 r2 sin2 d2 (r+ C

3r2 + 1)2d2 (3.36)


B(r) for r= 13

28


35/58

Again, to check whether this singularity is a coordinate or real singularity,

we need to calculate RR for this metric.

We start by calculating the Christoffel symbols using equation (1.3). The non-zero

components turn out to be:

001 = 010=

1

r

100 = 1

3(3r2 + 1)r

111 = 3r

3r2 + 1

122 = 1

3(3r2 + 1)r

133 = 1

3(3r2 + 1)r sin2

144 = 1

3

3r2 + 1

r+ C

3r2 + 1

3r2 + 1 + 3Cr

212 = 212=

1

r

233 =

sin cos

313 = 331=

1

r

332 = 323=

cos

sin

414 = 441=

3r2 + 1 + 3Cr

(r+ C

3r2 + 1)

3r2 + 1(3.37)

(3.38)

The components of the Reimann curvature tensor are calculated from

equation (1.4). The non-zero components are given by:

R0101 = 3r2

3r2 + 1

R0202 = 13

(3r2 + 1)r2

R0303 = 13

(3r2 + 1)r2 sin2

R0404 = 133r2 + 1r

r+ C3r2 + 1

3r2 + 1 + 3Cr

29


36/58

R1212 = 3r2

3r2 + 1

R1313 = 3r2

3r2 + 1sin2

R1414 =3

3r2 + 1r2 + 6r3C+ 2Cr+ 33r2 + 1C2r2 + 3r2 + 1C2(3r2 + 1)3/2

R2323 = 1

3r2 sin2 (2 + 3r2)

R2424 = 1

3

3r2 + 1r

r+ C

3r2 + 1

3r2 + 1 + 3Cr

R3434 = 1

3

3r2 + 1r

r+ C

3r2 + 1

3r2 + 1 + 3Cr

sin2

(3.39)

Taking= 1, We can now calculate RR. The result obtained is:

RR = 720r4

C2

(r+ C3r2 + 1)4 + 360r

4

C4

(r+ C3r2 + 1)4 + 40r

4

(r+ C3r2 + 1)4

480C3r33r2 + 1

(r+ C3r2 + 1)4 +

160Cr33r2 + 1

(r+ C3r2 + 1)4

312C2r2

(r+ C3r2 + 1)4

240C4r2

(r+ C3r2 + 1)4

8r2

(r+ C3r2 + 1)4 +

184C3r3r2 + 1

(r+ C3r2 + 1)4

24Cr3r2 + 1

(r+ C3r2 + 1)4

76C2

(r+ C3r2 + 1)4 +

4

(r+ C3r2 + 1)4

+

64C4

(r+ C3r2 + 1)4 40C3

3r2 + 1

(r+ C3r2 + 1)4r + 40C

3r2 + 1

3(r+ C3r2 + 1)4r+

52

3

C2

(r+ C3r2 + 1)4r2

16C4

(r+ C3r2 + 1)4r2 +

8

3

C4

(r+ C3r2 + 1)4r4

+32

3

C33r2 + 1

(r+ C3r2 + 1)4r3

We notice that the above expression does not blow up for r= 13

. Therefore, we

conclude that this is a coordinate singularity.

30


37/58


Let us derive the geodesic equations for this metric. Again, we derive the geodesic

equaitons by varying a Lagrangian. The Lagrangian L= gxx for this metric is

given by:

L= r2t2 33r2 + 1

r2 r22 r2 sin2 2 (r+ C

3r2 + 1)22 (3.40)


equations. They are given by:

d

d

2r2t

= 0

3r

3r2 + 1+ rt2 9r

(3r2 + 1)2r2 r(2 + sin2 2)

+(r+ C3r2 + 1)

1 + 3Cr3r2 + 1

2 = 0

+2r

r sin cos 2 = 0d

d

2r2 sin2

= 0

d

d

2(r+ C

3r2 + 1)2

= 0

(3.41)


r2t= k

3r

3r2 + 1+ rt2 9r

(3r2 + 1)2r2 r(2 + sin2 2)

+(r+ C3r2 + 1) 1 + 3Cr3r2 + 1

2 = 0

31


38/58

+2r

r sin cos 2 = 0

r2 sin2 = h

(r+ C3r2 + 1)2=l

(3.42)

where k has the significance of energy and hand l have the significance of angular

momenta along the and axes respectively.

Because of spherical symmetry, we can, without loss of generality, confine

our attention to particles moving in the equatorial plane, = 2 .

The remaining equations now become:

r2t= k

3r

3r2 + 1+ rt2 9r

(3r2 + 1)2r2 r2

+(r+ C

3r2 + 1) 1 + 3Cr

3r2

+ 1

2 = 0

r2= h

(r+ C

3r2 + 1)2=l

(3.43)


We are interested in trajectories of photons. Therefore, we replace the

r-equation by gxx = 0.

r2t= k

r2t2 33r2 + 1

r2 r22 r2 sin2 2 (r+ C

3r2 + 1)22 = 0

r2= h

(r+ C

3r2 + 1)2=l

32


39/58

(3.44)

Again, we are interested in the radial motion of photons. For radial

motion, we have = 0 and = 0.

The first integral reduces to:

r2t2 33r2 + 1

r2 = 0 (3.45)


dt=

3r2(3r2 + 1)

dr (3.46)

for = 1 we have:dt=

3

r2(3r2 + 1)dr (3.47)

After integrating, we get:

t=

3tanh1(

1 3r2) + constant (outgoing photons)

t=

3tanh1(

1 3r2) +constant (incoming photons) (3.48)


We now perform a series of coordinate transformations to get rid of the coordinate

singularity.

First, denote the above integration constants by u and v and take them as

new coordinates. We have:

u= t +

3tanh1(

1 3r2)

33


40/58

v= t

3tanh1(

1 3r2)

(3.49)

differentiating we get:

du= dt +

3

r2(3r2 + 1)dr

dv= dt

3

r2(3r2 + 1)

dr

(3.50)

Multiplying du and dv together by r2 we get:

r2dudv =r2dt2 31 3r2 dr

2 (3.51)


ds2 =r2dudv r2d2 r2 sin2 d2 (r+ C

3r2 + 1)2d2 (3.52)

Now let

T =u+ v

2

X= u

v

2 (3.53)

which gives dT2 dX2 =dudv.In terms of the new coordinates T and X, the line element takes the form

ds2 =r2dT2 r2dX2 r2d2 r2 sin2 d2 (r+ C

3r2 + 1)2d2 (3.54)

Notice that the line element has become well behaved at r= 13

.

34


41/58

The relation between the new coordinates T and X and the old

coordinates t and r is now given implicitly by:

T= U+ V

2 =t

X= U V

2 =

3tanh1(

1 3r2) (3.55)

(3.56)

We have got rid of the singularity at r= 13

by performing this series of

coordinate transformations.

3.3 Analyzing the Third Physical Solution

The third physical solution to be analyzed is solution 5. Recall that it is given by:

A(r) = A(r)

B(r) = 1r2 + 1

C(r) = C1

r2 + 1

(3.57)

where C1 is a constant of integration.


ds2 =A(r)2dt2 1r2 + 1

dr2 r2d2 r2 sin2 d2 C21 (r2 + 1)d2 (3.58)


C1.

35


42/58


ds2 =A(r)2dt2

1

r2

+ 1

dr2

r2d2

r2 sin2 d2

(r2 + 1)d2 (3.59)


ds2 =A(r)2dt2 dr2 r2d2 r2 sin2 d2 d2 (3.60)


ds2 =A(r)2dt2 1r2 + 1

dr2 r2d2 r2 sin2 d2 (r2 + 1)d2 (3.61)


ds2 =A(r)2dt2 11 r2 dr

2 r2d2 r2 sin2 d2 (1 r2)d2 (3.62)


B(r) for r= 1

To check whether this singularity is real or coordinate, we need to

calculate RR for this metric.

We start by calculating the Christoffel symbols from equation (1.3). The

non-zero components turn out to be:

001 = 010 =

A(r)

A(r)100 = (r

2 + 1)A(r)A(r)

36


43/58

111 = r

r2 + 1

122 = (r2 + 1)r

133 = (r2 + 1)r sin2

144 = (r2 + 1)r

212 = 212 =

1

r

233 = sin cos

313 = 331 =

1

r

3

32

= 3

23

=cos

sin

414 = 441 =

r

r2 + 1

(3.63)

We now calculate the components of the Reimann curvature tensor using

equation (1.4). The non-zero components are then given by:

R0101 = A(r) (A(r)r2 + A(r) + A(r)r)

r2 + 1

R0202 = (r2 + 1)A(r)A(r)r

R0303 = (r2 + 1)A(r)A(r)r sin2

R0404 = (r2 + 1)A(r)A(r)r

R1212 =

r2

r2 + 1

R1313 = r2

r2 + 1sin2

R1414 =

R2323 = r4 sin2

R2424 = (r2 + 1)r2

R3434 = (r2 + 1)r2 sin2

(3.64)

37


44/58

Taking= 1, We can calculate RR. The result obtained is:

RR = 24 +

16r2A(r)2

A(r)2 16A

(r)2

A(r)2 +

8A(r)2

A(r)2r2+

4r4A(r)2

A(r)2

8r2A(r)2

A(r)2 +

8r3A(r)A(r)

A(r)2 +

4A(r)2

A(r)2 8rA

(r)A(r)

A(r)2

We notice that the above expression does not blow up for r= 1 if A(r) is a

well-behaved function that has no roots. Therefore, we conclude that thissingularity is a coordinate singularity.


The Lagrangian L= gxx for this metric is given by:

L= A(r)2t2 1r2 + 1

r2 r22 r2 sin2 2 (r2 + 1)2 (3.65)


equations:

d

d

2A(r)2t

= 0

r

r2 + 1A(r)A(r)t2 r

(r2 + 1)2r2 r(2 sin2 2) r2 = 0

+2r

r sin cos 2 = 0d

d

2r2 sin2

= 0

d

d2(r

2 + 1)= 0(3.66)

38


45/58


A(r)2t= k

r

r2 + 1A(r)A(r)t2 r

(r2 + 1)2r2 r(2 sin2 2) r2 = 0

+2r

r sin cos 2 = 0

r2 sin2 = h

(r2 + 1)=l

(3.67)

where k has the significance of energy, and h and l have the significance of angular

momenta along the and axes respectively.

As in the previous two cases, we can confine our attention to particles

moving in the equatorial plane, = 2

The 4 remaining equations now become:

A(r)2t= k

r

r2 + 1A(r)A(r)t2 r

(r2 + 1)2r2 r2 r2 = 0

r2= h

(r2 + 1)=l

(3.68)


We are interested in trajectories of photons. Therefore, we replace the

r-equation by the first integral for null geodesics.

A(r)2t= k

39


46/58

A(r)2t2 1r2 + 1

r2 r22 r22 (r2 + 1)2 = 0

r2= h

(r2 + 1)=l

(3.69)

For radial motion, we have = 0 and = 0. The first integral reduces to:

A(r)2t2 1r2 + 1

r2 = 0 (3.70)


dt= drA(r)2(r2 + 1)

(3.71)

for = 1 we have:dt= dr

A(r)2(r2 + 1)

(3.72)

t=

drA(r)2(r2 + 1)

(3.73)

which of course cannot be solved unless A(r) is given.

However, let dr

A(r)2(r2+1)=F(r) + constant , then:

t= F(r) + constant (outgoing photons)

t= F(r) + constant (incoming photons) (3.74)


Denote the above integration constants by u and v and take them as new

coordinates. We have:

40


47/58

u= t + F(r)

v= t F(r)

(3.75)

Differentiating we get:

du= dt + dr

A(r)2(3r2 + 1)

dv= dt drA(r)2(3r2 + 1)

(3.76)

(3.77)

Multiplying du and dv together by A(r)2 we get:

A(r)2dudv =A(r)2dt2 11 r2 dr

2 (3.78)


ds2 =A(r)2dudv

r2d2

r2 sin2 d2

(1

r2)d2 (3.79)

Now let

T =u + v

2

X= u v

2

which gives dT2 dX2 =dudv.

41


48/58

In terms of the new coordinates T and X, the line element takes the form

ds2 =A(r)2dT2

A(r)2dX2

r2d2

r2 sin2 d2

(1

r2)d2 (3.80)

The relation between the new coordinates T and X and the old

coordinates t and r is now given implicitly by:

T =u+ v

2

=t

X= u v

2 =F(r)

(3.81)

Notice that the above line element still has a singularity at r= 1. This

singularity appears in g44.

To get rid of this singularity, we consider motion of photons in the-direction. Let us first rewrite the first integral of motion for photons in terms of

the new coordinates. This is given by:

A(r)2 T2 A(r)2 X2 r22 r2 sin2 2 (r2 + 1)2 = 0 (3.82)

We now consider motion in the equatorial plane and, in particular, motion along

the -direction. Therefore we have: = 2

, X= 0, and = 0.

Now, for = 1, the first integral becomes:

A(r)2 T2 (1 r2)2 = 0 (3.83)

which upon solving gives

42


49/58

T =

1 r2|A(r)| + constant (3.84)

Denote the integration constants by xand y respectively and take them as the new

coordinates.

x = T

1 r2|A(r)|

y = T+

1 r2

|A(r)|

(3.85)

Derive w.r.t to get:

dx

d

= dT

d

1 r2

|A(r)|dy

d =

dT

d +

1 r2|A(r)|

(3.86)

Multiplying the above equations together by A(r)2d2 and replacing inside the line

element, it becomes:

ds2 =A(r)2dxdy A(r)2dX2 r2d2 r2 sin 2d2 (3.87)

Now, let

U = y+ x

2

V = y x

2(3.88)

43


50/58

The line element then becomes:

ds2 =A(r)2dU2

A(r)2dX2

r2d2

r2 sin 2d2

A(r)2dV2 (3.89)

The relation between the new coordinate U and Vand the old coordinates T and

is given by:

U = x+ y

2 =T

V = y x2

= 1 r2|A(r)|(3.90)

We have got rid of the singularity at r= 1 by performing this series of

coordinate transformations. Note, however, that A(r) should be a well-behaved

function that has no roots.

44


51/58

Chapter 4

Dimensionally Reducing the

Action

In this chapter, we will reduce the action in 5-dimensional spacetime of part 1 into

an action in four dimensional spacetime plus a scalar field. We check that the

same set of solutions is still valid.

Recall that the action in five-dimensional spacetime is given by:

I5= 3

abcde(eaRbcRde 2

3eaebecRde +1

52eaebeced ee) (4.1)

which in component form becomes:

I5 = 3

abcde(eaR

bcR

de

2

3eae

be

cR

de+

1

52eae

be

ce

de

e) (4.2)

where of course the Greek indices are curved indices, and Latin indices are tangent

space indices.

This action is dimensionally reduced by Chamseddine [3] as follows:

First, let = 4, then

45


52/58

I4= 3k M4

d4xabcde(eaR

bcR

de4+ e

a4R

bcR

de

2eae

be

c4R

de

43

eaebe

cR

de4+

2eaebe

ce

de

e4) (4.3)

Second, rewrite the above expression in terms of the four-dimensional

fields. The action is then given by:

I4 = 3kM

4

d4xabcd(e4R

abR

cd4+ 2e

aR

bcR

d44+ 2e

aR

b4R

cd4

+e44RabR

cd+ 4e

a4R

bcR

d4

8

3eae

be

cR

d44

4eaebe44Rcd 4eaebe4Rcd4 4e4eaeb4Rcd 4eaebec4Rd4+2eae

be

crhoe

de

44+ 4

2eaebe

ce

4e

d4) (4.4)

(4.5)

where

Rab=ab ab + ak bk ak bk+ a4 b4 a4 b4 (4.6)

We use this formula to calculate the curvature tensor components from

the veilbeins and spin connections in chapter 1. For our choice of metric ansatz,

the curvature tensor components turn out to remain the same.

Replacing the values of the curvature tensors and veirbeins in the action,

we get an action in terms of the fields A(r), B(r), and C(r).

This system corresponds to gravity coupled to a vector and a scalar, where

C(r) =e44

is the scalar field and is called a dilaton. The vector field corresponds to

non-diagonal components in the metric and turns out to be zero in this case.

The Lagrangian for this action is then given by:

46


53/58

L=

8sin C(r)r2A(r)

B(r)

+12A(r)C(r)sin

B(r)3

12A(r)B(r)C(r)sin

B(r)4

+8sin rC(r)A(r)

B(r)3 +

4sin A(r)C(r)B(r)

B(r)2 + 8 sin B(r)C(r)A(r)

8sin C(r)A(r)B(r)

+8sin C(r)A(r)B(r)

B(r)2 +

8sin rA (r)C(r)

B(r)3

4sin A(r)C(r)

B(r) 4sin A(r)C

(r)

B(r) +

4sin A(r)C(r)

B(r)3

+8sin C(r)A(r)

B(r)3 24sin C(r)A

(r)B(r)

B(r)4 +

16 sin C(r)A(r)rB (r)

B(r)2

+24A(r)B(r)r2 sin 2C(r) +16 sin r2

A(r)C(r)B(r)B(r)2 8sin C(r)A(r)B(r)+

8sin C(r)r2A(r)B(r)

B(r)2 24rB

(r)sin A(r)C(r)

B(r)4 16sin C(r)rA

(r)

B(r)

16sin r2A(r)C(r)

B(r) 16sin r

2A(r)C(r)

B(r) 32 sin rA(r)C

(r)

B(r)

(4.7)

The equations of motion are then obtained by using the Euler-Lagrange

equations for A(r), B(r), and C(r).

Since the Lagrangian contains second derivatives, the generalized

Euler-Lagrange equations are given by:

L

r

L

(r)

+

2

r2

L

(2r)

= 0 (4.8)

where (A(r), B(r), C(r)).The equations of motion for A,B, and C respectively are then given by:

24B(r)r2 sin 2C(r) 8sin r2C(r)

B(r) 16 sin rC

(r)

B(r) 24B

(r)C(r)sin

B(r)4

+8sin C(r)B(r)

B(r)2 + 8 sin B(r)C(r) 8sin C(r)

B(r) +

8sin r2C(r)B(r)

B(r)2

+16sin C(r)rB(r)

B(r)2 +8sin C

(r)B(r)3

8sin C(r)B(r)

= 0

47


54/58

(4.9)

8sin r2A(r)C(r)

B(r)2 + 8 sin C(r)A(r) 8 sin A

(r)C(r)

B(r)2

16sin C(r)rA(r)

B(r)2 +

24A(r)C(r)sin

B(r)4 16 sin rA(r)C

(r)

B(r)2

8sin C(r)A(r)B(r)2

+ 24A(r)r2 sin 2C(r) = 0

(4.10)

8sin r2A(r)

B(r) + 8 sin B(r)A(r) 8sin A(r)

B(r) +

8sin A(r)B(r)

B(r)2

8sin A(r)

B(r) +

8 sin A(r)

B(r)3 24A

(r)B(r)sin

B(r)4 + 24A(r)B(r)r2 sin 2

+16sin A(r)rB (r)

B(r)2 +

8 sin r2A(r)B(r)

B(r)2 16sin rA

(r)

B(r) = 0

(4.11)

After solving this set of differential using MAPLE, we found out that the

solutions turn out to be exactly the same as those obtained for the action in

five-dimensional spacetime. As a double check, the solutions, mentioned in chapter

2 above, were inserted in the above equations of motion in equation 4.9, 4.10, and

4.11. It was verified that all solutions satisfy the above equations.

48


55/58

Chapter 5

Conclusion and Future Work

In the problem above, we considered a topological action for gravity in five

dimensions from which equations of motion are derived. The action contained a

quadratic term in R, the Gauss-Bonnet term; a linear term in R, an Einstein term;

and a cosmological constant. We then took as an ansatz a static, spherically

symmetric metric in five dimensions. The metric contained three unknown

functions A(r), B(r), and C(r). After calculating the curvatures from the spin

connections, we plugged into the equations of motion to get three coupled,

independent differential equations in terms of A(r), B(r), and C(r). This system is

completely solvable and turns out to have seven independent solutions that were

found using MAPLE. Three of these solutions were analyzed for singularities.

Neither of the analyzed solutions contained a singularity except for a coordinatesingularity in the deSitter space for each. After deriving the geodesic equations

that describe the radial motion of photons, we perform coordinate transformations

to get rid of the coordinate singularities. These transformations are analogous to

the Eddington-Finkelstein coordinate transformations applied Schwarzschild

geometry. Next, we dimensionally reduced the action in five dimensions to an

action in four dimensions that contained, in addition to the four dimensional

metric components, a scalar field. The scalar field was taken to be C(r). The

49


56/58

scalar curvatures were then plugged in to obtain a lagrangian in terms of the three

unknown functions A, B, and C. The equations of motion were then obtained by

considering the generalized Euler-Lagrange equations for these three fields. After

plugging the equations into MAPLE, they turned out to have the same set of

solutions as the equations considered in five-dimensions.

As future work, we might work on finding solutions for these equations of

motion by taking a non-static spherical metric rather than a static one.

50


57/58

Bibliography

[1] M. Alcubierre, Introduction to 3+1 Numerical Relativity, International Series of

Monographs on Physics, Oxford Science Publications, pp. 1 32.

[2] A.H. Chamseddine, Applications of the Gauge Principle to Gravitational

Interactions, Int.J.Geom.Meth.Mod.Phys.3:149-176, 2006.

[3] A.H. Chamseddine, Topological Gravity and Supergravity in Various

Dimensions,Nuclear Physics B, Volume 346, Issue 1, p. 213-234, 1990.

[4] A.E. Blechman, A Mathematics Primer for Physics Graduate Students , 2007.

[5] R. Utiyama, Invariant Theoretical Interpretation of Interaction, Phys. Rev.

101, 15971607, 1956.

[6] T.W.B. Kibble Lorentz Invariance and the Gravitational Field, Journal of

Mathematical Physics, Vol. 2, p.212-221, 1960.

[7] A.H. Chamseddine, Topological Gauge Theory of Gravity in Five and All Odd

Dimensions, Physics Letters B, Volume 233, Issue 3-4, p. 291-294,1989.

[8] R.M. Wald,General RelativityThe University of Chicago Press, Chicago and

London, 1984, pp. 148 160.

[9] M.P. Hobson, G. Efstathiou, and A.N. Lasenby, General Relativity, An

Introduction for Physicists, Cambridge University Press, New York, 2006.

51


58/58

[10] C.W Misner, K.S. Thorne, and J.A. Wheeler,Gravitation, W.H. Freeman and

Company, U.S.A, 1973.

52

spherically symmetric solutions for five dimensional topological gravity

Documents