Bianchi type I solutions to Einstein s vacuum equations. Oliver Lindblad Petersen,

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1 Bianchi type I solutions to Einstein s vacuum equations. Oliver Lindblad Petersen, May 21, 2013

2 Abstract A natural question in general relativity is whether there exist singularities, like the Big Bang and black holes, in the universe. Albert Einstein did not in the beginning believe that singularities in general relativity are generic ([2], [3]. He claimed that the existence of singularities is due to symmetry assumptions. The symmetry assumptions are usually spatial isotropy and spatial homogeneity. Spatial isotropy means intuitively that, for a fixed time, universe looks the same at all points and in all spatial directions. In the present paper, we will show the following: If we solve Einstein s vacuum equations with a certain type of initial data, called the Bianchi type I, the resulting space-time will either be the Minkowski space or an anisotropic space-time equipped with a so called Kasner metric. We show that, in the anisotropic case, the space-time will contain a certain singularity: the Big Bang. We distinguish between two different classes of a Kasner metrics; the Flat Kasner metric and the Non-flat Kasner metric. In the case of a Flat Kasner metric, we show that it is possible to isometrically embed the entire space-time into Minkowski space. In the case of the Non-flat Kasner metric, the space-time is not extendible and the gravity goes to infinity approaching the time of the Big Bang. In addition we show, using any Kasner metric, that the universe expands proportional to the time passed since the Big Bang. This happens even though some directions will shrink or not change. The conclusion is: We have found two natural classes of anisotropic space-times, that include a Big Bang and expand. These results supports the idea that singularities are generic, i.e. are not due to the assumptions of symmetry of the universe. 1

3 Contents 1 Introduction 3 2 Formulating Einstein s vacuum equations The mathematical formulation of Einstein s equations Defining the manifold, the metric and the orthonormal frame Formulating the initial value problem in the case of the Bianchi type I metrics Einstein s equations as an initial value problem Initial conditions in the present case The evolution equation Ric(e i, e i = Solving Einstein s vacuum equations Defining the s through Einstein s equations The Minkowski space as a particular case The non-minkowski case Timelike geodesics in (K, g Using killing fields Interval of existence The Kretschmann scalar The Kretschmann scalar for (K, g Extendibility of (K, g Embedding of (K, g into Minkowski space for the Flat Kasner metric Non-extendibility for the Non-flat Kasner metric The expansion of (K, g (K, g is anisotropic The 3-torus as a submanifold Volume of a the torus, at a fixed time Length of one circle in the torus, at a fixed time Appendix Basic properties of the frame Ricci and Scalar curvature

4 Chapter 1 Introduction Albert Einstein invented the theory of general relativity in 1915 ([2], [3]. The theory is essentially a set of equations, with a space-time as the outcome. We start with a class of metrics, satisfying some common symmetry assumptions. Einstein s equations give a restriction on the class of metrics, telling us the possible form of the metrics. Each metric in the resulting class, together with the corresponding Lorentz manifold, constitutes a space-time, i.e. a model of the universe. After the publication of the equations, researchers found singularities in some of the space-times obtained through Einstein s equations. A typical example is the so called Schwarzschild metric. But Albert Einstein did not believe that the universe contains any singularities. He claimed that the existence of singularities in general relativity, was due to symmetry assumptions on symmetry of the universe. The symmetry assumptions he was referring to are called spatial isotropy and spatial homogeneity. The first means intuitively: For a fixed time, at each point in the universe, all spatial directions look the same. The latter means intuitively: For a fixed time, at each point in the universe, the surrounding looks the same. In the current paper, we prove the following: There are space-times that are spatially anisotropic (i.e. not spatially isotropic, that contain singularities. This is a first argument against the idea that the singularities are due to assumptions on the symmetry. We solve Einstein s vacuum equations with a certain class of metrics, invariant under spatial translations in R 3. The class of metrics is called Bianchi type I. As solutions, depending on choice of initial values, we will obtain three classes of Lorentz manifolds. The classes of corresponding metrics are: The trivial solution: Minkowski metric. The first anisotropic class: the Flat Kasner metrics. The second anisotropic class: the Non-flat Kasner metrics. For the Non-flat Kasner metrics, we obtain the following characterization of t = 0: A point in time where the geodesics (i.e. the particles started to exist. A point in time where the curvature becomes infinite (the gravity explodes at t = 0. We cannot extend the space-time into a larger space-time that includes the point t = 0 (nothing has happened before time t = 0. The space-time expands. For the Flat-Kasner metrics, we have a more absurd situation at t = 0: A point in time where the geodesics (i.e. the particles started to exist. 3

5 CHAPTER 1. INTRODUCTION 4 The curvature stays bounded also close to t = 0 (the gravity does not explode close to t = 0. We can extend the space-time into a larger space-time that includes t = 0. We can in fact send a light particle from the time before t = 0, that will never enter the smaller space-time. The space-time expands. Since the two last classes of space-times contain a Big Bang and expand, even though they are anisotropic, the results imply that singularities can exist in space-times without the standard symmetry assumptions.

6 Chapter 2 Formulating Einstein s vacuum equations. 2.1 The mathematical formulation of Einstein s equations. Assume that (M, g is a Lorentz manifold. We start by defining the Einstein gravity tensor (tensor field on (M, g: Definition 1. The Einstein gravity tensor G is defined as G = Ric 1 2 Sg where Ric is the Ricci curvature tensor and S is the scalar curvature. Having defined this, we can now formulate Einstein s equations: Definition 2. Einstein s equations are formulated as G + Λg = T where G is the Einstein tensor, Λ the cosmological constant and T is the stress energy tensor. All involved tensors are 2-tensor fields on M. The intention of this paper is not to motivate the equation in any greater detail, but rather to use it for its purpose, to obtain a class of space-times. The tensor T can be seen as containing the information about the matter. We will consider vacuum solutions, and hence we can assume that T = 0. Also the cosmological constant is assumed to be 0 in this model. This leads to the following equation: G = 0 ( Defining the manifold, the metric and the orthonormal frame. In the theory of special relativity, one considers the manifold R 4 with the Minkowski metric, g M = dt 2 + dx i dx i. It is natural to let the coefficients in the metric depend on one variable. Then the initial value problem of Einstein s equations translate to ODE s, for which the theory is more clear than in the 5

7 CHAPTER 2. FORMULATING EINSTEIN S VACUUM EQUATIONS. 6 case of PDE s. In general relativity it is hence customary to consider metrics of the form g = dt 2 + a(t 2 dx i dx i. (2.2 But these metrics are all spatially isotropic, i.e. the Lorentz manifold looks the same in all spatial directions. That the model of the universe is spatially isotropic is therefore a simplification. Hence it is of interest to study spatially anisotropic metrics. The natural generalization of (2.2 is hence to solve Einstein s equations with the following class of metrics: Definition 3. Consider a metric of the form g = dt 2 + (t 2 dx i dx i on K := I R 3, where : I (0, is smooth. We assume that I is the maximal interval of existence for (a 1 (t, a 2 (t, a 3 (t. This class of metrics is called the Bianchi type I metrics. Remark 4. Note that the metrics in the Bianchi type I still satisfy some symmetry. The metrics are, for a fixed time, invariant under translations in R 3. It is customary to formulate the equations in terms of an orthonormal frame. The natural orthonormal frame for (K, g is the following: Definition 5. We define an orthonormal frame for the manifold (K, g as e 0 = t, e i = 1, i = 1, 2, 3. xi 2.3 Formulating the initial value problem in the case of the Bianchi type I metrics. We will from now on use the following convention: α, β = 0, 1, 2, 3, i, j = 1, 2, 3. In other words: Every time we use greek letters, we will run over {0, 1, 2, 3}, every time we use latin letters, we will run over {1, 2, 3}. In this section, we will formulate Einstein s with the Bianchi class I as initial data. impose conditions on the coefficients { }, in terms of ODE s. This will Einstein s equations as an initial value problem. Assume that {e i } 3 i=0 is a local orthonormal frame on (M, g. In terms of the orthonormal frame, Einsteins equation s are formulated as G(e α, e β (t = 0 t I. (2.3 It is customary to view this as an initial value problem. Assume initially: G(e 0, e i (t 0 = 0, i = 1, 2, 3 (2.4 G(e 0, e 0 (t 0 = 0 (2.5 We also want a so called evolution equation. This is some subset of Einstein s equations that we start solving. After solving the evolution equation, one either proves that the rest of the equations are satisfied, or continues solving the rest of the equations. In the case of Einstein s vacuum equations, it is natural to solve Ric(e α, e β = 0 for some choice of α, β. The following Lemma shows that Ric(e α, e β = 0 must be satisfied if the vacuum equation is satisfied.

8 CHAPTER 2. FORMULATING EINSTEIN S VACUUM EQUATIONS. 7 Lemma 6. G(e α, e β = 0, α, β Ric(e α, e β = 0, α, β Proof. By definition of G, G(e α, e β = Ric(e α, e β 1 2 Sg(e α, e β. We use the notation G(e α, e β = G αβ, Ric(e α, e β = Ric αβ and g(e α, e b = g αβ. Assume first that Ric αβ = 0, α, β. Since the scalar curvature is defined as S = tr g Ric this is zero, since Ric αβ = 0. Assume instead that G αβ = 0, α, β. Then the trace of G vanishes and we obtain 0 = tr g (G = g αβ (Ric αβ 1 2 Sg αβ = g αβ Ric αβ 1 2 Sgαβ g αβ We see that S = 0, and = tr g (Ric 1 S 4 = S 2S = S 2 G αβ = Ric αβ Hence if Ric αβ = 0, α, β, then G αβ = 0, α, β. This lemma proves, in particular, that if G = 0, then Ric(e α, e β = 0 α, β. We define the evolution equation as: Ric(e i, e i (t = 0, t I and i = 1, 2, 3 (2.6 The reason why a solution of the evolution equation satisfying the initial conditions (2.4 and (2.5, solves G = 0 is at this point not obvious at all. It will later be shown that, G = 0 is satisfied for all times Initial conditions in the present case. Now we want to translate the initial value problem into the present case of Bianchi type I metrics. Assume the Lorentz manifold is (K, g, i.e. the manifold I R 3 equipped with a metric in the class Bianchi type 1. We start by treating equation (2.4: This can be reformulated as G(e 0, e i (t 0 = 0, i = 1, 2, 3. G(e 0, e i (t 0 = Ric(e 0, e i (t S(t 0g(e 0, e i (t 0 = Ric(e 0, e i (t 0. Note by Lemma 50 in the Appendix, we know that Ric(e 0, e i = 0. Hence the equation (2.4 is always satisfied and does not impose any conditions on the s. Now we treat equation (2.5: This can be reformulated as G(e 0, e 0 (t 0 = 0. G(e 0, e 0 (t 0 = Ric(e 0, e 0 (t S(t 0g(e 0, e 0 (t 0 = Ric(e 0, e 0 (t S(t 0.

9 CHAPTER 2. FORMULATING EINSTEIN S VACUUM EQUATIONS. 8 Note that by Lemma 49 in the Appendix, By Lemma 53 in the Appendix, Ric(e 0, e 0 = S = θ 2 + ( 2 (ȧi (ȧi + t. ( 2 (ȧi (ȧi + 2 t, where Hence is the same as and we simplify the condition to θ(t = θ(t G(e 0, e 0 (t 0 = 0 ȧ i (t (t. (2.7 2 (ȧi (t 0 = 0 (t 0 θ(t The evolution equation Ric(e i, e i = 0. 2 (ȧi (t 0 = 0. (2.8 (t 0 Recall that the goal is to solve equation (2.6. By Lemma 51 in the Appendix, (ȧi Ric(e i, e i (t = t (t + ȧi(t (t θ(t, where θ is defined in equation (2.7. Hence equation (2.6 imposes the following condition on the s: (ȧi t (t + ȧi(t θ(t = 0. (2.9 (t

10 Chapter 3 Solving Einstein s vacuum equations. We are now in shape to solve Einstein s vacuum equations. We define the s through the equations (2.8 and (2.9. We solve the equations for { } and insert them in the definition of Bianchi type I metrics. Hence we obtain a subclass of metrics, that will satisfy Einstein s equations. 3.1 Defining the s through Einstein s equations. In the preceding chapter we defined the class of BIanchi type I metrics as the metrics of the form g = dt 2 + (t 2 dx i dx i. The preceding chapter motivates the following definition: Definition 7. Define (t for i = 1, 2, 3 through: For all t I we define (t to satisfy equation (2.9, i.e. (ȧi t + θ ȧi = 0, (3.1 with initial conditions including equation (2.8 θ(t (ȧi (t 0 = 0 (3.2 (t 0 (t 0 = α i (3.3 ȧ i (t 0 = β i, (3.4 where t 0 R, α i (0, and β i R. Let I denote the maximum existence interval for the solution (a 1 (t, a 2 (t, a 3 (t. We define the manifold K := I R 3. We identify (K, g as this manifold equipped with the corresponding metric. Also, to simplify notation, define θ i (t := ȧi(t (t. (3.5 9

11 CHAPTER 3. SOLVING EINSTEIN S VACUUM EQUATIONS. 10 Note that θ(t = θ i (t. We will now remark how to rewrite the equations above in a way that will be more convenient for the coming lemmas. Remark 8. Equation (3.1 can be rewritten as By linearity of the derivative, we get 0 = t (θ i (t + θ(tθ i (t = t ( t (θ i (t + θ(tθ i (t = 0. (3.6 θ i (t + θ(t 3.2 The Minkowski space as a particular case. θ i (t = θ(t + θ(t 2. (3.7 Assume the initial dats such that θ(t 0 = 0. We will now show that in this particular case, we actually get the Minkowski space. Lemma 9. Assume θ(t 0 = 0. Then I = (t, t + = R. Furthermore (K, g is isometric to Minkowski space R 4 1. Proof. We start by solving equation (3.7 for the given initial data. The uniqueness theorem for first order ODE s applies, hence θ(t = 0. Knowing this, we can rewrite (3.6 as follows: θ i (t = θ(tθ i (t = 0 The uniqueness theorem of ODE s applies again, and θ i (t = A i solves the equation. Note now that equation (3.2 can be rewritten as θ(t 0 2 θ i (t 0 2 = 0. Since θ(t 0 = 0 we get θ i (t 0 = 0 θ i (t = 0. Hence ȧ i (t = 0, and the solution is given by (t = C i R for i = 1, 2, 3. Hence (t, t + = R. We conclude that the metric becomes g = dt 2 + Ci 2 dx 2 i. This makes (K, g isometric to Minkowski space R 4 1 under the explicit isometry: f : (K, g R 4 1 (t, x, y, z (t, C 1 x, C 2 y, C 3 z Since Einstein s equations were to be fulfilled not only initially, but for all times, we need to check that the choices of the s satisfy also the equation (2.1. Proposition 10. Assume the initial condition G(e 0, e β (t 0 = 0, and θ(t 0 = 0. Then G(e α, e β (t = 0, α, β and t I Proof. By Lemma 6, this is equivalent to the Ricci curvature to vanish for all times. But the Ricci curvature does indeed vanish, since all s are constant by Lemma 9, and by Lemmas 49, 50, 51, 52 in the Appendix. Remark 11. We have hence solved Einstein s equations for the initial value θ(t 0 = 0. solution is (up to isometry the Minkowski space. The

12 CHAPTER 3. SOLVING EINSTEIN S VACUUM EQUATIONS The non-minkowski case In this section we consider the case when the initial data for θ is nonzero. We will first prove that, up to a change of time orientation, we have a maximal existence interval (t a, for θ. Lemma 12. Assume that θ(t 0 > 0. (This covers all cases, because if θ(t 0 < 0, we can change the time orientation. Then the existence interval of θ is of the form (t a, where t a >. Furthermore, θ(t = 1 t t a and Proof. Equation (3.7 has θ(t = 1 t+c This gives the solution lim θ(t =, t t a θ(t 0 = lim θ(t = 0. t as its solution. Hence 1 t 0 + C C = 1 θ(t 0 t 0. θ(t = 1 t + 1 θ(t 0 t 0 Using now that θ(t 0 > 0 gives a positive interval of existence. Hence, for all t > C, θ(t is continuously differentiable and we have uniqueness on the existence interval ( C,. Choose t a := C, then the limits in the lemma are obvious and do also imply that (t a, is the maximal existence interval. The next step is to prove that there is nothing special about the constant t a. After a suitable time coordinate translation, we can assume it to be 0. The time t a will be considered to be the point in time were the space-time starts to exists. This moment is what we usually call the Big Bang and hence it is convenient if we can assume that it happened at t a = 0. Lemma 13. Assume that θ(t 0 > 0. After a suitable time coordinate translation, the solution can be written as θ(t = 1 t, t R +, and θ(t 0 > 0, for some t 0 R +. Proof. Assume that we have a solution Define: Θ(s = 1 s + 1 Θ(s s, Θ(s 0 > ( θ(t = Θ t 1 Θ(s 0 + s 0 = 1 t. If we define t 0 := 1 Θ(s 0, we have θ(t 0 = Θ(s 0 > 0. Hence so far we know that the solution for θ can be written as θ(t = 1 t. Since the goal is to determine the functions (t, we need now to compute the functions θ i = ȧi. Lemma 14. Assume that θ(t = 1 t, t R +, and θ(t 0 > 0, for some t 0 R +. Then θ i (t = p i t, t R + for i = 1, 2, 3, where p i is a real constant. Furthermore, p i = 1.

13 CHAPTER 3. SOLVING EINSTEIN S VACUUM EQUATIONS. 12 Proof. Equation (3.6 becomes θ i (t + θ i(t = 0. t The solution is given by θ i (t = p i t, and the interval of existence is the same as for θ(t, i.e. (0,. Uniqueness is clear for t 0. The existence interval needs to contain t 0 > 0, hence the maximal existence interval is (0,. In addition we have the following: 1 t = θ(t = θ i (t = p i t = 1 t p i. This result makes it possible to compute the s. Lemma 15. Assume that θ(t = 1 t t R + and that θ(t 0 > 0. For i = 1, 2, 3, we have the solution (t = C i t pi, t (0,, where C i is a positive real constant. Proof. The differential equation for (t is given by: This gives the solution θ i (t = p i, t (0,, (3.8 t ȧ i (t p i (t t = 0, t (0,. (t = C i t pi. The existence interval will be the same as for θ i, namely (0,. Uniqueness is clear and since (t is non-zero for all times, the solution is also well defined. By assumption θ(t 0 = 1 t 0 > 0 t 0 > 0. But by definition, (t 0 = C i t 0 p i = α i > 0. Hence the constants C i are all positive. Hence the metric is of the form g = dt 2 + Ci 2 t 2pi dx i dx i Remark 16. A similar argument to the proof in Lemma 9, shows that this case is up to isometry, equivalent to the case when C i = 1, i.e. g = dt 2 + t 2pi dx i dx i. As in the Minkowski case, we need to show that the equation (2.1 is satisfied. Proposition 17. Assume that θ(t = 1 t and θ(t 0 0 holds, then G(e 0, e 0 (t = 0, t R +. Furthermore, p 1 p 2 + p 2 p 3 + p 1 p 3 = 0.

14 CHAPTER 3. SOLVING EINSTEIN S VACUUM EQUATIONS. 13 Proof. G(e 0, e 0 = Ric(e 0, e Sg(e 0, e 0 = Ric(e 0, e S = Ric(e 0, e Ric(e 0, e Ric(e i, e i = Ric(e 0, e Ric(e i, e i 2 ( { } = Lemma 49 and 51 = 1 (θ 2 i + 2 θ i + 1 ( θi + θ i θ = θ 2 i + 1 θ i = θ 1 θ 2 + θ 1 θ 3 + θ 2 θ 3 By Lemma 14, θ i (t = pi t. We now use the initial assumption that G(e 0, e 0 (t 0 = 0. The initial condition implies Hence we get 0 = G(e 0, e 0 (t 0 = p 1p 2 t p 2p 3 t p 1p 3 t 2 0 p 1 p 2 + p 2 p 3 + p 1 p 3 = 0. G(e 0, e 0 (t = p 1p 2 t 2 + p 2p 3 t 2 + p 1p 3 t 2 = 1 t 2 (p 1p 2 + p 2 p 3 + p 1 p 3 = 0. The following result will restrict the choices of the p i s even further. Lemma 18. Assume that θ(t = 1 t t R + and that θ(t 0 > 0. Then p 2 i = 1. Proof. We have already proven the following p i = 1 and p 1 p 2 + p 2 p 3 + p 1 p 3 = 0. Hence we have ( 2 1 = p i = p 2 i + 2(p 1 p 2 + p 2 p 3 + p 1 p 3 = p 2 i. Proposition 19. Assume that θ(t = 1 t t R + and that θ(t 0 > 0. Then G(e i, e i (t = 0, Proof. We know that Ric(e i, e i = 0. Hence = 1 2 Ric(e 0, e 0 (t = 1 2 t (0,. G(e i, e i (t = Ric(e i, e i (t 1 2 S(tg(e i, e i (t = 1 2 S(t = 1 2 = 1 2t 2 ( Ric(e 0, e 0 (t + ( θ i (t 2 + θ i (t = 1 2 ( 2 p i p i Ric(e i, e i (t ( 2 pi t 2 = 1 (1 1 = 0. 2t2 p i t 2 = 1 2t 2 ( pi p 2 i

15 CHAPTER 3. SOLVING EINSTEIN S VACUUM EQUATIONS. 14 Theorem 20. Assume θ(t 0 0. Then Einstein s vacuum equation G = 0 is satisfied. Proof. We have shown that this is, up to change of time coordinate and time orientation, θ(t = 1 t and θ(t 0 > 0. In the Propositions 17 and 19, we showed that G(e 0, e 0 = 0 and that G(e i, e i = 0. Note that, by Lemmas 50 and 52, Ric(e α, e β = 0 and g(e α, e β = 0 if α β. Hence G(e α, e β = 0 if α β. This implies that the Einstein tensor vanishes. We are now in ready to summarize the results in a remark. Remark 21. We have solved Einstein s vacuum equations for the initial value θ(t 0 0. We know that up to isometry, time coordinate translation and time orientation, a space-time, in this paper, is a manifold K and a metric g given by: where K = I R 3 = R + R 3. (3.9 g = dt 2 + t 2pi dx i dx i. (3.10 p i = 1, (3.11 p 2 i = 1. (3.12 Definition 22. The class of metrics of the form (3.10 on the manifold (3.9, with p i s satisfying equations (3.11 and (3.12, is called the class of Kasner metrics. A metric in this class is called a Kasner metric. We split the class of Kasner metrics into two subclasses of metrics. Definition 23. The class of Kasner metrics where some p i = 1, is called the class of Flat Kasner metrics. The class of Kasner metrics where all p i metrics. 1, is called the class of Non-flat Kasner We characterize the Flat and Non-flat Kasner metrics further: Lemma 24. The p i s must satisfy the following: Flat metric: Two of them equals 0, the third equals 1. Non-flat metric: Two of them are positive, the third is negative and has absolute value strictly less than 1 2. Proof. We know from 3 p2 i = 1 that the absolute value of each p i must be less than 1. Hence it follows immediately from 3 p i = 1 that maximum one of the indices can be negative. The first case to consider is when all are nonnegative: Since p 1 p 2 + p 2 p 3 + p 1 p 3 = 0 p i p j = 0. This implies that 2 of them are 0, hence the 3rd equals 1. The second case is when 1 of the p i s is negative. Through basic geometry calculations of the intersection of the plane 3 p i = 1 and the sphere 3 p2 i = 1, one realizes that the negative p i must have absolute value strictly less than 1/2. Since p i 1 and 3 p i = 1 the others must be positive.

16 Chapter 4 Timelike geodesics in (K, g. The theory for the case of the Minkowski space is clear, it is the theory of special relativity. Hence in this chapter and from now on we will only consider the solutions for initial conditions satisfying θ(t 0 0. We have shown in the previous chapter, that the manifold (K, g is R + R 3 equipped with a Kasner metric. In this chapter we will prove that all future pointing timelike geodesics can be extended to an existence interval of (a,, where a > and the timelike future pointing geodesic approaches the boundary t = 0 as the parameter approaches a. We will also prove that the existence interval cannot be extended to (,. We will use the following notation for a parametrized geodesic γ: 4.1 Using killing fields γ(s = (t(s, x 1 (s, x 2 (s, x 3 (s. There is a special type of vector fields, called Killing vector fields. We will show that the vector fields i are Killing vector fields, and use them to find the maximal existence intervals of the geodesics. Definition 25. A vector field X on a semi-riemannian manifold is called a Killing vector field if the Lie derivative of the metric tensor, with respect to X, vanishes: L X g = 0. Lemma 26. i are Killing fields for i = 1, 2, 3. Proof. By [1, Proposition 9.5], i is a killing field if and only if: for all vector fields on V, W on (K, g. i V, W = [ i, V ], W + V, [ i, W ] In the remaining part of the proof, we will use the Einstein summation convention. Let V = V α α and W = W β β. It follows that i V, W = i ( W α V β g αβ = {gαβ = 0 if α β, i (g αα = 0} = ( i (W α V α + W α i (V α g αα, [ i, V ], W = i (V α α, W = i (V α W β g αβ = {g αβ = 0 if α β} = i (V α W α g αα, V, [ i, W ] = i (W α V α g αα, similarily. Hence the lemms proved. 15

17 CHAPTER 4. TIMELIKE GEODESICS IN (K, G. 16 Lemma 27. The derivatives of the space component of the geodesic is given by x i(s = C i t(s 2pi, where C i R. This means that if γ exists on I, then the above equality holds on I. Proof. By [1, Lemma 9.26] we know that, since i is a Killing field, γ, i (s = C i R. Written in coordinates: C i = γ, i (s = x i(s t(s 2pi, x i(s = C i t(s 2pi. 4.2 Interval of existence Remark 28. (Local existence of geodesics The local existence of a geodesic on a manifold is a standard result in semi-riemannian geometry. Hence we know that the geodesics in (K, g exist locally. Hence we know that for each geodesic, we have a maximal interval of existence, say I := (s min, s max R. The next step is to show that this interval actually is of the form (a,. The following lemms mainly for later use, but already reveals some of the geometry of the geodesic. Lemma 29. Let I := (s min, s max R be the maximal interval of existence of the future pointing timelike geodesic γ. Then t is monotone increasing on I. Proof. By definition of a timelike geodesic, γ (s, γ (s = C 0 < 0. This implies Since C 0 > 0, we see that t (s 2 + t(s 2pi x (s 2 = C 0, C 0 > 0. 0 < t(s 2pi x (s 2 + C 0 = t (s 2, s I. We assumed initially that t (s 0 > 0. Then, since t (s is continuous, and nonzero for all s, it follows that t (s > 0 for all s I. The next lemma states that the time component of any geodesic defined only on a finite interval, tend to infinity. Lemma 30. Assume that γ : I := [0, s max M, s max <, is a timelike future pointing geodesic that can not be extended to any bigger interval of the form [0, s max + ɛ, where ɛ > 0. Assume also that t 0 := t(0 > 0. Then M > 0, s I such that t(s > M.

18 CHAPTER 4. TIMELIKE GEODESICS IN (K, G. 17 Proof. Assume to obtain a contradiction, that the proposition is not true. Then there exists an M 0 R +, such that t(s M 0, s I. Then [t(0, M 0 ] is a compact interval which contains t(i. By Lemma 22, t is monotone in I and hence t(s converges, to say t 1 R, i.e. We define lim t(s = t 1. s s max t(s max := t 1. We note that t(i [t 0, t 1 ] with t 0 > 0 and t 1 <. By Lemma 20, and by integrating, we achieve x i (s x i (s = s s x i(udu = s s C i t(u 2pi du. This implies (since t 2pi is monotone as a function of t s x i (s x i (s = C i t(u 2pi du C i (s s max(t 2pi 0, t 2pi 1. In particular, By assumption s max <. Hence s x i (s x i (0 C i (s s 0 max(t 2pi 0, t 2pi 1. x i (s x i (0 C i (s max s 0 max(t 2pi 0, t 2pi 1. and we have a uniform bound. Also, x i (s is monotone on I, since x i (s = C it(s 2pi has the same sign as C i, and is nonzero. This means that x i (s converges as s s max. We define Hence we have implicitly defined x i (s max = γ(s max = lim x i (s. s s max lim γ(s. s s max By [1, Lemma 5.8], a geodesic is extendible as a geodesic if and only if it is continuously extendible. Since we proved that it is continuously extendible, the geodesic is extendible as a geodesic and we have reached a contradiction. Having done this, we are finally in shape for one of the two theorems of this chapter: to prove that all geodesics can be extended to an infinite open right ray (a,. Theorem 31. All timelike future pointing geodesics γ : I := [0, a M can be extended to a timelike future pointing geodesic α : [0, M. Proof. By assumption t(0 > 0 and since γ is future pointing, t (0 > 0. By Lemma 29, t is monotone and hence t(s t(0 s I. Hence t(s never goes to zero, and we don t get a problem there. We know that one necessary condition for γ to be a geodesic, is that γ (s, γ (s = t (s 2 + t(s 2pi x i(s 2 is constant. By Lemma 27, we can substitute x i (s = C it(s 2pi, and see that t (s 2 + C 2 i t(s 2pi

19 CHAPTER 4. TIMELIKE GEODESICS IN (K, G. 18 is constant. Hence the derivative must vanish, i.e. t (s + This gives the following bound on t (s: t (s p i C 2 i t(s 2pi 1 = 0 p i C 2 i t(s 2pi 1 Now, by Lemma 24, i, p i > 1 2. This implies that i, 2p i 1 < 0. Since t is monotone increasing on I, we realize that t(s 2pi 1 decreases. In particular it is bounded. Hence there exists a constant A such that: t (s A, s I It is now clear that t(s is bounded on I if I is a finite interval. Hence by Lemma 30, all geodesics can be extended to positive infinity. Now that we know that all geodesics can be extended to (a,, we want to prove that no geodesic can be extended to negative infinity, i.e. that (a, actually is the form of the maximal interval of existence. Theorem 32. There are no timelike future pointing geodesics of the form γ : I := (, 0] K. Proof. Assume we have a future pointing timelike geodesic γ : (, 0] K. As before, we denote it by γ(s = (t(s, x 1 (s, x 2 (s, x 3 (s. Hence the equation can be rewritten as Rearranging gives t (s 2 + t (s 2 = γ, γ = C 0, C 0 > 0 t(s 2pi x i(s 2 = C 0. t(s 2pi x i(s 2 + C 0 C 0 > 0. Since t : I R is continuous, the sign of t (0 determines the sign of t (s for all s. Since it is future pointing, t (0 > 0. Hence t (s > 0 for all s I. We get Note now that This implies t(0 t(s = 0 s t (s > C 0 t (udu 0 C 0 du = C 0 (0 s = C 0 s t(s t(0 + C 0 s Since t(s > 0 for all s I, this contradicts the assumption that the existence interval I of the geodesic is (, 0]. s

20 CHAPTER 4. TIMELIKE GEODESICS IN (K, G. 19 We have proven the following result: Every future pointing timelike geodesic γ : (c, d K can be extended to a geodesic α : (a, K, where a > and γ = α (c,d. We have also proved that the maximal interval of a future pointing timelike geodesic is of that form. It will also be necessary to characterize the behavior as the parameter s approaches the boundary of the maximal interval of existence: Corollary 33. If γ(s is a timelike future pointing geodesic with (a, as its maximal interval of existence, then lim t(s = 0. s a Proof. Assume that there exists δ > 0 such that t(s > δ, s (a,. We know that t (s > 0 on this interval. Hence a similar argument to the proof of Lemma??, shows that the geodesic is extendible to the point γ(a. This contradicts the maximality of the interval of existence.

21 Chapter 5 The Kretschmann scalar An interesting coordinate invariant scalar function on the a semi-riemannian manifold is the Kretschmann scalar. We will in this chapter compute the Kretchmann scalar for (K, g. 5.1 The Kretschmann scalar for (K, g Definition 34. The Kretschmann scalar is defined R αβγδ R αβγδ. Lemma 35. The Kretschmann scalar on (K, g is Proof. Note: R αβγδ R αβγδ = 4 t 4 (p 2 1p p 2 1p p 2 2p R αβγδ R αβγδ = ( p 2 2 i p i g αβ = g αβ = g αα αβγδα β γ δ = αβγδ g αα g ββ g γγ g δδ R 2 αβγδ g ββ g γγ g δδ R αβγδ R α β γ δ Case 1: All indices are nonzero. R(e i, e j e k = D [ei,e j]e k [D ei, D ej ]e k = D ei (D ej e k + D ej (D ei e k ȧ = D ei (δ jk e k ȧ 0 a k + D ej (δ ik e k ȧ 0 a k = δ k ȧ jk a k D ei (e 0 + δ k ik a k D ej (e 0 ȧ = δ k ȧ i ȧ jk a k e i + k ȧ j ik a k a j e j, ȧ R(e i, e j e k, e l = δ k jk = (δ ik δ jl δ il δ jk ȧk ȧ l a k a l. a k where ik is the Kroneker delta. ȧ i ȧ e i, e l + δ k ȧ j ȧ ik a k a j e j, e l = δ il δ k ȧ i ȧ jk a k + δ ik δ k ȧ j jl a k a j We only calculate the non-zero components. tensor, this means that (i j and (k l. Note that by symmetries of the Riemann 1. (l i (δ ik δ jl δ il δ jk = δ ik δ jl 0 (i = k and (j = l : R ijij = R(e i, e j e i, e j = ȧi ȧ j a j 2. (l = i (δ ik δ jl δ il δ jk = (δ ik δ ij δ jk 0 (j = k i = l : R ijji = R(e i, e j e j, e i = ȧi ȧ j a j 20

22 CHAPTER 5. THE KRETSCHMANN SCALAR 21 Case 2: 1 index is zero. o l = 0 R(e i, e j e k e 0 by the calculations in Case 1. This means that R ijk0 = 0. o o k = 0 R ij0l = R ijl0 = 0 by the previous. i = 0 or j = 0 R 0jkl = R j0kl = R klj0 = 0 by the previous. Case 3: 3 indices are zero. o R(e 0, e 0 e 0 = 0 R 000l = 0. Hence from the identities of the Riemannian tensor we get: R 000l = R 00l0 = R 0l00 = R l000 = 0. Case 4: All indices are zero. o In this case the component obviously vanishes. Case 5: 2 indices are zero. o R(e 0, e 0 e i = R(e 0, e 0 e i R(e 0, e 0 e i = 0 R 00ij = R ij00 = 0. o R(e i, e 0 e j = D [ei,e 0]e j [D ei, D e0 ]e j = D ȧ i a e i (e j (D ei (D e0 e j D e0 (D ei e j ( ( ( i 2 ( = ȧi ȧ δ ij D ei (e i + δ ij D i ȧ e0 e 0 = δ i ȧ ij a e0 i + i t e 0 R i0i0 = R i00i = R 0ii0 = R 0i0i = Summary: The non-zero terms are (i j, i, j = 1, 2, 3 R ijij = R ijji = ȧiȧj a j, ( ȧ i 2 t ( R i0i0 = R i00i = R 0ii0 = R 0i0i = ( ȧ i ȧ i. 2 t ( ȧ i. Note now that g αα g ββ g γγ g δδ = 1 an even number of indices are zero. This is the case for all non-zero terms above. Hence the Kretchmann scalar is: ( R 2 ijij + Rijji 2 ( + R 2 i0i0 + Ri00i 2 + R0ii0 2 + R0i0i 2 R αβγδ R αβγδ = i j Now, using we obtain = 2 i j R 2 ijij (ȧ1 ȧ 2 (ȧ1 ȧ 3 (ȧ2 ȧ 3 = a 1 a 2 a 1 a 3 a 2 a 3 ( p 2 = 4 1 p 2 2 t 4 + p2 1p 2 3 t 4 + p2 2p 2 3 t ( pi t = p i t t 2 R 2 i0i0 ( 2 (ȧi (ȧi 2 + t R αβγδ R αβγδ = 4 t 4 (p 2 1p p 2 1p p 2 2p ( p 2 i t 2 + t ( pi 2 t ( p 2 2 i p i

23 CHAPTER 5. THE KRETSCHMANN SCALAR 22 Remark 36. The Kretchmann scalar is 0 if and only if one p i = 1. Hence if all p i s are different from zero, the Kretchmann scalar approaches infinity when t 0.

24 Chapter 6 Extendibility of (K, g. In this chapter, we will discuss the possibility of finding an extension of (K, g. In other words, another Lorentz manifold ( K, ḡ and an isometric embedding from (K, g into ( K, ḡ. This is the point where the Flat and Non-flat cases give very different results. 6.1 Embedding of (K, g into Minkowski space for the Flat Kasner metric. In this section we assume that g is the Flat Kasner metric. Proposition 37. Assume p 1 = 1 and p i = 0 if i = 2, 3. Then there is an isometric immersion of (K, g into the Minkowski space R 4 1, given by f : (K, g (R 4 1, g M f(t, x 1, x 2, x 3 = (t cosh(x 1, t sinh(x 1, x 2, x 3 Proof. It is clear that f is a diffeomorphism onto its image. Since we have chosen coordinates, we can represent df as a Jacobian matrix Jf: cosh(x 1 t sinh(x Jf(t, x 1, x 2, x 3 = sinh(x 1 t cosh(x Hence, if v = (v 0, v 1, v 2, v 3 and w = (w 0, w 1, w 2, w 3 in some tangent space T p K, where p = (t, x 1, x 2, x 3, we get v 0 cosh(x 1 + v 1 t sinh(x 1 g M (df(v, df(w = v 0 sinh(x 1 + v 1 t cosh(x 1 v 2 v w 0 cosh(x 1 + w 1 t sinh(x w 0 sinh(x 1 + w 1 t cosh(x v 0 cosh(x 1 + v 1 t sinh(x 1 = v 0 sinh(x 1 + v 1 t cosh(x 1 v 2 v 3 w 2 w 3 w 0 cosh(x 1 w 1 t sinh(x 1 w 0 sinh(x 1 + w 1 t cosh(x 1 w 2 w 3 23

25 CHAPTER 6. EXTENDIBILITY OF (K, G. 24 = v 0 w 0 cosh 2 (x 1 (v 0 w 1 + w 0 v 1 t cosh(x 1 sinh(x 1 v 1 w 1 t 2 sinh 2 (x 1 +v 0 w 0 sinh 2 (x 1 + (v 0 w 1 + v 1 w 0 t sinh(x 1 cosh(x 1 + v 1 w 1 t 2 cosh 2 (x 1 +v 2 w 2 + v 3 w 3 = v 0 w 0 (sinh 2 (x 1 cosh 2 (x 1 + v 1 w 1 t 2 (cosh 2 (x 1 sinh 2 (x 1 + v 2 w 2 + v 3 w 3 Now, using the identity cosh 2 (x sinh 2 (x = 1, we get g M (df(v, df(w = v 0 w 0 + t 2 v 1 w 1 + v 2 w 2 + v 3 w 3 = g(v, w and, since the point p and the vectors v and w were arbitrary, the proof is done. 6.2 Non-extendibility for the Non-flat Kasner metric. In this section we assume that g is a Non-flat Kasner metric. Lemma 38. Assume (K, g is extendible, i.e. that there exists a non-surjective isometric embedding ι : (K, g ( K, ḡ. where ( K, ḡ is a connected Lorentz manifold of the same dimension as K. Then there exists a future pointing timelike geodesic γ such that Im(γ ι(k, Im(γ (ι(k c. Proof. By assumption there exists a Lorentz manifold ( K, ḡ and an isometric embedding ι : (K, g ( K, ḡ. Note that this implies that timelike geodesics in (K, g are timelike geodesics in ( K, ḡ. Since ι is a homeomorphism, ι(k is an open subset of K. Hence ι(k has a boundary in K, we call it K. Let x 0 K. Let γ be a timelike geodesic starting from x 0. If this geodesic now touches ι(k, we are done, so assume it stays in ι(k c. Choose a convex neighborhood U around x 0. Let x 1 Im(γ U. Hence x 0 and x 1 are connected by the timelike geodesic γ. Hence there exists a timelike tangent vector v T x0 K, such that expx0 (v = x 1. Since U is a convex neighborhood, the exponential map is actually defined on the entire U. Define now the function: v x1 : U K T K, v x1 (p = (p, exp 1 p (x 1. By [1, Lemma 5.9], the function v x1 is continuous. Now define the function f : T K R, f(p, v = ḡ p (v, v. Since the function v x1 and the metric ḡ is continuous on both arguments, we know that f is continuous. Thus (f v x1 1 (R is an open set W U containing x 0 such that at each point p W, the vector exp 1 p (x 1 is timelike. Hence W is an open neighborhood of x 0 K. By definition of boundary there exists a point y ι(k such that exp 1 y (x 1 is timelike, and exp y (exp 1 y (x 1 = x 1. Define the geodesic α through α(s = exp y (s exp 1 y (x 1. By definition of the exponential map, the geodesic is well-defined. Since exp 1 y (x 1 is timelike, the geodesic is also timelike. Note that α(0 = exp y (0 = y, α(1 = exp y (exp 1 y (x 1 = x 1. Since α is timelike, y ι(k and x 1 ι(k c, the proof is done.

26 CHAPTER 6. EXTENDIBILITY OF (K, G. 25 Theorem 39. (K, g is not extendible, i.e. there exists no isometric embedding ι : (K, g ( K, ḡ where ( K, ḡ is a Lorentz manifold and ι is not surjective. Proof. Assume, to reach a contradiction, that (K, g is extendible. By the previous lemma, we know that there exists a future pointing timelike geodesic γ : [0, h] K such that it starts in ι(k and ends in ι(k c or the other way around. Assume now that the geodesic is future pointing. This is no loss of generality, because if it is not future pointing, α(s := γ( s would be. Then we have the following two cases: Case 1: γ(0 ι(k and γ(h ι(k c In this case, since ι(k is open in K, we can define the following geodesic in ι(k: α 1 : [0, ɛ ι(k α 1 = γ [0,ɛ But because ι is an isometry onto its image, this geodesic can be lifted through ι to a geodesic α 1 K : [0, ɛ K such that α = ι α K. But from Theorem 31 we know that α K can be extended to a geodesic β K on an interval [0, and be well-defined in (K, g. Hence we can define β := ι β K : [0, ι(k. Note now that by construction β(s = γ(s s [0, ɛ But γ leaves ι(k in finite time, and β does not leave ι(k. This is a contradiction to the assumption that γ is a geodesic. Case 2: γ(0 ι(k c and γ(h ι(k Define similar to case 1: α 2 : (δ, h] ι(k, α 2 = γ (δ,h]. where (δ, h] is the maximal connected interval where γ is in ι(k. Hence the lift α 2 K : (δ, h] K is well-defined. Note now that by the assumptions on (δ, h] and ι, we cannot extend the interval of α 2 K beyond δ. We also know, by Corollary 33, that if t α is the time component of α 2 K, then t α (s 0 as s δ +. Since the connection is preserved under an isometric embedding ([1, Proposition 3.59], the Kretschmann scalar R αβγδ R αβγδ is also preserved. If we denote the Kretschmann scalar on K by R αβγδ Rαβγδ, we can write this as R αβγδ R αβγδ (p = R αβγδ Rαβγδ (ι(p. Since t α (s 0 as s δ +, we see by Lemma 35, that as s δ +. This means that R αβγδ R αβγδ (α 2 K(s R αβγδ Rαβγδ (ι(α 2 K(s as s δ +. By definition of αk 2, this is the same as R Rαβγδ αβγδ (γ(s as s δ +. But, by assumption, γ is also defined on γ(δ, which contradicts that the Kretschmann scalar is smooth on K.

27 Chapter 7 The expansion of (K, g. The customary opinion today is that the universe is expanding ever since the Big Bang. Also this model will show this phenomenon. The volume will grow proportional to the time since the Big Bang. We prove that (K, g is anisotropic. The conclusion is hence that (K, g contains a Big Bang (chapter 4 and chapter 6 and expand, but is not isotropic. An interesting consequence of anisotropy is that different directions in the space-time may not change or even shrink (section 7.4. As usual, we assume that K = R + R 3 and that g is a Kasner metric. 7.1 (K, g is anisotropic. We start by proving that (K, g is anisotropic. Definition 40. Let (M, g be a Lorentz manifold, with a preferred foliation M = I S where I corresponds to time, and S to space. (M, g is called (spatially isotropic if: p = (t, x I S and v p, w p T p M such that v p, w p T p ({t} S g(v p, v p = g(w p, w p there exists a local isometry φ defined on U p such that dφ(v p = w p (t 1, x 1 U φ(t 1, x 1 {t 1 } S Proposition 41. (K, g is not spatially isotropic. Remark 42. The preferred foliation that will be used is I = R + and S = R 3. Proof. Assume, to reach a contradiction, that (K, g is spatially isotropic. Fix p = (t 1, x 1 K. We know that the frame is defined everywhere on K, hence in particular at p. We define v p := e 1 (p and w p := e 2 (p. Then v p, w p T p ({t} R 3 and g(v p, v p = 1 = g(w p, w p. By assumption, there exists a local isometry φ defined on U p such that it dφ(v p = w p. By [1, Lemma 4.7], the second fundamental form is preserved locally, in particular dφ (II(v p, v p = II(dφ(v p, dφ(v p = II(w p, w p (7.1 26

28 CHAPTER 7. THE EXPANSION OF (K, G. 27 As usual, we let D denote the Levi-Civita connection on (K, g. By definition, II(V, V is the, to the hypersurface {t} R 3, normal component of D V V. Note, by Lemma 34, D e1 e 1 = ȧ1 a 1 e t D e2 e 2 = ȧ2 a 2 e t Now, T p ({t} R 3 is spanned by e 1 (p, e 2 (p and e 3 (p. Hence both D e1 e 1 and D e2 e 2 are already normal. This gives: II(v p, v p = ȧ1 a 1 e t II(w p, w p = ȧ2 a 2 e t Now, by equation 7.1, together with dφ(ii(v p, v p = dφ( ȧ1 a 1 e t and II(w p, w p = ȧ2 a 2 e t, we get (ȧ1 dφ e t = ȧ2 e t a 2 Since φ is a local isometry, (ȧ1 g e t, ȧ1 e t a 1 a t But Hence ( (ȧ1 (ȧ1 (ȧ2 = g dφ e t, dφ e t = g e t, ȧ2 e t a 1 a 1 a 2 a 2 (ȧ1 (ȧ1 g e t, ȧ1 e t = a 1 a t (ȧ2 g e t, ȧ2 e t a 2 a 2 p 1 2 t 2 = a (ȧ2 =. a (ȧ1 (ȧ2 = = p 2 2 This gives p 1 2 = p 2 2 and an analogous argument for the other indices implies: Now, implies that p 2 i = 1 3. This contradicts a 1 a 2 p 1 2 = p 2 2 = p 3 2 p 2 i = 1 p i = 1. Hence this is a contradiction to the assumption that (K, g is spatially isotropic. t 2

29 CHAPTER 7. THE EXPANSION OF (K, G The 3-torus as a submanifold. Define the action a : Z 3 R 3 R 3 as a(z, r = r + z. This induces naturally a quotient Lie group T := R 3 /Z 3. Definition torus and 3-torus space-time: The Lie group T := R 3 /Z 3 above is called the 3-torus. The space-time R + T is called the 3-torus space time. Note that for any time t > 0, the rest space M t := {t} T R + T is naturally a Lie subgroup. It has a natural induced metric g t = t 2pi dx 2 i. 7.3 Volume of a the torus, at a fixed time. The next lemma shows that the space-time actually expands proportional to the time passed since the Big Bang. Lemma 44. Define the natural volume form as The volume of M t is then given by Proof. Now, p 1 + p 2 + p 3 = 1, and we get: And the volume form will be: ω t = det g t dx 1 dx 2 dx 3. ω t (M t = ω t = t M t t 2p1 0 0 det g t = det 0 t 2p2 0 = t 2(p1+p2+p3 0 0 t 2p3 det g t = t 2 = t 2 ω t = t 2 dx 1 dx 2 dx 3 = t dx 1 dx 2 dx 3 Hence the total volume at time t ω t = t dx 1 dx 2 dx 3 = t M t T since the volume of the torus with the Euclidean metric is 1. T dx 1 dx 2 dx 3 = t

30 CHAPTER 7. THE EXPANSION OF (K, G Length of one circle in the torus, at a fixed time. We consider now, again for fixed t, the 3 Lie subgroups given by the following injective maps: ι i : S 1 M t = {t} T, ι i (s = (t, δ ij s, i = 1, 2, 3. We denote the respective obtained submanifold as S t,i, i = 1, 2, 3. On S t,i, we get the induced metric g t,i = t 2pi dx i 2. The next lemma shows that even though the universe expands proportional to the time (Lemma 44, the length of one side of the torus in some particular direction can both shrink (Non-flat Kasner or be unchanged (Flat Kasner. Lemma 45. Define the length form (volume form in 1 dimension as ω t,i = det g t,i dx i = t pi dx i. Then the length of S t,i is ω t,i (S t,i = ω t,i = t pi S t,i Proof. The length of the i:th circle in the torus for a fixed time is: ω t,i (S t,i = ω t,i = S t,i t pi dx i = t pi S t,i dx i = t pi S t,i since the length of S 1 with the Euclidean metric is 1.

31 Chapter 8 Appendix Recall that we use the following convention: α, β = 0, 1, 2, 3, i, j = 1, 2, Basic properties of the frame In this section we append the straightforward, but tedious, calculations of the bracket of two frame components and the Levi-Civita connection of a frame component with respect to another frame component. Lemma 46. The only nonzero brackets of two frame components is Proof. We have (define a 0 = 1 [e α, e β ] = e α e β e β e α = 1 a α = 1 a β a α a 2 β x α [e 0, e i ] = ȧi e i [e i, e 0 ] = ȧi e i x α ( 1 a β x β 1 a β x β + 1 a α a β a 2 α x β x α x β ( 1 a α x α But a β x = 0 if α 0 or β = 0. Hence the only non-zero cases are when when the index in the α numerator equals zero and the index in the numerator does not equal. [e 0, e i ] = 1 a 0 a 2 i ȧ i x i = ȧi x i = ȧi e i a 2 i [e i, e 0 ] = ȧi e i Lemma 47. The only non-zero components of the Levi-Civita connection with respect to this frame is D ei e 0 = ȧi e i D ei e i = ȧi e 0 30

32 CHAPTER 8. APPENDIX 31 Proof. 2 D eα e β, e γ = e α e β, e γ + e β e γ, e α > e γ e α, e β e β, [e β, e γ ] + e β, [e γ, e α ] + e γ, [e α, e β ] = e α, [e β, e γ ] + e β, [e γ, e α ] + e γ, [e α, e β ] Case 1: α, β, γ 0 Brackets are zero by Lemma 1, hence D eα e β, e γ = 0 Case 2: One index is not equal to zero, say α 0, the other cases are analogous. [e β, e γ ] = 0, [e γ, e α ]e α e α e β, and [e α, e β ] e α e γ D eα e β, e γ = 0 Case 3: γ = α = β = 0 All brackets are zero D eα e β, e γ = 0 Case 4: 1 index equals zero. The same check as in Case 2 shows that the 2 other indices must be equal. 1. α = 0, β = γ = i 0 2. β = 0, α = γ = i 0 D e0 e i, e i = 1 2( ei, [e i, e 0 ] + e i, [e 0, e i ] = 0 D ei e 0, e i = 1 2( ei, [e 0, e i ] + e i, [e i, e 0 ] = 3. γ = 0, β = α = i 0 = e i, [e 0, e i ] = e i, ȧi e i = ȧi D ei e i, e 0 = 1 2( ei, [e i e 0 ] + e i, [e 0, e i ] = e i, [e 0, e i ] Conclusion: (Only nonzero components. = e i, ȧi e i = ȧi D ei e 0 = ȧi e i D ei e i = ȧi e Ricci and Scalar curvature In this section the straightforward Ricci curvature tensor calculations are appended. Definition 48. The Ricci curvature is given by Ric(e α, e β = where α, β {0, 1, 2, 3}. ɛ γ R(e α, e γ e β, e γ γ=0

33 CHAPTER 8. APPENDIX 32 Lemma 49. Ric(e 0, e 0 = 3 Proof. Case 1: α = 0 Case 2: α = i 0 where we have: ( ( ȧi 2 ( + ȧ t i R(e 0, e α e 0 = D [e0,e α] [D e0, D eα ]e 0 R(e 0, e 0 = 0 R(e 0, e i e 0 = D [e0,e i] [D e0, D ei ]e 0 (ȧi D [e0,e i]e 0 = D ȧ i a e i e 0 = ȧi D ei e 0 = i [D e0, D ei ]e 0 = D e0 (D ei e 0 D ei (D e0 e 0 = D e0 (ȧi e i result. = e 0 (ȧi 2 e i ( 2 (ȧi (ȧi R(e 0, e i e 0 = t e i ( 2 (ȧi (ȧi R(e 0, e i e 0, e i = t Inserting in the formula for Ric(e 0, e 0, noting that ɛ α = e i ȧi D e0 e i = t { 1, if α 0 1, if α = 0 (ȧi e i } gives the desired Lemma 50. For i {1, 2, 3} Ric(e 0, e i = 0 Proof. Analogous to the previous lemma, we check: D [e0,e α]e i and [D e0, D eα ]e i One realizes immediately both would equal zero if α i. Hence let α = i: Hence the lemms proved. Lemma 51. For i {1, 2, 3} where Proof. 2 (ȧi D [e0,e i]e i = D ȧ i a e i e i = ȧi D ei e i = e 0 i [D e0, D ei ]e i = D e0 (D ei e i D ei (D e0 e i = D e0 (e 0 = 0 (ȧi R(e 0, e i e i = e 0 R(e 0, e i e i, e i = 0 (ȧi Ric(e i, e i = t + ȧi θ θ = ȧ i R(e i, e α e i = D [ei,e α]e i [D ei, D eα ]e i

34 CHAPTER 8. APPENDIX 33 D [ei,e α]e i : [D ei, D e0 ]e i : [D ei, D ei ]e i : [D ei, D ej ]e i, ȧ i D [ei,e α]e i = δ α0 D ȧ i a e i = δ α0 D ei e i = δ α0 i 0 j i 2 (ȧi e 0 [D ei, D e0 ]e i = D e0 (D ei (e i D ei (D e0 (e i (ȧi (ȧi = D e0 e 0 = t e 0 [D ei, D ei ]e i = 0 (ȧi [D ei, D ej ]e i = D ej (D ei (e i D ei (D ej (e i = D ej e 0 (ȧi = e j e 0 + ȧi ȧ [ ] j e j = = (t = ȧi ȧ j e j a j a j Hence the terms of the Ricci curvature tensor are computed. We obtain the Ricci tensor: Ric(e i, e i = 2 (ȧi (ȧi ɛ α R(e i, e α e i, e α = + t + ȧi ȧ j = t a j j i α=0 (ȧi + ȧi θ Lemma 52. For i, j {1, 2, 3} and j, Ric(e i, e j = 0 Proof. D [ei,e α]e j : [D ei, D e0 ]e j : [D ei, D ei ]e j : [D ei, D ej ]e j R(e i, e α e j = D [ei,e α]e j [D ei, D eα ]e j D [ei,e α]e j = δ 0α ȧ i D ei e j = 0 [D ei, D e0 ]e j = D e0 (D ei (e j D ei (D e0 (e j = 0 [D ei, D ei ]e j = 0 (ȧj [D ei, D ej ]e j = D ej (D ei (e j D ei (D ej (e j = D ei e 0 a j ȧ i = ȧj D ei (e 0 = a ȧj e i j a j [D ei, D ek ]e j, i k j [D ei, D ek ]e j = D ek (D ei (e j D ei (D ek (e j = 0

35 CHAPTER 8. APPENDIX 34 So far, the only potential non-zero term would be the one containing [D ei, D ej ]e j. would be the term: (ȧj ȧ i R(e i, e j e j, e j = D [ei,e j]e j [D ei, D ej ]e j, e j = e i, e j = 0 a j That Hence all terms are zero and the lemms proved. We end the section by deriving the scalar curvature. Lemma 53. The scalar curvature S is given by S(t = θ(t (ȧi (t (ȧi + 2 t (t (t Proof. We write the proof in coordinates given by the orthonormal frame. S = g αβ Ric αβ = Ric 00 + Ric ii = θ (ȧi (ȧi + 2 t

36 Bibliography [1] B. O Neill, Semi-Riemannian geometry: With applications to general relativity. California: Academic Press. (1983 [2] A. Einstein, Zur allgemeinen Relativitätstheorie. Preuss. Akad. Wiss. Berlin, Sitzber. (1915 [3] A. Einstein, Die Feldgleichungen der Gravitation. Preuss. Akad. Wiss. Berlin, Sitzber. (