On the asymptotics of models for a spatially homogeneous ...kth.diva-portal.org/smash/get/diva2:561554/FULLTEXT01.pdfCosmology, which is a part of general relativity, is a eld in which

The Royal Institute of Technology (KTH)

Department of Mathematics

Bachelor Thesis, SA104X

On the asymptotics of models fora spatially homogeneous and

isotropic spacetime

Authors:Ludvig Hult,[email protected]

Eric Larsson,[email protected]

Supervisor:Hans Ringstrom

May 21, 2012

Abstract

In this thesis, we present an analysis of two cosmological models.

We begin by determining the asymptotics for solutions to Einstein’s equation inan isotropic, homogeneous and flat spacetime under weak assumptions on thebehaviour of matter. Then, we examine how two different matter models affectthe asymptotics. One matter model considers dust and radiation as perfectfluids, and the other involves Vlasov matter. Finally, we consider the differencesand similarities of the resulting asymptotics.

We conclude by discussing some physical consequences of the results.

The thesis also contains supplementary chapters for readers not familiar workingwith asymptotics of solutions to ordinary differential equations.

Contents

1 Introduction 31.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Cosmology . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.2 Ordinary differential equations . . . . . . . . . . . . . . . 7

1.2 Summary of the thesis . . . . . . . . . . . . . . . . . . . . . . . . 71.2.1 Problem statement and motivation . . . . . . . . . . . . . 71.2.2 Results and conclusions . . . . . . . . . . . . . . . . . . . 8

I Preliminaries 9

2 Elementary theory of ordinary differential equations 102.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 Local existence and uniqueness . . . . . . . . . . . . . . . . . . . 112.3 Maximal existence intervals . . . . . . . . . . . . . . . . . . . . . 11

3 Analysis of a model problem 133.1 Problem statement and some observations . . . . . . . . . . . . . 13

3.1.1 Existence and interval of existence . . . . . . . . . . . . . 133.2 Asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2.1 Non-zero first order derivative . . . . . . . . . . . . . . . . 153.2.2 Non-zero higher order derivatives . . . . . . . . . . . . . . 21

II Cosmology 29

4 Einstein’s equation in spatially flat, homogeneous and isotropicspacetime 304.1 Problem statement . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.1.1 The Hubble constant . . . . . . . . . . . . . . . . . . . . . 314.2 Conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . 314.3 Existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . 324.4 Asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5 Dynamics for a model with fluids 385.1 Problem statement . . . . . . . . . . . . . . . . . . . . . . . . . . 385.2 Conservation laws and connections to the general case . . . . . . 395.3 Future asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . 41

1

5.3.1 Asymptotics of a . . . . . . . . . . . . . . . . . . . . . . . 415.3.2 Asymptotics of H . . . . . . . . . . . . . . . . . . . . . . 42

5.4 Past asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6 Dynamics for a model with Vlasov matter 486.1 Problem statement . . . . . . . . . . . . . . . . . . . . . . . . . . 486.2 Conservation laws and connections to the general case . . . . . . 496.3 Future asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . 51

6.3.1 Asymptotics of ρV l and pV l . . . . . . . . . . . . . . . . . 516.3.2 Asymptotics of a . . . . . . . . . . . . . . . . . . . . . . . 526.3.3 Asymptotics of H . . . . . . . . . . . . . . . . . . . . . . 52

6.4 Past asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.4.1 Asymptotics of ρV l and pV l . . . . . . . . . . . . . . . . . 536.4.2 Asymptotics of a and H . . . . . . . . . . . . . . . . . . . 54

7 Conclusions and discussion 617.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

7.1.1 Geometric consequences . . . . . . . . . . . . . . . . . . . 627.2 Matter models . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

7.2.1 Future asymptotics . . . . . . . . . . . . . . . . . . . . . . 637.2.2 Past asymptotics . . . . . . . . . . . . . . . . . . . . . . . 63

Acknowledgments

We would like to thank our supervisor, Hans Ringstrom, for support and inspi-ration.

2

Chapter 1

Introduction

This chapter provides a background for the mathematical and cosmological mod-els in this thesis. It also contains a summary describing the purpose and objec-tives of the thesis as well as the main results and conclusions. The rest of thethesis is divided into two parts.

Part I, Preliminaries, is intended to help readers not familiar with determin-ing asymptotics for solutions to ordinary differential equations. It contains areference for some of the most fundamental theorems in the theory of ordinarydifferential equations. and an analysis of a model problem. The intention isto provide a complete and accessible example that will introduce the methodsused later in the thesis.

Part II concerns the cosmological models and is the core of the thesis. It startsout with some results that hold more or less independently of the matter model.Then follows a derivation of the asymptotics for the solution to Einstein’s equa-tion when the matter is modelled by fluids and Vlasov matter respectively, anda chapter relating the models to each other with discussion and conclusions.

1.1 Background

1.1.1 Cosmology

The following is a very brief summary of the cosmological background for thisthesis. A more complete (and considerably more technical) account can befound in, for instance, chapter 5 of [1].

Cosmology, which is a part of general relativity, is a field in which one studiesthe universe as a whole. The central concept is that of a spacetime, and its ge-ometry. The most well known spacetime model is the Minkowski space occuringin special relativity, which is described in introductory modern physics courses.The Minkowski spacetime happens to be a vector space, but in general relativ-ity the spacetimes under considerations are generally more exotic constructions,not being vector spaces and lacking global coordinate systems. A spacetime,

3

Introduction Background

together with a matter model, forms a model of the universe. Matter modelsare discussed further below. The fundamental equation of general relativity iscalled Einstein’s equation and describes how the matter affects the geometry ofthe spacetime. In turn, the geometry affects the behaviour of the matter.

Einstein’s equation is very general and one typically introduces further assump-tions to simplify the model. Cosmology studies the universe on very large scales,where galaxies are considered small. A natural first approximation is to assumethat the universe is spatially homogeneous (one cannot distinguish between dif-ferent points) and isotropic (one cannot distinguish between different direc-tions). A priori this does not make sense, since a spacetime does not have anatural decomposition into space and time. To be able to speak of homogeneousand isotropic spacetime one needs to introduce a notion of a global time, i. e.a distinguished set of reference frames that agree on time measurements. Withsuch a set of reference frames, the spacetime can be decomposed into a spatialpart and a temporal part. Under these assumptions, there are standard choicesof spacetimes among which we will confine our attention to what is known as aflat spacetime. In this class of spacetimes, the spatial part of the spacetime is,at each point in time, the three-dimensional euclidean space R3.

Apart from the choice of spacetime, one needs to choose a suitable model forthe contents of the universe. Typically one wants to have some matter, with orwithout mass, that can model radiation and galaxies. Under the conditions ofisotropy, homogeneity and flatness one can show that the only two propertiesaffecting the spacetime are the pressure and the energy density of the mat-ter. Therefore, with the specific spacetime and assumptions discussed earlier,specifying the pressure and energy density yields a unique cosmological model.

Einstein’s equation in a spatially isotropic, homogeneous and flatspacetime

Einstein’s equation postulates the interaction between spacetime and matter.Under the assumptions of homogeneity and isotropy, the equation reduces to asystem of ordinary differential equations stated below.

In what follows, a(t) is a scale factor in the metric of spacetime. Loosely speak-ing, it describes the time evolution of the distance between ”nearby” galaxies. Itis often convenient to use the Hubble constant H(t), defined byH(t) = a(t)/a(t).Note that H is not constant in time. The energy density ρ(t) describes howtightly packed the matter is. It is related to the pressure p(t). Later on, we willconsider different examples of such functions ρ(t) and p(t).

The parameter Λ ∈ R is called a cosmological constant and is, in this model,necessary for the expansion of the universe to be consistent with observations.

With the preceding notation, Einstein’s equation takes the following form:

3

(a(t0)

a(t0)

)2

= ρ(t0) + Λ,

2a(t)

a(t)+

(a(t)

a(t)

)2

= Λ− p(t).

4


We shall only be concerned with the case where ρ(t) > 0 and p(t) ≥ 0, cor-responding to positive energy densities and nonnegative pressures. For similarphysical reasons, we will only consider the case where Λ > 0, though we remarkthat this parameter is only important for the future behaviour of the universe,and not the past.

It is necessary to demand that a(t) is never zero for the spacetime metric tobe nondegenerate1. Since a is twice differentiable, the sign of a is preserved.Changing the sign of a does not change the Einstein equation in any way sincethis changes the signs of the derivatives as well, so we can freely decide whatsign to consider. We will let a be positive. The sign of a is also arbitrary;changing the sign of a and changing the direction of time will leave the equationsunchanged. It seems natural to denote increasing t as forwards in time. Withthat assumption, one can calculate the Hubble constant from observed empiricaldata and find it to be positive [2]. Therefore, we will take a to be positive.

Next we will consider two matter models, giving rise to different ρ(t) and p(t).

Perfect fluids

The most common matter model is the fluid model. Different fluids are char-acterized by the quotients between pressure and energy density. The two mostnotable special cases are dust, that gives rise to no pressure at all, and radia-tion, where the pressure is a third of the energy density. This simple relationlets us completely describe the matter by the energy density function ρ(t). Thesubscript m denotes dust terms whilst rad denotes radiation. This is expressedas follows.

prad(t) =1

3ρrad(t),

pm(t) = 0.

We will consider the dust and the radiation case side by side, using a totalpressure and energy density of the form

ρ(t) = ρm(t) + ρrad(t),

p(t) = pm(t) + prad(t).

These functions are assumed to satisfy the following equations:

ρrad + 3a(t)

a(t)(ρrad + prad) = 0, (1.1.1)

ρm + 3a(t)

a(t)(ρm + pm) = 0. (1.1.2)

1This is a technical condition that roughly describes that distances make sense.

5


Vlasov matter

We will also consider a slightly less common model using Vlasov matter. Thismodel differs from the dust and radiation model in that the latter does not keeptrack of the momentum of matter, while the Vlasov matter model describesparticles with positions and momenta. The model was originally used to de-scribe the dynamics of particles interacting electromagnetically in a plasma, bya particle density function. We will not use the classical model described belowdirectly, but it provides some understanding for how relativistic Vlasov mat-ter behaves. While the Lorentz force will be dominating in a electromagneticplasma, we will substitute this with a gravitational force.

The classical description is centered around the density function

Φ: I × R3 × R3 → [0,∞).

This function provides a full description of the system via the (non-negative)density of particles with momentum p at each time t and point r in space. Iis some time interval; I ⊆ R. From the distribution Φ, one can obtain physicalproperties such as the density of mass at some point x ∈ R3 and time t ∈ I.This is defined by ρ : I × R3 → R, such that

ρ(t,x) =

∫R3

Φ(t,x,p) dp.

The mass distribution ρ gives rise to a gravitational potential V such that

∆V = c0ρ

for a constant c0, whose value will not be of any importance in what follows.Note that this is an alternate form of Newton’s gravitational law. Since weassume that the particles interact only via the potential V , and not in anyother way, their motion is described by

x(t) = p(t),

p(t) = c1∇V.

We also postulate that the particles do not collide with each other. This isachieved by demanding that Φ is constant along orbits γ(t) = (t, x(t), p(t)).

In general relativity, the formalism for Vlasov matter is different but the under-lying idea is the same. For a full discussion of this system, see for instance [3]and [4]. Below we will use the simplified version that holds for flat, homogeneousand isotropic spacetimes. The energy density and pressure can be expressed as

ρV l(t) =

∫R3

f(a(t) q)(1 + |q|2)1/2 dq,

pV l(t) =1

3

∫R3

f(a(t) q)|q|2

(1 + |q|2)1/2dq.

We will demand that f : R3 → R is non-negative, smooth and has compactsupport. For the model to be consistent with the assumption of isotropy, wealso demand that f is invariant under rotation, that is

f(p) = g(|p|)

6

Introduction Summary of the thesis

for some function g : R→ R.

We remark that some of these conditions on f may be relaxed, but this resultsonly in slightly more technical proofs.

1.1.2 Ordinary differential equations

The equations considered in this thesis are the ordinary differential equationsstated above. There is a comprehensive theory of ordinary differential equations,the fundamental results of which concern existence and uniqueness for solutionsunder certain regularity conditions. Less known are theorems concerning exis-tence intervals for solutions. It can be shown that a solution will either becomeunbounded within finite time or exist for all time. Such theorems are needed inthis thesis and are therefore stated in chapter 2.

When the exact solutions are not known, or impossible to express explicitly,it may be of interest to see how they behave at the boundary of the existenceinterval. A convenient way to express such asymptotics is by the use of o andO. Definitions of these symbols are stated in chapter 2 for reference.

We will frequently use auxiliary functions as a method of obtaining asymptoticsin this thesis. As an example, imagine we want to determine the behaviour ofsome function φ. By intuition for the equations, or some other reason, we havecome to the conclusion that φ should decline like a power function. Then, wemay examine functions of the form t 7→ tnφ(t). If it can be determined, forinstance, that t2φ(t) → 0 as t → ∞, this tells us that φ decays to zero fasterthan 1/t2. If we on the other hand suspected that φ declined like a exponentialwe would examine functions like t 7→ φ(t)eαt and so on.

Another method that will be of use in this thesis is iteration. Imagine onederives an identity of the form φ(t) =

∫ t0f(s, φ(s)) ds for some function f . If φ

is known to have some form of regularity, this integral may give the conclusionthat φ is even more regular. By iterating the argument, one gets increasinglyspecific information about the function φ.

1.2 Summary of the thesis

1.2.1 Problem statement and motivation

When determining how the universe behaves, one encounters the problem offinding suitable spacetimes and matter models, and examining how they behave.Eventually one wants to perform measurements to determine the validity of thechosen model. Therefore, it is interesting to relate the different models to eachother to see if there is a measurable difference between them. For instance, onemay compare different matter models in a fixed class of spacetimes.

We will in this thesis determine the asymptotics for the solutions to Einstein’sequation in an isotropic, homogeneous and flat spacetime. First we do this in acase with a minimum of assumptions on the behaviour of matter, and then with

7

Introduction Summary of the thesis

two different matter models. The first model consists of perfect fluids and thesecond model consists of Vlasov matter. We will also consider differences andsimilarities in the asymptotics for the two matter models to determine whetheror not the models can be tweaked to give arbitrarily similar behaviour.

1.2.2 Results and conclusions

In our analysis of Einstein’s equation with only mild conditions on the mattermodel, we have shown that the scale factor of the metric goes to zero in finitetime when going backwards in time. This implies that the existence intervals ofsolutions are bounded from below, by a point that we call the Big Bang. Closeto that point in time, the universe is very compressed, and expands arbitrarilyfast. When investigating asymptotics forward in time we have, with some con-ditions on the matter, shown that the scale factor a grows exponentially, whilstthe Hubble constant tends to the constant H∞ =

√Λ/3, corresponding to an

expanding universe.

Our analysis of the two matter models yields somewhat similar results, mosteasily stated in the Hubble constant. Given a specific fluid model, one canconstruct a Vlasov matter model that will agree in the two leading terms whengoing forwards in time. The difference between the two solutions is bounded byan exponential function approaching zero. When going backwards in time theHubble constant will be the same to first order in the two models. The relativedifference in Hubble constant between the models approaches zero like

√t, when

t→ 0.

These results imply that the two models can be adjusted to fit each other par-tially. There are ”missing” terms in the asymptotics that must be introduced ifone wants to achieve greater agreement between the two models. One way to dothis could be to introduce other types of fluids with other quotients p/ρ, or usemassless Vlasov matter where the integrals for energy and pressure take otherforms than for the massive Vlasov matter described in this thesis. It would beinteresting to further investigate these possibilities.

8

Part I

Preliminaries

9

Chapter 2

Elementary theory ofordinary differentialequations

The theorems of this chapter are standard. For the omitted proofs, we refer thereader to chapters 1 and 2 of Hartman’s Ordinary differential equations [5].

2.1 Notation

The symbols o and O are commonly used to express the asymptotics of a func-tion.Definition 2.1.1. The expression

f(t) = g(t) +O(h(t))

will be interpreted as

lim supt→a

∣∣∣∣f(t)− g(t)

h(t)

∣∣∣∣ <∞when h(t) 6= 0, and

f(t)− g(t) = 0 whenever h(t) = 0.

Definition 2.1.2. The expression

f(t) = g(t) + o(h(t))

will be interpreted as

lim supt→a

∣∣∣∣f(t)− g(t)

h(t)

∣∣∣∣ = 0.

when h(t) 6= 0, and

f(t)− g(t) = 0 whenever h(t) = 0.

10

Ordinary differential equations Local existence and uniqueness

2.2 Local existence and uniqueness

The following theorem settles the question of existence of solutions, at leastlocally.Theorem 2.2.1. Let f : Rn+1 → Rn be continuous, and assume that the partialderivatives ∂f

∂xi for i = 1, 2, . . . , n exist and are continuous. Consider the problemdxdt (t) = f(t, x(t)),

x(t0) = x0

where t0 ∈ R and x0 ∈ Rn. There is an ε > 0 and a unique continuouslydifferentiable function x : (t0 − ε, t0 + ε)→ Rn satisfying the above problem.Theorem 2.2.2. Let f : Rn+1 → Rn be continuous, and assume that the partialderivatives ∂f


x(t0) = x0

where t0 ∈ R and x0 ∈ Rn.Suppose that t0 ∈ (t−, t+) and that x, y : (t−, t+)→ Rn are continuously differ-entiable solutions to the above problem. Then x(t) = y(t) for all t ∈ (t−, t+).

2.3 Maximal existence intervals

The two theorems above allow us to define a maximal existence interval forsolutions to an initial value problem.Definition 2.3.1. The maximal existence interval to the problem

dxdt (t) = f(t, x(t)),

x(t0) = x0

is the union of all intervals I ⊂ R containing t0 such that there is a functionx : I → R which is a solution to the problem.Proposition 2.3.1. The maximal existence interval J of solutions to the initialvalue problem discussed in this chapter is nonempty and unique. Further, thereis a solution, defined on J , to the problem.

Proof. Let Iαα∈A be the collection of existence intervals of the problem, in-dexed by some set A, and let xα be the corresponding solutions. By defini-tion, J =

⋃α∈A Iα. Obviously, J is connected, and hence an interval. Define

x : J → R by letting x(t) = xα(t) where t ∈ Iα. This is well-defined, since foreach α, β ∈ A, it holds that xα(t) = xβ(t) whenever t ∈ Iα ∩ Iβ . Close to eacht ∈ J , the function x (being identical to some xα) is a solution to the initialvalue problem. Hence J is an existence interval of the problem. Since J is theunion of all existence intervals, it is also maximal.

11

Ordinary differential equations Maximal existence intervals

Theorem 2.3.2. Let f : Rn+1 → Rn be continuous, and assume that the partialderivatives ∂f


x(t0) = x0

where t0 ∈ R and x0 ∈ Rn. Let (t−, t+) be the maximal existence interval ofsolutions to the problem. If t+ <∞, then x is unbounded on [t0, t+). Similarly,if t− > −∞, then x is unbounded on (t−, t0].Corollary 2.3.3. Let I ⊂ R and I 6= R. If x : I → R is a solution to theproblem, bounded on I, then I is not the maximal existence interval.

Proof. This is simply the contrapositive of Theorem 2.3.2.

12

Chapter 3

Analysis of a modelproblem

In this chapter, we will consider the problem of determining the asymptotics forsolutions to an ordinary differential equation. This will act as a model problemfor the cosmological problems considered in later chapters. No part of the thesislogically requires this chapter.

3.1 Problem statement and some observations

Consider a smooth function f : R −→ R. Suppose that there is a nonemptyinterval I = (x−, x+) ⊆ R such that

f(x−) = f(x+) = 0,

∀x ∈ I f(x) > 0.(3.1.1)

We will discuss the initial value problem

dφ

dt(t) = f(φ(t)),

φ(0) = x0,(3.1.2)

where x0 ∈ I.

3.1.1 Existence and interval of existence

Remark 3.1.1. By a direct application of Theorems 2.2.1 and 2.2.2, we see thatthe initial value problem has a unique solution in a neighborhood of 0.Lemma 3.1.2. Any solution to the initial value problem, defined on an openand connected set, is bounded from above by x+ and from below by x−.

13

Analysis of a model problem Asymptotics

Proof. Assume, to get a contradiction, that there is a solution φu defined onsome open connected set J ⊆ R such that φu(t) /∈ I for some t ∈ J . Bydefinition, solutions to the initial value problem are differentiable and hencecontinuous. Since φu(0) = x0, it holds by the intermediate value property thatthere is some t ∈ J such that φu(t) = x+ or φu(t) = x−. We prove the result inthe first case; the second is completely analogous. Since f is positive on I, sothat φu is increasing, such a t is positive.

Let A = t ≥ 0: φu(t) = x+ By continuity of φu, the set A is closed. As seenabove, A is nonempty. Let t∗ = inf A. Obviously 0 /∈ A, so t∗ 6= 0. Considerthe initial value problem

dφ

dt(t) = f(φ(t)),

φ(t∗) = x+.(3.1.3)

Of course, φu is a solution to this problem, as is the constant function φc(t) =x+. By Theorem 2.2.2, these two solutions agree on a neighborhood of t∗. This,however, contradicts the fact that t∗ is the infimum of A. Hence φu takes itsvalues in I.

Proposition 3.1.3. The initial value problem has a solution φ : R −→ I definedon all of R.

Proof. We have now shown that all solutions to the problem, defined on openconnected sets, are bounded. By the corollary to Theorem 2.3.2, this shows thatthe maximal existence interval for solutions to the problem is R, and the proofis complete.

3.2 Asymptotics

Lemma 3.2.1. Let φ : J → R be a solution to the initial value problem consid-ered. Then φ is strictly increasing.

Proof. Since φ(t) ∈ I for all t ∈ J and f(x) > 0 for all x ∈ I, it holds that

φ(t) = f(φ(t)) > 0

and φ is strictly increasing.

Proposition 3.2.2. Let φ be a solution to the initial value problem considered.Then

limt→∞

φ(t) = x+

andlim

t→−∞φ(t) = x−.

Proof. We begin by proving the first statement.

By Lemma 3.1.2, φ takes its values in the interval I. By Lemma 3.2.1, φ isstrictly increasing. Hence the limit

χ = limt→∞

φ(t)

14


exists, and χ ≤ x+. If χ = x+ we are done, so suppose not.

Since φ is increasing, it holds that φ(t) ∈ [x0, χ] for all t > 0. The function f ,being continuous, has some minimum value m on [x0, χ]. Since f(x) > 0 for allx ∈ [x0, χ], it holds that m > 0. Hence, for t > 0,

φ(t) = f(φ(t)) ≥ m.

This shows that φ is unbounded, which contradicts Lemma 3.1.2. Hence χ = x+.

By a similar argument, we see that limt→−∞ φ(t) = x−.

It is now natural to ask in what manner φ converges to its limits. We will seethat this depends on the order of the first nonvanishing derivative of f in x+

and x−.Remark 3.2.3. We will only consider the asymptotics of solutions in the limitt → ∞, since the proofs of the corresponding results in the limit t → −∞ arevery similar.

3.2.1 Non-zero first order derivative

Suppose that f ′(x+) = −α 6= 0, and let φ be a solution to the correspondinginitial value problem. We will keep f and φ fixed throughout this section. Theauxiliary functions δ and Ψ defined for t ≥ 0 by

δ(t) = x+ − φ(t),

Ψ(t) = eαtδ(t)

will be useful. We will use the notation

Ψ0 = Ψ(0)

andΨ∞ = lim

t→∞Ψ(t)

once we have shown that the latter limit exists.Remark 3.2.4. The reason for considering Ψ is that we suspect that δ(t) ≈ Ae−αtfor some A. One way to make this precise is to state it as Ψ being convergent.We will later prove that this is indeed the case. The function δ is introducedpurely for notational convenience.Proposition 3.2.5. It holds that f ′(x+) < 0, so that

α > 0.

Proof. The function f is smooth, which implies that

f ′(x+) = limh→0+

f(x+ − h)− f(x+)

−h.

Since f(x) > 0 for x ∈ I and f(x+) = 0, this limit is nonpositive:

limh→0

f(x+)− f(x+ − h)

h= limh→0−f(x+ − h)

h≤ 0.

We have assumed that f ′(x+) 6= 0, and hence f ′(x+) < 0.

15


The following lemma is the crucial one in obtaining the results of this section.Using this lemma together with a bound for δ typically gives rise to a betterbound for δ, and the argument may then be repeated to obtain an even betterbound. This will be illustrated in the proofs of the remaining propositions inthis section.Lemma 3.2.6. There is a bounded, continuous function ρ : [0,∞) → R suchthat

ln

(Ψ(t)

Ψ0

)=

∫ t

0

δ(s)ρ(s) ds.

Further, limt→∞ ρ(t) = ρ∞ for some real number ρ∞.

Proof. Note thatδ(t) = −f(x+ − δ(t)).

By series expanding f , we see that

−f(x+ − δ(t)) = −f(x+) + f ′(x+)δ(t)− f ′′(θ(δ(t)))

2δ2(t)

where θ is a function such that limx→0 θ(x) = x+. Define

ρ(t) = −f′′(θ(δ(t)))

2

for all t ≥ 0. With this notation

δ(t) = −δ(t)α+ δ(t)2ρ(t).

Since δ(t) → 0 as t → ∞, by Proposition 3.2.2, and f ′′ is continuous, we seethat ρ is convergent as t → ∞, and hence bounded on some interval (t∗,∞).Since δ and δ are continuous, and δ(t) 6= 0 for all t, ρ is continuous. Hence ρis bounded on [0, t∗], which combined with the boundedness on (t∗,∞) showsthat ρ is bounded.

Recall thatΨ(t) = eαtδ(t).

It holds that

Ψ(t) = αeαtδ(t) + eαtδ(t) = αeαtδ(t)− eαtδ(t)α+ eαtδ(t)2ρ(t) =

= Ψ(t) (α− α+ δ(t)ρ(t)) = Ψ(t)δ(t)ρ(t).

Hence

ln

(Ψ(t)

Ψ0

)=

∫ t

0

Ψ(s)

Ψ(s)ds =

∫ t

0

δ(s)ρ(s) ds (3.2.1)

and we have proved the lemma.

Proposition 3.2.7. For each β > 0, there is a real number Dβ such that

δ(t) ≤ Dβet(β−α).

16


Proof. Choose some β > 0, and let R be a bound of |ρ|. Proposition 3.2.2guarantees that δ(t)→ 0, so there is a constant Aβ such that δ(t) < β/R for allt > Aβ . Hence∫ t

0

δ(s)ρ(s) ds ≤ R∫ t

0

δ(s) ds = R

∫ Aβ

0

δ(s) ds+R

∫ t

Aβ

δ(s) ds

≤ R∫ Aβ

0

δ(s) ds+

∫ t

Aβ

β ds = Cβ + βt

where Cβ = R∫ Aβ

0δ(s) ds−Aββ. Using this and Lemma 3.2.6, we see that

Ψ(t) ≤ Ψ0 exp(Cβ + βt).

Let Dβ = Ψ0eCβ . By the definition of Ψ

δ(t) ≤ Dβet(β−α),

and the proof is complete.

We have shown that δ is bounded by exponentials arbitrarily close to e−αt. Inparticular, if we choose β < α, the exponentials will be decaying. With thisresult, we have rough control over the solution φ, and we wish to refine this.This will be accomplished by returning to Lemma 3.2.6, and using the boundobtained in Proposition 3.2.7. This is an example of the iterative nature of themethod we use to determine the asymptotics of φ.Lemma 3.2.8. Let ρ be as in Lemma 3.2.6. Then the integral

g(t) =

∫ ∞t

δ(s)ρ(s) ds

is convergent for all t ≥ 0, and there is a constant C such that

|g(t)| ≤ Ce−αt.

Further, Ψ is convergent to some Ψ∞ and

Ψ(t) = Ψ∞e−g(t).

Proof. The result can be obtained by combining Lemma 3.2.6 and Proposition3.2.7. As before, let R be a bound for ρ. Fix an arbitrary positive β < α. Wesee that

ln

(Ψ(t)

Ψ0

)=

∫ t

0

δ(s)ρ(s) ds

≤∫ t

0

RDβet(β−α) ds

=RDβ

α− β

(1− et(β−α)

)= C1 + C2e

t(β−α),

17


This inequality gives a bound

ln

(Ψ(t)

Ψ0

)≤ C1

for Ψ, so thatΨ(t) ≤ C3

for a constant C3. Thenδ(t) ≤ C3e

−αt. (3.2.2)

To obtain a lower bound for δ, we perform the following trick: We know fromthe bound (3.2.2) for δ that the integral

∫∞0δ(s)ρ(s) ds is convergent. Denote

it by M =∫∞

0δ(s)ρ(s) ds. Then the equation in the conclusion of Lemma 3.2.6

may be written as

ln

(Ψ(t)

Ψ0

)=

∫ t

0

δ(s)ρ(s) ds = M −∫ ∞t

δ(s)ρ(s) ds. (3.2.3)

For notational convenience, we introduce the function

g(t) =

∫ ∞t

δ(s)ρ(s) ds.

Using the bound (3.2.2), we see that

|g(t)| =∣∣∣∣∫ ∞t

δ(s)ρ(s) ds

∣∣∣∣ ≤ C4

∫ ∞t

e−αs ds =C4

αe−αt, (3.2.4)

for some constant C4, proving the first part of the lemma.

Expressing Ψ in terms of g, by using equation (3.2.3), we have

Ψ(t) = Ψ0eMe−g(t).

The bound for g implies in particular that Ψ is convergent, with

limt→∞

Ψ(t) = Ψ0eM .

Recall that we use the notation Ψ∞ to denote this limit. Hence

Ψ(t) = Ψ∞e−g(t),

proving the lemma.

Proposition 3.2.9. It holds that

δ(t) = Ψ∞e−αt +O(e−2αt).

Proof. Let g be as in Lemma 3.2.8, so that

Ψ(t) = Ψ∞e−g(t).

18


By a series expansion of e−g(t), considering some technicalities similar to thosein the first part of the proof of Lemma 3.2.6, we obtain

Ψ(t) = Ψ∞ (1− g(t)ρ(g(t))) ,

where ρ is bounded. Rewriting this in terms of δ, we obtain∣∣δ(t)−Ψ∞e−αt∣∣ =

∣∣−Ψ∞e−αtg(t)ρ(g(t))

∣∣The boundedness of ρ and the bound for g from Lemma 3.2.8 give an estimate∣∣δ(t)−Ψ∞e

−αt∣∣ ≤ ∣∣Ce−2αt∣∣ ,

where C is some constant, so that∣∣∣∣δ(t)−Ψ∞e−αt

e−2αt

∣∣∣∣ ≤ |C| .Hence

δ(t) = Ψ∞e−αt +O(e−2αt),

which proves the lemma.

The previous proposition specifies the asymptotics of φ rather well, but it maybe strengthened further.Theorem 3.2.10. It holds that

δ(t) = Ψ∞e−αt +

ρ∞Ψ2∞

αe−2αt + o

(e−2αt

).

Proof. Recall that the conclusion of Lemma 3.2.6 may be written as

ln

(Ψ(t)

Ψ0

)= M −

∫ ∞t

δ(s)ρ(s) ds,

where

M =

∫ ∞0

δ(s)ρ(s) ds.

Rewriting this slightly, using the definition Ψ(t) = eαtδ(t) of Ψ, and adding theintegral

∫∞te−αsρ∞Ψ∞ ds to both sides, we obtain

ln

(Ψ(t)

Ψ0

)−M +

∫ ∞t

e−αsρ∞Ψ∞ ds =

∫ ∞t

e−αs [ρ∞Ψ∞ − ρ(s)Ψ(s)] ds.

We will now show that this expression is o (e−αt). We see that∣∣∣∣ln(Ψ(t)

Ψ0

)− M +

∫ ∞t

e−αsρ∞Ψ∞ ds

∣∣∣∣ =

∣∣∣∣∫ ∞t

e−αs [ρ∞Ψ∞ − ρ(s)Ψ(s)] ds

∣∣∣∣=

∣∣∣∣∫ ∞t

e−αs [ρ∞Ψ∞ −Ψ∞ρ(s) + Ψ∞ρ(s)− ρ(s)Ψ(s)] ds

∣∣∣∣=

∣∣∣∣∫ ∞t

e−αs [Ψ∞(ρ∞ − ρ(s)) + ρ(s)(Ψ∞ −Ψ(s))] ds

∣∣∣∣≤∫ ∞t

e−αs [Ψ∞|ρ∞ − ρ(s)|+ |ρ(s)||Ψ∞ −Ψ(s)|] ds

19


Proposition 3.2.9 tells us that there is a constant C such that

|Ψ(t)−Ψ∞| ≤ Ce−αt

for sufficiently large t. Fix some ε > 0. Since ρ is convergent to ρ∞,

|ρ(t)− ρ∞| < ε

for sufficiently large t. For such large t,∫ ∞t

e−αs [Ψ∞|ρ∞ − ρ(s)|+ |ρ(s)||Ψ∞ −Ψ(s)|] ds

≤∫ ∞t

e−αs[Ψ∞ε+ |ρ(s)|Ce−αs

]ds

≤∫ ∞t

e−αs[Ψ∞ε+ |ρ(s)− ρ∞|Ce−αs + |ρ∞|Ce−αs

]ds

≤∫ ∞t

e−αs[ε(Ψ∞ + Ce−αs

)+ |ρ∞|Ce−αs

]ds

=ε

α

(Ψ∞e

−αt +C

2e−2αt

)+ |ρ∞|

C

2αe−2αt.

This long calculation shows that∣∣∣∣∣∣ln(

Ψ(t)Ψ0

)−M +

∫∞te−αsρ∞Ψ∞ ds

e−αt

∣∣∣∣∣∣ ≤ εΨ∞α

(1 +

C

2e−αt

)+ |ρ∞|

C

2αe−αt

for all t > t1(ε) and ε > 0. Accordingly,

lim supt→∞

∣∣∣∣∣∣ln(

Ψ(t)Ψ0

)−M +

∫∞te−αsρ∞Ψ∞ ds

e−αt

∣∣∣∣∣∣ = 0

so that

ln

(Ψ(t)

Ψ0

)= M −

∫ ∞t

e−αsρ∞Ψ∞ ds+ o(e−αt

)= M − ρ∞Ψ∞

αe−αt + o

(e−αt

).

Hence

Ψ(t) = Ψ0eM exp

[ρ∞Ψ∞α

e−αt]

exp[o(e−αt

)].

Series expanding this to second order, and using the fact that Ψ∞ = Ψ0eM ,

yields

Ψ(t) = Ψ∞

(1 +

ρ∞Ψ∞α

e−αt +O(e−2αt

))(1 + o

(e−αt

)+O

((o(e−αt

))2)),

so that

Ψ(t) = Ψ∞

(1 +

ρ∞Ψ∞α

e−αt + o(e−αt

))

20


Using the definition of Ψ(t), we arrive at the desired conclusion

δ(t) = Ψ∞e−αt +

ρ∞Ψ2∞

αe−2αt + o

(e−2αt

).

We have now specified the asymptotics of φ using the constant Ψ∞. It turnsout that the solutions φ to the initial value problem are uniquely determinedby the value of this parameter Ψ∞, in the following sense.Proposition 3.2.11. Let φ1 = x+ − Ψ∞,1e

−αt + O(e−2αt) and φ2 = x+ −Ψ∞,2e

−αt + O(e−2αt) be distinct solutions to the initial value problem. ThenΨ∞,1 6= Ψ∞,2.

Proof. Since φ1 and φ2 are solutions to the initial value problem, we know thatφ1(0) ∈ I and φ2(0) ∈ I. Obviously φ1(0) 6= φ2(0), since otherwise φ1 and φ2

would be identical. Let us assume without loss of generality that

φ1(0) < φ2(0).

The function φ1 is strictly increasing, converging to x+ > φ2(0) so by the in-termediate value theorem there is a t0 ∈ R+ such that φ1(t0) = φ2(0). ByTheorem 2.2.2, regarding uniqueness of solutions to ordinary differential equa-tions, it holds that

φ1(t0 + t) = φ2(t)

for all t ∈ R. This means that

Ψ∞,2 = limt→∞

(x+ − φ2(t))eαt = limt→∞

(x+ − φ1(t0 + t))eαt =

= limt→∞

(x+ − φ1(t))eα(t−t0) = Ψ∞,1e−αt0 .

Since by assumption we have α 6= 0 and t0 6= 0, the conclusion follows.

Remark 3.2.12. The conclusion of Proposition 3.2.11 may be interpreted ineither of the following equivalent ways:

• The mapping φ 7→ limt→∞(x+ − φ(t))eαt from the solution space of theinitial value problem to the real numbers is injective.

• Every solution is uniquley identified by the second coefficient in the expo-nential expansion.

3.2.2 Non-zero higher order derivatives

In the previous section, we determined the future asymptotics of solutions tothe initial value problem when the derivative f ′(x+) is nonzero. In this section,we will assume that f (k)(x+) = 0 for all k < n and f (n)(x+) 6= 0, for someinteger n ≥ 2.

Introduce the real number α such that f (n)(x+) = α n!n−1 (−1)n, and let φ be a

solution to the initial value problem corresponding to f . We will keep f and φfixed throughout this section.

21


Definition 3.2.1. For convenience of notation, introduce the constant

λ =1

n− 1.

Definition 3.2.2 (Auxiliary functions). Let the functions η, δ and Ψ be definedby

η(t) = x+ − φ(t),

δ(t) = ηn−1(t),

Ψ(t) =1

δ(t)− αt.

These functions will be fixed throughout the section.Lemma 3.2.13. It holds that (−1)nf (n)(x+) > 0, so that

α > 0.

Proof. By a series expansion of f

f(x+ − ζ) = f(x+)− f ′(x+)ζ + · · ·

+f (n)(x+)

n!(−ζ)n +O

(ζn+1

).

By the condition that the first n− 1 derivatives of f at x+ are zero, the aboveexpression can be written

f(x+ − ζ) =f (n)(x+)

n!(−ζ)n +O

(ζn+1

).

For sufficiently small ζ > 0, the right hand side has the same sign as f(n)(x+)n! (−ζ)n,

and the left hand side is positive. Hence

(−1)nf (n)(x+) > 0.

Lemma 3.2.14. There is a bounded, continuous function ρ : [0,∞) → R suchthat

δ(t) = −αδ2(t)− ρ(t)δ2+λ(t)

for t ≥ 0.

Proof. Recall that η(t) = x+ − φ(t), so that

η(t) = −φ(t) = −f(φ(t)) = −f(x+ − η(t)).

By a series expansion of f ,

f(x+ − η(t)) = f(x+)− f ′(x+)η(t) + · · ·

+f (n)(x+)

n!(−η(t))n +

f (n+1)(θ(−η(t)))

(n+ 1)!(−η(t))n+1

22


where θ is a function such that limx→0 θ(x) = x+. By the condition that thefirst n − 1 derivatives of f at x+ are zero, and the definition of α, the aboveexpression can be written

f(x+ − η(t)) =α

n− 1η(t)n +

f (n+1)(θ(−η(t)))

(n+ 1)!(−η(t))n+1.

Define

ρ(t) = (n− 1)(−1)n+1 f(n+1)(θ(−η(t)))

(n+ 1)!

for all t ≥ 0. With this notation

η(t) = − α

n− 1ηn(t)− ρ(t)

n− 1ηn+1(t).

Recall that η(t) → 0 as t → ∞ by Proposition 3.2.2, that limx→0 θ(x) = x+

and that f (n+1) is continuous. Hence ρ is bounded on some interval (t∗,∞).Since η and η are continuous, and η(t) 6= 0 for all t, ρ is continuous. Hence ρis bounded on [0, t∗], which combined with the boundedness on (t∗,∞) showsthat ρ is bounded.

Recall that δ(t) = ηn−1(t) and λ = 1/(n− 1). Hence

δ(t) = (n− 1)ηn−2(t)η(t) = −αδ2(t)− ρ(t)δ2+ 1n−1 (t) = −αδ2(t)− ρ(t)δ2+λ(t).

This proves the lemma.

Lemma 3.2.15. There is a constant A > 0 such that

δ(t) ≥ 1

At

for all sufficiently large t.

Proof. Recall that we have introduced

Ψ(t) =1

δ(t)− αt.

Let Ψ0 = Ψ(0). By Lemma 3.2.14

Ψ(t) = − δ(t)

δ2(t)− α = ρ(t)δλ(t).

Equivalently

Ψ(t)−Ψ0 =

∫ t

0

ρ(s)δλ(s) ds. (3.2.5)

Let R be a bound for |ρ|. Since δ(t)→ 0 when t→∞ we can utilize that therefor each β > 0 is a Cβ > 0 such that

Ψ(t)−Ψ0 =

∫ t

0

ρ(s)δλ(s) ds ≤ R∫ t

0

δλ(s) ds ≤ CβR+Rβt.

23


Fix such a β. Using the definition of Ψ, we can rewrite the above inequality as

δ(t) ≥ 1

Ψ0 +RCβ + (Rβ + α)t.

Let A = 2(α+Rβ). We see that if t > (RCβ + Ψ0)/(α+Rβ), then

δ(t) ≥ 1

At,

and the proof is complete.

Lemma 3.2.16. Suppose that A is a real number such that

δ(t) ≥ 1

At

for sufficiently large t. Let γ > 0 be a real number such that γ ≤ λ and γ < α/A.Then it holds that

δ(t) =1

αt+O

(1

t1+γ

)for all sufficiently large t.

Proof. We introduce the auxiliary functions

Ω(t) = δ(t)t− 1

α

and

E(t) =Ω(t)2

2.

By differentiating and using that tδ(t) = Ω(t) + 1/α we obtain

Ω(t) = δ(t) + t(−αδ(t)2 − ρ(t)δ(t)2+λ

)= δ(t) + Ω(t)

(−αδ(t)− ρ(t)δ1+λ

)− δ(t)− ρ(t)δ(t)1+λ

α

= Ω(t)(−αδ(t)− ρ(t)δ(t)1+λ

)− ρ(t)δ(t)1+λ

α

andE(t) =Ω(t)Ω(t)

=Ω(t)2δ(t)(−α− ρ(t)δλ(t)

)− Ω(t)

ρ(t)δ(t)1+λ

α

=− E(t)δ(t)2[α+ ρ(t)δλ(t)

]− δ(t)1+λE(t)1/2

√2ρ(t)

α.

We want to show that E is in some sense small, so we want to bound E fromabove. Since α > 0, ρ is bounded and δ(t)→ 0 as t→∞, there is for each ε > 0a constant Cε > 2α− ε such that Cε ≤ 2

[α+ ρ(t)δλ(t)

]for large t. We will fix

a particular ε later. Since ρ is bounded, so is |√

2ρ(t)/α|. Hence

E(t) ≤ −δ(t)[CεE(t)−Dδλ(t)E(t)1/2

](3.2.6)

24


for some constant D and all sufficiently large t. Note that we may choose Cεand D positive. This inequality implies that E is bounded. To see this, notethat Cε−Dδλ(t) > 0 for large t, and hence −δ(t)

[CεE(t)−Dδλ(t)E(t)1/2

]< 0

whenever E(t) > 1 and t is large enough. This means that E is bounded forlarge t, and by continuity it is also bounded on every compact.

That E is bounded implies that Ω, and hence also tδ(t), is bounded. Let a boundfor tδ(t) be F 1/(1+λ). Using this bound allows us to conclude from (3.2.6) that

E(t) ≤ −CεE(t)δ(t) +DFE(t)1/2

t1+λ.

Using the fact that δ(t) ≥ 1/(At), we see that

E(t) ≤ −CεE(t)

At+DF

E(t)1/2

t1+λ

for large t.

Introduce the functionE(t) = t2γE(t)

Using the above bound for E we obtain, for large t,

dE

dt(t) = 2γ

t2γE(t)

t+ t2γE(t)

≤ 2γE(t)

t− Cε

t2γE(t)

At+DFtγ

(t2γE(t))1/2

t1+λ

= − E(t)

t

(CεA− 2γ

)+DF

E(t)1/2

t1+λ−γ (3.2.7)

By hypothesis, γ ≤ λ. Recall that we may choose Cε > 2α − ε, for each ε > 0Choosing a sufficiently small value of ε, we can ensure that 2γ ≤ Cε/A. Thisimplies that E is bounded, by an argument analogous to the one proving thatE is bounded. Let a bound for E be 1

2G2. We have now shown that

E(t) ≤ G2

2t2γ

for all sufficiently large t. Hence∣∣∣∣tδ(t)− 1

α

∣∣∣∣ = |Ω(t)| ≤ G

tγ,

proving the lemma.

Proposition 3.2.17. There is a γ > 0 such that

δ(t) =1

αt+O

(1

t1+γ

).

Proof. Lemma 3.2.15 guarantees that there is a constant A fulfilling the condi-tion of Lemma 3.2.16, which then gives the desired conclusion.

25


Lemma 3.2.18. For every a such that 0 ≤ a < 1, there is a t∗ such that

δ(t)[α+ ρ(t)δλ(t)

]≥ a

t∀t > t∗.

Proof. Since ρ(t)δλ(t)→ 0, there is for every ε1 > 0 a t1 such that

δ(t) [α+ ρ(t)δ(t)] ≥ δ(t) (α− ε1) ∀t > t1.

Using Proposition 3.2.17 we see that there are constants G and γ > 0 such that

δ(t) (α− ε1) ≥ (α− ε1)

(1

αt− G

t1+γ

)=α− ε1αt

(1− Gα

tγ

)for large t.

Further note thatGα

tγ→ 0 as t→∞

since γ > 0, which means that for every ε2 > 0 there is a t2 such that

1− Gα

tγ≥ (1− ε2) ∀t > t2.

Take t∗ = maxt1, t2 and note that the observations above lead to


]≥ α− ε1

αt(1− ε2) ∀t > t∗.

Given any 0 ≤ a < 1, we can choose ε1 and ε2 such that

1

α(α− ε1)(1− ε2) ≥ a.

With this choice it holds that


]≥ a

t∀t > t∗,

proving the lemma.

Theorem 3.2.19. Let n > 1 be an integer. Let f (k)(x+) = 0 for each 0 < k < nand f (n)(x+) = α n!

n−1 (−1)n 6= 0. Then the solution φ is of the form

φ(t) =

x+ −

1

αt+O

(t−(1+γ)

)for all 0 < γ < 1 if n = 2,

x+ −1

(αt)1/(n−1)+O

(t−2/(n−1)

)if n > 2.

26


Proof. The theorem follows from Lemma 3.2.16 if we can choose the parameterA arbitrarily close to α.

Lemma 3.2.18 shows that for each 0 < a < 1 and all sufficiently large t

1

tδ(t)≤ α+ ρ(t)δλ(t)

a.

Since a may be chosen arbitrarily close to 1, ρ is bounded and δ(t)→ 0, we seethat for each ε it holds for sufficiently large t that

1

tδ(t)≤ α− ε.

Hence

δ(t) ≥ 1

At

for each A < α, and Lemma 3.2.16 shows that

δ(t) =1

αt+O

(1

t1+γ

)for all γ such that γ ≤ λ and γ < 1.

We consider two separate cases.

n=2

In this case, λ = 1. Hence the functions δ and η are identical and

δ(t) =1

αt+O

(1

t1+γ

)for all γ < 1. Recall that x+ − η(t) = φ(t), so that

φ(t) = x+ −1

αt+O

(1

t1+γ

), for all 0 < γ < 1.

This proves the theorem in the case n = 2.

n>2

In this case, 0 < λ < 1. Hence

δ(t) =1

αt+O

(1

t1+γ

)for all γ ≤ λ, and in particular for γ = λ.

In terms of the function η, this result can be written

η(t) =

(1

αt+O

(1

t1+λ

))λ=

1

(αt)λ

(1 +O

(1

tλ

))λ=

1

(αt)λ

(1 +O

(1

tλ

)).

27


Hence

η(t) =1

(αt)λ+O

(1

t2λ

).

Recall that λ = 1/(n− 1) and that x+ − η(t) = φ(t). In this notation

φ(t) = x+ −1

(αt)1/(n−1)+O

(t−2/(n−1)

).

This proves the theorem in the case n > 2.

28

Part II

Cosmology

29

Chapter 4

Einstein’s equation inspatially flat, homogeneousand isotropic spacetime

4.1 Problem statement

In this chapter, we will derive the main results by analysing Einstein’s equationin a spatially flat, homogeneous and isotropic spacetime. The physical settingand the motivation is described in Section 1.1. Mathematically we have thefollowing set of equations:

3

(a(t0)

a(t0)

)2

= ρ(t0) + Λ, (4.1.1)

2a(t)

a(t)+

(a(t)

a(t)

)2

= Λ− p(t), (4.1.2)

p(t) ≥ 0, ρ(t), Λ, a(t), a(t0) > 0.

Remark 4.1.1. One may also study the case when ρ is allowed to be zero. How-ever, we will consider the case when this is not so.

The purpose of the chapter is to derive some results that are not dependenton the particular choice of matter model. Nevertheless, we need some weakassumptions on ρ and p. These are formalized in the following definition.Definition 4.1.1. A matter model (ρ, p) is well-behaved if it satisfies the fol-lowing conditions.

• There are functions ρ and p such that ρ(t) = ρ(a(t)) and p(t) = p(a(t))for all t in the common domain of a, ρ and p.

• The function ρ is unbounded on an interval J if and only if a takes valuesarbitrarily close to 0 on J .

30

Einstein’s equation Conservation laws

• Let J be an interval. Then the following are equivalent.

– a/a is bounded on J .

– ρ is bounded on J .

– p is bounded on J .

• It holds that ρ(t) = −3H(t)(ρ(t) − p(t)) for all t in the common domainof H, ρ and p.

Remark 4.1.2. The condition

ρ(t) = −3H(t)(ρ(t) + p(t)) (4.1.3)

can be interpreted as a physical continuity equation. Consider an expandingcubic box filled with energy. As it expands, it performs mechanical work onits surroundings, corresponding to to the pressure it is subjected to. This workequals a decrease in the internal energy of the box. Expressing this in ourvariables yields the equation above.

4.1.1 The Hubble constant

Introduction of the Hubble constant will simplify the expressions a bit. We

define it as H(t) = a(t)a(t) . Please note that, contrary to what the name implies,

it is not constant. Introducing it into the equations above gives rise to thefollowing set of equations:

3H2(t0) = ρ(t0) + Λ, (4.1.4)

2H(t) + 3H(t)2 = Λ− p(t). (4.1.5)

Note that information about a and H is almost equivalent. Any solution a tothe former system lets us construct H from the definition, since it has to holdthat a is everywhere nonzero for a to be a solution to the system. Conversely,if a solution H to the latter system is given, we can construct a solution a tothe former system via the equation

a(t) = a(t0) exp

∫ t

t0

H(s)ds.

4.2 Conservation laws

Let the maximal existence interval for solutions to (4.1.1)-(4.1.2) be I ⊆ R. Atheorem about the form of I is found in the next section.Lemma 4.2.1. If

ρ(t) = −3H(t)(ρ(t) + p(t)),

then the initial condition (4.1.4) is conserved for all time:

3 (H(t))2

= ρ(t) + Λ ∀t ∈ I.

31

Einstein’s equation Existence of solutions

Proof. Consider the difference function g defined by

g(t) = 3H(t)2 − ρ(t)− Λ.

Recall that the matter model being well-behaved implies that ρ(t) = −3H(ρ+p).Differentiate g with respect to time, and use this hypothesis to obtain

g(t) = 6H(t)H(t)− ρ(t)

= 3H(t)(−3H(t)2 + Λ− p(t)

)+ 3H(ρ(t) + p(t))

= 3H(t)(−3H(t)2 + Λ + ρ(t)

)= −3H(t)g(t).

We see that given any C1 function H on the interval I, this is a well poseddifferential equation for g. One solution satisfying the condition g(t0) = 0 isthe zero function. By the uniqueness theorem of ordinary differential equations,there are no other solutions. Hence g(t) = 0 for all t ∈ I, which is what wewanted to prove.

Remark 4.2.2. Note that if the matter model is well-behaved, then the hypoth-esis of Lemma 4.2.1 is satisfied.

4.3 Existence of solutions

As noted above, the systems (4.1.1)-(4.1.2) and (4.1.4)-(4.1.5) give very simi-lar information. In particular, the maximal existence interval for solutions to(4.1.1)-(4.1.2) is equal to the maximal existence interval for solutions to (4.1.4)-(4.1.5).Lemma 4.3.1. If the matter model is well-behaved, then H is strictly decreasingand bounded from below by

√Λ/3.

Proof. By equation (4.1.5)

2H(t) = −3H(t)2 − p(t) + Λ ≤ −3H(t)2 + Λ.

Hence H is strictly decreasing at t whenever H(t) >√

Λ/3. By the assumptionsof our model, ρ(t) is positive. This, together with Lemma 4.2.1, leads to theconclusion that

H2(t) > Λ/3

for all t ∈ I. Since Λ is positive this means that H is never zero. Hence H hasthe same sign throughout I. Since H(t0) > 0, we have now shown that

H(t) >√

Λ/3

for all t ∈ I, so H is strictly decreasing and bounded from below.

Corollary 4.3.2. Suppose that the matter model is well-behaved. Then themaximal existence interval of solutions to (4.1.4)-(4.1.5) contains [t0,∞).

32

Einstein’s equation Existence of solutions

Proof. We have shown that H is bounded on I ∩ [t0,∞). By hypothesis, ρ andp are also bounded on this set. Theorem 2.3.2 then tells us that if t+ = sup I isfinite, then the solution can be extended beyond t+. This would contradict themaximality of I, so [t0,∞) ⊆ I.

Corollary 4.3.3. Suppose that the matter model is well-behaved. Then a isstrictly increasing and unbounded.

Proof. From the relation

a(t) = a(t0) exp

∫ t

t0

H(s)ds

it is clear that the positivity of H implies that a is strictly increasing for all t.Using the lower bound

√Λ/3 for H yields

a(t) ≥ a(t0) exp

∫ t

t0

√Λ

3ds.

If ρ and p are such that they are bounded on each interval where H is bounded,then the previous Corollary shows that I contains arbitrarily large elements. Inthis case, a is unbounded on I.

Lemma 4.3.4. Suppose that the matter model is well-behaved. If (−∞, t0] ⊆ I,then H is unbounded on (−∞, t0].

Proof. Using equation (4.1.5) and the fact that H is decreasing, we get

2H(t) = −3H2(t)− p(t) + Λ ≤ −3H2(t) + Λ ≤ −3H2(t0) + Λ

for t < t0.

Denote −C = −3H2(t0) + Λ. By the initial condition, C > 0. Hence

H(t) ≤ −C2

for t < t0, showing that

H(t) ≥ H(t0) +C

2(t0 − t)

for t < t0. If I ∩ (−∞, t0] contains arbitrarily small elements, then H is un-bounded on I ∩ (−∞, t0], and the proof is complete.

Theorem 4.3.5. Suppose that the matter model is well-behaved. Then themaximal existence interval of solutions to (4.1.4)-(4.1.5) is (t−,∞) for somet− ∈ R.

Proof. We have already shown that [t0,∞) ⊆ I under these hypotheses. If(−∞, t0] 6⊆ I, then the existence interval is bounded from below and we aredone, so suppose not. In that case Lemma 4.3.4 tells us that H is unbounded.

As before,2H(t) ≤ −3H2(t0) + Λ.

33

Einstein’s equation Asymptotics

Since H is unbounded and strictly decreasing, there exists a t1 ∈ I and aconstant D > 0 such that

−3H2(t) + Λ ≤ −2H2(t)

for t < t1. This means thatH(t)

H2(t)≤ −1.

Integrating this inequality, we have

1

H(t)− 1

H(t1)=

∫ t1

t

H(s)

H2(s)ds ≤ −(t1 − t)

so that

H(t) ≥ 11

H(t1) − (t1 − t)

for t < t1. For t close to t− = t1 − 1DH(t1) (which is certainly less than t1) the

right hand side of the inequality is unbounded, proving that

t1 −1

H(t1)6∈ I.

Hence the infimum of I is some real number t−, and the proof is complete.

The above theorem shows that H becomes unbounded close to some t− < t0. Ifthe matter model is well-behaved, then the system of equations we are studyingis autonomous, and so remains unchanged under the change of variable t 7→t − t−. Hence we may choose t− = 0. For simplicity, this is the convention wewill adopt for the remainder of the text.

4.4 Asymptotics

We conclude this chapter with some notes concerning the asymptotics of solu-tions.Lemma 4.4.1. Let H∞ =

√Λ/3, and suppose that the matter model is well-

behaved. Thenlimt→∞

H(t) = H∞.

Proof. To see this, one can use the result of Lemma 4.3.1 which states that H isdecreasing and bounded from below by H∞. This means that H is convergent.Since H is decreasing, this implies that

lim supt→∞

H(t) = 0.

By combining equation (4.1.4) with the conclusion of Lemma 4.2.1, we obtain

2H(t) + ρ(t) + p(t) = 0.

34


Hence

− lim inft→∞

(ρ(t) + p(t)) = lim supt→∞

(−ρ(t)− p(t)) = lim supt→∞

2H(t) = 0.

Since ρ and p are nonnegative, it holds that lim inft→∞ ρ(t) ≥ 0 and

lim inft→∞

ρ(t) ≤ lim inft→∞

(ρ(t) + p(t)) = 0.

We have now shown thatlim inft→∞

ρ(t) = 0.

By Lemma 4.2.1, this implies that

lim inft→∞

3H2(t) = lim inft→∞

ρ(t) + Λ = Λ.

Since H is convergent, limt→∞H(t) = lim inft→∞H(t). Since H is everywherepositive, this proves the lemma.

Proposition 4.4.2. Suppose that the matter model is well-behaved. Then

limt→0+

a(t) = 0.

Proof. We have seen that limt→0+ H(t) =∞. Hence, by Lemma 4.2.1,

limt→0+

ρ(t) =∞.

Recall that ρ is unbounded on an interval J if and only if a takes values arbi-trarily close to 0 on J . This implies that a takes values arbitrarily close to zeroon each interval (0, ε). Since a is increasing, we have now shown that

limt→0+

a(t) = 0,

proving the proposition.

Lemma 4.4.3. Suppose that the matter model is well-behaved. Suppose furtherthat the energy density can be written ρ(t) = O(a−3(t)). Let H∞ =

√Λ/3,

M =∫∞t0

(−1 +

√1 + ρ(s)/Λ

)ds and α = a(t0)eH∞(M−t0). Then

a(t) = αeH∞t +O(e−2H∞t)

in the limit t→∞.

Proof. By Lemma 4.2.1, equation (4.1.1) holds for all t ∈ I. Hence

a(t) = ±a(t)H∞

√1 +

ρ(t)

Λ.

By Corollary 4.3.3, it holds that a(t) > 0 for all t ∈ I. It is now easy to get alower bound for a. Recall that ρ(t) is positive. This means that

a(t) ≥ a(t)H∞

35


and hencea(t) ≥ eH∞t. (4.4.1)

The error made in this underestimation seems to tend to zero with time, andthis inspires to the idea that a may be written as the sum of an exponentiallyincreasing term, as in the bound above, and a term tending to zero. Let usintroduce an auxiliary function A that will help us prove this.

A(t) = a(t)e−H∞t.

We will compute the time derivative of A and construct an ordinary differentialequation for A. By solving this equation we will have a new expression that willlead us to a better estimate of a.

A(t) = a(t)e−H∞t − a(t)H∞e−H∞t

=

[H∞

√1 +

ρ(t)

Λ−H∞

]a(t)e−H∞t

=

[−1 +

√1 +

ρ(t)

Λ

]H∞A(t).

Integrating this, one obtains an expression for A:

A(t) = A(t0) exp

[H∞

∫ t

t0

(−1 +

√1 +

ρ(s)

Λ

)ds

]

We need to learn more about the integral inside the exponential to be able tocontinue. The square root in the integrand can be series expanded for small ρ,and if we combine the assumption that ρ(t) = O(1/a3(t)) and equation (4.4.1),we see that the integrand falls off at least exponentially, which implies that theintegral converges.

This allows us to express the integral using two terms M and −g(t), where

M =

∫ ∞t0

(−1 +

√1 +

ρ(s)

Λ

)ds

and

g(t) =

∫ ∞t

(−1 +

√1 +

ρ(s)

Λ

)ds.

Note that this is the constant M in the hypothesis. With this notation

A(t) = A(t0) exp [H∞(M − g(t))].

This relates to a according to

a(t) = A(t0)eH∞teH∞Me−H∞g(t). (4.4.2)

If g(t) is small and approaches 0 as t → ∞, we can series expand the lastexponential factor. After doing this, we can finally retrieve a new bound for a.

36


The following steps consist of showing that g(t) is bounded by something de-caying fast, series expanding e−H∞g(t), and solving everything for an explicitbound on a.

We will once again use the assumption that ρ(t) = O(a−3) and equation (4.4.1).

|g(t)| =

∣∣∣∣∣∫ ∞t

(−1 +

√1 +

ρ(s)

Λ

)ds

∣∣∣∣∣=

∣∣∣∣∫ ∞t

(−1 + 1 +

1

2

ρ(s)

Λr

(ρ(s)

Λ

))ds

∣∣∣∣ where r is bounded close to 0

=

∣∣∣∣∫ ∞t

ρ(s)

Λ

1

2r

(ρ(s)

Λ

)ds

∣∣∣∣≤ 1

Λ

∫ ∞t

∣∣∣∣ρ(s)

2r

(ρ(s)

Λ

)∣∣∣∣ ds.

There are constants K1, K2 and K3 such that for sufficiently large t it holdsthat

1

Λ

∫ ∞t

∣∣∣∣ρ(s)

2r

(ρ(s)

Λ

)∣∣∣∣ ds ≤ K1

∫ ∞t

∣∣∣∣a(s)−3r

(ρ(s)

Λ

)∣∣∣∣ ds

≤ K1

∫ ∞t

∣∣∣∣e−3H∞sr

(ρ(s)

Λ

)∣∣∣∣ ds

≤ K2

∫ ∞t

∣∣e−3H∞s∣∣ ds

= K2

∫ ∞t

e−3H∞s ds

= K3e−3H∞t.

This is the nice bound for g. A series expansion of (4.4.2) now yields

a(t) = A(t0)eH∞teH∞M (1−H∞g(t)r(|g(t)|))

where r(|g(t)|) is bounded by some R. Recall that

α = a(t0)e−H∞t0eH∞M

so that the above equation may be written as

a(t) = αeH∞t (1−H∞g(t)r(|g(t)|)) .

We now see that there is a constant C1 such that∣∣a(t)− αeH∞t∣∣ =

∣∣C1eH∞tg(t)r(|g(t)|)

∣∣Using the bounds for g and r, we see that there is a constant C2 such that∣∣a(t)− αeH∞t

∣∣ ≤ C2e−3H∞teH∞t.

Hence ∣∣∣∣a(t)− αeH∞t

e−2H∞t

∣∣∣∣ ≤ C2,

which proves the lemma.

37

Chapter 5

Dynamics for a model withfluids


In this chapter we will investigate how matter and radiation modelled by fluidsinfluence the asymptotics of Einstein’s equation. Fluids were described andmotivated in the introduction, section 1.1. Mathematically, we will consider theproblem

3

(a(t0)

a(t0)

)2

= ρ(t0) + Λ, (5.1.1)

2a(t)

a(t)+

(a(t)

a(t)

)2

= Λ− p, (5.1.2)

whereρ(t) = ρm(t) + ρrad(t),

p(t) = pm(t) + prad(t).

The equations governing the time evolution of the energy and pressure functionsare

prad(t) = 3ρrad(t), pm(t) = 0,

ρrad(t) + 3a(t)

a(t)(ρrad(t) + prad(t)) = 0, (5.1.3)

ρm(t) + 3a(t)

a(t)(ρm(t) + pm(t)) = 0. (5.1.4)

We will demand that

ρm, ρrad > 0, pm, prad ≥ 0, Λ, a, a(t0) > 0.

38

Dynamics for a model with fluids Conservation laws

The aim of this chapter is to find the asymptotics of solutions to the systemabove. We want to reuse the general results of chapter 4, and so we need toprove that the equations above imply that the matter model is well-behaved inthe sense of Definition 4.1.1. This will be done in Lemma 5.2.6.

The equations can be written in terms of the Hubble constant H(t) = a(t)/a(t)as follows:

3 (H(t0))2

= ρrad(t0) + ρm(t0) + Λ, (5.1.5)

2H(t) + 3H2(t) = Λ− prad(t), (5.1.6)

ρrad(t) + 3H(t) (ρrad(t) + prad(t)) = 0, (5.1.7)

ρm(t) + 3H(t) (ρm(t) + pm(t)) = 0. (5.1.8)

Using the definition of the pressures prad and pm we can reformulate the equa-tions to yield the following form:

2H(t) + 3H2(t) = Λ− ρrad(t)/3, (5.1.9)

ρrad(t) + 4H(t)ρrad(t) = 0, (5.1.10)

ρm(t) + 3H(t)ρm(t) = 0. (5.1.11)

5.2 Conservation laws and connections to thegeneral case

This section will establish some conserved properties of the system. First, wewill see that the total energy is conserved. Second, we will see that the initialcondition holds and that the results in chapter 4 are applicable on this system.

Let I be the maximal existence interval for solutions to the system under con-sideration. After proving Lemma 5.2.6 below, we will have a result on the formof I.Proposition 5.2.1. The total matter energy is conserved. More rigorously,there exists an Rm ∈ R such that

ρm(t)a3(t) = Rm

for all t ∈ R in the interior of the common domain of a and ρm.

Further, there is a conserved radiative energy Rrad ∈ R such that

ρrad(t)a4(t) = Rrad

for all t ∈ R in the interior of the common domain of a and ρrad.

Proof. The two cases are extremely similar; we will only prove the former case.

Substituting pm = 0 in (5.1.4) yields

ρm + 3a

aρm = 0.

39

Dynamics for a model with fluids Conservation laws

Hencea3ρm + 3a2aρm = 0

andd

dt(a3ρm) = 0.

This completes the proof.

Remark 5.2.2. We will keep the constants Rm and Rrad (which are obviouslyunique) fixed. Note that Rm and Rrad are positive, since we assumed thatρm(t0), ρm(t0) and a(t0) are all positive.Proposition 5.2.3. It holds that ρ(t) = −3H(t)(ρ(t) + p(t), so that the initialcondition (5.1.5) holds for all time:

3 (H(t))2

= ρrad(t) + ρm(t) + Λ ∀t ∈ I.

Proof. Note that

ρ(t) = ρrad(t) + ρm(t) = −4Hρrad(t)− 3Hρm(t)

and

−3H(t)(ρ(t) + p(t)) = −3H(ρrad + ρm + prad) = −4Hρrad − 3Hρm(t).

Hence ρ(t) = −3H(t)(ρ(t)+p(t), and by Lemma 4.2.1 the initial condition holdsfor all time, as desired.

Corollary 5.2.4. It holds for all t ∈ I that

H(t) = −2H2(t) +ρm(t)

6+

2Λ

3.

Proof. This is a simple consequence of (5.1.9) and Proposition 5.2.3.

Lemma 5.2.5. Consider a solution to the above system, defined on some in-terval J . The following are equivalent.

• H is bounded on J .

• ρ is bounded on J .

• p is bounded on J .

Proof. It holds that pm(t) = 0 and prad = ρrad(t)/3, and all the functionsinvolved are nonnegative. This shows that ρ is bounded on J if and only if p isbounded on J .

Proposition 5.2.3 tells us that for all t ∈ J , it holds that

ρ(t) = ρrad(t) + ρm(t) = 3 (H(t))2 − Λ.

Hence ρ is bounded if an only if H is bounded.

Lemma 5.2.6. The fluid matter model is well-behaved in the sense of Definition4.1.1.

40

Dynamics for a model with fluids Future asymptotics

Proof. We have already shown two of the conditions in the definition, in Propo-sition 5.2.3 and Lemma 5.2.5.

By letting

ρ(x) =Rmx3

+Rradx4

and p(x) =Rrad3x4

we see that ρ(t) = ρ(a(t)) and p(t) = p(a(t)) for all t ∈ I.

Further, we see from the explicit expression for ρ obtained in Proposition 5.2.1that ρ is unbounded on an interval J if and only if a takes values arbitrarilyclose to 0 on J .


Remark 5.2.7. Lemma 5.2.6 shows that all results of chapter 4 are applicable.In particular, the maximal existence interval for solutions to (5.1.1)-(5.1.4) isI = (0,∞), by Theorem 4.3.5 and the remark following it.Corollary 5.2.8. It holds that

limt→0+

a2(t)H(t) =

√Rrad

3.

Proof. By Proposition 5.2.3,

a2(t)H(t) =1√3

√Rrad +Rma(t) + Λa4(t)

for all positive t. By Proposition 4.4.2, limt→0+ a(t) = 0, and the proof iscomplete.

5.3 Future asymptotics

This section concerns the asymptotics of solutions when t→∞. We will exam-ine both a and H.

5.3.1 Asymptotics of a

Theorem 5.3.1. Let H∞ =√

Λ/3, M =∫∞t0

(−1 +

√1 + ρ(s)/Λ

)ds and

α = a(t0)eH∞(M−t0). Then

a(t) = αeH∞t +O(e−2H∞t)

when t→∞.

Proof. Proposition 5.2.3 states that the initial condition holds for all time. Itfollows from Proposition 5.2.1 that the energy density obeys ρ(t) = O(a(t)−3).These two conditions allow us to use Lemma 4.4.3, and we have proven ourtheorem.

41

Dynamics for a model with fluids Future asymptotics

5.3.2 Asymptotics of H

Theorem 5.3.2. It holds that

H(t) = H∞ +RmH∞2Λα3

e−3H∞t +RradH∞

2Λα4e−4H∞t +O(e−6H∞t).

Proof. From Proposition 5.2.3 and the fact that H is positive, we have that

H(t) = H∞

√1 +

ρ(t)

Λ.

Series expanding the square root and rearranging yields

H(t)−H∞ =H∞

2

ρ(t)

Λ+O(ρ(t)2).

Using ρ(t) = Rma3(t) + Rrad

a4(t) gives

H(t)−H∞ =H∞2Λ

[Rma

−3(t) +Rrada−4(t)

]+O(a−6(t)).

Apply Theorem 5.3.1 to find

H(t)−H∞ =H∞2Λ

Rm(αeH∞t +O(e−2H∞t)

)−3

+H∞2Λ

Rrad(αeH∞t +O(e−2H∞t)

)−4

+O(e−6H∞t),

so that

H(t)−H∞ =RmH∞

2Λ (αeH∞t)3

(1 +O(e−3H∞t)

)−3

+RradH∞

2Λ (αeH∞t)4

(1 +O(e−3H∞t)

)−4

+O(e−6H∞t).

By expanding 1(1+k−3t)n = 1 +O(k−3t) we find

H(t)−H∞ =RmH∞

2Λ (αeH∞t)3 +O(e−6H∞t)

+RradH∞

2Λ (αeH∞t)4 +O(e−7H∞t)

+O(e−6H∞t).

This proves our theorem.

42

Dynamics for a model with fluids Past asymptotics

5.4 Past asymptotics

In this section, we will be concerned with the asymptotics of the solutions inthe limit t→ 0+.Remark 5.4.1. We will find the auxiliary function

Ω(t) = a2(t)H(t)

useful. By Corollary 5.2.8, Ω(t) has a limit Ω0 =√

Rrad3 as t → 0+. Hence it

is natural to extend Ω by Ω(0) = Ω0. Note that Ω0 is positive, since Rrad ispositive.Lemma 5.4.2. It holds for all sufficiently small t > 0 that

1

H(t)= 2t−

∫ t

0

(2

3

Λ

H2(s)+ a(s)

η

H2(s)a4(s)

)ds,

where η = Rm/6.

Proof. By equation (5.1.6) and Proposition 5.2.3

6H(t) + 9H2(t) = 3Λ− ρrad(t),

0 = ρrad(t) + ρm(t)− 3H2 + Λ.

Adding these yields

6H(t) = −12H2(t) + 4Λ + ρm(t).

From Proposition 5.2.1, we know that

ρm(t) =Rma3(t)

.

With

η =Rm6

we have

H(t) = −2H2(t) +2

3Λ +

η

a3(t)

so that

− H(t)

H2(t)= 2− 2

3

Λ

H2(t)− a(t)

η

H2(t)a4(t).

Integrating this equality from t1 to t yields

1

H(t)− 1

H(t1)= 2(t− t1)−

∫ t

t1

(2

3

Λ

H2(s)+ a(s)

η

H2(s)a4(s)

)ds.

We have already seen that a4(s)H2(s), and hence the integrand above, remainsbounded close to zero, so that

limt1→0+

∫ t

t1

(2

3

Λ

H2(s)+ a(s)

η

H2(s)a4(s)

)ds =∫ t

0

(2

3

Λ

H2(s)+ a(s)

η

H2(s)a4(s)

)ds.

43


Letting t1 → 0 and using the fact that H(t1)→∞ as t1 → 0 we obtain

1

H(t)= 2t−

∫ t

0

(2

3

Λ

H2(s)+ a(s)

η

H2(s)a4(s)

)ds. (5.4.1)

Remark 5.4.3. It holds that

lims→0+

(2

3

Λ

H2(s)+ a(s)

η

H2(s)a4(s)

)= 0.

Lemma 5.4.4.

limt→0+

a(t)√t

=4

√4

3Rrad.

Proof. Choose an arbitrary 0 < ε < 1 and let t2 be such that the expression inRemark 5.4.3 does not exceed ε for t < t2. Then Lemma 5.4.2 tells us that, fort < t2,

2t− εt ≤ 1

H(t)≤ 2t+ εt.

Thus we have shown that

limt→0+

tH(t) =1

2.

To see what this implies for a, we use the auxiliary function Ω(t) = a2(t)H(t).We can express a as

a(t) =

√Ω(t)

H(t).

While t is small enough, the inequalities for H imply√Ω(t)(2− ε) ≤ a(t)√

t≤√

Ω(t)(2 + ε).

Recall that Ω(t) has a nonzero limit√

13Rrad as t→ 0. Using this and the fact

that ε was arbitrary, we get the conclusion that

limt→0+

a(t)√t

=√

2Ω0 =4

√4

3Rrad

as desired.

Lemma 5.4.5. The limit

limt→0+

√2Ω(t)−

√2Ω0√

t

exists.

Proof. We begin by noting that√2Ω(t)−

√2Ω0√

t=

Ω(t)− Ω0√t

√2√

Ω(t) +√

Ω0

.

44


Since both Ω0 is positive and Ω(t) is positive with a positive limit as t → 0+,the limit

limt→0+

√2Ω(t)−

√2Ω0√

t

exists if and only if the limit

limt→0+

Ω(t)− Ω0√t

exists. By the same trick, the existence of this limit is equivalent to the existenceof the limit

limt→0+

3Ω2(t)− 3Ω20√

t.

We now note that

3Ω2(t) = 3a4(t)H2(t)

= a4(t)ρrad(t) + a4(t)ρm(t) + Λa4(t)

= Rrad +Rma(t) + Λa4(t)

and3Ω2

0 = Rrad

so that3Ω2(t)− 3Ω2

0√t

=3a(t)√

t(Rm + Λa3(t)).

Using Lemma 5.4.4 we conclude that

limt→0+

3Ω2(t)− 3Ω20√

t= Rm

4

√4

3Rrad.

Lemma 5.4.6. Let η = Rm/6, as in Lemma 5.4.2. Then

limt→0+

(2

3

Λ

t2H2(t)t3/2 +

a(t)√t

η

H2(t)a4(t)

)=Rm6

(3

Rrad

)3/4

.

Proof. Recall that

limt→0

a2(t)H(t) = Ω0 =

√Rrad

3.

By Lemma 5.4.4

limt→0

a2(t)

t= 2

√Rrad

3.

In the course of the proof of Lemma 5.4.4, we saw that

limt→0

tH(t) =1

2.

Combining these observations yields

limt→0

(2

3

Λ

t2H2(t)t3/2 +

a(t)√t

η

H2(t)a4(t)

)=Rm6

(3

Rrad

)3/4

.

45


Definition 5.4.1. For notational reasons, we introduce the constant

Ξ =Rm9

(3

Rrad

)3/4

.

Lemma 5.4.7. For each ε > 0 there is a t∗ such that for all t < t∗∣∣∣∣ 1

H(t)− 2t+ Ξt3/2

∣∣∣∣ ≤ εt3/2.Proof. Rewriting the conclusion of Lemma 5.4.2 slightly, we see that

1

H(t)= 2t−

∫ t

0

√s

(2

3

Λ

s2H2(s)s3/2 +

a(s)√s

η

H2(s)a4(s)

)ds.

By Lemma 5.4.6, it holds that

lims→0

(2

3

Λ

s2H2(s)s3/2 +

a(s)√s

η

H2(s)a4(s)

)=

3

2Ξ.

Fix some 0 < ε < 1 and let t2 be such that∣∣∣∣23 Λ

s2H2(s)s3/2 +

a(s)√s

η

H2(s)a4(s)− 3

2Ξ

∣∣∣∣ < 3

2ε

for t < t2. Then ∣∣∣∣ 1

H(t)− 2t+ Ξt3/2

∣∣∣∣ ≤ ∫ t

0

3

2

√sεds = εt3/2. (5.4.2)

Theorem 5.4.8.

a(t) =4

√4

3Rradt

1/2 +O(t).

Proof. As before,

a(t) =

√Ω(t)

H(t).

The inequalities for H given by Lemma 5.4.7 yield√Ω(t)(2t− (Ξ + ε)t3/2) ≤ a(t) ≤

√Ω(t)(2t− (Ξ− ε)t3/2).

By series expanding the square roots in the bounds, we obtain√2Ω(t)− (Ξ + ε)

√Ω(t)O(

√t) ≤ a(t)√

t≤√

2Ω(t)− (Ξ− ε)√

Ω(t)O(√t).

Rewriting this slightly gives us√2Ω(t)−

√2Ω0√

t− (Ξ + ε)

√Ω(t)O(1)

≤ a(t)−√t√

2Ω0

t≤√

2Ω(t)−√

2Ω0√t

− (Ξ− ε)√

Ω(t)O(1)

46


for all t smaller than some t∗ depending on the arbitrary ε > 0. We can nowuse Lemma 5.4.5 to see that all terms in both the upper and lower bound arebounded close to t = 0. Hence we have shown that

a(t) =√

2Ω0t1/2 +O(t)

and the conclusion follows after noting that Ω0 =√

13Rrad.

Theorem 5.4.9.

H(t) =1

2t+

Ξ

4+ o

(1√t

).

Proof. Lemma 5.4.7 states that∣∣∣∣ 1

H(t)− 2t+ Ξt3/2

∣∣∣∣ ≤ εt3/2for each ε > 0 and all t sufficiently close to zero. Using this, we see that

H(t) ≤ 1

2t− Ξt3/2 − εt3/2

which means that

H(t)− 1

2t≤ 1

2t− Ξt3/2 − εt3/2− 1

2t=

Ξ + ε

2(2√t− Ξt− εt)

.

Hence

H(t)− 1

2t− Ξ

4√t≤ Ξ + ε

2(2√t− Ξt− εt)

− Ξ

4√t

=2ε+ Ξ2

√t+ εΞ

√t

4(2√t− Ξt− εt)

.

Analogously,

H(t)− 1

2t− Ξ

4√t≥ −2ε+ Ξ2

√t− εΞ

√t

4(2√t− Ξt+ εt)

.

We have shown that ∣∣∣∣∣H(t)− 12t −

Ξ4

1√t

∣∣∣∣∣is bounded by something which tends to zero as t → 0. This completes theproof.

47

Chapter 6

Dynamics for a model withVlasov matter


In this section we will determine the asymptotic behavior of solutions to Ein-stein’s equation in spatially flat, homogeneous and isotropic spacetime when thematter is modelled by Vlasov matter. The physical motivation and setting forthe model is given in section 1.1. The mathematical problem is:

3

(a(t0)

a(t0)

)2

= ρV l(t0) + Λ, (6.1.1)

2a(t)

a(t)+

(a(t)

a(t)

)2

= Λ− pV l(t), (6.1.2)

ρV l(t) =

∫R3

f(a(t) q)(1 + |q|2)1/2 dq,

pV l(t) =1

3

∫R3

f(a(t) q)|q|2

(1 + |q|2)1/2dq.

We will demand that f is nonnegative, smooth and has compact support. Sincewe work with isotropic models, we also demand that there is a function g : R→ Rsuch that f(p) = g(|p|). Further, we will demand that f is not identically zero,so that ρ is positive.

We will prove shortly (in Lemma 6.2.5) that the general results in chapter 4 areapplicable to this system. When this has been shown we can use Theorem 4.3.5to conclude that the solution so this system has an existence interval which isbounded from below and unbounded from above. As before, we can withoutloss of generality let the infimum of the existence interval be 0. The problemthat concerns us is then to determine the behaviour of solutions to the systemas t→ 0 or t→∞.

48

Vlasov matter Conservation laws

As in chapter 4, the equations (6.1.1) and (6.1.2) may be rewritten with theHubble constant as follows.

3 (H(t0))2

= ρV l(t0) + Λ (6.1.3)

2H(t) + 3H2(t) = Λ− pV l(t) (6.1.4)

We introduce for each n ∈ Z the shorthand notation

In =

∫R3

f(q)|q|n dq.

Note that these integrals are guaranteed to converge for n ≥ −2.

6.2 Conservation laws and connections to thegeneral case

There is a nice way to express ρV l and pV l using the rotational symmetry off . This will often be used without further comment. We derive the expressionsbelow.Remark 6.2.1. It holds that

ρV l(t) =1

a4(t)

∫ ∞0

4πx2g(x)(a2(t) + x2)1/2 dx

and

pV l(t) =1

3

1

a4(t)

∫ ∞0

4πx2g(x)x2

(a2(t) + x2)1/2dx.

Proof. This is shown by a simple change of variables. Recall that we haveassumed that

f(p) = g(|p|)for some function g : R→ R. Rewriting the integral, we have

ρV l(t) =

∫R3

f(a(t) q)(1 + |q|2)1/2 dq =

∫R3

g(a(t) |q|)(1 + |q|2)1/2 dq =

=

∫ ∞0

4πx2g(a(t)x)(1 + x2)1/2 dx =1

a4(t)

∫ ∞0

4πx2g(x)(a2(t) + x2)1/2 dx

and

pV l(t) =1

3

∫R3

f(a(t) q)|q|2

(1 + |q|2)1/2dq =

1

3

∫R3

g(a(t) |q|) |q|2

(1 + |q|2)1/2dq =

=1

3

∫ ∞0

4πx2g(a(t)x)x2

(1 + x2)1/2dx =

1

3

1

a4(t)

∫ ∞0

4πx2g(x)x2

(a2(t) + x2)1/2dx

as desired.

Let I be the maximal existence interval for solutions to the system under con-sideration. After proving Lemma 6.2.5 below, we will have a result on the formof I.

49

Vlasov matter Conservation laws

Proposition 6.2.2. It holds that ˙ρV l(t) = −3H(t)(ρV l(t) + pV l(t), so that theinitial condition (6.1.3) holds for all time:

3 (H(t))2

= ρV l(t) + Λ ∀t ∈ I.

Proof. We wish to use Lemma 4.2.1, and hence need to prove that

ρV l(t) = −3H(t)(ρV l(t) + pV l(t)).

First,

ρV l(t)− 3pV l(t) =1

a2(t)

∫ ∞0

4πx2g(x)√a2(t) + x2

dx.

Then, note that the derivative of ρV l is

˙ρV l(t) =− 4a(t)

a5(t)

∫ ∞0

4πx2g(x)√a2(t) + x2 dx

+1

a4(t)

∫ ∞0

4πx2g(x)a(t)a(t)√a2(t) + x2

dx.

Hence

˙ρV l(t)

H(t)=− 4ρV l(t) +

1

a2(t)

∫ ∞0

4πx2g(x)√a2(t) + x2

dx

=− 4ρV l(t) + (ρV l(t)− 3pV l(t))

and a use of Lemma 4.2.1 proves the proposition.

Proposition 6.2.3. It holds that

ρV l(t) ≥ 3pV l(t) ∀t ∈ I

Proof. Note that

ρV l(t)− 3pV l(t) =

∫R3

f (a(t)q)

(√1 + |q|2 − |q|2√

1 + |q|2

)dq.

Hence the proposition is true if√1 + |q|2 ≥ |q|2√

1 + |q|2

everywhere. Form the quotient of the two terms. We see that√1 + |q|2|q|2√1+|q|2

=1 + |q|2

|q|2= 1 +

1

|q|2,

proving the proposition.

Lemma 6.2.4. Consider a solution to the above system, defined on some in-terval J . The following are equivalent.

50

Vlasov matter Future asymptotics

• H is bounded on J .

• ρV l is bounded on J .

• pV l is bounded on J .

Proof. Proposition 6.2.2 tells us that for all t ∈ J , it holds that

ρV l(t) = 3H2(t)− Λ.

It is immediate from this equation that ρV l is bounded if and only if H isbounded.

By Proposition 6.2.3, it holds that pV l is bounded if ρV l is bounded.

Suppose now that pV l is bounded. From the definition of pV l, we see that thisimplies that a does not take values arbitrarily close to 0, which in turn impliesthat ρV l is bounded.


Lemma 6.2.5. The Vlasov matter model is well-behaved in the sense of Defi-nition 4.1.1.

Proof. We have already shown two of the conditions in the definition, in Propo-sition 6.2.2 and Lemma 6.2.4.

It is immediate from the definition of ρV l and pV l that there are functions ρand p such that ρV l(t) = ρ(a(t)) and pV l(t) = p(a(t)) for all t ∈ I.

Further, we see from the definition that ρV l is unbounded on an interval J ifand only if a takes values arbitrarily close to 0 on J .


Lemma 6.2.5 shows that all results of chapter 4 are applicable. In particular,the maximal existence interval for solutions to the system is I = (0,∞), byTheorem 4.3.5 and the remark following it.

6.3 Future asymptotics

This section examines the dynamics of the system in the limit t→∞.

6.3.1 Asymptotics of ρV l and pV l

Lemma 6.3.1. The pressure and energy density satisfy

pV l(t) =I2

3a(t)−5 + o(a(t)−5),

ρV l(t) = I0a(t)−3 + o(a(t)−3)

when t→∞.

51

Vlasov matter Future asymptotics

Proof. The proofs of the two claims are almost identical. We will only provethe former. It holds that

pV l =1

3

∫R3

f(a(t)q)|q|2

(1 + |q|2)1/2dq = p = a(t)q =

=1

3

1

a(t)5

∫R3

f(p)|p|2(

1 +(|p|a(t)

)2)1/2

dp

We can take limits inside this integral, and since Corollary 4.3.3 tells us that agoes to infinity, we find that

a(t)5pV l →I2

3as t→∞.

Hence

pV l =I2

3a(t)−5 + o(a(t)−5).


6.3.2 Asymptotics of a

Theorem 6.3.2. It holds that

a(t) = αeH∞t +O(e−2H∞t).

Proof. Proposition 6.2.2 states that the initial condition holds for all time. Itfollows from Lemma 6.3.1 that the energy density obeys ρ(t) = O(a(t)−3).These two conditions allow Lemma 4.4.3 to be used, and we have proven ourtheorem.

6.3.3 Asymptotics of H

Theorem 6.3.3.

H(t) = H∞ +I0H∞2Λα3

e−3H∞t +I2H∞4Λα5

e−5H∞t +O(e−6H∞t)

Proof. From Proposition 6.2.2 and the fact that H is positive, we have that

H(t) = H∞

√1 +

ρV l(t)

Λ

Series expanding the square root and rearranging yields

H(t)−H∞ =H∞

2

ρV l(t)

Λ+O(ρV l(t)

2). (6.3.1)

52

Vlasov matter Past asymptotics

We need to determine the asymptotics of ρV l to second order, to preserve theinformation in the above equation. Series expanding ρV l will suffice:

ρV l(t) =

∫R3

f(a(t)q)(1 + |q|2)1/2dq = p = qa(t) =

=1

a(t)3

∫R3

f(p)

(1 +

(|p|a(t)

)2)1/2

dp

=1

a(t)3

∫R3

f(p)

[1 +

1

2

|p|2

a2(t)+O

(|p|4

a4(t)

)]dp

= I0a−3(t) +

1

2I2a−5(t) +O(a−7(t)).

Insert this approximation into equation (6.3.1), and use Theorem 6.3.2. Finally,follow up with a simple calculation identical to the one in the proof of Theorem5.3.2, to prove the theorem.

6.4 Past asymptotics

In this section, we will be concerned with the asymptotics of the solutions inthe limit t→ 0+.

6.4.1 Asymptotics of ρV l and pV l

Lemma 6.4.1. It holds that

limt→0

a4(t)ρV l(t) = I1.

Proof. By Remark 6.2.1,

a4(t)ρV l(t) =

∫ ∞0

4πx2g(x)(a2(t) + x2)1/2 dx

We have already seen that limt→0 a(t) = 0. Using Lebesgues dominated conver-gence theorem (where the integrand is bounded by say 4πx2g(x)

√1 + x2)

limt→0+

a4(t)ρV l(t) =

∫ ∞0

4πx3g(x) dx = I1.

Corollary 6.4.2. It holds that

limt→0+

a2(t)H(t) =

√I1

3.

Proof. By Proposition 6.2.2

a4(t)H2(t) =a4(t)

3(ρV l(t) + Λ) .

From Lemma 6.4.1 and the fact that limt→0 a(t) = 0, the conclusion followsimmediately.

53



limt→0+

a4(t)pV l(t) =1

3I1.

Proof. By Remark 6.2.1,

a4(t)pV l(t) =1

3

∫ ∞0

4πx2g(x)x2

(a2(t) + x2)1/2dx.

We have already seen that limt→0 a(t) = 0. Using Lebesgues dominated conver-gence theorem

limt→0+

a4(t)pV l(t) =1

3

∫ ∞0

4πx3g(x) dx =1

3I1.


limt→0+

a2(t)(ρV l(t)− 3pV l(t)) = I−1.

Proof. By Remark 6.2.1 about our standard change of variables

a4(t)(ρV l(t)− 3pV l(t)) =

∫ ∞0

4πx2g(x)a2

√a2 + x2

dx.

Hence

a2(t)(ρV l(t)− 3pV l(t)) =

∫ ∞0

4πx2g(x)1√

a2 + x2dx.

Note that the integrand is bounded by the integrable function x 7→ 4πxg(x).Since a(t)→ 0 as t→ 0, Lebesgue’s dominated convergence theorem gives us

limt→0+

a2(t)(ρV l(t)− 3pV l(t)) =

∫ ∞0

4πxg(x) dx = I−1.

6.4.2 Asymptotics of a and H

Lemma 6.4.5. There is a function η : R+ → R+ such that

limt→0

η(t) =I−1

6

and1

H(t)= 2t−

∫ t

0

(2

3

Λ

H2(s)+ a2(s)

η(s)

H2(s)a4(s)

)ds

for all sufficiently small t > 0. Further,

lims→0+

(2

3

Λ

H2(s)+ a2(s)

η(s)

H2(s)a4(s)

)= 0.

54


Proof. By equation (6.1.4) and Proposition 6.2.2

6H(t) + 9H2(t) = 3(Λ− pV l(t)),

0 = ρV l(t)− 3H(t)2 + Λ.

Adding these yields

6H(t) = −12H2(t) + 4Λ + (ρV l(t)− 3pV l(t)).

From Lemma 6.4.4, we know that

ρV l(t)− 3pV l(t) = 6η(t)

a2(t)

where η is a function that tends to I−1/6 as t→ 0. Hence

H(t) = −2H2(t) +2

3Λ +

η(t)

a2(t)

so that

− H(t)

H2(t)= 2− 2

3

Λ

H2(t)− a2(t)

η(t)

H2(t)a4(t).

Integrating this equality from t1 to t yields

1

H(t)− 1

H(t1)= 2(t− t1)−

∫ t

t1

(2

3

Λ

H2(s)+ a2(s)

η(s)

H2(s)a4(s)

)ds

Since 3H2a4 = a4ρV l+a4Λ and a4(t)ρV l(t) are bounded from below by a positive

number, both terms in the integrand tend to zero as s→ 0. Letting t1 → 0 andusing the fact that H(t1)→∞ as t1 → 0 we obtain

1

H(t)= 2t−

∫ t

0

(2

3

Λ

H2(s)+ a2(s)

η(s)

H2(s)a4(s)

)ds. (6.4.1)

Remark 6.4.6. We will find the auxiliary function

Ω(t) = a2(t)H(t)

useful. By Corollary 6.4.2, Ω(t) has a nonzero limit Ω0 =√

13I1 as t → 0+.

Hence it is natural to extend Ω by Ω(0) = Ω0.Lemma 6.4.7.

limt→0

a(t)√t

=4

√4

3I1.

Proof. Choose an arbitrary 0 < ε < 1 and let t2 be such that the integrand in(6.4.1) does not exceed ε for t < t2. Then

2t− εt ≤ 1

H(t)≤ 2t+ εt

55


for t < t2. Thus we have shown that

limt→0+

tH(t) =1

2.

To see what this implies for a, we use the auxiliary function Ω(t) = a2(t)H(t).Using this function, we can express a as

a(t) =

√Ω(t)

H(t).

Hence the inequalities for H yield√Ω(t)(2− ε) ≤ a(t)√

t≤√

Ω(t)(2 + ε).

Recall that Ω(t) has a nonzero limit√

13I1 as t → 0. Using this and the fact

that ε was arbitrary, we get the conclusion that

limt→0+

a(t)√t

=√

2Ω0 =4

√4

3I1

as desired.

Lemma 6.4.8. There is a real number t∗ > 0 and a constant ΩB ∈ R such that∣∣∣∣∣√

2Ω(t)−√

2Ω0

t

∣∣∣∣∣ ≤ ΩB

for all t ∈ (0, t∗].

Proof. We begin by noting that√2Ω(t)−

√2Ω0

t=

Ω(t)− Ω0

t

√2√

Ω(t) +√

Ω0

.

Since both Ω0 and Ω(t) are positive for all t, the boundedness of√2Ω(t)−

√2Ω0

t

on some set is equivalent to the boundedness of

Ω(t)− Ω0

t

on the same set. By the same trick, it is enough to show that

3Ω2(t)− 3Ω20

t

is bounded.

56


By the definition of Ω0,

3Ω20 = I1 =

∫ ∞0

4πx3g(x) dx.

By Proposition 6.2.2,

3Ω2(t) = a4(t)ρV l(t) + a4(t)Λ = a4(t)Λ +

∫ ∞0

4πx2g(x)√a2(t) + x2 dx.

The expression under consideration can now be written as

3Ω2(t)− 3Ω20

t=a4(t)Λ

t+

4π

t

∫ ∞0

(x2g(x)

√a2(t) + x2 − x3g(x)

)dx.

Slightly rewritten, the second term becomes

4π

t

∫ ∞0

x2g(x)(√

a2(t) + x2 − x)

dx.

To simplify matters further, note that

√a2(t) + x2 − x =

a2(t) + x2 − x2√a2(t) + x2 + x

=a2(t)√

a2(t) + x2 + x.

With this, we want to determine if the limit

limt→0+

(a4(t)Λ

t+

4π

t

∫ ∞0

x2g(x)a2(t)√

a2(t) + x2 + xdx

)

is finite.

Apply Lemma 6.4.7, to find that the above limit is

limt→0+

(a2(t)

√4

3I1 + 4π

4

√4

3I1

∫ ∞0

x2g(x)√a2(t) + x2 + x

dx

).

The first term goes to zero as t → 0+, since a(t) → 0. Taking the limit insidethe integral with the help of Lebesgue’s dominated convergence theorem tellsus that the limit is

4π4

√4

3I1

∫ ∞0

x2g(x)

2xdx.

This integral is bounded since g(x) has compact support.

We have now shown that3Ω2(t)− 3Ω2

0

t

is bounded for sufficiently small t, which by the previous remarks is equivalentto what we wanted to prove.

57


Lemma 6.4.9. Let η be the function from Lemma 6.4.5. Then

limt→0+

(2

3

Λ

t2H2(t)t+

a2(t)

t

η(t)

H2(t)a4(t)

)=I−1√3I1

.

Proof. Recall that

limt→0+

η(t) =I−1

6

and that

limt→0+

a2(t)H(t) = Ω0 =

√I1

3.

By Lemma 6.4.7

limt→0+

a2(t)

t=

√4

3I1.

In the course of the proof of Lemma 6.4.7, we saw that

limt→0+

tH(t) =1

2.

Combining these yields

limt→0+

(2

3

Λ

t2H2(t)t+

a2(t)

t

η(t)

H2(t)a4(t)

)=

√4

3I1I−1

6I13=I−1√3I1

.

Lemma 6.4.10. For each ε > 0 there is a t∗ such that for all t < t∗∣∣∣∣ 1

H(t)− 2t+

I−1

2√

3I1t2∣∣∣∣ ≤ εt2.

Proof. Rewriting the conclusion of Lemma 6.4.5 slightly, we see that

1

H(t)= 2t−

∫ t

0

s

(2

3

Λ

s2H2(s)s+

a2(s)

s

η(s)

H2(s)a4(s)

)ds.

Let K = I−1

2√

3I1. By Lemma 6.4.9, it holds that

lims→0+

(2

3

Λ

s2H2(s)s+

a2(s)

s

η(s)

H2(s)a4(s)

)= 2K.

Fix some 0 < ε < 1 and let t2 be such that∣∣∣∣23 Λ

s2H2(s)s+

a2(s)

s

η(s)

H2(s)a4(s)− 2K

∣∣∣∣ < 2ε

for t < t2. Then ∣∣∣∣ 1

H(t)− 2t+Kt2

∣∣∣∣ ≤ ∫ t

0

sεds = εt2,

proving the result.

58


Theorem 6.4.11.

a(t) =4

√4

3I1t

1/2 +O(t3/2).

Proof. As before,

a(t) =

√Ω(t)

H(t).

Let K = I−1

2√

3I1. The inequalities for H given by Lemma 6.4.10 yield√

Ω(t)(2t− (K + ε)t2) ≤ a(t) ≤√

Ω(t)(2t− (K − ε)t2).

By series expanding the square roots in the bounds, we obtain√2Ω(t)− (K + ε)

√Ω(t)O(t) ≤ a(t)√

t≤√

2Ω(t)− (K − ε)√

Ω(t)O(t).

Rewriting this slightly gives us√2Ω(t)−

√2Ω0

t− (K + ε)

√Ω(t)O(1)

≤ a(t)−√t√

2Ω0

t3/2≤√

2Ω(t)−√

2Ω0

t− (K − ε)

√Ω(t)O(1)

for all t smaller than some t∗ depending on an arbitrary ε > 0.

We can now use Lemma 6.4.8 to see that all terms in both the upper and lowerbound are bounded close to t = 0. Hence we have shown that

a(t) =√

2Ω0t1/2 +O(t3/2)

and the conclusion follows after noting that Ω0 =√

13I1.

Theorem 6.4.12.

H(t) =1

2t+I−1

8√

3I1

+ o(1).

Proof. Let K = I−1

2√

3I1. Lemma 6.4.10 states that∣∣∣∣ 1

H(t)− 2t+Kt2

∣∣∣∣ ≤ εt2for each ε > 0 and all t sufficiently close to zero. Using this, we see that

H(t) ≤ 1

2t−Kt2 − εt2

which means that

H(t)− 1

2t≤ 1

2t−Kt2 − εt2− 1

2t=

K + ε

2(2−Kt− εt).

59


Hence

H(t)− 1

2t− K

4≤ K + ε

2(2−Kt− εt)− K

4=

2ε+K2t+ εKt

4(2−Kt− εt).

Analogously,

H(t)− 1

2t− K

4≥ −2ε+K2t− εKt

4(2−Kt+ εt).

We now have bounds for∣∣H(t)− 1

2t −K4

∣∣ which both tend to zero as t → 0.This completes the proof.

60

Chapter 7

Conclusions and discussion

In the preceding chapters we have studied Einstein’s equation in three differentsettings; first with minimal restrictions on the matter, and then with two dif-ferent matter models. The general setting allowed only a few things to be saidabout the solutions, but provided some insight into the approximate behaviourof solutions. With only that information, a great deal can be said about howthe spacetime behaves. In the later chapters, the asymptotics of the solutionscould be established in more detail.

In this chapter we will recapitulate some of the findings and provide supplemen-tary comments to illustrate some of the features in the results.

7.1 Generalities

Under the assumptions of spatial homogeneity, isotropy and flatness, Einstein’sequation boiled down to an initial condition and an equation for time evolution.These contained the pressure and energy density of the matter contained inthe universe. To derive the results of that chapter, we made some assumptionsabout the matter. To begin with, we assumed that the energy density wasnonzero, which corresponds to having something in our universe. This case ismore interesting than the vacuum case, when the energy density and pressure arezero. Since the matter models under consideration later on had nonzero energydensity, we imposed this condition in the general case. Secondly, we assumedthat the energy density ρ and the pressure p satisfied certain conditions, such asbeing bounded when the Hubble constant H was bounded. Both matter modelsunder consideration satisfy these conditions.

With these assumptions we found that solutions to the Einstein equation werevalid on an interval ranging from some initial time and infinitely forward intime. When approaching the initial time, the Hubble constant diverges and thescale factor a tends to zero; these are some of the properties associated with aBig Bang.

With the further assumption that the energy density approaches zero at least

61

Conclusions and discussion Generalities

Figure 7.1: Schematic behaviour of solutions to Einstein’s equation under mildassumptions on the matter model.

as fast as 1/a3 when t→∞, we could establish the first order asymptotics, anexponential, for a. This fits well with the leading term in asympotics for H, aconstant, since H = a/a.

These findings are summarized in Figure 7.1.

7.1.1 Geometric consequences

Cosmology is formalized in the language of differential geometry, in which onecan express the general Einstein’s equation. One of the central concepts is thecurvature of the spacetime and by calculating it one can tell, for instance, howfree falling particles behave when acted upon by the gravitational fields. In ourspecific case, the scale factor a determines the curvature, and the results of thisthesis will thus have implications for the geometry of the spacetime and howfree falling particles behave.

When approaching t = 0, whilst going backwards in time, one can interpret theresults as the gravitational fields becoming arbitrarily strong. Hence we have aphysical singularity at that point. This means that a free falling particle willnot have a space-time trajectory that extends infinitely backwards; there willbe a ”beginning” to the trajectory. One way to express this is to say that theparticle ”leaves” the universe in finite time.

On the other hand, a particle in free fall can continue to exist for all time,without ”leaving the universe”, when going forwards in time.

62

Conclusions and discussion Matter models

7.2 Matter models

The matter models under consideration gave rise to asymptotics that were deter-mined in the preceeding chapters. We will here discuss behaviour of the Hubbleconstant, since it relates things that can be calculated from observations.

The first thing that should be noted is that the leading term in both modelsagree with what was found in the more general analysis. We will now examineto what extent the higher order terms agree.

7.2.1 Future asymptotics

In the two preceding chapters, we found that

Hf (t) = H∞ +RmH∞2Λα3

f

e−3H∞t +RradH∞

2Λα4f


and

HV l(t) = H∞ +I0H∞2Λα3

V l

e−3H∞t +I2H∞4Λα5

V l


where subscript V l denotes the Vlasov model and f denotes the fluid modelwith dust and radiation. The constant α arises in the proof of Lemma 4.4.3 andincludes an integral over ρ.

It is easy to see that if I0/α3V l = Rm/α

3f , then two first terms in the asymptotics

are identical. However, if Rrad 6= 0, it holds that

Hf (t)−HV l(t) = O(e−4H∞t

)independent of the values of other parameters. This signifies that Vlasov mattermodels, while emulating dust well, fail to emulate radiation in the limit t→∞.

If one desires better agreement between the models, one could introduce anotherform of matter in the Vlasov model. It is possible that massless Vlasov matterwould introduce the terms needed to achieve this. We have not considered thistype of matter model in the thesis, though it may be a fruitful subject for furtherresearch.

7.2.2 Past asymptotics

The preceding chapters gave

Hf (t) =1

2t+

Ξ

4

1√t

+ o

(1√t

)and

HV l(t) =1

2t+I−1

8√

3I1

+ o(1),

63

Conclusions and discussion Matter models

where Ξ is a model parameter depending on Rm and Rrad. This means that, aslong as Ξ 6= 0, no matter how we choose the values of Rm, Rrad, I1 and I−1, itwill hold that

Hf (t)−HV l(t) = O

(1√t

).

From the definition of Ξ (Definition 5.4.1), we see that Ξ = 0 if and only if theenergy originating from dust is zero, so that Rm = 0. We can interpret this asa failure of the Vlasov model to simulate dust in the limit t→ 0.

It is interesting to consider the relative difference between the models. Since thesolutions diverge faster than the difference, the relative difference will in facttend to zero:

Hf (t)−HV l(t)Hf (t)

= O(√

t).

64

Bibliography

[1] Robert M. Wald, General Relativity, The University of Chicago Press, 1984.

[2] John P. Huchra, Science, New Series, Vol. 256, No. 5055 (Apr. 17, 1992),pp. 321-325,

[3] H. Andreasson, The Einstein–Vlasov System/Kinetic Theory, Liv-ing Rev. Relativity 14, (2011) 4. URL (cited on 2012-05-17):http://www.livingreviews.org/lrr-2011-4

[4] A. D. Rendall, An introduction to the Einstein-Vlasov system,arXiv:gr-qc/9604001v1.

[5] Philip Hartman Ordinary Diffrential Equations, John Wiley & Sons, Inc,1964.

65

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