A Short Course in Modern Cosmology

Lecture notes for Physics 555, University of Washington, Spring 2004

David B. Kaplan

April 2,2004

Abstract

These notes comprise a 15-lecture course introducing graduate students to current theoriesin cosmology. Background in thermodynamics, particle physics, astronomy and general rela-tivity are all helpful, but I do not assume expertise in any of these fields beyond that givenby a good undergraduate education in physics. The goal of the course is to impart some ofthe current excitement in the field, largely due to recent observations of the cosmic microwaveanisotropy, and of distant type IA supernovae. Throughout I will emphasize connections be-tween cosmology and particle physics. A major topic in cosmology omitted in these notes isthe study of galaxy formation.

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Contents

1 The Universe we see 11.1 Visible structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Hubble expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Age of the universe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 The cosmological principle, and the hot big bang theory . . . . . . . . . . . . . 5

2 Cosmology and GR 82.1 The metric and and coordinate transformations . . . . . . . . . . . . . . . . . . 82.2 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Two dimensional isotropic metrics . . . . . . . . . . . . . . . . . . . . . . . . . 112.4 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.5 Two dimensional isotropic and homogeneous metrics . . . . . . . . . . . . . . . 12

3 The Friedmann-Robertson-Walker metric 143.1 Three dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.2 Four dimensions: The Friedmann-Robertson-Walker metric . . . . . . . . . . . 153.3 The redshift of light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.4 Redshift of a thermal spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.5 Measures of distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.5.1 Parallax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.5.2 Angular distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.5.3 Luminosity distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4 The dynamical FRW universe 204.1 The Einstein equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4.1.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.2 Solutions for the equation of state p = wρ . . . . . . . . . . . . . . . . . . . . . 21

5 Thermodynamics 255.1 ρ, p, and n for a thermalized species of particle. . . . . . . . . . . . . . . . . . . 255.2 An inventory of energy in the universe. . . . . . . . . . . . . . . . . . . . . . . . 26

5.2.1 Standard model particles . . . . . . . . . . . . . . . . . . . . . . . . . . 265.2.2 Dark matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5.3 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.4 Matter-radiation equality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

6 Departure from thermal equilibrium 33

7 Nucleosynthesis 37

8 Recombination 41

9 Baryogenesis 42

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10 Inflation 4710.1 Problems with the conventional big bang . . . . . . . . . . . . . . . . . . . . . . 47

10.1.1 The flatness problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4710.1.2 The horizon problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

10.2 How inflation solves the problems . . . . . . . . . . . . . . . . . . . . . . . . . . 4910.3 Scalar field models for inflation . . . . . . . . . . . . . . . . . . . . . . . . . . . 5110.4 Fluctuations in the inflaton field . . . . . . . . . . . . . . . . . . . . . . . . . . 54

10.4.1 Quantum fluctuations of a free field . . . . . . . . . . . . . . . . . . . . 5510.4.2 Quantum fluctuations in the inflaton field . . . . . . . . . . . . . . . . . 56

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1 The Universe we see

Cosmology has become a very exciting science in the past decade or so. The field has beendriven by advances in the quality and quantity of observational data, driven by advances intechnology. An important factor has been the ability to view the universe in frequencies of lightother than optical: gamma-ray, X-ray, infrared, radio and microwave. Since 1987, the field ofneutrino astronomy has developed as well. Not only are we receptive to more messengers fromouter space, but the nature of observation has been changed with the advent of computers,which allow astronomers to process huge quantities of data, using techniques more familiarto high energy particle physics than astronomy. Computers have also allowed highly detailedsimulations of structure formation in the early universe, which offer important clues aboutproperties of the dark matter in the universe.

As a result of all these advances we now have a far more detailed picture of the evolution ofthe universe than was available a couple of decades ago. What is exciting is that this detailedpicture has a lot of mysteries remaining to be understood! That is why cosmology is in agolden age right now – there is lots of data, lots of synthesis, and lots of work to be donebefore we can claim to have a deep understanding of the evolution of the universe.

The aim of this course is to teach first year graduate students about some of the basicfacts about the current paradigm about how the universe has evolved, as well as to introducethem to some of the current outstanding questions. Cosmology is a deeply satisfying sciencesince it brings together so many different topics in physics, such as general relativity, quantumfield theory, classical mechanics of complex systems, hydrodynamics, thermodynamics, non-equilibrium statistical mechanics, all aspects of particle physics, and more. In these lectures,I do not assume that you are masters of all these disciplines, nor do I claim that you will beexperts when the course is over. Rather, I intend to introduce the relevant tools needed fromthese fields in order to answer cosmological questions. If the you want deeper explanations orderivations, you will have to go elsewhere; if I have sustained your curiosity, I will consider thecourse a success.

1.1 Visible structure

Let’s start with a basic inventory of what we see, and what the relevant length scales are fordescribing the universe we see today. Since stars produce light, they are the most obviouscomponent of matter in the universe, even though in fact they only constitute a small fractionof it. Our sun is a typical star, and has a mass M = 2× 1030 kg. There are about 1011 starsin the Milky Way galaxy. Stars in our galaxy have masses in the range 10−1 − 10+1 solarmasses. The nearest stars to the sun are several light-years away, or roughly one parsec —the unit of distance used by astronomers, where

One parsec = 3.261 light-years

The Milky Way is a spiral galaxy with a bulge in the center; it roughly forms a disk which is12.5 kpc in radius (where 1 kpc = 103 parsecs), and 0.3 kpc thick. The sun is situated about2/3 of the way out from the center of the galaxy. At this radius, the galaxy is rotating onceevery 2× 108 years.

The Milky Way is surrounded by small clumps of stars called globular clusters, thoughtto be left over remnants from the creation of the galaxy. They are distributed roughly sym-metrically about the galactic bulge at a distance of 5 − 30 kpc from the center of the galaxy,and they contain roughly 106 stars. About 130 globular clusters are known in the Milky Way.

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The closest galaxy to our own is a small irregular one called the Large Magellanic Cloud(LMC) which is about 50 kpc away from us. The nearest large galaxy is the spiral galaxyAndromeda, which is 770 kpc = 0.77 Mpc (where one Mpc is a “mega-parsec”) away fromus.

Galaxies are seen to form clusters on a scale of several cubic Mpc, while on a scale ofseveral hundred Mpc one sees structures called “super-clusters” of galaxies. An exampleof a super-cluster is the Coma cluster at a distance of 100Mpc, containing 104 galaxies. Onsimilar scales on also sees voids (empty regions), as well as filaments and sheets of galaxies.There is no evidence for structure on scales much longer than 100Mpc. Thus the universelooks isotropic on these large scales.

(See, for example, this image produced by the 2df Galaxy Redshift Survey of over 200,000galaxies).

There is even stronger evidence for isotropy in the microwave spectrum, where one seesa background thermal spectrum corresponding to about a temperature of about 2.7 K. Thesolar system is moving relative to this cosmic microwave background (CMB), as evident froma dipole Doppler shift in the microwave spectrum; when this dipole is removed, the spectrumlooks exceedingly isotropic, to one part in 105. As we will discuss later, these microwaves aretelling us information about the universe about 5 Gpc away, when it was about 100,000 yearsold. Look at images of the sky as seen in microwaves by the COBE satellite, with temperatureresolution scales of ∆T = 4K (isotropic), ∆T = 3.3mK (with evidence of the dipole Dopplershift), and ∆T = 18µK (where evidence for anisotropy appears).

1.2 Hubble expansion

In the 1910’s and 1920’s, Slipher examined the spectra of 41 spiral nebulae and discoveredthe atomic transition lines to be shifted relative to those on earth. We define the redshiftparameter

1 + z =ν1

ν0(1)

where ν1 is the emitted frequency of light (which is known for a particular atomic transition)and ν0 is the frequency of the light detected on earth. A positive value for z implies ν0 < ν1,or that the light received is redder than the light emitted; a negative z implies a blue-shift.Slipher found that his galaxies had redshifts in the range −0.001 < z < +0.006, but with themajority having positive z. This could be interpreted as a Doppler shift, if most of thesenearby nebulae were moving away from us.

The Doppler shift is an important phenomenon, so here I will give a derivation of the effectusing Special Relativity (SR), revisiting it later in the context of General Relativity (GR).

A photon traveling in the n direction, where n is a unit three-vector, , has a 4-momentumof the form pµ = (E, ~p) = hν(1, n), where I have used the fact that the photon is massless, sothat pµpµ = 0, and that a photon’s energy E is related to its frequency ν by E = hν.

Now suppose there is a galaxy at some distance from us along the x-axis, and that it isreceding from us at velocity v in the x direction. It emits a photon with frequency ν1 andmomentum p′µ = (E, px) = hν0(1,−1) (restricting our attention to the time and x-componentsof the four vector) which we receive as a photon of frequency ν0 and four momentum pµ =hν0(1,−1). We can relate the two frequencies by using the Lorentz transformation to relate

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the two momenta, pα = Λ βα p′β, or in matrix notation,(

hν0

−hν0

)=

(γ βγβγ γ

) (hν1

−hν1

), β ≡ v

c, γ ≡ 1√

1− β2. (2)

From this we find

ν0 = ν1γ(1− β) = ν1

√1− β

1 + β. (3)

We can then define the redshift z:

z =ν1

ν0− 1 =

√1 + β

1− β− 1 . (4)

Note that (i) z > 0 for β > 0 (source moving away from us); (ii) for very small β, z ' β; and(iii) for β → 1, we have z →∞.

Wirtz (1918) suggested that galaxies are receding from us, with the furthest galaxies re-ceding the fastest. However, it was Hubble in 1929 who performed a systematic study of theredshift of nearby galaxies, plotting the velocity versus the distance to the galaxy. Here is theoriginal figure from Hubble’s 1929 paper.

Since then data has been collected to much larger distances. The most recent and most dis-tant data comes from observations of type IA supernovae, which serve as very bright standardcandles. A recent compilation of this data is shown in Fig. 1.

The expansion rate observed today is called the Hubble constant H0, and is convention-ally written as

H0 = 100hkm− s−1 −Mpc−1 , h = 0.72± 0.08 . (5)

The present value is much smaller than Hubble’s initial determination, due to an error on hispart in correctly determining the distance to the sources he was looking at.

Note that H0 times a distance equals a velocity. For example, to find the Hubble velocityof the Coma cluster at a distance of 100 Mpc, one multiplies H0 by 100 Mpc to get a velocityof 7 × 103 km/s. This velocity is actually comparable to the relative velocities (“peculiarvelocities”) of the galaxies within the cluster. However, the farther away one looks, the lessimportant are the peculiar velocities compared to the Hubble flow. This motion discovered byHubble is consistent with the effects of an expanding Universe. If one puts dots on a balloonwith radius a(t), then two dots on the balloon at angular separation θ will be separated by adistance s = a(t)θ along the surface of the balloon. If a(t) = eH0t, then it follows that s = H0s,the linear relationship noted by Hubble. A universe expanding in this exponential manner iscalled a de Sitter universe, and Hubble thought he had found evidence for such. In fact, wenow know that the universe is expanding, but with H = a/a being a function of time, insteadof constant as supposed by Hubble.

1.3 Age of the universe

If the universe is expanding, then tracing it back in time would seem to lead to an initialsingularity when it all started — the big bang. Therefore the age of the universe must befinite...what is it?

3

20

21

22

23

24

25

0.01 0.02 0.04 0.114

16

18

20

22

24

26

0.40.2 0.6 1.0

0.40.2 0.6 1.0

mag

nit

ud

e

redshift

Type Ia Supernovae

Calan/Tololo

Supernova Survey

High-Z Supernova SearchSupernova Cosmology Project

fain

ter

DeceleratingUniverse

AcceleratingUniverse

without vacuum energywith vacuum energy

empty

mas

sde

nsity

0

1

Perlmutter, Physics Today (2003)

0.1

1

0.01

0.001

0.0001

Rel

ativ

e b

rig

htn

ess

0.70.8 0.6 0.5

Scale of the Universe[relative to today's scale]

Figure 1: Redshift/luminosity plot for distant receding type 1A supernovae

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H0 has dimensions 1/time, so the age as determined from the expansion will be T ∼ 1/H0 ∼1010 years (to compute this from eq. (5), it is convenient to remember that there are ∼ 3×107

seconds in a year). Other age determinations exist:

1. From dating the earth’s crust using radioactive decay, one finds it to be at least 3.7×109

years old.

2. Dating the age of heavy elements, one finds that they were created (perhaps in a super-nova) between 8× 109 and 15× 109 years ago.

3. One can also estimate the ages of globular clusters. One can measure the color (andhence the temperature) and luminosity of stars, constructing the Hertzsprung-Russelldiagram. Most stars lie on a diagonal strip called the “main sequence”. Two otherpopulations that exist are white dwarfs (hot and dim) and the red giants (cool andbright). Stars typically evolve along the main sequence, eventually departing to form redgiants or white dwarfs. The time spent on the main sequence is

tm.s. ∼ 1.1× 1010 M

M

L

L(6)

where M and L are the mass and luminosity respectively, empirically related by

L

L∼

(M/M)4 L > L

(M/M)2.8 L < L(7)

Therefore, if one looks at a globular cluster and identifies the stars just leaving the mainsequence, then their luminosity may be used to determine their age; assuming that thestars in the globular cluster all formed at the same time, then on has determined the ageof the cluster. One finds in this manner globular clusters whose ages are 14.9± 3.5× 109

years.

It used to be rather embarrassing that the age of globular clusters kept on coming outolder than the age of the universe as determined from its expansion. The recent evidence thatthe universe is accelerating solves that problem, as an accelerating universe has been aroundlonger than the current expansion rate would suggest, as the expansion used to be slower inthe past. The present estimate for the age of the universe is about 14× 109 years.

1.4 The cosmological principle, and the hot big bang theory

Clearly, on “short” distance scales (< 100 Mpc) the universe is very complicated. We haveseen, though, that at longer distance scales it looks increasingly isotropic. Therefore it isplausible that the early history, and present-day large scale structure of the universe couldbe described by a simple model based one what is referred to as the “cosmological principle”,which entails three assumptions:

1. The universe is isotropic (observationally true on large distance scales);

2. The universe is homogeneous (this is not easily testable, but is equivalent to the Coper-nican idea that we do not occupy a special point in the universe, so that if the universeis isotropic about us, it is isotropic about every point in space; this in turn impliesinvariance under translations);

5

+5

Electroweakphase transition begins

Nucleosynthesis Today

Quarkconfinement

Recombinationof hydrogen

Galaxy formation

Time

1 eV1 MeV

1 sec.

1 GeV

10 sec.−1010 sec. 10 yr. +1010 yr.

100 GeV Temperature−4

10 eV

−6

Figure 2: A partial time-line of the big bang

3. The large scale dynamics of cosmological evolution are described by general relativity(GR has been tested to some degree locally, but we do not have any direct experimentalinformation to support the idea that it is the correct theory over very large distances).

Most of the course will be involved in examining the hot big bang theory following fromthese assumptions. We will end up with the “concordance model”, which is an model for theevolution of the universe which seems to fit all present data. One of the exciting features ofcosmology as a field to work in is that the concordance model is clearly incomplete invokingingredients which have yet to be directly discovered or understood (eg, dark matter, darkenergy and the existence of an “inflationary epoch”). Yet at the same time, the amount ofdata it can explain satisfactorily is quite extensive.

The basic idea of the hot big bang theory — which we will revise extensively before arrivingat the concordance model – is that the universe was once very small and in thermal equilibriumand hot. We can trace its thermal history back in time with some degree of certainty to theepoch when its temperature was about 100 GeV 1, since we have probed energies up to the100 GeV scale in particle accelerators. Some of the epochs we can identify are shown in thetime-line pictured in Fig. 2. We will learn how temperature and time are related, why theseepochs occur where they appear on the time-line.

Furthermore, we will see that for the later epochs, there is strong observational evidencesupporting the big bang theory. I will just sketch here the most dramatic of these pieces ofevidence: the cosmic microwave background (CMB). Since the ionization energy of hydrogenis 13.6 eV, by the time the temperature of the universe is down to 1 eV, most of the electronsand protons in the universe have combined to form neutral hydrogen. As a result, the uni-verse, which was a very opaque plasma, becomes transparent to light. The universe thereforeconsists of neutral hydrogen, and a black body distribution of photons corresponding to atemperature of T = 1 eV, which travel from then on without interacting with anything (toa good approximation). This occurs when the universe is about 105 years old. As we willsee, GR predicts that from that epoch to today, the universe expands at a rate R(t) ∝ t2/3.Now, the number density of photons per “co-moving volume” doesn’t change as the volumeexpands; rather, the number density dilutes: N = nR3 = constant, implying that n ∝ t−2.Therefore the number density today n0 is related to the number density n1 at recombinationby n0 = n1(t1/t0)2 = 10−10n1. The photons were originally in a black body distribution, andwe will later show that they retain such a distribution even as the universe expands, although

1I will usually set Boltzmann’s constant kB to unity, as well as the speed of light c and Planck’s constant ~. ThuskBT = T = energy, which will be quoted in electron-volts (eV). For reference, the relation between eV and degreesKelvin (K) is roughly 1eV ' 104K.

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at a cooler temperature. The number density of blackbody photons at temperature T is givenby the Planck formula to be ∝ T 3. Therefore the temperature of photons today (T0) is relatedthe the temperature at recombination T1 = 1 eV by

T0 = T1(t1/t0)2/3 ' 1 eV× (105 years/1010 years)2/3 ' 1/3× 10−3 eV ∼ 3K .

So the basic prediction of the big bang theory is that we are bathed today in 3K photons. Thiswas observed by Penzias and Wilson in the 1960’s, and received spectacular confirmation bythe COBE satellite.

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2 Cosmology and GR

The first step toward a cosmological theory, following what we called the “cosmological prin-ciple” is to implement the assumptions of isotropy and homogeneity withing the context ofgeneral relativity (GR). I will assume that you are familiar with special relativity; some ac-quaintance with GR would be helpful but not necessary for this course. Good introductionscan be found in the recent book Gravity: An Introduction to Einstein’s General Relativity byJames Hartle, or at Sean Caroll’s web site.

General Relativity (GR) is the theory that controls the large scale evolution of the universe.There are two parts to the theory, which are generalizations the Poisson equation and Newton’slaw:

∇2φ = 4πGρ , (8)a = −∇φ . (9)

The first of the above equations tells us how a mass distribution ρ gives rise to a gravitationalpotential φ, while the second tells us how a particle accelerates in the resulting potential. InGR, the Poisson equation eq. (8) is replaced by Einstein’s equation

Gµν = 8πGNTµν , (10)

where Gµν on the left hand side is called the Einstein tensor, and which describes the geometryof spacetime, while Tµν on the right hand side is the energy-momentum tensor, describing thedistribution of energy and momentum. Tµν acts as a source for spacetime curvature, just as ρserves as a source for the gravitational potential φ in Poisson’s equation eq. (8).

Given a particular geometry, the trajectory that particles follow is given by a geodesic,which is path of shortest distance between two points, with the concept of distance suitablydefined. Therefore the geodesic equation replaces Newton’s law, eq. (9). I will begin by brieflydiscussing differential geometry and geodesics, then turning to Einstein’s equations and someof their solutions.

2.1 The metric and and coordinate transformations

Consider some manifold in d dimensions, such as a curved two-dimensional surface, or fourdimensional Minkowski space. We can refer to points on the surface by labeling them eachwith a set of d coordinates xµ where µ = 1, . . . , d. The geometry of the manifold can bespecified in terms of a metric gµν which allows us to compute the distance ds between thepoint at xµ and one nearby at xµ + dxµ, where

ds2 = gµνdxµdxν . (11)

In general, gµν is a function of xµ; it is symmetric, gµν = gνµ. As is the convention in specialrelativity, an upper and a lower index which are identical are assumed to be summed over(“contracted”).

If we change coordinate systems from xµ to xµ, then in terms of the new coordinates

ds2 = gµν

(∂xµ

∂xα

) (∂xν

∂xβ

)dxαdxβ = gαβdx

αdxβ , (12)

so we see that in the new coordinate system, the metric is given by

gαβ = gµν∂xµ

∂xα∂xν

∂xβ(13)

8

If we think of gµν as a matrix, then we can write the above transformation as a matrix equation

g(x) → g(x) = ΛT (x)g(x(x))Λ(x) , Λαβ ≡∂xα

∂xβ. (14)

We can also define the inverse of the metric, writing it with upper indices:

gαβgβγ ≡ gαγ =

1 α = γ,

0 otherwise.. (15)

In general, tensors are objects with upper and/or lower indices that behave in the followingway under coordinate transformations x → x: for each lower index, the tensor T ······α··· getsmultiplied by dxα

dxβwith the α’s contracted; while for each upper index, the tensor T ···α······ gets

multiplied bydxβ

dxα with the α’s contracted. The metric gµν , its inverse gµν and the partial derivatives∂µ ≡ ∂

∂xµ are examples of tensors; the coordinate xµ is not a tensor, but the differential dxµ is.Indices on tensors can be raised and lowered by means of the metric tensor. A string of

tensors with some indices contracted is also a tensor. A tensor with no indices, or equivalentlya string of tensors with all their indices contracted is a scalar, invariant under all coordinatetransformations.

Here are some simple examples of metrics and coordinates in two dimensions:

plane (Cartesian) plane (polar) 2-sphere

xµ x, y r, θ θ, φ

ds2 dx2 + dy2 dr2 + r2dθ2 a2(dθ2 + sin2 θdφ2

)gµν

(1

1

) (1

r2

)a2

(1

sin2 θ

)

gµν(

11

) (1

r−2

)a−2

(1

sin−2 θ

)xµ x, y r, r2θ a2θ, a2φ sin2 θ

(16)

One can easily arrive at the above metric in polar coordinates by making the change ofvariables

x = r cos θ , y = r sin θ (17)

and following the prescription in eq. (14). A more amusing change of variables is to take thesphere, expressed in terms of polar and azimuthal angles θ, φ, and to define ρ = sin θ. Thendρ = cos θdθ, or dθ2 = dρ2/(1− ρ2). Thus in the new variables ρ, φ we have

ds2 = a2

(dρ2

1− ρ2+ ρ2dφ2

). (sphere) (18)

This is still the metric for a sphere. Note that it looks similar to the metric for a plane withr ≡ aρ.

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2.2 Geodesics

Given a metric, one can compute the shortest path between two points; such paths are calledgeodesics. It is interesting to learn to compute them for two reasons: (i) objects moving ina gravitational field follow geodesics; (ii) computing geodesics is an efficient way to computeChristoffel symbols, an important object in differential geometry out of which on constructscurvature.

The equation for the geodesics can be simply derived using variational calculus. Weparametrize a path xµ(λ) by the parameter λ with, for example, ξµ(0) = xµi being the startof the path and xµ(1) = xµf being the end of the path. Then we wish to stationarize the pathlength L:

0 = δL = δ

∫ xµf

xµi

ds = δ

∫ 1

0

√d2s

dλ2dλ = δ

∫ 1

0

√gµν

dxµ

dλ

dxν

dλdλ . (19)

We can make our task slightly easier by noting that the path that extremizes the integral of√ds2/d2λ also extremizes the integral of 1

2 d2s/dλ2, so we solve

0 = δ

∫ 1

0

(12 gµν

dxµ

dλ

dxν

dλ

)dλ , (20)

which yields the “Lagrangian”

L = gµν(x)xµxν (21)

which obeys the Euler Lagrange equation (where gµν,α ≡ ∂gµν/∂xα)

0 =d

dλ

∂L

∂xµ− ∂L

∂xµ

=d

dλ

(gαν

dxν

dλ

)− 1

2gµν,αdxµ

dλ

dxν

dλ

= gαν,βdxβ

dλ

dxν

dλ+ gαν

d2xν

dλ2− 1

2gµν,αdxµ

dλ

dxν

dλ, (22)

where gµν,α ≡ ∂gµν/∂xα. By renaming dummy (contracted) indices, the above expressionmay be rewritten in the form

0 =d2xα

dλ2+ Γαµν

dxµ

dλ

dxν

dλ, (23)

Γαµν ≡ 12gαβ (gβµ,ν + gβµ,ν − gµν,β) . (24)

Γαµν is called a Christoffel symbol; it is symmetric in its lower indices,

Γαβγ = Γαγβ . (25)

One can show that under coordinate transformations iΓ transforms inhomogeneously, and istherefore not a tensor. Often the simplest way to compute the Christoffel symbols is not to usethe formula eq. (24), but rather the variational equation eq. (20) in conjunction with eq. (23).

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2.3 Two dimensional isotropic metrics

As an example it is interesting to compute the Christoffel symbols for the most general isotropicmetric in two dimensions. We take as our coordinates r, θ, in which case the most generalline element is

ds2 = A(r, θ)dr2 +B(r, θ)dθ2 + C(r, θ)dθdr (26)

By making a change of variables, it is possible to cancel the mixed term proportional to C(r, θ).If we now insist that the metric be isotropic (independent of θ) then the line element takes theform

ds2 = A(r)dr2 + B(r)dθ2 . (27)

I will assume that A and B are positive functions. Then we can make the change of variablesa2ρ2 = B(r), where a is a number of dimension length. We arrive at element

ds2 = h(ρ)dρ2 + ρ2dθ2 , (28)

where h(ρ) is an arbitrary, dimensionless function. Since we do not want a conical singularityat the origin (if you roll a piece of paper into a cone, the tip of the cone is called a conicalsingularity) it follows that for a very small circle about the origin, we had better recover theflat space relation between the circumference C and the radius R of the circle: C = 2πR.With the above metric we find that for the circle dρ, 0 ≤ θ < 2π, the length of the radiusis R = h(0)dρ, while the length of the circumference is C = 2πdρ. So we have the additionalconstraint

h(0) = 1 . (29)

This is the most general isotropic metric in two dimensions.Now let us compute the Christoffel symbols. From ds2 we ‘construct the “Lagrangian”

L =(h(ρ)ρ2 + ρ2θ2

)(30)

with the Euler-Lagrange equations

0 =d

dλ(2hρ)−

(h′ρ2 + 2ρθ2

),

0 =d

dλ

(2ρ2θ

), (31)

which can be rewritten as

ρ+h′

2hρ2 − ρ

hθ2 = 0 , (32)

(33)

θ +2ρρθ = 0 . (34)

By comparing eq. (34) with eq. (23), we find that the nonzero Christoffel symbols for thismetric eq. (28) are just

Γ111 =

h′

2h, Γ1

22 = −ρh, Γ2

12 = Γ221 =

1ρ. (35)

11

2.4 Curvature

Next let us ask what the most general isotropic, homogeneous metric is in two dimensions. Tosee if the metric describes a homogeneous space, we need to define something that is supposedto remain constant over the manifold. The quantity should not be coordinate dependent, orwhat looks constant in in set of coordinates does not in another, and by “homogeneity” we referto an intrinsic property of the space independent of coordinate choice. The scalar curvaturesuch a quantity.

Suppose you started walking East along the earth’s equator holding an arrow in you handpointed East. After walking a quarter of the way around the world, you turn at right anglesheading North, but you are careful to not turn the arrow; as you march North, the arrow ispointing at 90 to your right.. When you reach the North Pole, you make a 90 left turn andhead South, again careful not to rotate the arrow, which now points directly behind you asyou proceed back to the equator. When you reach the equator, you are back at your startingpoint; however the arrow which you were careful to never rotate relative to your path (paralleltransport) is now pointing North, whereas it was pointing East when you began your journey.This rotation of the arrow is a result of the curvature of the Earth; nothing analogous occurswhen tracing a triangular path on a flat surface.

By considering infinitesimal closed paths on a manifold and the effect on the orientation ofarrows (vectors) one can make precise a definition of curvature. One arrives at the definitionof three useful and related tensors for describing curvature, The Riemann tensor, the Riccitensor, the Ricci scalar, and the Einstein tensor. Their definitions are

Rαβγδ =∂Γαβδ∂xγ

−∂Γαβγ∂xδ

+ ΓαγεΓεβδ − ΓαδεΓ

εβγ Riemann tensor

Rαβ = Rγαγβ =∂Γγαβ∂xγ

− ∂Γγαγ∂xβ

+ ΓγαβΓδγδ − ΓγαδΓ

δβγ Ricci tensor

R = Rαα Ricci scalar

(36)

The Riemann tensor has some simplifying symmetries:

• Antisymmetry: Rαβγδ = −Rβαγδ = −Rαβδγ ;• Symmetry: Rαβγδ = Rγδαβ ;

• A cyclic property: Rαβγδ +Rαγδβ +Rαδβγ = 0 .

As a result of these symmetries, one finds that the Riemann tensor in d dimensions has ingeneral Cd = d2(d2 − 1)/12 independent components. For d = 2, 3, 4 dimensions we findC2 = 1, C3 = 6 and C4 = 10. Metrics with special symmetry can have far fewer independentcomponents.

The Ricci scalar R is something that must be constant everywhere on our two dimensionalmanifold if it is to describe a homogeneous space.

2.5 Two dimensional isotropic and homogeneous metrics

So as a warm-up for four dimensional cosmology, let’s compute the Ricci scalar for the isotropicmetric eq. (28), demand it be constant, and determine the allowed functions h(ρ) in themetric eq. (28). Since we are in two dimensions, there is only one independent component of

12

the Riemann tensor. For the two-dimensional isotropic metric with non-vanishing Christoffelsymbols given by eq. (35), the Riemann tensor is given by

R1212 = ∂1Γ1

22 − ∂2Γ121 + Γ1

11Γ122 − Γ1

22Γ221 =

ρh′

2h2. (37)

Then, using the symmetries of the Riemann tensor,

R2121 = g22g11R

1212 =

h′

2ρh. (38)

It follows that the Ricci tensor is

R11 = R2121 =

h′

2ρh, R22 = R1

212 =ρh′

2h2, R12 = R21 = 0 , (39)

or

Rαβ =h′

2ρh2

(h

ρ2

)=

h′

2ρh2gαβ . (40)

Finally this implies that the Ricci scalar is given by

R = Rαα = gαβRαβ =h′

ρh2. (41)

Requiring that R equal a constant c everywhere gives us the differential equation for thefunction h(ρ) : h′/h2 = cρ, with solution

h =1

1− cρ2/2, (42)

where I have implemented the boundary condition on h, eq. (29). We can now rescale ρ →√2/|c|ρ, to obtain the metric

ds2 = a2

(dρ2

1− kρ2+ ρ2dθ2

), (43)

with k = c/|c| = 0,±1, and a2 = 2/|c|. The parameter a has dimension of length and iscalled the scale factor; the curvature R = c is given by R = 2k/a2. Note that the sign of thecurvature appears as the parameter k: a flat space corresponds to k = 0, and the metric isjust that for the plane in polar coordinates; if k = 1 we have constant positive curvature andget the metric for the sphere that we derived in eq. (18). The case k = −1, a surface withconstant negative curvature, is the surface AdS2, two dimensional anti-de Sitter space. It canbe described as a surface embedded in three dimensional Minkowski space, namely the surfacesatisfying

z2 − x2 − y2 = 0 (44)

in a three dimensional space with line element

ds2 = dx2 + dy2 − dz2 . (45)

As we will see, the metric eq. (43) is similar to what we will encounter when describing theisotropic, homogeneous universe.

13

3 The Friedmann-Robertson-Walker metric

3.1 Three dimensions

The most general isotropic and homogeneous metric in three dimensions is similar to the twodimensional result of eq. (43):

ds2 = a2

(dr2

1− kr2+ r2dΩ2

), dΩ2 = dθ2 + sin2 θdφ2 , k = 0,±1 . (46)

The angles φ and θ are the usual azimuthal and polar angles of spherical coordinates, withθ ∈ [0, π], φ ∈ [0, 2π). As before, the parameter k can take on three different values: for k = 0,the above line element describes ordinary flat space in spherical coordinates; k = 1 yieldsthe metric for S3, with constant positive curvature, while k = −1 is AdS3 and has constantnegative curvature. As in the two dimensional case, the change of variables r = sinχ (k = 1) orr = sinhχ (k = −1) makes the global nature of these manifolds more apparent. For example,for the k = 1 case, after defining r = sinχ, the line element becomes

ds2 = a2(dχ2 + sin2 χdθ2 + sin2 χ sin2 θdφ2

). (47)

This is equivalent to writing

ds2 = dX2 + dY 2 + dZ2 + dW 2 , (48)

where

X = a sinχ sin θ cosφ ,Y = a sinχ sin θ sinφ ,Z = a sinχ cos θ ,W = a cosχ , (49)

which satisfy X2 + Y 2 + Z2 + W 2 = a2. So we see that the k = 1 metric corresponds toa 3-sphere of radius a embedded in 4-dimensional Euclidean space. One also sees a problemwith the r = sinχ coordinate: it does not cover the whole sphere. In going from the northpole (χ = 0) to the equator (χ = π/2), the variable r ranges from r = 0 to r = 1; however, inproceeding from the equator to the south pole of the sphere (χ = π), r sinχ runs back fromr = 1 to r = 0. So the full space has coordinates which are not single valued in r.

The proper distance dbetween the origin and an object at radial coordinate r is given by

d =∫ r

0a

dr′√1− kr′2

dr′ = a×

sin−1 r k = 1r k = 0sinh−1 k = −1

(50)

For the case k = 1 we see that d is not defined for r > 1, and that for r = 1, d = aπ/2. Thisis the distance along the sphere from the north pole to the equator; to compute the distancefrom the north pole to the south pole in these coordinates, you need to double the result, toget d = aπ.

14

3.2 Four dimensions: The Friedmann-Robertson-Walker met-ric

It is simple to go to the case of interest: four dimensional spacetime. Because of the homogene-ity, we can choose the same time coordinate for each point in space, and at each time slice, wemust have the isotropic and homogeneous three dimensional metric eq. (46). However, thereis no constraint relating the scale factor a at different time slices, which can therefore be afunction of time. Aside from isotropy and homogeneity, general relativity requires that locally(eg, near the origin) the line element be invariant under Lorentz transformations:

ds2 = dt2 − d~x2 . (51)

Thus we arrive at the Friedmann-Robertson-Walker (FRW) metric, which is the most generalmetric (up to coordinate transformations) fulfilling the cosmological principle:

ds2 = dt2 − a(t)2(

dr2√1− kr2

+ r2dΩ2

). (52)

These coordinates t, r, θ, φ are called co-moving coordinates. The reason is becausetwo objects at different spatial coordinates can remain at those coordinates at all times, whilethe proper distance between them changes with time according to how the scale factor a(t)changes with time. Picture to dots on a balloon whose coordinates are fixed, while the radiusof the balloon changes with time. That the proper distance will change with time is evident— one needs merely to replace a in the three dimensional example eq. (50) by a(t). However,one needs to show that in fact an object can remain at rest at fixed spatial coordinate. Theworld line of such an object, parametrized by λ = t, satisfies

dt

dλ= 1 ,

dxi

dλ= 0 , (53)

and we have to show that this is indeed a geodesic for our metric. Since the geodesic equationeq. (23) is

0 =d2xα

dλ2+ Γαµν

dxµ

dλ

dxν

dλ, (54)

it follows that the world line eq. (53) satisfies this equation provided that Γαtt vanishes. However,that is evident if you think of how the geodesic equation is derived from the metric by theEuler-Lagrange equations: since the coefficient of dt2 in eq. (52) is coordinate independent,there can be no terms proportional to (dt/dλ)2 in the geodesic equation.

3.3 The redshift of light

Now I will demonstrate that light gets red-shifted in the FRW universe due to the time de-pendent scale factor. Suppose there is a galaxy at rest in the co-moving coordinates, withradial coordinate r1, and we are at the origin. Light is emitted by this galaxy at time t1 withfrequency ν1 and received on earth at time t0 with frequency ν0. We wish to determine therelation between ν0 and ν1.

Light travels on ‘null geodesics — that is, on geodesics with ds = 0. Note that in flatspace where ds2 = dt2−d~x2, ds = 0 simply means that the speed of light is c = 1. However, it

15

is possible to choose locally flat coordinates about any point in curved space, so having lighttravel on null geodesics means that any observer along its path sees it go by with velocityc = 1. Thus along a radial light path in the FRW universe,

dt

a(t)=

dr√1− kr2

. (55)

Now consider the emission of two subsequent crests of a light wave:

crest #1 t r

emitted: t1 r1received: t0 0

crest #2 t r

emitted: t1 + δt1 r1received: t0 + δt0 0

(56)

It follows from eq. (55) that∫ t0

t1

dt

a(t)=

∫ t0+δt0

t1+δt1

dt

a(t)=

∫ 0

r1

dr√1− kr2

. (57)

Subtracting the first integral from the second, and assuming δt0,1 aa , we get

δt0a(t0)

=δt1a(t1)

(58)

or, since δt0,1 = 1/ν0,1, the redshift z is given by

1 + z =ν1

ν0=a(t0)a(t1)

. (59)

Note that the redshift only depends on the ratio of the scale factor at reception to the scalefactor at emission. It is not simply a function of the relative motion of the source at the timeof emission; in fact if the universe were to contract for a while, and then expand, it is possiblefor a source which is approaching us to emit light which we see as shifted to the red.

It is useful to consider eq. (59) in terms of the wavelength λ instead of the frequency ν:

1 + z =λ0

λ1=a(t0)a(t1)

. (60)

We see that the wavelength of light just contracts and stretches with the scale factor, and thatfact explains the whole redshift phenomenon. In this view it is not much like the Doppler shiftdescribed in §1.2 in the context of special relativity.

3.4 Redshift of a thermal spectrum

Suppose we view a thermal source with temperature T1 with absolute luminosity L. The poweremitted per unit area per unit frequency is given by the Planck formula

Sν =2πhc2

ν31

ehν1/kT1 − 1. (61)

Integrating Sν over frequency ν1 and multiplying by the area A of the source gives

L = Aπ4

15h4(kT )4 , (62)

16

so the total energy emitted by the object per time dt1 in a frequency interval [ν1, ν1 + dν1] is

dE1 = ASνdν1dt1 = L15π4

(hν1

kT1

)3 1ehν1/kT1 − 1

(h

kT1

)dν1 dt1 . (63)

The energy received on earth per unit area in a time dt0 in a frequency interval [ν0, ν0 + dν0]is then obtained from dE1 above in the following way:

1. One must divide by the area of the spherical shell of light, when received; the relevantpart of the metric is ds2 = a(t)2r21dΩ

2, and so the area of the light pulse is (4πa20r

21),

where a0 = a(t0).

2. One must account for the fact that the photons emitted with frequency ν1 are receivedwith frequency ν0 = ν1/(1 + z) and have correspondingly lower energy

3. We have also seen that photons emitted over a time dt1 will be received over a timeinterval dt0 = dt1(1 + z).

Therefore the energy we receive on earth per unit area is

dE0 =(

11 + z

)L

4π(a0r1)215π4

(hν1

kT1

)3 1ehν1/kT1 − 1

(h

kT1

)dν0 dt0

=(

11 + z

)2 L

4π(a0r1)215π4

(hν0

kT0

)3 1ehν1/kT0 − 1

(h

kT0

)dν0 dt0

= `15π4

(hν0

kT0

)3 1ehν1/kT0 − 1

(h

kT0

)dν0 dt0 (64)

where we have defined the apparent luminosity

` =(

11 + z

)2 L

4π(a0r1)2(65)

and

T0 ≡T1

1 + z. (66)

Note that the expressions for dE0dt0dν0

and dE1dt1dν1

take the same functional form, except for theoverall normalization, the substitution of ν0 for ν1, and of T0 for T1. This means that thespectrum observed still looks like blackbody radiation, but with a temperature red-shifted bya factor of 1/(1 + z).

3.5 Measures of distance

To observationally determine the parameters of the FRW metric, k and a(t), one important toolwill be to compare the redshift of an object with its distance. However, while it is theoreticallystraightforward to compute the proper distance to an object∫ r

0

dr′√1− kr′2

, (67)

that is not a quantity readily measured. Instead one usually measures distances in one of thethree following ways.

17

d

b

2θ

Figure 3: Determining a distance by parallax

3.5.1 Parallax

For relatively nearby objects (. 50kpc) one can use parallax to measure the distance of anobject. The idea is to look at the same object from opposite sides of the earth’s orbit about thesun, and record the angle 2θ by which the orientation of the telescope must be shifted betweenthe two observations. If the radius of the earth’s orbit projected onto the plane perpendicularto the starlight propagation x1is b, then in Euclidean space, the distance to a distant object atr = r1 is given by d = b/θ (see Fig. 3). Therefore one defines the parallax distance dp = b/θ.In the FRW space one can show that

dp = a(t0)r1√

1− kr21. (68)

This looks rather strange; for example, with k = r1 = 1, this distance diverges, even thoughwe know that the k = 1 space is compact, and that the greatest [proper distance one can getfrom any other point in that space is aπ! However, the above result makes sense. Recall thatfor k = 1, the spatial slices of our space are 3-spheres. If the star is at the north pole, and weare observing from two point on the equator, then our telescope always points due north, andθ = 0. It follows that dp = ∞ even though the star is a finite proper distance away. I will notderive the above result, but you can find it in Weinberg’s gravitation book, and other sources.

3.5.2 Angular distance

Another way to measure distance is to measure the angular extension of a distant object whosesize is known, as in Fig. 4. If the object has diameter D, then a suitable definition of angulardistance for distant objects is dA = D/θ. In the FRW metric, D is the proper width of theobject at the time of the emission of light. Using ds2 = . . . − a(t)2r2dθ2 − . . ., this gives us

D θ

Figure 4: Determining a distance by angular extension

18

D = a(t1)r1θ, for an object of angular extension θ emitting light at time t1, so that

dA = r1a(t1) . (69)

3.5.3 Luminosity distance

Suppose we have a source whose absolute luminosity L is known, defined as the total energyoutput of the source per unit time. If we draw an imaginary sphere about the source inter-secting our position, then in Minkowski space, the power received on earth is ` = L/A, whereA = 4πd2, d being the distance to the source. Therefore a sensible definition of luminositydistance is

dL =

√L

4π`. (70)

With the FRW metric, we use the apparent luminosity for `, given in eq. (65). As a resultthe luminosity distance is given by

dL = r1a(t0)(1 + z) (71)

19

4 The dynamical FRW universe

4.1 The Einstein equations

Einstein’s equations

Gµν = 8πGNTµν (72)

relate the expansion rate a(t) to energy distribution in the universe. On the left hand side isthe Einstein tensor which can be determined from the metric, and is defined as

Gµν = Rµν − 12Rgµν . (73)

where Rµν and R are the Ricci tensor and Ricci scalar respectively. On the right side of theequation is Newton’s constant GN and the energy momentum tensor of matter, Tµν . Thecomponent T00 tells us about the energy density, while Tij is related to the flow of material inthe j direction with nonzero momentum pointing in the i direction.

For the FRW metric one finds

R00 = −3a

a, Rij =

[a

a+ 2

(a

a

)2

+ 2k

a2

]gij , R0i = 0 , (74)

where gij is the spatial part of the FRW metric. Noting that gij = δij , it is easy to constructthe Einstein tensor with one upper and one lower index,

G00 = 3

[(a

a

)2

+k

a2

]

Gij =

[2a

a+

(a

a

)2

+k

a2

]δij . (75)

Our isotropy and homogeneity assumptions for the metric are justified if the energy-momentum tensor Tµν has a similar structure: independent of spatial coordinates, and withT ij ∝ δij with T 0

i = T i0 = 0. Throughout this course we will be treating matter to be close toan ideal fluid with energy density ρ and pressure p, for which

Tµν =

ρ−p

−p−p

(76)

Einstein’s equations then read: (a

a

)2

+k

a2=

8πGN3

ρ , (77)

2a

a+

(a

a

)2

+k

a2= −8πGNp . (78)

It is useful to manipulate the equations to simplify the second one. By subtracting eq. (77)from eq. (78) we get an expression fro the cosmic acceleration:

a

a= −4πGN

3(ρ+ 3p) . (79)

20

It may seem a little disturbing to find two differential equations to solve for one function,a(t). However, we can manipulate our equations again to show that one combination of thetwo Einstein equations is simply equivalent to energy conservation. If we differentiate eq. (77)with respect to time, and then make use of both equations we find

8πGN3

ρ =(a

a

) [2a

a− 2

(a

a

)2

− 2k

a2

]= −8πGN

(a

a

)(ρ+ p) , (80)

or

ρ = −3(a

a

)(ρ+ p) . (81)

Note that this equation has no factor of GN , and so is not telling us about gravity. In fact itcan be rewritten as

d(a3ρ

)dt

= −pda3

dt(82)

which is the energy conservation equation

dE = −p dV (83)

since a small cube of co-moving space has proper volume dV = a3(r2 sin θ/√

1− kr2)drdθdφ.For most applications it is convenient to use the Friedmann equation eq. (77) as our first

equation, and then either the equation for the acceleration eq. (79) or for energy conservationeq. (81) as our second independent equation.

4.1.1 Summary

If we define

H =(a

a

), (84)

then our two equations to solve are:

H2 + ka2 = 8πGN

3 ρ , Friedmann equation

ρ = −3H(ρ+ p) , Energy conservation(85)

4.2 Solutions for the equation of state p = wρ

To proceed, we need to know the equation of state which relates ρ and p. We will assume fornow that the matter of the universe is a single material which satisfies

p = wρ . (86)

Various cases of particular interest are:

• Non-relativistic, pressureless matter: w = 0. If the universe is dominated by a dilute gasof non-relativistic particles, or a less dilute gas of weakly interacting particles, then itspressure is negligible, and we have w = 0.

21

• Highly relativistic particles: w = 1/3. When the universe is dominated by particleswhose energy is so much greater than their mass that they can be treated as massless,then one finds w = 1/3. Note that for this special value, the energy momentum tensoreq. (76) is traceless, Tµµ = 0. There is a reason for this: if all your particles are massless,then they move at the speed of light and there is no special frame (unlike for massiveparticles, where there is a rest frame). But Tµµ has dimensions of mass, and is invariantunder Lorentz transformations. The only dimensionful quantity around is the energy ofthe particles, but that is not Lorentz invariant, so Tµµ must vanish.

• Cosmological constant (vacuum energy): w = −1. In non-gravitational physics one canalways add a constant to the Lagrangian without consequence. It shifts the value of theenergy, but since only energy differences are measured, there is no innate definition of zeroenergy. However, gravity couples to energy, and shifting the Lagrangian by a constantshifts Tµν . The constant term is called the cosmological constant; it is a divergent quantityin quantum field theory, and so naively one would think it should be very big, but as wewill see, observations of our cosmos tell us it is very small. Since adding a constant to theLagrangian is Lorentz invariant, it follows that the change in Tµν must be proportionalto gµν −δµν . Thus it must be that such a form of “matter” must have −p = ρ, or w = −1.Note that when w < 0, the pressure is negative, which implies tension...the materialwould want to contract in the absence of gravity.

• Matter with w = −1/3. Although there is no reason to expect matter with w = −1/3, itis interesting to note from eq. (79) that for w < −1/3, the expansion of the universe isaccelerating (a > 0) while for w > −1/3, it is decelerating (a < 0).

From the equation for energy conservation eq. (85) we find ρ = −3H(1 + w)ρ or

dρ

ρ= −3(1 + w)

da

a(87)

with solution

ρ ∝ a−3(1+w) . (88)

Note that non-relativistic matter has ρ ∝ a−3, which accounts for dilution of the particlesdues to the expansion of the volume they are in; for relativistic matter, ρ ∝ a−4, where theextra factor of a−1 accounts for the redshift of the energy; and a cosmological constant has anenergy density that remains constant as the universe expands, where the increase of energycomes from the universe expanding against the negative pressure.

Next we look at the Friedmann equation, eq. (85). A static solution is possible with H = 0provided that k 6= 0 and

k

a2=

8πGN3

ρ . (89)

This solution appealed to Einstein on philosophical grounds before he learned of Hubble’sdiscovery of cosmological expansion. It is not a stable configuration — small perturbationsabout this solution cause it to collapse or expand.

If H 6= 0 then we can rewrite the Friedmann equation as

k

a2= H2 (Ω− 1) , (90)

22

where

Ω(t) =ρ(t)ρcrit(t)

, ρcrit(t) ≡3H2(t)8πGN

(91)

ρc ≡ ρcrit(t0) =3H2

0

8πGN= 1.88h2 × 10−29 gm cm−3 = 1.1× 10−5h2 GeV cm−3 .

In principle then we see how to tell what k is: we measure the total energy density ρ of theuniverse, we determine ρc by measuring the Hubble constant and then by constructing theratio we find Ω. If Ω > 1 we have k = 1, and if Ω < 1 then k = −1, while if Ω = 1, k = 0.Of course, if a 1, the right hand side may be close to zero, even if k 6= 0; this is the ideabehind the theory of inflation. As we will see later, we have evidence that Ω is very close orequal to one. That means that the curvature term in our universe is negligible.

Now suppose the universe contains all three types of energy: relativistic (ρR), nonrelativistic(ρM , “M” is for “matter”), and cosmological constant (ρΛ); we have seen that these scale asa−4, a−3, and a0 respectively. Then the total energy density, as a function of the scale factora, is given by

ρ = ρR + ρM + ρΛ

= ρc

(ΩR

(a0

a

)4+ ΩM

(a0

a

)3+ ΩΛ

). (92)

We see that the early universe (small a) will be radiation dominated, while the late universeis dominated by the cosmological constant, if nonzero, or matter if it is. Everything with thesubscript 0 on ρ and H refers to today’s value; the terms ΩM , ΩR and ΩΛ always refer totoday’s values. Using this formula we can find a useful formula for the evolution of Ω. Fromeq. (89) we have

Ω− 1 = (Ω0 − 1)a2H2

0

a20H

2. (93)

We can rewrite

H2

H20

=ρcrit(t)ρc

=ρcrit(t)ρ

ρ

ρc

=1Ω

(ΩR

(a0

a

)4+ ΩM

(a0

a

)3+ ΩΛ

). (94)

Plugging the above result into the previous one, we get

Ω− 1 =Ω0 − 1

1− Ω0 + ΩR

(a0a

)2 + ΩM

(a0a

)+ ΩΛ

(aa0

)2 . (95)

We see that in the early universe (Ω − 1) ∝ (Ω0 − 1)(a/a0)2, so that if Ω0 . 1 then Ω musthave satisfied Ω = 1 to high accuracy in the early universe, and will evolve to Ω = 0 in the farfuture if ΩΛ = 0. In contrast, if Ω0 > 1, then we must also have had Ω = 1 to high accuracyin the early universe, but in the future Ω will grow and eventually diverge.

If we set k = 0, it is simple to solve for for the evolution of the universe for a simpleequation of state p = wρ. By taking the time derivative of the Friedmann equation we get

2HH =8πGN

3ρ =

8πGN3

[−3H(1 + w)ρ] = −3H3(1 + w) . (96)

23

Thus

dH

H2= −3(1 + w)

2dt , (97)

with solution

H =2

3(1 + w)t. (98)

I have chosen the integration constant such that H = ∞ at t = 0. Writing H = a/a andintegrating again, we find

a(t) =

a0e

±λt w = −1,a0(t/t0)2/3(1+w) w 6= −1

(99)

24

5 Thermodynamics

In the big bang theory, one starts from the assumption that particles in the early universe were(mostly) in thermal equilibrium. This obviously can’t be exactly true in an expanding universe,but it suffices that the expansion be adiabatic, so that particles can readjust their distributionsthrough interactions quickly enough to track the slowly changing temperature. We will re-examine this assumption later, but begin by assuming that particles have short enough meanfree paths that they have thermal distributions appropriate for Minkowski spacetime,

f(p) =g

e(Ep−µ)/T ∓ 1,

− bosons+ fermions

(100)

where g counts the number of degrees of freedom of the particle (e.g. 2 for spin, 3 for color,etc).

5.1 ρ, p, and n for a thermalized species of particle.

The energy-momentum tensor for a collection of particles is

Tαβ(x) =∑n

pαnpβn

Enδ3(x− xn(t)) . (101)

Therefore

ρ = 〈T 00〉 = 〈E〉 = g

∫d3p

(2π)3f(p)Ep =

g

2π2

∫ ∞

m

(E2 −m2)1/2E2

e(Ep−µ)/T,

p = −13〈T ii〉 = g

∫d3p

(2π)3f(p)

p2

3Ep=

g

6π2

∫ ∞

m

(E2 −m2)3/2

e(Ep−µ)/T,

n = 〈1〉 =g

2π2

∫ ∞

m

(E2 −m2)1/2Ee(Ep−µ)/T

. (102)

These integrals take on simple values in the ultra-relativistic and nonrelativistic limits:

Nondegenerate, relativistic matter: T m, T µ.

n =ζ(3)π2

gT 3 ×

1 bosons34 fermions

ρ =π2

30gT 4 ×

1 bosons78 fermions

p =ρ

3(103)

When there are several relativistic species at the same temperature, we write

ρR =π2

30g∗T

4 , (104)

where

g∗ =∑

bosons

gi +78

∑fermions

gi , (105)

25

the sums being over all relativistic species in thermal equilibrium at temperature T . If inaddition there are decoupled relativistic species at a different temperature Ti, then we mustreplace gi → gi(Ti/T )4:

g∗ =∑

bosons

gi

(TiT

)4

+78

∑fermions

gi

(TiT

)4

. (106)

Nonrelativistic matter: T m

n = g

(mT

2π

)3/2

e−(m−µ)/T

ρ = mnp = nT ρ . (107)

5.2 An inventory of energy in the universe.

In order to develop a realistic picture for how the universe evolves, we need to know what sortof particles are important for the energy and pressure at different epochs. At earlier times,when the universe was hotter, more types of particles will be in thermal equilibrium. Whatparticles were these? All we know about right now are the particles of the standard model,and those only up to energies of ∼ 100 GeV. However, we do know that there is also darkmatter in the universe, which is believed to be something entirely different. Finally, there hasbeen recent evidence for a cosmological constant, or dark energy. Here I will discuss whatwe know about the standard model constituents of the universe, and a little about the darkmatter. We will discuss dark energy later.

A particle with mass m will count as relativistic particle (or radiation in the standardparlance) for T & m) but will become nonrelativistic (called matter) once the temperaturedrops well below its mass. If there is no chemical potential, thermal equilibrium requires thenumber density to drop precipitously at the cross-over between the two regimes. This canhappen via annihilation or decay, or inelastic collisions, provided they are fast enough to keepup with the cooling of the universe. If they can’t keep up, then we have a departure fromthermal equilibrium, which is discussed in §6. First we turn to the question of what sort ofparticles there are which contribute to ρ and p in the early universe.

5.2.1 Standard model particles

The standard model consists of three families of fermions, conveniently described as fifteentypes of two-component fermions (Weyl fermion) per family. They can interact through threeforces: the strong force, which couples to a quantum number called color and is mediated byeight massless spin-one particles called the gluons; the electromagnetic force, mediated by themassless photon; and the weak force, mediated by the massive W± and Z0 spin-one bosons.The fermions are:

Family I: uL, uR, dL, dR, eL, eR, νe ,Family II: cL, cR, sL, sR, µL, µR, νµ ,

Family III: tL, tR, bL, bR, τL, τR, ντ . (108)

The L and the R refer to left- and right-chirality Weyl fermions (chirality refers to the ±eigenvalues of the Dirac matrix γ5), with two components per Weyl fermion; this includes allthe spins as well as both particle and antiparticle. All neutrinos have left chirality.

26

All fermions participate in the neutral current weak interactions (Z exchange), while onlythe L fermions interaction via charged current weak interactions (W± exchange). All of thequarks come in three colors each and interact through both the strong and electromagnetic in-teractions; the charged leptons (e, µ, τ) are neutral under color and so do not interact strongly,but they do interact electromagnetically; finally the neutrinos (νe, νµ, ντ ) can only participatein the weak interactions.

The eight gluons and the one photon each come in two helicity states, while the massiveW+, W− and Z0 bosons come in three spin states. In addition the standard model predictsthe existence of a single spin-zero boson, the Higgs, which has yet to be seen.

The masses of the standard model particles may be found in the particle data book, orthe web page of the Particle Data Group. The heaviest known particle is the top quark atmt = 174± 5 GeV, followed by the Z (MZ = 91 GeV) and W± (MW = 80 GeV). The Higgsboson mass, if it exists, is mH & 120 GeV. There could be many other particles with massesM & 100 GeV which we would not know about; there could also be lighter particles whichinteract with our world only very weakly.

Assuming that there are no particles beyond the standard model, then the scenario for theuniverse as it cools in the first few minutes is as follows:

• At T 100 GeV, all particles in the standard model are relativistic and in thermalequilibrium, and g∗ (eq. (105))is given by

g∗ = (2 + 8× 2 + 3× 3 + 1) +78

(3× 15× 2) = 106.75 , (109)

where the bosons are the photon, gluons, W± and Z, and the Higgs; the fermions arethe three families of 15 Weyl doublets each.

• As the universe cools to T ' 100 GeV the electroweak phase transition occurs, where theHiggs field develops a vacuum expectation value and the W and Z and fermions acquirea mass.

• As the temperature drops, the top quarks, Higgs, W ’s and Z’s annihilate and decay. AtT . 1 GeV, the b and c quarks are nonrelativistic and are annihilating and decaying. Theremaining quarks and gluons bind up into a complicated spectrum of strongly interactingcomposite states (hadrons) called glueballs, or mesons (qq states) or baryons (qqq states).All of the heavier states, including the baryons, are annihilating and decaying.

• For T . 100 MeV, the hadrons are all disappearing, except for a small number of non-relativistic protons and neutrons. Strangely enough, there has been an excess of baryonnumber (B = (number of baryons−number of anti-baryons) and so this small number ofresidual nucleons have nothing to annihilate with. In thermal equilibrium are the chargedleptons, photons and neutrinos.

• When 1 MeV & T & 0.1 MeV, the weak interactions go out of equilibrium: at lowenergies the weak interactions are too slow to keep up with the cooling rate of theuniverse. So neutrinos decouple from everything else, and neutrons start to decay. Atthe same time, neutrons and protons bind up into light nuclei, in a process called BigBang Nucleosynthesis, or BBN.

• Below T ∼ 100 keV the charged leptons are all gone, except for one electron for everyresidual proton, to maintain charge neutrality.

27

• At about T ' 1 eV two important things happen: the energy density of the universebegins to be dominated by nonrelativistic matter (baryons and dark matter) instead ofrelativistic matter (the photons and neutrinos). Also, protons and electrons bind to formneutral hydrogen, an era known as recombination. At this point, the thermal photonsdecouple and continue to redshift; they are seen today as the CMB.

• As the universe continues to evolve, small density fluctuations present at recombinationevolve and gravitationally collapse to form the structure we see today.

• Today we have residual baryons and dark matter, as well as a background of photonsand neutrinos.

For T . 1 MeV the nonzero baryon density in the universe becomes important. So whatis Ωb today? Most visible baryons are in the form of inter-galactic gas (mostly hydrogen)within galactic clusters. This gas can be observed by examining the absorption spectra oflight from distant quasars (quasi-stellar objects, which are exceedingly bright) as it passesthrough the intervening gas. These measurements give Ωbh

3/2 ∼ 0.02. From examining thespectrum of fluctuations in the CMB, one comes up with a more precise and similar value,Ωbh

2 = 0.024+0.004−0.003. Finally, the theory of nucleosynthesis predicts that the ratio of primordial

deuterium to hydrogen ratio depends sensitively on the baryon density, and one finds Ωbh2 =

0.0205 ± 0.0018. As we will discuss later, the existence of a nonzero baryon density todayimplies a tiny (O(10−10)) asymmetry between the number of baryons and anti-baryons in theearly universe. It is widely believed that this asymmetry evolved from microphysical processesduring the early expansion of the universe, through a process called baryogenesis, which wewill discuss later.

5.2.2 Dark matter

There is now strong evidence that there is more matter in the universe than what meets the eye.There exists some sort of non-baryonic dark matter which is observed by its gravitationaleffects on luminous matter, as opposed to being observed directly. It is currently believed thatnonrelativistic matter in the universe today has energy density ΩM ' 0.27 on cosmologicalscales, while ordinary baryonic matter contributes only about Ωb ' 0.04 to the total.

Most striking is the evidence at the galactic scale from rotation curves, measurements ofhow fast visible matter in galaxies is rotating about the galactic center. For a circular orbit ofradius r about the center, Newtonian mechanics implies a orbital velocity of

v(r) =

√GNM(r)

r, M(r) ∼ 4π

∫ r

0ρ(r′)r′2dr′ , (110)

where ρ(r) is the mass density. The evidence from numerous galaxies is that v(r) rises in thecore of the galaxy, but then becomes constant for r larger than a few kpc, staying constantout to as large a radius as can be observed (∼ 20 kpc). This implies that M(r) ∝ r andρ(r) ∝ 1/r2. In contrast, if the visible disk faithfully described the mass density of a galaxy,one would find v(r) ∝ 1/

√r. There are numerous additional observations at the galactic scale

which point to the existence of a spherical dark matter halo about galaxies. Since M(r) ∝ r,and where the halo ends is hard to determine (since there are few light sources far out from thegalactic center) it is difficult to determine the total dark matter mass clustered about a galaxy.For some time, it was expected that this dark matter could be in the form of Jupiters, or lowmass, low luminosity stars. However experiments searching for such objects (eg, MACHO)

28

Figure 5: The rotation curves of various galaxies, from this source.

by seeing their gravitational lensing effects on stars in our galaxy or the LMC ruled out thatpossibility in our galactic neighborhood.

The first evidence for the existence of dark matter at the cluster scale was made in 1933(Zwicky) who determined that the mass-to-light ratio in the Coma cluster, M/L, was about400 times greater than M/L, roughly 102 times what M/L is in our neighborhood. Present-day measurements at the cluster scale reveal ΩM ' 0.2 − 0.3, where the mass distribution inclusters can be determined by observing the velocities of galaxies withing the cluster. Anindependent check of the matter distribution at the cluster scale can be made by observingthe gravitational lensing of light from distant quasars by galaxy clusters.

These measurements do not tell you whether the matter being observed is ordinary (bary-onic) gas, or something new. However, the baryonic component is severely constrained by thetheory of nucleosynthesis, and the observed abundances of light elements, especially deuterium(see §7). A lower bound on the size of the ordinary baryonic component of the inter-galactic gascomes from the study of X-ray emission from these regions, giving Ωb/Ω/M ≥ 0.06h−3/2 + .02.When combined with the bound from theories of primordial nucleosynthesis, which put anupper bound on Ωbh

2 this yields ΩM < 0.30 for h = 0.75. A similar number results from stud-ies of the Sunyaev-Zeldovitch effect, the spectral distortion of the CMB due to Comptonscattering of CMB photons off free electrons in the ionized inter-galactic medium.

Finally, there is evidence for Ωm ' 0.23 at cosmological scales from studies of the fluctuationspectrum of the CMB. This will be discussed later. With the current value for the criticaldensity

ρc =3H2

0

8πGN= 1.88h2 × 10−29 gm cm−3 = 1.1× 10−5h2 GeV cm−3 , (111)

we see that the dark matter density is roughly equivalent to the mass of a single proton forevery cubic meter of space.

Aside from observational constraints for the existence of dark matter, there also exists moreindirect but compelling evidence for the existence of cold dark matter (CDM) from galaxyformation simulations which we will discuss later.

There are many attempts to detect dark matter in terrestrial experiments, assuming thatthe dark matter comprises of some exotic, weakly interacting particle which is a relic of thebig bang. Since dark matter clusters in galaxies, it is thought that the local density of darkmatter is much higher than the cosmological average, with estimates in the 0.2−0.8 GeV cm−3

29

range. By virial arguments, it is thought that the typical rms velocity of dark matter particlesin our neighborhood will be about

√〈v2〉 ' 270 km/s.

5.3 Entropy

How does temperature vary with time, or scale factor a? To answer this question, it is helpfulto realize that if the expansions is slow enough that particles remain in thermal equilibrium(i.e. the expansion is adiabatic), then the entropy is conserved. Assuming that all chemicalpotentials vanish, we have the thermodynamic relation

TdS = dE + pdV (112)

where S is the entropy, and dV corresponds to a co-moving volume element. But we havealready seen that one of Einstein’s equations gives energy conservation, in the form of E =−pV , which is seen to yield S = 0, which means that entropy is conserved. we see that

∂S

∂E=

1T,

∂S

∂V=p

T, (113)

which can be integrated to give

S =E + pV

T. (114)

Therefore the entropy density is

s =S

V=ρ+ p

T. (115)

From eq. (103) and eq. (107) we see that nonrelativistic matter contributes negligibly to theentropy density, proportional to e−m/T , while for relativistic matter

s =2π2

45g∗T

3 . (116)

If we have several relativistic species at different temperatures, we define

s =2π2

45g∗sT

3 , g∗s =∑

bosons

gi

(TiT

)3

+78

∑fermions

gi

(TiT

)3

. (117)

The statement that S is constant implies that sa3 is constant, where a(t) is the evolving scalefactor. So long as g∗ doesn’t change, this means that the product aT is constant. However, asthe temperature drops, eventually the relativistic assumption T m will fail for any massivespecies, and therefore g∗ will change in time. For this reason aT is not a constant.

Some uses for entropy:

The temperature of light, decoupled particles. As we will discuss shortly, neutrinosdecouple at a temperature dec ' 1 MeV, below which they are essentially noninteracting andrelativistic. At Tdec the neutrinos are in a thermal distribution, and we have seen in § 3.4 thatas the universe continues to expand, they will stay in a thermal distribution with temperatureTdec(adec/a), where adec is the scale factor at the time of decoupling. Below Tdec, the relativisticparticles are the photon, three neutrinos (at the same temperature), electron and positron;

30

there also exists a small number of protons and light nuclei, but they do not contributesignificantly to the entropy. However, at a temperature several times below the electronmass (me = 0.511 MeV), the electrons and positrons have become nonrelativistic and haveessentially all annihilated into photons. The photons are still in thermal equilibrium due tothe small number of residual protons and electrons that make up the baryonic matter of theuniverse. Therefore, while at T = Tdec the neutrinos and photons are at the same temperature,at T me the photons have been replenished and are hotter than the neutrinos. To find therelative temperatures after cooling below the electron mass, we invoke the constancy of entropy,and the fact that since neutrinos are decoupled, their entropy, and the entropy of the photonsand electrons, are separately conserved. So we can just compute he two values for g∗ above anbelow the electron mass for the photon-electron system

T> & me : g>∗ = 2 + 78 × (4) = 11

2T< me : g<∗ = 2 ,

(118)

where we are only keeping particles in thermal equilibrium (eg, not the neutrinos) — photons,electrons and positrons for T>, but only photons for T<. In the above formulas, the 2 countsthe two polarizations of photon, while the 4 counts the two spin states of the electron and thetwo spin states of the positron. Since a3s is constant, it follows that the photon temperaturetoday Tγ,0 satisfies

(Tγ,0)3 = (Tdec)3

(g>∗g<∗

) (adec

a0

)3

. (119)

On the other hand, the neutrino temperature has simply red-shifted since Tdec, and so theneutrino temperature today is Tν,0 which satisfies

(Tν,0)3 = (Tdec)3(adec

a0

)3

. (120)

Comparing the above two expressions, we find

Tν,0Tγ,0

=(g>∗g<∗

)1/3

=(

411

)1/3

= 0.7 . (121)

As a fiducial volume. Since the S is constant, 1/s serves as a definition of volume. Forexample one parameter of interest in the universe is the baryon number density, nb. As far aswe know experimentally, baryon number is conserved, and so nb − nb only changes due to thedilution of baryons as the universe expands. In order to describe the baryon asymmetry by aconstant, one considers instead the ratio

nbs,

which is constant, as it measures the baryon number in a co-moving volume. What is usuallyquoted, however, is the baryon-to-photon ratio today, which is called η. We know that η ' 5×10−10, and its value will come up later when we discuss both nucleosynthesis and baryogenesis.Note that the ratio nγ/s is not a constant, as the number of photons in a co-moving volumeneed not be conserved. To compute the ratio today, we use the fact that the entropy todayresides in photons and three species of neutrino, so that today

s =2π2

45

(2 +

78× 3× 2×

(411

))T 3

0 'nγ7

(122)

Thus nb/s ' η/7 today.

31

5.4 Matter-radiation equality

Today one finds that ΩM = ρM/ρc = 0.23 where ρM is the nonrelativistic (“matter”) compo-nent of energy today, and ρc is the critical density defined in eq. (91). This is much greaterthan ΩR due to photons and neutrinos, and so we say that the universe is matter dominatedtoday. Yet we know from eq. (92) that radiation must have dominated earlier in the historyof the universe, and that the expansion rate was correspondingly different. To compute thetemperature when matter and radiation contributed equally, we equate ρR = ρM at photontemperature Teq, and solve for Teq, where

ρR =π2

30g∗(Teq)4 ,

ρM = ΩMρc

(a0

a

)3= ΩMρc

(Teq

T0

)3

, (123)

where we have assumed in the second line that Tdec < me so that entropy conservation impliesaT has been constant between the time of matter-radiation equality and today (we will seethat this assumption is justified). Equating the two above quantities, and using

T0 =(

T0

2.75K

)× 2.4× 10−4 eV , (124)

ρc = 8.1h2 × 10−11 eV4 = 4.2(

h

0.72

)2

× 10−11 eV4 , (125)

yields

Teq =30π2g∗

ΩMρcT 3

0

=(

T0

2.75K

)−3 (h

0.72

)2 (ΩM

0.23

)× 0.63 eV , (126)

where we computed g∗ from eq. (106)

g∗ = 2 +78× 3× 2×

(411

)4/3

= 3.4 , (127)

appropriate for two polarizations of photon, and three flavors of neutrinos with two helicitystates each, and a temperature given by eq. (121). We see that Teq me and so we werejustified in our calculation of g∗. Note that matter-radiation equality corresponds to a redshift

zeq =Teq

T0− 1 ' 2700 (128)

and that this epoch will coincide pretty closely (and coincidentally) with the time when elec-trons and protons bind into neutral hydrogen.

32

6 Departure from thermal equilibrium

Entropy conservation implies that the rate of change of the temperature is given by the expan-sion rate, T /T ' −a/a = H, except when the temperature in near the mass of some particle.Thus thermal equilibrium can be maintained only so long as particle interactions are rapidcompared to the Hubble expansion H. Suppose we have a scattering or annihilation processa + b −→ c + d, which has a cross section σ (which has dimensions of area, or in units with~ = c = 1, mass−2). Then the rate per particle to experience such a collision is roughly givenby Γ ' nσv, where n is the density of initial particles (either a or b) and v is their relativevelocity. You can check that Γ has dimension of inverse time, or mass. Thus for thermalequilibrium to be maintained, we must have Γ > H.

Just on dimensional grounds it is easy to see that there will be various sorts of cases thermalequilibrium will not be maintained:

1. In a radiation dominated universe, H ∝ T 2/Mp, where

Mp = G−1/2N = 1.221× 1019 GeV

is the Planck mass. If the scattering process only involves massless particles, includingin intermediate states, then by dimensional grounds it must go as Γ ∼ α2T , whereα = g2/4π and g is a typical coupling constant in the process. Thus in the very earlyuniverse, when T/Mp & α2, this scattering process will be out of thermal equilibrium.

2. It can happen that only interaction light particles can experience involves the mediation ofa heavy virtual state with mass M . Neutrinos serve as a particularly important example,as they only interact through the weak interactions, involving the exchange of massiveW or Z bosons. At temperatures T M the quantum mechanical amplitude for such aprocess is proportional to 1/M2, and Γ, which is a probability, must be proportional to1/M4. Since Γ has dimension of mass, it follows on dimensional grounds that Γ ∝ T 5/M4.Thus at low enough temperature, it will always happen that Γ < H and the processes goout of thermal equilibrium.

3. There are other ways to depart from thermal equilibrium involving phase transitions. Inquantum field theory, the vacuum we live in as very complex, and like any ground statein a condensed matter system, it can undergo phase transitions as the temperature ischanged. Examples of phase transitions we believe exist include electroweak symmetrybreaking, associated with a temperature Teq ' 100 GeV, where the Higgs gets its vacuumexpectation value and gives mass to W , Z, and fermions in the standard model. Thechiral symmetry breaking phase transition where quarks and anti-quarks condense isassociated with the strong interactions, at a temperature Tstrong ' 1 GeV. One caninvent many other theories containing phase transitions, such as axion models or GrandUnified Theories (GUTS). If any of these phase transitions are first order, then theuniverse can experience supercooling as the universe expands, followed by an explosiveboiling period when the true ground state supplants the metastable state that the universefinds itself in. Phase transitions are also associated with the production of topologicaldefects, which behave like heavy particles or extended objects, and which are often outof thermal equilibrium.

4. Coherent oscillations of scalar fields are sources of possible departure from thermal equi-librium in the early universe. A scalar field takes on a value not at the minimum of

33

its potential, and oscillates about the minimum; provided that the field is weakly cou-pled, its oscillations will decay slowly as its energy is dissipated in particle production.Examples of such fields that have been contemplated include the axion (a dark mattercandidate), the inflaton (the cause of an epoch of exponential growth in the scale fac-tor a), quintessence (a theory for the dark energy), as well as various fields invented toexplain baryogenesis.

Let us do a rough calculation for the temperature Tdec where neutrinos decouple. Any lowenergy 2 ↔ 2 type of interaction involving neutrinos must have an amplitude proportional tog2/M2 ' GF where g is the weak boson coupling and M is its mass; GF is the Fermi constant,given by

GF = 1.1664× 10−5 GeV−2 . (129)

Therefore Γ ∝ G2F , and on dimensional grounds, Γ ' G2

FT5. Equating this to H ' T 2/MP at

T = Tdec we get

Tdec '(MPG

2F

)1/3 ' 1 MeV . (130)

This estimate is actually very close to the correct answer. The value of Tdec is quite importantfor the theory of nucleosynthesis, which we will discuss shortly.

Another type of non-equilibrium phenomenon of interest is the abundance of relic massiveparticles. Suppose that the temperature drops below the mass of a stable particle. Theequilibrium number distribution is then ∝ e−m/T , but in order to reduce their population, theparticles and anti-particles must find each other to annihilate. If the annihilation rate is tooslow, then the particles go out of equilibrium, and there will be a residual population of theseheavy particles around today. This is called relic freeze-out, and could be a problem or avirtue, depending on whether or not the energy density in these particles is similar to what isrequired to explain the dark matter we indirectly observe.

Relic freeze-out can be studied by means of the Boltzmann equation. The non-relativisticBoltzmann equation is given by Lf = C[f ], where f(x,v, t) is the particle distribution inphase space, L is the differential operator

L =ddt

=∂

∂t+ v · ∇x +

Fm· ∇v , (131)

and C is a collision term that describes how particles are scattered into or out of a particularcell in phase space. We need to come up with a version of this equation which is locally Lorentzinvariant, and which properly accounts for gravitational forces in the L operator. We take fto be a function of momentum instead of velocity, f = f(xµ, pν), and replace the terms in mL,by

m

(∂

∂t+ v · ∇x

)−→ pα

∂

∂xα, F · ∇v −→ −Γαβγp

βpγ∂

∂pα. (132)

The second substitution follows by replacing the acceleration d2xi/dt2 in the non-relativisticexpression for L by d2xα/dλ2 in the geodesic equation eq. (23), where the parameter λ is takento be the proper time τ along the path of the particle, using mdxα/dτ = pα. Recall that atenet of GR is that the geodesic equation accounts for the effect of all gravitational forces onthe particle.

34

Specializing to the isotropic and homogeneous FRW metric, f becomes a function of onlyt and p0 = E, with

mLf =(E∂

∂t−H|p2| ∂

∂E

)f . (133)

Here H = a/a arises from the FRW value Γ0µν = −Hgij , where gij is the spatial part of the

FRW metric eq. (52) (which comes with a negative signature). This can be simplified furtherby considering an equation for the density na of particles of type a:

ga

∫d3p

(2π)3maLfa = ga

∫d3p

(2π)3

(E∂

∂t−H|p2| ∂

∂E

)fa

=dnadt

+ 3Hna , (134)

where to get the result I integrated by parts and used the expression for the density

na(t) = g

∫d3p

(2π)3fa . (135)

For the right-hand side of the Boltzmann equation for the process a+ b→ c+ d we write

ga∫ d3p

(2π)3maCnr[fa] −→ −

∫ ∏i=a,b,c,dgi

(d3pi

2Ei(2π)3

)(2π)4δ4(ptot)

×[∣∣Mab→cd

∣∣2fafb(1± fc)(1± fd)−∣∣Mcd→ab

∣∣2fcfd(1± fa)(1± fb)] (136)

where we have replaced the non-relativistic collision term Cnr by its covariant version. Thescattering amplitude M is to be computed from quantum field theory; we have assumed herethat the mean-free-path of particles is short enough here that the amplitudes in flat space canbe used. The (1 + f) or (1 − f) terms are the stimulated emission or Pauli blocking termsrelevant for bosons or fermions respectively.

Now I will restrict the discussion to annihilation processes XX → ψψ and consider relicX particles. Consider the X’s to be massive, weakly interacting particles, for example, whichdecay into states that are not weakly interacting I will also make a number of simplifyingassumptions:

• time reversal invariance is assumed (equivalent to CP -invariance, in a relativistic fieldtheory) so that

∣∣Mcd→ab

∣∣2 =∣∣Mab→cd

∣∣2;• no chemical potentials for any of the species, and Maxwell-Boltzmann statistics so that

we can replace (1± f) → 1.

• the ψ, ψ products are in thermal equilibrium.

• the X particles are in kinetic equilibrium, but not necessarily in chemical equilibrium.This means that their distribution fX is proportional to the equilibrium distribution, buthas the wrong normalization. This can occur if X-number conserving interactions are alot faster than annihilations, which will always be the case if X density is small, since thelatter process will go as the density of X particles to the second power, since the scarceX and X particles have to find each other to annihilate, but not to exchange energy withother particles in the medium.

35

With these simplifications we get

dnXdt

+ 3HnX = −∫ ∏

i=X,X,ψ,ψ

gi

(d3pi

2Ei(2π)3

)(2π)4δ4(ptot) |MXX→ψψ

∣∣2 (fXfX − f eq

ψ feq

ψ

)(137)

Because of energy conservation enforced by the δ-function,

f eqψ f

eq

ψ= e−(Eψ+Eψ)/T = e−(EX+EX)/T = f eq

X feqX, (138)

while by our assumption of kinetic equilibrium for the X particles and anti-particles, we have

fXfX =(nXneqX

)2

f eqX f

eqX. (139)

Thus our equation can be written as

dnXdt

+ 3HnX = −〈σav〉[(nX)2 − (neq

X )2], (140)

where the thermally averaged annihilation cross section is defined to be

〈σav〉 ≡1

(neqX )2

∫ ∏i=X,X,ψ,ψ

gi

(d3pi

2Ei(2π)3

)(2π)4δ4(ptot) |MXX→ψψ|2 e−(EX+EX)/T . (141)

Eq. 140 can be massaged into a more useful form. First, define

YX = nX/s , (142)

where s is the entropy density, and use the conservation of entropy d(a3s)/dt to show that

dnXdt

+ 3HnX = sYX . (143)

Second, define the variable

ξ = mX/T ; (144)

assuming radiation domination, we know that H = 1/(2t) ∝ T 2, so that

d/dt = Hξd/dξ . (145)

Finally, define the annihilation rate

Γa ≡ neqX 〈σav〉 . (146)

Then our Boltzmann equation eq. (140) can be expressed as

ξ

Y eqX

dYXdξ

= −ΓaH

[(YXY eqX

)2

− 1

]. (147)

You will analyze this equation in your next problem set.

36

7 Nucleosynthesis

The theory of nucleosynthesis is one of the cornerstones of cosmology, since it makes testablepredictions about primordial nuclear abundances, and because it places strong constraints onthe number of relativistic species at a temperature T ' 1 MeV, as well as on η, the baryon tophoton ratio.

The equilibrium number density of a nucleus with atomic number A is

nA = gA

(MAT

2π

)3/2

eµA−MA)/T . (148)

Since such a nucleus has Z protons and (A − Z) neutrons and a binding energy BA, we canwrite

µA = Zµp + (A− Z)µn , MA = ZMp + (A− Z)Mn −BA ., (149)

Therefore

e(µA−MA)/T = e(Z(µp−Mp)+(A−Z)(µn−MN )+BA)/T

= nZp n(A−Z)n eBA/T 2−A

(2π

MNT

)3A/2

, (150)

where I am approximating here Mp = Mn ≡ MN here, and I used gN = 2 for the two spinstates of the proton or neutron. Defining

XA =AnAnB

, nB = np + nn , (151)

we get

XA = gAA5/22−AXZ

p X(A−Z)n

(2πTMN

)3(A−1)/2 (nBT 3

)A−1eBA/T

= gAA5/22−AXZ

p X(A−Z)n (2π)3(A−1)/2

(2ζ(3)π2

)A−1 (T

MN

)3(A−1)/2 (nBnγ

)A−1

eBA/T .

(152)

The quantity nB/nγ is constant for T . me (about when electrons and positrons annihilated)and equals a few times 10−10. The factor of (nB/nγ)A−1 suppresses the formation of big nuclei,even when the eBA/T factor is large. This just reflects the large entropy then, and the factthat there are many photons around that can ionize the nucleus, even at temperatures wellbelow the binding energy BA. Note that for 56Fe η55 ' 10−550!

We can define the temperature TA when XA ' 1. This temperature is found by setting theexponent in NA to zero:

0 ' (A− 1)[

BAT (A− 1)

− ln η − 32

lnT/M]

(153)

or

TA 'BA/(A− 1)

− ln η − (3/2) lnT/MN. (154)

37

Some values one finds are

A TA ( MeV)2 0.073 0.114 0.2812 0.25

(155)

4He (A = 4) is the most favored of the light nuclei because it is much more tightly boundthan 3He, 3H or D =2 H. Evidently the temperature has to drop to T ' 0.1 MeV before theequilibrium density of nuclei can become significant.

Nucleosynthesis then proceeds as follows:

• T ∼ 10 MeV. At this temperature before nucleosynthesis starts, the weak interactionsare in thermal equilibrium, Xn ' Xp ' 0.5, while the heavier nuclei are extremely rare(X2 ' 10−11, X3 ' 10−23, etc.

• T = 1.0 − 0.3 MeV. At this stage the weak interaction processes freeze out (see §6) atTf ' 0.8 MeV, and so protons can no longer be converted into neutrons. At T = Tf theneutron-proton ratio is

Xn

Xp

∣∣∣Tf

= e(Mn−Mp)/Tf ' 16. (156)

and it begins to fall as neutrons decay with a lifetime τn = 886 sec. At the same time,electrons and positrons begin to significantly annihilate since me = 0.511 MeV. SinceT > TA, there are still no appreciable densities of nuclei.

• T = 0.3 − 0.1 MeV. Coincidentally, the age of the universe when T = 0.1 MeV, theuniverse is about one minute old, a comparable time scale to the neutron lifetime, soby now the neutron-proton ratio has been reduced slightly to Xn

Xp' 1/7. Around T '

0.2 MeV the equilibrium value for 4He reaches unity, but in fact X4 cannot keep up withits equilibrium value. By far the fastest production rates involve sequential addition ofnucleons; the rate for more than two particles fusing simultaneously is suppressed by apower of η for each initial particle needed. However since D, 3He and 3H are weaklybound, they have not formed in large numbers yet. In addition, the Coulomb barrierwhich disfavors fusion is becoming significant, entering the rate for the fusion of twonuclei as a factor of

e−2

(1A1

+ 1A2

)(Z1Z2)2/3T

−1/3MeV . (157)

Even for the process D + D → 4He the above factor gives a suppression of 4 × 10−3

at T = 0.1 MeV, and it is much smaller for the fusion of larger nuclei. Eventually atT ' 0.1 MeV almost all available neutrons have been gobbled up by 4He, but fusioncannot proceed further efficiently because of the growing Coulomb barrier, and becausethere are no strongly bound nuclei with A = 5 or A = 8 (since the only quick way to getto heavier elements would be via the fusion of an 4He nucleus with a nucleon or another4He).

• T < 0.1 MeV. Almost all baryon number is locked up in the form of protons or α particles(4He nuclei); fusion processes freeze out. Since all remaining neutrons ended up in 4He,

38

Figure 6: Nuclear abundances during nucleosynthesis

which contains two neutrons per nucleus, the ratio X4 at the end of nucleosynthesis isgiven by

X4 =4n4

nB' 4(nn/2)nn + np

=2(nn/np)

1 + (nn/np)' 2/7

1 + 1/7= 0.25 . (158)

So a basic prediction of nucleosynthesis theory is that a quarter of the primordial baryonnumber should be in the form of 4He, and the rest in hydrogen. A detailed analysis givesthe abundances as a function of time as shown in Fig. 6

Much more detailed analyses than the above estimate are performed, involving all possiblenuclear reaction networks and measured cross sections, integrated without all of the above ap-proximations. The results depend on two cosmological parameters that cannot be determinedexperimentally: the value for η, and the value of g∗ at the time of nucleosynthesis. The latteris important, as it controls the expansion rate of the universe, and hence affects the weakfreezeout temperature Tf , and hence also the neutron to proton ratio at T = Tf . One doesnot know η since observations only detect baryons which are interacting with photons, eitherby emitting or absorbing light. We do not know a priori how much dark baryon number isin the universe, which we cannot see, but which affects the outcome of nucleosynthesis. Wealso do not know the value of g∗ a priori since there could be additional decoupled relativisticparticles which have very weak interactions and have not been detected.

The dependence of X4 on g∗ is simple to estimate. The weak interaction rate for equili-brating protons and neutrons scales with temperature as Γw ∼ G2

FT5 for T larger than several

MeV. If at T = Tf we equate Γw = H, we find Tf ∝ g1/6∗ , and so a small change in g∗ leads

to a small change in the neutron to proton ratio:

δXn

Xp= δe−∆M/Tf =

δTfTf

∆MTf

Xn

Xp=

16g∗

∆MTf

Xn

Xp. (159)

39

Since Tf ' ∆M , we find that the fractional change in Xn/X6, and hence in X4, is aboutδg∗/6g∗. Suppose we had Nν species of neutrinos, instead of the three in the standard model.Then we would have

g∗ = 10.25 + 278(Nν − 3) = 10.25 +

74(Nν − 3) , (160)

and so δg∗/6g∗ ' .03(Nν − 3). So we see that every additional neutrino species changes theprediction for the helium abundance by about 3%. Throughout the universe, there appears tobe about one helium atom for every ten hydrogen atoms. Stellar fusion processes are calculatedto be responsible for only about ten percent of this abundance, which makes the primordialabundance about X4 ∼ 0.9× 4/14 = 0.26. A more precise determination leads on to concludethat

X4 = 0.234± 0.002± 0.005 , (161)

in excellent agreement with our rather crude estimate. In fact the precision of the observationand theory of stellar production of helium means that we can conclude that Nν ≤ 4, a remark-able conclusion, considering that we are ruling out particles with only very weak interactionsfrom fossil evidence from when the universe was a few minutes old! We presently know fromthe measured decay width of the Z boson that Nν = 3; however, the nucleosynthesis bound isnot really on Nν but on g∗, and so the helium abundance tells us that there is little room forexotic relativistic particles, even if noninteracting, at the time of nucleosynthesis. The boundis very restrictive since it only depends on the expansion rate of the universe at the time, andnot on the interactions of the hypothetical particle.

The dependence of the nuclear abundances on η is more complicated. As we have seen, X4

is very insensitive to η. However, the deuterium abundance X2 ends up being proportionalto η−1.6. The reason for this is because rates for processes such as pD → 3He, which destroythe weakly bound (B2 = 2.2 MeV) and disfavored deuterium are proportional to η. So thesmaller η is, the less efficient is the destruction of deuterium, and the larger is X2. Anothernice thing about deuterium is that, because of its weak binding, it is not created in starsor supernovae, but only destroyed. Therefore the deuterium one observes is primordial inorigin, provides a lower bound on X2. Deuterium is measured by looking at absorption linesof light emitted by quasars from redshifts z ∼ 1 − 5. One sees the Lyman-alpha forest, adense series of absorption lines as the light passes through inter-galactic hydrogen at variousredshifts. Observers are able to detect slightly shifted absorption lines from deuterium, andthereby deduce its abundance from the strength of the lines.

The current status of the theory and observation of nuclear abundances is summarized inFig. ??. One concludes that

4(3)× 10−10 . η . 7(10)× 10−10 . (162)

or

0.015(0.011) . Ωbh2 . 0.026(0.038) (163)

where the most conservative bounds are in parentheses. Again, a remarkable deduction, sinceit tells us about the total baryon number content of the universe, visible today or not.

40

8 Recombination

Another epoch of great interest is the time of recombination, when electrons and protons bindinto hydrogen, and the universe becomes transparent to light. The equilibrium abundances ofelectrons, protons, and hydrogen are given by

ne = ge

(MeT

2π

)3/2

e(µe−Me)/T ,

np = gp

(MpT

2π

)3/2

e(µp−Mp)/T ,

nH = gH

(MpT

2π

)3/2

e(µp−Mp+µe−Me+BH)/T ,

= nenp

(2πMeT

)3/2

nenpeBH/T . (164)

where by charge neutrality (ignoring the small fraction of helium) we must have ne = np. Thequantity BH = 13.6 eV is the ground state binding energy of hydrogen; I have assumed we areonly interested in the ground state of hydrogen, with gH = 4 = gpge.

Define the fraction of baryon number in the form of free protons as

Xe =npnB

, (165)

and then, with nB ' (np + nH) (again, ignoring the primordial helium), compute the equilib-rium value for

1−Xeqe

(Xeqe )2

=nBnHn2p

= ηnγ

(2πMeT

)3/2

eBH/T

= η

(2πTMe

)3/2 (2ζ(3)π2

)= ζ(3)

√32πη

(T

Me

)3/2

eBH/T . (166)

As was the case for nucleosynthesis, the high value of the entropy (small η prefactor) ensuresthat T has to get far below BH before an appreciable amount of hydrogen is produced. Buteventually Xe becomes small in equilibrium, at which point (1 − Xe)/X2

e ' 1/X2e , at which

point Xe becomes

Xeqe '

(π

32ζ(3)2

)1/4

η−1/2

(Me

T

)3/4

e−B/2T . (167)

To find when e+ p↔ H goes out of equilibrium, one equates the recombination rate Γrecto the Hubble expansion rate, using the averaged cross section for electron capture to the nth

level of hydrogen

〈σrec|v|〉 =4παM2e

BH/n

(3MeT )1/2. (168)

with n = 1, and one finds that equilibrium can be maintained to about z = 1100, while forlower Z the ionization fraction is at a few percent level. Performing the computations will bea homework problem.

41

9 Baryogenesis

We now turn to a more speculative subject, baryogenesis, providing an explanation for theorigin of matter in the universe. Recall that from nucleosynthesis one derives

4(3)× 10−10 . η . 7(10)× 10−10 . (169)

Assuming that there are no anti-baryons 2, this is equivalent to the bounds on η = nB/nγ :

0.015(0.011) . Ωbh2 . 0.026(0.038) (170)

where the figures in parentheses are the most conservative bounds one can imagine. Weactually see fewer baryons than that, with Ωluminous

b h2 . 0.01. There are several pieces ofevidence that the baryons we see are truly baryons and not sequestered anti-baryons. For onething, cosmic ray anti-protons are only seen at the 10−4 level, consistent with their productionin the atmosphere from proton collision, and anti-nuclei have not been discovered in cosmicrays. Secondly, we know that in the intergalactic space a lot of gas resides, which means thatgalaxies are not totally sequestered from each other. Thus is there were regions of matter andanti-matter, we would have expected to see annihilation processes at their interfaces, in theform of high energy photons.

Of course, it is possible that the universe has zero net baryon number, but that withinour horizon there is an excess of baryons, compensated for by an excess of antimatter in someother horizon. This idea cannot work without invoking inflation however. Suppose therewas a mechanism at early time to produce regions where nB = ±nγ with random sign. Themaximum size for these regions from causality arguments would be the horizon size at thattime. Our current horizon would therefore contain N = (H/H0)3 such domains, and thereforewe would find today η ' 1/

√N . Even if this epoch which produced the huge baryon number

fluctuations was as low as T = 1 TeV, we would find N ' 1042, which would lead to a valuefor η over ten orders of magnitude too small too explain the baryon asymmetry we see.

Recall from eq. (122) that nB/s ' η/7 is a constant during the adiabatic expansion ofthe universe. Therefore nB/s was about 10−10 when the temperature was far above thenucleon mass. At that epoch, there would have been quarks and anti-quarks in thermalequilibrium, with densities (nb + nb) ' s (up to some not too large factor), so that theearly universe would have to exhibit the tiny asymmetry between matter and anti-matter of(nb − nb)/(nb + nb) ∼ 10−10.

Sakharov was the first to consider the possibility that the matter/anti-matter asymmetrycould arise from micro-physical processes during the big bang. He discovered three conditionsthat would have to be fulfilled for this to be the case:

1. Baryon number violating processes must exist;

2. C and CP violation must exist;

3. There must have been an epoch when baryon violating processes went out of thermalequilibrium.

2Recall that protons and neutrons have baryon number B = 1; anti-nucleons have baryon number B = −1.Quarks and anti-quarks have baryon number + 1

3 and − 13 respectively. I will use nb and nb to represent the density

of baryons and anti-baryons respectively, while nB = (nb − nb) is the net baryon number density; Ωb will represent(ρb + ρb)/ρc.

42

The reason for the first rule is obvious: if the universe begins with no net baryon number,and ends up with nonzero baryon number, baryon number must not be conserved. In theSU(3) × SU(2) × U(1) standard model (SM), the lowest dimension operators one can writedown which violate baryon number are dimension six, taking the form

qqq`

Λ2(171)

where the q’s are quark fields, ` is a lepton, and Λ has dimension of mass. The lifetime of theproton (τ & 1034 years) forces Λ to be a very large scale, since the above operator can lead tothe decay p → π0e+, or p → K+ν, etc. Since τ ∼ m5

p/Λ4, one finds Λ & 1016 GeV, which is

the grand unified theory (GUT) scale. Operators such as eq. (171) can be generated, in grandunified theories, for example, by the exchange of bosons with mass M and coupling g, withg2/M2 ' 1/Λ2. The large Λ scale implies that if baryogenesis is to exploit this hypothesizedsource of baryon violation, it would have to occur at very high temperature, and is referred toas “GUT-scale” baryogenesis.

Another possible source of baryon number violation is the standard model, surprisinglyenough. ’t Hooft showed that because of an anomaly, neither baryon number B nor leptonnumber L are conserved by the weak interactions, although the combination (B − L) is. Atzero temperature, he found that

∆B = ∆L ∝ Nfe−4π/αw , (172)

where Nf is the number of families (3) and αw is the weak fine structure constant, αw ' 1/25,and so the exponential suppression is fantastically small. However, it has been shown thatat finite temperature T > 100 GeV (the scale of the W and Z boson masses) the electroweakB violation rate is much more rapid, Γ /B ' 30 × α5

wT . This opens the possibility for havingbaryogenesis occur at the electroweak scale. It also presents a challenge for GUT scale baryo-genesis, as electroweak baryon violation can wash out a baryon asymmetry produced earlier.A solution to this quandary is to produce nonzero (B − L) charge at the GUT scale, sincethis quantum number is conserved by the electroweak interactions. Note that the operatoreq. (171) won’t do for generating nonzero (B − L), however, as it also conserves (B − L).

The second of Sakharov’s rules follow similar reasoning. A baryon charge is odd under bothof the discrete symmetries C, which interchanges quarks with anti-quarks, and CP , whichinterchanges quarks with anti-quarks and followed by a parity flip. Since we wish to have theuniverse start off initially any net charges, the production of a B asymmetry must necessarilyinvolve both C and CP violation. It is worth commenting that the SM violates C maximallysince C would exchange a left-handed neutrino with a left-handed anti-neutrino, but the latterdoes not exist in the SM. In contrast, CP would exchange a left-handed neutrino with a right-handed anti-neutrino, which does exist. Nevertheless, it is known that CP violation exists innature, probably in the SM, as is observed in the kaon and B meson systems. Nevertheless,the CP violation in the standard model appears to be much too small to account for thebaryon asymmetry we see. The effects are large in the B and K meson systems because wehave nearly degenerate states there, which have large mixing effects induced by very smallinteractions. However, at finite temperature, all states are broad, and there should not be anysuch enhancement of CP violation. Then a reparametrization invariant estimate of SM CPviolation gives the dimensionless measure δCP ' 10−20, much too small to explain η ∼ 10−10.Therefore it is a pretty robust prediction of baryogenesis that there must exist new sources

43

of CP violation, which could possibly be measurable in experiments search for electric dipolemoments in quarks and leptons 3.

The third of Sakharov’s rules mandates departure from thermal equilibrium, and is aconsequence of CPT symmetry. The idea is that CPT symmetry implies that for every statewith baryon number B and energy E, there exists a state with baryon number −B and identicalenergy E. (For example, CPT symmetry implies that proton and anti-proton have the samemass). Thus in thermal equilibrium, states with baryon number B and −B will be equallypopulate4d, and the net baryon number will have to be zero, even if there are B- and C-and CP -violating processes. Formally, if Ω is a CPT transformation, then ΩBΩ−1=−B, andΩHΩ−1 = H, so that

〈B〉 = Tr Be−βH

= Tr Ω−1ΩBΩ−1Ωe−βH

= Tr(ΩBΩ−1

) (Ωe−βHΩ−1

)= Tr −Be−βH= −〈B〉 (173)

implying 〈B〉 = 0.We have seen that departure from equilibrium can occur whenever there are relic particles

which cannot annihilate or decay fast enough as the temperature drops. Annihilation processesquench themselves, as the number density of the relic drops, since the rate is proportional to therelic density. Therefore one will typically depart from thermal equilibrium whenever the life-time of the relic is long compared to the Hubble time, 1/H. Note that 1/H = 3MPl/8π

√g∗T

2

which is a very short time scale at the GUT scale (8π/3)√g∗T ∼Mpl/5, since the universe is

expanding so quickly then. However, at the electroweak scale, one has (8π/3)√g∗T ∼ M−16

pl

and so only incredibly weakly interacting particles can go out of thermal equilibrium. However,one can also go out of thermal equilibrium during a first order phase transition (when thereis typically supercooling, and then explosive formation of bubbles of the new low temperaturephase). For this reason, electroweak baryogenesis is associated with a strong first order elec-troweak phase transition. Other possibilities exist for nonequilibrium epochs, such as duringthe period of reheating after inflation.

GUT-scale baryogenesis Now I will sketch several scenarios for baryogenesis, startingwith a GUT scale model. Suppose there are X and X bosons which can either decay into 2quarks or an anti-quark and a lepton.

particle final state Bfinal branching ratioX qq 2/3 rX q ¯ −1/3 1− rX qq −2/3 rX q` 1/3 1− r

(174)

We can define a measure of the asymmetry in baryon number production

εX =∑

final state f

BfΓ(X → f) + Γ(X → f)

ΓX(175)

3Electric dipole moments violate T (time reversal symmetry) since the spin of a particle reverses under T , but itscharge distribution does not. However, in a Lorentz-invariant and unitary (probability conserving) theory, CPT isan automatic symmetry, and so T violation implies CP violation, which is harder to measure directly.

44

where ΓX is the total width and Bf is the baryon number of the state f . By CPT, the totalwidths of the X and its antiparticle X are equal; their branching ratios into a particularfinal state f may be different, provided there is C, CP violation. In the present example,εX = (r − r). The scenario then is that X and X bosons are massive, and that at the timewhen T ∼MX , their annihilation is not efficient, and more importantly, their lifetime is muchlonger than the Hubble time. Suppose their width is given by Γ = α

4πMX , where α = g2/4πand g is a typical coupling of the X. Then we are requiring that α/4π >

√g∗MX/MPl, or

MX >[α/(4π

√g∗)

]MPl. We know that G∗ ' 100 in the SM, and could be ∼ 1000 in a

GUT. So for α/4π ∼ 10−3 it is reasonable to have MX be as low as 1014− 1015 GeV. Ignoringannihilation then, which will shut off as the particle density gets lower anyway,, the X’s andX’s will hang around for a while, both with number density nX = nX ' nγ until they decayat a time t ∼ Γ−1

X . They decay with their unequal branching ratios, and produce a baryonasymmetry nB/nγ ' εX . The reverse processes which remove baryon number, f → X andf → X, are out of thermal equilibrium and do not proceed, as the particles in the final statef now have an energy T mX and the process will have an e−mX/T suppression.

It is the baryon to entropy ratio that is conserved, rather than the baryon to photon ratio,so what one finds today is η ' εXg

s∗(today)/gs∗(T = MX) which could give another 10−2

suppression factor.This difference (r−r) is naturally small, since it cannot arise in leading order in perturbation

theory, but must arise as an interference term between a tree-level and a one-loop amplitude.At tree level, for example, the amplitude for X → qq could equal g, but then the amplitudefor X → qq would equal g∗. If g 6= g∗, that is a sign of CP violation; however, even if thereis CP violation present, one finds at tree level ΓX→qq/ΓX→qq = |g|2/|g∗|2 = 1. In fact, thisfeature would appear to persist to higher order perturbation theory: for every graph X → f ,there is an analogous graph for X → f with all the arrows on the propagators reversed andall couplings replaced by their complex conjugates. Since the widths depend only on theabsolute value of the amplitude, they can differ only if there is some source of complex phasein the amplitudes which is not due to complex coupling constants. There is: the iε in thepropagators of internal particles in the Feynman diagrams. So to get r 6= r it is necessary tohave (i) loops involving internal propagation of particles, and (ii) the kinematics must be suchthat the internal particles in the diagram could be on shell, so that the amplitude is sensitiveto the iε pole prescription. We see that at the very least, a nonzero value for εX requiresan interference between a tree level and one-loop amplitude, and so there will be typically afactor of α/4π suppression, where α is a coupling constant of the theory, on top of suppressionfactors from the measure of CP violation (often quite small) and from kinematic factors. Notethat α has to be small, otherwise the X’s could decay quickly and would not go out of thermalequilibrium in the first place.

Various problems to overcome with GUT-scale baryogenesis: (i) If there is inflation, thebaryon asymmetry one has produced will typically be inflated away; (ii) If one does not producea (B − L) asymmetry (and the above example does not), then the anomalous electroweakprocesses will equilibrate the asymmetry to zero. (This problem is turned into a feature inmodels of leptogenesis where a lepton asymmetry is produced at the GUT scale, which isthen equilibrated into nonzero baryon asymmetry by the electroweak anomaly).

Electroweak baryogenesis As mentioned above, there is a phase transition at a temper-ature of several hundred GeV where the Higgs field H gets an expectation value and breaksthe SU(2)×U(1) gauge symmetry, Phase transitions are typically first order, or second order.

45

In second order phase transitions, the order parameter (〈H〉 in this case) changes continuouslyat the phase transition, although its derivative with respect to temperature is discontinuous.In the case of a first order phase transition, however, the order parameter has to jump discon-tinuously, tunneling through a barrier in the free energy. Typically this process will involvesuper-cooling, where the universe gets trapped in the symmetric minimum H = 0 below thetemperature where it ceased to be the point that minimizes the free energy. Eventually then thevacuum tunnels to the correct ground state, y nucleating a bubble of the true vacuum, whichthen expands and takes over, releasing latent heat in the process. This is a nonequilibriumprocess.

Whether or not the electroweak phase transition is first order depends on details of thHiggs interaction about which we have know direct knowledge. If the phase transition was firstorder, then a rather complicated scenario has been developed for how the baryon asymmetrycan be generated. As the bubble walls expand, the “snow-plow” particles in front of them —particularly the top quark, which is massless outside the bubble (symmetric phase) and heavyinside the bubble (broken phase). With CP violating interactions between the top quark andthe Higgs field, it is possible to have more anti-top quarks than top quarks pushed in front ofthe expanding wall. In this symmetric phase, baryon violation is rapid, and tries to equilibratethe anti-top excess by producing net baryon number. As soon as the baryons produced getengulfed by the expanding bubble, they are in the broken symmetry phase where the baryonviolating processes are exponentially suppressed, and so the baryon asymmetry is preserved.

Electroweak baryogenesis has the advantage of being late, so that inflation can be wellover with before the baryon asymmetry is produced. It is also appealing because extra CPviolation at the weak scale may be testable, and because having the transition be first orderplaces constraints on the Higgs sector, that can also be experimentally relevant.

Other scenarios There are several other popular scenarios for baryogenesis, which I willjust mention. Affleck-0ine baryogenesis exploits the fact that in supersymmetric theories,the potentials for scalar fields often have flat directions. In this scenario a large coherentfield carrying baryon number develops, and which preferentially decays into quarks. In lep-togenesis, as mentioned above, a lepton number asymmetry is produced, and subsequentlyturned into a baryon asymmetry by anomalous weak interactions. There are also theorieswhere baryogenesis is associated with the reheating phase after inflation.

46

10 Inflation

10.1 Problems with the conventional big bang

Inflation is a theory introduced by Guth in 1981 to address several serious problems with thestandard big bang model. The first is the flatness problem, why |Ω − 1| is so close to unitytoday, implying that k/a2 is very small. The second is the horizon problem, which is really aproblem of initial conditions: how did causally disconnected regions end up having the sametemperature (to high precision). The third issue, which may or may not be a problem, isthat in most theories of grand unification, the universe undergoes a phase transition whenthe temperature is not too far below the Planck scale; as the universe cool and goes throughthis phase transition, topological defects tend to be produced, such as magnetic monopoles, orcosmic strings, or domain walls which would be left to day as heavy relic particles, dominatingthe energy density of the universe today. Guth’s solution is quite simple it turns out: one justrequires there to be an epoch when the scale factor of the universe is accelerating. We will seethat not only can the three above problems be solved, but inflation also provides the densityperturbations required to explain how structure was able to form out of such a homogeneousbeginning. First: a little more discussion of the flatness and horizon problems:

10.1.1 The flatness problem

The measure of curvature in the FRW universe is

k

a2= H2 (Ω− 1) . (176)

As usual

Ω =ρ

ρcrit, ρcrit =

38πH2M2

Pl , ρc ≡ ρcrit,0 = 2.775× 1011 MM3Pl

h2

=(3.00× 10−3 eV

)4h2 . (177)

We have seen in eq. (95) that

Ω− 1 =Ω0 − 1

1− Ω0 + ΩR

(a0a

)2 + ΩM

(a0a

)+ ΩΛ

(aa0

)2 , (178)

where ΩR, ΩM , ΩΛ refer to energy today in the form of relativistic matter (w = 1/3), nonrel-ativistic matter (w = 0), and cosmological constant (w = −1) respectively. So let us imaginehow Ω would evolve in the future for various types of energy.

i. We see that if Ω0 = 1, then Ω = 1 forever.

ii. If ΩΛ = 0, and Ω0 < 1: At a = a0 the denominator begins at the value 1, and then as aincreases, Ωm quickly dominates, and the denominator drops monotonically to 1−Ω0 > 0.Thus one finds Ω goes to zero as Ωm/(1− Ω0)× (a0/a).

iii. If ΩΛ = 0, and Ω0 > 1: The denominator evolves as before, but now (1−Ω0) < 0, and sothe denominator vanishes at finite a, and so Ω →∞. This does not mean that ρ→∞ —rather H → 0 as the universe stops expanding and starts to collapse. At the turn-aroundpoint, ρcrit = 0.

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iv. If ΩΛ > 0 and Ω0 > 1 then again the universe stops expanding at some point andrecollapses.

v. If ΩΛ > 0 and Ω0 < 1, then the denominator is positive and steadily increasing, andΩ → 1.

Instead of looking into our future, we can imagine how we got where we are today, with Ω0 ' 1.A simple calculation shows that at the time of nucleosynthesis, the universe must have had|Ω− 1| ∼ 10−16. This looks like an extreme fine-tuning, and it is much worse if one traces thestandard big bang back to the Planck scale.

However, the formula eq. (178) also provides an answer for how this apparent fine tuningof Ω could come about dynamically: we see that during a cosmological constant dominatedepoch, Ω is driven toward 1. Inflation is basically a scenario in which the universe spends sometime cosmological constant dominated. This solution works more quickly than is apparent ineq. (178), since the scale factor a grows exponentially fast during a Λ-dominated epoch.

This explanation is really very simple: it says that if there was a period when the scalefactor a grew exponentially fast in the past, then the k/a2 term in the Friedmann equation isnegligible today.

The flatness problem can be expressed in a different way, which makes it clearer froma geometric point of view how flat is flat enough. We can define a curvature length scaleLc =

√a2/|k|, assuming k 6= 0, and the horizon scale Lh = 1/H. Then (Ω− 1) = η(Lh/Lc)2,

where η = Sign(k) = ±1. So “flat” means that the curvature length scale is much largerthan the horizon length scale. Suppose we go to comoving coordinates instead: then matteris at rest, the horizon scale is Lh/a and the curvature scale is Lc/a = 1, independent oftime. Under the normal evolution of the universe, Lh/a is expanding (as t1/3 or t1/2), sothat even if one starts off with Lh/a << Lc/a, eventually Lh/a grows to be comparable, and|Ω − 1| = (Lh/a)/(Lc/a) becomes O(1). In inflation this problem is solved by having Lh/abecome very small, so that it takes much longer for than the age of the universe for it tosubsequently grow to be O(Lc/a).

10.1.2 The horizon problem

The second problem is why the universe looks so isotropic. It is ironic that the assumption ofisotropy leads to the FRW solution, and then we find that due to its expansion the universe iscomprised of causally disconnected regions that constant merge as the universe evolves. Theproblem is that when we look back in the microwaves to the surface of last scattering, we see ablackbody with a temperature that has variation δT/T . 10−5 (after the dipole doppler shiftdue to the Earth’s motion motion is subtracted). What is astonishing about this is that thephotons we see were only in causal contact over the horizon size at recombination. The horizonat recombination only subtends about one degree now. To see this, we compute the ratio ofthe horizon size at decoupling, divided by the distance from the surface of last scattering tothe earth, at the time of last scattering:

∆θ =H−1

1

a1r1. (179)

Since the universe has been matter dominated since recombination, so that H = 2/3t anda = (t/t0)2/3, we have

a1r1 = a1

∫ t0

t1

dt′

a(t′)=

(t1t0

)2/3

3(t1/30 − t

1/31

)t2/30

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' 3t2/31 t1/30

= 2H−11

√a0/a1 (180)

so that

∆θ =12

√a1

a0=

12(1 + zdec)−1/2 = 0.015 rad = 0.86 deg . (181)

In the above calculation, I used zdec ' 1100. Therefore one would only expect δT/T to besmall over angular scales less than a degree. In fact, the temperature of the CMB is extremelyuniform over all angles...how did matter not in causal contact come to the same temperature?Inflation will explain this by having all the visible universe in causal contact in the far past,later to be flung apart faster than the speed of light, and then still later to come back intoour view. It is because of the early period of causal contact that all we see is at the sametemperature today.

10.2 How inflation solves the problems

The idea of inflation is that there is a period when a > 0. This is equivalent to saying

d

dt

1aH

= − a

a2< 0 . (182)

This means that the horizon in comoving coordinates is shrinking during inflation. So we getthe following scenario:

1. The universe is initially radiation dominated, and there is some expanding comovinghorizon (Fig. 7). Within this horizon, everything is thermally equilibrated, but notwithout. Furthermore, the universe does not look very flat, with Lh/a not necessarilymuch smaller than the comoving curvature radius Lc/a.

2. There is a period of inflation, during which the comoving horizon Lh/a is shrinking,according to eq. (182) — Fig. 8. At the end of inflation Lh/a Lc/a so the universelooks very flat. There are also very few relic particles within the horizon.

3. At the end of inflation, energy is dumped back into the universe (and epoch calledreheating), and the horizon starts to expand again (Fig. 9). As it does, what is cominginto the horizon are regions of space that had all been in thermal contact before inflationbegan.

The only problem with this scenario is apparently that at the end of inflation, not only relicparticles will be removed from the horizon, but very little ordinary matter or radiation willremain as well. So inflation is only viable if it is followed by a period of reheating, at a lowenough temperature that the undesirable relic particles are not reintroduced. Inflation thenis a a way of manufacturing the desired initial conditions for the standard big bang scenario.As we will see later though, inflation provides a bonus: it explains the origin of the densityfluctuations (departures from homogeneity) which can eventually grow into the structure wesee today in baryons and dark matter.

How much inflation is needed? To solve the horizon problem, the entire surface of lastscattering that we see today must have been within a single causal horizon before inflation

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c

Comoving horizon,L /ah

L /a

Before inflation

T

T’

Figure 7: A possible view before inflation: The comoving horizon size Lh/a may be similar in size tothe comoving curvature radius Lc/a (so that Ω can differ significantly from unity), and is expanding.Temperatures can differ by a lot between causally disconnected regions. Nasty relic particles (dots)exist within the horizon.

c

Comoving horizon,L /ah

L /a

During inflation

Figure 8: During inflation: The comoving horizon size Lh/a is shrinking rapidly.

cL /a

Comoving horizon,L /ahAfter inflation

T

T’

horizonPre−inflation comoving

Figure 9: After inflation: The comoving horizon size Lc/a is now tiny, so that (Lh/a)/(Lc/a) is verysmall and Ω ' 1; after reheating it begins to expand again. The horizon expands into a region withthe same temperature, as the larger region was in thermal contact before inflation began. Thereare very few relic particles (or none) with in the horizon now.

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started. Since recombination occurred when the universe was very young, we might as well saythat our entire horizon was contained within a single pre-inflation causal region. Consider theamount of entropy S within our horizon today. It is carried by photons and (cooler) neutrinos,and assuming three flavors of neutrinos, it is given by

Shor =4π3

(1H0

)3 2π2

45T 3

0 ×(

2 +78× 3× 2× 4

11

). (183)

Using the numbers

H−10 T0 =

(2h× 10−42 GeV

)−1 (2× 10−13 GeV

)= h−1 × 1029 , (184)

we get for the entropy in our horizon

Shor ' 1088h−3 . (185)

All of this entropy must come from a patch that was inflated by a factor of eN . The physical sizeof the patch of universe that eventually became our horizon today started off with size H−1,where H was the expansion rate during inflation. At the end of the inflationary period, thepatch had stretched to size eNH−1. At this point the the field begins to oscillate, the universeis matter dominated and begins to reheat. By the time the reheating epoch is finished, withscale factor is aRH , the temperature is TRH , and the universe is radiation dominated. Nowthe size of the patch is L = H−1eN (aRH/aosc ' eNH−1

(V0/T

4RH

)1/3, since during the matterdominated era, a ∝ ρ−1/3. Into this patch goes the entropy associated with TRH :

S ∼ L3T 3RH = e3NH−3V0/TRH ' e3NM3

Pl/(TRH√V0) (186)

using H '√V0/MPl. Taking the log, and insisting S ' 1088 yields

N ≥ 13

ln(

1088TRH√V0

M3Pl

)= 53 +

23

ln(M/1014 GeV) +13

ln(TRH/1010 GeV) . (187)

10.3 Scalar field models for inflation

Consider a scalar field whose Lagrangian is

L = 12∂µφ∂

µφ− V (φ) , (188)

and its equation of motion i s ∂2φ+ V ′(φ) = 0, where ∂2 is the general covariant laplacian,

∂2 =1√−g

∂α√−ggαβ∂β . (189)

In an FRW background, this becomes

φ+ 3Hφ− ∇2

a2φ+ V ′(φ) = 0 . (190)

To find out the effect of this field on gravity, we need to compute its contribution to theenergy-momentum tensor, which is defined to be

Tµν =2√−g

δS

δgµν, (191)

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where S is the action

S =∫d4x

√−gL . (192)

Computing Tµν is easy so long as one knows how to take the variation of√−g with respect

to the metric. Note that for a matrix M ,

δ detM = det(M + δM)− detM = detM[det(1 +M−1δM)− 1

]= detM Tr M−1δM +O(δM2) , (193)

so that

∂ detM∂Mij

=(M−1

)ji. (194)

Armed with the above identity and assuming a spatially homogeneous field φ(t), it is straightforward to compute

T 00 = 1

2 φ2 + V , T ij =

(−1

2 φ2 + V

)δij , (195)

and T 0i = T i0 = 0. From this we see

ρ = 12 φ

2 + V (φ) , p = 12 φ

2 − V (φ) . (196)

For example, a simple harmonic oscillator in flat space will have V = 12m

2φ2, and a classicalsolution φ(t) = φ0 sinmt; this will lead to ρ = m2φ2

0 and p = 12m

2φ20 cos 2mt. The pressure

averages to zero over time, and one can interpret the classical field configuration as a collectionof zero momentum particles of massm and density n = mφ2

0, behaving as nonrelativistic matterwith w = 0.

However, note that if V (φ) = 12m

2φ2 + V0, where V0 is a constant, then we still havethe classical solution φ(t) = φ0 sinmt, but now Tµν describes nonrelativistic matter plus acosmological constant. Shifting V (φ) by a constant, which has no effect on classical or quantummechanics, has a profound effect on a gravitating system.

So one idea for implementing Guth’s idea for inflation is to introduce a scalar field (theinflaton with the potential shown in Fig. 10. There is a very flat segment with vacuum energyV0, where the field φ(t) rolls slowly, so that ρ ∼ −p ∼ V0, leading to a Λ dominated universewith accelerating scale factor. After sufficient inflation, the scalar field rolls into a well whoseminimum has been tuned to zero vacuum energy density, and the field begins to oscillate.During this epoch, φ oscillates and the universe becomes matter dominated. The φ field isassumed to decay at this point, turning the kinetic energy in the inflaton field into ordinaryparticles which thermalize and return the universe to the conventional, radiation dominatedbig bang scenario. By now there have been a plethora of variations of this basic scenario.

The Friedmann equation and equation of motion for φ are

K

a2+H2 =

8πρ3M2

Pl

=8π

3M2Pl

(12 φ

2 + V)

φ+ 3Hφ = −V ′ (197)

where in this section I will write the curvature term asK instead of k, which will be reserved forthe wavenumber of a plane wave. We will analyze this model in the slow roll approximation:

52

V0

Figure 10: A prototypical slow-roll scalar potential for inflation.

i. φ Hφ

ii. φ2 V

We also assume that k/a2 is small, which is true not long after inflation has begun. With theseassumptions the equations of motion become

H2 ' 8πV3M2

pl

, 3Hφ ' −V ′ . (198)

Now we have to see whether these equations are consistent with our assumptions. We can putconstraints on two parameters we will define, ε and η, which depend on the potential V andits derivatives. Using the equations eq. (198) we have

φ2 '(V ′

3H

)2

'M2Pl

24π

(V ′

V

)2

V , (199)

so that our assumption that φ2 V implies that

ε(φ) ≡ MP l2

16π

(V ′

V

)2

1 . (200)

Note that by our equations,

ε = − H

H2, (201)

so ε 1 implies that H is almost constant.

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Next, take the time derivative of the second of eq. (198), and then divide by 9H2φ, arrivingat

φ

3Hφ+

H

3H2' − V ′′

9H2' −

M2Pl

24πV ′′

V. (202)

Given that |H/H2| = ε 1 and that we are assuming φ Hφ, it follows that

η(φ ≡3M2

Pl

8π

(V ′′

V

) 1 . (203)

Assuming that we have a potential satisfying the slow roll constraints ε 1 and η 1,then it is easy to see that the potential can lead to inflation. We know that

a

a= H +H2 = H2 (1− ε) > 0 , (204)

since ε 1. Therefore the scale factor is accelerating and we have inflation.At the end of inflation comes the time when φ starts to oscillate, and decays, bumping

some of its kinetic energy back into the universe, rethermalizing it. During this period, in thesimplest models, one has

ρφ + (3H + Γφ)ρφ = 0 , ρrad + (4H − Γφ)ρrad = 0 , (205)

where ρrad is the energy density in the decay products, and Γφ is the width of the φ particle.The initial conditions at t = tasc are ρφ ' Vφosc , the potential energy just before oscillationsbegin, and ρrad ' 0. During the oscillation period, the universe is matter dominated. AttRH ' Γ−1

φ the φ particles decay and the world is once again radiation dominated, with anenergy density roughly given by ρ(tRH) ' V (tosc)(tosc/tRH)3 approximating the exponentialdecay of φ by a sudden decay at t = tRH .

One of the odd features of inflation is that the inflaton has to have a very flat potentialduring inflation, but cannot be very weakly coupled during the reheat epoch, or else the energydensity stored in the inflaton field will dilute away during the expansion of the oscillation periodbefore it decays and reheats the universe. There are a very large number of inflation modelsconsidered in the literature.

10.4 Fluctuations in the inflaton field

One should think of inflation as being a theory of initial conditions for the conventional bigbang theory. We have seen how the standard (non-accelerating) FRW universe is constructedon the assumption of homogeneity and isotropy, but then ironically leads to solutions thatappear to be inconsistent with homogeneity and isotropy without additional assumptions thatappear to violate causality. However, we can’t have the universe become too homogeneous,or else the spatial structures we see in the sky would never have formed! Amazingly enough,inflation is such a good theory of initial conditions, that it also offers an explanation of smalldeviations from homogeneity, which can subsequently, through gravitational instabilities, growinto the structure we see today. And amazingly enough, the perturbations that end up beinggalaxies and clusters of galaxies begin their life as quantum fluctuations.

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10.4.1 Quantum fluctuations of a free field

First consider a free field in flat space. It is described by a Lagrange density

L = 12∂µφ∂

µφ (206)

leading to the equation of motion

∂2φ = 0 . (207)

An arbitrary classical solution can be written in a Fourier decomposition as

φ(x, t) =∫

d3k2ωk(2π)3

ake−i(ωk−k·x) + c.c. , ωk = |k| . (208)

One can also compute the canonical momentum

Π(x, t) =∂L∂φ

= φ (209)

and impose canonical quantization conditions

[φ(x, t), Π(y, t)] = δ3(x− y) . (210)

It is easy to show that the Fourier coefficients ak in the classical expansion become operatorssatisfying

[ak, a†q] = 2ωk(2π)3δ3(k− q) , [ak, aq] = [a†k, a†q] = 0 . (211)

The as and a†s are just creation and annihilation operators similar to the ladder operatorsof the simp[le harmonic oscillator in quantum mechanics, although with a perhaps unfamiliarnormalization.

Next one defines the vacuum |0〉 to be the state annihilated by all of the aq operators:aq|0〉 = 0.

Although the vacuum contains no particles, the field does fluctuate in the vacuum. Supposewe write φ = φcl + δφ, where φcl is a classical solution to the equations of motion, and wequantize δφ. Note that δφ obeys the same equation of motion as φ and so it is quantized asabove. Define

δφp(t) =∫

d3x δφ(x, t)e−ip·x =1

2ωp

(ape

−iωpt + a†−peiωpt

). (212)

Then by using the commutation relations eq. (221) it is straightforward to compute

〈0| δφ†k(t)δφq(t) |0〉 =1

2ωk

12ωq

ei(ωk+ωq)〈0| a−ka†−q |0〉 =

12ωk

(2π)3δ3(k− q) . (213)

We define the spectrum Pφ(k) of δφ fluctuations by

〈0| δφ†k(t)δφq(t) |0〉 = Pφ(k)2π2

k3(2π)3δ3(k− q) , (214)

so that for the free field we have the spectrum

Pφ(k) =k3

4π2ωk=

(k

2π

)2

. (215)

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10.4.2 Quantum fluctuations in the inflaton field

Now consider the inflaton field, which we again write as φ = φcl + δφ. Plugging this into theequation of motion eq. (190) and expanding to first order in δφ we get

δφ+ 3Hδφ− ∇2

a2δφ+ V ′′(φcl)δφ = 0 . (216)

We now assume that V ′′(φcl) is ignorable and that H is constant, so that a ∝ eHt. In thatcase, it is possible to show that

δφ(x, t) =∫

d3k2ωk(2π)3

Nkck

(i+

k

aH

)ei(k/aH+k·x) + c.c. (217)

is a solution. Note that k is the comoving wave number, so that 2π/k is its comoving wave-length. The physical wavenumber is k/a.

Now consider the following expansion: I assume that (i) k aH (we consider a wavewell within the horizon); (ii) I expand about t = t, the time inflation begins, assuming that(t − t)H 1 (that is, we look at the wave for much less than the Hubble time, althoughperhaps for many oscillation times). I also set a(t) = 1 for convenience. The rationale fordoing this expansion is that the solution eq. (217) ought to look like the free particle in flatspacetime we considered in the previous section, and so we ought to be able to quantize it ina similar fashion; we need only to fix the normalization constant N so that the creation andannihilation operators c†, c have the same commutation relations as a† and a did. Under theexpansion

k

aH=

k

H− (t− t)k +O((t− t)2) (218)

and so in this regime,

δφ(x, t) '∫

d3k2ωk(2π)3

Nkckk

Hei(k/H−(t−t)k+k·x) . (219)

Note that we can drop the subleading term in the expansion of the prefactor, but not in theexponent...note that e1000 is not a good approximation to e1000+π even though π 1000. Ifwe define the normalization constant

Nk =H

ke−ik(t+1/H (220)

then δφ above looks like the free field solution for φ in eq. (208), and the quantization goesthrough as before, with

[ck, c†q] = 2ωk(2π)3δ3(k− q) , [ck, cq] = [c†k, c†q] = 0 , ωk = k . (221)

and we can define the vacuum as the state annihilated by c: ck|0〉 = 0.However, now look at the fluctuation after it has gone outside the horizon, at a time t when

k aH. Now we have

δφ(x, t) '∫

d3k2ωk(2π)3

iNkckei(k/aH+k·x) + h.c. (222)

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where Nk is given by eq. (220). If we define

δφp(t) =∫d3x δφ(x, t)e−ip·x (223)

as before, then when we compute the fluctuation spectrum we get

〈0| δφ†k(t)δφq(t) |0〉 =H2

2k2(2π)3δ3(k− q) , (224)

so that the spectrum as defined in eq. (214) is

Pφ(k) =(H

2π

)2

, (225)

which is evidently scale invariant.One can relax the rigid assumptions we made about V ′′ being negligible, H being constant,

etc, and calculate corrections to the spectrum in terms of properties of the scalar potential V .

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