1 Statistical Physics and Cosmology

The original aim of these notes was to provide a brief summary of the essentialpoints of each of the sixteen lectures. In previous years a handwritten versionwas circulated after each lecture and this is reflected in the rather informalstructure of the current typscript. The notes are not intended as a systematicaccount of statistical physics, nor as a treatise on cosmology. For that thestudent is directed to the recommended textbooks.

It should be borne in mind that these notes do not contain all the ex-planations and motivational remarks, nor all the steps which were spelt inthe lectures. The required algebraic manipulations are almost always ratherstraightforward and should cause no great difficulty. What is more difficultto acquire is the physical intuition behind the calculations. A major aim ofthe lectures as delivered is to attempt to impart that intuition.

1.1 Pre-requisites

The study of statistical physics and cosmology is not one which can be em-barked upon with absolutely no pre-requistites. It calls upon quite a widerange of facts and ideas drawn from various sources. However as far as thepresent course is concerned, the small number of astrophysical facts neededare contained herein. One also needs some familiarity with elementary spe-cial relativity and quantum mechanics. The course has been deliberatelydesigned so that no knowledge of general relativity is required.

1.2 Web Pages

Students wishing to find out more about modern cosmology including prettypictures of galaxies etc might like to consult the DAMTP Relativity Groupweb pages. The URL is www.damtp.cam.ac.uk/user/gr/public/

1.3 Suitable textbooks

The following is a list of suitable textbooks at the level of the course. Unfor-tunately Pointon is out of print but it should be available in college libraries.Students report that they have found Guenault most helpful. For cosmology,For comsology I think that Roos or the newcomer, Linder, are probably thebooks closest to the present level. They do however contain some generalrelativity Weinberg’s book is outstanding and usually found very helpful. Itsucceeds in conveying all the main ideas in an an entirely non-technical wayand yet contains many of the main formulae in a technical appendix. The

1

older books by Sciama and by Bondi are very good on the basic facts. Inparticular that by Bondi discusses the Newtonian approach to cosmologyadopted in this course. Those by Kolb and Turner and by Borner will takeyou well into the frontiers of the subject but are probaly to advanced for afirst look at the subject.

Statistical Physics

1 ? A J Pointon: Introduction to Statistical Physics for Students

2 T Guenault: Statistical Physics

3 M G Bowler: Lectures on Statistical Mechanics

Cosmology

1 M Berry: Principles of Cosmology and Gravitation

2 ? S Weinberg The First Three Minutes

3 J N Islam An Introduction to Mathematical Relativity

4 M Roos Introduction to Cosmology

5 E W Kolb and M S Turner The Early Universe

6 D W Sciama Modern Cosmology

7 G Borner The Early Universe

8 H Bondi Cosmology

9 E V Linder First Principles of Cosmology

Background Reading on Modern Physics

1 F J Blatt Modern Physics

Contents

1 Statistical Physics and Cosmology 11.1 Pre-requisites . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Web Pages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Suitable textbooks . . . . . . . . . . . . . . . . . . . . . . . . 1

2

2 Introduction 4

3 Statistical Physics 53.1 Kinetic Theory of Gases . . . . . . . . . . . . . . . . . . . . . 53.2 Statistical Mechanics . . . . . . . . . . . . . . . . . . . . . . . 9

3.2.1 The Most Probable Distribtion . . . . . . . . . . . . . 93.2.2 Consistency and Interpretation of Lagrange Multipliers 113.2.3 Interpretation of α and β . . . . . . . . . . . . . . . . . 113.2.4 Boltzmann Distribution with Degeneracy . . . . . . . . 15

3.3 Non-Relativistic Maxwell-Boltzmann Gas . . . . . . . . . . . . 153.3.1 Gibbs Paradox . . . . . . . . . . . . . . . . . . . . . . 193.3.2 Application to Cosmology: Ionisation . . . . . . . . . . 20

3.4 Relativistic Maxwell-Boltzmann Gas . . . . . . . . . . . . . . 223.5 Bosons and Fermions . . . . . . . . . . . . . . . . . . . . . . . 243.6 Bose-Einstein distribution . . . . . . . . . . . . . . . . . . . . 25

3.6.1 The photon gas . . . . . . . . . . . . . . . . . . . . . . 253.7 Fermi-Dirac distribution . . . . . . . . . . . . . . . . . . . . . 28

3.7.1 Massless Neutrinos . . . . . . . . . . . . . . . . . . . . 283.7.2 Degenerate Fermi Gas . . . . . . . . . . . . . . . . . . 29

3.8 White Dwarfs, Neutron Stars and Black Holes . . . . . . . . . 313.8.1 White Dwarfs . . . . . . . . . . . . . . . . . . . . . . . 313.8.2 Neutron Stars . . . . . . . . . . . . . . . . . . . . . . . 323.8.3 Black Holes . . . . . . . . . . . . . . . . . . . . . . . . 323.8.4 Stellar Structure . . . . . . . . . . . . . . . . . . . . . 323.8.5 Virial Theorem . . . . . . . . . . . . . . . . . . . . . . 333.8.6 Order of Magnitude Estimates . . . . . . . . . . . . . . 33

4 Cosmology 354.1 Isotropy & Homogeneity of the Universe . . . . . . . . . . . . 35

4.1.1 Cosmic Distance Ladder . . . . . . . . . . . . . . . . . 354.1.2 Hubble Expansion . . . . . . . . . . . . . . . . . . . . 364.1.3 Cosmic microwave background . . . . . . . . . . . . . . 37

4.2 Kinematics of Homogeneous Isotropic Expansion . . . . . . . . 374.2.1 Homogeneity and the Cosmological Principle . . . . . . 384.2.2 Behaviour of Distances, Areas, and Volumes . . . . . . 39

4.3 Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.3.1 Dynamics of Isotropic Homogeneous Pressure-Free Fluid 404.3.2 Dynamics of Homogeneous Isotropic Fluid with Pressure 444.3.3 General Upper Bound on the Age of the Universe . . . 46

4.4 Redshifting & Motion of Light . . . . . . . . . . . . . . . . . . 474.4.1 Proof of Redshift Formula . . . . . . . . . . . . . . . . 48

3

4.4.2 Singularity Theorems (Hawking & Penrose) . . . . . . 504.5 Thermal History of the Universe . . . . . . . . . . . . . . . . . 50

4.5.1 Decoupling . . . . . . . . . . . . . . . . . . . . . . . . 514.5.2 Decoupling Temperature & Temperature at End of Ra-

diation Era . . . . . . . . . . . . . . . . . . . . . . . . 524.5.3 Nucleosynthesis . . . . . . . . . . . . . . . . . . . . . . 534.5.4 Calculation of Relic Neutrino Number Density . . . . . 544.5.5 Comparison with observations . . . . . . . . . . . . . . 55

4.6 Current problems . . . . . . . . . . . . . . . . . . . . . . . . . 564.6.1 Initial Conditions . . . . . . . . . . . . . . . . . . . . . 564.6.2 The solution? . . . . . . . . . . . . . . . . . . . . . . . 58

2 Introduction

Modern cosmology is based on two basic observational facts.

The Hubble expansion The observed luminous universe is made up ofsystems of stars like our own Milky Way called galaxies . Each galaxy con-tains up to 1012 stars.

In 1929 Edwin Hubble discovered that the light coming towards us fromthese galaxies is systematically ”redshifted”, the observed wavelengths λo

are increased by a factor (1 + z) over their emitted values λe. The (positive)quantity z is called the ”redshift”. He interpreted this as being due to theDoppler effect. The galaxies he claimed are systematically moving awy fromus. Special relativity gives the recession speed (assuming they move alongthe line of sight) from the formula

1 + z =λo

λe

=

√c + v

c− v.

For small z and hence small vc

we have

z ≈ v

c.

Hubble obtained an estimate of the distances r of the galaxies by selectingthose he thought had the same intrinsic total luminosity L , that is, the sametotal energy emitted per unit time in their rest frame. Nowadays such objectsare referred to as ”standard candles”. He actually measured their observedbrightnesses S (the energy we receive per unit are per unit time). If the usuallaws of geometry and optics hold and we ignore the redshifting, we have

S =L

4πr2.

4

Hubble plotted logz against −logS and obtained a straight line with slope12. He deduced the Hubble Law

v = Hr,

where the constant of proportionality H is called ”Hubble’s constant”.

The Cosmic Microwave Background In 1965 two engineers at the BellTelephone Labs in New Jersey, (Penzias and Wilson) were investigating in-terference in the propagation of microwaves (i.e. electromagnetic radiationof about a centimetre wavelength). They realized that the earth is bathedin an isotropic bath of microwaves with wavelengths peaked at about 3 cmwavelength. The spectrum is ”Planckian”, that is it is characteristic of a”black body” with temperature T ≈ 3K. Another way to say this is thatthe earth is imerssed in a gas of about 400 photons per cm3 , with a thermaldistribution.

These two facts lead, almost immediately to the

Hot Big Bang Theory which states that the universe we see today hasexpanded from a previous state of high density and high temperature.

To understand the universe therefore we must understand matter at highdensity and high pressure. In order to do so we need first to develop thekinetic theory of gases and the elements of statistical mechanics. Havingdone that we shall return to apply our knowledge to cosmology.

3 Statistical Physics

3.1 Kinetic Theory of Gases

The aim of kinetic theory of gases is to describe their macroscopic proper-ties , such as pressure P , temperature T , and volume V and the relationshipsbewteen them in terms of microscopic dynamical properties of the molecules,atoms, or other particles making up the gas. The theory was developed orig-inally in the eighteenth century to treat such familar gases as oxygen andhydrogen at ordinary pressures and temperatures. Later it was extended to

5

cover more exotic gases in which the particles move relativistically includ-ing photons or neutrinos. It is these cases which are important for cosmology.

Initially we focus on n(v)d3v =, the average number of particles per unitvolume with velocity vectors ~v inside a cell d3v in velocity space. Later, whenwe consider photons or neutrinos which always move at the speed of light, itwill turn out to be more convenient to work in terms of the distribution oftheir momenta ~p.

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v

vz

vy

vx

Let’s introduce polar coordinates in velocity space:

d3v = v2dv dΩ = v2dv sin θ dθ dφ, (1)

where dΩ is the element of solid angle.We now assume an isotropic distribution:

n(~v) = n(|~v|). (2)

The number of particles with speed v → v + dv (v = |~v|) per unit volume is

f(v)dv =

∫ 2π

0

dφ

∫ π

0

dθ n(v)v2 sin θ dθ dφ = 4πn(v)v2dv. (3)

If we assume that the only energy is kinetic energy (i.e. we have a gas of”free particles”), then the energy density is

U =

∫ ∫ ∫

ε(v)n(~v)d3v, (4)

where

ε(v) =1

2mv2 non-relativistic case

ε(v) =mc2

√

1− v2/c2relativistic case.

(5)

6

Thus

U =

∫ ∞

0

f(v)mv2

2dv non-relativistic (6)

U =

∫ c

0

f(v)mc2

√

1− v2/c2(7)

relativistic (n.b. includes rest mass energy).

Number flux The number of particles striking the surface per unit areaper unit time in the direction (θ, φ) is

f(v)dv v cos θsin θ dθ dφ

4π(8)

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θ

since

sin θ dθ dφ

4πf(v)dv (9)

is the number/unit volume in solid angle sin θ dθ dφ, and all particles strikingthe surface in unit time arrived from inside the cylinder whose volume isv cos θ.

The total numbers striking the surface per unit area, per unit time is

∫ π/2

0

dθ

∫ 2π

0

dφf(v)

4πv dv cos θ sinθ =

f(v)v dv

4. (10)

Pressure: This is the momentum reversed per unit area per unit time indirection (θ, φ),

f(v)dv 2pv cos θ cos θsin θ dθ dφ

4π. (11)

7

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The total pressure is

P =

∫

dv

∫ π/2

0

dθ

∫ 2π

0

dφ f(v)2pv

4πcos2 θ sin θ

P =1

3

∫

pf(v)vdv.

In the non-relativistic case p = mv, so P = 13

∫∞

0mv2f(v)dv,

P =2

3U (12)

This does not include the rest mass energy.

In the relativistic case p = mv/√

1− v2/c2, so in the extreme relativisticcase as v → c, ε→ pc and

P → 1

3U (13)

This includes the rest mass energy. However if the particles are ultra-relativistic the contribution from the rest mass will be negligible.

The previous results followed without knowing the detailed form of n(~v).Maxwell guessed that

i) n(~v) = n(v2

x + v2y + v2

z

)

ii) n(~v) = g(vx)g(vy)g(vz)

(i.e. vx, vy, vz are independently distributed) and deduced that n(~v) is aGaussian or Normal distribution:

n(~v) = constant× exp[−constant′

(v2

x + v2y + v2

z

)]. (14)

We shall derive this (in the non-relativistic case) using statistical mechanics.

8

3.2 Statistical Mechanics

The idea is that for given values of the macroscopic variables there are manypossible microscopic configurations. In order to calculate the macroscopicvariables we need to know how the total energy is distributed among themany possible microstates of the system. We specify a distribution by givingthe occupation number ni telling us how many particles are in state i withenergy Ei. Each distribution occurs with a certain weight W (n1, n2, . . .)equal to the number of different microscopic configurations giving rise to thesame distribution of energies.

3.2.1 The Most Probable Distribtion

The idea is that we get a good estimate of the behaviour of the system byconfining attention to the most probable distribution which is obtained bymaximising log W subject to the constraints:

∑

i

ni = N, total number of particles (15)

∑

i

Eini = E, total energy of particles. (16)

Because N is very large (of the order of 1023 per cm3 for a gas at room tem-perature and atmospheric pressure) we treat the ni’s as continuous variablesand extremize

log W − β

(∑

i

Eini − E

)

− α

(∑

i

ni −N

)

(17)

with respect to the ni where α and β are Lagrange multipliers.

1

W

∂W

∂ni

∣∣∣∣ni=ni

= βEi + α (18)

Example: Maxwell-Boltzmann statistics for N distinguishable particles.We can imagine picking out the particles from a bag containing N parti-cles in N ! ways and assigning them to the k possible energy states. Theorder in which we assign the ni particles to the i’th state does not matterand therefore:

W =N !

n1!n2! · · ·nk!(19)

1

W

∂W

∂ni= − log ni. (20)

9

Using the Stirling approximation log x ' x log x− x,

ni = exp−(βEi + α). (21)

In fact (see later) β = 1/(kT ) where k is Boltzmann’s constant, T = tem-perature, and α = −βµ, where µ is the chemical potential.

If we are given N and E we can calculate T and µ by substituting theBoltzmann distribution into the constraints (15, 16).

Thus for a two-level system, k = 2,

n1

n2

= exp−∆E

kT, ∆E = E1 − E2. (22)

An application to Cosmology

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•

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star orgalaxy

photon

gas at temperature Tcapable of being in two levels

••

•

•

•

∗•

•

•

•

The observed light intensity shows absorption lines.

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intensity

EE1 E2

〈∆E〉

Assume the rate of absorption ∝ n1/n2, then knowing ∆E and n1/n2 weestimate T , e.g.

i) Cyanogen molecules (CN) in our galaxy were observed before Penzias& Wilson in 1965 to have T ∼ 3K.

10

ii) In 1994, observations of a cloud of neutral carbon atoms seen in frontof the quasar Q1331+170 which has a redshift z ' 1.776 were observedto have T ' 7.4± 0.8K. At redshift z the microwave background had

T = (1 + z) 2.726± 0.010 (23)

(we shall prove this later).

3.2.2 Consistency and Interpretation of Lagrange Multipliers

To complete our calculation we need:

i) to check that W (n1, n2, . . . , nk) is indeed highly peaked at its mostprobable value.

ii) to give an interpretation of α and β.

Sharpness of Maximum Use Taylor expansion for log W :

∂2log W

∂ni2

= − 1

ni. (24)

This implies

W = W exp−∑

i

(ni − ni)2

2ni(Gaussian). (25)

The standard deviation of (ni− ni)/ni is 1/√

ni, which is very small for largeni.

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pppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppp

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

W

1 ni/ni

width ∼ 1/√

ni

3.2.3 Interpretation of α and β

There are 3 steps.

11

I: Thermal Equilibrium of Independent Systems

1 2

W1 W2

Assume W = W1W2, log W = log W1 + log W2.

i) Allow the exchange of energy, but not particles. There are three con-straints and therefore three Lagrange multipliers (not four):

Extremize:

log W1 + log W2 − α1

(∑

i

n(1)i −N (1)

)

− α2

(∑

i

n(2)i −N (2)

)

−β

(∑

i

n(1)i E

(1)1 +

∑

i

n(2)i E

(2)1

)

. (26)

Then

1

W1

∂W1

∂n(1)i

= β(

E(1)i − µ1

)

, α1 = −βµ1

1

W2

∂W2

∂n(2)i

= β(

E(2)i − µ2

)

, α2 = −βµ2

(27)

e.g. Maxwell-Boltzmann case

n(1)i = exp−β

(

E(1)i − µ1

)

(28)

n(2)i = exp−β

(

E(2)i − µ2

)

. (29)

Thus, two systems in thermal contact at equilibrium have the same

value of β (cf. temperature).

ii) Remove the partition and allow free exchange of both energy and par-ticles, then there are two constraints and two Lagrange multipliers.

... .

....

. ..

...

. ..

.

. . .

...

12

Extremize:

log W1 + log W2 − α

(∑

i

n(1)i +

∑

i

n(2)i −N

)

−β

(∑

i

n(1)i E

(1)i +

∑

i

n(2)i E

(2)i

)

. (30)

This implies

1

W1

∂W1

∂n(1)i

= β(

E(1)i − µ

)

, α = −βµ (31)

1

W2

∂W2

∂n(2)i

= β(

E(2)i − µ

)

. (32)

Therefore, two systems in thermal contact that are allowed to exchange

particles have the save values of µ (cf. chemical potential).

II: Definition of Entropy

S = k log W = k max log W (33)

S is a measure of the likelihood or probability of the equilibrium configura-tions.

Consider two independent systems allowed to come into equilibrium.

S1 S2

S1 = maxn

(1)i

log W1, S2 = maxn

(2)i

log W2. (34)

This system is subject to four constraints.The combined entropy S3,

S3 = maxn

(1)i ,n

(2)i

(log W1 + log W2) (35)

is subject to fewer constraints (i.e. 3 or 2).

S3 ≥ S1 + S2. (36)

Entropy cannot decrease when systems come into equilibrium (2nd Law ofThermodynamics).

13

III: Quasi-Static Variation of Equilibria We have seen that µ and βhave properties resembling chemical potential and 1/(kT ), but to make thecorrespondence quantitative we consider changing (very slowly) N , E, andV through a sequence of equilibrium configurations. We know (or. lookup)from thermodynamics that we have the Thermodynamic Relation,

dE = −PdV + µdN + TdS.change in work gain in increase

internal done potential in heat

energy energy energy

(37)

[It is not obvious that the increase in heat energy can be written as TdS.This is established in thermodynamics books. However we shall be able toprove this using statistical mechanics.]

Regarding S = k log W as a function of E, N , and V .

1

kdS =

∑

i

1

W

(∂W

∂ni

)

n=ni

dni =∑

i

(βEi + α)dni, (38)

where

E =∑

i

niEi, N =∑

i

ni. (39)

Then1

kdS = βdE + αdN −

∑

i

βinidEi. (40)

We interpret∑

nidEi = dW = −PdV as the work done on the systemassuming that no particles or energy are added (i.e. “adiabatically”). Wefind that if β = 1/(kT ), α = −βµ, equation (40) becomes

dE = TdS + µdN − PdV (41)

Conclusion: We have shown that:

1) The quantities T and µ are equal for two systems in equilibrium.

2) For quasi-static variations of equilibrium configurations

dE = TdS + µdN − PdV.

3) For two systems allowed to come into equilibrium

S3 ≥ S1 + S2.

These are the “zero’th, first, and second laws of thermodynamics” respec-tively. They justify calling β = 1/(kT ) the temperature and µ = −α/β thechemical potential.

14

3.2.4 Boltzmann Distribution with Degeneracy

If there are gi distinct energy states with energy Ei then

W = N !∏

i

gni

i

ni!. (42)

Going through the same steps as before, we find that

ni = gi exp [−β(Ei − µ)] (43)

e.g.

i) spin degeneracy of an electron at rest: g = 2, ↑ or ↓.In absence of magnetic fields there exist two states with angular mo-mentum projection ±~/2.

ii) a photon with fixed momentum has two polarisation states (plane po-larised or circularly polarised). For circular polarisation angular mo-mentum along the direction of motion is ±~:

“two helicity states”

pppppppppppppppppppppppppppppppppppppp

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.......................................

.....................................

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...

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...........................................................................................................................................................

......................................

left-handedright-handed

iii) any particle with fixed energy E is degenerate w.r.t. the direction ofthe momentum ~p.

3.3 Non-Relativistic Maxwell-Boltzmann Gas

To find the states we solve the non-relativistic Schrodinger equation for aparticle of mass M in a box of sides (a, b, c) subject to periodic boundaryconditions:

− ~2

2M∇2Ψi = EiΨi. (44)

The solutions are

Ψi = exp 2πi

(xl

a+

ym

b+

zn

c

)

= exp i~k · ~x, (45)

with i↔ (l, m, n) ∈ Z× Z× Z, and

Ei =~

2

2M(2π)2

(l2

a2+

m2

b2+

n2

c2

)

=h2

2M

(l2

a2+

m2

b2+

n2

c2

)

, (46)

where h = 2π~.

15

Allowed momenta These are given by

~p = ~~k = h

(l

a,m

b,n

c

)

, (47)

and are uniformly distributed on a lattice in momentum space with density

abc

h3=

V

h3, (48)

where V = abc is the volume of the box.Now if V is large the levels are very closely spaced and we can pass from

a summation over l, m, n to an integral by the rule:

∑

i

↔∑

l,m,n

↔∫

R3

V d3p

h3. (49)

Thus

n(~p)d3~p =d3p

h3exp

(

−p2

2M− µ

kT

)

, (50)

which implies

n(~v) ∝ exp

(

−M~v 2

2kT

)

, (51)

which is indeed a Gaussian distribution as Maxwell predicted.The total number of particles in volume V is given by

N = V

∫

n(p)d3p =V

h3

∫ ∫ ∫

d3p exp

(

− p2

2MkT

)

expµ

kT,

=V

h3exp

µ

kT

[∫ +∞

−∞

dpx exp

(

− p2x

2MkT

)]3

,

N =V

h3(2πMkT )3/2 exp

µ

kT(52)

To calculate the total energy E one can differentiate N w.r.t. M:

∂N

∂M=

V

MkT

∫p2

2Mn(p)d3p (53)

Then

E = MkT∂N

∂M, (54)

16

which gives

E =3

2kT

V

h3(2πMkT )3/2 exp µ/(kT ) (55)

or using the expression for N above,

E =3

2NkT (56)

A note on boundary conditions Students encountering the density ofstates formula for the first time are often worried that it depends upon thechoice of periodic boundary conditions. For example if one imposes Dirichletboundary conditions, Ψ = 0 on the sides of the box, as one often does inquantum mechanics courses one would get (un-normalized)-solutions of theform

Ψ = sin(lπx

a) sin(

mπx

b) sin(

nπx

c),

with (l, m, n) positive integers. These are of course not momentum eigen-states but they are energy eigen-states. One may label these energy eigensstates by a lattice of points lying in the positive octant of momentum spacebut now with 2×2×2 = 8 times the density we found before. The net resultis that the number of states having energies between E and E + dE is un-changed. In practice it is usually most convenient to use periodic boundaryconditions. The number of states with energies between E and E + dE (inthe limit of large volume V ) are found not to depend on the precise boundaryconditions. To check this one may repeat the calculation when one imposesthe Neumann bouundary condition that the normal derivative of Ψ vanisheson the sides of the box. This amounts to replacing the sine function with thecosine function. If you are still sceptical try various combinations of periodic,Neumann and Dirichlet boundary conditions.

Pressure

P =2

3U =

2

3

E

V=

NkT

V. (57)

This is Boyle-Charles Law

PV = NkT (58)

17

Entropy

S

k= log W = N log N −N +

∑

ni log gi −∑

(ni log ni − ni), (59)

but log gi − log ni = β(Ei − µ), therefore,

S = Nk log N + kβ(E − µN). (60)

Substitution from the expressions for E gives

S = Nk log

[V

h3(2πMkT )3/2

]

+3

2Nk (61)

Application For an adiabatic expansion the added heta energy δQ = 0which implies that the change in entropy dS = 0. Thus an adiabatic processis one for which S = constant at constant N . This implies

V T 3/2 = constant (62)

Using Boyle-Charles Law

PV 5/3 = constant (63)

Alternative Derivation of Pressure Using the Thermodynamic Re-lation

S = S(E, T, N), (64)

TdS = dE − µdN + PdV. (65)

Therefore

P = T

(∂S

∂V

)

E,N

. (66)

Now use the expression E = 23NkT to eliminate T from (61) and get

S = Nk log

[

V

h2

(

2πM2

3E

)3/2]

+3

2Nk, (67)

which implies

P =NkT

V. (68)

18

3.3.1 Gibbs Paradox

We used earlier that for independent systems at some T and µ:

1 2

i) W3 = W1W2.

ii) S3 = S1 + S2, entropy is extensive.

Both are false in our case!. This is esaily seen using the formula for theweight

W = N !∏

i

gni

i

ni!. (69)

As a consequence one arrives at certain paradoxes, e.g. the entropy of a gascan change by removing a fictitious partition. Sackur & Tetrode suggested(before quantum mechanics) the replacement:

W →∏

i

gni

i

ni!(70)

(removing N !) since (N1 + N2) 6= N1!N2!.This alters the entropy to

S =5

2Nk + Nk log

[V

Nh3(2πMkT )3/2

]

(71)

Now S is “extensive”, i.e. sending N → 2N and V → 2V sends S → 2S.The old formula has the effect that if N → 2N and V → 2V then S →2S + Nk log 2 !

Note however that both formulae give the same answer for the pressureP and other physically measurable quantities (as long as N is held fixed).The origin of this problem is that we have assumed our particles are distin-

guishable. This can only be corrected by using quantum statistics. This wewill do shortly.

19

3.3.2 Application to Cosmology: Ionisation

The following calculation is representative of how we use the ideas we havedeveloped above are used to study gases with more than one species of par-ticle. We think of each species of particle as defining a system. They areallowed to exchange energy and if chemical reactions are allowed they arealso allowed to change their number, subject to the constraints imposed byconservation laws.

A neutral gas of hydrogen atoms will heat up on adiabatic compression(T ∝ V −2/3) and eventually become ionised:

H → e + p − Eion

hydrogen electron proton

atom g = 12 g = 1

2

−e +e

(72)

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p

e

••

e+••

−

p

where Eion is the ionisation energy ' 13.6 eV .At low temperature H is neutral ⇒ transparent.At high temperature H is ionised ⇒ opaque, because it is an

electrically conducting “plasma”.

Calculation of the Amount of Ionisation: Saha’s Equation We max-imise

log W = log W+ + log W− + log W0, (73)

subject to three constraints, where+ ↔ proton− ↔ electron0 ↔ H atom

.

During ionisation,

N+ → N+ + 1N− → N− + 1N0 → N0 − 1

⇒ N+ + N0 = constantN− + N0 = constant

20

Let

E+ =p2

+

2M+, E− =

p2−

2M−

, E0 =p2

0

2M0, (74)

be kinetic energies. The hydrogen atom’s energy is potential plus kinetic,

EH =p2

0

2M0

− Eion, (Eion > 0). (75)

Energy conservation implies that

n+E+ + n−E− + n0(E0 − Eion) = constant, (76)

so we extremize

log W − β(n+E+ + n−E− + n0(E0 − Eion)) (77)

−α+(n+ + n0)− α−(n− + n0).

This is as if we had three independent systems, but with all three tempera-tures equal,

β+ = β− = β0, (78)

and

α0 = α+ + α− − βEion, (79)

or chemical potentials,

µ+ = −α+/β,µ− = −α−/β,µ0 = µ+ + µ− + Eion.

(80)

Therefore

N±

V=

g+

h3(2πM±kT )3/2 exp βµ±, (81)

N0

V=

g0

h3(2πM0kT )3/2 exp β(µ+ + µ− + Eion), (82)

n+n−

n0

=g+g−g0

(2πMekT

h2

)3/2

exp (−βEion) (83)

where we have used the fact that M+ 'M0 = Mp, the proton mass, and Me

is the electron mass.Note that

n+n−

n0 1, unless

kT ≥ Eion

i.e. T >∼ 3× 103K.

21

Application to Cosmology At around t ' 105 yrs after the Big Bang theredshifted temperature of the universe, T = 3K(1+ z) was around 3× 103K(z ∼ 103) and all hydrogen would have been ionised and hence opaque. Wewill study this in detail later.

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.......... .......... .......... .......... .......... .......... .......... ..........

............................................................................................................................................................................................................................ .....

.......................................................................................................................................................................................................................

105 yrstime

Big Bang

TRANSPARENTT < 3× 103K

T > 3× 103K OPAQUE

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..

1.5× 1010 yrs (now)

3.4 Relativistic Maxwell-Boltzmann Gas

To find the allowed energy levels we need the analogue of Shrodinger for arelativistic particle. Strictly speaking, this takes us a little beyond elementaryquantum mechanics. However we can follow the pattern which should befamilar. We make the replacements:

E → −~

i

∂

∂t, ~p→ ~

i~∇, (84)

in the special relativity relation between energy and momentum

E2 = ~p 2c2 + M2c4, (85)

to get what is called the Klein-Gordon Equation,

1

c2

∂2

∂t2φ = ~∇ 2φ− M2c2

~2φ (86)

n.b. if M = 0 we recover the wave equation.We solve this by separation of variables in a box of sides (a, b, c) subject

to periodic boundary conditions:

φ = e−iEt/~ei~k·~x, ~k = 2π

(l

a,m

b,n

c

)

, (87)

as before, and we conclude that the allowed values of the momentum ~p arethe same as in the non-relativistic case. This is true both for massive andfor massless particles.

22

Thus we obtain the Juttner distribution:

n(~p)d3p =d3p

h3exp

[

−√

~p 2c2 + M2c4 − µ

kT

]

, (88)

N =V

h3

∫

d3p exp(

−√

~p 2c2 + M2c4)

expµ

kT. (89)

This is difficult to evaluate exactly unless M = 0 (zero rest mass case). Weshall proceed indirectly in order to illustrate the use of the thermodynamicrelation. Define the Helmholtz Free Energy F = F (T, V, N) by

F ≡ E − TS ⇒ dF = −PdV − SdT + µdN, (90)

P = − ∂F

∂V

∣∣∣∣T,N

(91)

Now

N =V

h3exp

µ

kTf(T, M, c), (92)

for some function f(T, M, c). Moreover we showed earlier(60) that in general,for the Maxwell Boltzmann distribution,

S = Nk log N +1

T(E − µN). (93)

Therefore,

F = N(µ− kT log N), (94)

= −kTN log

(V

h3f(T, M, c)

)

. (95)

Using (91), P = NkT/V or

PV = NkT (96)

i.e. the Juttner gas obeys the familiar Boyle-Charles Law.

23

Order of Magnitudes The Juttner distribution differs significantly fromthe Maxwell distribution when Mc2 ' kT . In the early universe the lowesttemperature when this happens is T = mec

2/k ' 1010K since the electronwith mass me is the lightest known massive particle:

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.................................................................................................................................................................................................................................................................................................Big Bang

Electrons bound inhydrogen atoms

Relativistic electrons

Ionized plasma, free electrons, non-relativistic

now, T ' 3K, t ∼ 15 Gyrs

T ∼ 3000K, t ∼ 105 yrs

T ∼ 1010K, t ∼ 1 sec

3.5 Bosons and Fermions

Swapping the position of two particles which are indistinguishable shouldmake no physical difference,

|Ψ(~x1, ~x2)| = |Ψ(~x2, ~x1)| .We may realize this in two ways:

Bosons: Ψ(~x1, ~x2) = Ψ(~x2, ~x1) orFermions: Ψ(~x1, ~x2) = −Ψ(~x2, ~x1)

The spin statistics theorem states that:

Bosons have integer spinss

~∈ Z

Fermions have 12 integer spins

s

~∈ Z + 1

2

Examples

electron s = 12~ fermion

photon s = ~ bosonpion s = 0 bosonproton s = 1

2~ fermionneutron s = 1

2~ fermionquark s = 1

2~ fermionneutrino s = 1

2~ fermion

24

3.6 Bose-Einstein distribution

For Bosons the counting problem amounts to putting ni indistingushableobjects in gi boxes. We lay out the ni objects in a line, separated by gi − 1partitions. We have (ni + gi− 1)! ways of laying them out in order. Howeverthe order in which we select the objects and the order in which we select thepartitions does not matter. Therefore:

W = WBE =∏

i

(gi + ni − 1)!

(gi − 1)!ni!, (97)

assume ni 1

∂log WBE

∂ni= log(gi + ni − 1)− log ni (98)

Then

ni =gi − 1

eβ(Ei−µ) − 1,

assume gi 1. This is the Bose-Einstein distribution

ni =gi

eβ(Ei−µ) − 1(99)

To get the classical, Maxwell-Boltzmann limit assume gi ni⇔ β(Ei−µ)1, then

ni ' gi exp [−β(Ei − µ)] (100)

3.6.1 The photon gas

E = pc = ~ω, µ = 0

n(p)d3p = 2× V

h3

d3p

exp(βpc)− 1(101)

(n.b. the factor of 2 comes from two polarisation states). We set µ = 0because photon number is not conserved.

Planck spectrum: The number of photons with energy ω → ω + dω perunit volume is

1

c3

1

π2

ω2dω

exp β~ω − 1(102)

Wien Displacement law : This expression is of the form

ω2dωf

(~ω

kT

)

for some function f .

25

Consequences of the Wien Displacement Law

i) the maximum scales with T as ωmax ∝ T , i.e. λmaxT = constant.

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...............

ωωmax

low Thigh T

ii) if we slowly (i.e. adiabatically) expand a box of thermal photons suchthat

V → V (1 + z)3,ω → ω/(1 + z),T → T/(1 + z),

the shape of the distribution will not change and the number of photonsin the interval ω → ωdω will equal the number in the interval

ω

1 + z→ ω + dω

1 + z

This happens to the microwave background as the universe expands.

Note that from the observed energy flux we can estimate the tem-perature (T ∼ 3K) and the number density (∼ 400 cm−3), and theenergy density (∼ 10−35 g/cm3). Since there is less than one protonand electron per m3, photons are by far the most plentiful particles in

the universe.

N =

∫ ∞

0

n(ω)dω =

∫ ∞

0

V

c3

1

π2

ω2dω

exp(β~ω)− 1, (103)

= 8π

(kT

hc

)3

V

∫ ∞

0

x2dx

ex − 1. (104)

Using

∫xndx

ex − 1= n!ζ(n + 1), where ζ(s) =

∞∑

n=1

1

ns(105)

26

we obtain

N = 16π

(kT

hc

)3

V ζ(3) (106)

E =

∫ ∞

0

~ωn(ω) dω, (107)

=8πh

c3

(kT

h

)4 ∫ ∞

0

x3dx

ex − 1. (108)

Using ζ(4) =∑∞

1 1/n4 = π4/90, we find

E =8π5

15h3

k4

c3T 4V (109)

Entropy

E = aT 4V, a =8π5k4

1h3c3. (110)

µ = 0 ⇒ dE = TdS = −PdV , therefore

∂S

∂E

∣∣∣∣V

=1

T=

(aV

E

)1/4

⇒ S =4

3(aV )1/4E3/4 + f(V ). (111)

S = 0 as T ↓ 0 implies

S =4a

3T 3V (112)

Pressure

F = E − TS = −a

3T 4V, P = −∂F

∂V

∣∣∣∣T

. (113)

P =1

3

E

V=

1

3energy density (114)

Adiabatic Expansion

S = constant ⇒ V T 3 = constant⇒ PV 4/3 = constant

27

3.7 Fermi-Dirac distribution

If no more than one of gi states may be occupied, then given ni particles, wemust choose ni of the states to fill

WFD =∏

i

(gi

ni

)

=∏

i

gi!

n!(gi − ni)!, (115)

therefore

∂log WFD

∂ni

= − log ni + log(gi − ni) (116)

and

ni =gi

exp β(Ei − µ) + 1(117)

3.7.1 Massless Neutrinos

n(~p)d3p = 2× V

h3

d3p

exp βpc + 1(118)

(n.b. the factor of 2 comes from the two helicity states).

N = 8π

(kT

hc

)3

V

∫ ∞

0

x2dx

ex + 1, (119)

E =8πh

c3

(kT

h

)4 ∫ ∞

0

x3dx

ex + 1, (120)

but∫

∞

0

xndx

ex + 1=

(

1− 1

2n

)∫∞

0

xndx

ex − 1, ‘ (121)

therefore

Nneutrinos =3

4Nphotons

Eneutrinos =7

8Ephotons

(122)

28

3.7.2 Degenerate Fermi Gas

Let µ = EF , “Fermi energy”.

limβ→∞

1

exp β(E − EF ) + 1= 1 E < EF (123)

= 0 E > EF

Therefore, if βEF 1 all states are uniformly populated inside a sphere ofradius

PF =

√

E2F −M2c4

c2, (124)

in momentum space.

N

V≡ n =

4π

3h3gsP

3F (125)

where gs is spin degeneracy.

E

V=

4πgs

h3

∫ PF

0

P 2√

P 2c2 + M2c4dP (126)

Calculation of pressure as a function of density dS = 0 impliesP = − ∂E

∂V|S. The answer depends on n( ~

Mc)3.

For n (Mc~

)3, the particles are non-relativistic, and

E

V= gsMc2 +

4πgs

h3

P 5F

10M(127)

= gsMc2 +

(4πgs

h3

)2/3 (3N

V

)5/31

10M, (128)

P =2

3

(E

V− gsMc2

)

. (129)

For n (Mc~

)3, the particles are relativistic and

E

V=

4πgs

h3

P 4F c

4=

3c

4

(4πgs

3h3

)−1/3 (N

V

)4/3

(130)

P =1

3

(E

V

)

(131)

29

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......................................................................................................................................................................................................................................................................................................................................................................................................................................................................

P

∝ n4/3

∝ n5/3

n = N/Vrelativisticnon-rel

(Mc

~

)3

Neutrinos Neutrinos were postulated by Fermi in the 1930’s to explainmissing energy in β-decay in nuclei. The basic reaction is

n→ p + e− + νe ∼ 13 minutes (132)

(where νe is an (anti)-neutrino). Neutrinos:

i) are electrically neutral,

ii) move at the velocity of light,

iii) have helicity or spin = ~/2.

Experiments carried out in 1957 by Wu at the suggestion of Lee & Yangshowed that neutrinos emitted in β-decay spin in a right-handed sense (i.e.parity or left-right symmetry is violated).

Particles & Anti-Particles According to Dirac, to every particle thereis an antiparticle, e.g.

i) electron ←→ positron massive, M 6= 02 spin states 2 spin states

ii) a photon is its own antiparticle massless, M = 0and it has 2 spin states

iii) neutrino ←→ anti-neutrino massless, M = 0νe νe

left-handed right-handed1 state 1 state

30

Three Families

In addition to the electron-neutrinos associated to the electronνe e

there are muon-neutrinos associated to the muonνµ µ

and tau-neutrinos associated to the tau particleντ τ

3.8 White Dwarfs, Neutron Stars and Black Holes

Once a star has exhausted it’s nuclear fuel it can only be supported againstthe inward forces due to its own gravitional field by ”degeneracy pressure”resulting from the Pauli Exclusion Principle. Depending upon which particleis degenerate we get a white dwarf or a neutron star. If degeneracy pressureis insufficient then nothing else can hold the star up and it must collapseto form a black hole. The discussion given below is mainly qualitative andconcentrates on orders of magnitude.

3.8.1 White Dwarfs

In this case we have degenerate assembly of electrons, with non-degenerateprotons to give overall charge neutrality. A typical radius is comparablewith that of the earth, but the mass is comparable with that of the sun. Thedensity is ∼ 105 g/cm3.

Example: Sirius Bessel observed elliptical orbit in 1844. Alvan Clarkobserved a faint companion in 1862. Adams observed it to be white hot in1914.

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.

..........................................................................................

.............................................................................

...........................................

.

...............................

.

...............................A B

star white dwarf

••

Using the theory of Newtonian orbits one may deduce that the faint com-panion has mass ∼ 4

5M. The observed luminosity tohether with the black

body emission allows one to estimate that radius < R/19 (where M andR are solar mass and radius). This gives a density of 60, 000 g/cm3.

In 1924 R.H. Fowler suggested it was made of a superperfect gas (i.e. whtawe now call degenerate). S. Chandrasehkar showed that no such configurationexists for M > 1.5M, Chandrasehkar Limit .

31

3.8.2 Neutron Stars

Landau suggested that for more massive stars it is possible that inverse β-decay takes place:

e− + p→ νe + n. (133)

The dense degenerate assembly of neutrons is called a neutron star. Landauobserved that there is also an upper bound for the mass of a neutron starM/M < ∼ 6.

3.8.3 Black Holes

If

GM

R>

c2

2⇒ R < RS =

2GM

c2(134)

not even light can escape from the collapsing star (where RS is the Schwarzschildradius).

3.8.4 Stellar Structure

Each element in the static star is in equilibrium under two forces, a pressuregradient and gravity,

~∇P + ρ~∇Φ = 0 (135)

where ρ is the density, and Φ is the Newtonian potential defines such that−ρ~∇Φ is the gravitaional force on a unit volume of material of density ρ.

Assume spherical symmetry: m(r) is the mass inside a sphere of radius r

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dP 4πr2dr

G4πr2ρdrm

r2

dP

dr= −Gρ

m(r)

r2 ,

dm

dr= 4πr2ρ.

(136)

32

Boundary conditions

∂P

∂r

∣∣∣∣r=0

= 0, P |r=R = 0. (137)

If we assume a relation P = P (ρ) between pressure P and density ρ , referrredto as an ”equation of state” we may integrate these equations.

3.8.5 Virial Theorem

We can get a feel for the solutions by establishing a general relationship ,valid for all Newtonian stars, relating their kinetic and potential energy. Infact a result like this hold for any gravitationally bound system.

∫ R

0

4πr3dP = −∫ R

0

Gm(r)

rdm(r) . (138)

Integrating by parts,

3

∫ R

0

P4πr2dr =

∫ R

0

Gm(r)dm(r)

r(139)

E.g., P = (γ − 1)ε where ε = energy density

3(γ − 1)

∫ R

0

4πr2εdr =

∫ R

0

Gm(r)dm(r)

r(140)

Therefore

3(γ − 1) Thermal Energy = −Gravitational Potential EnergyThermal + Potential Energy = (4− 3γ) Thermal Energy

(141)

Now any bound sytem must have negative total energy and we thereforededuce that

A star cannot be bound stably if γ ≤ 4

3

The final conclusion is that if the density increases so that material be-comes sufficiently relativistic that γ ≤ 4

3, the star becomes unstable.

3.8.6 Order of Magnitude Estimates

(factors of order unity ignored)

Pressure support implies,

P

R' GM

R2ρ ⇒ P

ρ' GM

R(142)

33

White Dwarfs

i) nucleons dominate the mass, therefore ρ ' ncmN , where mN is nucleonmass and

nc = critical number density of electrons= number density of protons

ii) electrons provide the pressure,

P ' ~2

men5/3

c , nc '(mec

~

)3

, (143)

therefore

P

ρ' ~

2

memNn2/3

c , (144)

and

GM

c2R' me

mN(145)

The Chandrasehkar mass M ' 4π3

R3ρ gives

M =1

m2N

(~c

G

)3/2

(146)

Neutron Stars In free space,

n→ p + e− + νe (β-decay). (147)

In the deep potential well of the star, the inverse is favoured,

p + e→ n + νe (inverse β-decay). (148)

Pressure is now provided by neutrons. Replace me by mN in the above:

GM

c2R' O(1) (149)

M =1

m2N

(~c

G

)3/2

(150)

Therefore,

escape velocity ' light velocity.

34

4 Cosmology

4.1 Isotropy & Homogeneity of the Universe

We begin our study of cosmology by summarising some of the data about itsbasic construction.

4.1.1 Cosmic Distance Ladder

• distances of stars in our galaxy are determined by trigonometry.

• distances to nearby galaxies are determined by period-luminosity rela-tion of cepheid variable stars.

• distances to more distant galaxies may also be estimated using Tully-Fisher relation between luminosity L and rotation speed v: L ∝ v4.

• distances to yet more distant galaxies are estimated by assuming bright-est galaxy in a cluster has some intrinsic luminosity for all clusters.

35

4.1.2 Hubble Expansion

Deduce velocity v from Doppler shift:

λobs

λem= 1 + z =

√c + v

c− v(151)

where z is redshift.

Last year it was true thatthe most distant galaxy observed had z = 4.38the most distant quasar observed had z = 4.9

In March 1998 a paper appeared claiming to have observed a galaxy witha redshift z = 5.34.

Hubble’s Law: isotropic expansion about us,

~v = H0~r (152)

where H0 is the Hubble constant, which is “probably” 50 < H0 < 50 Km/sec/Mpc.H0 = 50 implies

Hubble time =1

H0

= 19.5 Gyrs

Hubble radius =c

H0

= 6000 Mpc.

(153)

Useful Orders of Magnitude Age of earth 4.5 Gyrs.Age of globular clusters 8− 15 Gyrs.c = 300 Mpc/Gyr.

1 parsec ' 3.09× 1018 cm ' 3.26 light years,= distance at which earth-sun distance

︸ ︷︷ ︸

1 astronomical unit

subtends 1 arcsec.

Distance to Andromeda Galaxy ' 0.5 Mpc.

36

4.1.3 Cosmic microwave background

is isotropic and thermal.

T = 2.735± 0.06 K, as measured by COBE satellite.

There exist dipole variations due to Doppler shift of our galaxy towardsVirgo cluster and Hydra Centauraus Supercluster ∼ 10−3 c. In fact COBEsees Doppler shift due to the earth’s motion about the sun (∼ 30 Km/sec).

There exist quadrupole variations ∼ 10−5 first seen by COBE. Thesecorrespond to density fluctuations which grew to form galaxies, stars, planet,etc.

Energy Density

T ∼ 2.7K ≡ 1 eV/cm3

∼= 400 photons/cm3 (154)

Microwave photons have the same total energy as starlight but are muchmore numerous.

4.2 Kinematics of Homogeneous Isotropic Expansion

• Each galaxy is assumed to satisfy

~vi = H(t)~ri(t), (155)

i.e. isotropy about us.

• Present value of Hubble’s constant H0 = H(t0), where t0 is time now.

~vi =d~ri

dt= H(t)~ri (156)

where ~ri is proper or physical distance.

~ri(t) = a(t)~xi

H(t) =a(t)

a(t)

(157)

and ~xi is independent of time. Therefore,

a(t) = a(t0) exp

∫ t

t0

H(t′)dt′ (158)

37

• a(t) is called the scale factor .

• ~xi is called the comoving coordinate of the ith galaxy.

4.2.1 Homogeneity and the Cosmological Principle

Choose galaxy 1 as new origin:

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0

~r2~r1

1

2

~ri = ~ri − ~r1

= a(t)~xi − a(t)~x1(159)

~ri = a(t) ~xi (160)

where ~xi = ~xi − ~x1.Thus, the expansion is isotropic relative to any galaxy participating in

the Hubble flow.

Copernican Principle The earth is not in a specially favoured positionin the universe.

Cosmological Principle The universe presents the same aspect from ev-ery point except for local irregularities, i.e. the universe seems homogeneousand isotropic when averaged over a suitably large scale.

Therefore, there exists some overall mass density ρ which depends onlyon time

ρ = ρ(t) (161)

38

4.2.2 Behaviour of Distances, Areas, and Volumes

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. ......................................

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.......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .

.......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .

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...

t t

t0

te

~r ~x

physical distance comoving coordinate

~r = a(t)~x,d~r = a(t)d~x,dl = |d~r| = a(t) |d~x| .

(162)

Thus,all distances scale ∝ a(t)all areas scale ∝ a2(t)all volumes scale ∝ a3(t).

Example

i) galaxies: number is constant in a comoving volume

ρ(t)dV = constantρ(t)a3(t)d3~x = constant

(163)

Therefore,

ρa3 = constant (164)

ii) pressure free fluid, “dust”: same argument ρ ∝ a3

iii) fluid with pressure: assume expansion is adiabatic,

sdV = constant, (165)

where s = S/V is the entropy per unit volume. Then

sa3 = constant (166)

39

e.g. microwave photons, s ∝ T 3,

Ta = constant (167)

Since nγ = number of photons per unit volume ∝ s ∝ T 3, we alsodeduce that

nγa3 = constant (168)

Energy density ∝ T 4 ∝ 1/a4.

4.3 Dynamics

The equations governing the bahaviour of the scale fcator a(t) cna be derivedusing simple Newtonian reasoning, subject only to an additional term in theNewtonian force law which enters when the pressure is significant comparedwith ρc2. The resulting (consistent) set of equations are identical to thosederived from Einstein’s theory of General Relativity.

4.3.1 Dynamics of Isotropic Homogeneous Pressure-Free Fluid

Consider a small sphere of radius r = |~r| = a(t) |~x| and apply Newton’s Lawto a small shell of thickness dr and mass δm = 4πr2 dr

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....................

r

δm r = −Gm

r2δm, (169)

δm a |~x| = −Gm

a2

δm

|~x|2. (170)

The mass inside radius r is

m =4π

3ρr3 =

4π

3ρa3 |~x|3 = constant. (171)

Then,

δm |~x| a = −4π

3

Gρ

aδm |~x| , (172)

40

which gives the Raychaudhuri equation,

a

a= −4πGρ

3(173)

n.b.

i) arbitrary radius of sphere cancels

ii) we can ignore GR because

GM

c2r=

4π

3

Gρ

c2r2

can be made as small as we like by taking r sufficiently small.

First Integral

a = −GM

a, M =

4π

3ρa3 = constant. (174)

Integrating we obtain

a2

2− GM

a= constant = −k

2(175)

which is the Friedman equation,

a2

a2=

8πGρ

3− k

a2(176)

By scaling a→ λa we may take:k = +1 bound ⇒ universe recollapsesk = 0 marginalk = −1 unbound ⇒ universe expands forever

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k = −1

k = 0

k = +1

r

a(t)

41

Solutions

a2 − 2GM

a= −k (177)

Einstein-DeSitter: k = 0

a = a0

(t

t0

)2/3

, (178)

implies,

a

a=

2

3

1

t0= H0. (179)

The age of the universe is

t0 =2

3

1

H0(180)

H0 ' 50 Km/sec/Mpc gives t0 ≈ 13 Gyrs.

Redshift

(1 + z) =

(t0t

)2/3

, (181)

t

t0=

1

(1 + z)3/2(182)

E.g., (typical quasar) z = 4.5 gives t ≈ 1 Gyr

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t

...

..................................................................................................................................................................................................................................................................................................................................... Big Bang

t = 13 Gyr

t = 1 Gyr

now

The density of quasars is bigger by a factor of (5.5)3.

42

The Age Problem

• Stars in the disc of our galaxy are young and rich in metals (“PopulationI”).

• Stars in the halo, in global clusters are old and poor in metals (“Pop-ulation II”). Globular clusters are believed to have ages: 12 ≤ age ≤16 Gyrs.

Therefore, if k = 0 then Hubble’s constant H0 cannot be too large.

Recollapsing solutions: k = +1

a2 =GM

a− 1. (183)

To solve,

a12 da√

2GM − a= dt (184)

substitute a = 2GM sin2 θ, then

2GM 2 sin2 θ dθ = dt (185)

2GM

(

θ − sin 2θ

2

)

= t (186)

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tt0

cycloidcurve

BigBang

BigCrunch

a(t)

max scale factor amax = 2GMtotal age ttotal = 2GMπ

43

Expanding solutions: k = −1 Substitute a = 2GM sinh2 θ,

2GM

(sinh 2θ

2− θ

)

= t, (187)

a ∼ t as t→∞.

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t

linear

a(t)

Limiting forms

• as t ↓ 0, a(t) ∼ t2/3, ∀k.

• as t ↑ ∞, a(t) ∼ t, k = −1 (“Milne” model).

• the k = 0 case is an attractor to the past, but a repellor to the future.

The flatness problem Today we are close to the Einstein-DeSitter model,which is a repellor. This must have required very special initial conditions.

4.3.2 Dynamics of Homogeneous Isotropic Fluid with Pressure

We take from GR the result that we must replace ρ by (ρ + 3P/c2) in theRaychaudhuri equation:

a

a= −4πG

3

(

ρ +3P

c2

)

(188)

First Integral is the same as before:

a2

a2=

8πGρ

3− k

a2(189)

Proof

d

dt

(1

2a2 − 4πG

3ρa2

)

= aa− 4πG

3ρa2 − 8πG

3ρaa, (190)

= −4πGa2

3

ρ +

(

ρ +P

c2

)3a

a

, on substitution of eq. (188)

= 0, by Thermodynamic relation.

44

Example: Radiation Dominated Universe Pc2

= 13ρ; γ = 4

3. Assume

k = 0, then

(a

a

)2

∝ 1

a4⇒ a ∝ t1/2, (191)

ρ =3

32πG

1

t2(192)

Age a ∝ t1/2, therefore

a

a=

1

2t= H, (193)

t0 =1

2H0

(194)

E.g., if H0 = 50 Km/Mpc/sec, then t0 ≈ 9.25 Gyrs which is too short. Thus,probably the universe is not dominated by massless particles.

Temperature Time Relation

ρ =

(8π5k4

30h3c5

)

T 4g(T ) (195)

where

g(T ) = 2 photons

=7

8neutrinos, etc.

(196)

Therefore

T ∝ t−1/2. (197)

E.g.,T ∼ 109 K ⇒ t ∼ 100 sec nucleosynthesisT ∼ 1010 sec ⇒ t ∼ 1 sec electron-pair annihilation

45

4.3.3 General Upper Bound on the Age of the Universe

The result which follows is most easily seen by noting that if a is negative thenthe graph of a(t) against t must lie below its tangent. The argument givenbelow however is actually the essential idea behind the famous Hawking-Penrose singularity theorems which show that classical physics implies thatthe universe had a singularity at the Big bang. Because it is so simple I willgive a slightly longer and more formal argument which generalizes.

a = −4πG

3

(

ρ +3P

c2

)

a. (198)

Then(

a

a

)

= −4πG

3

(

ρ +3P

c2

)

−(

a

a

)2

, (199)

H = −H2 − 4πG

3

(

ρ +3P

c2

)

. (200)

Assume ρ + 3P/c2:

H < −H2, (201)

dH

H2< −dt, (202)

1

H− 1

H0< t− t0, (203)

H ≥ H0

1 + H0(t− t0)(204)

Then

age ≤ 1

H0

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.

H

t0

actual Hubble

constant

our lower

bound

t = t0 −1

H0

46

4.4 Redshifting & Motion of Light

In a time dt a photon travels towards us a comoving distance dx and physicaldistance a(t) dx. The physical velocity = a(t) dx

dt= c, i.e. the motion is given

by

dx = − c dt

a(t)(205)

Example

a(t) =

(t

t0

) 23

(we use units s.t. a0 = 1), (206)

−dx = ct230

dt

t23

, (207)

x = 3ct230

(

t130 − t

13e

)

(since x = 0 at t = t0), (208)

x = 3ct0

(

1−(

tet0

) 13

)

, (209)

but

1 + z =

(t0te

) 23

, (210)

therefore,

x = 3ct0

(

1− 1

(1 + z)12

)

(211)

Limiting values:

i) z small, implies (Hubble’s Law)

x ' 3

2ct0z '

c

H0z. (212)

ii) z →∞,

x→ 3ct0 =2c

H0

. (213)

47

Horizon Light coming from galaxies outside a co-moving radius 2c/H0 hasnot yet had time to reach us.

..

..............................................................................................................................................................................................................................................................

......................................

. .. .. .. .. .

...

..

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t = 0

t = t0

horizon radius

4.4.1 Proof of Redshift Formula

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....................

................................................................................................................................................................................................

xe x0

t3

t1

t4

t2

1

c(xe − x0) =

1

c

∫ xe

x0

dx , (214)

=

∫ t2

t1

dt′

a(t′), (215)

=

∫ t4

t3

dt′

a(t′). (216)

Set,

t3 = t1 + ∆te,t4 = t2 + ∆t0.

Then∫ t2

t1

dt′

a(t′)=

∫ t2+∆t0

t1+∆te

dt′

a(t′). (217)

Now in the limit as ∆t0, ∆te ↓ 0,

∆t0a(t0)

=∆tea(te)

, (218)

48

∆t0∆te

=λ0

λe

=νe

ν0

= 1 + z =a(t0)

a(te), (219)

1 + z =a(t0)

a(te)(220)

Example: angular sizes A galaxy of size l0, i.e. physical or proper lengthl0, will subtend an angle δθ

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δx

δθ

x

such that,

a(te)δx = le,

δθ =δx

x,

(221)

implies,

δθ =le

a(te)x. (222)

If we assume a(t) ∝ t2/3 (Einstein-DeSitter case), we find

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................................................................

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δθ

‘zz ' 1.25

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... Big Bang

our past light cone

z = 1.25 ≡ t = 9.15 Gyrsif H0 = 50 Km/sec/Mpc

49

4.4.2 Singularity Theorems (Hawking & Penrose)

Since all particles travel no faster than light they must be confined insidethe light cone. Following them back into the past we see that if the pastlight cone reconverges the density will become infinite when the past lightcone converges to a point at the big bang. This is the basis of a rigourousmathematical theorem showing that classical general relativity must breakdown at the big bang singularity.

4.5 Thermal History of the Universe

• The thermal evolution of the the universe depends on the dimensionlessrate η = nB/nγ , where nB = number density of baryons, nγ = thenumber density of photons.

• nB ∝ 1/a3, nγ ∝ 1/a3, implies

η = constant

• since the photon entropy ∝ nγ ,

1

η' entropy

baryon

• nγ ' 400 /cm3

• nB ≥ 3× 10−8 /cm3 in luminous matter

• nB < 3× 10−7 /cm3 from theory of nucleosynthesis

• nB = 3× 10−6(H0/50)2

Thus

10−10 <nB

nγ

< 10−9 (223)

50

4.5.1 Decoupling

• At high temperatures hydrogen was ionised to form a “plasma” whichis opaque to electromagnetic radiation.

• At low temperatures protons and electrons combine to form Hydrogen

p + e− → H

which is transparent to radiation.

• The transition occurs at the ionisation temperature

TD '1

2

mec2

kα2 ' 4, 533 K ' 13.6 eV (224)

where α =(

e2

4π~c

)

' 1/137,

T ' 4.5× 103 ⇔ z ' 1.5× 103 (225)

.......... .......... .......... .......... .......... .......... .......... ....................

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T

TD

topaque transparent

T ∝ 1

a

T ∝ 1

a2 (hydrogen)

tD

• The collisions between photons and electrons maintain thermal equi-librium before tD.

• After tD the protons retain their Planckian spectrum without collisionsjust by redshifting as explained earlier

• After tD the hydrogen cools adiabatically as a non-relativistic gas,PV 5/3 = constant, or V T 3/2 = constant

T ∝ 1

a2(226)

Thus, the hydrogen cools more rapidly than photons.

51

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•

microwave photons

transparent

surface of last scattering

opaque

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• Using z ≈ 1.5× 103 and (1 + z) = (t0/te)2/3, if k = 0 we deduce that

t ≈ 105 yrs.

4.5.2 Decoupling Temperature & Temperature at End of Radia-tion Era

• Let fB be the fraction of total matter density in baryons,

ρmatter =1

fBmNnB. (227)

• photon energy density is given by

ργ =3!

2!

ζ(4)

J(3)

kT

c2nγ (228)

• ρmatter = ργ at T = TR, implies

TR =mNc2

fBk

2!

3!

ζ(3)

ζ(4)

(nB

nγ

)

(229)

TR

TD' mN

me

1

2α2

ζ(3)

ζ(4)

(nB

nγ

)1

fB(230)

mN

me

' 1836; α ' 1

137; fB = O

(1

10

)

;nB

nγ

= O(10−8

).

Therefore, TR/TD = O(1), the two temperatures roughly coincide.

52

4.5.3 Nucleosynthesis

If T > 1010 K, the reactions

n + e+ p + νe,

p + e− n + νe,

n p + e− + νe,

are rapid and maintain n and p in equilibrium.Since kT mNc2 we use non-relativistic formulae:

nn =

(2πmNkT

h3

)3/2

expµn

kT, (231)

np =

(2πmpkT

h3

)3/2

expµp

kT. (232)

Now µn − µp is the difference in energy increase due to the addition ofone neutron or one proton,

µn − µp = (mn −mp)c2 (233)

np

nn=

(mp

mN

)3/2

exp

[

−(mn −mp)c2

kT

]

, (234)

i.e.,

np

nn' exp

[

−(mn −mp)c2

kT

]

(235)

since mp/mn ' 1.As T dropped below ∼ 1010 K the electron neutrinos νe are no longer in

equilibrium (they “decouple”) and henceforth their abundance drops withoutreference to anything else (see later). As T dropped to ∼ 109 K the neutronsreacted slowly to form deuterium,

p + n→ D + γ, D = 2H.

Deuterium then reacted rapidly to form Helium-4, 4He

D + D → 3He + n,D + D → 3H + p,

3He + n → 3H + p,3H + D → 4He + n,

53

(where 3H is tritium).As a result almost all the neutrinos reacted to form 4He.

The abundance by mass Y is defined by:

Y =4n4He

np + nn

, (236)

but n4He ≈ 12nn (because almost all nn’s formed 4He), therefore

Y =2nn

np + nn=

2

1 + np

nn

=2

(

1 + exp(

− (mn−mp)c2

kT∗

)) (237)

where T∗ is called the “freeze out” temperature.The freeze out temperature depends on the rate of expansion of the uni-

verse which in turn depends on

i) η = nB/nγ,

ii) the number of neutrino species

ρrad = ργ + ρνe+ ρνµ

+ ρντ,

= ργ

(1 + 3× 7

8

).

(238)

4.5.4 Calculation of Relic Neutrino Number Density

The electron neutrinos, νe, went out of equilibrium at T ' 1 MeV ∼ 1 sec.Electron-positron pairs annihilated at T ' 0.5 MeV ∼ 4 secs. This meantthat the temperature rose sharply.

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T

Tt

t1 sec 4 secs

Tν

neutrinos redshift Tν

independently of photons

T ∝ t−12

1 MeV

Ti

To calculate the reheat temperature we assume that entropy does notdecrease. Before:

S = (1 +7

8× 2) T 3

i × constant (239)

54

(where 1 comes from the photons and 7/8 is for the electrons and positrons).After:

S = (1 ) T 3f × constant (240)

Thus,

Ti

Tf

=

(4

11

) 13

, (241)

nγ(Ti)

nγ(Tf)=

(4

11

)

. (242)

nνe/nγ = 3/4 at some temperature, so

nνe

nγ=

4

11× 3

4=

3

11. (243)

4.5.5 Comparison with observations

Astronomers have found that

Y ' 0.23± 0.2, (244)

which in turn implies that

i) there are probably no more than three neutrino species,

ii) 3× 10−10 < η < 6× 10−10.

At the same time that the primordial Helium was made (about 25% byweight) 3He, 7Li, and 7Be were also formed. These give good agreementwith observations, with k = 0 and

ρB

ρmatter' 1

This indicates that not all the matter in the universe has yet been seen, theremust be “dark matter” which is probably not baryons but rather some exoticnew type of matter.

There is futher evidence for ”dark matter” which is based on the observedrotation curves of galaxies and the velocities of galaxies in of clusters ofgalaxies. In both cases one may use the Virial Theorem to estimate the massof these systems assuming that they are bound. The estimated mass exceedsthe mass of the observed luminous matter, such as stars. the nature of themissing matter is one of the greatest puzzles in astrophysics today.

55

4.6 Current problems

The cosmological theory developed above is usually thought of as the ”stan-dard model”. It seems to be well supported by the observations and isaccepted at least in outline by most people working in the field. The greatestuncertainty concerns the question of the value of k and whether the cosmo-logical constant is non-zero. In the final lecture of the course I will give a verybrief account of some ideas which attempt to take us beyond the standardmodel and into unknown territory.

4.6.1 Initial Conditions

The hot big bang model is able to account for the present appearance of theuniverse but it fails to provide an explanation for certain initial conditions.It also leaves unexplained some puzzling questions including:

i) Why is η ' 10−10, i.e. why is there so much entropy per baryon? Whyis the total number of baryons non-zero?

ii) Why does the universe have almost zero kinetic energy?

iii) What about “horizons”?

iv) What accounts for the initial fluctuations in density δρ/ρ ∼ 10−5 onthe last scattering surface which grew to make galaxies?

In this lecture we will discuss (ii) (and (iii)).

The Horizon Problem The co-moving radius of the last scattering surfaceis given by

x

c=

∫ t0

tR

dt

a(t)=

∫ t0

tR

(t0t

)2/3

dt, (245)

= 3t0

1−(

tRt0

)1/3

= 3t0

1− 1

(1 + zR)12

' 3t0, (246)

56

x

c' 2

H0, x ' 2c

H0(247)

During the period from the big bang,

a(t) =

(t

tR

)1/2 (tRt0

)2/3

. (248)

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.......................................................................................................................................................................................................................................................................................................................................

a(t)

ttR

t0

Thus since the big bang light can have travelled a total comoving distancegiven by

x

c=

∫ tR

0

(tRt

)1/2

dt

(tRt0

)−2/3

, (249)

= 2tR

(tRt0

)−2/3

= 2t0

(tRt0

)1/3

= 2t01

(1 + zR)1/2, (250)

x

c' 2t0

z1/2R

⇒ x ' 4

3

c

H0

1

z1/2R

(251)

The ratio

2

3

1

z1/2R

' O(10−2). (252)

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The Kinetic Energy Problem (Also called the horizon problem). Recallthat,

Ω =8πGρ

3H2, H =

a

a, (253)

57

tells us how close to the Einstein-DeSitter model we are. The Friedmanequation may be re-expressed as

1

Ω= 1− 3k

8πGρ

1

a2(254)

Ω > 1 ⇔ k = 1,Ω = 1 ⇔ k = 0,Ω < 1 ⇔ k = −1.

(255)

At present, 1Ω0−1 = ε0, say with ε0 = O(1). Let’s calculate 1

ΩR−1, the value

at the beginning of the matter dominated era. Since ρ ∝ 1/a3,

1

ΩR

− 1 = ε0a(tR)

a(t0)= ε0(1 + z)−1, (256)

1

ΩR

− 1 = ε0 × 10−3 = O(10−3), (257)

which requires “fine tuning”. The problem gets worse the earlier we go, e.g.at t ∼ 1 sec, T ∼ 1010 K

1

Ω1 sec

− 1 = εR

(a(1 sec)

a(tR)

)2

= εR

(TR

1010 K

)2

(258)

1

Ω1 sec− 1 = ε0 × 10−17 = O(10−17) !!! (259)

4.6.2 The solution?

We must replace the assumption a(t) ∝ t1/2 by a faster rate of expansion,eg.

a(t) ∝ tn; n > 1,or a(t) ∝ eHt; H = constant.

Flatness Problem

a(t) ∝ tn ⇔ ρ ∝ 1

a2n(260)

i.e. γ = 2n/3. Now,

1

Ω− 1 ∝ 1

ρa2∝ a2n−2 (261)

which decreases with a as long as n > 1.

58

Horizon Problem∫ tR

0

dt

a(t)∝∫ tR

0

dt

tn→∞, if n ≥ 1. (262)

Therefore, light may travel arbitrary distances since the big bang.An equally acceptable solution is to set a(t) ∝ eHt since then ρ = con-

stant. This corresponds to

P

c2= −ρ (263)

(i.e. tension and not pressure) and is called a cosmological term.m

59

Index

adiabatic, 9, 33

black hole, 26Bose-Einstein distribution, 19Boyle-Charles Law, 12, 18

cepheid, 29Chandrasehkar Limit, 26comoving coordinate, 31cosmic microwave background, 30cosmological term, 52

dark matter, 49dipole variations, 30Doppler shift, 29

Einstein-DeSitter, 35energy density, 2

Fermi-Dirac distribution, 22fine tuning, 51freeze out, 47Friedman equation, 34

Helmholtz Free Energy, 18Hubble

expansion, 29law, 30

isotropic distribution, 1

Juttner distribution, 17

Klein-Gordon Equation, 17

Maxwell-Boltzmann, 4microwave background, 21

neutrinos, 23, 24neutron star, 26

occupation number, 4

Planck spectrum, 20plasma, 44

quadrupole variations, 30

Raychaudhuri equation, 34, 37

scale factor, 31Schwarzschild radius, 26singularity theorems, 43spin statistics theorem, 19Stirling approximation, 5

virial theorem, 27

wave equation, 17white dwarf, 25Wien Displacement law, 20

60