lecture notes from fys5130 - cosmological physicsfolk.uio.no/finnr/notes/fys5130.pdf · these...

Lecture notes fromFYS5130 - Cosmological Physics

Spring 2005

Lectures by Finn RavndalDepartment of Physics

University of Oslo

Transcribed by:Øystein Rudjord

Gorm Krogh JohnsenJostein Riiser Kristiansen

Last updated: September 29, 2005

Preface

These lecture notes are taken from the course FYS5130 - Cosmological Physicsgiven by prof. Finn Ravndal at the University of Oslo in the spring semester 2005.The notes are mainly transcribed for our personal use, but we are of course happyif someone else also find them useful.

If you find something in these notes that make no sense to you, it could be dueto one of the following reasons:

1. It is some kind of an internal joke.

2. We totally misunderstood what was said in the lectures.

3. It is a typo.

4. You are reading this in year> 2050 and your notions of physics have changedradically.

Most probably it will be either number 2 or 3. In those cases we would be morethan happy to hear from you, such that the mistake can be corrected.

We want to acknowledge Finn Ravndal both for giving lectures on such an in-teresting subject, and also for taking the time to read through this manuscript andprovide us with numerous corretcions.

Oslo, September 28, 2005

Øystein Rudjord ([email protected])Gorm Krogh Johnsen ([email protected])Jostein Riiser Kristiansen ([email protected])

Contents

1 Lecture 1 71.1 Classical introduction . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Quantum field theory . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Lecture 2 132.1 2nd quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Classical field theory . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Fermions and bosons . . . . . . . . . . . . . . . . . . . . . . . . 152.4 Approximation to point particles . . . . . . . . . . . . . . . . . . 18

2.4.1 Interactions between point particles . . . . . . . . . . . . 18

3 Lecture 3 213.1 Relativistic fields . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1.1 Relativistic scalar field theory . . . . . . . . . . . . . . . 223.1.2 External scalar field . . . . . . . . . . . . . . . . . . . . . 24

3.2 Vacuum energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2.1 Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.3 The Casimir effect . . . . . . . . . . . . . . . . . . . . . . . . . 273.4 Bernoulli numbers . . . . . . . . . . . . . . . . . . . . . . . . . 30

4 Lecture 4 334.1 Regularization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.2 Dimensional regularization . . . . . . . . . . . . . . . . . . . . . 344.3 Quantum statistical mechanics . . . . . . . . . . . . . . . . . . . 37

4.3.1 Example with relativistic scalar particles . . . . . . . . . 37

5 Lecture 5 415.1 Path integral formalism . . . . . . . . . . . . . . . . . . . . . . . 41

5.1.1 The partition function for the harmonic oscillator . . . . . 445.2 Divergent products and functional determinants . . . . . . . . . . 455.3 Generalized zeta-function regularization . . . . . . . . . . . . . . 46

5.3.1 Some examples . . . . . . . . . . . . . . . . . . . . . . . 47

1

2 CONTENTS

6 Lecture 6 516.1 Free scalar fields . . . . . . . . . . . . . . . . . . . . . . . . . . 526.2 Radiation with massless particles and free energy . . . . . . . . . 54

6.2.1 Pressure in arbitrary dimensions . . . . . . . . . . . . . . 566.2.2 Example with zero temperature . . . . . . . . . . . . . . 56

6.3 Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576.3.1 Example with fermionic Z . . . . . . . . . . . . . . . . . 596.3.2 Example from QED . . . . . . . . . . . . . . . . . . . . 60

6.4 Interacting fields . . . . . . . . . . . . . . . . . . . . . . . . . . 616.4.1 Dimensional analysis . . . . . . . . . . . . . . . . . . . . 62

7 Lecture 7 657.1 Theory of free fields . . . . . . . . . . . . . . . . . . . . . . . . . 657.2 Wick’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

7.2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 697.3 Wick’s theorem outside field theory . . . . . . . . . . . . . . . . 71

7.3.1 Example: Stirling’s formula for n! . . . . . . . . . . . . . 71

8 Lecture 8 738.1 Correlators with interactions at finite temperature . . . . . . . . . 738.2 Phase transitions . . . . . . . . . . . . . . . . . . . . . . . . . . 75

8.2.1 The Higgs model . . . . . . . . . . . . . . . . . . . . . . 758.2.2 The partition function in the Higgs model . . . . . . . . . 768.2.3 Veff and an early-universe phase transition . . . . . . . . . 77

8.3 Particle creation in an external field at zero temperature . . . . . . 79

9 Lecture 9 819.1 External fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 819.2 Toy model: Harmonic oscillator . . . . . . . . . . . . . . . . . . 829.3 Oscillator with constant frequency . . . . . . . . . . . . . . . . . 859.4 Oscillator with time dependent frequency . . . . . . . . . . . . . 87

9.4.1 The relation between the vacuum states |0a〉 and |0b〉 . . . 909.5 Quantum fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

9.5.1 Lorentz invariant norm . . . . . . . . . . . . . . . . . . . 93

10 Lecture 10 9510.1 Effective action . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

10.1.1 Example with Hint = −Lint . . . . . . . . . . . . . . . 9610.1.2 Example with Hint 6= −Lint . . . . . . . . . . . . . . . 96

10.2 Quantization of external fields . . . . . . . . . . . . . . . . . . . 9710.3 Particle production in constant electric field . . . . . . . . . . . . 99

10.3.1 QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

CONTENTS 3

11 Lecture 11 10311.1 Massless scalar field in a general space-time . . . . . . . . . . . . 103

11.1.1 Example with particle creation in accelerated systems . . . 10311.1.2 Scalar fields in a Robertson-Walker background . . . . . . 10611.1.3 Introduction of the d’Alembertian . . . . . . . . . . . . . 10811.1.4 Variation of the metric . . . . . . . . . . . . . . . . . . . 108

11.2 The Einstein-Hilbert action . . . . . . . . . . . . . . . . . . . . . 11011.3 The energy-momentum tensor . . . . . . . . . . . . . . . . . . . 111

11.3.1 Example with a Maxwell field . . . . . . . . . . . . . . . 11111.3.2 Example with a scalar field . . . . . . . . . . . . . . . . . 112

12 Lecture 12 11512.1 Conservation of the energy-momentum tensor . . . . . . . . . . . 115

12.1.1 Generalized Bianchi identity . . . . . . . . . . . . . . . . 11912.2 Weyl transformations . . . . . . . . . . . . . . . . . . . . . . . . 120

12.2.1 Example with a free scalar field . . . . . . . . . . . . . . 12112.2.2 Example with Maxwell theory . . . . . . . . . . . . . . . 12212.2.3 Example with coupled Maxwell and scalar theory . . . . . 123

12.3 Quantization of scalar fields . . . . . . . . . . . . . . . . . . . . 12312.3.1 Scalar field in a RW background . . . . . . . . . . . . . . 124

13 Lecture 13 12913.1 The Dirac equation in curved spacetime . . . . . . . . . . . . . . 12913.2 Gravitons in flat spacetime . . . . . . . . . . . . . . . . . . . . . 13013.3 Gravitational waves . . . . . . . . . . . . . . . . . . . . . . . . . 132

13.3.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13213.3.2 Particle hit by a wave . . . . . . . . . . . . . . . . . . . . 13413.3.3 Energy contents in a wave . . . . . . . . . . . . . . . . . 135

14 Lecture 14 13714.1 Quantum fluctuations in a classical background (repetition) . . . . 13714.2 Propagators from using the Einstein-Hilbert action . . . . . . . . 13814.3 Canonical normalization of the gravitational field . . . . . . . . . 13914.4 The energy-momentum tensor for the graviton field . . . . . . . . 14014.5 Higher order terms and renormalizability . . . . . . . . . . . . . . 14114.6 Green functions and propagators . . . . . . . . . . . . . . . . . . 14114.7 Example with gravitational radiation . . . . . . . . . . . . . . . . 145

15 Lecture 15 14715.1 Gravitational radiation . . . . . . . . . . . . . . . . . . . . . . . 14715.2 Emitted energy and momenta . . . . . . . . . . . . . . . . . . . . 14915.3 Detecting gravitational waves . . . . . . . . . . . . . . . . . . . . 150

15.3.1 Indirect observations . . . . . . . . . . . . . . . . . . . . 15015.3.2 Direct observations . . . . . . . . . . . . . . . . . . . . . 150

4 CONTENTS

15.4 Gravitational radiation in a RW metric . . . . . . . . . . . . . . . 15215.4.1 Tensor fluctuations . . . . . . . . . . . . . . . . . . . . . 15215.4.2 Calculating the Ricci tensor . . . . . . . . . . . . . . . . 15315.4.3 The Friedmann equations . . . . . . . . . . . . . . . . . . 15415.4.4 Scalar field in the RW metric . . . . . . . . . . . . . . . . 156

16 Lecture 16 15916.1 Cosmology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

16.1.1 Important equations and numbers . . . . . . . . . . . . . 15916.1.2 The horizon and cosmological distances . . . . . . . . . . 16116.1.3 The early universe . . . . . . . . . . . . . . . . . . . . . 162

16.2 Inflation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

17 Lecture 17 16717.1 Inflation driven by a classical scalar field . . . . . . . . . . . . . . 167

17.1.1 The slow-roll approximation . . . . . . . . . . . . . . . . 16817.1.2 Example with SRA using a quadratic potential . . . . . . 170

17.2 Fluctuations of φ . . . . . . . . . . . . . . . . . . . . . . . . . . 17117.2.1 Quantum fluctuations . . . . . . . . . . . . . . . . . . . . 174

18 Lecture 18 17918.1 Primordial spectrum . . . . . . . . . . . . . . . . . . . . . . . . . 17918.2 Vacuum fluctuations of φ . . . . . . . . . . . . . . . . . . . . . . 17918.3 Inflation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18118.4 Metric fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . 182

18.4.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . 18218.4.2 The fluctuations . . . . . . . . . . . . . . . . . . . . . . . 18418.4.3 Scalar perturbations . . . . . . . . . . . . . . . . . . . . 18618.4.4 Gauge invariant variables . . . . . . . . . . . . . . . . . . 18818.4.5 Choice of gauge . . . . . . . . . . . . . . . . . . . . . . 188

19 Lecture 19 19119.1 Tensor fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . 191

19.1.1 The power spectrum of the tensor fluctuations . . . . . . . 193

20 Lecture 20 19720.1 Scalar perturbations . . . . . . . . . . . . . . . . . . . . . . . . . 197

20.1.1 Coupling the equations . . . . . . . . . . . . . . . . . . . 19820.1.2 A gauge invariant curvature scalar . . . . . . . . . . . . . 20020.1.3 The power spectrum for scalar perturbations . . . . . . . . 203

21 Lecture 21 20521.1 Newtonian theory . . . . . . . . . . . . . . . . . . . . . . . . . . 20621.2 Perfect fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

21.2.1 Perfect perturbations . . . . . . . . . . . . . . . . . . . . 208

CONTENTS 5

21.2.2 Non-perfect fluids . . . . . . . . . . . . . . . . . . . . . 20921.3 The relativistic perturbation equations . . . . . . . . . . . . . . . 210

21.3.1 Example with matter domination . . . . . . . . . . . . . . 21121.3.2 Example with radiation domination . . . . . . . . . . . . 211

21.4 Adiabatic perturbations and sound velocity . . . . . . . . . . . . 21221.5 Verifying the constancy ofR . . . . . . . . . . . . . . . . . . . . 213

22 Lecture 22 21722.1 More perturbations . . . . . . . . . . . . . . . . . . . . . . . . . 21722.2 The Boltzmann equation . . . . . . . . . . . . . . . . . . . . . . 222

23 Lecture 23 22523.1 The Boltzmann equation for photons . . . . . . . . . . . . . . . . 22523.2 The brightness equation . . . . . . . . . . . . . . . . . . . . . . . 22723.3 Properties of Θ and its multipole expansion . . . . . . . . . . . . 229

23.3.1 Line-of-sight integration . . . . . . . . . . . . . . . . . . 232

24 Lecture 24 23524.1 Degree of ionization . . . . . . . . . . . . . . . . . . . . . . . . 23624.2 The temperature fluctuations in CMB . . . . . . . . . . . . . . . 238

24.2.1 Solution of simplified case . . . . . . . . . . . . . . . . . 23824.3 Exact solution of δT

T . . . . . . . . . . . . . . . . . . . . . . . . 23924.3.1 Correlation function to lowest order . . . . . . . . . . . . 24024.3.2 Solution to (24.4) by line-of-sight integration . . . . . . . 24224.3.3 Simplifications due to instantaneous recombination . . . . 24424.3.4 Sachs-Wolfe . . . . . . . . . . . . . . . . . . . . . . . . 245

A The Riemann tensor and gauge fields 249

B de Sitter space 251

6 CONTENTS

Chapter 1

Lecture 1

1.1 Classical introduction

From Newtonian mechanics we have F = ma. We may express this using thecoordinate q, and by the change of the potential V (q)

F = mq = −V ′(q) = −dV

dq= −∂V (q, t)

∂q

The last expression is valid in the case of a time-dependent potential. We may writethe energy as a sum of the kinetic and the potential energy.

E = T + V =1

2mq2 + V

Energy conservation gives:

dE

dt= mqq +

∂V

∂qq +

∂V

∂t=

(mq +

∂V

∂q

)q +

∂V

∂t= 0

when V is time independent. In Lagrangian mechanics we define a Lagrangian L.

L = L(q, q) = T − V =1

2mq2 − V (q)

We define the action as the time integral of the Lagrangian from a time ti to a timetf

S =

∫ tf

ti

dtL(q, q)

Hamilton’s principle states that the action should have a stationary value, e.g. δS =0 when q(t)→ q(t) + δq(t). We get

δS =

∫ tf

ti

dt

(∂L

∂qδq +

∂L

∂qδq

)= 0 (1.1)

7

8 CHAPTER 1. LECTURE 1

The multiplication rule for differentiation gives

∂

∂t

∂L

∂qδq =

∂L

∂qδq + δq

∂

∂t

∂L

∂q⇒ ∂L

∂qδq =

∂

∂t

∂L

∂qδq − δq ∂

∂t

∂L

∂q

Inserted in (1.1) this gives

δS =

∫ tf

ti

dtδq

(∂L

∂q− ∂

∂t

∂L

∂q

)+∂L

∂qδq

∣∣∣∣tf

ti

= 0

Since δq vanishes at the end points, δq(ti) = δq(tf ) = 0, the last term will vanishand we are left with the Euler-Lagrange equation

∂L

∂q− d

dt

∂L

∂q= 0 (1.2)

For the Lagrangian L(q, q) = 12mq

2 − V (q) we get

∂V

∂q= −V ′(q)

∂V

∂q= mq

⇒ mq = −V ′(q)

which we recognize from Newtonian mechanics.We define the canonical momentum:

p =∂L

∂q

and from Euler-Lagrange we get:

p =∂L

∂q

We define the Hamiltonian H = H(q, p) as

H(q, p) = qp− L(q, q)

which is a Legendre transformation. This gives

q =∂H

∂p(1.3)

p = −∂H∂q

(1.4)

These are Hamilton’s equations.

1.2. QUANTUM MECHANICS 9

1.2 Quantum mechanics

Canonical quantization

We want p and q to become operators.

q → q

p→ p

These operators do not commute. We define the commutator

[q, p] = ih

In the same way the Hamiltonian will now become an operator H → H =H(q, p). The state of the system is described by a vector |Ψ, t〉 in Hilbert spaceand Schrödinger’s equation

ih∂

∂t|Ψ, t〉 = H|Ψ, t〉

Many-body systems

We have N free particles that can move in a external potential U(x). They aredescribed by the Hamiltonian

H =N∑

n=1

[1

2mp2n + U(xn)

]=

N∑

n=1

hn

where hn = 12m p2

n + U(xn) is the Hamiltonian for one particle. We then have

“single-particle eigenvalues” h|k〉 = εk|k〉, and the single-particle states |k〉 satisfythe completeness relation

∑∞k=0 |k〉〈k| = 1.

In coordinate representation we let x → x and p → −ih∇ which give us thesingle-particle Hamiltonian h = − h2

2m∇2 + U(x) and we get the single-particle

eigenvalue equation huk(x) = εkuk(x) where uk(x) = 〈x|k〉 is a single-particlewave function that satisfy the completeness relation.

∞∑

k=0

uk(x)u∗k(x′) = δ(x − x′)

We now have eigenstates E for many-body systems H|ΨN 〉 = E|ΨN 〉 where|ΨN 〉 = |k1〉|k2〉 . . . |kN 〉 is a many-body state. A bosonic pair is symmetric andhas a state given by

|k1, k2〉S =

√1

2(|k1〉|k2〉+ |k2〉|k1〉)


while a pair of fermions has an antisymmetric state given by

|k1, k2〉A =

√1

2(|k1〉|k2〉 − |k2〉|k1〉)

The total energy is just the sum over all the single-particle energies E = εk1 + εk1 .Instead of describing every single particle it is more convenient to use occupationnumbers for the number of particles in every energy level.

E =

∞∑

k=0

nkεk

where the occupation number nk is the number and particles in a single-particlestate |k〉. For an arbitrary number of particles we will get a Fock space, which is atensor product of Hilbert spaces

F = H0 ⊗H1 ⊗ . . .⊗HN

From the quantization of the harmonic oscillator we have the lowering, raisingand number operators, a†, a and N respectively, where N = a†a and we have thecommutators

[a, a†

]= 1

[N , a

]=[a†a, a

]= a† [a, a] +

[a†, a

]a = −a

[N , a†

]= a†

We have the number operators Nk = a†kak for particles in a state |k〉, and a and a†

are annihilation and creation operators respectively.E.g.

Nk|nk〉 = nk|nk〉ak|nk〉 =

√nk|nk − 1〉

a†k|nk〉 =√nk + 1|nk + 1〉

We now get the Hamiltonian H =∑∞

k=0 Nkεk with energy eigenstates H|Ψ〉 =E|Ψ〉 where E =

∑∞k=0 nkεk is the total energy, and |Ψ〉 = |n0, n1, . . .〉 is the

many-body state. We define the vacuum state |0〉 by

ak|0〉 = 0

and a single-particle state |k〉 as

|k〉 = a†k|0〉

1.3. QUANTUM FIELD THEORY 11

1.3 Quantum field theory

A particle in position x is now represented by

|x〉 =

∞∑

k=0

|k〉〈k|x〉 =

∞∑

k=0

u∗k(x)|k〉 =

∞∑

k=0

u∗k(x)a†k|0〉 = Ψ†(x)|0〉

where Ψ†(x) =∑∞

k=0 u∗k(x)a†k or Ψ(x) =

∑∞k=0 uk(x)ak where Ψ(x) is a

quantum field operator for many-body systems. We introduce time variation us-ing the time propagator

Ψ(x, t) = eihHtΨ(x)e−

ihHt

=∞∑

k=0

uk(x)eihHktake

− ihHkt

=∞∑

k=0

akuk(x)e−ihεkt

Here we have written the Hamiltonian as H =∑∞

k=0 Hk where Hk = Nkεk.We have the commutators between the creation and annihilation operators

[ak, ak′ ] = 0[ak, a

†k′

]= δkk′

which give us the commutation relations between the field operators

[Ψ(x, t), Ψ(x′, t)

]= 0

[Ψ(x, t), Ψ†(x′, t)

]=∑

kk′uk(x)u∗k′(x)

[ak, a

†k′

]e−

ih

(εk−εk′)t

=∞∑

k=0

uk(x)u∗k(x′) = δ(x − x′)

(1.5)

Now we will study free particles in a volume V = L3. We have the Hamilto-nian h = − h2

2m∇2 and the eigenvalue equation huk(x) = εkuk(x) where uk(x) =√1V e

ih

k·x and εk = k2

2m where k is the momentum. Our volume gives peri-

odic boundary conditions u(x, y, z) = u(x + L, y, z) when ki = 2πhL ni where

ni = 0,±1,±2, . . .. Normalization gives

∫d3xuk(x)u∗k′(x) =

1

V

∫ L2

−L2

dx

∫ L2

−L2

dy

∫ L2

−L2

dzeih

(k−k′)·x = δkk′


and we have

1

V

∑

k

eik·(x−x′) = δ(x− x′) =1

V

∑

n

∆nx∆ny∆nzeih

k·(x−x′)

=

∫d3k

(2πh)3eih

k·(x−x′) = δ(x − x′)

where ∆ni = L2πh∆ki and we have used the fact that 1

V

∑k →

∫d3k

(2πh)3 whenV →∞.

Chapter 2

Lecture 2

2.1 2nd quantization

The step that we performed last lecture, when we obtained a field operator froma wave function, Ψ(x, t) → Ψ(x, t), is often called 2nd quantization. In field-description we will have a new interpretation of the creation and annihilation oper-ators, a† and a. In the field theoretical description a will annihilate a particle in onestate and a† will create a new particle in another state. This justifies the use of theterms “creation” and “annihilation” operators, while one in conventional quantummechanics rather should use the terms “raising” and “lowering” operators.

We can rewrite our Hamiltonian as

H =∑

k

a†kakεk =

∫dxΨ†(x)

[− h2

2m∇2 + U

]Ψ(x)

since[− h2

2m∇2 + U

]= h

huk(x) = εkuk(x)

Ψ(x) =∑

k

akuk(x)

Similarly for our number operator we get

N =∑

k

a†kak

=

∫d3xΨ†Ψ

=

∫d3x

∑

kk′a†kak′ u

∗kuk′︸︷︷︸δkk′

=∑

k

a†kak

13


Now we want to study the time evolution of Ψ(x, t). Using the Heisenberg equa-tion of motion we find

ih∂

∂tΨ(x, t) =

[Ψ(x, t), H

]

=

∫d3x′

[Ψ(x, t), Ψ†(x′)hΨ(x′, t)

]

=

∫d3x′δ(x− x′)h(x′, t)Ψ(x′, t)

= h(x)Ψ(x, t)

So we end up with the equation[− h2

2m∇2 + U

]Ψ(x, t) = ih

∂Ψ

∂t(2.1)

This equation really looks like the good old Scrödinger equation (SE), and ofcourse this is not just a coincidence. Formally it is of course the same equation.But the fact is that also in this case the interpretation is somewhat different fromstandard quantum mechanics. In contrast to the one-particle description that weare used to with the ordinary SL, here we are using a field operator that in prin-ciple may describe an arbitrary number of particles. This is what is called 2ndquantization.

2.2 Classical field theory

We now introduce the Hamilton density H, given by

H =

∫d3xH

H = Ψ†(x, t)[− h2

2m∇2 + U

]Ψ(x, t)

We want to introduce a Lagrange formalism since that is a more fundamental the-ory than the Hamiltonian. We use what we know from classical field theory 1.Classically we have

H = Ψ∗(− h2

2m∇2 + U

)Ψ

where the field operator Ψ(x) has become a complex Schrödinger field. Havingthis Hamiltonian density, we know that we can construct a Lagrangian density

L = πΨ−H = ihΨ∗Ψ−Ψ∗(− h2

2m∇2 + U

)Ψ

= ihΨ∗Ψ +h2

2m|∇Ψ|2 −Ψ∗UΨ (2.2)

1That is, a field theory where we have not performed a 2nd quantization, and where we usecomplex functions instead of field operators

2.3. FERMIONS AND BOSONS 15

Here Π is the canonical field momentum

Π ≡ ∂L∂Ψ

and we see that (2.2) satisfies

H = πΨ−L

Let us see how we can use the canonical momentum in the quantum theory. Inthe quantum theory we had the non-trivial commutator

[Ψ(x, t), Ψ†(x′, t)

]= δ(x − x′)

Using the canonical momentum we get

Ψ → Ψ

Π → Π

⇒[Ψ(x, t), Π(x′, t)

]= ihδ(x− x′)

Back to the classical theory, the Lagrangian can be expressed as

L =

∫d3xL.

That gives us the action

S =

∫dt L =

∫d4xL

Using Hamilton’s principle, δS = 0, we now get a new Euler-Lagrange equation:

∂L∂Ψ− ∂

∂t

∂L∂Ψ−∇ · ∂L

∂(∇Ψ)= 0.

Using the Lagrangian density from (2.2) and solving the complex conjugated ver-sion of this equation we obtain:

ih∂Ψ

∂t=

[− h2

2m+ U(x)

]Ψ(x, t)

which is a “classical field Schrödinger equation”.

2.3 Fermions and bosons

Now we will have a closer look at bosons and fermions. For particles that obeyBose-Einstein statistics we have

Ψ(x) =∑

k

akuk(x) and[ak, a

†k′

]= δkk′


which gives [Ψ(x, t), Ψ†(x′, t)

]= δ(x− x′)

Classically we haveΨ =

∑

k

akuk(x)

where Ψ, ak and uk are complex. For particles that obey Fermi-Dirac statistics wewant to comply with the Pauli principle. That will be the case if we substitute allthe commutators in the Bose-Einstein theory with anti-commutators in the Fermi-Dirac theory.

[ , ]︸︷︷︸Bose−Einstein

→ , ︸︷︷︸Fermi−Dirac

where the anitcommutator is defined

A,B = AB +BA.

Then, for fermions we have:

Ψ(x) =∑

k

bkuk(x)

bk, b†k′ = δkk′

bk, bk′ = 0

H =∑

k

b†kbk︸︷︷︸bNk

εk

The Pauli principle implies that Nk for fermions only has the eigenvalues 0 and 1.We want to find out if this is true in this formalism.

N2k = b†k bkb

†k bk = b†k(1− b

†k bk)bk = b†k bk = Nk

which givesNk(Nk − 1) = 0 → nk(1− nk) = 0

Thus nk can only take the values nk = 0, 1, which means that our formalismwith anti-commutators obeys the Pauli principle.

In classical theory the fermion operator will become

Ψ(x) =∑

k

bkuk(x)

where uk(x) is a complex function and bk is a Grassmann number (Γ) (see Table2.1). We see that a classical fermion field must be a Grassmann field. Using theDirac equation we get:

(i 6 ∂ −m)Ψ(x) = 0

2.3. FERMIONS AND BOSONS 17

—————————————————————————————-

Grassmann numbers (Γ) have certain properties that make them well suited fordescribing fermionic fields. Consider two Grassmann numbers α and β. They willanti-commute:

αβ + βα = 0

which implies thatαβ = −βα ⇒ α2 = β2 = 0

If you multiply a Grassmann number by a complex number you get a Grassmannnumber. You can have complex Grassmann numbers, and they will have the fol-lowing properties:

ψ = α+ iβ

ψ∗ = α− iβψ2 = (α− iβ)(α + iβ)

= α2 − β2 + i (αβ + βα)︸︷︷︸=0

= 0

ψ∗ψ = (α− iβ)(α + iβ)

= α2 + β2 + i (αβ − βα)︸︷︷︸6=0

= 2iαβ 6= 0

This last expression, stating that complex Grassmann numbers will have anon-vanishing ψ∗ψ is good, since this will allow us to have non-vanishing massterms on the form mψ∗ψ in our theory.

—————————————————————————————-

Table 2.1: Grassmann numbers


whereΨ =

∑

k

(bkuk + d†kuvk) and 6 ∂ = γµ∂µ

Here bk and d†k are Grassmann variables, and d† describes antiparticles.Note that although a quantity like (Ψ∗Ψ)2 = 0 in Grassmann theory, this is not

necessarily true for real fermions. This is due to the existence of the two spin statesof the fermions, such that

(Ψ∗Ψ)2 → (Ψ∗↑Ψ↓)(Ψ∗↓Ψ↑) = (Ψ∗↑Ψ

∗↓)(Ψ↑Ψ↓)

where the pairs in the last expression are called Cooper pairs.

2.4 Approximation to point particles

It is interesting to notice that in this formulation one is able to have a Fermi-Diraccondensate in much the same manner that we have Bose-Einstein condensates. Inboth cases the atoms will be described using a local field Ψ(x) where the atoms aretreated as point particles. Whether the use of point particles is a valid approxima-tion or not depends of course on the energy range in which we are working. Theapproximation is valid as long as the typical de Broglie wavelength λB of the atomis much larger than the typical size of the atom. In room temperature this is alwaysthe case.

In the case of atoms we are used to work with well defined energy bands. Herewe can treat each band as being one Ψ-field. This is valid as long as the energydifferences between the states inside an energy band are much smaller than thetypical temperature that we are considering. A typical difference between energylevels in an atom is ∼ 1eV. That corresponds to a temperature around 104K. Thatmeans that the point particle approximation is a very good approximation whenworking in room temperature.

Analogously, when working with for example protons, a local field descriptionis valid as long as we do not have any excitations of the quarks in the proton. Later,when talking about interactions, we will have the same situation. The scatteringlength a of a typical potential will have a finite extension, and using a point particletreatment is valid as long as λB >> a. The point here is that whatever theory youare using, it has a finite range of validity, and it is important not to use the theoryoutside this range. This is what mathematicians never understand when workingwith physics.

2.4.1 Interactions between point particles

For an interaction between a point particle and an external field we have the Hamilto-nian density

H = Ψ†(x)

(−h2

2m∇2 + U

)Ψ(x).

2.4. APPROXIMATION TO POINT PARTICLES 19

Here the first term on the right hand side is the free particle part, while the last termis taking care of the interactions. U(x) is annihilating a particle in x and creating anew particle in the same point.

An internal interaction between two particles in different points can be de-scribed by a potential V (x − x′). This interaction can then be described by aHamiltonian density

HI =

∫d3x

∫d3x′

1

2Ψ†(x)Ψ(x)V (x− x′)Ψ†(x′)Ψ(x′)

For our approximation of a point particle, this interaction will be a δ-potential:

V (x− x′)→ λδ(x − x′)

λ ∼ a ∼∫

d3xV (x).

The a is still the scattering length. Do not confuse this λ with the de Broglie wave-length λB . Using this delta function we get the Hamiltonian

H =

∫d3x

λ

2(Ψ†(x)Ψ(x))2

and the interaction hamiltonian density then yields

H =λ

2(Ψ†(x)Ψ(x))2

To find our new L, including the interaction term, we use (2.2) and include our newinteraction term. This gives

L = ihΨ∗Ψ +h2

2m|∇Ψ|2 −Ψ∗UΨ− λ

2(Ψ∗Ψ)2

This Lagrangian density is frequently used in particle physics, nuclear physics,super-conductivity etc.. The last term makes is nonlinear and complicates the the-ory a lot. Using the complex conjugated Euler-Lagrange equations we obtain thefield equation

ih∂Ψ

∂t= − h2

2m∇2Ψ + UΨ + λ(Ψ∗Ψ)Ψ

This equation can be viewed as a “non-linear Schrödinger equation” for classicalfields. When used in Bose-Einstein theory it is called the Gross-Pitaevskij equa-tion. It is used to determine the distribution of particles in a potential. There havealso been attempts to include a similar non-linear term in standard quantum mech-anics, but without great success.

Chapter 3

Lecture 3

3.1 Relativistic fields

We will now like to consider relativistic fields describing particles with rest massm. For such particles energy and momentum are related by

E2 = m2c2 + c2p2

From non-relativistic quantum mechanics we know that if the particles may bedescribed by a single scalar wave function, φ(x) say, then momentum and energysatisfy the following :

E → i∂

∂tp→ −ih∇

The resulting and relativistic wave-equation is called the Klein-Gordon equationand is given by:

− ∂2

∂t2φ(x, t) = −∇2φ+m2φ (3.1)

This equation was actually first described by Erwin Schrödinger, but is still mostcommonly referred to as the Klein-Gordon equation (KG). The equation, since itis relativistic, can be written covariantly. To do so, there must be no doubt aboutwhat metric that is being used. For our purpose we choose a metric with positivetime component:

ηµν =

+1 0 0 00 −1 0 00 0 −1 00 0 0 −1

The Klein-Gordon equation may now be more compactly written in covariant formas

(∂2 +m2)φ = 0 (3.2)

Where the shorthand notation ∂2 ≡ ∂µ∂µ has been used.

21


3.1.1 Relativistic scalar field theory

We now want to consider a relativistic field theory describing scalar fields. Toconstruct such a theory the field φ turns into a field operator.

φ(x)→ φ(x)

However, and as seen earlier in lecture 2 in the case of the Schroedinger equationin (2.1), the field operator is still satisfying the same equation as the field itself, thatis the KG-equation

(∂2 +m2)φ = 0

Thus the field operator φ has a classical counterpart in the scalar field φ, satisfying(3.2). In the case of considering the field as a classical quantity, we know that theequation of motion can be derived from the classical Lagrangian density of the fieldby applying the variational principle. The Lagrangian density is in this particularcase given by:

L =1

2ηµν∂µ∂νφ−

1

2m2φ2 =

1

2φ2 − 1

2(∇φ)2 − 1

2m2φ2 (3.3)

The Euler-Lagrange equation with respect to the field φ:

∂L∂φ− ∂

∂xµ∂L∂φ,µ

= 0

results in equation (3.2) as expected.The energy density of the field is given by the Hamiltonian density H

H = φπ −L

where the canonical momentum in this case is π ≡ ∂L∂φ

= φ. With the Lagrangian

density as in (3.3), the Hamiltonian density turns out to be:

H =1

2φ2 +

1

2(∇φ)2 +

1

2m2φ2

To construct a quantum theory of these classical quantities, there are (at least)two approaches. These are the canonical quantization or quantization by Fourierexpanding the field in its plane waves. The first procedure we saw in the secondlecture, and as a short reminder it looks like:

φ → φπ → π

⇒[φ(x, t), π(x′, t)

]= iδ(x − x′)

where we assign the well-known constraint, represented by the δ-function to thegeneralized coordinate and its conjugated momentum. The other way to quantizeis to Fourier expand the field in a finite volume. The field and its adjoint editionthen takes form as:

φ(x, t) =

√1

V

∑

k

φk(t)eik·x

3.1. RELATIVISTIC FIELDS 23

φ∗(x, t) =

√1

V

∑

k

φ∗k(t)e−ik·x

here k takes discrete values. The constraint on the field φ that it is real valuedimplies the following:

φk = φ∗−k (3.4)

This is clearly seen by letting k → −k in the expression of the adjoint field andthen finally demanding it to equal the field itself:

φ(x, t) = φ∗(x, t) =

√1

V

∑

k

φ∗−k, (t)e+ik·x

We are allowed to perform this substitution in wave number since the wave numberk is arbitrary in the sense that whether we sum over all the positive or all thenegative values, the result should be the same in both cases. The next step is tofind the Lagrangian. To do so, we will make use of the following relation whenintegrating the Lagrangian density to get the Lagrangian:

∫

Vd3xeik·xe−ik

′·x = V δkk′

here V is the volume of the box in which we are considering the Fourier modesof the field. The Lagrangian now follows directly from the expression in (3.3) byintegrating over the whole volume of the box.

L =

∫d3xL =

1

2

∫d3x

√1

V

√1

V

∑

kk′φkφ

∗k′e

ik·xe−ik′·x + . . .

=1

2

∑

k

φkφ−k + . . .

=1

2

∑

k

[φkφ

∗k −

1

2(k2 +m2)φ∗kφk

](3.5)

here the k2 and the m2 arise from the gradient term and the mass term in theLagrangian density as given in (3.3), respectively. So from the above expression wesee that the Lagrangian may be written as a sum over all Fourier modes k. We alsonotice that the Lagrangian consists of a kinetic term and a quadratic potential term,and thus is similar to the harmonic oscillator. To see this similarity more explicitwe define the oscillation frequency to be ωk ≡

√k2 +m2 and the Lagrangian then

turns into a harmonic oscillator for every mode k:

L =1

2

∑

k

[φkφ

∗k −

1

2ω2

kφ∗kφk

]

The field operator:

φk =

√1

2ωk(ak + a†−k) and φk(t) =

√1

2ωk[ake

−iωkt + a†−keiωkt ] (3.6)


and the adjoint operator:

φ†k =

√1

2ωk(ak − a†−k) = φ−k

here the last equality follows from the real-value constraint that is assigned to thefield φ as seen in equation (3.4). The operators are also set to satisfy the canonicalcommutation relation [ak, a

†k′ ] = δkk′ of the harmonic oscillator. We can now find

the Hamiltonian by:

H =

∫d3xH =

1

2

∑

k

ωk(a†kak + aka†k)

=∑

k

ωk(a†kak +1

2)

where, in the last step, the number operator Nk = a†kak has been used. The vacuumstate is |0〉 with the annihilation operator acting on it as ak|0〉 = 0. This fact leadsto a profound problem in quantum field theory that is yet to be solved. If we take acloser look at the vacuum energy, E0 of the field as it is given by the Hamiltonian,it is clearly seen that the energy diverges:

H|0〉 = E0|0〉 ⇒ E0 =1

2

∑

k

ωk = +∞!

This is so since all frequencies ωk are positive, that is, ωk =√

k2 +m2 > 0, and,when summing over all possible wave numbers, the sum diverges.The time dependent field operator now looks like:

φ(x, t) =

√1

V

∑

k

φk(t)eik·x

=∑

k

√1

2ωkV

[ake

i(k·x−ωt) + a†ke−i(k·x−ωt)

]

=∑

k

√1

2ωkV

[ake−i(k·x) + a†ke

+i(k·x)

](3.7)

In the last step the exponents have been expressed in terms of the 4-vector identityk · x = ωkt− k · x.

3.1.2 External scalar field

We can expand the formal quantization process of the field to also account foran external scalar field. In this case the external field is chosen to be spatially

3.2. VACUUM ENERGY 25

dependent U = U(x). When the scalar field φ is in the presence of U(x) theLagrangian density changes from the free Lagrangian density

L0 =1

2(∂µφ)2 − 1

2m2φ2

to a density containing an additional term determined by the external field U

L =1

2(∂µφ)2 − 1

2m2φ2 − 1

2Uφ2

The Lagrangian density is a classical quantity and we retrieve the (classical) equa-tion of motion by applying the Euler-Lagrange equation.

[∂2 +m2 + U(x)]φ(x) = 0 (3.8)

The similarity with the Klein-Gordon equation that we found earlier is striking.This motivates us to introduce an effective and position dependent mass M(x) thatnow replaces the mass m in the ordinary Klein-Gordon case as seen in (3.1). If wedefine the effective mass as M 2(x) = m2 + U(x), equation (3.8) takes form as

[∂2 +M2(x)]φ(x) = 0

The question is now how to construct a quantized theory in this case. On theassumption that the particular Klein-Gordon equation has solutions of the formu = up(x), we may find the field operator. The functions u(x), often referred to asmodes, are classical functions. The field operator is then given by

φ(x) =∑

p

[apup + a†pu∗p(x)] (3.9)

and is normalized by the commutator identity

[ap, a†p′ ] = δpp′

This particular example of how to quantize a theory with a non-trivial Lagrangianis made to illustrate the fact that the field operator always will be of the form asgiven in (3.9).

3.2 Vacuum energy

If the volume that we Fourier expand over is large enough, the sum in the expres-sion of the vacuum energy can be replaced by an integral over all modes k.

E0 =1

2

∑

k

ωk =1

2V

∫d3k

(2π)3(k2 +m2)1/2

=1

2V

∫ ∞

0

4πk2dk

8π3(k2 +m2)1/2

⇒ ε0 =E0

V=

1

4π2

∫ ∞

0dkk2

√k2 +m2


which clearly diverges. A suitable comment in this manner should be that all ef-fective field theories are only valid up to some given value of the wavenumber.Another way to view this is that one cannot construct a field theory with infinitelylarge resolution, that is with a wavenumber that approaches infinity. It is thereforecommon to introduce a so called cut-off K in the frequency. The maximal rangeof validity of the theory is therefore found by letting the cut-off frequency be of theorder of the Planck scale. Hence:

kmax = K = MPlanck ≈ 1018 GeV

There are several ways to make the integral of the energy density ε0 finite. Thisprocess is called regularization and will be dealt with in more detail later. However,and this is the important issue, is that for a physically consistent theory the choiceof method of regularize cannot influence the final answer. As an illustration ofdifferent types of regularization, let us consider two specific examples:

ε0 =

14π2

∫ K0 dkk2

√k2 +m2

14π2

∫∞0 dkk2

√k2 +m2e−k/K

The first case has the cut-off explicit as the upper limit of the integral, while theother introduces an exponentially damping term in the integrand to prevent diver-gence. The vacuum energy in the special case of a massless scalar field is to firstorder given by:

ε0 =

116π2K

4 + . . .

34π2K

4 + . . .

illustrating the problem that the vacuum energy can not be properly described sincehere two different regularizations give different answers. The vacuum energy prob-lem is explicitly shown if we try to compare the predicted value of the vacuum en-ergy with that of today’s best fit to observations. In this case we use as the predictedvalue a term ∼ K4. The observed vacuum energy density ρΛ (the cosmologicalconstant) is

ρΛ = µ4

where µ ≈ 10−3 eV. We then get

ε0ρΛ∼ K4

µ4∼(

1018GeV

10−3eV

)4

= 10120

This is a huge error! Hence, the field theoretical description in this manner isno way near of giving a correct estimate of the value of the vacuum energy asit is observed in the universe. Actually this misconception between theory andobservations is the largest ever observed.

3.3. THE CASIMIR EFFECT 27

3.2.1 Fermions

From the second lecture we saw how a field theory for fermions could be extractedsimply by substituting the commutator with anticommutators

[ , ]︸︷︷︸Bose−Einstein

→ , ︸︷︷︸Fermi−Dirac

and vice versa. The anticommutator is then defined to obey the canonical commut-ation relation, only now in anticommutator form

bk, b†k′ = δkk′ = [ak, a†k′ ]

The Hamiltonian in these two cases is

Hbos =1

2

∑

k

ωka†k, ak =∑

k

ωk(a†kak +1

2)

Hferm =1

2

∑

k

ωk[b†k, bk] =∑

k

ωk(b†kbk −1

2)

We observe that the vacuum energies of the fermions and bosons are equally largewith opposite sign and hence will cancel. For this to be fulfilled there must how-ever be a special symmetry between fermions and bosons called supersymmetry(SUSY). There is yet no physical evidence of such a symmetry as a fundamentalsymmetry of nature. Nevertheless, the idea has been pursued since supersymmetrygives, as a general feature, a vacuum energy which is 0, due to the cancellationsof the contributions from fermions and bosons. Such a cancellation is importantin supersymmetric quantum field theories, where the divergent contributions to thevacuum energy may be avoided. Signs of SUSY are to be searched for at energiesof the order of TeV at CERN. Even if supersymmetry is found at these high ener-gies, it is not necessarily the final solution to the vacuum energy problem. If wenow assign the vacuum energy with ε0 = K4

SUSY the fraction between predictedand observed values is

ε0ρΛ

=

(1TeV

10−3 eV

)4

= 1060

so even with SUSY at energies E ∼ TeV there seem to be problems estimatingthe vacuum energy.

3.3 The Casimir effect

The Casimir effect arises when two (parallel) plates are separated by, in this par-ticular case a scalar field. There is then a measurable, attractive force between thetwo plates that may, for instance be conductive iron plates. This is shown in figure(3.1).


Z

A

L

Figure 3.1: The Casimir effect: Two plates are separated at a distance L. A repres-ents the area of the plate in the transverse directions x and y. An attractive forcebetween the two plates, depending on the distance between them, is observed.

We will consider the situation where the space between the plates is “filled”with a massless scalar field. We then have:

∂2µφ = 0⇒ ωnkT =

√k2T + n2K2, K =

π

L

The boundary conditions are as given by Dirichlet and Neumann, and look like

• Dirichlet: φ(z = 0) = φ(z = L) = 0.

• Neumann: ∂zφ|z=0 = ∂z|z=L = 0.

A closer look at the Dirichlet conditions yields

u(xT, z, t) ∼ sin(kzz)eikT ·xT e−iωt

with kz taking discrete values, kz = πLn, n = 1, 2, 3, . . ., and xT describing the

two transverse directions x and y. The vacuum energy can now be expressed interms of a two-dimensional integral over transverse momenta as:

E0 = A

∫d2kT(2π)2

∞∑

k=1

1

2

√k2T + n2K2 (3.10)

3.3. THE CASIMIR EFFECT 29

Here, A, is the area of the plates. The integral is divergent, so a regularization isneeded, that is, we make the integral finite:

φ2 = φ(x)φ(x′)

The x′ can be expressed in terms of x, t − iε by a Schwinger time split whicheffectively is to introduce an exponential damping. This results in

φφ ∼ eiωte−iωt′ ∼ e−ωε

To proceed with the calculation of the vacuum energy, we perform a substitutionby ω2 = k2

T + (nK)2 ⇒ ωdω = kTdkT .Thus (3.10) can be written as:

E0 = A

∫2πkT dkT

4π2

∞∑

k=1

1

2ω e−εω

=A

4π

∞∑

n=1

∫ ∞

nKdωω2e−εω

∣∣∣∣ε→0

=A

4π

∞∑

n=1

∂2

∂ε2

∫ ∞

nKdωe−εω

︸︷︷︸1εe−εKn

=A

4π

∂2

∂ε21

ε

∞∑

n=1

e−εKn

=A

4π

∂2

∂ε21

ε

1

eKε − 1(3.11)

To deal with the last fraction in the above expression, it is necessary to consider arelation derived by Euler

t

et − 1=

∞∑

n=0

Bnn!tn

Bn are known as the Bernoulli numbers and will be described later in this lecture.The numbers we need for this pupose are B0 = 1, B1 = −1

2 , B2 = 16 , and

B3 = 0, Thus the last fraction in the expression of the vacuum energy can thennow be expanded, so that the regularized vacuum energy, E reg

0 takes the form:

Ereg0 =

A

4π

∂2

∂ε21

ε

(1

Kε− 1

2+Kε

12− (Kε)3

720+ ...

)

ε→0

The vacuum energy density is further found simply by dividing with the Casimir-volume between the two plates, that is the spatial range of the field.

εreg0 =

Ereg0

AL=

3

2π2ε4− π2

1440L4


If the plates are largely separated, hence L→∞, the vacuum energy density is notaffected by the Casimir-energy, and therefore looks like:

ε0 =3

2π2ε4=

3K4

2π2, with K =

1

ε

This vacuum energy density is often referred to as the open or free vacuum energydensity. The renormalized vacuum energy is finally found by subtracting the freeenergy density, ε0 from the regularized energy density εreg

0 .

εren = εreg0 − ε0 = − π2

1440L4

Here the minus corresponds to an attractive force between the two plates in figure(3.1). So effectively, the plates will contribute to the vacuum energy in such a waythat the vacuum energy between the plates reduces. This means that we have anattractive force between the plates.

Before we end this (long) lecture, let us take a brief look at the mathematicswith Bernoulli numbers.

3.4 Bernoulli numbers

Bernoulli numbers are defined, as we have seen, by the Euler expansion:

t

et − 1=∞∑

n=0

Bnn!tn (3.12)

The sum is quite similar to the power expansion of the natural exponential functionex. If we treat the B’s as ordinary numbers, Bn → Bn, this allows us to rewriteequation (3.12) as

t

et − 1=∞∑

n=0

Bn

n!tn = eBt

Solving for t results in

t = (et − 1)eBt = e(B+1)t − eBt

= (B + 1)t−Bt+

∞∑

n=2

(B + 1)n −Bn

n!tn

︸︷︷︸!=0

And finally the Bernoulli numbers are given by the explicit formula

Bn = (B + 1)n, n ≥ 2

Examples of Bernoulli numbers are

3.4. BERNOULLI NUMBERS 31

• n=2:

B2 = (B + 1)2 = B2 + 2B + 1⇒ B1 = −1

2

• n=3:

B3 = (B + 1)3 = B3 + 3B2︸︷︷︸3B2

+ 3B︸︷︷︸3(−1/2)

+1 ⇒ B2 =1

6

We can also expand this description to account for Bernoulli polynomials wherenow Bn → Bn(x):

t

et − 1ext =

∞∑

n=0

Bn(x)

n!tn

= eBtext = e(B+x)t =

∞∑

n=0

(B + x)n

n!tn (3.13)

So from the relation 3.13) above, we retrieve the explicit formula of the Bernoullipolynomials.

Bn(x) = (B+x)n

Actually, the Bernoulli numbers are just a special case of the Bernoulli polynomi-als.

Chapter 4

Lecture 4

We will in this lecture study different ways to regularize the divergent field. Wewill also see how the divergent integral stated below can be made finite.

E0

A=

∫d2kT(2π)2

∞∑

k=1

1

2

√k2T + n2K2 (4.1)

4.1 Regularization

Last lecture we were looking at the Casimir effect. To summarize we consideredtwo metal plates with an observable attractive force between them, caused by amassless scalar field with Lagrangian density L = 1

2 (∂µφ)2. We found the renor-

malized energy density εrenφ = − π2

1440L4 .If we now consider the 1-dimensional Casimir effect we get the energy

E0 =1

2

π

L

∞∑

n=1

n =π

2L(1 + 2 + 3 + . . .) = S =∞ (4.2)

Here we will need to regularize. We use the exponential damping method and getthe finite sum

Sreg = limε→0

∞∑

n=1

ne−εn =

(− ∂

∂ε

) ∞∑

n=1

e−εn =

(− ∂

∂ε

)1

eε − 1

and here we use our Bernoulli trick again. We can write the last factor as

1

eε − 1=

1

ε

ε

eε − 1=

1

ε

∞∑

n=0

Bnn!εn =

1

ε− 1

2+

ε

12− ε3

720+ . . .

This gives us the regularized sum

Sreg =

(− ∂

∂ε

)1

eε − 1

∣∣∣∣ε→0

=1

ε2+ 0− 1

12+

ε2

240+ . . .

∣∣∣∣ε→0

=1

ε2− 1

12

33


Where the first term is the divergent vacuum energy and the second term is thefinite, renormalized contribution. The interesting physics is described by this last,finite term. We will now try to isolate this term by applying zeta-function regular-ization.

At this point we should familiarize ourselves with Riemann’s zeta-function.(Actually it was initially introduced by Euler, and later further studied by Riemann.)It can be written on the form

ζ(s) =

∞∑

n=1

1

ns

where s > 1. It can also be expressed with help of the Γ-function

ζ(s) =1

Γ(s)

∫ ∞

0dx

xs−1

ex − 1(4.3)

where still s > 1. Here we have used that

1

ex − 1=

e−x

1− e−x = e−x + e−2x + . . .

and the integral definition of the Γ-function

Γ(s) =

∫ ∞

0dxe−xxs−1

The (non-regularized) sum we had earlier can now be written

S = 1 + 2 + 3 + . . . = ζ(−1)

As we have the relation between the zeta-function (with odd-numbered argument)and the Bernoulli numbers

ζ(1− 2n) = −B2n

2n(4.4)

we get our sum

ζ(−1) = −B2

2= −1

2· 1

6= − 1

12

which is the same as we got above. Funny.

4.2 Dimensional regularization

Now we will consider the same effect, but with extra dimensions. We will denotethe number of spatial dimensions d, and the total number of dimensions D, that iswith one time dimension we have D = d + 1. In this case we will have to solvethe following integral to find the energy

I =

∫ddk

(2π)d(k2 +m2

)−N

4.2. DIMENSIONAL REGULARIZATION 35

—————————————————————————————-

One funny thing that is not really relevant at this point is that the solid angles canbe written as a series of integrals in the different dimensions. This is best shownwith some examples

Ω2 = 2π

∫ π

0dθ sin θ = 4π

Ω3 = 2π

∫ π

0dθ1 sin θ1

∫ π

0dθ2 sin2 θ2 = 2π2

Ω4 = 2π

∫ π

0dθ1 sin θ1

∫ π

0dθ2 sin2 θ2

∫ π

0dθ3 sin3 θ3 =

8

3π2

and so on.

—————————————————————————————-

Table 4.1: Solid angles

but we can rewrite this with the help of the Γ-function

I =

∫ddk

(2π)d1

Γ(N)

∫ ∞

0dssN−1e−s(k

2+m2)

=1

Γ(N)

∫ ∞

0dssN−1e−sm

2

∫ddk

(2π)de−sk

2

If we assume spherical symmetry, we can rewrite this with help of the solid angleas ddk = Ωd−1k

d−1dk. Now we can rewrite the last integral

I ′ =∫

ddk

(2π)de−sk

2=

1

(2π)dΩd−1

∫ ∞

0dkkd−1e−sk

2

And if we substitute with t = sk2 we recognize the Γ-function

I ′ =1

(2π)dΩd−1

1

2sd2−1

Γ

(d

2

)(4.5)

Now we need to find Ωd−1. As we know we can rewrite a Gaussian integral in ddimensions as a product of the solid angle and the Γ function

∫ddke−k

2= Ωd−1

∫ ∞

0dkkd−1e−k

2= Ωd−1

Γ(d2

)

2

but we can also imagine that just raising a one dimensional Gaussian integral to thepower of d would do the same job

∫ddke−k

2=

(∫ ∞

−∞dke−k

2

)d= π

d2


And if we combine these expressions we find the solid angle

Ωd−1 =2π

d2

Γ(d2

)

we can insert this into (4.5) and we get

I ′ =1

2dπd2

s1− d2

Great! Now let’s insert this in our original integral

I =1

2dπd2

1

Γ(N)

∫ ∞

0dssN−

d2 e−sm

2

and once again we recognize the Γ-function and finally get

I =1

(4π)d2

Γ(N − d

2

)

Γ(N)(m2)

d2−N (4.6)

Well, if we now try our ordinary two dimensions in this integral we get

I(d = 2, N = −1

2) =

1

4π

Γ(−3

2

)

Γ(−1

2

)(m2)32 = −m

3

6π

where we have used that Γ(z + 1) = zΓ(z). By comparing with the originalexpression for E0

A that we had in equation (4.1), we also get the energy per area

E0

A= − 1

6π

1

2

∞∑

n=1

m3(n) = − 1

6π

1

2

∞∑

n=1

(πL

)3n3

and the energy density

E0

AL= − 1

6π

π3

2L4

∞∑

n=1

n3

where we recognize the sum as the zeta-function∑∞

n=1 n3 = ζ(−3) = − 1

120 andwe get

E0

AL= − π2

1440L4

as before.

4.3. QUANTUM STATISTICAL MECHANICS 37

4.3 Quantum statistical mechanics

We all know the eigenvalue equation for the Hamiltonian H|n〉 = En|n〉, andthe probability for a specific state , Pn = 1

Z e−βEn where β = 1

kBT. The sum of

probabilities should be unity∑

n Pn = 1 and that leads us to the partition function,

Z =∑

n

e−βEn =∑

n

〈n|e−βH |n〉 =∑

n

ρnn = Trρ (4.7)

Where ρ is the density operator and ρmn is the corresponding density matrix. Fromthis we get the expectation value of an observable A

〈A〉 =∑

n

Pn〈n|A|n〉 =1

Z

∑

n

e−βEn〈n|A|n〉

=1

Z

∑

n

〈n|e−βHA|n〉 =1

ZTrρA =

1

ZTrAρ

So we can find the energy, as the expectation value of the Hamiltonian

E = 〈H〉 =1

ZTrρH =

1

ZTr(e−βHH)

=1

Z

(− ∂

∂β

)Tre−βH =

1

Z

(− ∂

∂β

)Z = − ∂

∂βlnZ

Where Z = e−βF and F is Helmholtz free energy. This gives us

E =∂

∂β(βF ) = F + β

∂F

∂β= F − T ∂F

∂T= F + TS

Where S = −∂F∂T is the entropy.

4.3.1 Example with relativistic scalar particles

Let us now say that we have a relativistic gas of scalar particles. We have theLagrangian density

L =1

2(∂µφ)2

and this gives us the following Hamiltonian

H =∑

k

ωk

(Nk +

1

2

)=∑

k

Hk

Where Nk = a†kak. We get the energy of each state

Ek = 〈Hk〉 =1

Z

∞∑

nk=0

ωk

(nk +

1

2

)e−β(nk+ 1

2)ωk


where

Zk =

∞∑

nk=0

e−β(nk+ 12)ωk =

e−12βωk

1− e−βωk

so that we have

Ek = − ∂

∂βlnZk =

1

2ωk +

ωk

eβωk − 1

where the first term is the zero-point energy, and the second term is the thermalenergy. This gives us the total energy for the gas

E =∑

k

(1

2ωk +

ωk

eβωk − 1

)= E0 +

∑

k

ωk

eβωk − 1

where E0 is the vacuum energy. If we neglect the vacuum energy, the expressionof the energy can be approximated as an integral (in arbitrary dimensions) by

E = V

∫ddk

(2π)dk

eβk − 1

Then we find the energy density, ε

ε =E

V=

Ωd−1

(2π)d

∫ ∞

0

dkkd

eβk − 1

and if we substitute with x = βk we get

ε =Ωd−1

(2π)d1

βd+1

∫ ∞

0dx

xd

ex − 1︸︷︷︸Γ(d+1)ζ(d+1)

where we see that the integral of the energy density can be written as a product ofthe Γ-function and the Riemann-zeta function (see (4.3)).

ε =Ωd−1

(2π)dΓ(D)ζ(D)(kBT )D

where D = d+ 1. Now, for D = 4 (ordinary spacetime) we get

Ω2 = 4π Γ(4) = 3! = 6 ζ(4) =π4

90

and this gives us energy density

ε =π2

30(kBT )4 ≡ aT 4 (4.8)

4.3. QUANTUM STATISTICAL MECHANICS 39

which we recognize as the Stefan-Boltzmann law. This gives us the entropy densitys = S

V (∂s

∂T

)

V

=1

T

(∂ε

∂T

)

V

= 4aT 2 ⇒ s =4

3aT 3

Then we find the pressure by this Maxwell relation(∂p

∂T

)

V

=

(∂S

∂V

)

T

= s =4

3aT 3

which gives

p =1

3aT 4 =

1

3ε

For D 6= 4 we get p = 1dε, and we will get the energy-momentum tensor

〈Tµν〉 =

ε 0 0 0 . . .0 ε

d 0 0 . . .0 0 ε

d 0 . . .0 0 0 ε

d . . ....

......

.... . .

and this is traceless〈T µµ〉 = gµν〈Tµν〉 = ε− ε = 0

The expectation value of the trace is always zero, regardless of the number ofdimensions. We remember that we had the Lagrangian density L = 1

2 (∂φ)2, andthis gives us the energy-momentum tensor

Tµν = ∂µφ∂νφ− ηµν1

2(∂λφ)2

with the trace

T µµ = (∂µφ)2 − D

2(∂λφ)2 =

(1− D

2

)(∂µφ)2

so the trace itself is equal to zero only for D = 2, while, as we have seen, theexpectation value of the trace is zero for all dimensions. This is very strange.

Chapter 5

Lecture 5

So far we have been considering only free fields. In the real world no fields arereally free, so it is time to get some interactions into our formalism. To do this, wewill introduce a field theory with non-zero temperature, and use a formalism basedon path integrals with imaginary time.

5.1 Path integral formalism

Recovering from last lecture, we know that we may write the partition function, Z ,in energy representation as

Z = Tre−βH =∑

n

〈n|e−βH |n〉

In coordinate basis, where q|q〉 = q|q〉 and 〈q|q ′〉 = δ(q − q′) we can in a similarway write Z as

Z = Tre−βH =

∫dq〈q|e−βH |q〉. (5.1)

This is leading us to the path integral formalism. We are used to a time propagatoron the form e−itH , but this is formally the same structure as eβH when it → β,where β is what is going to be called imaginary time. This is a very useful concept.We are later going to see that statistical mechanics can be viewed upon as themovement of a quantum mechanical system in imaginary time.

Remembering the density operator introduced in (4.7), we now define a densitymatrix in coordinate basis as

ρ(qb, qa;β) = 〈qb|e−βH |qa〉= 〈qb| e−εHe−εH ...e−εH︸︷︷︸

N :Nε=β

|qa〉

Later we will (of course) take the limit (ε → 0, N → ∞). As if this doesn’t looknasty enough, we will now insert a complete set of states,

∫dq|q〉〈q| = 1, between

41


all the e−εHs in the above expression, giving us a product of N − 1 integrals:

ρ(qb, qa;β) =

N−1∏

n=1

∫ +∞

−∞dqn〈qb|e−εH |qn−1〉〈qn−1|e−εH |qn−2〉. . . 〈q2|e−εH |q1〉〈q1|e−εH |qa〉

(5.2)All these matrix elements have the same form, so it is sufficient to calculate onlyone of them. We will try to find a general expression for this element. Using thewell-known forms of the Lagrangian and Hamiltonian

L =m

2q2 − V (q)

H =1

2mp2 + V (q)

we have (to first order in ε) that:

〈q′|e−εH |q〉 = 〈q′|1− εH|q〉= 〈q′|q〉 − ε〈q′|H(p, q)|q〉

= δ(q′ − q)︸︷︷︸R dp2πei(q′−q)p

−ε∫

dp2π〈q′|H(p, q)|p〉︸︷︷︸“p2

2m+V (q′)

”〈q′|p〉

e−iqp︷︸︸︷〈p|q〉

=

∫ ∞

−∞

dp2πei(q

′−q)p [1− εH(q′, p)]

=

∫ ∞

−∞

dp2πei(q

′−q)pe−ε»p2

2m+V (q′)

–

=

√m

2πεe−ε»m2

“q′−qε

”2+V (q′)

–

(5.3)

Actually, Dirac obtained this matrix element already in the 30s, but Feynman wasthe first one to use it when considering the product from (5.2). This product nowyields

ρ(qb, qa;β) =( m

2πε

)N/2 N−1∏

n=1

∫ ∞

−∞dqn exp

−

N∑

n=1

ε

[m

2

(qn − qn−1

ε

)2

+ V (qn)

]

See figure 5.1 to get an idea of how the path integrals are working. We aregetting the integral in the limit ε→ 0 and N →∞. Using the definitions

qn ≡ q(τn) , τn = nε , β = Nε

we haveqn − qn−1

ε

∣∣∣∣ε→0

=dqndτ≡ qn.

5.1. PATH INTEGRAL FORMALISM 43

qb q

τ

β

N

ε

q2 q1 qa

Figure 5.1: Keeping qa and qb constant we sum over all possible positions inbetween

Then, in the same limitN−1∑

n=1

ε→∫ β

0dτ

and we get that the density matrix now looks like

ρ(qb, qa;β) =

∫D′qe−

R β0

dτLE(q,q) (5.4)

where

D′q =( m

2πε

)N/2 N−1∏

n=1

dqn

∣∣∣∣∣N→∞

and LE = m2 q

2 + V (q) is the Euclidean Lagrangian, which is the continuous limitof the sum in the former exponents. Now, introducing the imaginary time, τ , byt → −iτ and comparing with our good, old Lagrangian L = m

2 q2 − V (q) this

yields

L→ −[m

2

(dq

dτ

)2

+ V (q)

]

and we see that L → −LE , which shows that we are now working in a Euclideanmetric (ds2 = dτ2 + dx2) instead of a Minkowski metric (ds2 = dt2− dx2). Thisfield theory is therefore called an Euclidean field theory.

The integral in (5.4) is of course not trivial, but it can be calculated analyticallyin some cases, and we will soon consider the case of a harmonic oscillator. Whatwe really are interested in, though, is not the matrix element, but the partitionfunction Z . We assume the integral to be over a periodic path, and thus we have


qa = qb = q0 = qβ ≡ q. The partition function as seen in (5.1) now yields

Z =

∫ ∞

−∞dqρ(q, q;β)

=

∫Dqe−

R β0

dτLE(q,q)

where ∫Dq =

( m

2πε

)N/2 ∫ N∏

n=1

dqn.

5.1.1 The partition function for the harmonic oscillator

Considering a harmonic oscillator potential we have

LE =1

2q2 +

1

2ω2q2

L =1

2q2 − 1

2ω2q2

and our partition function yields

Z =

∫Dqe

− 12

R β0 dτ

h( dq

dτ )2+ω2q2

i.

And since we have periodic paths, q(0) = q(β):

∫ β

0

(dq

dτ

)2

dτ = qq|β0︸︷︷︸=0

−∫ β

0dτq

d2

dτ2q

Which gives

Z =

∫Dqe−

12

R β0

dτq(τ)[−∂/tau2+ω2]q(τ)

Now, taking the periodic boundary conditions explicitly into account we have:

q(τ) =

√1

β

∑

n

qneiωnτ

q(τ + β) =

√1

β

∑

n

qneiωn(τ+β) !

= q(τ)

This tells us that eiωnβ = 1, and thus

ωn = 2πβ n , n = 0,±1,±2, . . . (5.5)

5.2. DIVERGENT PRODUCTS AND FUNCTIONAL DETERMINANTS 45

The frequencies in (5.5) are called Matsubara frequencies. Originally this form-alism was introduced in solid state physics, but this periodicity will occur in allquantum mechanical systems at a finite temperature.

Earlier we have performed a normalization over the volume. Analogously, herewe will perform a normalization over the τ -coordinate, and this will lead us to

∫ β

0dτeiωnτeiωmτ = βδn,−m

This gives:

∫ β

0dτq2 =

∑

n

q∗nqn

∫ β

0dτ q2 =

∑

n

qnqnω2n

and our partition function becomes

Z =

∫Dqe−

12

P∞n=−∞ q∗n(ω2

n+ω2)qn

=∏

n

∫ ∞

−∞

dqn(√

2π)Ne−

12

Pn q∗n(ω2

n+ω2)qn

where

qn =

√1

2(xn + iyn).

Remembering that q∗n(ω2n + ω2)qn = q∗−n(ω2

n + ω2)q−n, when summing over alln (both positive and negative), the 1/2-factor in the exponent vanishes. If we alsoinsert for qn, we have

Z =∏

n

∫ ∞

−∞

dxn(√

2π)N

∫ ∞

−∞

dyn(√

2π)Ne−(x2

n+y2n)λn

=

∞∏

n=−∞

√1

λn

=

∞∏

n=−∞

1√ω2n + ω2

(5.6)

which of course is a divergent product.

5.2 Divergent products and functional determinants

Earlier we have dealt divergent sums. Now we will develop a method for solvingdivergent products using functional determinants. We have encountered integrals


on the form

Z =∏

i

∫ ∞

−∞

dxi(√

2π)Ne−

12xiAijxj (5.7)

where Aij = Aji. Now we will use a rotation matrix, R, to transform x into a newvariable y. R is orthogonal, such that R−1 = RT . We have:

xiAijxj = xTAx = yT RAR−1︸︷︷︸

AD(diagonal)

y = yTADy = yiaiδijyj = aiy2i .

Using this, (5.7) yields

Z =∏

i

∫ ∞

−∞

dyi(√

2π)Ne−

12aiy

2i

=

(1∏i ai

)1/2

= det−12AD = det−

12A

Here, the last step follows fromAD = RAR−1 ⇒ detAD = |AD| = |R||A||R−1| =|A|.

Now, using the expressions already obtained for Z in the case of a harmonicoscillator, we have:

Z =∏

n

∫ ∞

−∞

dqn√2πe−

12

Pn q∗n(ω2

n+ω2)qn

=

∞∏

n=−∞

1√ω2n + ω2

=

∫Dqe−

12

R∞0 dτq(−∂2

τ+ω2)q

= det−12(−∂2

τ + ω2)

(5.8)

5.3 Generalized zeta-function regularization

When regularizing our divergent sums, we made use of Riemann’s ζ-function. Forthe divergent products we will use a method called generalized zeta-function reg-ularization. Physicists tend to claim that this method was invented by StephenHawking, but in fact the mathematicians had already been playing around with itfor a long time when the physicists finally found it and saved it from the ocean ofuselessness.

Now, we have encountered problems on the form

detM =∏

n

λn =?

5.3. GENERALIZED ZETA-FUNCTION REGULARIZATION 47

We define our generalized zeta-function, f(s) by1

f(s) =∞∑

n=1

1

λsn, s > 1

Again it is possible to extend the definition to include all values of s. We thenrewrite the definition

f(s) =

∞∑

n=1

e−s logλn .

Differentiation gives

f ′(s) = −∞∑

n=1

logλn e−s logλn

e.g.

f ′(0) = −∞∑

n=1

logλn.

Thus we see that

log detM =∑

n

logλn = −f ′(0)

detM = e−f′(0) (5.9)

5.3.1 Some examples

Let us now consider a couple of simple applications. First, a product of a constant:

∞∏

n=1

a = a · a · a · . . .

= a1+1+1+...

= aζ(0) = a−12 =

√1

a

That was easy! Let us take a small step further:

∞∏

n=1

n = 1 · 2 · 3 · . . .

= e−ζ′(0) = e

12ln(2π) =

√2π

1In the lectures it was denoted Z(s), but if we stick to the f(s) notation used by Petter Callin, weavoid confusing it with the partition function.


In this example λn = 1 and f(s) = ζ(s). As a last example, let us consider thepartition function for the harmonic oscillator that we saw in (5.6):

Z =∏

n

(1

ω2n + ω2

) 12

=1

ω

∞∏

n=1

1

ω2n︸︷︷︸

“2πβn”2

+ω2

=1

ω

1∞∏

n=1

(2π

βn

)2

︸︷︷︸= β

2π2π=β

∞∏

n=1

(1 +

ω2

ω2n

)

︸︷︷︸Euler

(5.10)

The part of the denominator embraced by Euler is a known Euler product

∞∏

n=1

(1 +

ω2

ω2n

)=

sinh βω2

βω2

.

which gives us

Z =1

2 sinh βω2

=e−βω/2

1− e−βω (5.11)

Of course it is silly to use such dirty techniques to obtain such well-known results.But in more advanced problems, we have to use this method to obtain any sensibleanswer.

From last lecture we remember that we expressed the partition function usingthe free energy F as Z = e−βF . Let us now use this to check the result (5.11). In

5.3. GENERALIZED ZETA-FUNCTION REGULARIZATION 49

the harmonic oscillator case F now yields

F = − 1

βlogZ

=1

2β

∑

n

log(ω2n + ω2

)

=1

2β

∞∑

n=−∞log

[(2π

βn

)2

+ ω2

]

∂F

∂ω=

1

2β

∞∑

−∞

2ω

ω2n + ω2

=1

2coth

βω

2

=1

2+

1

eβω − 1(5.12)

In the last equality we have used that

∞∑

−∞

1

n2 + a2=π

acoth a.

Integration yields

F =1

2ω +

1

βlog(

1− e−βω)

+ const. (5.13)

where this constant formally is infinite. This expression is in agreement with ourresult in (5.11).

A standard example where these techniques have to be used is for a gas ofmassless particles with a Lagrangian:

L =1

2(∂µφ)2 − λ

4φ4

which does not have any exact solutions in quantum mechanics or in classical fieldtheory, thus one has to apply perturbation theory. In standard quantum field theory(QFT) the temperature is set to zero. Finite temperature field theory will provideus with the same Feynman diagrams and integrals in QFT, but we will now workin a Euclidean space instead of a Minkowski space.

In QFT we are used to working with propagators of the form

〈0|Tφ(x)φ(x′)|0〉.

In the Euclidean theory the propagators will be the correlation function

〈φ(x)φ(x′)〉

where the φs are classical fields. Later we will calculate these propagators.

Chapter 6

Lecture 6

In this lecture we will continue our investigation of the partition function Z andthe free energy F for a free scalar field given by a free Lagrangian density L0 andthen finally turn the attention towards interacting scalar fields that are of physicallymore interest. To go from free theory to an interacting one, we assign interactingterms to the free Lagrangian. But before we do just that, let us take a short reminderof what we arrived at in the end of the last lecture.

When we considered a harmonic oscillator at finite temperature, then the par-tition function was given by

Z = Tre−βH =

∫Dqe−

12

R β0

dτq(−∂2τ+ω2)q

Here all the paths that we integrated over are periodic with the conditions q(τ +β) = q(τ). Then we are allowed to write q(τ) as a power expansion in τ leavingus with :

q(τ) =

√1

β

∞∑

n=−∞qne

iωnτ (6.1)

with ωn = 2πβ n, n = 0,±1,±2, . . . We then further derived the partition function

of the harmonic oscillator:

Z =∞∏

n=−∞

1√ω2n + ω2

which results in the free energy a’ la Helmholtz

F =1

2hω︸︷︷︸vacuum

+hω

eβhω − 1︸︷︷︸thermal

(6.2)

The first term corresponds to the vacuum energy, while the second term gives theenergy contribution due to thermal excitation. We found a more compact and ab-stract representation of the partition function that looked like

Z = det−1/2(−∂2τ + ω2)

51


and then of course

−βF = logZ =1

2log det(−∂2

τ + ω2) (6.3)

The point of considering the βF term again is that it can be rewritten into a slightlydifferent form that is more convenient in some calculations. To see how this isdone, let us take a look at a diagonal matrix

AD =

a1 0 . . .0 a2 . . ....

.... . .

We then have the well-known formulas for the determinant and the trace of thematrix

detAD =∏

i

ai

TrAD =∑

i

ai

Hence, the relation between trace and the determinant is log detA = Tr logAresulting in that (6.3) can be expressed as

βF =1

2Tr log(−∂2

τ + ω2) (6.4)

Now we turn our attention towards free scalar fields and take a closer look athow we can find expressions of physical quantities such as energy and pressure.

6.1 Free scalar fields

The Lagrangian density of a free scalar field with mass m is given by:

L =1

2(∂µφ)2 − 1

2m2φ2

and the partition function is as always

Z = Tre−βH

The scalar field is now considered not to be a function of time and space, butinstead a function of imaginary (Euclidean) time and space. This is the same shiftof variable as we have done earlier when we looked at the Feynman path integral

6.1. FREE SCALAR FIELDS 53

formalism in lecture 5. The formal shift from real to imaginary time will thereforebe:

φ(t,x)→ φ(τ,x)

The scalar field is also constrained with the boundary conditions as we have seenearlier, only now the field is periodic with respect to the Euclidean time τ. We canthen again expand the variable in powers of τ just as we saw in (6.1).Hence:

φ(τ,x) = φ(τ + β,x)

=

√1

β

∞∑

n=−∞φn(x)eiωnτ

=

√1

βV

∑

k,n

φnkei(k·x+ωnτ)

The main point is now that what we have earlier done for one harmonic oscillatorcan be applied to all the modes labeled with the index k. The partition function forthe free scalar field can therefore be expressed as:

Z =∏

k,n

1√ω2n + ω2

k

and the free energy can be seen as a sum over all modes of the (one-dimensional)harmonic oscillator in equation (6.2).

F =∑

k

[1

2hωk +

hωk

eβhωk − 1

]

Here each harmonic oscillator has frequency ωk =√

k2 +m2. So then the freeenergy can be written as

βF =1

2Tr log(−∂2 +m2)

The partition function can be rewritten as:

Z = Tre−βH =

∫Dφe−

R β0 dτ

Rd3xLE (6.5)

where LE is the Euclidean Lagrangian density which is found by letting t→ iτ asseen in lecture 6. In this shifting from real time to imaginary time, the Lagrangiandensity of the scalar field changes somehow. We remember that the Lagrangianwas given by

L =1

2(∂µφ)2 − 1

2m2φ2 =

1

2φ2 − 1

2(∇φ)2 − 1

2m2φ2


Now t → iτ implies that ∂φ∂t → −i

∂φ∂τ so then the Euclidean Lagrangian density

in this particular case is:

−L → LE =1

2

(∂φ

∂τ

)2

+1

2(∇φ)2 +

1

2m2φ2 =

1

2(∂µφ)2 +

1

2m2φ2

where, in the last step, we have used the shorthand notation ∂µ = (∂τ ,∇)⇒ ∂2 =∂2

∂τ2 +∇2. So once again we return to the partition function and find that (6.5) nowcan be expressed as

Z =

∫Dφe−

12

Rd4x[(∂µφ)2+m2φ2

]

=

∫Dφe−

12

Rd4xφ(−∂2+m2)φ

= det−12 (−∂2 +m2)

So far so good. We have been through some piece of derivation to finallyretrieve an expression for the partition function of the (massive) scalar field φ, allmotivated by extending our knowledge of the harmonic oscillator to account formore than just the one-particle description that a single harmonic oscillator gives.Let us therefore look at one particular example where we want to find the freeenergy of the radiation of a massless particle.

6.2 Radiation with massless particles and free energy

The free energy is related to the partition function in the following way

βF = − logZ =1

2

∑

k,n

log(ω2n + ω2

k)

with each frequency ωk given as previous in this lecture and is representing theenergy of each particle with wavenumber k.

We have in the previous lectures seen that in statistical mechanics the free en-ergy was found by summing over all modes in the expression of the one-dimensionalharmonic oscillator. Then by considering equation (6.2), the free energy will be

βF = β∑

k

[1

2ωk +

ωk

eβhωk − 1

]

But there is also another possible approach which is to first sum over k and thenover n. The expression for the free energy then turns into

βF =1

2V∑

n

∫d3k

(2π)3log(k2 + m2

︸︷︷︸= 0

+ω2n) (6.6)

6.2. RADIATION WITH MASSLESS PARTICLES AND FREE ENERGY 55

since the radiation is massless. It should be clear to the reader that this integralis divergent. So in order to receive any physically meaningful information aboutthe radiation system, the integral must be made finite in some way. The method ofchanging divergence into convergence is the well-known regularization. We preferdimensional regularization and, to keep the result as general as possible, we applyregularization in arbitrary dimensions (as we did in lecture 4). Hence:

∫ddk

(2π)d1

(k2 +m2)N=

1

(4π)d/2Γ(N − d/2)

Γ(N)(m2)d/2−N .

We choose N = 1. Furthermore:

∫dm2 :

∫ddk

(2π)dlog(k2 +m2) =

Γ(−d/2)

(4π)d/2(m2)d/2

The above equation is very useful. We also notice that it is correct dimensionally.In our case, with m2 = 0 and D = d+ 1 = 4:

βF =1

2V

∞∑

n=−∞

(− Γ(−3/2)

(4π)3/2|ωn|3

)

and finally we have an answer of what we were looking for. Here the frequencyωn is the Matsubara-frequency given by ωn = 2π

β n with the n′s taking on discretevalues n = 0,±1,±2, . . . The free energy density is now simple to find, just bydividing by the volume. If we write out the Γ′s it looks like

F =F

V= − 1

2β

43

√π√

4π4π

(2π

β

)3 ∞∑

n=−∞|n|3 (6.7)

The last sum can be rewritten as a sum of the form of the zeta-function and then beevaluated. This is seen by

∞∑

n=−∞|n|3 = 2

∞∑

n=1

|n|3 = 2ζ(−3) = 21

120

The pressure in the gas of massless particles is

P = −(∂F

∂V

)

T

= −F

⇒ P =π2

90(kBT )4

which we recognize as the ordinary Stefan-Boltzmann pressure.


6.2.1 Pressure in arbitrary dimensions

However, there is nothing stopping us from calculating the radiation pressure alsowith arbitrary dimension number. An interesting question to ask would be abouthow the expression of the pressure is given in, for instance, extra dimensions. Thespace time then has a dimension D that represents both the (one) temporal dimen-sion and all the spatial dimensions. (For extra dimensions D is of course greaterthan four). The pressure is, according to (6.7) :

PD = − 1

2β

∞∑

n=−∞

Γ(−d/2)

(4π)d/2

(2π

β

)d|n|d (6.8)

Again the sum over n can be evaluated by the zeta-function as ζ(−d). It is possibleto write the expression of the radiation pressure in d dimensions in a more compactform as

PD =Γ(D/2)

(π)D/2ζ(D)(kBT )D (6.9)

The important point to notice in the above expression, is that the Γ′s and ζ ′s in-volve the total dimension number D and not the spatial dimension number d. Theexplanation is that we want a more convenient formula and that ζ-functions withdifferent arguments D and − d can be related by the mathematical identity:

Γ(s/2)π−s/2ζ(s) = Γ(1− s

2

)π−( 1−s

2)ζ(1− s)

We observe that if s = D = d+ 1 is being substituted into the above formula, thenthe relation between the two pressures in (6.8) and (6.9) is clearly seen.

6.2.2 Example with zero temperature

We can also apply this formalism to do calculations with zero temperatures. If asimilar problem is considered, that is to find the free energy only now in Casimirform, then we will observe that the Casimir description is the same as that ofStefan-Boltzmann at zero temperature. The two situations are:

• Casimir: The Casimir energy (at zero temperature) can be found by consid-ering (6.6) and perform the substitutions (as β →∞)

∑

n

→ β

∫ ∞

−∞

dkD2π

∑

n

→ V

∫ddk

(2π)d

6.3. FERMIONS 57

• Stefan-Boltzmann If the temperature is zero, then the free energy equals theusual energy (F = E − TS) so that:

βF = βE =1

2Vd−1

∞∑

n=1

dd−1k

(2π)d−1β

∫dkD2π︸︷︷︸

βR

ddk

(2π)d

log(k2 +

(πLn)2

+ k2D

)

=1

2Vd−1β

∞∑

n=1

∫ddk

(2π)dlog

(k2 +

(πLn)2)

We can then find the energy density as

ρD =E

V= −Γ(D/2)

(4π)D/2ζ(D)

LD

which is very similar to the expression for the pressure in (6.8), only nowwith an opposite sign. This is due to

P = −(∂F

∂V

)

T

= −F → ρD

what we have studied here, is the bosonic case with scalar particles. Now, we takea look at the fermionic case.

6.3 Fermions

If we want to construct a Lagrangian formulation that describes fermions, then thefermion field must necessarily be a Grassmann variable. Some of the properties ofGrassmann numbers are listed in table 1. The spinor describing the fermion fieldmust be antisymmetric, that is:

ΨiΨj = −ΨjΨi

But before we take a deeper look at the properties of the fermions and try to findthe partition function, let us first remind ourselves of what we already know in thebosonic case. The partition function of the bosons is

Z =

∫Dφe−

12φTAφ

here A is a symmetric matrix with Aij = Aji and φ is a vector

φ =

φ1

φ2...φN


By means of the Gaussian integrals seen earlier, the partition function therefore canbe rewritten

Z =

∫Dφe−

12φTAφ = det−

12A

If φ is a complex field, it can be expressed as a linear combination of two real fields

as φ =√

12(φ1 + iφ2) 6= φ∗. The partition function will then take form as

Z =

∫Dφ

∫Dφ∗e−φ

∗Aφ

= det−1A

with the matrix in the exponent being hermitean A = A† and symmetric.Now back to the fermions. The fermion integrals of the partition function will

be of the same form as in the bosonic case:

Z =

∫DΨe−

12

ΨTDΨ (6.10)

Here Ψ is (still) a Grassmann variable, but now the matrix is antisymmetric, Dij =−Dji and the vector Ψ is given by

Ψ =

Ψ1

Ψ2...

ΨN

It is then possible to show that the partition function in the fermionic case can bewritten as

Z =

∫DΨe−

12

ΨTDΨ = det−12D

The Grassmann variable and its properties can give us some new and interestinginformation about integrating with respect to the fermion field Ψ (which we willhave to do in order to find the partition function).

If we assume that f is a function of one Grassmann variable Ψ, a Taylor ex-pansion gives:

f(Ψ) = f(0) + f ′(0)Ψ +1

2f ′′(0) Ψ2

︸︷︷︸= 0

+ . . .

We notice that the second order term, and all higher order terms in Ψ is zero. Thisis due to the properties of Grassmann variables and the only two interesting termsare a constant term f(0) and a term to first order in Ψ.We can then write down twobasis integrals where the integrands are a constant or Ψ, respectively.

I0 =

∫dΨ

I1 =

∫dΨΨ

6.3. FERMIONS 59

We will further assign some constraints to these integrals. The two integrals mustbe invariant under a change of variable Ψ→ Ψ′ = Ψ + α. The added term, α is aconstant Grassmann number, so then the differential is unchanged dΨ′ = dΨ. Wecan now write the integrals as

I0 =

∫dΨ′ =

∫dΨ

I1 =

∫dΨΨ′ =

∫dΨΨ

⇒ I1 =

∫dΨ(Ψ + α) =

∫dΨΨ +

∫dΨα

︸︷︷︸= 0

!= 0

So due to the last term in the equation above, it is clear that the first integral, I0 isequal to zero. Hence

I0 =

∫dΨ = 0

I1 =

∫dΨiΨj = δij

The properties of these two basic integrals will be of great importance when welater in this lecture will calculate the fermionic partition function (See the ex-ample). However, we will first consider a different change of variable given byΨ → Ψ′ = αΨ and still demand the integral to be invariant. We can then re-trieve some properties about the differential dΨ. (So far these properties have beenunknown, and we have only used it as an ordinary differential)

I1 =

∫dΨ′Ψ′ = 1

=

∫d(αΨ)αΨ = α

∫d(αΨ)Ψ

!=

∫dΨΨ

⇒ d(αΨ) =1

αdΨ

If we on the other hand consider bosonic variables, then the differential behaves as

d(αx) = αdx

So the differentials in the two cases of fermionic and bosonic variables are in somesense opposite in their behavior. Now, let us take a look at the fermionic determ-inant, and hence the partition function in the situation with N = 2 where theformalism we just derived will be useful.

6.3.1 Example with fermionic Z

The partition function for fermions with N = 2 is given by (according to equation(6.10) )

Z =

∫dΨ1

∫dΨ2e

− 12

ΨT1 DΨ2


The matrix D is antisymmetric, and given by

D =

(0 a−a 0

)

So then it is easy to calculate the exponent in the expression of the partitionfunction and find Z:

ΨTDΨ = 2aΨ1Ψ2

furthermore

Z =

∫dΨ1

∫dΨ2e

−aΨ1Ψ2 =

∫dΨ1

∫dΨ2[1− aΨ1Ψ2 +

1

2( )2 + . . .]

And here comes our little trick in. We remember from our derivation of the twobasic integrals that the only non-vanishing integral was of the form

∫dΨΨ. In the

above expression, we recognize the integral as a sum of I0, I1 and higher orderterms in Ψ, respectively. The only contribution to the partition function comesfrom the I1-like integral, and we write

Z =

∫dΨ1

∫dΨ2(−aΨ1Ψ2)

In order to calculate Z , we will have to perform one permutation between the(neighboring) fermion fields Ψ1 and Ψ2. From the Feynman rules in QFT, we knowthat each interchange of fermion fields correspond to a change in the sign:

Z =

∫dΨ1

∫dΨ2(+aΨ2Ψ1)

= a =√a2

= det12D

since detD = a2. If we instead have complex fields, the value of the partition

function changes. We write the complex field as Ψ =√

12(Ψ1 + iΨ2) 6= Ψ∗ and

then find the following partition function:

Z =

∫dΨ

∫dΨ∗e−Ψ∗DΨ = detD

We will take a briefer look at these functional determinants later in this course.

6.3.2 Example from QED

One of the leading physicists in taking advantage of the functional determinantswas Schwinger. He developed some of the theory when considering QuantumElectro Dynamics (QED). The Lagrangian density in QED is

L = Ψ(iγµ∂µ −m− eγµAµ)Ψ− 1

4F 2µν

6.4. INTERACTING FIELDS 61

We can again find the partition function (now with respect to real time t)

Z =

∫dΨ

∫dΨ

∫DAµ exp

i

∫d4x[Ψ(iγµ∂µ −m− eγµAµ)Ψ− 1

4F 2

µν ]

=

∫DAµ det(iγµ∂µ −m− eγµAµ)ei

Rd4x(− 1

4F 2µν)

The determinant in the expression above can be rewritten. We have earlier seen thatthere is a relation between determinants and traces given by elog det( ) = eTr log( )

so then the partition function is

Z =

∫DAµe

iRdx[− 1

4F 2µν+tr log(iγµ∂µ−m−eγµAµ)]

where now the effective Lagrangian of the fermion field is written as

−1

4F 2µν + tr log(iγµ∂µ −m− eγµAµ)

by means of the special trace tr. The two variants of trace, Tr and tr differ in thesense that

• Tr: does also involve a summation over the integral∫d4x

• tr: does not sum over∫d4x

6.4 Interacting fields

In this course we will mainly consider scalar fields. When interactions are included,the Lagrangian density of the field looks like

L =1

2(∂µφ)2 − U(φ) (6.11)

The first term describes the non-interacting, free part of the field and U(φ) givesthe interacting potential of order three or higher. When only considering the freepart, we have seen that the partition function can be analytically solved by meansof Gaussian functional integrals. If interactions are included, the expression of thepartition function is often so complicated that numerical methods must be used orperturbation theory applied in order to solve for it. Consider a potential, leading tointeractions, and given by

U(φ) =1

2m2φ2 + gφp (6.12)

Here g is a coupling constant which, as we will see later, is directly related towhether a theory is renormalizable or not. If p > 2 and we want to calculate thefunctional integral by

Z =

∫Dφe−

Rd4x[ 1

2(∂µφ)2+U ], (6.13)

this cannot be done analytically.


6.4.1 Dimensional analysis

It was stated in the previous section that the dimension of the coupling constant isof importance in understanding renormalization. To see this, there will be a needof a quick review of what is called dimensional analysis. If the constants h and care set to unity, then all quantities will be measured in length or mass. We will hereillustrate dimensional analysis with respect to mass-dimensions:

dim[m] = +1

dim[E] = +1

dim[pµ] = +1

(6.14)

with pµ = i ∂dxµ we get that

dim[xµ] = −1

The dimension of an exponent (or any other mathematical argument) must be zero.We can then find the dimension of the scalar field in arbitrary dimensions by con-sidering the dimension of the free Lagrangian density (which often ends up in theexponent) :

dim[dDx(∂µφ)2]!

= 0

⇒ −D + 2dim[∂µφ] = 0

⇒ dim[∂µφ] =D

2= dim[φ] + 1

⇒ dim[φ] =D − 2

2

So in our common space time with D = 4, we have that dim[φ] = +1. Thefield can also be a dimensionless quantity, then with D = 2. This is the case infor instance string theory. If we are to find the connection between the couplingconstant g and renormalization, we must also include the potential. The potentialforms a separate term in the exponent, and since dim[d4x] = −4 in four space-time dimensions, then the dimension of the potential must be four. With the samepotential as in (6.12) we get

dim[g] + p dim[φ] = 4

⇒ dim[g] = 4− p⇒ dim[g] = 0 , if p = 4

In a renormalizable theory, the coupling constants are dimensionless. Hence, theφ4 potential is somewhat special when D = 4.

D = 6 :

6.4. INTERACTING FIELDS 63

The dimension of the field, with 5 spatial dimensions, is dim[φ] = 2. In this casegφ3 gives a renormalizable theory. Therefore the Lagrangian density must be

L =1

2(∂µφ)2 − gφ3

Actually this theory is also asymptotically free, but that seems to be another story.

We now return to the expression for the partition function as seen in (6.13). Allthe interesting physics is contained in the partition function, and we rewrite it as

Z =

∫Dφe−

Rd4x[ 1

2(∂µφ)2+U ] =

∫Dφe−

Rd4x(L0+LI)

here the total Lagrangian density (as seen in (6.11)) is separated into a free part L0

and an interacting part LI containing the term ∼ φ4. Such an integral cannot besolved to find an exact solution. However, if we rewrite the partition function

Z =

∫ ∞

−∞dxe−

12x2−gx4

and if the coupling constant is small, then the partition function can be written asan expansion in x4. This yields:

Z =

∫ ∞

−∞dx exp

− 1

2x2

[1− gx4 +

(gx4)2

2!− . . .

]

So we see that the partition function written on this form is nothing but a seriesof different Gaussian integrals that each and one of them can be calculated separ-ately. The partition function can therefore be expressed as a series in the couplingconstant g.

Z = C0 + gC1 + g2C2 + . . .

This perturbative expansion of Z is only a good approximation as long as g is verysmall. In the following lectures this is the approach we will follow to find thepartition function. We will then see that the integrals in (6.13), giving Z , can betreated and interpreted by means of simple Feynman diagrams.

Chapter 7

Lecture 7

We will now end the part with field theory by constructing perturbation theoryby means of Wicks theorem. From earlier we remember that we could find theexpectation values with help of the density operator, ρ, given by

ρ = e−βH

and we saw that we could express the partition function as the trace of this operator,and that this could again be expressed as an integral

Z = Trρ =

∫Dφe−

R β0dτRd3xLE(φ,φ) = e−βF (7.1)

where LE is the Euclidean Lagrangian density, and F is the free energy. Now ifwe have an operator A = A(φ), we can use this to find its expectation value asfollows

〈A〉 =1

ZTrρA =

1

ZTrA(φ)e−βH =

1

Z

∫DφA(φ)e−

R β0dτRd3xLE (7.2)

where the last equality follows from the same procedure as we used to find thefunctional integral for Z .

7.1 Theory of free fields

For free fields the Lagrangian, LE , doesn’t have any interaction terms, so it turnsout simply

LE = L0 =1

2(∂µφ)2 +

1

2m2φ2

and from (7.1) we have that the corresponding free field partition function becomes

Z0 =

∫Dφe−

12

RdτRd3xφ(−∂2

τ+m2)φ

65


Now we want to express the the integral in the exponent as a matrix product, bywriting the variables as matrix/vector indices as φ(x) = φ(τ,x) = φτ,x. Theseindices are continuous, just as when we write ψx = 〈x|ψ〉 = ψ(x) in ordinaryquantum mechanics. Now the free field partition function can be written

Z0 =

∫Dφe−

12φ†Aφ

=

∫Dφe−

12φiAijφj

= det−12A = det−

12 (∂2

τ +m)

Where Dφ =∏i dφi.

We will now find some of the simplest expectation values, first in free fieldtheory. We find the expectation value for φ by using (7.2).

〈φ(x)〉 =1

Z0

∫Dφφ(x)e−

12

R β0dτRd3xφ(∂2

T+m2)φ

The exponential is an even, Gaussian function, while φ(x) is an odd function. Thisgives that the integrand is odd and the integral vanishes

〈φ(x)〉 = 0

〈φ(x)〉 is called a 1-point function. A 2-point function on the other hand is

〈φ(x)φ(x′)〉 =1

Z0

∫Dφφ(x)φ(x′)e−

12

R β0 dτ

Rd3xφ(∂2

T+m2)φ

and this is also called a correlation function, or a correlator for short. It can beinterpreted as a quantum mechanical propagator in imaginary time.

If we now make a substitution φ → φ′ = Rφ, (on component form φ′i =Rijφj) where R−1 = RT so that

∫Dφ =

∫Dφ′. This gives us the partition

function

Z0 =

∫Dφ′e−

12φ′iA′ijφ′j

where A′ = RAR−1 is diagonal. For non-free fields we will consider an additionalterm in the exponent

Z =

∫Dφe−

12φTAφ+BT φ

=

∫Dφe−

12φiAijφj+Biφi

where Bi is called a source function. This gives the expectation value for the fieldφi

〈φi〉 =1

Z

∫Dφφie

− 12φTAφ+BT φ

=1

Z

∂

∂BiZ =

∂

∂BilnZ

7.1. THEORY OF FREE FIELDS 67

Now we should be able to find the free field expectation value (the 1-point function)by letting B → 0 in the previous expression.

〈φi〉0 =1

Z0

∫Dφφie

− 12φTAφ =

1

Z0

∂

∂BiZ

∣∣∣∣B→0

(7.3)

and we can use the same trick on the correlator (the 2-point function)

〈φiφj〉0 =1

Z0

∫Dφφiφje

− 12φTAφ =

1

Z0

∂

∂Bi

∂

∂BjZ

∣∣∣∣B→0

(7.4)

If we again substitute φ → φ′ = Rφ, we get the non-free field partition func-tion

Z =

∫Dφ′e−

12φ′iA′ijφ′j+B

′jφ′j

where B′j = BR−1. If we write out the integral we get

Z =∏

i

∫ ∞

−∞

dφ′i√2πe−

12φ′iAiφ

′i+B

′iφ′i

= det−12Ae

12B′TA′−1B′

= det−12Ae

12B′ia−1i B′i

= det−12Ae

12BTA−1B

= Z0e12BiA

−1ij Bj

where ai are the diagonal components of A′ (remember that A′ is a diagonal mat-rix). Now, we have found the full partition function, and can then find the two-pointcorrelation function. If we go back to the correlation in equation (7.4) function withthis new expression for Z we get

〈φiφj〉0 =1

Z0

∫Dφφiφje

− 12φiAijφj =

1

Z0

∂

∂Bi

∂

∂BjZ

∣∣∣∣B→0

=1

Z0

∂

∂Bi

∂

∂BjZ0e

12BiA

−1ij Bj

∣∣∣∣B→0

= A−1ij

So we see that the expectation value of the two-point correlation function is theinverse of the matrix, Aij that is in the exponent. This is not a coincidence. If welook at the propagator ∆ for the scalar field, then

∆ ∼ 1

−∂2 +m2→ 1

p2 +m2(7.5)

The propagator should be compared with Aij ∼ ∂2 +m2 as seen earlier. The termin the exponent, when calcualating the partition function, is usually the Lagrangiandensity. The Lagrangian density of the scalar field is L0 = 1

2φ(−∂2τ + m2)φ and


should be compared with the propagator given above in (7.5). This is also illus-trated by considering the Dirac propagator, D and the corresponding Lagrangian.These are

D ∼ 1

iγµ∂µ −mwith L0 = Ψ(iγµ∂µ −m)Ψ

To be a little more specific, we look at the partition function given by

Z =

∫Dφk,ne

− 12φ∗k,n(k2+m2)φk,n

where

k2 =

(2π

βn

)2

+ k2

We have the correlation function between two modes in momentum space

〈φ∗k,nφk′,n′〉 =1

k2 +m2δkk′δnn′ =

1

k2 + ω2n +m2

δkk′δnn′

where ωn = 2πβ n is the Matsubara frequency. We can express the φ(x, τ) function

as a Fourier sum of modes.

φ(x, τ) =

√1

βV

∑

kn

φk,nei(k·x+iωnτ)

and the correlation function becomes

〈φ(x, τ)φ(x′, τ ′)〉 =1

βV

∑

k,n

∑

k′,n′〈φk,nφk′,n′〉ei(k·x+ωnτ)ei(k

′·x′+ω′nτ ′)

we have 〈φk,nφk′,n′〉 from above and we get

〈φ(x, τ)φ(x′, τ ′)〉 =1

βV

∑

k,n

1

k2 + ω2n +m2

ei(x−x′)·k+i(τ−τ ′)ωn

and if we let V →∞ we get

〈φ(x, τ)φ(x′, τ ′)〉 =1

β

∫d3k

(2π)3

1

k2 + ω2n +m2

eik·(x−x′)

and in the limit T → 0 we return to ordinary 0-temperature field theory

〈φ(x, τ)φ(x′, τ ′)〉 =

∫d4k

(2π)4

1

k2 +m2eik·(x−x

′)

7.2. WICK’S THEOREM 69

7.2 Wick’s theorem

We will now derive Wick’s theorem. This is really an old mathematical theorem,and not only used in quantum field theory.

We can write the correlation function between two field operators as a contrac-tion

〈φiφj〉 = φiφj = A−1ij = Kij

And also with several operators

〈φiφjφk〉 =1

Z0

∫Dφφiφjφke

− 12φiAijφj = 0

〈φiφjφkφl〉 =∂

∂Bi

∂

∂Bj

∂

∂Bk

∂

∂Ble

12BiA

−1ij Bj

∣∣∣∣B→0

=

(∂

∂Bi. . .

∂

∂Bl

)(1 +

1

2BTA−1B +

1

2!(1

2BTA−1B)2 + . . .)

)

= A−1ij A

−1kl +A−1

ik A−1jl +A−1

il A−1jk

= φiφjφkφl + φiφjφkφl + φiφjφkφl

=∑

(contractions of all fields)

The rule is that we can write a general correlator as the sum of all possible 2-pointcorrelators of the relevant field operators. From here we can move around theoperators, but if we are dealing with a fermion field we must remember that theyanticommute

φiφjφkφl = (−φiφk)φjφl

7.2.1 Example

Let us look at a Lagrangian density with an interaction term

LE =1

2(∂µφ)2 +

λ

4!φ4

First we find the interaction partition function

Z =

∫Dφe−

R β0dτRd3x[ 1

2(∂µφ)2+ λ

4!φ4]

=

∫Dφe−

Rd4x λ

4!φ4e−

12

Rd4xφ(−∂2+m2)φ (7.6)

and as we remember that we could write the expectation value of an free fieldoperator

〈A〉0 =1

Z0

∫DφA(φ)e−

Rd4xL0


we can use this on (7.6) to get

Z = Z0〈e−Rd4x λ

4!φ4〉0

= Z0

[1− λ

4!

∫d4x〈φ4(x)〉0 +

1

2!

(λ

4!

)2 ∫d4xd4y〈φ4

xφ4y〉+ . . .

]

Here we can use Wick’s theorem to find that

〈φ4(x)〉0 = 〈φxφxφxφx〉0= φxφx φxφx ·3= 3K(x− x)K(x− x) = 3K(0)

Now we shall draw some Feynman diagrams. For each contraction between twofield operators we will draw a line between two points in space.

〈φiφj〉0 = 〈φ(xi)φ(xj)〉0 = φ(xi)φ(xj) = xi × × xj

and for each integral in spacetime we will draw a fat dot

−λ∫d4x = •

This gives that our partition function is actually a series of Feynman diagrams

Z

Z0= 1 +

1

8©•©

+1

128©•©©•©

+1

16©•©•©

+1

48©()••

+ . . . =∣∣∣e−β(F−F0)

∣∣∣

and we get

−β∆F =1

8©•©

+1

16©•©•©

+1

48©()••

+ . . .

this is only disconnected diagrams. This expression is independent of dimension.We find the K function

K(x− x′) =1

β

∑

n

∫d3k

(2π)3

eik·(x−x′)

ω2n + k2

We have K(0) (loop curve)

K(0) =1

β

∫d3k

(2π)3

∞∑

n=−∞

1

k + ω2n

and through dimension regularization we get

K(0) = − 1

2πβ

∞∑

n=1

2π

βn =

1

12β2

7.3. WICK’S THEOREM OUTSIDE FIELD THEORY 71

and this gives

−β∆F =1

8

(1

128β2

)2

λ

∫ β

0dβ

∫d3x =

1

8

(1

128β2

)2

λβV

And this gives us the pressure

p =π2

90

(1− 5λ

64π2

)(kBT )4 =

π2

90(kBT )4 − ∆F

V

And this result is independent of dimension.

7.3 Wick’s theorem outside field theory

As Wick’s theorem is a general mathematical theorem it is also in use outside fieldtheory. In general, if we have an integral

I =

∫ b

a

dφ√2πe−A(φ)

To treat this we must use the sadelpoint approximation, where we find a globalminimum Ac = A(φc) of A(φ) and make an series expansion around this point

A(φ) = A(φc) +A′c(φ− φc) +1

2!A′′c (φ− φc)2 +

1

3!A′′′c (φ− φc)3 + . . .

where A′c = dAdφ

∣∣∣∣φ0

= 0. We can rewrite the integral as

I ≈∫ ∞

−∞

dφ√2πe−Ac−

12A′′cϕ

2− 13!A′′′c ϕ

3+...

and now we can think of K = 1A′′c

as the propagator, and −A(3)c and −A(4)

c as the3 and 4 line vertices respectively.

7.3.1 Example: Stirling’s formula for n!

We can write n! with help of the Γ-function

n! =

∫ ∞

0dφφne−φ

If we just rewrite this a little bit we have

n!√2π

=

∫ ∞

0

dφ√2πe−φ+n lnφ


That makes our A-function

A(φ) = φ− n lnφ

and its derivatives

A′(φ) = 1− n

φ⇒ φc = n

Ac = n− n lnn = − ln(ne

)n

A′′c =1

n⇒ K = n

A(3)c = − 2

n2

A(4)c =

4

n3

If we associate our partition function with n! as

Z =n!√2π

=

∫ ∞

−∞

dφ√2πe−Ac−

12A′′c φ

2+... = e−W

where −W is a series of Feynman diagrams

−W =

AC︷︸︸︷• +

log A′′C︷︸︸︷1

2©

︸︷︷︸1−loop

+1

8©•©


+1

16©•©•©


+1

48©()••


+1

8©• •©


+1

12©|••


+ . . .

So we get

n! =√

2πn(ne

)n [1 +

1

12n+ . . .

]≈√

2π(n+1

6)(ne

)n

and this is a very good approximation for n! for large n. Even for small n it isreally close. Examples are 0! ≈ 1.023 and 4! ≈ 23.99.

Chapter 8

Lecture 8

We will begin this lecture with finishing our study of finite temperature field theory.

8.1 Correlators with interactions at finite temperature

We remember from last lecture that when including an interaction term in our Lag-rangian density for a massive scalar field, we would get a correlator on the form

〈φ(x)φ(x′)〉 =1

Z

∫Dφφ(x)φ(x′)e

−R β0 dτ

Rdx3

LE︷︸︸︷[1

2φ(−∂2 +m2)φ︸︷︷︸

Z0

+λ

4!φ4

]

(8.1)This correlator must be solved either numerically or perturbatively. Using perturb-ation theory we separate the free part and the interaction part for Z in (8.1), andapplying Wick’s theorem we have

Z =

∫Dφe−

R β0 dτ

Rd3xLE

= Z0〈e−R β0

dτR

d3x λ4!φ4〉0

= Z0

[1 +

1

8©•©

+ . . .

](8.2)

The integral in (8.1) yields


RdτR

d3xLE =


RdτR

d3x 12

(∂2+m2)φe−R

dτR

d3x λ4!φ4

= Z0 〈φ(x)φ(x′)e−R β0

Rd3x λ

4!φ4〉0 (8.3)

73


where the 0 subscript means that this is the expectation value for the free Lag-rangian. Now, inserting the results from (8.2) and (8.3) into (8.1) yields

〈φ(x)φ(x′)〉 =〈φ(x)φ(x′)e−

R β0

Rd3x λ

4!φ4〉0

〈e−R β0

dτR

d3x λ4!φ4〉0

(8.4)

This exact expression for the correlator is known as the Gell-Mann-Low theorem.To the first order, and when applying Wick’s theorem, we have

〈φ(x)φ(x′)[1− λ

4!

∫d4yφ4(y)

]〉0 = 〈φ(x)φ(x′)〉0−

λ

4!

∫d4y φ(x)φ(x′)φ(y)φ(y)φ(y)φ(y) + . . .

where we are supposed to make all possible contractions. The contractions shownin the above expression do not lead to any connection between the points x and yand are thus represented by what is called disconnected diagrams. These contrac-tions may be illustrated by Feynman diagrams on the form

x× × x′ 1

8©•©y

But of course we will also have contractions on the form

φ(x)φ(x′)φ(y)φ(y)φ(y)φ(y)

which will correspond to connected diagrams on the form

x× •© × x′

Using the Gell-Mann-Low theorem (8.4) we now have

〈φ(x)φ(x′)〉 =× ×+× × 1

8©•©

+ 12 × •© ×

1 + 18©•©

=× ×

(1 + 1

8©•©)

+ 12 × •© ×

1 +1

8©•©

︸︷︷︸Z

= × ×+1

2× •© ×+O(λ2) (8.5)

In the last line the first term represents the free part and the second term is a firstorder mass correction or what is often called self-energy. This result, where thediagram(s) in the denominator cancels the disconnected diagrams in the numerator,is general. Therefore we may write:

〈φ(x)φ(x′)〉 = 〈φ(x)φ(x′)e−R β0 dτ

Rd3x λ

4!φ4〉connected

0

In the context of cosmology this is interesting because it may be possible to makea theory where the cosmological constant Λ can be described in this way.

8.2. PHASE TRANSITIONS 75

φC

V

φ

Figure 8.1: The Higgs potential as a function of φ. The minima are at φC .

8.2 Phase transitions

Phase transitions play an important role in the early universe, and understandingphase transitions is probably crucial to understand important fields like the baryonasymmetry. First we will study spontaneous symmetry breaking with a Higgs-potential and see how this will lead to a phase transition at a finite temperature.

8.2.1 The Higgs model

In the Higgs model the mass of the particles is given by the expectation value of ascalar field with the Lagrangian

L =1

2(∂µφ)2 +

1

2µ2φ2 − λ

4φ4

The first term is a kinetic term, while the two last terms represents a potential

V (φ) = −1

2µ2φ2 +

λ

4φ4.

See figure (8.1) 1. We find the minima of the potential:

dV

dφ= −µ2φ+ λφ3 = 0

φ = 0 or φ = ±φC where φC =

√µ2

λ= const.

And we define

VC ≡ V (φC) = −1

2µ2µ

2

λ+λ

4

(µ2

λ

)2

= −µ4

4λ

1If we had a negative sign in the second term of our Lagrangian, we would have particles withm2 < 0. These are mathematical particles called tachyons that sometimes occur in theories withcertain instabilities.


As is easily seen, the potential is symmetric under the change of sign for the φ-field.When a particle “choose” a position φ, the symmetry is broken. In this model theVC will represent a cosmological constant, and we notice that even classical fieldswill contribute to a Λ. Inserting typical values for µ and λwill give a Λ that is about50 orders of magnitude bigger than the observed one. So why don’t we observethis contribution to the cosmological constant? We don’t have a clue.

8.2.2 The partition function in the Higgs model

Now we will study the Higgs model in the early universe. We consider the partitionfunction

Z =

∫Dφe−

R β0 dτ

Rd3xLE(φ,∂µφ)

where

φ(x) =

√1

βV

∑

k,n

φknei(k·x+ωnτ)

= φ+ ϕ(x)

where the first term, φ, represents the k = n = 0-mode of the scalar field. Sinceall k = 0-modes are in the φ-part, we have that

∫d3xϕ = 0. Now we will follow

much the same procedure as we used when working with Stirling’s formula lastweek, and we expand Z around φ.

Z =

∫dφ∫Dφe−

R β0 dτ

Rd3xLE(φ+ϕ)

where

LE(φ+ ϕ) =1

2(∂µϕ)2 − 1

2µ2(φ+ ϕ)2 +

λ

4(φ+ ϕ)4

=1

2(∂µϕ)2 − 1

2µ2(φ2 + 2φϕ︸︷︷︸

=0

+ϕ2) +λ

4(φ4 + 4φ3ϕ︸︷︷︸

=0

+6φ2ϕ2 + 4φϕ3 + ϕ4)

where the underbraced terms will vanish since they are linear in ϕ and∫

d3xϕ = 0.Inserting this, our partition function now yields

Z =

∫dφe−

R β0 dτ

Rd3x(− 1

2µ2φ2+λ

4φ4)

∫Dϕ e

−R β0 dτ

Rd3x

[free part︷︸︸︷

1

2(∂µϕ)2 +

1

2m2ϕ2

+

interaction part︷︸︸︷λφϕ3 +

λ

4ϕ4

]

=

∫dφe−

R β0 dτ

Rd3xVeff(φ,T ) (8.6)

where m2 = −µ2 + 3λφ2. We see that the free part can be solved exactly sinceit depends on ϕ2. Veff is called an effective potential and its form determines whatkind of symmetries we have in our theory. Notice that Veff has the same form aswe have seen for the free energy in earlier lectures.

8.2. PHASE TRANSITIONS 77

8.2.3 Veff and an early-universe phase transition

The effective potential can now be expanded like

Veff(φ, T ) = V(0)

eff + V(1)

eff + V(2)

eff + . . .

where the first term is the classical term, the second a 1-loop correction, the third a2-loop-correction and so on. The first term looks like

V(0)

eff = −1

2µ2φ2 +

λ

4φ4

Here, we will only consider the 1-loop corrections. Using m2 = −µ2 + 3λφ2 andω =√

k2 + m2 we have:

V(1)

eff = −1

2Tr log(−∂2 +m2)

=

∫d3k

(2π)3T log(1− e− ωT )

=T 4

2π2

∫ ∞

0dx x2 log

(1− e−

√x2+y2

)

= −π2

90T 4 +

m2T 2

24− m3T

12π+ . . .

Inserting for m2 in the second term and neglecting the third term we now get

Veff(φ, T ) =1

2

(−µ2 +

λ

4T 2

)φ2 +

λ

4φ4 + . . .

We see that when T grows large enough, the effective potential will become posit-ive. Thus, considering the Higgs model, the spontaneous symmetry breaking willbe ehh. . . broken. Then there is now way for giving our particles their mass. So,if we believe in the Higgs model, the particles will have no mass in the very earlyuniverse, when the temperature is above some critical temperature TC . Veff(φ) isplotted for different temperatures in figure (8.2). Solving Veff = 0 we see that

TC =2µ√λ≈ O(100GeV) , kB = 1

In figure (8.3) you can see φC as a function of T. The inflation epoch is assumedto have happened at a temperature around 1015GeV, that is, a long time beforethe particles got their mass. As we don’t know much about the physics aboveT ∼ 100GeV , a lot of unknown fun can have happened in the cooling phase afterthe inflation. The transition at T ≈ 100GeV is called the electroweak transition. Itis probable that the baryon asymmetry was generated during this phase transition.


T=T_CT=0T>T_C

Legend

Veff T = 0

φ

T >> TCT = TC

Figure 8.2: The effective potential Veff(φ) for different temperatures

1

TT

φC

TC

Figure 8.3: φC(T ). It is easy to see how the φC suddenly breaks down when thetemperature reaches a critical value TC

8.3. PARTICLE CREATION IN AN EXTERNAL FIELD AT ZERO TEMPERATURE79

8.3 Particle creation in an external field at zero temperat-ure

Let us consider a good, old harmonic oscillator with the Lagrangian

L =1

2q2 − 1

2ω2q2

where ω is a time-dependent function ω = ω(t). Let us first consider a very simpletime dependence where

ω =

ωa for t < 0ωb for t > 0

The ground state of the harmonic oscillator is then given by

ψ0(x) =

ψa0 = |0in〉 = |0a〉 ∼ e−

12ω2ax

2for t < 0

ψb0 = |0out〉 = |0b〉 ∼ e−12ω2bx

2for t > 0

Particle production may occur in the transition t < 0 → t > 0. The amplitude forexciting the oscillator to a state n in this transition is given by

An =

∫dxψ∗a0(x)ψbn(x)

The amplitude for the transition between the two ground states (without any excit-ation at all) is called vacuum persistence amplitude, and is given by

Z = 〈0b|0a〉

That this magnitude is denoted Z just like the partition function is of course notdone for confusing, but rather for enlightening. This Z can be shown to be equi-valent to a partition function defined as a functional integral over real time:

Z =

∫Dφei

Rd4xL(φ,∂µφ)

This Z then denotes the amplitude for the vacuum to remain vacuum during thet < 0 → t > 0 transition.

This simple model with a step-function-like change of ω is just the simplestapproximation to a theory describing what we call reality. Other models have amore smooth transition where the values for ω are approaching ωa when t→ −∞and ωb as t → ∞. If this transition is sufficiently smooth, the oscillator willremain in its ground state, Z = 1, and no particles will be created. This is calledan adiabatic transition. Next lecture we will consider such smooth transitions usingcreation and annihilation operators and S-matrices.

Chapter 9

Lecture 9

In the last part of the previous lecture, we considered particle creation in externalfields. When we investigated the harmonic oscillator in such environments, wesaw that there were two possible frequencies ωa and ωb describing the oscillationdepending on whether the time twas smaller or greater than zero, respectively. Theparticles were created as the frequency of an harmonic oscillator changes when thetime changes from negative to positive. We will continue this investigation of theharmonic oscillator in external fields and see how we can extract quantum fieldsfrom their well known classical counterparts. The oscillator will also be equippedwith raising and lowering operators.

9.1 External fields

The fundamental quantity of any physical system is the Lagrangian density L.An example of a Lagrangian in an external field is for instance to consider anelectrically charged particle with mass m:

L =∣∣(∂µ − ieAµ(x)

)∣∣2 −m2φ∗(x)φ(x) (9.1)

Here, Aµ(x) is the electromagnetic field representing a classical background. Asomewhat simpler example is the Lagrangian of a field in a scalar background:

L = |∂µφ|2 −m2φ∗φ−A(x)φ∗φ (9.2)

or of a neutral scalar field in a non-trivial space-time metric gµν :

L =1

2gµν∂µφ∂νφ−

1

2m2φ2 (9.3)

Now, the metric tensor is a function of the space-time coordinate with gµν =gµν(x).What we want to take a further look at, is a Robertson-Walker metric (RW)

81


with a metric tensor given as:

gµν =

1 0 0 00 −a2 0 00 0 −a2 00 0 0 −a2

so that equation (9.3) now turns into

L =1

2φ2 − 1

2a2(∇φ)2 − 1

2m2φ2 (9.4)

As an easy start of external fields, and to get used to the formalism, let us thereforeconsider a toy model, that is, an ordinary harmonic oscillator.

9.2 Toy model: Harmonic oscillator

In its most general case, the Lagrangian of the harmonic oscillator is :

L =1

2q2 − 1

2ω2q2 (9.5)

The frequency is then generally a function of time ω = ω(t) and the system cannotbe expected to respect energy conservation. To proceed, we make the followingsimplification: The frequency is assumed to be a constant of the time, either ωa orωb in intervals where the time is negative or positive, respectively. This is illustratedin figure (9.1). In the areas where the frequency is constant, energy is conservedand the system can be described by ωa or ωb. The two time intervals are usuallyreferred to as in-physics (I) or out-physics (II).

This simplification is being used quite often to solve QM oscillator problems.

• In introductory courses in Quantum Mechanics (fys3110) the simplificationis even stronger as the frequency is assumed to be a step function with respectto time. This is shown in figure (9.2)

• Radiative decay. In a neutral atom there is an equal number of protons andelectrons. The electrons are orbiting the nuclei and a wave-function of theatom is then ΨA(r, Z). Here r is the radius of the atom and Z is the protonnumber. If the nucleus experiences a β-decay, then the total charge changesinstantaneously. Electrons orbiting the nuclei will possess the same wavefunction (from the nuclei), but the charge of the nuclei changes instantan-eously with one elementary charge. This behaviour corresponds to figure(9.2).

So in order to solve the oscillator problem described by the Lagrangian in (9.5) onemust first consider the classical motion:

(∂2t + ω2)q(t) = 0 (9.6)

9.2. TOY MODEL: HARMONIC OSCILLATOR 83

I II

ωa

ωb

ω

t

Figure 9.1: Simplification of the oscillator. The frequency is assumed to be aconstant, ωa for a long time interval (I), then changing during short time to anotherconstant ωb (II)

with general solutionsq(t) = C1f1(t) + C2f2(t) (9.7)

The coefficients are depending on the initial conditions that are assigned to thespecific problem. The classical solutions are more conveniently expressed in termsof complex functions as

q(t) = au(t) + a∗u∗(t). (9.8)

In the Heisenberg picture, the quantum mechanical description is represented byoperators, so the transition from classical to Q.M. description is

q(t)→ q(t)

The solution is then written in operator form

q(t) = au(t) + a†u∗(t) (9.9)

Here u(t) and u∗(t) are called mode-functions of the generalized coordinate. Again,we see that the question arises of how to define the initial conditions of the systemin order to find exact solutions. Generalized momenta is defined in the usual way,and in this particular case it is given by:

p(t) =d

dtq(t) = au(t) + a†u∗(t) (9.10)


ω

t

ωa

ωb

Figure 9.2: Further simplification of the oscillator. The frequency is assumed tobehave as a step-function changing instantaneously from ωa to ωb.

The canonical commutation relation is what actually quantizes the solutions and isthe usual [

q(t), p(t)]

= i (9.11)

We are free to insert the expressions of q and p into the commutator. The result is:[q(t), p(t)

]= [a, a†](uu∗ − uu∗) + [a, a]︸︷︷︸

=0

−like terms.

= [a, a†](uu∗ − uu∗) = i (9.12)

It will now be the need of introducing an inner product of the classical solution.This is done in order to get a unique description of the commutator [a, a†] in equa-tion (9.12). The inner-product is defined as

(u, u) ≡ i(u∗u− u∗u) (9.13)

Actually this inner product is a normalization product that is independent of thetime. This is seen by differentiating the product :

d

dt(u, u) = i(u∗u+ u∗u− u∗u− u∗u) (9.14)

From the Klein-Gordon-like equation in (9.6) the u is constrained by

u = −ω2(t)u

u∗ = −ω2(t)u∗

Hence, it is obvious that the inner product is time independent:

d

dt(u, u) = 0 (9.15)

The inner product is also a real quantity. This is seen by

(u, u)∗ = −i(uu∗ − uu∗) = +i(u∗u− u∗u) = (u, u) (9.16)

9.3. OSCILLATOR WITH CONSTANT FREQUENCY 85

So the inner product can be used as a norm. (We will in the coming lectures see thatthe inner product is describing bosons and that for bosons it is not always positive.For fermions on the other hand, it is always positive).

We are now going to make a choice. The solution, represented by u is chosento have a positive norm. That is:

(u, u) = +1 (9.17)

It is now possible to describe the commutator [a, a†] seen in (9.12) uniquely by

[q(t), p(t)] = i(u, u)[a, a†] = i

⇒ [a, a†] = 1 (9.18)

When this commutator identity is established, commutator algebra can be doneconstructing an infinite number of states. This is done by applying the numberoperator, N = a†a, and all the states are actually eigenstates of N . In order toconstruct all the states, we need a zero-state |0〉 of the Hilbert space where theinner product is defined. The zero-state is defined by:

a|0〉 = 0 (9.19)

So then a complete set of states of the Hilbert space can be established, and theseare generally given by:

|n〉 =(a†)n√n!|0〉 (9.20)

What we have done so far (and that differs from common oscillator physics) isvalid for an arbitrary frequency ω = ω(t). Then the state |0〉 is not necessarily theground state. We continue our investigation of the oscillator by first considering thesimplified case where the frequency is assumed constant (step-function like) andthen the more complicated situation where the frequaency ω is time-dependent.

9.3 Oscillator with constant frequency

As stated earlier, the energy is conserved when ω(t) = ω = const. We will nowfind the solutions of the (9.6) in terms of the mode functions u. The Hamiltonianoperator with constant frequency is

H =1

2p2 +

1

2ω2q2 (9.21)

And the position and momentum operators are as in (9.9) and (9.10):

q = au+ a†u∗

p =d

dtq = au+ a†u∗


Inserting into the expression of the Hamiltonian results in:

H =1

2(u2 + ω2u2)aa+

1

2(u2 + ω2u2)∗a†a† +

1

2(|u|2 + ω|u|2)(a†a+ aa†)

(This is a general result and is also valid when ω is time dependent). We furtherinvestigate when |0〉 is an eigenstate of the Hamiltonian operator. That is:

H|0〉 = 0 +1

2(u2 + ω2u2)∗a†a†|0〉 +

1

2(|u|2 + ω2|u|2)|0〉 (9.22)

Here the zero term follows from the definition of the zero-state with a|0〉 = 0. Ifthe zero-state |0〉 is supposed to be an eigenstate of H , then the following musthold:

H|0〉 = number× |0〉

So also the second term in (9.22) must vanish, resulting in

u2 + ω2u2 = 0 ⇒ u(t) ∼ e±iωt (9.23)

So there seem to be two solutions for the mode u(t). However, due to the constraintassigned to the inner product that it is positive with (u, u) = +1 there is only onepossible solution for u(t). It is found by combining the solution in (9.23) and thedefinition of the inner product:

(u, u) = i(u∗u− u∗u)!= 1

⇒ u(t) =

√1

2ωe−iωt (9.24)

So the entire solution to (9.6), which is the position operator in the Heisenbergpicture, can then be further written as

q(t) = au(t) + a†u∗(t)

=

√1

2ω(ae−iωt + a†eiωt) (9.25)

and finally the Hamiltonian operator regains the well-known form as

H =1

2ω(a†a+ aa†)

= ω(aa† +1

2) (9.26)

So in words we are back to normal with our common oscillator description whenwe have assumed the frequency to be time independent.

9.4. OSCILLATOR WITH TIME DEPENDENT FREQUENCY 87

9.4 Oscillator with time dependent frequency

Now the general situation with ω = ω(t) as seen in figure (9.1) is being considered.The frequency is no longer a constant with time, but evolves from ωa (early times)towards ωb (later times). The modes will in this case satisfy the classical solutionsto the equation

[∂2t + ω2(t)]u(t) = 0

with the general solution

q(t) = au(t) + a†u∗(t) (9.27)

Specifying the initial conditions when the oscillator frequency is a varying quantitywith time demands a little precaution. The modes u(t) will depend on whether theinitial conditions are defined at early times, that is, with ω = ωa, or at late timeswhen the frequency is ω = ωb. These two occasions are both equally correct, andit is common to distinguish between in-modes and out-modes.

• In-modes: The in-modes are uniquely defined at early times, that is at t =−∞ as

u(t→ −∞) =

√1

2ωae−iωat

We already know that the inner product with respect to the modes u isconstant with time. This is seen from the solution of the mode u(t) by(u, u) = i(u∗u − u∗u) = 1. It also follows from the definition of thein-modes that the commutator identity is preserved [a, a†] = 1 as seen in(9.18). So when time is taken to large, positive values, the in-mode turnsinto a linear combination of exponentials

u(t)

∣∣∣∣t→∞

∼ ( )e−iωbt + ( )e+iωbt

Here, the parenthesis correspond to the so called Bogoliubov coefficients.These coefficients will later be shown to be the connection between the in-modes and the out-modes.

• Out-modes: The out-modes are defined to have a unique behavior at latetimes, t→ +∞, in the same way as the in-modes at early times. The modes,now denoted by v(t) are given by:

v(t→ +∞) =

√1

2ωbe−iωbt

However, the modes are valid also when time is very small. Then the out-mode looks like:

v(t)

∣∣∣∣t→−∞

= ( )e−iωat + ( )e+iωat (9.28)


So to summarize, we have that

The in-modes are uniquely defined by their behavior at early times, t→ −∞, and theout-modes show a unique behavior at times approaching positive infinite, t→ +∞.

We have earlier in this lecture seen that the coordinate solution, q(t), of theclassical Klein-Gordon equation (9.6) can be expressed in terms of the in-modesu(t). This was shown in equation (9.25 ). As discussed when defining the in- andout modes, they both serve satisfactory solutions, only depending on the choice ofinitial conditions. Therefore there is nothing stopping us from expressing q(t) interms of also the out-modes v(t) and define the inner-product for them analogouslyto what was done for the in-modes earlier. The solution then looks like:

q(t) = bv(t) + b†v∗(t) (9.29)

with the inner product with respect to v as

(v, v) = +1 (9.30)

One should take notice that the coefficients in front of the modes now are differentthan in the in-mode situation in (9.9). The new (raising and lowering) operatorsare easily shown to satisfy the commutator identity [b, b†] = 1. (exercise!). Hence,there seems to be more than just one description of the solution q(t). Actually, theHilbert space containing the quantum mechanical states of the oscillator, is equallydescribed by the in-operators (a and a†) or the out-operators (b and b†). We willin our continuation establish a connection between the two sets of operators of theoscillator. There exist two vacuum states denoted by |0a〉 and |0b〉 and defined inthe usual way as:

a|0a〉 = b|0b〉 = 0

They represent the two different situations of the initial conditions

• |0a〉: The ground state of the oscillator at t→ −∞.

• |0b〉: The ground state of the oscillator at t→ +∞.

From the two ground states new and excited states can be constructed by applyingthe two different sets of raising and lowering operators. From the relation in (9.29)between the field-operator and the second set of raising and lowering operators, thefollowing holds:

b = (v, q)

this is so, since (v, v) = +1 and (v, v∗) = i(v∗v∗ − v∗v∗) = 0. A more compactformalism of the inner product will be further used. This is in general written as:

(f, g) = i(f ∗g − f ∗ g)

= if∗(

→∂

∂t−←∂

∂t)g

= if∗↔∂t g

9.4. OSCILLATOR WITH TIME DEPENDENT FREQUENCY 89

Here we have used a notation of the form←∂∂t where the arrow indicates the direction

of action of the derivative. The operator b can now, if we insert for the field operatorq(t) in (9.27), be expressed as:

b = (v, q) = a(v, u) + a†(v, u∗) (9.31)

We will now seek to find relations between the two sets of operators a and b. Todo so, there will be the need of introducing coefficients describing inner productsbetween the in-modes and the out modes u and v given by:

A = (u, v)⇒ A∗ = (v, u)

B = (u, v∗)⇒ B∗ = (v∗, u)

So then the b-operator in (9.31) is written

b = A∗a−Ba† (9.32)

Hence,A andB are both Bogoliubov coefficients. We can derive some of the prop-erties of these coefficients if we consider the commutator between the b-operators.Then we will need the adjoint operator:

b† = Aa† −B∗a (9.33)

and the commutator is:

[b, b†] = AA∗ [a, a†]︸︷︷︸=1

+BB∗ [a†, a]︸︷︷︸=−1

= 1 (9.34)

So the Bogoliubov coefficients satisfy

|A|2 − |B|2 = 1

This relation holds only for bosons. Formally the same procedure can be runthrough for fermions, only now in anticommutator form. Due to the formal shift

[ , ]Bose−Einstein → , ︸︷︷︸Fermi−Dirac

,

the same relation between the Bogoliubov coefficients holds only now by changingsign

|A|2 + |B|2 = 1

The relations between the two sets of operators are

b = b = A∗a−Ba† (9.35)

a = Ab+Bb† (9.36)


Combining with the Bogoliubov coefficients, the relations lead to an expression ofthe out-mode in terms of the in-mode

v(t) = Au(t) +B∗u∗(t) (9.37)

So when the time satisfies t → −∞, that is, in the in-area, then the out-mode(defined at t→ +∞) evolves as

v(t) = Au(t) +B∗u∗(t)→ Ae−iωat +B∗eiωat (9.38)

which is the same answer that we found in equation (9.28), only now with theBogoliubov coefficients properly established. So now the out-modes are expressedin terms of the in-modes u(t).

9.4.1 The relation between the vacuum states |0a〉 and |0b〉One situation that we will meet quite often later in this course, is the one wherewe prepare the system in the vacuum state by choosing the initial conditions atearly time, t → −∞. This state, which is the easiest state, will therefore be thein-vacuum |0a〉 defined by

a|0a〉 = 0 (9.39)

The system is viewed in the Heisenberg picture where all states are time independ-ent. This means that if we start out with the in- vacuum state |0a〉, then the systemwill always be in this state. However, as time goes by and t → +∞, then theoperators to be used now, b and b† are not necessarily annihilation and creationoperators of this state any longer. So the state that was vacuum state at early timesis no longer such a state as t→ +∞:

b|0a〉 6= 0 (9.40)

Since the state |0a〉 is no longer a ground state, there is (another) physical inter-pretation of this state as an excited state. As the state is being excited somewherein the transition t = −∞ → t =∞, quanta are created. The number of quanta isrelated to the Bogoliubov coefficients in the following way

n = 〈0a|b†b|0a〉= 〈0a|(Aa† −B∗a)(A∗a−Ba†)|0a〉= |B|2〈0a|aa∗|0a〉 = |B|2 (9.41)

where theB is a Bogoliubov coefficient. Now it is time to find the relation betweenthe states |0a〉 and |0b〉. We start out with the definition of the “early” vacuum stateand insert for the operator a and find:

a|0a〉 = (Ab+Bb†)|0a〉 = 0

⇒ b|0a〉 = −BAb†|0a〉 (9.42)

9.5. QUANTUM FIELDS 91

We will now perform a hand waving derivation of the relation between the twovacuum states |0a〉 and |0b〉motivated by the coordinate representation of the Q.M.operators q and p. In the commutator [q, p] = i in the coordinate representation,the momentum operator is given by p = −i ∂∂q . In the similar way, the b-operatoris stated to be

b =∂

∂b†

So inserted in the equation (9.42), we find

b|0a〉 =∂

∂b†|0a〉 = −B

Ab†|0a〉

⇒ |0a〉 = Ce−B2Ab†b† |0b〉

The relation between the two vacuum states is commonly written in terms of anS-operator as:

|0a〉 = S|0b〉

We end our investigation of the oscillator by considering some of the propertiesof the S-operator:

• The S-operator is unitary S†S = SS† = 1 and the a and b operators areconnected by a = SbS†.

• The matrix elements of the S-matrix are defined by Smn = 〈mb||na〉. Thezero-element of the matrix was seen in the end of the last lecture when theadiabatic transition was considered, and is defined by

S00 = 〈0b||0a〉 = Z = 〈0a|S|0a〉. (9.43)

This matrix element is the amplitude that if the system starts out in theground state, it remains a ground state as t approaches infinity. If Z = 1there is an adiabatic transition called vacuum-persistence.

9.5 Quantum fields

We will now consider a classical field with a classical equation of motion and itssolutions. The point in viewing the problem (the equations) classically, is that weknow the solutions, and when we have these solutions, then for each of them thereexists an operator that either creates or annihilates a particle. And eventually wehave reached our goal which is a quantum field description (with a well-definedquantum field operator). So let us therefore first take a look at the (classical) Lag-rangian density of a free field:

L =1

2ηµν∂µφ∂νφ−

1

2m2φ2 (9.44)


The classical equation of motion is

(∂2 +m2)u(x) = 0 (9.45)

with the solutions for each wavenumber of the mode function as

uk(x) =

√1

2ωkVe−ik·x =

√1

2ωkVei(k·x−ωkt), (9.46)

where the frequency is the usual ωk =√

k2 +m2 and V is the volume of the boxwhere the system is expected to show periodic boundary conditions. The modefunctions uk form a complete set and therefore a field-operator can be established.It is given by

φ(x) =∑

k

(akuk + a†ku∗k). (9.47)

By means of the canonical commutator identity [φ, π] = i, the commutator relationbetween the creation and annihilation operators is found to be

[ak, a†k′ ] = δkk′ (9.48)

In the deduction above, the system is expected to show periodic boundary condi-tions.

The classical modes are further defined in terms of an inner product

(uk, uk′) = i

∫d3xu∗k(

←∂t −

→∂t)uk′

= i

∫d3xu∗k

↔∂t uk′ (9.49)

To calculate this inner product, we will need to compute the spatial integral of twomode functions. This integral is given by the relation

∫d3xu∗kuk′ =

1

2ωkδkk′

The inner product is then(uk, uk′) = δkk′ (9.50)

This is so since the derivative↔∂t brings down one ωk’s from the exponent of the

mode functions seen in (9.46). Similarly, the complex conjugated inner product is

(u∗k, u∗k′) = −δkk′ (9.51)

The structure and shape of the quantum field operator as we have found in thissection is valid for all fields. This is so since we started out with the classical viewof the problem and found its solution, and at the same time we know that theseclassical solutions have a well-defined and consistent inner product.

9.5. QUANTUM FIELDS 93

9.5.1 Lorentz invariant norm

One other interesting property of the norm is that it is invariant under a Lorentztransformation. This is seen by

(uk, uk′) = i

∫

dΩdΣµ(u∗k

↔∂µ uk′) (9.52)

Here, Σµ is a vector perpendicular to the 3-dimensional hypersurface x. This isshown in figure (9.3).

x

t

dΣµ = d3x

Figure 9.3: Minkowski diagram showing the volume element before the Lorentztransformation.

The Lorentz invariance is seen if we perform the same calculation in anotherspace-time x′. This is illustrated in figure (9.4) where the spatial surface is tilteddue to the Lorentz transformation. The integration volume perpendicular to x ′ isdifferent from that of x. After the transformation, the inner product is

(uk, uk′) = i

∫

dΩdΣ′µ(u∗k

↔∂µ uk′) + i

∫

Ωd4x∂µ(u∗k

↔∂µ uk′) (9.53)

In order for the inner product to be Lorentz invariant, the last term has to be zero.This is shown by

∂µ(u∗k↔∂µ uk′) = (∂2u∗k)︸︷︷︸

=m2u∗k

uk′ − u∗k ∂2uk′︸︷︷︸=−m2uk

= 0 (9.54)

Hence, the two expressions of the inner product are equal and the inner product isLorentz invariant as stated:

∫

dΩdΣµ(u∗k

↔∂µ uk′) =

∫

dΩdΣ′µ(u∗k

↔∂µ uk′) (9.55)


t

x′

x

Ω

dΣ′µ

Figure 9.4: Minkowski diagram showing the space time after the Lorentz trans-formation. The integration is now performed not only perpendicular to the spatialaxis, x′, but also over the area Ω.

Chapter 10

Lecture 10

10.1 Effective action

We looked at an ordinary free field where an external field is turned on, so that ωvaries as by figure (9.1). The action is defined by

S =

∫d4xL(φ, ∂µφ) (10.1)

We have the Lagrangian density

L =1

2(∂µφ)2 − 1

2m2φ2 − 1

2Aφ2 (10.2)

where A is an external field. We find the classical modes by solving the Euler-Lagrange equation

∂L∂φ− ∂

∂xµ∂L∂φ,µ

= 0 (10.3)

From this we get two solutions. We get the in-solutions uk(x) which gives the fieldoperator

φ =∑

k

[akuk(x) + a†u∗k(x)

]

and the out-solutions (funny word in Norwegian) vk(x) which gives

φ =∑

k

[bkvk(x) + b†v∗k(x)

]

and we define the in-vacuum |0a〉 by ak|0a〉 = 0 and the out-vacuum |0b〉 bybk|0b〉 = 0. We have the survival amplitude Z , which is the amplitude that the fieldstays in the vacuum state forever.

Z = 〈0a|0b〉 = Z[A] = eiW [A] (10.4)

where W [A] is the effective action. The probability for the field to stay in thevacuum state is then |Z|2 = e−2ImW [A].

95


We remember from the finite temperature field theory that we used a similarnotation for the partition function

Z[A] =

∫Dφe−

R β0 dτ

Rd3xLE(φ,∂µφ) = e−βF [A] (10.5)

Where F is Helmholtz free energy. This is no coincidence. We remember fromrelativistic quantum field theory that we could write the propagator as

Z = 〈0|Te−iR

d4xHint(φ)|0〉 (10.6)

where Hint is the Hamiltonian density in the interaction picture. Based on thatL = L0 + Lint and that H = H0 +Hint it can be shown that this is equal to

Z =

∫Dφei

Rd4xL(φ,∂µφ) = 〈ei

Rd4xLint(φ,∂µφ)〉0 (10.7)

where we only considered the connected Feynman diagrams. We recognize herethe partition function.

From thermodynamics we remember that the free energy is given as F = E −TS, and when T → 0 we can associate F with vacuum energy.

Here we prefer the expression with the Lagrangian density. Very often we havethat Hint = −Lint.

10.1.1 Example withHint = −LintFor example if we have the Lagrangian density

L = L0 + Lint =1

2(∂µφ)2 − λ

4φ4

which gives that π = ∂L∂φ

= φ. From this we find the Hamiltonian density

H = H0 +Hint= πφ−L = φ2 − 1

2φ2 − 1

2(∇φ)2 +

λ

4φ4

=1

2π2 +

1

2(∇φ)2 +

λ

4φ4 (10.8)

We see that Hint = −Lint.

10.1.2 Example withHint 6= −LintBut this is not always the case. If we for example have the Lagrangian density

L =1

2(∂µφ)2 − g

4φ2(∂µφ)2 (10.9)

10.2. QUANTIZATION OF EXTERNAL FIELDS 97

where the last term is the interaction Lagrangian density. We find the generalizedmomentum (or whatever it is called in field theory) to be

π =∂L∂φ

= φ− g

4φ2φ (10.10)

and if we use this to find the Hamiltonian density H = H0 +Hint = πφ − L wesee that for this case

Hint 6= −Lint. (10.11)

From this follows that the expression 〈eiR

d4xHint〉0 is not Lorentz invariant (andperhaps not gauge invariant), but 〈e−i

Rd4xLint〉0 is correct. This gives a hint that

the path integral formalism is the most fundamental. This was a long digression.Now back to the quantization of external fields.

10.2 Quantization of external fields

We define orthonormality by the inner product

(uk, uk′) = (vk, vk′) = δkk′ (10.12)

and(uk, u

∗k′) = 0

and this gives the commutation relations

[ak, a†k′ ] = [bk, b

†k′ ] = δkk′

and we can express ak by

ak = (uk, φ) =∑

k′

(Akk′ bk′ +Bkk′ b

†k′

)(10.13)

WhereAkk′ = (uk, vk) andBkk′ = (uk, v∗k) are the Bogoliubov coefficients. From

[ak, a†k′ ] = δkk′ it follows that

∑

l

= (AklA∗k′l −BklB∗k′l) = δkk′ (10.14)

Usually the Bogoliubov coefficients are diagonal

Akk′ = Akδkk′

Bkk′ = Bkδkk′

We have defined vacuum by ak|0a〉 = 0 and as ak = Ak bk +Bk b†k we can find

the average number of particles, n, produced

nk = 〈0a|b†k bk|0a〉 = |Bk|2 (10.15)


And we can find the amplitude to produce a pair of particles. In that case the in-state is |0a〉 and the out-state is b†k b

†k|0b〉. As we know that bk|0a〉 = −Bk

Akb†k|0a〉

and that the vacuum condition gives 〈0b|b†k = 0, we get the amplitude

〈0b|bk bk|0a〉 = −BkAk〈0b|bk b†k|0a〉

= −BkAk〈0b|0a〉

= −BkAk

Z (10.16)

Thus, the probability amplitude for producing one pair of particles is given by thefraction −Bk

Akmultiplied with the probability that vacuum remains vacuum.

Unitarity gives that 〈0a|0a〉 = 1 and the completeness relation allowes us toexpand this in two particle states

∑∞n=0 |2nb〉〈2nb| = 1. And these together with

our newfound amplitude gives us

〈0a|0a〉 = |Z|2 + |Z|2∣∣∣∣B

A

∣∣∣∣2

+ |Z|2∣∣∣∣B

A

∣∣∣∣4

+ . . . = 1 (10.17)

We factorize out |Z|2 and find the sum

|Z|2(

1 +

∣∣∣∣B

A

∣∣∣∣2

+ . . .

)= |Z|2 1

1−∣∣BA

∣∣2 = |Z|2 |A|2|A|2 − |B|2 = |Z|2|A|2 = 1

where we have used the relation |A|2 − |B|2 = 1 (valid for bosons only!). Thisresult gives us an expression for |Z|

|Z| = 1

|A| (10.18)

Just as a check, if we now return to the average number of particles produced, thisshould be

n = |Z|2∞∑

n=1

n

∣∣∣∣B

A

∣∣∣∣2n

= |Z|2∣∣BA

∣∣2(

1−∣∣BA

∣∣2)2 =

1

|A|2

∣∣BA

∣∣2(

1−∣∣BA

∣∣2)2 = |B|2

(10.19)The same as we got above in equation (10.15).

For fermions, things are a little different. If we do the same operations, butwith commutator relations replaced with anticommutator relations, we will get that|A|2 + |B|2 = 1 (the point here is the plus sign). This leads to another amplitude|Z| = |A|, but the average number of produced particles stays the same n = |B|2as in the boson case. What we have shown in this section is the standard formalismon how particles are produced in an external field.

10.3. PARTICLE PRODUCTION IN CONSTANT ELECTRIC FIELD 99

m

−m

Figure 10.1: The ordinary Dirac sea

10.3 Particle production in constant electric field

As before we will consider a field with varying ω, but now it is caused by an electricfield being turned on, left constant for a while (infinitely long actually), and thenturned off again.

If we look at the old Dirac-sea where we have a large potential gap between thepositive and negative area. The production of a particle would be accompanied withproduction of corresponding particle with equal but negative energy (figure (10.1)).We can imagine that if we introduce an electric field we will tilt this potential gap.In this case we could imagine that a particle tunnels through the gap, with the resultthat two particles are produced, both with positive energy (figure (10.2)). That is,we would have a net production of particles.

We now want to find the tunneling amplitude. The potential of the electric fieldis given by

V (z) = eEz (10.20)

The transverse momentum, kT = (kx, ky), will here be conserved. The same istrue for the energy (here denoted ω) except for the energy from the electric field.That gives us the energy equation

(ω − eEz)2 = k2T + k2

z +m2 (10.21)

We will find the position where the particles enter and leave the forbidden region.


Figure 10.2: “Tilted” Dirac sea, due to a electric field

It is defined by kz = 0, turning (10.21) into

(ω − eEz)2 = k2T +m2 = m2

T (10.22)

where m2T = k2

T +m2. This gives two solutions

ω − eEz1 = −mT

ω − eEz2 = +mT

Now we get the tunnelling current

n = e−2R z2z1

dzkz = e−2R z2z1

dz

r1−“ω−eEzmT

”2

(10.23)

substituting xmT = ω − eEz yields

n = e−2m2

TeE

R 1−1 dx

√1−x2

= e−πm2

TeE (10.24)

To see what kind of fields we are talking about here we find an approximate fieldstrength when producing particles. In order to achieve such production, the expo-nent in (10.24) must be smaller than unity

E =πm2

e∼ 1016 V

m(10.25)

and this is a terribly strong field.

10.3. PARTICLE PRODUCTION IN CONSTANT ELECTRIC FIELD 101

As we earlier found that n = |B|2 we get for bosons that

|Zk|2 = e−2ImWk =1

|Ak|2=

1

1 + |Bk|2=

1

1 + nk(10.26)

and the total partition function

|Z|2 = e−2ImW =∏

k

|Zk|2 = e−2Pk ImWk =

∏

k

(1 + nk)−1 (10.27)

and this gives

2ImW =∑

k

ln(1 + nk) =∑

k

ln

(1 + e−

πm2T

eE

)(10.28)

and as we know, when the volume grows large we can write the sum as an integral

∑

k

= V

∫d3k

(2π)3=

∫d3x

∫d3k

(2π)3

So we write

2ImW [A] = 2

∫d4xLeff (A)

=∑

k

ln(1 + nk)

=

∫dωdt

2π

∫d2kT(2π)2

∫d2xT ln (1 + n(ω,kT ))

=eE

(2π)3

∫dtdzdxdy

∫d2kT ln (1 + n(ω, kT ))

=eE

2π

∫d4x

(2π)2

∫d2kT ln (1 + n(ω, kT )) (10.29)

where we have made the substitution dω = eEdz. Now we use our previous result

that n = e−πm2

TeE , and thus we get

2ImLeff =eE

2π

∫d2kT(2π)2

ln

(1 + e−

πm2T

eE

)(10.30)

A series expansion of the logarithm function yields

ln(1 + x) =

∞∑

n=1

(−1)n+1

nxn

and we can use that m2T = k2

T +m2 and then find the integral∫ ∞

0πdk2

T e−πneEk2T =

eE

n(10.31)


So our expression turns out

2ImLeff =1

8π

(eE

π

)2 ∞∑

n=1

(−1)n

n2e−

nπm2

eE (10.32)

we know that the free particle Lagrangian density yields

L =1

2E2 − i

16π

(eE

π

)2

e−πm2

eE + . . . (10.33)

All the particle production so far has been with bosons. We now end this lectureby investigating the fermion case as in QED.

10.3.1 QED

From QED we know that

L = ψ(iγµ∂µ −m− eγµA)ψ − 1

4F 2µν (10.34)

and

Z[A] = eiW [A] =

∫Dψ

∫Dψei

Rd4xL = ei

Rd4x[− 1

4F 2µν+Tr log(iγµ∂µ−m−eγµ∂µA)]

(10.35)Schwinger looked at the case where Fµν = const and calculated the exact expres-sion for Leff

Leff =

∫ ∞

0

ds

8π2s

[e2ab

cosh(eas) cos(ebs)

sinh(eas) sin(ebs)− 1

s2

]e−ism

2(10.36)

where

a2 − b2 = E2 −B2 =1

2FµνF

µν

and

ab = E ·B =1

4FµνF

µν

If we expand Leff in a and b we get

Leff = −1

4F 2µν +

2α2

45m4

[(E2 −B2)2 + 7(E ·B)2

]+ . . . (10.37)

The second term was first found by Euler and Heisenberg in 1933, but Schwingercould expand this to any order.

Chapter 11

Lecture 11

This lecture we are going to continue the study of classical fields in non-trivialbackgrounds. We are going to focus on the use of the variation principle, andapply this principle on the metric. We will end up studying the Einstein-Hilbertaction and the energy-momentum tensor.

11.1 Massless scalar field in a general space-time

We assume a scalar field φ = φ(x) in a space-time with a metric

gµν ⇒ ds2 = gµν︸︷︷︸=gµν(x)

dxµdxν .

Earlier we have used the principle of extremal action on a scalar field living in aMinkowski metric, ηµν , using the action

S =

∫d4xL =

∫d4x

[1

2ηµν∂µφ∂νφ− V (φ)

](11.1)

The transition to a general metric is as simple as we could hope for, using what iscalled “minimal coupling” we just substitute the Minkowski metric with a generalmetric and obtain

S =

∫d4x√−g

[1

2gµν∂µφ∂νφ− V (φ)

](11.2)

whereg = det(gµν).

11.1.1 Example with particle creation in accelerated systems

Here we are going to see how particle creation occurs for an observer acceleratedrelative to a particle-less vacuum. Assume a system Σ in 1 + 1 dimensions with

103


(x, t)

x′

t′

Figure 11.1: An accelerated system Σ shown in an inertial system Σ′

local coordinates (x, t) moving with constant rest acceleration a relative to an iner-tial system Σ′ with coordinates (x′, t′) see figure 11.2. The primed coordinates aregiven in terms of the un-primed coordinates using what is called Rindler coordin-ates1:

x′ =1

aeaxcosh(at) (11.3)

t′ =1

aeaxsinh(at) (11.4)

Now, assume a massless scalar field φ′(x′, t′) in Σ′. The corresponding field in Σis denoted φ(x, t). Σ′ has an action defined by

S =

∫d2x′

[1

2

(∂φ′

∂t′

)2

− 1

2

(∂φ′

∂x′

)2]

(11.5)

Using (11.3) we see that our line element is given by

ds2 = dt′2 − dx′2 = e2ax(dt2 − dx2) (11.6)

where e2ax is called a conformal factor. So, in our accelerated system the metricbecomes

gµν =

(e2ax 0

0 −e2ax

)⇒ gµν =

(e−2ax 0

0 e−2ax

)

since gαλ gλβ = δβα. This gives us√−g = e2ax. In Σ the action thus yields

S =

∫d2x

1

2e2ax

[(∂φ

∂t

)2

−(∂φ

∂x

)2]

(11.7)

1These are easily derived from Lorentz transformations. See for example chap. 3.2 in “Lecturenotes on general relativity” by Øyvind Grøn.

11.1. MASSLESS SCALAR FIELD IN A GENERAL SPACE-TIME 105

In both systems we now write out the fields as integrals over all Fourier modes.

φ(x) =

∫dk2π

√1

2ω

[ake

ik·x + a†ke−ik·x

]

φ′(x′) =

∫dk′

2π

√1

2ω

[ak′e

ik′·x′ + a†k′e−ik′·x′

]

where the a-operators of course satisfy our beloved commutation relation[aki , a

†kj

]= δkikj

As known from a general treatment of hyperbolically accelerated systems, suchsystems will always create a horizon. See figure 11.2. As time increases the world

t′

t = +∞

t = −∞

x′

Figure 11.2: The creation of a horizon. The world-lines of the Σ-system willconverge to this horizon as t→∞.

lines in Σ will approach the edge of the light cone in Σ′, but never cross this line.The result is that, as t → ∞, no particle inside the colored area of figure 11.2can ever be observed by an observer in Σ′. That means that even if the space-timeaccording to an observer in Σ is non-empty, an observer in Σ′ may experiencevacuum. Remembering the definition of Bogoliubov coefficients from earlier lec-tures, we know that we may write the a-operators in Σ expressed in terms of the aoperators in Σ′:

ak =∑

k′

[Akk′ ak′ +Bkk′ a

†k′

]. (11.8)

Let us assume that we have a vacuum in the inertial system Σ′. That means that

ak′ |0′〉 = 0 (11.9)


The corresponding expectation value for an observer in the accelerated system Σis then

〈0′|a†kak|0′〉 = |Bkk′ |2

where

Bkk′ = (uk, v∗k′) = i

∫ ∞

−∞dx[u∗kv

∗k′ − u∗kv∗k′ ]

uk =

√1

2ωei(k

′x′−ωt′)

vk′ =

√1

2ω′ei(k

′x′−ω′t′) (11.10)

Since we are dealing with massless fields, we have that ω = |k| and ω ′ = |k′|.Combining the three lines in (11.10) one should obtain that

〈0′|a†kak|0′〉 = |Bkk′ |2

∼ e−πωa

sinhπωa

∼ 1

e2πωa − 1

∼ 1

eωT − 1

, (11.11)

where T = a2π . This is a remarkable result! An observer in accelerated motion

in an inertial system with vacuum will, according to this, experience a thermalbackground with temperature T = a

2π2. A common physical explanation for this

effect is that on the horizon, virtual particle-antiparticle-pairs are being createdand annihilated all the time. Some times these pairs will be created in such away that the pair get separated by the horizon before they annihilate, and that thiscreates a net creation of particles inside the horizon. Remembering the equivalenceprinciple, this is closely related to the creation of Hawking radiation on the surfaceof black holes. Systems like the one studied in this example is still subject forseveral problems when it comes to physical interpretation.

11.1.2 Scalar fields in a Robertson-Walker background

Now we will use the variation principle when having a special metric, namely theRobertson-Walker (RW) one, as a background for a scalar field. The RW metric is

2A similar effect has been investigated by Jon Magne Leinaas and John Bell at CERN. Theyconsidered a particle in a circular path and found that it would experience being in a thermal bath.Studying the distribution of spin, this effect is claimed to be found in particle accelerators.


given by

ds2 = dt2 − a2(t)dx2

gµν =

1 0 0 00 −a2(t) 0 00 0 −a2(t) 00 0 0 −a2(t)

gµν =

1 0 0 00 −a−2(t) 0 00 0 −a−2(t) 00 0 0 −a−2(t)

g = det(gµν) = −a6 (11.12)

where a is the scale factor. Using (11.2), the action for a scalar field in a RW-background is now given by

S =

∫d4xa3

[1

2φ2 − 1

2a2(∇φ)2 − V (φ)

].

Now, let us study a small variation φ → φ + δφ in the scalar fields and apply theprinciple of stationary action to obtain the equations of motion for φ:

δS =

∫d4xa3

[φδφ− 1

a2(∇φ) · δ(∇φ)︸︷︷︸

=∇δφ

−V ′δφ]

=

∫d4x

[− d

dt(a3φ) + a∇2φ− a3V ′

]δφ

=

∫d4xa3

[−3

a

aφ− φ+

1

a2∇2φ− V ′

]δφ

!= 0

Now we demand the expression in the brackets to vanish, and thus we obtain theequation of motion

φ+ 3Hφ− 1

a2∇2φ+ V ′ = 0 (11.13)

where H = aa is the Hubble parameter. This is a frequently used equation. Often

it appears without the third term. This is usually a very good approximation sincea grows rapidly in most universe models and thus makes this third term neglect-able. But of course one may also encounter problems where this term indeed isimportant.


11.1.3 Introduction of the d’Alembertian

Varying the scalar field in (11.2) using a general metric gµν one obtains

δS =

∫d4x√−g

[gµν∂µφ∂νδφ − V ′δφ

]

=

∫d4x√−g

[1√−g ∂ν(

√−g gνµ∂µφ)− V ′]δφ

!= 0 (11.14)

We now introduce the d’Alembertian operator given by

2 =1√−g∂µ

(√−g gµν∂ν)

= ∂ν∂ν = ∂2 (11.15)

This is easily seen if you let gµν raise the index of ∂µ. The 2 operator can beconsidered as a covariant version of the Laplacian ∇2.

2φ = ∇µ∇µφ

where ∇µ is a covariant derivative given in terms of the Christoffel symbols as

∇µAν = ∂µAν +AλΓνλµ (11.16)

Using the d’Alembertian we can now write (11.14) as

2φ+V ′(φ) = 0

11.1.4 Variation of the metric

Until now we have been studying variations of a scalar field. Now we will see whathappens if we vary the metric. This will lead us to a very useful quantity called theEinstein-Hilbert action, which will be introduced in the next section.

Recalling some basic linear algebra we know that

g = det(gµν) =∑

ν

gµνγµν

where the µ in the last expression can be chosen arbitrarily, but should not besummed over. Here γµν is the cofactor, which can be defined as

γij = (−1)i+jdet(gij)

where gij is the matrix which is left when one removes the ith row and the jthcolumn from the full g-matrix. The cofactors can be used to express the contrav-ariant of the metric since

gµνgνλ = δλµ ⇒ gνλ =

1

gγλν


Using this result we can write the derivative of the determinant as

∂λg =∂g

∂xλ=

∂g

∂gµν︸︷︷︸=γµν

∂gµν∂xλ

= g gµν︸︷︷︸=γµν

∂λgµν (11.17)

and similarly we can write the variation of the metric as

δgµν =∂gµν∂x

δx ⇒ δg = g gµνδgµν

Later we will also need to know δ√−g, which now yields

δ√−g =

1

2√−g δ(−g)

=1

2√−g (−g)gµνδgµν

=1

2

√−g gµνδgµν (11.18)

Enough about variation of the metric for a while . . . Let us go back to (11.16),where we defined the covariant derivative of a vector in terms of Christoffel sym-bols. As we know from kindergarten the Christoffel symbols can be expressed interms of the derivatives of the metric as

Γµλν =1

2gµρ [gρλ,ν + gρν,λ − gλν,ρ] (11.19)

From this definition it is obvious that the Christoffel symbols are symmetric in thelower indices. We are now interested in the divergence of a vector given as

∇µAµ = ∂µAµ +AνΓµνµ

so we are only interested in the Christoffel symbols on the form Γµλµ. Then, givena symmetric metric, the first and third term in the brackets in (11.19) will canceleach other out since now µ = ν and

gνρ [gρλ,ν − gλν,ρ] = δνλ,ν − δρλ,ρ = 0

Then we have that

Γµλµ =1

2gµρgµρ,λ

=1

2

1

−g ∂λ(−g)

=1

2∂λln(−g)

= ∂λln√−g


where we have used (11.17) to get from the first to the second line. Using this, wecan now write ∇µAµ as

∇µAµ = ∂µAµ +Aλ∂λln

√−g

=1√−g

(∂µ√−gAµ

)(11.20)

So if Aµ can be derived from a scalar potential

Aµ = ∇µφ = ∂µφ

we see that

∇µ∂µφ =1√−g

(∂µ√−g∂µφ

)

=1√−g∂µ

(√−ggµν∂νφ)

which is exactly the 2-operator that we found in (11.15), using variation calculus.We see that this 2-operator has the same meaning as the divergence of a gradient.

11.2 The Einstein-Hilbert action

We start this section by stating the famous Einstein equations:

Eµν = Rµν −1

2gµνR = 8πGTµν

These equations should result from a variation principle for a given action. Hilbertwas the first to express such an action in a general coordinate system. The Einstein-Hilbert action is then given by

S =

∫d4x√−g

[− 1

2κR+ L(φ, ∂µφ)

](11.21)

where R = Rµνgµν is the Ricci scalar which is a contraction of the Ricci tensor,

and κ = 8πG. Now we introduce the Lagrangian for a massless scalar field and avariation in the metric.

gµν → gµν + δgµν

L =1

2gµν∂µφ∂νφ− V

11.3. THE ENERGY-MOMENTUM TENSOR 111

So δS yields

δS =

∫d4x

[1

2κδ(√−gR) + δ(

√−gL)

]

=

∫d4x

[− 1

2κ

((δ√−g)R+

√−gRµνδgµν +√−g gµνδRµν

)+δ(√−gL)

δgµνδgµν

]

The third term is a surface-term that will vanish. Using (11.18) we have for thesecond term that:

(δ√−g)R = −1

2

√−g gµνδgµνR (11.22)

and we can rewrite δS as

δS =

∫d4x√−g

[− 1

2κ(Rµν −

1

2gµνR) +

1√−gδ(√−gL)

δgµν

]δgµν = 0

(11.23)We see that we obtain the Einstein equations if we demand that the last term in thebrackets is the energy momentum tensor Tµν .

11.3 The energy-momentum tensor

Given a metric and a Lagrangian you can always find the energy momentum tensorusing the expression

Tµν =2√−g

δ(√−gL)

δgµν

= 2∂L∂gµν

− gµνL (11.24)

since using (11.18) we have

δ√−gδgµν

=1

2

√−g gµν . (11.25)

(11.24) is a very important and extremely useful equation, and we will show a fewexamples on how it can be used for different Lagrangians.

11.3.1 Example with a Maxwell field

First we will calculate the energy-momentum tensor for a Maxwell field with Lag-rangian

L = −1

4F 2µν =

1

4FαβFρσg

αρgβσ

The derivative with respect to the metric yields

∂L∂gµν

= −1

2FµνF

αν =

1

2FµαF

αν


Inserting into (11.24) gives the energy-momentum tensor

Tµν =1

2FµνF

αν +

1

4gµνF

2αβ

Notice the familiar Maxwellian form when using a Minkowski metric, gµν = ηµν .

11.3.2 Example with a scalar field

Assume a massless scalar field with Lagrangian

L =1


∂L∂gµν

=1

2∂µφ∂νφ

Which give the energy-momentum tensor

Tµν = ∂µφ∂νφ− gµν[

1

2(∂λφ)2 − V

](11.26)

Let us now examine the simple but interesting case of a constant φ.

φ = const. = φ0 → Tµν = gµν V (φ0)︸︷︷︸=ρ0

= ρ0gµν

Inserting this into the Einstein equations yields

Eµν = 8πGρ0gµν ≡ Λgµν

where Λ = 8πGρ0. So we see, not too surprisingly, that this scalar field willcorrespond to Einstein’s cosmological constant. Inserting this into (11.21) andadding a matter Lagrangian Lm, we have the action

S =

∫d4x√−g

[− 1

2κR− ρ0 + Lm

]

=

∫d4x√−g

[− 1

2κ(R+ 2Λ) + Lm

]

and the Einstein equations now yield

Eµν = 8πGTµν + gµνΛ

Let us study another case which is more general in the sense that we no longerdemand our scalar field to be constant, but less general in the sense that we choosea special metric, namely the RW-metric:


11.3. THE ENERGY-MOMENTUM TENSOR 113

Using (11.26) we will now calculate some components of the energy-momentumtensor and see how our well known relations between the time-derivatives of thescalar field and the energy density and pressure fall out of our formalism. Let usstart with the 00-component:

T00 = φ2 −[

1

2φ2 − 1

2a2(∇φ)2 − V

]

=1

2φ2 + V +

1

2a(∇φ)2 (11.27)

Now we calculate the 11-component:

T11 = (∂xφ)2 + a2

[1

2φ2 +

1

2a2(∇φ)2 − V

]

= a2

[1

2φ2 +

1

2a2

((∂xφ)2 − (∂yφ)2 − (∂zφ)2

)− V

]

Raising one of the indices we have

T 11 = g11T11 =

1

a2T11 =

1

2φ2 +

1

2a2

((∂xφ)2 − (∂yφ)2 − (∂zφ)2

)− V

The 22 and 33 components will of course be completely analog. It is common todefine a pressure p like

p ≡ 1

3

(T 1

1 + T 22 + T 3

3

)

=1

2φ2 − V (φ) − 1

6a2(∇φ)2

In cosmology one usually omits the last term because the scale factor a growsrapidly to a large value and we are thus left with the well-known expression forpressure from a scalar field:

p =1

2φ2 − V (φ) (11.28)

Notice that in the most general case one also has non-diagonal components of theform

T01 = φ∂xφ , T12 = ∂xφ∂yφ

But usually when working in cosmology one assumes a hydrodynamical modelwhere the non-diagonal components vanish because of homogeneity and isotropy:

〈∂µφ∂νφ〉 = 0 for µ 6= ν

Then the last term in the expression (11.27) for T00 vanish, and we recognize the00-component as the well-known expression for energy density from a scalar field:

ρ =1

2φ2 + V (11.29)


And we can write the energy-momentum tensor on the form

〈T µν〉 =

ρ 0 0 00 −p 0 00 0 −p 00 0 0 −p

Chapter 12

Lecture 12

In this lecture, we will finish our investigation of the energy-momentum tensor andthen turn the attention towards the concept of how to quantize curved spaces. Inparticular we will consider the Robertson-Walker metric (RW) in this manner.

12.1 Conservation of the energy-momentum tensor

First, a reconsideration of the definition of the energy-momentum tensor as seen inequation (11.24):

Tµν =2√−g

δ(√−gL)

δgµν

The definition above is actually the one and only correct definition of the energy-momentum tensor. We will in the following take a look at the energy-momentumtensor and investigate under what kind of circumstances it is conserved. We willstart out with the easy situation of a flat space and then turn to the more complicatedcurved spaces thereafter:

• Flat space-time: From relativistic quantum field theory and Noether’s the-orem we know that quantities are conserved due to invariance or symmetriesof the Lagrangian density under translations or rotations. Hence, the energy-momentum tensor is conserved due to translations with respect to time andspace. The conservation of momenta is a direct fact of spatial invariance,while energy is conserved from temporal (time) invariance. This is stated as:

∂µTµν = 0

• Curved space-time: In the transition from flat to curved spaces, the (usual)partial derivative, ∂µ changes into a covariant derivative, ∇µ. And insteadof considering translations (in time and space), we look at invariance dueto local coordinate transformations. From the flat case, a qualified guess istherefore that the energy-momentum tensor is conserved as

∇µTµν = 0 (12.1)

115


and where the local coordinate transformation is the small shift in position

xµ → xµ(x) = xµ + ξµ(x) (12.2)

We now want to convince ourselves that the conservation of the energy-momentumtensor in (12.1) is correct. To do so we will perform such a local coordinate trans-formation which now also includes a coordinate transformation of the metric gµν .Then we further demand that the action of the scalar field, Sφ, is invariant in thetransformation, and then finally minimizing the action and applying the variationalprinciple.

First let us notice that the Bianchi identity automatically leads to conservationof the energy-momentum tensor. This is seen by combining the Einstein equations

Eµν = 8πGTµν

with the Bianchi identity 1

∇µEµν = 0

Now we go on with investigating the Tµν under a local coordinate transforma-tion with respect to the metric:

gµν(x) → gµν(x)

= gµν(x+ ξ) = ξλ∂λgµν(x) + gµν(x)

= ξλ∂λgµν + higher order terms (12.3)

where in the first step, the relation xµ → xµ(x) = xµ + ξµ(x) in (12.2) has beeninserted for. To find the desired variation in the metric, δgµν , we need to use therelation:

gµν →∂xρ

∂xµ∂xσ

∂xνgρσ(x) (12.4)

where the derivatives are∂xρ

∂xµ= δρµ + ∂µξ

ρ

∂xσ

∂xν= δσν + ∂νξ

σ

If we now insert these derivatives into equation (12.4), we find an expression ofgµν as:

gµν = (δρµ + ∂µξρ)(δσν + ∂νξ

σ)gρσ(x) = gµν(x) + ξλ∂λgµν

which finally gives us the change or variation in the metric by subtracting gµν(x)from gµν(x):

δgµν = gµν(x)− gµν(x)

= −ξλ ∂λgµν︸︷︷︸Christoffel

−gµλ∂νξλ − gνλ∂µξλ (12.5)

1Another way to interpret this identity is ∂∂Σ = 0, and is described by Wheeler as “the boundaryof a boundary is zero”.

12.1. CONSERVATION OF THE ENERGY-MOMENTUM TENSOR 117

The variation in the metric can be expressed in terms of the covariant derivativesas:

δgµν = −∇µξν −∇νξµ (12.6)

We will now apply the action Sφ of a scalar field and then the variational principlewith stationary action to derive the conservation of the energy-momentum tensorin curved space-time. With curvature (and 4 space-time dimensions), the actionintegral is the generalized expression of (11.2) and looks like:

Sφ =

∫d4x√−gL(φ, ∂µφ) (12.7)

and where the curvature normalization term is the usual g = det(gµν). To apply thevariational principle with stationary action, it is now time to consider the specificcoordinate transformation xµ → xµ(x) = xµ + ξµ(x) in (12.2). We have alreadyshown that the metric then transforms as

gµν → gµν = gµν + δgµν (12.8)

However, not only the metric changes in the coordinate transformation, but alsothe scalar field. The change in the scalar field, δφ is:

φ(x) = φ(x) = φ(x+ ξ)

= φ(x) + ξλ∂λφ (12.9)

Here, the term ξλ∂λφ is the (small) change in the scalar field. Then the variationof the field due to the transformation (12.2) is:

δφ = φ(x)− φ(x) = −ξλ∂λφ (12.10)

So in order to have coordinate invariance (which we have assumed), the actionintegral Sφ must be invariant under such a transformation. Hence, if the Lagrangiandensity is invariant under the transformation:

xµ → xµ + ξµ (12.11)

then applying the variational principle and demanding the action to be stationaryyields:

δSφ = 0 =

∫d4x√−g

[1

2Tµνδg

µν +1√−g

∂(√−gL)

∂φδφ

](12.12)

The second term in the expression above can be shown to be of no contribution.We recall that the Lagrangian density of a general scalar field is given by:

L =1



if we compare with the equation in (11.14) we see that this second term in (12.12)satisfies the Euler-Lagrange equations and therefore vanishes. With only one con-tributing term, the variation of the action now is:

δSφ =

∫d4x√−g(1

2Tµνδg

µν)

=

∫d4x√−g(Tµν(∇µξν)) (12.13)

The last step follows from tha relation between the variation of the metric tensorand the covariant derivatives and is given by

δgµν = ∇µξν +∇νξµ (12.14)

This relation right above can be justified from its covariant relative in (12.6) andthen use the metric tensor inverse identity to find:

gµνgµν = δµµ

⇒ = gµν(δgµν) = −(δgµν)gµν (12.15)

So we notice the change in sign compared to (12.6). It is therefore possible towrite the action integral in the desired form for investigating the conservation ofthe energy-momentum tensor, that is, the term ∇µTµν .

δSφ =

∫d4x√−g

[∇µ(Tµνξ

ν)− (∇µTµν)ξν] !

= 0 (12.16)

To find a solution of the second term in the integrand, let us take a look at the firstone. This term is a so-called total divergence and it equals zero (when integrated)by means of Gauss’ 4-dimensional divergence theorem. However, stating this factdemands some argumentation. Gauss’ divergence theorem in its usual form in-volves partial derivatives ∂µ and not, as we have in (12.16), covariant derivatives,∇µ. The common Gauss’ theorem is given by:

∫d4x(∂µAµ) =

∫d3ΣµAµ = 0 (12.17)

where Σµ is a surface term on the considered surface. So in order to show that thefirst term (in (12.16)) is zero, and hence also the second term, we must convinceourselves that Gauss’ theorem also is valid when dealing with ∇µ’s. That this isthe fact is seen from the definition:

∇µgρσ = 0 (def.)

⇒∇µ√−g = 0 (12.18)

So the√−g-term in the action integral makes no influence when changing from

partial derivatives to Christoffel symbols. A more direct way to see this fact, is toconsider

∇µAµ =1√−g∂

µ(√−gAµ)

12.1. CONSERVATION OF THE ENERGY-MOMENTUM TENSOR 119

So we see that the√−g-terms cancel and we are left with only partial derivatives.

So finally, with stationary action, we write (12.16) as:

δSφ =

∫d4x√−g

[∇µ(Tµνξ

ν)︸︷︷︸=0

−(∇µTµν)ξν] !

= 0 (12.19)

which gives the important result of the conservation of the energy-momentumtensor:

∇µTµν = 0

In the (long) derivation, we have illustrated the following important fact:

Conservation of physical quantities in flat space-time due to invariance withrespect to translations is in the case of curved spaces replaced by invariance dueto coordinate transformations.

12.1.1 Generalized Bianchi identity

We will now, very short and in no detail, consider how we can find a generalizedBianchi identity which results in a generalized conservation of Tµν . The Einsteinequation is given by:

Eµν = 8πGTµν (12.20)

and the Bianchi identity∇µEµν = 0 (12.21)

which leads to conservation of the energy-momentum tensor, ∇µTµν = 0. TheEinstein tensor expressed in terms of the Ricci scalar and the Ricci tensor:

Eµν = Rµν −1

2gµνR (12.22)

From the Einstein-Hilbert action, we find the following action:

S =

∫d4x√−g

[− 1

16πGR+ . . .+ L

](12.23)

We assume the “rest”-terms after the Ricci scalar also to be a function of the Ricciscalar, that is with + . . .+ = f(R) in the expression above. The question is whendoes the conservation-identity ∇µTµν = 0 hold. (Actually, we very much wouldlike this important to hold in most cases). It can be shown 2 that the Bianchi identityholds for any function f(R). Actually, this is a generalized Bianchi identity and asevery such identity leading to the conservation ∇µTµν = 0.

2Tomi Koivisto has shown this. Ask him if you have any further questions


12.2 Weyl transformations

In this section, we will consider scaling transformations for both flat and curvedspaces. When the global scale-properties of flat space are attempted to be trans-ferred to curved spaces, problems occur. The solution to these problems is theintroduction of Weyl-transformations. Let us therefore take a look at the propertiesof flatness and curvature with respect to scaling transformations.

• Flat space: We perform the following scaling transformation in Minkowskispace:

xµ → xµ = λxµ

This means that the metric transforms as:

ds2 = dxµdxµ → λ2ds2 (12.24)

An example of a scale invariant theory is one describing scalar and masslessparticles given with the Lagrangian density

L =1

2(∂µφ)2

• Curved space: In opposition to the case of flat space, scale invariance isnot satisfied when applying such a global scale transformation in a generalcurved space. There will be the need of limiting to a local scale transforma-tion in order to achieve scale invariance. An example of such a local one istransforming the metric into (compare with that of flat space (12.24)) :

ds2 → λ2(x)ds2 (12.25)

or with respect to the metric tensor

gµν(x)→ λ2(x)gµν(x)

Hence, in every point the metric is changed locally, that is independentlyof any point in the neighborhood. It is this local transformation which iscalled a Weyl-transformation. (In some occasions the Weyl-transformationsare referred to as conformal transformations. This is not always a good idea,since conformal transformations normally are used to describe flat spaces). Itshould be obvious that the local scale transformation not necessarily prevailsglobal curvature. This can be illustrated by a curved object, for instance aballoon. If a local transformation is applied to the balloon, then for examplethe point at the top of the balloon may be enlarged, resulting in a “bump”when no other points are enlarged at the same time. So the global shape ofthe balloon is not prevailed. In the case of local scale transformations anypoint may then be enlarged independently of the neighboring areas. If wenow assign a physical interpretation to the balloon, say as the universe, then

12.2. WEYL TRANSFORMATIONS 121

a global scale transformation, for instance by enlarging the entire balloon,that is, every point simultaneously, would be a pretty tough constraint to theuniverse. Then all points in the universe would have to expand equally andsimultaneously. This illustrates that global scale transformations in curvedspace are non-trivial.

Let us follow up the theme and investigate whether or not Weyl invariance can beassigned to every kind of transformations. A local Weyl-transformation may looklike:

λ(x) = 1 +1

2ε(x)

⇒ λ2(x) = 1 + ε(x)−O2(ε2)

The last term can be omitted when ε is small. Similarly, the transformation of themetric tensor is

gµν → (1 + ε)gµν = gµν + εgµν

So the variation in the metric tensor is proportional to the tensor itself. This is seenby:

δgµν = (1 + ε)gµν − gµν = εgµν

If we have Weyl-invariance, the variation of the action should be zero with

δSW = 0

=1

2

∫dx√−gTµνgµνε(x) (12.26)

So Weyl invariance is found when Tµνgµν = 0, which leads to the following fact:

Tµνgµν = T µµ = 0

Hence, the trace T µµ vanishes due to the Weyl-invariance. We will in the follow-ing, for illustration, consider different physical theories and see if they are Weyl-invariant or not.

12.2.1 Example with a free scalar field

The Lagrangian describing a free and massless scalar field is the well-known

L =1

2gµν∂µφ∂νφ


From the definition of the Energy-momentum tensor seen in (11.24), we find inthis particular case that:

Tµν =2√−g

δ(√−gL)

δgµν= 2

∂L∂gµν

− gµνL

= ∂µφ∂νφ− gµν1

2(∂λφ)2

The trace was given by the relation T µµ = Tµνgµν which results in

T µµ = Tµνgµν = (∂λφ)2 − 1

2D(∂λφ)2

=1

2(2−D)(∂λφ)2

Here D is the total number of both spatial and temporal dimensions. We observethat the trace is zero only for one spatial dimension, D = 1 + 1 = 2. A mass-less scalar field is therefore only Weyl invariant in 2 space-time dimensions. Anexample of such a theory is String theory since strings only live in one spatialdimension.

12.2.2 Example with Maxwell theory

The Lagrangian of the photon theory is

L =1

4F 2µν

Again, by using the definition, the energy-momentum tensor now is

Tµν = FµλFλν +

1

4gµνF

2αβ .

The trace is then

T µµ =1

4(D − 4)F 2

αβ .

We see that the trace in the case of Maxwell theory is zero only when the number ofspace-time dimensions, D, equals four, and Weyl invariance is only satisfied whenthis is fulfilled. If now electromagnetic photon theory is considered in dimensionshigher than four, it will be quite different. Let us therefore take a short look at thissituation

• D>4: The expectation value of the trace is now given by:

〈T µν〉 =

ρ 0 0 . . .0 −p 0 . . .0 0 −p . . ....

......

. . .

(12.27)

12.3. QUANTIZATION OF SCALAR FIELDS 123

We know from standard (D = 4) statistical mechanics that the pressure and energydensity in the hydro dynamical approximation is related by p = 1

3ρ. However,in this particular case where the number of dimensions is arbitrary, the relationchanges somehow

p =1

3ρ→ p =

1

D − 1ρ. (12.28)

The general relation between pressure and energy follows, as in the usual D = 4-case, from kinematics. It can be derived by considering the radiation pressure thatis caused by massless photons hitting a wall. Then the expectation value of thetrace of the higher dimensional Maxwell theory is the sum of all diagonal elementsin (12.27), and by means of relation (12.28) it is found to be zero for all dimensions:

〈T µµ〉 = 0

This was actually seen in the last part of lecture 4 when relations between pressureand energy density in quantum statistical mechanics were derived. Hence, that theexpectation value of the tensor is zero is not so surprising, but that the tensor itselfvanishes when D = 4 is a bit more remarkable. 3

12.2.3 Example with coupled Maxwell and scalar theory

An interesting question to ask in this sense, is whether or not it is possible toachieve Weyl invariance when Maxwell fields are coupled to scalar fields. We havein the previous two examples seen that the theory describing scalar theory only isWeyl invariant when D = 2, and on the other hand electromagnetic photon theoryis only Weyl invariant for a total dimension satisfying D = 4. At first sight theanswer to this question seems to be no, but this is actually not so. It can be shownthat if a term ∼ 1

12Rφ2 is added to the Lagrangian of the scalar field, that is

L =1

2gµν∂µφ∂νφ+

1

12Rφ2

then the coupled theory is Weyl-invariant. This is so since the added term in thescalar Lagrangian causes an extra term in Tµν which results in that the (scalar)trace vanishes after all:

T µν = 0

It should also be noted that if the R is a constant, the added term 112Rφ

2 acts as amass term of the scalar field.

12.3 Quantization of scalar fields

We are now about to perform a quantization with respect to the scalar field in aRW metric (as was considered in lecture 11). This quantization is very important

3See article by Finn, Håvard and Ingunn. finn artikkelen


in cosmology and will be of great use later in this course. We will see that it leadsto, among others, the primordial spectrum. However, that will be subject of somelater lecture.

12.3.1 Scalar field in a RW background

The line element in the Robertson-Walker metric is given by


and the metric itself is

gµν =

1 0 0 00 −a2(t) 0 00 0 −a2(t) 00 0 0 −a2(t)

For later use it is convenient to define the spatial determinant, h, in analogy withthat of (11.12) as the determinant of gµν when the first row and column are re-moved. 4

So far we have not assigned any constraints to the scale factor, a(t), but ratherconsidered an arbitrary background. Later, when inflation is being considered, thescale factor will be exponential or a power function with respect to time t, but sofar the scale factor is assumed to be arbitrary. The scalar field for massive particlesis described by the Lagrangian

L =1

2gµν∂µφ∂νφ−

1

2m2φ2

the solutions, we have earlier seen, satisfies the classical wave equation given by

2φ+ V ′(φ) = 0→ 2φ+m2φ = 0

which is also called the (general) equation of motion of the scalar field. And forthe RW case, this equation turns into the earlier seen

φ+ 3Hφ− 1

a2∇2φ+ V ′ = 0. (12.29)

The scalar field is now rewritten

φ(x, t) = ϕ(t)eik·x

which results in that the RW equation of motion of the scalar field in (12.29) canbe written as

ϕ+ 3Hϕ+

(k2

a2+m2

)ϕ = 0. (12.30)

4To avoid any confusion, the first row and column that are to be removed before we find thespatial determinant, are the ones involving time components so that only the a2-s are left.


We observe that the effective mass-term now also contains the scale factor, a andthe wave number k. The equation above is a classical equation and therefore thesolution is the classical mode-functions

uk(x) = ϕk(t)eik·x.

The mode-functions are normalized by means of the inner product as we saw in(9.49) in lecture 9.

(uk, uk′) = i

∫d3x√hu∗k

↔∂t uk′

where h is the spatial determinant, in this case given by h =√a6 = a3.The

spatial determinant ensures that the integral is normalized. The inner product nowbecomes

(uk, uk′) = ia3(2π)3δ(k− k′)(ϕ∗kϕk′ − ϕ∗kϕk′).

If we now require the normalization

(uk, uk) = (2π)3δ(k − k′), (12.31)

then the normalization condition satisfies

ia3(ϕ∗kϕk − ϕ∗kϕk) = 1 (12.32)

The field operator with standard normalization now looks like

φ(x) =

∫d3k

(2π)3[akuk(x) + h.c.]. (12.33)

Here, h.c. corresponds to hermitean conjugated terms. We know that the annihil-ation and creation operators are defined in the usual way by means of the innerproduct (uk, uk′), (lecture 9, equation(9.18)). This yields

(uk, uk′) = (2π)3δ(k − k′)

= [ak, a†k′ ].

We will now proceed with simplifying the field equation of the scalar field in(12.30). The simplified equation will then in turn be comparable with that of anoscillator with time dependent frequency.

1. We substitute time, t with conformal time τ in the line element. That is:

ds2 = a2(τ)[dτ 2 − dx2]

The derivatives of the scalar field with respect to t now changes into:

ϕ =dϕ

dt=dϕ

dτ

dτ

dt=

1

aϕ′

ϕ =1

a2ϕ′′ − a′

a3ϕ′


It is convenient also to write the Hubble parameter in terms of conformaltime:

H =a

a=

1

a2a′ =

1

aH

(The conformal Hubble parameter must not be confused with the Hamilto-nian density H). We have H = a′

a . So equation (12.30) is written to theform:

ϕ′′k + 2Hϕ′k + (k2 +m2a2)ϕk = 0 (12.34)

2. We observe that the field equation still has a friction term ∼ ϕ′k when inter-preting it as an oscillator. This term complicates the whole story and it canbe removed by rewriting ϕk = 1

aψk(t). Equation (12.34) can then be writtenas

ψ′′k + ω2k(τ)ψk = 0

where ω2k = k2 +m2a2 − a′′

a = ω2k(τ)

We see that after the two small steps, the field equation of the scalar field inan RW background is precisely the same problem that we considered in lecture 9where an oscillator with time dependent frequency was considered as an exampleof quantization of an external field. We can also find, from (12.32) , that

ψ′∗k ψk − ψ∗kψ′k = i (12.35)

The quantum field in terms of its fieldoperator in (12.33) can now be re-expressedas:

φ(x) =1

a

∫d3k

(2π)3[akψk(τ)eik·x + h.c.] (12.36)

where the normalization is given by

[ak, a†k′ ] = (2π)3δ(k− k′)

However, there are some difficulties when applying the scalar field theory tophysical problems such as describing the universe at an early stage. We have re-written the theory to a form containing an oscillator. For a given value of the scalefactor a(τ), we would like the oscillator frequency to be well-defined an (nearly)constant for early values of τ . The frequency should therefore be of the form asshown in figure (12.1) We would in that case be able to repeat the process we didwhen simplifying the oscillator with time dependent frequency in the previous lec-tures (see figure 9.1) and then be able to describe the vacuum consistently. Thevacuum is given by the in-modes when time is taken to the far past. Different


modes (which may be a result of a non-constant frequency at small τ ) result indifferent vacuums and different sets of operators that again are connected with theBogoliubov coefficients. Hence, what acts as vacuum at some set of initial condi-tions, will not necessarily be so at different initial conditions when the modes aredefined. Instead we may have a state where particles are present. Different kindsof vacuums will decide whether there is a possibility of particle creation or not. Soif the frequency has a nastier shape than that in figure 12.1 , we can not be sureof what kind of vacuum our physical system possesses. Such problems are typicalwhen looking at inflation.

In the coming lectures, we will learn more about gravitons and gravitationalwaves in flat space.

t

ωk

I

Figure 12.1: To define the vacuum in the early universe in a proper way, the oscil-lator frequency should satisfy a constant value at early times (I).

Chapter 13

Lecture 13

13.1 The Dirac equation in curved spacetime

First we will have just a quick look at the Dirac equation in curved spacetime. Weshall not use it further in this course, but it is useful to have just seen it.

We are used to the Dirac equation in flat spacetime on the form

(γµ∂µ −m)ψ(x) = 0

where the anticommutator of the γ matrices gives the metric

γµγν + γνγµ = 2gµν(x)

The metric can be expressed as a transformed Minkowski metric

gµν(x) = V µµ V

νν ηµν

And thus this transformation tensor can be used to transform the γ matrices as well

γµ(x) = V µµ (x)γµ

where

γ0 =

[1 00 −1

]

and

γk =

[0 σk−σk 0

]

We want this to be invariant under local Lorentz transformations

ψ(x)→ ψ′(x) = S(x)ψ(x)

WhereS = e−

i4σµνθ

µν(x)

129


and σµν = i2 [γµ, γν ] . We ensure this invariance if we replace the differential op-

erator

∂µ →∇µ = ∂µ −1

2Ωαβµ σαβ,

where Ωαβµ is called the gravitational field . This is similar to the ordinary gauge-

field Aµ in the ordinary covariant derivative

∂µ → Dµ = ∂µ − iAaµTa

where T fulfills the commutation relation

[Ta, Tb] = ifabcTc.

13.2 Gravitons in flat spacetime

Usually when we go from flat to curved spacetime we exchange the Minkowskimetric with the general metric, and we will also do so here, but now we only haveslightly curved spacetime. We will write

ηµν → gµν(x) = ηµν + hµν(x)

where |hµν | 1. As gµλgλν = δµν we get that

gµν = ηµν − hµν +O(h2)

Now we introduce a coordinate transformation

xµ → xµ = xµ + ξµ(x)

and the metric transforms as

gµν(x)→ gµν(x) =∂xα

∂xµ∂xβ

∂xνgαβ(x) = ηµν + hµν(x)

as we saw in the last lecture we can now express hµν

hµν = hµν − ∂µξν − ∂νξµ

We have the Christoffel symbols . . .

Γσµν =1

2

(∂µh

σν + ∂νh

σµ − ∂σhµν

)

. . . and the Riemann curvature tensor.

Rσρµν =1

2(∂ρ∂µh

σν − ∂ρ∂νhσµ + ∂σ∂νhρµ − ∂σ∂µhρν)

13.2. GRAVITONS IN FLAT SPACETIME 131

Contraction gives the Ricci tensor

Rµν = Rλµλν =1

2(∂λ∂µh

λν + ∂λ∂νh

λµ − ∂µ∂νhλλ − ∂λ∂λhµν

Both the Riemann and the Ricci tensor are invariant under this transformation.We introduce now hλλ, and the transformation

hµν = hµν −1

2ηµνh

and this has nothing to do with the previous hµν . We now choose Hilbert gauge1

∂µhµν = 0

This is analog to Lorentz2-gauge in electromagnetism ∂µAµ = 0. hµν has its tracereversed under the transformation

hµµ = h− 2h = −h

another transformation returns the ordinary trace

hµµ = hµµ −1

2ηµµh = h

In Hilbert-gauge we get the Ricci curvature tensor

Rµν = −1

2∂2hµν

and the Ricci scalar becomes R = − 12∂

2h. This leads to the Einstein tensor

Eµν = Rµν −1

2ηµνR

= −1

2∂2

(hµν −

1

2ηµνh

)

= 8πGTµν

we recognize hµν in the expression and this leads us to

∂2hµν = −16πGTµν (13.1)

again this is analog to electromagnetism where we have ∂2Aµ = Jµ.As usual we are interested in finding the Lagrangian density. Here it is

L =1

2(∂νhαβ)2 − 1

2(∂αh)2 + ∂αh

αν∂νh− ∂αhαν∂βhβν1This gauge has many names. Einstein called it Harmonic gauge, some people call it de Donder

gauge, Finn Ravndal calls it Hilbert-gauge and so will we in this course.2Actually this is named after a Danish physicist named Lorenz, but enough people have mistaken

this for the famous Lorentz so it is common to wrongfully use the name Lorentz-gauge.


The trace reversed energy momentum tensor is now

Tµν = Tµν −1

2ηµνT

λλ

and this leads us to the relation

∂2hµν = −16πGTµν .

We can write this as

hµν = 16πG1

−∂2Tµν

where the differential operator can be thought of as the momentum operator 1−∂2 ∼

1k2 . The change of action is

δS = −1

2

∫d4x√−gT ′µνhµν =

∫d4x√−gLint

where we have the interaction Lagrangian density

Lint = −1

2T ′µνh

µν = −1

2T ′µν16πG

1

−∂2T µν = −8πGT ′µν

1

−∂2T µν =

1

M2p

T ′µν1

−∂2T µν

where the Planck mass is

M2p =

1

8πG

Now we want to express Tµν with help of a polarization matrix

Tµν = PµνρσTρσ

where

Pµνρσ =1

2(ηµρηνσ + ηµσηνρ − ηµνηρσ) (13.2)

if we insert this in our Lagrangian density we get

Lint =1

M2p

T ′µνDµνσρTρσ

where Dµνσρ(k) =Pµνσρk2+iε is a graviton propagator in momentum space.

13.3 Gravitational waves

13.3.1 Basics

First, if we again look back at electromagnetism (that is, QED) we see that thephoton field Aµ has four degrees of freedom. Then we impose a Lorentz-gauge∂µA

µ = 0 which removes one degree of freedom. Now if we have a gauge trans-formation Aµ → Aµ

′= Aµ+∂χ. And we demand that ∂µA′µ = ∂µAµ+∂2χ = 0.

13.3. GRAVITATIONAL WAVES 133

This gives us that ∂2χ = 0. Now we are down to two degrees of freedom. We canwrite

Aµ(x) =

∫d3k

(2π)3

1

2ω

∑

λ=R,L

[eµ(λ, k)aλ(k)e−ik·x + h.c.

]

where

kµ = ω(1, 0, 0, 1)

εµR =

√1

2(0, 1, i, 0)

εµL =

√1

2(0, 1,−i, 0)

(This is only so when the vector k is along the z-axis.)

Now, let’s do the same to gravitation. We start with the tensor hµν(x) with 16components. It is symmetric, so we have only ten degrees of freedom. The Hilbert-gauge ∂µhµν = 0 removes four degrees of freedom. Then to have invariance underthe transformation hµν → h′µν = hµν+∂µχν+∂νχµ we must have ∂2χµ = 0. Andthis puts us down another four degrees of freedom so we end up with 10−4−4 = 2degrees of freedom, just as in QED.

But we have not yet specified the constraints that removes the last four degreesof freedom. We choose these to be

hµµ = −h = 0

hµ0 = 0 (13.3)

this is really five equations, but they are not totally independent, so they will re-move only four degrees of freedom. This is called TT-gauge where the T’s standsfor transverse and traceless.

Now we can write out

hµν(x) =

∫d3k

(2π)3

1

2ω

∑

λ=+,×

[eµν(λ, k)aλ(k)e−ik·x + h.c.

]

we can write this in matrix form

hµν(z, t) =

0 0 0 00 h+ h× 00 h× −h+ 00 0 0 0

e

i(kz−ωt) = (h+e+ + h×e×)ei(kz−ωt) (13.4)


where the “basis” matrices are

e+ =

0 0 0 00 1 0 00 0 −1 00 0 0 0

= eµν(+, k)

e× =

0 0 0 00 0 1 00 1 0 00 0 0 0

= eµν(×, k)

and the propagator can be written

iDµνρσ = 〈0|Thµν(x)hρσ(x′)|0〉

13.3.2 Particle hit by a wave

Now we shall see what happens if a particle is hit by a gravitational wave. Atfirst the particle is at rest, with a 4-velocity uµ(t = 0) = (1, 0, 0, 0). As the4-acceleration for this particle is zero, the geodesic equation for this particle is

duµ

ds+ Γµαβu

αuβ = 0

as only the time component of the velocity vectors are nonzero, we get the Chris-toffel symbols

Γµ00 =1

2ηµν(hν0,0 + h0ν,0 − h00,ν)

we see from (13.3) that all these h-components are zero in TT-gauge and we get

Γµ00 = 0⇒ duµ

ds= 0

We now look at two test particles separated with a spatial distance along the x-axis, dxµ = (0, ξ, 0, 0). The line element gives us (we choose the positive solution)

ds =√gµνdxµdxν = ξ

√−g11 = ξ

[1− 1

2h+ cos(kz − ωt)

]

and two particles separated in the y-direction, dxµ = (0, 0, ξ, 0) leads to the dis-tance

ds = ξ

[1 +

1

2h+ cos(kz − ωt)

]

so, if we have four particles they will oscillate as in figure (13.1). For the cross-components we will get the same oscillations, but along the diagonals.

13.3. GRAVITATIONAL WAVES 135

Figure 13.1: The oscillations of four particles hit by a gravitational wave

13.3.3 Energy contents in a wave

In vacuum the Einstein tensor is zero

Rµν −1

2gµνR = 0

which gives that the Ricci tensor is also zero Rµν = 0, but if we expand the Riccitensor to second order

Rµν = R(0)µν +R(1)

µν +R(2)µν + . . .

∼ 0 + hµν + h2µν + . . .

The expectation value of the second order term is not generally zero

〈R(2)µν 〉 6= 0

This term gives geometry to the background. The energy-momentum tensor for thebackground is then given from the Einstein equations

〈R(2)µν 〉 −

1

2ηµν〈R(2)〉 = 8πGtµν

and from here it can be shown that

tµν =1

32πG〈∂µhmn∂νhmn〉

This gives in TT-gauge that

t00 =1

16πG〈(∂0h+)2 + (∂0h×)2〉

Chapter 14

Lecture 14

This lecture we will continue our study of gravitational waves and we will see howsuch waves may be quantized.

14.1 Quantum fluctuations in a classical background (re-petition)

In lecture 11 we introduced in (11.21) the Einstein-Hilbert action for a gravitationalfield

S =

∫d4x√−g

[− 1

16πGR+ Lm

](14.1)

where Lm is a common Lagrangian density for matter. Applying the principle ofstationary action gives us the Einstein field equations as classical solutions:

Eµν = Rµν −1

2gµνR = 8πGTmµν (14.2)

which gives as a classical solution the background metric gµν(x). Including quantumfluctuations to the metric on the form hµν(x), we obtain a total metric

gµν(x) = gµν(x) + hµν(x) (14.3)

or assuming a Minkowski background:

gµν(x) = ηµν + hµν(x) (14.4)

Assuming |hµν | small, gµν is usually given as a series expansion around gµν . Lastlecture we introduced the Hilbert gauge

∂µhµν = 0 (14.5)

where

hµν = hµν −1

2ηµνh (14.6)

137


Now, enforcing this Hilbert gauge, the Einstein equations (14.2) reduce to

∂2hµν = −16GTµν (14.7)

or∂2hµν = −16GTµν (14.8)

where Tµν = Tµν− 12ηµνT . Actually, Einstein himself found this expression before

he found his full equations.In lecture 13 we used these equations to find the interaction between matter

through graviton interchange:

hµν = −16πG1

∂2Tµν (14.9)

In the quantum mechanical description the inverse differential operator 1∂2 will give

us a Feynman propagator on the form

Dµνρσ(x− x′) =

∫d4k

(2π)4

Pµνρσk2 + iε

e−ik(x−x′) (14.10)

where Pµνρσ is the projection operator introduced in (13.2), and provided that weare dealing with two electric charges this will give us the common Newtonian po-tential e2

4πr in the non-relativistic limit. Now, classically we have to view the 1∂2 -

operator a bit different, and instead of a Feynman propagator we will obtain whatis called a retarded operator. This will be studied in some detail at the end of thissection using Green’s functions, so be patient.

14.2 Propagators from using the Einstein-Hilbert action

Early in the course we learned how to obtain propagators from an action usingtechniques with partial integration and so on. Let’s do this with the Einstein-Hilbertaction!

Starting with an Einstein-Hilbert action given by

S =

∫d4x√−g

(− 1

16πGR

), gµν = ηµν + hµν (14.11)

andgµν = ηµν − hµν + hµαh

αν +O(h3) (14.12)

and expanding also R to the second order we obtain1

S =1

32πG

∫d4x

[1

2(∂µhαβ)2 − 1

2(∂µh)2 + (∂αh

αν)(∂νh)− (∂αhαν)(∂βh

βµν )

]

(14.13)

1We don’t bother to do the detailed calculations here. Talk to Petter Callin. He has done it.

14.3. CANONICAL NORMALIZATION OF THE GRAVITATIONAL FIELD139

We haven’t made any choice of gauge yet. Choosing Hilbert gauge yields

∂µhµν

!=

1

2∂νh (14.14)

two of the terms will cancel out and we are left with

S =1

32πG

∫d4x

[1

2(∂µhαβ)2 − 1

2(∂µh)2

](14.15)

Applying the techniques from the first lectures, for example that∫

d4x(∂µφ)2 →−∫

d4xφ∂2φ when using partial integration we have

S =1

32πG

∫d4x

1

2hµν

1

2[−ηµαηνβ + ηµβηνα + ηµνηαβ ]

︸︷︷︸=Pµναβ

∂2hαβ

Here Pµναβ is the projection operator defined in (13.2) and keeping in mind that∂2 → 1

k2 when Fourier transforming we see that our propagator is

Dµναβ =Pµναβk2 + iε

(14.16)

14.3 Canonical normalization of the gravitational field

Last lecture we saw that Hilbert gauge corresponds to Lorentz gauge in QED, butthat we still were left with two degrees of freedom. We introduced the TT gaugegiven by the two conditions

i) hµµ = h = 0

ii) hµ0 = 0 ⇒ ∂ihij = 0

The last condition corresponds to the coulomb gauge from QED, ∂iAi = 0. Weremember from (13.4) that we were left with two independent amplitudes given byh+ and h×. In this TT gauge our action becomes

S =1

32πG

∫d4x

1

2(∂µhij)

2 (14.17)

If we compare (14.17) with the action for a massless scalar field

Sφ =

∫d4x

1

2(∂µφ)2 (14.18)

we see that

hij ∼2

MPφ

h2ij ∼

4

M2P

φ2 = 32πGφ2

where M 2P = 1

8πG . So if we want to normalize the gravitational field such that thecoefficient in front of (∂ )2 is 1

2 , we must impose the canonical normalization


hµν =2

MPHµν

on the gravitational field.

14.4 The energy-momentum tensor for the graviton field

As mentioned at the end of last lecture, Isaacson derived an energy momentumtensor for gravitational waves on the form (in TT-gauge)

tµν =1

32πG〈∂µhij∂νhij〉 (14.19)

The reason why we have to use the mean value over a volume of some finite exten-sion, is that the equivalence principle makes it possible to locally transform awayany gravitational effect, which would make our calculations tremendously boringand uninteresting if using a local theory. The original derivation of (14.19) madeby Isaacson himself (which can be found in Misner, Thorne, Wheeler as well) ispainfully long and boring. Luckily Finn, Håvard and Ingunn have done it in a muchmore compact and pleasant (but less stringent?) way which we will repeat here.

For a scalar field we have the energy-momentum tensor

T φµν = ∂µφ∂νφ−1

2ηµν(∂λφ)2

The mean value of the energy-momentum tensor in a volume V4 (which is small,but still large enough to contain lots of fluctuations) is given by

〈T φµν〉 =1

V4

∫d4xT φµν

= 〈∂µφ∂νφ〉 −1

2ηµν〈(∂λφ)2〉 (14.20)

The mean value in the last term can be written

〈(∂λφ)2〉 =1

V4

∫d4x(∂λφ)2

=1

V4

∫d4xφ∂2φ = 0 (14.21)

because the surface term vanish since we are free to impose periodic boundaryconditions. Then we have, by comparing (14.17) and (14.18), that

〈T φµν〉 = 〈∂µφ∂νφ〉

=1


which is just the same as Isaacson’s expression. Note that we assumed flat spacein this derivation, but the result is also valid in a general background.

14.5. HIGHER ORDER TERMS AND RENORMALIZABILITY 141

14.5 Higher order terms and renormalizability

From the simple relation gµν = ηµν + hµν we see that the dimension of hµν is

dim[hµν ] = 0

If we had expanded the action in (14.15) to a higher order it would look like

S =1

32πG

∫d4x

[1

2(∂µhαβ)2 − 1

2(∂µh)2 +

a

4(∂µh)2h2

αβ +O(h5)

]

Introducing our physical Hµν given by hµν = 2MP

Hµν this action becomes

S =

∫d4x

[1

2(∂µHαβ)2 − 1

2(∂µH)2 +

a

M2P

(∂µH)2H2αβ +O(H5)

]

The last term is describing an interaction between four gravitons. Since the coup-ling is of the order∼ 1

M2P

, this is not a renormalizable interaction (that is the case of

all quantum field theories with couplings on the form ∼ 1mass ). Many people have

used this fact to argue that one may not quantize the gravitational field. Formallythis is of course true, but at the same time the question on renormalizability is un-interesting for a physicist working in a real world. As mentioned earlier, it is just aquestion of staying inside the range of validity for your theory. In the theory terms

on the form ∼∣∣∣ kMP

∣∣∣4

that cannot be absorbed will appear, but in practical prob-

lems in astrophysics where k is small, these problems can safely be swept underthe rug, and you can treat it as an effective field theory. The theory will howeverbreak down if you go sufficiently far back in the inflationary phase of the Universe.The problem with such a quantized theory of gravitation is that the effects are toosmall to be observed in objects like quasars, so you have to go all the way back tothe inflationary phase to observe them.

Historically the first physicist to work with this kind of quantized gravitationwas John F. Donoghue. Both he and Cliff Burgess later put it into an effective fieldtheory. The Dane Emil Bjerrum-Bohr showed that there were some mistakes inwhat these people had done. Today, the theory that we use are based on the worksof these people.

14.6 Green functions and propagators

Both in quantum and classical theory we have expressions on the form

∂2hµν = −16πGTµν

∂2Aµ = −eJµwhich can be interpreted as

D︸︷︷︸diff. operator

A(x)︸︷︷︸field

= J(x)︸︷︷︸source


So formally we can write the solution as

A(x) = D−1J(x)

The further procedure is known from electromagnetism where we introduce aGreen function which is a solution for a point source, such that

D G(x− x′) = δ(x − x′) (14.23)

per definition of a Green function. A more fancy way to write the δ-function is:

δ(x − x′) =

∫d4k

(2π)4e−ik(x−x′)

and using an inverse Fourier transform of the Green function we can write it as

G(x− x′) =

∫d4k

(2π)4G(k)e−ik(x−x′) (14.24)

To choose a specific example, we choose the differential operator D = ∂2. Re-membering that ∂2 → −k2 when Fourier transforming we demand that the Greenfunction (14.24) should obey (14.23):

∫d4k

(2π)4

[−k2G(k)

]e−ik(x−x′) !

=

∫d4k

(2π)4eik(x−x′)

And thus we have that

G(k) = − 1

k2

and

G(x− x′) = −∫

d4k

(2π)4

1

k2e−ik(x−x′) (14.25)

Now, since G(x-x’) is a solution for a point source we have that

A(x) =

∫d4x′G(x− x′)J(x′)

and

D A(x) =

∫d4x′D G(x− x′)︸︷︷︸

δ(x−x′)

J(x′) = J(x)

Now, let’s try to calculate G(x− x′) from (14.25). We have

G(x− x′) = −∫

d3k

(2π)3

∫ ∞

−∞

dk0

2π

1

k20 − k2

e−ik0(t−t′)+ik(x−x′)

We have poles at±k (see figure 14.1). What we do now is of course to integratearound the poles in the complex plane. We will obtain different Green functionsdepending on how we choose our path of integration.

14.6. GREEN FUNCTIONS AND PROPAGATORS 143

Im(k0)

k Re(k0)−k

Figure 14.1: The Green function has poles for in ±k

To obtain the Feynman propagator one choose to go “below” one of the polesand “above” the other, or equivalently shifting one pole ε in positive imaginarydirection and the other one ε in negative imaginary direction (see figure 14.2). Thepoles for k0 then shift like

1

k2→ 1

k2 + iε=

1

k20 − k2 + iε

k0 = ±√

k2 − iεThere are two different contours that both satisfy these requirements, one that isclosed in the upper half plane and one closed in the lower half plane correspondingto a particle with positive energy going forward in time (t > t′) and the other onecorresponding to a anti-particle going backwards in time (t < t′). Again, see figure14.2.

In classical physics this negative energy solution is considered unphysical, soone has to deal with the poles in a somewhat different way. What one does is tomake a shift of the kind

1

k2→ 1

k2 + iεk0=

1

k20 − k2 + iεk0

k0 = ±√

k2 − iεk0

which shifts both of the poles in negative imaginary direction. As one can seefrom figure 14.3, we have to possible contours. One is in the lower plane andis the physical t > t′ solution which gives us the non-zero part of the retardedpropagator. The other contour is in the upper half plane and gives the zero partof the retarded propagator. Similarly we can construct an advanced propagator by


Re(k0)

Im(k0)

Re(k0)

k0 = −k + iε

k0 = k − iε

t′ > tIm(k0)

k0 = −k + iε

k0 = k − iε

t > t′

Figure 14.2: There are two possible contours of integration. The one to the leftcorresponding to an anti-particle with negative energy going backwards in time,and the one to the right corresponding to a particle with positive energy goingforward in time.

letting the parameter ε be negative. In classical physics only the retarded solutionis used.

We have seen that in quantum mechanics and classical physics we have thesame expression for the Green function, it is just a question of how we deal withthe poles2.

Applying residue calculations we will now sketch how one may calculate theclassical retarded propagator from the Green function.

Gk(x− x′) =

∫d3k

(2π)3eik(x−x′) sin k(t− t′)

k

=1

8π3

∫ ∞

0dkk22π

∫ 1

−1dµeikµ(x−x′) sin k(t− t′)

k

=δ(t− t′ − (x− x′))

4π(x− x′)

where µ = cosθ.

2“Those terms are not as harsh as they might have been. The Germans fig-ure that they can deal with the Poles at any time once they deal with the Soviets”http://members.aol.com/althist1/PODMar99/UkraineOption2.htm

14.7. EXAMPLE WITH GRAVITATIONAL RADIATION 145

t > t′

Im(k0)

Re(k0)

t < t′

Figure 14.3: Integrating along the solid line in the lower half plane corresponds tofinding the non-zero part of the retarded propagator, while an integration along thedashed line in the upper half plane yields the zero part of the retarded operator

14.7 Example with gravitational radiation

We can use this result for calculating gravitational radiation from (14.9) if we sub-stitute the 1

∂2 operator with Gk. Then we have

∂2hµν = −16πGGk Tµν

and we have

hµν(x) = −16πG

4π

∫d4x′

δ(t− t′ − (x− x′))(x− x′)

Tµν(x′)

= −4G

∫d3x′

Tµν(x′, t− (x− x′))(x− x′)

Next lecture we will finish this calculation and use the result on a binary star systemand on inflation.

Chapter 15

Lecture 15

We ended the last lecture in the middle of a calculation of the gravitational radiationfrom a general source of gravity. For convenience, the entire calculation is writtenbelow. An illustration of gravitational radiation is seen in figure (15.1). The sourcesof the gravitational waves are matter distributions such as for instance galaxies.

x

Observer

x′

x− x′Source

Tµν(x)

Figure 15.1: The figure shows a general matter (and energy) distribution causinggravitational radiation.

15.1 Gravitational radiation

The wave equation of gravitational waves was found in (13.1) in lecture 13 and inTT-gauge, it can be written as

∂2hµν = −16πGTµν

⇒ hµν = −16πG1

∂2Tµν

If we instead of the operator 1∂2 apply the general Green’s function as calculated in

(14.26), the wave equation turns into

147


hµν(x) = −16πG

4π

∫d4x′

δ(t− t′ − |x− x′|)|x− x′| Tµν(x′)

= −4G

∫d3x′

Tµν(x′, t− |x− x′|)|x− x′|

If the size of the source is small compared to the distance to the observer, that iswith |x| = r >> |x′|, the expression simplifies somewhat

hµν(x) = −4G

r

∫d3x′Tµν(x′, t− r)

This will be the situation if gravitational waves are observed from (distant) galax-ies. We also notice from the expression above that the gravitational emission hassimilar structure as electromagnetic radiation from an antenna, say.

To proceed with the calculation, we will have to do some small tricks.

• Conservation of energy and momenta is a fact of the following property ofthe energy -momentum tensor

∂µTµν = 0 (15.1)

In our calculation, we are only interested in the ij-components of h and T ,that is Hij and Tij . So if the relation in (15.1) is applied twice and combinedwith the fact that ∂2 = ∂2

t −∇2, the following holds

∂2T00

∂t2=

∂2Tij∂xi∂xj

. (15.2)

• We start out with the expression

∂2

∂t2

∫d3xT00x

kxl

Here, the partial differentiation ∂2

∂t2only acts on the term T00. By the relation

in (15.2) this yields

∂2

∂t2

∫d3xT00x

kxl =

∫d3x

∂2Tij∂xi∂xj

xkxl

= 2

∫d3xTkl. (15.3)

The last step follows by applying two partial integrations.

Hence, the calculation proceeds with

hij(x, t) =2G

r

∂2

∂t2

∫d3xT00(x′, t− r)x′ix′j

=2G

rIij(t− r) (15.4)

15.2. EMITTED ENERGY AND MOMENTA 149

where the moment of inertia, defined as

Iij(t− r) ≡∫d3xT00(x′, t− r)x′ix′j (15.5)

has been inserted for. To find the final answer to the calculation, we will apply TT-gauge. This traceless-transverse gauge demands both sides being traceless. Above,the righthand side is not so, and the trace therefore has to be subtracted in RHS (Wealready know that the lefthand side, hij is traceless in TT-gauge). This gives

hij(x, t) =2G

r

[Iij(t− r)−

1

3δij Ikk

](15.6)

A more compact answer can be found by inserting for the the quadrupole moment,Q of the source. Thus

hij =2G

3rQij (15.7)

The quadrupole moment is defined asQij ≡∫d3xT00(3xixj−δijx2). The expres-

sion found in (15.7) is useful in calculations of gravitational waves from varioussources. We also notice by comparing the equations (15.6) and (15.7) that thequadrupole moment simply is the traceless moment of inertia of the mass/energydistribution.

15.2 Emitted energy and momenta

We would now like to find out just how much energy and momenta that is beingemitted in the gravitational radiation when TT-gauge is applied. These quantitiesare both given by the Isaacson-tensor of energy and momenta, tµν as found inlecture 13

tµν =1

32πG〈∂µhij∂νhij〉.

The energy density of the radiation is given by the zeroth component of the Isaac-son tensor

t00 =1

32πG〈hij hij〉

=1

32πG〈h2ij〉

By applying the equation in (15.7), an expression for emitted energy, − dEdt , can be

derived

−dEdt

=G

45〈

...Qij

2〉 (15.8)


15.3 Detecting gravitational waves

Gravitational radiation and the various attempts and set-ups to detect it is a topicreceiving an increasing interest from physicists. The radiation is predicted in Ein-stein’s General Theory of Relativity, but is yet to be properly investigated throughexperiments. It is common to differ between indirect and direct observations ofgravitational radiation.

15.3.1 Indirect observations

The indirect observations are in a sense more vague than the direct ones, but there-fore also easier to perform. Gravitational radiation has been detected indirectly bylooking at binary pulsars. One example is the binary pulsar system PSR1913+16.In this system two stars are orbiting each other. The way the gravitational radiationhas been detected, is that the orbiting period of the pulsar has been detected over atime interval of more than 20 years, and the period is observed to be changing withtime. This is due to energy loss which can be explained as emitted gravitationalwaves. The first to observe these effects were Joseph H. Taylor Jr. and Russel A.Hulse who measured the orbiting time between the seventies and the early nineties.The observed values corresponded very satisfactory to Einstein’s predictions, andtheir long effort eventually paid off when receiving the Nobel prize in physics,1993.1

15.3.2 Direct observations

The main interest today is to try to measure gravitational radiation directly. Thisis among other places done in large observatories in the USA. The set-up is quitesimilar to that legendary of Michelson and Morley with detectors perpendicular toeach other and of the size L ∼ 3km in each direction. When gravitational wavesenter the detectors, there will be a small shift

L → L(1 + h) (15.9)

From the interference pattern, this shift, h can in principle be detected and wouldbe a confirmation that gravitational waves are present. However, to perform suchan observation, a lot of technical problems, such as disturbing noise from the sur-roundings have to be solved. To get a feeling with just how fine tuned the ob-servationally set-up has to be, let us estimate the magnitude of the shift, h due togravitational radiation from a binary system analog to the situation of indirect de-tection. A sketch of such a system is shown in figure (15.2). For simplicity, weassume the two stars in the binary system to be of equal mass, M .

1There is more to read, and satisfactory plots to view in the article written by Taylor, “Binarypulsars and relativistic gravity”. Rev. Mod.Phys. 66, 711-719 (1994). Of specific interest is figure10 in this article.

15.3. DETECTING GRAVITATIONAL WAVES 151

M

M

Figure 15.2: Detecting gravitational radiation directly: A binary system with equalmass, M causes gravitational radiation.

From (15.7) we find that

hij ∼G

rQij ∼

GM

rP 2b

(GMP 2b )2/3

≈ 10−21

(M

Msun

)5/3(1h

Pb

)2/3 100pc

r. (15.10)

Here, Pb is the period of the double-star system, Msun is the mass of the sun, pc isparsec and h is one hour. So for gravitational radiation emitted from a system of thesun mass, with orbiting time of about 1h and with distance to the observer of a fewhundred parsec (typically for galaxies?), the shift to be observed is of magnitude10−21. This is an extremely small number, but still seems to be the best we canhope for at the time being. (Actually, a lot of systems give rise to gravitationalradiation smaller than this). If we now look at the actual shift in length, δL, we findwith L ∼ 103m and h ∼ 10−21 that the scale of the observations is of magnitude

δL = Lh = 103 × 10−21 = 10−16cm

This length is about 11000 of the radius of an atom. In order to detect gravitational

radiation, the set-ups thus involves scale smaller than atoms. The sensitivity insuch measurements is in (near?) future expected to be as good as 10−23, so that thewaves eventually can be detected.


15.4 Gravitational radiation in a RW metric

We have earlier investigated perturbations of the Minkowski metric, ηµν when con-sidering gravitational radiation. The perturbed metric was then written as the flatMinkowski metric with a small perturbation, hµν on top

gµν = ηµν + hµν (15.11)

This is the simplest possible case of perturbations of a metric. In order to proceedwith cosmology, the Minkowski metric is replaced by the Robertson-Walker (RW)metric. The RW metric is a much better choice than the flat Minkowski metricas the background metric in cosmology and is still quite easy to deal with. Theprocedure will be similar to the one we saw for the case of a flat metric in the twoprevious lectures. The perturbations will cause tensor fluctuations on the metric.These fluctuations will be of great importance when the lectures proceed with in-flation. We will also see (in the end of this lecture) that the tensor fluctuations arevery similar to the simpler scalar fields and that the tensor fluctuations can be founddirectly from the scalar fluctuations simply by scaling a prefactor. This will be themain topic in the rest of this lecture and we will also derive the two Friedmannequations. There will be quite a lot of mathematics (and even more omitted) on ourway, but please hang in.

15.4.1 Tensor fluctuations

The metric that we will use now with the RW background is

ds2 = dt2 − a2(t)dx2 = a2(τ)(dτ 2 − dx2) (15.12)

Here, we have applied conformal time, τ in the last term. Conformal time is definedby dτ = dt

a . (One of the advantages of conformal time, is that with such a choice ofcoordinates, light always travels at 45 degrees on a space-time diagram, regardlessof the behavior of the scale factor). The metric we are going to use is the RW:

gµν = a2(τ)ηµν (15.13)

The perturbed metric, which we will investigate further is then found by adding asmall term, hµν to the Minkowski metric.

gµν = a2(τ)(ηµν + hµν) (15.14)

The perturbations in the metric can be interpreted as the tensor fluctuations that weare interested in. The equation of motion of the tensor fluctuations is now foundby means of the Einstein-Hilbert action

S = − 1

16πG

∫d4x√−g(R+ 2Λ). (15.15)

15.4. GRAVITATIONAL RADIATION IN A RW METRIC 153

Here, R is the perturbed Ricci scalar. To proceed, we will need the scalar curvature,R. It is found from the Ricci-tensor, Rµν . Before we calculate the Ricci-tensor (andthen in turn the Ricci-scalar), let us just remind ourselves of the Riemann-tensor,which is the basis for all calculations in this matter.

Rµναβ = ∂αΓµνβ − ∂βΓµνα + ΓρνβΓµρα − ΓρναΓµρβ (15.16)

For the interested reader, there will be some short comments on the similaritiesbetween the Riemann tensor and gauge-fields in Appendix A. The Ricci tensor isnow calculated, also leading to the two Friedmann equations. These results will beneeded to continue with the tensor-fluctuations, but first, let us calculate the Riccitensor.

15.4.2 Calculating the Ricci tensor

From its definition, the Ricci tensor is

Rµν = Rαµαν = ∂αΓαµν − ∂νΓβµβ + ΓαµνΓβαβ − ΓβµαΓαβν (15.17)

From the above expression, we see that if we find the Christoffel symbols, Γ, thenthe Ricci tensor is also found. The contributing (non-vanishing) components of theChristoffels are as follows

Γ000 =

a′

a≡ H (15.18)

Γ0ij = Hδij (15.19)

Γi0j = Hδij . (15.20)

Here, we have defined the conformal Hubble parameter, H and the primes denotedifferentiation with respect to conformal time, such as a′ = da

dτ . If the Γ’s areinserted in the expression for the Ricci tensor in (15.17), we find the componentsof Rµν to be

R00 = −3H′

⇒ R00 = g00R00 = − 3

a2H′ (15.21)

Rij = (H′ + 2H2)δij

⇒ Rij = giiRij = − 1

a2(H′ + 2H2)δij . (15.22)

In the above calculations, we have used the metric gµν = 1a2(τ)

to raise one index.The Ricci scalar is therefore

R = R00 +Rii = − 6

a2(H′ +H2). (15.23)

We eventually want to find the perturbed scalar, R in the expression of the actionfor the tensor fluctuations. This scalar will be found in a minute, but since we areso close to deriving the Friedmann equation, let us allow us to do just that beforewe proceed.


15.4.3 The Friedmann equations

The Friedmann equations can now easily be found from the Einstein equationsEµν = 8πGTµν . The first equation is found by applying the zeroth component(the energy density) of the energy-momentum tensor and using the definition ofthe Einstein tensor

E00 ≡ R00 −1

2g00R = 8πGT00 = 8πGρa2

⇒ −3H′ − 1

2a2

[− 6

a2(H′ +H2)

]= 8πGρa2

⇒H2 =8πG

3ρa2 (15.24)

which is the first of the Friedmann equations. The second Friedmann equation isderived from the spatial components of the Einstein tensor. From these we find that

H2 + 2H′ = −8πGpa2

If the expression above is combined with the first Friedmann equation, then thesecond follows directly as

H′ = −4πG

3(ρ+ 3p)a2 (15.25)

We will now proceed with calculating the curvature scalar, R of the fluctuatingmetric, gµν . This scalar is one of the terms in the Einstein-Hilbert action that wesaw in (15.15) and given by

S = − 1

16πG

∫d4x√−g(R+ 2Λ). (15.26)

The perturbed metric can be written as

gµν = a2(τ)gµν (15.27)

where gµν = ηµν + hµν . In the expression above the perturbed Ricci scalar,R isfound by some neat formula when gµν is known. Ingunn is the expert with thisformula and it looks like

R =R

a2− 6

2a

a3(15.28)

The (Einstein-Hilbert) action is now calculated to second order in the perturbations.From equation (15.15) we see that we also need to calculate

√−g. This is wherePetter Callin comes in (again). By the fact that log detM = Tr logM , the termcan be found 2

√−g = a4

[1 +

1

2h+

1

8h2 − 1

4(hαβ) + . . .

](15.29)

2For the complete calculation, see the Master thesis of Petter Callin


In order to find the physical degrees of freedom and simplify the expression, weassign the TT gauge. By this choice of gauge, the trace is zero, h = 0, and allthe zero components are also zero, hµ0 = 0 as seen earlier. The formula for thecurvature scalar in (15.28) involves the “box” -or d’Alembertian operator 2. Thisoperator was defined in (11.15) as

2 =1√−g ∂µ

(√−g gµν∂ν)

In a flat background, the box operator acting on the scale factor is 2a = a′′ sincethe scale factor only depends on the time 3.

If the expressions of R and√g are inserted in the action integral and if we do

remember the constraints in the TT-gauge (h=0, for instance), we find

S = − 1

16πG

∫d4xa4

(1− 1

4h2αβ

)[R

a2− 6

a′′

a3+ 2Λ

]. (15.30)

The second term in the brackets is actually equal to the background Ricci scalarthat we found earlier in (15.23), that is

Rbackground = R = − 6

a2(H′ +H2)

This will be shown in the following. The derivative of the conformal Hubble para-meter with respect to conformal time is

H′ =d

dτ

(a′

a

)=a′′

a−(a′

a

)2

=a′′

a−H2 (15.31)

and

R = − 6

a2(H′ +H2) = − 6

a2

a′′

a

= −6a′′

a3(15.32)

This is the same term that we saw in the action in (15.30). This background Ricciscalar is the one fluctuating and can be expressed as

R = hαβ2hαβ −1

2hαβ,νh

νβ,α +3

4hαβ,νh

αβ,ν

3According to calculations done by Mr. Callin, the box-operator also contains terms dependingon hµν . Hence, the calculation involves additional terms and is more complicated. However, the neteffect of these extra terms sum up to zero, so the result is the same in both situations, and since weare most interested in the result, we proceed with clean conscience although Petter most probably isright.


If we insert for R in the action, S, the result will be

S = − 1

16πG

∫d4xa2

[hαβ2hαβ + . . .

]

The action can be rewritten into a more convenient form in order to state our point,namely the similarity between tensor fluctuations and scalar fluctuations. This isdone by partial integrating, applying the two Friedmann equations and write it inthe TT gauge. The action of the tensor fluctuations then looks like

S =1

32πG

∫d4xa2 1

2

(∂µhij

)2 (15.33)

We notice that the action integral of the tensor fluctuations is of the same form asthat of a massless scalar field that we have seen many times before. To illustratethis point and investigate even further, let us take a look at a massless scalar fieldin a RW metric.

15.4.4 Scalar field in the RW metric

The action of the scalar field from which we in turn find the classical Euler-Lagrange equations of motion is given by

S =

∫d4x√−gL

=

∫d4x√−g1

2gµν∂µφ∂νφ

=

∫d4xa4 1

2

1

a2ηµν∂µφ∂νφ

=

∫d4xa2 1

2ηµν∂µφ∂νφ (15.34)

which should be compared with (15.33). The similarity between the tensor fluc-tuations and the scalar ones can be very helpful in calculations since now a tensormode, hij can be translated into a scalar mode, φ and vice versa by rescaling theprefactor of the integrals. Thus, the tensor fluctuations can be found directly fromthe scalar fluctuations. This will be very useful when we proceed to inflation andwant to find the tensor fluctuations then. This kind of “duality” also applies to theexpectation values of the energy-momentum tensors in the two situations. For ascalar field the energy-momentum tensor is (Here, the upper φ is not an index, butdenotes the Tµν of the scalar field)

T φµν = ∂µφ∂νφ−1

2gµν(∂λφ)2

The expectation value is found, remembering that the last (surface) term now van-ishes due to periodic boundary conditions (As seen in equation (14.21) in lecture14).

〈T φµν〉 = 〈∂µφ∂νφ〉 − 0 (15.35)


Now, we can apply our important result to easily find the expectation value of theenergy-momentum tensor in the situation of tensor fluctuations simply by rescalingthe prefactor by 1

32πG . Hence:

〈T hµν〉 =1


This is the Isaacson relation in a RW metric.4

We will now end the lecture by looking at the classical equation of motion of ageneral massive scalar field with respect to conformal time, τ . Earlier we havefound this equation with cosmic time in (11.13) to be

φ+ 3Hφ− 1

a2∇2φ+ V ′ = 0 (15.37)

Now, we will see that the conformal edition is equivalent. By means of the principleof stationary action and varying the metric φ → φ + δφ, the variation principleyields

δS = 0 =

∫d4xa2(τ)(ηµν∂µφ∂νδφ−m2a2φδφ)

=

∫d4x

[− ∂

∂τ(a2φ′) + a2∇2φ−m2a4φ

]δφ

=

∫d4xa2

[− φ′′ +∇2φ− 2Hφ′ −m2a2φ

]δφ (15.38)

This is valid for all variations in the field, hence the brackets must be zero

φ′′ −∇2φ+ 2Hφ′ +m2a2φ = 0 (15.39)

This is the same equation of motion that we found in (12.34) and is seen by swap-ping from conformal to cosmic time.

Next lecture will give a phenomenological introduction to inflation and earlyuniverse physics.

4There may be a possibility that the partial derivatives, ∂ should be replaced by covariant deriv-atives in (15.36) . This is probably not so since, when applying the TT gauge one chooses a specificcoordinate system. So then if one changes coordinate system (which would motivate the need of acovariant derivative), these changes are most likely zero due to the choice of gauge. Since there stillseems to be some uncertainty about this topic, sources such as the original article by Isaacson or“Gravitation” by Misner et. al. should be consulted.

Chapter 16

Lecture 16

16.1 Cosmology

As this is a course in cosmological physics we shall go briefly through some basiccosmology before we start on inflation. The point here is to get an overview overthe most important concepts in modern cosmology.

16.1.1 Important equations and numbers

The Robertson-Walker spacetime is given by the line element


And Einstein’s field equations are

Eµν = 8πGTµν

Where the energy momentum tensor is

T µν =

ρ 0 0 00 p 0 00 0 p 00 0 0 p

and as we saw earlier the Einstein tensor can be found from the Christoffel symbolswhich are

Γ0ij = aaδij

Γk0j =a

aδkj

The Ricci tensor is thus given by

R00 = −3a

a

Rij = (aa+ 2a2)δij

159


which gives the Ricci scalar

R = −6

[a

a+

(a

a

)2]

This leaves us the Einstein equations. The time (00) component becomes(a

a

)2

= H2 =8πG

3ρ (16.1)

which we recognize as the first Friedmann equation. The spatial (ij) component is

2a

a+a2

a2= −8πGp

Combination of these two equations leads us to the second Friedmann equation

a

a= −4πG

3(ρ+ 3p) (16.2)

By differentiating the Friedmann’s first equation one can obtain the first law ofthermodynamics (for adiabatic expansion) as the equation ρ + 3H(ρ + p) = 0.This can also be found by covariant differentiating the energy-momentum tensor∇µT µν 1.

We assume the equation of state p = wρ, and by integrating the adiabaticequation we get that

ρ(a) = ρ0a−3(1+w)

If we assume that the scale factor evolves as a(t) ∼ tp where 0 < p < 1 and bycombining this with the Friedmann equations and integrating we get

a(t) =

(t

t0

) 23(1+w)

Here we let the scale factor today be a0 = a(t0) = 1. We have the Hubbleparameter

H0 =a

a

∣∣∣∣t0

=2

3(1 + w)t−10

If we assume matter-dominated universe (w=0) we get the age of the universet0 = 2

3H−10 . The ΛCDM model predicts the age t0 = 0.9H−1

0 . The value of the

Hubble parameter today is given as H0 = 100hkm/sMpc where h is measured to be

h ≈ 0.7. This gives

H−10 = 1010h−1yr =

10

hGyr ≈ 14Gyr

In the ΛCDM model this will give the age t0 ≈ 13Gyr. If we insert the speed oflight c = 3 · 105 km

s we get the distance 3h103Mpc = 4200Mpc. This gives a clue

about the size of the observable universe.1Calculating this would be a good exercise.

16.1. COSMOLOGY 161

δθ

t0

Dp(t0)

δLp ≈ 100Mpc

Figure 16.1: A supercluster far away

16.1.2 The horizon and cosmological distances

Let us look at a photon emitted from a galaxy at time te that is observed here,on earth at time to. Light follows lightlike worldlines, so ds2 = 0 which givesdt = a(t)dr. Integration gives us

∫ ro

re

dr =

∫ to

te

dt

a(t)

The physical distance to the galaxy is then

Dp(to) =

∫ to

te

dt

a(t)

That makes the physical distance to a source that emitted a photon at time te = 0

Dh(t0) =

∫ to

0

dt

a(t)

which is the distance to the horizon. It will be of order of magnitude Dh(t0) ∼ 1H0

,

where 1H0

= DH(t) is called the Hubble radius. If we assume a matter dominated

universe where a(t) =(tt0

) 23

we get the size of the observable universe

Dh(t0) = 2H−10 = 104Mpc

To get an idea of cosmological distances we can remember that the size ofa galaxy is approximately 0.1Mpc = 100kpc ≈ 300000ly, the size of a galaxycluster is about 30 times as large 0.1Mpc × 30 = 3Mpc. Superclusters are againabout 30 times as large as a cluster, 3Mpc× 30 = 90Mpc ∼ 100Mpc.


We must remember that e.g. superclusters evolve as the light is traveling to-wards us. If the “angular size” of a supercluster (or other structure) is δθ we havethat

δθ =δLpDp(tp)

=(1 + ze)δL0

Dp(t0)=

δL0

DA(t0)

Where ze is the redshift and δL0 is the size of the galaxy at the time of emission.We have here introduced the angular distance DA(t0) = 1

1+zeDp(t0).

16.1.3 The early universe

The early universe was dominated by radiation and the energy density was givenby

ρ = ρr =π2

30T 4g

where g is the degrees of freedom of the particles, and is derived from the particlephysics. The universe expanded as a(t) ∼ t

12 . This gives the Hubble parameter

H = aa = 1

2t . And from the first Friedmann equation we get

H2 =8πG

3ρ

1

4t2=

8πG

3ρ

t2 =3

32πGρ

and if we insert the expression for ρ the relation between temperature and time(and degrees of freedom) is

(tT 2)2 =45

16π3Gg

If we insert all the constants we get the practical formula

tT 2 =2.42√g

Where t is expressed in seconds, and T is expressed in MeV.The factor g must of course be found. This varies as particles gains mass. When

the universe was one second old, and the nucleosynthesis started, g consisted of thedegrees of freedom for the photon (2), three types of massless neutrinos and theirantiparticles (3 · 2) and the electron and positron with spin up and down (2 · 2), thatis

gNS = 2 +7

8(3 · 2 + 2 · 2) =

43

4

Here the factor 78 comes from statistical mechanics as neutrinos and electrons (and

positrons) follows Fermi-Dirac distribution.

16.2. INFLATION 163

1015GeV

tGUT tEW tQCD

10−5s 1s

tNS

104yr

trm tLS t0

1010yr

103GeV 200MeV 1MeV

T

t

Figure 16.2: timeline

Even earlier, at T > 1TeV all the particles (almost) were massless, and g wasmuch higher. We had in addition gluons with different spin (8 · 2), the W ± andZ0 bosons with three degrees of freedom each (they are still massive), the Higgsboson (1), the six quarks and their antiparticles with color and spin (6 · 3 · 2 · 2) andwe also get the µ and τ particles. That gives us g

g = 2 + 8 · 2 + 3 · 3 + 1 +7

8(6 · 3 · 2 · 2 + 3 · 2 · 2 + 3 · 2) =

427

4≈ 100

If one extrapolates the coupling constants of the elementary forces one expectthat they will meet at a point around 1015GeV (figure 16.3) and the physics willbe described by a grand unifying theory (GUT). At even higher energies (Planckenergy, 1018GeV) this too will break down, and we will need an universal theoryof everything (TOE).

16.2 Inflation

Inflation was introduced by Alan Guth in 1981 and later refined by Andrej Linde.It solves to important problems in cosmology, namely the flatness problem (why isthe universe so extremely flat?) and the horizon problem (Why is the universe soisotropic when it is causally isolated by horizons?).

The horizon problem can easily be seen if one looks at our local part of the uni-verse (the observable universe). If we go backwards in time we see that it shrinks(matter dominated, a ∼ t2/3, and earlier radiation dominated a ∼ t1/2), but itshrinks slower than the horizon (H−1 ∼ t) (figure 16.4). This means that at anearlier time some of our local universe was outside the horizon. Yet when we seeit today it is surprisingly isotropic even if it has never been in causal contact.

Now if we introduce inflation as a vacuum dominated epoch in the early uni-verse where the universe expanded exponentially, a(t) ∼ eHI t. If we then plot thescale of our local universe in comoving coordinates (Lp = 1

aLp so that the size


10

30

100

102 10151018

E(GeV)

TOEGUT

α−13

α−11

α−12

Figure 16.3: At 1015GeV the coupling constants are of equal size and we get GUT-physics

t

H−1

104

102

100

10−2

∼ t

∼ t2/3

∼ t1/2

trm

Figure 16.4: It looks like all our observable universe cannot have been in thermalcontact until today.

16.2. INFLATION 165

∼ t1/3

∼ t1/2

∼ e−HI t

t

(aH)−1 Mpc

104

102

100

10−2

trmtfti

Figure 16.5: With inflation the local universe can have been in thermal contactbefore.

of the comoving horizon is (aH)−1) together with our current (constant) horizonwe see that they cross at an earlier time (figure 16.5). This makes it possible fordifferent parts of the universe to have been in contact before.

To have inflation we must have

ddt

(1aH

)< 0

⇒ ddt

1a = − a

a2 < 0

⇒ a > 0

And if it is supposed to solve the horizon problem we must demand that the localuniverse is within the horizon before the inflation starts. That is

(1

aH

)

ti

> H−10

and this gives us how much the universe must expand during inflation

eHI(tf−ti) ≥ e55

Chapter 17

Lecture 17

Last lecture we saw why we need inflation. In this lecture we will see how wecan produce an inflationary epoch in the very early universe using a classical scalarfield and the so-called slow-roll approximation (SRA). The model we will study iscalled “chaotic inflation”. It was first introduced by A. Linde in 1983, and although(or should we say ’because’?) it is simple, it seems to explain most of propertiesof the universe that we observe today.

At the end of the lecture we will study quantum perturbations on the top of thisscalar field and see how these fluctuations evolve.

17.1 Inflation driven by a classical scalar field

We remember from last lecture that inflation is defined to be an epoch with acceler-ated expansion, that is a > 0. Assuming an equation of state relating pressure andenergy, p = wρ we have from the second Friedmann equation (16.2) that inflationrequires w < − 1

3 . Standard models of inflation has w = −1.

The scalar field φ that is responsible for inflation in our model is called aninflaton and has the well-known scalar field Lagrangian

L =1

2gµν∂µφ∂νφ− V (φ) (17.1)

Early on many thought that this scalar field was the Higgs field, but that idea hasbeen abandoned. The Higgs field has too many constraints from the standard modelof particle physics that doesn’t fit into the requirements of an inflaton field. Wedon’t know what the inflaton field is and where it comes from, but it could explain alot in a simple way, so we assume that it is there and try to deduce some observablesthat might confirm its existence.

In lecture 11 we showed that the pressure (11.28) and energy density (11.29)

167


in a scalar field is given by

ρ =1

2φ2 + V (φ)

p =1

2φ2 − V (φ)

where V (φ) is a potential. Here we have omitted terms on the form 16a2 (∇φ)2

that will vanish rapidly as a grows. For inflation we must have w < − 13 which

corresponds to

1

2φ2 − V < −1

3

(1

2φ2 + V

)

2

3φ2 − 2

3V < 0

V > φ2

For a scalar field the first Friedmann equation (16.1) will take the form

(a

a

)2

=1

3M2P

(1

2φ2 + V

), MP =

√1

8πG≈ 1018GeV (17.2)

In a RW metricds2 = dt2 − a2dx2

we have the equation of motion for a scalar field given in lecture 11 (11.13):

φ+ 3Hφ+ V ′ = 0 (17.3)

The first Friedman equation and (17.3) are coupled and with a given potential theymay be integrated numerically 1. Such an integration will include a choice of initialconditions for φ and φ. Hopefully the result will look something like figure 17.1where one has an “attraction pool” around the initial state such that the evolutionon the field is not critical dependent on the initial conditions. Hopefully, in thisattraction pool we will have that V φ2, such that the SRA (soon to be defined)can be used.

17.1.1 The slow-roll approximation

Solving the above equations is non-trivial. The most common approximation todo is the slow-roll approximation (SRA). In this approximation φ is changing veryslowly, such that V φ2 and φ ≈ 0. Inserted in (17.3) this gives the equation ofmotion

3Hφ+ V ′ = 0 (17.4)

1Gorm will do this in his master thesis. Ask him for results next year.

17.1. INFLATION DRIVEN BY A CLASSICAL SCALAR FIELD 169

attraction pool

φ

φ

φ ≈ const.

Figure 17.1: Hopefully a numerical solution of the equations will show a behaviorlike this. Here the system is not depending critically on the initial conditions, andfor a long time φ is almost constant.

An arbitrary chosen potential will of course not satisfy the above conditions. Thepotential must be sufficiently large and have a small gradient. From (17.4) we havethat φ must satisfy

φ = − V ′

3H

Under the assumption φ2 < V the first Friedmann equation (17.2) gives

H2 =V

3M2P

(17.5)

and we have (V ′

3H

)2

< V ⇒ 1

3

(V ′

V

)2

< M−2P

So a constraint on V is that V ′ cannot be too big to have inflation in the SRA. Thenit is natural to define a slow-roll parameter

ε ≡ 1

2M2P

(V ′

V

)2

(17.6)

We say that we have inflation when ε < 1, and that inflation ends when ε = 1.This assumed V -domination will give an equation of state parameter w = −1.

From solving the Friedmann equations we know that such an equation of state willgive an exponential evolution of the scale factor, a ∼ eHt. This is not exactlycorrect since H is not a constant, but it works well in a low order approximation.


17.1.2 Example with SRA using a quadratic potential

For some choices of V the SRA equations may be solved exactly. The most com-mon example, which also is frequently used by Linde, is

V =1

2m2φ2

V ′ = m2φ

ε =1

2M2P

m2φ12m

2φ2= 2

(MP

φ

)2 !≤ 1

The last condition tells us that φ ≥√

2MP . This looks nasty. To be able to useSRA we need a field larger than the Planck mass, and in this range it seems a bitsuspicious to use a classical field. What saves us is that even if φ > MP , theenergy density does not have to be that large. The energy density in SRA is givenby

V =1

2m2φ2 ≥ m2M2

P = M4P

(m

MP

)2

< M4P

if m MP . So if the mass-term of the field is not too big, we avoid the Planckproblem, and we may safely use classical gravitation.

With this potential (17.5) yields

H2 =1

3M2P

V =1

6M2P

m2φ2

⇒ H =

√1

6

m

MPφ (17.7)

Inserting this into (17.4) gives

3

√1

6

m

Mpφφ+m2φ = 0

⇒ φ = −√

2

3mMP (17.8)

That is, the velocity of the field is constant (not that strange actually, since wealready have assumed that φ ≈ 0). Integration gives

φ(t) = φi = −√

2

3MPmt (17.9)

Now, let us study how the scale factor evolves. We have

d

dφln a =

ddt ln a

dφdt

=a

aφ=H

φ

17.2. FLUCTUATIONS OF φ 171

which is a very useful equation. Inserting for H and φ we have

d

dφln a =

√16mMP

φ

−√

23mMP

= − φ

2M2P

lnafai

=

∫ φf

φi

dφφ

2M2P

= − 1

4M2P

(φ2f − φ2

i

)

=1

4M2P

(φ2i − 2M2

P

)=

φ2i

4M2P

− 1

2(17.10)

Since we know that the scale factor will grow exponentially, we may define aN by

afai≡ eN

where N is called the number of e-foldings given by the inflation. We know thatwe should have N > 55 for the inflation to solve the problems that we constructedit to solve. Then we must have

lnafai

=φ2i

4MP− 1

2= N ≥ 55

φi = 2√NMP ∼ 14MP (17.11)

17.2 Fluctuations of φ

Now we will study how fluctuations of the inflaton will be affected by the inflation.Much of the calculations involved have been performed earlier in this course inmore detail than we will do here.

We assume a fluctuating inflaton on the form

φ(x, t) = φ0(t)︸︷︷︸classical evolution

+ δφ(x, t)︸︷︷︸fluctuation

(17.12)

We know from lecture 11 (11.13) that a scalar field in RW-metric will obey

φ+ 3Hφ− 1

a2∇2φ+ V ′ = 0

Inserting (17.12) gives

φ0 +∂2

∂t2δφ+ 3H(φ0 +

∂

∂tδφ) − 1

a2(∇δφ) + V ′(φ0 + δφ) = 0 (17.13)


when remembering that φ0 only depends on time. For a scalar field we have thewell-known Lagrangian

L =1

2(∂µφ)2 − V (φ)

The potential V has a minimum for φ = φ0 = 0, so we may expand V like

V (φ) = V (0) +1

2V ′′φ2 + . . . (17.14)

and we have that

L =1

2(∂µφ)2 − 1

2V ′′φ2 + . . .

We see that the second term in the Lagrangian has the form of a mass term withm2φ = V ′′, so the second derivative of the potential may be interpreted as an effect-

ive mass term. So for the last term in (17.13) we have

V ′(φ0 + δφ) = V ′(φ0) + V ′′(φ0) + δφ

= V ′(φ0) +m2φδφ

Subtracting the classical φ0-terms from (17.13) we are left with an equation for thefluctuations given by

(∂2

∂t2− 1

a2∇2 + 3H

∂

∂t+m2

φ

)δφ = 0

And as always when working with a field, we assume a finite volume and expandit in Fourier modes:

δφ(x, t) =

√1

V

∑

k

ϕk(t)eik·x

where k is a comoving wave number. Since the physical distance is given by ax,the physical momentum is p = 1

ak, Performing the same steps as when deriving(11.13) in lecture 11, we obtain

ϕk + 3Hϕk +

(k2

a2+m2

φ

)ϕk = 0 (17.15)

This is basically the same equation of motion as obtained for a scalar field inRW-background earlier, so the fluctuations will follow the same equation as thebackground scalar field. The only difference is the mφ-term which really is not aconstant. But to the lowest order we set this mass-term to be mφ = 0. To motivate

this, we can study the ε-parameter ε = 12M

2P

(V ′V

)2. Assuming ε ≈ constant we

have

V ′2 = ε2

M2P

V 2



2V ′V ′′ = ε4

M2P

V V ′ ⇒ ε ≈M2P

V ′′

V

This motivates the introduction of a new parameter

η = M2P

V ′′

V

so V ′′ = η VM2P

, and with a quadratic potential we had H2 = 13M

2PV . This gives us

V ′′ =3M2

PH2

M2P

= 3ηH2

and we have that m2φ = 3ηH2 and η 1. Since mφ → 0 when η → 0 we may

omit the mass-term in (17.15) to the lowest order, and we have

ϕk + 3Hϕk +k2

a2ϕk = 0 (17.16)

Now we introduce conformal time τ given by

dτ =dta(t)

• If we now have power-law behaviour of the scale factor a(t) ∼ tp, then τwill evolve like

τ ∼∫

d tt−p =t1−p

1− pt ∼ τ

11−p , a(τ) ∼ τ

p1−p

• In the inflationary epoch we have a(t) ∼ eHt such that

τ ∼∫

dt e−Ht =e−Ht

H+ τC

a(τ) = − 1

H(τ − τC)⇒ a(τ) = − 1

Hτ

when we choose τC = 0. Applying the condition a > 0 for inflation, we seethat we will have inflation for τ < 0 and that inflation ends when τ → 0.

Now, in a similar way as we derived (12.34) in lecture 12, we can now write(17.16), using conformal time, as

ϕ′′k + 2Hϕ′k + k2ϕk = 0 (17.17)

where H = a′a and a prime as usual denotes differentiation with respect to con-

formal time τ . Then, again as in lecture 11, we rewrite the field as ϕk(τ) = 1auk(τ)

to get rid of the friction term. Equation (17.17) can then be written as


u′′k +

(k2 − a′′

a

)uk = 0 (17.18)

We have

a′ =1

Hτ2and a′′ =

−2

Hτ3

which givesa′′

a=

2

τ2

and thus we have that

u′′k +

(k2 − 2

τ2

)uk = 0 (17.19)

This equation may be solved using Bessel functions j1(kτ) and y1(kτ). The solu-tion then becomes

uk(τ) = Ake−ikτ

(1− i

kτ

)+Bke

ikτ

(1 +

i

kτ

)(17.20)

where Ak and Bk are Bogoliubov coefficients which we do not know, but thathave to be determined from some principle. (17.20) is the general solution fornon-vacuum fluctuations.

17.2.1 Quantum fluctuations

Here follows just a quick review on how the quantized fluctuations will behave.Quantum fluctuations will be on the form

δφ(x, τ) =1

a

∫d3k

(2h)3

[akuk(τ)eik·x + a†ku

∗k(τ)e−ik·x

]

Normalizing the modes we have the Wronskian

u′∗k uk − u∗ku′k = i

then one see at once that the a-operator has the usual canonical commutator:[ak, a

†k′

]= (2h)3δ(k − k′)

with a vacuum defined by ak|0〉 = 0. The final solution will also in this casedepend on the determination of the Bogoliubov coefficients Ak and Bk.

When working with fluctuations it can be very useful to have a figure like figure17.2 in mind. When we use comoving coordinates along the y-axis a fluctuationmode may be represented by a horizontal line, an it is easy to see how a specific


t0

(aH)−1

t

mode outside horizon

VD RD MD

tf teq

Figure 17.2: A nice figure to have in mind. In comoving coordinates the modes aresimply horizontal lines. Here, one of the modes re-enters the horizon in the matterdominated epoch (MD), while the other re-enters in the radiation dominated (RD)epoch. Notice how the vacuum dominated (VD) epoch ensures that all the modesare in causal contact early on.


mode leaves and enters the horizon. The comoving scale is λ = 2πk and a fluctu-

ation mode will be inside the horizon provided

k−1 = λ < (aH)−1 ⇒ k > aH

During inflation a = − 1Hτ , so in this epoch the mode is inside the horizon when

k > − 1

HτH ⇒ |kτ | > 1

since τ < 0 during inflation. Similarly a mode will be outside the horizon when|kτ | < 1 and pass through the horizon when |τ | = 1

k . Very early in the inflationepoch we have |kτ | 1, and (17.19) yields

u′′k + k2uk = 0

that is, we have effectively a free field. We choose an in- mode very early, e.g. we

set Bk = 0 and Ak =√

12k , and we have the solution

uk =

√1

2ke−ikτ

The√

12k -factor comes from the harmonic oscillator where we have

H =1

2p2 +

1

2ω2q2 , q =

√1

2ω(a+ a†)

This choice of solution is called a Bunch-Davies (BD) vacuum, and is the mostcommon solution to work with in inflationary cosmology.

Some people, e.g. Ulf Danielson, have commented that when enforcing sucha boundary condition very early you will get a physical momentum p = k

a whichis much larger than the Planck-mass, and thus you do not have a valid field theory.Ulf Danielson is demanding p < pP and uses boundary conditions that dependson k, such that the Bogoliubov coefficients will vary with k. That gives somesmall ripples on the CMB power spectrum. These ripples are too small to observewith the current WMAP-data, but they might be large enough to be detectable withfuture CMB-surveys like the Planck satellite.

Anyway, we continue using the BD-vacuum, and we study one mode:

δφk = ϕk(t)eik·x =1

auk(τ)eik·x

Inside the horizon we only have the Ak-term, which gives

|δφk| =∣∣∣uk

a

∣∣∣ =

√1

2k

1Hτ

=

√1

2k(Hτ)


log|δφk|

k2 > k1

logτ

k1

Figure 17.3: The quantum fluctuations freezes as they leave the horizon.

Outside the horizon we have both the Ak and Bk-part, which give

|δφk| =∣∣∣uk

a

∣∣∣ =

√12k

1kτ

1Hτ

=H√2k3

(17.21)

So outside the horizon the fluctuations will be constant in time, while the modeswill decay when inside the horizon (see figure 17.3). Notice from the figure howthe quantum fluctuations get “frozen” and behave like classical fluctuations whenmoving past the horizon.

Chapter 18

Lecture 18

18.1 Primordial spectrum

The primordial spectrum is due to quantum fluctuations and are given by the two-point correlation function. The potential, V we earlier in the last lecture Taylor-expanded in (17.14). This combined with the fluctuation in the scalar field as seenin (17.12) yields

V (φ0 + δφ) = V (φ0) +1

2V ′′(φ0)δφ2

︸︷︷︸Gaussian

+1

6V ′′′(φ0)δφ3 + . . .︸︷︷︸

non−Gaussian

(18.1)

The fluctuations in the equation above will in overall be Gaussian. From quantumfield theory we know that the higher order terms, that is of order 3 or higher giverise to interactions. (One may think of these interactions as some vertex in a generalFeynman diagram). The higher order terms will spoil the Gaussian shape of thefluctuations. Therefore it will often be important to know the size of these extraterms relatively to the Gaussian ones. In our case, however, we will neglect thecorrections and assume Gaussian fluctuations. We want to derive an expressionfor the power spectrum of the fluctuations in the field, φ. The power spectrum issometimes also referred to as a correlation function. To do so we will need thevacuum expectation value of the field operator φ(x).

18.2 Vacuum fluctuations of φ

We will now consider a free field in Minkowski space. To find the vacuum expect-ation value of φ, we need the operator itself (of a massless field).

φ(x) =

√1

V

∑

k

[akϕk(t)eik·x + a†kϕ

∗k(t)e−ik·x

]. (18.2)

Here, ϕk(t) =√

12ke

ikt and the frequency ωk =√

k2 +m2 = k since the field is

massless. The quantum field, φ fluctuates in a vacuum described by the Minkowski

179


metric so then φ0 = 0. The vacuum fluctuation of the scalar field is given by

〈φ2(x)〉 = 〈0|φ(x)φ(x)|0〉

We calculate this by applying the definition of the vacuum state, ak|0〉 = 0 and theconjugated vacuum 〈0|a†k = 0. Thus inserting for the field operator in (18.2), theexpectation value then takes form as

〈φ2(x)〉 =1

V

∑

k,k′〈0|ak′ a

†k|0〉ϕk′ϕ

∗kei(k′−k)x (18.3)

If we apply the canonical commutator relation that[ak′ , a

†k

]= δk,k′ , the above

expression can be written into a single sum

〈φ2(x)〉 =1

V

∑

k

∣∣ϕk

∣∣2

=

∫d3k

(2π)3

∣∣ϕk(t)∣∣2. (18.4)

The last step follows under the assumption that the integration volume is largeso that the sum can be very well approximated to an integral by the earlier seensubstitution 1

V

∑k →

∫d3k

(2πh)3 when V →∞. The power spectrum is now found

by writing the integral in spherical coordinates (d3k = 4πk2dk) and then justrewriting the expression slightly into the desired form. That is

〈φ2(x)〉 =

∫4πk2dk

(2π)2

∣∣ϕk(t)∣∣2

=

∫dk

k

k3

2π2|ϕk|2

︸︷︷︸powerspectrum

(18.5)

We now have the power spectrum or correlation function, ∆2φ(k) defined as

∆2φ(k) =

k3

2π2|ϕk|2

In some literature, only the last term, |ϕk|2 = Pφ(k) is referred to as the powerspectrum. We will use only one of them, namely

∆2φ(k) =

k3

2π2Pφ (18.6)

The term 〈φ2(x)〉 is not only a step in finding the power spectrum, but has alsoan interpretation itself. If we plot the power spectrum, ∆2

φ(k) as a function of the

18.3. INFLATION 181

ln k

∆2φ(k)

Figure 18.1: The figure shows some power spectrum as a function of the wavenum-ber, k. The total area beneath the curve of the power spectrum function is thevacuum expectation value 〈φ2(x)〉.

wavenumber, k, then the vacuum expectation value 〈φ2〉 is the total area beneaththe curve. This is so since from (18.5) we have

∫dk

k=

∫d ln k →

∫dk

k∆2

φ = 〈φ2(x)〉 = Tot. area

This is shown schematically in figure (18.1).The power spectrum in a Minkowski space-time can now easily be calculated

as

∆2φ(k) =

k3

2π2Pφ =

k3

2π2

1

2k

=k2

4π2(18.7)

To further find the vacuum expectation value of the scalar field, one has to integrateover all momentas (wavenumbers, see fig (18.1)) and this integral clearly diverges.The infinity can be removed by a regularization similar to the one seen in the Cas-simir situation, section (3.3).

18.3 Inflation

In the last lecture, we wrote the Fourier functions ϕk(τ) in terms of new mode-functions in order to get rid of the friction term in the equation of motion,ϕk(τ) = 1

auk(τ). This in turn lead to the conclusion that the fluctuations will


be constant with respect to time outside the horizon and time dependent inside.Outside the horizon the absolute value of the mode function therefore looks like(see (17.21))

|ϕk| =H√2k3

We are now considering inflation where the dynamics is described by a scalar fieldφ. We want to calculate the power spectrum in the inflationary epoch by applyingthe expression of the (scalar) fluctuations outside the horizon. In this manner oneshould notice that the result only is valid in the inflationary time interval [ti, tf ]since after tf the scalar field vanishes. (The usual way to think of this is thataround tf the scalar field dumps its kinetic energy driving the inflation process intoconventional radiation and matter. This is referred to as the reheating process).The power spectrum now is

∆2φ(k) =

k3

2π2|ϕk|2 =

k3

2π2

H2

2k3

=

(H

2π

)2

(18.8)

Thus, the power spectrum (outside the horizon) is independent of wave numberwhen H is constant. This is called Harrison-Zeldovich (HZ) scaling and is actuallyonly correct in the limit of infinitely slow rolling of the inflaton, φ. The reasonthat the HZ scaling is not quite exact, is due to the fact that modes with differentwavenumbers, k will leave the (comoving) horizon at different times. So the scalarpotential, V will be of different values when modes are crossing the horizon. SinceV ∼ H2 in the SRA, different values of V will give a power spectrum ∆2

φ(k)depending on the wavenumber. This is sketched in figure (18.2). However thiseffect is very small in the slow-roll approximation, since then the scalar potentialis assumed to be slowly changing. So to lowest order the HZ scaling is a very goodapproximation.

18.4 Metric fluctuations

We will now take a brief look at fluctuations in the metric and motivate why metricfluctuations are of interest in various subjects in cosmology.

18.4.1 Motivation

In inflation the energy-momentum tensor is dependent of the scalar field drivingthe dynamics. This is so since the Lagrangian, L contains φ and Tµν , as seen in(11.24), in turn depends on L. Hence, as the scalar field fluctuates as

φ→ φ+ δφ,

18.4. METRIC FLUCTUATIONS 183

t

Com.coordinates

k2

k1

Figure 18.2: Modes with different wavenumbers k1 and k2 leave the horizon atdifferent times, resulting in a k-dependent power spectrum. To lowest order, how-ever, the power spectrum is independent of wavenumber and the HZ scaling can beassumed.

then the energy-momentum tensor will be fluctuating

Tµν → Tµν + δTµν .

From the Einstein equation it is then clear that the Einstein tensor and thus also thegeometry will fluctuate. From

Eµν → Eµν + δEµν

one therefore finds an expression of the fluctuation of the Einstein tensor similar tothe Einstein equation as

δEµν = 8πGδTµν (18.9)

The fluctuation of the Einstein tensor will perturb the metric and so metric (tensor)fluctuations will be of great interest. The reason for writing tensor fluctuations inthe parenthesis, is that they will be the variants of the metric fluctuations of mostinterest in the following. The metric fluctuations are in general written as

gµν → gµν + δgµν

An important fact in this manner is that an arbitrary perturbation to the metriccan be expressed as a sum of scalar, vector and tensor fluctuations, depending onhow each of them behave under coordinate transformations. This is known as Thedecomposition theorem and yields

δgµν = δgtensorµν + δgscalar

µν + δgvectorµν


The three components of the fluctuation can be expanded in terms of sphericalcoordinates. They will be orthogonal to each other and can be solved separatelysince they evolve independently. That is, if some physical process in the earlyuniverse sets up tensor perturbations, these do not include scalar perturbations.So when we want to determine the evolution of for instance tensor fluctuations,we don’t have to worry about the two other components. It is also possible toshow that the vector modes usually do not contribute in inflation. This is so sincethere is no way to source such modes with a single scalar field. Hence, in simplemodels of inflation, the vector modes vanish. We are then left with only two of thecomponents and the easiest of them to calculate are the tensor modes. We will findthat the tensor fluctuations are on the same form as the power spectrum that wesaw in (18.8) in the case of inflation

∆2T ∼

(H

2π

)2

The similarity between tensor and scalar fluctuations was seen also in lecture 15.In a RW metric and with a TT choice of gauge the action integrals of tensor andscalar perturbations were proportional

S =1

32πG

∫d4xa2 1

2

(∂µhij

)2 ∼∫d4xa2 1

2(∂λφ)2

Knowing one of them, we can therefore easily find the other. We will also inthe following see that the tensor fluctuations, δgtensor

µν do not change during theexpansion of the universe. They can in principle be observed in the polarization ofthe CMB, but they are very small, much less than the scalar ones.

18.4.2 The fluctuations

When we now will study the metric fluctuations, we will use RW as the backgroundmetric. The perturbation to the metric will be parameterized and this new metricwe will split into the three basic components mentioned earlier. We will then in-vestigate some of these terms in more detail, also applying gauge transformationsto them. The RW background metric in terms of conformal coordinates is given by

ds2 = gµνdxµdxν

= a2(τ)ηµνdxµdxν

= a2(dτ2 − dx2)

= a2(dτ2 − δijdxidxj) (18.10)

A perturbation in the metric can therefore be written

gµν → gµν + δgµν = a2(ηµν + hµν)


where hµν is the common, small perturbation. In the following, we will only cal-culate to the first order in the perturbations and omit all higher order corrections.That is

gµν = a−2(ηµν − hµν + . . .)

In order to separate out the vector, tensor and scalar perturbations from the fluctu-ation in the metric, we will rewrite it by parameterizing the Minkowski metric, ηµνand the perturbation, hµν as follows

ηµν =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

hµν =

(2A BiBi −Cij

)

(18.11)

The hµν matrix is also a 4 × 4 matrix with 10 degrees of freedom (DOF). All theterms in this matrix are small and their respective DOFs are: A : 1, Bi : 3 andCij : 6 with ij running from 1 to 3. To find the three different components of themetric fluctuation, we rewrite the line element in terms of the parameterizations as

ds2 = a2[(1 + 2A)dτ 2 + 2Bidxidτ − (δij + Cij)dx

idxj]

A is a scalar since it has one DOF, but the Bi and Cij terms can be separatedfurther.We write out the Bi and Cij terms to find the tensor, vector and scalarconstituents

Bi = −∂iB + Vi. (18.12)

Here, the B term is a scalar potential and V is a pure vector. The expression aboveis actually the same as when a three-vector (Bi) is decomposed into a divergenceof a scalar field (∂iB) and a curl (Vi). From calculus, we know that a vector field(in the three-dimensional case) always can be separated into two constituents: Onewith zero divergence and the other with zero curl. The latter can therefore beexpressed as the gradient of some scalar field (B). The Cij term is given by

Cij︸︷︷︸6

= −2Dδij︸︷︷︸1

+ 2∂i∂jE︸︷︷︸1

+ ∂iEj + ∂jEi︸︷︷︸2

+ hij︸︷︷︸2

Here, the numbers beneath each term gives the DOF, and we see that it sums up to6. Also Cij is a symmetric tensor of rank 3, the D and E terms are scalars and theEi-field in ∂iEj +∂jEi is a symmetric, divergence free vector field. (Actually, onewould expect 3 DOFs for the term ∂iEj + ∂jEi, but due to the divergence relation∂iEi = 0, it reduces to only 2 DOFs). The last term, hij is the (spatially) tensorfluctuations constrained with the TT gauge hii = 0 and ∂ihij = 0. It is now timeto characterize the three constituents of the metric perturbations:


• The contributions to the vector perturbations come from the terms Vi and∂iEj + ∂jEi, but as we have mentioned earlier, these will decay rapidly asthe universe expands exponentially during inflation.

• The tensor perturbations are

ds2 = a2[dτ2 − (δij + hij)dxidxj ]

• The scalar perturbations are

ds2 = a2(1 + 2A)dτ 2 − 2∂iBdxidτ − [(1− 2D)δij + 2E,ij]dx

idxj(18.13)

18.4.3 Scalar perturbations

We will now consider the scalar perturbations in more detail. These perturbationsare very interesting since they are responsible for the structure of the universe, thatis, for the deviation from the FRW universe which is isotropic and homogenous.The line element of the scalar perturbations contains 4 scalar fields A, B, D andE. They all depend on space and time. In terms of the metric tensor, the scalarperturbations can be written as

g00 = a2(1 + 2A)

g0i = −a2∂iB

gij = −a2[(1− 2D)δij + 2E,ij ] (18.14)

This is found from the line element ds2 given in (18.13. We will later in thislecture see that two of the scalar functions can be omitted due to a suitable choiceof gauge. This gauge is called Newtonian gauge. We will now investigate how themetric changes under a coordinate transformation

xµ → xµ = xµ + ξµ(x)

We write ξµ = (ξ0, ξi). The components ξ0 and ξi are small perturbations of thesame order as the variables characterizing the perturbations. The component ξ i isa three-vector. By the same principle as seen in (18.12), it can be separated into ascalar part (a gradient) and a (transverse) vector part

ξi = ∂iξ + ξitr

This splitting will be very useful when we a little later are going to calculate howthe four scalar fields change under a gauge transformation. The metric now trans-forms as

δgµν → δgµν −∇µξν −∇νξµ (18.15)


where the ∇s are covariant derivatives. To find the perturbations in the metric, wewill need the Christoffel symbols and the contravariant ξµ. It is given by

ξµ = gµνξν = a2(ξ0, ξi)

when inserting for gµν . The covariant derivatives are found by

∇µξν = ∂µξν − Γλµνξλ

With the background metric defined as ds2 = a2(dτ2 − dx2) and the Christof-fel symbols given by Γλµν = 1

2gαλ[gαµ,ν + gαν,µ − gµν,α] the components of the

Christoffel symbols are

Γ000 = H

Γ0ij = Hδij

Γi0j = HδijLet us examine how the scalar part of the metric perturbation changes under such acoordinate transformation. We will consider the 00-perturbation, δg00. The othersare left as an excercise for the reader. From the equation in (18.13) we find that thetemporal components of the perturbation of the metric is

δg00 = 2Aa2

By means of the coordinate transformation, this component now is

δg00 → δg00 = 2Aa2

= 2Aa2 − 2(∇0ξ0)

= 2Aa2 − 2

[∂

∂τ(a2ξ0)−Ha2ξ0

](18.16)

where in the last step, the index of ξ is raised resulting in an a2 term. We furtherget

A = A− 2a′

aξ0 − ξ0′ +Hξ0

= A−Hξ0 − ξ0′ (18.17)

So this is how the scalar potential, A transforms in this coordinate transformation.In a similar way, we can find how the other components ( δg0i and δgij ) of themetric transform and then finally receive all the four scalar perturbations

A = A−Hξ0 − ξ0′

B = B + ξ0 + ξ′

D = D +Hξ0

E = E + ξ (18.18)


The equations above do not say how the four fields themselves change under the(general) coordinate transformation. This is so since the individual functions, Afor example, transform as a scalar and do not change under a spatial coordinatetransformation. We have transformed the metric and accommodated the resultingchanges in new definitions of A, B, D and E. This is not the same as seeing howA by itself changes under a transformation.

Although there are four functions characterizing the scalar fluctuations, thesecan be manipulated so that two of them vanish. In the coordinate (gauge) trans-formation, there are two parameters, ξ0 and ξ that can be chosen on a free basis.This is the choice of a gauge. For example, starting with a metric in which E 6= 0 itis trivial to make a transformation to eliminate E: simply choose ξ = −E resultingin E = 0. Hence, there are not four, but 4 − 2 = 2 functions which give physicalmeaning. Motivated by this fact one would expect there are two variables that areleft unchanged under a general coordinate transformation. Actually, there are, andJ. Bardeen was the first to identify them.

18.4.4 Gauge invariant variables

We will now in short look at the two potentials introduced by Bardeen. We willshow explicitly that one of the potentials is invariant and leave the other as anexercise. The potentials are

Φ = A+H(B −E ′) + (B −E′)′Ψ = D −H(B −E ′) (18.19)

That the second is gauge invariant, is seen by

Ψ → Ψ = D −H(B − E′)= D +Hξ0 −H[(B + ξ0 + ξ′)−E′ − ξ′]= D −H(B −E ′) = Ψ (18.20)

These invariants are very useful. If equations simplify in a particular gauge, thenone can do calculations in that gauge, form the gauge-invariant variables, and thenfinally turn these into the perturbations in any other gauge. We will probably dothat in some later lecture. We will now end this lecture by looking at differentchoices of gauge.

18.4.5 Choice of gauge

• Newtonian gauge. We already saw one example of the Newtonian gaugeearlier when the number of scalar functions was reduced from four to two.The most common is to eliminate E and B by transforming them to vanish.In that case, Ψ = D and Φ = A. The line element of the scalar perturbations


now takes form as

ds2 = a2(1 + 2A)dτ 2 − (1− 2D)δijdxidxj

= a2(1 + 2Φ)dτ 2 − (1− 2Ψ)δijdxidxj (18.21)

The reason why this gauge is called Newtonian is its similarity to the New-tonian limit in General Relativity: Consider a Schwarzschild line element inthe weak field limit given by

ds2 = (1 + 2φ)dt2 − (1− 2φ)dx2

with φ = −GMr . If now φ = Ψ = Φ, the line element of the scalar fluctu-

ations in this gauge is the Schwarzschild line element at large distances. Inthe limit of large distance to the mass M we get Newtonian physics. Thisgauge is sometimes also referred to as longitudinal gauge.

• Synchronous gauge. This gauge was first brought up by Lifshitz andsome of his students. It is often preferred when doing calculations withcomputers since the equations are better behaved numerically in this gaugewhere now B = A = 0. The line element then is

ds2 = a2dτ2 − [(1− 2D)δij +E,ij]dxidxj

• Comoving gauge. This gauge is defined so that the perturbation of thematter (energy) component of the energy-momentum tensor vanishes, that is

δT 0i = 0

This is so since the observer is comoving and hence has no velocity relativeto the matter.

In the coming lectures we will make the approximation Φ = Ψ in the Newtoniangauge. We then have established the metric in this gauge. What is left before wecan investigate the fluctuations in this metric, is to calculate the fluctuations of theEinstein tensor which we will need in equation (18.9) given by


Chapter 19

Lecture 19

19.1 Tensor fluctuations

Last lecture we considered scalar fluctuations of the metric. For tensor fluctuationswe had the metric

ds2 = a2[dτ2 − (δij + hij) dx

2]

and for scalar fluctuations

ds2 = a2[(1 + 2A)dτ − (1− 2D)dx2

]= a2

[(1 + 2Φ)dτ − (1− 2Ψ)dx2

]

where we have used the Bardeen potentials Φ = A and Ψ = D1, and chosenNewtonian (also called longitudinal) gauge.

We got a relation between fluctuations similar to the field equations

δEµν = 8πGδTµν

and we had the tensor fluctuations of the metric as

δgij = −a2hij ⇒ δgij = +1

a2hij

From this we can find the fluctuations of the Christoffel symbols, δΓλµν and fromthat we can find the fluctuations of the Ricci tensor δRµν and then we can find thefluctuations of the Einstein tensor δEµν . The Christoffel symbols are not tensors,but they can be expressed with only low components, Γµαβ = 1

2gµνΓναβ and this is

given asΓναβ = ∂αgβν + ∂βgαν − ∂νgαβ

this gives us the fluctuations

2δΓµαβ = δgµνΓναβ + gµνδΓναβ

1Be careful, some people use different definitions of Φ and Ψ.

191


whereδΓναβ = ∂αδgβν + ∂βδgαν − ∂νδgαβ

If we insert for the metric we get

δΓ0ij =

1

2

(h′ij + 2Hhij

)

δΓi0j =1

2hij′

δΓkij =1

2

(∂jh

ki + ∂ih

kj − ∂khij

)

The Ricci tensor is given from the Christoffel symbols as

Rµν = ∂αΓαµν − ∂νΓβµβ + ΓΓ− ΓΓ

so we get the fluctuations of the Ricci scalar

δRµν = ∂αδΓαµν − ∂νδΓβµβ + δΓΓ + ΓδΓ− δΓΓ− ΓδΓ

(We haven’t bothered writing out all the contractions in the last terms) In our casethe fluctuations of the Ricci tensor becomes

δRij = − 1

2a2

(h′′ij + 2Hh′ij −∇2hij

)

And then the fluctuations of the Ricci scalar, R = gµνRµν

δR = δgµνRµν + gµνδRµν =1

a2

(hijRij − δijδRij

)

In TT-gauge hµν should be traceless, and as Rµν ∝ hµν the last term must be zero.We also get that the the first term is zero, but this is less obvious. The result is thatthe Ricci scalar don’t fluctuate, δR = 0.

As the Einstein tensor is given by E ij = Rij− 1

2δijR, we find the fluctuations of

the this to be

δEij = δRij −1

2δijδR = δRij

Now we have derived the lefthand side of the perturbed Einstein equation, δEµν =8πGδTµν , so then it is natural to continue with rhs.

The energy-momentum tensor for a perfect fluidis given as

T µν =

ρ 0 0 00 −p 0 00 0 −p 00 0 0 −p

=

(ρ 00 T ij

)

and the fluctuations of the spatial components are

δT ij = −δpδij −Πij

19.1. TENSOR FLUCTUATIONS 193

where we have included the possibility for anisotropic stress through the term−Πi

j . We can then express the tensor fluctuations (for example caused by neut-rinos) through the field equations for fluctuations δE i

j = 8πGδT ij .

h′′ij + 2Hh′ij −∇2hij = 16πGa2ΠTij

Here we have used that δpδij = 0 since Rij is traceless. Weinberg found in 2004that the anisotropy would have only small effect, so we can neglect it, and we get

h′′ij + 2Hh′ij −∇2hij = 0 (19.1)

This is the wave equation for a free scalar field. This can be derived from the action

Sh =1

32πG

∫d4x

1

2a2 (∂µhij)

2 =

∫d4xa2 1

2(∂µφ)2 (19.2)

where the last equality is shown in chapter 15.

19.1.1 The power spectrum of the tensor fluctuations

We had the power spectrum for the scalar field (from the inflation), ∆2φ =

(H2π

)2.

From (19.2) we see that the relation between the scalar an tensor spectrums are

∆2T = ∆2

h(k) = 2 · 32πG∆2φ(k) =

8

M2p

∆2φ =

2

π2

(H

Mp

)2

where we have used the Planck mass M 2p = 1

8πG , and the factor 2 comes from thespin of the graviton. We see that also the tensor fluctuations are constant outsidethe horizon and that the result is independent of k. This is (again) an approximationthat is very good to the lowest order, since then HZ scaling is respected. This isverified by the following argument: In the Slow-roll approximation (SRA), H 2 ∼V . Thus, if the inflaton field V is slowly evolving, then H ∼ const. At the sametime, the fluctuations, when crossing the horizon, satisfy k−1 ∼ (aH)−1 (Again,see figure (18.2)). So to the lowest order, the tensor power spectrum is independentof k. Higher order terms will, however, give a k-dependent power spectrum. It isusual to write this as

∆2T (k) ∼ knT (k) (19.3)

so thatnT ln k = ln ∆2

T + . . .

or we can also write

nT =d(ln ∆2

T )

d(ln k)(19.4)


We now want to find nT as given in the tensor power spectrum in (19.3). As themodes cross the horizon, then k = aH . This gives

ln k = ln a+ lnH

Differentiation with respect to time gives

d ln k

dt=

d ln a

dt+

d lnH

dt=a

a+H

H

Where the first term is large and the last term is small due to the assumption thatthe Hubble parameter is (almost) constant in the de-Sitter like inflation dynamics.The last term is therefore ignored. As a

a = H we then get

d(lnk) = Hdt (19.5)

In lecture 17 we learned that in the slow-roll approximation we have

φ = − V′

3H=

dφ

dt

So for a general function F and applying (19.5) we then get

dF

d ln k=

1

H

dF

dt=

φ

H

dF

dφ= − V ′

3H2

dF

dφ= −M2

p

V ′

V

dF

dφ

In our case F = ln ∆2T ∼ lnH2 ∼ lnV and from equation (19.4) we find the

desired result

nT = −M2p

V ′

V

d lnV

dφ= −M2

p

(V ′

V

)2

= −2ε

so that∆2T ∼ k−2ε (19.6)

where ε 1 during inflation. If we are able to measure the power spectrum ofthe tensor fluctuations of the tensor fluctuations we can determine the form of theinflaton potential. But this is hard, as this power spectrum is much smaller than thepower spectrum of the scalar fluctuations so one must measure it very precisely.

We can assume that the universe expands as a(t) ∼ tp where p = 12 for a

radiation dominated universe, and p = 23 for a matter dominated universe. Using

conformal time

τ =

∫dt

a(t)=

∫dt

tp=

1

1− pt1−p

so we say that t ∼ τ1

1−p and thus

a(τ) ∼ τp

1−p =

∼ τ radiation dominated∼ τ2 matter dominated

19.1. TENSOR FLUCTUATIONS 195

∆2T

ττ0τf τ∗τeq

Figure 19.1: The tensor fluctuations decrease rapidly inside the horizon

The conformal Hubble parameter is H = a′a = 1

τ in both radiation and matterdominated universes (but H = − 1

τ in a vacuum dominated universe).In the next lecture, we will study the scalar fluctuations. We will then find a

similar equation of motion to (19.1). However, the situation will be a little morecomplicated due to the coupling between the scalar fluctuations and the inflatonfield.

Chapter 20

Lecture 20

In lecture 18 we split the metric perturbations into three parts, namely scalar, vec-tor and tensor perturbations. Vector perturbations decay quickly and are of littlecosmological interest. The tensor perturbations are the simplest ones to treat math-ematically, and we reviewed some of their properties last lecture. But the tensorperturbations are small, and even if they may tell us about some really interestingproperties of the early universe, they have no deep impact on the structures andanisotropies that we observe in the universe today. Those structures are caused bythe scalar perturbations.

20.1 Scalar perturbations

In (18.21) we introduced the longitudinal (or conformal Newtonian) gauge givenby

ds2 = a(τ)[(1 + 2A)dτ 2 − (1− 2D)dx2

](20.1)

What we want to do now is to find the equations of motion (EOM) for the scalarpotentials A and D. Then, when having these EOM we may couple this perturbedmetric to the perturbations of the inflaton field, as found earlier. The procedurewhen treating the scalar perturbations is very similar to what we did last lecturewhen studying tensor perturbations. The perturbations to the RW-metric are givenby

δg00 = 2a2A

δgij = 2a2Dδij ⇒ δgij = − 2

a2Dδij (20.2)

This perturbed metric is then used to obtain the perturbed Einstein tensor in theperturbed Einstein equation

δEµν = 8πGδT µν

Using the perturbed metric to calculate the Christoffel symbols, the Riemann tensor,the Ricci tensor and the Ricci scalar in this gauge, one hopefully ends up with a

197


perturbed Einstein tensor looking like this:

δE00 =

2

a2

[∇2D − 3H(D′ +HA)

]

δEij = − 1

a2

[D′′ + (H2 + 2H′)A+H(A′ + 2D′)

]δij −

1

a2

[∇2(A−D)δij − ∂i∂j(A−D)

]

δE0i =

2

a2∂i(D

′ +HA) (20.3)

where a prime as usual denotes derivation with respect to conformal time τ .

20.1.1 Coupling the equations

For a scalar field we have the energy momentum tensor

Tµν = ∂µφ∂νφ− gµνL

which gives the perturbed energy momentum tensor

δTµν = ∂µδφ∂νφ+ ∂µφ∂νδφ− δgµνL− gµνδL (20.4)

The explicit calculation of the components of the perturbed energy momentumtensor will not be done here, but should be straightforward using the techniquespresented in this course. The results, inserted in the perturbed Einstein equations,yield

δE00 = 8πG

1

a2

(−Aφ′2 +A2V ′δφ+ φ′δφ′

)(20.5)

δEij = 8πG1

a2(Aφ′2 + a2V ′δφ − φ′δφ′)δij (20.6)

δE0i = 8πG

1

a2φ′∂iδφ (20.7)

where V is the potential. These equations combined with the expressions for theperturbation of the Einstein tensor given in (20.3), give us the three equations thatwe want to solve. They look pretty nasty, but solving them will give a lot of can-cellation of terms and the results will be rather simple. The explicit derivation,although very beautiful, will be left as an exercise, and only the results will bestated.

The first result comes from a mere inspection of (20.6). For i 6= j we haveδij = 0 and thus ∂i∂j(A−D) = 0 which gives the important result that

A = D

The reason for why (A − D) cannot be a constant is that the metric should re-duce to the Minkowski metric in the limit where A and D approach 0. Either can(A −D) possess a first order coordinate dependence, since that would violate ourassumptions of isotropy and homogeneity. In longitudinal gauge we know that the

20.1. SCALAR PERTURBATIONS 199

two gauge invariant variables are given by Ψ = D + [. . .] and Φ = A + [. . .] andthus Ψ = Φ, which is very useful to know.

Then, simply by inserting in the perturbed Einstein equation, we have

∇2D − 3H(D′ +HA) = 4πG(−Aφ′2 +A2V ′δφ+ φ′δφ′

)

(20.8)

D′ +HA = 4πGφ′δφ(20.9)

D′′ + (H2 + 2H′)A+H(A′ + 2D′) = 4πG(−Aφ′2 + a2V ′δφ− φ′δφ′)(20.10)

Here φ is the usual inflaton background evolving like

φ′′ + 2Hφ′ + a2V ′ = 0

Subtracting (20.8) from (20.10) and inserting for δφ from (20.9) one obtains

D′′−∇2D+2

(H− φ′′

φ

)D′+2

(H′ +Hφ

′′

φ

)D = 0 (20.11)

Now we will use this equation to study how the perturbations evolve outside thehorizon. During inflation we have

H = aH = eHtH , a(τ) = − 1

Hτ(20.12)

The Dk-modes are as usual found by the Fourier transformation

D(x, τ) =

∫d3k

(2π3)Dk(τ)eik·x

In first order perturbation theory the equations are linear and the modes evolveindependent of one another. This simplifies the equations enormously, but is alsoan extremely good approximation in the early universe. Using SRA as we did inlecture 17 we end up with an equation for Dk

D′′k + k2Dk +O(ε)D′k +O(ε)Dk = 0

When ε→ 0 in SRA the last two terms may safely be omitted. Studying k-modesoutside the horizon, also the second term becomes small, and the first term will bethe leading one, that is D′′k = 0. Integration yields

Dk(τ) = c1(k) + c2(k)τ

From (20.12) we have that −Hτ = e−Ht. Since we know that τ → 0 as inflationis approaching its end we must have c1(k) = const. and c2(k) → 0. That meansthat D = A = const. outside the horizon.


20.1.2 A gauge invariant curvature scalar

Now we want to find the magnitude of the A and D potentials. When τ is constant,the metric gives us a 3D hyper-surface with scalar curvature R(3) = 4

a2∇2D whereD is called a curvature perturbation. Our results should be expressed in a gauge-invariant way, but we remember that under a gauge transformation xµ → xµ =xµ + ξµ(x) we had

D → D +Hξ0

such that D is not a gauge invariant quantity. But it is possible to construct a gaugeinvariant curvature perturbation R given by D plus an additional part. To find thisR we first recall some of the properties of the scalar inflaton field φ. Since φ is ascalar field it transforms like φ(x) = φ(x). For the fluctuation δφ we have

δφ(x) = φ(x)− φ0(x)

= φ(x)− φ0(x+ ξ)

= φ(x)− φ0(x)︸︷︷︸δφ(x)

−ξµ∂µφ0

where φ0 is the background field. So δφ will transform like

δφ→ δφ = δφ− ξ0φ′0 = δφ− ξ0φ′

where the last equality simply is a change of notation. So we try to express ourgauge invariant curvature scalar like

R = D +Hδφφ′→ R = D +H δφ

φ′

= D +Hξ0 +Hφ′

(δφ − ξ0φ′) = R

SoR may safely be defined as

R ≡ D +Hδφφ′

(20.13)

This quantity is extremely important in inflationary cosmology. We will later showthat R is constant from the inflationary epoch until today, so it can be used as amessenger that brings us information from the very early quantum universe. Rcontains the inflaton field φ which is a meaningless quantity in the late universe.So after the inflation, R has to be redefined, but we postpone that for the nextlecture. What we will do now is to show that R is constant during inflation formodes outside the horizon. Using (20.9) we write R as

R = D + 2HM 2P

D′ +HAφ′2



R′ = D′ + 2M2P

[H′(D′ +HA

φ′2

)+H

(D′ +HA

φ′2

)′]

Calculating this takes a lot of spacetime, but after using (20.8)-(20.10) one is leftwith

R′ = 2M2P

(H∇2D

φ′2

)+

2M2P

φ′2

[φ′2

2M2P

+H′ −H2

]

︸︷︷︸=0

That the underbraced term vanishes one can see from the Friedmann equations.They say that

H2 =1

3M2P

a2ρ and H2 −H′ = 1

2M2P

a2(ρ+ p)

Using that ρ = 12a2φ

′2 + V and p = 12a2φ

′2 − V we have

φ′2

2M2P

= H2 −H′ (20.14)

and thus the underbraced term indeed vanishes, and we are left with

R′ = 2M2P

(H∇2D

φ′2

)

(20.15)

Taking a little step to the side, we play around a little with φ′2

2M2P

. We start with

the slow-roll equation 3Hφ = − dVdφ . By definition of conformal time we have

ddt = 1

addτ . Inserted in the slow-roll equation this gives

3Hφ = 3dadt

a

dφdt

= 3a′

a3φ′ = −dV

dφ

and thus we have that φ′2 = − a4

9H2

(dVdφ

)2, where H = a′

a . Remembering the

definition of the slow-roll parameter ε ≡ 12M

2P

(V ′V

)2we have

φ′2

2M2P

=a4

9H2

(dVdφ

)2 1

2M2P

= εV 2a4

9M4PH2

From the first Friedmann equation we have H2 = a2

3M2PV which gives us that

φ′2

2M2P

= εH2 = H2 −H′


where the last equality comes from (20.14). This means that

ε = 1− H′

H2

Using this result in (20.15) and remembering that ∇2 → k2 when Fourier trans-forming we have

R′kH =

aRk

aH=Rk

H=

1

ε

(k

H

)2

|Dk|2

when only considering one mode. From the definition ofR we have

Rk = Dk +Hδφk

φ′

= Dk +H

φδφk

From (20.9) we haveD′ +HA = 4πGφ′δφ

We have already shown that D is a constant and thus that D ′ vanishes. Consideringonly one k-mode we then have

Dk = Ak = 4πGφ′

Hδφ

=1

2M2P

φ

Hδφk

Here we see the coupling between the perturbation in the metric and the perturb-ation in the inflaton field very explicitly. But this relation is not general, and isonly valid in longitudinal gauge. Then of course the next step is to calculate thecorresponding relation with the gauge independent R instead of D.

Rk = Dk +H

φδφk

=1

2M2P

φ

Hδφk +

H

φδφk

Playing the same ε-game as above, we can write this as

Rk = (1 + ε)H

φδφk

Since ε is really small we have to a very good approximation:

Rk =H

φδφk

(20.16)


Now, that is a nice equation! In figure 20.1 one can see how R is constant in thedifferent epochs while Dk = Ψk vary.

1aH

τeq

Rk = 1

τ

Ψk

23Rk

35Rk

τf

Figure 20.1: Outside the horizon R will be constant while Ψk = Dk will vary inthe different epochs, being 0 during inflation.

20.1.3 The power spectrum for scalar perturbations

In (17.21) we saw that during inflation we have

δφk =H√2k3

Combining this with (20.16) we can find the power spectrum for R:

∆2R(k) =

k3

2π2|Rk|2

=k3

2π2

(H

φ

)2

|δφk|2

=

(H2

2πφ

)2

(20.17)

From the first Friedmann equation we have H2 ∼ V and from the slow-roll ana-lysis we have φ2 ∼ ε. Including the prefactors we may express the power spectrumas

∆2R(k) =

1

24π2ε

(V

M4P

)(20.18)


This quantity is measured directly through the Sachs-Wolfe effect in the CMB. Sowe may observe the fraction V

ε , but not these two quantities separately. Now wehave been studying scalar fluctuations. For tensor fluctuations we had

∆2T (k) =

2

3π2

(V

M4P

)

and thus∆2T

∆2S

= 16ε 1

In lecture 19 we expressed the power spectrum for the tensor fluctuations like

∆2T (k) = knT where nT = −2ε

Similarly is common to express the power spectrum for the scalar fluctuations as

∆2S ∼ knS−1 ⇒ nS − 1 = −2ε

From this definition we have that

nS − 1 =dln∆2

S

dln k

=dln Vdln k

− dlnεdln k

= −2ε− (4ε− 2η)

= 2η − 6ε (20.19)

This gives a small deviation from nS = 1. The data from WMAP is not giving anunambiguous answer to how large this quantity is.

Chapter 21

Lecture 21

In the previous lectures, we have considered perturbation theory in the inflationaryepoch. We then found the equations that couple the inflaton field to the geometryof the universe by means of the Einstein equations


The perturbations of the energy-momentum tensor give rise to fluctuations in themetric. These fluctuations were called curvature perturbations, R. We showedthat the curvature fluctuations are constant outside the horizon. This is due tothe fact that the modes of the fluctuations are outside the horizon provided theirwavelength, λ ∼ 1

k , is larger than the comoving horizon scale, H. That is, R =const outside horizon when k << H. This is shown in figure (20.1). However,we actually only showed that the curvature fluctuations are constant in the infla-tionary era. Therefore it remains to show thatR is constant also in the latercomingradiation dominated (RD) and matter dominated (MD) epochs of the history of theuniverse. (Actually, this is true only with adiabatic perturbations when no entropyis produced). All this is shown towards the end of this lecture. So in some sense theR acts as a messenger from very early times and up until today. One of the mes-sages that we can measure is the non-zero fluctuations in density and temperature,δT 6= 0 and δρm 6= 0. On our way we will derive the three relativistic general-izations of the fundamental equations that we have in non-relativistic perturbationtheory. So far we have considered fluctuations in the inflationary epoch describedclassically by the inflaton field φ. Now we will need to extend our description toalso account for relativistic perturbations in order to describe the late universe. Wefirst take a look at the Newtonian perturbation theory.

205


21.1 Newtonian theory

The non-perturbative theory starts with the perturbations in pressure, energy dens-ity, gravitational potential and velocity given by1

ρ → ρ+ δρ

p → p+ δp

Φ → Φ + δΦ

0 → v. (21.2)

The last equation describes that the perturbed velocity is nonzero although we haveassumed comoving coordinates (i.e. zero velocity). The Newtonian equations forperturbations in the gravitation field are

∇2δΦ = 4πGa2δ (Poisson)

∂δ

∂t+

1

a∇v = 0 (continuity)

∂v

∂t+Hv = −1

a∇δφ− 1

aρ∇δp (Euler) (21.3)

In the Poisson equation we have used the shorthand notation, δ ≡ δρρ . The con-

tinuity equation ensures particle conservation. These three equations are valid forperturbations inside the horizon, but their validity ends when we enter and crossthe horizon. When we now go to the non-Newtonian theory things get a little morecomplicated. However, we will see that in simple universe models the equationsdon’t get that nasty after all. One very useful simplification in models describing ahomogeneous and isotropic universe is the concept of perfect fluid.

21.2 Perfect fluids

We start this section by looking at some of the criterias for being a perfect fluid.These are:

• The temperature and the entropy of the fluid have to be unique so that thefluid is in thermal equilibrium. The fluid should not experience shear forcessince such forces, when currents are present, will produce heat and thenagain entropy.

• The particles in the perfect fluid should be close to each other and frequentlyinteracting. This justifies the hydrodynamical description as a fluid. Thisis not a good approach if the density is low, or particles move freely. Oneexample of a non-perfect fluid, which we will look at in the coming lectures,is the photons and neutrinos decoupled. After decoupling from matter, thesemove freely and can no longer be described hydrodynamical.

1For more details about the derivation of the Newtonian perturbation equations, see the note usedin Ast 4220.

21.2. PERFECT FLUIDS 207

We now want to find the perturbed velocities uµ and uµ and then in turn the fluc-tuations in the energy-momentum tensor, δTµν . This will be needed in order toproceed with the perturbed Einstein equation we saw in (21.1) in the beginning ofthis lecture. We start out with the energy-momentum tensor for a perfect fluid

Tµν = (p+ ρ)uµuν − gµνp (21.4)

the metric we use is the usual one

ds2 = a2(dτ2 − dx2)

The four velocity identity makes us find the zero-components of u by

u2 = 1 = gµνuµuν = (u0)2a2.

The last step follows from uµ = (u0,u = 0) since we have chosen comovingcoordinates, thus the fluid is at rest in our frame. The components are u0 = 1

a andu0 = a. We now perturb the 4-velocity and the metric

uµ → uµ = uµ + δuµ

(Here, we still demand the perturbed velocity to be normalized by the identityuu = 1.) The perturbed metric is the same as we have seen a hundred timesbefore, namely

gµν → ˜gµν = gµν + δgµν

and the perturbations are the 4× 4 matrix given by the four potentials

gµν = a2

(1 + 2A −Bi−Bi −(1− 2D)δij −Eij

)

All the terms in the matrix are small, so any product of these terms will be neglectedin the following. We can then find uµ and uµ. We write

uµ = (u0 + δu0, δui) = (1

a+ δu0, δui)

= (1

a+ δu0,

1

avi) (21.5)

Here, we have introduced peculiar velocity, vi = aδui in the last step. We will nowsketch some of the derivations of the components of the perturbed velocities

u0 = g0µuµ = a2(1 + 2A)u0 − a2Biδu

i

= a2(1 + 2A)(1

a+ δu0)

= a+ a2δu0 + 2aA = u0 + a2δu0 + 2aA

= u0 + δu0

⇒ δu0 = a2δu0 + 2aA (21.6)


Another component is

δui = giµuµ

= a(vi −Bi) + higher order terms (21.7)

From the 4-velocity identity uu = 1, we find

δu0 = −1

aA+ . . .

So with all the components inserted for, the perturbed velocities are

uµ =1

a(1−A, vi)

uµ = a(1 +A, vi −Bi) (21.8)

21.2.1 Perfect perturbations

We now turn our attention towards the perturbations in the energy-momentumtensor given as

Tµν → Tµν = Tµν + δTµν

The energy-momentum tensor is a symmetric 4× 4 matrix which means 10 DOFs.We will see that the perturbed energy-momentum tensor can be separated into twoparts; perfect and non-perfect perturbations, each of them having 5 DOFs. The per-fect perturbations of Tµν will be written on the same form as the energy-momentumtensor for a perfect fluid seen in (21.4), so the shape of the tensor will be main-tained.

Tµν = (p+ ρ)uµuν − gµν pHere, the perturbations to the pressure and energy density are as given in equation(21.2). Hence, we can find the perfect perturbation to the energy-momentum tensorby calculating Tµν and then subtract the unperturbed tensor to find

δT µν = T µν − T µνBy means of the components

T µν =

(T 0

0 T i0T 0j T ij

)

we can express the perfect perturbations of the energy-momentum tensor as

δT µν =

(δρ (ρ+ p)vi

(ρ+ p)(vi −Bi) −δρδij

)(21.9)

There are altogether 5 DOFs that conserve the perfect shape of Tµν . These areδρ(1), δp(1) and vi(3) since the spatial index runs over three components. Theother 5 DOFs must therefore be in the non-perfect part of the perturbation of theenergy-momentum tensor described by a non-perfect fluid.

21.2. PERFECT FLUIDS 209

21.2.2 Non-perfect fluids

Perfect fluids have an equation of state (EOS), relating pressure and density as

p = p(ρ) = wρ

We saw earlier that if the fluid depends on entropy, σ, then it is not perfect. Ageneral fluid will therefore have an EOS

p = p(ρ, σ)

Here, σ = Sn where S is the entropy density and n is the density if the particles

in the fluid. We want to find the perturbed energy-momentum tensor that consistof a perfect and a non-perfect part. The perfect perturbations were seen in (21.9),but the non-perfect ones are yet to be found in order to get the components ofthe perturbed Einstein equations. The usual way to account for also non-perfectcontributions to the fluid, is to introduce an anisotropic part to the perturbed energy-momentum tensor by writing

δTij = δijδp+ Σij (21.10)

Here, the first term is an isotropic part. It contains the trace of Tij and is a partof the perfect Tµν . (δijδp is often called bulk-viscosity and corresponds to the−δpδij-term of δT µν in equation (21.9)). The second term, Σij , is the anisotropicpart and the physical interpretation is anisotropic pressure or stress. It is tracelessand contains the 5 remaining DOFs of the entire δTµν .This description of non-perfect fluids also has its range of validity and eventu-ally breaks down. Then the Boltzmann equations are needed to further describethe fluid. This will be investigated in a future lecture. So to find the non-perfectcomponents to the perturbation, we neglect the δijδp-term and concentrate on theanisotropic part. As seen earlier, perturbations can always be separated into scalar,vector and tensor components.

Σij = ΣSij + ΣV

ij + ΣTij (21.11)

Here, the tensor part contains the perturbation hij , and the vector part looks likeΣvij = ∂ivj+∂jvi−the trace. We will need the scalar perturbations of the isotropic

pressure. These are:

ΣSij = (∂i∂j −

1

3δij∇2)Σ

The gradient of the scalar velocity potential, V can be written as

vi = ∂iV = −∂iV⇒ v = −∇V (21.12)

We then finally have both the perfect and non-perfect perturbations of the energy-momentum tensor and can find the components of the perturbed Einstein equations.We will from the Einstein equations derive the generalizations of the three Newto-nian perturbation equations in (21.3).


21.3 The relativistic perturbation equations

The perturbed Einstein equations are given by δEµν = 8πGδTµν . We find thecomponents of the equation by following the same procedure as we used in theinflaton-case earlier. We choose longitudinal (or conformal Newtonian) gauge sothat A = Φ and D = Ψ and get 4 equations

∇2Ψ− 3H(Ψ′ +HΨ) = 4πGa2δρ , (00) (21.13)

Ψ′ +HΦ = −4πGa2(ρ+ p)V , (0i) (21.14)

Ψ−Φ = 8πGa2Σ , (i = j) (21.15)

Ψ′′ +H+H(Φ′+ 2Ψ′) + (2H′ +H2)Φ +1

3∇2(Φ−Ψ) = 4πGa2δp , (i 6= j)

(21.16)We notice from the third equation that the two Bardeen potentials, Ψ and Φ areequal if the fluid is perfect, that is, if there are no anisotropic terms ∼ Σ present.In the beginning of this lecture we saw the three Newtonian perturbation equations(Euler, continuity and Poisson) in (21.3). These can now be generalized to alsoaccount for relativistic perturbations. The Poisson equation is found by combiningthe first and the second equation above. This yields

∇2Ψ = 4πGa2[ δρ︸︷︷︸Newtonian

− 3H(ρ+ p)V︸︷︷︸relativistic

] (21.17)

We see that the new Poisson equation for the perturbations is the former and New-tonian term with an additional relativistic correction. In order to find also the twoother perturbation equations, we will use the equation of motion (EOM) for mat-ter. This equation shows how perturbations in the metric depend on perturbationsin matter and vice versa. The relativistic Euler equation will then follow frommomentum conservation, while the continuity equation comes from energy (orparticle) conservation. The EOM for matter is

∇µT µν = 0 = ∂µTµν + ΓµνλT

λν − ΓλµνT

µλ

The variation of the above equation is

δ(∇µT µν) = 0 = ∂µδTµν + δΓµνλT

λν + ΓµνλδT

λν − δΓλµνT µλ − ΓλµνδT

µλ

We have calculated the Γ’s and the δΓ’s earlier. If we plug in for all these terms,we find the two desired equations depending on the index ν.

ν = 0 energy conservation

ν = i momentum conservation

For ν = 0, the equation is

δ′ + 3H(δp

δρ− w)δ = (1 +w)(3Ψ′ +∇2V ) (21.18)

21.3. THE RELATIVISTIC PERTURBATION EQUATIONS 211

This equation replaces the Newtonian continuity equation seen in (21.3). It de-scribes the fluctuations in density and gives conservation of particles (i.e. energy).One should also notice that the term ∼ ( δpδρ − w) = 0 when the fluid follows anequation of state p = wρ as for perfect fluids. When ν = j, we get

V ′ + (1− 3w)HV +w′

1 + w= Φ +

δp

p+ ρ+

2

3

∇2Σ

1 + w(21.19)

The equation is the generalization of the Newtonian Euler equation. We notice alsothe derivative term of the equation of state parameter, w. In general, this parameteris not necessarily a constant, but may be a function of time. An example is thetransition from RD to MD epoch when w changes continuously from 1

3 to 0. Thethree relativistic equations we have found are

∇2Ψ = 4πGa2[δρ − 3H(ρ+ p)V ]

δ′ + 3H(δp

δρ−w)δ = (1 + w)(3Ψ′ +∇2V )

V ′ + (1− 3w)HV +w′

1 + w= Φ +

δp

p+ ρ+

2

3

∇2Σ

1 + w(21.20)

They look a bit nasty, and in their most general form, they are. However, whenconsidering our usual universe models, we often have fluids with simple equationof state, so that the equations simplify significantly. We will now apply the two lastequations to the MD and RD epochs

21.3.1 Example with matter domination

When the universe is matter dominated, the equation of state parameter is w = 0.The continuity and Euler equations then turn into

δ′m = ∇2Vm + 3Ψ′

V ′m +HVm = Φ (21.21)

21.3.2 Example with radiation domination

In a RD universe the parameter is w = 13 . Then the pressure density term in the

Euler equation isδp

p+ ρ=

13δρ

ρ+ 13ρ

=1

4

δργργ

=1

4δγ

So now the continuity and Euler equations are

δ′γ =4

3∇2Vγ + 4Ψ′

V ′γ =1

4δγ + Φ (21.22)


One short comment to the equations for RD should be that the photons here areassumed not to be decoupled. The photons then contribute to the perfect fluid.

Before we end this section by looking at the curvature perturbation,R, showingthat it is constant outside the horizon in all epochs of the universe history, we takea short look at adiabatic perturbations and the concept of sound velocity in thecosmological fluid.

21.4 Adiabatic perturbations and sound velocity

We know that in general the equation of state is not like p = wρ , but rather givenby p = p(ρ, σ) since the pressure may depend on entropy. Since pressure is aphysical quantity, the differential, δp is exact and we can write it as

δp =

(∂p

∂ρ

)

σ

δρ+

(∂p

∂σ

)

ρ

δσ

The first term represents the adiabatic or isentropic perturbations, while the secondgives the isocurvature (or entropy) perturbations. The adiabatic term motivates fordefining the sound velocity, cs in the fluid by

(∂p

∂ρ

)

σ

= c2s = w (21.23)

where the last equality is valid in the case of a perfect fluid. And, thus, the sound

velocity in the (perfect) fluid applied in the RD and MD epochs are cs =√

13 and

cs = 0, respectively.One way to define adiabatic fluctuations is by demanding that

δx

x− δy

y= 0

Another way to explain adiabatic fluctuations is that the entropy stays constant sothat the equation of state is p = p(ρ, σ) = p(ρ) = wρ. In a general fluid it istherefore convenient to define a measure of the entropy fluctuations, S by

S = H(δp

p′− δρ

ρ′

)= H

(δp

p− δρ

ρ

)(21.24)

If then S 6= 0, entropy is being produced in the fluctuation and it is not adiabatic.From equation (21.24) we can rewrite the condition for adiabatic perturbations into

∂p

∂ρ=p′

ρ′=δp

δρ

21.5. VERIFYING THE CONSTANCY OFR 213

21.5 Verifying the constancy ofRWe showed in the previous lecture that the gauge invariant curvature perturbation,R, is constant during inflation for modes outside the horizon. We also claimed thatR is constant from the inflation epoch and up until today. This is yet to be verified.We have found the perturbation equations for VD and MD and we therefore needto establish contact with the inflation era by constructing an R that is valid alsowhen we enter the fluid description that is being used in the RD and MD epochs.During inflation, the curvature perturbation was

R = D +Hδφφ′

= Ψ +Hδφφ′

where the last equality is a consequence of our longitudinal choice of gauge. Themetric that will be used is

ds2 = a2[(1 + 2A)dτ 2 − (1− 2D)dx2]

(We actually introduced the curvature perturbation by setting dτ = 0 and obtainingan 3D hypersurface. That this surface is curved comes from the 2D-term in theline element.). The original curvature perturbation we had (in section 20.1.2) wasD, which we found was not in general gauge invariant, but transformed (under acoordinate transformation) as

D → D = D +Hξ0

We now define D = Dc = R, where Dc is the curvature perturbation, D ina comoving gauge. We want that R is invariant in longitudinal gauge, and nowfurther investigate what it takes to achieve just that. In the gauge transformation,Dc transformed as

Dc = D +Hξ0

The ξ0 is yet unknown, but soon to be found due to constraints in the longitudinalgauge and the velocity potential. Although the potential B is zero in the longitud-inal gauge, a “new” B occurs when a coordinate transformation is performed.

B → B = B + ξ0 − ξ′ = ξ0 − ξ

If the transformed B is still to be zero, then ξ0 = ξ′. In the transformation, thevelocity potential, V , will also change

V → V = V − ξ′ = Vc!= 0

The last step follows automatically since the gauge is comoving and so there shouldbe no local velocity. Thus, we seem to have found ξ0 which now must satisfy

ξ0 = V = ξ′


It is now time to see just what the curvature perturbation, R, looks like. We findthat

D = D +Hξ0 = D +HV⇒R = D +HV = Ψ +HV (21.25)

The last step is due to the longitudinal gauge. We can manipulate the expressionof R a little more in order to verify that it is all-time constant outside the horizon.The curvature perturbation must be valid for all epochs in the history of the uni-verse. We have considered both a classical field description (inflation) and a fluiddescription (RD and MD) for the perturbations in the energy-momentum tensor.By setting them equal, we have

δT 0i =

1

a2φ′∂iδφ

︸︷︷︸field

= (ρ+ p)∂iV︸︷︷︸fluid

(21.26)

Here, φ is the usual inflaton field. We remember from scalar field theory that energydensity and pressure were given by

ρ =1

2a2φ′2 + V

p =1

2a2φ′2 − V

Inserting this into equation (21.26) leads to

1

a2φ′∂iδφ = (ρ+ p)∂iV =

1

a2φ′2∂iV

⇒ V =δφ

φ′(21.27)

The curvature perturbation will therefore now be

R = Ψ +HV = Ψ +Hδφφ′

By means of the (0i) Einstein equation

Ψ′ +HΦ = 4πGa2(ρ+ p)V,

we can derive an equation for the curvature perturbation

R = Ψ +2

3

Φ + Ψ′/H1 + w

(21.28)

21.5. VERIFYING THE CONSTANCY OFR 215

This expression gives us the curvature perturbations for RD and MD. In order toshow that R is a constant, we differentiate the expression and then find

3

2(1 + w)

R′H =

1

H2∇2[c2sΨ +

1

3(Φ−Ψ)] +

3

2c2s(1 + w)S (21.29)

If we assume adiabatic expansion of the universe, then there is per definition no in-crease in entropy, so the last term above vanishes. When we also take into accountthat ∇2 → −k2, equation (21.29) reduces to

R′H ∼

k2

H2→ 0 (21.30)

This is valid for all modes outside the horizon. This is true since modes are outsidethe (comoving) horizon provided their wavelengths λ ∼ 1

k are greater than thehorizon scale ∼ 1

H . Thus, k2 H and therefore R′ → 0. So to sum up, thecurvature perturbation is constant for all times, provided that the modes are outsidethe horizon and that the perturbation is adiabatic.

R′ = 0

i) k H , (ouside horizon)

ii) adiabatic

Chapter 22

Lecture 22

22.1 More perturbations

From (21.28) we had the curvature perturbation, R

R = Ψ +2

3

Φ + Ψ′/H1 + w

We get R′ = 0 where we have assumed that we are outside the horizon, k H,and that we have adiabatic fluctuations. We have the pressure, p = p(ρ, S) withthe perturbation

δp =

(∂p

∂ρ

)

S

δρ+

(∂p

∂S

)

ρ

δS

Where the first term can be called curvature fluctuation, adiabatic fluctuation orisentropic fluctuation according to taste, and the second term can be called iso-curvature fluctuation, or entropy fluctuation.

Let us first look at isocurvature fluctuations. In this case ρ = const and δρ = 0.We can express the density as a sum of the energy density from matter and fromradiation

ρ = ρr + ρm

δρ = δρr + δρm = 0

which gives that the fluctuations of matter and radiation is equal with opposite sign(figure (22.1))

δρr = −δρm

Now to the adiabatic case where δS = 0. We have entropy per mass particle

σm =Sγnm

217


δρδρm

δρr = −δρm

Figure 22.1: For isocurvature perturbations the matter and radiation fluctuationshave opposite sign.

Here Sγ is the radiation entropy so that Sγ ∼ T 3 ∼ a−3 and nm ∼ a−3 so thatσm is constant during the evolution of the universe. In the adiabatic case we getδσm = 0. We can express this vanishing fluctuation as

δσm =δSγnm− Sγn2m

δnm = Sγ1

nm

(δSγSγ− δnm

nm

)

As the particle density is nm = ρmm and the radiation entropy evolves as Sγ ∼

T 3 ∼ ρ3/4γ we get

δσm =(

constants)(3

4

δργργ− δρm

ρm

)

as we had that δσm = 0 this gives us

3

4

δργργ

= δm (22.1)

We can picture adiabatic expansion with the traditional particle in a one di-mensional box with size L. Assume that the particle is in the ground state. Nowif the box expands slowly (adiabaticly) to size 2L the wave function will be in thenew ground state with twice as large wavelength. But if the box expand quickly(non-adiabaticly), the expansion is to quick for the wave function to follow, so itwill retain the old wavelength. And this is not a ground state in the new box, it is asuperposition of several excited energy eigenstates (figure (22.3)).

We continue to look at adiabatic expansion and we write

δρ = ρδt

δP = P δt

22.1. MORE PERTURBATIONS 219

δρ

δρm ∼ δργ

Figure 22.2: For adiabatic perturbations matter and radiation perturbations behavesimilarly

L 2L

2L

ψ adiabatic

non-adiabatic

Figure 22.3: In the non-adiabatic case, the expansion is too quick for the wavefunction to follow, but in the new box this is no longer the ground state, it is asuperposition of excited states. In the adiabatic expansion on the other hand, thewave function change, so that the system is still in the ground state.


so that

δt =δρ

ρ=δp

p=δργργ

=δρcρc

So we getδργργ

=δρmρm

(22.2)

We remember the equations

ργ + 3H(ργ + pγ) = 0

ρm + 3H(ρm + pm) = 0

and as pγ = 13ργ and pm = 0 we get

ργ = −4Hργ

ρm = −3Hρm

and equation (22.2) can then be written

δργ4Hργ

=δρm

3Hρm

We can rewrite equation (21.28) and insert for Φ = Ψ to get

2

3

Ψ′

H +5 + 3w

3Ψ = (1 +w)R (22.3)

Here R is constant, and w is constant in the radiation dominated and the matterdominated epoch (but with different value). We have the conformal Hubble para-meter H ∼ 1

τ and this allows us to write equation (22.3)

Ψ′ +n

τΨ =

p

τ

where n and p are constants. This is a simple differential equation with solution

Ψ(τ) = const+ τ−n

as τ increases the last term will be small and we get Ψ ≈ const if we now insertΨ′ = 0 into equation (22.3) we get

Ψ =3 + 3w

5 + 3wR =

23R in RD35R in MD

If we assume no anisotropic stress, Σ = 0 and Ψ = Φ we have the equationsfrom (21.21) and (21.22)

δ′m −∇2Vm = 3Ψ′ (22.4)

V ′m +HVm = Ψ (22.5)

δ′γ −4

3∇2Vγ = 4Ψ′ (22.6)

V ′γ −1

4δγ = Ψ (22.7)

22.1. MORE PERTURBATIONS 221

R

ττf τeq τ0

35R

23R

Ψ

Figure 22.4: The evolution of Ψ in the different epochs

The perturbed Einstein equations give

δEi0 = 8πGδT i0

so we get

Ψ′ +HΨ =3

2H2(1 +w)V = (ργ + pγ)Vγ + (ρm + pm)Vm (22.8)

we Fourier expand

δ = δ(x, t) =∑

k

δk(t)eik·x

V = V (x, t) =∑

k

Vk(t)eik·x

Ψ = Ψ(x, t) =∑

k

Ψk(t)eik·x

and as ∇2 → −k2 equation (22.4) turns out

δ′mk + k2Vmk = 3Ψ′k

or without the indicesδ′m + k2Vm = 3Ψ′

Let us look at the initial conditions early in the radiation dominated epoch.Here w = 1

3 and

Ψ =2

3R = const =

3

2H4

3Vr


which leads us to

Vr =Ψ

2HSInce adiabatic expansion requires these velocities to be equal, this gives us Vr =Vγ and

V ′γ =−Ψ

2H2H′ (22.9)

The adiabatic fluctuations become

Vγ = Vm = Vr =Ψ

2HAnd with the Friedmann equation in the radiation dominated epoch

H′ = −4πG

3(ρ+ 3p)a2 = −8πG

3ρa2 = −H2

and if we insert this into (22.9) we get

V ′γ =Ψ

2

If we insert into (22.8) we find

Ψ

2− 1

4δγ = Ψ

and further

δγ = −2Ψ =4

3δm (22.10)

where the last equality is due to the adiabatic rquirement in (22.1).

22.2 The Boltzmann equation

Classically the Hamilton function of a system of particles is the sum of kinetic andpotential energy

H = K + U =N∑

a=1

p2a

2m+ U(q1, . . .qN ; p1, . . .pN )

We will for simplicity write p = (p1,p2, . . .pN ) = (pi) and likewise q = (qi).The Hamilton equations are

qi =∂H

∂pi⇒ qi = qi(t) (22.11)

pi = −∂H∂qi⇒ pi = pi(t) (22.12)

22.2. THE BOLTZMANN EQUATION 223

qi

pi

Figure 22.5: The ensemble of systems behave as a gas of non-interacting particles

So the system of particles is described as a single point in a 6N -dimensionalphase space.An ensemble is a set of such identical systems but with different initialconditions. This ensemble will then behave as a gas of free particles (figure 22.5).We introduce the density in the phase space, ρ = ρ(q,p, t). The conservation of acurrent of system particles is expressed by

∂ρ

∂t+∇ · J = 0 (22.13)

where ∇ =(∂∂q ,

∂∂p

)and J = ρ (q, p). (22.13) then turns out

∂ρ

∂t+

∂

∂qi(ρqi) +

∂

∂pi(ρpi) = 0

or even better∂ρ

∂t+ qi

∂ρ

∂qi+ pi

∂ρ

∂pi= 0 (22.14)

But we can use the comoving derivative (or the Lagrange derivative)

d

dt=

∂

∂t+ v · ∇

where ∂∂t is called the Euler derivative. In our case the comoving derivative is

d

dt=

∂

∂t+ qi

∂

∂q i+ pi

∂

∂p i(22.15)

Combining (22.14) and (22.15) we find Liouville’s theorem

dρ

dt= 0


and if we apply this together with (22.15) and the Hamilton equations, (22.11) and(22.12), we get

dρ

dt=∂ρ

∂t+∂H

∂pi

∂ρ

∂qi− ∂H

∂qi

∂ρ

∂pi= 0 (22.16)

Now we define the Poisson bracket

[A,B] =∂A

∂pi

∂B

∂qi− ∂B

∂pi

∂A

∂qi

so (22.16) becomesdρ

dt=∂ρ

∂t+ [H, ρ] = 0

In quantum mechanics q and p becomes operators

qi → qi

pi → pi

and the Poisson bracket turns to Heisenberg brackets

[A,B]Poisson → [A, B]Heisenberg

The expectation value of an observable A = A(q, p) can be expressed

〈A〉 =1

Z

∫d3Nqd3Npρ(q, p, t)A(q, p, t)

where Z =∫

d3Nq∫

d3Npρ is the partition function. In quantum mechanics ρbecomes a density operator ρ→ ρ where

Z = Trρ

We also get the reduced density functions

ρnm = ρn (q1,q2, . . . qn; p1,p2, . . .pm)

=

∫d3qn+1 . . .

∫d3qN

∫d3pm+1 . . .

∫d3pNρ(q, p, t)

In particular we have the 2-particle distribution function

ρ20 = (x1,x2) = g(x1,x2)

and the 1-particle density function

ρ11(x1,p1) = f(x1,p1)

and the comoving derivative of this

df

dt=∂f

∂t+ x

∂f

∂x+ p

∂f

∂p= C[f ] (22.17)

This is the Boltzmann equation. We now have a 6-dimensional phase space withinteractions, and the interactions are given by the collision integral C[f ] definedhere.

Chapter 23

Lecture 23

In this lecture we will study temperature fluctuations, that is, the perturbation equa-tions for photons. In equilibrium the photon distribution is Bose-Einstein and isgiven by

f(x,p, t) =1

ep/T (t) − 1But as discussed earlier, the equilibrium assumption, although simple, is not agood model for the photon distribution in the universe today. For non-equilibriumsystems we have to apply the more general Boltzmann formalism. Luckily thisformalism was introduced last lecture in section (22.2).

23.1 The Boltzmann equation for photons

The photon distribution f(x,p, t) satisfies the Boltzmann equation

df

dt=∂f

∂t+

dxi

dt

∂f

∂xi+

dpi

dt

∂f

∂pi= C[f ] (23.1)

Here pi denotes the physical momentum1 and C[f ] on RHS is a collision term tobe determined later. We now write the momentum as p = np where n is a unitvector in the direction of p satisfying n · n = 1. Using this, we write (23.1) as

df

dt=∂f

∂t+

dxi

dt

∂f

∂xi+

dp

dt

∂f

∂p+

dni

dt

∂f

∂ni(23.2)

Without any interactions n will be a constant, so dni

dt is a first order term. Thelast term in (23.2) is therefore of second order and will be omitted in our furtheranalysis. We now make the usual decomposition of f into an equilibrium part anda perturbation part:

f(x,p, t) = f0(x,p, t)︸︷︷︸equilibrium

+ f1(x,p, t)︸︷︷︸perturbation=δf

1In curved space time one may also define for example a comoving momentum that is differentfrom the local, measurable and physical momentum.

225


We express our perturbed metric in longitudinal gauge as

ds2 = a2[(1 + 2Φ)dτ 2 − (1− 2Ψ)dx2

](23.3)

Now we want to find an expression for the dpdt -term in (23.2). First we introduce

the covariant momentum pµ defined by

pµ =dxµ

dλ

From general relativity we have the geodesic equation

dpµ

dλ+ Γµαβ p

αpβ = 0 (23.4)

The physical energy and physical momentum are

p =√g00p

0 (physical energy)

pi =√|gii|pi = nip (physical momentum)

(23.5)

Using the metric defined in (23.3) these quantities become

p = a(1 + Φ)p0

pi = a(1−Ψ)pi

(23.6)

For us it will be sufficient to study the µ = 0-equation in (23.4). Calculating theChristoffel symbols for µ = 0 in our metric yields

Γ000 = H+ Φ′

Γ00i = ∂iΦ

Γ0ij = [H− 2H(Φ + Ψ)] δij

(23.7)

Inserting this into (23.4), using p0 = 1a (1− Φ)p and remembering that

dΦ

dτ=∂Φ

∂τ+

dxi

dτ

∂Φ

∂xi= Φ′ + ui∂iΦ

one hopefully finds that

1

p

dp

dτ= −H+ Ψ′ − n · ∇Ψ (23.8)

23.2. THE BRIGHTNESS EQUATION 227

This is actually the only result that we will need from gravitational theory. To thelowest order, when not considering the effects of the perturbed metric, this gives

1

p

dp

dt= −H = − a

a

d

dt(ln pa) = 0 ⇒ pa = const. ⇒ p ∼ 1

a

which is the familiar relation for redshifting of free photons with the expansion ofthe universe. Inserting (23.8) into the Boltzmann equation (23.2) we get the finalversion of our Boltzmann equation for photons

df

dτ=∂f

∂τ+ n · ∇f + p

∂f

∂p

(−H+ Ψ′ − n · ∇Φ

)= C[f ] (23.9)

23.2 The brightness equation

In (23.9) we still have the mysterious C[f ] collision term on the RHS. So let’ssee what kind of interactions we have for T 1MeV. The particles present atthis temperature are photons, neutrinos, electrons, protons and α-particles. Exceptfrom the neutrinos, which are decoupled at this temperature, all these particles areinteracting through electromagnetic interactions. In addition the electrons, pro-tons and α-particles are coupled through Coulomb interaction. At this temperaturethe electrons are non-relativistic and interact with the photons through Thompsonscattering. The cross-section for this process is given by

dσ

dΩ=

1

2(1 + cos2θ)r2

0 (23.10)

where θ is the deflection angle for the photon and r0 is the “classical electronradius” r0 = α h

mec= 2.8× 10−13cm. From the expression for r0 one notices that

the cross section is inversely proportional to the electron mass. This is why thecorresponding cross-sections for Thompson scattering on protons and α-particlesare neglectable compared to the electron scattering. The Thompson scattering iselastic so |p| = |p′| = p, that is, the photon doesn’t change its energy, only itsdirection through this process. We now may write the total photon distributionfunction as a perturbed Bose-Einstein distribution with a temperature that dependson n rather than p.

f(x,p, t) =1

ep

T (x,n,t) − 1= f0(x,p, t) + δf

where

δf =∂f

∂TδT = − p

T

∂f0

∂pδT


This last expression makes it tempting to define a quantity that looks like

Θ(x,n, t) =1

TδT (x,n, t) (23.11)

and we have

δf = −p∂f0

∂pΘ (23.12)

Inserting this expression for f into the Boltzmann equation (23.9) one should endup with

−p∂f0

∂p

[Θ′ + n · ∇(Θ + Φ)−Ψ′

]= aC[f ] (23.13)

Now we almost have the Boltzmann equation for θ. What is left is to find a moreexplicit expression for C[f ]. A detailed calculation of this term requires a bit ofwork. Luckily these details are given in section 4.3 in Modern Cosmology by ScottDodelson. Here we will just scratch the surface of classical kinetic theory to shedsome light on the origin of the C[f ]-term, and then state the final result.

While statistical mechanics is dealing with large systems in equilibrium, kinetictheory is dealing with large systems out of equilibrium. The variables that we workwith in kinetic theory is the distribution function f(x,v, t) and the momentump = mv. As Boltzmann showed, one may write

(∂

∂t+ v

∂

∂x+ a

∂

∂v

)f =

∫d3v1

∫dσ|v − v′|(f ′f ′1 − ff1) (23.14)

where v = x, a = v and dσ is the differential cross-section given by dσ =dσdΩdΩ. Here one is considering scattering processes where two particles with initialvalues v and v1 are scattered to the new values v′ and v′1. (23.14) is an exactequation2 . We are considering Thompson scattering where dσ is the given by theThompson cross-section, so our collision term will have the form σTneΘ, wherene is the electron density and σT is the Thompson cross-section. Doing the detailedcalculations and including all terms and factors one is left with

aC[f ] = aNeσT p∂f0

∂p(Θ

︸︷︷︸δf

−Θ0 − n · ve) (23.15)

where the meaning of Θ0 is soon to be defined. Inserting this into (23.13) one findsthe Boltzmann equation for Θ which is called the brightness equation:

2From this equation Boltzmann found his famous H-theorem

23.3. PROPERTIES OF Θ AND ITS MULTIPOLE EXPANSION 229

Θ′ −Ψ′ + n · ∇(Θ + Φ) = −κ′(Θ0 −Θ + n · ve) (23.16)

where κ is the optical depth defined as κ = −aneσT . Θ is commonly expanded inits multipoles

Θ = Θ(x,n, t) = Σ multipoles

The monopole

Θ0(x, t) =1

4π

∫dΩΘ(x,n, t)

can be interpreted as the average temperature fluctuation in all directions. Later wewill couple the temperature fluctuations to fluctuations in the gravitational potentialto determine the observable temperature fluctuations today.

23.3 Properties of Θ and its multipole expansion

We want to relate the Θ-multipoles to the perturbations in the energy density forphotons. We use the perturbed distribution function to find the perturbed energymomentum tensor. In flat space we have

T µν(x, t) = 2

∫d3p

(2π)3Epµpνf(x,p, t) (23.17)

We know that d3pE is Lorentz invariant, so T µν must be Lorentz invariant. The

00-component in the equilibrium case yields

T 00 = ργ = 2

∫dp3

(2π)3

pp

pf0(x,p, t)

since E = p for photons. If the photon gas is in equilibrium, the distributionfunction is Bose-Einstein and the energy density obeys the Stefan-Boltzmann lawργ = π2

15T4. In the general case we have

T 00 = ργ + δργ = 2

∫d3p

(2π)3p(f0 + δf) (23.18)


and thus, using (23.12), we have

δργ = 2

∫d3p

(2π)3pδf

= −2

∫d3p

(2π)3pp∂f0

∂pΘ

= −2

∫dp p2

(2π)3p2∂f0

∂p×∫

dΩΘ(x,n, t)

= 2

∫dp 4p3

(2π)3f0 × 4πΘ0(x, t)

= 4× 2

∫d3p

(2π)3pf0

︸︷︷︸ργ

×Θ0

= 4ργΘ0 (23.19)

So we have the beautiful relation

δγ =δργργ

= 4Θ0 (23.20)

Another similar calculation is to study the i0-component of the energy-momentumtensor. We have

T i0 = (ργ + pγ)viγ

= 2

∫d3p

(2π)3piδf

where vi denotes a peculiar velocity. pi = nip and pγ = 13ργ . We then have

4

3ργv

iγ = −2

∫d3p

(2π)3pp∂f0

∂pni

︸︷︷︸4ργ

Θ(x,n, t) (23.21)

In (21.12) in lecture 21 we defined vi as the derivative of a potential:

vi = ∂iVγ = −∂iVγ∂iv

iγ = −∇2Vγ

where Vγ = Vγ(x, t). Now we want to Fourier transform our x-dependencies intok-space. We get

Θ(x,n, t) =

∫d3k

(2π)3Θ(k,n, t)eik·x

Vγ(x, t) =

∫d3k

(2π)3Vγ(k, t)eik·x (23.22)


and we have

−∇2Vγ ⇒4

3ργk

2Vγ = 4ργi

4π

∫dΩ(k · n)Θ (23.23)

Θ = Θ(k,n, t) is a scalar function which we now want to expand in multipoles.We define a quantity µ = k · n = cos θ and write Θ = Θ(k, µ, t). We then

havedΩ = 2π sin θdθ = 2πd cos θ = 2πdµ

We now write Θ as a sum of Legendre polynomials

Θ(k, µ, t) =

∞∑

l=0

(−i)l(2l + 1)Θl(k, t)Pl(µ) (23.24)

We have the orthogonality relation for Legendre polynomials

∫ 1

−1dµPl(µ)Pl′(µ) =

2

2l + 1δll′

So we have that1

4π

∫dΩPlPl′ =

δll′

2l + 1

and thus we can write a single multipole of Θ as

Θl(k, t) =il

4π

∫dΩ Θ(k,n, t)Pl(k · n︸︷︷︸

µ

) (23.25)

Some values for the first Legendre polynomials are

P0(µ) = 1 , P1(µ) = µ , P2(µ) =1

2(3µ2 − 1)

If we now look back to (23.23) we find that

4

3ργk

2Vγ = 4ργi

4π

∫dΩ(k · n)Θ

= 4ργkΘ1

and we have thatkVγ = 3Θ1 (23.26)

Analogously to what we have done for Θ0 and Θ1 we can find that Θ2 is connectedto the anisotropic stress Πγ ∼ Φ−Ψ like

Πγ = 12Θ2

So we see that for the lowest multipoles in the temperature fluctuations, the amp-litude of each of the multipoles has a direct physical interpretation.


23.3.1 Line-of-sight integration

From the brightness equation (23.16) we now want to determine the magnitudeof Θ today given its value, say at the last scattering surface (LSS)3. Solving thisproblem can be very nasty when working with higher order terms that include Θ1,Θ2, Θ3 etc., so it was a big relief when Seljak and Zaldarriaga in 1996 presen-ted a technique called line-of-sight integration as a very direct way to solve theproblem. This method is used in all modern computer programs working withCMB anisotropies. As a very simple example we will consider the case whereΨ′ = Φ′ = H′ = 0. Then the brightness equation (23.16) yields

Θ′ + ik · nΘ = 0

Θ′ + ikµΘ = 0dΘ

Θ= −ikµdτ

ln Θ = −ikµτ + ln Θ∗(23.27)

where a subscript ∗ denotes evaluation at LSS. So Θ today is given by

Θ(τ0) = e−ikµ(τ0−τ∗)Θ(τ∗) (23.28)

In the next lecture we will see that even if we include all the potentials, the solu-tion will be on the same, simple form. A well-known technique from quantummechanics when one has such a plane wave solution, is to expand it as a product ofspherical Bessel functions, jl, and Legendre polynomials.

eik·r =

∞∑

l=0

il(2l + 1)jl(k · r)Pl(k · r)

Of course, since Θ is on this form, we may use the same technique here. AssumingΘ∗ = Θ(k, τ∗) = Θ0(k, τ∗) (that is, we only have a monopole at LSS), expandingand projecting out the multipoles using (23.25) we will find that

Θl(k, τ0) = Θ0(k, τ∗)jl(kD) (23.29)

3One of the less pleasant facts about modern cosmology is that “last scattering surface” has ex-actly the same acronym as “large scale structure”.


where D is the comoving angular diameter distance to LSS given by

D =

∫ t0

t∗

dta(t)

=

∫ τ0

τ∗dτ = τ0 − τ∗ ≈ τ0

We see that we get a wide spectrum of Θ-multipoles by only starting with a mono-pole at LSS. This is just the Sachs-Wolfe effect. We also notice how the sphericalBessel function, jl, enters as a propagator for the temperature fluctuations. Nextlecture we will solve the full equation, and the result will be almost as beautiful asthis one.

Chapter 24

Lecture 24

In this lecture, we will derive the final equation that determines the fluctuations intemperature of the background radiation. We will first find the simplified equa-tion, under the assumption that there are no interactions, gravitational potentialor perturbations, and then finally find the exact equation of the temperature fluc-tuations by applying the method of line of sight integration, first developed bySeljak and Zaldarriaga in 1996. Thus we receive an equation that fully describesthe fluctuating properties of the CMB-spectra. However, before we go on withthese derivations, we will in short remind ourselves with what we found in the lastlecture.

The temperature fluctuations, δTT were given by

Θ(x,n, τ) =δT

T(x,n, τ)

We saw that it is more convenient to work with these functions in Fourier space, soFourier transforming the Θ-function leads to

Θ(x,n, τ) =

∫d3k

(2π)3Θ(k,n, τ)eik·x (24.1)

where the Θ(k,n, τ)-function can be expressed in terms by the Legendre polyno-mials, Pl, as a sum by

Θ(k,n, τ) =

∞∑

l=0

(−i)l(2l + 1)Θl(k, τ)Pl(k · n) (24.2)

And lastly plane waves are expanded in terms of the spherical harmonic functionsby

eik·x =∞∑

l=0

il(2l + 1)jl(kx)Pl(k · x) (24.3)

All these expansions listed above are standard and so easy to use (although hard toderive) that all master students should be familiar with them. We will now take a

235


deep look at the equation governing how the Θ-function (and then again the CMBtemperature fluctuations) vary with time, direction and wavenumber. This is ourstarting point for the further derivations. The equation was given by

Θ′ + ikµΘ− κ′Θ = Ψ′ − ikµΦ− κ′(Θ0 − iµkVb) (24.4)

Here, Θ = Θ(k,n, τ), µ = k · n and −κ′ = aneσT ⇒ −κ = neσT wherea is the scale factor, ne is the number of free electrons and σT is the Thompsoncross-section seen earlier in the previous lecture. Thus, in equation (24.4), onlyscattering on free electrons contribute to the last term. Before we proceed to ourmain topic, let us take just a short look at free electrons, degree of ionization, andthe conformal Hubble parameter. We will investigate how the conformal Hubbleparameter evolves with the redshift.

24.1 Degree of ionization

Due to conservation of electric charge, the number of free electrons equal the num-ber of free protons before recombination. That is

ne = np

The number of free baryons is given by

nb = np + nH

The degree of ionization, X is defined as the fraction between free electrons andfree baryons

X =nenb

and is equal to zero when no free electrons are present. A schematic illustration ofthe degree of ionization as a function of the redshift, z is shown in figure (24.1 ) andis based on observations by WMAP. We see that the degree of ionization is equal to1 in the early universe, but decreases rapidly to zero at the time of recombinationwhen neutral atoms formed and the universe became transparent. The fact that theionization increases again at later times is a bit controversial. This reionizationitself, towards the value of one, is predicted by the models of stellar formations,but is expected to happen at z ∼ 6 and not at z ∼ 20 as the WMAP-data suggestsin the figure. Some speculate that the WMAP- data are not precise enough. Onlytime will show.

We can also find how the intensity of light (=luminosity?) changes with theconformal Hubble-parameter. This is due to the fact that when photons are scattered,the intensity in the original direction of the photon is smaller after the scattering

24.1. DEGREE OF IONIZATION 237

X

Zz∗ ∼ 1100 20 0

1

Figure 24.1: The degree of ionization: The figure shows how the degree of ioniza-tion, X depends on the redshift. We observe that X is equal to one until recombin-ation begins at ∼ 1000, then dropping rapidly towards zero thereafter. The valueofX rises to 1 again at a redshift z ≈ 20. This is due to stellar production resultingin ionizing UV -radiation. The data are from WMAP.

than before, that is I(t0) < I(t). We can find the damping factor by

dI = −IneσTdt⇒ dI

I= −neσTdt

⇒ I(t0) = I(t)e−R t0t neσT dt

= I(t)e−κ (24.5)

So the luminosity is reduced by a factor κ, which plays the role as an optical depthin this manner. The conformal Hubble-parameter can be written as

κ =

∫ t

t0

κdt = −∫ t0

tκdt

If we use that the number of free electrons decrease with expansion of space, ne ∼1a3 , we can find −κ as a function of z. This is shown in figure (24.2). If the curveof −κ is integrated, we find the optical depth, κ. Figure (24.3) shows how κ varieswith the redshift. Per definition κ(z∗) = 1 at LSS, and this is also seen in thefigure.


1100 20 z

−H

Figure 24.2: The figure shows the redshift dependence of the time derivative of theoptical depth.

24.2 The temperature fluctuations in CMB

Now it is time to solve the equation seen in (24.4) governing the temperature fluc-tuations. We will do this in two different situations; in the first, we assume a verysimplified equation, corresponding to free electron travels. The last occasion is therealistic and exact situation, yet more complicated due to interactions and gravita-tion perturbations. But before we do that, the full equation is written again

Θ′ + ikµΘ− κ′Θ = Ψ− ikµΦ− κ′(Θ0 − iµkVb)

24.2.1 Solution of simplified case

We now solve the equation for the temperature fluctuation, making a lot of sim-plifications. If we assume that there are no interactions, then κ′ = 0 since nowκ′ ∼ σT = 0. Furthermore Ψ = Φ = 0 when no perturbations to the gravitationalpotential are present. This gives an equation

Θ′ + ik · nΘ = 0 (24.6)

which is easy to solve

Θ(k,n, τ0) = Θ(k,n, τ∗)e−ik·n(τ0−τ∗) (24.7)

Here, τ∗ denotes time when the recombination took place, giving a transparentuniverse. The exponential term in the solution to the temperature fluctuations de-scribes a plane wave from the last scattering surface (LSS). This is not so surprisingsince we have already assumed no interactions that may influence the wave. If thereare only monopoles, Θ0 on LSS, the expression of the temperature fluctuations are

Θ(k,n, τ∗) = Θ0(k, τ∗)

24.3. EXACT SOLUTION OF δTT 239

1100 20 z

1

H

Figure 24.3: The figure shows how the optical depth evolves with respect to theredshift. We see that the value of κ is large in the early universe, then decreasingrather rapid at recombination. function plotted here, is the integral of −κ andshould therefore be compared with figure (24.2).

This means that all the angular dependence (in the sky) of the temperature fluc-tuations is given by the exponential term in (24.7) since Θ0 has the same valuein all directions per definition of a monopole. This makes it possible to write thefluctuations as

Θl(k, τ0) = Θ0jl(kD)

where D is the distance to the last scattering surface and is given by

D =

∫dt

a=

∫ τ0

τ∗= τ0 − τ∗

≈ τ0. (24.8)

The last step is due to the fact that τ0 >> τ∗. This was the simplified situation. Wewill now find the exact solution of the temperature fluctuations in the CMB.

24.3 Exact solution of δTT

We will now derive the precise expression of the temperature fluctuations in theCMB, given in (24.4), by following the line of sight integration approach. But firstwe find an expression for a very useful quantity in CMB physics, the correlationfunction. There are several versions of the correlation function, depending on howmany separate points in the sky that are correlated. We will consider the simplestone which is the two-point correlation function gaining information from only twoseparate points. This is shown schematically in figure (24.4).


β

LSS

n′

n

γ

γ

Figure 24.4: The two-point correlation: The figure shows two pieces of informa-tion, represented by the vectors n and n′ arriving from the last scattering surface(LSS). The arrows indicate the directions of the radiation carrying the temperaturefluctuations. The angular distribution between the (two) photons is given by theangle β.

24.3.1 Correlation function to lowest order

The two-point correlation function, C(β) is the expectation value of the product oftwo separate temperature fluctuations and is defined by

C(β) = 〈Θ(x,n, τ0)Θ(x,n′, τ0)〉 (24.9)

The two-point correlation function can be written into a more convenient form byexpanding in terms of the Legendre polynomials

C(β) =1

4π

∞∑

l=0

(2l + 1)ClPl(cos β) (24.10)

Here, cosβ = n · n′ being the angle between the two points correlated. Thefunction Cl in the above expression is the one we are going to find. In some sensethis function “codes” the correlation function giving the temperature fluctuations.We will see that Cl depends on Θl that we are going to find later in this lecture. Byinserting the Fourier expansion of Θ given in (24.1) in (24.9), C(β) takes form as

C(β) =

∫d3k

(2π)3

∫d3k′

(2π)3〈Θ(k,n, τ0)Θ(k′,n′, τ0)〉ei(k+k′)·x

=

∫d3k

(2π)3

∫d3k′

(2π)3

∞∑

l,l′=0

(−i)l+l′(2l + 1)(2l′ + 1)〈Θl(k, τ0)Θl(k′, τ0)〉(24.11)

Pl(k · n)Pl(k′ · n′)ei(k+k′)·x


In the last step, we have used the equations (24.2). The Θl−terms in the bracketsare fluctuating temperature amplitudes. These fluctuations are Gaussian as pre-dicted by inflation. Therefore the fluctuations are orthogonal and this fact makes itpossible to simplify the expression for C(β). The orthogonality is expressed by

〈Θl(k, τ0)Θl(k′, τ0)〉 = (2π)3δ(k + k′)δll′〈|Θl(k, τ0)|2〉

Thus when we integrate over all modes, we see that∫

dk′

(2π)3δ(k + k′) ⇒ k′ = −k

Similarly, summation over l gives the relation∑

l′δll′ ⇒ l′ = l

which again leads to

Pl′(k′ · n′)→ Pl(−k · n′) = (−1)lPl(k · n′)(−i)2l(−1)l = +1 (24.12)

We can show that the sign of the expression of C(β) now is positive. The expres-sion will contain a (−i)2l−term from the main expression and (−1)l from (24.12),and it is easy to verify that (−i)2l(−1)l = +1 always holds. So what we are leftwith now, after the simplifications, is an expression that looks like

C(β) =

∞∑

l=0

∫d3k

(2π)3(2l + 1)2〈|Θl(k, τ0)|2〉Pl(k · n)Pl(k · n′) (24.13)

The last step we have to in order to find the Cl function, is to simplify the productof the Legendre polynomials in expression (24.13). Remembering that

∫d3k =∫

dkk2dΩk, we have the following∫dΩkPl(k · n)Pl(k · n′) =

4π

2l + 1Pl(n · n′)

Now, the expression for C(β) turns into

C(β) =∞∑

l=0

∫d3k

(2π)3(2l + 1)〈|Θl(k, τ0)|2〉Pl(n · n′)

(here, we have used that∫dkk2 × 4π =

∫dkk2dΩ =

∫d3k since 4π is the solid

angle). To find Cl we compare with equation (24.10), which was given by

C(β) =1

4π

∞∑

l=0

(2l + 1)ClPl(cos β)

=1

4π

∞∑

l=0

(2l + 1)ClPl(n · n′) (24.14)

And then finally we find that


Cl = 4π

∫d3k

(2π)3〈|Θl(k, τ0)|2〉 (24.15)

This is an important expression. If we know Cl, we saw in equation (24.10) that thetwo point correlation function, C(β) can be calculated. The two point correlationfunction is also an observable so now theory and observations can be compared.We saw in the simplified equation of the temperature fluctuations that Θl ( andthen again Cl ) could be found quite easily. We will now consider the general casewhere interactions are allowed and solve equation (24.4) exactly by line-of-sightintegration.

24.3.2 Solution to (24.4) by line-of-sight integration

The method we are about to use was first proposed by Seljak and Zaldarriaga.They showed that it is possible to solve the equation governing the temperaturefluctuations by only performing one integration. Earlier, one had to solve an infiniteset of coupled equations. This was due to the fact that Θ =

∑l Θl, giving one

equation for each value of l.The equation that we will solve is

Θ′ + ikµΘ− κ′Θ = Ψ′ − ikµΦ− κ′(Θ0 − iµkVb) (24.16)

The left hand side of the previous equation can be written as

Θ′ + ikµΘ− κ′Θ = e−iµkτ eκd

dτ(eiµkτ−κΘ)

So then the equation turns into

d

dτ(eiµkτ−κΘ) = [Ψ′ − ikµΦ− κ′(Θ0 − iµkVb)]e+iµkτ e−κ (24.17)

By integrating this equation from zero to infinity, we find that the temperaturefluctuations look like

Θ(k,n, τ0) =

∫ ∞

0dτeiµk(τ−τ0)−κ[Ψ′ − iµkΦ− κ′(Θ0 − iµkVb)]

The term containing iµk in the previous expression contains an angular depend-ence. We can rewrite this term as a derivative d

dτ eiµkτ = iµkeiµkτ and then per-

form a partial integration. This leads to a term of the form

d

dτe−κ = −κ′e−κ ≡ g(τ).

The function g(τ), is often referred to as a “visibility function”. It tells us what theprobability is that photons experienced their last scattering at the time τ . From the


g

z1100

Figure 24.5: The visibility function: The figure sketches the visibility function, gas a function of the cosmological redshift. The function gives the probability thatthe photons had their last scattering at time τ (or the corresponding redshift value,z). We observe that the probability for the photons having their last scattering hasits maximum value at z = 1100.

definition of the visibility function, −κ′e−κ ≡ g(τ), the term−κ′ is the probabilitythat photons are scattered and e−κ is the probability that a photon is unscattereduntil the time τ . An illustration of the visibility function is shown in figure (24.5)where we see that the photons are most likely to have experienced their last scat-tering around the time when the universe became transparent.

We in some sense avoided the angular dependence by rewriting the iµk-termand then perform a partial integration. From the expression in (24.17) we see thatwe have two terms ∼ iµk. This means that two partial integrations are needed inorder to find the expression for the temperature fluctuations which now looks like

Θ(k,n, τ0) =

∫ τ0

0dτeiµk(τ−τ0)−κ[Ψ′ − iµkΦ− κ′(Θ0 − iµkVb)]

=

∫ τ0

0dτe−ik·n(τ0−τ)[g(Θ0 + Ψ)

+(gVb)′ + e−κ(Φ′ + Ψ′)].

Here, the g(Θ0 + Ψ)-term comes from interactions and Vb represents the velocityof the scattering objects of the photons. We have earlier seen that we need theamplitude, Θl, of the fluctuations in the expression for Cl (see equation (24.15) ).So in order to separate out the amplitude of these CMB fluctuations, we expand theexponential term in equation (24.18) in spherical harmonics and end up with

Θl(k, τ) =

∫ τ0

0dτ [g(Θ0 + Ψ) + (gVb)

′ + e−κ(Φ′ + Ψ′)]jl(kτ0 − kτ ). (24.18)


θπ2

dσdΩ

π

r20

r20

2

Figure 24.6: The Thomson cross-section: the figure shows how the cross-sectionchanges with the angle θ.

This equation is pretty much the exact answer to the fluctuations in the temperatureof the CMB as described by equation (24.4). However, there is one small detail thatis worth mentioning: The Thomson cross-section has an angular dependence seenby dσ

dΩ = r02 (1+cos2 θ). The variation of the cross-section with respect to the angle,

θ is sketched in figure (24.6). We see that there is a little more scattering forwardsand backwards than in between. In the derivation of the Boltzmann equation, weassumed that there was no angular dependence of the cross-section. The smalldeviation from this assumption due to the angular dependence, manifests itself asa polarization of the temperature fluctuations. The Planck satellite is supposed toobserve this polarization which is expected to be small.

24.3.3 Simplifications due to instantaneous recombination

The solution to the temperature fluctuations found in equation (24.18) can be sim-plified by assuming that the recombination occurs instantaneously. In that case thefraction of ionization, X as seen in figure (24.1), will turn into a step function withrespect to the redshift, z. This is shown in figure (24.7). With this simplification,the temperature fluctuations satisfy

X ∼ ne ∼ Θ(τ∗ − τ).

where Θ here denotes the step-function. The point in this manner is that the visib-ility function, g turns into a delta function, being equal to zero for all values of theredshift except when z = z∗ (or τ0 = τ∗) at last scattering. This is shown in figure(24.8). The equation (24.18) is therefore written, substituting g → δ(τ − τ∗), as


X

Zz∗ ∼ 1100 20 0

1 1

1100

X

z

Figure 24.7: Instantaneous recombination: If the recombination is assumed to beinstantaneous, then the fraction of ionization turns into a step function.

g

z1100

g

z1100

Figure 24.8: The visibility function, g turns into a delta function under the assump-tion that the recombination takes place instantaneously.

Θl(k, τ0) = (Θ0 + Ψ)jl(kD)︸︷︷︸S.W.

+ kVbj′l(kD)︸︷︷︸

doppler

+

∫ τ

τ∗dτe−κ(τ)(Φ′ + Ψ′)jl(kτ0 − kτ)

︸︷︷︸I.S.W.

(24.19)Here, the first term describes the Sachs-Wolfe effect, the middle term is the dopplereffect due to the velocity of baryons, while the last term gives the Integrated Sachs-Wolfe. Actually, since we performed an integration over a delta function, all valuesin the last expression should be taken at the value τ = τ∗. Thus, the equation moreprecisely is

Θl(k, τ0) = (Θ0+Ψ)

∣∣∣∣τ∗

jl(kD)+kVbj′l(kD)+

∫ τ

τ∗dτe−κ(τ)(Φ′+Ψ′)jl(kτ0−kτ)

Finally in this lecture, we take a look at the Sachs-Wolfe effect since this gives usan important connection to inflation.

24.3.4 Sachs-Wolfe

The first term in equation (24.19) can be written as

(Θ0 + Ψ)∗ = (1

4δγ + Ψ)∗


In the very end of lecture 22, equation (22.10), we found an expression of δγ .Hence

(Θ0 + Ψ)∗ = (−2

3Ψ + Ψ)∗ =

1

3Ψ∗

=1

3Ψm (24.20)

In the last step, the Bardeen potential, Ψ∗ has been replaced with Ψm, representingmatter domination, since the last scattering occurs in the matter dominated epoch.We know from the previous lectures that the initial value of Ψm = 3

5 ( figure (20.1)shows this very clearly). If we insert equation (24.20) into the expression for theamplitude, Cl, we get the following

CS.W.l = 4π

∫d3k

(2π)3〈|Θl(k, τ0)|2〉j2

l (kD)

=4π

9

∫ ∞

0

dk

k

k3

2π2〈|Ψm|2〉

︸︷︷︸∆2

Ψ

j2l (kD) (24.21)

From the expression above, we recognize the underbraced term as the Power spec-trum for the potential Ψ. We can rewrite the expression for the amplitudes, C S.W.

l

of the temperature fluctuations by using

Ψm =3

5R ⇒ ∆2

Ψ =9

25∆2R.

This gives us

CS.W.l =

4π

9

∫dk

k∆2R

9

25j2l (kD)

=4π

25

∫ ∞ dx

xj2l (x)

︸︷︷︸= 1

2l(l+1)

∆2R (24.22)

The final expression is usually written on a form so that the we see the connectionto the power spectrum, ∆2

R more clearly

l(l + 1)

2πCS.W.l =

1

25∆2R (24.23)

The Power spectrum is shown in figure (24.9). The Sachs-Wolfe effect comesin for l ≤ 100, and from the figure, we observe that this is when the value of thePower spectrum is approximately constant and equal to 2× 10−10.


2

4

6

200 1000

[10−10]

l(l+1)2π Cl

l100

Figure 24.9: The Power spectrum: The graph shows the Power spectrum as afunction of l.

We end this course by estimating the value of the inflaton potential, V . Whenwe considered inflation in the previous lectures, we found that

∆2R =

1

24π2ε

(V

M4p

)≈ 50× 10−10

This makes it possible to find the fraction between V and ε that looks like

(V

ε

)1/4

≈ 6× 1016GeV (24.24)

During inflation the parameter ε is small in order for the potential to be sufficientlyshallow to allow a slow rolling. If we assume ε ∼ 10−4, then the value of thepotential driving inflation is

V 1/4 ∼ 1016GeV

It is a bit remarkable that this energy is also of the same order as we expect to havein the GUT regime. Whether or not this is a coincidence is yet an open question,but do not despair; the truth is out there...

Appendix A

The Riemann tensor and gaugefields

We will now take a look at the similarities between the Riemann tensor and gaugefields. The Riemann tensor is given by

Rµναβ = ∂αΓµνβ − ∂βΓµνα + ΓρνβΓµρα − ΓρναΓµρβ

We observe that the Riemann tensor is of the same form as a gauge field

Fαβ = ∂αAβ − ∂βAα + [Aα, Aβ ]

= ∂αAβ − ∂βAα +AαAβ −AβAα (A.1)

Here Aα = Aαα(Tα) with Tα being the generator of the field and the upper index,α being the group index.. This gives us the indices (Aα)ij = A α

α (Tα)ij . The pointis that the upper µ and lower ν indices of the Riemann tensor correspond to theij-indices in gauge theory.

249

250 APPENDIX A. THE RIEMANN TENSOR AND GAUGE FIELDS

Appendix B

de Sitter space

The metric for the de Sitter space is

ds2 = dt2 − e2Htdx2

But this only covers half of the de Sitter space. To cover the whole de Sitter spaceyou have to use the metric

ds2 = dt2 − cosh2(Ht)

(dr2

1−H2r2+ r2dΩ2

)

This can also be written

ds2 =(1−H2r2

)dt2 −

(dr2

1−H2r2+ r2dΩ2

)

Lots of different metrics were tried before the Robertson-Walker metric wasfound. Originally Friedmann used Poincaré coordinates with the metric

dσ2 =dx2

1 + dx22 + dx2

3

x23

But this was difficult. Einstein used spherical coordinates with the metric

ds2 = dt2 − a2

(dr2

1−H2r2+ r2dΩ2

)

this can also be written (with redefined r)

ds2 = dt2 − a2 dr2 + r2dΩ2

(1 + 1

4H2r2)2 = dt2 − a2

(dχ2 + sin2 χdΩ2

)

And in 1918 de Sitter used

ds2 =ηµνdxµdxν(1 + 1

4H2x2)2

where x = x2 − t2. It should be noted that all the components of this metricapproaches zero, ds2 → 0 as |x| → ∞. Finally it was Lemaitre who finally usedthe metric

ds2 = dt2 − e2Htdx2

251

lecture notes from fys5130 - cosmological physicsfolk.uio.no/finnr/notes/fys5130.pdf · these...

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