Lecture notes from
FYS5130 - Cosmological Physics
Spring 2005
Lectures by Finn Ravndal
Department of Physics
University of Oslo
Transcribed by:
Øystein Rudjord
Gorm Krogh Johnsen
Jostein Riiser Kristiansen
Last updated: September 29, 2005
Preface
These lecture notes are taken from the course FYS5130 - Cosmological Physics
given 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 happy
if someone else also find them useful.
If you find something in these notes that make no sense to you, it could be due
to 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 changed
radically.
Most probably it will be either number 2 or 3. In those cases we would be more
than 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 interesting subject, and also for taking the time to read through this manuscript and
provide us with numerous corretcions.
Oslo, September 28, 2005
Øystein Rudjord (oysteir@fys.uio.no)
Gorm Krogh Johnsen (gormj@fys.uio.no)
Jostein Riiser Kristiansen (j.r.kristiansen@astro.uio.no)
Contents
1 Lecture 1
1.1 Classical introduction . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Quantum field theory . . . . . . . . . . . . . . . . . . . . . . . .
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7
9
11
2 Lecture 2
2.1 2nd quantization . . . . . . . . . . . . . . .
2.2 Classical field theory . . . . . . . . . . . .
2.3 Fermions and bosons . . . . . . . . . . . .
2.4 Approximation to point particles . . . . . .
2.4.1 Interactions between point particles
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13
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14
15
18
18
3 Lecture 3
3.1 Relativistic fields . . . . . . . . . . .
3.1.1 Relativistic scalar field theory
3.1.2 External scalar field . . . . . .
3.2 Vacuum energy . . . . . . . . . . . .
3.2.1 Fermions . . . . . . . . . . .
3.3 The Casimir effect . . . . . . . . . .
3.4 Bernoulli numbers . . . . . . . . . .
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21
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22
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27
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4 Lecture 4
4.1 Regularization . . . . . . . . . . . . . . . . . . .
4.2 Dimensional regularization . . . . . . . . . . . .
4.3 Quantum statistical mechanics . . . . . . . . . .
4.3.1 Example with relativistic scalar particles
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33
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37
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5 Lecture 5
5.1 Path integral formalism . . . . . . . . . . . . . . . . . .
5.1.1 The partition function for the harmonic oscillator
5.2 Divergent products and functional determinants . . . . .
5.3 Generalized zeta-function regularization . . . . . . . . .
5.3.1 Some examples . . . . . . . . . . . . . . . . . .
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41
41
44
45
46
47
1
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CONTENTS
2
6
7
8
9
Lecture 6
6.1 Free scalar fields . . . . . . . . . . . . . . . . .
6.2 Radiation with massless particles and free energy
6.2.1 Pressure in arbitrary dimensions . . . . .
6.2.2 Example with zero temperature . . . . .
6.3 Fermions . . . . . . . . . . . . . . . . . . . . .
6.3.1 Example with fermionic Z . . . . . . . .
6.3.2 Example from QED . . . . . . . . . . .
6.4 Interacting fields . . . . . . . . . . . . . . . . .
6.4.1 Dimensional analysis . . . . . . . . . . .
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51
52
54
56
56
57
59
60
61
62
Lecture 7
7.1 Theory of free fields . . . . . . . . . . . .
7.2 Wick’s theorem . . . . . . . . . . . . . .
7.2.1 Example . . . . . . . . . . . . .
7.3 Wick’s theorem outside field theory . . .
7.3.1 Example: Stirling’s formula for n!
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65
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71
71
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Lecture 8
8.1 Correlators with interactions at finite temperature . . .
8.2 Phase transitions . . . . . . . . . . . . . . . . . . . .
8.2.1 The Higgs model . . . . . . . . . . . . . . . .
8.2.2 The partition function in the Higgs model . . .
8.2.3 Veff and an early-universe phase transition . . .
8.3 Particle creation in an external field at zero temperature
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73
73
75
75
76
77
79
Lecture 9
9.1 External fields . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 Toy model: Harmonic oscillator . . . . . . . . . . . . . . .
9.3 Oscillator with constant frequency . . . . . . . . . . . . . .
9.4 Oscillator with time dependent frequency . . . . . . . . . .
9.4.1 The relation between the vacuum states |0a i and |0b i
9.5 Quantum fields . . . . . . . . . . . . . . . . . . . . . . . .
9.5.1 Lorentz invariant norm . . . . . . . . . . . . . . . .
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81
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90
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95
. 95
. 96
. 96
. 97
. 99
. 102
10 Lecture 10
10.1 Effective action . . . . . . . . . . . . . . .
10.1.1 Example with Hint = −Lint . . .
10.1.2 Example with Hint 6= −Lint . . .
10.2 Quantization of external fields . . . . . . .
10.3 Particle production in constant electric field
10.3.1 QED . . . . . . . . . . . . . . . .
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CONTENTS
3
11 Lecture 11
11.1 Massless scalar field in a general space-time . . . . . . . . . .
11.1.1 Example with particle creation in accelerated systems .
11.1.2 Scalar fields in a Robertson-Walker background . . . .
11.1.3 Introduction of the d’Alembertian . . . . . . . . . . .
11.1.4 Variation of the metric . . . . . . . . . . . . . . . . .
11.2 The Einstein-Hilbert action . . . . . . . . . . . . . . . . . . .
11.3 The energy-momentum tensor . . . . . . . . . . . . . . . . .
11.3.1 Example with a Maxwell field . . . . . . . . . . . . .
11.3.2 Example with a scalar field . . . . . . . . . . . . . . .
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103
103
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106
108
108
110
111
111
112
12 Lecture 12
12.1 Conservation of the energy-momentum tensor . . . . . .
12.1.1 Generalized Bianchi identity . . . . . . . . . . .
12.2 Weyl transformations . . . . . . . . . . . . . . . . . . .
12.2.1 Example with a free scalar field . . . . . . . . .
12.2.2 Example with Maxwell theory . . . . . . . . . .
12.2.3 Example with coupled Maxwell and scalar theory
12.3 Quantization of scalar fields . . . . . . . . . . . . . . .
12.3.1 Scalar field in a RW background . . . . . . . . .
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115
115
119
120
121
122
123
123
124
13 Lecture 13
13.1 The Dirac equation in curved spacetime
13.2 Gravitons in flat spacetime . . . . . . .
13.3 Gravitational waves . . . . . . . . . . .
13.3.1 Basics . . . . . . . . . . . . . .
13.3.2 Particle hit by a wave . . . . . .
13.3.3 Energy contents in a wave . . .
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129
129
130
132
132
134
135
14 Lecture 14
14.1 Quantum fluctuations in a classical background (repetition)
14.2 Propagators from using the Einstein-Hilbert action . . . .
14.3 Canonical normalization of the gravitational field . . . . .
14.4 The energy-momentum tensor for the graviton field . . . .
14.5 Higher order terms and renormalizability . . . . . . . . . .
14.6 Green functions and propagators . . . . . . . . . . . . . .
14.7 Example with gravitational radiation . . . . . . . . . . . .
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137
137
138
139
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141
141
145
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147
147
149
150
150
150
15 Lecture 15
15.1 Gravitational radiation . . .
15.2 Emitted energy and momenta
15.3 Detecting gravitational waves
15.3.1 Indirect observations
15.3.2 Direct observations .
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CONTENTS
4
15.4 Gravitational radiation in a RW metric
15.4.1 Tensor fluctuations . . . . . .
15.4.2 Calculating the Ricci tensor .
15.4.3 The Friedmann equations . . .
15.4.4 Scalar field in the RW metric .
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16 Lecture 16
16.1 Cosmology . . . . . . . . . . . . . . . . . . .
16.1.1 Important equations and numbers . . .
16.1.2 The horizon and cosmological distances
16.1.3 The early universe . . . . . . . . . . .
16.2 Inflation . . . . . . . . . . . . . . . . . . . . .
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152
152
153
154
156
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159
159
159
161
162
163
17 Lecture 17
17.1 Inflation driven by a classical scalar field . . . . . . . .
17.1.1 The slow-roll approximation . . . . . . . . . .
17.1.2 Example with SRA using a quadratic potential
17.2 Fluctuations of φ . . . . . . . . . . . . . . . . . . . .
17.2.1 Quantum fluctuations . . . . . . . . . . . . . .
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167
167
168
170
171
174
18 Lecture 18
18.1 Primordial spectrum . . . . . . .
18.2 Vacuum fluctuations of φ̂ . . . .
18.3 Inflation . . . . . . . . . . . . .
18.4 Metric fluctuations . . . . . . .
18.4.1 Motivation . . . . . . .
18.4.2 The fluctuations . . . . .
18.4.3 Scalar perturbations . .
18.4.4 Gauge invariant variables
18.4.5 Choice of gauge . . . .
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179
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182
182
184
186
188
188
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19 Lecture 19
191
19.1 Tensor fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . 191
19.1.1 The power spectrum of the tensor fluctuations . . . . . . . 193
20 Lecture 20
20.1 Scalar perturbations . . . . . . . . . . . . . . . . .
20.1.1 Coupling the equations . . . . . . . . . . .
20.1.2 A gauge invariant curvature scalar . . . . .
20.1.3 The power spectrum for scalar perturbations
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197
197
198
200
203
21 Lecture 21
205
21.1 Newtonian theory . . . . . . . . . . . . . . . . . . . . . . . . . . 206
21.2 Perfect fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
21.2.1 Perfect perturbations . . . . . . . . . . . . . . . . . . . . 208
CONTENTS
21.2.2 Non-perfect fluids . . . . . . . . .
21.3 The relativistic perturbation equations . . .
21.3.1 Example with matter domination . .
21.3.2 Example with radiation domination
21.4 Adiabatic perturbations and sound velocity
21.5 Verifying the constancy of R . . . . . . . .
5
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209
210
211
211
212
213
22 Lecture 22
217
22.1 More perturbations . . . . . . . . . . . . . . . . . . . . . . . . . 217
22.2 The Boltzmann equation . . . . . . . . . . . . . . . . . . . . . . 222
23 Lecture 23
23.1 The Boltzmann equation for photons . . . .
23.2 The brightness equation . . . . . . . . . . .
23.3 Properties of Θ and its multipole expansion
23.3.1 Line-of-sight integration . . . . . .
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24 Lecture 24
24.1 Degree of ionization . . . . . . . . . . . . . . . . . . . .
24.2 The temperature fluctuations in CMB . . . . . . . . . . .
24.2.1 Solution of simplified case . . . . . . . . . . . . .
24.3 Exact solution of δT
. . . . . . . . . . . . . . . . . . . .
T
24.3.1 Correlation function to lowest order . . . . . . . .
24.3.2 Solution to (24.4) by line-of-sight integration . . .
24.3.3 Simplifications due to instantaneous recombination
24.3.4 Sachs-Wolfe . . . . . . . . . . . . . . . . . . . .
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225
225
227
229
232
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235
236
238
238
239
240
242
244
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 the
coordinate q, and by the change of the potential V (q)
F = mq̈ = −V ′ (q) = −
dV
∂V (q, t)
=−
dq
∂q
The last expression is valid in the case of a time-dependent potential. We may write
the energy as a sum of the kinetic and the potential energy.
1
E = T + V = mq̇ 2 + V
2
Energy conservation gives:
∂V
∂V
dE
= mq̇ q̈ +
q̇ +
=
dt
∂q
∂t
mq̈ +
∂V
∂q
q̇ +
∂V
=0
∂t
when V is time independent. In Lagrangian mechanics we define a Lagrangian L.
1
L = L(q, q̇) = T − V = mq̇ 2 − V (q)
2
We define the action as the time integral of the Lagrangian from a time ti to a time
tf
Z tf
dtL(q, q̇)
S=
ti
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 =
Z
tf
dt
ti
∂L
∂L
δq +
δq̇
∂q
∂ q̇
7
=0
(1.1)
CHAPTER 1. LECTURE 1
8
The multiplication rule for differentiation gives
∂L
∂ ∂L
∂L
∂ ∂L
∂ ∂L
∂ ∂L
δq =
δq̇ + δq
⇒
δq̇ =
δq − δq
∂t ∂ q̇
∂ q̇
∂t ∂ q̇
∂ q̇
∂t ∂ q̇
∂t ∂ q̇
Inserted in (1.1) this gives
δS =
Z
tf
dtδq
ti
∂L
∂ ∂L
−
∂q
∂t ∂ q̇
∂L tf
δq = 0
+
∂ q̇ ti
Since δq vanishes at the end points, δq(ti ) = δq(tf ) = 0, the last term will vanish
and we are left with the Euler-Lagrange equation
d ∂L
∂L
−
=0
∂q
dt ∂ q̇
(1.2)
For the Lagrangian L(q, q̇) = 12 mq̇ 2 − V (q) we get
∂V
= −V ′ (q)
∂q
∂V
= mq̇
∂ q̇
⇒ mq̈ = −V ′ (q)
which we recognize from Newtonian mechanics.
We define the canonical momentum:
p=
∂L
∂ q̇
ṗ =
∂L
∂q
and from Euler-Lagrange we get:
We define the Hamiltonian H = H(q, p) as
H(q, p) = q̇p − L(q, q̇)
which is a Legendre transformation. This gives
∂H
∂p
∂H
ṗ = −
∂q
q̇ =
These are Hamilton’s equations.
(1.3)
(1.4)
1.2. QUANTUM MECHANICS
9
1.2 Quantum mechanics
Canonical quantization
We want p and q to become operators.
q → q̂
p → p̂
These operators do not commute. We define the commutator
[q̂, p̂] = ih̄
In the same way the Hamiltonian will now become an operator H → Ĥ =
H(q̂, p̂). The state of the system is described by a vector |Ψ, ti in Hilbert space
and Schrödinger’s equation
ih̄
∂
|Ψ, ti = Ĥ|Ψ, ti
∂t
Many-body systems
We have N free particles that can move in a external potential U (x). They are
described by the Hamiltonian
X
N
N X
1 2
p̂n + U (x̂n ) =
ĥn
Ĥ =
2m
n=1
n=1
1
where ĥn = 2m
p̂2n + U (x̂n ) is the Hamiltonian for one particle. We then have
“single-particle eigenvalues”
ĥ|ki = εk |ki, and the single-particle states |ki satisfy
P∞
the completeness relation k=0 |kihk| = 1.
In coordinate representation we let x̂ → x and p̂ → −ih̄∇ which give us the
h̄2
∇2 + U (x) and we get the single-particle
single-particle Hamiltonian ĥ = − 2m
eigenvalue equation ĥuk (x) = εk uk (x) where uk (x) = hx|ki is a single-particle
wave function that satisfy the completeness relation.
∞
X
k=0
uk (x)u∗k (x′ ) = δ(x − x′ )
We now have eigenstates E for many-body systems Ĥ|ΨN i = E|ΨN i where
|ΨN i = |k1 i|k2 i . . . |kN i is a many-body state. A bosonic pair is symmetric and
has a state given by
r
1
|k1 , k2 iS =
(|k1 i|k2 i + |k2 i|k1 i)
2
CHAPTER 1. LECTURE 1
10
while a pair of fermions has an antisymmetric state given by
|k1 , k2 iA =
r
1
(|k1 i|k2 i − |k2 i|k1 i)
2
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 occupation
numbers for the number of particles in every energy level.
E=
∞
X
nk εk
k=0
where the occupation number nk is the number and particles in a single-particle
state |ki. For an arbitrary number of particles we will get a Fock space, which is a
tensor product of Hilbert spaces
F = H0 ⊗ H1 ⊗ . . . ⊗ HN
From the quantization of the harmonic oscillator we have the lowering, raising
and number operators, ↠, â and N̂ respectively, where N̂ = ↠â and we have the
commutators
i
h
â, ↠= 1
h
i h
i
h
i
N̂ , â = ↠â, â = ↠[â, â] + ↠, â â = −â
h
i
N̂ , ↠= â†
We have the number operators N̂k = â†k âk for particles in a state |ki, and â and â†
are annihilation and creation operators respectively.E.g.
N̂k |nk i = nk |nk i
√
âk |nk i = nk |nk − 1i
√
â†k |nk i = nk + 1|nk + 1i
P
We now get the Hamiltonian
Ĥ = ∞
k=0 N̂k εk with energy eigenstates Ĥ|Ψi =
P∞
E|Ψi where E = k=0 nk εk is the total energy, and |Ψi = |n0 , n1 , . . .i is the
many-body state. We define the vacuum state |0i by
âk |0i = 0
and a single-particle state |ki as
|ki = â†k |0i
1.3. QUANTUM FIELD THEORY
11
1.3 Quantum field theory
A particle in position x is now represented by
|xi =
∞
X
k=0
|kihk|xi =
∞
X
u∗k (x)|ki =
k=0
∞
X
k=0
u∗k (x)â†k |0i = Ψ̂† (x)|0i
P∞
P∞ ∗
†
where Ψ̂† (x) =
k=0 uk (x)âk or Ψ̂(x) =
k=0 uk (x)âk where Ψ̂(x) is a
quantum field operator for many-body systems. We introduce time variation using the time propagator
i
i
Ψ̂(x, t) = e h̄ Ĥt Ψ̂(x)e− h̄ Ĥt
∞
X
i
i
=
uk (x)e h̄ Ĥk t âk e− h̄ Ĥk t
k=0
=
∞
X
i
âk uk (x)e− h̄ εk t
k=0
P
Here we have written the Hamiltonian as Ĥ = ∞
k=0 Ĥk where Ĥk = N̂k εk .
We have the commutators between the creation and annihilation operators
h
[âk , âk′ ] = 0
i
âk , â†k′ = δkk′
which give us the commutation relations between the field operators
h
i
Ψ̂(x, t), Ψ̂(x′ , t) = 0
h
i X
h
i i
†
† ′
∗
Ψ̂(x, t), Ψ̂ (x , t) =
uk (x)uk′ (x) âk , âk′ e− h̄ (εk −εk′ )t
=
kk ′
∞
X
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 Hamiltoh̄2
∇2 and the eigenvalue equation ĥuk (x) = εk uk (x) where uk (x) =
nian ĥ = − 2m
q i
1 h̄ k·x
k2
and εk = 2m
where k is the momentum. Our volume gives periVe
odic boundary conditions u(x, y, z) = u(x + L, y, z) when ki =
ni = 0, ±1, ±2, . . .. Normalization gives
Z
d3 xuk (x)u∗k′ (x) =
1
V
Z
L
2
−L
2
dx
Z
L
2
−L
2
dy
Z
L
2
−L
2
i
′
2πh̄
L ni
dze h̄ (k−k )·x = δkk′
where
CHAPTER 1. LECTURE 1
12
and we have
1 X
V
k
′
eik·(x−x ) = δ(x − x′ ) =
=
where ∆ni =
V → ∞.
L
2πh̄ ∆ki
Z
i
′
1 X
∆nx ∆ny ∆nz e h̄ k·(x−x )
V n
d3 k i k·(x−x′ )
= δ(x − x′ )
e h̄
(2πh̄)3
and we have used the fact that
1
V
P
k
→
R
d3 k
(2πh̄)3
when
Chapter 2
Lecture 2
2.1 2nd quantization
The step that we performed last lecture, when we obtained a field operator from
a wave function, Ψ(x, t) → Ψ̂(x, t), is often called 2nd quantization. In fielddescription we will have a new interpretation of the creation and annihilation operators, ↠and â. In the field theoretical description â will annihilate a particle in one
state and ↠will create a new particle in another state. This justifies the use of the
terms “creation” and “annihilation” operators, while one in conventional quantum
mechanics rather should use the terms “raising” and “lowering” operators.
We can rewrite our Hamiltonian as
Z
X †
h̄2 2
b =
∇ + U Ψ̂(x)
H
âk âk εk = dxΨ̂† (x) −
2m
k
since
h̄2 2
−
∇ +U
2m
= ĥ
ĥuk (x) = εk uk (x)
X
Ψ̂(x) =
âk uk (x)
k
Similarly for our number operator we get
X †
b =
N
âk âk
=
=
=
Zk
Z
d3 xΨ̂† Ψ̂
d3 x
X
X
kk ′
â†k âk
k
13
â†k âk′ u∗k uk′
| {z }
δkk′
CHAPTER 2. LECTURE 2
14
Now we want to study the time evolution of Ψ̂(x, t). Using the Heisenberg equation of motion we find
h
i
∂
b
ih̄ Ψ̂(x, t) = Ψ̂(x, t), H
∂t
Z
h
i
=
d3 x′ Ψ̂(x, t), Ψ̂† (x′ )ĥΨ̂(x′ , t)
Z
=
d3 x′ δ(x − x′ )ĥ(x′ , t)Ψ̂(x′ , t)
= ĥ(x)Ψ̂(x, t)
So we end up with the equation
∂ Ψ̂
h̄2 2
∇ + U Ψ̂(x, t) = ih̄
−
2m
∂t
(2.1)
This equation really looks like the good old Scrödinger equation (SE), and of
course 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 from
standard quantum mechanics. In contrast to the one-particle description that we
are used to with the ordinary SL, here we are using a field operator that in principle may describe an arbitrary number of particles. This is what is called 2nd
quantization.
2.2 Classical field theory
We now introduce the Hamilton density H, given by
Z
b
b
H =
d3 xH
h̄2 2
†
b
∇ + U Ψ̂(x, t)
H = Ψ̂ (x, t) −
2m
We want to introduce a Lagrange formalism since that is a more fundamental theory than the Hamiltonian. We use what we know from classical field theory 1 .
Classically we have
h̄2 2
∗
∇ +U Ψ
H=Ψ −
2m
where the field operator Ψ̂(x) has become a complex Schrödinger field. Having
this Hamiltonian density, we know that we can construct a Lagrangian density
h̄2 2
∗
∗
L = π Ψ̇ − H = ih̄Ψ Ψ̇ − Ψ −
∇ +U Ψ
2m
= ih̄Ψ∗ Ψ̇ +
h̄2
|∇Ψ|2 − Ψ∗ U Ψ
2m
(2.2)
1
That is, a field theory where we have not performed a 2nd quantization, and where we use
complex functions instead of field operators
2.3. FERMIONS AND BOSONS
15
Here Π is the canonical field momentum
Π≡
∂L
∂ Ψ̇
and we see that (2.2) satisfies
H = π Ψ̇ − L
Let us see how we can use the canonical momentum in the quantum theory. In
the quantum theory we had the non-trivial commutator
h
i
Ψ̂(x, t), Ψ̂† (x′ , t) = δ(x − x′ )
Using the canonical momentum we get
h
i
Ψ → Ψ̂
⇒ Ψ̂(x, t), Π̂(x′ , t) = ih̄δ(x − x′ )
Π → Π̂
Back to the classical theory, the Lagrangian can be expressed as
Z
L = d3 xL.
That gives us the action
S=
Z
dt L =
Z
d4 xL
Using Hamilton’s principle, δS = 0, we now get a new Euler-Lagrange equation:
∂L
∂L
∂ ∂L
−∇·
−
= 0.
∂Ψ ∂t ∂ Ψ̇
∂(∇Ψ)
Using the Lagrangian density from (2.2) and solving the complex conjugated version of this equation we obtain:
h̄2
∂Ψ
= −
+ U (x) Ψ(x, t)
ih̄
∂t
2m
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 obey
Bose-Einstein statistics we have
h
i
X
Ψ̂(x) =
âk uk (x) and âk , â†k′ = δkk′
k
CHAPTER 2. LECTURE 2
16
which gives
Classically we have
h
i
Ψ̂(x, t), Ψ̂† (x′ , t) = δ(x − x′ )
Ψ=
X
ak uk (x)
k
where Ψ, ak and uk are complex. For particles that obey Fermi-Dirac statistics we
want to comply with the Pauli principle. That will be the case if we substitute all
the commutators in the Bose-Einstein theory with anti-commutators in the FermiDirac theory.
→
{, }
[, ]
| {z }
|{z}
F ermi−Dirac
Bose−Einstein
where the anitcommutator is defined
{A, B} = AB + BA.
Then, for fermions we have:
Ψ̂(x) =
X
b̂k uk (x)
k
{b̂k , b̂†k′ } = δkk′
{b̂k , b̂k′ } = 0
X †
b =
H
b̂k b̂k εk
|{z}
k
bk
N
b k for fermions only has the eigenvalues 0 and 1.
The Pauli principle implies that N
We want to find out if this is true in this formalism.
which gives
bk
b 2 = b̂† b̂k b̂† b̂k = b̂† (1 − b̂† b̂k )b̂k = b̂† b̂k = N
N
k
k
k
k
k
k
b k (N
b k − 1) = 0
N
→
nk (1 − nk ) = 0
Thus nk can only take the values nk = {0, 1}, which means that our formalism
with anti-commutators obeys the Pauli principle.
In classical theory the fermion operator will become
X
Ψ(x) =
bk uk (x)
k
where uk (x) is a complex function and bk is a Grassmann number (Γ) (see Table
2.1). We see that a classical fermion field must be a Grassmann field. Using the
Dirac equation we get:
(i 6 ∂ − m)Ψ̂(x) = 0
2.3. FERMIONS AND BOSONS
17
—————————————————————————————Grassmann numbers (Γ) have certain properties that make them well suited for
describing fermionic fields. Consider two Grassmann numbers α and β. They will
anti-commute:
αβ + βα = 0
which implies that
αβ = −βα ⇒ α2 = β 2 = 0
If you multiply a Grassmann number by a complex number you get a Grassmann
number. You can have complex Grassmann numbers, and they will have the following properties:
ψ = α + iβ
ψ ∗ = α − iβ
ψ 2 = (α − iβ)(α + iβ)
= α2 − β 2 + i (αβ + βα) = 0
| {z }
=0
ψ ∗ ψ = (α − iβ)(α + iβ)
= α2 + β 2 + i (αβ − βα)
| {z }
= 2iαβ 6= 0
6=0
This last expression, stating that complex Grassmann numbers will have a
non-vanishing ψ ∗ ψ is good, since this will allow us to have non-vanishing mass
terms on the form mψ ∗ ψ in our theory.
—————————————————————————————Table 2.1: Grassmann numbers
CHAPTER 2. LECTURE 2
18
where
Ψ̂ =
X
(b̂k uk + dˆ†k uvk)
and
k
6 ∂ = γ µ ∂µ
Here b̂k and dˆ†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 states
of 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-Dirac
condensate in much the same manner that we have Bose-Einstein condensates. In
both cases the atoms will be described using a local field Ψ(x) where the atoms are
treated as point particles. Whether the use of point particles is a valid approximation or not depends of course on the energy range in which we are working. The
approximation is valid as long as the typical de Broglie wavelength λB of the atom
is much larger than the typical size of the atom. In room temperature this is always
the case.
In the case of atoms we are used to work with well defined energy bands. Here
we can treat each band as being one Ψ-field. This is valid as long as the energy
differences between the states inside an energy band are much smaller than the
typical temperature that we are considering. A typical difference between energy
levels in an atom is ∼ 1eV. That corresponds to a temperature around 104 K. That
means that the point particle approximation is a very good approximation when
working in room temperature.
Analogously, when working with for example protons, a local field description
is 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 scattering
length a of a typical potential will have a finite extension, and using a point particle
treatment is valid as long as λB >> a. The point here is that whatever theory you
are using, it has a finite range of validity, and it is important not to use the theory
outside this range. This is what mathematicians never understand when working
with physics.
2.4.1 Interactions between point particles
For an interaction between a point particle and an external field we have the Hamiltonian density
2
−h̄ 2
†
b
∇ + U Ψ̂(x).
H = Ψ̂ (x)
2m
2.4. APPROXIMATION TO POINT PARTICLES
19
Here the first term on the right hand side is the free particle part, while the last term
is taking care of the interactions. U (x) is annihilating a particle in x and creating a
new particle in the same point.
An internal interaction between two particles in different points can be described by a potential V (x − x′ ). This interaction can then be described by a
Hamiltonian density
Z
Z
1
3
b
HI = d x d3 x′ Ψ̂† (x)Ψ̂(x)V (x − x′ )Ψ̂† (x′ )Ψ̂(x′ )
2
For our approximation of a point particle, this interaction will be a δ-potential:
V (x − x′ ) → λδ(x − x′ )
Z
λ ∼ a ∼ d3 xV (x).
The a is still the scattering length. Do not confuse this λ with the de Broglie wavelength λB . Using this delta function we get the Hamiltonian
Z
b = d3 x λ (Ψ̂† (x)Ψ̂(x))2
H
2
and the interaction hamiltonian density then yields
b = λ (Ψ̂† (x)Ψ̂(x))2
H
2
To find our new L, including the interaction term, we use (2.2) and include our new
interaction term. This gives
L = ih̄Ψ∗ Ψ̇ +
h̄2
λ
|∇Ψ|2 − Ψ∗ U Ψ − (Ψ∗ Ψ)2
2m
2
This Lagrangian density is frequently used in particle physics, nuclear physics,
super-conductivity etc.. The last term makes is nonlinear and complicates the theory a lot. Using the complex conjugated Euler-Lagrange equations we obtain the
field equation
h̄2 2
∂Ψ
=−
∇ Ψ + U Ψ + λ(Ψ∗ Ψ)Ψ
ih̄
∂t
2m
This equation can be viewed as a “non-linear Schrödinger equation” for classical
fields. When used in Bose-Einstein theory it is called the Gross-Pitaevskij equation. It is used to determine the distribution of particles in a potential. There have
also been attempts to include a similar non-linear term in standard quantum mechanics, but without great success.
20
CHAPTER 2. LECTURE 2
Chapter 3
Lecture 3
3.1 Relativistic fields
We will now like to consider relativistic fields describing particles with rest mass
m. For such particles energy and momentum are related by
E 2 = m2 c2 + c2 p2
From non-relativistic quantum mechanics we know that if the particles may be
described by a single scalar wave function, φ(x) say, then momentum and energy
satisfy the following :
∂
p → −ih̄∇
E→i
∂t
The resulting and relativistic wave-equation is called the Klein-Gordon equation
and is given by:
∂2
(3.1)
− 2 φ(x, t) = −∇2 φ + m2 φ
∂t
This equation was actually first described by Erwin Schrödinger, but is still most
commonly referred to as the Klein-Gordon equation (KG). The equation, since it
is relativistic, can be written covariantly. To do so, there must be no doubt about
what metric that is being used. For our purpose we choose a metric with positive
time component:


+1 0
0
0
 0 −1 0
0 

η µν = 
 0
0 −1 0 
0
0
0 −1
The Klein-Gordon equation may now be more compactly written in covariant form
as
(∂ 2 + m2 )φ = 0
(3.2)
Where the shorthand notation ∂ 2 ≡ ∂ µ ∂µ has been used.
21
CHAPTER 3. LECTURE 3
22
3.1.1 Relativistic scalar field theory
We now want to consider a relativistic field theory describing scalar fields. To
construct such a theory the field φ turns into a field operator.
b
φ(x) → φ(x)
However, and as seen earlier in lecture 2 in the case of the Schroedinger equation
in (2.1), the field operator is still satisfying the same equation as the field itself, that
is the KG-equation
(∂ 2 + m2 )φb = 0
Thus the field operator φb 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 the
equation of motion can be derived from the classical Lagrangian density of the field
by applying the variational principle. The Lagrangian density is in this particular
case given by:
1
1
1
1
1
L = η µν ∂µ ∂ν φ − m2 φ2 = φ̇2 − (∇φ)2 − m2 φ2
2
2
2
2
2
(3.3)
The Euler-Lagrange equation with respect to the field φ:
∂L
∂ ∂L
−
=0
∂φ ∂xµ ∂φ,µ
results in equation (3.2) as expected.
The energy density of the field is given by the Hamiltonian density H
H = φ̇π − L
= φ̇. With the Lagrangian
where the canonical momentum in this case is π ≡ ∂L
∂ φ̇
density as in (3.3), the Hamiltonian density turns out to be:
1
1
1
H = φ̇2 + (∇φ)2 + m2 φ2
2
2
2
To construct a quantum theory of these classical quantities, there are (at least)
two approaches. These are the canonical quantization or quantization by Fourier
expanding the field in its plane waves. The first procedure we saw in the second
lecture, and as a short reminder it looks like:
h
i
φ → φ̂
b t), π̂(x′ , t) = iδ(x − x′ )
⇒ φ(x,
π → π̂
where we assign the well-known constraint, represented by the δ-function to the
generalized coordinate and its conjugated momentum. The other way to quantize
is to Fourier expand the field in a finite volume. The field and its adjoint edition
then takes form as:
r
1 X
φk (t)eik·x
φ(x, t) =
V
k
3.1. RELATIVISTIC FIELDS
23
∗
φ (x, t) =
r
1 X ∗
φk (t)e−ik·x
V
k
here k takes discrete values. The constraint on the field φ that it is real valued
implies the following:
(3.4)
φk = φ∗−k
This is clearly seen by letting k → −k in the expression of the adjoint field and
then finally demanding it to equal the field itself:
r
1 X ∗
∗
φ−k , (t)e+ik·x
φ(x, t) = φ (x, t) =
V
k
We are allowed to perform this substitution in wave number since the wave number
k is arbitrary in the sense that whether we sum over all the positive or all the
negative values, the result should be the same in both cases. The next step is to
find the Lagrangian. To do so, we will make use of the following relation when
integrating the Lagrangian density to get the Lagrangian:
Z
′
d3 xeik·x e−ik ·x = V δkk′
V
here V is the volume of the box in which we are considering the Fourier modes
of the field. The Lagrangian now follows directly from the expression in (3.3) by
integrating over the whole volume of the box.
r r
Z
Z
1
1 X
′
1
3
3
φ̇k φ̇∗k′ eik·x e−ik ·x + . . .
L =
d xL =
d x
2
V V
kk′
X
1
φ̇k φ̇−k + . . .
=
2
k
1 2
1X
2 ∗
∗
(3.5)
φ̇k φ̇k − (k + m )φk φk
=
2
2
k
here the k2 and the m2 arise from the gradient term and the mass term in the
Lagrangian density as given in (3.3), respectively. So from the above expression we
see that the Lagrangian may be written as a sum over all Fourier modes k. We also
notice 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 explicit
we define the oscillation frequency to be ωk ≡ k2 + m2 and the Lagrangian then
turns into a harmonic oscillator for every mode k:
1 2 ∗
1X
∗
φ̇k φ̇k − ωk φk φk
L=
2
2
k
The field operator:
r
r
1
1
†
(âk + â−k ) and φ̂k (t) =
[âk e−iωkt + â†−k eiωkt ] (3.6)
φ̂k =
2ωk
2ωk
CHAPTER 3. LECTURE 3
24
and the adjoint operator:
φ̂†k
=
r
1
(âk − â†−k ) = φ̂−k
2ωk
here the last equality follows from the real-value constraint that is assigned to the
field φ as seen in equation (3.4). The operators are also set to satisfy the canonical
commutation relation [âk , â†k′ ] = δkk′ of the harmonic oscillator. We can now find
the Hamiltonian by:
Z
1X
Ĥ =
d3 xĤ =
ωk (â†k âk + âk â†k )
2
k
X
1
=
ωk (â†k âk + )
2
k
where, in the last step, the number operator N̂k = â†k âk has been used. The vacuum
state is |0i with the annihilation operator acting on it as âk |0i = 0. This fact leads
to a profound problem in quantum field theory that is yet to be solved. If we take a
closer look at the vacuum energy, E0 of the field as it is given by the Hamiltonian,
it is clearly seen that the energy diverges:
Ĥ|0i = E0 |0i ⇒ E0 =
1X
ωk = +∞!
2
k
√
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:
r
1 X
φ̂k (t)eik·x
V
k
r
X
1
† −i(k·x−ωt)
i(k·x−ωt)
=
âk e
+ âk e
2ωk V
k
r
X
1
† +i(k·x)
−i(k·x)
=
âk e
+ âk e
2ωk V
φ̂(x, t) =
(3.7)
k
In the last step the exponents have been expressed in terms of the 4-vector identity
k · x = ωk t − k · x.
3.1.2 External scalar field
We can expand the formal quantization process of the field to also account for
an 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) the
Lagrangian density changes from the free Lagrangian density
1
1
L0 = (∂µ φ)2 − m2 φ2
2
2
to a density containing an additional term determined by the external field U
1
1
1
L = (∂µ φ)2 − m2 φ2 − U φ2
2
2
2
The Lagrangian density is a classical quantity and we retrieve the (classical) equation 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) that
now replaces the mass m in the ordinary Klein-Gordon case as seen in (3.1). If we
define the effective mass as M 2 (x) = m2 + U (x), equation (3.8) takes form as
[∂ 2 + M 2 (x)]φ(x) = 0
The question is now how to construct a quantized theory in this case. On the
assumption that the particular Klein-Gordon equation has solutions of the form
u = up (x), we may find the field operator. The functions u(x), often referred to as
modes, are classical functions. The field operator is then given by
X
(3.9)
φ̂(x) =
[âp up + â†p u∗p (x)]
p
and is normalized by the commutator identity
[âp , â†p′ ] = δpp′
This particular example of how to quantize a theory with a non-trivial Lagrangian
is made to illustrate the fact that the field operator always will be of the form as
given in (3.9).
3.2 Vacuum energy
If the volume that we Fourier expand over is large enough, the sum in the expression of the vacuum energy can be replaced by an integral over all modes k.
Z
1X
d3 k
1
E0 =
(k2 + m2 )1/2
ωk = V
2
2
(2π)3
k
Z ∞
4πk2 dk 2
1
V
(k + m2 )1/2
=
2
8π 3
0
Z ∞
p
E0
1
⇒ ǫ0 =
= 2
dkk2 k2 + m2
V
4π 0
CHAPTER 3. LECTURE 3
26
which clearly diverges. A suitable comment in this manner should be that all effective 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 infinitely
large resolution, that is with a wavenumber that approaches infinity. It is therefore
common to introduce a so called cut-off K in the frequency. The maximal range
of validity of the theory is therefore found by letting the cut-off frequency be of the
order of the Planck scale. Hence:
kmax = K = MPlanck ≈ 1018 GeV
There are several ways to make the integral of the energy density ǫ0 finite. This
process 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 choice
of method of regularize cannot influence the final answer. As an illustration of
different types of regularization, let us consider two specific examples:
ǫ0 =



1
4π 2
1
4π 2
RK
0
R∞
0
√
dkk2 k2 + m2
√
dkk2 k2 + m2 e−k/K
The first case has the cut-off explicit as the upper limit of the integral, while the
other introduces an exponentially damping term in the integrand to prevent divergence. The vacuum energy in the special case of a massless scalar field is to first
order given by:
ǫ0 =



1
4
16π 2 K
3
K4
4π 2
+ ...
+ ...
illustrating the problem that the vacuum energy can not be properly described since
here two different regularizations give different answers. The vacuum energy problem is explicitly shown if we try to compare the predicted value of the vacuum energy with that of today’s best fit to observations. In this case we use as the predicted
value a term ∼ K 4 . The observed vacuum energy density ρΛ (the cosmological
constant) is
ρΛ = µ4
where µ ≈ 10−3 eV. We then get
K4
ǫ0
∼ 4 ∼
ρΛ
µ
1018 GeV
10−3 eV
4
= 10120
This is a huge error! Hence, the field theoretical description in this manner is
no way near of giving a correct estimate of the value of the vacuum energy as
it is observed in the universe. Actually this misconception between theory and
observations is the largest ever observed.
3.3. THE CASIMIR EFFECT
3.2.1
27
Fermions
From the second lecture we saw how a field theory for fermions could be extracted
simply by substituting the commutator with anticommutators
[, ]
|{z}
Bose−Einstein
→
{, }
| {z }
F ermi−Dirac
and vice versa. The anticommutator is then defined to obey the canonical commutation relation, only now in anticommutator form
{b̂k , b̂†k′ } = δkk′ = [âk , â†k′ ]
The Hamiltonian in these two cases is
X
1X
1
Ĥbos =
ωk {â†k , âk } =
ωk (â†k âk + )
2
2
k
k
X
1X
1
ωk [b̂†k , b̂k ] =
Ĥferm =
ωk (b̂†k b̂k − )
2
2
k
k
We observe that the vacuum energies of the fermions and bosons are equally large
with opposite sign and hence will cancel. For this to be fulfilled there must however be a special symmetry between fermions and bosons called supersymmetry
(SUSY). There is yet no physical evidence of such a symmetry as a fundamental
symmetry of nature. Nevertheless, the idea has been pursued since supersymmetry
gives, as a general feature, a vacuum energy which is 0, due to the cancellations
of the contributions from fermions and bosons. Such a cancellation is important
in supersymmetric quantum field theories, where the divergent contributions to the
vacuum energy may be avoided. Signs of SUSY are to be searched for at energies
of the order of TeV at CERN. Even if supersymmetry is found at these high energies, it is not necessarily the final solution to the vacuum energy problem. If we
4
now assign the vacuum energy with ǫ0 = KSUSY
the fraction between predicted
and observed values is
1TeV 4
ǫ0
=
= 1060
ρΛ
10−3 eV
so even with SUSY at energies E ∼ T eV there seem to be problems estimating
the vacuum energy.
3.3 The Casimir effect
The Casimir effect arises when two (parallel) plates are separated by, in this particular case a scalar field. There is then a measurable, attractive force between the
two plates that may, for instance be conductive iron plates. This is shown in figure
(3.1).
CHAPTER 3. LECTURE 3
28
A
Z
L
Figure 3.1: The Casimir effect: Two plates are separated at a distance L. A represents the area of the plate in the transverse directions x and y. An attractive force
between 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:
q
π
∂µ2 φ = 0 ⇒ ωnkT = kT2 + n2 K 2 , K =
L
The boundary conditions are as given by Dirichlet and Neumann, and look like
• Dirichlet: φ(z = 0) = φ(z = L) = 0.
• Neumann: ∂z φ|z=0 = ∂z |z=L = 0.
A closer look at the Dirichlet conditions yields
u(xT , z, t) ∼ sin(kz z)eikT ·xT e−iωt
π
with kz taking discrete values, kz = L
n, n = 1, 2, 3, . . ., and xT describing the
two transverse directions x and y. The vacuum energy can now be expressed in
terms of a two-dimensional integral over transverse momenta as:
E0 = A
Z
∞
q
d2 kT X 1
kT2 + n2 K 2
(2π)2
2
k=1
(3.10)
3.3. THE CASIMIR EFFECT
29
Here, A, is the area of the plates. The integral is divergent, so a regularization is
needed, that is, we make the integral finite:
φ̂2 = φ̂(x)φ̂(x′ )
The x′ can be expressed in terms of x, t − iǫ by a Schwinger time split which
effectively is to introduce an exponential damping. This results in
′
φ̂φ̂ ∼ eiωt e−iωt ∼ e−ωǫ
To proceed with the calculation of the vacuum energy, we perform a substitution
by ω 2 = kT2 + (nK)2 ⇒ ωdω = kT dkT .
Thus (3.10) can be written as:
E0
Z
∞
2πkT dkT X 1 −ǫω
ωe
= A
4π 2
2
k=1
∞ Z
A X ∞
2 −ǫω =
dωω e 4π
ǫ→0
n=1 nK
Z
∞
∞
A X ∂2
=
dωe−ǫω
4π n=1 ∂ǫ2 nK
{z
}
|
1 −ǫKn
e
ǫ
=
=
∞
A ∂ 2 1 X −ǫKn
e
4π ∂ǫ2 ǫ n=1
1
A ∂2 1
4π ∂ǫ2 ǫ eKǫ − 1
(3.11)
To deal with the last fraction in the above expression, it is necessary to consider a
relation derived by Euler
∞
X
t
Bn n
=
t
t
e −1
n!
n=0
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 = − 12 , B2 = 16 , and
B3 = 0, Thus the last fraction in the expression of the vacuum energy can then
now be expanded, so that the regularized vacuum energy, E0reg takes the form:
A ∂2 1 1
1 Kǫ (Kǫ)3
reg
E0 =
− +
−
+ ...
4π ∂ǫ2 ǫ Kǫ 2
12
720
ǫ→0
The vacuum energy density is further found simply by dividing with the Casimirvolume between the two plates, that is the spatial range of the field.
ǫreg
0 =
3
π2
E0reg
= 2 4−
AL
2π ǫ
1440L4
CHAPTER 3. LECTURE 3
30
If the plates are largely separated, hence L → ∞, the vacuum energy density is not
affected by the Casimir-energy, and therefore looks like:
ǫ0 =
3K 4
1
3
=
, with K =
2
4
2
2π ǫ
2π
ǫ
This vacuum energy density is often referred to as the open or free vacuum energy
density. The renormalized vacuum energy is finally found by subtracting the free
energy density, ǫ0 from the regularized energy density ǫreg
0 .
ǫren = ǫreg
0 − ǫ0 = −
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 way
that the vacuum energy between the plates reduces. This means that we have an
attractive force between the plates.
Before we end this (long) lecture, let us take a brief look at the mathematics
with Bernoulli numbers.
3.4 Bernoulli numbers
Bernoulli numbers are defined, as we have seen, by the Euler expansion:
∞
X Bn
t
=
tn
et − 1 n=0 n!
(3.12)
The sum is quite similar to the power expansion of the natural exponential function
ex . If we treat the B’s as ordinary numbers, Bn → B n , this allows us to rewrite
equation (3.12) as
∞
X
Bn n
t
=
t = eBt
et − 1 n=0 n!
Solving for t results in
t = (et − 1)eBt = e(B+1)t − eBt
∞
X
(B + 1)n − B n n
t
= (B + 1)t − Bt +
n!{z
|
}
n=2
!
=0
And finally the Bernoulli numbers are given by the explicit formula
B n = (B + 1)n , n ≥ 2
Examples of Bernoulli numbers are
3.4. BERNOULLI NUMBERS
• n=2:
• n=3:
31
B 2 = (B + 1)2 = B 2 + 2B + 1 ⇒ B1 = −
1
2
1
B 3 = (B + 1)3 = B 3 + |{z}
3B 2 + |{z}
3B +1 ⇒ B2 =
6
3B2
3(−1/2)
We can also expand this description to account for Bernoulli polynomials where
now Bn → Bn (x):
∞
X
Bn (x) n
t
xt
e
=
t
t
e −1
n!
n=0
= eBt ext = e(B+x)t =
∞
X
(B + x)n
n=0
n!
tn
(3.13)
So from the relation 3.13) above, we retrieve the explicit formula of the Bernoulli
polynomials.
B n (x) = (B + x)n
Actually, the Bernoulli numbers are just a special case of the Bernoulli polynomials.
32
CHAPTER 3. LECTURE 3
Chapter 4
Lecture 4
We will in this lecture study different ways to regularize the divergent field. We
will also see how the divergent integral stated below can be made finite.
E0
=
A
Z
∞
q
d2 kT X 1
kT2 + n2 K 2
(2π)2
2
(4.1)
k=1
4.1 Regularization
Last lecture we were looking at the Casimir effect. To summarize we considered
two metal plates with an observable attractive force between them, caused by a
massless scalar field with Lagrangian density L = 12 (∂µ φ)2 . We found the renorπ2
malized energy density εren
φ = − 1440L4 .
If we now consider the 1-dimensional Casimir effect we get the energy
E0 =
∞
1π X
π
n=
(1 + 2 + 3 + . . .) = S = ∞
2L
2L
(4.2)
n=1
Here we will need to regularize. We use the exponential damping method and get
the finite sum
∞
∞
X
∂
1
∂ X −εn
reg
−εn
e
= −
S
= lim
ne
= −
ε
ε→0
∂ε n=1
∂ε e − 1
n=1
and here we use our Bernoulli trick again. We can write the last factor as
∞
1
1 ε
1 X Bn n 1 1
ε
ε3
=
=
ε
=
−
+
−
+ ...
eε − 1
ε eε − 1
ε
n!
ε 2 12 720
n=0
This gives us the regularized sum
1
1 1
ε2
1
1
∂
reg
= 2 +0−
+
+ . . . = 2−
S
= −
ε
∂ε e − 1 ǫ→0 ε
12 240
ε
12
ε→0
33
CHAPTER 4. LECTURE 4
34
Where the first term is the divergent vacuum energy and the second term is the
finite, renormalized contribution. The interesting physics is described by this last,
finite term. We will now try to isolate this term by applying zeta-function regularization.
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
∞
X
1
ζ(s) =
ns
n=1
where s > 1. It can also be expressed with help of the Γ-function
Z ∞
1
xs−1
ζ(s) =
dx x
Γ(s) 0
e −1
(4.3)
where still s > 1. Here we have used that
1
e−x
=
= e−x + e−2x + . . .
ex − 1
1 − e−x
and the integral definition of the Γ-function
Z ∞
dxe−x xs−1
Γ(s) =
0
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
B2
1 1
1
=− · =−
2
2 6
12
which is the same as we got above. Funny.
ζ(−1) = −
4.2 Dimensional regularization
Now we will consider the same effect, but with extra dimensions. We will denote
the number of spatial dimensions d, and the total number of dimensions D, that is
with one time dimension we have D = d + 1. In this case we will have to solve
the following integral to find the energy
Z
−N
dd k
I=
k 2 + m2
d
(2π)
4.2. DIMENSIONAL REGULARIZATION
35
—————————————————————————————One funny thing that is not really relevant at this point is that the solid angles can
be written as a series of integrals in the different dimensions. This is best shown
with some examples
Z π
Ω2 = 2π
dθ sin θ = 4π
0
Z π
Z π
dθ2 sin2 θ2 = 2π 2
dθ1 sin θ1
Ω3 = 2π
Z0 π
Z0 π
Z π
8
2
dθ1 sin θ1
Ω4 = 2π
dθ2 sin θ2
dθ3 sin3 θ3 = π 2
3
0
0
0
and so on.
—————————————————————————————Table 4.1: Solid angles
but we can rewrite this with the help of the Γ-function
Z ∞
Z
1
dd k
2
2
dssN −1 e−s(k +m )
I =
d
(2π) Γ(N ) 0
Z
Z ∞
1
dd k −sk2
N −1 −sm2
=
e
dss
e
Γ(N ) 0
(2π)d
If we assume spherical symmetry, we can rewrite this with help of the solid angle
as dd k = Ωd−1 kd−1 dk. Now we can rewrite the last integral
Z ∞
Z
2
1
dd k −sk2
′
dkkd−1 e−sk
e
=
Ωd−1
I =
d
d
(2π)
(2π)
0
And if we substitute with t = sk2 we recognize the Γ-function
d
1
1
Ω
Γ
I′ =
d−1
d
−1
(2π)d
2
2s 2
(4.5)
Now we need to find Ωd−1 . As we know we can rewrite a Gaussian integral in d
dimensions as a product of the solid angle and the Γ function
Z
Z ∞
Γ d2
d−1 −k 2
d −k 2
dkk e
= Ωd−1
d ke
= Ωd−1
2
0
but we can also imagine that just raising a one dimensional Gaussian integral to the
power of d would do the same job
Z ∞
d
Z
d
−k 2
d −k 2
dke
= π2
=
d ke
−∞
CHAPTER 4. LECTURE 4
36
And if we combine these expressions we find the solid angle
d
Ωd−1
we can insert this into (4.5) and we get
I′ =
2π 2
=
Γ d2
1
2d π
d
d
2
s1− 2
Great! Now let’s insert this in our original integral
I=
1
d
d
Γ(N
)
2 π2
1
Z
∞
d
2
dssN − 2 e−sm
0
and once again we recognize the Γ-function and finally get
Γ N − d2
d
(m2 ) 2 −N
I=
d
(4π) 2 Γ(N )
1
(4.6)
Well, if we now try our ordinary two dimensions in this integral we get
1 Γ − 32
1
m3
2 32
I(d = 2, N = − ) =
=
−
(m
)
2
4π Γ − 12
6π
where we have used that Γ(z + 1) = zΓ(z). By comparing with the original
expression for EA0 that we had in equation (4.1), we also get the energy per area
∞
∞
n=1
n=1
1 1 X π 3 3
1 1X 3
E0
n
m (n) = −
=−
A
6π 2
6π 2
L
and the energy density
∞
E0
1 π3 X 3
n
=−
AL
6π 2L4 n=1
where we recognize the sum as the zeta-function
we get
π2
E0
=−
AL
1440L4
as before.
P∞
n=1 n
3
1
= ζ(−3) = − 120
and
4.3. QUANTUM STATISTICAL MECHANICS
37
4.3 Quantum statistical mechanics
We all know the eigenvalue equation for the Hamiltonian Ĥ|ni = En |ni, and
the probability for a specific P
state , Pn = Z1 e−βEn where β = kB1T . The sum of
probabilities should be unity n Pn = 1 and that leads us to the partition function,
Z=
X
e−βEn =
X
X
hn|e−β Ĥ |ni =
ρnn = T r ρ̂
n
n
(4.7)
n
Where ρ̂ is the density operator and ρmn is the corresponding density matrix. From
this we get the expectation value of an observable A
hÂi =
=
X
n
Pn hn|Â|ni =
1 X −βEn
hn|Â|ni
e
Z n
1
1
1 X
hn|e−β Ĥ Â|ni = T r ρ̂Â = T r Âρ̂
Z n
Z
Z
So we can find the energy, as the expectation value of the Hamiltonian
1
1
E = hĤi = T r ρ̂Ĥ = T r(e−β Ĥ Ĥ)
Z
Z
∂
1
∂
∂
1
−β Ĥ
ln Z
−
T re
=
−
Z=−
=
Z
∂β
Z
∂β
∂β
Where Z = e−βF and F is Helmholtz free energy. This gives us
E=
∂
∂F
∂F
(βF ) = F + β
=F −T
= F + TS
∂β
∂β
∂T
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 the
Lagrangian density
1
L = (∂µ φ)2
2
and this gives us the following Hamiltonian
X
X 1
=
Ĥk
Ĥ =
ωk N̂k +
2
k
k
Where N̂k = â†k âk . We get the energy of each state
∞
1 −β (nk + 1 )ωk
1 X
ˆ
2
ωk nk +
e
Ek = hHk i =
Z
2
nk =0
CHAPTER 4. LECTURE 4
38
where
∞
X
Zk =
1
−β (nk + 12 )ωk
e
nk =0
so that we have
Ek = −
e− 2 βωk
=
1 − e−βωk
1
ωk
∂
ln Zk = ωk + βω
∂β
2
e k −1
where the first term is the zero-point energy, and the second term is the thermal
energy. This gives us the total energy for the gas
X
X 1
ωk
ωk
ωk + βω
= E0 +
E=
βω
2
e k −1
e k −1
k
k
where E0 is the vacuum energy. If we neglect the vacuum energy, the expression
of the energy can be approximated as an integral (in arbitrary dimensions) by
E=V
Z
dd k
k
d
βk
(2π) e − 1
Then we find the energy density, ε
Ωd−1
E
=
ε=
V
(2π)d
Z
∞
0
dkkd
eβk − 1
and if we substitute with x = βk we get
ε=
Ωd−1 1
(2π)d β d+1
Z
∞
|0
xd
dx x
e −1
{z
}
Γ(d+1)ζ(d+1)
where we see that the integral of the energy density can be written as a product of
the Γ-function and the Riemann-zeta function (see (4.3)).
ε=
Ωd−1
Γ(D)ζ(D)(kB T )D
(2π)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
(kB T )4 ≡ aT 4
30
(4.8)
4.3. QUANTUM STATISTICAL MECHANICS
39
which we recognize as the Stefan-Boltzmann law. This gives us the entropy density
s = VS
∂s
1 ∂ε
4
=
= 4aT 2 ⇒ s = aT 3
∂T V
T ∂T V
3
Then we find the pressure by this Maxwell relation
∂p
∂S
4
=
= s = aT 3
∂T V
∂V T
3
which gives
1
1
p = aT 4 = ε
3
3
For D 6= 4 we get p = 1d ε, and we will get the energy-momentum tensor

and this is traceless
ε 0
0 ε

d

hTµν i = 0 0
0 0

.. ..
. .
0
0
ε
d
0
..
.

0 ...
0 . . .

0 . . .

ε

.
.
.

d
.. . .
.
.
hT µµ i = gµν hTµν i = ǫ − ǫ = 0
The expectation value of the trace is always zero, regardless of the number of
dimensions. We remember that we had the Lagrangian density L = 21 (∂φ)2 , and
this gives us the energy-momentum tensor
1
Tµν = ∂µ φ∂ν φ − ηµν (∂λ φ)2
2
with the trace
T µµ = (∂µ φ)2 −
D
(∂λ φ)2 =
2
1−
D
2
(∂µ φ)2
so the trace itself is equal to zero only for D = 2, while, as we have seen, the
expectation value of the trace is zero for all dimensions. This is very strange.
40
CHAPTER 4. LECTURE 4
Chapter 5
Lecture 5
So far we have been considering only free fields. In the real world no fields are
really free, so it is time to get some interactions into our formalism. To do this, we
will introduce a field theory with non-zero temperature, and use a formalism based
on 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
X
Z = T re−β Ĥ =
hn|e−β Ĥ |ni
n
In coordinate basis, where q̂|qi = q|qi and hq|q ′ i = δ(q − q ′ ) we can in a similar
way write Z as
Z
Z = T re−β Ĥ = dqhq|e−β Ĥ |qi.
(5.1)
This is leading us to the path integral formalism. We are used to a time propagator
on the form e−itĤ , but this is formally the same structure as eβ Ĥ 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 the
movement of a quantum mechanical system in imaginary time.
Remembering the density operator introduced in (4.7), we now define a density
matrix in coordinate basis as
ρ(qb , qa ; β) =
hqb |e−β Ĥ |qa i
Ĥ
= hqb | |e−εĤ e−ε{z
...e−εĤ} |qa i
N :N ε=β
Later we will (of course) take the limit (ε → 0, N → ∞).R As if this doesn’t look
nasty enough, we will now insert a complete set of states, dq|qihq| = 1, between
41
CHAPTER 5. LECTURE 5
42
all the e−εĤ s in the above expression, giving us a product of N − 1 integrals:
ρ(qb , qa ; β) =
N
−1 Z +∞
Y
n=1
−∞
dqn hqb |e−εĤ |qn−1 ihqn−1 |e−εĤ |qn−2 i. . . hq2 |e−εĤ |q1 ihq1 |e−εĤ |qa i
(5.2)
All these matrix elements have the same form, so it is sufficient to calculate only
one of them. We will try to find a general expression for this element. Using the
well-known forms of the Lagrangian and Hamiltonian
L =
Ĥ =
m 2
q̇ − V (q)
2
1 2
p̂ + V (q̂)
2m
we have (to first order in ǫ) that:
hq ′ |e−εĤ |qi = hq ′ |1 − εĤ|qi
=
hq ′ |qi − εhq ′ |H(p̂, q̂)|qi
=
R
=
=
=
δ(q ′ − q) −ε
| {z }
Z
Z
dp i(q ′ −q)p
e
2π
∞
−∞
∞
−∞
r
Z
e−iqp
z }| {
dp
hq ′ |H(p̂, q̂)|pi hp|qi
{z ” }
2π “|
p2
+V (q ′ )
2m
hq ′ |pi
dp i(q′ −q)p e
1 − εH(q ′ , p)
2π
dp i(q′ −q)p −ε
e
e
2π
m −ε
e
2πε
»
m
2
“
q ′ −q
ε
»
”2
p2
+V
2m
–
(q ′ )
–
+V (q ′ )
(5.3)
Actually, Dirac obtained this matrix element already in the 30s, but Feynman was
the first one to use it when considering the product from (5.2). This product now
yields
#)
( N " −1 Z ∞
m N/2 NY
X m qn − qn−1 2
ρ(qb , qa ; β) =
ε
+ V (qn )
dqn exp −
2πε
2
ε
−∞
n=1
n=1
See figure 5.1 to get an idea of how the path integrals are working. We are
getting the integral in the limit ε → 0 and N → ∞. Using the definitions
qn ≡ q(τn ) , τn = nε , β = N ε
we have
dqn
qn − qn−1 =
≡ q̇n .
ε
dτ
ε→0
5.1. PATH INTEGRAL FORMALISM
τ
β
N
ǫ
43
1
0
0
1
0
1
0
1
0
1
0
1
1111111111111111111
0000000000000000000
0
1
0
1
0
1
0
1
1111111111111111111
0000000000000000000
0
1
0
1
0
1
0
1
0
1
1111111111111111111
0000000000000000000
0
1
0
1
0q q q
1
q
q
2
1
b
a
Figure 5.1: Keeping qa and qb constant we sum over all possible positions in
between
Then, in the same limit
N
−1
X
n=1
ε→
Z
β
dτ
0
and we get that the density matrix now looks like
ρ(qb , qa ; β) =
where
Z
D ′ qe−
Rβ
0
dτ LE (q,q̇)
(5.4)
−1
m N/2 NY
dqn Dq=
2πǫ
n=1
′
N →∞
2
and LE = m
2 q̇ + V (q) is the Euclidean Lagrangian, which is the continuous limit
of the sum in the former exponents. Now, introducing the imaginary time, τ , by
2
t → −iτ and comparing with our good, old Lagrangian L = m
2 q̇ − V (q) this
yields
" #
m dq 2
L→−
+ V (q)
2 dτ
and we see that L → −LE , which shows that we are now working in a Euclidean
metric (ds2 = dτ 2 + dx2 ) instead of a Minkowski metric (ds2 = dt2 − dx2 ). This
field theory is therefore called an Euclidean field theory.
The integral in (5.4) is of course not trivial, but it can be calculated analytically
in some cases, and we will soon consider the case of a harmonic oscillator. What
we really are interested in, though, is not the matrix element, but the partition
function Z. We assume the integral to be over a periodic path, and thus we have
CHAPTER 5. LECTURE 5
44
qa = qb = q0 = qβ ≡ q. The partition function as seen in (5.1) now yields
Z ∞
dqρ(q, q; β)
Z =
−∞
Z
Rβ
=
Dqe− 0 dτ LE (q,q̇)
where
Z
Dq =
N
m N/2 Z Y
dqn .
2πε
n=1
5.1.1 The partition function for the harmonic oscillator
Considering a harmonic oscillator potential we have
LE
1 2 1 2 2
q̇ + ω q
2
2
1 2 1 2 2
q̇ − ω q
2
2
=
L =
and our partition function yields
h
i
Z
R
dq 2
− 21 0β dτ ( dτ
) +ω2 q2
.
Z = Dqe
And since we have periodic paths, q(0) = q(β):
Z
0
β
Which gives
Z=
dq
dτ
Z
2
dτ = q q̇|β0 −
|{z}
=0
1
Dqe− 2
Rβ
0
Z
0
β
dτ q
d2
q
dτ 2
dτ q(τ )[−∂/tau2 +ω 2 ]q(τ )
Now, taking the periodic boundary conditions explicitly into account we have:
r
1X
q(τ ) =
qn eiωn τ
β n
r
1X
!
qn eiωn (τ +β) = q(τ )
q(τ + β) =
β n
This tells us that eiωn β = 1, and thus
ωn =
2π
β 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 formalism was introduced in solid state physics, but this periodicity will occur in all
quantum mechanical systems at a finite temperature.
Earlier we have performed a normalization over the volume. Analogously, here
we will perform a normalization over the τ -coordinate, and this will lead us to
Z
β
dτ eiωn τ eiωm τ = βδn,−m
0
This gives:
Z
Z
β
dτ q 2 =
0
X
qn∗ qn
n
β
dτ q̇ 2 =
0
X
q̇n q̇n ωn2
n
and our partition function becomes
Z
1 P∞
∗
2
2
Z =
Dqe− 2 n=−∞ qn (ωn +ω )qn
Y Z ∞ dqn
1 P
∗
2
2
√
=
e− 2 n qn (ωn +ω )qn
N
−∞ ( 2π)
n
where
qn =
r
1
(xn + iyn ).
2
∗ (ω 2 + ω 2 )q
Remembering that qn∗ (ωn2 + ω 2 )qn = q−n
−n , when summing over all
n
n (both positive and negative), the 1/2-factor in the exponent vanishes. If we also
insert for qn , we have
Y Z ∞ dxn Z ∞ dyn
2
2
√
√
e−(xn +yn )λn
Z =
N
N
−∞ ( 2π)
−∞ ( 2π)
n
r
∞
Y
1
=
λ
n
n=−∞
=
∞
Y
n=−∞
p
1
ωn2
+ 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 solving
divergent products using functional determinants. We have encountered integrals
CHAPTER 5. LECTURE 5
46
on the form
YZ
Z=
∞
1
dx
√ i e− 2 xi Aij xj
N
( 2π)
−∞
i
(5.7)
where Aij = Aji . Now we will use a rotation matrix, R, to transform x into a new
variable y. R is orthogonal, such that R−1 = RT . We have:
xi Aij xj = xT Ax = y T
T
2
−1
RAR
| {z } y = y AD y = yi ai δij yj = ai yi .
AD (diagonal)
Using this, (5.7) yields
YZ
∞
1
dy
2
√ i e− 2 ai yi
N
−∞ ( 2π)
i
1/2
1
Q
=
i ai
Z =
=
1
1
det− 2 AD = det− 2 A
Here, the last step follows from AD = RAR−1 ⇒ detAD = |AD | = |R||A||R−1 | =
|A|.
Now, using the expressions already obtained for Z in the case of a harmonic
oscillator, we have:
Y Z ∞ dqn
1 P
∗
2
2
√ e− 2 n qn (ωn +ω )qn
Z =
2π
−∞
n
=
=
∞
Y
p
1
ωn2 + ω 2
n=−∞
Z
R
− 12 0∞ dτ q (−∂τ2 +ω 2 )q
Dqe
1
= det− 2 −∂τ2 + ω 2
(5.8)
5.3 Generalized zeta-function regularization
When regularizing our divergent sums, we made use of Riemann’s ζ-function. For
the divergent products we will use a method called generalized zeta-function regularization. Physicists tend to claim that this method was invented by Stephen
Hawking, but in fact the mathematicians had already been playing around with it
for a long time when the physicists finally found it and saved it from the ocean of
uselessness.
Now, we have encountered problems on the form
Y
detM =
λn =?
n
5.3. GENERALIZED ZETA-FUNCTION REGULARIZATION
47
We define our generalized zeta-function, f (s) by1
f (s) =
∞
X
1
s
λ
n=1 n
,
s>1
Again it is possible to extend the definition to include all values of s. We then
rewrite the definition
∞
X
e−s logλn .
f (s) =
n=1
Differentiation gives
∞
X
′
f (s) = −
logλn e−s logλn
n=1
e.g.
f ′ (0) = −
Thus we see that
log detM =
X
n
∞
X
logλn .
n=1
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:
∞
Y
n=1
a = a · a · a · ...
= a1+1+1+...
=
ζ(0)
a
− 21
= a
=
r
1
a
That was easy! Let us take a small step further:
∞
Y
n=1
n = 1 · 2 · 3 · ...
′
1
= e−ζ (0) = e 2 ln(2π) =
√
2π
1
In the lectures it was denoted Z(s), but if we stick to the f(s) notation used by Petter Callin, we
avoid confusing it with the partition function.
CHAPTER 5. LECTURE 5
48
In this example λn = 1 and f (s) = ζ(s). As a last example, let us consider the
partition function for the harmonic oscillator that we saw in (5.6):
Z =
Y
n
=
1
2
ωn + ω 2
∞
1 Y
ω
2
1
ωn2
n=1
=
1
“
|{z}
2π
n
β
+ω 2
”2
1
1
2 Y
∞ ∞ Y
ω
ω2
2π
1+ 2
n
β
ωn
n=1
n=1
|
{z
}|
{z
}
β
2π=β
= 2π
(5.10)
Euler
The part of the denominator embraced by Euler is a known Euler product
∞ Y
n=1
ω2
1+ 2
ωn
=
sinh βω
2
βω
2
.
which gives us
Z =
=
1
2 sinh βω
2
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 sensible
answer.
From last lecture we remember that we expressed the partition function using
the 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
=
=
∂F
∂ω
1
log Z
β
1 X
log ωn2 + ω 2
2β n
#
"
∞
2π 2
1 X
2
log
n +ω
2β n=−∞
β
= −
=
=
=
∞
1 X 2ω
2β −∞ ωn2 + ω 2
βω
1
coth
2
2
1
1
+
2 eβω − 1
(5.12)
In the last equality we have used that
∞
X
−∞
n2
π
1
= coth a.
2
+a
a
Integration yields
F =
1
1
ω + log 1 − e−βω + const.
2
β
(5.13)
where this constant formally is infinite. This expression is in agreement with our
result in (5.11).
A standard example where these techniques have to be used is for a gas of
massless particles with a Lagrangian:
λ
1
L = (∂µ φ)2 − φ4
2
4
which does not have any exact solutions in quantum mechanics or in classical field
theory, thus one has to apply perturbation theory. In standard quantum field theory
(QFT) the temperature is set to zero. Finite temperature field theory will provide
us with the same Feynman diagrams and integrals in QFT, but we will now work
in a Euclidean space instead of a Minkowski space.
In QFT we are used to working with propagators of the form
h0|T {φ̂(x)φ̂(x′ )}|0i.
In the Euclidean theory the propagators will be the correlation function
hφ(x)φ(x′ )i
where the φs are classical fields. Later we will calculate these propagators.
50
CHAPTER 5. LECTURE 5
Chapter 6
Lecture 6
In this lecture we will continue our investigation of the partition function Z and
the free energy F for a free scalar field given by a free Lagrangian density L0 and
then finally turn the attention towards interacting scalar fields that are of physically
more interest. To go from free theory to an interacting one, we assign interacting
terms to the free Lagrangian. But before we do just that, let us take a short reminder
of what we arrived at in the end of the last lecture.
When we considered a harmonic oscillator at finite temperature, then the partition function was given by
Z
R
1 β
2
2
−β Ĥ
Z = T re
= Dqe− 2 0 dτ q(−∂τ +ω )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 leaving
us with :
r
∞
1 X
qn eiωn τ
(6.1)
q(τ ) =
β n=−∞
with ωn = 2π
β n, n = 0, ±1, ±2, . . . We then further derived the partition function
of the harmonic oscillator:
∞
Y
1
p
Z=
2
ωn + ω 2
n=−∞
which results in the free energy a’ la Helmholtz
F =
1
h̄ω
h̄ω + βh̄ω
2
e
|{z}
| {z− 1}
vacuum
(6.2)
thermal
The first term corresponds to the vacuum energy, while the second term gives the
energy contribution due to thermal excitation. We found a more compact and abstract representation of the partition function that looked like
Z = det−1/2 (−∂τ2 + ω 2 )
51
CHAPTER 6. LECTURE 6
52
and then of course
−βF = log Z =
1
log det(−∂τ2 + ω 2 )
2
(6.3)
The point of considering the βF term again is that it can be rewritten into a slightly
different form that is more convenient in some calculations. To see how this is
done, let us take a look at a diagonal matrix


a1 0 . . .


AD =  0 a2 . . . 
.. . .
..
.
.
.
We then have the well-known formulas for the determinant and the trace of the
matrix
Y
det AD =
ai
i
T rAD =
X
ai
i
Hence, the relation between trace and the determinant is log det A = T r log A
resulting in that (6.3) can be expressed as
1
βF = T r log(−∂τ2 + ω 2 )
2
(6.4)
Now we turn our attention towards free scalar fields and take a closer look at
how 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:
1
1
L = (∂µ φ)2 − m2 φ2
2
2
and the partition function is as always
Z = T re−β Ĥ
The scalar field is now considered not to be a function of time and space, but
instead a function of imaginary (Euclidean) time and space. This is the same shift
of 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 therefore
be:
φ(t, x) → φ(τ, x)
The scalar field is also constrained with the boundary conditions as we have seen
earlier, only now the field is periodic with respect to the Euclidean time τ. We can
then again expand the variable in powers of τ just as we saw in (6.1).
Hence:
φ(τ, x) = φ(τ + β, x)
r
∞
1 X
φn (x)eiωn τ
=
β n=−∞
r
1 X
=
φnk ei(k·x+ωn τ )
βV
k,n
The main point is now that what we have earlier done for one harmonic oscillator
can be applied to all the modes labeled with the index k. The partition function for
the free scalar field can therefore be expressed as:
Z=
Y
k,n
q
1
ωn2 + ωk2
and the free energy can be seen as a sum over all modes of the (one-dimensional)
harmonic oscillator in equation (6.2).
X 1
h̄ωk
h̄ωk + βh̄ω
F =
2
e k −1
k
Here each harmonic oscillator has frequency ωk =
energy can be written as
βF =
√
k2 + m2 . So then the free
1
T r log(−∂ 2 + m2 )
2
The partition function can be rewritten as:
Z
R 3
Rβ
Z = T re−β Ĥ = Dφe− 0 dτ d xLE
(6.5)
where LE is the Euclidean Lagrangian density which is found by letting t → iτ as
seen in lecture 6. In this shifting from real time to imaginary time, the Lagrangian
density of the scalar field changes somehow. We remember that the Lagrangian
was given by
1
1
1
1
1
L = (∂µ φ)2 − m2 φ2 = φ̇2 − (∇φ)2 − m2 φ2
2
2
2
2
2
CHAPTER 6. LECTURE 6
54
Now t → iτ implies that
in this particular case is:
∂φ
∂t
→ −i ∂φ
∂τ so then the Euclidean Lagrangian density
1
1
1
1 ∂φ 2 1
+ (∇φ)2 + m2 φ2 = (∂µ φ)2 + m2 φ2
−L → LE =
2 ∂τ
2
2
2
2
where, in the last step, we have used the shorthand notation ∂µ = (∂τ , ∇) ⇒ ∂ 2 =
∂2
+ ∇2 . So once again we return to the partition function and find that (6.5) now
∂τ 2
can be expressed as
Z
R
− 12 d4 x (∂µ φ)2 +m2 φ2
Z =
Dφe
Z
R 4
1
2
2
=
Dφe− 2 d xφ(−∂ +m )φ
1
= det− 2 (−∂ 2 + m2 )
So far so good. We have been through some piece of derivation to finally
retrieve an expression for the partition function of the (massive) scalar field φ, all
motivated by extending our knowledge of the harmonic oscillator to account for
more 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 free
energy 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 = − log Z =
1X
log(ωn2 + ωk2 )
2
k,n
with each frequency ωk given as previous in this lecture and is representing the
energy of each particle with wavenumber k.
We have in the previous lectures seen that in statistical mechanics the free energy was found by summing over all modes in the expression of the one-dimensional
harmonic oscillator. Then by considering equation (6.2), the free energy will be
X 1
ωk
ωk + βh̄ω
βF = β
2
e k −1
k
But there is also another possible approach which is to first sum over k and then
over n. The expression for the free energy then turns into
Z
d3 k
1 X
log(k2 + |{z}
m2 +ωn2 )
βF = V
3
2
(2π)
n
=0
(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 integral
is divergent. So in order to receive any physically meaningful information about
the radiation system, the integral must be made finite in some way. The method of
changing divergence into convergence is the well-known regularization. We prefer
dimensional regularization and, to keep the result as general as possible, we apply
regularization in arbitrary dimensions (as we did in lecture 4). Hence:
Z
1
1
Γ(N − d/2) 2 d/2−N
dd k
=
(m )
.
d
2
2
N
d/2
(2π) (k + m )
Γ(N )
(4π)
We choose N = 1. Furthermore:
Z
dm
2
:
Z
dd k
Γ(−d/2) 2 d/2
(m )
log(k2 + m2 ) =
d
(2π)
(4π)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:
∞ X
Γ(−3/2)
1
3
|ωn |
−
βF = V
2 n=−∞
(4π)3/2
and finally we have an answer of what we were looking for. Here the frequency
′
ωn is the Matsubara-frequency given by ωn = 2π
β n with the n s taking on discrete
values n = 0, ±1, ±2, . . . The free energy density is now simple to find, just by
dividing by the volume. If we write out the Γ′ s it looks like
√ ∞
1 43 π 2π 3 X
F
|n|3
=− √
F=
V
2β 4π4π β
n=−∞
(6.7)
The last sum can be rewritten as a sum of the form of the zeta-function and then be
evaluated. This is seen by
∞
X
n=−∞
|n|3 = 2
∞
X
n=1
|n|3 = 2ζ(−3) = 2
1
120
The pressure in the gas of massless particles is
P
⇒P
∂F
= −
∂V
=
T
= −F
π2
(kB T )4
90
which we recognize as the ordinary Stefan-Boltzmann pressure.
CHAPTER 6. LECTURE 6
56
6.2.1 Pressure in arbitrary dimensions
However, there is nothing stopping us from calculating the radiation pressure also
with arbitrary dimension number. An interesting question to ask would be about
how the expression of the pressure is given in, for instance, extra dimensions. The
space time then has a dimension D that represents both the (one) temporal dimension and all the spatial dimensions. (For extra dimensions D is of course greater
than four). The pressure is, according to (6.7) :
PD =
∞
1 X Γ(−d/2) 2π d d
|n|
2β n=−∞ (4π)d/2
β
(6.8)
Again the sum over n can be evaluated by the zeta-function as ζ(−d). It is possible
to write the expression of the radiation pressure in d dimensions in a more compact
form as
Γ(D/2)
ζ(D)(kB T )D
(6.9)
PD =
(π)D/2
The important point to notice in the above expression, is that the Γ′ s and ζ ′ s involve the total dimension number D and not the spatial dimension number d. The
explanation is that we want a more convenient formula and that ζ-functions with
different arguments D and − d can be related by the mathematical identity:
Γ(s/2)π −s/2 ζ(s) = Γ
1 − s −( 1−s )
π 2 ζ(1 − s)
2
We observe that if s = D = d + 1 is being substituted into the above formula, then
the 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 a
similar problem is considered, that is to find the free energy only now in Casimir
form, then we will observe that the Casimir description is the same as that of
Stefan-Boltzmann at zero temperature. The two situations are:
• Casimir: The Casimir energy (at zero temperature) can be found by considering (6.6) and perform the substitutions (as β → ∞)
X
n
X
n
→ β
→ V
Z
∞
−∞
Z
dkD
2π
dd k
(2π)d
6.3. FERMIONS
57
• Stefan-Boltzmann If the temperature is zero, then the free energy equals the
usual energy (F = E − T S) so that:
βF
Z
∞
X
π 2
dd−1 k
1
dkD
2
log k2 +
= βE = Vd−1
n + kD
β
d−1
2
(2π)
2π
L
n=1 |
{z
}
β
=
∞
X
1
Vd−1 β
2
n=1
Z
R
dd k
(2π)d
π 2
dd k
2
log
k
+
n
(2π)d
L
We can then find the energy density as
ρD =
Γ(D/2) ζ(D)
E
=−
V
(4π)D/2 LD
which is very similar to the expression for the pressure in (6.8), only now
with an opposite sign. This is due to
∂F
P =−
= −F → ρD
∂V T
what we have studied here, is the bosonic case with scalar particles. Now, we take
a look at the fermionic case.
6.3 Fermions
If we want to construct a Lagrangian formulation that describes fermions, then the
fermion field must necessarily be a Grassmann variable. Some of the properties of
Grassmann numbers are listed in table 1. The spinor describing the fermion field
must be antisymmetric, that is:
Ψi Ψj = −Ψj Ψi
But before we take a deeper look at the properties of the fermions and try to find
the partition function, let us first remind ourselves of what we already know in the
bosonic case. The partition function of the bosons is
Z
1 T
Z = Dφe− 2 φ Aφ
here A is a symmetric matrix with Aij = Aji and φ is a vector


φ1
 φ2 


φ= . 
.
 . 
φN
CHAPTER 6. LECTURE 6
58
By means of the Gaussian integrals seen earlier, the partition function therefore can
be rewritten
Z
1
1 T
Z = Dφe− 2 φ Aφ = det− 2 A
If φ is aq
complex field, it can be expressed as a linear combination of two real fields
as φ =
1
2 (φ1
+ iφ2 ) 6= φ∗ . The partition function will then take form as
Z
Z
∗
Z =
Dφ Dφ∗ e−φ Aφ
= det−1 A
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
1 T
(6.10)
Z = DΨe− 2 Ψ DΨ
Here Ψ is (still) a Grassmann variable, but now the matrix is antisymmetric, Dij =
−Dji and the vector Ψ is given by


Ψ1
 Ψ2 


Ψ= . 
 .. 
ΨN
It is then possible to show that the partition function in the fermionic case can be
written as
Z
1
1 T
Z = DΨe− 2 Ψ DΨ = det− 2 D
The Grassmann variable and its properties can give us some new and interesting
information about integrating with respect to the fermion field Ψ (which we will
have to do in order to find the partition function).
If we assume that f is a function of one Grassmann variable Ψ, a Taylor expansion gives:
1
Ψ2 + . . .
f (Ψ) = f (0) + f ′ (0)Ψ + f ′′ (0) |{z}
2
=0
We notice that the second order term, and all higher order terms in Ψ is zero. This
is due to the properties of Grassmann variables and the only two interesting terms
are a constant term f (0) and a term to first order in Ψ. We can then write down two
basis integrals where the integrands are a constant or Ψ, respectively.
Z
I0 =
dΨ
Z
I1 =
dΨΨ
6.3. FERMIONS
59
We will further assign some constraints to these integrals. The two integrals must
be invariant under a change of variable Ψ → Ψ′ = Ψ + α. The added term, α is a
constant Grassmann number, so then the differential is unchanged dΨ′ = dΨ. We
can now write the integrals as
Z
Z
′
I0 =
dΨ = dΨ
Z
Z
′
I1 =
dΨΨ = dΨΨ
Z
Z
Z
!
⇒ I1 =
dΨ(Ψ + α) = dΨΨ + dΨα = 0
| {z }
=0
So due to the last term in the equation above, it is clear that the first integral, I0 is
equal to zero. Hence
Z
I0 =
dΨ = 0
Z
I1 =
dΨi Ψj = δij
The properties of these two basic integrals will be of great importance when we
later in this lecture will calculate the fermionic partition function (See the example). However, we will first consider a different change of variable given by
Ψ → Ψ′ = αΨ and still demand the integral to be invariant. We can then retrieve some properties about the differential dΨ. (So far these properties have been
unknown, and we have only used it as an ordinary differential)
Z
I1 =
dΨ′ Ψ′ = 1
Z
Z
Z
!
=
d(αΨ)αΨ = α d(αΨ)Ψ =
dΨΨ
1
dΨ
α
If we on the other hand consider bosonic variables, then the differential behaves as
⇒ d(αΨ) =
d(αx) = αdx
So the differentials in the two cases of fermionic and bosonic variables are in some
sense opposite in their behavior. Now, let us take a look at the fermionic determinant, and hence the partition function in the situation with N = 2 where the
formalism 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
Z
Z=
dΨ1
1
T
dΨ2 e− 2 Ψ1 DΨ2
CHAPTER 6. LECTURE 6
60
The matrix D is antisymmetric, and given by
0 a
D=
−a 0
So then it is easy to calculate the exponent in the expression of the partition
function and find Z:
ΨT DΨ = 2aΨ1 Ψ2
furthermore
Z
Z
Z
Z
1
Z = dΨ1 dΨ2 e−aΨ1 Ψ2 = dΨ1 dΨ2 [1 − aΨ1 Ψ2 + ( )2 + . . .]
2
And here comes our little trick in. We remember from our derivation
R of the two
basic 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 order
terms in Ψ, respectively. The only contribution to the partition function comes
from the I1 -like integral, and we write
Z
Z
Z = dΨ1 dΨ2 (−aΨ1 Ψ2 )
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 know
that each interchange of fermion fields correspond to a change in the sign:
Z
Z
Z =
dΨ1 dΨ2 (+aΨ2 Ψ1 )
√
= a = a2
1
= det 2 D
since det D = a2 . If we instead have complex fields,qthe 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
Z
∗
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 determinants
was Schwinger. He developed some of the theory when considering Quantum
Electro Dynamics (QED). The Lagrangian density in QED is
1 2
L = Ψ̄(iγ µ ∂µ − m − eγµ Aµ )Ψ − Fµν
4
6.4. INTERACTING FIELDS
61
We can again find the partition function (now with respect to real time t)
Z
Z
Z
Z
1 2
4
µ
Z =
dΨ dΨ̄ DAµ exp i d x[Ψ̄(iγ ∂µ − m − eγµ Aµ )Ψ − F µν ]
4
Z
R 4
1 2
=
DAµ det(iγ µ ∂µ − m − eγµ Aµ )ei d x(− 4 Fµν )
The determinant in the expression above can be rewritten. We have earlier seen that
there is a relation between determinants and traces given by elog det( ) = eT r log( )
so then the partition function is
Z
R
1 2
µ
Z = DAµ ei dx[− 4 Fµν +tr log(iγ ∂µ −m−eγµ Aµ )]
where now the effective Lagrangian of the fermion field is written as
1 2
+ tr log(iγ µ ∂µ − m − eγµ Aµ )
− Fµν
4
by means of the special trace tr. The two variants of trace, T r and tr differ in the
sense that
R
• Tr: does also involve a summation over the integral d4 x
R
• tr: does not sum over d4 x
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
1
L = (∂µ φ)2 − U (φ)
(6.11)
2
The first term describes the non-interacting, free part of the field and U (φ) gives
the interacting potential of order three or higher. When only considering the free
part, we have seen that the partition function can be analytically solved by means
of Gaussian functional integrals. If interactions are included, the expression of the
partition function is often so complicated that numerical methods must be used or
perturbation theory applied in order to solve for it. Consider a potential, leading to
interactions, and given by
1 2 2
m φ + gφp
(6.12)
2
Here g is a coupling constant which, as we will see later, is directly related to
whether a theory is renormalizable or not. If p > 2 and we want to calculate the
functional integral by
Z
R 4 1
2
Z = Dφe− d x[ 2 (∂µ φ) +U ] ,
(6.13)
U (φ) =
this cannot be done analytically.
CHAPTER 6. LECTURE 6
62
6.4.1 Dimensional analysis
It was stated in the previous section that the dimension of the coupling constant is
of importance in understanding renormalization. To see this, there will be a need
of a quick review of what is called dimensional analysis. If the constants h̄ and c
are set to unity, then all quantities will be measured in length or mass. We will here
illustrate dimensional analysis with respect to mass-dimensions:
dim[m] = +1
dim[E] = +1
dim[pµ ] = +1
(6.14)
with pµ = i dx∂µ 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 considering the dimension of the free Lagrangian density (which often ends up in the
exponent) :
!
dim[dD x(∂µ φ)2 ]
=
0
⇒ −D + 2dim[∂µ φ]
=
⇒ dim[∂µ φ]
=
⇒ dim[φ]
=
0
D
= dim[φ] + 1
2
D−2
2
So in our common space time with D = 4, we have that dim[φ] = +1. The
field can also be a dimensionless quantity, then with D = 2. This is the case in
for instance string theory. If we are to find the connection between the coupling
constant g and renormalization, we must also include the potential. The potential
forms a separate term in the exponent, and since dim[d4 x] = −4 in four spacetime dimensions, then the dimension of the potential must be four. With the same
potential 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, the
φ4 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 case
gφ3 gives a renormalizable theory. Therefore the Lagrangian density must be
1
L = (∂µ φ)2 − gφ3
2
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). All
the interesting physics is contained in the partition function, and we rewrite it as
Z
Z
R 4
R
− d4 x[ 12 (∂µ φ)2 +U ]
= Dφe− d x(L0 +LI )
Z = Dφe
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 be
solved to find an exact solution. However, if we rewrite the partition function
Z ∞
1 2
4
dxe− 2 x −gx
Z=
−∞
and if the coupling constant is small, then the partition function can be written as
an expansion in x4 . This yields:
Z ∞
1 2
(gx4 )2
4
− ...
dx exp − x
1 − gx +
Z=
2
2!
−∞
So we see that the partition function written on this form is nothing but a series
of different Gaussian integrals that each and one of them can be calculated separately. The partition function can therefore be expressed as a series in the coupling
constant g.
Z = C0 + gC1 + g2 C2 + . . .
This perturbative expansion of Z is only a good approximation as long as g is very
small. In the following lectures this is the approach we will follow to find the
partition function. We will then see that the integrals in (6.13), giving Z, can be
treated and interpreted by means of simple Feynman diagrams.
64
CHAPTER 6. LECTURE 6
Chapter 7
Lecture 7
We will now end the part with field theory by constructing perturbation theory
by means of Wicks theorem. From earlier we remember that we could find the
expectation values with help of the density operator, ρ̂, given by
ρ̂ = e−β Ĥ
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
Rβ
R 3
Z = T r ρ̂ = Dφe− 0 dτ d xLE (φ,φ̇) = e−βF
(7.1)
where LE is the Euclidean Lagrangian density, and F is the free energy. Now if
we have an operator  = A(φ̂), we can use this to find its expectation value as
follows
Z
Rβ
R 3
1
1
1
hAi = T r ρ̂Â = T rA(φ̂)e−β Ĥ =
(7.2)
DφA(φ)e− 0 dτ d xLE
Z
Z
Z
where the last equality follows from the same procedure as we used to find the
functional integral for Z.
7.1 Theory of free fields
For free fields the Lagrangian, LE , doesn’t have any interaction terms, so it turns
out simply
1
1
LE = L0 = (∂µ φ)2 + m2 φ2
2
2
and from (7.1) we have that the corresponding free field partition function becomes
Z
R
R 3
1
2
2
Z0 = Dφe− 2 dτ d xφ(−∂τ +m )φ
65
CHAPTER 7. LECTURE 7
66
Now we want to express the the integral in the exponent as a matrix product, by
writing the variables as matrix/vector indices as φ(x) = φ(τ, x) = φτ,x . These
indices are continuous, just as when we write ψx = hx|ψi = ψ(x) in ordinary
quantum mechanics. Now the free field partition function can be written
Z
1 †
Z0 =
Dφe− 2 φ Aφ
Z
1
=
Dφe− 2 φi Aij φj
1
1
= det− 2 A = det− 2 (∂τ2 + m)
Q
Where Dφ = i dφi .
We will now find some of the simplest expectation values, first in free field
theory. We find the expectation value for φ by using (7.2).
Z
R
R 3
1 β
1
2
2
Dφφ(x)e− 2 0 dτ d xφ(∂T +m )φ
hφ(x)i =
Z0
The exponential is an even, Gaussian function, while φ(x) is an odd function. This
gives that the integrand is odd and the integral vanishes
hφ(x)i = 0
hφ(x)i is called a 1-point function. A 2-point function on the other hand is
Z
R 3
R
1 β
1
2
2
′
hφ(x)φ(x )i =
Dφφ(x)φ(x′ )e− 2 0 dτ d xφ(∂T +m )φ
Z0
and this is also called a correlation function, or a correlator for short. It can be
interpreted as a quantum mechanical propagator in imaginary time.
If we now make a substitution Rφ → φ′ R= Rφ, (on component form φ′i =
Rij φj ) where R−1 = RT so that Dφ = Dφ′ . This gives us the partition
function
Z
1 ′ ′
′
Z0 = Dφ′ e− 2 φi Aij φj
where A′ = RAR−1 is diagonal. For non-free fields we will consider an additional
term in the exponent
Z
1 T
T
Z =
Dφe− 2 φ Aφ+B φ
Z
1
=
Dφe− 2 φi Aij φj +Bi φi
where Bi is called a source function. This gives the expectation value for the field
φi
Z
1 T
T
1
Dφφi e− 2 φ Aφ+B φ
hφi i =
Z
∂
1 ∂
Z=
ln Z
=
Z ∂Bi
∂Bi
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.
Z
1 ∂ 1
− 12 φT Aφ
=
hφi i0 =
Z
Dφφi e
(7.3)
Z0
Z0 ∂Bi B→0
and we can use the same trick on the correlator (the 2-point function)
Z
1 ∂ ∂
1
− 12 φT Aφ
=
Z Dφφi φj e
hφi φj i0 =
Z0
Z0 ∂Bi ∂Bj B→0
(7.4)
If we again substitute φ → φ′ = Rφ, we get the non-free field partition function
Z
1 ′ ′
′
′ ′
Z = Dφ′ e− 2 φi Aij φj +Bj φj
where Bj′ = BR−1 . If we write out the integral we get
Z =
YZ
i
∞
−∞
1 ′
′
′ ′
dφ′
√ i e− 2 φi Ai φi +Bi φi
2π
1
1
′ T A′ −1 B ′
1
1
′ −1 B ′
i
1
1
T A−1 B
= det− 2 Ae 2 B
= det− 2 Ae 2 Bi ai
= det− 2 Ae 2 B
1
−1
= Z0 e 2 Bi Aij
Bj
where ai are the diagonal components of A′ (remember that A′ is a diagonal matrix). Now, we have found the full partition function, and can then find the two-point
correlation function. If we go back to the correlation in equation (7.4) function with
this new expression for Z we get
Z
1
1 ∂ ∂
− 21 φi Aij φj
Z hφi φj i0 =
Dφφi φj e
=
Z0
Z0 ∂Bi ∂Bj B→0
−1
1
1 ∂ ∂
Z0 e 2 Bi Aij Bj =
= A−1
ij
Z0 ∂Bi ∂Bj
B→0
So we see that the expectation value of the two-point correlation function is the
inverse of the matrix, Aij that is in the exponent. This is not a coincidence. If we
look at the propagator ∆ for the scalar field, then
∆∼
−∂ 2
1
1
→ 2
2
+m
p + m2
(7.5)
The propagator should be compared with Aij ∼ ∂ 2 + m2 as seen earlier. The term
in the exponent, when calcualating the partition function, is usually the Lagrangian
density. The Lagrangian density of the scalar field is L0 = 12 φ(−∂τ2 + m2 )φ and
CHAPTER 7. LECTURE 7
68
should be compared with the propagator given above in (7.5). This is also illustrated by considering the Dirac propagator, D and the corresponding Lagrangian.
These are
1
with L0 = Ψ̄(iγ µ ∂µ − m)Ψ
D∼ µ
iγ ∂µ − m
To be a little more specific, we look at the partition function given by
Z
Z=
1
∗
Dφk,n e− 2 φk,n (k
where
k2 =
2π
n
β
2
2 +m2 )φ
k,n
+ k2
We have the correlation function between two modes in momentum space
hφ∗k,n φk′ ,n′ i =
k2
1
1
δkk′ δnn′ = 2
δkk′ δnn′
2
+m
k + ωn2 + m2
where ωn = 2π
β n is the Matsubara frequency. We can express the φ(x, τ ) function
as a Fourier sum of modes.
r
1 X
φk,n ei(k·x+iωn τ )
φ(x, τ ) =
βV
kn
and the correlation function becomes
hφ(x, τ )φ(x′ , τ ′ )i =
1 XX
′ ′
′ ′
hφk,n φk′ ,n′ iei(k·x+ωn τ ) ei(k ·x +ωn τ )
βV
′ ′
k,n k ,n
we have hφk,n φk′ ,n′ i from above and we get
hφ(x, τ )φ(x′ , τ ′ )i =
1
1 X
′
′
ei(x−x )·k+i(τ −τ )ωn
βV
k2 + ωn2 + m2
k,n
and if we let V → ∞ we get
hφ(x, τ )φ(x′ , τ ′ )i =
1
β
Z
1
d3 k
′
eik·(x−x )
3
2
2
2
(2π) k + ωn + m
and in the limit T → 0 we return to ordinary 0-temperature field theory
′
′
hφ(x, τ )φ(x , τ )i =
Z
d4 k
1
′
eik·(x−x )
4
2
2
(2π) k + m
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 contraction
hφi φj i = φi φj = A−1
ij = Kij
And also with several operators
Z
1
1
hφi φj φk i =
Dφφi φj φk e− 2 φi Aij φj = 0
Z0
∂ ∂ ∂ ∂ 1 Bi A−1
B
j
ij
e2
hφi φj φk φl i =
∂Bi ∂Bj ∂Bk ∂Bl
B→0
∂
1 1 T −1 2
1 T −1
∂
...
1 + B A B + ( B A B) + . . .)
=
∂Bi
∂Bl
2
2! 2
−1
−1 −1
−1 −1
= A−1
ij Akl + Aik Ajl + Ail Ajk
= φi φj φk φl + φi φj φk φl + φi φj φk φl
X
=
(contractions of all fields)
The rule is that we can write a general correlator as the sum of all possible 2-point
correlators of the relevant field operators. From here we can move around the
operators, but if we are dealing with a fermion field we must remember that they
anticommute
φi φj φk φl = (− φi φk ) φj φl
7.2.1 Example
Let us look at a Lagrangian density with an interaction term
1
λ
LE = (∂µ φ)2 + φ4
2
4!
First we find the interaction partition function
Z
Rβ
R 3 1
2 λ 4
Z =
Dφe− 0 dτ d x[ 2 (∂µ φ) + 4! φ ]
Z
R 4
R 4 λ 4
1
2
2
=
Dφe− d x 4! φ e− 2 d xφ(−∂ +m )φ
(7.6)
and as we remember that we could write the expectation value of an free field
operator
Z
R 4
1
hAi0 =
DφA(φ)e− d xL0
Z0
CHAPTER 7. LECTURE 7
70
we can use this on (7.6) to get
Z = Z0 he−
"
= Z0
R
λ 4
d4 x 4!
φ
λ
1−
4!
Z
i0
1
d xhφ (x)i0 +
2!
4
4
#
2 Z
λ
4
4
4 4
d xd yhφx φy i + . . .
4!
Here we can use Wick’s theorem to find that
hφ4 (x)i0 = hφx φx φx φx i0
= φx φx φx φx ·3
= 3K(x − x)K(x − x) = 3K(0)
Now we shall draw some Feynman diagrams. For each contraction between two
field operators we will draw a line between two points in space.
hφi φj i0 = hφ(xi )φ(xj )i0 = φ(xi )φ(xj ) = xi ×
× xj
and for each integral in spacetime we will draw a fat dot
Z
−λ d4 x = •
This gives that our partition function is actually a series of Feynman diagrams
Z
1 1
1
1 •
•
=1+ • +
• • + +
() + . . . = e−β(F −F0 ) •
•
Z0
8
128
16 48
and we get
−β∆F =
1
1 •
1
•
() + . . .
• + +
• 48 •
8 16 this is only disconnected diagrams. This expression is independent of dimension.
We find the K function
Z
′
d3 k eik·(x−x )
1X
′
K(x − x ) =
β n
(2π)3 ωn2 + k2
We have K(0) (loop curve)
1
K(0) =
β
Z
∞
1
d3 k X
3
(2π) n=−∞ k + ωn2
and through dimension regularization we get
K(0) = −
∞
1
1 X 2π
n=
2πβ n=1 β
12β 2
7.3. WICK’S THEOREM OUTSIDE FIELD THEORY
71
and this gives
1
−β∆F =
8
1
128β 2
2 Z
λ
β
dβ
0
Z
1
d x=
8
3
1
128β 2
2
λβV
And this gives us the pressure
π2
∆F
π2
5λ
4
p=
(kB T )4 −
(k
T
)
=
1−
B
90
64π 2
90
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 field
theory. In general, if we have an integral
I=
Z
b
a
dφ
√ e−A(φ)
2π
To treat this we must use the sadelpoint approximation, where we find a global
minimum Ac = A(φc ) of A(φ) and make an series expansion around this point
A(φ) = A(φc ) + A′c (φ − φc ) +
where
A′c
=
dA dφ φ0
1 ′′
1
Ac (φ − φc )2 + A′′′
(φ − φc )3 + . . .
2!
3! c
= 0. We can rewrite the integral as
I≈
Z
∞
−∞
1 ′′ 2
1 ′′′ 3
dφ
√ e−Ac − 2 Ac ϕ − 3! Ac ϕ +...
2π
(3)
(4)
and now we can think of K = A1′′ as the propagator, and −Ac and −Ac as the
c
3 and 4 line vertices respectively.
7.3.1 Example: Stirling’s formula for n!
We can write n! with help of the Γ-function
Z ∞
dφφn e−φ
n! =
0
If we just rewrite this a little bit we have
Z ∞
dφ
n!
√ =
√ e−φ+n ln φ
2π
2π
0
CHAPTER 7. LECTURE 7
72
That makes our A-function
A(φ) = φ − n ln φ
and its derivatives
n
⇒ φc = n
φ
n n
Ac = n − n ln n = − ln
e
1
′′
Ac = ⇒ K = n
n
2
A(3)
c =− 2
n
4
A(4)
c = 3
n
A′ (φ) = 1 −
If we associate our partition function with n! as
Z ∞
1 ′′ 2
dφ
n!
√ e−Ac − 2 Ac φ +... = e−W
Z=√ =
2π
2π
−∞
where −W is a series of Feynman diagrams
log A′′
C
z}|{
AC
z}|{
1
1 • 1
1 •
1
1
•
() + • • +
| +...
−W = • + + • + +
•
• 8
•
2
8
16
48
12
| {z } | {z } | {z }
| {z } |{z}
| {z } 3−loop
1−loop
2−loop
2−loop
2−loop
3−loop
So we get
n! =
√
r
n n 1 n n
1
2πn
+ . . . ≈ 2π(n + )
1+
e
12n
6 e
and this is a very good approximation for n! for large n. Even for small n it is
really 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 Lagrangian density for a massive scalar field, we would get a correlator on the form
1
Z
hφ(x)φ(x′ )i =
Z
−
Rβ
0
dτ
R
dx3
Dφφ(x)φ(x′ )e
z
LE
}|
{
1
λ
φ(−∂ 2 + m2 )φ + φ4
2
|
{z
} 4!
Z0
(8.1)
This correlator must be solved either numerically or perturbatively. Using perturbation theory we separate the free part and the interaction part for Z in (8.1), and
applying Wick’s theorem we have
Z =
Z
Dφe−
Rβ
Rβ
0
dτ
R
R
3
d3 xLE
λ
4
= Z0 he− 0 dτ d x 4! φ i0
1
= Z0 1 + • + . . .
8
(8.2)
The integral in (8.1) yields
Z
′
−
Dφφ(x)φ(x )e
R
dτ
R
d3 xLE
=
Z
Dφφ(x)φ(x′ )e−
= Z0 hφ(x)φ(x′ )e−
73
R
dτ
Rβ R
0
R
d3 x 21 (∂ 2 +m2 )φ −
λ 4
d3 x 4!
φ
e
i0
R
dτ
R
λ 4
d3 x 4!
φ
(8.3)
CHAPTER 8. LECTURE 8
74
where the 0 subscript means that this is the expectation value for the free Lagrangian. Now, inserting the results from (8.2) and (8.3) into (8.1) yields
′
hφ(x)φ(x )i =
hφ(x)φ(x′ )e−
−
he
Rβ
0
dτ
R
Rβ R
0
λ 4
φ
d3 x 4!
λ 4
d3 x 4!
φ
i0
i0
(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
Z
Z
λ
λ
′
4
4
′
hφ(x)φ(x ) 1 −
d yφ (y) i0 = hφ(x)φ(x )i0 −
d4 y φ(x)φ(x′ ) φ(y)φ(y) φ(y)φ(y) + . . .
4!
4!
where we are supposed to make all possible contractions. The contractions shown
in the above expression do not lead to any connection between the points x and y
and are thus represented by what is called disconnected diagrams. These contractions may be illustrated by Feynman diagrams on the form
1
•y
8
But of course we will also have contractions on the form
× x′
x×
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
′
hφ(x)φ(x )i =
=
×
×
×+×
×
1
•
8
1 + 18 •
× 1 + 81 • + 12 ×
1
1+ •
8
| {z }
+
1
2
•
×
•
×
×
Z
1
×+ ×
•
× +O(λ2 )
(8.5)
= ×
2
In the last line the first term represents the free part and the second term is a first
order mass correction or what is often called self-energy. This result, where the
diagram(s) in the denominator cancels the disconnected diagrams in the numerator,
is general. Therefore we may write:
hφ(x)φ(x′ )i = hφ(x)φ(x′ )e−
Rβ
0
dτ
R
λ 4
d3 x 4!
φ connected
i0
In the context of cosmology this is interesting because it may be possible to make
a theory where the cosmological constant Λ can be described in this way.
8.2. PHASE TRANSITIONS
75
V
φC
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 understanding
phase transitions is probably crucial to understand important fields like the baryon
asymmetry. First we will study spontaneous symmetry breaking with a Higgspotential 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 a
scalar field with the Lagrangian
1
1
λ
L = (∂µ φ)2 + µ2 φ2 − φ4
2
2
4
The first term is a kinetic term, while the two last terms represents a potential
1
λ
V (φ) = − µ2 φ2 + φ4 .
2
4
See figure (8.1) 1 . We find the minima of the potential:
dV
= −µ2 φ + λφ3 = 0
dφ
φ=0
or
φ = ±φC
φC =
where
And we define
1 µ2 λ
VC ≡ V (φC ) = − µ2
+
2 λ
4
1
µ2
λ
2
r
µ2
= const.
λ
=−
µ4
4λ
If we had a negative sign in the second term of our Lagrangian, we would have particles with
m < 0. These are mathematical particles called tachyons that sometimes occur in theories with
certain instabilities.
2
CHAPTER 8. LECTURE 8
76
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 the
VC will represent a cosmological constant, and we notice that even classical fields
will contribute to a Λ. Inserting typical values for µ and λ will give a Λ that is about
50 orders of magnitude bigger than the observed one. So why don’t we observe
this 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 partition
function
Z
R 3
Rβ
Z = Dφe− 0 dτ d xLE (φ,∂µ φ)
where
φ(x) =
r
1 X
φkn ei(k·x+ωn τ )
βV
k,n
= φ̄ + ϕ(x)
where the first term, φ̄, represents the k = n = R0-mode of the scalar field. Since
all k = 0-modes are in the φ̄-part, we have that d3 xϕ = 0. Now we will follow
much the same procedure as we used when working with Stirling’s formula last
week, and we expand Z around φ̄.
Z
Z
R 3
Rβ
Z = dφ̄ Dφe− 0 dτ d xLE (φ̄+ϕ)
where
LE (φ̄ + ϕ) =
=
1
(∂µ ϕ)2 −
2
1
(∂µ ϕ)2 −
2
1 2
λ
µ (φ̄ + ϕ)2 + (φ̄ + ϕ)4
2
4
1 2 2
λ
µ (φ̄ + 2φ̄ϕ +ϕ2 ) + (φ̄4 + 4φ̄3 ϕ +6φ̄2 ϕ2 + 4φ̄ϕ3 + ϕ4 )
|{z}
| {z }
2
4
=0
=0
where the underbraced terms will vanish since they are linear in ϕ and
Inserting this, our partition function now yields
R
d3 xϕ = 0.
free part
interaction part
}|
{
}|
{ z
z
1
λ
1
R
R
3
2
2
2
Z
Z
− 0β dτ d3 x
(∂µ ϕ) + m ϕ + λφ̄ϕ + ϕ4
R
R
4)
− 0β dτ d3 x(− 12 µ2 φ̄2 + λ
φ̄
2
2
4
4
Z =
dφ̄e
Dϕ e
Z
R 3
Rβ
(8.6)
=
dφ̄e− 0 dτ d xVeff (φ̄,T )
where m2 = −µ2 + 3λφ̄2 . We see that the free part can be solved exactly since
it depends on ϕ2 . Veff is called an effective potential and its form determines what
kind of symmetries we have in our theory. Notice that Veff has the same form as
we 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
(0)
(1)
(2)
Veff (φ̄, T ) = Veff + Veff + Veff + . . .
where the first term is the classical term, the second a 1-loop correction, the third a
2-loop-correction and so on. The first term looks like
λ
1
(0)
Veff = − µ2 φ̄2 + φ̄4
2
4
Here,√we will only consider the 1-loop corrections. Using m2 = −µ2 + 3λφ̄2 and
ω = k2 + m2 we have:
(1)
Veff
1
= − T r log(−∂ 2 + m2 )
2
Z
ω
d3 k
=
T log(1 − e− T )
3
(2π)
Z ∞
√
4
T
2
− x2 +y 2
dx
x
log
1
−
e
=
2π 2 0
m2 T 2 m3 T
π2
−
+ ...
= − T4 +
90
24
12π
Inserting for m2 in the second term and neglecting the third term we now get
1
Veff (φ̄, T ) =
2
λ
λ 2
−µ + T φ̄2 + φ̄4 + . . .
4
4
2
We see that when T grows large enough, the effective potential will become positive. Thus, considering the Higgs model, the spontaneous symmetry breaking will
be 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 early
universe, when the temperature is above some critical temperature TC . Veff (φ̄) is
plotted for different temperatures in figure (8.2). Solving Veff = 0 we see that
2µ
TC = √
λ
≈ O(100GeV) , kB = 1
In figure (8.3) you can see φC as a function of T. The inflation epoch is assumed
to have happened at a temperature around 1015 GeV, that is, a long time before
the particles got their mass. As we don’t know much about the physics above
T ∼ 100GeV , a lot of unknown fun can have happened in the cooling phase after
the inflation. The transition at T ≈ 100GeV is called the electroweak transition. It
is probable that the baryon asymmetry was generated during this phase transition.
CHAPTER 8. LECTURE 8
78
Vef f
T =0
T = TC
T >> TC
φ̄
Legend
T=T_C
T=0
T>T_C
Figure 8.2: The effective potential Veff (φ̄) for different temperatures
φC
1
TC
T
T
Figure 8.3: φC (T ). It is easy to see how the φC suddenly breaks down when the
temperature 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 temperature
Let us consider a good, old harmonic oscillator with the Lagrangian
1
1
L = q̇ 2 − ω 2 q 2
2
2
where ω is a time-dependent function ω = ω(t). Let us first consider a very simple
time dependence where
ωa for t < 0
ω=
ωb for t > 0
The ground state of the harmonic oscillator is then given by
(
1 2 2
for t < 0
ψa0 = |0in i = |0a i ∼ e− 2 ωa x
ψ0 (x) =
− 21 ωb2 x2
for t > 0
ψb0 = |0out i = |0b i ∼ e
Particle production may occur in the transition t < 0 → t > 0. The amplitude for
exciting the oscillator to a state n in this transition is given by
Z
∗
An = dxψa0
(x)ψbn (x)
The amplitude for the transition between the two ground states (without any excitation at all) is called vacuum persistence amplitude, and is given by
Z = h0b |0a i
That this magnitude is denoted Z just like the partition function is of course not
done for confusing, but rather for enlightening. This Z can be shown to be equivalent to a partition function defined as a functional integral over real time:
Z
R 4
Z = Dφei d xL(φ,∂µ φ)
This Z then denotes the amplitude for the vacuum to remain vacuum during the
t < 0 → t > 0 transition.
This simple model with a step-function-like change of ω is just the simplest
approximation to a theory describing what we call reality. Other models have a
more smooth transition where the values for ω are approaching ωa when t → −∞
and ωb as t → ∞. If this transition is sufficiently smooth, the oscillator will
remain in its ground state, Z = 1, and no particles will be created. This is called
an adiabatic transition. Next lecture we will consider such smooth transitions using
creation and annihilation operators and S-matrices.
80
CHAPTER 8. LECTURE 8
Chapter 9
Lecture 9
In the last part of the previous lecture, we considered particle creation in external
fields. When we investigated the harmonic oscillator in such environments, we
saw that there were two possible frequencies ωa and ωb describing the oscillation
depending on whether the time t was smaller or greater than zero, respectively. The
particles were created as the frequency of an harmonic oscillator changes when the
time changes from negative to positive. We will continue this investigation of the
harmonic oscillator in external fields and see how we can extract quantum fields
from their well known classical counterparts. The oscillator will also be equipped
with 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 an
electrically charged particle with mass m:
2
L = ∂µ − ieAµ (x) − m2 φ∗ (x)φ(x)
(9.1)
Here, Aµ (x) is the electromagnetic field representing a classical background. A
somewhat simpler example is the Lagrangian of a field in a scalar background:
L = |∂µ φ|2 − m2 φ∗ φ − A(x)φ∗ φ
(9.2)
or of a neutral scalar field in a non-trivial space-time metric gµν :
L=
1
1 µν
g ∂µ φ∂ν φ − m2 φ2
2
2
(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
CHAPTER 9. LECTURE 9
82
with a metric tensor given as:
gµν


1
0
0
0
 0 −a2
0
0 

=
2
 0
0
−a
0 
0
0
0
−a2
so that equation (9.3) now turns into
L=
1 2
1
1
φ̇ − 2 (∇φ)2 − m2 φ2
2
2a
2
(9.4)
As an easy start of external fields, and to get used to the formalism, let us therefore
consider 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 2 1 2 2
q̇ − ω q
2
2
(9.5)
The frequency is then generally a function of time ω = ω(t) and the system cannot
be expected to respect energy conservation. To proceed, we make the following
simplification: The frequency is assumed to be a constant of the time, either ωa or
ωb in intervals where the time is negative or positive, respectively. This is illustrated
in figure (9.1). In the areas where the frequency is constant, energy is conserved
and the system can be described by ωa or ωb . The two time intervals are usually
referred 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 simplification
is even stronger as the frequency is assumed to be a step function with respect
to time. This is shown in figure (9.2)
• Radiative decay. In a neutral atom there is an equal number of protons and
electrons. The electrons are orbiting the nuclei and a wave-function of the
atom is then ΨA (r, Z). Here r is the radius of the atom and Z is the proton
number. If the nucleus experiences a β-decay, then the total charge changes
instantaneously. Electrons orbiting the nuclei will possess the same wave
function (from the nuclei), but the charge of the nuclei changes instantaneously 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) one
must first consider the classical motion:
(∂t2 + ω 2 )q(t) = 0
(9.6)
9.2. TOY MODEL: HARMONIC OSCILLATOR
83
ω
ωb
ωa
I
t
II
Figure 9.1: Simplification of the oscillator. The frequency is assumed to be a
constant, ωa for a long time interval (I), then changing during short time to another
constant ωb (II)
with general solutions
q(t) = C1 f1 (t) + C2 f2 (t)
(9.7)
The coefficients are depending on the initial conditions that are assigned to the
specific problem. The classical solutions are more conveniently expressed in terms
of complex functions as
q(t) = au(t) + a∗ u∗ (t).
(9.8)
In the Heisenberg picture, the quantum mechanical description is represented by
operators, so the transition from classical to Q.M. description is
q(t) → q̂(t)
The solution is then written in operator form
q̂(t) = âu(t) + ↠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 system
in order to find exact solutions. Generalized momenta is defined in the usual way,
and in this particular case it is given by:
p̂(t) =
d
q̂(t) = âu̇(t) + ↠u̇∗ (t)
dt
(9.10)
CHAPTER 9. LECTURE 9
84
ω
ωb
ωa
t
Figure 9.2: Further simplification of the oscillator. The frequency is assumed to
behave as a step-function changing instantaneously from ωa to ωb .
The canonical commutation relation is what actually quantizes the solutions and is
the usual
q̂(t), p̂(t) = i
(9.11)
We are free to insert the expressions of q̂ and p̂ into the commutator. The result is:
q̂(t), p̂(t) = [â, ↠](uu̇∗ − u̇u∗ ) + [â, â] −like terms.
| {z }
=0
†
∗
∗
= [â, â ](uu̇ − u̇u ) = 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 equation (9.12). The inner-product is defined as
(u, u) ≡ i(u∗ u̇ − u̇∗ u)
(9.13)
Actually this inner product is a normalization product that is independent of the
time. This is seen by differentiating the product :
d
(u, u) = i(u̇∗ u̇ + u∗ ü − ü∗ u − u̇∗ u̇)
dt
(9.14)
From the Klein-Gordon-like equation in (9.6) the ü is constrained by
ü = −ω 2 (t)u
ü∗ = −ω 2 (t)u∗
Hence, it is obvious that the inner product is time independent:
d
(u, u) = 0
dt
(9.15)
The inner product is also a real quantity. This is seen by
(u, u)∗ = −i(uu̇∗ − u̇u∗ ) = +i(u∗ u̇ − u̇∗ 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 that
the 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 chosen
to 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
[q̂(t), p̂(t)] = i(u, u)[â, ↠] = i
⇒ [â, ↠] = 1
(9.18)
When this commutator identity is established, commutator algebra can be done
constructing an infinite number of states. This is done by applying the number
operator, N̂ = ↠â, and all the states are actually eigenstates of N̂ . In order to
construct all the states, we need a zero-state |0i of the Hilbert space where the
inner product is defined. The zero-state is defined by:
â|0i = 0
(9.19)
So then a complete set of states of the Hilbert space can be established, and these
are generally given by:
(a† )n
(9.20)
|ni = √ |0i
n!
What we have done so far (and that differs from common oscillator physics) is
valid for an arbitrary frequency ω = ω(t). Then the state |0i is not necessarily the
ground state. We continue our investigation of the oscillator by first considering the
simplified case where the frequency is assumed constant (step-function like) and
then 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 now
find the solutions of the (9.6) in terms of the mode functions u. The Hamiltonian
operator with constant frequency is
Ĥ =
1 2 1 2 2
p̂ + ω q̂
2
2
And the position and momentum operators are as in (9.9) and (9.10):
q̂ = âu + ↠u∗
d
q̂ = âu̇ + ↠u̇∗
p̂ =
dt
(9.21)
CHAPTER 9. LECTURE 9
86
Inserting into the expression of the Hamiltonian results in:
1
1
1
Ĥ = (u̇2 + ω 2 u2 )ââ + (u̇2 + ω 2 u2 )∗ ↠↠+ (|u̇|2 + ω|u|2 )(↠â + â↠)
2
2
2
(This is a general result and is also valid when ω is time dependent). We further
investigate when |0i is an eigenstate of the Hamiltonian operator. That is:
1
1
Ĥ|0i = 0 + (u̇2 + ω 2 u2 )∗ ↠↠|0i + (|u̇|2 + ω 2 |u|2 )|0i
2
2
(9.22)
Here the zero term follows from the definition of the zero-state with â|0i = 0. If
the zero-state |0i is supposed to be an eigenstate of Ĥ, then the following must
hold:
Ĥ|0i = number × |0i
So also the second term in (9.22) must vanish, resulting in
u̇2 + ω 2 u2 = 0 ⇒ u(t) ∼ e±iωt
(9.23)
So there seem to be two solutions for the mode u(t). However, due to the constraint
assigned to the inner product that it is positive with (u, u) = +1 there is only one
possible solution for u(t). It is found by combining the solution in (9.23) and the
definition of the inner product:
!
(u, u) = i(u∗ u̇ − u̇∗ u) = 1
r
1 −iωt
e
⇒ u(t) =
2ω
(9.24)
So the entire solution to (9.6), which is the position operator in the Heisenberg
picture, can then be further written as
q̂(t) = âu(t) + ↠u∗ (t)
r
1
(âe−iωt + ↠eiωt )
=
2ω
(9.25)
and finally the Hamiltonian operator regains the well-known form as
1
ω(↠â + â↠)
2
1
= ω(â↠+ )
2
Ĥ =
(9.26)
So in words we are back to normal with our common oscillator description when
we 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 solutions
to the equation
[∂t2 + ω 2 (t)]u(t) = 0
with the general solution
q̂(t) = âu(t) + ↠u∗ (t)
(9.27)
Specifying the initial conditions when the oscillator frequency is a varying quantity
with time demands a little precaution. The modes u(t) will depend on whether the
initial conditions are defined at early times, that is, with ω = ωa , or at late times
when the frequency is ω = ωb . These two occasions are both equally correct, and
it 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
r
1 −iωa t
e
u(t → −∞) =
2ωa
We already know that the inner product with respect to the modes u is
constant with time. This is seen from the solution of the mode u(t) by
(u, u) = i(u∗ u̇ − u̇∗ u) = 1. It also follows from the definition of the
in-modes that the commutator identity is preserved [â, ↠] = 1 as seen in
(9.18). So when time is taken to large, positive values, the in-mode turns
into a linear combination of exponentials
u(t)
∼ ( )e−iωb t + ( )e+iωb t
t→∞
Here, the parenthesis correspond to the so called Bogoliubov coefficients.
These coefficients will later be shown to be the connection between the inmodes and the out-modes.
• Out-modes: The out-modes are defined to have a unique behavior at late
times, t → +∞, in the same way as the in-modes at early times. The modes,
now denoted by v(t) are given by:
r
1 −iωb t
v(t → +∞) =
e
2ωb
However, the modes are valid also when time is very small. Then the outmode looks like:
v(t)
= ( )e−iωa t + ( )e+iωa t
(9.28)
t→−∞
CHAPTER 9. LECTURE 9
88
So to summarize, we have that
The in-modes are uniquely defined by their behavior at early times, t → −∞, and the
out-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 the
classical Klein-Gordon equation (9.6) can be expressed in terms of the in-modes
u(t). This was shown in equation (9.25 ). As discussed when defining the in- and
out modes, they both serve satisfactory solutions, only depending on the choice of
initial conditions. Therefore there is nothing stopping us from expressing q(t) in
terms of also the out-modes v(t) and define the inner-product for them analogously
to what was done for the in-modes earlier. The solution then looks like:
q̂(t) = b̂v(t) + b̂† 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 different
than in the in-mode situation in (9.9). The new (raising and lowering) operators
are easily shown to satisfy the commutator identity [b̂, b̂† ] = 1. (exercise!). Hence,
there seems to be more than just one description of the solution q(t). Actually, the
Hilbert space containing the quantum mechanical states of the oscillator, is equally
described by the in-operators (â and ↠) or the out-operators (b̂ and b̂† ). We will
in our continuation establish a connection between the two sets of operators of the
oscillator. There exist two vacuum states denoted by |0a i and |0b i and defined in
the usual way as:
â|0a i = b̂|0b i = 0
They represent the two different situations of the initial conditions
• |0a i: The ground state of the oscillator at t → −∞.
• |0b i: The ground state of the oscillator at t → +∞.
From the two ground states new and excited states can be constructed by applying
the 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, the
following holds:
b̂ = (v, q̂)
this is so, since (v, v) = +1 and (v, v ∗ ) = i(v ∗ v̇ ∗ − v̇ ∗ v ∗ ) = 0. A more compact
formalism of the inner product will be further used. This is in general written as:
(f, g) = i(f ∗ ġ − f˙ ∗ g)
→
←
∂
∂
= if ( − )g
∂t
∂t
∗
↔
= if ∗ ∂t g
9.4. OSCILLATOR WITH TIME DEPENDENT FREQUENCY
89
←
∂
where the arrow indicates the direction
Here we have used a notation of the form ∂t
of action of the derivative. The operator b̂ can now, if we insert for the field operator
q(t) in (9.27), be expressed as:
b̂ = (v, q̂) = â(v, u) + ↠(v, u∗ )
(9.31)
We will now seek to find relations between the two sets of operators â and b̂. To
do so, there will be the need of introducing coefficients describing inner products
between 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
b̂ = A∗ â − Bâ†
(9.32)
Hence, A and B are both Bogoliubov coefficients. We can derive some of the properties of these coefficients if we consider the commutator between the b-operators.
Then we will need the adjoint operator:
b̂† = A↠− B ∗ â
(9.33)
[b̂, b̂† ] = AA∗ [â, ↠] +BB ∗ [↠, â] = 1
| {z }
| {z }
(9.34)
and the commutator is:
=1
=−1
So the Bogoliubov coefficients satisfy
|A|2 − |B|2 = 1
This relation holds only for bosons. Formally the same procedure can be run
through for fermions, only now in anticommutator form. Due to the formal shift
[ , ]Bose−Einstein →
{, }
| {z }
,
F ermi−Dirac
the same relation between the Bogoliubov coefficients holds only now by changing
sign
|A|2 + |B|2 = 1
The relations between the two sets of operators are
b̂ = b̂ = A∗ â − Bâ†
â = Ab̂ + B b̂
†
(9.35)
(9.36)
CHAPTER 9. LECTURE 9
90
Combining with the Bogoliubov coefficients, the relations lead to an expression of
the 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−iωa t + B ∗ eiωa t
(9.38)
which is the same answer that we found in equation (9.28), only now with the
Bogoliubov coefficients properly established. So now the out-modes are expressed
in terms of the in-modes u(t).
9.4.1 The relation between the vacuum states |0a i and |0b i
One situation that we will meet quite often later in this course, is the one where
we prepare the system in the vacuum state by choosing the initial conditions at
early time, t → −∞. This state, which is the easiest state, will therefore be the
in-vacuum |0a i defined by
â|0a i = 0
(9.39)
The system is viewed in the Heisenberg picture where all states are time independent. This means that if we start out with the in- vacuum state |0a i, then the system
will always be in this state. However, as time goes by and t → +∞, then the
operators to be used now, b̂ and b̂† are not necessarily annihilation and creation
operators of this state any longer. So the state that was vacuum state at early times
is no longer such a state as t → +∞:
b̂|0a i =
6 0
(9.40)
Since the state |0a i is no longer a ground state, there is (another) physical interpretation of this state as an excited state. As the state is being excited somewhere
in the transition t = −∞ → t = ∞, quanta are created. The number of quanta is
related to the Bogoliubov coefficients in the following way
n̄ = h0a |b̂† b̂|0a i
= h0a |(A↠− B ∗ â)(A∗ â − B↠)|0a i
2
= |B|2 h0a |ââ∗ |0a i = |B|
(9.41)
where the B is a Bogoliubov coefficient. Now it is time to find the relation between
the states |0a i and |0b i. We start out with the definition of the “early” vacuum state
and insert for the operator â and find:
â|0a i = (Ab̂ + B b̂† )|0a i = 0
B
⇒ b̂|0a i = − b̂† |0a i
A
(9.42)
9.5. QUANTUM FIELDS
91
We will now perform a hand waving derivation of the relation between the two
vacuum states |0a i and |0b i motivated by the coordinate representation of the Q.M.
operators q̂ and p̂. In the commutator [q̂, p̂] = i in the coordinate representation,
∂
. In the similar way, the b-operator
the momentum operator is given by p̂ = −i ∂q
is stated to be
∂
b̂ =
∂ b̂†
So inserted in the equation (9.42), we find
b̂|0a i =
∂
∂ b̂†
|0a i = −
B
† b̂†
⇒ |0a i = Ce− 2A b̂
B †
b̂ |0a i
A
|0b i
The relation between the two vacuum states is commonly written in terms of an
S-operator as:
|0a i = Ŝ|0b i
We end our investigation of the oscillator by considering some of the properties
of the S-operator:
• The S-operator is unitary Ŝ † Ŝ = Ŝ Ŝ † = 1 and the â and b̂ operators are
connected by â = Ŝ b̂Ŝ † .
• The matrix elements of the S-matrix are defined by Smn = hmb ||na i. The
zero-element of the matrix was seen in the end of the last lecture when the
adiabatic transition was considered, and is defined by
S00 = h0b ||0a i = Z = h0a |Ŝ|0a i.
(9.43)
This matrix element is the amplitude that if the system starts out in the
ground state, it remains a ground state as t approaches infinity. If Z = 1
there 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 its
solutions. The point in viewing the problem (the equations) classically, is that we
know the solutions, and when we have these solutions, then for each of them there
exists an operator that either creates or annihilates a particle. And eventually we
have reached our goal which is a quantum field description (with a well-defined
quantum field operator). So let us therefore first take a look at the (classical) Lagrangian density of a free field:
L=
1
1 µν
η ∂µ φ∂ν φ − m2 φ2
2
2
(9.44)
CHAPTER 9. LECTURE 9
92
The classical equation of motion is
(∂ 2 + m2 )u(x) = 0
(9.45)
with the solutions for each wavenumber of the mode function as
r
r
1
1
−ik·x
uk (x) =
e
=
ei(k·x−ωk t) ,
(9.46)
2ωk V
2ωk V
√
where the frequency is the usual ωk = k2 + m2 and V is the volume of the box
where the system is expected to show periodic boundary conditions. The mode
functions uk form a complete set and therefore a field-operator can be established.
It is given by
X
φ̂(x) =
(âk uk + â†k u∗k ).
(9.47)
k
By means of the canonical commutator identity [φ̂, π̂] = i, the commutator relation
between the creation and annihilation operators is found to be
[âk , â†k′ ] = δkk′
(9.48)
In the deduction above, the system is expected to show periodic boundary conditions.
The classical modes are further defined in terms of an inner product
Z
←
→
(uk , uk′ ) = i d3 xu∗k (∂t − ∂t )uk′
Z
↔
(9.49)
= i d3 xu∗k ∂t uk′
To calculate this inner product, we will need to compute the spatial integral of two
mode functions. This integral is given by the relation
Z
1
δkk′
d3 xu∗k uk′ =
2ωk
The inner product is then
(uk , uk′ ) = δkk′
(9.50)
↔
∂t
This is so since the derivative
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 this
section is valid for all fields. This is so since we started out with the classical view
of the problem and found its solution, and at the same time we know that these
classical 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 Lorentz
transformation. This is seen by
Z
↔
′
dΣµ (u∗k ∂ µ uk′ )
(uk , uk ) = i
(9.52)
dΩ
Here, Σµ is a vector perpendicular to the 3-dimensional hypersurface x. This is
shown in figure (9.3).
t
dΣµ = d3 x
x
Figure 9.3: Minkowski diagram showing the volume element before the Lorentz
transformation.
The Lorentz invariance is seen if we perform the same calculation in another
space-time x′ . This is illustrated in figure (9.4) where the spatial surface is tilted
due to the Lorentz transformation. The integration volume perpendicular to x′ is
different from that of x. After the transformation, the inner product is
Z
Z
↔
↔
′
∗ µ
d4 x∂µ (u∗k ∂ µ uk′ )
dΣµ (uk ∂ uk′ ) + i
(9.53)
(uk , uk′ ) = i
Ω
dΩ
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′ ) = (∂ 2 u∗k ) uk′ − u∗k ∂ 2 uk′ = 0
| {z }
| {z }
(9.54)
=−m2 uk
=m2 u∗k
Hence, the two expressions of the inner product are equal and the inner product is
Lorentz invariant as stated:
Z
Z
↔
↔
dΣµ (u∗k ∂ µ uk′ ) =
dΣ′µ (u∗k ∂ µ uk′ )
(9.55)
dΩ
dΩ
CHAPTER 9. LECTURE 9
94
t
dΣ′µ
x′
Ω
x
Figure 9.4: Minkowski diagram showing the space time after the Lorentz transformation. The integration is now performed not only perpendicular to the spatial
axis, 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 ω
varies as by figure (9.1). The action is defined by
Z
S = d4 xL(φ, ∂µ φ)
(10.1)
We have the Lagrangian density
L=
1
1
1
(∂µ φ)2 − m2 φ2 − Aφ2
2
2
2
(10.2)
where A is an external field. We find the classical modes by solving the EulerLagrange equation
∂L
∂ ∂L
− µ
=0
(10.3)
∂φ ∂x ∂φ,µ
From this we get two solutions. We get the in-solutions uk (x) which gives the field
operator
i
Xh
φ̂ =
âk uk (x) + ↠u∗k (x)
k
and the out-solutions (funny word in Norwegian) vk (x) which gives
i
Xh
φ̂ =
b̂k vk (x) + b̂† vk∗ (x)
k
and we define the in-vacuum |0a i by âk |0a i = 0 and the out-vacuum |0b i by
b̂k |0b i = 0. We have the survival amplitude Z, which is the amplitude that the field
stays in the vacuum state forever.
Z = h0a |0b i = Z[A] = eiW [A]
(10.4)
where W [A] is the effective action. The probability for the field to stay in the
vacuum state is then |Z|2 = e−2ImW [A] .
95
CHAPTER 10. LECTURE 10
96
We remember from the finite temperature field theory that we used a similar
notation for the partition function
Z
R 3
Rβ
(10.5)
Z[A] = Dφe− 0 dτ d xLE (φ,∂µ φ) = e−βF [A]
Where F is Helmholtz free energy. This is no coincidence. We remember from
relativistic quantum field theory that we could write the propagator as
Z = h0|T e−i
R
d4 xHint (φ̂)
|0i
(10.6)
where Hint is the Hamiltonian density in the interaction picture. Based on that
L = L0 + Lint and that H = H0 + Hint it can be shown that this is equal to
Z
R 4
R 4
Z = Dφei d xL(φ,∂µ φ) = hei d xLint (φ,∂µ φ) i0
(10.7)
where we only considered the connected Feynman diagrams. We recognize here
the partition function.
From thermodynamics we remember that the free energy is given as F = E −
T S, and when T → 0 we can associate F with vacuum energy.
Here we prefer the expression with the Lagrangian density. Very often we have
that Hint = −Lint .
10.1.1 Example with Hint = −Lint
For example if we have the Lagrangian density
L = L0 + Lint =
which gives that π =
∂L
∂ φ̇
λ
1
(∂µ φ)2 − φ4
2
4
= φ̇. From this we find the Hamiltonian density
H = H0 + Hint
1
λ
1
= π φ̇ − L = φ̇2 − φ̇2 − (∇φ)2 + φ4
2
2
4
λ
1 2 1
π + (∇φ)2 + φ4
=
2
2
4
(10.8)
We see that Hint = −Lint .
10.1.2 Example with Hint 6= −Lint
But this is not always the case. If we for example have the Lagrangian density
g
1
L = (∂µ φ)2 − φ2 (∂µ φ)2
2
4
(10.9)
10.2. QUANTIZATION OF EXTERNAL FIELDS
97
where the last term is the interaction Lagrangian density. We find the generalized
momentum (or whatever it is called in field theory) to be
π=
∂L
g
= φ̇ − φ2 φ̇
4
∂ φ̇
(10.10)
and if we use this to find the Hamiltonian density H = H0 + Hint = π φ̇ − L we
see that for this case
Hint 6= −Lint .
(10.11)
R
4
i d xHint i is not Lorentz invariant (and
From this follows that the expression he
0
R 4
−i
d
xLint i is correct. This gives a hint that
perhaps not gauge invariant), but he
0
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
[âk , â†k′ ] = [b̂k , b̂†k′ ] = δkk′
and we can express âk by
âk = (uk , φ̂) =
X
Akk′ b̂k′ + Bkk′ b̂†k′
(10.13)
k′
Where Akk′ = (uk , vk ) and Bkk′ = (uk , vk∗ ) are the Bogoliubov coefficients. From
[âk , â†k′ ] = δkk′ it follows that
X
(10.14)
= (Akl A∗k′ l − Bkl Bk∗′ l ) = δkk′
l
Usually the Bogoliubov coefficients are diagonal
Akk′ = Ak δkk′
Bkk′ = Bk δkk′
We have defined vacuum by âk |0a i = 0 and as âk = Ak b̂k + Bk b̂†k we can find
the average number of particles, n̄, produced
n̄k = h0a |b̂†k b̂k |0a i = |Bk |2
(10.15)
CHAPTER 10. LECTURE 10
98
And we can find the amplitude to produce a pair of particles. In that case the inBk †
b̂ |0a i
state is |0a i and the out-state is b̂†k b̂†k |0b i. As we know that b̂k |0a i = − A
k k
and that the vacuum condition gives h0b |b̂†k = 0, we get the amplitude
Bk
h0b |b̂k b̂†k |0a i
Ak
Bk
= − h0b |0a i
Ak
Bk
= − Z
Ak
h0b |b̂k b̂k |0a i = −
(10.16)
Thus, the probability amplitude for producing one pair of particles is given by the
Bk
fraction − A
multiplied with the probability that vacuum remains vacuum.
k
Unitarity gives that h0a |0a i =P1 and the completeness relation allowes us to
expand this in two particle states ∞
n=0 |2nb ih2nb | = 1. And these together with
our newfound amplitude gives us
2
2
2 B 4
2 B h0a |0a i = |Z| + |Z| + |Z| + . . . = 1
A
A
(10.17)
We factorize out |Z|2 and find the sum
!
2
B |A|2
1
2
2
= |Z|2 |A|2 = 1
|Z| 1 + + . . . = |Z|2
=
|Z|
B 2
2 − |B|2
A
|A|
1− A
where we have used the relation |A|2 − |B|2 = 1 (valid for bosons only!). This
result 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, this
should be
B 2
B 2
2n
∞
X
B 1
2
2
2
A
A
n = |Z| n̄ = |Z|
2 =
2 = |B|
2
2
2
A
|A|
n=1
1 − B
1 − B
A
A
(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, but
with 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 n̄ = |B|2
as in the boson case. What we have shown in this section is the standard formalism
on how particles are produced in an external field.
10.3. PARTICLE PRODUCTION IN CONSTANT ELECTRIC FIELD
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−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 electric
field being turned on, left constant for a while (infinitely long actually), and then
turned off again.
If we look at the old Dirac-sea where we have a large potential gap between the
positive and negative area. The production of a particle would be accompanied with
production 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 result
that 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 field
is given by
V (z) = eEz
(10.20)
The transverse momentum, kT = (kx , ky ), will here be conserved. The same is
true for the energy (here denoted ω) except for the energy from the electric field.
That gives us the energy equation
(ω − eEz)2 = k2T + kz2 + m2
(10.21)
We will find the position where the particles enter and leave the forbidden region.
CHAPTER 10. LECTURE 10
100
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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 = m2T
(10.22)
where m2T = k2T + m2 . This gives two solutions
ω − eEz1 = −mT
ω − eEz2 = +mT
Now we get the tunnelling current
−2
n̄ = e
Rz
2
z1
−2
dzkz
=e
Rz
2
z1
dz
r
1−
“
ω−eEz
mT
”2
(10.23)
substituting xmT = ω − eEz yields
n̄ = e−
2m2
T
eE
R1
−1
√
dx 1−x2
= e−
πm2
T
eE
(10.24)
To see what kind of fields we are talking about here we find an approximate field
strength when producing particles. In order to achieve such production, the exponent in (10.24) must be smaller than unity
E=
and this is a terribly strong field.
V
πm2
∼ 1016
e
m
(10.25)
10.3. PARTICLE PRODUCTION IN CONSTANT ELECTRIC FIELD
101
As we earlier found that n̄ = |B|2 we get for bosons that
|Zk |2 = e−2ImWk =
1
1
1
=
=
|Ak |2
1 + |Bk |2
1 + n̄k
and the total partition function
P
Y
Y
|Z|2 = e−2ImW =
|Zk |2 = e−2 k ImWk =
(1 + n̄k )−1
k
(10.26)
(10.27)
k
and this gives
2ImW =
X
ln(1 + n̄k ) =
k
X
−
ln 1 + e
k
πm2
T
eE
(10.28)
and as we know, when the volume grows large we can write the sum as an integral
Z
Z
Z
X
d3 k
d3 k
3
=V
=
d
x
3
(2π)
(2π)3
k
So we write
Z
2ImW [A] = 2 d4 xLef f (A)
X
=
ln(1 + n̄k )
k
=
=
=
Z
Z 2
d kT
dωdt
d2 xT ln (1 + n(ω, kT ))
2π
(2π)2
Z
Z
eE
dtdzdxdy d2 kT ln (1 + n(ω, kT ))
(2π)3
Z
Z
eE
d4 x
d2 kT ln (1 + n(ω, kT ))
(10.29)
2π
(2π)2
Z
where we have made the substitution dω = eEdz. Now we use our previous result
−
that n̄ = e
πm2
T
eE
, and thus we get
2ImLef f
eE
=
2π
Z
πm2
d2 kT
− eET
ln 1 + e
(2π)2
(10.30)
A series expansion of the logarithm function yields
ln(1 + x) =
∞
X
(−1)n+1
n=1
n
xn
and we can use that m2T = k2T + m2 and then find the integral
Z ∞
πn 2
eE
πdkT2 e− eE kT =
n
0
(10.31)
CHAPTER 10. LECTURE 10
102
So our expression turns out
2ImLef f
1
=
8π
eE
π
2 X
∞
(−1)n − nπm2
e eE
2
n
n=1
(10.32)
we know that the free particle Lagrangian density yields
1
i
L = E2 −
2
16π
eE
π
2
e−
πm2
eE
+ ...
(10.33)
All the particle production so far has been with bosons. We now end this lecture
by investigating the fermion case as in QED.
10.3.1 QED
From QED we know that
1 2
L = ψ̄(iγ µ ∂µ − m − eγ µ A)ψ − Fµν
4
(10.34)
and
iW [A]
Z[A] = e
=
Z
Dψ
Z
D ψ̄ei
R
d4 xL
= ei
R
2 +T r log(iγ µ ∂ −m−eγ µ ∂ A)]
d4 x[− 14 Fµν
µ
µ
(10.35)
Schwinger looked at the case where Fµν = const and calculated the exact expression for Lef f
Z ∞
1 −ism2
cosh(eas) cos(ebs)
ds
2
− 2 e
(10.36)
e ab
Lef f =
8π 2 s
sinh(eas) sin(ebs)
s
0
where
1
a2 − b2 = E2 − B2 = Fµν F µν
2
and
If we expand Lef f
1
ab = E · B = Fµν F µν
4
in a and b we get
2α2 2
1 2
2 2
2
+
(E
−
B
)
+
7(E
·
B)
+ ...
Lef f = − Fµν
4
45m4
(10.37)
The second term was first found by Euler and Heisenberg in 1933, but Schwinger
could expand this to any order.
Chapter 11
Lecture 11
This lecture we are going to continue the study of classical fields in non-trivial
backgrounds. We are going to focus on the use of the variation principle, and
apply this principle on the metric. We will end up studying the Einstein-Hilbert
action 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 dxµ dxν .
|{z}
=gµν (x)
Earlier we have used the principle of extremal action on a scalar field living in a
Minkowski metric, ηµν , using the action
Z
Z
1 µν
4
4
(11.1)
S = d xL = d x η ∂µ φ∂ν φ − V (φ)
2
The transition to a general metric is as simple as we could hope for, using what is
called “minimal coupling” we just substitute the Minkowski metric with a general
metric and obtain
Z
1 µν
4 √
(11.2)
S = d x −g g ∂µ φ∂ν φ − V (φ)
2
where
g = 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 accelerated
relative to a particle-less vacuum. Assume a system Σ in 1 + 1 dimensions with
103
CHAPTER 11. LECTURE 11
104
t′
(x, t)
x′
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 inertial system Σ′ with coordinates (x′ , t′ ) see figure 11.2. The primed coordinates are
given in terms of the un-primed coordinates using what is called Rindler coordinates1 :
1 ax
x′ =
e cosh(at)
(11.3)
a
1 ax
t′ =
e sinh(at)
(11.4)
a
Now, assume a massless scalar field φ′ (x′ , t′ ) in Σ′ . The corresponding field in Σ
is denoted φ(x, t). Σ′ has an action defined by
" #
Z
∂φ′ 2 1 ∂φ′ 2
2 ′ 1
−
(11.5)
S= d x
2 ∂t′
2 ∂x′
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 metric
becomes
2ax
−2ax
e
0
e
0
µν
gµν =
⇒ g =
0
−e2ax
0
e−2ax
√
since gαλ gλβ = δαβ . This gives us −g = e2ax . In Σ the action thus yields
" 2 #
Z
2
∂φ
1
∂φ
S = d2 x e2ax
−
(11.7)
2
∂t
∂x
1
These are easily derived from Lorentz transformations. See for example chap. 3.2 in “Lecture
notes 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.
r
Z
i
1 h ik·x
dk
âk e
+ â†k e−ik·x
φ̂(x) =
2π 2ω
r
Z
i
dk′
′ ′
′ ′
1 h
′ ′
φ̂ (x ) =
âk′ eik ·x + â†k′ e−ik ·x
2π 2ω
where the a-operators of course satisfy our beloved commutation relation
h
i
âki , â†kj = δki kj
As known from a general treatment of hyperbolically accelerated systems, such
systems will always create a horizon. See figure 11.2. As time increases the world
t′
t = +∞
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′
Figure 11.2: The creation of a horizon. The world-lines of the Σ-system will
converge 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 11.2
can ever be observed by an observer in Σ′ . That means that even if the space-time
according to an observer in Σ is non-empty, an observer in Σ′ may experience
vacuum. Remembering the definition of Bogoliubov coefficients from earlier lectures, we know that we may write the a-operators in Σ expressed in terms of the a
operators in Σ′ :
i
Xh
âk =
Akk′ âk′ + Bkk′ â†k′ .
(11.8)
k′
Let us assume that we have a vacuum in the inertial system Σ′ . That means that
âk′ |0′ i = 0
(11.9)
CHAPTER 11. LECTURE 11
106
The corresponding expectation value for an observer in the accelerated system Σ
is then
h0′ |â†k âk |0′ i = |Bkk′ |2
where
Bkk′ =
uk
vk′
(uk , vk∗′ )
r
=i
Z
∞
−∞
dx[u̇∗k vk∗′ − u∗k v̇k∗′ ]
1 i(k′ x′ −ωt′ )
=
e
2ω
r
1 i(k′ x′ −ω′ t′ )
e
=
2ω ′
(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
h0′ |â†k âk |0′ i = |Bkk′ |2
πω
∼
∼
∼
e− a
sinh πω
a
1
2πω
a
−1
1
,
ω
eT − 1
e
(11.11)
a
. This is a remarkable result! An observer in accelerated motion
where T = 2π
in an inertial system with vacuum will, according to this, experience a thermal
a 2
. A common physical explanation for this
background with temperature T = 2π
effect is that on the horizon, virtual particle-antiparticle-pairs are being created
and annihilated all the time. Some times these pairs will be created in such a
way that the pair get separated by the horizon before they annihilate, and that this
creates a net creation of particles inside the horizon. Remembering the equivalence
principle, this is closely related to the creation of Hawking radiation on the surface
of black holes. Systems like the one studied in this example is still subject for
several 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 the
Robertson-Walker (RW) one, as a background for a scalar field. The RW metric is
2
A similar effect has been investigated by Jon Magne Leinaas and John Bell at CERN. They
considered 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.
11.1. MASSLESS SCALAR FIELD IN A GENERAL SPACE-TIME
107
given by
ds2 = dt2 − a2 (t)dx2


1
0
0
0
 0 −a2 (t)

0
0

gµν = 
2
 0

0
−a (t)
0
2
0
0
0
−a (t)

1
0
0
0
−2 (t)

0
−a
0
0
gµν = 
−2
 0
0
−a (t)
0
0
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 RWbackground is now given by
S=
Z
4
3
d xa
1
1 2
2
φ̇ − 2 (∇φ) − V (φ) .
2
2a
Now, let us study a small variation φ → φ + δφ in the scalar fields and apply the
principle of stationary action to obtain the equations of motion for φ:
δS =
Z
i
h
1
d4 xa3 φ̇δφ̇ − 2 (∇φ) · δ(∇φ) −V ′ δφ
| {z }
a
=∇δφ
d 3
2
3 ′
=
d x − (a φ̇) + a∇ φ − a V δφ
dt
Z
ȧ
1
!
=
d4 xa3 −3 φ̇ − φ̈ + 2 ∇2 φ − V ′ δφ = 0
a
a
Z
4
Now we demand the expression in the brackets to vanish, and thus we obtain the
equation of motion
φ̈ + 3H φ̇ −
1 2
∇ φ+V′ =0
a2
(11.13)
where H = aȧ is the Hubble parameter. This is a frequently used equation. Often
it appears without the third term. This is usually a very good approximation since
a grows rapidly in most universe models and thus makes this third term neglectable. But of course one may also encounter problems where this term indeed is
important.
CHAPTER 11. LECTURE 11
108
11.1.3 Introduction of the d’Alembertian
Varying the scalar field in (11.2) using a general metric gµν one obtains
Z
√ δS =
d4 x −g gµν ∂µ φ∂ν δφ − V ′ δφ
Z
√
1
!
νµ
′
4 √
=
d x −g √ ∂ν ( −g g ∂µ φ) − V δφ = 0
−g
(11.14)
We now introduce the d’Alembertian operator given by
√
1
2 = √ ∂µ
−g gµν ∂ν
−g
=
∂ ν ∂ν = ∂ 2
(11.15)
This is easily seen if you let gµν raise the index of ∂µ . The 2 operator can be
considered as a covariant version of the Laplacian ∇2 .
2φ = ∇µ ∇µ φ
where ∇µ is a covariant derivative given in terms of the Christoffel symbols as
∇µ Aν = ∂µ Aν + Aλ Γνλµ
(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 what
happens if we vary the metric. This will lead us to a very useful quantity called the
Einstein-Hilbert action, which will be introduced in the next section.
Recalling some basic linear algebra we know that
X
g = det(gµν ) =
gµν γ µν
ν
where the µ in the last expression can be chosen arbitrarily, but should not be
summed over. Here γ µν is the cofactor, which can be defined as
γij = (−1)i+j det(gij )
where gij is the matrix which is left when one removes the ith row and the jth
column from the full g-matrix. The cofactors can be used to express the contravariant of the metric since
gµν gνλ = δλµ
⇒
gνλ =
1 λν
γ
g
11.1. MASSLESS SCALAR FIELD IN A GENERAL SPACE-TIME
109
Using this result we can write the derivative of the determinant as
∂λ g =
∂g
∂g ∂gµν
=
= g gµν ∂λ gµν
λ
| {z }
∂x
∂gµν ∂xλ
| {z }
=γ µν
(11.17)
=γ µν
and similarly we can write the variation of the metric as
∂gµν
δx ⇒ δg = g gµν δgµν
∂x
√
Later we will also need to know δ −g, which now yields
δgµν =
√
δ −g =
=
=
1
√ δ(−g)
2 −g
1
√ (−g)gµν δgµν
2 −g
1√
−g gµν δgµν
2
(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 symbols. As we know from kindergarten the Christoffel symbols can be expressed in
terms of the derivatives of the metric as
1
Γµλν = gµρ [gρλ,ν + gρν,λ − gλν,ρ ]
2
(11.19)
From this definition it is obvious that the Christoffel symbols are symmetric in the
lower indices. We are now interested in the divergence of a vector given as
∇µ Aµ = ∂µ Aµ + Aν Γµνµ
so we are only interested in the Christoffel symbols on the form Γµλµ . Then, given
a symmetric metric, the first and third term in the brackets in (11.19) will cancel
each other out since now µ = ν and
gνρ [gρλ,ν − gλν,ρ ] = δνλ,ν − δρλ,ρ = 0
Then we have that
1 µρ
g gµρ,λ
2
1 1
=
∂λ (−g)
2 −g
1
∂λ ln(−g)
=
2 √
= ∂λ ln −g
Γµλµ =
CHAPTER 11. LECTURE 11
110
where we have used (11.17) to get from the first to the second line. Using this, we
can now write ∇µ Aµ as
√
∇µ Aµ = ∂µ Aµ + Aλ ∂λ ln −g
√
1
∂µ −gAµ
= √
−g
(11.20)
So if Aµ can be derived from a scalar potential
Aµ = ∇µ φ = ∂ µ φ
we see that
∇µ ∂ µ φ =
=
√
1
√
∂µ −g∂ µ φ
−g
√
1
√ ∂µ
−ggµν ∂ν φ
−g
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:
1
Eµν = Rµν − gµν R = 8πGTµν
2
These equations should result from a variation principle for a given action. Hilbert
was the first to express such an action in a general coordinate system. The EinsteinHilbert action is then given by
S=
Z
1
d x −g − R + L(φ, ∂µ φ)
2κ
4
√
(11.21)
where R = Rµν gµν 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 a
variation in the metric.
gµν
→ gµν + δgµν
1 µν
g ∂µ φ∂ν φ − V
L =
2
11.3. THE ENERGY-MOMENTUM TENSOR
111
So δS yields
Z
√
1 √
4
δS =
d x
δ( −gR) + δ( −gL)
2κ
√
Z
δ( −gL) µν
√
√
√
1
µν
µν
4
δg
(δ −g)R + −gRµν δg + −g g δRµν +
=
d x −
2κ
δgµν
The third term is a surface-term that will vanish. Using (11.18) we have for the
second term that:
√
1√
(11.22)
(δ −g)R = −
−g gµν δgµν R
2
and we can rewrite δS as
√
Z
1
1 δ( −gL)
1
4 √
δS = d x −g − (Rµν − gµν R) + √
δgµν = 0
2κ
2
−g δgµν
(11.23)
We see that we obtain the Einstein equations if we demand that the last term in the
brackets 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 tensor
using the expression
√
2 δ( −gL)
Tµν = √
−g δgµν
∂L
= 2 µν − gµν L
(11.24)
∂g
since using (11.18) we have
√
1√
δ −g
−g gµν .
=
µν
δg
2
(11.25)
(11.24) is a very important and extremely useful equation, and we will show a few
examples 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 Lagrangian
1
1 2
= Fαβ Fρσ gαρ gβσ
L = − Fµν
4
4
The derivative with respect to the metric yields
∂L
1
1
= − Fµν F αν = Fµα F αν
µν
∂g
2
2
CHAPTER 11. LECTURE 11
112
Inserting into (11.24) gives the energy-momentum tensor
Tµν =
1
1
2
Fµν F αν + gµν Fαβ
2
4
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
1
L = gµν ∂µ φ∂ν φ − V (φ)
2
∂L
1
= ∂µ φ∂ν φ
µν
∂g
2
Which give the energy-momentum tensor
Tµν = ∂µ φ∂ν φ − gµν
1
(∂λ φ)2 − V
2
(11.26)
Let us now examine the simple but interesting case of a constant φ.
φ = const. = φ0
→
Tµν = gµν V (φ0 ) = ρ0 gµν
| {z }
=ρ0
Inserting this into the Einstein equations yields
Eµν = 8πGρ0 gµν ≡ Λgµν
where Λ = 8πGρ0 . So we see, not too surprisingly, that this scalar field will
correspond to Einstein’s cosmological constant. Inserting this into (11.21) and
adding a matter Lagrangian Lm , we have the action
Z
1
4 √
S =
d x −g − R − ρ0 + Lm
2κ
Z
1
4 √
=
d x −g − (R + 2Λ) + Lm
2κ
and the Einstein equations now yield
Eµν = 8πGTµν + gµν Λ
Let us study another case which is more general in the sense that we no longer
demand our scalar field to be constant, but less general in the sense that we choose
a special metric, namely the RW-metric:
ds2 = dt2 − a2 (t)dx2
11.3. THE ENERGY-MOMENTUM TENSOR
113
Using (11.26) we will now calculate some components of the energy-momentum
tensor and see how our well known relations between the time-derivatives of the
scalar field and the energy density and pressure fall out of our formalism. Let us
start with the 00-component:
1
1 2
2
2
T00 = φ̇ − φ̇ − 2 (∇φ) − V
2
2a
1 2
1
(11.27)
=
φ̇ + V + (∇φ)2
2
2a
Now we calculate the 11-component:
1
2
2 1 2
2
T11 = (∂x φ) + a
φ̇ + 2 (∇φ) − V
2
2a
1
1
2
2
2
2
2
φ̇ + 2 (∂x φ) − (∂y φ) − (∂z φ) − V
= a
2
2a
Raising one of the indices we have
T 11 = g11 T11 =
1
1
1
T11 = φ̇2 + 2 (∂x φ)2 − (∂y φ)2 − (∂z φ)2 − V
2
a
2
2a
The 22 and 33 components will of course be completely analog. It is common to
define a pressure p like
p ≡
=
1 1
T 1 + T 22 + T 33
3
1 2
1
φ̇ − V (φ) − 2 (∇φ)2
2
6a
In cosmology one usually omits the last term because the scale factor a grows
rapidly to a large value and we are thus left with the well-known expression for
pressure from a scalar field:
p=
1 2
φ̇ − V (φ)
2
(11.28)
Notice that in the most general case one also has non-diagonal components of the
form
T01 = φ̇∂x φ , T12 = ∂x φ∂y φ
But usually when working in cosmology one assumes a hydrodynamical model
where the non-diagonal components vanish because of homogeneity and isotropy:
h∂µ φ∂ν φi = 0
for
µ 6= ν
Then the last term in the expression (11.27) for T00 vanish, and we recognize the
00-component as the well-known expression for energy density from a scalar field:
1
ρ = φ̇2 + V
2
(11.29)
114
CHAPTER 11. LECTURE 11
And we can write the energy-momentum tensor on the form


ρ 0
0
0
 0 −p 0
0 

hT µν i = 
 0 0 −p 0 
0 0
0 −p
Chapter 12
Lecture 12
In this lecture, we will finish our investigation of the energy-momentum tensor and
then turn the attention towards the concept of how to quantize curved spaces. In
particular 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 in
equation (11.24):
√
2 δ( −gL)
Tµν = √
−g δgµν
The definition above is actually the one and only correct definition of the energymomentum tensor. We will in the following take a look at the energy-momentum
tensor and investigate under what kind of circumstances it is conserved. We will
start out with the easy situation of a flat space and then turn to the more complicated
curved spaces thereafter:
• Flat space-time: From relativistic quantum field theory and Noether’s theorem we know that quantities are conserved due to invariance or symmetries
of the Lagrangian density under translations or rotations. Hence, the energymomentum tensor is conserved due to translations with respect to time and
space. The conservation of momenta is a direct fact of spatial invariance,
while energy is conserved from temporal (time) invariance. This is stated as:
∂ µ Tµν = 0
• Curved space-time: In the transition from flat to curved spaces, the (usual)
partial derivative, ∂ µ changes into a covariant derivative, ∇µ . And instead
of considering translations (in time and space), we look at invariance due
to local coordinate transformations. From the flat case, a qualified guess is
therefore that the energy-momentum tensor is conserved as
∇µ Tµν = 0
115
(12.1)
CHAPTER 12. LECTURE 12
116
and where the local coordinate transformation is the small shift in position
xµ → x̄µ (x) = xµ + ξµ (x)
(12.2)
We now want to convince ourselves that the conservation of the energy-momentum
tensor in (12.1) is correct. To do so we will perform such a local coordinate transformation 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 the
transformation, and then finally minimizing the action and applying the variational
principle.
First let us notice that the Bianchi identity automatically leads to conservation
of the energy-momentum tensor. This is seen by combining the Einstein equations
Eµν = 8πGTµν
with the Bianchi identity 1
∇µ Eµν = 0
Now we go on with investigating the Tµν under a local coordinate transformation with respect to the metric:
gµν (x) → ḡµν (x̄)
=
ḡµν (x + ξ) = ξ λ ∂λ ḡµν (x) + ḡµν (x)
=
ξ λ ∂λ gµν + higher order terms
(12.3)
where in the first step, the relation xµ → x̄µ (x) = xµ + ξµ (x) in (12.2) has been
inserted for. To find the desired variation in the metric, δgµν , we need to use the
relation:
∂ x̄ρ ∂ x̄σ
ḡρσ (x̄)
(12.4)
gµν →
∂xµ ∂xν
where the derivatives are
∂ x̄ρ
= δµρ + ∂µ ξ ρ
∂xµ
∂ x̄σ
= δνσ + ∂ν ξ σ
∂xν
If we now insert these derivatives into equation (12.4), we find an expression of
gµν as:
gµν = (δµρ + ∂µ ξ ρ )(δνσ + ∂ν ξ σ )ḡρσ (x) = ḡµν (x) + ξ λ ∂λ gµν
which finally gives us the change or variation in the metric by subtracting gµν (x)
from ḡµν (x):
δgµν
= ḡµν (x) − gµν (x)
= −ξ λ
∂λ gµν
| {z }
−gµλ ∂ν ξ λ − gνλ ∂µ ξ λ
(12.5)
Christof f el
1
Another way to interpret this identity is ∂∂Σ = 0, and is described by Wheeler as “the boundary
of 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 derivatives
as:
δgµν = −∇µ ξν − ∇ν ξµ
(12.6)
We will now apply the action Sφ of a scalar field and then the variational principle
with stationary action to derive the conservation of the energy-momentum tensor
in curved space-time. With curvature (and 4 space-time dimensions), the action
integral is the generalized expression of (11.2) and looks like:
Z
√
(12.7)
Sφ = d4 x −gL(φ, ∂µ φ)
and where the curvature normalization term is the usual g = det(gµν ). To apply the
variational principle with stationary action, it is now time to consider the specific
coordinate transformation xµ → x̄µ (x) = xµ + ξµ (x) in (12.2). We have already
shown that the metric then transforms as
gµν → ḡµν = gµν + δgµν
(12.8)
However, not only the metric changes in the coordinate transformation, but also
the scalar field. The change in the scalar field, δφ is:
φ(x) = φ̄(x̄) = φ̄(x + ξ)
= φ̄(x) + ξ λ ∂λ φ
(12.9)
Here, the term ξ λ ∂λ φ is the (small) change in the scalar field. Then the variation
of the field due to the transformation (12.2) is:
δφ = φ̄(x) − φ(x) = −ξ λ ∂λ φ
(12.10)
So in order to have coordinate invariance (which we have assumed), the action
integral Sφ must be invariant under such a transformation. Hence, if the Lagrangian
density is invariant under the transformation:
xµ → xµ + ξµ
(12.11)
then applying the variational principle and demanding the action to be stationary
yields:
√
Z
√
1
1 ∂( −gL)
δSφ = 0 = d4 x −g Tµν δgµν + √
δφ
(12.12)
2
−g
∂φ
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:
1
L = gµν ∂µ φ∂ν φ − V (φ)
2
CHAPTER 12. LECTURE 12
118
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 contributing term, the variation of the action now is:
Z
√
1
δSφ =
d4 x −g( Tµν δgµν )
2
Z
√
(12.13)
=
d4 x −g(Tµν (∇µ ξ ν ))
The last step follows from tha relation between the variation of the metric tensor
and the covariant derivatives and is given by
δgµν = ∇µ ξ ν + ∇ν ξ µ
(12.14)
This relation right above can be justified from its covariant relative in (12.6) and
then use the metric tensor inverse identity to find:
gµν gµν
⇒
= δµµ
= gµν (δgµν ) = −(δgµν )gµν
(12.15)
So we notice the change in sign compared to (12.6). It is therefore possible to
write the action integral in the desired form for investigating the conservation of
the energy-momentum tensor, that is, the term ∇µ Tµν .
Z
!
√ (12.16)
δSφ = d4 x −g ∇µ (Tµν ξ ν ) − (∇µ Tµν )ξ ν = 0
To find a solution of the second term in the integrand, let us take a look at the first
one. 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 fact
demands some argumentation. Gauss’ divergence theorem in its usual form involves partial derivatives ∂ µ and not, as we have in (12.16), covariant derivatives,
∇µ . The common Gauss’ theorem is given by:
Z
Z
4
µ
d x(∂ Aµ ) = d3 Σµ Aµ = 0
(12.17)
where Σµ is a surface term on the considered surface. So in order to show that the
first term (in (12.16)) is zero, and hence also the second term, we must convince
ourselves that Gauss’ theorem also is valid when dealing with ∇µ ’s. That this is
the fact is seen from the definition:
∇µ gρσ = 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 to
consider
√
1
∇µ Aµ = √ ∂ µ ( −gAµ )
−g
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:
Z
!
√ (12.19)
δSφ =
d4 x −g ∇µ (Tµν ξ ν ) −(∇µ Tµν )ξ ν = 0
| {z }
=0
which gives the important result of the conservation of the energy-momentum
tensor:
∇µ Tµν = 0
In the (long) derivation, we have illustrated the following important fact:
Conservation of physical quantities in flat space-time due to invariance with
respect to translations is in the case of curved spaces replaced by invariance due
to coordinate transformations.
12.1.1 Generalized Bianchi identity
We will now, very short and in no detail, consider how we can find a generalized
Bianchi identity which results in a generalized conservation of Tµν . The Einstein
equation is given by:
Eµν = 8πGTµν
(12.20)
and the Bianchi identity
∇µ Eµν = 0
(12.21)
1
Eµν = Rµν − gµν R
2
(12.22)
which leads to conservation of the energy-momentum tensor, ∇µ Tµν = 0. The
Einstein tensor expressed in terms of the Ricci scalar and the Ricci tensor:
From the Einstein-Hilbert action, we find the following action:
Z
√ 1
R + ... + L
S = d4 x −g −
16πG
(12.23)
We assume the “rest”-terms after the Ricci scalar also to be a function of the Ricci
scalar, that is with + . . . + = f (R) in the expression above. The question is when
does the conservation-identity ∇µ Tµν = 0 hold. (Actually, we very much would
like this important to hold in most cases). It can be shown 2 that the Bianchi identity
holds for any function f (R). Actually, this is a generalized Bianchi identity and as
every such identity leading to the conservation ∇µ Tµν = 0.
2
Tomi Koivisto has shown this. Ask him if you have any further questions
CHAPTER 12. LECTURE 12
120
12.2 Weyl transformations
In this section, we will consider scaling transformations for both flat and curved
spaces. When the global scale-properties of flat space are attempted to be transferred to curved spaces, problems occur. The solution to these problems is the
introduction of Weyl-transformations. Let us therefore take a look at the properties
of flatness and curvature with respect to scaling transformations.
• Flat space: We perform the following scaling transformation in Minkowski
space:
xµ → x̄µ = λxµ
This means that the metric transforms as:
ds2 = dxµ dxµ → λ2 ds2
(12.24)
An example of a scale invariant theory is one describing scalar and massless
particles given with the Lagrangian density
1
L = (∂µ φ)2
2
• Curved space: In opposition to the case of flat space, scale invariance is
not satisfied when applying such a global scale transformation in a general
curved space. There will be the need of limiting to a local scale transformation in order to achieve scale invariance. An example of such a local one is
transforming 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 independently
of any point in the neighborhood. It is this local transformation which is
called a Weyl-transformation. (In some occasions the Weyl-transformations
are referred to as conformal transformations. This is not always a good idea,
since conformal transformations normally are used to describe flat spaces). It
should be obvious that the local scale transformation not necessarily prevails
global curvature. This can be illustrated by a curved object, for instance a
balloon. If a local transformation is applied to the balloon, then for example
the 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 of
the balloon is not prevailed. In the case of local scale transformations any
point may then be enlarged independently of the neighboring areas. If we
now 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 the
universe. Then all points in the universe would have to expand equally and
simultaneously. This illustrates that global scale transformations in curved
space are non-trivial.
Let us follow up the theme and investigate whether or not Weyl invariance can be
assigned to every kind of transformations. A local Weyl-transformation may look
like:
1
λ(x) = 1 + ǫ(x)
2
⇒ λ2 (x) = 1 + ǫ(x) − O2 (ǫ2 )
The last term can be omitted when ǫ is small. Similarly, the transformation of the
metric tensor is
gµν → (1 + ǫ)gµν = gµν + ǫgµν
So the variation in the metric tensor is proportional to the tensor itself. This is seen
by:
δgµν = (1 + ǫ)gµν − gµν = ǫgµν
If we have Weyl-invariance, the variation of the action should be zero with
δSW
= 0
Z
√
1
=
dx −gTµν gµν ǫ(x)
2
(12.26)
So Weyl invariance is found when Tµν gµν = 0, which leads to the following fact:
Tµν gµν = Tµµ = 0
Hence, the trace Tµµ vanishes due to the Weyl-invariance. We will in the following, for illustration, consider different physical theories and see if they are Weylinvariant or not.
12.2.1 Example with a free scalar field
The Lagrangian describing a free and massless scalar field is the well-known
1
L = gµν ∂µ φ∂ν φ
2
CHAPTER 12. LECTURE 12
122
From the definition of the Energy-momentum tensor seen in (11.24), we find in
this particular case that:
√
2 δ( −gL)
∂L
Tµν = √
= 2 µν − gµν L
−g δgµν
∂g
1
= ∂µ φ∂ν φ − gµν (∂λ φ)2
2
The trace was given by the relation Tµµ = Tµν gµν which results in
Tµµ = Tµν gµν
1
= (∂λ φ)2 − D(∂λ φ)2
2
1
(2 − D)(∂λ φ)2
=
2
Here D is the total number of both spatial and temporal dimensions. We observe
that the trace is zero only for one spatial dimension, D = 1 + 1 = 2. A massless scalar field is therefore only Weyl invariant in 2 space-time dimensions. An
example of such a theory is String theory since strings only live in one spatial
dimension.
12.2.2 Example with Maxwell theory
The Lagrangian of the photon theory is
1 2
L = Fµν
4
Again, by using the definition, the energy-momentum tensor now is
1
2
Tµν = Fµλ Fνλ + gµν Fαβ
.
4
The trace is then
1
2
.
T µµ = (D − 4)Fαβ
4
We see that the trace in the case of Maxwell theory is zero only when the number of
space-time dimensions, D, equals four, and Weyl invariance is only satisfied when
this is fulfilled. If now electromagnetic photon theory is considered in dimensions
higher than four, it will be quite different. Let us therefore take a short look at this
situation
• D>4: The expectation value of the trace is now given by:


ρ 0
0 ...
 0 −p 0 . . . 


µ
hT ν i =  0 0 −p . . . 


..
..
..
..
.
.
.
.
(12.27)
12.3. QUANTIZATION OF SCALAR FIELDS
123
We know from standard (D = 4) statistical mechanics that the pressure and energy
density in the hydro dynamical approximation is related by p = 31 ρ. However,
in this particular case where the number of dimensions is arbitrary, the relation
changes somehow
1
1
ρ.
(12.28)
p= ρ→p=
3
D−1
The general relation between pressure and energy follows, as in the usual D = 4case, from kinematics. It can be derived by considering the radiation pressure that
is caused by massless photons hitting a wall. Then the expectation value of the
trace of the higher dimensional Maxwell theory is the sum of all diagonal elements
in (12.27), and by means of relation (12.28) it is found to be zero for all dimensions:
hT µµ i = 0
This was actually seen in the last part of lecture 4 when relations between pressure
and energy density in quantum statistical mechanics were derived. Hence, that the
expectation value of the tensor is zero is not so surprising, but that the tensor itself
vanishes 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 to
achieve Weyl invariance when Maxwell fields are coupled to scalar fields. We have
in the previous two examples seen that the theory describing scalar theory only is
Weyl invariant when D = 2, and on the other hand electromagnetic photon theory
is only Weyl invariant for a total dimension satisfying D = 4. At first sight the
answer to this question seems to be no, but this is actually not so. It can be shown
1
Rφ2 is added to the Lagrangian of the scalar field, that is
that if a term ∼ 12
L=
1
1 µν
g ∂µ φ∂ν φ + Rφ2
2
12
then the coupled theory is Weyl-invariant. This is so since the added term in the
scalar 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
mass term of the scalar field.
1
2
12 Rφ
acts as a
12.3 Quantization of scalar fields
We are now about to perform a quantization with respect to the scalar field in a
RW metric (as was considered in lecture 11). This quantization is very important
3
See article by Finn, Håvard and Ingunn. finn artikkelen
CHAPTER 12. LECTURE 12
124
in cosmology and will be of great use later in this course. We will see that it leads
to, among others, the primordial spectrum. However, that will be subject of some
later lecture.
12.3.1 Scalar field in a RW background
The line element in the Robertson-Walker metric is given by
2
ds2 = dt − a2 (t)dx2
and the metric itself is
gµν


1
0
0
0
 0 −a2 (t)

0
0

=
2
 0

0
−a (t)
0
2
0
0
0
−a (t)
For later use it is convenient to define the spatial determinant, h, in analogy with
that of (11.12) as the determinant of gµν when the first row and column are removed. 4
So far we have not assigned any constraints to the scale factor, a(t), but rather
considered an arbitrary background. Later, when inflation is being considered, the
scale factor will be exponential or a power function with respect to time t, but so
far the scale factor is assumed to be arbitrary. The scalar field for massive particles
is described by the Lagrangian
1
1
L = gµν ∂µ φ∂ν φ − m2 φ2
2
2
the solutions, we have earlier seen, satisfies the classical wave equation given by
2φ + V ′ (φ) = 0 → 2φ + m2 φ = 0
which is also called the (general) equation of motion of the scalar field. And for
the RW case, this equation turns into the earlier seen
φ̈ + 3H φ̇ −
1 2
∇ φ + V ′ = 0.
a2
(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) can
be written as
2
k
2
ϕ̈ + 3H ϕ̇ +
+ m ϕ = 0.
(12.30)
a2
4
To avoid any confusion, the first row and column that are to be removed before we find the
spatial determinant, are the ones involving time components so that only the a2 -s are left.
12.3. QUANTIZATION OF SCALAR FIELDS
125
We observe that the effective mass-term now also contains the scale factor, a and
the wave number k. The equation above is a classical equation and therefore the
solution 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.
Z
√
↔
′
(uk , uk ) = i d3 x 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 now
becomes
(uk , uk′ ) = ia3 (2π)3 δ(k − k′ )(ϕ∗k ϕ̇k′ − ϕ̇∗k ϕk′ ).
If we now require the normalization
(uk , uk ) = (2π)3 δ(k − k′ ),
(12.31)
then the normalization condition satisfies
ia3 (ϕ∗k ϕ̇k − ϕ̇∗k ϕk ) = 1
The field operator with standard normalization now looks like
Z
d3 k
φ̂(x) =
[âk uk (x) + h.c.].
(2π)3
(12.32)
(12.33)
Here, h.c. corresponds to hermitean conjugated terms. We know that the annihilation and creation operators are defined in the usual way by means of the inner
product (uk , uk′ ), (lecture 9, equation(9.18)). This yields
(uk , uk′ ) = (2π)3 δ(k − k′ )
= [âk , â†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 an
oscillator 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:
ϕ̇ =
ϕ̈ =
dϕ
dϕ dτ
1
=
= ϕ′
dt
dτ dt
a
1 ′′ a′ ′
ϕ − 3ϕ
a2
a
CHAPTER 12. LECTURE 12
126
It is convenient also to write the Hubble parameter in terms of conformal
time:
ȧ
1
1
H = = 2 a′ = H
a
a
a
(The conformal Hubble parameter must not be confused with the Hamilto′
nian density H). We have H = aa . So equation (12.30) is written to the
form:
(12.34)
ϕ′′k + 2Hϕ′k + (k2 + m2 a2 )ϕk = 0
2. We observe that the field equation still has a friction term ∼ ϕ′k when interpreting it as an oscillator. This term complicates the whole story and it can
be removed by rewriting ϕk = a1 ψk (t). Equation (12.34) can then be written
as
ψk′′ + ωk2 (τ )ψk = 0
where ωk2 = k2 + m2 a2 −
a′′
a
= ωk2 (τ )
We see that after the two small steps, the field equation of the scalar field in
an RW background is precisely the same problem that we considered in lecture 9
where an oscillator with time dependent frequency was considered as an example
of quantization of an external field. We can also find, from (12.32) , that
ψk′∗ ψk − ψk∗ ψk′ = i
(12.35)
The quantum field in terms of its fieldoperator in (12.33) can now be re-expressed
as:
Z
d3 k
1
[âk ψk (τ )eik·x + h.c.]
(12.36)
φ̂(x) =
a
(2π)3
where the normalization is given by
[âk , â†k′ ] = (2π)3 δ(k − k′ )
However, there are some difficulties when applying the scalar field theory to
physical problems such as describing the universe at an early stage. We have rewritten the theory to a form containing an oscillator. For a given value of the scale
factor 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 as
shown in figure (12.1) We would in that case be able to repeat the process we did
when simplifying the oscillator with time dependent frequency in the previous lectures (see figure 9.1) and then be able to describe the vacuum consistently. The
vacuum is given by the in-modes when time is taken to the far past. Different
12.3. QUANTIZATION OF SCALAR FIELDS
127
modes (which may be a result of a non-constant frequency at small τ ) result in
different vacuums and different sets of operators that again are connected with the
Bogoliubov coefficients. Hence, what acts as vacuum at some set of initial conditions, will not necessarily be so at different initial conditions when the modes are
defined. Instead we may have a state where particles are present. Different kinds
of vacuums will decide whether there is a possibility of particle creation or not. So
if the frequency has a nastier shape than that in figure 12.1 , we can not be sure
of what kind of vacuum our physical system possesses. Such problems are typical
when looking at inflation.
In the coming lectures, we will learn more about gravitons and gravitational
waves in flat space.
ωk
I
t
Figure 12.1: To define the vacuum in the early universe in a proper way, the oscillator frequency should satisfy a constant value at early times (I).
128
CHAPTER 12. LECTURE 12
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. We
shall 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
(γ µ ∂µ − m) ψ(x) = 0
where the anticommutator of the γ matrices gives the metric
γµ γν + γν γµ = 2gµν (x)
The metric can be expressed as a transformed Minkowski metric
gµν (x) = Vµµ̂ Vνν̂ ηµ̂ν̂
And thus this transformation tensor can be used to transform the γ matrices as well
γµ (x) = Vµµ̂ (x)γµ̂
where
1 0
γ =
0 −1
0̂
and
0
σk
γ =
−σk 0
k̂
We want this to be invariant under local Lorentz transformations
ψ(x) → ψ ′ (x) = S(x)ψ(x)
Where
i
S = e− 4 σµ̂ν̂ θ
129
µ̂ν̂ (x)
CHAPTER 13. LECTURE 13
130
and σµ̂ν̂ =
erator
i
2
[γµ̂ , γν̂ ] . We ensure this invariance if we replace the differential op1
∂µ → ∇µ = ∂µ − Ωα̂µβ̂ σα̂β̂ ,
2
where Ωµα̂β̂ is called the gravitational field . This is similar to the ordinary gaugefield Aµ in the ordinary covariant derivative
∂µ → Dµ = ∂µ − iAaµ Ta
where T fulfills the commutation relation
[Ta , Tb ] = ifabc Tc .
13.2 Gravitons in flat spacetime
Usually when we go from flat to curved spacetime we exchange the Minkowski
metric with the general metric, and we will also do so here, but now we only have
slightly curved spacetime. We will write
ηµν → gµν (x) = ηµν + hµν (x)
where |hµν | ≪ 1. As gµλ gλν = δνµ we get that
gµν = η µν − hµν + O(h2 )
Now we introduce a coordinate transformation
xµ → x̄µ = xµ + ξµ (x)
and the metric transforms as
gµν (x) → ḡµν (x̄) =
∂xα ∂xβ
gαβ (x) = ηµν + h̄µν (x)
∂ x̄µ ∂ x̄ν
as we saw in the last lecture we can now express h̄µν
h̄µν = hµν − ∂µ ξν − ∂ν ξµ
We have the Christoffel symbols . . .
Γσµν =
1
∂µ hσν + ∂ν hσµ − ∂ σ hµν
2
. . . and the Riemann curvature tensor.
Rσρµν =
1
(∂ρ ∂µ hσν − ∂ρ ∂ν hσµ + ∂ σ ∂ν hρµ − ∂ σ ∂µ hρν )
2
13.2. GRAVITONS IN FLAT SPACETIME
131
Contraction gives the Ricci tensor
1
Rµν = Rλµλν = (∂λ ∂µ hλν + ∂λ ∂ν hλµ − ∂µ ∂ν hλλ − ∂ λ ∂λ hµν
2
Both the Riemann and the Ricci tensor are invariant under this transformation.
We introduce now hλλ , and the transformation
1
h̄µν = hµν − ηµν h
2
and this has nothing to do with the previous h̄µν . We now choose Hilbert gauge1
∂ µ h̄µν = 0
This is analog to Lorentz2 -gauge in electromagnetism ∂ µ Aµ = 0. h̄µν has its trace
reversed under the transformation
h̄µµ = h − 2h = −h
another transformation returns the ordinary trace
1
hµµ = h̄µµ − η µµ h̄ = h
2
In Hilbert-gauge we get the Ricci curvature tensor
1
Rµν = − ∂ 2 hµν
2
and the Ricci scalar becomes R = − 12 ∂ 2 h. This leads to the Einstein tensor
Eµν
1
= Rµν − ηµν R
2
1 2
1
= − ∂ hµν − ηµν h
2
2
= 8πGTµν
we recognize h̄µν in the expression and this leads us to
∂ 2 h̄µν = −16πGTµν
(13.1)
again this is analog to electromagnetism where we have ∂ 2 Aµ = Jµ .
As usual we are interested in finding the Lagrangian density. Here it is
1
1
L = (∂ν hαβ )2 − (∂α h)2 + ∂α hαν ∂ν h − ∂α hαν ∂β hβν
2
2
1
This 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.
2
Actually 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.
CHAPTER 13. LECTURE 13
132
The trace reversed energy momentum tensor is now
1
T̄µν = Tµν − ηµν T λλ
2
and this leads us to the relation
∂ 2 hµν = −16πGT̄µν .
We can write this as
hµν = 16πG
1
Tµν
−∂ 2
where the differential operator can be thought of as the momentum operator
1
k 2 . The change of action is
Z
Z
√
1
′
µν
4 √
δS = −
d x −gTµν h = d4 x −gLint
2
1
−∂ 2
∼
where we have the interaction Lagrangian density
1 µν
1 ′
1
1 ′ 1 µν
1 ′ µν
′
h = − Tµν
16πG 2 T̄ µν = −8πGTµν
T̄ = 2 Tµν
T̄
Lint = − Tµν
2
2
2
−∂
−∂
Mp
−∂ 2
where the Planck mass is
1
8πG
with help of a polarization matrix
Mp2 =
Now we want to express T̄µν
T̄µν = Pµνρσ T ρσ
where
1
Pµνρσ = (ηµρ ηνσ + ηµσ ηνρ − ηµν ηρσ )
2
if we insert this in our Lagrangian density we get
Lint =
where Dµνσρ (k) =
Pµνσρ
k 2 +iε
(13.2)
1 ′ µνσρ
T D
Tρσ
Mp2 µν
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 the
photon field Aµ has four degrees of freedom. Then we impose a Lorentz-gauge
∂µ Aµ = 0 which removes one degree of freedom. Now if we have a gauge trans′
formation Aµ → Aµ = Aµ +∂χ. And we demand that ∂ µ A′µ = ∂ µ Aµ +∂ 2 χ = 0.
13.3. GRAVITATIONAL WAVES
133
This gives us that ∂ 2 χ = 0. Now we are down to two degrees of freedom. We can
write
Z
i
d3 k 1 X h
−ik·x
e
(λ,
k)a
(k)e
+
h.c.
Aµ (x) =
µ
λ
(2π)3 2ω
λ=R,L
where
kµ = ω(1, 0, 0, 1)
r
1
µ
(0, 1, i, 0)
εR =
2
r
1
µ
εL =
(0, 1, −i, 0)
2
(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 16
components. It is symmetric, so we have only ten degrees of freedom. The Hilbertgauge ∂ µ h̄µν = 0 removes four degrees of freedom. Then to have invariance under
the transformation hµν → h′µν = hµν +∂µ χν +∂ν χµ we must have ∂ 2 χµ = 0. And
this puts us down another four degrees of freedom so we end up with 10−4−4 = 2
degrees of freedom, just as in QED.
But we have not yet specified the constraints that removes the last four degrees
of freedom. We choose these to be
h̄µµ = −h = 0
hµ0 = 0
(13.3)
this is really five equations, but they are not totally independent, so they will remove only four degrees of freedom. This is called TT-gauge where the T’s stands
for transverse and traceless.
Now we can write out
hµν (x) =
Z
i
d3 k 1 X h
−ik·x
e
(λ,
k)a
(k)e
+
h.c.
µν
λ
(2π)3 2ω
λ=+,×
we can write this in matrix form

0 0
0
0 h+ h×
hµν (z, t) = 
0 h× −h+
0 0
0

0
0
 ei(kz−ωt) = (h+ e+ + h× e× )ei(kz−ωt) (13.4)
0
0
CHAPTER 13. LECTURE 13
134
where the “basis” matrices are
e+
e×

0
0
= 
0
0

0
0
= 
0
0

0 0 0
1 0 0
 = eµν (+, k)
0 −1 0
0 0 0

0 0 0
0 1 0
 = eµν (×, k)
1 0 0
0 0 0
and the propagator can be written
iDµνρσ = h0|T hµν (x)hρσ (x′ )|0i
13.3.2 Particle hit by a wave
Now we shall see what happens if a particle is hit by a gravitational wave. At
first the particle is at rest, with a 4-velocity uµ (t = 0) = (1, 0, 0, 0). As the
4-acceleration for this particle is zero, the geodesic equation for this particle is
duµ
+ Γµαβ uα uβ = 0
ds
as only the time component of the velocity vectors are nonzero, we get the Christoffel symbols
1
Γµ00 = η µν (hν0,0 + h0ν,0 − h00,ν )
2
we see from (13.3) that all these h-components are zero in TT-gauge and we get
Γµ00 = 0 ⇒
duµ
=0
ds
We now look at two test particles separated with a spatial distance along the xaxis, dxµ = (0, ξ, 0, 0). The line element gives us (we choose the positive solution)
ds =
p
gµν
dxµ dxν
√
= ξ −g11
1
= ξ 1 − h+ cos(kz − ωt)
2
and two particles separated in the y-direction, dxµ = (0, 0, ξ, 0) leads to the distance
1
ds = ξ 1 + h+ cos(kz − ωt)
2
so, if we have four particles they will oscillate as in figure (13.1). For the crosscomponents 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
1
Rµν − gµν R = 0
2
which gives that the Ricci tensor is also zero Rµν = 0, but if we expand the Ricci
tensor to second order
Rµν
(0)
(1)
(2)
= Rµν
+ Rµν
+ Rµν
+ ...
∼ 0 + hµν + h2µν + . . .
The expectation value of the second order term is not generally zero
(2)
hRµν
i=
6 0
This term gives geometry to the background. The energy-momentum tensor for the
background is then given from the Einstein equations
1
(2)
hRµν
i − ηµν hR(2) i = 8πGtµν
2
and from here it can be shown that
tµν =
1
h∂µ hmn ∂ν hmn i
32πG
This gives in TT-gauge that
t00 =
1
h(∂0 h+ )2 + (∂0 h× )2 i
16πG
136
CHAPTER 13. LECTURE 13
Chapter 14
Lecture 14
This lecture we will continue our study of gravitational waves and we will see how
such waves may be quantized.
14.1 Quantum fluctuations in a classical background (repetition)
In lecture 11 we introduced in (11.21) the Einstein-Hilbert action for a gravitational
field
Z
1
4 √
(14.1)
R + Lm
S = d x −g −
16πG
where Lm is a common Lagrangian density for matter. Applying the principle of
stationary action gives us the Einstein field equations as classical solutions:
1
m
Eµν = Rµν − ḡµν R = 8πGTµν
2
(14.2)
which gives as a classical solution the background metric ḡµν (x). Including quantum
fluctuations to the metric on the form hµν (x), we obtain a total metric
gµν (x) = ḡµν (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 ḡµν . Last
lecture we introduced the Hilbert gauge
where
∂µ h̄µν = 0
(14.5)
1
h̄µν = hµν − ηµν h
2
(14.6)
137
CHAPTER 14. LECTURE 14
138
Now, enforcing this Hilbert gauge, the Einstein equations (14.2) reduce to
∂ 2 h̄µν = −16GTµν
(14.7)
∂ 2 hµν = −16GT̄µν
(14.8)
or
where T̄µν = Tµν − 21 ηµν T . 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 = −16πG
1
T̄µν
∂2
In the quantum mechanical description the inverse differential operator
us a Feynman propagator on the form
Z
d4 k Pµνρσ −ik(x−x′ )
′
Dµνρσ (x − x ) =
e
(2π)4 k2 + iε
(14.9)
1
∂2
will give
(14.10)
where Pµνρσ is the projection operator introduced in (13.2), and provided that we
are dealing with two electric charges this will give us the common Newtonian poe2
in the non-relativistic limit. Now, classically we have to view the ∂12 tential 4πr
operator a bit different, and instead of a Feynman propagator we will obtain what
is called a retarded operator. This will be studied in some detail at the end of this
section 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 using
techniques with partial integration and so on. Let’s do this with the Einstein-Hilbert
action!
Starting with an Einstein-Hilbert action given by
Z
1
4 √
R
,
gµν = ηµν + hµν
(14.11)
S = d x −g −
16πG
and
gµν = η µν − hµν + hµα hαν + O(h3 )
(14.12)
and expanding also R to the second order we obtain1
Z
1
1
1
2
2
αν
αν
βµ
4
S=
d x (∂µ hαβ ) − (∂µ h) + (∂α h )(∂ν h) − (∂α h )(∂β hν )
32πG
2
2
(14.13)
1
We 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
! 1
∂µ hµν = ∂ν h
2
two of the terms will cancel out and we are left with
Z
1
1
1
2
2
4
S=
d x (∂µ hαβ ) − (∂µ h)
32πG
2
2
(14.14)
(14.15)
R
Applying
techniques from the first lectures, for example that d4 x(∂µ φ)2 →
R 4 the
− d xφ∂ 2 φ when using partial integration we have
Z
1
1
1
d4 x hµν [−ηµα ηνβ + ηµβ ηνα + ηµν ηαβ ] ∂ 2 hαβ
S=
32πG
2
2
|
{z
}
=Pµναβ
Here Pµναβ is the projection operator defined in (13.2) and keeping in mind that
∂ 2 → k12 when Fourier transforming we see that our propagator is
Dµναβ =
Pµναβ
k2 + iε
(14.16)
14.3 Canonical normalization of the gravitational field
Last lecture we saw that Hilbert gauge corresponds to Lorentz gauge in QED, but
that we still were left with two degrees of freedom. We introduced the TT gauge
given by the two conditions
i)
hµµ
=h=0
ii) hµ0 = 0
⇒ ∂i hij = 0
The last condition corresponds to the coulomb gauge from QED, ∂i Ai = 0. We
remember from (13.4) that we were left with two independent amplitudes given by
h+ and h× . In this TT gauge our action becomes
Z
1
1
(14.17)
d4 x (∂µ hij )2
S=
32πG
2
If we compare (14.17) with the action for a massless scalar field
Z
1
Sφ = d4 x (∂µ φ)2
2
(14.18)
we see that
hij
∼
h2ij
∼
2
φ
MP
4 2
φ = 32πGφ2
MP2
1
where MP2 = 8πG
. So if we want to normalize the gravitational field such that the
coefficient in front of (∂ )2 is 12 , we must impose the canonical normalization
CHAPTER 14. LECTURE 14
140
hµν =
2
Hµν
MP
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 momentum
tensor for gravitational waves on the form (in TT-gauge)
1
h∂µ hij ∂ν hij i
(14.19)
32πG
The reason why we have to use the mean value over a volume of some finite extension, is that the equivalence principle makes it possible to locally transform away
any gravitational effect, which would make our calculations tremendously boring
and uninteresting if using a local theory. The original derivation of (14.19) made
by Isaacson himself (which can be found in Misner, Thorne, Wheeler as well) is
painfully long and boring. Luckily Finn, Håvard and Ingunn have done it in a much
more compact and pleasant (but less stringent?) way which we will repeat here.
For a scalar field we have the energy-momentum tensor
tµν =
1
φ
= ∂µ φ∂ν φ − ηµν (∂λ φ)2
Tµν
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
Z
1
φ
φ
d4 xTµν
hTµν
i =
V4
1
= h∂µ φ∂ν φi − ηµν h(∂λ φ)2 i
(14.20)
2
The mean value in the last term can be written
Z
1
d4 x(∂λ φ)2
h(∂λ φ)2 i =
V4
Z
1
=
d4 xφ∂ 2 φ = 0
(14.21)
V4
because the surface term vanish since we are free to impose periodic boundary
conditions. Then we have, by comparing (14.17) and (14.18), that
φ
hTµν
i = h∂µ φ∂ν φi
1
h∂µ hij ∂ν hij i
(14.22)
=
32πG
which is just the same as Isaacson’s expression. Note that we assumed flat space
in 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
Z
1
a
1
1
5
2
2
2
4
d x (∂µ hαβ ) − (∂µ h) + (∂µ h)2hαβ + O(h )
S=
32πG
2
2
4
Introducing our physical Hµν given by hµν = M2P Hµν this action becomes
Z
1
1
a
4
2 2
5
2
2
S = d x (∂µ Hαβ ) − (∂µ H) + 2 (∂µ H) Hαβ + O(H )
2
2
MP
The last term is describing an interaction between four gravitons. Since the coupling is of the order ∼ M12 , this is not a renormalizable interaction (that is the case of
P
1
all quantum field theories with couplings on the form ∼ mass
). Many people have
used this fact to argue that one may not quantize the gravitational field. Formally
this is of course true, but at the same time the question on renormalizability is uninteresting for a physicist working in a real world. As mentioned earlier, it is just a
question of staying
the range of validity for your theory. In the theory terms
inside
k 4
on the form ∼ MP that cannot be absorbed will appear, but in practical problems in astrophysics where k is small, these problems can safely be swept under
the rug, and you can treat it as an effective field theory. The theory will however
break 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 too
small to be observed in objects like quasars, so you have to go all the way back to
the inflationary phase to observe them.
Historically the first physicist to work with this kind of quantized gravitation
was John F. Donoghue. Both he and Cliff Burgess later put it into an effective field
theory. The Dane Emil Bjerrum-Bohr showed that there were some mistakes in
what these people had done. Today, the theory that we use are based on the works
of these people.
14.6 Green functions and propagators
Both in quantum and classical theory we have expressions on the form
∂ 2 hµν
2
= −16πGTµν
∂ Aµ = −eJµ
which can be interpreted as
D
|{z}
A(x) = J(x)
| {z } | {z }
diff. operator field
source
CHAPTER 14. LECTURE 14
142
So formally we can write the solution as
A(x) = D −1 J(x)
The further procedure is known from electromagnetism where we introduce a
Green 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:
Z
d4 k −ik(x−x′ )
′
δ(x − x ) =
e
(2π)4
and using an inverse Fourier transform of the Green function we can write it as
Z
d4 k
′
G(k)e−ik(x−x )
(14.24)
G(x − x′ ) =
4
(2π)
To choose a specific example, we choose the differential operator D = ∂ 2 . Remembering that ∂ 2 → −k2 when Fourier transforming we demand that the Green
function (14.24) should obey (14.23):
Z
Z
−ik(x−x′ ) !
d4 k ik(x−x′ )
d4 k 2
−k
G(k)
e
=
e
(2π)4
(2π)4
And thus we have that
G(k) = −
and
′
G(x − x ) = −
Z
1
k2
d4 k 1 −ik(x−x′ )
e
(2π)4 k2
(14.25)
Now, since G(x-x’) is a solution for a point source we have that
Z
A(x) = d4 x′ G(x − x′ )J(x′ )
and
D A(x) =
Z
d4 x′ D G(x − x′ ) J(x′ ) = J(x)
|
{z
}
δ(x−x′ )
Now, let’s try to calculate G(x − x′ ) from (14.25). We have
Z ∞
Z
dk0
1
d3 k
′
′
e−ik0 (t−t )+ik(x−x )
G(x − x′ ) = −
2
3
2
(2π) −∞ 2π k0 − k
We have poles at ±k (see figure 14.1). What we do now is of course to integrate
around the poles in the complex plane. We will obtain different Green functions
depending on how we choose our path of integration.
14.6. GREEN FUNCTIONS AND PROPAGATORS
143
Im(k0 )
−k
k
Re(k0 )
Figure 14.1: The Green function has poles for in ±k
To obtain the Feynman propagator one choose to go “below” one of the poles
and “above” the other, or equivalently shifting one pole ε in positive imaginary
direction and the other one ε in negative imaginary direction (see figure 14.2). The
poles for k0 then shift like
1
1
1
= 2
→ 2
2
k
k + iε
k0 − k2 + iε
p
k0 = ± k2 − iε
There are two different contours that both satisfy these requirements, one that is
closed in the upper half plane and one closed in the lower half plane corresponding
to a particle with positive energy going forward in time (t > t′ ) and the other one
corresponding to a anti-particle going backwards in time (t < t′ ). Again, see figure
14.2.
In classical physics this negative energy solution is considered unphysical, so
one has to deal with the poles in a somewhat different way. What one does is to
make a shift of the kind
1
1
1
→ 2
= 2
2
2
k
k + iεk0
k0 − k + iεk0
k0 = ±
p
k2 − iεk0
which shifts both of the poles in negative imaginary direction. As one can see
from figure 14.3, we have to possible contours. One is in the lower plane and
is the physical t > t′ solution which gives us the non-zero part of the retarded
propagator. The other contour is in the upper half plane and gives the zero part
of the retarded propagator. Similarly we can construct an advanced propagator by
CHAPTER 14. LECTURE 14
144
t′ > t
Im(k0 )
t > t′
Im(k0 )
k0 = −k + iε
k0 = −k + iε
Re(k0 )
k0 = k − iε
Re(k0
k0 = k − iε
Figure 14.2: There are two possible contours of integration. The one to the left
corresponding to an anti-particle with negative energy going backwards in time,
and the one to the right corresponding to a particle with positive energy going
forward in time.
letting the parameter ǫ be negative. In classical physics only the retarded solution
is used.
We have seen that in quantum mechanics and classical physics we have the
same expression for the Green function, it is just a question of how we deal with
the poles2 .
Applying residue calculations we will now sketch how one may calculate the
classical retarded propagator from the Green function.
′
Gk (x − x ) =
=
=
Z
d3 k ik(x−x′ ) sin k(t − t′ )
e
(2π)3
k
Z 1
Z ∞
′
1
ikµ(x−x′ ) sin k(t − t )
2
dµe
dkk
2π
8π 3 0
k
−1
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 figure 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
Im(k0 )
t < t′
Re(k0 )
t > t′
Figure 14.3: Integrating along the solid line in the lower half plane corresponds to
finding the non-zero part of the retarded propagator, while an integration along the
dashed 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 substitute the ∂12 operator with Gk . Then we have
∂ 2 hµν = −16πGGk T̄µν
and we have
Z
16πG
δ(t − t′ − (x − x′ ))
T̄µν (x′ )
d 4 x′
4π
(x − x′ )
Z
T̄µν (x′ , t − (x − x′ ))
= −4G d3 x′
(x − x′ )
hµν (x) = −
Next lecture we will finish this calculation and use the result on a binary star system
and on inflation.
146
CHAPTER 14. LECTURE 14
Chapter 15
Lecture 15
We ended the last lecture in the middle of a calculation of the gravitational radiation
from a general source of gravity. For convenience, the entire calculation is written
below. An illustration of gravitational radiation is seen in figure (15.1). The sources
of the gravitational waves are matter distributions such as for instance galaxies.
Observer
Source
x − x′
Tµν (x)
x
x′
Figure 15.1: The figure shows a general matter (and energy) distribution causing
gravitational radiation.
15.1 Gravitational radiation
The wave equation of gravitational waves was found in (13.1) in lecture 13 and in
TT-gauge, it can be written as
∂ 2 hµν
⇒ hµν
= −16πGT̄µν
1
= −16πG 2 T̄µν
∂
If we instead of the operator ∂12 apply the general Green’s function as calculated in
(14.26), the wave equation turns into
147
CHAPTER 15. LECTURE 15
148
Z
δ(t − t′ − |x − x′ |)
16πG
T̄µν (x′ )
d4 x′
hµν (x) = −
4π
|x − x′ |
Z
T̄µν (x′ , t − |x − x′ |)
= −4G d3 x′
|x − x′ |
If the size of the source is small compared to the distance to the observer, that is
with |x| = r >> |x′ |, the expression simplifies somewhat
Z
4G
hµν (x) = −
d3 x′ Tµν (x′ , t − r)
r
This will be the situation if gravitational waves are observed from (distant) galaxies. We also notice from the expression above that the gravitational emission has
similar 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 of
the energy -momentum tensor
∂ µ Tµν = 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 combined
with the fact that ∂ 2 = ∂t2 − ∇2 , the following holds
∂ 2 T00
∂ 2 Tij
=
.
∂t2
∂xi ∂xj
• We start out with the expression
∂2
∂t2
Z
(15.2)
d3 xT00 xk xl
2
∂
Here, the partial differentiation ∂t
2 only acts on the term T00 . By the relation
in (15.2) this yields
Z
Z
∂ 2 Tij k l
∂2
3
k l
3
x x
d
xT
x
x
=
d
x
00
∂t2
∂xi ∂xj
Z
= 2 d3 xTkl .
(15.3)
The last step follows by applying two partial integrations.
Hence, the calculation proceeds with
hij (x, t) =
=
Z
2G ∂ 2
d3 xT00 (x′ , t − r)x′i x′j
r ∂t2
2G ¨
Iij (t − r)
r
(15.4)
15.2. EMITTED ENERGY AND MOMENTA
149
where the moment of inertia, defined as
Z
Iij (t − r) ≡ d3 xT00 (x′ , t − r)x′i x′j
(15.5)
has been inserted for. To find the final answer to the calculation, we will apply TTgauge. 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 (We
already know that the lefthand side, hij is traceless in TT-gauge). This gives
1 ¨
2G ¨
Iij (t − r) − δij Ikk
(15.6)
hij (x, t) =
r
3
A more compact answer can be found by inserting for the the quadrupole moment,
Q of the source. Thus
hij =
2G
Q̈ij
3r
(15.7)
R
The quadrupole moment is defined as Qij ≡ d3 xT00 (3xi xj −δij x2 ). The expression found in (15.7) is useful in calculations of gravitational waves from various
sources. We also notice by comparing the equations (15.6) and (15.7) that the
quadrupole moment simply is the traceless moment of inertia of the mass/energy
distribution.
15.2 Emitted energy and momenta
We would now like to find out just how much energy and momenta that is being
emitted in the gravitational radiation when TT-gauge is applied. These quantities
are both given by the Isaacson-tensor of energy and momenta, tµν as found in
lecture 13
1
h∂µ hij ∂ν hij i.
tµν =
32πG
The energy density of the radiation is given by the zeroth component of the Isaacson tensor
1
t00 =
hḣij ḣij i
32πG
1
hḣ2 i
=
32πG ij
By applying the equation in (15.7), an expression for emitted energy, − dE
dt , can be
derived
−
dE
G ... 2
= hQij i
dt
45
(15.8)
CHAPTER 15. LECTURE 15
150
15.3 Detecting gravitational waves
Gravitational radiation and the various attempts and set-ups to detect it is a topic
receiving an increasing interest from physicists. The radiation is predicted in Einstein’s General Theory of Relativity, but is yet to be properly investigated through
experiments. It is common to differ between indirect and direct observations of
gravitational radiation.
15.3.1 Indirect observations
The indirect observations are in a sense more vague than the direct ones, but therefore also easier to perform. Gravitational radiation has been detected indirectly by
looking at binary pulsars. One example is the binary pulsar system P SR1913+16.
In this system two stars are orbiting each other. The way the gravitational radiation
has been detected, is that the orbiting period of the pulsar has been detected over a
time interval of more than 20 years, and the period is observed to be changing with
time. This is due to energy loss which can be explained as emitted gravitational
waves. 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, and
their 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. This
is among other places done in large observatories in the USA. The set-up is quite
similar to that legendary of Michelson and Morley with detectors perpendicular to
each other and of the size L ∼ 3km in each direction. When gravitational waves
enter 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 would
be a confirmation that gravitational waves are present. However, to perform such
an observation, a lot of technical problems, such as disturbing noise from the surroundings have to be solved. To get a feeling with just how fine tuned the observationally set-up has to be, let us estimate the magnitude of the shift, h due to
gravitational radiation from a binary system analog to the situation of indirect detection. A sketch of such a system is shown in figure (15.2). For simplicity, we
assume the two stars in the binary system to be of equal mass, M .
1
There is more to read, and satisfactory plots to view in the article written by Taylor, “Binary
pulsars and relativistic gravity”. Rev. Mod.Phys. 66, 711-719 (1994). Of specific interest is figure
10 in this article.
15.3. DETECTING GRAVITATIONAL WAVES
151
M
M
Figure 15.2: Detecting gravitational radiation directly: A binary system with equal
mass, M causes gravitational radiation.
From (15.7) we find that
hij
GM
G
Q̈ij ∼
(GM Pb2 )2/3
r
rPb2
5/3 2/3
M
1h
100pc
−21
≈ 10
.
Msun
Pb
r
∼
(15.10)
Here, Pb is the period of the double-star system, Msun is the mass of the sun, pc is
parsec and h is one hour. So for gravitational radiation emitted from a system of the
sun mass, with orbiting time of about 1h and with distance to the observer of a few
hundred parsec (typically for galaxies?), the shift to be observed is of magnitude
10−21 . This is an extremely small number, but still seems to be the best we can
hope for at the time being. (Actually, a lot of systems give rise to gravitational
radiation smaller than this). If we now look at the actual shift in length, δL, we find
with L ∼ 103 m and h ∼ 10−21 that the scale of the observations is of magnitude
δL = Lh = 103 × 10−21 = 10−16 cm
1
of the radius of an atom. In order to detect gravitational
This length is about 1000
radiation, the set-ups thus involves scale smaller than atoms. The sensitivity in
such measurements is in (near?) future expected to be as good as 10−23 , so that the
waves eventually can be detected.
CHAPTER 15. LECTURE 15
152
15.4 Gravitational radiation in a RW metric
We have earlier investigated perturbations of the Minkowski metric, ηµν when considering gravitational radiation. The perturbed metric was then written as the flat
Minkowski 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 proceed
with cosmology, the Minkowski metric is replaced by the Robertson-Walker (RW)
metric. The RW metric is a much better choice than the flat Minkowski metric
as the background metric in cosmology and is still quite easy to deal with. The
procedure will be similar to the one we saw for the case of a flat metric in the two
previous lectures. The perturbations will cause tensor fluctuations on the metric.
These fluctuations will be of great importance when the lectures proceed with inflation. We will also see (in the end of this lecture) that the tensor fluctuations are
very similar to the simpler scalar fields and that the tensor fluctuations can be found
directly from the scalar fluctuations simply by scaling a prefactor. This will be the
main topic in the rest of this lecture and we will also derive the two Friedmann
equations. There will be quite a lot of mathematics (and even more omitted) on our
way, 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 defined
by dτ = dt
a . (One of the advantages of conformal time, is that with such a choice of
coordinates, light always travels at 45 degrees on a space-time diagram, regardless
of 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 a
small term, hµν to the Minkowski metric.
g̃µν = a2 (τ )(ηµν + hµν )
(15.14)
The perturbations in the metric can be interpreted as the tensor fluctuations that we
are interested in. The equation of motion of the tensor fluctuations is now found
by means of the Einstein-Hilbert action
Z
p
1
d4 x −g̃(R̃ + 2Λ).
(15.15)
S=−
16πG
15.4. GRAVITATIONAL RADIATION IN A RW METRIC
153
Here, R̃ is the perturbed Ricci scalar. To proceed, we will need the scalar curvature,
R̃. It is found from the Ricci-tensor, Rµν . Before we calculate the Ricci-tensor (and
then 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 = ∂α Γµνβ − ∂β Γµνα + Γρνβ Γµρα − Γρνα Γµρβ
(15.16)
For the interested reader, there will be some short comments on the similarities
between the Riemann tensor and gauge-fields in Appendix A. The Ricci tensor is
now calculated, also leading to the two Friedmann equations. These results will be
needed to continue with the tensor-fluctuations, but first, let us calculate the Ricci
tensor.
15.4.2 Calculating the Ricci tensor
From its definition, the Ricci tensor is
Rµν = Rαµαν = ∂α Γαµν − ∂ν Γβµβ + Γαµν Γβαβ − Γβµα Γαβν
(15.17)
From the above expression, we see that if we find the Christoffel symbols, Γ, then
the Ricci tensor is also found. The contributing (non-vanishing) components of the
Christoffels are as follows
a′
≡H
(15.18)
Γ000 =
a
Γ0ij = Hδij
(15.19)
Γi0j
= Hδji .
(15.20)
Here, we have defined the conformal Hubble parameter, H and the primes denote
da
. If the Γ’s are
differentiation with respect to conformal time, such as a′ = dτ
inserted in the expression for the Ricci tensor in (15.17), we find the components
of Rµν to be
R00 = −3H′
⇒
Rij
⇒
3 ′
H
a2
= (H′ + 2H2 )δij
1
= gii Rij = − 2 (H′ + 2H2 )δij .
a
R00 = g00 R00 = −
Rij
(15.21)
(15.22)
In the above calculations, we have used the metric gµν = a21(τ ) to raise one index.
The Ricci scalar is therefore
6
(15.23)
R = R00 + Rii = − 2 (H′ + H2 ).
a
We eventually want to find the perturbed scalar, R̃ in the expression of the action
for the tensor fluctuations. This scalar will be found in a minute, but since we are
so close to deriving the Friedmann equation, let us allow us to do just that before
we proceed.
CHAPTER 15. LECTURE 15
154
15.4.3 The Friedmann equations
The Friedmann equations can now easily be found from the Einstein equations
Eµν = 8πGTµν . The first equation is found by applying the zeroth component
(the energy density) of the energy-momentum tensor and using the definition of
the Einstein tensor
1
E00 ≡ R00 − g00 R = 8πGT00 = 8πGρa2
2
6
1
2
′
2
′
⇒ −3H − a − 2 (H + H ) = 8πGρa2
2
a
8πG 2
⇒ H2 =
ρa
3
(15.24)
which is the first of the Friedmann equations. The second Friedmann equation is
derived 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 the
second follows directly as
H′ = −
4πG
(ρ + 3p)a2
3
(15.25)
We will now proceed with calculating the curvature scalar, R̃ of the fluctuating
metric, g̃µν . This scalar is one of the terms in the Einstein-Hilbert action that we
saw in (15.15) and given by
Z
p
1
d4 x −g̃(R̃ + 2Λ).
(15.26)
S=−
16πG
The perturbed metric can be written as
g̃µν = a2 (τ )gµν
(15.27)
where gµν = ηµν + hµν . In the expression above the perturbed Ricci scalar,R̃ is
found by some neat formula when gµν is known. Ingunn is the expert with this
formula and it looks like
2a
R
(15.28)
R̃ = 2 − 6 3
a
a
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 −g̃. This is where
Petter Callin comes in (again). By the fact that log det M = T r log M , the term
can be found 2
p
1
1
1
−g̃ = a4 1 + h + h2 − (hαβ ) + . . .
(15.29)
2
8
4
2
For the complete calculation, see the Master thesis of Petter Callin
15.4. GRAVITATIONAL RADIATION IN A RW METRIC
155
In order to find the physical degrees of freedom and simplify the expression, we
assign the T T gauge. By this choice of gauge, the trace is zero, h = 0, and all
the zero components are also zero, hµ0 = 0 as seen earlier. The formula for the
curvature scalar in (15.28) involves the “box” -or d’Alembertian operator 2. This
operator was defined in (11.15) as
√
1
2 = √ ∂µ
−g gµν ∂ν
−g
In a flat background, the box operator acting on the scale factor is 2a = a′′ since
the scale factor only depends on the time 3 .
√
If the expressions of R̃ and g̃ are inserted in the action integral and if we do
remember the constraints in the TT-gauge (h=0, for instance), we find
Z
a′′
R
1 2
1
4
4
− 6 3 + 2Λ .
(15.30)
d xa 1 − hαβ
S=−
16πG
4
a2
a
The second term in the brackets is actually equal to the background Ricci scalar
that we found earlier in (15.23), that is
Rbackground = R = −
6
(H′ + H2 )
a2
This will be shown in the following. The derivative of the conformal Hubble parameter with respect to conformal time is
H′ =
=
′ 2
d a′
a
a′′
−
=
dτ a
a
a
′′
a
− H2
a
(15.31)
and
6
6 a′′
′
2
(H
+
H
)
=
−
a2
a2 a
′′
6a
= − 3
a
R = −
(15.32)
This is the same term that we saw in the action in (15.30). This background Ricci
scalar is the one fluctuating and can be expressed as
1
3
R = hαβ 2hαβ − hαβ,ν hνβ,α + hαβ,ν hαβ,ν
2
4
3
According to calculations done by Mr. Callin, the box-operator also contains terms depending
on hµν . Hence, the calculation involves additional terms and is more complicated. However, the net
effect of these extra terms sum up to zero, so the result is the same in both situations, and since we
are most interested in the result, we proceed with clean conscience although Petter most probably is
right.
156
CHAPTER 15. LECTURE 15
If we insert for R in the action, S, the result will be
Z
1
S=−
d4 xa2 hαβ 2hαβ + . . .
16πG
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 is
done by partial integrating, applying the two Friedmann equations and write it in
the T T gauge. The action of the tensor fluctuations then looks like
Z
2
1
1
d4 xa2 ∂µ hij
(15.33)
S=
32πG
2
We notice that the action integral of the tensor fluctuations is of the same form as
that of a massless scalar field that we have seen many times before. To illustrate
this point and investigate even further, let us take a look at a massless scalar field
in 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 EulerLagrange equations of motion is given by
Z
√
S =
d4 x −gL
Z
√ 1
=
d4 x −g gµν ∂µ φ∂ν φ
2
Z
1
1
=
d4 xa4 2 η µν ∂µ φ∂ν φ
2a
Z
1
=
d4 xa2 η µν ∂µ φ∂ν φ
(15.34)
2
which should be compared with (15.33). The similarity between the tensor fluctuations and the scalar ones can be very helpful in calculations since now a tensor
mode, hij can be translated into a scalar mode, φ and vice versa by rescaling the
prefactor of the integrals. Thus, the tensor fluctuations can be found directly from
the scalar fluctuations. This will be very useful when we proceed to inflation and
want to find the tensor fluctuations then. This kind of “duality” also applies to the
expectation values of the energy-momentum tensors in the two situations. For a
scalar field the energy-momentum tensor is (Here, the upper φ is not an index, but
denotes the Tµν of the scalar field)
1
φ
Tµν
= ∂µ φ∂ν φ − gµν (∂λ φ)2
2
The expectation value is found, remembering that the last (surface) term now vanishes due to periodic boundary conditions (As seen in equation (14.21) in lecture
14).
φ
hTµν
i = h∂µ φ∂ν φi − 0
(15.35)
15.4. GRAVITATIONAL RADIATION IN A RW METRIC
157
Now, we can apply our important result to easily find the expectation value of the
energy-momentum tensor in the situation of tensor fluctuations simply by rescaling
1
. Hence:
the prefactor by 32πG
h
hTµν
i=
1
h∂µ hij ∂ν hij i
32πG
(15.36)
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 a
general massive scalar field with respect to conformal time, τ . Earlier we have
found this equation with cosmic time in (11.13) to be
1 2
∇ φ+V′ =0
(15.37)
a2
Now, we will see that the conformal edition is equivalent. By means of the principle
of stationary action and varying the metric φ → φ + δφ, the variation principle
yields
Z
δS = 0 =
d4 xa2 (τ )(ηµν ∂µ φ∂ν δφ − m2 a2 φδφ)
Z
∂ 2 ′
2 2
2 4
4
(a φ ) + a ∇ φ − m a φ δφ
=
d x −
∂τ
Z
=
d4 xa2 − φ′′ + ∇2 φ − 2Hφ′ − m2 a2 φ δφ (15.38)
φ̈ + 3H φ̇ −
This is valid for all variations in the field, hence the brackets must be zero
φ′′ − ∇2 φ + 2Hφ′ + m2 a2 φ = 0
(15.39)
This is the same equation of motion that we found in (12.34) and is seen by swapping from conformal to cosmic time.
Next lecture will give a phenomenological introduction to inflation and early
universe physics.
4
There may be a possibility that the partial derivatives, ∂ should be replaced by covariant derivatives in (15.36) . This is probably not so since, when applying the T T gauge one chooses a specific
coordinate system. So then if one changes coordinate system (which would motivate the need of a
covariant derivative), these changes are most likely zero due to the choice of gauge. Since there still
seems to be some uncertainty about this topic, sources such as the original article by Isaacson or
“Gravitation” by Misner et. al. should be consulted.
158
CHAPTER 15. LECTURE 15
Chapter 16
Lecture 16
16.1 Cosmology
As this is a course in cosmological physics we shall go briefly through some basic
cosmology before we start on inflation. The point here is to get an overview over
the most important concepts in modern cosmology.
16.1.1 Important equations and numbers
The Robertson-Walker spacetime is given by the line element
ds2 = dt2 − a2 (t)dx2
And Einstein’s field equations are
Eµν = 8πGTµν
Where the energy momentum tensor is

ρ

0
T µν = 
0
0
0
p
0
0
0
0
p
0

0
0

0
p
and as we saw earlier the Einstein tensor can be found from the Christoffel symbols
which are
Γ0ij = aȧδij
ȧ
Γk0j = δkj
a
The Ricci tensor is thus given by
ä
a
= (aä + 2ȧ2 )δij
R00 = −3
Rij
159
CHAPTER 16. LECTURE 16
160
which gives the Ricci scalar
"
ä
+
R = −6
a
2 #
ȧ
a
This leaves us the Einstein equations. The time (00) component becomes
2
8πG
ȧ
= H2 =
ρ
a
3
(16.1)
which we recognize as the first Friedmann equation. The spatial (ij) component is
ä ȧ2
2 + 2 = −8πGp
a a
Combination of these two equations leads us to the second Friedmann equation
4πG
ä
=−
(ρ + 3p)
(16.2)
a
3
By differentiating the Friedmann’s first equation one can obtain the first law of
thermodynamics (for adiabatic expansion) as the equation ρ̇ + 3H(ρ + p) = 0.
This can also be found by covariant differentiating the energy-momentum tensor
∇µ T µν 1 .
We assume the equation of state p = wρ, and by integrating the adiabatic
equation we get that
ρ(a) = ρ0 a−3(1+w)
If we assume that the scale factor evolves as a(t) ∼ tp where 0 < p < 1 and by
combining this with the Friedmann equations and integrating we get
2
t 3(1+w)
a(t) =
t0
Here we let the scale factor today be a0 = a(t0 ) = 1. We have the Hubble
parameter
2
ȧ t−1
H0 = =
a
3(1 + w) 0
t0
If we assume matter-dominated universe (w=0) we get the age of the universe
t0 = 32 H0−1 . The ΛCDM model predicts the age t0 = 0.9H0−1 . The value of the
Hubble parameter today is given as H0 = 100h km/s
Mpc where h is measured to be
h ≈ 0.7. This gives
10
Gyr ≈ 14Gyr
h
In the ΛCDM model this will give the age t0 ≈ 13Gyr. If we insert the speed of
3
3
light c = 3 · 105 km
s we get the distance h 10 Mpc = 4200Mpc. This gives a clue
about the size of the observable universe.
H0−1 = 1010 h−1 yr =
1
Calculating this would be a good exercise.
16.1. COSMOLOGY
161
t0
δLp ≈ 100Mpc
δθ
Dp (t0 )
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 gives
dt = a(t)dr. Integration gives us
Z
ro
dr =
re
Z
to
te
dt
a(t)
The physical distance to the galaxy is then
Dp (to ) =
Z
to
te
dt
a(t)
That makes the physical distance to a source that emitted a photon at time te = 0
Dh (t0 ) =
Z
to
0
dt
a(t)
which is the distance to the horizon. It will be of order of magnitude Dh (t0 ) ∼ H10 ,
where H10 = DH (t) is called the Hubble radius. If we assume a matter dominated
2
3
universe where a(t) = tt0
we get the size of the observable universe
Dh (t0 ) = 2H0−1 = 104 Mpc
To get an idea of cosmological distances we can remember that the size of
a galaxy is approximately 0.1Mpc = 100kpc ≈ 300000ly, the size of a galaxy
cluster is about 30 times as large 0.1Mpc × 30 = 3Mpc. Superclusters are again
about 30 times as large as a cluster, 3Mpc × 30 = 90Mpc ∼ 100Mpc.
CHAPTER 16. LECTURE 16
162
We must remember that e.g. superclusters evolve as the light is traveling towards us. If the “angular size” of a supercluster (or other structure) is δθ we have
that
δLp
(1 + ze )δL0
δL0
δθ =
=
=
Dp (tp )
Dp (t0 )
DA (t0 )
Where ze is the redshift and δL0 is the size of the galaxy at the time of emission.
1
Dp (t0 ).
We have here introduced the angular distance DA (t0 ) = 1+z
e
16.1.3 The early universe
The early universe was dominated by radiation and the energy density was given
by
π2 4
ρ = ρr =
T g
30
where g is the degrees of freedom of the particles, and is derived from the particle
1
physics. The universe expanded as a(t) ∼ t 2 . This gives the Hubble parameter
1
. And from the first Friedmann equation we get
H = aȧ = 2t
H2 =
1
4t2
=
t2 =
8πG
ρ
3
8πG
ρ
3
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
16π 3 Gg
If we insert all the constants we get the practical formula
2.42
tT 2 = √
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 the
degrees of freedom for the photon (2), three types of massless neutrinos and their
antiparticles (3 · 2) and the electron and positron with spin up and down (2 · 2), that
is
43
7
gN S = 2 + (3 · 2 + 2 · 2) =
8
4
Here the factor 78 comes from statistical mechanics as neutrinos and electrons (and
positrons) follows Fermi-Dirac distribution.
16.2. INFLATION
163
10−5 s
1s
1010 yr
104 yr
t
tGU T
tEW
tQCD
tN S
trm
tLS
t0
T
1015 GeV 103 GeV
200MeV 1MeV
Figure 16.2: timeline
Even earlier, at T > 1TeV all the particles (almost) were massless, and g was
much higher. We had in addition gluons with different spin (8 · 2), the W ± and
Z 0 bosons with three degrees of freedom each (they are still massive), the Higgs
boson (1), the six quarks and their antiparticles with color and spin (6 · 3 · 2 · 2) and
we also get the µ and τ particles. That gives us g
7
427
g = 2 + 8 · 2 + 3 · 3 + 1 + (6 · 3 · 2 · 2 + 3 · 2 · 2 + 3 · 2) =
≈ 100
8
4
If one extrapolates the coupling constants of the elementary forces one expect
that they will meet at a point around 1015 GeV (figure 16.3) and the physics will
be described by a grand unifying theory (GUT). At even higher energies (Planck
energy, 1018 GeV) this too will break down, and we will need an universal theory
of 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 is
the universe so extremely flat?) and the horizon problem (Why is the universe so
isotropic when it is causally isolated by horizons?).
The horizon problem can easily be seen if one looks at our local part of the universe (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 it
shrinks slower than the horizon (H −1 ∼ t) (figure 16.4). This means that at an
earlier time some of our local universe was outside the horizon. Yet when we see
it 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 universe where the universe expanded exponentially, a(t) ∼ eHI t . If we then plot the
scale of our local universe in comoving coordinates (Lp = a1 Lp so that the size
CHAPTER 16. LECTURE 16
164
100
30
α1−1
TOE
GUT
α2−1
α3−1
10
E(GeV)
10
2
10
15
1018
Figure 16.3: At 1015 GeV the coupling constants are of equal size and we get GUTphysics
H −1
104
102
100
∼ t2/3
10−2
∼ t1/2
∼t
t
trm
Figure 16.4: It looks like all our observable universe cannot have been in thermal
contact until today.
16.2. INFLATION
165
(aH)−1
Mpc
104
102
∼ e−HI t
100
∼ t1/3
10−2
∼ t1/2
ti
tf
t
trm
Figure 16.5: With inflation the local universe can have been in thermal contact
before.
of the comoving horizon is (aH)−1 ) together with our current (constant) horizon
we see that they cross at an earlier time (figure 16.5). This makes it possible for
different parts of the universe to have been in contact before.
To have inflation we must have
1
d
dt aH < 0
⇒
⇒
d 1
dt ȧ
= − ȧä2 < 0
ä > 0
And if it is supposed to solve the horizon problem we must demand that the local
universe is within the horizon before the inflation starts. That is
1
> H0−1
aH ti
and this gives us how much the universe must expand during inflation
eHI (tf −ti ) ≥ e55
166
CHAPTER 16. LECTURE 16
Chapter 17
Lecture 17
Last lecture we saw why we need inflation. In this lecture we will see how we
can produce an inflationary epoch in the very early universe using a classical scalar
field and the so-called slow-roll approximation (SRA). The model we will study is
called “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 properties
of the universe that we observe today.
At the end of the lecture we will study quantum perturbations on the top of this
scalar 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 accelerated expansion, that is ä > 0. Assuming an equation of state relating pressure and
energy, p = wρ we have from the second Friedmann equation (16.2) that inflation
requires w < − 31 . Standard models of inflation has w = −1.
The scalar field φ that is responsible for inflation in our model is called an
inflaton and has the well-known scalar field Lagrangian
L=
1 µν
g ∂µ φ∂ν φ − V (φ)
2
(17.1)
Early on many thought that this scalar field was the Higgs field, but that idea has
been abandoned. The Higgs field has too many constraints from the standard model
of particle physics that doesn’t fit into the requirements of an inflaton field. We
don’t know what the inflaton field is and where it comes from, but it could explain a
lot in a simple way, so we assume that it is there and try to deduce some observables
that might confirm its existence.
In lecture 11 we showed that the pressure (11.28) and energy density (11.29)
167
CHAPTER 17. LECTURE 17
168
in a scalar field is given by
ρ =
p =
1 2
φ̇ + V (φ)
2
1 2
φ̇ − V (φ)
2
where V (φ) is a potential. Here we have omitted terms on the form 6a12 (∇φ)2
that will vanish rapidly as a grows. For inflation we must have w < − 31 which
corresponds to
1 1 2
1 2
φ̇ − V < −
φ̇ + V
2
3 2
2 2 2
φ̇ − V < 0
3
3
V > φ̇2
For a scalar field the first Friedmann equation (16.1) will take the form
r
2
1
ȧ
1
1 2
=
≈ 1018 GeV
φ̇ + V
,
MP =
2
a
8πG
3MP 2
(17.2)
In a RW metric
ds2 = dt2 − a2 dx2
we have the equation of motion for a scalar field given in lecture 11 (11.13):
φ̈ + 3H φ̇ + V ′ = 0
(17.3)
The first Friedman equation and (17.3) are coupled and with a given potential they
may be integrated numerically 1 . Such an integration will include a choice of initial
conditions for φ and φ̇. Hopefully the result will look something like figure 17.1
where one has an “attraction pool” around the initial state such that the evolution
on the field is not critical dependent on the initial conditions. Hopefully, in this
attraction pool we will have that V ≫ φ̇2 , 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 to
do is the slow-roll approximation (SRA). In this approximation φ is changing very
slowly, such that V ≫ φ̇2 and φ̈ ≈ 0. Inserted in (17.3) this gives the equation of
motion
3H φ̇ + V ′ = 0
(17.4)
1
Gorm will do this in his master thesis. Ask him for results next year.
17.1. INFLATION DRIVEN BY A CLASSICAL SCALAR FIELD
169
111111111
000000000
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
φ̇
attraction pool
φ̇ ≈ const.
φ
Figure 17.1: Hopefully a numerical solution of the equations will show a behavior
like this. Here the system is not depending critically on the initial conditions, and
for a long time φ̇ is almost constant.
An arbitrary chosen potential will of course not satisfy the above conditions. The
potential must be sufficiently large and have a small gradient. From (17.4) we have
that φ̇ must satisfy
V′
φ̇ = −
3H
Under the assumption φ̇2 < V the first Friedmann equation (17.2) gives
H2 =
and we have
V′
3H
2
<V
V
3MP2
1
3
⇒
(17.5)
V′
V
2
< MP−2
So a constraint on V is that V ′ cannot be too big to have inflation in the SRA. Then
it is natural to define a slow-roll parameter
1
ε ≡ MP2
2
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 will
give an exponential evolution of the scale factor, a ∼ eHt . This is not exactly
correct since H is not a constant, but it works well in a low order approximation.
CHAPTER 17. LECTURE 17
170
17.1.2 Example with SRA using a quadratic potential
For some choices of V the SRA equations may be solved exactly. The most common example, which also is frequently used by Linde, is
V
V′
1 2 2
m φ
2
= m2 φ
=
ε =
1 2 m2 φ
M
=2
2 P 12 m2 φ2
MP
φ
2
!
≤1
√
The last condition tells us that φ ≥ 2MP . This looks nasty. To be able to use
SRA we need a field larger than the Planck mass, and in this range it seems a bit
suspicious to use a classical field. What saves us is that even if φ > MP , the
energy density does not have to be that large. The energy density in SRA is given
by
m 2
1 2 2
2
2
4
< MP4
V = m φ ≥ m MP = MP
2
MP
if m ≪ MP . So if the mass-term of the field is not too big, we avoid the Planck
problem, and we may safely use classical gravitation.
With this potential (17.5) yields
1
1
V =
m2 φ2
2
2
3M
6MP
r P
1 m
φ
⇒H =
6 MP
H2 =
Inserting this into (17.4) gives
r
3
(17.7)
1 m
φφ̇ + m2 φ = 0
6 Mp
⇒ φ̇ = −
r
2
mMP
3
(17.8)
That is, the velocity of the field is constant (not that strange actually, since we
already have assumed that φ̈ ≈ 0). Integration gives
r
2
MP mt
(17.9)
φ(t) = φi = −
3
Now, let us study how the scale factor evolves. We have
d
ln a =
dφ
d
dt ln
dφ
dt
a
=
ȧ
aφ̇
=
H
φ̇
17.2. FLUCTUATIONS OF φ
171
which is a very useful equation. Inserting for H and φ̇ we have
q
1 m
φ
d
6 MP φ
ln a = q
=−
dφ
2MP2
− 23 mMP
af
ln
ai
=
Z
φf
dφ
φi
φ
2MP2
1
φ2f − φ2i
2
4MP
1
1
φ2i
−
φ2i − 2MP2 =
2
2
4MP
4MP2
= −
=
(17.10)
Since we know that the scale factor will grow exponentially, we may define a N by
af
≡ eN
ai
where N is called the number of e-foldings given by the inflation. We know that
we should have N > 55 for the inflation to solve the problems that we constructed
it to solve. Then we must have
ln
af
φ2i
1
=
− = N ≥ 55
ai
4MP
2
√
φi = 2 N MP ∼ 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 in
more detail than we will do here.
We assume a fluctuating inflaton on the form
φ(x, t) =
φ (t)
| 0{z }
classical evolution
+ δφ(x, t)
| {z }
(17.12)
fluctuation
We know from lecture 11 (11.13) that a scalar field in RW-metric will obey
φ̈ + 3H φ̇ −
1 2
∇ φ+V′ =0
a2
Inserting (17.12) gives
φ̈0 +
∂
1
∂2
δφ + 3H(φ̇0 + δφ) − 2 (∇δφ) + V ′ (φ0 + δφ) = 0
2
∂t
∂t
a
(17.13)
CHAPTER 17. LECTURE 17
172
when remembering that φ0 only depends on time. For a scalar field we have the
well-known Lagrangian
1
L = (∂µ φ)2 − V (φ)
2
The potential V has a minimum for φ = φ0 = 0, so we may expand V like
1
V (φ) = V (0) + V ′′ φ2 + . . .
2
(17.14)
and we have that
1
1
L = (∂µ φ)2 − V ′′ φ2 + . . .
2
2
We see that the second term in the Lagrangian has the form of a mass term with
m2φ = V ′′ , so the second derivative of the potential may be interpreted as an effective mass term. So for the last term in (17.13) we have
V ′ (φ0 + δφ) = V ′ (φ0 ) + V ′′ (φ0 ) + δφ
= V ′ (φ0 ) + m2φ δφ
Subtracting the classical φ0 -terms from (17.13) we are left with an equation for the
fluctuations given by
2
∂
1 2
∂
2
− ∇ + 3H + mφ δφ = 0
∂t2 a2
∂t
And as always when working with a field, we assume a finite volume and expand
it in Fourier modes:
r
1 X
ϕk (t)eik·x
δφ(x, t) =
V
k
where k is a comoving wave number. Since the physical distance is given by ax,
the physical momentum is p = a1 k, Performing the same steps as when deriving
(11.13) in lecture 11, we obtain
ϕ̈k + 3H ϕ̇k +
k2
2
+ mφ ϕk = 0
a2
(17.15)
This is basically the same equation of motion as obtained for a scalar field in
RW-background earlier, so the fluctuations will follow the same equation as the
background scalar field. The only difference is the mφ -term which really is not a
constant. But to the lowest order we set this mass-term to be mφ = 0. To motivate
′ 2
this, we can study the ε-parameter ε = 21 MP2 VV . Assuming ε ≈ constant we
have
2
V ′2 = ε 2 V 2
MP
17.2. FLUCTUATIONS OF φ
173
Differentiation gives
2V ′ V ′′ = ε
4
VV′
2
MP
ε ≈ MP2
⇒
V ′′
V
This motivates the introduction of a new parameter
η = MP2
V ′′
V
so V ′′ = η MV 2 , and with a quadratic potential we had H 2 = 13 MP2 V . This gives us
P
3MP2 H 2
= 3ηH 2
MP2
V ′′ =
and we have that m2φ = 3ηH 2 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
ϕ̈k + 3H ϕ̇k +
k2
ϕk = 0
a2
(17.16)
Now we introduce conformal time τ given by
dτ =
dt
a(t)
• If we now have power-law behaviour of the scale factor a(t) ∼ tp , then τ
will evolve like
Z
t1−p
τ ∼ d tt−p =
1−p
1
t ∼ τ 1−p
,
p
a(τ ) ∼ τ 1−p
• In the inflationary epoch we have a(t) ∼ eHt such that
Z
e−Ht
+ τC
τ ∼ dt e−Ht =
H
1
1
⇒
a(τ ) = −
H(τ − τC )
Hτ
when we choose τC = 0. Applying the condition ä > 0 for inflation, we see
that we will have inflation for τ < 0 and that inflation ends when τ → 0.
a(τ ) = −
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 + k2 ϕk = 0
′
(17.17)
where H = aa and a prime as usual denotes differentiation with respect to conformal time τ . Then, again as in lecture 11, we rewrite the field as ϕk (τ ) = a1 uk (τ )
to get rid of the friction term. Equation (17.17) can then be written as
CHAPTER 17. LECTURE 17
174
a′′
uk = 0 (17.18)
u′′k + k2 −
a
We have
a′ =
1
Hτ 2
a′′ =
and
which gives
−2
Hτ 3
2
a′′
= 2
a
τ
and thus we have that
u′′k
2
+ k − 2
τ
2
uk = 0
(17.19)
This equation may be solved using Bessel functions j1 (kτ ) and y1 (kτ ). The solution then becomes
i
i
−ikτ
ikτ
uk (τ ) = Ak e
1−
+ Bk e
(17.20)
1+
kτ
kτ
where Ak and Bk are Bogoliubov coefficients which we do not know, but that
have to be determined from some principle. (17.20) is the general solution for
non-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
Z
i
d3 k h
1
† ∗
−ik·x
ik·x
(τ
)e
â
u
(τ
)e
+
â
u
δφ̂(x, τ ) =
k k
k k
a
(2h̄)3
Normalizing the modes we have the Wronskian
′
uk∗ uk − u∗k u′k = i
then one see at once that the a-operator has the usual canonical commutator:
h
i
âk , â†k′ = (2h̄)3 δ(k − k′ )
with a vacuum defined by âk |0i = 0. The final solution will also in this case
depend on the determination of the Bogoliubov coefficients Ak and Bk .
When working with fluctuations it can be very useful to have a figure like figure
17.2 in mind. When we use comoving coordinates along the y-axis a fluctuation
mode may be represented by a horizontal line, an it is easy to see how a specific
17.2. FLUCTUATIONS OF φ
175
(aH)−1
mode outside horizon
tf
VD
t0
teq
RD
t
MD
Figure 17.2: A nice figure to have in mind. In comoving coordinates the modes are
simply horizontal lines. Here, one of the modes re-enters the horizon in the matter
dominated epoch (MD), while the other re-enters in the radiation dominated (RD)
epoch. Notice how the vacuum dominated (VD) epoch ensures that all the modes
are in causal contact early on.
CHAPTER 17. LECTURE 17
176
mode leaves and enters the horizon. The comoving scale is λ =
ation mode will be inside the horizon provided
k−1 = λ < (aH)−1
⇒
2π
k
and a fluctu-
k > aH
1
, so in this epoch the mode is inside the horizon when
During inflation a = − Hτ
k>−
1
H
Hτ
⇒
|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 | = k1 . Very early in the inflation
epoch we have |kτ | ≫ 1, and (17.19) yields
u′′k + k2 uk = 0
that is, we have effectively
q a free field. We choose an in- mode very early, e.g. we
set Bk = 0 and Ak =
1
2k ,
and we have the solution
uk =
The
q
1
2k -factor
r
1 −ikτ
e
2k
comes from the harmonic oscillator where we have
1
1
H = p2 + ω 2 q 2
2
2
,
q̂ =
r
1
(â + ↠)
2ω
This choice of solution is called a Bunch-Davies (BD) vacuum, and is the most
common solution to work with in inflationary cosmology.
Some people, e.g. Ulf Danielson, have commented that when enforcing such
a boundary condition very early you will get a physical momentum p = ka which
is 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 depends
on k, such that the Bogoliubov coefficients will vary with k. That gives some
small ripples on the CMB power spectrum. These ripples are too small to observe
with the current WMAP-data, but they might be large enough to be detectable with
future CMB-surveys like the Planck satellite.
Anyway, we continue using the BD-vacuum, and we study one mode:
δφk = ϕk (t)eik·x =
1
uk (τ )eik·x
a
Inside the horizon we only have the Ak -term, which gives
q
r
1
u 1
2k
k
|δφk | = = 1 =
(Hτ )
a
2k
Hτ
17.2. FLUCTUATIONS OF φ
177
log|δφk |
k1
k2 > k1
logτ
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
q
1 1
u H
2k kτ
k
|δφk | = =
=√
1
a
2k3
Hτ
(17.21)
So outside the horizon the fluctuations will be constant in time, while the modes
will decay when inside the horizon (see figure 17.3). Notice from the figure how
the quantum fluctuations get “frozen” and behave like classical fluctuations when
moving past the horizon.
178
CHAPTER 17. LECTURE 17
Chapter 18
Lecture 18
18.1 Primordial spectrum
The primordial spectrum is due to quantum fluctuations and are given by the twopoint correlation function. The potential, V we earlier in the last lecture Taylorexpanded in (17.14). This combined with the fluctuation in the scalar field as seen
in (17.12) yields
1
1
(18.1)
V (φ0 + δφ) = V (φ0 ) + V ′′ (φ0 )δφ2 + V ′′′ (φ0 )δφ3 + . . .
2
6
|
{z
} |
{z
}
Gaussian
non−Gaussian
The fluctuations in the equation above will in overall be Gaussian. From quantum
field theory we know that the higher order terms, that is of order 3 or higher give
rise to interactions. (One may think of these interactions as some vertex in a general
Feynman diagram). The higher order terms will spoil the Gaussian shape of the
fluctuations. Therefore it will often be important to know the size of these extra
terms relatively to the Gaussian ones. In our case, however, we will neglect the
corrections and assume Gaussian fluctuations. We want to derive an expression
for the power spectrum of the fluctuations in the field, φ. The power spectrum is
sometimes also referred to as a correlation function. To do so we will need the
b
vacuum expectation value of the field operator φ(x).
18.2 Vacuum fluctuations of φ̂
We will now consider a free field in Minkowski space. To find the vacuum expectb we need the operator itself (of a massless field).
ation value of φ,
r
X
1
†
−ik·x
ik·x
∗
b
b
ak ϕk (t)e
+b
ak ϕk (t)e
.
(18.2)
φ(x)
=
V
k
q
√
1 ikt
Here, ϕk (t) = 2k
e and the frequency ωk = k2 + m2 = k since the field is
massless. The quantum field, φ̂ fluctuates in a vacuum described by the Minkowski
179
CHAPTER 18. LECTURE 18
180
metric so then φ0 = 0. The vacuum fluctuation of the scalar field is given by
hφ2 (x)i = h0|φ̂(x)φ̂(x)|0i
We calculate this by applying the definition of the vacuum state, âk |0i = 0 and the
conjugated vacuum h0|â†k = 0. Thus inserting for the field operator in (18.2), the
expectation value then takes form as
hφ2 (x)i =
1 X
′
h0|âk′ â†k |0iϕk′ ϕ∗k ei(k −k)x
V
′
(18.3)
k,k
If we apply the canonical commutator relation that âk′ , â†k = δk,k′ , the above
expression can be written into a single sum
1 X 2
ϕk
V
k
Z
2
d3 k .
=
ϕ
(t)
k
(2π)3
hφ2 (x)i =
(18.4)
The last step follows under the assumption that the integration volume is large
so that the sumPcan beRvery well approximated to an integral by the earlier seen
d3 k
substitution V1 k → (2πh̄)
3 when V → ∞. The power spectrum is now found
by writing the integral in spherical coordinates (d3 k = 4πk2 dk) and then just
rewriting the expression slightly into the desired form. That is
Z
2
4πk2 dk 2
ϕ
(t)
hφ (x)i =
k
(2π)2
Z
dk k3
=
(18.5)
|ϕk |2
k |2π 2{z }
powerspectrum
We now have the power spectrum or correlation function, ∆2φ (k) defined as
∆2φ (k) =
k3
|ϕk |2
2π 2
In some literature, only the last term, |ϕk |2 = Pφ (k) is referred to as the power
spectrum. We will use only one of them, namely
∆2φ (k) =
k3
Pφ
2π 2
(18.6)
The term hφ2 (x)i is not only a step in finding the power spectrum, but has also
an interpretation itself. If we plot the power spectrum, ∆2φ (k) as a function of the
18.3. INFLATION
181
∆2φ (k)
ln k
Figure 18.1: The figure shows some power spectrum as a function of the wavenumber, k. The total area beneath the curve of the power spectrum function is the
vacuum expectation value hφ2 (x)i.
wavenumber, k, then the vacuum expectation value hφ2 i is the total area beneath
the curve. This is so since from (18.5) we have
Z
Z
Z
dk 2
dk
= d ln k →
∆ φ = hφ2 (x)i = Tot. area
k
k
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
k3 1
P
=
φ
2π 2
2π 2 2k
2
k
4π 2
(18.7)
To further find the vacuum expectation value of the scalar field, one has to integrate
over 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 Cassimir situation, section (3.3).
18.3 Inflation
In the last lecture, we wrote the Fourier functions ϕk (τ ) in terms of new modefunctions in order to get rid of the friction term in the equation of motion,
ϕk (τ ) = a1 uk (τ ). This in turn lead to the conclusion that the fluctuations will
CHAPTER 18. LECTURE 18
182
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))
H
|ϕk | = √
2k3
We are now considering inflation where the dynamics is described by a scalar field
φ. We want to calculate the power spectrum in the inflationary epoch by applying
the expression of the (scalar) fluctuations outside the horizon. In this manner one
should 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 that
around tf the scalar field dumps its kinetic energy driving the inflation process into
conventional radiation and matter. This is referred to as the reheating process).
The power spectrum now is
k3 H 2
k3
2
|ϕ
|
=
k
2π 2
2π 2 2k3
2
H
=
2π
∆2φ (k) =
(18.8)
Thus, the power spectrum (outside the horizon) is independent of wave number
when H is constant. This is called Harrison-Zeldovich (HZ) scaling and is actually
only correct in the limit of infinitely slow rolling of the inflaton, φ. The reason
that the HZ scaling is not quite exact, is due to the fact that modes with different
wavenumbers, k will leave the (comoving) horizon at different times. So the scalar
potential, V will be of different values when modes are crossing the horizon. Since
V ∼ H 2 in the SRA, different values of V will give a power spectrum ∆2φ (k)
depending on the wavenumber. This is sketched in figure (18.2). However this
effect is very small in the slow-roll approximation, since then the scalar potential
is assumed to be slowly changing. So to lowest order the HZ scaling is a very good
approximation.
18.4 Metric fluctuations
We will now take a brief look at fluctuations in the metric and motivate why metric
fluctuations are of interest in various subjects in cosmology.
18.4.1 Motivation
In inflation the energy-momentum tensor is dependent of the scalar field driving
the 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
Com.coordinates
k2
k1
t
Figure 18.2: Modes with different wavenumbers k1 and k2 leave the horizon at
different times, resulting in a k-dependent power spectrum. To lowest order, however, the power spectrum is independent of wavenumber and the HZ scaling can be
assumed.
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 the
geometry will fluctuate. From
Eµν → Eµν + δEµν
one therefore finds an expression of the fluctuation of the Einstein tensor similar to
the 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 in
the parenthesis, is that they will be the variants of the metric fluctuations of most
interest 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 metric
can be expressed as a sum of scalar, vector and tensor fluctuations, depending on
how each of them behave under coordinate transformations. This is known as The
decomposition theorem and yields
tensor
scalar
vector
δgµν = δgµν
+ δgµν
+ δgµν
CHAPTER 18. LECTURE 18
184
The three components of the fluctuation can be expanded in terms of spherical
coordinates. They will be orthogonal to each other and can be solved separately
since they evolve independently. That is, if some physical process in the early
universe 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 to
show that the vector modes usually do not contribute in inflation. This is so since
there is no way to source such modes with a single scalar field. Hence, in simple
models of inflation, the vector modes vanish. We are then left with only two of the
components and the easiest of them to calculate are the tensor modes. We will find
that the tensor fluctuations are on the same form as the power spectrum that we
saw 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 and
scalar perturbations were proportional
Z
Z
2
1
1
4
21
S=
∂µ hij ∼ d4 xa2 (∂λ φ)2
d xa
32πG
2
2
Knowing one of them, we can therefore easily find the other. We will also in
tensor do not change during the
the following see that the tensor fluctuations, δgµν
expansion of the universe. They can in principle be observed in the polarization of
the 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 background
metric. The perturbation to the metric will be parameterized and this new metric
we will split into the three basic components mentioned earlier. We will then investigate some of these terms in more detail, also applying gauge transformations
to them. The RW background metric in terms of conformal coordinates is given by
ds2 = gµν dxµ dxν
= a2 (τ )ηµν dxµ dxν
= a2 (dτ 2 − dx2 )
= a2 (dτ 2 − δij dxi dxj )
A perturbation in the metric can therefore be written
gµν → gµν + δgµν = a2 (ηµν + hµν )
(18.10)
18.4. METRIC FLUCTUATIONS
185
where hµν is the common, small perturbation. In the following, we will only calculate to the first order in the perturbations and omit all higher order corrections.
That is
gµν = a−2 (η µν − hµν + . . .)
In order to separate out the vector, tensor and scalar perturbations from the fluctuation in the metric, we will rewrite it by parameterizing the Minkowski metric, ηµν
and the perturbation, hµν as follows


1 0
0
0
0 −1 0
0

ηµν = 
0 0 −1 0 
0 0
0 −1
2A Bi
hµν =
Bi −Cij
(18.11)
The hµν matrix is also a 4 × 4 matrix with 10 degrees of freedom (DOF). All the
terms in this matrix are small and their respective DOFs are: A : 1, Bi : 3 and
Cij : 6 with ij running from 1 to 3. To find the three different components of the
metric fluctuation, we rewrite the line element in terms of the parameterizations as
ds2 = a2 [(1 + 2A)dτ 2 + 2Bi dxi dτ − (δij + Cij )dxi dxj ]
A is a scalar since it has one DOF, but the Bi and Cij terms can be separated
further.We write out the Bi and Cij terms to find the tensor, vector and scalar
constituents
Bi = −∂i B + Vi .
(18.12)
Here, the B term is a scalar potential and V is a pure vector. The expression above
is actually the same as when a three-vector (Bi ) is decomposed into a divergence
of a scalar field (∂i B) and a curl (Vi ). From calculus, we know that a vector field
(in the three-dimensional case) always can be separated into two constituents: One
with zero divergence and the other with zero curl. The latter can therefore be
expressed as the gradient of some scalar field (B). The Cij term is given by
Cij = −2Dδij + 2∂i ∂j E + ∂i Ej + ∂j Ei + hij
{z
} |{z}
|{z} | {z } | {z } |
6
1
1
2
2
Here, the numbers beneath each term gives the DOF, and we see that it sums up to
6. Also Cij is a symmetric tensor of rank 3, the D and E terms are scalars and the
Ei -field in ∂i Ej + ∂j Ei is a symmetric, divergence free vector field. (Actually, one
would expect 3 DOFs for the term ∂i Ej + ∂j Ei , but due to the divergence relation
∂i Ei = 0, it reduces to only 2 DOFs). The last term, hij is the (spatially) tensor
fluctuations constrained with the T T gauge hii = 0 and ∂i hij = 0. It is now time
to characterize the three constituents of the metric perturbations:
CHAPTER 18. LECTURE 18
186
• The contributions to the vector perturbations come from the terms Vi and
∂i Ej + ∂j Ei , but as we have mentioned earlier, these will decay rapidly as
the universe expands exponentially during inflation.
• The tensor perturbations are
ds2 = a2 [dτ 2 − (δij + hij )dxi dxj ]
• The scalar perturbations are
ds2 = a2 {(1 + 2A)dτ 2 − 2∂i Bdxi dτ − [(1 − 2D)δij + 2E,ij ]dxi dxj }
(18.13)
18.4.3 Scalar perturbations
We will now consider the scalar perturbations in more detail. These perturbations
are very interesting since they are responsible for the structure of the universe, that
is, 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 and
E. They all depend on space and time. In terms of the metric tensor, the scalar
perturbations can be written as
g00 = a2 (1 + 2A)
g0i = −a2 ∂i B
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 this
lecture see that two of the scalar functions can be omitted due to a suitable choice
of gauge. This gauge is called Newtonian gauge. We will now investigate how the
metric changes under a coordinate transformation
xµ → x̄µ = xµ + ξ µ (x)
We write ξ µ = (ξ 0 , ξ i ). The components ξ 0 and ξ i are small perturbations of the
same order as the variables characterizing the perturbations. The component ξ i is
a three-vector. By the same principle as seen in (18.12), it can be separated into a
scalar part (a gradient) and a (transverse) vector part
i
ξ i = ∂i ξ + ξtr
This splitting will be very useful when we a little later are going to calculate how
the four scalar fields change under a gauge transformation. The metric now transforms as
δgµν → δgµν − ∇µ ξν − ∇ν ξµ
(18.15)
18.4. METRIC FLUCTUATIONS
187
where the ∇s are covariant derivatives. To find the perturbations in the metric, we
will need the Christoffel symbols and the contravariant ξµ . It is given by
ξµ = gµν ξ ν = a2 (ξ 0 , ξ i )
when inserting for gµν . The covariant derivatives are found by
∇µ ξν = ∂µ ξν − Γλµν ξλ
With the background metric defined as ds2 = a2 (dτ 2 − dx2 ) and the Christoffel symbols given by Γλµν = 21 gαλ [gαµ,ν + gαν,µ − gµν,α ] the components of the
Christoffel symbols are
Γ000 = H
Γ0ij = Hδij
Γi0j = Hδji
Let us examine how the scalar part of the metric perturbation changes under such a
coordinate transformation. We will consider the 00-perturbation, δg00 . The others
are left as an excercise for the reader. From the equation in (18.13) we find that the
temporal components of the perturbation of the metric is
δg00 = 2Aa2
By means of the coordinate transformation, this component now is
δg00 → δḡ00 = 2Āa2
=
=
2Aa2 − 2(∇0 ξ0 )
∂ 2 0
2 0
2
(a ξ ) − Ha ξ
2Aa − 2
∂τ
(18.16)
where in the last step, the index of ξ is raised resulting in an a2 term. We further
get
a′ 0
′
ξ − ξ 0 + Hξ 0
a
′
= A − Hξ 0 − ξ 0
Ā = A − 2
(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 the
metric transform and then finally receive all the four scalar perturbations
Ā = A − Hξ 0 − ξ 0
′
B̄ = B + ξ 0 − ξ ′
D̄ = D + Hξ 0
Ē = E − ξ
(18.18)
CHAPTER 18. LECTURE 18
188
The equations above do not say how the four fields themselves change under the
(general) coordinate transformation. This is so since the individual functions, A
for example, transform as a scalar and do not change under a spatial coordinate
transformation. We have transformed the metric and accommodated the resulting
changes in new definitions of A, B, D and E. This is not the same as seeing how
A by itself changes under a transformation.
Although there are four functions characterizing the scalar fluctuations, these
can be manipulated so that two of them vanish. In the coordinate (gauge) transformation, 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 it
is trivial to make a transformation to eliminate E: simply choose ξ = −E resulting
in Ē = 0. Hence, there are not four, but 4 − 2 = 2 functions which give physical
meaning. Motivated by this fact one would expect there are two variables that are
left unchanged under a general coordinate transformation. Actually, there are, and
J. 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 will
show explicitly that one of the potentials is invariant and leave the other as an
exercise. 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 → Ψ̄ = D̄ − H(B̄ − Ē ′ )
=
=
D + Hξ 0 − H[(B + ξ 0 + ξ ′ ) − E ′ − ξ ′ ]
D − H(B − E ′ ) = Ψ
(18.20)
These invariants are very useful. If equations simplify in a particular gauge, then
one can do calculations in that gauge, form the gauge-invariant variables, and then
finally turn these into the perturbations in any other gauge. We will probably do
that in some later lecture. We will now end this lecture by looking at different
choices of gauge.
18.4.5 Choice of gauge
• Newtonian gauge. We already saw one example of the Newtonian gauge
earlier 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
18.4. METRIC FLUCTUATIONS
189
now takes form as
ds2 = a2 {(1 + 2A)dτ 2 − (1 − 2D)δij dxi dxj }
= a2 {(1 + 2Φ)dτ 2 − (1 − 2Ψ)δij dxi dxj }
(18.21)
The reason why this gauge is called Newtonian is its similarity to the Newtonian limit in General Relativity: Consider a Schwarzschild line element in
the weak field limit given by
ds2 = (1 + 2φ)dt2 − (1 − 2φ)dx2
with φ = − GM
r . If now φ = Ψ = Φ, the line element of the scalar fluctuations in this gauge is the Schwarzschild line element at large distances. In
the limit of large distance to the mass M we get Newtonian physics. This
gauge is sometimes also referred to as longitudinal gauge.
• Synchronous gauge. This gauge was first brought up by Lifshitz and
some of his students. It is often preferred when doing calculations with
computers since the equations are better behaved numerically in this gauge
where now B = A = 0. The line element then is
ds2 = a2 {dτ 2 − [(1 − 2D)δij + E,ij ]dxi dxj }
• Comoving gauge. This gauge is defined so that the perturbation of the
matter (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 relative
to the matter.
In the coming lectures we will make the approximation Φ = Ψ in the Newtonian
gauge. We then have established the metric in this gauge. What is left before we
can investigate the fluctuations in this metric, is to calculate the fluctuations of the
Einstein tensor which we will need in equation (18.9) given by
δEµν = 8πGδTµν
(18.22)
190
CHAPTER 18. LECTURE 18
Chapter 19
Lecture 19
19.1 Tensor fluctuations
Last lecture we considered scalar fluctuations of the metric. For tensor fluctuations
we had the metric
ds2 = a2 dτ 2 − (δij + hij ) dx2
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 = D 1 , and chosen
Newtonian (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 = −a2 hij ⇒ δgij = +
1 ij
h
a2
From this we can find the fluctuations of the Christoffel symbols, δΓλµν and from
that we can find the fluctuations of the Ricci tensor δRµν and then we can find the
fluctuations of the Einstein tensor δEµν . The Christoffel symbols are not tensors,
but they can be expressed with only low components, Γµαβ = 12 gµν Γναβ and this is
given as
Γναβ = ∂α gβν + ∂β gαν − ∂ν gαβ
this gives us the fluctuations
2δΓµαβ = δgµν Γναβ + gµν δΓναβ
1
Be careful, some people use different definitions of Φ and Ψ.
191
CHAPTER 19. LECTURE 19
192
where
δΓναβ = ∂α δgβν + ∂β δgαν − ∂ν δgαβ
If we insert for the metric we get
δΓ0ij
=
δΓi0j
=
δΓkij
=
1 ′
hij + 2Hhij
2
1 i′
h
2 j
1
∂j hki + ∂i hkj − ∂k hij
2
The Ricci tensor is given from the Christoffel symbols as
Rµν = ∂α Γαµν − ∂ν Γβµβ + ΓΓ − ΓΓ
so we get the fluctuations of the Ricci scalar
δRµν = ∂α δΓαµν − ∂ν δΓβµβ + δΓΓ + ΓδΓ − δΓΓ − ΓδΓ
(We haven’t bothered writing out all the contractions in the last terms) In our case
the fluctuations of the Ricci tensor becomes
δRji = −
1
h′′ij + 2Hh′ij − ∇2 hij
2
2a
And then the fluctuations of the Ricci scalar, R = gµν Rµν
δR = δgµν Rµν + gµν δRµν =
1
ij
ij
h
R
−
δ
δR
ij
ij
a2
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 that
the Ricci scalar don’t fluctuate, δR = 0.
As the Einstein tensor is given by Eji = Rji − 21 δji R, we find the fluctuations of
the this to be
1
δEji = δRji − δji δR = δRji
2
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


ρ 0
0
0
0 −p 0
ρ 0
0
µ


Tν = 
=
0 Tji
0 0 −p 0 
0 0
0 −p
and the fluctuations of the spatial components are
δTji = −δpδji − Πij
19.1. TENSOR FLUCTUATIONS
193
where we have included the possibility for anisotropic stress through the term
−Πij . We can then express the tensor fluctuations (for example caused by neutrinos) through the field equations for fluctuations δEji = 8πGδTji .
h′′ij + 2Hh′ij − ∇2 hij = 16πGa2 ΠTij
Here we have used that δpδji = 0 since Rji is traceless. Weinberg found in 2004
that the anisotropy would have only small effect, so we can neglect it, and we get
h′′ij + 2Hh′ij − ∇2 hij = 0
(19.1)
This is the wave equation for a free scalar field. This can be derived from the action
Z
Z
1
1
2
4 1 2
Sh =
(19.2)
d x a (∂µ hij ) = d4 xa2 (∂µ φ)2
32πG
2
2
where the last equality is shown in chapter 15.
19.1.1 The power spectrum of the tensor fluctuations
H
We had the power spectrum for the scalar field (from the inflation), ∆2φ = 2π
From (19.2) we see that the relation between the scalar an tensor spectrums are
H 2
2
8
∆2T = ∆2h (k) = 2 · 32πG∆2φ (k) = 2 ∆2φ = 2
Mp
π
Mp
2
.
1
where we have used the Planck mass Mp2 = 8πG
, and the factor 2 comes from the
spin of the graviton. We see that also the tensor fluctuations are constant outside
the horizon and that the result is independent of k. This is (again) an approximation
that is very good to the lowest order, since then HZ scaling is respected. This is
verified 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 same
time, 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 independent
of k. Higher order terms will, however, give a k-dependent power spectrum. It is
usual to write this as
∆2T (k) ∼ knT (k)
(19.3)
so that
nT ln k = ln ∆2T + . . .
or we can also write
nT =
d(ln ∆2T )
d(ln k)
(19.4)
CHAPTER 19. LECTURE 19
194
We now want to find nT as given in the tensor power spectrum in (19.3). As the
modes cross the horizon, then k = aH. This gives
ln k = ln a + ln H
Differentiation with respect to time gives
d ln k
d ln a d ln H
ȧ Ḣ
=
+
= +
dt
dt
dt
a H
Where the first term is large and the last term is small due to the assumption that
the Hubble parameter is (almost) constant in the de-Sitter like inflation dynamics.
The last term is therefore ignored. As ȧa = H we then get
d(ln k) = Hdt
(19.5)
In lecture 17 we learned that in the slow-roll approximation we have
φ̇ = −
V′
dφ
=
3H
dt
So for a general function F and applying (19.5) we then get
′
dF
1 dF
φ̇ dF
V ′ dF
2 V dF
=
=
=−
=
−M
p
d ln k
H dt
H dφ
3H 2 dφ
V dφ
In our case F = ln ∆2T ∼ ln H 2 ∼ ln V and from equation (19.4) we find the
desired result
′ 2
′
2 V
2 V d ln V
= −Mp
= −2ε
nT = −Mp
V dφ
V
so that
∆2T ∼ k−2ε
(19.6)
where ε ≪ 1 during inflation. If we are able to measure the power spectrum of
the tensor fluctuations of the tensor fluctuations we can determine the form of the
inflaton potential. But this is hard, as this power spectrum is much smaller than the
power 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 = 32 for a matter dominated universe. Using
conformal time
Z
Z
dt
1 1−p
dt
=
=
t
τ=
p
a(t)
t
1−p
1
so we say that t ∼ τ 1−p and thus
p
∼ τ radiation dominated
1−p
=
a(τ ) ∼ τ
∼ τ 2 matter dominated
19.1. TENSOR FLUCTUATIONS
195
∆2T
τf
τeq
τ∗
τ0
Figure 19.1: The tensor fluctuations decrease rapidly inside the horizon
′
The conformal Hubble parameter is H = aa = τ1 in both radiation and matter
dominated universes (but H = − τ1 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 more
complicated due to the coupling between the scalar fluctuations and the inflaton
field.
τ
196
CHAPTER 19. LECTURE 19
Chapter 20
Lecture 20
In lecture 18 we split the metric perturbations into three parts, namely scalar, vector and tensor perturbations. Vector perturbations decay quickly and are of little
cosmological interest. The tensor perturbations are the simplest ones to treat mathematically, and we reviewed some of their properties last lecture. But the tensor
perturbations are small, and even if they may tell us about some really interesting
properties of the early universe, they have no deep impact on the structures and
anisotropies that we observe in the universe today. Those structures are caused by
the scalar perturbations.
20.1 Scalar perturbations
In (18.21) we introduced the longitudinal (or conformal Newtonian) gauge given
by
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 scalar
potentials A and D. Then, when having these EOM we may couple this perturbed
metric to the perturbations of the inflaton field, as found earlier. The procedure
when treating the scalar perturbations is very similar to what we did last lecture
when studying tensor perturbations. The perturbations to the RW-metric are given
by
δg00 = 2a2 A
δgij
= 2a2 Dδij
⇒
δgij = −
2
Dδij
a2
(20.2)
This perturbed metric is then used to obtain the perturbed Einstein tensor in the
perturbed Einstein equation
δE µν = 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
CHAPTER 20. LECTURE 20
198
perturbed Einstein tensor looking like this:
2 2
′
∇
D
−
3H(D
+
HA)
a2
1 1 = − 2 D ′′ + (H2 + 2H′ )A + H(A′ + 2D′ ) δij − 2 ∇2 (A − D)δij − ∂ i ∂j (A − D)
a
a
2
=
∂i (D ′ + HA)
(20.3)
a2
δE 00 =
δE ij
δE 0i
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 = ∂µ φ∂ν φ − gµν L
which gives the perturbed energy momentum tensor
δTµν = ∂µ δφ∂ν φ + ∂µ φ∂ν δφ − δgµν L − gµν δL
(20.4)
The explicit calculation of the components of the perturbed energy momentum
tensor will not be done here, but should be straightforward using the techniques
presented in this course. The results, inserted in the perturbed Einstein equations,
yield
1
−Aφ′2 + A2 V ′ δφ + φ′ δφ′
2
a
1
= 8πG 2 (Aφ′2 + a2 V ′ δφ − φ′ δφ′ )δij
a
1
= 8πG 2 φ′ ∂i δφ
a
δE 00 = 8πG
(20.5)
δE ij
(20.6)
δE 0i
(20.7)
where V is the potential. These equations combined with the expressions for the
perturbation of the Einstein tensor given in (20.3), give us the three equations that
we want to solve. They look pretty nasty, but solving them will give a lot of cancellation 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 be
stated.
The first result comes from a mere inspection of (20.6). For i 6= j we have
δij = 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 reduce
to the Minkowski metric in the limit where A and D approach 0. Neither can
(A − D) possess a first order coordinate dependence, since that would violate our
assumptions 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 + [. . .] and
thus Ψ = Φ, which is very useful to know.
Then, simply by inserting in the perturbed Einstein equation, we have
∇2 D − 3H(D′ + HA) = 4πG −Aφ′2 + A2 V ′ δφ + φ′ δφ′
(20.8)
D ′ + HA =
4πGφ′ δφ
(20.9)
D ′′ + (H2 + 2H′ )A + H(A′ + 2D ′ ) =
4πG(−Aφ′2 + a2 V ′ δφ − φ′ δφ′ )
(20.10)
Here φ is the usual inflaton background evolving like
φ′′ + 2Hφ′ + a2 V ′ = 0
Subtracting (20.8) from (20.10) and inserting for δφ from (20.9) one obtains
φ′′
D −∇ D+2 H −
φ
′′
2
φ′′
D +2 H + H
φ
′
′
D = 0 (20.11)
Now we will use this equation to study how the perturbations evolve outside the
horizon. During inflation we have
H = aH = eHt H
,
a(τ ) = −
1
Hτ
(20.12)
The Dk -modes are as usual found by the Fourier transformation
Z
d3 k
Dk (τ )eik·x
D(x, τ ) =
(2π 3 )
In first order perturbation theory the equations are linear and the modes evolve
independent of one another. This simplifies the equations enormously, but is also
an extremely good approximation in the early universe. Using SRA as we did in
lecture 17 we end up with an equation for Dk
Dk′′ + k2 Dk + O(ε)Dk′ + O(ǫ)Dk = 0
When ε → 0 in SRA the last two terms may safely be omitted. Studying k-modes
outside the horizon, also the second term becomes small, and the first term will be
the leading one, that is Dk′′ = 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 inflation
is approaching its end we must have c1 (k) = const. and c2 (k) → 0. That means
that D = A = const. outside the horizon.
CHAPTER 20. LECTURE 20
200
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) = a42 ∇2 D where
D is called a curvature perturbation. Our results should be expressed in a gaugeinvariant way, but we remember that under a gauge transformation xµ → x̄µ =
xµ + ξ µ (x) we had
D → D + Hξ 0
such that D is not a gauge invariant quantity. But it is possible to construct a gauge
invariant curvature perturbation R given by D plus an additional part. To find this
R we first recall some of the properties of the scalar inflaton field φ. Since φ is a
scalar field it transforms like φ̄(x̄) = φ(x). For the fluctuation δφ we have
¯
δφ(x̄)
= φ̄(x̄) − φ0 (x̄)
= φ(x) − φ0 (x + ξ)
= φ(x) − φ0 (x) −ξ µ ∂µ φ0
{z
}
|
δφ(x)
where φ0 is the background field. So δφ will transform like
¯ = δφ − ξ 0 φ′ = δφ − ξ 0 φ′
δφ → δφ
0
where the last equality simply is a change of notation. So we try to express our
gauge invariant curvature scalar like
¯
δφ
δφ
→
R̄
=
D̄
+
H
φ′
φ′
H
= D + Hξ 0 + ′ (δφ − ξ 0 φ′ ) = R
φ
R = D+H
So R may safely be defined as
R≡D+H
δφ
φ′
(20.13)
This quantity is extremely important in inflationary cosmology. We will later show
that R is constant from the inflationary epoch until today, so it can be used as a
messenger that brings us information from the very early quantum universe. R
contains 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 next
lecture. What we will do now is to show that R is constant during inflation for
modes outside the horizon. Using (20.9) we write R as
R = D + 2HMP2
D ′ + HA
φ′2
20.1. SCALAR PERTURBATIONS
201
Differentiation gives
′
′
R =D +
2MP2
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 left
with
′2
2MP2
φ
H∇2 D
′
2
+
R′ = 2MP2
+
H
−
H
φ′2
φ′2 2MP2
{z
}
|
=0
That the underbraced term vanishes one can see from the Friedmann equations.
They say that
H2 =
Using that ρ =
1 2
a ρ
3MP2
1
φ′2
2a2
H2 − H′ =
and
+ V and p =
1
φ′2
2a2
1 2
a (ρ + p)
2MP2
− V we have
φ′2
= H2 − H′
2MP2
(20.14)
and thus the underbraced term indeed vanishes, and we are left with
′
R =
2MP2
H∇2 D
φ′2
(20.15)
Taking a little step to the side, we play around a little with
φ′2
2MP2
. We start with
the slow-roll equation 3H φ̇ = − dV
dφ . By definition of conformal time we have
1 d
d
dt = a dτ . Inserted in the slow-roll equation this gives
da
dφ
a′
dV
= 3 3 φ′ = −
3H φ̇ = 3 dt
a dt
a
dφ
2
′
a4
dV
and thus we have that φ′2 = − 9H
, where H = aa . Remembering the
2
dφ
′ 2
we have
definition of the slow-roll parameter ε ≡ 12 MP2 VV
φ′2
a4
=
9H2
2MP2
dV
dφ
2
1
V 2 a4
=
ε
2MP2
9MP4 H2
From the first Friedmann equation we have H2 =
a2
3MP2
φ′2
= εH2 = H2 − H′
2MP2
V which gives us that
CHAPTER 20. LECTURE 20
202
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 transforming we have
R′k
Ṙk
aṘk
1 k 2
=
=
=
|Dk |2
H
aH
H
ε H
when only considering one mode. From the definition of R we have
Rk = Dk + H
= Dk +
δφk
φ′
H
δφk
φ̇
From (20.9) we have
D ′ + HA = 4πGφ′ δφ
We have already shown that D is a constant and thus that D ′ vanishes. Considering
only one k-mode we then have
Dk = Ak = 4πG
=
φ′
δφ
H
1 φ̇
δφk
2MP2 H
Here we see the coupling between the perturbation in the metric and the perturbation in the inflaton field very explicitly. But this relation is not general, and is
only valid in longitudinal gauge. Then of course the next step is to calculate the
corresponding relation with the gauge independent R instead of D.
H
δφk
φ̇
H
1 φ̇
δφk + δφk
2
2MP H
φ̇
Rk = Dk +
=
Playing the same ε-game as above, we can write this as
Rk = (1 + ε)
H
δφk
φ̇
Since ε is really small we have to a very good approximation:
Rk =
H
φ̇
δφk
(20.16)
20.1. SCALAR PERTURBATIONS
203
Now, that is a nice equation! In figure 20.1 one can see how R is constant in the
different epochs while Dk = Ψk vary.
1
aH
Rk = 1
2
R
3 k
3
R
5 k
Ψk
τf
τeq
τ
Figure 20.1: Outside the horizon R will be constant while Ψk = Dk will vary in
the 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
H
δφk = √
2k3
Combining this with (20.16) we can find the power spectrum for R:
k3
|Rk |2
2π 2
k3 H 2
=
|δφk |2
2π 2 φ̇
2 2
H
=
2π φ̇
∆2R (k) =
(20.17)
From the first Friedmann equation we have H 2 ∼ V and from the slow-roll analysis we have φ̇2 ∼ ǫ. Including the prefactors we may express the power spectrum
as
∆2R (k) =
1
24π 2 ε
V
MP4
(20.18)
CHAPTER 20. LECTURE 20
204
This quantity is measured directly through the Sachs-Wolfe effect in the CMB. So
we may observe the fraction Vε , but not these two quantities separately. Now we
have been studying scalar fluctuations. For tensor fluctuations we had
2
V
2
∆T (k) = 2
3π
MP4
and thus
∆2T
= 16ε ≪ 1
∆2S
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
dln∆2S
dln k
dlnε
dln V
−
=
dln k
dln k
= −2ε − (4ε − 2η)
nS − 1 =
= 2η − 6ε
(20.19)
This gives a small deviation from nS = 1. The data from WMAP is not giving an
unambiguous answer to how large this quantity is.
Chapter 21
Lecture 21
In the previous lectures, we have considered perturbation theory in the inflationary
epoch. We then found the equations that couple the inflaton field to the geometry
of the universe by means of the Einstein equations
δEµν = 8πGδTµν
(21.1)
The perturbations of the energy-momentum tensor give rise to fluctuations in the
metric. These fluctuations were called curvature perturbations, R. We showed
that the curvature fluctuations are constant outside the horizon. This is due to
the fact that the modes of the fluctuations are outside the horizon provided their
wavelength, λ ∼ k1 , 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 inflationary era. Therefore it remains to show that R is constant also in the latercoming
radiation dominated (RD) and matter dominated (MD) epochs of the history of the
universe. (Actually, this is true only with adiabatic perturbations when no entropy
is produced). All this is shown towards the end of this lecture. So in some sense the
R acts as a messenger from very early times and up until today. One of the messages 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 generalizations of the fundamental equations that we have in non-relativistic perturbation
theory. So far we have considered fluctuations in the inflationary epoch described
classically by the inflaton field φ. Now we will need to extend our description to
also account for relativistic perturbations in order to describe the late universe. We
first take a look at the Newtonian perturbation theory.
205
CHAPTER 21. LECTURE 21
206
21.1 Newtonian theory
The non-perturbative theory starts with the perturbations in pressure, energy density, 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 have
assumed comoving coordinates (i.e. zero velocity). The Newtonian equations for
perturbations in the gravitation field are
∇2 δΦ = 4πGa2 δ
(Poisson)
1
∂δ
+ ∇v = 0
(continuity)
∂t
a
1
1
∂v
+ Hv = − ∇δφ − ∇δp (Euler)
∂t
a
aρ
(21.3)
In the Poisson equation we have used the shorthand notation, δ ≡ δρ
ρ . The continuity equation ensures particle conservation. These three equations are valid for
perturbations inside the horizon, but their validity ends when we enter and cross
the horizon. When we now go to the non-Newtonian theory things get a little more
complicated. However, we will see that in simple universe models the equations
don’t get that nasty after all. One very useful simplification in models describing a
homogeneous 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 the
fluid is in thermal equilibrium. The fluid should not experience shear forces
since such forces, when currents are present, will produce heat and then
again entropy.
• The particles in the perfect fluid should be close to each other and frequently
interacting. This justifies the hydrodynamical description as a fluid. This
is not a good approach if the density is low, or particles move freely. One
example of a non-perfect fluid, which we will look at in the coming lectures,
is the photons and neutrinos decoupled. After decoupling from matter, these
move freely and can no longer be described hydrodynamical.
1
For more details about the derivation of the Newtonian perturbation equations, see the note used
in Ast 4220.
21.2. PERFECT FLUIDS
207
We now want to find the perturbed velocities ũµ and ũµ and then in turn the fluctuations in the energy-momentum tensor, δTµν . This will be needed in order to
proceed with the perturbed Einstein equation we saw in (21.1) in the beginning of
this lecture. We start out with the energy-momentum tensor for a perfect fluid
Tµν = (p + ρ)uµ uν − gµν p
(21.4)
the metric we use is the usual one
ds2 = a2 (dτ 2 − dx2 )
The four velocity identity makes us find the zero-components of u by
u2 = 1 = gµν uµ uν = (u0 )2 a2 .
The last step follows from uµ = (u0 , u = 0) since we have chosen comoving
coordinates, thus the fluid is at rest in our frame. The components are u0 = a1 and
u0 = a. We now perturb the 4-velocity and the metric
uµ → ũµ = uµ + δuµ
(Here, we still demand the perturbed velocity to be normalized by the identity
ũũ = 1.) The perturbed metric is the same as we have seen a hundred times
before, namely
gµν → g˜µν = gµν + δgµν
and the perturbations are the 4 × 4 matrix given by the four potentials
−Bi
2 1 + 2A
g̃µν = a
−Bi
−(1 − 2D)δij − Eij
All the terms in the matrix are small, so any product of these terms will be neglected
in the following. We can then find ũµ and ũµ . We write
1
ũµ = (u0 + δu0 , δui ) = ( + δu0 , δui )
a
1
0 1 i
= ( + δu , v )
a
a
(21.5)
Here, we have introduced peculiar velocity, v i = aδui in the last step. We will now
sketch some of the derivations of the components of the perturbed velocities
ũ0 = g̃0µ ũµ = a2 (1 + 2A)u0 − a2 Bi δui
1
= a2 (1 + 2A)( + δu0 )
a
= a + a2 δu0 + 2aA = u0 + a2 δu0 + 2aA
= u0 + δu0
⇒ δu0 = a2 δu0 + 2aA
(21.6)
CHAPTER 21. LECTURE 21
208
Another component is
δui = g̃iµ ũµ
= a(vi − Bi ) + higher order terms
(21.7)
From the 4-velocity identity ũũ = 1, we find
1
δu0 = − A + . . .
a
So with all the components inserted for, the perturbed velocities are
1
(1 − A, v i )
a
= a(1 + A, vi − Bi )
ũµ =
ũµ
(21.8)
21.2.1 Perfect perturbations
We now turn our attention towards the perturbations in the energy-momentum
tensor given as
Tµν → T̃µν = 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 two
parts; perfect and non-perfect perturbations, each of them having 5 DOFs. The perfect perturbations of Tµν will be written on the same form as the energy-momentum
tensor for a perfect fluid seen in (21.4), so the shape of the tensor will be maintained.
T̃µν = (p̃ + ρ̃)ũµ ũν − g̃µν p̃
Here, 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 tensor
by calculating T̃µν and then subtract the unperturbed tensor to find
δT µν = T̃ µν − T µν
By means of the components
T µν
=
T 00 T i0
T 0j T ij
we can express the perfect perturbations of the energy-momentum tensor as
δρ
(ρ + p)v i
µ
(21.9)
δT ν =
(ρ + p)(vi − Bi )
−δpδji
There are altogether 5 DOFs that conserve the perfect shape of Tµν . These are
δρ(1), δp(1) and v i (3) since the spatial index runs over three components. The
other 5 DOFs must therefore be in the non-perfect part of the perturbation of the
energy-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. A
general 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 consist
of 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 of
the perturbed Einstein equations. The usual way to account for also non-perfect
contributions to the fluid, is to introduce an anisotropic part to the perturbed energymomentum tensor by writing
δTij = δij δp + Σij
(21.10)
Here, the first term is an isotropic part. It contains the trace of Tij and is a part
of the perfect Tµν . (δij δp is often called bulk-viscosity and corresponds to the
−δpδji -term of δT µν in equation (21.9)). The second term, Σij , is the anisotropic
part and the physical interpretation is anisotropic pressure or stress. It is traceless
and contains the 5 remaining DOFs of the entire δTµν .
This description of non-perfect fluids also has its range of validity and eventually breaks down. Then the Boltzmann equations are needed to further describe
the fluid. This will be investigated in a future lecture. So to find the non-perfect
components to the perturbation, we neglect the δij δp-term and concentrate on the
anisotropic part. As seen earlier, perturbations can always be separated into scalar,
vector and tensor components.
Σij = ΣSij + ΣVij + ΣTij
(21.11)
Here, the tensor part contains the perturbation hij , and the vector part looks like
Σvij = ∂i vj +∂j vi −the trace. We will need the scalar perturbations of the isotropic
pressure. These are:
1
ΣSij = (∂i ∂j − δij ∇2 )Σ
3
The gradient of the scalar velocity potential, V can be written as
v i = ∂ i V = −∂i V
⇒ v = −∇V
(21.12)
We then finally have both the perfect and non-perfect perturbations of the energymomentum tensor and can find the components of the perturbed Einstein equations.
We will from the Einstein equations derive the generalizations of the three Newtonian perturbation equations in (21.3).
CHAPTER 21. LECTURE 21
210
21.3 The relativistic perturbation equations
The perturbed Einstein equations are given by δEµν = 8πGδTµν . We find the
components of the equation by following the same procedure as we used in the
inflaton-case earlier. We choose longitudinal (or conformal Newtonian) gauge so
that A = Φ and D = Ψ and get 4 equations
∇2 Ψ − 3H(Ψ′ + HΨ) = 4πGa2 δρ
, (00)
Ψ′ + HΦ = −4πGa2 (ρ + p)V , (0i)
Ψ − Φ = 8πGa2 Σ , (i = j)
1
Ψ′′ + H + H(Φ′ + 2Ψ′ ) + (2H′ + H2 )Φ + ∇2 (Φ − Ψ) = 4πGa2 δp
3
(21.13)
(21.14)
(21.15)
, (i 6= j)
(21.16)
We notice from the third equation that the two Bardeen potentials, Ψ and Φ are
equal 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 also
account for relativistic perturbations. The Poisson equation is found by combining
the first and the second equation above. This yields
∇2 Ψ = 4πGa2 [
δρ
|{z}
N ewtonian
− 3H(ρ + p)V ]
{z
}
|
(21.17)
relativistic
We see that the new Poisson equation for the perturbations is the former and Newtonian term with an additional relativistic correction. In order to find also the two
other perturbation equations, we will use the equation of motion (EOM) for matter. This equation shows how perturbations in the metric depend on perturbations
in matter and vice versa. The relativistic Euler equation will then follow from
momentum conservation, while the continuity equation comes from energy (or
particle) conservation. The EOM for matter is
∇µ T µν = 0 = ∂µ T µν + Γµνλ T λν − Γλµν T µλ
The variation of the above equation is
δ(∇µ T µν ) = 0 = ∂µ δT µν + δΓµνλ T λν + Γµνλ δT λν − δΓλµν T µλ − Γλµν δT 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
− w)δ = (1 + w)(3Ψ′ + ∇2 V )
δρ
(21.18)
21.3. THE RELATIVISTIC PERTURBATION EQUATIONS
211
This equation replaces the Newtonian continuity equation seen in (21.3). It describes the fluctuations in density and gives conservation of particles (i.e. energy).
δp
− w) = 0 when the fluid follows an
One should also notice that the term ∼ ( δρ
equation of state p = wρ as for perfect fluids. When ν = j, we get
V ′ + (1 − 3w)HV +
δp
2 ∇2 Σ
w′
= Φ+
+
1+w
p+ρ 31+w
(21.19)
The equation is the generalization of the Newtonian Euler equation. We notice also
the derivative term of the equation of state parameter, w. In general, this parameter
is not necessarily a constant, but may be a function of time. An example is the
transition from RD to MD epoch when w changes continuously from 13 to 0. The
three relativistic equations we have found are
∇2 Ψ = 4πGa2 [δρ − 3H(ρ + p)V ]
δp
− w)δ = (1 + w)(3Ψ′ + ∇2 V )
δρ
w′
δp
2 ∇2 Σ
V ′ + (1 − 3w)HV +
= Φ+
+
1+w
p+ρ 31+w
δ′ + 3H(
(21.20)
They look a bit nasty, and in their most general form, they are. However, when
considering our usual universe models, we often have fluids with simple equation
of state, so that the equations simplify significantly. We will now apply the two last
equations 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
′
δm
= ∇2 Vm + 3Ψ′
Vm′ + HVm = Φ
(21.21)
21.3.2 Example with radiation domination
In a RD universe the parameter is w = 31 . Then the pressure density term in the
Euler equation is
1
δρ
1
δp
1 δργ
= 3 1 =
= δγ
p+ρ
4 ργ
4
ρ + 3ρ
So now the continuity and Euler equations are
δγ′
=
Vγ′ =
4 2
∇ Vγ + 4Ψ′
3
1
δγ + Φ
4
(21.22)
CHAPTER 21. LECTURE 21
212
One short comment to the equations for RD should be that the photons here are
assumed not to be decoupled. The photons then contribute to the perfect fluid.
Before we end this section by looking at the curvature perturbation, R, showing
that it is constant outside the horizon in all epochs of the universe history, we take
a short look at adiabatic perturbations and the concept of sound velocity in the
cosmological fluid.
21.4 Adiabatic perturbations and sound velocity
We know that in general the equation of state is not like p = wρ , but rather given
by p = p(ρ, σ) since the pressure may depend on entropy. Since pressure is a
physical quantity, the differential, δp is exact and we can write it as
δp =
∂p
∂ρ
δρ +
σ
∂p
∂σ
δσ
ρ
The first term represents the adiabatic or isentropic perturbations, while the second
gives the isocurvature (or entropy) perturbations. The adiabatic term motivates for
defining the sound velocity, cs in the fluid by
∂p
∂ρ
= c2s = w
(21.23)
σ
where the last equality is valid in the case of a perfect fluid. And, thus, the
qsound
velocity in the (perfect) fluid applied in the RD and MD epochs are cs =
cs = 0, respectively.
One way to define adiabatic fluctuations is by demanding that
1
3
and
δx δy
−
=0
ẋ
ẏ
Another way to explain adiabatic fluctuations is that the entropy stays constant so
that the equation of state is p = p(ρ, σ) = p(ρ) = wρ. In a general fluid it is
therefore convenient to define a measure of the entropy fluctuations, S by
δp δρ
S=H ′ − ′
p
ρ
δp δρ
−
=H
ṗ
ρ̇
(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′
δp
∂p
= ′ =
∂ρ
ρ
δρ
21.5. VERIFYING THE CONSTANCY OF R
213
21.5 Verifying the constancy of R
We showed in the previous lecture that the gauge invariant curvature perturbation,
R, is constant during inflation for modes outside the horizon. We also claimed that
R 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 need
to establish contact with the inflation era by constructing an R that is valid also
when we enter the fluid description that is being used in the RD and MD epochs.
During inflation, the curvature perturbation was
R=D+H
δφ
δφ
=Ψ+H ′
φ′
φ
where the last equality is a consequence of our longitudinal choice of gauge. The
metric 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 obtaining
an 3D hypersurface. That this surface is curved comes from the 2D-term in the
line element.). The original curvature perturbation we had (in section 20.1.2) was
D, which we found was not in general gauge invariant, but transformed (under a
coordinate transformation) as
D → D̄ = D + Hξ0
We now define D̄ = Dc = R, where Dc is the curvature perturbation, D in
a comoving gauge. We want that R is invariant in longitudinal gauge, and now
further 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 longitudinal
gauge and the velocity potential. Although the potential B is zero in the longitudinal gauge, a “new” B occurs when a coordinate transformation is performed.
B → B̄ = B + ξ0 − ξ ′ = ξ0 − ξ
If the transformed B is still to be zero, then ξ0 = ξ ′ . In the transformation, the
velocity potential, V , will also change
!
V → V̄ = V − ξ ′ = Vc = 0
The last step follows automatically since the gauge is comoving and so there should
be no local velocity. Thus, we seem to have found ξ0 which now must satisfy
ξ0 = V = ξ ′
CHAPTER 21. LECTURE 21
214
It is now time to see just what the curvature perturbation, R, looks like. We find
that
D̄ = D + Hξ0 = D + HV
⇒ R = D + HV = Ψ + HV
(21.25)
The last step is due to the longitudinal gauge. We can manipulate the expression
of 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 universe. We have considered both a classical field description (inflation) and a fluid
description (RD and MD) for the perturbations in the energy-momentum tensor.
By setting them equal, we have
δT 0i =
1 ′ i
φ ∂ δφ = (ρ + p)∂ i V
2
a
| {z } | {z }
field
(21.26)
fluid
Here, φ is the usual inflaton field. We remember from scalar field theory that energy
density and pressure were given by
ρ =
p =
1 ′2
φ +V
2a2
1 ′2
φ −V
2a2
Inserting this into equation (21.26) leads to
1 ′ i
1
φ ∂ δφ = (ρ + p)∂ i V = 2 φ′2 ∂ i V
2
a
a
δφ
⇒V =
φ′
The curvature perturbation will therefore now be
R = Ψ + HV = Ψ + H
δφ
φ′
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 Φ + Ψ′ /H
3 1+w
(21.28)
(21.27)
21.5. VERIFYING THE CONSTANCY OF R
215
This expression gives us the curvature perturbations for RD and MD. In order to
show that R is a constant, we differentiate the expression and then find
R′
1
3
1
3
(1 + w)
= 2 ∇2 c2s Ψ + (Φ − Ψ)] + c2s (1 + w)S
2
H
H
3
2
(21.29)
If we assume adiabatic expansion of the universe, then there is per definition no increase in entropy, so the last term above vanishes. When we also take into account
that ∇2 → −k2 , equation (21.29) reduces to
k2
R′
∼ 2 → 0 (21.30)
H
H
This is valid for all modes outside the horizon. This is true since modes are outside
the (comoving) horizon provided their wavelengths λ ∼ k1 are greater than the
1
horizon scale ∼ H
. Thus, k2 ≪ H and therefore R′ → 0. So to sum up, the
curvature perturbation is constant for all times, provided that the modes are outside
the horizon and that the perturbation is adiabatic.
(
i) k ≪ H , (ouside horizon)
′
R =0
ii) adiabatic
216
CHAPTER 21. LECTURE 21
Chapter 22
Lecture 22
22.1 More perturbations
From (21.28) we had the curvature perturbation, R
R=Ψ+
2 Φ + Ψ′ /H
3 1+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) with
the perturbation
∂p
∂p
δp =
δρ +
δS
∂ρ S
∂S ρ
Where the first term can be called curvature fluctuation, adiabatic fluctuation or
isentropic fluctuation according to taste, and the second term can be called isocurvature 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 from
radiation
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 =
Sγ
nm
217
CHAPTER 22. LECTURE 22
218
δρ
δρm
δρr = −δρm
Figure 22.1: For isocurvature perturbations the matter and radiation fluctuations
have opposite sign.
Here Sγ is the radiation entropy so that Sγ ∼ T 3 ∼ a−3 and nm ∼ a−3 so that
σm is constant during the evolution of the universe. In the adiabatic case we get
δσm = 0. We can express this vanishing fluctuation as
δSγ
δSγ
Sγ
1
δnm
δσm =
− 2 δnm = Sγ
−
nm
nm
n m Sγ
nm
As the particle density is nm =
T3
∼
3/4
ργ
ρm
m
we get
δσm
and the radiation entropy evolves as Sγ ∼
3 δρ
δρm
γ
= constants
−
4 ργ
ρm
as we had that δσm = 0 this gives us
3 δργ
= δm
4 ργ
(22.1)
We can picture adiabatic expansion with the traditional particle in a one dimensional box with size L. Assume that the particle is in the ground state. Now
if the box expands slowly (adiabaticly) to size 2L the wave function will be in the
new 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 it
will retain the old wavelength. And this is not a ground state in the new box, it is a
superposition of several excited energy eigenstates (figure (22.3)).
We continue to look at adiabatic expansion and we write
δρ = ρ̇δt
δP = Ṗ δt
22.1. MORE PERTURBATIONS
219
δρ
δρm ∼ δργ
Figure 22.2: For adiabatic perturbations matter and radiation perturbations behave
similarly
ψ
L
adiabatic
2L
non-adiabatic
2L
Figure 22.3: In the non-adiabatic case, the expansion is too quick for the wave
function to follow, but in the new box this is no longer the ground state, it is a
superposition of excited states. In the adiabatic expansion on the other hand, the
wave function change, so that the system is still in the ground state.
CHAPTER 22. LECTURE 22
220
so that
δt =
δρ
δp
δργ
δρc
=
=
=
ρ̇
ṗ
ρ̇γ
ρ̇c
So we get
δργ
δρm
=
ρ̇γ
ρ̇m
(22.2)
We remember the equations
ρ̇γ + 3H(ργ + pγ ) = 0
ρ̇m + 3H(ρm + pm ) = 0
and as pγ = 31 ργ and pm = 0 we get
ρ̇γ = −4Hργ
ρ̇m = −3Hρm
and equation (22.2) can then be written
δργ
δρm
=
4Hργ
3Hρm
We can rewrite equation (21.28) and insert for Φ = Ψ to get
2 Ψ′ 5 + 3w
+
Ψ = (1 + w)R
(22.3)
3H
3
Here R is constant, and w is constant in the radiation dominated and the matter
dominated epoch (but with different value). We have the conformal Hubble parameter H ∼ τ1 and this allows us to write equation (22.3)
p
n
Ψ′ + Ψ =
τ
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 insert
Ψ′ = 0 into equation (22.3) we get
2
3 + 3w
R in RD
Ψ=
R = 33
5 + 3w
5 R in MD
If we assume no anisotropic stress, Σ = 0 and Ψ = Φ we have the equations
from (21.21) and (21.22)
′
δm
− ∇2 Vm = 3Ψ′
Vm′
+ HVm = Ψ
4
δγ′ − ∇2 Vγ = 4Ψ′
3
1
Vγ′ − δγ = Ψ
4
(22.4)
(22.5)
(22.6)
(22.7)
22.1. MORE PERTURBATIONS
221
Ψ
R
2
R
3
3
R
5
τf
τeq
τ0
τ
Figure 22.4: The evolution of Ψ in the different epochs
The perturbed Einstein equations give
δE0i = 8πGδT0i
so we get
3
Ψ′ + HΨ = H2 (1 + w)V = (ργ + pγ )Vγ + (ρm + pm )Vm
2
(22.8)
we Fourier expand
δ = δ(x, t) =
X
δk (t)eik·x
k
V = V (x, t) =
X
Vk (t)eik·x
k
Ψ = Ψ(x, t) =
X
Ψk (t)eik·x
k
and as ∇2 → −k2 equation (22.4) turns out
′
+ k2 Vmk = 3Ψ′k
δmk
or without the indices
′
δm
+ k2 Vm = 3Ψ′
Let us look at the initial conditions early in the radiation dominated epoch.
Here w = 13 and
2
3 4
Ψ = R = const = H Vr
3
2 3
CHAPTER 22. LECTURE 22
222
which leads us to
Ψ
2H
SInce adiabatic expansion requires these velocities to be equal, this gives us Vr =
Vγ and
−Ψ ′
Vγ′ =
H
(22.9)
2H2
The adiabatic fluctuations become
Vr =
Vγ = Vm = Vr =
Ψ
2H
And with the Friedmann equation in the radiation dominated epoch
H′ = −
4πG
8πG 2
(ρ + 3p)a2 = −
ρa = −H2
3
3
and if we insert this into (22.9) we get
Vγ′ =
Ψ
2
If we insert into (22.8) we find
Ψ 1
− δγ = Ψ
2
4
and further
4
δm
3
where the last equality is due to the adiabatic rquirement in (22.1).
δγ = −2Ψ =
(22.10)
22.2 The Boltzmann equation
Classically the Hamilton function of a system of particles is the sum of kinetic and
potential energy
H =K +U =
N
X
p2a
+ U (q1 , . . . qN ; p1 , . . . pN )
2m
a=1
We will for simplicity write p = (p1 , p2 , . . . pN ) = (pi ) and likewise q = (qi ).
The Hamilton equations are
∂H
⇒ qi = qi (t)
∂pi
∂H
ṗi = −
⇒ pi = pi (t)
∂qi
q̇i =
(22.11)
(22.12)
22.2. THE BOLTZMANN EQUATION
223
pi
qi
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 -dimensional
phase space.An ensemble is a set of such identical systems but with different initial
conditions. 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 a
current of system particles is expressed by
where ∇ =
∂
∂
∂q , ∂p
∂ρ
+∇·J=0
∂t
(22.13)
and J = ρ (q̇, ṗ). (22.13) then turns out
∂
∂
∂ρ
+
(ρq̇i ) +
(ρṗi ) = 0
∂t
∂qi
∂pi
or even better
∂ρ
∂ρ
∂ρ
+ q̇i
+ ṗi
=0
∂t
∂qi
∂pi
But we can use the comoving derivative (or the Lagrange derivative)
(22.14)
d
∂
=
+v·∇
dt
∂t
where
∂
∂t
is called the Euler derivative. In our case the comoving derivative is
∂
∂
∂
d
+ ṗi
=
+ q̇i
dt
∂t
∂q i
∂p i
Combining (22.14) and (22.15) we find Liouville’s theorem
dρ
=0
dt
(22.15)
CHAPTER 22. LECTURE 22
224
and if we apply this together with (22.15) and the Hamilton equations, (22.11) and
(22.12), we get
dρ
∂ρ ∂H ∂ρ
∂H ∂ρ
=
+
−
=0
(22.16)
dt
∂t
∂pi ∂qi
∂qi ∂pi
Now we define the Poisson bracket
∂A ∂B
∂B ∂A
[A, B] =
−
∂pi ∂qi
∂pi ∂qi
so (22.16) becomes
∂ρ
dρ
=
+ [H, ρ] = 0
dt
∂t
In quantum mechanics q and p becomes operators
qi → q̂i
pi → p̂i
and the Poisson bracket turns to Heisenberg brackets
[A, B]P oisson → [Â, B̂]Heisenberg
The expectation value of an observable A = A(q, p) can be expressed
Z
1
hAi =
d3N qd3N pρ(q, p, t)A(q, p, t)
Z
R
R
where Z = d3N q d3N pρ is the partition function. In quantum mechanics ρ
becomes a density operator ρ → ρ̂ where
Z = T r ρ̂
We also get the reduced density functions
ρnm = ρn (q1 , q2 , . . . qn ; p1 , p2 , . . . pm )
Z
Z
Z
Z
=
d3 qn+1 . . . d3 qN d3 pm+1 . . . d3 pN ρ(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
∂f
∂f
∂f
=
+ ẋ
+ ṗ
= C[f ]
dt
∂t
∂x
∂p
(22.17)
This is the Boltzmann equation. We now have a 6-dimensional phase space with
interactions, and the interactions are given by the collision integral C[f ] defined
here.
Chapter 23
Lecture 23
In this lecture we will study temperature fluctuations, that is, the perturbation equations for photons. In equilibrium the photon distribution is Bose-Einstein and is
given by
1
f (x, p, t) = p/T (t)
e
−1
But as discussed earlier, the equilibrium assumption, although simple, is not a
good model for the photon distribution in the universe today. For non-equilibrium
systems we have to apply the more general Boltzmann formalism. Luckily this
formalism 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
∂f
dxi ∂f
dpi ∂f
=
+
+
= C[f ]
dt
∂t
dt ∂xi
dt ∂pi
(23.1)
Here pi denotes the physical momentum1 and C[f ] on RHS is a collision term to
be determined later. We now write the momentum as p = np where n is a unit
vector in the direction of p satisfying n · n = 1. Using this, we write (23.1) as
∂f
dxi ∂f
dp ∂f
dni ∂f
df
=
+
+
+
dt
∂t
dt ∂xi
dt ∂p
dt ∂ni
(23.2)
i
Without any interactions n will be a constant, so dn
dt is a first order term. The
last term in (23.2) is therefore of second order and will be omitted in our further
analysis. We now make the usual decomposition of f into an equilibrium part and
a perturbation part:
f (x, p, t) = f0 (x, p, t) + f1 (x, p, t)
| {z }
| {z }
equilibrium
perturbation=δf
1
In curved space time one may also define for example a comoving momentum that is different
from the local, measurable and physical momentum.
225
CHAPTER 23. LECTURE 23
226
We express our perturbed metric in longitudinal gauge as
ds2 = a2 (1 + 2Φ)dτ 2 − (1 − 2Ψ)dx2
Now we want to find an expression for the
the covariant momentum p̃µ defined by
p̃µ =
dp
dt -term in (23.2).
(23.3)
First we introduce
dxµ
dλ
From general relativity we have the geodesic equation
dp̃µ
+ Γµαβ p̃α p̃β = 0
dλ
(23.4)
The physical energy and physical momentum are
√
g00 p̃0
(physical energy)
p
|gii |p̃i = ni p
=
(physical momentum)
p =
pi
(23.5)
Using the metric defined in (23.3) these quantities become
p = a(1 + Φ)p̃0
pi = a(1 − Ψ)p̃i
(23.6)
For us it will be sufficient to study the µ = 0-equation in (23.4). Calculating the
Christoffel symbols for µ = 0 in our metric yields
Γ000 = H + Φ′
Γ00i = ∂i Φ
Γ0ij
= [H − 2H(Φ + Ψ)] δij
Inserting this into (23.4), using p̃0 = a1 (1 − Φ)p and remembering that
∂Φ dxi ∂Φ
dΦ
=
+
= Φ′ + ui ∂i Φ
dτ
∂τ
dτ ∂xi
one hopefully finds that
1 dp
= −H + Ψ′ − n · ∇Ψ (23.8)
p dτ
(23.7)
23.2. THE BRIGHTNESS EQUATION
227
This is actually the only result that we will need from gravitational theory. To the
lowest order, when not considering the effects of the perturbed metric, this gives
ȧ
1 dp
= −H = −
p dt
a
1
d
(ln pa) = 0 ⇒ pa = const. ⇒ p ∼
dt
a
which is the familiar relation for redshifting of free photons with the expansion of
the universe. Inserting (23.8) into the Boltzmann equation (23.2) we get the final
version of our Boltzmann equation for photons
df
∂f
∂f
=
+ n · ∇f + p
−H + Ψ′ − n · ∇Φ = C[f ]
dτ
∂τ
∂p
(23.9)
23.2 The brightness equation
In (23.9) we still have the mysterious C[f ] collision term on the RHS. So let’s
see what kind of interactions we have for T ≪ 1MeV. The particles present at
this temperature are photons, neutrinos, electrons, protons and α-particles. Except
from the neutrinos, which are decoupled at this temperature, all these particles are
interacting through electromagnetic interactions. In addition the electrons, protons and α-particles are coupled through Coulomb interaction. At this temperature
the electrons are non-relativistic and interact with the photons through Thompson
scattering. The cross-section for this process is given by
1
dσ
= (1 + cos2 θ)r02
dΩ
2
(23.10)
where θ is the deflection angle for the photon and r0 is the “classical electron
radius” r0 = α mh̄e c = 2.8 × 10−13 cm. From the expression for r0 one notices that
the cross section is inversely proportional to the electron mass. This is why the
corresponding cross-sections for Thompson scattering on protons and α-particles
are neglectable compared to the electron scattering. The Thompson scattering is
elastic so |p| = |p′ | = p, that is, the photon doesn’t change its energy, only its
direction through this process. We now may write the total photon distribution
function as a perturbed Bose-Einstein distribution with a temperature that depends
on n rather than p.
f (x, p, t) =
=
where
δf =
1
p
T (x,n,t)
−1
e
f0 (x, p, t) + δf
∂f
p ∂f0
δT = −
δT
∂T
T ∂p
CHAPTER 23. LECTURE 23
228
This last expression makes it tempting to define a quantity that looks like
Θ(x, n, t) =
1
δT (x, n, t)
T
(23.11)
and we have
δf = −p
∂f0
Θ (23.12)
∂p
Inserting this expression for f into the Boltzmann equation (23.9) one should end
up with
∂f0 ′
Θ + n · ∇(Θ + Φ) − Ψ′ = aC[f ]
(23.13)
−p
∂p
Now we almost have the Boltzmann equation for θ. What is left is to find a more
explicit expression for C[f ]. A detailed calculation of this term requires a bit of
work. Luckily these details are given in section 4.3 in Modern Cosmology by Scott
Dodelson. Here we will just scratch the surface of classical kinetic theory to shed
some 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, kinetic
theory is dealing with large systems out of equilibrium. The variables that we work
with in kinetic theory is the distribution function f (x, v, t) and the momentum
p = mv. As Boltzmann showed, one may write
Z
Z
∂
∂
∂
3
+v
+a
(23.14)
f = d v1 dσ|v − v′ |(f ′ f1′ − f f1 )
∂t
∂x
∂v
where v = ẋ, a = v̇ and dσ is the differential cross-section given by dσ =
dσ
dΩ dΩ. Here one is considering scattering processes where two particles with initial
values v and v1 are scattered to the new values v′ and v1′ . (23.14) is an exact
equation2 . We are considering Thompson scattering where dσ is the given by the
Thompson cross-section, so our collision term will have the form σT ne Θ, where
ne is the electron density and σT is the Thompson cross-section. Doing the detailed
calculations and including all terms and factors one is left with
∂f0
(Θ −Θ0 − n · ve )
aC[f ] = aNe σT p
∂p
| {z }
(23.15)
δf
where the meaning of Θ0 is soon to be defined. Inserting this into (23.13) one finds
the Boltzmann equation for Θ which is called the brightness equation:
2
From 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 = −ane σT . Θ is commonly expanded in
its multipoles
Θ = Θ(x, n, t) = Σ multipoles
The monopole
Θ0 (x, t) =
1
4π
Z
dΩΘ(x, n, t)
can be interpreted as the average temperature fluctuation in all directions. Later we
will couple the temperature fluctuations to fluctuations in the gravitational potential
to 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 for
photons. We use the perturbed distribution function to find the perturbed energy
momentum tensor. In flat space we have
T µν (x, t) = 2
Z
d3 p µ ν
p p f (x, p, t)
(2π)3 E
(23.17)
3
We know that dEp is Lorentz invariant, so T µν must be Lorentz invariant. The
00-component in the equilibrium case yields
T
00
= ργ = 2
Z
dp3 pp
f0 (x, p, t)
(2π)3 p
since E = p for photons. If the photon gas is in equilibrium, the distribution
function is Bose-Einstein and the energy density obeys the Stefan-Boltzmann law
2
ργ = π15 T 4 . In the general case we have
T
00
= ργ + δργ = 2
Z
d3 p
p(f0 + δf )
(2π)3
(23.18)
CHAPTER 23. LECTURE 23
230
and thus, using (23.12), we have
Z
d3 p
δργ = 2
pδf
(2π)3
Z
∂f0
d3 p
Θ
pp
= −2
3
(2π)
∂p
Z
Z
dp p2 2 ∂f0
= −2
p
× dΩΘ(x, n, t)
(2π)3 ∂p
Z
dp 4p3
= 2
f0 × 4πΘ0 (x, t)
(2π)3
Z
d3 p
pf0 ×Θ0
= 4×2
(2π)3
|
{z
}
ργ
= 4ργ Θ0
(23.19)
So we have the beautiful relation
δγ =
δργ
= 4Θ0 (23.20)
ργ
Another similar calculation is to study the i0-component of the energy-momentum
tensor. We have
T0i = (ργ + pγ )vγi
Z
d3 p i
p δf
= 2
(2π)3
where v i denotes a peculiar velocity. pi = ni p and pγ = 13 ργ . We then have
Z
d3 p
∂f0 i
4
(23.21)
ργ vγi = −2
n Θ(x, n, t)
pp
3
(2π)3 ∂p
{z
}
|
4ργ
In (21.12) in lecture 21 we defined v i as the derivative of a potential:
v i = ∂ i Vγ = −∂i Vγ
∂i vγi
= −∇2 Vγ
where Vγ = Vγ (x, t). Now we want to Fourier transform our x-dependencies into
k-space. We get
Z
d3 k
Θ(k, n, t)eik·x
Θ(x, n, t) =
(2π)3
Z
d3 k
Vγ (x, t) =
Vγ (k, t)eik·x
(23.22)
(2π)3
23.3. PROPERTIES OF Θ AND ITS MULTIPOLE EXPANSION
231
and we have
−∇2 Vγ ⇒
i
4
ργ k2 Vγ = 4ργ
3
4π
Z
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 = k̂ · n = cos θ and write Θ = Θ(k, µ, t). We then
have
dΩ = 2π sin θdθ = 2πd cos θ = 2πdµ
We now write Θ as a sum of Legendre polynomials
Θ(k, µ, t) =
∞
X
(−i)l (2l + 1)Θl (k, t)Pl (µ)
(23.24)
l=0
We have the orthogonality relation for Legendre polynomials
Z 1
2
δll′
dµPl (µ)Pl′ (µ) =
2l
+1
−1
So we have that
1
4π
Z
dΩPl Pl′ =
δll′
2l + 1
and thus we can write a single multipole of Θ as
Z
il
dΩ Θ(k, n, t)Pl (k̂
· n)
Θl (k, t) =
|{z}
4π
(23.25)
µ
Some values for the first Legendre polynomials are
1
P0 (µ) = 1 , P1 (µ) = µ , P2 (µ) = (3µ2 − 1)
2
If we now look back to (23.23) we find that
4
ργ k2 Vγ
3
Z
i
= 4ργ
dΩ(k · n)Θ
4π
= 4ργ kΘ1
and we have that
kVγ = 3Θ1
(23.26)
Analogously to what we have done for Θ0 and Θ1 we can find that Θ2 is connected
to the anisotropic stress Πγ ∼ Φ − Ψ like
Πγ = 12Θ2
So we see that for the lowest multipoles in the temperature fluctuations, the amplitude of each of the multipoles has a direct physical interpretation.
CHAPTER 23. LECTURE 23
232
23.3.1 Line-of-sight integration
From the brightness equation (23.16) we now want to determine the magnitude
of Θ today given its value, say at the last scattering surface (LSS)3 . Solving this
problem 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 presented a technique called line-of-sight integration as a very direct way to solve the
problem. This method is used in all modern computer programs working with
CMB anisotropies. As a very simple example we will consider the case where
Ψ′ = Φ′ = H′ = 0. Then the brightness equation (23.16) yields
Θ′ + ik · nΘ = 0
Θ′ + ikµΘ = 0
dΘ
= −ikµdτ
Θ
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 solution will be on the same, simple form. A well-known technique from quantum
mechanics when one has such a plane wave solution, is to expand it as a product of
spherical Bessel functions, jl , and Legendre polynomials.
ik·r
e
=
∞
X
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. Assuming
Θ∗ = Θ(k, τ∗ ) = Θ0 (k, τ∗ ) (that is, we only have a monopole at LSS), expanding
and projecting out the multipoles using (23.25) we will find that
Θl (k, τ0 ) = Θ0 (k, τ∗ )jl (kD)
(23.29)
3
One of the less pleasant facts about modern cosmology is that “last scattering surface” has exactly the same acronym as “large scale structure”.
23.3. PROPERTIES OF Θ AND ITS MULTIPOLE EXPANSION
233
where D is the comoving angular diameter distance to LSS given by
Z τ0
Z t0
dt
dτ = τ0 − τ∗ ≈ τ0
=
D=
τ∗
t∗ a(t)
We see that we get a wide spectrum of Θ-multipoles by only starting with a monopole at LSS. This is just the Sachs-Wolfe effect. We also notice how the spherical
Bessel function, jl , enters as a propagator for the temperature fluctuations. Next
lecture we will solve the full equation, and the result will be almost as beautiful as
this one.
234
CHAPTER 23. LECTURE 23
Chapter 24
Lecture 24
In this lecture, we will derive the final equation that determines the fluctuations in
temperature of the background radiation. We will first find the simplified equation, under the assumption that there are no interactions, gravitational potential
or perturbations, and then finally find the exact equation of the temperature fluctuations by applying the method of line of sight integration, first developed by
Seljak and Zaldarriaga in 1996. Thus we receive an equation that fully describes
the fluctuating properties of the CMB-spectra. However, before we go on with
these derivations, we will in short remind ourselves with what we found in the last
lecture.
The temperature fluctuations, δT
T were given by
Θ(x, n, τ ) =
δT
(x, n, τ )
T
We saw that it is more convenient to work with these functions in Fourier space, so
Fourier transforming the Θ-function leads to
Z
d3 k
Θ(k, n, τ )eik·x
(24.1)
Θ(x, n, τ ) =
(2π)3
where the Θ(k, n, τ )-function can be expressed in terms by the Legendre polynomials, Pl , as a sum by
∞
X
Θ(k, n, τ ) =
(−i)l (2l + 1)Θl (k, τ )Pl (k̂ · n)
(24.2)
l=0
And lastly plane waves are expanded in terms of the spherical harmonic functions
by
∞
X
ik·x
il (2l + 1)jl (kx)Pl (k̂ · x̂)
(24.3)
e
=
l=0
All these expansions listed above are standard and so easy to use (although hard to
derive) that all master students should be familiar with them. We will now take a
235
CHAPTER 24. LECTURE 24
236
deep look at the equation governing how the Θ-function (and then again the CMB
temperature fluctuations) vary with time, direction and wavenumber. This is our
starting 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 = k̂ · n and −κ′ = ane σT ⇒ −κ̇ = ne σT where
a is the scale factor, ne is the number of free electrons and σT is the Thompson
cross-section seen earlier in the previous lecture. Thus, in equation (24.4), only
scattering on free electrons contribute to the last term. Before we proceed to our
main topic, let us take just a short look at free electrons, degree of ionization, and
the conformal Hubble parameter. We will investigate how the conformal Hubble
parameter evolves with the redshift.
24.1 Degree of ionization
Due to conservation of electric charge, the number of free electrons equal the number 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 and
free baryons
ne
X=
nb
and is equal to zero when no free electrons are present. A schematic illustration of
the degree of ionization as a function of the redshift, z is shown in figure (24.1 ) and
is based on observations by WMAP. We see that the degree of ionization is equal to
1 in the early universe, but decreases rapidly to zero at the time of recombination
when neutral atoms formed and the universe became transparent. The fact that the
ionization increases again at later times is a bit controversial. This reionization
itself, 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 suggests
in the figure. Some speculate that the WMAP- data are not precise enough. Only
time will show.
We can also find how the intensity of light (=luminosity?) changes with the
conformal 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
1
z∗ ∼ 1100
20
0
Z
Figure 24.1: The degree of ionization: The figure shows how the degree of ionization, X depends on the redshift. We observe that X is equal to one until recombination begins at ∼ 1000, then dropping rapidly towards zero thereafter. The value
of X rises to 1 again at a redshift z ≈ 20. This is due to stellar production resulting
in ionizing U V -radiation. The data are from WMAP.
than before, that is I(t0 ) < I(t). We can find the damping factor by
dI
dI
⇒
I
= −Ine σT dt
= −ne σT dt
⇒ I(t0 ) = I(t)e−
R t0
t
ne σT dt
= I(t)e−κ
(24.5)
So the luminosity is reduced by a factor κ, which plays the role as an optical depth
in this manner. The conformal Hubble-parameter can be written as
κ=
Z
t
t0
κ̇dt = −
Z
t0
κ̇dt
t
If we use that the number of free electrons decrease with expansion of space, ne ∼
1
, we can find −κ̇ as a function of z. This is shown in figure (24.2). If the curve
a3
of −κ̇ is integrated, we find the optical depth, κ. Figure (24.3) shows how κ varies
with the redshift. Per definition κ(z∗ ) = 1 at LSS, and this is also seen in the
figure.
CHAPTER 24. LECTURE 24
238
−Ḣ
1100
20
z
Figure 24.2: The figure shows the redshift dependence of the time derivative of the
optical depth.
24.2 The temperature fluctuations in CMB
Now it is time to solve the equation seen in (24.4) governing the temperature fluctuations. We will do this in two different situations; in the first, we assume a very
simplified equation, corresponding to free electron travels. The last occasion is the
realistic and exact situation, yet more complicated due to interactions and gravitation 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 simplifications. If we assume that there are no interactions, then κ′ = 0 since now
κ′ ∼ σT = 0. Furthermore Ψ = Φ = 0 when no perturbations to the gravitational
potential are present. This gives an equation
Θ′ + ik · nΘ = 0
(24.6)
Θ(k, n, τ0 ) = Θ(k, n, τ∗ )e−ik·n(τ0 −τ∗ )
(24.7)
which is easy to solve
Here, τ∗ denotes time when the recombination took place, giving a transparent
universe. The exponential term in the solution to the temperature fluctuations describes a plane wave from the last scattering surface (LSS). This is not so surprising
since we have already assumed no interactions that may influence the wave. If there
are only monopoles, Θ0 on LSS, the expression of the temperature fluctuations are
Θ(k, n, τ∗ ) = Θ0 (k, τ∗ )
24.3. EXACT SOLUTION OF
δT
T
239
H
1
20
1100
z
Figure 24.3: The figure shows how the optical depth evolves with respect to the
redshift. We see that the value of κ is large in the early universe, then decreasing
rather rapid at recombination. function plotted here, is the integral of −κ and
should therefore be compared with figure (24.2).
This means that all the angular dependence (in the sky) of the temperature fluctuations is given by the exponential term in (24.7) since Θ0 has the same value
in all directions per definition of a monopole. This makes it possible to write the
fluctuations as
Θl (k, τ0 ) = Θ0 jl (kD)
where D is the distance to the last scattering surface and is given by
D =
Z
≈ τ0 .
dt
=
a
Z
τ0
τ∗
= τ0 − τ∗
(24.8)
The last step is due to the fact that τ0 >> τ∗ . This was the simplified situation. We
will now find the exact solution of the temperature fluctuations in the CMB.
24.3 Exact solution of
δT
T
We will now derive the precise expression of the temperature fluctuations in the
CMB, given in (24.4), by following the line of sight integration approach. But first
we find an expression for a very useful quantity in CMB physics, the correlation
function. There are several versions of the correlation function, depending on how
many separate points in the sky that are correlated. We will consider the simplest
one which is the two-point correlation function gaining information from only two
separate points. This is shown schematically in figure (24.4).
CHAPTER 24. LECTURE 24
240
γ
n′
β
LSS
γ
n
Figure 24.4: The two-point correlation: The figure shows two pieces of information, represented by the vectors n and n′ arriving from the last scattering surface
(LSS). The arrows indicate the directions of the radiation carrying the temperature
fluctuations. The angular distribution between the (two) photons is given by the
angle β.
24.3.1 Correlation function to lowest order
The two-point correlation function, C(β) is the expectation value of the product of
two separate temperature fluctuations and is defined by
C(β) = hΘ(x, n, τ0 )Θ(x, n′ , τ0 )i
(24.9)
The two-point correlation function can be written into a more convenient form by
expanding in terms of the Legendre polynomials
∞
1 X
(2l + 1)Cl Pl (cos β)
4π
C(β) =
(24.10)
l=0
Here, cos β = n · n′ being the angle between the two points correlated. The
function Cl in the above expression is the one we are going to find. In some sense
this 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. By
inserting the Fourier expansion of Θ given in (24.1) in (24.9), C(β) takes form as
Z
Z
d3 k′
d3 k
′
hΘ(k, n, τ0 )Θ(k′ , n′ , τ0 )iei(k+k )·x
C(β) =
(2π)3
(2π)3
Z
Z
∞
d3 k′ X
d3 k
′
′
=
(−i)l+l (2l + 1)(2l′ + 1)hΘl (k, τ0 )Θl (k
(24.11)
, τ0 )i
3
3
(2π)
(2π) ′
l,l =0
′
′
Pl (k̂ · n)Pl (k̂ · n′ )ei(k+k )·x
24.3. EXACT SOLUTION OF
δT
T
241
In the last step, we have used the equations (24.2). The Θl −terms in the brackets
are fluctuating temperature amplitudes. These fluctuations are Gaussian as predicted by inflation. Therefore the fluctuations are orthogonal and this fact makes it
possible to simplify the expression for C(β). The orthogonality is expressed by
hΘl (k, τ0 )Θl (k′ , τ0 )i = (2π)3 δ(k + k′ )δll′ h|Θl (k, τ0 )|2 i
Thus when we integrate over all modes, we see that
Z
dk′
δ(k + k′ ) ⇒ k′ = −k
(2π)3
Similarly, summation over l gives the relation
X
δll′ ⇒ l′ = l
l′
which again leads to
ˆ · n′ ) = (−1)l Pl (k̂ · n′ )(−i)2l (−1)l = +1
Pl′ (k̂′ · n′ ) → Pl (−k
(24.12)
We can show that the sign of the expression of C(β) now is positive. The expression 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 left
with now, after the simplifications, is an expression that looks like
∞ Z
X
d3 k
(2l + 1)2 h|Θl (k, τ0 )|2 iPl (k̂ · n)Pl (k̂ · n′ )
(24.13)
C(β) =
(2π)3
l=0
The last step we have to in order to find the Cl function, is to simplify theRproduct
3
of
R the 2Legendre polynomials in expression (24.13). Remembering that d k =
dkk dΩk , we have the following
Z
4π
dΩk Pl (k̂ · n)Pl (k̂ · n′ ) =
Pl (n · n′ )
2l + 1
Now, the expression for C(β) turns into
∞ Z
X
d3 k
(2l + 1)h|Θl (k, τ0 )|2 iPl (n · n′ )
C(β) =
(2π)3
l=0
R
R
R
(here, we have used that dkk2 × 4π = dkk2 dΩ = d3 k since 4π is the solid
angle). To find Cl we compare with equation (24.10), which was given by
C(β) =
∞
1 X
(2l + 1)Cl Pl (cos β)
4π
l=0
=
∞
1 X
(2l + 1)Cl Pl (n · n′ )
4π
l=0
And then finally we find that
(24.14)
CHAPTER 24. LECTURE 24
242
Cl = 4π
Z
d3 k
h|Θl (k, τ0 )|2 i
(2π)3
(24.15)
This is an important expression. If we know Cl , we saw in equation (24.10) that the
two point correlation function, C(β) can be calculated. The two point correlation
function is also an observable so now theory and observations can be compared.
We saw in the simplified equation of the temperature fluctuations that Θl ( and
then again Cl ) could be found quite easily. We will now consider the general case
where interactions are allowed and solve equation (24.4) exactly by line-of-sight
integration.
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 temperature
fluctuations by only performing one integration. Earlier, one hadP
to solve an infinite
set 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 eκ
d iµkτ −κ
(e
Θ)
dτ
So then the equation turns into
d iµkτ −κ
(e
Θ) = [Ψ′ − ikµΦ − κ′ (Θ0 − iµkVb )]e+iµkτ e−κ
dτ
(24.17)
By integrating this equation from zero to infinity, we find that the temperature
fluctuations look like
Z ∞
dτ eiµk(τ −τ0 )−κ [Ψ′ − iµkΦ − κ′ (Θ0 − iµkVb )]
Θ(k, n, τ0 ) =
0
The term containing iµk in the previous expression contains an angular dependd iµkτ
e
= iµkeiµkτ and then perence. We can rewrite this term as a derivative dτ
form a partial integration. This leads to a term of the form
d −κ
e = −κ′ e−κ ≡ g(τ ).
dτ
The function g(τ ), is often referred to as a “visibility function”. It tells us what the
probability is that photons experienced their last scattering at the time τ . From the
24.3. EXACT SOLUTION OF
δT
T
243
g
1100
z
Figure 24.5: The visibility function: The figure sketches the visibility function, g
as a function of the cosmological redshift. The function gives the probability that
the 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 has
its maximum value at z = 1100.
definition of the visibility function, −κ′ e−κ ≡ g(τ ), the term −κ′ is the probability
that photons are scattered and e−κ is the probability that a photon is unscattered
until 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 scattering around the time when the universe became transparent.
We in some sense avoided the angular dependence by rewriting the iµk-term
and then perform a partial integration. From the expression in (24.17) we see that
we have two terms ∼ iµk. This means that two partial integrations are needed in
order to find the expression for the temperature fluctuations which now looks like
Z τ0
dτ eiµk(τ −τ0 )−κ [Ψ′ − iµkΦ − κ′ (Θ0 − iµkVb )]
Θ(k, n, τ0 ) =
Z0 τ0
dτ e−ik̂·n(τ0 −τ ) [g(Θ0 + Ψ)
=
0
+(gVb )′ + e−κ (Φ′ + Ψ′ )].
Here, the g(Θ0 + Ψ)-term comes from interactions and Vb represents the velocity
of the scattering objects of the photons. We have earlier seen that we need the
amplitude, 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 the
exponential term in equation (24.18) in spherical harmonics and end up with
Z τ0
dτ [g(Θ0 + Ψ) + (gVb )′ + e−κ (Φ′ + Ψ′ )]jl (kτ0 − kτ ). (24.18)
Θl (k, τ ) =
0
CHAPTER 24. LECTURE 24
244
dσ
dΩ
r02
r02
2
π
2
π
θ
Figure 24.6: The Thomson cross-section: the figure shows how the cross-section
changes with the angle θ.
This equation is pretty much the exact answer to the fluctuations in the temperature
of the CMB as described by equation (24.4). However, there is one small detail that
is worth mentioning: The Thomson cross-section has an angular dependence seen
dσ
by dΩ
= r20 (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 forwards
and backwards than in between. In the derivation of the Boltzmann equation, we
assumed that there was no angular dependence of the cross-section. The small
deviation from this assumption due to the angular dependence, manifests itself as
a polarization of the temperature fluctuations. The Planck satellite is supposed to
observe 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 simplified by assuming that the recombination occurs instantaneously. In that case the
fraction of ionization, X as seen in figure (24.1), will turn into a step function with
respect 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 visibility function, g turns into a delta function, being equal to zero for all values of the
redshift 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
24.3. EXACT SOLUTION OF
δT
T
245
X
X
1
1
z∗ ∼ 1100
0
20
1100
Z
z
Figure 24.7: Instantaneous recombination: If the recombination is assumed to be
instantaneous, then the fraction of ionization turns into a step function.
g
g
1100
1100
z
z
Figure 24.8: The visibility function, g turns into a delta function under the assumption that the recombination takes place instantaneously.
Θl (k, τ0 ) = (Θ0 + Ψ)jl (kD) + kVb jl′ (kD) +
|
{z
} | {z }
S.W.
doppler
Z
τ
| τ∗
dτ e−κ(τ ) (Φ′ + Ψ′ )jl (kτ0 − kτ )
{z
}
I.S.W.
(24.19)
Here, the first term describes the Sachs-Wolfe effect, the middle term is the doppler
effect due to the velocity of baryons, while the last term gives the Integrated SachsWolfe. Actually, since we performed an integration over a delta function, all values
in the last expression should be taken at the value τ = τ∗ . Thus, the equation more
precisely is
Z τ
Θl (k, τ0 ) = (Θ0 +Ψ) jl (kD)+kVb jl′ (kD)+
dτ e−κ(τ ) (Φ′ +Ψ′ )jl (kτ0 −kτ )
τ∗
τ∗
Finally in this lecture, we take a look at the Sachs-Wolfe effect since this gives us
an important connection to inflation.
24.3.4 Sachs-Wolfe
The first term in equation (24.19) can be written as
1
(Θ0 + Ψ)∗ = ( δγ + Ψ)∗
4
CHAPTER 24. LECTURE 24
246
In the very end of lecture 22, equation (22.10), we found an expression of δγ .
Hence
2
1
(Θ0 + Ψ)∗ = (− Ψ + Ψ)∗ = Ψ∗
3
3
1
=
Ψm
3
(24.20)
In the last step, the Bardeen potential, Ψ∗ has been replaced with Ψm , representing
matter domination, since the last scattering occurs in the matter dominated epoch.
We know from the previous lectures that the initial value of Ψm = 35 ( figure (20.1)
shows this very clearly). If we insert equation (24.20) into the expression for the
amplitude, Cl , we get the following
ClS.W.
d3 k
h|Θl (k, τ0 )|2 ijl2 (kD)
(2π)3
Z
4π ∞ dk k3
h|Ψm |2 i jl2 (kD)
9 0 k |2π 2 {z
}
= 4π
=
Z
(24.21)
∆2Ψ
From the expression above, we recognize the underbraced term as the Power spectrum for the potential Ψ. We can rewrite the expression for the amplitudes, ClS.W.
of the temperature fluctuations by using
9
3
Ψm = R ⇒ ∆2Ψ = ∆2R .
5
25
This gives us
ClS.W.
=
=
Z
4π
dk 2 9 2
∆
j (kD)
9
k R 25 l
Z ∞
4π
dx 2
j (x) ∆2R
25
x l
{z
}
|
(24.22)
1
= 2l(l+1)
The final expression is usually written on a form so that the we see the connection
to the power spectrum, ∆2R more clearly
l(l + 1) S.W.
1
= ∆2R
Cl
2π
25
(24.23)
The Power spectrum is shown in figure (24.9). The Sachs-Wolfe effect comes
in for l ≤ 100, and from the figure, we observe that this is when the value of the
Power spectrum is approximately constant and equal to 2 × 10−10 .
24.3. EXACT SOLUTION OF
δT
T
247
6
l(l+1)
2π Cl
[10−10 ]
4
2
200
100
1000
l
Figure 24.9: The Power spectrum: The graph shows the Power spectrum as a
function of l.
We end this course by estimating the value of the inflaton potential, V . When
we considered inflation in the previous lectures, we found that
V
1
2
≈ 50 × 10−10
∆R =
24π 2 ǫ Mp4
This makes it possible to find the fraction between V and ǫ that looks like
V
ǫ
1/4
≈ 6 × 1016 GeV
(24.24)
During inflation the parameter ǫ is small in order for the potential to be sufficiently
shallow to allow a slow rolling. If we assume ǫ ∼ 10−4 , then the value of the
potential driving inflation is
V 1/4 ∼ 1016 GeV
It is a bit remarkable that this energy is also of the same order as we expect to have
in the GUT regime. Whether or not this is a coincidence is yet an open question,
but do not despair; the truth is out there...
248
CHAPTER 24. LECTURE 24
Appendix A
The Riemann tensor and gauge
fields
We will now take a look at the similarities between the Riemann tensor and gauge
fields. The Riemann tensor is given by
Rµναβ = ∂α Γµνβ − ∂β Γµνα + Γρνβ Γµρα − Γρνα Γµρβ
We observe that the Riemann tensor is of the same form as a gauge field
Fαβ
= ∂α Aβ − ∂β Aα + [Aα , Aβ ]
= ∂α Aβ − ∂β Aα + Aα Aβ − Aβ Aα
(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 = Aαα (Tα )ij . The point
is that the upper µ and lower ν indices of the Riemann tensor correspond to the
ij-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 − e2Ht dx2
But this only covers half of the de Sitter space. To cover the whole de Sitter space
you have to use the metric
dr 2
2
2
2
2
2
+ r dΩ
ds = dt − cosh (Ht)
1 − H 2r2
This can also be written
2
dr 2
2
2 2
2
2
ds = 1 − H r dt −
+ r dΩ
1 − H 2 r2
Lots of different metrics were tried before the Robertson-Walker metric was
found. Originally Friedmann used Poincaré coordinates with the metric
dσ 2 =
dx21 + dx22 + dx23
x23
But this was difficult. Einstein used spherical coordinates with the metric
dr 2
2
2
2
2
2
+ r dΩ
ds = dt − a
1 − H 2r2
this can also be written (with redefined r)
dr 2 + r 2 dΩ2
2
2
2
2
2
ds2 = dt2 − a2
=
dt
−
a
dχ
+
sin
χdΩ
2
1 + 41 H 2 r 2
And in 1918 de Sitter used
ds2 =
ηµν dxµ dxν
2
1 + 41 H 2 x2
where x = x2 − t2 . It should be noted that all the components of this metric
approaches zero, ds2 → 0 as |x| → ∞. Finally it was Lemaitre who finally used
the metric
ds2 = dt2 − e2Ht dx2
251