Chapter 2.
INFLATION
Why is the universe homogeneous and isotropic?
Why is the CMB so uniform?
2.1. THE HORIZON PROBLEM
Consider the propagation of light in the FRW spacetime
h
i
2
2
2
2
2
2
ds = a (τ ) dτ − dχ − Sk (χ)dΩ .
Because of isotropy, we can focus on purely radial geodesics (dθ = dφ = 0):
h
i
2
2
2
2
ds = a (τ ) dτ − dχ .
Photons travel on null geodesics,
ds2 = 0 ⇒ ∆χ = ±∆τ
(straight lines)
Consider the FRW universe in these coordinates:
comoving particle outside
the particle horizon at p
event horizon at p
p
particle horizon at p
It is the particle horizon that is relevant for the horizon problem:
Z t
Z a
Z ln a
dt
da
χph (τ ) = τ − τi =
=
=
(aH)−1 d ln a (?)
ti a(t)
ai aȧ
ln ai
↑
comoving Hubble radius
e.g. for a fluid with w = P/ρ:
1
(aH)−1 = H0−1 a 2 (1+3w) .
Ordinary matter satisfies the Strong Energy Condition (SEC): 1 + 3w > 0.
⇒ (aH)−1 increases and (?) is dominated by late times.
For the fluid, we have
i ai →0 , w>− 1
1
2H0−1 1 (1+3w)
2H0−1 h 1 (1+3w)
3
2 (1+3w)
2
− ai
a
−−−−−−−−−→
a2
χph (a) =
1 + 3w
1 + 3w
2
(aH)−1
=
1 + 3w
[NB: In the standard cosmology: particle horizon ∼ Hubble radius ∼ “horizon”]
The singularity is at τi ≡
2H0−1 12 (1+3w)
a
= 0 and
1 + 3w i
the particle horizon is finite and grows as χph = τ .
2
This has an important consequence: most parts of the cosmic microwave background (CMB) have never been in causal contact.
D
TODAY
CMB PHOTON
q
p
RECOMBINATION
SINGULARITY
p
q
D
SURFACE OF LAST-SCATTERING
The CMB is made of 104 causally disconnected regions,
= horizon problem
yet it is observed to be almost perfectly uniform!?
3
2.2. A SHRINKING HUBBLE SPHERE
A simple solution to the horizon problem is
to postulate a shrinking Hubble sphere
d
(aH)−1 < 0
dt
⇔
SEC-violating fluid
1 + 3w < 0
2H0−1 12 (1+3w) w<− 13
a
−−−−−→ −∞
This implies that τi ≡
1 + 3w i
“There was more (conformal) time between the
singularity and recombination than we had thought!”
D
BIG BANG
q
p
RECOMBINATION
END OF INFLATION
INFLATION
CAUSAL
CONTACT
SINGULARITY
4
HUBBLE RADIUS
vs.
PARTICLE HORIZON
(aH)−1
χph
↑
↑
distance over which particles
can travel in one expansion time
distance over which particles can
travel in the history of the universe
Consider two particles separated by a distance λ:
λ > (aH)−1
λ > χph
particles cannot talk
to each other now
particles could never
have communicated
Conservative solution to the horizon problem:
⇒
The observable universe was inside the
Hubble radius at the beginning of inflation
(a0 H0 )−1 < (aI HI )−1
scales
inflation
standard Big Bang
inflation
reheating
“ Big Bang ”
time
How much inflation do we need? (aI HI )−1 > (a0 H0 )−1 ∼ e60 (aE HE )−1
During inflation H ≈ const., so we find aE > e60 aI
“60 e-folds”
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CONDITIONS FOR INFLATION
The shrinking Hubble sphere is equivalent to other ways of describing inflation:
Proof
• Accelerated expansion
ä > 0
d(aH)−1
d
ä
= (ȧ)−1 = − 2 < 0 ⇒
dt
dt
(ȧ)
• Slowly-varying Hubble
Ḣ
ε≡− 2 <1
H
ȧH + aḢ
(ε − 1)
d(aH)−1
=−
=
<0 ⇒
dt
(aH)2
a
• Exponential expansion
2
2
2Ht
ds ≈ dt − e
dx
2
ε1 → H=
ȧ
≈ const. ⇒
a
• Negative pressure
1
P
w≡ <−
ρ
3
ä
ρ + 3P
=−
>0 ⇒
a
6Mpl2
• Constant density
d ln ρ d ln a = 2ε < 1
H 2 ∝ ρ → 2H Ḣ ∝ ρ̇ ⇒
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ä > 0
a(t) = eHt
ρ + 3P < 0
ε=−
1 ρ̇
2 Hρ
ε<1
2.3 THE PHYSICS OF INFLATION
Inflation occurs: ε = −
Inflation lasts:
η=
d ln H
Ḣ
=
−
< 1, where dN ≡ d ln a = Hdt.
H2
dN
d ln ε
ε̇
=
< 1
dN
Hε
What microphyscis leads to {ε, |η|} < 1?
Scalar Field Dynamics
Inflation is often modelled by the evolution of a scalar field φ (the inflaton)
with energy density V (φ) (the inflaton potential):
The stress-energy tensor associated with the inflaton is
1 αβ
Tµν = ∂µ φ∂ν φ − gµν
g ∂α φ∂β φ − V (φ)
2
Let us evaluate this for a homogeneous field φ = φ(t):
1 2
φ̇ + V (φ) (= KE + PE)
2
1
1
Pφ ≡ − T i i = φ̇2 − V (φ) (= KE − PE)
3
2
ρφ ≡ T 0 0
=
7
We then feed this into the Friedmann equations
r
1
ρφ
1 2
~c
=
φ̇ + V
(F1 )
Mpl ≡
H2 =
2
2
3Mpl
3Mpl 2
8πG
1 φ̇2
ρφ + Pφ
= −
Ḣ = −
2Mpl2
2 Mpl2
(F2 )
Take a time derivative of (F1 )
i
1 h
0
φ̇φ̈ + V φ̇ ,
2H Ḣ =
3Mpl2
where V 0 ≡
dV
dφ
and use (F2 ) to get the Klein-Gordon equation:
φ̈
+
ACCELERATION
FRICTION
The ratio of (F2 ) and (F1 ) gives
V0
= −
3H φ̇
(KG)
FORCE
ε=
1 2
2 φ̇
Mpl2 H 2
< 1.
Inflation occurs if the KE is small = slow-roll inflation
For inflation to last the acceleration should be small: δ ≡ −
It is easy to show that η =
ε̇
= 2(ε − δ).
Hε
The conditions {ε, |δ|} 1 imply {ε, |η|} 1.
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φ̈
< 1
H φ̇
So far, this was exact. Now, we make the slow-roll approximation:
1) ε =
1 2
2 φ̇
Mpl2 H 2
2) |δ| =
V
3Mpl2
1
⇒
H2 ≈
|φ̈|
1
H|φ̇|
⇒
3H φ̇ ≈ −V 0
(F)SR
(KG)SR
(KG)SR
↓
0 2
1 2
φ̇
2
M
V
pl
• Hence, we find ε = 22 2 ≈
≡ v
Mpl H
2
V
↑
(F)SR
• Next, we consider
d
(KG)SR ⇒ 3Ḣ φ̇ + 3H φ̈ = −V 00 φ̇, which leads to
dt
ε + δ ≈ Mpl2
V 00
≡ ηv
V
Successful SR inflation occurs when {v , |ηv |} 1 :
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The total amount of inflation is
Z
Ntot =
aE
Z
d ln a =
aI
Using Hdt =
tE
Hdt
tI
1 dφ
H
1 |dφ|
dφ = ± √
≈√
, we find
2v Mpl
2ε Mpl
φ̇
Z
φE
Ntot =
φI
1 |dφ|
√
& 60
2v Mpl
to solve the horizon problem.
Inflation ends when the energy becomes dominated by kinetic energy:
The field starts oscillating
hPφ i ≈ 0 ⇒ hρφ i ∝ a−3
and behaves like matter:
This energy needs to be converted to
= reheating
Standard Model degrees of freedom
This initiates the Hot Big Bang.
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