An Introduction to Inflation

Page 1

String Journal Club An Introduction To Inflation Johar M. Ashfaque 1

The Standard Model of Cosmology

1.1

Einstein’s Equations In Empty Space

Cosmological Principle − On large distances the universe is homogeneous and isotropic. The FRW metric reads

ds2

=

gµν dxµ dxν

=

−dt2 + a2 (t)

dr2 2 2 2 2 + r dθ + sin θdφ 1 − Kr2

where t is the cosmic time, a(t) is the scale factor and K is the spatial curvature which defines the geometry of the universe where positive corresponds to the closed universe, zero curvature leads to the flat universe and the negative corresponds to the open one. Einstein equations 1 Gµν ≡ Rµν − gµν R = 8πGTµν − Λgµν 2 where Rµν is the Ricci tensor, R is the Ricci scalar, Tµν is the energy-momentum tensor, G is the gravitational constant, and Λ is the cosmological constant. In empty space, the energy-momentum tensor vanishes and we have that 1 Rµν − gµν R + Λgµν = 0 2 which on multiplying with g µν and using the fact that g µν gµν = n for n is the dimension of space-time, we obtain 2n R= Λ. n−2 Note.

nR + nΛ = 0 ⇒ R = R− 2

2n 2n − Λ⇒R= Λ. 2−n n−2

Now going back to the equation 1 Rµν − gµν R + Λgµν = 0 2 and substituting in R=

2n Λ n−2

gives

Rµν

n Λ−Λ = gµν n−2 nΛ − nΛ + 2Λ = gµν n−2 2Λ = gµν . n−2 1

1.2

The Perfect Fluid

The energy-momentum tensor of a perfect fluid takes the form T 00

= ρ

T ij

= pgij

or more precisely 

T µν

ρ 0  0 −p =  0 0 0 0

0 0 −p 0

 0 0  . 0  −p

The fully covariant expression for the components of the energy-momentum tensor of a perfect fluid in an arbitrary coordinate system reads p µν T = ρ + 2 uµ uν − pg µν . c Note. The energy-momentum tensor T µν is symmetric and is made up from two scalar fields ρ and p and the vector field u that characterise the perfect fluid.

1.3

FRW Elements

The cosmological principle states that any particular time, the universe looks the same from all positions in space and all the directions in space at any point are equivalent.

1.4

Maximally Symmetric 3-Space

A maximally symmetric space is specified by just one number, the curvature κ which is independent of the coordinates. Such a constant curvature space is clearly homogeneous and isotropic. The curvature tensor Rijkl has six independent components defined as Rijkl = κ(gik gjl − gil gjk )

1.5

The FRW Metric

The Ricci tensor is Rjk

= g il Rijkl = −2κgjk

with the Ricci scalar −2κg jk gjk = −2κ · (3) = −6κ. Consider the line element dσ 2 = B(r)dr2 + r2 dθ2 + r2 sin2 θdφ2 where B(r) is an arbitrary function of r and gθθ = r2 .

grr = B(r), Then given

Rij = −2κgij and Rrr = −

1 dB(r) , rB(r) dr

Rθθ = 2

1 r dB(r) − −1 2 B(r) 2B (r) dr

we have

1 dB(r) 1 = 2κB(r) ⇒ B = rB(r) dr C − κr2

and 1+

r dB(r) 1 r − = 2κr2 ⇒ 1 + (2κr) − C = κr2 ⇒ C = 1, 2B 2 (r) dr B(r) 2

where by making use of the quotient rule we had 2κr dB(r) . = dr (C − κr2 )2 Hence dσ 2 = B(r)dr2 + r2 dθ2 + r2 sin2 θdφ2 ⇒ dσ 2 =

1.6

Computing Rrr Rrr

1.7

= ∂r Γkrk − ∂k Γkrr + Γlrk Γklr − Γlrr Γklk 1 1 2 1 dB 1 dB 2 1 dB 1 dB = ∂r + + − ∂r + 2− + 2B dr r r 2B dr r 2B dr 2B dr r 1 dB 2 = − 2B dr r 1 dB = − rB dr

Computing Rθθ Rθθ

1.8

1 dr2 + r2 dθ2 + r2 sin2 θdφ2 . 1 − κr2

= ∂θ Γkθk − ∂k Γkθθ + Γlθk Γklθ − Γlθθ Γklk 1 2 r 2 r r 1 dB = ∂θ + cot θ − ∂r − + + cot2 θ + − − r B r B B 2B dr r r dB 2 r dB 2 1 + cot2 θ − + + = − cosec2 θ + − 2 2 dr B B dr B 2B B 1 r − 2r dB = −1+ B 2B 2 dr 1 r dB = −1− B 2B 2 dr

The Inflaton Potential & The Klein-Gordon Equation

The Friedmann equation reads H2 =

8πG ρ 3

H2 =

ρ 3MP2 l

which simplifies to

after making use of the reduced Planck mass r MP l = instead of the Newton’s constant G.

3

~c 8πG

Combining Friedmann’s equation and the fluid equation (continuity equation) H2

=

ρ̇ + 3H(ρ + P )

=

ρ 3MP2 l 0

we have the second Friedmann equation Ḣ + H 2

1 (ρ + 3P ) 6MP2 l 3P H2 = − 1+ . 2 ρ

= −

For ρ

=

P

=

1 2 φ̇ + V (φ) 2 1 2 φ̇ − V (φ) 2

Substituting ρ into the Friedmann equation yields H2 =

1 2 1 φ̇ + V (φ) 3MP2 l 2

and taking the time derivative gives 1 0 2ḢH = φ̇φ̈ + V φ̇ . 3MP2 l Substituting ρ and P into the second Friedmann equation, we obtain Ḣ =

−(ρ + P ) −(φ̇2 ) = . 2MP2 l 2MP2 l

Ḣ =

−(ρ + P ) −(φ̇2 ) = . 2 2MP l 2MP2 l

By substitution of

into

1 0 2ḢH = φ̇φ̈ + V φ̇ . 3MP2 l

gives us the Klein-Gordon equation φ̈ + 3H φ̇ + V 0 (φ) = 0.

1.9

The Slow-Roll Regime

The potential must dominate over kinetic energy 1 2 φ̇ < V (φ). 2 In the limit of vanishing kinetic energy the acceleration is exponential and never ending.

4

1.9.1

Derivations

We consider

1 2 φ̇ V (φ). 2 Moreover, if the damping the KG equation is large, the acceleration term can be neglected when compared to the friction term Ď†Ěˆ 3H φ̇. This enforces the slow-roll regime where 2

=

8πG 1 2 φ̇ + V (φ) 3 2

Ď†Ěˆ + 3H φ̇ + V 0 (φ)

=

0

H

simplify to H2

8Ď€G V (φ) 3 −V 0 (φ).

'

3H φ̇ '

We now wish to recast these equations. We begin by replacing φ̇ ' −

1 0 V (φ) 3H

in the condition 12 φ̇2 V (φ) to obtain 2 1 V 0 (φ) V 2 3H

1 V 0 (φ)2 V 2 9( 8πG 3 V (φ)) 2 MP2 l V 0 (φ) 1. 6 V (φ)

�⇒ �⇒

Similarly, the condition on acceleration Ď†Ěˆ 3H φ̇ is rewritten as

Ď†Ěˆ = ∂t (φ̇) ' ∂t

V 0 (φ) 3H

=

V 00 (φ) φ̇ 3H

yielding V 00 (φ) φ̇ 3H φ̇ 3H

1 V 00 (φ) 1 9H 2 1 V 00 (φ) 1 3(8πG)V (φ) MP2 l V 00 (φ) 1 3 V (φ)

�⇒ �⇒ �⇒

Defining the potential slow-roll parameters M2 V ≥ P l 6 and ΡV ≥

V 0 (φ) V (φ)

2

MP2 l V 00 (φ) 3 V (φ) 5

then the conditions determining the range of validity of the slow-roll regime are simply V 1,

|ΡV | 1.

The parameters constrain the potential to be almost flat such that the inflaton field slowly rolls down the potential. As long as the these parameters are small the space-time inflates (technically speaking the universe will inflate if the condition V 1 is met) and is approximately de Sitter a(t) âˆź eHt .

1.10

The Conditions of Slow-Roll Inflation Summarized

1.10.1

The First Condition

Motion of the inflaton is overdamped so that the force V 0 (φ) in Ď†Ěˆ + 3H φ̇ + V 0 (φ) = 0,

Fluid Equation

balances the friction term 3H φ̇ such that φ̇ ' −

1 0 V (φ). 3H

This condition is referred to as the attractor solution. 1.10.2

The Second Condition M2 V ≥ P l 6

V 0 (φ) V (φ)

2 1

which means that φ̇ < V (φ) is well satisfied and H2 '

1.10.3

κ2 V (φ). 3

The Third Condition |ΡV | 1

for ΡV ≥

1.11

MP2 l V 00 (φ) . 3 V (φ)

The Boltzmann Equation ȧ dn + 3Hn = 0, H = dt a

which for 1+2 3+4 can be expressed as 1 d(na3 ) = âˆ’Îąn1 n2 + βn3 n4 a3 dt

6

2

The Relativistic Limit: T m or x → 0

At early times

µ 1 T

for all particles. Set µ ≡ 0. Then n

=

ρ

=

∞

Z

1 2π 2

dp 0

exp ∞

Z

1 2π 2

√

p2 p2 +m2 T

±1

p

dp 0

p2 p2 + m2 √ p2 +m2 exp ±1 T

where +1 is used for fermions and −1 is used for bosons. Define x≡

m , T

so that n= where

1 3 T I± (x) 2π 2

∞

Z I± (x) ≡

dξ 0

and ρ= where

∞

Z J± (x) ≡ 0

p T

ξ≡

ξ2 p exp[ ξ 2 + x2 ] ± 1

1 4 T J± (x) 2π 2 p ξ 2 ξ 2 + x2 p dξ . exp[ ξ 2 + x2 ] ± 1

The functions I± (x) and J± (x) have analytic expansions in certain limits.

2.1

Bosons In The Relativistic Limit: Number Density

For bosons, we have

∞

Z I− (0) =

dξ 0

eξ

ξ2 −1

which is a standard integral giving I− (0) = 2ζ(3) since ζ(x) =

1 Γ(x)

Z

∞

du 0

ux−1 eu − 1

and making use of the fact that Γ(3) = (3 − 1)! = 2.

2.2

Fermions In The Relativistic Limit: Number Density

For fermions, we have

∞

Z I+ (0) =

dξ 0

where we will have to use

eξ

ξ2 +1

1 1 2 = ξ − eξ + 1 e − 1 e2ξ − 1

7

in order to obtain I+ (0)

2.3

=

2ζ(3) −

=

3 ζ(3) 2

1 4

Z

∞

u2 du −1

eu

0

Bosons In The Relativistic Limit: Energy Density

For bosons, we have

∞

Z J− (0) ≡

dξ 0

eξ

ξ3 . −1

Again, this is a standard integral so J− (0) =

2.4

π4 15

Fermions In The Relativistic Limit: Energy Density

For fermions, we have

∞

Z J+ (0) ≡

dξ 0

ξ3 eξ − 1

where we yet again use the fact that 1 1 2 = ξ − eξ + 1 e − 1 e2ξ − 1 in order to obtain J+ (0)

Z π4 1 ∞ u3 − du 15 8 0 eu − 1 4 7 π 8 15

= =

2.5

Number Density, Energy Density and Pressure: Radiation

2.6

Entropy Density

In the relativistic limit, T m and the energy density takes the form π2 g∗S T 4 ρ= 30

(

1 7 8

Bosons Fermions

where g∗S denotes the effective number of relativistic degrees of freedom. Pressure is then given by P

= =

π2 g∗S T 4 90 ζ(4) g∗S T 4 π2

Then entropy density s is given by the relation s≡

S ρ+p = V T 8

which leads to s = = =

π 2 4 g∗S T 4 30 3 T 4 ζ(4)g∗S T 3 π2 2 2 ζ(4)g∗S T 3 π

where g∗S =

X bosons

gi

Ti T

3 +

9

7 8

X f ermions

gi

Ti T

3 .

3

The Early Universe & Cosmological Phase Transitions

3.1

Shortcomings of Standard Cosmology

The standard model of cosmology explains the Hubble expansion law successfully, the cosmic background radiation and the abundance of the light elements whilst providing us with a framework to understand the large-scale structure formation. However, if the standard model of cosmology is extrapolated all the way back to the singularity at time t = 0 we run into difficulties as we are compelled to assume a very special initial state. Here, we give a list of outstanding problems • The Horizon Problem - This problem stems from the large-scale homogeneity and isotropy of the universe and in particular of the background cosmic radiation. • The Flatness Problem - The density of the present universe is within one order of magnitude of the critical density. • The Density Fluctuation Problem - It is now generally believed that galaxies and clusters of galaxies evolved by gravitational instability from small density fluctuations in the early universe. The origin of these tiny fluctuations is mysterious in the standard cosmology and they simply have to be postulated. • The Thermal State Problem - The initial thermal state postulated in the standard model is also inexplicable because for temperatures T > 1016 GeV the expansion of the universe is too fast for thermal equilibrium to be established. • The Singularity Problem - The cosmological singularity at t = 0 corresponds to a state of infinite energy density and signals the breakdown of classical general relativity. • The Cosmological Constant Problem - Empirical evidence shows that Λ is over 10120 times smaller than the Planck scale. In addition to these problems, we have a few more concerning the matter content of the universe • The Baryon Asymmetry Problem - The creation of more matter than anti-matter in the early universe requires some out-of-equilibrium process with baryon-number violating interactions. • The Dark Matter Problem - Many candidates have been suggested but the exact nature of dark matter remains unknown. • The Exotic Relics Problem - Phase transitions and other processes in the early universe can be expected to create topological defects, exotic particles and small black holes. To give an example, look at the large overproduction of monopoles which appears to be an inevitable predictions of the GUT models.

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