computational physics and astrophysicslaguna.gatech.edu/compphys/lectures/inflation.pdfthe cosmic...

Computational Physics and AstrophysicsCosmological Inflation

Kostas KokkotasUniversity of Tubingen, Germany

and

Pablo LagunaGeorgia Institute of Technology, USA

Spring 2012

Kokkotas & Laguna Computational Physics and Astrophysics

Our Universe


Cosmic Expansion

Co-moving coordinates expand at exactly the same rate as theuniverse.To a good approximation, galaxies following the cosmologicalflow do not change their co-moving coordinate location.Therefore, the co-moving distance between galaxies is constant.Physical distances on the other hand change because of thecosmological expansion.


Cosmic Expansion

Co-moving distance between A and B: (~xA − ~xB)t = (~xA − ~xB)t+∆t

Physical distance between A and B:(~rA −~rB

)t 6=

(~rA −~rB

)t+∆t

From homogeneity and isotropy in the universe, we have that:

~rA(t)−~rB(t) = a(t) (~xA − ~xB)

Thus, the scale or expansion factor a(t) encapsulates thedynamics of the universe.


Evolution of the Scale Factor

Einstein Equations: Gµν = 8π Tµν that is Geometry =Matter-Energy.Conservation of Matter-Energy: ∇νTµν = 0From Einstein’s equations(

aa

)2

+ka2 =

8π3ρ Friedmann equation

with ρ the energy density and k the spatial curvature of theuniverse, respectively.From conservation of matter-energy

ρ = −3aa

(ρ+ p)

with p the pressure in the universe.For matter dominated ρ ∝ a−3 and for radiation dominatedρ ∝ a−4


Curvature

The second term in theFriedmann equation is acurvature term

H2 +ka2 =

8π3ρ

where H ≡ a/a.k = 0 (flat): Flat Euclideanspace.k = +1 (closed): Geometryof a three-sphere.k = −1 (open): Geometryof a three-hyperboloid


Friedmann Equation

Recall

H2 +ka2 =

8π3ρ

If k = 0, then

H2 =8π3ρc

with ρc called the critical densityThen

H2 +ka2 =

8π3ρ

1 +k

a2H2 =8π

3 H2 ρ

1 +k

a2H2 =ρ

ρc= Ω density parameter


The Cosmic Microwave Background Radiation


Cosmic Microwave Background Anisotropies

∆TT

= 10−5

Ω = 1


Inflation

The universe seems to have emerged from a very special set ofinitial conditionsA set of initial conditions fined tuned to be patially flat Ω = 1 andhighly homogeneous/isotropic ∆T/T = 10−5

Is there a mechanism that could take a wide spectrum of initialconditions and evolve them toward flatness andhomogeneity/isotropy?The answers is YES.The inflationary universe scenario provides such mechanism.


Flatness Problem

The density of the universe ρ seems to be finely tuned to beequal to the critical density ρc , that is Ω = 1.In other words, the universe seems to be fine-tuned to be flat,that is k = 0.How natural are these values?Recall Friedmann equation

1 +ka2

1H2 = Ω

1 +ka2

38πρc

= Ω

1 +ka2

38πρ

ρ

ρc= Ω

1 +3 k8π

1a2ρ

Ω = Ω

(Ω−1 − 1)ρa2 = −3 k8π

= const


Flatness Problem

Given

(Ω−1 − 1)ρa2 = −3 k8π

= const

|Ω−10 − 1|ρ0 a2

0 = |Ω−1 − 1|ρa2 = const

|Ω−10 − 1| = |Ω−1 − 1| ρ

ρ0

a2

a20

But

ρ = ρ0

(a0

a

)3

thus

|Ω−10 − 1| = |Ω−1 − 1|a0

a


Flatness Problem

Consider near the Big Bang ( a0/aBB = 1062 at the Planck epoch) asmall deviation of ΩBB from unity; that is, |Ω−1

BB − 1| = ε with |ε| 1.Thus

|Ω−10 − 1| = |Ω−1

BB − 1| a0

aBB

|Ω−10 − 1| = ε

a0

aBB

|Ω−10 − 1| = 1062ε

In order to have today a small deviation δ = |Ω−10 − 1| 1, we require

ε = 10−62δ


Particle Horizon

Horizons exist because there is finite amount of time since theBig Bang.The particle horizon is the maximum, finite distance from whichparticles (or photons) could have traveled to the observer in thistime.It represents the boundary between the observable and theunobservable regions of the universe.


The particle horizon is calculated from

rh(t) =

∫ t

0

dta

Therefore, it depends on the scale factor and thus the matter contentof the universe.


The particle horizon now is much larger than the particle horizonwhen the CMB photons where emitted.That is, two widely separated parts of the CMB will havenon-overlapping horizons.Horizon Problem: How come then we see them at almost thesame temperature.We need a much larger particle horizon when CMB photons areemitted to bring the entire visible universe in causal contact.


Flatness and Horizon Problems Solution

In the very early universe,We need to drive Ω→ 1

Ω = 1 +k

H2 a2

We need to increase the way particle horizon grows.

rh(t) =

∫ t

0

dta

Therefore, we change how H and thus a evolve.Recall that from Friedmann equation

H2 =8π3ρ ∝ a−n

with n = 3,4As the universe expands, H2 a2 decays, and thus 1/H2 a2 grows(Flatness problem).


A possible solution (Guth) is that H = a/a = constant > 0. That isa ∝ eH t . Then

Ω = 1 +k

H2 e2 H t → 1

and

rh(inflation) =

∫ tend

rinit

dta∝ eH ∆t rh(today)

physical coordinates (left), co-moving coordinates (right)


Scalar Fields and Inflation

Consider a Universe filled with a scalar field φ.Ignoring the curvature term, Friedmann equation reads

H2 =8π3ρ with ρ =

12φ2 + V (φ)

and V (φ) a potential to be determined.The equation for the dynamics of the scalar field is (assuming ahomogenous field)

φ+ 3 H φ+dVdφ

= 0

That is, the equations for a and φ are:

φ+ 3 H φ+dVdφ

= 0

H2 =8π3

(12φ2 + V

)Kokkotas & Laguna Computational Physics and Astrophysics

The onset of Inflation

Recall that we need a type of matter such that ρ ≈ constant, soH ≈ constant and then a ∝ eHt

Therefore, we require that

φ2 V

|φ| |3 H φ|,∣∣∣∣dVdφ

∣∣∣∣which is equivalent to requiring that the potential energydominates over the kinetic energy (slow-roll approximation;).


The end of Inflation

At the bottom of the potential V ≈ 0

The field relaxes, converting the energy in the inflation potentialinto a thermalized gas of matter and radiation (reheating)


Phase Transitions in the Early Universe

First-order phase transition via bubble nucleation (i.e. boiling water)


Phase Transitions in the Early Universe

Second-order phase transition, the old phase transforms itself into thenew phase in a continuous manner.


Domain WallsRecall (one-dimensional case)

φ+ 3 H φ− ∂2xφ+

dVdφ

= 0

H2 =8π3

(12φ2 + V

)

with a Mexican Sombreropotential

V (φ) =λ

8(φ2 − η2)2


Domain Walls

Consider the static case solution

d2

dx2φ =λ

2φ(φ2 − η2)

which has the following solution

φ(x) = η tanh

[√λη

2(x − x0)

]


Domain Walls

Domain Wall: the boundary between regions with φ = ±η. Theenergy at the wall is V (0) = λ

8 η4


Project

Solve the equation

φ+ 3 H φ− ∂2xφ+

dVdφ

= 0

with

H2 =8π3

(12φ2 + V

)V (φ) =

λ

8(φ2 − η2)2

Impose periodic boundary conditions in a computation domain oflength L = 1024 and grid-spacing dx = 1.Set the values of the parameters λ and η to unity.In calculating H2, use the average values of φ and φ over theentire computational domain.Initial conditions:

φ = 0.01 ξ and φ = 0.01 ξ

with ξ ∈ [−1,1] a random number.Kokkotas & Laguna Computational Physics and Astrophysics

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