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EL UNIVERSO INFLACIONARIO

César Alonso Valenzuela Toledo

[email protected]

Universidad del Valle

Barranquilla - 26/10/2015

(2)

Problemas del Modelo Cosmológico Estándar

¿Por qué es plano?

Problema de Planitud:

2.

d

dt (aH)

1

> 0

(Materia ordinaria)

|⌦

BBN

1 |  10

16

|⌦

GUT

1 |  10

55

|⌦

pl

1 |  10

61

Actualmente el Universo es

prácticamente plano:

|⌦ 1 | ⇠ 10

2

|⌦ 1 | / t

2(1 n)

n = 2/3 (MD) n = 1/2 (RD)

{

Ecuación de

Friedmann:

⌦ 1 = K

a

2

H

2

(3)

Problemas del Modelo Cosmológico Estándar

Problema de Horizonte:

2.

r

p

(⌧ ) =

Z

t ti

dt

0

a(t

0

) =

Z

ln a ln ai

(aH)

1

d ln a

(aH)

1

= H

0 1

a

12 (1+3!)

Para un universo dominado por

un fluido con ecuación de estado

: P = !⇢

Se obtiene:

Radio comóvil de Hubble (Horizonte)

r

p

(t) = 2H

0 1

1 + 3! a(t)

12(1+3!)

= 2

1 + 3! (aH)

1

r

p

(a =) = 2H

0 1

1 + 3!

⇣ a

12(1+3!)

a

i12(1+3!)

⌘ ⌧ ⌧

i i ⌘ 2H0 1

1 + 3! ai12(1+3!)

i ! 0 Si ti ! 0

a ! 0, ! > 1

3

(4)

En el desacople habían regiones causalmente desconectadas y por lo tanto nunca estuvieron en contacto

⇠ 104

10 1. Classical Dynamics of Inflation

1.2.3 Shock in the CMB

A moment’s thought will convince the reader that the finiteness of the conformal time elapsed between ti = 0 and the time of CMB decoupling trec implies a serious problem: most spots in the CMB have non-overlapping past light-cones and hence never were in causal contact (see fig. 1.1). Why aren’t there order-one fluctuations in the CMB temperature?

Past Light-Cone

Recombination

Particle Horizon Conformal Time

Last-Scattering Surface

Big Bang Singularity

rec

0

i = 0

Figure 1.1: Conformal diagram for the standard FRW cosmology.

CMB correlations. Let us compute the angle subtended by the comoving horizon at recombination.

This is defined as the ratio of the comoving particle horizon at recombination and the comoving angular diameter distance from us (an observer at redshift z = 0) to recombination (z ' 1090) (cf. fig. 1.1)

hor = dhor

dA . (1.2.14)

A fundamental quantity is the comoving distance between redshifts z1 and z2

2 1 = Z z2

z1

dz

H(z) ⌘ I(z1, z2) . (1.2.15)

The comoving particle horizon at recombination is

dhor = ⌧rec i ⇡ I(zrec,1) . (1.2.16) In a flat universe, the comoving angular diameter distance from us to recombination is

dA = ⌧0 rec =I(0, zrec) . (1.2.17) The angular scale of the horizon at recombination therefore is

hor dhor

dA = I(zrec,1)

I(0, zrec) . (1.2.18)

Using

H(z) = H0 q

m(1 + z)3+ ⌦ (1 + z)4+ ⌦, (1.2.19)

Problemas del Modelo Cosmológico Estándar

✓ ⇠ 2.3

o

No hay correlación.

Anisotropías:

T /T ' 10

5

T = 2.725 ± 0.002 K

¿Por qué el Universo

es tan uniforme?

(5)

¿Quién disparó la expansión?

LSS + RCF :

dec

⇠ T

T

CMB

⇠ 10

5

¿Cuál es el origen de ? ⇢

Problemas del Modelo Cosmológico Estándar

Reliquias no deseadas:

3.

Densidad inicial:

4.

Impulso inicial:

5.

Monopolos magnéticos y otros defectos topológicos.

¿Por qué no se observan?

Filosófico: Condiciones iniciales.

(6)

Inflación

Hasta el momento hemos considerado un universo cuyo contenido energético cumple con la condición fuerte de energía, es decir materia ordinaria.

! > 1/3 (aH)

1

> 0 d

2

a

dt

2

< 0

¡Puede ser la causa de nuestros problemas!

¿Qué pasa si:

d

2

a

dt

2

> 0 (aH)

1

< 0

! < 1/3 ?

i

! 1

i ⌘ 2H0 1

1 + 3! ai12(1+3!)

(7)

Inflación

12 1. Classical Dynamics of Inflation

Past Light-Cone

Recombination

Particle Horizon

Conformal Time

Last-Scattering Surface

Big Bang Singularity Reheating

causal contact

Inflation

rec

0

0

i = 1

Figure 1.2: Conformal diagram for inflationary cosmology.

This implies that there was much more conformal time between the singularity and decoupling than we had thought! Fig. 1.2 shows the new conformal diagram. The past light cones of widely separated points in the CMB now had time to intersect before the time ⌧ = 0. In inflationary cosmology, ⌧ = 0 isn’t the initial singularity, but instead becomes the time of reheating. There is time both before and after ⌧ = 0.

A decreasing comoving horizon means that large scales entering the present universe were inside the horizon before inflation (see fig. 1.3). Causal physics before inflation therefore had time to establish spatial homogeneity. With a period of inflation, the uniformity of the CMB is not a mystery anymore.

1.3.2 Solution of the Flatness Problem

In foonote 3, I advertised that any solution to the horizon problem also solves the other Big Bang puzzles. Let me therefore demonstrate that a shrinking Hubble sphere indeed solves the flatness problem.

Consider adding spatial curvature to the Friedmann equation (1.2.2), H2 =

3Mpl2

k

a2 . (1.3.25)

Dividing both sides by the Hubble parameter, we can write this as 1 ⌦(a) = k

(aH)2 , (1.3.26)

Solución problema de horizonte:

1.

(8)

Inflación 34 2. Inflation

• if > ph, then the particles could never have communicated.

• if > (aH) 1, then the particles cannot talk to each other now.

Inflation is a mechanism to achieve ph (aH) 1. This means that particles can’t communi- cate now (or when the CMB was created), but were in causal contact early on. In particular, the shrinking Hubble sphere means that particles which were initially in causal contact with another—i.e. separated by a distance < (aIHI) 1—can no longer communicate after a suf- ficiently long period of inflation: > (aH) 1; see fig. 2.4. However, at any moment before horizon exit (careful: I really mean exit of the Hubble radius!) the particles could still talk to each other and establish similar conditions. Everything within the Hubble sphere at the beginning of inflation, (aIHI) 1, was causally connected.

Since the Hubble radius is easier to calculate than the particle horizon it is common to use the Hubble radius as a means of judging the horizon problem. If the entire observable universe was within the comoving Hubble radius at the beginning of inflation—i.e. (aIHI) 1 was larger than the comoving radius of the observable universe (a0H0) 1—then there is no horizon problem.

Notice that this is more conservative than using the particle horizon since ph(t) is always bigger than (aH) 1(t). Moreover, using (aIHI) 1 as a measure of the horizon problem means that we don’t have to assume anything about earlier times t < tI.

time scales

reheating

inflation Big Bang

standard Big Bang inflation

Figure 2.4: Scales of cosmological interest were larger than the Hubble radius until a ⇡ 10 5 (where today is at a(t0) ⌘ 1). However, at very early times, before inflation operated, all scales of interest were smaller than the Hubble radius and therefore susceptible to microphysical processing. Similarly, at very late times, the scales of cosmological interest are back within the Hubble radius. Notice the symmetry of the inflationary solution. Scales just entering the horizon today, 60 e-folds after the end of inflation, left the horizon 60 e-folds before the end of inflation.

Duration of inflation.—How much inflation do we need to solve the horizon problem? At the very least, we require that the observable universe today fits in the comoving Hubble radius at the begin- ning of inflation,

(a0H0) 1 < (aIHI) 1 . (2.2.12) Let us assume that the universe was radiation dominated since the end of inflation and ignore the relatively recent matter- and dark energy-dominated epochs. Remembering that H / a 2 during radiation domination, we have

a0H0

aEHE a0 aE

aE a0

2

= aE

a0 T0

TE ⇠ 10 28 , (2.2.13)

Las regiones causalmente conectadas

crecen exponencialmente.

(9)

Inflación

Expansión cuasi-de Sitter :

H ' cte ! a(t) ↵ e

Ht

|⌦

FI

1 |↵ exp( 2H t) ! |⌦ 1 | ! 0

Entonces la solución es un atractor. ⌦ = 1

Solución problema de planitud:

2.

Los problemas del modelo cosmológico estándar

se solucionan si en un principio incluimos una época de expansión acelerada. Este periodo recibe el nombre de Inflación.

Conclusión:

(10)

Inflación

Condiciones para Inflación:

2. Expansión acelerada:

3. Parámetro de Hubble lentamente variable:

1. Radio de Hubble decreciente:

d

dt (aH)

1

= 1

a (1 ✏), ✏ ⌘ H ˙ H

2

⌘ ⌘ ˙✏

H✏ , |⌘| < 1 N ⇠ 40 60 e-folds

N ⌘ d ln a = Hdt Número de e-folds

d

dt (aH)

1

= a ¨

˙a

2

a > 0 ¨ d

dt (aH)

1

< 0

✏ = H ˙

H

2

= d ln H

dN < 1

(11)

Inflación

4. Expansion cuasi-de Sitter:

5. Presión Negativa:

¨ a

a = ˙ H + H

2

= 4⇡G

3 (⇢ + 3P ) = H

2

2

1 + 3 P

! = P

⇢ < 1 3 H ⇠ const.

a = a 0 e Ht

✏ = 3 2

1 + P

< 1 ,

P < 0

(12)

La Física de Inflación

S = Z

d

4

x p

g

✓ 1

2 M

pl2

R 1

2 g

µ⌫

@

µ

@

V ( )

Escalar de Ricci Potencial Inflatón:

= (t)

Modelo simple:

37 2. Inflation

Figure 2.5: Example of a slow-roll potential. Inflation occurs in the shaded parts of the potential.

The stress-energy tensor of the scalar field is

4

T

µ⌫

= @

µ

@

g

µ⌫

✓ 1

2 g

@

@ V ( )

. (2.3.26)

Consistency with the symmetries of the FRW spacetime requires that the background value of the inflaton only depends on time, = (t). From the time-time component T

00

= ⇢ , we infer that

⇢ = 1

2 ˙

2

+ V ( ) . (2.3.27)

We see that the total energy density, ⇢ , is simply the sum of the kinetic energy density,

12

˙

2

, and the potential energy density, V ( ). From the space-space component T

ij

= P

ji

, we find that the pressure is the di↵erence of kinetic and potential energy densities,

P = 1

2 ˙

2

V ( ) . (2.3.28)

We see that a field configuration leads to inflation, P <

13

⇢ , if the potential energy dominates over the kinetic energy.

Next, we look in more detail at the evolution of the inflaton (t) and the FRW scale factor a(t). Substituting ⇢ from (2.3.27) into the Friedmann equation, H

2

= ⇢ /(3M

pl2

), we get

H

2

= 1 3M

pl2

 1

2 ˙

2

+ V . (F)

Taking a time derivative, we find

2H ˙ H = 1 3M

pl2

h ˙ ¨ + V

0

˙ i

, (2.3.29)

where V

0

⌘ dV/d . Substituting ⇢ and P into the second Friedmann equation ( 2.2.21), H = ˙ (⇢ + P )/(2M

pl2

), we get

H = ˙ 1 2

˙

2

M

pl2

. (2.3.30)

Notice that ˙ H is sourced by the kinetic energy density. Combining (2.3.30) with (2.3.29) leads to the Klein-Gordon equation

¨ + 3H ˙ + V

0

= 0 . (KG)

4You can derive this stress-energy tensor either from Noether’s theorem or from the action of a scalar field.

You will see those derivations in the QFT course: David Tong, Part III Quantum Field Theory.

M

pl

=

r 1

Masa de Planck:

8⇡G

(13)

La Física de Inflación

T

µ⌫

= @

µ

@

g

µ⌫

✓ 1

2 g

@

@ V ( )

Variando la acción respecto al tensor métrico:

T

µ⌫

= 2 p g

@( p

g L

mat

)

@g

µ⌫

R

µ⌫

1

2 Rg

µ⌫

⇤g

µ⌫

= 8GT

µ⌫

L

mat

= L

mat

(g

µ⌫

, x

µ

,

i

, @

µ i

, A

iµ

, @

A

iµ

, . . .)

S

g

µ⌫

= 0 :

Dado que el campo es homogéneo

⇢ = 1

2 ˙

2

+ V ( ) + ( r )

2

2a

2

P = 1

2 ˙

2

V ( ) ( r )

2

6a

2

}

P = 1

2 ˙

2

V ( )

⇢ = 1

2 ˙

2

+ V ( )

(14)

La Física de Inflación

V ( ) ˙

2

Si P ' ⇢

! = 1

{ cuasi-exponencial Inflación

Conclusión:

Un campo escalar cuya energía es dominante en el universo y cuya energía potencial domina sobre la energía cinética produce Inflación.

Ecuaciones de

Friedmann : { H H = ˙

2

= 3M 1 2 M 1 ˙

pl22pl2

1 2 ˙ + V

Si V ( ) > ˙

2

! < 1 3 Inflación

! = P

⇢ =

1

2

˙

2

V ( )

1

2

˙

2

+ V ( )

(15)

La Física de Inflación

S = 0

Variando la acción respecto al campo:

¨ + 3H ˙ + V

0

( ) = 0

Ecuación de Klein-Gordon:

Aproximación de evolución lenta (slow-roll):

Inflación requiere:

✏ = H ˙

H

2

< 1 ✏ =

1

2

˙

2

M

pl2

H

2

< 1

⇢ = 3M

pl2

H

2

V ( ) > ˙

2

Se cumple si la energía cinética, , sólo hace una pequeña contribución a la densidad de energía total:

1 2 ˙2

(16)

La Física de Inflación

Para el caso de inflación cuasi-exponencial:

V ( ) ˙

2

✏ ⌧ 1 }

Este caso particular se denomina inflación de evolución lenta.

37 2. Inflation

Figure 2.5: Example of a slow-roll potential. Inflation occurs in the shaded parts of the potential.

The stress-energy tensor of the scalar field is 4 Tµ⌫ = @µ @ gµ⌫

1

2g @ @ V ( )

. (2.3.26)

Consistency with the symmetries of the FRW spacetime requires that the background value of the inflaton only depends on time, = (t). From the time-time component T00 = ⇢ , we infer that

⇢ = 1

2 ˙2 + V ( ) . (2.3.27)

We see that the total energy density, ⇢ , is simply the sum of the kinetic energy density, 12 ˙2, and the potential energy density, V ( ). From the space-space component Tij = P ji, we find that the pressure is the di↵erence of kinetic and potential energy densities,

P = 1

2 ˙2 V ( ) . (2.3.28)

We see that a field configuration leads to inflation, P < 13⇢ , if the potential energy dominates over the kinetic energy.

Next, we look in more detail at the evolution of the inflaton (t) and the FRW scale factor a(t). Substituting ⇢ from (2.3.27) into the Friedmann equation, H2 = ⇢ /(3Mpl2 ), we get

H2 = 1 3Mpl2

1

2 ˙2 + V . (F)

Taking a time derivative, we find

2H ˙H = 1 3Mpl2

h ˙ ¨ + V 0 ˙i

, (2.3.29)

where V 0 ⌘ dV/d . Substituting ⇢ and P into the second Friedmann equation (2.2.21), H =˙ (⇢ + P )/(2Mpl2 ), we get

H =˙ 1 2

˙2

Mpl2 . (2.3.30)

Notice that ˙H is sourced by the kinetic energy density. Combining (2.3.30) with (2.3.29) leads to the Klein-Gordon equation

¨ + 3H ˙ + V 0 = 0 . (KG)

4You can derive this stress-energy tensor either from Noether’s theorem or from the action of a scalar field.

You will see those derivations in the QFT course: David Tong, Part III Quantum Field Theory.

¨

Para mantener la anterior condición también es necesario garantizar que el potencial es lo suficientemente plano, esto se logra si es también despreciable.

Ecuación de Friedmann:

H

2

' 1

3M

pl2

V ( )

3H ˙ ' V

0

¨ ⌧ 3H ˙

{

⌘ ¨

⌘ = ˙✏

H ˙

H✏ = 2(✏ ), | | ⌧ 1

(17)

La Física de Inflación

Conclusión:

La inflación de evolución

lenta ocurre si

: {✏, |⌘|} ⌧ 1 { ✏, ⌘

Parámetros de evolución lenta

:

Las ecuaciones dinámicas se simplifican.

{

En términos del potencial:

✏ = 1

2 M

pl2

✓ V

0

V

|⌘| = M

pl2

|V

00

| V

Número de e-folds:

N =

Z

tf

ti

Hdt =

Z

e

i

H

˙ d ' 1 M

pl2

Z

e

i

V

V

0

d

Evolución lenta

(18)

La Física de Inflación

Inflación con campos vectoriales:

S = Z

d

4

x p

g

✓ 1

2 M

pl2

R 1

4 F

µ⌫

F

µ⌫

+ 1 2

m

2

+ R 6

A

µ

A

µ

p 1

g

@

@x

µ

p gF

µ⌫

+

m

2

+ R 6

A

= 0

Variando la acción respecto al campo :

Considerando un campo

cuasi homogéneo : @

i

A

= 0 B ¨

i2

+ 3HB

i

+ m

2

B

i

= 0

B

i

= A

i

a A

0

= 0

{

Mukhanov et al., JCAP 0806 (2006) 009

(19)

La Física de Inflación

Variando la acción respecto a la métrica

se obtiene el tensor energía -impulso T

µ⌫

= . . .

Considerando ( ) vectores aleatoriamente orientados o tres vectores mutuamente ortogonales

N ⇠ 10

2

:

⇢ = 3 2

⇣ B ˙

k2

+ m

2

B

k2

P = 3 2

⇣ B ˙

k2

m

2

B

k2

{

! = B ˙

k2

m

2

B

k2

B ˙

k2

+ m

2

B

k2

! < 1 3 P ' ⇢

{

B ˙

k2

⌧ m

2

B

k2

si

Inflación

Referencias

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