Gravitation: Cosmology - An Introduction to General...

Gravitation: CosmologyAn Introduction to General Relativity

Pablo Laguna

Center for Relativistic AstrophysicsSchool of Physics

Georgia Institute of Technology

Notes based on textbook: Spacetime and Geometry by S.M. CarrollSpring 2013

Pablo Laguna Gravitation: Cosmology

Copernican Principle

Copernican principle:

The universe is pretty much the same everywhere.

There should be no special observers.

The Earth is not in a central, specially position.The Copernican principle applies on the very largest scales, where local variations in density are averagedover.The Copernican principle has been validated by a number of different observations, such as

Number counts of galaxies

Oservations of diffuse X-ray and γ-ray backgrounds.

Cosmic microwave background radiation.

The cosmic microwave background is not perfectly smooth, the deviations from regularity are on the order of10−5


Formally, the Copernican principle is related to isotropy and homogeneity of the universe.

Isotropy: a manifold is iotropic at a given point if it looks the same in every direction. More formally, amanifold M is isotropic around a point p if, for any two vectors V and W in TpM, there is an isometry of Msuch that the pushforward of W under the isometry is parallel with V (not pushed forward).

Homogeneity: the metric is the same throughout the space. That is, given any two points p and q in M,there is an isometry which takes p into q.

Notice a manifold can be homogeneous but nowhere isotropic (such as R× S2 in the usual metric), or itcan be isotropic around a point without being homogeneous (such as a cone).

If a space is isotropic everywhere then it is homogeneous. Likewise, if it is isotropic around one point andalso homogeneous, it will be isotropic around every point.

Observational evidence for isotropy + the Copernican principle implies homogeneity and isotropyeverywhere in space in the universe.

Since isotropy can be viewed as rotational invariance and homogeneity as translational invariance, thus thespace is maximally symmetric; that is, it has the maximum number of Killing vectors.

Notice that the expansion of the universe tells us that the space-time is not homogeneous and isotropic.


Cosmological Models

Observations of distant galaxies show to be receding from us; the universe is apparently not static, butchanging with time.

We need then a cosmological models homogeneous and isotropic in space, but not in time.

That is, we need to foliate the spacetime manifold representing the universe into spacelike slices such thateach slice is homogeneous and isotropic.

That is, we consider our spacetime to be R× Σ, where R represents the time direction and Σ is ahomogeneous and isotropic three-manifold.

Σ must be a maximally symmetric space. That is space has its maximum possible number of Killing vectors.

Therefore the metric takes the form

ds2 = −dt2 + a2(t)γij (u)dui duj.

where t is the timelike coordinate, and (u1, u2, u3) are the spatial coordinates on Σ.

γij is the maximally symmetric metric on Σ.

The function a(t) is known as the scale factor.

The coordinates used here are known as comoving coordinates. That is, the metric is free of cross termsdt dui and the spacelike components are proportional to a single function of t ,

An observer who stays at constant ui is called comoving.

Only a comoving observer will think that the universe looks isotropic.


Recall, for maximally symmetric Euclidean three-metrics γij

(3)Rijkl = k(γikγjl − γilγjk ) ,

with k a constant. Thus the Ricci tensor is then (3)Rjl = 2kγjl and (3)R = 6 k

the metric can be put in the form

dσ2 = γij dui duj = e2β(r)dr2 + r2(dθ2 + sin2θ dφ2)

The components of the Ricci tensor are

(3)R11 =2

r∂1β

(3)R22 = e−2β (r∂1β − 1) + 1

(3)R33 = [e−2β (r∂1β − 1) + 1] sin2θ

Using (3)Rjl = 2kγjl and (3)R = 6 k , one gets

β = −1

2ln(1− kr2) .

This gives the Friedmann-Robertson-Walker metric.

ds2 = −dt2 + a2(t)

[dr2

1− kr2+ r2(dθ2 + sin2

θ dφ2)

]


The Friedmann-Robertson-Walker metric can then be rewritten as ds2 = −dt2 + a2(t) dσ2 where

dσ2 =dr2

1− kr2+ r2(dθ2 + sin2

θ dφ2)

is the metric on Σ.

One still needs to find from the Einstein’s equations an equation that determines the behavior of the scalefactor a(t).

Notice that the substitutions

k → k|k|

r →√|k| r

a → a√|k|

leave the metric invariant. Therefore the only relevant parameter is k/|k|, and there are three cases ofinterest: k = −1, 0, 1.


Case k = 0 (flat): This case corresponds tovanishing curvature on Σ

dσ2 = dr2 + r2dΩ2

which is simply flat Euclidean space. Globally,it could describe R3 or a more complicatedmanifold, such as the three-torusS1 × S1 × S1.

Case k = +1 (closed): This casecorresponds to constant positive curvature onΣ. Defining r = sinχ one gets

dσ2 = dχ2 + sin2χ dΩ2

which is the metric of a three-sphere. Theonly possible global structure is actually thethree-sphere.

Case k = −1 (open): This case correspondsto constant negative curvature on Σ Definer = sinhψ to obtain

dσ2 = dψ2 + sinh2ψ dΩ2


Setting a ≡ da/dt , the Christoffel symbols are given by

Γ011 =

aa

1− kr2Γ0

22 = aar2 Γ033 = aar2 sin2

θ

Γ101 = Γ1

10 = Γ202 = Γ2

20 = Γ303 = Γ3

30 =a

aΓ1

22 = −r(1− kr2) Γ133 = −r(1− kr2) sin2

θ

Γ212 = Γ2

21 = Γ313 = Γ3

31 =1

rΓ2

33 = − sin θ cos θ Γ323 = Γ3

32 = cot θ .

The nonzero components of the Ricci tensor are

R00 = −3a

a

R11 =aa + 2a2 + 2k

1− kr2

R22 = r2(aa + 2a2 + 2k)

R33 = r2(aa + 2a2 + 2k) sin2θ ,

and the Ricci scalar is then

R = 6

a

a+

(a

a

)2

+k

a2


Energy-momentum Tensor

The universe is not empty. We model its matter and energy as a perfect fluid.

The energy-momentum tensor for a perfect fluid is given by

Tµν = (p + ρ)UµUν + pgµν ,

where ρ and p are the energy density and pressure, respectively, as measured in the rest frame of anobserver with 4-velocity Uµ.

In comoving coordiantes the fluid is at rest; thus, Uµ = (1, 0, 0, 0) and the energy-momentum tensor is

Tµν =

ρ 0 0 000 gij p0

or Tµν = diag(−ρ, p, p, p)

Note that T = Tµµ = −ρ + 3p


Conservation of Energy and Equation of State

Consider the zero component of the conservation of energy equation:

0 = ∇µTµ0

= ∂µTµ0 + Γµµ0T 0

0 − Γλµ0Tµλ

= −ρ− 3a

a(ρ + p) .

Introduce the equation of state p = wρ where w is a constant independent of time.

The conservation of energy equation becomes

ρ

ρ= −3(1 + w)

a

a,

which can be integrated to obtainρ ∝ a−3(1+w)

.


Cosmological Models

DUST or MATTER-DOMINATED:

Dust is collisionless, nonrelativistic matter, which obeys w = 0.

Examples include ordinary stars and galaxies, for which the pressure is negligible in comparison with theenergy density.

The energy density in matter falls off as ρ ∝ a−3

Notice that this is just the decrease in the number density of particles as the universe expands since for dustthe energy density is dominated by the rest energy, which is proportional to the number density.

RADIATION-DOMINATED:

Radiation includes actual electromagnetic radiation, or massive particles moving at relative velocitiessufficiently close to the speed of light that they become indistinguishable from photons.

Recall that

Tµν =1

4π(FµλFνλ −

1

4gµνFλσFλσ) .

with the trace given by

Tµµ =1

4π

[FµλFµλ −

1

4(4)FλσFλσ

]= 0 .

Thus, T = Tµµ = −ρ + 3p implies that p = ρ/3 or w = 1/3

Thus, the energy density in radiation falls off as ρ ∝ a−4

Today the energy density of the universe is dominated by matter, with ρmat/ρrad ∼ 106.


VACUUM-DOMINATED:

Introducing energy into the vacuum is equivalent to introducing a cosmological constant.

Einstein’s equations with a cosmological constant are

Gµν = 8πTµν − Λgµν

This is equivalent to the equations with no cosmological constant but an energy-momentum tensor for thevacuum,

T (vac)µν = −

Λ

8πgµν

Recall thatTµν = (p + ρ)UµUν + p gµν

Thus

ρ = −p =Λ

8π

or equivalently w = −1.

Fromρ ∝ a−3(1+w)

.

one has that ρ ∝ constant

Since the energy density in matter and radiation decreases as the universe expands, if there is a nonzerovacuum energy it tends to win out over the long term.


Friedmann Equation and Hubble Parameter

Starting from

Rµν = 8π(

Tµν −1

2gµνT

).

The µν = 00 equation is

−3a

a= 4π(ρ + 3p)

and the µν = ij equations give

a

a+ 2

(a

a

)2

+ 2k

a2= 4π(ρ− p)

It is not difficult to show that these equations yield

Friedmann Equations

a

a= −

4π

3(ρ + 3p)

(a

a

)2

=8π

3ρ−

k

a2

The rate of expansion is characterized by the Hubble parameter, H = a/a

The value of the Hubble parameter at the present epoch is the Hubble constant, H0. Measurements indicatethat H0 = 70± 10 km/sec/Mpc.

One parametrizes H0 as H0 = 100 h km/sec/Mpc so that h ≈ 0.7


More cosmological parameters and scales

Hubble lengthdH = H−1

0 c = 3, 000 h−1 Mpc

Hubble timetH = H−1

0 = 9.8× 109 h−1 yr

Deceleration parameter,

q = −aa

a2

Density parameter,

Ω =8π

3H2ρ =

ρ

ρcrit

where the critical density is defined by

ρcrit =3H2

8π


Density Parameter Ω

Thus

(a

a

)2

=8π

3ρ−

k

a2

H2 =8π

3ρ−

k

a2

1 =8π

H2ρ−

k

a2 H2

1 =ρ

ρcrit−

k

a2 H2

Thus

Ω− 1 =k

H2a2.

The sign of k is therefore determined by whether Ω is greater than, equal to, or less than one. We have

ρ < ρcrit ↔ Ω < 1 ↔ k = −1 ↔ openρ = ρcrit ↔ Ω = 1 ↔ k = 0 ↔ flatρ > ρcrit ↔ Ω > 1 ↔ k = +1 ↔ closed .

The density parameter, then, tells us which of the three Robertson-Walker geometries describes ouruniverse.

From the cosmic microwave background observations, we believe that Ω = 1


Evolution of the Scale Factor

Recall that from ρ = −3(ρ + p)a/a and p = w ρ, we have that ρ ∝ a−3(1+w) which can be rewritten asρ = ρ0a−n if w = n/3− 1

Define the curvature “density” and its density parameter as

ρc = −3 k

8π a2and Ωc = −

k

H2 a2

Therefore, from Friedmann equation

H =a

a=

( 8π

3ρ

)1/2 ( 8π

3ρ0a−n

)1/2

we get that a ∝ a1−n/2 which yields a ∝ t2/n

w n amatter 0 3 t2/3

radiation 13 4 t1/2

curvature − 13 2 t

vacuum −1 0 eHt

Notice that all the non-vacuum solutions have a = 0, the Big Bang singularity

The matter dominated solution is also called the Einstein-de Sitter universe and the vacuum dominatedsolution the de Sitter universe


From the Friedmann equation

H2 =8π

3

∑i

ρi

Dividing by H2 one gets1 =

∑i

Ωi or 1 = Ωc + Ω

where Ω is the sum of all the non-curvature densities.

Taking the time derivative of the Hubble parameter

H =a

a−(

a

a

)2

= −4π∑

i

(1 + wi )ρi

Since |wi | ≤ 1 then H ≤ 0. That is, the rate of expansion of the universe decreases.

The acceleration of the universe refers to a.


From

H2 =8π

3

(ρM0 a−3 + ρR0 a−4 + ρΛ0 + ρc0 a−2

)we see that

As a→ 0 the universe becomes first dominated by matter and then by radiation

As a→∞ the universe is dominated by Λ

If Λ < 0 the universe will eventually re-collapse.

If Λ > 0 the universe expands for ever or re-collapse depending on the value of ΩM

Let’s denote by a∗ the value of the scale factor at the turnaround point between perpetual expansion and eventualre-collapse. Then

0 = H2∗ =

8π

3

(ρM0 a−3

∗ + ρΛ0 + ρc0 a−2∗

)0 =

H2∗

H20

= ΩM0 a−3∗ + ΩΛ0 + Ωc0 a−2

∗

0 = ΩM0 a−3∗ + ΩΛ0 + (1− ΩM0 + ΩΛ0) a−2

∗

0 = ΩM0 + ΩΛ0 a3∗ + (1− ΩM0 + ΩΛ0) a∗

Solving this equation for a∗ and finding the value of ΩΛ0 given ΩM0 for which real solutions exists, one gets thecondition for the turnaround point.


Current observations favor ΩΛ0 ∼ 0.7 given ΩM0 ∼ 0.3


Redshift and Distances

Recall H0 is related to the age of the universe

We would like to also know Ω since among other things determines the curvature k of the universe.

In a FRW universe, there are no timelike Killing vectors, but there is a Killing tensor

Kµν = a2(gµν + UµUν )

where Kµν satisfies∇(σKµν) = 0.

Consider a particle with 4-velocity Vµ = dxµ/dλ, thus

K 2 = KµνVµVν = a2[VµVµ + (UµVµ)2]

will be a constant along geodesics.

For massive particles, VµVµ = −1, or (V 0)2 = 1 + |~V |2 where |~V |2 = gij Vi V j . So

|~V | =K

a.

The particle therefore slows down with respect to the comoving coordinates as the universe expands.

Thats is, a gas of particles will cool down as the universe expands.


For null geodesics VµVµ = 0, and

UµVµ =K

a.

Recall that the frequency of the photon as measured by a comoving observer is ω = −UµVµ.

The frequency of the photon emitted with frequency ωem will therefore be observed with a lower frequencyωobs as the universe expands:

ωobs

ωem=

aem

aobs

The redshift z between two events is defined to be

z =λobs − λem

λem=

aobs

aem− 1

Notice that this redshift is not the same as the conventional Doppler effect; it is the expansion of space.

To measure z, we know the rest-frame wavelengths and thus λem and we observe λobs .


Instantaneous Physical Distance and Hubble Law

Recall the FRW metricds2 = −dt2 + a2

[dχ2 + S2

k (χ) dΩ2]

where Sk = sinχ, χ, sinhχ for k = +1, 0,−1, respectively.

The instantaneous distance isdP (t) = a(t)χ

The observed velocity is then

v = dP (t) = a(t)χ =a

adP

thus

Hubble law

v = H0 dP


Luminosity Distance

We define the luminosity distance as

d2L =

L

4πF

where L is the absolute luminosity of the source and F is the flux measured by the observer.

This is motivated by the fact that for a source at distance d , the flux over the luminosity is

F

L=

1

A(d)

with A(d) the area of a sphere center at the source.

In an FRW universe, A = 4πa20 S2

k , where a0 is the scale factor when the photons are observed.

The flux is diluted by two additional effects: (1) the individual photons redshift by a factor (1 + z), and (2)the photons hit the sphere less frequently, since two photons emitted a time δt apart will be measured at atime (1 + z)δt apart.

ThereforeF

L=

1

4πa20S2

k (1 + z)2,

ordL = a0 Sk (χ)(1 + z)

dL is in principle measurable, since there are some astrophysical sources whose absolute luminosities areknown standard candles

The redshift z is an observable, but χ and a0 are not.


Consider a null radial geodesic0 = ds2 = −dt2 + a2 dχ2

thus

χ =

∫ dt

a(t)=

∫ da

a2 H=

∫ z

0

dz′

H(z′)

On the other hand, from Friedmann equation,

H2 =8π

3

∑i

ρi

where to simplify our existence we assume ρi = ρi0 a−ni = ρi0(1 + z)ni

Thus, we rewrite H(z) = H0 E(z) where

E(z) =

∑i

Ωi0(1 + z)ni

1/2

So the luminosity distance isdL(z) = a0 Sk (χ) (1 + z)

where

χ =1

a0 H0

∫ z

0

dz′

E(z′)


Recall, Ωc0 = −k/(a0 H0)2 therefore a0 = H−10

√−k/Ωc0 = H−1

0 /√|Ωc0|

Thus,

dL(z) = (1 + z)H−1

0√|Ωc0|

Sk (χ)

where

χ =√|Ωc0|

∫ z

0

dz′

E(z′)

and it is understood that Ωc0 = 1− Ω0

Thus to calculate dL we need the observables Ω0 and H0.

There seems to be an issue with the above expression for a flat Universe since in that case k = 0 andΩc0 = 0.

However, in this case Sk (χ) = χ and

dL(z) = (1 + z) H−10

∫ z

0

dz′

E(z′)


Lookback time

If the age of the Universe today is t0 and the age when a photon was emitted is t∗. the lookback time is

t0 − t∗ =

∫ t0

t∗dt

=

∫ a0

a∗

da

a H= H−1

0

∫ a0

a∗

(H0

H

)( a0

a

)d

(a

a0

)

= H−10

∫ 0

z∗

dz

(1 + z) E(z)

where we have used a/a0 = 1 + z and H = H0 E(z)

Consider a flat (k = 0) matter-dominated (ρ = ρM = ρM0 a−3) universe. Then with E(z) = (1 + z)3/2

one gets

t0 − t∗ = H−10

∫ 0

z∗

dz′

(1 + z)5/2

=2

3H−1

0

[1− (1 + z∗)−3/2

]

Let t∗ → 0 or z∗ →∞

Age of the Universe in a Matter Dominated Universe

t0 =2

3H−1

0 where tH = H−10 the Hubble time


The Cosmic Microwave Background Radiation


Black Body Radiation



Cosmic Microwave Background Anisotropies

Temperature Anisotropies

Θ(n) ≡∆T

T(n) =

∑lm

almYlm(n)

Power Spectrum

〈Θ(n),Θ(n′)〉 =∑lm

Cl Ylm(n)Y∗lm(n′)

=∑

l

Cl(2l + 1)

4πPl (n · n′)

where 〈〉 denotes average over all directions and “positions” with 〈alma∗l′m′ 〉 = δll′δmm′Cl

We cannot average over all positions, so the observed Cl is an average over m

Cobsl ≡

1

2 l + 1

∑m

alma∗lm =1

4π

∫d2n d2n′ Pl (n · n′)Θ(n),Θ(n′)

Cosmic Variance ⟨(Cl − Cobsl

Cl

)2 ⟩=

2

2 l + 1





The CMBR spectrum support the view that density perturbations are imprinted on all scales at extremelyearly times.

The observations support support also the view that the Universe is spatially flat; thus, Ω = 1 and thenΩΛ0 = 0.7± 01 or equivalently ρvac ≈ 10−8erg/cm3. Which is consistent with the Type Ia SN results.


Polarization of the Cosmic Microwave BackgroundRadiation


The Matter Content in the Universe

Matter is equivalent to non-relativistic particles and isapproximately ΩM0 = 0.3± 0.1

Baryonic matter: stars, planes, gas, dust (i.e. protons andneutrons) is accounts for Ωb = 0.04± 0.02

Dark or non-baryonic matter: accounts for ΩM0 − Ωb

We do not know yet the the origin of dark matter.

Coincidence problem: we have that

ΩΛ

ΩM∝ a3

Since the vacuum and matter energy are comparable today,the vacuum energy is undetectably small in the past!


Inflation

The universe seems to have merged from one set of initial conditions

A set of initial conditions fined tuned to be spatially flat and highly homogeneous and isotropic

Is there a mechanism that could take a wide spectrum of initial conditions and evolve them toward flatnessand homogeneity/isotropy?

The answers is YES. The inflationary universe scenario provides such mechanism.


Flatness Problem

The densiy of the universe seems to be finely tuned to be equal to the critical density ρc . Thus, the universeis fine-tuned to be flat

Recall Friedmann equation

H2 =8π

3ρ−

κ

a2

3 H2 a2

8π= ρ a2 −

3κ

8π

ρc a2 = ρ a2 −3κ

8π

(ρc − ρ) a2 = −3κ

8π

(Ω−1 − 1)ρ a2 = −3κ

8π= const

At a ≈ 0, the term ρ a2 ≈ 1060ρ0 a20

Therefore, at a ≈ 0, the factor (Ω−1 − 1) ≈ 10−60(Ω−10 − 1)

Thus, any small deviation from Ω = 1 will yield a huge value of Ω0


Particle Horizon

Horizons exist because there is finite amount of time since the Big Bang.

The photons travel only a finite distance during this time.

Consider a photon along a radial null geodesic in a matter dominated, flat universe. Thus, from0 = ds2 = −dt2 + a2 dr2 with H2 = H2

0 a−3, the comoving distance traveled by this photon in an intervalof time [t1, t2] is

r =

∫ t2

t1

dt

a=

∫ a2

a1

da

H a2=

∫ a2

a1

da

H0 a1/2= 2 H−1

0 (√

a2 −√

a1)

The comoving horizon distance traveled since the Big Bang to a time in which a = a∗ isrhor (a∗) = 2 H−1

0√

a∗

The physical horizon distance is then the proper distance traveled by the photon, namelydhor (a∗) = a∗ rhor (a∗) = 2 H−1

∗One can show that in general dhor (a∗) ≈ H−1

∗ = dH (a∗) where dH is the Hubble distance.


Horizon ProblemGiven

r = 2 H−10 (√

a2 −√

a1)

Consider a photon originated at the CMB. The comoving distance between a point in the CMB(a1 = aCMB ≈ 1/1200) and us (a2 = a0 = 1) is

r = 2 H−10 (1−

√aCMB) ≈ 2 H−1

0

The comoving horizon distance between a point in the CMB and the Big Bang

rhor (ACMB) = 2 H−10√

aCMB ≈ 6× 10−2 H−10

Thus, two widely separated parts of the CMB will have non-overlapping horizons. How come then we seethem at almost the same temperature.


Inflation

Recall, Friedmann equation

H2 =8π

3ρ−

κ

a2

For matter dominated ρ ∝ a−3 and for radiation dominated ρ ∝ a−4

Thus, as the Universe expands (i.e. a grows), the curvature term becomes more dominant.

We need a type of matter for which 8π3 ρ κ

a2 , for instance ρ = constant

If ρ = constant with 8π3 ρ κ

a2 , one has that H = a/a = constant > 0 and thus a ∝ eH t

Notice that in this case one would have a period of accelerated expansion (a > 0).

Since Ωc = 1− Ω = − κ

H2 a2 , with a period of inflation (i.e. a ∝ eH t ), one can potentially drive Ω→ 1

and solve the flatness problem


The same acceleration could also grow our horizon in such a way that all of the observable universe todaywas in causal contact at the time when the CMB was generated. That is,

r(inf → dec) =

∫ tdec

tinf

dt

a=

∫ adec

ainf

da

H a2

= H−1inf (a−1

inf − a−1dec ) ≈ H−1

inf a−1inf

r(dec → today) =

∫ t0

tdec

dt

a≈ 2 H−1

0

One can then select Hinf such that

r(inf → dec) ≈ H−1inf a−1

inf r(dec → today) ≈ 2 H−10


How to start and stop InflationConsider a Universe filled with a scalar field φ. Then, Friedmann equation reads

H2 =8π

3ρ +

κ

a2with ρ =

1

2φ

2 + V (φ)

and V a potential to be determined.

The equation for the dynamics of the scalar field is (assuming a homogenous field)

φ + 3 H φ + V ′(φ) = 0

Ignoring for the moment the curvature term, the equations for a and φ are then:

φ + 3 H φ + V ′(φ) = 0

H2 =8π

3

( 1

2φ

2 + V (φ)

)

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

Therefore, we require that

φ2 V

|φ| |3 H φ|, |V ′|

which is equivalent to requiring that the potential energy dominates over the kinetic energy (slow-rollapproximation;).


Therefore

3 H φ ≈ −V ′ H2 ≈8π

3V

Define

ε =1

16π

(V ′

V

)2

η =1

8π

(V ′′

V

)

where ε measures the slope of the potential and η its curvature

We will see that the necessary conditions for inflation (slow-roll parameters) requires ε < 1, |η| < 1

Inflation also requires acceleration; that is, a > 0.

Notice thata

a= H + H2

> 0 ⇐⇒ −H

H2< 1

But

H =(H2)˙

2 H=

4π

3

(V ′ φ

H

)= −

4π

9

(V ′

H

)2

= −1

6

(V ′)2

V

Thus

−H

H2< 1 ⇐⇒

1

16π

(V ′

V

)2

= ε < 1


e-foldings

The amount of inflation is given by the number of e-foldings

N ≡ lna(te)

a(ti )=

∫ te

tiH dt ≈ −8π

∫ φe

φi

V

V ′dφ

The minimum amount of inflation required to solve the various cosmological problems is about 70 e-foldings.

Inflation ends when V ≈ 0


Example

Consider the case of power-law inflation, V ∝ φn

Then

N = −8π∫ φe

φi

V

V ′dφ = −8π

∫ φe

φi

φ

ndφ = −

8π

n

(φ

2e − φ

2i

)=

8π

nφ

2i

where we set φe = 0, so V (φe) ≈ 0

In order to get N = 70, one needs

φi ≈

√70 n

8π≈ 1.7

√n

From the slow-roll parameters one also needs,

ε =1

16π

(n

φi

)2

< 1 η =1

8π

[n (n − 1)

φi

]< 1

or

φi >n

√16π

φi >n (n − 1)

8π


Revisit Horizon Problem

Recall, particle horizon size

dH (inf → dec) = adec

∫ tdec

tinf

dt

a= adec

∫ adec

ainf

da

H a2

≈ H−1inf

(adec

ainf− 1

)≈ H−1

inf

(eHinf ∆t − 1

)

dH (dec → today) = a0

∫ t0

tdec

dt

a≈ 2 H−1

0

ThendH (inf → dec) ≈ H−1

inf eHinf ∆t r(dec → today) ≈ 2 H−10


Relic Fluctuations from Inflation

Let’s consider the following perturbation of the scalar field

φ(t)→ φ0(t) + δφ(t,~x)

The perturbation δφ will induce a perturbation in the stress-energy tensor, and therefore, through Einstein’sequation, there will also be a perturbation in the metric.

In a flat FRW model, the perturbed metric will read

ds2 = −(1 + 2 Φ) dt2 + (1− 2 Φ) a2(t) (dx2 + dy2 + dz2)

with Φ(t,~x) the metric perturbation associated with the scalar perturbation δφ.

The perturbed part of the Einstein equations reads

δG00 = 2(∇2Φ− 3 H Φ− 3 H2Φ)

δG0i = 2∂i (Φ + H Φ)

δGii = −2[Φ + 4 Φ H + (2 H + 3 H2) Φ]

Notice that from 3 H φ0 = −V ′ and H2 = 8π V/3 one has that H = −4πφ20


The perturbed part of the stress-energy tensor reads

δT 00 = −φ2

0 Φ + φ0 δφ + V ′δφ

δT 0i = φ0∂i (δφ)

δT ii = φ

20 Φ− φ0 δφ + V ′δφ

From the δGµν = 8π δTµν (in units 8π = 1)

∇2Φ− 3 H Φ− 3 H2Φ = 4π(−φ20 Φ + φ0 δφ + V ′δφ)

Φ + H Φ = 4π φ0 δφ

Φ + 4 Φ H + (2 H + 3 H2) Φ = −4π(φ20 Φ− φ0 δφ + V ′δφ)

Using the equations for φ0 and H, one can show that the third equation can be obtained from the timederivative of the first equation.


The equations for Φ and δφ are then

Φ + H Φ = 4π φ0 δφ(4π φ2

0 + ~∇2)

Φ = 4π φ20

d

dt

(δφ

φ0

)

In Fourier space, these equations read

Φk + H Φk = 4π φ0 δφk(1 +

1

4π φ20

k2

a2

)Φk =

d

dt

(δφk

φ0

)

with k the wave number.

Important to point out that Φk and δφk are not gauge invariant.


Introduce the following gauge invariant perturbation

ξk ≡ Φk + Hδφk

φ0

Which leads to

ξk +

(φ2

H+

2 φ0

φ0+ 3 H

)ξk +

k2

a2ξk = 0

This equation and the equations

H2 =8π

3

( 1

2φ

20 + V

)

φ0 + 3 H φ0 + V ′ = 0

are sufficient to obtain the evolution of a, φ0 and ξk .


If we convert from cosmic time t to conformal time η such that dt = a dη, the metric becomes

ds2 = a2(−dη2 + dx2 + dy2 + dz2)

If we introduce the definitions α = a φ0/H and χk = α ξk , the equation for ξk becomes

χ′′k +

(k2 −

α′′

α

)χk = 0

where primes denote derivatives with respect to conformal time.

If k2 α′′/α, one can integrate the equation to get χk ∝ α or equivalently ξk = constant.

In the slow-roll approximation, α′′/α ≈ 2 a2 H2 and

χ′′k +

(k2 − 2 a2 H2

)χk = 0

If k a H (sub-horizon scale perturbations), the perturbations χk oscillate. The equation becomes that ofa harmonic oscillator.

If k a H (super-horizon scale perturbations), the perturbations χk exponentially grow.



Date post:	08-May-2020
Category:	Documents
Upload:	others
View:	4 times
Download:	0 times

Gravitation: Cosmology - An Introduction to General...

Documents