Standard Cosmology II - Istituto Nazionale di Fisica Nucleare

Astroparticle Course 1

Standard CosmologyStandard Cosmology

The fate of the universeThe fate of the universe

Red shiftRed shift

Hubble lawHubble law

Age of the universeAge of the universe

Problems of standard cosmologyProblems of standard cosmology


LongLong--term behaviour of a(t) (I)term behaviour of a(t) (I)

The full solutions to the Friedmann equations can be quite

complicated. However, it is possible to study a few generic case, in

particular starting from the present time (⇒ radiation plays no role)

and considering t→∞. In case Λ=0,

3

3

0

0a

am ρρρ == k

a

aGa N −=

3

8 0302 ρπ

&

0=k3/2

)( tta ≈ Einstein-de Sitter universe

1−=k ttat

≈∞→

)(lim

since at the end the matter term becomes

negligible

1+=k3

8 030

max

ρπ aGa N=

imposing that the right member be 0


LongLong--term behaviour of a(t) (II)term behaviour of a(t) (II)

In case Λ≠0,33

82

0302 a

ka

aGa N Λ

+−=ρπ

&

0<Λ

)(3/

00)()(

ttetata

−Λ≅

Λ contribution is negligible for small a(t) and dominates for large a(t). This is puzzling: why is it that we live at a so special epoch,

when ρΛ is of the order of ρm?

033

82max

max

030 =

Λ+−

ak

a

aGN ρπ

oscillating universe

1,00 −=>Λ k

exponential expansion

10 =>Λ kthere exists Λc giving a static solution,

Λ>Λc ⇒ expansion forever, Λ<Λc ⇒ forbidden range for a(t)

Problem of

coincidence


Solutions of the Friedmann equationSolutions of the Friedmann equation

Λ=0, k=-1

Λ=0, k=0

Λ=0, k=+1

Λ>0, k=0

tH=H0-1


Redshift (I)Redshift (I)

Let us consider the geodesic motion of a particle that is not

necessarily massless,

0=Γ+λλ

ανµ

να

µ

d

dxu

d

du

where uµ=dxµ/ds and λ is some affine parameter. If λ is the proper length, s, the geodesic equation becomes

01

0=+ u

a

a

ds

ud

u

r&r


Redshift (II)Redshift (II)

Since u0=dt/ds, we get

that is the three-momentum of a freely-propagating particle in an expanding universe red shifts. This occurs for photons too,

a

a

u

u &r

r

−=

•

1−∝ aur 1−∝= aump

rr

zta

ta

t

t+== 1

)(

)(

)(

)(

1

0

1

0

λ

λ λ(t0)=received wave length λ(t1)=emitted wave length

z=redshift

The redshift z is often used in place of the scale factor.


Redshift (III)Redshift (III)

Consider a fixed comoving volume of the universe, i.e. a volume

specified by fixed values of the coordinates, from which one

may obtain the physical volume at a given time t by multiplying

by a(t)3. Given a fixed number of dust particles (of mass m) within this comoving volume, the energy density will then scale

just as the physical volume, i.e. as a(t)−3, in agreement with the

equation of state with w = 0.

In the case of radiation, because of the redshift, the energy

density of a fixed number of photons in a fixed comoving

volume drops with the physical volume (as for dust) and by an

extra factor of the scale factor, as the expansion of the universe stretches the wavelengths of light. Thus, the energy density of

radiation will scale as a(t)−4, once again in agreement with the

equation of state with w = 1/3.


Linear HubbleLinear Hubble’’s laws law

In a uniformly expanding

universe, in first approxi-

mation, Hubble’s law is valid,

Hdv =

Let us consider a bidimensional analogy. The distance between

two points on a 2-sphere at t1[t2] is d1=a(t1)s[d2=a(t2)s], where s is the arclength distance between the points. Then

Hdsastt

tata

tt

ddv

tttt==

−

−=

−

−=

→→&

12

12

12

12 )()(limlim

1212


Deviations from linear HubbleDeviations from linear Hubble’’s laws law

Actually, acceleration equation

implies that, if matter drives the

expansion, the rate is presently

decelerating.

+−−−+= ...)(

2

1)(1)()(

2

0

2

00000 ttHqttHtata

)3(3

4p

G

a

a N +−= ρπ&&

However, observations indicate that vacuum energy is

dominating and the expansion is accelerating. In any case, we

may expand the scale factor in a Taylor series around the present time, t0,

which defines the deceleration parameter, q0. 2

00

0

0Ha

aq

&&−=

ρ>0 p∼0

a0=a(t0)


Deceleration parameterDeceleration parameter

From the acceleration equation, the expression of ρc, and the equation of state for each different contribution (matter, radiation, and vacuum) to the universe energy density, we get

∑∑ +Ω=+=i

ii

i

iiN p

H

Gq )31(

2

1)3(

3

8

2

10002

0

0ωρ

π

Since a cosmological constant has ω=-1, q0 can be negative if

the energy density of the universe is dominated by ρΛ. In

general, acceleration of the expansion is possible for ωi=-1/3.


RedRed--shift expansionshift expansion

Using the definition of red-shift, the expansion of a(t) gives

1

2

0

2

00000 ...)(

2

1)(1

)(1

−

+−−−+==+ ttHqttH

ta

az

...)(2

1)(112

0

2

00

00 +−

++−−=+ ttH

qttHz

...)()(2

1)(11

2

0

2

0

2

0

2

0000 +−+−+−−=+ ttHttHqttHz

...)(2

1)(2

0

2

0

0

00 +−

++−−= ttH

qttHz


Particle horizonParticle horizonA light ray that moves radially (θ,φ=const) toward us, from r=re at time

te to r=0 at time t, along a geodesic

(ds2=0 ⇒ dt=a dr/(1-kr2)1/2), will satisfy∫∫ =

−

t

t

r

e

e

ta

dt

kr

dr

)'(

'

10 2

and the physical distance will be

∫∫ =−

=t

t

r

e

e

ta

dtta

kr

drtatd

)'(

')(

1)()(

0 2

Particle horizon is the farthest physical

distance we can observe in the past. It

is given by the limit te → 0, unless the integral does not converge (this does

∫=t

pta

dttatd

0 )'(

')()(

not happen in conventional cosmological theories). In a RD era,

dp=2t, while in a MD era dp≅3t.


Event horizonEvent horizon

∫∫∞

=− t

r

ta

dt

kr

dro

)'(

'

10 2 ∫∞

=t

eta

dttatd

)'(

')()(

While particle horizon is the maximum distance wherefrom

signals can reach the observer, event horizon is the maximum

distance that signals we send can reach. It is determined by

In a matter dominated era, the integral diverges, so that there is no event horizon. But, in presence of a cosmological constant,

a(t) will eventually grow exponentially, giving a finite event

horizon. This means that, as time passes, all sources of light

outside our Local Group will move beyond de and become

unosservable.


HorizonsHorizons

Let us remember the definition of particle and event horizon:

∫=t

pta

dttatd

0 )'(

')()( ∫

∞

=t

eta

dttatd

)'(

')()(

If a(t)∼tα, then∞=

−= )(

1)( td

ttd ep

α1<α

1)()(

−=∞=

α

ttdtd ep1>α

If a(t)∼eHt, thenH

tdH

etd e

Ht

p

1)(

1)( =

−=


Distance expansionDistance expansion

Expanding left and right member of previous equation gives

...2

11

12

2++=

−

kr

kr...

61

3

0 2++=

−∫ ee

r

rk

rkr

dre

+−

++−−== ...)(

21)(1

1

)(

1

)(

1 2

0

2

0

0

00

0

0

0

ttHq

ttHata

a

ata

+−+−=∫ ...)(

2)

1

)(

20

0

eoeo

t

ttt

Htt

ata

dto

e

+−+−= ...)(

2)

1 20

0

eoeoe ttH

tta

rand finally


Luminosity distance (I)Luminosity distance (I)

To obtain information on the cosmological parameters we use

measurements like the intensity of a source of known intrinsic

strangth, a standard candle. Given a source of absolute

luminosity L, its apparent luminosity at a distance d is l=L/(4πd2). At large distance this expression needs modification. Suppose

that Nγ photons was emitted isotropically at te. If there were no

expansion, a telescope of area A, at distance a(te)r, would see a

fraction of photons,

( )2

det

)(4 rta

A

N

N

eπγ

=

Due to the expansion, at detection time, to,( )2

det

)(4 rta

A

N

N

oπγ

=


Luminosity distance (II)Luminosity distance (II)

Two additional effects have to be taken into account. First, all

photons will have energy red-shifted by the factor 1+z. Second,

the rate of arrival of individual photons is lower than the rate at

which they are emitted, since the time interval will be increased by the same factor, 1+z. This means that the flux measured at the

telescope will be

( ) 2224)1()(4 Lo

d

L

zrta

L

ππ=

+=F

which defines the luminosity distance, dL,

)1()( zrtad oL +=


HubbleHubble’’s laws law

After some arithmetics,

...)(2

1)(22

0

0

0 +−

++−= eoeo ttH

qttHz

+

+−=− ...

21

1 20

0

zq

zH

tt eo

+−+−= ...)(

2

1 20

0

eoeoe ttH

tta

r

++−= ...)1(

2

11 2

0

00

zqzHa

re

+−+= ...)1(

2

11 2

0

0

zqzH

dL

deviation from the linear Hubble law

z∼v


Standard candlesStandard candles

Establishing the distances to

other galaxies has been one

of the main efforts of

observational cosmology. For minimizing the

uncertainties introduced by

peculiar velocities, one needs to reach the Hubble flow, that is the

distance range where the velocities of the galaxies are dominated

by the cosmological expansion. To do that, one of the most

common methods exploits the homogeneity of the peak brightness of Type Ia supernovae (SNIa), which has about a 20%

intrinsic dispersion. Connecting the distance of the galaxy hosting

the supernova with the recession velocity (given by the red-shift),

one builds a Hubble diagram.

SNIa: a small star (white dwarf) which, accreting matter from a companion, overcomes the Chandrasekar limit


Hubble diagramHubble diagram

Hubble diagram for high red-shift SNIa from the Supernova

Cosmology Project and low red-shift SNIa from the CfA and

Calan/Tololo Supernova Survey.

apparent magnitude

The data favour a

universe with a positive cosmological constant.

The analysis of SNIa

gives:

15.065.0

)05.0(02.065.0

0 ±−=

±±=

q

h

( )ΛΩ−Ω= 22

10 Mq


Degeneracy in the SNIa analysisDegeneracy in the SNIa analysis

The SNIa data are

insufficient to break the degeneracy of the density

terms. The results can be

approximated by the linear

combination

1.02.06.08.0 ±−=Ω−Ω ΛM

Such a degeneracy can be

break only with the inclusion

of other kind of observation.


Hubble parameter evolutionHubble parameter evolution

Remembering the rate of change of ρM and ρΛ with a(t), the definition of red-shift and curvature parameter,

( )( )03000

3

001 ΛΛΛ Ω++Ω=+

=+= z

a

aMcMM ρρρρρρ

( ) ( )( )200302

0

211 zzHH kM +Ω+Ω++Ω= Λ

( )202

0

2

0

2

0

21 zH

a

a

a

k

a

kk +Ω=

−=−


Age of universeAge of universe

From the definition of the Hubble parameter and red-shift,

( )dt

dz

zz

dt

d

a

a

dt

dH

+−=+−=

=

1

1)1ln(ln

0

( ) ( )( ) 2/120030

0

1

11

)1(1

1

1

zzH

z

Hzdz

dt

kM +Ω+Ω++Ω

+−=

+−=

Λ

−

( ) ( )( )∫+Ω+Ω++Ω+

=

Λ

∞→

z

kM

zzzz

dz

Ht

0 2/1200300

0

'1'1)'1(

'lim

1

and the present age of the universe is


Age of a flat universeAge of a flat universe

001 =Ω=Ω=Ω Λ kM ( ) 3

2

10 2/500 =+

= ∫∞

z

dztH

For H0=65 km s-1 Mpc-1, this gives an age of about 10 billion years.

The age of the universe for a

given ΩM is always larger in the flat case (with a

cosmological constant). Also,

the age is larger for smaller

values of ΩM.

Einstein-de Sitter universe


The flatness problemThe flatness problem

From the Friedmann equation,

2

2

3

8

a

kGH N −= ρ

π2

2

3

81

8

3

aG

k

G

H

NN ρπρπ

−=

cρ

2

2

1

8

31

−− ≈−=−Ω α

ρπa

aG

k

N

α−a

This shows that at early times (small a) Ω must have been very close to unity. For example, at the time of nucleosynthesis:

Ω(t=1s)=1±10-16 and at the Planck time Ω(t=10-43s)=1±10-60.

60

2

0

2

0

1101

−− ≈

=

≈−Ω

PlT T

T

a

a

Pl

How is it possible a How is it possible a similar finesimilar fine--tuning?tuning?


The horizon problemThe horizon problem

Let us consider two photons

which arrive to us today from

two different directions in the

sky. It is possible to consider

directions corresponding to regions causally decoupled

at the time of decoupling.

How is it possible How is it possible that the two that the two photons have the photons have the same T?same T?


The The ΛΛ problemproblem

How is it possible to have a so How is it possible to have a so small dimensionless number?small dimensionless number?

Going back to the complete Friedmann equation, we see that the

comparison between the matter term and the cosmological

constant term is given by their ratio, κ,

323

82 Λ+−=

a

kGH N ρ

π

ρπκ

NG8

Λ=

Considering the value that κ has today, one obtains that at the Planck time it should have been 10-122!!!!

Standard Cosmology II - Istituto Nazionale di Fisica Nucleare

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