Quintessence and the Quintessence and the Accelerating UniverseAccelerating Universe

Jérôme MartinJérôme MartinInstitut d’Astrophysique de ParisInstitut d’Astrophysique de Paris

BibliographyBibliography

2) “Cosmological constant vs. quintessence”, P. Binétruy, hep-ph/0005037.

3) “The cosmological constant and dark energy”, P. Peebles and B. Ratra, astro-ph/0207347.

4) “The cosmological constant”, S. Weinberg, Rev. Mod. Phys. 61, 1 (1989).

5) B. Ratra and P. Peebles, Phys. Rev. D 37, 3406 (1988).

6) I. Zlatev, L. Wang and P.J. Steinhardt, Phys. Rev. Lett. 82, 896 (1999), astro-ph/9807002.

7) P. Brax and J. Martin, Phys. Lett 468B, 40 (1999).

1) “The case for a positive cosmological Lambda term”, V. Sahni and A. Starobinsky, astro-ph/9904398.

PlanPlan

I) The accelerating universe:

II) The cosmological constant problem:

III) Quintessence

SNIa, CMB

Why the cosmological constant is not a satisfactory candidate for dark energy

The notion of tracking fields

The accelerating universeThe accelerating universe

II

The luminosity distanceThe luminosity distance (I)(I)

ÐR = É SÉ t

nRöh÷

ÐR = 4ùd2L

L

The flux received from the source is

É S

dL

nR or L

is the distance to the

source

The luminosity distance (II)The luminosity distance (II)Let us now consider the same physical situation but in a FLRW curved spacetime:

ds2=à dt2+a2(t)(dr2+r2dÒ22)If we define dL as dL ñ 4ùÐ

R

Lr

, then the luminosity distance takes the form:

dL(z) = aR(1+z)RtEtRa(ü)dü

For small redshifts, one has

dL(z) = HR

1ôz+ 2

1(qR à 1)z2+áááõ

with,

H = aaç

q= à aç2aa�

Hubble parameter

Acceleration parameter

Equations of motion (I)Equations of motion (I)The dynamics of the scale factor can be calculated from the Einstein equations:

Rö÷à 21Rí ö÷+Ëí ö÷= ôTö÷

For a FLRW universe:

H2= 3ôP

i=1N úi+ 3

Ë ; àà2aa� + a2

aç2á= ô

Pi=1N pi à Ë

energy density pressure

pË= à ú

Ë= à Ë=ô

The equation of state of a cosmological constant is given by:

!Ëñ cst = à 1

Equations of motion (II)Equations of motion (II)The Einstein equations can also be re-written as:

H2=3ôX

i=1

N+1

úi ; àà2aa� + a2

aç2á= ô

Pi=1N+1pi

with "N +1= Ë"These equations can be combined to get an expression for the acceleration of the scale factor:

aa� = à 6

ôPi=1N+1(úi+3pi)

aa� > 0) ú

T+3p

T< 0

In particular this is the case for a cosmological constant

Acceleration: basic mechanismAcceleration: basic mechanism

A phase of acceleration can be obtained if two basic principles of general relativity and field theory are combined :

General relativity: “any form of energy weighs”

Field theory: “the pressure can be negative”

aa� = à 6

ô ( ú+ )3p p< à ú=3) a� >0

Relativistic term

The acceleration parameterThe acceleration parameter

qñ à aç2aa� = à H2

1aa�

Friedmann equation

Equation giving the acceleration of the scale factor

q= 21Òm+Òr à ÒË

Òm= úT

úm

Òr = úT

úr

ÒË = úT

úË

Matter

Radiation

Vacuum energy

pm= 0

pr= 3

1úr

pË= à ú

Ë

òúT= 8ùG

3H20 ' 10à47GeV 4

ó

Critical energy density

SNIa as standard candles (I)SNIa as standard candles (I)

the width of the light curve is linked to the absolute luminosity

dL(z) = (4ù`

L )1=2The luminosity distance is

`L

: apparent luminosity

: absolute luminositywhere

Clearly, the main difficulty lies in the measurement of the absolute luminosity

SNIa:

SNIa as standard candles (II)SNIa as standard candles (II)

dLL

z

The Hubble diagramThe Hubble diagramHubble diagram: luminosity distance (standard candles) vs. redshift in a

FLRW Universe:

q0= àaç20

a0a� 0 ' 21Òmà ÒË;

Òi ñ úcriúi

úcri = 8ùG

3H20 ' 10à47 GeV 4

q0< 0

The universe is accelerating

The CMB anisotropy The CMB anisotropy measurementsmeasurements

= hTîT(e1) T

îT(e2)iP

`=01 C`P (̀cosî )

COBE has shown that there are temperature fluctuations at the level

TîT ' 10à5

The two-point correlation function is

The position of the first peak depends on

Òm+ÒË

The cosmological parameters The cosmological parameters

Òm ' 0:3; ÒË ' 0:7

The universe is accelerating !

q0<0) a� 0>0

SNIa: 21Òmà ÒË

CMB: Òm+ÒË

The cosmological constant problemThe cosmological constant problem

IIII

The cosmological constant (I)The cosmological constant (I)

Rö÷à 21Rí ö÷+Ëí ö÷= ôTö÷

Bare cosmological constant

ú= 21R

0kmax

(2ù)3dk öh! k

' 16ù2öh k4max

úvac ' 1074 GeV 4 ' 10122úcri

Contribution from the vacuumT vacö÷ = à úvací ö÷

The cosmological constant (II)The cosmological constant (II)The Einstein equations can be re-written under the following form

Rö÷à 21Rí ö÷= à ô

àôË +úvac

áí ö÷+ááá

The cosmological constant problem is : ôË +úvac< úcri

“ Answer “: because there is a deep (unknown!) principle such that the cancellation is exact (SUSY?? …) .

However, the recent measurements of the Hubble diagram indicate

ôË +úvac6=0' 0:7úcri

The cosmological constant (III)The cosmological constant (III)

Maybe super-symmetry can play a crucial role in this unknown principle ?

fQr;Qösg= 2í örsPö

H = P0= 41P

rQ2r

Qrj0i = 0) h0jHj0i = 0

The SUSY algebra

yields the following relation between the Hamiltonian and the super-symmetry generators

but SUSY has to be broken … M S ' 1TeV

The cosmological constant (IV)The cosmological constant (IV)Since a cosmological constant has a constant energy density, this means that its initial value was extremely small in comparison with the energy densities

of the other form of matter

Coincidence problem, fine-tuning of the initial conditions

Radiation

Matter

Cosmological constant

' 100 orders of magnitude

The cosmological constant (V)The cosmological constant (V)

The vacuum has the correct equation of state:

! Ëñ p=ú= à 1

It is important to realize that the cosmological constant problem is a “theoretical” problem. So far a cosmological constant is still compatible with the observations

IIIIII

QuintessenceQuintessence

Quintessence: the main idea (I)Quintessence: the main idea (I)

1) One assumes that the cosmological constant vanishes due to some (so far) unknown principle.

2) The acceleration is due to a new type of fluid with a negative equation of state which, today, represents 70% of the matter content of the universe.

This is the fifth component (the others being baryons, cdm, photons and neutrinos) and the most important one … hence its name

Plato

Scalar fieldsScalar fields A simple way to realize the previous program is to consider a scalar field

S = àRd4x à í

pô

21í ö÷@öQ@÷Q+V(Q)

õ

The stress-energy tensor is defined by: Tö÷= à àíp2

î í ö÷î S

Tö÷=@öQ@÷Q à í ö÷

ô

21í ëì@ëQ@ìQ+V(Q)

õ

The conservation of the stress-energy tensor implies

úç+3aaç(ú+p) = 0 Q� +3a

açQç+ dQdV = 0

Quintessence: the main idea (II)Quintessence: the main idea (II)A scalar field Q can be a candidate for dark energy. Indeed, the time-time and space-space components of the stress-energy tensor are given by:

ú= 21Qç2+V(Q) p= 2

1Qç2à V(Q)

V(Q) ý 21Qç2 ) ! Q = ú

p ' à 1

This is a well-known mechanism in the theory of inflation at very high redshifts. The theoretical surprise is that this kind of exotic

matter could dominate at small redshifts, i.e. now.

A generic property of this kind of model is that the equation

of state is now redshift-dependent ! = ! (z)

The proto-typical modelThe proto-typical model

A typical model where all the main properties of quintessencecan be discussed is given by

V(Q) =M4+ëQàë ë > 0

Two free parameters:

M

ë: energy scale

: power index

Evolution of the quintessence field Evolution of the quintessence field

a21(aa0)2=

m2

P`

8ù (úQ+úB)

The equations of motion controlling the evolution of the system are (in conformal time):

ú0B+3aa0(úB+pB) = 0; ! B = 3

1;0

Q00+2aa0Q0+a2 dQ

dV(Q) =0

! Q = (Q0)2=(2a2)+V(Q)(Q0)2=(2a2)àV(Q)

c2SQ ñ ú0

Q

p0

Q =à 31(HQ02Q00

+1); H ñ aa0

! 0Q = à 3H(1+! Q)(c2SQ à ! Q)

1) Friedmann equation:

quintessenceBackground: radiation or matter

2) Conservation equation for the background :

3) Conservation equation for the quintessence field:

Using the equation of state parameter and the “sound velocity”,

the Klein-Gordon equation can be re-written as

Initial conditionsInitial conditions

ë = 6;M = 106 GeV

ÒQ(z = 1028) = 10à4ÒR

1) The initial conditions are fixed after inflation

2) One assumes that the quintessence field is subdominant

initially.

zi ' 1028

Equipartition

a21H2 ' m2

Pl

8ùúB a(ñ) = a0ñ2=(1+3! B)

úQ ü úB

Quintessence is a test field

The free parameters are chosen to be (see below)

Kinetic eraKinetic era

! 0Q = à 3H(1+! Q)(c2SQ à ! Q)

! Q = (Q0)2=(2a2)+V(Q)(Q0)2=(2a2)àV(Q) ! 1

ú0Q+3aa0(1+! Q)úQ =0) úQ / 1=a6

c2SQ = ! Q = 1

Q02=(2a2) ý V(Q)

2a2Q02

/ ñà6 ) Q = Qf à a(ñ)A

The potential energy becomes constant even if the kinetic one still dominates!

Transition eraTransition era

Q02=(2a2) ü V(Q)

! Q = (Q0)2=(2a2)+V(Q)(Q0)2=(2a2)àV(Q) ! à 1

ú0Q+3aa0(1+! Q)úQ =0) úQ / cst

! 0Q = à 3H(1+! Q)(c2SQ à ! Q)

c2SQ = 16=! Q

But the kinetic energy still redshifts as úKQ / 1=a6

Potential eraPotential era

c2sQ à 1/ a5 The sound velocity has to change

3H22dQ2d2V = H

1(c2sQ)0à 3(c2sQ à 1)(!

B+c2sQ+2) ü 1

c2sQ = à 2à ! B = à 7=3

Q =Qf+Ba4 ) úKQ / a4

The potential era cannot last forever

The potential energy still dominates

The attractor (I)The attractor (I)At this point, the kinetic and potential energy become comparable

Q =Q0ñ4=(ë+2)

If the quintessence field is a test field, then the Klein-Gordon equation with the inverse power –law potential has the solution

úQ / aà3ë(1+! B)=(2+ë)

! Q = à 2+ë2àë! B

Q00+2aa0Q0+a2 dQ

dV(Q) =0

a(ñ) = a0ñ2=(1+3! B)

Redshifts more slowly than the background and therefore is going to dominate

The equation of state tracks the background equation of state

The equation of state is negative!

The attractor (II)The attractor (II)

Equivalence between radiation and matter

One can see the change in the quintessence equation of state when the background equation of state evolves

The attractor (III)The attractor (III)ññ eüü

Q =Qpu p= düdu

(p;u)(0;1) î u; î p

õæ= à 2(ë+2)ë+10 æ2(ë+2)

i 15ë2+108ë +92p

düd î p

î u

ò ó= à ë+2

ë+10 à ë+24(ë+6)

1 0

ò óî pî u

ò ó

Let us introduce a new time defined by and define and by u p

Particular solution

The Klein-Gordon equation, viewed as a dynamical system in the plane , possesses a critical point and small perturbations around this point, , obey

Solutions to the equation are det(M à õI ) = 0

Re(õæ) < 0 The particular solution is an attractor

The attractor (IV)The attractor (IV)This solution is an attractor and is therefore insensible to the initial conditions

Different initial conditions

! Q < 0

The equation of state obtained is negative as required

Consequences for the free parametersConsequences for the free parameters

In order to have ÒQ ' 0:7 M ' 10(19ëà47)=(4+ë) GeVone must choose

ë = 10) M = 1010 GeVFor example

High energy physics !

Q =Q0ñ4=(ë+2) ) dQ2d2V = 2

9H2ëë+1(1à ! 2Q)

' úQ=Q2 ' úQ=m

2P` Q ' mP`

(valid when the quintessence field is about to dominate)

SuperGravity is going to play an important role in the model building problem

A note of the model building problemA note of the model building problem

The fact that, at small redshifts, the value of the quintessence field is the Planck mass means that supergravity must be used for model building. A model gives

A potential V(Q) =M4+ëQàë arises in supersymmetry in the study of gaugino condensation.

V(Q) = e4ùQ2=m

2

P`M4+ëQàë

usual term Sugra correction

At small redshifts, the exponential factor pushes the equation of state towards –1 independently of ë . The model predicts ! Q ' à 0:82

Problems with quintessenceProblems with quintessence

m' dQ2d2VììììQ' mP`

' 10à33eV

The mass of the quintessence field at very small redshift (i.e. now)

The quintessence field must be ultra-light (but this comes “naturally” from the value of M)

This field must therefore be very weakly coupled to matter (this is bad)

Quintessential cosmological Quintessential cosmological perturbationsperturbations

The main question is: can the quintessence field be clumpy?

At the linear level, one writes Q(ñ) + îQ(ñ;xi) and one has to solve the perturbed Klein-Gordon equation:

î Q00+2HîQ0+ôk2+a2dQ2

d2VõîQ +

2Q0

(3h0à h`) = 0

Coupling with the perturbed metric tensor

No growing mode for úQ

î úQ

NB: there is also an attractor for the perturbed quantities, i.e. the final result does not depend on the initial conditions.

ConclusionsConclusionsQuintessence can solve the coincidence and (maybe) the fine tuning problem: the clue to these problems is the concept of tracking field.

There are still important open questions: model building, clustering properties, etc …

A crucial test: the measurement of the equation of state and of its evolution ! Q6=à 1; dz

d! Q 6=0

SNAP ! Q(z) = ! 0+! 1z+ááá