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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 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:
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 (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
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