Transcript Quintessence and the Accelerating Universe

Quintessence and the
Accelerating Universe
Jérôme Martin
Institut d’Astrophysique de Paris
Bibliography
1) “The case for a positive cosmological Lambda term”, V. Sahni and A. Starobinsky,
astro-ph/9904398.
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), astroph/9807002.
7) P. Brax and J. Martin, Phys. Lett 468B, 40 (1999).
Plan
I) The accelerating universe:
SNIa, CMB
II) The cosmological constant problem:
Why the cosmological constant is not a satisfactory candidate for dark energy
III) Quintessence
The notion of tracking fields
I
The accelerating universe
The luminosity distance (I)
The flux received
from the source is
nR or L
ÐR =
ÐR =
n ö
h÷
R
É SÉ t
L
4ùd2
L
dL is the distance to the
source
ÉS
The luminosity distance (II)
Let us now consider the same physical situation but in a FLRW curved spacetime:
2
2
2
2
ds = à r dt + a ( t )(dr + r
If we define
dL
as
L
4ùÐR
dL ñ
2
dÒ2)
2
, then the luminosity distance takes the form:
dL( z) = aR(1 + z)
Rt R
tE
For small redshifts, one has
dü
a(ü)
ô
õ
dL(z) = H1R z + 12(qR à 1)z2 + ááá
with,
H=
aç
a
q= à
aa
•
aç2
Hubble parameter
Acceleration parameter
Equations of motion (I)
The dynamics of the scale factor can be calculated from the Einstein equations:
Rö÷ à
1
2Rí ö÷ +
Ëí ö÷ = ôTö÷
For a FLRW universe:
2
H =
ôP N
i= 1 úi
3
+
Ë
3;
à a•
à 2a +
á
aç2
a2
= ô
P
pË = à úË = à Ë=ô
The equation of state of a cosmological constant is given by:
Ë
àË
pressure
energy density
!
N
i= 1 pi
ñ cst = à 1
Equations of motion (II)
The Einstein equations can also be re-written as:
N+ 1
ôX
H =
úi ;
3 i= 1
2
with
à a•
à 2a +
á
aç2
a2
= ô
P
N+ 1
i= 1 pi
"N + 1 = Ë"
These equations can be combined to get an expression for the acceleration of the scale
factor:
ô P N+ 1
i= 1 (úi
6
a
•
a=
à
a
•
a>
0 ) úT + 3pT < 0
+ 3pi )
In particular this is the case for a cosmological constant
Acceleration: basic mechanism
•
a
=
a
à
ô
6
( ú + 3p )
p< à ú=3 ) a
•> 0
Relativistic term
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”
The acceleration parameter
qñ à
aa
•
aç2 =
à
1a
•
H 2a
Equation giving the acceleration of the
scale factor
Friedmann equation
q = 12Òm + Òr à ÒË
Òm =
úm
úT
Òr =
úr
úT
ÒË =
úË
úT
ò
Matter
pm = 0
Radiation
pr = 13úr
Vacuum energy
pË = à ú Ë
úT =
3H 20
à 47
GeV
8ùG ' 10
ó
4
Critical energy density
SNIa as standard candles (I)
The luminosity distance is
dL(z) = (4Lù` )1=2
where
L
`
: absolute luminosity
: apparent luminosity
Clearly, the main difficulty lies in the measurement of the absolute luminosity
SNIa:
the width of the light curve is linked to the absolute luminosity
SNIa as standard candles (II)
z
L
dL
The Hubble diagram
Hubble diagram: luminosity distance (standard candles) vs. redshift in a
FLRW Universe:
•0
a0a
2
ç0
a
q0 = à
Òi ñ
úi
úcri
úcri =
3H 0
8ùG
'
1
à
2Òm
ÒË;
2
' 10à 47 GeV
4
q0 < 0
The universe is accelerating
The CMB anisotropy measurements
COBE has shown that there are temperature fluctuations at
the level îTT ' 10à 5
The two-point correlation function is
P
1
C P (co sî )
`= 0 ` `
= hîTT(e1) îTT(e2)i
The position of the first peak depends on
Òm + ÒË
The cosmological parameters
SNIa: 12Òm à ÒË
CM B: Òm + ÒË
Òm ' 0:3;
ÒË ' 0:7
q0 < 0 ) a
•0 > 0
The universe is accelerating !
II
The cosmological constant problem
The cosmological constant (I)
Rö÷ à
1
2Rí ö÷ +
Ëí ö÷ = ôTö÷
Bare cosmological constant
ú=
'
R
Contribution from the
vacuum
1 kmax dk
h
ö! k
2 0
(2ù) 3
vac
Tö÷
= à úvací ö÷
ö
4
h
2 k max
16ù
úvac ' 1074 GeV
4
' 10122úcri
The cosmological constant (II)
The Einstein equations can be re-written under the following form
Rö÷ à
1
2Rí ö÷ =
àË
á
à ô ô + úvac í ö÷ + ááá
The cosmological constant problem is :
Ë
ô
+ 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
Ë
ô
+ úvac6
= 0 ' 0:7úcri
The cosmological constant (III)
Maybe super-symmetry can play a crucial role in this unknown principle ?
The SUSY algebra
ö sg = 2í örsPö
f Qr ; Q
yields the following relation between the Hamiltonian and the supersymmetry generators
H = P0 =
1P
r
4
Q2r
Q r j0i = 0 ) h0jHj0i = 0
but SUSY has to be broken …
M S ' 1 TeV
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
' 100 orders of magnitude
Cosmological constant
The cosmological constant (V)
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
The vacuum has the correct equation of state:
! Ë ñ p=ú = à 1
III
Quintessence
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 fields
A simple way to realize the previous program is to consider a scalar field
ô
õ
R 4 p
S = à d x à í 12í ö÷@öQ @÷Q + V(Q)
The stress-energy tensor is defined by:
p 2 î Sö÷
àí îí
Tö÷ = à
ô
õ
Tö÷ = @öQ@÷Q à í ö÷ 12 í ëì @ë Q@ì Q + V(Q)
The conservation of the stress-energy tensor implies
úç+
aç
3a(ú +
p) = 0
• + 3açQ
ç
Q
a +
dV
dQ
= 0
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=
1 ç2
2Q +
V(Q)
V(Q) ý
1 ç2
2Q
p=
)
!
1 ç2
2Q
Q
=
à V(Q)
p
ú
'
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 model
A typical model where all the main properties of quintessence
can be discussed is given by
V(Q) = M 4+ ë Qà ë
ë> 0
Two free parameters:
M
: energy scale
ë : power index
Evolution of the quintessence field
The equations of motion controlling the evolution of the system are (in
conformal time):
1 a0 2
( )
a2 a
1) Friedmann equation:
=
8ù
2 (úQ
mP`
+ úB)
Background: radiation or
matter
quintessence
2) Conservation equation for the background :
ú0B
+
0
3aa(úB
+ pB) = 0;
!
B
= 13; 0
0
3) Conservation equation for the quintessence field: Q00+ 2 a Q0+ a2dVd( Q) = 0
a
Q
Using the equation of state parameter and the “sound velocity”,
0
00
p
( Q0) 2=(2a2)à V( Q)
c2SQ ñ Q0 = à 1( 2Q 0 +
! Q = ( Q0) 2=(2a2)+ V( Q)
úQ
3 HQ
the Klein-Gordon equation can be re-written as
!
0
Q
= à 3H (1 + ! Q)(c2SQ à ! Q)
1);
Hñ
a0
a
Initial conditions
1) The initial conditions are fixed after inflation
zi ' 1028
2) One assumes that the quintessence field is subdominant
initially.
Ò (z = 1028) = 10à4ÒR
Q
Equipartition
1 2
a2H
'
8ù
m2Pl úB
úQ ü úB
a(ñ) = a0ñ2=(1+ 3! B)
Quintessence is a test field
The free parameters are chosen to be
ë = 6; M = 106 GeV(see
below)
Kinetic era
Q 02=(2a2) ý V(Q)
!Q=
!
0
Q
ú0Q
( Q0) 2=(2a2)à V( Q)
( Q0) 2=(2a2)+ V( Q)
! 1
= à 3H (1 + ! Q)(c2SQ à ! Q)
+
Q 02
2a2
0
3aa(1 +
! Q)úQ = 0 )
/ ñà 6 ) Q = Q f à
úQ / 1=a6
A
a(ñ)
The potential energy becomes constant even if
the kinetic one still dominates!
c2SQ = !
Q
= 1
Transition era
Q 02=(2a2) ü V(Q)
!
!
0
Q
Q=
( Q0) 2=(2a2)à V( Q)
( Q0) 2=(2a2)+ V( Q)
= à 3H (1 + !
c2SQ = 16
=!
ú0Q
+
0
3aa(1 +
!
2
Q)(cSQ
à 1
à !
Q)
Q
! Q)úQ = 0 )
úQ /
But the kinetic energy still redshifts as
úKQ / 1=a6
cst
Potential era
c2sQ à 1 / a5
2 d2V
3H 2dQ 2
=
The sound velocity has to change
1 2 0
2
H (csQ) à 3(csQ à 1)(!
c2sQ = à 2 à !
B
B
The potential energy still dominates
+ c2sQ + 2) ü 1
= à 7=3
Q = Qf + Ba4 ) úKQ / a4
The potential era cannot last forever
The attractor (I)
At this point, the kinetic and potential energy become
comparable
If the quintessence field is a test field, then the Klein-Gordon
equation with the inverse power –law potential has the solution
Q00+
0
a
2 a Q0+
a2dVd(QQ) = 0
a(ñ) = a0ñ2=(1+ 3! B)
Q = Q0ñ4=( ë+ 2)
úQ / aà 3ë(1+ ! B)=(2+ ë)
!
Q
= à
2à ë! B
2+ ë
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)
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)
Let us introduce a new time
ü defined by ñ ñ
eü and define u and p by
Q = Qpu
p=
du
dü
Particular solution
The Klein-Gordon equation, viewed as a dynamical system in the plane (p; u) , possesses a
critical point (0; 1) and small perturbations around this point, î u; î p, obey
d
dü
ò
ò
ó
îp
îu
=
à
ë+ 10
ë+ 2
1
à
4(ë+ 6)
ë+ 2
óò
0
ó
îp
îu
det (M à õI ) = 0 are
p
ë+ 10
i
õæ = à 2(ë+ 2) æ 2(ë+ 2) 15ë 2 + 108ë + 92
Solutions to the equation
Re (õæ) < 0
The particular solution is an attractor
The attractor (IV)
This solution is an attractor and is therefore insensible to the initial conditions
The equation of state obtained is negative as required
Different
initial conditions
!
Q
< 0
Consequences for the free parameters
Q=
Q0ñ4=( ë+ 2)
2
dV
dQ2
)
=
9 2 ë+ 1
H ë (1 à
2
!
2
)
Q
(valid when the quintessence field is about to dominate)
' úQ=Q2
' úQ=m 2P
`
Q ' mP`
SuperGravity is going to play an important role in the model building problem
In order to have
For example
ÒQ ' 0:7
one must choose
M ' 10(19ëà 47) =(4+ ë) GeV
ë = 10 ) M = 1010 GeV
High energy physics !
A note of the model building problem
A potential V(Q) = M 4+ ëQà ë arises in supersymmetry in the
study of gaugino condensation.
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
2
4ùQ2=mP`
V(Q) = e
Sugra correction
M 4+ ëQà ë
usual term
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 quintessence
The mass of the quintessence field at very small redshift (i.e. now)
ì
ì
d Vì
m ' dQ 2ì
2
' 10à 33 eV
Q' m P`
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
perturbations
The main question is: can the quintessence field be clumpy?
At the linear level, one writes
Q(ñ) + î Q(ñ; x i )
and one has to solve the
perturbed Klein-Gordon equation:
ô
õ
0
2
Q
0
dV
+
(3h
à h` ) = 0
î Q00+ 2Hî Q0+ k 2 + a2dQ
î
Q
2
2
Coupling with the perturbed metric
tensor
No growing mode for
î úQ
úQ
NB: there is also an attractor for the perturbed quantities, i.e. the final result does
not depend on the initial conditions.
Conclusions
Quintessence 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
d
and of its evolution ! Q6
= à 1; d! z 6
=0
Q
SNAP
! Q(z) = ! 0 + ! 1z + ááá