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Cargese - 6 th July 2010 Cargese - 06/07/10 PPInstitut d’Astrophysique de Paris G R ! C O Bouncing alternatives to inflation Cargese - 6 th July 2010 2 Problems with standard model: Singularity Horizon Flatness Homogeneity Perturbations Dark matter Dark energy / cosmological constant Baryogenesis ... Topological defects (monopoles) Accepted solution = INFLATION (Linde’s book)
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Transcript of Bouncing alternatives to inflation P!#$% P&'( - UMR 7332cosmo/EcoleCosmologie/Dossier... ·...

  • Cargese - 6th July 2010

    Cargese - 06/07/10Patri% Pe'r

    Institut dAstrophysique de Paris

    GR!CO

    Bouncing alternatives to inflation

    Cargese - 6th July 2010 2

    Problems with standard model:

    Singularity

    Horizon

    Flatness

    Homogeneity

    Perturbations

    Dark matter

    Dark energy / cosmological constant

    Baryogenesis

    ...

    Topological defects (monopoles)

    Accepted solution = INFLATION

    (Lindes book)

  • Cargese - 6th July 2010

    Alternative model???

    string based ideas (PBB, other brane models, string gas, )

    singularity, initial conditions & homogeneity

    provide challengers!

    Inflation:

    solves cosmological puzzlesuses GR + scalar fields [(semi-)classical]can be implemented in high energy theoriesmakes falsifiable predictions ...... consistent with all known observationsstring based ideas (brane inflation, ...)

    Quantum gravity / cosmology

    bounces

    3

    Cargese - 6th July 2010 4

    Problems of Inflation 3

    t

    xtR

    t0

    ti

    tf (k)

    ti (k)

    k

    H!1

    Fig. 1. Space-time diagram (sketch) showing the evolution of scales in inflationarycosmology. The vertical axis is time, and the period of inflation lasts between ti andtR, and is followed by the radiation-dominated phase of standard big bang cosmol-ogy. During exponential inflation, the Hubble radius H1 is constant in physicalspatial coordinates (the horizontal axis), whereas it increases linearly in time aftertR. The physical length corresponding to a fixed comoving length scale labelled byits wavenumber k increases exponentially during inflation but increases less fast thanthe Hubble radius (namely as t1/2), after inflation.

    From R. Brandenberger, in M. Lemoine, J. Martin & P. P. (Eds.), Inflationary cosmology,Lect. Notes Phys. 738 (Springer, Berlin, 2007).

    Scalar field origin?

    Trans-Planckian

    Hierarchy (amplitude)

    Singularity

    Validity of GR?

    V ()

    4 1012

    1

    t; !(t) = !0a0

    a(t) !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    1

    t; !(t) = !0a0

    a(t) !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    Einf $ 103M

    Pl

    1

    t; !(t) = !0a(t)

    a0 !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    Einf $ 103M

    Pl

    S3 > S2 > S1

    a()

    +g

    +d

    = Tij(k)

    g

    d

    vk eicSk

    2cSk

    1

  • Cargese - 6th July 2010

    A brief history of bouncing cosmology

    R. C. Tolman, On the Theoretical Requirements for a Periodic Behaviour of the Universe, PRD 38, 1758 (1931) G. Lematre, LUnivers en expansion, Ann. Soc. Sci. Bruxelles (1933)

    t; !(t) = !0a0

    a(t) !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    Einf $ 103M

    Pl

    S1

    1

    t; !(t) = !0a0

    a(t) !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    Einf $ 103M

    Pl

    S2 > S1

    1

    t; !(t) = !0a0

    a(t) !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    Einf $ 103M

    Pl

    S3 > S2 > S1

    1

    R. Durrer & J. Laukerman, The oscillating Universe: an alternative to inflation, Class. Quantum Grav. 13, 1069 (1996)

    Quantum nucleation?

    Penrose: BH formation

    PBB - Ekpyrotic - Modified gravity - Quantum cosmology - Quintom - Horava-Lifshitz - Lee-Wick - ...

    A. A. Starobinsky, On one non-singular isotropic cosmological model, Sov. Astron. Lett. 4, 82 (1978)

    5

    Cargese - 6th July 2010 66

    Pre Big Bang scenario: (cf. M.Gasperini & G. Veneziano, arXiv: hep-th/0703055)

  • Cargese - 6th July 2010

    Pre Big Bang scenario: (cf. M.Gasperini & G. Veneziano, arXiv: hep-th/0703055)

    20 M. Gasperini and G. Veneziano

    tions [44, 45]. With such potential V = V () the string cosmology equationscan be rewritten in terms of a, , = a3 and p = pa3 as follows [36, 37, 38]:

    2 3H2 V () = 22se ,

    H H = 2s e p,

    2 2 3H2 + V ()

    V

    = 0. (12)

    These equations are still invariant under the duality transformations (4),(7) but, differently from Eq. (5), they admit regular and self-dual solutions.We can also obtain exact analytical integrations for appropriate forms of thepotential V (), and for equations of state such that p/ can be written asintegrable function of a suitable time parameter [15].

    Let us consider, as a simple example, the exponential potential V =V0 exp(2) (with V0 > 0), to be regarded here only as an effective, low-energy description of the quantum-loop backreaction, possibly computable athigher orders. Let us use, in addition, an equation of state (motivated by an-alytical simulations concerning the equation of state of a string gas in back-grounds with rolling horizons [46]) evolving between the asymptotic valuesp = /3 at t and p = /3 at t +, so as to match the low-energypre-and post-big bang solutions (10) and (8), respectively. The plot of thecorresponding solution (see [15] for the exact analytic form) is illustrated inFig. 2.

    t

    H2

    gS

    e

    pe

    PRE#BIGBANG POST#BIGBANG

    V

    Fig. 2. Example of smooth transition between a phase of pre-big bang inflation andthe standard radiation-dominated evolution.

    The solution smoothly interpolates between the string perturbative vac-uum at t and the standard, radiation-dominated phase at constantdilaton (described by Eq. (8)) at t +, after a pre-big bang phase of grow-ing curvature and growing dilaton described by Eq. (10). The dashed curves

    22 M. Gasperini and G. Veneziano

    HE

    gSE

    aE

    PRE"BIGBANG POST"BIGBANG

    Fig. 3. Example of pre-big bang evolution represented in the E-frame, where thescale factor is shrinking and the Hubble parameter HE is negative. The plots areobtained from Eq. (14) with a0 = 0.8, 0 = 0, 0 = 1, 0 = 1.

    strong coupling, in a marked quantum regime. Nevertheless, an epoch of pre-big bang inflation is able to solve the kinematical problems of the standardscenario starting from different initial conditions which are not necessarilyunnatural [49] or unlikely [50] (see also [51] for a detailed comparison of thepre-big bang versus post-big bang inflationary kinematics). A possible excep-tion concerns the presence of primordial shear, which is not automaticallyinflated away during the phase of pre-big bang evolution: the isotropizationof the three-dimensional spatial sections might require some specific post-bigbang mechanism (see e.g. the discussion of [52]), differently from the standardinflationary scenario where the dilution of shear is automatic.

    Quantum effects, in the pre-big bang scenario, can become important to-wards the end of the inflationary regime. We can say, in particular, that themonotonic growth of the curvature and of the string coupling automaticallyprepares the onset of a typically stringy epoch at strong coupling. Thisepoch could be characterized by the production of a gas of heavy objects(such as winding strings [53, 54] or mini-black holes [55]) as well as light,higher-dimensional branes [56]. In such a context the interaction (and/ or theeventual collision) of two branes can drive a phase of slow-roll inflation [26],as discussed in Sect. 3.

    At this point of the cosmological evolution there are two possible alterna-tives.

    i) The phase of string/brane dominated inflation is long enough to diluteall effects of the preceding phase of dilaton inflation, and to give rise toan epoch of slow-roll inflation able to prepare the subsequent evolution,according to the conventional inflationary picture.

    ii) The back-reaction of the quantum fluctuations, amplified by the phase ofpre-big bang inflation, induces a bounce as soon as the Universe reaches

    string frame Einstein frame

    7

    Cargese - 6th July 2010 8

    Ekpyrotic scenario:

    3

    inspiration in the extra dimensional scenarios, a la Ran-dall Sundrum [4], and can be motivated by compact-ifying the action of 11 dimensional supergravity on anS1/Z2 orbifold, compactified on a CalabiYau three-fold.This results in an effectively five dimensional action read-ing

    S5

    M5d5x

    g5[

    R(5)

    1

    2()2

    3

    2

    e2F2

    5 !

    ]

    , (1)

    where is the scalar modulus, and F the field strength ofa four-form gauge field. Two fourdimensional boundarybranes (orbifold fixed planes), one of which to be lateridentified with our universe, are separated by a finite gap.Both are BPS states [13], i.e., they can be described atlow energy by an effective N = 1 supersymmetric model,so that their curvature vanishes. This is how the flatnessproblem is addressed in the ekpyrotic model.

    Rorb

    Bulk

    K=0

    Visible

    Hidden

    4 D

    bra

    ne,

    Qua

    si B

    PS

    FIG. 1: Schematic representation of the old ekpyrotic modelas a bulk boundary branes in an effective five dimensionaltheory. Our Universe is to be identified with the visible brane,and a bulk brane is spontaneously nucleated near the hiddenbrane, moving towards our universe to produce the Big-Bangsingularity and primordial perturbations. In the new ekpy-rotic scenario, the bulk brane is absent and it is the hiddenbrane that collides with the visible one, generating the hotBig Bang singularity.

    In the old scenario [8], the five dimensional bulk isalso assumed to contain various fields not described here,whose excitations can lead to the spontaneous nucleationof yet another, much lighter, freely moving, brane. Inthe so-called new scenario [9], and its cyclic exten-sion [23], it is the hidden boundary brane that is ableto move in the bulk. In both cases, this extra brane, ifassumed BPS (as demanded by minimization of the ac-tion) is flat, parallel to the boundary branes and initially

    at rest. Non perturbative effects yield an interaction po-tential between the visible and the bulk brane. The dis-tance of the former to the latter can be regarded as ascalar field living on the four dimensional visible bound-ary brane whose effective action is thus that of four di-mensional GR together with a scalar field evolving inan exponential potential, namely

    S4 =

    M4d4x

    g4[

    R(4)

    2

    1

    2()2 V ()

    ]

    , (2)

    with

    V () = Vi exp[

    4

    mPl( i)

    ]

    , (3)

    where is a constant and = 8G = 8/m2Pl

    . Apartfrom the sign, the potential is the one that leads to thewell known power-law inflation model if the value of lies in a given range [24].

    The interaction between the two branes results in one(bulk or hidden) brane moving towards the other (vis-ible) boundary until they collide. This impact time isthen identified with the Big-Bang of standard cosmol-ogy. Slightly before that time, the exponential potentialabruptly goes to zero so the boundary brane is led to asingular transition at which the kinetic energy of the bulkbrane is converted into radiation. The result is, from thispoint on, exactly similar to standard big bang cosmology,with the difference that the flatness problem is claimedto be solved by saying our Universe originated as a BPSstate (see however [23]).

    FIG. 2: Scale factor in the new ekpyrotic scenario. TheUniverse starts its evolution with a slow contraction phasea ()1+ with = 0.9 on the figure. The bounce itselfis explicitly associated with a singularity which is approachedby the scalar field kinetic term domination phase, and theexpansion then connects to the standard Big-Bang radiationdominated phase.

    3

    inspiration in the extra dimensional scenarios, a la Ran-dall Sundrum [4], and can be motivated by compact-ifying the action of 11 dimensional supergravity on anS1/Z2 orbifold, compactified on a CalabiYau three-fold.This results in an effectively five dimensional action read-ing

    S5

    M5d5x

    g5[

    R(5)

    1

    2()2

    3

    2

    e2F2

    5 !

    ]

    , (1)

    where is the scalar modulus, and F the field strength ofa four-form gauge field. Two fourdimensional boundarybranes (orbifold fixed planes), one of which to be lateridentified with our universe, are separated by a finite gap.Both are BPS states [13], i.e., they can be described atlow energy by an effective N = 1 supersymmetric model,so that their curvature vanishes. This is how the flatnessproblem is addressed in the ekpyrotic model.

    Rorb

    Bulk

    K=0

    Visible

    Hidden

    4 D

    bra

    ne,

    Qua

    si B

    PS

    FIG. 1: Schematic representation of the old ekpyrotic modelas a bulk boundary branes in an effective five dimensionaltheory. Our Universe is to be identified with the visible brane,and a bulk brane is spontaneously nucleated near the hiddenbrane, moving towards our universe to produce the Big-Bangsingularity and primordial perturbations. In the new ekpy-rotic scenario, the bulk brane is absent and it is the hiddenbrane that collides with the visible one, generating the hotBig Bang singularity.

    In the old scenario [8], the five dimensional bulk isalso assumed to contain various fields not described here,whose excitations can lead to the spontaneous nucleationof yet another, much lighter, freely moving, brane. Inthe so-called new scenario [9], and its cyclic exten-sion [23], it is the hidden boundary brane that is ableto move in the bulk. In both cases, this extra brane, ifassumed BPS (as demanded by minimization of the ac-tion) is flat, parallel to the boundary branes and initially

    at rest. Non perturbative effects yield an interaction po-tential between the visible and the bulk brane. The dis-tance of the former to the latter can be regarded as ascalar field living on the four dimensional visible bound-ary brane whose effective action is thus that of four di-mensional GR together with a scalar field evolving inan exponential potential, namely

    S4 =

    M4d4x

    g4[

    R(4)

    2

    1

    2()2 V ()

    ]

    , (2)

    with

    V () = Vi exp[

    4

    mPl( i)

    ]

    , (3)

    where is a constant and = 8G = 8/m2Pl

    . Apartfrom the sign, the potential is the one that leads to thewell known power-law inflation model if the value of lies in a given range [24].

    The interaction between the two branes results in one(bulk or hidden) brane moving towards the other (vis-ible) boundary until they collide. This impact time isthen identified with the Big-Bang of standard cosmol-ogy. Slightly before that time, the exponential potentialabruptly goes to zero so the boundary brane is led to asingular transition at which the kinetic energy of the bulkbrane is converted into radiation. The result is, from thispoint on, exactly similar to standard big bang cosmology,with the difference that the flatness problem is claimedto be solved by saying our Universe originated as a BPSstate (see however [23]).

    FIG. 2: Scale factor in the new ekpyrotic scenario. TheUniverse starts its evolution with a slow contraction phasea ()1+ with = 0.9 on the figure. The bounce itselfis explicitly associated with a singularity which is approachedby the scalar field kinetic term domination phase, and theexpansion then connects to the standard Big-Bang radiationdominated phase.

    3

    inspiration in the extra dimensional scenarios, a la Ran-dall Sundrum [4], and can be motivated by compact-ifying the action of 11 dimensional supergravity on anS1/Z2 orbifold, compactified on a CalabiYau three-fold.This results in an effectively five dimensional action read-ing

    S5

    M5d5x

    g5[

    R(5)

    1

    2()2

    3

    2

    e2F2

    5 !

    ]

    , (1)

    where is the scalar modulus, and F the field strength ofa four-form gauge field. Two fourdimensional boundarybranes (orbifold fixed planes), one of which to be lateridentified with our universe, are separated by a finite gap.Both are BPS states [13], i.e., they can be described atlow energy by an effective N = 1 supersymmetric model,so that their curvature vanishes. This is how the flatnessproblem is addressed in the ekpyrotic model.

    Rorb

    Bulk

    K=0

    Visible

    Hidden

    4 D

    bra

    ne,

    Qua

    si B

    PS

    FIG. 1: Schematic representation of the old ekpyrotic modelas a bulk boundary branes in an effective five dimensionaltheory. Our Universe is to be identified with the visible brane,and a bulk brane is spontaneously nucleated near the hiddenbrane, moving towards our universe to produce the Big-Bangsingularity and primordial perturbations. In the new ekpy-rotic scenario, the bulk brane is absent and it is the hiddenbrane that collides with the visible one, generating the hotBig Bang singularity.

    In the old scenario [8], the five dimensional bulk isalso assumed to contain various fields not described here,whose excitations can lead to the spontaneous nucleationof yet another, much lighter, freely moving, brane. Inthe so-called new scenario [9], and its cyclic exten-sion [23], it is the hidden boundary brane that is ableto move in the bulk. In both cases, this extra brane, ifassumed BPS (as demanded by minimization of the ac-tion) is flat, parallel to the boundary branes and initially

    at rest. Non perturbative effects yield an interaction po-tential between the visible and the bulk brane. The dis-tance of the former to the latter can be regarded as ascalar field living on the four dimensional visible bound-ary brane whose effective action is thus that of four di-mensional GR together with a scalar field evolving inan exponential potential, namely

    S4 =

    M4d4x

    g4[

    R(4)

    2

    1

    2()2 V ()

    ]

    , (2)

    with

    V () = Vi exp[

    4

    mPl( i)

    ]

    , (3)

    where is a constant and = 8G = 8/m2Pl

    . Apartfrom the sign, the potential is the one that leads to thewell known power-law inflation model if the value of lies in a given range [24].

    The interaction between the two branes results in one(bulk or hidden) brane moving towards the other (vis-ible) boundary until they collide. This impact time isthen identified with the Big-Bang of standard cosmol-ogy. Slightly before that time, the exponential potentialabruptly goes to zero so the boundary brane is led to asingular transition at which the kinetic energy of the bulkbrane is converted into radiation. The result is, from thispoint on, exactly similar to standard big bang cosmology,with the difference that the flatness problem is claimedto be solved by saying our Universe originated as a BPSstate (see however [23]).

    FIG. 2: Scale factor in the new ekpyrotic scenario. TheUniverse starts its evolution with a slow contraction phasea ()1+ with = 0.9 on the figure. The bounce itselfis explicitly associated with a singularity which is approachedby the scalar field kinetic term domination phase, and theexpansion then connects to the standard Big-Bang radiationdominated phase.

    3

    inspiration in the extra dimensional scenarios, a la Ran-dall Sundrum [4], and can be motivated by compact-ifying the action of 11 dimensional supergravity on anS1/Z2 orbifold, compactified on a CalabiYau three-fold.This results in an effectively five dimensional action read-ing

    S5

    M5d5x

    g5[

    R(5)

    1

    2()2

    3

    2

    e2F2

    5 !

    ]

    , (1)

    where is the scalar modulus, and F the field strength ofa four-form gauge field. Two fourdimensional boundarybranes (orbifold fixed planes), one of which to be lateridentified with our universe, are separated by a finite gap.Both are BPS states [13], i.e., they can be described atlow energy by an effective N = 1 supersymmetric model,so that their curvature vanishes. This is how the flatnessproblem is addressed in the ekpyrotic model.

    Rorb

    Bulk

    K=0

    Visible

    Hidden

    4 D

    bra

    ne,

    Qua

    si B

    PS

    FIG. 1: Schematic representation of the old ekpyrotic modelas a bulk boundary branes in an effective five dimensionaltheory. Our Universe is to be identified with the visible brane,and a bulk brane is spontaneously nucleated near the hiddenbrane, moving towards our universe to produce the Big-Bangsingularity and primordial perturbations. In the new ekpy-rotic scenario, the bulk brane is absent and it is the hiddenbrane that collides with the visible one, generating the hotBig Bang singularity.

    In the old scenario [8], the five dimensional bulk isalso assumed to contain various fields not described here,whose excitations can lead to the spontaneous nucleationof yet another, much lighter, freely moving, brane. Inthe so-called new scenario [9], and its cyclic exten-sion [23], it is the hidden boundary brane that is ableto move in the bulk. In both cases, this extra brane, ifassumed BPS (as demanded by minimization of the ac-tion) is flat, parallel to the boundary branes and initially

    at rest. Non perturbative effects yield an interaction po-tential between the visible and the bulk brane. The dis-tance of the former to the latter can be regarded as ascalar field living on the four dimensional visible bound-ary brane whose effective action is thus that of four di-mensional GR together with a scalar field evolving inan exponential potential, namely

    S4 =

    M4d4x

    g4[

    R(4)

    2

    1

    2()2 V ()

    ]

    , (2)

    with

    V () = Vi exp[

    4

    mPl( i)

    ]

    , (3)

    where is a constant and = 8G = 8/m2Pl

    . Apartfrom the sign, the potential is the one that leads to thewell known power-law inflation model if the value of lies in a given range [24].

    The interaction between the two branes results in one(bulk or hidden) brane moving towards the other (vis-ible) boundary until they collide. This impact time isthen identified with the Big-Bang of standard cosmol-ogy. Slightly before that time, the exponential potentialabruptly goes to zero so the boundary brane is led to asingular transition at which the kinetic energy of the bulkbrane is converted into radiation. The result is, from thispoint on, exactly similar to standard big bang cosmology,with the difference that the flatness problem is claimedto be solved by saying our Universe originated as a BPSstate (see however [23]).

    FIG. 2: Scale factor in the new ekpyrotic scenario. TheUniverse starts its evolution with a slow contraction phasea ()1+ with = 0.9 on the figure. The bounce itselfis explicitly associated with a singularity which is approachedby the scalar field kinetic term domination phase, and theexpansion then connects to the standard Big-Bang radiationdominated phase.

    3

    inspiration in the extra dimensional scenarios, a la Ran-dall Sundrum [4], and can be motivated by compact-ifying the action of 11 dimensional supergravity on anS1/Z2 orbifold, compactified on a CalabiYau three-fold.This results in an effectively five dimensional action read-ing

    S5

    M5d5x

    g5[

    R(5)

    1

    2()2

    3

    2

    e2F2

    5 !

    ]

    , (1)

    where is the scalar modulus, and F the field strength ofa four-form gauge field. Two fourdimensional boundarybranes (orbifold fixed planes), one of which to be lateridentified with our universe, are separated by a finite gap.Both are BPS states [13], i.e., they can be described atlow energy by an effective N = 1 supersymmetric model,so that their curvature vanishes. This is how the flatnessproblem is addressed in the ekpyrotic model.

    Rorb

    Bulk

    K=0

    Visible

    Hidden

    4 D

    bra

    ne,

    Qua

    si B

    PS

    FIG. 1: Schematic representation of the old ekpyrotic modelas a bulk boundary branes in an effective five dimensionaltheory. Our Universe is to be identified with the visible brane,and a bulk brane is spontaneously nucleated near the hiddenbrane, moving towards our universe to produce the Big-Bangsingularity and primordial perturbations. In the new ekpy-rotic scenario, the bulk brane is absent and it is the hiddenbrane that collides with the visible one, generating the hotBig Bang singularity.

    In the old scenario [8], the five dimensional bulk isalso assumed to contain various fields not described here,whose excitations can lead to the spontaneous nucleationof yet another, much lighter, freely moving, brane. Inthe so-called new scenario [9], and its cyclic exten-sion [23], it is the hidden boundary brane that is ableto move in the bulk. In both cases, this extra brane, ifassumed BPS (as demanded by minimization of the ac-tion) is flat, parallel to the boundary branes and initially

    at rest. Non perturbative effects yield an interaction po-tential between the visible and the bulk brane. The dis-tance of the former to the latter can be regarded as ascalar field living on the four dimensional visible bound-ary brane whose effective action is thus that of four di-mensional GR together with a scalar field evolving inan exponential potential, namely

    S4 =

    M4d4x

    g4[

    R(4)

    2

    1

    2()2 V ()

    ]

    , (2)

    with

    V () = Vi exp[

    4

    mPl( i)

    ]

    , (3)

    where is a constant and = 8G = 8/m2Pl

    . Apartfrom the sign, the potential is the one that leads to thewell known power-law inflation model if the value of lies in a given range [24].

    The interaction between the two branes results in one(bulk or hidden) brane moving towards the other (vis-ible) boundary until they collide. This impact time isthen identified with the Big-Bang of standard cosmol-ogy. Slightly before that time, the exponential potentialabruptly goes to zero so the boundary brane is led to asingular transition at which the kinetic energy of the bulkbrane is converted into radiation. The result is, from thispoint on, exactly similar to standard big bang cosmology,with the difference that the flatness problem is claimedto be solved by saying our Universe originated as a BPSstate (see however [23]).

    FIG. 2: Scale factor in the new ekpyrotic scenario. TheUniverse starts its evolution with a slow contraction phasea ()1+ with = 0.9 on the figure. The bounce itselfis explicitly associated with a singularity which is approachedby the scalar field kinetic term domination phase, and theexpansion then connects to the standard Big-Bang radiationdominated phase.

  • Cargese - 6th July 2010 9

    Standard Failures and some solutions

    Singularity

    Horizon

    Flatness

    Homogeneity

    PerturbationsOthers

    Merely a non issue in the bounce case!

    see coming slides

    dark matter/energy, baryogenesis, ...

    T0a30 ! 1500!Pl

    nS = 0.96 0.02 = w < 8 104

    ! 0.62

    T

    S=

    C(T)10

    C(S)10= F (, )

    A2T

    A2S

    w

    T

    S! 4 102

    nS 1

    dH a(t) t

    ti

    d

    a()

    ti

    d

    dt| 1| = 2

    a

    a3

    6

    accelerated expansion (inflation) or decelerated contraction (bounce)

    LastScatteringsurface

    a < 0 & a < 0

    crit 3H2

    8GN

    totcrit

    = 1 = K = 0

    a

    a=

    1

    3[ 4GN ( + p)]

    H2 +Ka2

    =1

    3(8GN + )

    =1

    3

    p =

    1

    T0a30 ! 1500!Pl

    nS = 0.96 0.02 = nS < 8 104

    ! 0.62

    T

    S=

    C(T)10

    C(S)10= F (, )

    A2T

    A2S

    w

    T

    S! 4 102

    nS 1

    dH a(t) t

    ti

    d

    a()

    6

    can be made divergent easily if

    T0a30 ! 1500!Pl

    nS = 0.96 0.02 = nS < 8 104

    ! 0.62

    T

    S=

    C(T)10

    C(S)10= F (, )

    A2T

    A2S

    w

    T

    S! 4 102

    nS 1

    dH a(t) t

    ti

    d

    a()

    ti

    6

    Large & flat Universe + low initial density + diffusion

    T0a30 ! 1500!Pl

    nS = 0.96 0.02 = w < 8 104

    ! 0.62

    T

    S=

    C(T)10

    C(S)10= F (, )

    A2T

    A2S

    w

    T

    S! 4 102

    nS 1

    dH a(t) t

    ti

    d

    a()

    ti

    d

    dt| 1| = 2

    a

    a3

    tdiffusiontHubble

    R1/3H

    (

    1 +

    AR2H

    )

    =

    6

    enough time to dissipate any wavelength

    T0a30 ! 1500!Pl

    nS = 0.96 0.02 = w < 8 104

    ! 0.62

    T

    S=

    C(T)10

    C(S)10= F (, )

    A2T

    A2S

    w

    T

    S! 4 102

    nS 1

    dH a(t) t

    ti

    d

    a()

    ti

    d

    dt| 1| = 2

    a

    a3

    tdiffusiontHubble

    R1/3H

    (

    1 +

    AR2H

    )

    =

    6

    vacuum state!

    tdissipation

    0

    I = R

    3 (4RR R2)

    =dV

    d= I

    S =1

    16GN

    d4xg

    [

    R +N

    i=1

    iI(i) V ()

    ]

    s n

    max(s) =Aa sin

    (

    s

    1 + x2)

    + Ba cos(

    s

    1 + x2)

    sin s

    1

    Cargese - 6th July 2010 10

    Initial conditionsfixed in the contracting era

    t; !(t) = !0a0

    a(t) !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    Einf $ 103M

    Pl

    S3 > S2 > S1

    1

    t; !(t) = !0a0

    a(t) !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    Einf $ 103M

    Pl

    S3 > S2 > S1

    a()

    1

    t; !(t) = !0a0

    a(t) !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    Einf $ 103M

    Pl

    S3 > S2 > S1

    a()

    +g

    +d

    = Tij(k)

    g

    d

    1

    ???????

  • Cargese - 6th July 2010-20 0 20

    t

    0

    20

    40

    60

    80

    100

    N(t)

    -20 0 20-3

    -2

    -1

    0

    1

    2

    3

    H(t)

    S =

    d4xg

    [

    R

    6!2Pl

    1

    2

    V ()]

    7

    11

    Self consistent bounce:

    One d.o.f. + 4 dimensions G.R.

    S =

    d4xg

    [

    R

    6!2Pl

    1

    2

    V ()]

    ds2 = dt2 a2(t)(

    dr2

    1 Kr2+ r2d2

    )

    7

    Positive spatial curvature

    S =

    d4xg

    [

    R

    6!2Pl

    1

    2

    V ()]

    ds2 = dt2 a2(t)(

    dr2

    1 Kr2+ r2d2

    )

    H2 =1

    3

    (

    1

    22 + V

    )

    Ka2

    7

    F. Falciano, M. Lilley & P. P., Phys. Rev. D77, 083513 (2008)

    Cargese - 6th July 2010 12

    50 0 50 100

    0.3

    0.2

    0.1

    0.0

    0.1

    0.2

    h

    50 0 50 1001045

    1038

    1031

    1024

    1017

    1010

  • Cargese - 6th July 2010

    100 1000

    k or n

    10-6

    10-4

    10-2

    100

    102

    k3|!

    BS

    T|2

    -1.5 -1 -0.5 0 0.5 1 1.5

    t

    0

    50

    100

    150

    200

    Vu(t)

    -1 0 1-15

    -10

    -5

    0

    5

    10

    15

    !BST(t)

    13

    a = a0

    [

    1 +

    (

    T

    T0

    )2]

    13(1)

    K = +1

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T )(2) [v, T ; a (T )]

    = 3!2

    Pl

    2

    + p

    a

    d

    d

    (v

    a

    )

    ds2 = a2(){

    (1 + 2) d2 [(1 2) ij + hij ] dxidxj}

    d = a31dT

    i(2)

    =

    d3x

    (

    1

    2

    2

    v2+

    2v,iv

    ,i a

    a

    )

    (2)

    vk +

    (

    c2Sk2

    a

    a

    )

    vk = 0

    4

    Perturbations:

    S =

    d4xg

    [

    R

    6!2Pl

    1

    2

    V ()]

    ds2 = dt2 a2(t)(

    dr2

    1 Kr2+ r2d2

    )

    H2 =1

    3

    (

    1

    22 + V

    )

    Ka2

    =3Hu2a2

    1

    a

    + p

    (

    1 3K

    a2

    )

    7

    S =

    d4xg

    [

    R

    6!2Pl

    1

    2

    V ()]

    ds2 = dt2 a2(t)(

    dr2

    1 Kr2+ r2d2

    )

    H2 =1

    3

    (

    1

    22 + V

    )

    Ka2

    =3Hu2a2

    1

    a

    + p

    (

    1 3K

    a2

    )

    u +

    [

    k2

    3K

    (

    1 c2S

    )

    ]

    u = 0

    P = AknS1 cos(

    kphk#

    +

    )

    L = p(X, )

    X 1

    2g

    T = ( + p)uu pg

    2Xp

    X p

    u 2X

    8

    > 1

    3

    S =

    d4xg

    [

    R

    6"2Pl

    1

    2

    V ()]

    ds2 = dt2 a2(t)(

    dr2

    1 Kr2+ r2d2

    )

    H2 =1

    3

    (

    1

    22 + V

    )

    Ka2

    =3Hu2a2

    1

    a

    + p

    (

    1 3K

    a2

    )

    u +

    [

    k2

    3K

    (

    1 c2S

    )

    ]

    u = 0

    P = AknS1 cos2(

    kphk#

    +

    )

    L = p(X, )

    X 1

    2g

    T = ( + p)uu pg

    2Xp

    X p

    8

    Cargese - 6th July 2010 14

    10 20 50 100 200 500 1000

    1000

    2000

    3000

    4000

    5000

    6000

    l

    !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    !l"1"Cl

    2

    Data!

    105

    104

    1000

    100

    10

    1

    104 1000

    0.001 0.01 0.1 1 10

    100 10 1

    Wavelength (h1 Mpc)

    Wave number k (h / Mpc)

    Pow

    er s

    pect

    rum

    tod

    ay P

    (k)

    (h1

    Mpc

    )3

    CMBSDSS Galaxies Cluster abundanceGravitational lensingLyman forest

    No obvious oscillations ...

  • Cargese - 6th July 2010 15

    1 10 100 1000l

    02000

    4000

    6000

    l(l

    +1)C

    l /2!!!"!#"$%

    WMAP1WMAP3

    #=103 k

    *=10

    -2 Mpc

    -1

    Cargese - 6th July 2010

    T0a30 ! 1500!Pl

    nS = 0.96 0.02 = w < 8 104

    ! 0.62

    T

    S=

    C(T)10

    C(S)10= F (, )

    A2T

    A2S

    w

    T

    S! 4 102

    nS 1

    dH a(t) t

    ti

    d

    a()

    ti

    d

    dt| 1| = 2

    a

    a3

    tdiffusiontHubble

    R1/3H

    (

    1 +

    AR2H

    )

    =

    > 1

    3

    6

    16

    A specific model: 4D Quantum cosmology

    Perfect fluid: bounce

    no horizon problem if

    Results:

  • Cargese - 6th July 2010 17

    Digression: about QM

    Schrdinger

    Polar form of the wave function

    Hamilton-Jacobi

    quantumpotential

    Cargese - 6th July 2010 18

    Ontological interpretation (BdB)

    Trajectories satisfy

    Properties:

    classical limit well definedstate dependent intrinsic reality

    no need for external classical domain!

    strictly equivalent to Copenhagen QMprobability distribution (attractor)

    non local

    x(t)

    md2x(t)

    dt2= (V + Q)

    t0; (x, t0) = | (x, t0)|2

    Q 0

    ds2 = N2()d a2()ijdxidxj

    p = p0

    + sN(1 + )

    1+

    (, , s) =

    T = pses/s0p(1+) s00

    H =p

    2a

    4aKa + pT

    a3

    N

    a3

    2

  • Cargese - 6th July 2010 19

    The two-slit experiment:

    Cargese - 6th July 2010 20

    Trajectories in the two-slit experiment

  • Cargese - 6th July 2010

    H =

    (

    p2a4a

    Ka +pTa3

    )

    N

    a3

    H = 0

    i

    T=

    1

    4a3(1)/2

    a

    [

    a(31)/2

    a

    ]

    + Ka

    K = 0 = 2a3(1)/2

    3(1 )= i

    T=

    1

    4

    2

    2

    > 0

    =

    =

    eiET (E)E(T )dE

    e(ET0)2

    =

    [

    8T0

    (T 20 + T2)

    2

    ]14

    exp

    (

    T0

    2

    T 20 + T2

    )

    eiS(,T )

    S =T2

    T 20 + T2

    +1

    2arctan

    T0T

    4

    4

    Quantum cosmology+ canonical transformation+ rescaling (volume )+ units

    = a simple Hamiltonian:

    space defined by constraint

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1/2

    3(1 ) = iT

    =14

    22

    > 0

    =

    3

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1/2

    3(1 ) = iT

    =14

    22

    > 0

    =

    3

    Wheeler-De Witt

    x(t)

    md2x(t)

    dt2= (V + Q)

    t0; (x, t0) = | (x, t0)|2

    Q 0

    ds2 = N2()d a2()ijdxidxj

    p = p0

    + sN(1 + )

    1+

    (, , s) =

    T = pses/s0p(1+) s00

    H =p

    2a

    4aKa + pT

    a3

    N

    a3

    H = 0

    2

    + Technical trick:

    21

    Cargese - 6th July 2010

    a = {a, H}

    a = a0

    [

    1 +

    (

    T

    T0

    )2]

    13(1)

    K = +1

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T )(2) [v, T ; a (T )]

    = 3!2

    Pl

    2

    + p

    a

    d

    d

    (v

    a

    )

    ds2 = a2(){

    (1 + 2) d2 [(1 2) ij + hij ] dxidxj}

    d = a31dT

    i(2)

    =

    d3x

    (

    1

    2

    2

    v2+

    2v,iv

    ,i a

    a

    )

    (2)

    5

    22

    WKB exact superposition:alternative way of getting the solution:

    Gaussian wave packet

    phase

    Bohmian trajectory

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1/2

    3(1 ) = iT

    =14

    22

    > 0

    =

    =

    eiET (E)E(T )dE

    3

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1/2

    3(1 ) = iT

    =14

    22

    > 0

    =

    =

    eiET (E)E(T )dE

    e(ET0)2

    3

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1/2

    3(1 ) = iT

    =14

    22

    > 0

    =

    =

    eiET (E)E(T )dE

    e(ET0)2

    =

    8T0

    (T 20 + T 2)2

    14

    exp T0

    2

    T 20 + T 2

    eiS(,T )

    3

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1/2

    3(1 ) = iT

    =14

    22

    > 0

    =

    =

    eiET (E)E(T )dE

    e(ET0)2

    =

    8T0

    (T 20 + T 2)2

    14

    exp T0

    2

    T 20 + T 2

    eiS(,T )

    S =T2

    T 20 + T 2+

    12

    arctanT0T

    4

    3

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1

    /23(1 ) = i

    T

    =14

    22

    > 0

    =

    =

    eiET (E)E(T )dE

    e(ET0)2

    =

    8T0

    (T 20 + T 2)2

    14

    exp T0

    2

    T20 + T 2

    eiS(,T )

    S =T2

    T20 + T 2

    +12

    arctanT0

    T

    4

    a = {a,H}

    3

  • Cargese - 6th July 2010 23

    quantum potential

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1

    /23(1 ) = i

    T

    =14

    22

    > 0

    =

    =

    eiET (E)E(T )dE

    e(ET0)2

    =

    8T0

    (T 20 + T 2)2

    14

    exp T0

    2

    T20 + T 2

    eiS(,T )

    S =T2

    T20 + T 2

    +12

    arctanT0

    T

    4

    a = {a,H}

    a = a0

    1 +

    T

    T0

    2 13(1)

    K = 0

    3

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1

    /23(1 ) = i

    T

    =14

    22

    > 0

    =

    =

    eiET (E)E(T )dE

    e(ET0)2

    =

    8T0

    (T 20 + T 2)2

    14

    exp T0

    2

    T20 + T 2

    eiS(,T )

    S =T2

    T20 + T 2

    +12

    arctanT0

    T

    4

    a = {a,H}

    a = a0

    1 +

    T

    T0

    2 13(1)

    K = 1

    3

    iT

    =14a3(1)/2

    a

    a(31)/2

    a

    +Ka

    K = 0 = 2a2(1

    /23(1 ) = i

    T

    =14

    22

    > 0

    =

    =

    eiET (E)E(T )dE

    e(ET0)2

    =

    8T0

    (T 20 + T 2)2

    14

    exp T0

    2

    T20 + T 2

    eiS(,T )

    S =T2

    T20 + T 2

    +12

    arctanT0

    T

    4

    a = {a,H}

    a = a0

    1 +

    T

    T0

    2 13(1)

    K = +1

    3

    a(T )

    Q(T )

    4

    a(T )

    Q(T )

    4

    Cargese - 6th July 2010 24

    What about the perturbations?

    conformal time

    Hamiltonian up to 2nd order

    factorization of the wave function

    comes from 0th order

    Bardeen (Newton) gravitational potential

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2)) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    d = a31dT

    4

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2)) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    d = a31dT

    4

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2)) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    d = a31dT

    4

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2)) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    ds2 = a2()(1 + 2) d2 [(1 2) ij + hij ] dxidxj

    d = a31dT

    i(2)

    =

    d3x1

    22

    v2+

    2v,iv

    ,i a

    a

    (2)

    vk +

    c2Sk

    2 a

    a

    vk = 0

    c2S

    =

    = 0

    vk exp(icSk

    2cSk

    4

    K = +1

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    ds2 = a2()(1 + 2) d2 [(1 2) ij + hij ] dxidxj

    d = a31dT

    i(2)

    =

    d3x1

    22

    v2+

    2v,iv

    ,i a

    a

    (2)

    vk +

    c2Sk

    2 a

    a

    vk = 0

    c2S

    =

    = 0

    4

  • Cargese - 6th July 2010 25

    + canonical transformations:

    Fourier mode

    vacuum initial conditions

    + evolution (matchings and/or numerics)

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2)) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    d = a31dT

    i(2)

    =

    d3x1

    22

    v2+

    2v,iv

    ,i a

    a

    (2)

    4

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2)) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    d = a31dT

    i(2)

    =

    d3x1

    22

    v2+

    2v,iv

    ,i a

    a

    (2)

    vk +

    c2Sk

    2 a

    a

    vk = 0

    4

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2)) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    d = a31dT

    i(2)

    =

    d3x1

    22

    v2+

    2v,iv

    ,i a

    a

    (2)

    vk +

    c2Sk

    2 a

    a

    vk = 0

    c2S

    =

    = 0

    4

    t; !(t) = !0a0

    a(t) !

    Pl

    V ()

    4 1012

    t(); a(t) 0

    Einf $ 103M

    Pl

    S3 > S2 > S1

    a()

    +g

    +d

    = Tij(k)

    g

    d

    vk eicSk

    2cSk

    1

    Cargese - 6th July 2010 26

    4

    -1!108

    -8!107

    -6!107

    -4!107

    -2!107 0

    2!107

    4!107

    6!107

    8!107

    1!108

    x

    102

    104

    106

    108

    1010

    S(x

    )

    |"e S(x)|

    |#m S(x)||S(x)|

    ~k=10

    -3

    -1!108

    -5!107 0

    5!107

    1!108

    x

    100

    102

    104

    106

    108

    v(x

    )

    |"e v(x)||#m v(x)||v(x)|

    ~k=10

    -3

    FIG. 1: Time evolution of the scalar mode function for the equation of state = 0.1 in the one-fluid model of the bounce. Theleft panel shows the full time evolution which was computed, i.e., the function S(x), while the right panel shows v(x) itself,both plots having k = 103. For x < 0, there are oscillations only in the real and imaginary parts of the mode, so the amplitudeshown is a non oscillating function of time. It however acquires an oscillating piece after the bounce has taken place.

    (a, T ) = a(13)/2

    (0)(a, T )

    2. Its continuity equation

    coming from Eq. (17) reads

    T

    a

    [

    a(31)

    2

    S

    a

    ]

    = 0, (21)

    which implies in the Bohm interpretation that

    a = a(31)

    2

    S

    a, (22)

    in accordance with the classical relations a = {a, H} =a(31)Pa/2 and Pa = S/a.

    Inserting the phase of (20) into Eq. (22), we obtain theBohmian quantum trajectory for the scale factor:

    a(T ) = a0

    [

    1 +

    (

    T

    T0

    )2]

    13(1)

    . (23)

    Note that this solution has no singularities and tends tothe classical solution when T . Remember that weare in the gauge N = a3, and T is related to conformaltime through

    NdT = ad = d = [a(T )]31 dT. (24)

    The solution (23) can be obtained for other initial wavefunctions (see Ref. [8]).

    The Bohmian quantum trajectory a(T ) can be usedin Eq. (16). Indeed, since one can view a(T ) as a func-tion of T , it is possible to implement the time dependentcanonical transformation generated by

    U = exp

    [

    i

    (

    d3xav2

    2a

    )]

    (25)

    exp{

    i

    [

    d3x

    (

    v + v

    2

    )

    ln

    (

    1

    a

    )]}

    . (26)

    As a(T ) is a given quantum trajectory coming fromEq. (17), Eq. (25) must be viewed as the generator ofa time dependent canonical transformation to Eq. (17).It yields, in terms of conformal time, the equation for(2)[v, a(), ]

    i(2)

    =

    d3x

    (

    1

    2

    2

    v2+

    2v,iv

    ,i a

    2av2

    )

    (2).

    (27)

    This is the most simple form of the Schrodinger equa-tion which governs scalar perturbations in a quantumminisuperspace model with fluid matter source. The cor-responding time evolution equation for the operator v inthe Heisenberg picture is given by

    v v,i,i a

    av = 0, (28)

    where a prime means derivative with respect to confor-mal time. In terms of the normal modes vk, the aboveequation reads

    vk +

    (

    k2 a

    a

    )

    vk = 0. (29)

    These equations have the same form as the equations forscalar perturbations obtained in Ref. [1]. This is quitenatural since for a single fluid with constant equation ofstate , the pump function z/z obtained in Ref. [1] isexactly equal to a/a obtained here. The difference isthat the function a() is no longer a classical solutionof the background equations but a quantum Bohmiantrajectory of the quantized background, which may leadto different power spectra.

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2)) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    d = a31dT

    i(2)

    =

    d3x1

    22

    v2+

    2v,iv

    ,i a

    a

    (2)

    vk +

    c2Sk

    2 a

    a

    vk = 0

    c2S

    =

    = 0

    vk exp(icSk

    2cSk

    s(x) = a12 (13)

    v(x)T0

    4

    a(T )

    Q(T )

    H = H(0) + H(2) +

    = (0) (a, T ) (2)) [v, T ; a (T )]

    = 32

    Pl

    2

    + p

    a

    dd

    v

    a

    d = a31dT

    i(2)

    =

    d3x1

    22

    v2+

    2v,iv

    ,i a

    a

    (2)

    vk +

    c2Sk

    2 a

    a

    vk = 0

    c2S

    =

    = 0

    vk exp(icSk

    2cSk

    s(x) = a12 (13)

    v(x)T0

    x TT0

    4

    4

    -1!108

    -8!107

    -6!107

    -4!107

    -2!107 0

    2!107

    4!107

    6!107

    8!107

    1!108

    x

    102

    104

    106

    108

    1010

    S(x

    )

    |"e S(x)|

    |#m S(x)||S(x)|

    ~k=10

    -3

    -1!108

    -5!107 0

    5!107

    1!108

    x

    100

    102

    104

    106

    108

    v(x

    )

    |"e v(x)||#m v(x)||v(x)|

    ~k=10

    -3

    FIG. 1: Time evolution of the scalar mode function for the equation of state = 0.1 in the one-fluid model of the bounce. Theleft panel shows the full time evolution which was computed, i.e., the function S(x), while the right panel shows v(x) itself,both plots having k = 103. For x < 0, there are oscillations only in the real and imaginary parts of the mode, so the amplitudeshown is a non oscillating function of time. It however acquires an oscillating piece after the bounce has taken place.

    (a, T ) = a(13)/2

    (0)(a, T )

    2. Its continuity equation

    coming from Eq. (17) reads

    T

    a

    [

    a(31)

    2

    S

    a

    ]

    = 0, (21)

    which implies in the Bohm interpretation that

    a = a(31)

    2

    S

    a, (22)

    in accordance with the classical relations a = {a, H} =a(31)Pa/2 and Pa = S/a.

    Inserting the phase of (20) into Eq. (22), we obtain theBohmian quantum trajectory for the scale factor:

    a(T ) = a0

    [

    1 +

    (

    T

    T0

    )2]

    13(1)

    . (23)

    Note that this solution has no singularities and tends tothe classical solution when T . Remember that weare in the gauge N = a3, and T is related to conformaltime through

    NdT = ad = d = [a(T )]31 dT. (24)

    The solution (23) can be obtained for other initial wavefunctions (see Ref. [8]).

    The Bohmian quantum trajectory a(T ) can be usedin Eq. (16). Indeed, since one can view a(T ) as a func-tion of T , it is possible to implement the time dependentcanonical transformation generated by

    U = exp

    [

    i

    (

    d3xav2

    2a

    )]

    (25)

    exp{

    i

    [

    d3x

    (

    v + v

    2

    )

    ln

    (

    1

    a

    )]}

    . (26)

    As a(T ) is a given quantum trajectory coming fromEq. (17), Eq. (25) must be viewed as the generator ofa time dependent canonical transformation to Eq. (17).It yields, in terms of conformal time, the equation for(2)[v, a(), ]

    i(2)

    =

    d3x

    (

    1

    2

    2

    v2+

    2v,iv

    ,i a

    2av2

    )

    (2).

    (27)

    This is the most simple form of the Schrodinger equa-tion which governs scalar perturbations in a quantumminisuperspace model with fluid matter source. The cor-responding time evolution equation for the operator v inthe Heisenberg picture is given by

    v v,i,i a

    av = 0, (28)

    where a prime means derivative with respect to confor-mal time. In terms of the normal modes vk, the aboveequation reads

    vk +

    (

    k2 a

    a

    )

    vk = 0. (29)

    These equations have the same form as the equations forscalar perturbations obtained in Ref. [1]. This is quitenatural since for a single fluid with constant equation ofstate , the pump function z/z obtained in Ref. [1] isexactly equal to a/a obtained here. The difference isthat the function a() is no longer a classical solutionof the background equations but a quantum Bohmiantrajectory of the quantized background, which may leadto different power spectra.

  • Cargese - 6th July 2010 27

    7

    -100 -75 -50 -25 0 25 50 75 100~kx

    10-20

    10-15

    10-10

    10-5

    100

    105

    Spec

    tra

    P!

    (x)/ ~k

    nS-1

    , ~k=10

    -7

    Ph(x)/

    ~k

    nT,

    ~k=10

    -7

    P!

    (x)/ ~k

    nS-1

    , ~k=10

    -5

    Ph(x)/

    ~k

    nT,

    ~k=10

    -5

    "=0.01

    FIG. 2: Rescaled power spectra for scalar and tensor pertur-bations as functions of time for = 0.01 and two differentvalues of k. It is clear from the figure that not only bothspectra reach a constant mode, but also that this mode doesbehave as indicated in Eqs. (46) and (50). It is purely inci-dental that the actual constant value of both modes are veryclose for that particular value of . The constant values ob-tained in this figure are the one used to derive the spectrumbelow. In this figure and the following, the value of nS usedto rescale is the one derived in Eq. (46), thus proving thevalidity of the analytic calculation.

    Defining

    L0 = T0a30 , (53)

    the curvature scale at the bounce (the characteristicbounce length-scale), namely, L0 1/

    R0 where R0 is

    the scalar curvature at the bounce, from which one canwrite

    k =k

    a0L0

    L0bouncephys

    , (54)

    one obtains the scalar perturbation density spectrum asa function of time through

    P =2( + 1)

    2

    (1 ))2k|f(x)|2

    (

    $PlL0

    )2

    , (55)

    where

    f(x) (

    1 + x2) 1+33(1) dS

    dx x

    (

    1 + x2) 43(1) S,

    while the tensor power spectrum is

    Ph =2k3

    2|v|2

    1 + x2

    (

    $PlL0

    )2

    , (56)

    in which the rescaled function v, defined through [11]

    v a

    12 (13)

    $Pl

    T0,

    10-6

    10-5

    10-4

    10-3

    10-2

    10-1

    100

    101

    ~k

    10-4

    10-2

    100

    102

    P!

    ( ~ k)/

    ~ kn

    S-1

    , P

    h(

    ~ k)/

    ~ kn

    T a

    nd T

    /S

    P!

    ( ~k)/

    ~k

    nS-1

    Ph(

    ~k)/

    ~k

    nT

    T/S

    FIG. 3: Rescaled power spectra for = 8104, correspond-ing to the conservative maximum bound on the deviation froma scale invariant spectrum nS = 1.01, as function of k. Thescalar spectrum P is the full line, while the dashed line isthe gravitational wave spectrum Ph. Also shown is the ratioT/S (dotted); in this case, the T/S " 5.2 103, i.e. almosttwo orders of magnitude below the current limit. This casehas a typical bounce length-scale of L0 1.47 10

    3"Pl . Theamplitude of the modes is obtained as the constant part ofFig. 2.

    satisfies the same dynamical equation (51) with cs 1,with subject to initial condition

    ini =

    3

    k$Ple

    ik . (57)

    From the above defined spectra, one reads the ampli-tudes

    A2S

    4

    25P (58)

    and

    A2T

    1

    100Ph, (59)

    where we assume the classical relation between and thecurvature perturbation through

    =5 + 3

    3(1 + ) P =

    [

    5 + 3

    3(1 + )

    ]2

    P, (60)

    to obtain the observed spectrum. Since both spectra areidentical power laws, and indeed almost scale invariantpower laws, the tensor-to-scalar (T/S) ratio, defined bythe CMB multipoles C$ at $ = 10 as

    T

    S

    C(T)10

    C(S)10= F(, )

    A2T

    A2S

    , (61)

    can easily be computed (see, e.g., [16] and referencestherein). In Eq. (61), the function F depends entirely

    Cargese - 6th July 2010 28

    spectrum

    id. grav. waves:

    same dynamics + initial conditions same spectrum

    CMB normalisation

    bounce curvature

    x TT0

    P k3 |k|2 A2SknS1

    5

    x TT0

    P k3 |k|2 A2SknS1

    +

    k2 a

    a

    = 0

    ha

    Ph k3 |hk|2 A2TknT

    ini exp(ik)

    k

    nT = nS 1 =12

    1 + 3

    A2S

    = 2.08 1010

    T0a30 1500Pl

    5

    x TT0

    P k3 |k|2 A2SknS1

    +

    k2 a

    a

    = 0

    ha

    Ph k3 |hk|2 A2TknT

    ini exp(ik)

    k

    nT = nS 1 =12

    1 + 3

    A2S

    = 2.08 1010

    T0a30 1500Pl

    5

    x TT0

    P k3 |k|2 A2SknS1

    +

    k2 a

    a

    = 0

    ha

    Ph k3 |hk|2 A2TknT

    ini exp(ik)

    k

    nT = nS 1 =12

    1 + 3

    A2S

    = 2.08 1010

    T0a30 1500Pl

    5

    x TT0

    P k3 |k|2 A2SknS1

    +

    k2 a

    a

    = 0

    ha

    Ph k3 |hk|2 A2TknT

    ini exp(ik)

    k

    nT = nS 1 =12

    1 + 3

    A2S

    = 2.08 1010

    T0a30 1500Pl

    5

    x TT0

    P k3 |k|2 A2SknS1

    +

    k2 a

    a

    = 0

    ha

    Ph k3 |hk|2 A2TknT

    ini exp(ik)

    k

    nT = nS 1 =12

    1 + 3

    A2S

    = 2.08 1010

    T0a30 1500Pl

    5

    x TT0

    P k3 |k|2 A2SknS1

    +

    k2 a

    a

    = 0

    ha

    Ph k3 |hk|2 A2TknT

    ini exp(ik)

    k

    nT = nS 1 =12

    1 + 3

    A2S

    = 2.08 1010

    T0a30 1500Pl

    5

    x TT0

    P k3 |k|2 A2SknS1

    +

    k2 a

    a

    = 0

    ha

    Ph k3 |hk|2 A2TknT

    ini exp(ik)

    k

    nT = nS 1 =12

    1 + 3

    A2S

    = 2.08 1010

    T0a30 1500Pl

    5

  • Cargese - 6th July 2010 29

    WMAP constraint

    predictions

    spectrum slightly blue

    power-law + concordance

    x TT0

    P k3 |k|2 A2SknS1

    +

    k2 a

    a

    = 0

    ha

    Ph k3 |hk|2 A2TknT

    ini exp(ik)

    k

    nT = nS 1 =12

    1 + 3

    A2S

    = 2.08 1010

    T0a30 1500Pl

    nS = 0.96 0.02 = nS < 8 104

    0.62

    5

    T

    S=

    C(T)10

    C(S)10= F (, )

    A2T

    A2S

    w

    T

    S 4 102

    nS 1

    6

    T

    S=

    C(T)10

    C(S)10= F (, )

    A2T

    A2S

    w

    T

    S 4 102

    nS 1

    6

    T0a30 ! 1500!Pl

    nS = 0.96 0.02 = w < 8 104

    ! 0.62

    T

    S=

    C(T)10

    C(S)10= F (, )

    A2T

    A2S

    w

    T

    S! 4 102

    nS 1

    dH a(t) t

    ti

    d

    a()

    ti

    6

    P. P. & N. Pinto-Neto, Phys. Rev. D78, 063506 (2008)

    Cargese - 6th July 2010 30

    monopoles = ???

    Dark energy ...

    Cosmology without inflation?

    New solutions to old puzzles

    No singularity

    G.R. ...

    (oscillations, T/S ...)New predictions

    String implementation

    Cosmology without inflation!

    Model dependence

    Future

    Non gaussianities