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Cosmology and stability in scalarndashtensor bigravity

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2014 Class Quantum Grav 31 075016

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Classical and Quantum Gravity

Class Quantum Grav 31 (2014) 075016 (16pp) doi1010880264-9381317075016

Cosmology and stability in scalarndashtensorbigravity

Kazuharu Bamba1 Yusuke Kokusho2 Shinrsquoichi Nojiri12

and Norihito Shirai2

1 Kobayashi-Maskawa Institute for the Origin of Particles and the UniverseNagoya University Nagoya 464-8602 Japan2 Department of Physics Nagoya University Nagoya 464-8602 Japan

E-mail nojiriphysnagoya-uacjp

Received 8 January 2014 revised 13 February 2014Accepted for publication 19 February 2014Published 14 March 2014

AbstractThe bigravity models coupled with two scalar fields are constructed Weshow that a wide class of the expansion history of the universe especiallycorresponding to dark energy andor inflation can be described by a solutionof the bigravity model We discuss the stability of the solution and give theconditions for the stability We also explicitly construct a model which givesa stable solution By using the stable model for an arbitrary evolution of theuniverse expansion we construct the BransndashDicke like model which reproducesthe evolution

Keywords bigravity massive gravity dark energyPACS numbers 9536+x 1210minusg 1110Ef

1 Introduction

Although free massive gravity was established about seventy-five years ago in [1] (for a recentreview see [2]) it is difficult to obtain consistent interacting or nonlinear models for a longtime because such a model contains the BoulwarendashDeser ghost [3 4] in general and therealso appears the van DamndashVeltmanndashZakharov discontinuity [5] in the massless limit m rarr 0It is known that the discontinuity can be screened by the Vainshtein mechanism [6] (see forexample [7]) Note that the extra degrees of freedom are cancelled by the ghost degrees offreedom

Recently however the study of the nonlinear massive gravity has progressed remarkablyand the ghost-free models which are called the de Rham Gabadadze Tolley (dRGT) modelshave been constructed and found for non-dynamical background metric in [8ndash10] and for

0264-938114075016+16$3300 copy 2014 IOP Publishing Ltd Printed in the UK 1

Class Quantum Grav 31 (2014) 075016 K Bamba et al

dynamical metric [11] The so-called minimal model first appeared in [12] (for the generalproof of absence of ghost in massive gravity see [13]) Even in the dRGT models theVainshtein mechanism works although there does not appear ghost Since the models for thedynamical background metric contain two metric (or symmetric tensor fields) such modelsare called as bigravity models

The massive gravity models have been applied to cosmology in [14ndash17] For the bimetricgravity some cosmological solutions including the ones describing the accelerating universehave been investigated [18ndash25] In case of the dRGT model it became clear that the flatFriedmannndashRobertsonndashWalker (FRW) cosmology is absent [17] In this paper we considerthe bigravity models with scalar fields We show that there are models which admit the stablesolution describing the FRW solution with the spatially flat metric

The organization of the paper is as follows In the next section we show one of thedifficulties in massive gravity theories with a non-dynamical background metric to constructa model in which the non-trivial FRW cosmology is realized Then in section 3 we considerbigravity models where the background metric is also dynamical We construct bigravitymodels coupled with two scalar fields and show that a wide class of the expansion history ofthe universe can be described by a solution of the bigravity model The solution is howevernot always stable under the perturbation that is the small perturbation to the solution growsup in general In section 4 we investigate the stability of the solutions and give the conditionsfor the stability Furthermore we explicitly build a model which gives a stable solution Insection 5 by using the stable model obtained in section 4 for an arbitrary evolution of theuniverse expansion we construct the BransndashDicke like model which reproduces the evolutionThe last section is devoted to conclusions

2 Difficulties of cosmology by massive gravity with scalar field

In this section we show that it is very difficult to construct a massive gravity model coupledwith a scalar field Such a coupling could lead to a solution describing the FRW space-timewith the vanishing spatial curvature

The starting action is given by

Smg = M2g

intd4x

radicminus det gR(g) + 2m2M2

eff

intd4x

radicminus det g

4sumn=0

βnen(radic

gminus1 f ) (1)

where R(g) is the scalar curvature for gμν and fμν is a non-dynamical reference metric Thetensor

radicgminus1 f is defined by the square root of gμρ fρν namely (

radicgminus1 f )μ ρ (

radicgminus1 f )ρ ν =

gμρ fρν For general tensor Xμν en(X )rsquos are defined by

e0(X ) = 1 e1(X ) = [X] e2(X ) = 12 ([X]2 minus [X2])

e3(X ) = 16 ([X]3 minus 3[X][X2] + 2[X3])

e4(X ) = 124 ([X]4 minus 6[X]2[X2] + 3[X2]2 + 8[X][X3] minus 6[X4])

ek(X ) = 0 for k gt 4 (2)

Here [X] expresses the trace of arbitrary tensor Xμν [X] = Xμ

μWe add the following terms to the action (1)

Sφ = minusM2g

intd4x

radicminus det g

1

2gμνpartμφpartνφ + V (φ)

(3)

2

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By the conformal transformation gμν rarr eminusϕ(φ)gJμν the total action SBD = Smg + Sϕ is

transformed as

SBD = +M2g

intd4x

radicminus det gJ

eminusϕRJ(g) minus 1

2eminusϕ(φ)(1 minus 3ϕprime(φ)2)gμνpartμφpartνφ minus eminus2ϕV (ϕ)

+ 2m2M2eff

intd4x

radicminus det gJ

4sumn=0

βn e(n2 minus2)ϕen(

radicgJminus1 f ) (4)

Thus we obtain a BransndashDicke type model Then if we have a solution where the scalar fieldis not constant but depends on the time coordinate and the space-time is the arbitrary FRWbackground even if flat background we may obtain arbitrary history of the expansion by theconformal transformation As we will see in (14) later however the scalar field should beconstant which is the motivation why we consider the bigravity

In the following just for simplicity we only investigate the minimal case [12]

Smg = M2g

intd4x

radicminus det g R(g) + 2m2M2

eff

intd4x

radicminus det g (3 minus tr

radicgminus1 f + det

radicgminus1 f ) (5)

In terms of en in (2) we find 3 minus trradic

gminus1 f + detradic

gminus1 f = 3e0((radic

gminus1 f )μ ν ) minuse1((

radicgminus1 f )μ ν )+e4((

radicgminus1 f )μ ν ) The minimal case could be a simplest but non-trivial case

and proposed in [12] In the model the interaction between two metrics gμν and fμν is onlygiven by the trace of (

radicgminus1 f )μ ν When we consider non-minimal models the calculations

becomes rather complicated but the quantitative structure in the arguments in this paper couldnot be changed In order to evaluate δ

radicgminus1 f two matrices M and N which satisfy the relation

M2 = N are taken Since δMM + MδM = δN we have

tr δM = 12 tr(Mminus1δN) (6)

For a while we examine the Einstein frame action (5) with (3) but matter contribution isneglected Therefore by the variation over gμν we acquire

0 = M2g

(12 gμνR(g) minus R(g)

μν

) + m2M2eff

gμν (3 minus tr

radicgminus1 f ) + 1

2 fμρ(radic

gminus1 f )minus1ρν

+ 12 fνρ (

radicgminus1 f )minus1ρ

μ

+ M2g

[12

(13 gρσ partρφpartσφ + V (φ)

)gμν minus 1

2partμφpartνφ] (7)

We should note that detradic

gdetradic

gminus1 f = detradic

f in general The variation of the scalar field φ

yields

0 = minusgφ + V prime(φ) (8)

where g is the drsquoAlembertian with respect to the metric g By multiplying the covariantderivative nablaμ

g with respect to the metric g by equation (7) and using the Bianchi identity0 = nablaμ

g ( 12 gμνR(g) minus R(g)

μν ) and equation (8) we have

0 = minusgμνnablaμg (tr

radicgminus1 f ) + 1

2nablaμg

fμρ(

radicgminus1 f )minus1ρ

ν + fνρ (radic

gminus1 f )minus1ρμ

(9)

In case of the Einstein gravity the conservation law of the energyndashmomentum tensor dependson the Einstein equation It can be derived from the Bianchi identity In case of massivegravity however the conservation laws of the energyndashmomentum tensor of the scalar fieldsare derived from the scalar field equations These conservation laws are independent of theEinstein equation The Bianchi identities present the equation (9) independent of the Einsteinequation

We assume the FRW universe for the metric gμν and the flat Minkowski space-time forfμν and use the conformal time t = τ for the universe with metric gμν

3

Class Quantum Grav 31 (2014) 075016 K Bamba et al

ds2g =

3sumμν=0

gμν dxμ dxν = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

)

ds2f =

3sumμν=0

fμν dxμ dxν = minusdτ 2 +3sum

i=1

(dxi)2 (10)

The physical meaning of the metric fμν has not been clear although there are several conjecturesas in [26] The longitudinal scalar modes in the metric play the roles when we consider thebackground as in (10) but these modes do not propagate which may be found by consideringthe local Lorentz frame The propagating modes could be massless tensor (massless graviton)and the massive tensor (massive graviton) and any scalar mode does not propagate

The (τ τ ) component of (7) reads

0 = minus3M2gH2 minus 3m2M2

eff(a2 minus a) + (

14 φ2 + 1

2V (φ)a(τ )2)M2g (11)

and from (i j) components we find

0 = M2g (2H + H2) + 3m2M2

eff(a2 minus a) + (

12 φ2 minus 1

2V (φ)a(τ )2)M2

g (12)

with H = aa where the dot denotes the derivative with respect to t From equation (9) wehave the following equation

a

a= 0 (13)

Hence a should be a constant a = a0 This means that the only consistent solution for gμν isthe flat Minkowski space Furthermore by using (11) and (12) we find

φ = 0 0 = 3m2M2eff

(a2

0 minus a0) minus 1

2V0a20M2

g (14)

Since φ is a constant we cannot obtain the expanding universe

3 Bigravity with two scalar fields and cosmology

In the last section we have observed difficulties to construct the model which generates theexpanding universe In this section we build models of cosmology by using the bigravity withtwo scalar fields The bimetric gravity proposed in [11] includes two metric tensors gμν andfμν In addition to the massless spin-two field corresponding to graviton it contains massivespin-two field although massive gravity models only have the massive spin-two field TheBoulwarendashDeser ghost [3] does not appear in such a theory

31 Bigravity models with scalar fields

We add the term containing the scalar curvature R( f ) given by fμν to the action (1) as follows

Sbi = M2g

intd4x

radicminus det gR(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det g

4sumn=0

βnen(radic

gminus1 f ) (15)

Here Meff is defined by1

M2eff

= 1

M2g

+ 1

M2f

(16)

There is a conjecture that the two dynamical metric may correspond to manifolds with twometric [26]

4

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We also involve the following terms given by two scalar fields ϕ and χ in the action (15)

Sϕ = minusM2g

intd4x

radicminus det g

1

2gμνpartμϕpartνϕ + V (ϕ)

+

intd4xLmatter(gμνi) (17)

Sξ = minusM2f

intd4x

radicminus det f

1

2f μνpartμξpartνξ + U (ξ )

(18)

For simplicity we start from the minimal case again

Sbi = M2g

intd4x

radicminus det gR(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det g (3 minus tr

radicgminus1 f + det

radicgminus1 f ) (19)

For a while we neglect the contributions from matters By the variation over gμν we againfind (7) On the other hand through the variation over fμν we acquire

0 = M2f

(12 fμνR( f ) minus R( f )

μν

) + m2M2eff

radicdet( f minus1g)

minus 12 fμρ(

radicgminus1 f )ρ ν minus 1

2 fνρ (radic

gminus1 f )ρ μ

+ det(radic

gminus1 f ) fμν

+ M2f

[12

(13 f ρσ partρξpartσ ξ + U (ξ )

)fμν minus 3

2partμξpartνξ] (20)

The variations of the scalar fields ϕ and ξ lead to

0 = minusgϕ + V prime(ϕ) 0 = minus f ξ + U prime(ξ ) (21)

corresponding to (8) Here f is the drsquoAlembertian with respect to the metric f Fromequation (7) and the Bianchi identity again we acquire (9) Similarly by using the covariantderivative nablaμ

f with respect to the metric f from (20) we have

0 = nablaμ

f

[radicdet( f minus1g)

minus 12 (

radicgminus1 f )minus1ν

σ gσμ minus 12 (

radicgminus1 f )minus1μ

σ gσν + det(radic

gminus1 f ) f μν]

(22)

The identities (9) and (22) impose strong constraints on the solutions Especially weinvestigate the solutions describing the FRW universe in the next subsection

32 Reconstruction of bigravity models

We examine whether we can construct models describing the arbitrarily given evolution of theexpansion in the universe

We take the FRW universes for the metric gμν as in (10) and use the conformal time t = τ Moreover instead of (10) we suppose the form of the metric fμν as follows

ds2g =

3sumμν=0

gμν dxμ dxν = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

)

ds2f =

3sumμν=0

fμν dxμ dxν = minusc(τ )2 dτ 2 + b(τ )23sum

i=1

(dxi)2 (23)

We should note the assumption in (10) could be most general form if we assume the spatial partof the space-time is uniform homogeneous and flat The redefinition of the time-coordinatealways gives the form of ds2

g but there does not any more freedom to choose c(τ ) = 1 norc(τ ) = b(τ ) In this case from the (τ τ ) component of (7) we find

0 = minus3M2gH2 minus 3m2M2

eff(a2 minus ab) + (

14 ϕ2 + 1

2V (ϕ)a(τ )2)

M2g (24)

5

Class Quantum Grav 31 (2014) 075016 K Bamba et al

and (i j) components yield

0 = M2g (2H + H2) + m2M2

eff(3a2 minus 2ab minus ac) + (14 ϕ2 minus 1

2V (ϕ)a(τ )2)

M2g (25)

On the other hand the (τ τ ) component of (20) leads to

0 = minus3M2f K

2 + m2M2effc

2

(1 minus a3

b3

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (26)

and from (i j) components we find

0 = M2f (2K + 3K2 minus 2LK) + m2M2

eff

(a3c

b2minus c2

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (27)

with K = bb and L = cc Both equations (9) and (22) yield the identical equation

cH = bK orca

a= b (28)

The above equation is the constraint relating the two metrics imposed by the equations ofmotion If a = 0 we obtain c = aba On the other hand if a = 0 we find b = 0 that is aand b are constant and c can be arbitrary

Next we redefine scalars as ϕ = ϕ(η) and ξ = ξ (ζ ) and identify η and ζ with theconformal time t as η = ζ = τ Hence we acquire

ω(τ )M2g = minus 4M2

g (H minus H2) minus 2m2M2eff(ab minus ac) (29)

V (τ )a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a2 minus 5ab minus ac) (30)

σ (τ )M2f = minus4M2

f (K minus LK) minus 2m2M2eff

(minus c

b+ 1

) a3c

b2 (31)

U (τ )c(τ )2M2f = M2

f (2K + 6K2 minus 2LK)+ m2M2eff

(a3c

b2minus 2c2 + a3c2

b3

) (32)

with

ω(η) = ϕprime(η)2 V (η) = V (ϕ (η)) σ (ζ ) = ξ prime(ζ )2 U (ζ ) = U (ξ (ζ )) (33)

Consequently for arbitrary a(τ ) b(τ ) and c(τ ) if we choose ω(τ ) V (τ ) σ (τ ) and U (τ ) tosatisfy equations (29)ndash(32) the cosmological model with given evolutions of a(τ ) b(τ ) andc(τ ) can be reconstructed

A reason why we introduced two scalar fields instead of one is that there are three degreesof freedom a b and c in metrics (23) and it is not trivial to describe them by using only onescalar field which might not be impossible but we have not succeeded

33 Conformal description of the accelerating universe

In the following we use the conformal time We describe how the known cosmologies can beexpressed by using the conformal time

The conformally flat FRW universe metric is given by

ds2 = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

) (34)

In this equation when a(τ )2 = l2

τ 2 the metric (34) corresponds to the de Sitter universewhich may represent inflation or dark energy in the model under consideration On the otherhand if a(τ )2 = l2n

τ 2n with n = 1 by redefining the time coordinate as

dt = plusmn ln

τ ndτ (35)

6

Class Quantum Grav 31 (2014) 075016 K Bamba et al

ie

t = plusmn ln

n minus 1τ 1minusn (36)

the metric (34) can be rewritten as

ds2 = minusdt2 +[plusmn(n minus 1)

t

l

]minus 2n1minusn 3sum

i=1

(dxi)2 (37)

Equation (37) shows that if 0 lt n lt 1 the metric corresponds to the phantom universe [27]if n gt 1 to the quintessence universe and if n lt 0 to decelerating universe In case of thephantom universe (0 lt n lt 1) we should choose + sign in plusmn of (35) or (36) and shift tin (37) as t rarr t minus t0 The time t = t0 corresponds to the Big Rip and the present time ist lt t0 and the limit of τ rarr infin is equivalent to the infinite past (t rarr minusinfin) In case of thequintessence universe (n gt 1) we may again select + sign in plusmn of (35) or (36) The limit ofτ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infin to that of t rarr 0 which may beequivalent to the Big Bang In case of the decelerating universe (n lt 0) we may take minus signin plusmn of (35) or (36) The limit of τ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infinto that of t rarr 0 which may again be considered to be the Big Bang We should also note thatin case of the de Sitter universe (n = 1) the limit of τ rarr 0 corresponds to that of t rarr +infinand that of τ rarr plusmninfin to that of t rarr minusinfin

34 Dark energy universe with a(τ ) = b(τ ) = c(τ )

If the space-time described by the metric gμν represents the universe where we live thefunctions c(τ ) and b(τ ) are not directly related to the expansion of our universe because thefunctions c(τ ) and b(τ ) correspond to the degrees of freedom in the Einstein frame metric fμν Therefore we may choose c(τ ) and b(τ ) in the consistent way convenient for the calculationThis does not mean c(τ ) and b(τ ) are not relevant for the physics besides the expansionof our universe In this section we simply take a(τ ) = c(τ ) = b(τ ) which satisfy thecondition (28) and therefore H = K = L From (29) and (31) we find ω(τ ) = σ (τ ) andthus ϕ(τ ) = ξ (τ ) and also V (τ ) = U (τ ) from (30) and (32)

By choosing a(τ ) = c(τ ) = b(τ ) equations (29)ndash(32) are simplified as

ω(τ ) = σ (τ ) = 4(minusH + H2) V (τ )a(τ )2 = U (τ )a(τ )2 = (2H + 4H2) (38)

Let us construct the models where the scale factor squared is given by a(τ )2 = l2n

τ 2n In thiscase we find

ω(τ ) = σ (τ ) = 4n(n minus 1)

τ 2 V (τ ) = U (τ ) = (2n + 4n2)l2n

τ 2(1minusn) (39)

It should be cautioned that if 0 lt n lt 1 ω(τ ) and σ (τ ) become negative and this conflictswith the definition in (33) Hence the universe corresponding to the phantom cannot be realizedas in the standard scalarndashtensor model whose situation is different from the case of F(R)-bigravity [22] (for modified gravity including F(R) gravity and dark energy problem see eg[28ndash31]) In case of n = 1 in which the de Sitter universe is realized both ω(τ ) and σ (τ )

vanish and V (τ ) and U (τ ) become constants This is equivalent to the cosmological constant

4 Stability of solutions

As we have shown a wide class of expansions of the universe can be reproduced in thebigravity models coupled to scalar fields The desired solution is however only one of the

7

Class Quantum Grav 31 (2014) 075016 K Bamba et al

solutions If the solution is not stable under the perturbation such a solution cannot be realizedunless we perform very fine-tuning In this section we study the stability of the solution in thelast section For this purpose we rewrite (29)ndash(32) in the following form

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2eff(a(τ )b(τ ) minus a(τ )c(τ )) (40)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a(τ )2 minus 5a(τ )b(τ ) minus a(τ )c(τ )) (41)

σ (ζ )ζ 2M2f = minus 4M2

f (K minus LK) minus 2m2M2eff

(minus c (τ )

b (τ )+ 1

)a (τ )3 c (τ )

b (τ )2 (42)

U (ζ )c(τ )2M2f =M2

f (2K| +6K2 minus 2LK)+m2M2eff

(a (τ )3 c (τ )

b (τ )2 minus 2c (τ )2 + a (τ )3 c (τ )2

b (τ )3

)

(43)

On the other hand the scalar field equations (21) can be rewritten to

0 = 3

(ω(η)η + ωprime(η)

2η2 + 2Hω(η)η

)+ V prime(η)a2

0 = 3

(σ (ζ )ζ + σ prime(ζ )

2ζ 2 + (3K minus L) σ (ζ )ζ

)+ U prime(ζ )a2 (44)

Equations in (38) implies that with a function f (τ ) if we choose

ω(η) = 4(minus f primeprime(η) + f prime(η)2) σ (ζ ) = 4(minus f primeprime(ζ ) + f prime(ζ )2)

V (η) = eminus2 f (η)(2 f primeprime(η) + 4 f prime(η)2) U (ζ ) = eminus2 f (ζ )(2 f primeprime(ζ ) + 4 f prime(ζ )2) (45)

we find the following solution

a(τ ) = b(τ ) = c(τ ) = e f (τ ) η = ζ = τ (46)

We explore the stability of the solution in (46)We may consider the following perturbation

H rarr H + δH K rarr K + δK a rarr a (1 + δ fa) b rarr b (1 + δ fb)

η rarr η + δη ζ rarr ζ + δζ (47)

In what follows just for simplicity we take

M2f = M2

g = M2eff

2= M2 (48)

Thus we obtain

d

dτ

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

= M

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

M =

⎛⎜⎜⎜⎜⎜⎜⎝

2H 0 CminusDB minusD

B3

HB (B minus 1)

A E 2C minus DB

C+DB minus 2C 3

HB (B minus 1)

0 0 0 0 1

AH minusAH 2HC minus2HC 1(1 + D

3

)AH minusADH

3 2H(C minus 2BD

3

)43 BD minus4H

⎞⎟⎟⎟⎟⎟⎟⎠

(49)

8

Class Quantum Grav 31 (2014) 075016 K Bamba et al

where

A equiv H

H+ 2H minus 4

H3

H B equiv 1 minus H2

H C equiv 1 + 2

H2

H D equiv 3m2a2

H E equiv minus H

Hminus 4

H3

H

(50)

The derivation of equations (49) and (50) is given in appendix A We should note that wehave deleted δK in (47) by using (A13)

The eigenvalue equation has the following form

0 = λ5 + c4λ4 + c3λ

3 + c2λ2 + c1λ + c0 (51)

where λ is the eigenvalue of the matrix M In order that the solution (46) could be stable allthe eigenvalues should be negative Then all the eigenmodes corresponding to the eigenvaluesdecrease and therefore any perturbation damps It requires ci gt 0 (i = 1 4) Especiallyminusc4 is the trace of the matrix M and we find

minus c4 = minus H

Hminus 4H minus 8

H3

Hlt 0 (52)

For the power expanding model (39) where H = minusnτ if τ gt 0 equation (52) leads to

4n2 + 2n + 1 lt 0 (53)

Thus there is no real solution for n As a result there does not exist any stable solution for thepower expanding model (39) On the other hand suppose τ lt 0 equation (52) yields

4n2 + 2n + 1 gt 0 (54)

for which there is a possibility that the solution might be stableWhen H = minusnτ in (39) the matrix M in (49) has the following form

M =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1+2nminusD0τminus2n+2

1minusn minus D01minusnτminus2n+2 3τ

1minusn

minus2minus2n+4n2

τ2+4n2

τ2 + 4n minus D0τ

minus2n+2

1minusn

1+2n+D0τminus2n+2

1minusn

minus2 minus 4n3t

1minusn

0 0 0 0 1

minus n(minus2minus2n+4n2)τ 2

n(minus2minus2n+4n2)τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

(1 + D0τ

minus2n+2

3

) (minus2minus2n+4n2)nτ 3 minusD0(minus2minus2n+4n2)nτminus2nminus1

3

2n(1+2n)

τ 2

minus 4(1minusn)D0τminus2n

3

4(1minusn)D0τminus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(55)

where the scale factor a(τ ) is given by a = a0τminusn and D0 equiv 3m2a2

0n Note that a0 = ln in (39)

As an example we may investigate the case n = minus12 In this case the eigenvalueequation has the following form

0 = λ

(λ minus 1

τ

) (λ minus 3

τ

) (2D0τ

4 + D0τ2 minus 2λ minus λ2τ

) (56)

Since there always appear positive eigenvalue the solution is not stableWe redefine

δ fa = (1 minus n)δ fa δ fb = (1 minus n)δ fb δH = (1 minus n)δH (57)

9

Class Quantum Grav 31 (2014) 075016 K Bamba et al

The matrix M in (55) has the following form

M =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1 + 2n minus D0τminus2n+2 minusD0τ

minus2n+2 3t

minus 2(1+2n)(1minusn)

τ2+4n2

τ

2(1 + 2n)(1 minus n)

minusD0τminus2n+2

minus1 + 4n2

+D0τminus2n+2 3t

0 0 0 0 1n(1+2n)

τ 2 minus n(1+2n)

τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

minus(1 + D0τ

minus2n+2

3

)1+2n

τ

D0(1+2n)nτminus2nminus1

32n(1+2n)

τ 2 minus 4(1minusn)D0τminus2n

34(1minusn)D0τ

minus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(58)

In section 34 in the part below equation (39) we have shown that if 0 lt n lt 1 the model isinconsistent As another example we consider the limit of n rarr 1 + 0 The matrix M in (58)reduces to

M =

⎛⎜⎜⎜⎜⎜⎜⎝

minus 2τ

0 3 minus D0 minusD0 3t

0 6τ

minusD0 3 + D0 3t

0 0 0 0 13τ 2 minus 3

τ 2 minus 6τ

6τ

1

minus 3+D0τ 3

D0τ 3

6τ 2 0 4

τ

⎞⎟⎟⎟⎟⎟⎟⎠

(59)

For this matrix the eigenvalue equation has the following form

0 = λ5 minus 14

τλ4 + (6D0 + 64)

λ3

τ 2minus (

2D20 + 66D0 + 33

) λ2

τ 3+ 3

(8D2

0 + 86D0 minus 63) λ

τ 4

minus 45(2D2

0 + 3) 1

τ 5 (60)

If

τ lt 0 and D0 gt

radic2353 minus 43

8= 0688 466 417 817 452 middot middot middot (61)

all the eigenvalues are negative and the system becomes stableIn general the eigenvalue equations (51) for the matrices (55) and (58) are rather

complicated and the explicit forms are given in appendix B As a result anyway we have founda solution which is stable under the perturbation Then we have shown that for an arbitrarilygiven history of the expansion of the universe we can construct a model who has a solutiongenerating the expansion and the solution is stable that is attractor solution

5 BransndashDicke type model

We introduce a parameter ε which is a positive but sufficiently small value (0 lt ε 1)In the previous section we have found that the model where n = 1 + ε and both τ and D0

satisfy equation (61) is stable and that the limit ε rarr 0 (n rarr 1) corresponds to the de Sitterspace In this section by starting with a model where ε gt 0 is small enough but finite weconstruct a model which reproduces an arbitrary expansion history of the universe by usingthe BransndashDicke type model

We here explore an arbitrary scale factor a(τ ) for τ lt 0 The scale factor correspondingto n = 1 + ε is given by a0τ

minus1minusε Hence the metric gμν corresponding to the scale factor a(τ )

is expressed by multiplying the metric gεμν corresponding to n = 1 + ε by a(τ )2aminus2

0 τ 2(1+ε)Since η = τ we rescale the metric gμν in the actions (17) and (19) as follows

gμν rarr a(η)minus2a20η

minus2(1+ε)gμν (62)

10

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By the conformal transformation gμν rarr eminusϕ(φ)gJμν the total action SBD = Smg + Sϕ is

transformed as

SBD = +M2g

intd4x

radicminus det gJ

eminusϕRJ(g) minus 1

2eminusϕ(φ)(1 minus 3ϕprime(φ)2)gμνpartμφpartνφ minus eminus2ϕV (ϕ)

+ 2m2M2eff

intd4x

radicminus det gJ

4sumn=0

βn e(n2 minus2)ϕen(

radicgJminus1 f ) (4)

Thus we obtain a BransndashDicke type model Then if we have a solution where the scalar fieldis not constant but depends on the time coordinate and the space-time is the arbitrary FRWbackground even if flat background we may obtain arbitrary history of the expansion by theconformal transformation As we will see in (14) later however the scalar field should beconstant which is the motivation why we consider the bigravity

In the following just for simplicity we only investigate the minimal case [12]

Smg = M2g

intd4x

radicminus det g R(g) + 2m2M2

eff

intd4x

radicminus det g (3 minus tr

radicgminus1 f + det

radicgminus1 f ) (5)

In terms of en in (2) we find 3 minus trradic

gminus1 f + detradic

gminus1 f = 3e0((radic

gminus1 f )μ ν ) minuse1((

radicgminus1 f )μ ν )+e4((

radicgminus1 f )μ ν ) The minimal case could be a simplest but non-trivial case

and proposed in [12] In the model the interaction between two metrics gμν and fμν is onlygiven by the trace of (

radicgminus1 f )μ ν When we consider non-minimal models the calculations

becomes rather complicated but the quantitative structure in the arguments in this paper couldnot be changed In order to evaluate δ

radicgminus1 f two matrices M and N which satisfy the relation

M2 = N are taken Since δMM + MδM = δN we have

tr δM = 12 tr(Mminus1δN) (6)

For a while we examine the Einstein frame action (5) with (3) but matter contribution isneglected Therefore by the variation over gμν we acquire

0 = M2g

(12 gμνR(g) minus R(g)

μν

) + m2M2eff

gμν (3 minus tr

radicgminus1 f ) + 1

2 fμρ(radic

gminus1 f )minus1ρν

+ 12 fνρ (

radicgminus1 f )minus1ρ

μ

+ M2g

[12

(13 gρσ partρφpartσφ + V (φ)

)gμν minus 1

2partμφpartνφ] (7)

We should note that detradic

gdetradic

gminus1 f = detradic

f in general The variation of the scalar field φ

yields

0 = minusgφ + V prime(φ) (8)

where g is the drsquoAlembertian with respect to the metric g By multiplying the covariantderivative nablaμ

g with respect to the metric g by equation (7) and using the Bianchi identity0 = nablaμ

g ( 12 gμνR(g) minus R(g)

μν ) and equation (8) we have

0 = minusgμνnablaμg (tr

radicgminus1 f ) + 1

2nablaμg

fμρ(

radicgminus1 f )minus1ρ

ν + fνρ (radic

gminus1 f )minus1ρμ

(9)

In case of the Einstein gravity the conservation law of the energyndashmomentum tensor dependson the Einstein equation It can be derived from the Bianchi identity In case of massivegravity however the conservation laws of the energyndashmomentum tensor of the scalar fieldsare derived from the scalar field equations These conservation laws are independent of theEinstein equation The Bianchi identities present the equation (9) independent of the Einsteinequation

We assume the FRW universe for the metric gμν and the flat Minkowski space-time forfμν and use the conformal time t = τ for the universe with metric gμν

3

Class Quantum Grav 31 (2014) 075016 K Bamba et al

ds2g =

3sumμν=0

gμν dxμ dxν = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

)

ds2f =

3sumμν=0

fμν dxμ dxν = minusdτ 2 +3sum

i=1

(dxi)2 (10)

The physical meaning of the metric fμν has not been clear although there are several conjecturesas in [26] The longitudinal scalar modes in the metric play the roles when we consider thebackground as in (10) but these modes do not propagate which may be found by consideringthe local Lorentz frame The propagating modes could be massless tensor (massless graviton)and the massive tensor (massive graviton) and any scalar mode does not propagate

The (τ τ ) component of (7) reads

0 = minus3M2gH2 minus 3m2M2

eff(a2 minus a) + (

14 φ2 + 1

2V (φ)a(τ )2)M2g (11)

and from (i j) components we find

0 = M2g (2H + H2) + 3m2M2

eff(a2 minus a) + (

12 φ2 minus 1

2V (φ)a(τ )2)M2

g (12)

with H = aa where the dot denotes the derivative with respect to t From equation (9) wehave the following equation

a

a= 0 (13)

Hence a should be a constant a = a0 This means that the only consistent solution for gμν isthe flat Minkowski space Furthermore by using (11) and (12) we find

φ = 0 0 = 3m2M2eff

(a2

0 minus a0) minus 1

2V0a20M2

g (14)

Since φ is a constant we cannot obtain the expanding universe

3 Bigravity with two scalar fields and cosmology

In the last section we have observed difficulties to construct the model which generates theexpanding universe In this section we build models of cosmology by using the bigravity withtwo scalar fields The bimetric gravity proposed in [11] includes two metric tensors gμν andfμν In addition to the massless spin-two field corresponding to graviton it contains massivespin-two field although massive gravity models only have the massive spin-two field TheBoulwarendashDeser ghost [3] does not appear in such a theory

31 Bigravity models with scalar fields

We add the term containing the scalar curvature R( f ) given by fμν to the action (1) as follows

Sbi = M2g

intd4x

radicminus det gR(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det g

4sumn=0

βnen(radic

gminus1 f ) (15)

Here Meff is defined by1

M2eff

= 1

M2g

+ 1

M2f

(16)

There is a conjecture that the two dynamical metric may correspond to manifolds with twometric [26]

4

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We also involve the following terms given by two scalar fields ϕ and χ in the action (15)

Sϕ = minusM2g

intd4x

radicminus det g

1

2gμνpartμϕpartνϕ + V (ϕ)

+

intd4xLmatter(gμνi) (17)

Sξ = minusM2f

intd4x

radicminus det f

1

2f μνpartμξpartνξ + U (ξ )

(18)

For simplicity we start from the minimal case again

Sbi = M2g

intd4x

radicminus det gR(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det g (3 minus tr

radicgminus1 f + det

radicgminus1 f ) (19)

For a while we neglect the contributions from matters By the variation over gμν we againfind (7) On the other hand through the variation over fμν we acquire

0 = M2f

(12 fμνR( f ) minus R( f )

μν

) + m2M2eff

radicdet( f minus1g)

minus 12 fμρ(

radicgminus1 f )ρ ν minus 1

2 fνρ (radic

gminus1 f )ρ μ

+ det(radic

gminus1 f ) fμν

+ M2f

[12

(13 f ρσ partρξpartσ ξ + U (ξ )

)fμν minus 3

2partμξpartνξ] (20)

The variations of the scalar fields ϕ and ξ lead to

0 = minusgϕ + V prime(ϕ) 0 = minus f ξ + U prime(ξ ) (21)

corresponding to (8) Here f is the drsquoAlembertian with respect to the metric f Fromequation (7) and the Bianchi identity again we acquire (9) Similarly by using the covariantderivative nablaμ

f with respect to the metric f from (20) we have

0 = nablaμ

f

[radicdet( f minus1g)

minus 12 (

radicgminus1 f )minus1ν

σ gσμ minus 12 (

radicgminus1 f )minus1μ

σ gσν + det(radic

gminus1 f ) f μν]

(22)

The identities (9) and (22) impose strong constraints on the solutions Especially weinvestigate the solutions describing the FRW universe in the next subsection

32 Reconstruction of bigravity models

We examine whether we can construct models describing the arbitrarily given evolution of theexpansion in the universe

We take the FRW universes for the metric gμν as in (10) and use the conformal time t = τ Moreover instead of (10) we suppose the form of the metric fμν as follows

ds2g =

3sumμν=0

gμν dxμ dxν = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

)

ds2f =

3sumμν=0

fμν dxμ dxν = minusc(τ )2 dτ 2 + b(τ )23sum

i=1

(dxi)2 (23)

We should note the assumption in (10) could be most general form if we assume the spatial partof the space-time is uniform homogeneous and flat The redefinition of the time-coordinatealways gives the form of ds2

g but there does not any more freedom to choose c(τ ) = 1 norc(τ ) = b(τ ) In this case from the (τ τ ) component of (7) we find

0 = minus3M2gH2 minus 3m2M2

eff(a2 minus ab) + (

14 ϕ2 + 1

2V (ϕ)a(τ )2)

M2g (24)

5

Class Quantum Grav 31 (2014) 075016 K Bamba et al

and (i j) components yield

0 = M2g (2H + H2) + m2M2

eff(3a2 minus 2ab minus ac) + (14 ϕ2 minus 1

2V (ϕ)a(τ )2)

M2g (25)

On the other hand the (τ τ ) component of (20) leads to

0 = minus3M2f K

2 + m2M2effc

2

(1 minus a3

b3

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (26)

and from (i j) components we find

0 = M2f (2K + 3K2 minus 2LK) + m2M2

eff

(a3c

b2minus c2

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (27)

with K = bb and L = cc Both equations (9) and (22) yield the identical equation

cH = bK orca

a= b (28)

The above equation is the constraint relating the two metrics imposed by the equations ofmotion If a = 0 we obtain c = aba On the other hand if a = 0 we find b = 0 that is aand b are constant and c can be arbitrary

Next we redefine scalars as ϕ = ϕ(η) and ξ = ξ (ζ ) and identify η and ζ with theconformal time t as η = ζ = τ Hence we acquire

ω(τ )M2g = minus 4M2

g (H minus H2) minus 2m2M2eff(ab minus ac) (29)

V (τ )a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a2 minus 5ab minus ac) (30)

σ (τ )M2f = minus4M2

f (K minus LK) minus 2m2M2eff

(minus c

b+ 1

) a3c

b2 (31)

U (τ )c(τ )2M2f = M2

f (2K + 6K2 minus 2LK)+ m2M2eff

(a3c

b2minus 2c2 + a3c2

b3

) (32)

with

ω(η) = ϕprime(η)2 V (η) = V (ϕ (η)) σ (ζ ) = ξ prime(ζ )2 U (ζ ) = U (ξ (ζ )) (33)

Consequently for arbitrary a(τ ) b(τ ) and c(τ ) if we choose ω(τ ) V (τ ) σ (τ ) and U (τ ) tosatisfy equations (29)ndash(32) the cosmological model with given evolutions of a(τ ) b(τ ) andc(τ ) can be reconstructed

A reason why we introduced two scalar fields instead of one is that there are three degreesof freedom a b and c in metrics (23) and it is not trivial to describe them by using only onescalar field which might not be impossible but we have not succeeded

33 Conformal description of the accelerating universe

In the following we use the conformal time We describe how the known cosmologies can beexpressed by using the conformal time

The conformally flat FRW universe metric is given by

ds2 = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

) (34)

In this equation when a(τ )2 = l2

τ 2 the metric (34) corresponds to the de Sitter universewhich may represent inflation or dark energy in the model under consideration On the otherhand if a(τ )2 = l2n

τ 2n with n = 1 by redefining the time coordinate as

dt = plusmn ln

τ ndτ (35)

6

Class Quantum Grav 31 (2014) 075016 K Bamba et al

ie

t = plusmn ln

n minus 1τ 1minusn (36)

the metric (34) can be rewritten as

ds2 = minusdt2 +[plusmn(n minus 1)

t

l

]minus 2n1minusn 3sum

i=1

(dxi)2 (37)

Equation (37) shows that if 0 lt n lt 1 the metric corresponds to the phantom universe [27]if n gt 1 to the quintessence universe and if n lt 0 to decelerating universe In case of thephantom universe (0 lt n lt 1) we should choose + sign in plusmn of (35) or (36) and shift tin (37) as t rarr t minus t0 The time t = t0 corresponds to the Big Rip and the present time ist lt t0 and the limit of τ rarr infin is equivalent to the infinite past (t rarr minusinfin) In case of thequintessence universe (n gt 1) we may again select + sign in plusmn of (35) or (36) The limit ofτ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infin to that of t rarr 0 which may beequivalent to the Big Bang In case of the decelerating universe (n lt 0) we may take minus signin plusmn of (35) or (36) The limit of τ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infinto that of t rarr 0 which may again be considered to be the Big Bang We should also note thatin case of the de Sitter universe (n = 1) the limit of τ rarr 0 corresponds to that of t rarr +infinand that of τ rarr plusmninfin to that of t rarr minusinfin

34 Dark energy universe with a(τ ) = b(τ ) = c(τ )

If the space-time described by the metric gμν represents the universe where we live thefunctions c(τ ) and b(τ ) are not directly related to the expansion of our universe because thefunctions c(τ ) and b(τ ) correspond to the degrees of freedom in the Einstein frame metric fμν Therefore we may choose c(τ ) and b(τ ) in the consistent way convenient for the calculationThis does not mean c(τ ) and b(τ ) are not relevant for the physics besides the expansionof our universe In this section we simply take a(τ ) = c(τ ) = b(τ ) which satisfy thecondition (28) and therefore H = K = L From (29) and (31) we find ω(τ ) = σ (τ ) andthus ϕ(τ ) = ξ (τ ) and also V (τ ) = U (τ ) from (30) and (32)

By choosing a(τ ) = c(τ ) = b(τ ) equations (29)ndash(32) are simplified as

ω(τ ) = σ (τ ) = 4(minusH + H2) V (τ )a(τ )2 = U (τ )a(τ )2 = (2H + 4H2) (38)

Let us construct the models where the scale factor squared is given by a(τ )2 = l2n

τ 2n In thiscase we find

ω(τ ) = σ (τ ) = 4n(n minus 1)

τ 2 V (τ ) = U (τ ) = (2n + 4n2)l2n

τ 2(1minusn) (39)

It should be cautioned that if 0 lt n lt 1 ω(τ ) and σ (τ ) become negative and this conflictswith the definition in (33) Hence the universe corresponding to the phantom cannot be realizedas in the standard scalarndashtensor model whose situation is different from the case of F(R)-bigravity [22] (for modified gravity including F(R) gravity and dark energy problem see eg[28ndash31]) In case of n = 1 in which the de Sitter universe is realized both ω(τ ) and σ (τ )

vanish and V (τ ) and U (τ ) become constants This is equivalent to the cosmological constant

4 Stability of solutions

As we have shown a wide class of expansions of the universe can be reproduced in thebigravity models coupled to scalar fields The desired solution is however only one of the

7

Class Quantum Grav 31 (2014) 075016 K Bamba et al

solutions If the solution is not stable under the perturbation such a solution cannot be realizedunless we perform very fine-tuning In this section we study the stability of the solution in thelast section For this purpose we rewrite (29)ndash(32) in the following form

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2eff(a(τ )b(τ ) minus a(τ )c(τ )) (40)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a(τ )2 minus 5a(τ )b(τ ) minus a(τ )c(τ )) (41)

σ (ζ )ζ 2M2f = minus 4M2

f (K minus LK) minus 2m2M2eff

(minus c (τ )

b (τ )+ 1

)a (τ )3 c (τ )

b (τ )2 (42)

U (ζ )c(τ )2M2f =M2

f (2K| +6K2 minus 2LK)+m2M2eff

(a (τ )3 c (τ )

b (τ )2 minus 2c (τ )2 + a (τ )3 c (τ )2

b (τ )3

)

(43)

On the other hand the scalar field equations (21) can be rewritten to

0 = 3

(ω(η)η + ωprime(η)

2η2 + 2Hω(η)η

)+ V prime(η)a2

0 = 3

(σ (ζ )ζ + σ prime(ζ )

2ζ 2 + (3K minus L) σ (ζ )ζ

)+ U prime(ζ )a2 (44)

Equations in (38) implies that with a function f (τ ) if we choose

ω(η) = 4(minus f primeprime(η) + f prime(η)2) σ (ζ ) = 4(minus f primeprime(ζ ) + f prime(ζ )2)

V (η) = eminus2 f (η)(2 f primeprime(η) + 4 f prime(η)2) U (ζ ) = eminus2 f (ζ )(2 f primeprime(ζ ) + 4 f prime(ζ )2) (45)

we find the following solution

a(τ ) = b(τ ) = c(τ ) = e f (τ ) η = ζ = τ (46)

We explore the stability of the solution in (46)We may consider the following perturbation

H rarr H + δH K rarr K + δK a rarr a (1 + δ fa) b rarr b (1 + δ fb)

η rarr η + δη ζ rarr ζ + δζ (47)

In what follows just for simplicity we take

M2f = M2

g = M2eff

2= M2 (48)

Thus we obtain

d

dτ

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

= M

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

M =

⎛⎜⎜⎜⎜⎜⎜⎝

2H 0 CminusDB minusD

B3

HB (B minus 1)

A E 2C minus DB

C+DB minus 2C 3

HB (B minus 1)

0 0 0 0 1

AH minusAH 2HC minus2HC 1(1 + D

3

)AH minusADH

3 2H(C minus 2BD

3

)43 BD minus4H

⎞⎟⎟⎟⎟⎟⎟⎠

(49)

8

Class Quantum Grav 31 (2014) 075016 K Bamba et al

where

A equiv H

H+ 2H minus 4

H3

H B equiv 1 minus H2

H C equiv 1 + 2

H2

H D equiv 3m2a2

H E equiv minus H

Hminus 4

H3

H

(50)

The derivation of equations (49) and (50) is given in appendix A We should note that wehave deleted δK in (47) by using (A13)

The eigenvalue equation has the following form

0 = λ5 + c4λ4 + c3λ

3 + c2λ2 + c1λ + c0 (51)

where λ is the eigenvalue of the matrix M In order that the solution (46) could be stable allthe eigenvalues should be negative Then all the eigenmodes corresponding to the eigenvaluesdecrease and therefore any perturbation damps It requires ci gt 0 (i = 1 4) Especiallyminusc4 is the trace of the matrix M and we find

minus c4 = minus H

Hminus 4H minus 8

H3

Hlt 0 (52)

For the power expanding model (39) where H = minusnτ if τ gt 0 equation (52) leads to

4n2 + 2n + 1 lt 0 (53)

Thus there is no real solution for n As a result there does not exist any stable solution for thepower expanding model (39) On the other hand suppose τ lt 0 equation (52) yields

4n2 + 2n + 1 gt 0 (54)

for which there is a possibility that the solution might be stableWhen H = minusnτ in (39) the matrix M in (49) has the following form

M =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1+2nminusD0τminus2n+2

1minusn minus D01minusnτminus2n+2 3τ

1minusn

minus2minus2n+4n2

τ2+4n2

τ2 + 4n minus D0τ

minus2n+2

1minusn

1+2n+D0τminus2n+2

1minusn

minus2 minus 4n3t

1minusn

0 0 0 0 1

minus n(minus2minus2n+4n2)τ 2

n(minus2minus2n+4n2)τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

(1 + D0τ

minus2n+2

3

) (minus2minus2n+4n2)nτ 3 minusD0(minus2minus2n+4n2)nτminus2nminus1

3

2n(1+2n)

τ 2

minus 4(1minusn)D0τminus2n

3

4(1minusn)D0τminus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(55)

where the scale factor a(τ ) is given by a = a0τminusn and D0 equiv 3m2a2

0n Note that a0 = ln in (39)

As an example we may investigate the case n = minus12 In this case the eigenvalueequation has the following form

0 = λ

(λ minus 1

τ

) (λ minus 3

τ

) (2D0τ

4 + D0τ2 minus 2λ minus λ2τ

) (56)

Since there always appear positive eigenvalue the solution is not stableWe redefine

δ fa = (1 minus n)δ fa δ fb = (1 minus n)δ fb δH = (1 minus n)δH (57)

9

Class Quantum Grav 31 (2014) 075016 K Bamba et al

The matrix M in (55) has the following form

M =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1 + 2n minus D0τminus2n+2 minusD0τ

minus2n+2 3t

minus 2(1+2n)(1minusn)

τ2+4n2

τ

2(1 + 2n)(1 minus n)

minusD0τminus2n+2

minus1 + 4n2

+D0τminus2n+2 3t

0 0 0 0 1n(1+2n)

τ 2 minus n(1+2n)

τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

minus(1 + D0τ

minus2n+2

3

)1+2n

τ

D0(1+2n)nτminus2nminus1

32n(1+2n)

τ 2 minus 4(1minusn)D0τminus2n

34(1minusn)D0τ

minus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(58)

In section 34 in the part below equation (39) we have shown that if 0 lt n lt 1 the model isinconsistent As another example we consider the limit of n rarr 1 + 0 The matrix M in (58)reduces to

M =

⎛⎜⎜⎜⎜⎜⎜⎝

minus 2τ

0 3 minus D0 minusD0 3t

0 6τ

minusD0 3 + D0 3t

0 0 0 0 13τ 2 minus 3

τ 2 minus 6τ

6τ

1

minus 3+D0τ 3

D0τ 3

6τ 2 0 4

τ

⎞⎟⎟⎟⎟⎟⎟⎠

(59)

For this matrix the eigenvalue equation has the following form

0 = λ5 minus 14

τλ4 + (6D0 + 64)

λ3

τ 2minus (

2D20 + 66D0 + 33

) λ2

τ 3+ 3

(8D2

0 + 86D0 minus 63) λ

τ 4

minus 45(2D2

0 + 3) 1

τ 5 (60)

If

τ lt 0 and D0 gt

radic2353 minus 43

8= 0688 466 417 817 452 middot middot middot (61)

all the eigenvalues are negative and the system becomes stableIn general the eigenvalue equations (51) for the matrices (55) and (58) are rather

complicated and the explicit forms are given in appendix B As a result anyway we have founda solution which is stable under the perturbation Then we have shown that for an arbitrarilygiven history of the expansion of the universe we can construct a model who has a solutiongenerating the expansion and the solution is stable that is attractor solution

5 BransndashDicke type model

We introduce a parameter ε which is a positive but sufficiently small value (0 lt ε 1)In the previous section we have found that the model where n = 1 + ε and both τ and D0

satisfy equation (61) is stable and that the limit ε rarr 0 (n rarr 1) corresponds to the de Sitterspace In this section by starting with a model where ε gt 0 is small enough but finite weconstruct a model which reproduces an arbitrary expansion history of the universe by usingthe BransndashDicke type model

We here explore an arbitrary scale factor a(τ ) for τ lt 0 The scale factor correspondingto n = 1 + ε is given by a0τ

minus1minusε Hence the metric gμν corresponding to the scale factor a(τ )

is expressed by multiplying the metric gεμν corresponding to n = 1 + ε by a(τ )2aminus2

0 τ 2(1+ε)Since η = τ we rescale the metric gμν in the actions (17) and (19) as follows

gμν rarr a(η)minus2a20η

minus2(1+ε)gμν (62)

10

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

ds2g =

3sumμν=0

gμν dxμ dxν = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

)

ds2f =

3sumμν=0

fμν dxμ dxν = minusdτ 2 +3sum

i=1

(dxi)2 (10)

The physical meaning of the metric fμν has not been clear although there are several conjecturesas in [26] The longitudinal scalar modes in the metric play the roles when we consider thebackground as in (10) but these modes do not propagate which may be found by consideringthe local Lorentz frame The propagating modes could be massless tensor (massless graviton)and the massive tensor (massive graviton) and any scalar mode does not propagate

The (τ τ ) component of (7) reads

0 = minus3M2gH2 minus 3m2M2

eff(a2 minus a) + (

14 φ2 + 1

2V (φ)a(τ )2)M2g (11)

and from (i j) components we find

0 = M2g (2H + H2) + 3m2M2

eff(a2 minus a) + (

12 φ2 minus 1

2V (φ)a(τ )2)M2

g (12)

with H = aa where the dot denotes the derivative with respect to t From equation (9) wehave the following equation

a

a= 0 (13)

Hence a should be a constant a = a0 This means that the only consistent solution for gμν isthe flat Minkowski space Furthermore by using (11) and (12) we find

φ = 0 0 = 3m2M2eff

(a2

0 minus a0) minus 1

2V0a20M2

g (14)

Since φ is a constant we cannot obtain the expanding universe

3 Bigravity with two scalar fields and cosmology

In the last section we have observed difficulties to construct the model which generates theexpanding universe In this section we build models of cosmology by using the bigravity withtwo scalar fields The bimetric gravity proposed in [11] includes two metric tensors gμν andfμν In addition to the massless spin-two field corresponding to graviton it contains massivespin-two field although massive gravity models only have the massive spin-two field TheBoulwarendashDeser ghost [3] does not appear in such a theory

31 Bigravity models with scalar fields

We add the term containing the scalar curvature R( f ) given by fμν to the action (1) as follows

Sbi = M2g

intd4x

radicminus det gR(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det g

4sumn=0

βnen(radic

gminus1 f ) (15)

Here Meff is defined by1

M2eff

= 1

M2g

+ 1

M2f

(16)

There is a conjecture that the two dynamical metric may correspond to manifolds with twometric [26]

4

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We also involve the following terms given by two scalar fields ϕ and χ in the action (15)

Sϕ = minusM2g

intd4x

radicminus det g

1

2gμνpartμϕpartνϕ + V (ϕ)

+

intd4xLmatter(gμνi) (17)

Sξ = minusM2f

intd4x

radicminus det f

1

2f μνpartμξpartνξ + U (ξ )

(18)

For simplicity we start from the minimal case again

Sbi = M2g

intd4x

radicminus det gR(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det g (3 minus tr

radicgminus1 f + det

radicgminus1 f ) (19)

For a while we neglect the contributions from matters By the variation over gμν we againfind (7) On the other hand through the variation over fμν we acquire

0 = M2f

(12 fμνR( f ) minus R( f )

μν

) + m2M2eff

radicdet( f minus1g)

minus 12 fμρ(

radicgminus1 f )ρ ν minus 1

2 fνρ (radic

gminus1 f )ρ μ

+ det(radic

gminus1 f ) fμν

+ M2f

[12

(13 f ρσ partρξpartσ ξ + U (ξ )

)fμν minus 3

2partμξpartνξ] (20)

The variations of the scalar fields ϕ and ξ lead to

0 = minusgϕ + V prime(ϕ) 0 = minus f ξ + U prime(ξ ) (21)

corresponding to (8) Here f is the drsquoAlembertian with respect to the metric f Fromequation (7) and the Bianchi identity again we acquire (9) Similarly by using the covariantderivative nablaμ

f with respect to the metric f from (20) we have

0 = nablaμ

f

[radicdet( f minus1g)

minus 12 (

radicgminus1 f )minus1ν

σ gσμ minus 12 (

radicgminus1 f )minus1μ

σ gσν + det(radic

gminus1 f ) f μν]

(22)

The identities (9) and (22) impose strong constraints on the solutions Especially weinvestigate the solutions describing the FRW universe in the next subsection

32 Reconstruction of bigravity models

We examine whether we can construct models describing the arbitrarily given evolution of theexpansion in the universe

We take the FRW universes for the metric gμν as in (10) and use the conformal time t = τ Moreover instead of (10) we suppose the form of the metric fμν as follows

ds2g =

3sumμν=0

gμν dxμ dxν = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

)

ds2f =

3sumμν=0

fμν dxμ dxν = minusc(τ )2 dτ 2 + b(τ )23sum

i=1

(dxi)2 (23)

We should note the assumption in (10) could be most general form if we assume the spatial partof the space-time is uniform homogeneous and flat The redefinition of the time-coordinatealways gives the form of ds2

g but there does not any more freedom to choose c(τ ) = 1 norc(τ ) = b(τ ) In this case from the (τ τ ) component of (7) we find

0 = minus3M2gH2 minus 3m2M2

eff(a2 minus ab) + (

14 ϕ2 + 1

2V (ϕ)a(τ )2)

M2g (24)

5

Class Quantum Grav 31 (2014) 075016 K Bamba et al

and (i j) components yield

0 = M2g (2H + H2) + m2M2

eff(3a2 minus 2ab minus ac) + (14 ϕ2 minus 1

2V (ϕ)a(τ )2)

M2g (25)

On the other hand the (τ τ ) component of (20) leads to

0 = minus3M2f K

2 + m2M2effc

2

(1 minus a3

b3

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (26)

and from (i j) components we find

0 = M2f (2K + 3K2 minus 2LK) + m2M2

eff

(a3c

b2minus c2

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (27)

with K = bb and L = cc Both equations (9) and (22) yield the identical equation

cH = bK orca

a= b (28)

The above equation is the constraint relating the two metrics imposed by the equations ofmotion If a = 0 we obtain c = aba On the other hand if a = 0 we find b = 0 that is aand b are constant and c can be arbitrary

Next we redefine scalars as ϕ = ϕ(η) and ξ = ξ (ζ ) and identify η and ζ with theconformal time t as η = ζ = τ Hence we acquire

ω(τ )M2g = minus 4M2

g (H minus H2) minus 2m2M2eff(ab minus ac) (29)

V (τ )a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a2 minus 5ab minus ac) (30)

σ (τ )M2f = minus4M2

f (K minus LK) minus 2m2M2eff

(minus c

b+ 1

) a3c

b2 (31)

U (τ )c(τ )2M2f = M2

f (2K + 6K2 minus 2LK)+ m2M2eff

(a3c

b2minus 2c2 + a3c2

b3

) (32)

with

ω(η) = ϕprime(η)2 V (η) = V (ϕ (η)) σ (ζ ) = ξ prime(ζ )2 U (ζ ) = U (ξ (ζ )) (33)

Consequently for arbitrary a(τ ) b(τ ) and c(τ ) if we choose ω(τ ) V (τ ) σ (τ ) and U (τ ) tosatisfy equations (29)ndash(32) the cosmological model with given evolutions of a(τ ) b(τ ) andc(τ ) can be reconstructed

A reason why we introduced two scalar fields instead of one is that there are three degreesof freedom a b and c in metrics (23) and it is not trivial to describe them by using only onescalar field which might not be impossible but we have not succeeded

33 Conformal description of the accelerating universe

In the following we use the conformal time We describe how the known cosmologies can beexpressed by using the conformal time

The conformally flat FRW universe metric is given by

ds2 = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

) (34)

In this equation when a(τ )2 = l2

τ 2 the metric (34) corresponds to the de Sitter universewhich may represent inflation or dark energy in the model under consideration On the otherhand if a(τ )2 = l2n

τ 2n with n = 1 by redefining the time coordinate as

dt = plusmn ln

τ ndτ (35)

6

Class Quantum Grav 31 (2014) 075016 K Bamba et al

ie

t = plusmn ln

n minus 1τ 1minusn (36)

the metric (34) can be rewritten as

ds2 = minusdt2 +[plusmn(n minus 1)

t

l

]minus 2n1minusn 3sum

i=1

(dxi)2 (37)

Equation (37) shows that if 0 lt n lt 1 the metric corresponds to the phantom universe [27]if n gt 1 to the quintessence universe and if n lt 0 to decelerating universe In case of thephantom universe (0 lt n lt 1) we should choose + sign in plusmn of (35) or (36) and shift tin (37) as t rarr t minus t0 The time t = t0 corresponds to the Big Rip and the present time ist lt t0 and the limit of τ rarr infin is equivalent to the infinite past (t rarr minusinfin) In case of thequintessence universe (n gt 1) we may again select + sign in plusmn of (35) or (36) The limit ofτ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infin to that of t rarr 0 which may beequivalent to the Big Bang In case of the decelerating universe (n lt 0) we may take minus signin plusmn of (35) or (36) The limit of τ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infinto that of t rarr 0 which may again be considered to be the Big Bang We should also note thatin case of the de Sitter universe (n = 1) the limit of τ rarr 0 corresponds to that of t rarr +infinand that of τ rarr plusmninfin to that of t rarr minusinfin

34 Dark energy universe with a(τ ) = b(τ ) = c(τ )

If the space-time described by the metric gμν represents the universe where we live thefunctions c(τ ) and b(τ ) are not directly related to the expansion of our universe because thefunctions c(τ ) and b(τ ) correspond to the degrees of freedom in the Einstein frame metric fμν Therefore we may choose c(τ ) and b(τ ) in the consistent way convenient for the calculationThis does not mean c(τ ) and b(τ ) are not relevant for the physics besides the expansionof our universe In this section we simply take a(τ ) = c(τ ) = b(τ ) which satisfy thecondition (28) and therefore H = K = L From (29) and (31) we find ω(τ ) = σ (τ ) andthus ϕ(τ ) = ξ (τ ) and also V (τ ) = U (τ ) from (30) and (32)

By choosing a(τ ) = c(τ ) = b(τ ) equations (29)ndash(32) are simplified as

ω(τ ) = σ (τ ) = 4(minusH + H2) V (τ )a(τ )2 = U (τ )a(τ )2 = (2H + 4H2) (38)

Let us construct the models where the scale factor squared is given by a(τ )2 = l2n

τ 2n In thiscase we find

ω(τ ) = σ (τ ) = 4n(n minus 1)

τ 2 V (τ ) = U (τ ) = (2n + 4n2)l2n

τ 2(1minusn) (39)

It should be cautioned that if 0 lt n lt 1 ω(τ ) and σ (τ ) become negative and this conflictswith the definition in (33) Hence the universe corresponding to the phantom cannot be realizedas in the standard scalarndashtensor model whose situation is different from the case of F(R)-bigravity [22] (for modified gravity including F(R) gravity and dark energy problem see eg[28ndash31]) In case of n = 1 in which the de Sitter universe is realized both ω(τ ) and σ (τ )

vanish and V (τ ) and U (τ ) become constants This is equivalent to the cosmological constant

4 Stability of solutions

As we have shown a wide class of expansions of the universe can be reproduced in thebigravity models coupled to scalar fields The desired solution is however only one of the

7

Class Quantum Grav 31 (2014) 075016 K Bamba et al

solutions If the solution is not stable under the perturbation such a solution cannot be realizedunless we perform very fine-tuning In this section we study the stability of the solution in thelast section For this purpose we rewrite (29)ndash(32) in the following form

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2eff(a(τ )b(τ ) minus a(τ )c(τ )) (40)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a(τ )2 minus 5a(τ )b(τ ) minus a(τ )c(τ )) (41)

σ (ζ )ζ 2M2f = minus 4M2

f (K minus LK) minus 2m2M2eff

(minus c (τ )

b (τ )+ 1

)a (τ )3 c (τ )

b (τ )2 (42)

U (ζ )c(τ )2M2f =M2

f (2K| +6K2 minus 2LK)+m2M2eff

(a (τ )3 c (τ )

b (τ )2 minus 2c (τ )2 + a (τ )3 c (τ )2

b (τ )3

)

(43)

On the other hand the scalar field equations (21) can be rewritten to

0 = 3

(ω(η)η + ωprime(η)

2η2 + 2Hω(η)η

)+ V prime(η)a2

0 = 3

(σ (ζ )ζ + σ prime(ζ )

2ζ 2 + (3K minus L) σ (ζ )ζ

)+ U prime(ζ )a2 (44)

Equations in (38) implies that with a function f (τ ) if we choose

ω(η) = 4(minus f primeprime(η) + f prime(η)2) σ (ζ ) = 4(minus f primeprime(ζ ) + f prime(ζ )2)

V (η) = eminus2 f (η)(2 f primeprime(η) + 4 f prime(η)2) U (ζ ) = eminus2 f (ζ )(2 f primeprime(ζ ) + 4 f prime(ζ )2) (45)

we find the following solution

a(τ ) = b(τ ) = c(τ ) = e f (τ ) η = ζ = τ (46)

We explore the stability of the solution in (46)We may consider the following perturbation

H rarr H + δH K rarr K + δK a rarr a (1 + δ fa) b rarr b (1 + δ fb)

η rarr η + δη ζ rarr ζ + δζ (47)

In what follows just for simplicity we take

M2f = M2

g = M2eff

2= M2 (48)

Thus we obtain

d

dτ

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

= M

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

M =

⎛⎜⎜⎜⎜⎜⎜⎝

2H 0 CminusDB minusD

B3

HB (B minus 1)

A E 2C minus DB

C+DB minus 2C 3

HB (B minus 1)

0 0 0 0 1

AH minusAH 2HC minus2HC 1(1 + D

3

)AH minusADH

3 2H(C minus 2BD

3

)43 BD minus4H

⎞⎟⎟⎟⎟⎟⎟⎠

(49)

8

Class Quantum Grav 31 (2014) 075016 K Bamba et al

where

A equiv H

H+ 2H minus 4

H3

H B equiv 1 minus H2

H C equiv 1 + 2

H2

H D equiv 3m2a2

H E equiv minus H

Hminus 4

H3

H

(50)

The derivation of equations (49) and (50) is given in appendix A We should note that wehave deleted δK in (47) by using (A13)

The eigenvalue equation has the following form

0 = λ5 + c4λ4 + c3λ

3 + c2λ2 + c1λ + c0 (51)

where λ is the eigenvalue of the matrix M In order that the solution (46) could be stable allthe eigenvalues should be negative Then all the eigenmodes corresponding to the eigenvaluesdecrease and therefore any perturbation damps It requires ci gt 0 (i = 1 4) Especiallyminusc4 is the trace of the matrix M and we find

minus c4 = minus H

Hminus 4H minus 8

H3

Hlt 0 (52)

For the power expanding model (39) where H = minusnτ if τ gt 0 equation (52) leads to

4n2 + 2n + 1 lt 0 (53)

Thus there is no real solution for n As a result there does not exist any stable solution for thepower expanding model (39) On the other hand suppose τ lt 0 equation (52) yields

4n2 + 2n + 1 gt 0 (54)

for which there is a possibility that the solution might be stableWhen H = minusnτ in (39) the matrix M in (49) has the following form

M =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1+2nminusD0τminus2n+2

1minusn minus D01minusnτminus2n+2 3τ

1minusn

minus2minus2n+4n2

τ2+4n2

τ2 + 4n minus D0τ

minus2n+2

1minusn

1+2n+D0τminus2n+2

1minusn

minus2 minus 4n3t

1minusn

0 0 0 0 1

minus n(minus2minus2n+4n2)τ 2

n(minus2minus2n+4n2)τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

(1 + D0τ

minus2n+2

3

) (minus2minus2n+4n2)nτ 3 minusD0(minus2minus2n+4n2)nτminus2nminus1

3

2n(1+2n)

τ 2

minus 4(1minusn)D0τminus2n

3

4(1minusn)D0τminus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(55)

where the scale factor a(τ ) is given by a = a0τminusn and D0 equiv 3m2a2

0n Note that a0 = ln in (39)

As an example we may investigate the case n = minus12 In this case the eigenvalueequation has the following form

0 = λ

(λ minus 1

τ

) (λ minus 3

τ

) (2D0τ

4 + D0τ2 minus 2λ minus λ2τ

) (56)

Since there always appear positive eigenvalue the solution is not stableWe redefine

δ fa = (1 minus n)δ fa δ fb = (1 minus n)δ fb δH = (1 minus n)δH (57)

9

Class Quantum Grav 31 (2014) 075016 K Bamba et al

The matrix M in (55) has the following form

M =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1 + 2n minus D0τminus2n+2 minusD0τ

minus2n+2 3t

minus 2(1+2n)(1minusn)

τ2+4n2

τ

2(1 + 2n)(1 minus n)

minusD0τminus2n+2

minus1 + 4n2

+D0τminus2n+2 3t

0 0 0 0 1n(1+2n)

τ 2 minus n(1+2n)

τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

minus(1 + D0τ

minus2n+2

3

)1+2n

τ

D0(1+2n)nτminus2nminus1

32n(1+2n)

τ 2 minus 4(1minusn)D0τminus2n

34(1minusn)D0τ

minus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(58)

In section 34 in the part below equation (39) we have shown that if 0 lt n lt 1 the model isinconsistent As another example we consider the limit of n rarr 1 + 0 The matrix M in (58)reduces to

M =

⎛⎜⎜⎜⎜⎜⎜⎝

minus 2τ

0 3 minus D0 minusD0 3t

0 6τ

minusD0 3 + D0 3t

0 0 0 0 13τ 2 minus 3

τ 2 minus 6τ

6τ

1

minus 3+D0τ 3

D0τ 3

6τ 2 0 4

τ

⎞⎟⎟⎟⎟⎟⎟⎠

(59)

For this matrix the eigenvalue equation has the following form

0 = λ5 minus 14

τλ4 + (6D0 + 64)

λ3

τ 2minus (

2D20 + 66D0 + 33

) λ2

τ 3+ 3

(8D2

0 + 86D0 minus 63) λ

τ 4

minus 45(2D2

0 + 3) 1

τ 5 (60)

If

τ lt 0 and D0 gt

radic2353 minus 43

8= 0688 466 417 817 452 middot middot middot (61)

all the eigenvalues are negative and the system becomes stableIn general the eigenvalue equations (51) for the matrices (55) and (58) are rather

complicated and the explicit forms are given in appendix B As a result anyway we have founda solution which is stable under the perturbation Then we have shown that for an arbitrarilygiven history of the expansion of the universe we can construct a model who has a solutiongenerating the expansion and the solution is stable that is attractor solution

5 BransndashDicke type model

We introduce a parameter ε which is a positive but sufficiently small value (0 lt ε 1)In the previous section we have found that the model where n = 1 + ε and both τ and D0

satisfy equation (61) is stable and that the limit ε rarr 0 (n rarr 1) corresponds to the de Sitterspace In this section by starting with a model where ε gt 0 is small enough but finite weconstruct a model which reproduces an arbitrary expansion history of the universe by usingthe BransndashDicke type model

We here explore an arbitrary scale factor a(τ ) for τ lt 0 The scale factor correspondingto n = 1 + ε is given by a0τ

minus1minusε Hence the metric gμν corresponding to the scale factor a(τ )

is expressed by multiplying the metric gεμν corresponding to n = 1 + ε by a(τ )2aminus2

0 τ 2(1+ε)Since η = τ we rescale the metric gμν in the actions (17) and (19) as follows

gμν rarr a(η)minus2a20η

minus2(1+ε)gμν (62)

10

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We also involve the following terms given by two scalar fields ϕ and χ in the action (15)

Sϕ = minusM2g

intd4x

radicminus det g

1

2gμνpartμϕpartνϕ + V (ϕ)

+

intd4xLmatter(gμνi) (17)

Sξ = minusM2f

intd4x

radicminus det f

1

2f μνpartμξpartνξ + U (ξ )

(18)

For simplicity we start from the minimal case again

Sbi = M2g

intd4x

radicminus det gR(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det g (3 minus tr

radicgminus1 f + det

radicgminus1 f ) (19)

For a while we neglect the contributions from matters By the variation over gμν we againfind (7) On the other hand through the variation over fμν we acquire

0 = M2f

(12 fμνR( f ) minus R( f )

μν

) + m2M2eff

radicdet( f minus1g)

minus 12 fμρ(

radicgminus1 f )ρ ν minus 1

2 fνρ (radic

gminus1 f )ρ μ

+ det(radic

gminus1 f ) fμν

+ M2f

[12

(13 f ρσ partρξpartσ ξ + U (ξ )

)fμν minus 3

2partμξpartνξ] (20)

The variations of the scalar fields ϕ and ξ lead to

0 = minusgϕ + V prime(ϕ) 0 = minus f ξ + U prime(ξ ) (21)

corresponding to (8) Here f is the drsquoAlembertian with respect to the metric f Fromequation (7) and the Bianchi identity again we acquire (9) Similarly by using the covariantderivative nablaμ

f with respect to the metric f from (20) we have

0 = nablaμ

f

[radicdet( f minus1g)

minus 12 (

radicgminus1 f )minus1ν

σ gσμ minus 12 (

radicgminus1 f )minus1μ

σ gσν + det(radic

gminus1 f ) f μν]

(22)

The identities (9) and (22) impose strong constraints on the solutions Especially weinvestigate the solutions describing the FRW universe in the next subsection

32 Reconstruction of bigravity models

We examine whether we can construct models describing the arbitrarily given evolution of theexpansion in the universe

We take the FRW universes for the metric gμν as in (10) and use the conformal time t = τ Moreover instead of (10) we suppose the form of the metric fμν as follows

ds2g =

3sumμν=0

gμν dxμ dxν = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

)

ds2f =

3sumμν=0

fμν dxμ dxν = minusc(τ )2 dτ 2 + b(τ )23sum

i=1

(dxi)2 (23)

We should note the assumption in (10) could be most general form if we assume the spatial partof the space-time is uniform homogeneous and flat The redefinition of the time-coordinatealways gives the form of ds2

g but there does not any more freedom to choose c(τ ) = 1 norc(τ ) = b(τ ) In this case from the (τ τ ) component of (7) we find

0 = minus3M2gH2 minus 3m2M2

eff(a2 minus ab) + (

14 ϕ2 + 1

2V (ϕ)a(τ )2)

M2g (24)

5

Class Quantum Grav 31 (2014) 075016 K Bamba et al

and (i j) components yield

0 = M2g (2H + H2) + m2M2

eff(3a2 minus 2ab minus ac) + (14 ϕ2 minus 1

2V (ϕ)a(τ )2)

M2g (25)

On the other hand the (τ τ ) component of (20) leads to

0 = minus3M2f K

2 + m2M2effc

2

(1 minus a3

b3

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (26)

and from (i j) components we find

0 = M2f (2K + 3K2 minus 2LK) + m2M2

eff

(a3c

b2minus c2

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (27)

with K = bb and L = cc Both equations (9) and (22) yield the identical equation

cH = bK orca

a= b (28)

The above equation is the constraint relating the two metrics imposed by the equations ofmotion If a = 0 we obtain c = aba On the other hand if a = 0 we find b = 0 that is aand b are constant and c can be arbitrary

Next we redefine scalars as ϕ = ϕ(η) and ξ = ξ (ζ ) and identify η and ζ with theconformal time t as η = ζ = τ Hence we acquire

ω(τ )M2g = minus 4M2

g (H minus H2) minus 2m2M2eff(ab minus ac) (29)

V (τ )a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a2 minus 5ab minus ac) (30)

σ (τ )M2f = minus4M2

f (K minus LK) minus 2m2M2eff

(minus c

b+ 1

) a3c

b2 (31)

U (τ )c(τ )2M2f = M2

f (2K + 6K2 minus 2LK)+ m2M2eff

(a3c

b2minus 2c2 + a3c2

b3

) (32)

with

ω(η) = ϕprime(η)2 V (η) = V (ϕ (η)) σ (ζ ) = ξ prime(ζ )2 U (ζ ) = U (ξ (ζ )) (33)

Consequently for arbitrary a(τ ) b(τ ) and c(τ ) if we choose ω(τ ) V (τ ) σ (τ ) and U (τ ) tosatisfy equations (29)ndash(32) the cosmological model with given evolutions of a(τ ) b(τ ) andc(τ ) can be reconstructed

A reason why we introduced two scalar fields instead of one is that there are three degreesof freedom a b and c in metrics (23) and it is not trivial to describe them by using only onescalar field which might not be impossible but we have not succeeded

33 Conformal description of the accelerating universe

In the following we use the conformal time We describe how the known cosmologies can beexpressed by using the conformal time

The conformally flat FRW universe metric is given by

ds2 = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

) (34)

In this equation when a(τ )2 = l2

τ 2 the metric (34) corresponds to the de Sitter universewhich may represent inflation or dark energy in the model under consideration On the otherhand if a(τ )2 = l2n

τ 2n with n = 1 by redefining the time coordinate as

dt = plusmn ln

τ ndτ (35)

6

Class Quantum Grav 31 (2014) 075016 K Bamba et al

ie

t = plusmn ln

n minus 1τ 1minusn (36)

the metric (34) can be rewritten as

ds2 = minusdt2 +[plusmn(n minus 1)

t

l

]minus 2n1minusn 3sum

i=1

(dxi)2 (37)

Equation (37) shows that if 0 lt n lt 1 the metric corresponds to the phantom universe [27]if n gt 1 to the quintessence universe and if n lt 0 to decelerating universe In case of thephantom universe (0 lt n lt 1) we should choose + sign in plusmn of (35) or (36) and shift tin (37) as t rarr t minus t0 The time t = t0 corresponds to the Big Rip and the present time ist lt t0 and the limit of τ rarr infin is equivalent to the infinite past (t rarr minusinfin) In case of thequintessence universe (n gt 1) we may again select + sign in plusmn of (35) or (36) The limit ofτ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infin to that of t rarr 0 which may beequivalent to the Big Bang In case of the decelerating universe (n lt 0) we may take minus signin plusmn of (35) or (36) The limit of τ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infinto that of t rarr 0 which may again be considered to be the Big Bang We should also note thatin case of the de Sitter universe (n = 1) the limit of τ rarr 0 corresponds to that of t rarr +infinand that of τ rarr plusmninfin to that of t rarr minusinfin

34 Dark energy universe with a(τ ) = b(τ ) = c(τ )

If the space-time described by the metric gμν represents the universe where we live thefunctions c(τ ) and b(τ ) are not directly related to the expansion of our universe because thefunctions c(τ ) and b(τ ) correspond to the degrees of freedom in the Einstein frame metric fμν Therefore we may choose c(τ ) and b(τ ) in the consistent way convenient for the calculationThis does not mean c(τ ) and b(τ ) are not relevant for the physics besides the expansionof our universe In this section we simply take a(τ ) = c(τ ) = b(τ ) which satisfy thecondition (28) and therefore H = K = L From (29) and (31) we find ω(τ ) = σ (τ ) andthus ϕ(τ ) = ξ (τ ) and also V (τ ) = U (τ ) from (30) and (32)

By choosing a(τ ) = c(τ ) = b(τ ) equations (29)ndash(32) are simplified as

ω(τ ) = σ (τ ) = 4(minusH + H2) V (τ )a(τ )2 = U (τ )a(τ )2 = (2H + 4H2) (38)

Let us construct the models where the scale factor squared is given by a(τ )2 = l2n

τ 2n In thiscase we find

ω(τ ) = σ (τ ) = 4n(n minus 1)

τ 2 V (τ ) = U (τ ) = (2n + 4n2)l2n

τ 2(1minusn) (39)

It should be cautioned that if 0 lt n lt 1 ω(τ ) and σ (τ ) become negative and this conflictswith the definition in (33) Hence the universe corresponding to the phantom cannot be realizedas in the standard scalarndashtensor model whose situation is different from the case of F(R)-bigravity [22] (for modified gravity including F(R) gravity and dark energy problem see eg[28ndash31]) In case of n = 1 in which the de Sitter universe is realized both ω(τ ) and σ (τ )

vanish and V (τ ) and U (τ ) become constants This is equivalent to the cosmological constant

4 Stability of solutions

As we have shown a wide class of expansions of the universe can be reproduced in thebigravity models coupled to scalar fields The desired solution is however only one of the

7

Class Quantum Grav 31 (2014) 075016 K Bamba et al

solutions If the solution is not stable under the perturbation such a solution cannot be realizedunless we perform very fine-tuning In this section we study the stability of the solution in thelast section For this purpose we rewrite (29)ndash(32) in the following form

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2eff(a(τ )b(τ ) minus a(τ )c(τ )) (40)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a(τ )2 minus 5a(τ )b(τ ) minus a(τ )c(τ )) (41)

σ (ζ )ζ 2M2f = minus 4M2

f (K minus LK) minus 2m2M2eff

(minus c (τ )

b (τ )+ 1

)a (τ )3 c (τ )

b (τ )2 (42)

U (ζ )c(τ )2M2f =M2

f (2K| +6K2 minus 2LK)+m2M2eff

(a (τ )3 c (τ )

b (τ )2 minus 2c (τ )2 + a (τ )3 c (τ )2

b (τ )3

)

(43)

On the other hand the scalar field equations (21) can be rewritten to

0 = 3

(ω(η)η + ωprime(η)

2η2 + 2Hω(η)η

)+ V prime(η)a2

0 = 3

(σ (ζ )ζ + σ prime(ζ )

2ζ 2 + (3K minus L) σ (ζ )ζ

)+ U prime(ζ )a2 (44)

Equations in (38) implies that with a function f (τ ) if we choose

ω(η) = 4(minus f primeprime(η) + f prime(η)2) σ (ζ ) = 4(minus f primeprime(ζ ) + f prime(ζ )2)

V (η) = eminus2 f (η)(2 f primeprime(η) + 4 f prime(η)2) U (ζ ) = eminus2 f (ζ )(2 f primeprime(ζ ) + 4 f prime(ζ )2) (45)

we find the following solution

a(τ ) = b(τ ) = c(τ ) = e f (τ ) η = ζ = τ (46)

We explore the stability of the solution in (46)We may consider the following perturbation

H rarr H + δH K rarr K + δK a rarr a (1 + δ fa) b rarr b (1 + δ fb)

η rarr η + δη ζ rarr ζ + δζ (47)

In what follows just for simplicity we take

M2f = M2

g = M2eff

2= M2 (48)

Thus we obtain

d

dτ

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

= M

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

M =

⎛⎜⎜⎜⎜⎜⎜⎝

2H 0 CminusDB minusD

B3

HB (B minus 1)

A E 2C minus DB

C+DB minus 2C 3

HB (B minus 1)

0 0 0 0 1

AH minusAH 2HC minus2HC 1(1 + D

3

)AH minusADH

3 2H(C minus 2BD

3

)43 BD minus4H

⎞⎟⎟⎟⎟⎟⎟⎠

(49)

8

Class Quantum Grav 31 (2014) 075016 K Bamba et al

where

A equiv H

H+ 2H minus 4

H3

H B equiv 1 minus H2

H C equiv 1 + 2

H2

H D equiv 3m2a2

H E equiv minus H

Hminus 4

H3

H

(50)

The derivation of equations (49) and (50) is given in appendix A We should note that wehave deleted δK in (47) by using (A13)

The eigenvalue equation has the following form

0 = λ5 + c4λ4 + c3λ

3 + c2λ2 + c1λ + c0 (51)

where λ is the eigenvalue of the matrix M In order that the solution (46) could be stable allthe eigenvalues should be negative Then all the eigenmodes corresponding to the eigenvaluesdecrease and therefore any perturbation damps It requires ci gt 0 (i = 1 4) Especiallyminusc4 is the trace of the matrix M and we find

minus c4 = minus H

Hminus 4H minus 8

H3

Hlt 0 (52)

For the power expanding model (39) where H = minusnτ if τ gt 0 equation (52) leads to

4n2 + 2n + 1 lt 0 (53)

Thus there is no real solution for n As a result there does not exist any stable solution for thepower expanding model (39) On the other hand suppose τ lt 0 equation (52) yields

4n2 + 2n + 1 gt 0 (54)

for which there is a possibility that the solution might be stableWhen H = minusnτ in (39) the matrix M in (49) has the following form

M =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1+2nminusD0τminus2n+2

1minusn minus D01minusnτminus2n+2 3τ

1minusn

minus2minus2n+4n2

τ2+4n2

τ2 + 4n minus D0τ

minus2n+2

1minusn

1+2n+D0τminus2n+2

1minusn

minus2 minus 4n3t

1minusn

0 0 0 0 1

minus n(minus2minus2n+4n2)τ 2

n(minus2minus2n+4n2)τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

(1 + D0τ

minus2n+2

3

) (minus2minus2n+4n2)nτ 3 minusD0(minus2minus2n+4n2)nτminus2nminus1

3

2n(1+2n)

τ 2

minus 4(1minusn)D0τminus2n

3

4(1minusn)D0τminus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(55)

where the scale factor a(τ ) is given by a = a0τminusn and D0 equiv 3m2a2

0n Note that a0 = ln in (39)

As an example we may investigate the case n = minus12 In this case the eigenvalueequation has the following form

0 = λ

(λ minus 1

τ

) (λ minus 3

τ

) (2D0τ

4 + D0τ2 minus 2λ minus λ2τ

) (56)

Since there always appear positive eigenvalue the solution is not stableWe redefine

δ fa = (1 minus n)δ fa δ fb = (1 minus n)δ fb δH = (1 minus n)δH (57)

9

Class Quantum Grav 31 (2014) 075016 K Bamba et al

The matrix M in (55) has the following form

M =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1 + 2n minus D0τminus2n+2 minusD0τ

minus2n+2 3t

minus 2(1+2n)(1minusn)

τ2+4n2

τ

2(1 + 2n)(1 minus n)

minusD0τminus2n+2

minus1 + 4n2

+D0τminus2n+2 3t

0 0 0 0 1n(1+2n)

τ 2 minus n(1+2n)

τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

minus(1 + D0τ

minus2n+2

3

)1+2n

τ

D0(1+2n)nτminus2nminus1

32n(1+2n)

τ 2 minus 4(1minusn)D0τminus2n

34(1minusn)D0τ

minus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(58)

In section 34 in the part below equation (39) we have shown that if 0 lt n lt 1 the model isinconsistent As another example we consider the limit of n rarr 1 + 0 The matrix M in (58)reduces to

M =

⎛⎜⎜⎜⎜⎜⎜⎝

minus 2τ

0 3 minus D0 minusD0 3t

0 6τ

minusD0 3 + D0 3t

0 0 0 0 13τ 2 minus 3

τ 2 minus 6τ

6τ

1

minus 3+D0τ 3

D0τ 3

6τ 2 0 4

τ

⎞⎟⎟⎟⎟⎟⎟⎠

(59)

For this matrix the eigenvalue equation has the following form

0 = λ5 minus 14

τλ4 + (6D0 + 64)

λ3

τ 2minus (

2D20 + 66D0 + 33

) λ2

τ 3+ 3

(8D2

0 + 86D0 minus 63) λ

τ 4

minus 45(2D2

0 + 3) 1

τ 5 (60)

If

τ lt 0 and D0 gt

radic2353 minus 43

8= 0688 466 417 817 452 middot middot middot (61)

all the eigenvalues are negative and the system becomes stableIn general the eigenvalue equations (51) for the matrices (55) and (58) are rather

complicated and the explicit forms are given in appendix B As a result anyway we have founda solution which is stable under the perturbation Then we have shown that for an arbitrarilygiven history of the expansion of the universe we can construct a model who has a solutiongenerating the expansion and the solution is stable that is attractor solution

5 BransndashDicke type model

We introduce a parameter ε which is a positive but sufficiently small value (0 lt ε 1)In the previous section we have found that the model where n = 1 + ε and both τ and D0

satisfy equation (61) is stable and that the limit ε rarr 0 (n rarr 1) corresponds to the de Sitterspace In this section by starting with a model where ε gt 0 is small enough but finite weconstruct a model which reproduces an arbitrary expansion history of the universe by usingthe BransndashDicke type model

We here explore an arbitrary scale factor a(τ ) for τ lt 0 The scale factor correspondingto n = 1 + ε is given by a0τ

minus1minusε Hence the metric gμν corresponding to the scale factor a(τ )

is expressed by multiplying the metric gεμν corresponding to n = 1 + ε by a(τ )2aminus2

0 τ 2(1+ε)Since η = τ we rescale the metric gμν in the actions (17) and (19) as follows

gμν rarr a(η)minus2a20η

minus2(1+ε)gμν (62)

10

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

and (i j) components yield

0 = M2g (2H + H2) + m2M2

eff(3a2 minus 2ab minus ac) + (14 ϕ2 minus 1

2V (ϕ)a(τ )2)

M2g (25)

On the other hand the (τ τ ) component of (20) leads to

0 = minus3M2f K

2 + m2M2effc

2

(1 minus a3

b3

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (26)

and from (i j) components we find

0 = M2f (2K + 3K2 minus 2LK) + m2M2

eff

(a3c

b2minus c2

)+

(1

4ξ 2 minus 1

2U (ξ )c(τ )2

)M2

f (27)

with K = bb and L = cc Both equations (9) and (22) yield the identical equation

cH = bK orca

a= b (28)

The above equation is the constraint relating the two metrics imposed by the equations ofmotion If a = 0 we obtain c = aba On the other hand if a = 0 we find b = 0 that is aand b are constant and c can be arbitrary

Next we redefine scalars as ϕ = ϕ(η) and ξ = ξ (ζ ) and identify η and ζ with theconformal time t as η = ζ = τ Hence we acquire

ω(τ )M2g = minus 4M2

g (H minus H2) minus 2m2M2eff(ab minus ac) (29)

V (τ )a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a2 minus 5ab minus ac) (30)

σ (τ )M2f = minus4M2

f (K minus LK) minus 2m2M2eff

(minus c

b+ 1

) a3c

b2 (31)

U (τ )c(τ )2M2f = M2

f (2K + 6K2 minus 2LK)+ m2M2eff

(a3c

b2minus 2c2 + a3c2

b3

) (32)

with

ω(η) = ϕprime(η)2 V (η) = V (ϕ (η)) σ (ζ ) = ξ prime(ζ )2 U (ζ ) = U (ξ (ζ )) (33)

Consequently for arbitrary a(τ ) b(τ ) and c(τ ) if we choose ω(τ ) V (τ ) σ (τ ) and U (τ ) tosatisfy equations (29)ndash(32) the cosmological model with given evolutions of a(τ ) b(τ ) andc(τ ) can be reconstructed

A reason why we introduced two scalar fields instead of one is that there are three degreesof freedom a b and c in metrics (23) and it is not trivial to describe them by using only onescalar field which might not be impossible but we have not succeeded

33 Conformal description of the accelerating universe

In the following we use the conformal time We describe how the known cosmologies can beexpressed by using the conformal time

The conformally flat FRW universe metric is given by

ds2 = a(τ )2

(minusdτ 2 +

3sumi=1

(dxi)2

) (34)

In this equation when a(τ )2 = l2

τ 2 the metric (34) corresponds to the de Sitter universewhich may represent inflation or dark energy in the model under consideration On the otherhand if a(τ )2 = l2n

τ 2n with n = 1 by redefining the time coordinate as

dt = plusmn ln

τ ndτ (35)

6

Class Quantum Grav 31 (2014) 075016 K Bamba et al

ie

t = plusmn ln

n minus 1τ 1minusn (36)

the metric (34) can be rewritten as

ds2 = minusdt2 +[plusmn(n minus 1)

t

l

]minus 2n1minusn 3sum

i=1

(dxi)2 (37)

Equation (37) shows that if 0 lt n lt 1 the metric corresponds to the phantom universe [27]if n gt 1 to the quintessence universe and if n lt 0 to decelerating universe In case of thephantom universe (0 lt n lt 1) we should choose + sign in plusmn of (35) or (36) and shift tin (37) as t rarr t minus t0 The time t = t0 corresponds to the Big Rip and the present time ist lt t0 and the limit of τ rarr infin is equivalent to the infinite past (t rarr minusinfin) In case of thequintessence universe (n gt 1) we may again select + sign in plusmn of (35) or (36) The limit ofτ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infin to that of t rarr 0 which may beequivalent to the Big Bang In case of the decelerating universe (n lt 0) we may take minus signin plusmn of (35) or (36) The limit of τ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infinto that of t rarr 0 which may again be considered to be the Big Bang We should also note thatin case of the de Sitter universe (n = 1) the limit of τ rarr 0 corresponds to that of t rarr +infinand that of τ rarr plusmninfin to that of t rarr minusinfin

34 Dark energy universe with a(τ ) = b(τ ) = c(τ )

If the space-time described by the metric gμν represents the universe where we live thefunctions c(τ ) and b(τ ) are not directly related to the expansion of our universe because thefunctions c(τ ) and b(τ ) correspond to the degrees of freedom in the Einstein frame metric fμν Therefore we may choose c(τ ) and b(τ ) in the consistent way convenient for the calculationThis does not mean c(τ ) and b(τ ) are not relevant for the physics besides the expansionof our universe In this section we simply take a(τ ) = c(τ ) = b(τ ) which satisfy thecondition (28) and therefore H = K = L From (29) and (31) we find ω(τ ) = σ (τ ) andthus ϕ(τ ) = ξ (τ ) and also V (τ ) = U (τ ) from (30) and (32)

By choosing a(τ ) = c(τ ) = b(τ ) equations (29)ndash(32) are simplified as

ω(τ ) = σ (τ ) = 4(minusH + H2) V (τ )a(τ )2 = U (τ )a(τ )2 = (2H + 4H2) (38)

Let us construct the models where the scale factor squared is given by a(τ )2 = l2n

τ 2n In thiscase we find

ω(τ ) = σ (τ ) = 4n(n minus 1)

τ 2 V (τ ) = U (τ ) = (2n + 4n2)l2n

τ 2(1minusn) (39)

It should be cautioned that if 0 lt n lt 1 ω(τ ) and σ (τ ) become negative and this conflictswith the definition in (33) Hence the universe corresponding to the phantom cannot be realizedas in the standard scalarndashtensor model whose situation is different from the case of F(R)-bigravity [22] (for modified gravity including F(R) gravity and dark energy problem see eg[28ndash31]) In case of n = 1 in which the de Sitter universe is realized both ω(τ ) and σ (τ )

vanish and V (τ ) and U (τ ) become constants This is equivalent to the cosmological constant

4 Stability of solutions

As we have shown a wide class of expansions of the universe can be reproduced in thebigravity models coupled to scalar fields The desired solution is however only one of the

7

Class Quantum Grav 31 (2014) 075016 K Bamba et al

solutions If the solution is not stable under the perturbation such a solution cannot be realizedunless we perform very fine-tuning In this section we study the stability of the solution in thelast section For this purpose we rewrite (29)ndash(32) in the following form

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2eff(a(τ )b(τ ) minus a(τ )c(τ )) (40)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a(τ )2 minus 5a(τ )b(τ ) minus a(τ )c(τ )) (41)

σ (ζ )ζ 2M2f = minus 4M2

f (K minus LK) minus 2m2M2eff

(minus c (τ )

b (τ )+ 1

)a (τ )3 c (τ )

b (τ )2 (42)

U (ζ )c(τ )2M2f =M2

f (2K| +6K2 minus 2LK)+m2M2eff

(a (τ )3 c (τ )

b (τ )2 minus 2c (τ )2 + a (τ )3 c (τ )2

b (τ )3

)

(43)

On the other hand the scalar field equations (21) can be rewritten to

0 = 3

(ω(η)η + ωprime(η)

2η2 + 2Hω(η)η

)+ V prime(η)a2

0 = 3

(σ (ζ )ζ + σ prime(ζ )

2ζ 2 + (3K minus L) σ (ζ )ζ

)+ U prime(ζ )a2 (44)

Equations in (38) implies that with a function f (τ ) if we choose

ω(η) = 4(minus f primeprime(η) + f prime(η)2) σ (ζ ) = 4(minus f primeprime(ζ ) + f prime(ζ )2)

V (η) = eminus2 f (η)(2 f primeprime(η) + 4 f prime(η)2) U (ζ ) = eminus2 f (ζ )(2 f primeprime(ζ ) + 4 f prime(ζ )2) (45)

we find the following solution

a(τ ) = b(τ ) = c(τ ) = e f (τ ) η = ζ = τ (46)

We explore the stability of the solution in (46)We may consider the following perturbation

H rarr H + δH K rarr K + δK a rarr a (1 + δ fa) b rarr b (1 + δ fb)

η rarr η + δη ζ rarr ζ + δζ (47)

In what follows just for simplicity we take

M2f = M2

g = M2eff

2= M2 (48)

Thus we obtain

d

dτ

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

= M

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

M =

⎛⎜⎜⎜⎜⎜⎜⎝

2H 0 CminusDB minusD

B3

HB (B minus 1)

A E 2C minus DB

C+DB minus 2C 3

HB (B minus 1)

0 0 0 0 1

AH minusAH 2HC minus2HC 1(1 + D

3

)AH minusADH

3 2H(C minus 2BD

3

)43 BD minus4H

⎞⎟⎟⎟⎟⎟⎟⎠

(49)

8

Class Quantum Grav 31 (2014) 075016 K Bamba et al

where

A equiv H

H+ 2H minus 4

H3

H B equiv 1 minus H2

H C equiv 1 + 2

H2

H D equiv 3m2a2

H E equiv minus H

Hminus 4

H3

H

(50)

The derivation of equations (49) and (50) is given in appendix A We should note that wehave deleted δK in (47) by using (A13)

The eigenvalue equation has the following form

0 = λ5 + c4λ4 + c3λ

3 + c2λ2 + c1λ + c0 (51)

where λ is the eigenvalue of the matrix M In order that the solution (46) could be stable allthe eigenvalues should be negative Then all the eigenmodes corresponding to the eigenvaluesdecrease and therefore any perturbation damps It requires ci gt 0 (i = 1 4) Especiallyminusc4 is the trace of the matrix M and we find

minus c4 = minus H

Hminus 4H minus 8

H3

Hlt 0 (52)

For the power expanding model (39) where H = minusnτ if τ gt 0 equation (52) leads to

4n2 + 2n + 1 lt 0 (53)

Thus there is no real solution for n As a result there does not exist any stable solution for thepower expanding model (39) On the other hand suppose τ lt 0 equation (52) yields

4n2 + 2n + 1 gt 0 (54)

for which there is a possibility that the solution might be stableWhen H = minusnτ in (39) the matrix M in (49) has the following form

M =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1+2nminusD0τminus2n+2

1minusn minus D01minusnτminus2n+2 3τ

1minusn

minus2minus2n+4n2

τ2+4n2

τ2 + 4n minus D0τ

minus2n+2

1minusn

1+2n+D0τminus2n+2

1minusn

minus2 minus 4n3t

1minusn

0 0 0 0 1

minus n(minus2minus2n+4n2)τ 2

n(minus2minus2n+4n2)τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

(1 + D0τ

minus2n+2

3

) (minus2minus2n+4n2)nτ 3 minusD0(minus2minus2n+4n2)nτminus2nminus1

3

2n(1+2n)

τ 2

minus 4(1minusn)D0τminus2n

3

4(1minusn)D0τminus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(55)

where the scale factor a(τ ) is given by a = a0τminusn and D0 equiv 3m2a2

0n Note that a0 = ln in (39)

As an example we may investigate the case n = minus12 In this case the eigenvalueequation has the following form

0 = λ

(λ minus 1

τ

) (λ minus 3

τ

) (2D0τ

4 + D0τ2 minus 2λ minus λ2τ

) (56)

Since there always appear positive eigenvalue the solution is not stableWe redefine

δ fa = (1 minus n)δ fa δ fb = (1 minus n)δ fb δH = (1 minus n)δH (57)

9

Class Quantum Grav 31 (2014) 075016 K Bamba et al

The matrix M in (55) has the following form

M =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1 + 2n minus D0τminus2n+2 minusD0τ

minus2n+2 3t

minus 2(1+2n)(1minusn)

τ2+4n2

τ

2(1 + 2n)(1 minus n)

minusD0τminus2n+2

minus1 + 4n2

+D0τminus2n+2 3t

0 0 0 0 1n(1+2n)

τ 2 minus n(1+2n)

τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

minus(1 + D0τ

minus2n+2

3

)1+2n

τ

D0(1+2n)nτminus2nminus1

32n(1+2n)

τ 2 minus 4(1minusn)D0τminus2n

34(1minusn)D0τ

minus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(58)

In section 34 in the part below equation (39) we have shown that if 0 lt n lt 1 the model isinconsistent As another example we consider the limit of n rarr 1 + 0 The matrix M in (58)reduces to

M =

⎛⎜⎜⎜⎜⎜⎜⎝

minus 2τ

0 3 minus D0 minusD0 3t

0 6τ

minusD0 3 + D0 3t

0 0 0 0 13τ 2 minus 3

τ 2 minus 6τ

6τ

1

minus 3+D0τ 3

D0τ 3

6τ 2 0 4

τ

⎞⎟⎟⎟⎟⎟⎟⎠

(59)

For this matrix the eigenvalue equation has the following form

0 = λ5 minus 14

τλ4 + (6D0 + 64)

λ3

τ 2minus (

2D20 + 66D0 + 33

) λ2

τ 3+ 3

(8D2

0 + 86D0 minus 63) λ

τ 4

minus 45(2D2

0 + 3) 1

τ 5 (60)

If

τ lt 0 and D0 gt

radic2353 minus 43

8= 0688 466 417 817 452 middot middot middot (61)

all the eigenvalues are negative and the system becomes stableIn general the eigenvalue equations (51) for the matrices (55) and (58) are rather

complicated and the explicit forms are given in appendix B As a result anyway we have founda solution which is stable under the perturbation Then we have shown that for an arbitrarilygiven history of the expansion of the universe we can construct a model who has a solutiongenerating the expansion and the solution is stable that is attractor solution

5 BransndashDicke type model

We introduce a parameter ε which is a positive but sufficiently small value (0 lt ε 1)In the previous section we have found that the model where n = 1 + ε and both τ and D0

satisfy equation (61) is stable and that the limit ε rarr 0 (n rarr 1) corresponds to the de Sitterspace In this section by starting with a model where ε gt 0 is small enough but finite weconstruct a model which reproduces an arbitrary expansion history of the universe by usingthe BransndashDicke type model

We here explore an arbitrary scale factor a(τ ) for τ lt 0 The scale factor correspondingto n = 1 + ε is given by a0τ

minus1minusε Hence the metric gμν corresponding to the scale factor a(τ )

is expressed by multiplying the metric gεμν corresponding to n = 1 + ε by a(τ )2aminus2

0 τ 2(1+ε)Since η = τ we rescale the metric gμν in the actions (17) and (19) as follows

gμν rarr a(η)minus2a20η

minus2(1+ε)gμν (62)

10

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

ie

t = plusmn ln

n minus 1τ 1minusn (36)

the metric (34) can be rewritten as

ds2 = minusdt2 +[plusmn(n minus 1)

t

l

]minus 2n1minusn 3sum

i=1

(dxi)2 (37)

Equation (37) shows that if 0 lt n lt 1 the metric corresponds to the phantom universe [27]if n gt 1 to the quintessence universe and if n lt 0 to decelerating universe In case of thephantom universe (0 lt n lt 1) we should choose + sign in plusmn of (35) or (36) and shift tin (37) as t rarr t minus t0 The time t = t0 corresponds to the Big Rip and the present time ist lt t0 and the limit of τ rarr infin is equivalent to the infinite past (t rarr minusinfin) In case of thequintessence universe (n gt 1) we may again select + sign in plusmn of (35) or (36) The limit ofτ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infin to that of t rarr 0 which may beequivalent to the Big Bang In case of the decelerating universe (n lt 0) we may take minus signin plusmn of (35) or (36) The limit of τ rarr 0 corresponds to that of t rarr +infin and that of τ rarr +infinto that of t rarr 0 which may again be considered to be the Big Bang We should also note thatin case of the de Sitter universe (n = 1) the limit of τ rarr 0 corresponds to that of t rarr +infinand that of τ rarr plusmninfin to that of t rarr minusinfin

34 Dark energy universe with a(τ ) = b(τ ) = c(τ )

If the space-time described by the metric gμν represents the universe where we live thefunctions c(τ ) and b(τ ) are not directly related to the expansion of our universe because thefunctions c(τ ) and b(τ ) correspond to the degrees of freedom in the Einstein frame metric fμν Therefore we may choose c(τ ) and b(τ ) in the consistent way convenient for the calculationThis does not mean c(τ ) and b(τ ) are not relevant for the physics besides the expansionof our universe In this section we simply take a(τ ) = c(τ ) = b(τ ) which satisfy thecondition (28) and therefore H = K = L From (29) and (31) we find ω(τ ) = σ (τ ) andthus ϕ(τ ) = ξ (τ ) and also V (τ ) = U (τ ) from (30) and (32)

By choosing a(τ ) = c(τ ) = b(τ ) equations (29)ndash(32) are simplified as

ω(τ ) = σ (τ ) = 4(minusH + H2) V (τ )a(τ )2 = U (τ )a(τ )2 = (2H + 4H2) (38)

Let us construct the models where the scale factor squared is given by a(τ )2 = l2n

τ 2n In thiscase we find

ω(τ ) = σ (τ ) = 4n(n minus 1)

τ 2 V (τ ) = U (τ ) = (2n + 4n2)l2n

τ 2(1minusn) (39)

It should be cautioned that if 0 lt n lt 1 ω(τ ) and σ (τ ) become negative and this conflictswith the definition in (33) Hence the universe corresponding to the phantom cannot be realizedas in the standard scalarndashtensor model whose situation is different from the case of F(R)-bigravity [22] (for modified gravity including F(R) gravity and dark energy problem see eg[28ndash31]) In case of n = 1 in which the de Sitter universe is realized both ω(τ ) and σ (τ )

vanish and V (τ ) and U (τ ) become constants This is equivalent to the cosmological constant

4 Stability of solutions

As we have shown a wide class of expansions of the universe can be reproduced in thebigravity models coupled to scalar fields The desired solution is however only one of the

7

Class Quantum Grav 31 (2014) 075016 K Bamba et al

solutions If the solution is not stable under the perturbation such a solution cannot be realizedunless we perform very fine-tuning In this section we study the stability of the solution in thelast section For this purpose we rewrite (29)ndash(32) in the following form

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2eff(a(τ )b(τ ) minus a(τ )c(τ )) (40)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a(τ )2 minus 5a(τ )b(τ ) minus a(τ )c(τ )) (41)

σ (ζ )ζ 2M2f = minus 4M2

f (K minus LK) minus 2m2M2eff

(minus c (τ )

b (τ )+ 1

)a (τ )3 c (τ )

b (τ )2 (42)

U (ζ )c(τ )2M2f =M2

f (2K| +6K2 minus 2LK)+m2M2eff

(a (τ )3 c (τ )

b (τ )2 minus 2c (τ )2 + a (τ )3 c (τ )2

b (τ )3

)

(43)

On the other hand the scalar field equations (21) can be rewritten to

0 = 3

(ω(η)η + ωprime(η)

2η2 + 2Hω(η)η

)+ V prime(η)a2

0 = 3

(σ (ζ )ζ + σ prime(ζ )

2ζ 2 + (3K minus L) σ (ζ )ζ

)+ U prime(ζ )a2 (44)

Equations in (38) implies that with a function f (τ ) if we choose

ω(η) = 4(minus f primeprime(η) + f prime(η)2) σ (ζ ) = 4(minus f primeprime(ζ ) + f prime(ζ )2)

V (η) = eminus2 f (η)(2 f primeprime(η) + 4 f prime(η)2) U (ζ ) = eminus2 f (ζ )(2 f primeprime(ζ ) + 4 f prime(ζ )2) (45)

we find the following solution

a(τ ) = b(τ ) = c(τ ) = e f (τ ) η = ζ = τ (46)

We explore the stability of the solution in (46)We may consider the following perturbation

H rarr H + δH K rarr K + δK a rarr a (1 + δ fa) b rarr b (1 + δ fb)

η rarr η + δη ζ rarr ζ + δζ (47)

In what follows just for simplicity we take

M2f = M2

g = M2eff

2= M2 (48)

Thus we obtain

d

dτ

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

= M

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

M =

⎛⎜⎜⎜⎜⎜⎜⎝

2H 0 CminusDB minusD

B3

HB (B minus 1)

A E 2C minus DB

C+DB minus 2C 3

HB (B minus 1)

0 0 0 0 1

AH minusAH 2HC minus2HC 1(1 + D

3

)AH minusADH

3 2H(C minus 2BD

3

)43 BD minus4H

⎞⎟⎟⎟⎟⎟⎟⎠

(49)

8

Class Quantum Grav 31 (2014) 075016 K Bamba et al

where

A equiv H

H+ 2H minus 4

H3

H B equiv 1 minus H2

H C equiv 1 + 2

H2

H D equiv 3m2a2

H E equiv minus H

Hminus 4

H3

H

(50)

The derivation of equations (49) and (50) is given in appendix A We should note that wehave deleted δK in (47) by using (A13)

The eigenvalue equation has the following form

0 = λ5 + c4λ4 + c3λ

3 + c2λ2 + c1λ + c0 (51)

where λ is the eigenvalue of the matrix M In order that the solution (46) could be stable allthe eigenvalues should be negative Then all the eigenmodes corresponding to the eigenvaluesdecrease and therefore any perturbation damps It requires ci gt 0 (i = 1 4) Especiallyminusc4 is the trace of the matrix M and we find

minus c4 = minus H

Hminus 4H minus 8

H3

Hlt 0 (52)

For the power expanding model (39) where H = minusnτ if τ gt 0 equation (52) leads to

4n2 + 2n + 1 lt 0 (53)

Thus there is no real solution for n As a result there does not exist any stable solution for thepower expanding model (39) On the other hand suppose τ lt 0 equation (52) yields

4n2 + 2n + 1 gt 0 (54)

for which there is a possibility that the solution might be stableWhen H = minusnτ in (39) the matrix M in (49) has the following form

M =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1+2nminusD0τminus2n+2

1minusn minus D01minusnτminus2n+2 3τ

1minusn

minus2minus2n+4n2

τ2+4n2

τ2 + 4n minus D0τ

minus2n+2

1minusn

1+2n+D0τminus2n+2

1minusn

minus2 minus 4n3t

1minusn

0 0 0 0 1

minus n(minus2minus2n+4n2)τ 2

n(minus2minus2n+4n2)τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

(1 + D0τ

minus2n+2

3

) (minus2minus2n+4n2)nτ 3 minusD0(minus2minus2n+4n2)nτminus2nminus1

3

2n(1+2n)

τ 2

minus 4(1minusn)D0τminus2n

3

4(1minusn)D0τminus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(55)

where the scale factor a(τ ) is given by a = a0τminusn and D0 equiv 3m2a2

0n Note that a0 = ln in (39)

As an example we may investigate the case n = minus12 In this case the eigenvalueequation has the following form

0 = λ

(λ minus 1

τ

) (λ minus 3

τ

) (2D0τ

4 + D0τ2 minus 2λ minus λ2τ

) (56)

Since there always appear positive eigenvalue the solution is not stableWe redefine

δ fa = (1 minus n)δ fa δ fb = (1 minus n)δ fb δH = (1 minus n)δH (57)

9

Class Quantum Grav 31 (2014) 075016 K Bamba et al

The matrix M in (55) has the following form

M =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1 + 2n minus D0τminus2n+2 minusD0τ

minus2n+2 3t

minus 2(1+2n)(1minusn)

τ2+4n2

τ

2(1 + 2n)(1 minus n)

minusD0τminus2n+2

minus1 + 4n2

+D0τminus2n+2 3t

0 0 0 0 1n(1+2n)

τ 2 minus n(1+2n)

τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

minus(1 + D0τ

minus2n+2

3

)1+2n

τ

D0(1+2n)nτminus2nminus1

32n(1+2n)

τ 2 minus 4(1minusn)D0τminus2n

34(1minusn)D0τ

minus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(58)

In section 34 in the part below equation (39) we have shown that if 0 lt n lt 1 the model isinconsistent As another example we consider the limit of n rarr 1 + 0 The matrix M in (58)reduces to

M =

⎛⎜⎜⎜⎜⎜⎜⎝

minus 2τ

0 3 minus D0 minusD0 3t

0 6τ

minusD0 3 + D0 3t

0 0 0 0 13τ 2 minus 3

τ 2 minus 6τ

6τ

1

minus 3+D0τ 3

D0τ 3

6τ 2 0 4

τ

⎞⎟⎟⎟⎟⎟⎟⎠

(59)

For this matrix the eigenvalue equation has the following form

0 = λ5 minus 14

τλ4 + (6D0 + 64)

λ3

τ 2minus (

2D20 + 66D0 + 33

) λ2

τ 3+ 3

(8D2

0 + 86D0 minus 63) λ

τ 4

minus 45(2D2

0 + 3) 1

τ 5 (60)

If

τ lt 0 and D0 gt

radic2353 minus 43

8= 0688 466 417 817 452 middot middot middot (61)

all the eigenvalues are negative and the system becomes stableIn general the eigenvalue equations (51) for the matrices (55) and (58) are rather

complicated and the explicit forms are given in appendix B As a result anyway we have founda solution which is stable under the perturbation Then we have shown that for an arbitrarilygiven history of the expansion of the universe we can construct a model who has a solutiongenerating the expansion and the solution is stable that is attractor solution

5 BransndashDicke type model

We introduce a parameter ε which is a positive but sufficiently small value (0 lt ε 1)In the previous section we have found that the model where n = 1 + ε and both τ and D0

satisfy equation (61) is stable and that the limit ε rarr 0 (n rarr 1) corresponds to the de Sitterspace In this section by starting with a model where ε gt 0 is small enough but finite weconstruct a model which reproduces an arbitrary expansion history of the universe by usingthe BransndashDicke type model

We here explore an arbitrary scale factor a(τ ) for τ lt 0 The scale factor correspondingto n = 1 + ε is given by a0τ

minus1minusε Hence the metric gμν corresponding to the scale factor a(τ )

is expressed by multiplying the metric gεμν corresponding to n = 1 + ε by a(τ )2aminus2

0 τ 2(1+ε)Since η = τ we rescale the metric gμν in the actions (17) and (19) as follows

gμν rarr a(η)minus2a20η

minus2(1+ε)gμν (62)

10

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

solutions If the solution is not stable under the perturbation such a solution cannot be realizedunless we perform very fine-tuning In this section we study the stability of the solution in thelast section For this purpose we rewrite (29)ndash(32) in the following form

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2eff(a(τ )b(τ ) minus a(τ )c(τ )) (40)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff(6a(τ )2 minus 5a(τ )b(τ ) minus a(τ )c(τ )) (41)

σ (ζ )ζ 2M2f = minus 4M2

f (K minus LK) minus 2m2M2eff

(minus c (τ )

b (τ )+ 1

)a (τ )3 c (τ )

b (τ )2 (42)

U (ζ )c(τ )2M2f =M2

f (2K| +6K2 minus 2LK)+m2M2eff

(a (τ )3 c (τ )

b (τ )2 minus 2c (τ )2 + a (τ )3 c (τ )2

b (τ )3

)

(43)

On the other hand the scalar field equations (21) can be rewritten to

0 = 3

(ω(η)η + ωprime(η)

2η2 + 2Hω(η)η

)+ V prime(η)a2

0 = 3

(σ (ζ )ζ + σ prime(ζ )

2ζ 2 + (3K minus L) σ (ζ )ζ

)+ U prime(ζ )a2 (44)

Equations in (38) implies that with a function f (τ ) if we choose

ω(η) = 4(minus f primeprime(η) + f prime(η)2) σ (ζ ) = 4(minus f primeprime(ζ ) + f prime(ζ )2)

V (η) = eminus2 f (η)(2 f primeprime(η) + 4 f prime(η)2) U (ζ ) = eminus2 f (ζ )(2 f primeprime(ζ ) + 4 f prime(ζ )2) (45)

we find the following solution

a(τ ) = b(τ ) = c(τ ) = e f (τ ) η = ζ = τ (46)

We explore the stability of the solution in (46)We may consider the following perturbation

H rarr H + δH K rarr K + δK a rarr a (1 + δ fa) b rarr b (1 + δ fb)

η rarr η + δη ζ rarr ζ + δζ (47)

In what follows just for simplicity we take

M2f = M2

g = M2eff

2= M2 (48)

Thus we obtain

d

dτ

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

= M

⎛⎜⎜⎜⎜⎜⎜⎝

δη

δζ

δ fa

δ fb

δH

⎞⎟⎟⎟⎟⎟⎟⎠

M =

⎛⎜⎜⎜⎜⎜⎜⎝

2H 0 CminusDB minusD

B3

HB (B minus 1)

A E 2C minus DB

C+DB minus 2C 3

HB (B minus 1)

0 0 0 0 1

AH minusAH 2HC minus2HC 1(1 + D

3

)AH minusADH

3 2H(C minus 2BD

3

)43 BD minus4H

⎞⎟⎟⎟⎟⎟⎟⎠

(49)

8

Class Quantum Grav 31 (2014) 075016 K Bamba et al

where

A equiv H

H+ 2H minus 4

H3

H B equiv 1 minus H2

H C equiv 1 + 2

H2

H D equiv 3m2a2

H E equiv minus H

Hminus 4

H3

H

(50)

The derivation of equations (49) and (50) is given in appendix A We should note that wehave deleted δK in (47) by using (A13)

The eigenvalue equation has the following form

0 = λ5 + c4λ4 + c3λ

3 + c2λ2 + c1λ + c0 (51)

where λ is the eigenvalue of the matrix M In order that the solution (46) could be stable allthe eigenvalues should be negative Then all the eigenmodes corresponding to the eigenvaluesdecrease and therefore any perturbation damps It requires ci gt 0 (i = 1 4) Especiallyminusc4 is the trace of the matrix M and we find

minus c4 = minus H

Hminus 4H minus 8

H3

Hlt 0 (52)

For the power expanding model (39) where H = minusnτ if τ gt 0 equation (52) leads to

4n2 + 2n + 1 lt 0 (53)

Thus there is no real solution for n As a result there does not exist any stable solution for thepower expanding model (39) On the other hand suppose τ lt 0 equation (52) yields

4n2 + 2n + 1 gt 0 (54)

for which there is a possibility that the solution might be stableWhen H = minusnτ in (39) the matrix M in (49) has the following form

M =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1+2nminusD0τminus2n+2

1minusn minus D01minusnτminus2n+2 3τ

1minusn

minus2minus2n+4n2

τ2+4n2

τ2 + 4n minus D0τ

minus2n+2

1minusn

1+2n+D0τminus2n+2

1minusn

minus2 minus 4n3t

1minusn

0 0 0 0 1

minus n(minus2minus2n+4n2)τ 2

n(minus2minus2n+4n2)τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

(1 + D0τ

minus2n+2

3

) (minus2minus2n+4n2)nτ 3 minusD0(minus2minus2n+4n2)nτminus2nminus1

3

2n(1+2n)

τ 2

minus 4(1minusn)D0τminus2n

3

4(1minusn)D0τminus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(55)

where the scale factor a(τ ) is given by a = a0τminusn and D0 equiv 3m2a2

0n Note that a0 = ln in (39)

As an example we may investigate the case n = minus12 In this case the eigenvalueequation has the following form

0 = λ

(λ minus 1

τ

) (λ minus 3

τ

) (2D0τ

4 + D0τ2 minus 2λ minus λ2τ

) (56)

Since there always appear positive eigenvalue the solution is not stableWe redefine

δ fa = (1 minus n)δ fa δ fb = (1 minus n)δ fb δH = (1 minus n)δH (57)

9

Class Quantum Grav 31 (2014) 075016 K Bamba et al

The matrix M in (55) has the following form

M =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1 + 2n minus D0τminus2n+2 minusD0τ

minus2n+2 3t

minus 2(1+2n)(1minusn)

τ2+4n2

τ

2(1 + 2n)(1 minus n)

minusD0τminus2n+2

minus1 + 4n2

+D0τminus2n+2 3t

0 0 0 0 1n(1+2n)

τ 2 minus n(1+2n)

τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

minus(1 + D0τ

minus2n+2

3

)1+2n

τ

D0(1+2n)nτminus2nminus1

32n(1+2n)

τ 2 minus 4(1minusn)D0τminus2n

34(1minusn)D0τ

minus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(58)

In section 34 in the part below equation (39) we have shown that if 0 lt n lt 1 the model isinconsistent As another example we consider the limit of n rarr 1 + 0 The matrix M in (58)reduces to

M =

⎛⎜⎜⎜⎜⎜⎜⎝

minus 2τ

0 3 minus D0 minusD0 3t

0 6τ

minusD0 3 + D0 3t

0 0 0 0 13τ 2 minus 3

τ 2 minus 6τ

6τ

1

minus 3+D0τ 3

D0τ 3

6τ 2 0 4

τ

⎞⎟⎟⎟⎟⎟⎟⎠

(59)

For this matrix the eigenvalue equation has the following form

0 = λ5 minus 14

τλ4 + (6D0 + 64)

λ3

τ 2minus (

2D20 + 66D0 + 33

) λ2

τ 3+ 3

(8D2

0 + 86D0 minus 63) λ

τ 4

minus 45(2D2

0 + 3) 1

τ 5 (60)

If

τ lt 0 and D0 gt

radic2353 minus 43

8= 0688 466 417 817 452 middot middot middot (61)

all the eigenvalues are negative and the system becomes stableIn general the eigenvalue equations (51) for the matrices (55) and (58) are rather

complicated and the explicit forms are given in appendix B As a result anyway we have founda solution which is stable under the perturbation Then we have shown that for an arbitrarilygiven history of the expansion of the universe we can construct a model who has a solutiongenerating the expansion and the solution is stable that is attractor solution

5 BransndashDicke type model

We introduce a parameter ε which is a positive but sufficiently small value (0 lt ε 1)In the previous section we have found that the model where n = 1 + ε and both τ and D0

satisfy equation (61) is stable and that the limit ε rarr 0 (n rarr 1) corresponds to the de Sitterspace In this section by starting with a model where ε gt 0 is small enough but finite weconstruct a model which reproduces an arbitrary expansion history of the universe by usingthe BransndashDicke type model

We here explore an arbitrary scale factor a(τ ) for τ lt 0 The scale factor correspondingto n = 1 + ε is given by a0τ

minus1minusε Hence the metric gμν corresponding to the scale factor a(τ )

is expressed by multiplying the metric gεμν corresponding to n = 1 + ε by a(τ )2aminus2

0 τ 2(1+ε)Since η = τ we rescale the metric gμν in the actions (17) and (19) as follows

gμν rarr a(η)minus2a20η

minus2(1+ε)gμν (62)

10

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

where

A equiv H

H+ 2H minus 4

H3

H B equiv 1 minus H2

H C equiv 1 + 2

H2

H D equiv 3m2a2

H E equiv minus H

Hminus 4

H3

H

(50)

The derivation of equations (49) and (50) is given in appendix A We should note that wehave deleted δK in (47) by using (A13)

The eigenvalue equation has the following form

0 = λ5 + c4λ4 + c3λ

3 + c2λ2 + c1λ + c0 (51)

where λ is the eigenvalue of the matrix M In order that the solution (46) could be stable allthe eigenvalues should be negative Then all the eigenmodes corresponding to the eigenvaluesdecrease and therefore any perturbation damps It requires ci gt 0 (i = 1 4) Especiallyminusc4 is the trace of the matrix M and we find

minus c4 = minus H

Hminus 4H minus 8

H3

Hlt 0 (52)

For the power expanding model (39) where H = minusnτ if τ gt 0 equation (52) leads to

4n2 + 2n + 1 lt 0 (53)

Thus there is no real solution for n As a result there does not exist any stable solution for thepower expanding model (39) On the other hand suppose τ lt 0 equation (52) yields

4n2 + 2n + 1 gt 0 (54)

for which there is a possibility that the solution might be stableWhen H = minusnτ in (39) the matrix M in (49) has the following form

M =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1+2nminusD0τminus2n+2

1minusn minus D01minusnτminus2n+2 3τ

1minusn

minus2minus2n+4n2

τ2+4n2

τ2 + 4n minus D0τ

minus2n+2

1minusn

1+2n+D0τminus2n+2

1minusn

minus2 minus 4n3t

1minusn

0 0 0 0 1

minus n(minus2minus2n+4n2)τ 2

n(minus2minus2n+4n2)τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

(1 + D0τ

minus2n+2

3

) (minus2minus2n+4n2)nτ 3 minusD0(minus2minus2n+4n2)nτminus2nminus1

3

2n(1+2n)

τ 2

minus 4(1minusn)D0τminus2n

3

4(1minusn)D0τminus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(55)

where the scale factor a(τ ) is given by a = a0τminusn and D0 equiv 3m2a2

0n Note that a0 = ln in (39)

As an example we may investigate the case n = minus12 In this case the eigenvalueequation has the following form

0 = λ

(λ minus 1

τ

) (λ minus 3

τ

) (2D0τ

4 + D0τ2 minus 2λ minus λ2τ

) (56)

Since there always appear positive eigenvalue the solution is not stableWe redefine

δ fa = (1 minus n)δ fa δ fb = (1 minus n)δ fb δH = (1 minus n)δH (57)

9

Class Quantum Grav 31 (2014) 075016 K Bamba et al

The matrix M in (55) has the following form

M =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1 + 2n minus D0τminus2n+2 minusD0τ

minus2n+2 3t

minus 2(1+2n)(1minusn)

τ2+4n2

τ

2(1 + 2n)(1 minus n)

minusD0τminus2n+2

minus1 + 4n2

+D0τminus2n+2 3t

0 0 0 0 1n(1+2n)

τ 2 minus n(1+2n)

τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

minus(1 + D0τ

minus2n+2

3

)1+2n

τ

D0(1+2n)nτminus2nminus1

32n(1+2n)

τ 2 minus 4(1minusn)D0τminus2n

34(1minusn)D0τ

minus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(58)

In section 34 in the part below equation (39) we have shown that if 0 lt n lt 1 the model isinconsistent As another example we consider the limit of n rarr 1 + 0 The matrix M in (58)reduces to

M =

⎛⎜⎜⎜⎜⎜⎜⎝

minus 2τ

0 3 minus D0 minusD0 3t

0 6τ

minusD0 3 + D0 3t

0 0 0 0 13τ 2 minus 3

τ 2 minus 6τ

6τ

1

minus 3+D0τ 3

D0τ 3

6τ 2 0 4

τ

⎞⎟⎟⎟⎟⎟⎟⎠

(59)

For this matrix the eigenvalue equation has the following form

0 = λ5 minus 14

τλ4 + (6D0 + 64)

λ3

τ 2minus (

2D20 + 66D0 + 33

) λ2

τ 3+ 3

(8D2

0 + 86D0 minus 63) λ

τ 4

minus 45(2D2

0 + 3) 1

τ 5 (60)

If

τ lt 0 and D0 gt

radic2353 minus 43

8= 0688 466 417 817 452 middot middot middot (61)

all the eigenvalues are negative and the system becomes stableIn general the eigenvalue equations (51) for the matrices (55) and (58) are rather

complicated and the explicit forms are given in appendix B As a result anyway we have founda solution which is stable under the perturbation Then we have shown that for an arbitrarilygiven history of the expansion of the universe we can construct a model who has a solutiongenerating the expansion and the solution is stable that is attractor solution

5 BransndashDicke type model

We introduce a parameter ε which is a positive but sufficiently small value (0 lt ε 1)In the previous section we have found that the model where n = 1 + ε and both τ and D0

satisfy equation (61) is stable and that the limit ε rarr 0 (n rarr 1) corresponds to the de Sitterspace In this section by starting with a model where ε gt 0 is small enough but finite weconstruct a model which reproduces an arbitrary expansion history of the universe by usingthe BransndashDicke type model

We here explore an arbitrary scale factor a(τ ) for τ lt 0 The scale factor correspondingto n = 1 + ε is given by a0τ

minus1minusε Hence the metric gμν corresponding to the scale factor a(τ )

is expressed by multiplying the metric gεμν corresponding to n = 1 + ε by a(τ )2aminus2

0 τ 2(1+ε)Since η = τ we rescale the metric gμν in the actions (17) and (19) as follows

gμν rarr a(η)minus2a20η

minus2(1+ε)gμν (62)

10

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

The matrix M in (55) has the following form

M =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

minus 2nτ

0 1 + 2n minus D0τminus2n+2 minusD0τ

minus2n+2 3t

minus 2(1+2n)(1minusn)

τ2+4n2

τ

2(1 + 2n)(1 minus n)

minusD0τminus2n+2

minus1 + 4n2

+D0τminus2n+2 3t

0 0 0 0 1n(1+2n)

τ 2 minus n(1+2n)

τ 2 minus 2n(1+2n)

τ

2n(1+2n)

τ1

minus(1 + D0τ

minus2n+2

3

)1+2n

τ

D0(1+2n)nτminus2nminus1

32n(1+2n)

τ 2 minus 4(1minusn)D0τminus2n

34(1minusn)D0τ

minus2n+2

34nτ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(58)

In section 34 in the part below equation (39) we have shown that if 0 lt n lt 1 the model isinconsistent As another example we consider the limit of n rarr 1 + 0 The matrix M in (58)reduces to

M =

⎛⎜⎜⎜⎜⎜⎜⎝

minus 2τ

0 3 minus D0 minusD0 3t

0 6τ

minusD0 3 + D0 3t

0 0 0 0 13τ 2 minus 3

τ 2 minus 6τ

6τ

1

minus 3+D0τ 3

D0τ 3

6τ 2 0 4

τ

⎞⎟⎟⎟⎟⎟⎟⎠

(59)

For this matrix the eigenvalue equation has the following form

0 = λ5 minus 14

τλ4 + (6D0 + 64)

λ3

τ 2minus (

2D20 + 66D0 + 33

) λ2

τ 3+ 3

(8D2

0 + 86D0 minus 63) λ

τ 4

minus 45(2D2

0 + 3) 1

τ 5 (60)

If

τ lt 0 and D0 gt

radic2353 minus 43

8= 0688 466 417 817 452 middot middot middot (61)

all the eigenvalues are negative and the system becomes stableIn general the eigenvalue equations (51) for the matrices (55) and (58) are rather

complicated and the explicit forms are given in appendix B As a result anyway we have founda solution which is stable under the perturbation Then we have shown that for an arbitrarilygiven history of the expansion of the universe we can construct a model who has a solutiongenerating the expansion and the solution is stable that is attractor solution

5 BransndashDicke type model

We introduce a parameter ε which is a positive but sufficiently small value (0 lt ε 1)In the previous section we have found that the model where n = 1 + ε and both τ and D0

satisfy equation (61) is stable and that the limit ε rarr 0 (n rarr 1) corresponds to the de Sitterspace In this section by starting with a model where ε gt 0 is small enough but finite weconstruct a model which reproduces an arbitrary expansion history of the universe by usingthe BransndashDicke type model

We here explore an arbitrary scale factor a(τ ) for τ lt 0 The scale factor correspondingto n = 1 + ε is given by a0τ

minus1minusε Hence the metric gμν corresponding to the scale factor a(τ )

is expressed by multiplying the metric gεμν corresponding to n = 1 + ε by a(τ )2aminus2

0 τ 2(1+ε)Since η = τ we rescale the metric gμν in the actions (17) and (19) as follows

gμν rarr a(η)minus2a20η

minus2(1+ε)gμν (62)

10

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

By using η and ζ the total action Stotal = Sbi +Sϕ +Sχ in (17) (18) and (19) has the followingform

Stotal = M2g

intd4x

radicminus det ge(η)R(g) + M2

f

intd4x

radicminus det f R( f )

+ 2m2M2eff

intd4x

radicminus det ge2(η)(3 minus eminus (η)

2 trradic

gminus1 f + eminus2(η) detradic

gminus1 f )

minus M2g

intd4x

radicminus det g

1

2e(η)(ω(η) minus 3prime(η)2)gμνpartμηpartνη + e2(η)V (η)

minus M2f

intd4x

radicminus det f

1

2σ (ζ ) f μνpartμζpartνζ + U (ζ )

(63)

where

(η) equiv ln(a(η)minus2a2

0ηminus2(1+ε)

) (64)

Furthermore with (39) we have

ω(η) = 4 (1 + ε) ε

η2 σ (ζ ) = 4 (1 + ε) ε

ζ 2

V (ζ ) = 2 (1 + ε) (3 + 2ε) a20

ηminus2ε U (ζ ) = 2 (1 + ε) (3 + 2ε) a2

0

ζminus2ε (65)

We assume that for the Jordan frame of the action (63) the matters do not couple with thescalar fields η (ϕ) nor ζ (χ ) Thus we see that an arbitrary expansion history of the universecan be reproduced by the BransndashDicke type model and the solution is stable by a construction

6 Conclusions

In the present paper we have constructed bigravity models coupled with two scalar fields Ithas been shown that a wide class of the expansion history of the universe can be described by asolution of the bigravity model Especially inflation andor present accelerating expansion canbe described by this models This situation is very different from the models in the massivegravity where the reference metric is not dynamical In general it is very difficult to constructa model of the massive gravity which gives any non-trivial evolution of the expansion in theuniverse The solution obtained in the bigravity model is however unstable in general that isif we add a perturbation to the solution the perturbation grows up Accordingly we have foundthe conditions for the stability of the solution and explicitly constructed a model in whichthere exists a stable solution The stability can be checked from the eigenvalue equation forthe five times five matrix in (49) The stable model describes the universe whose expansion isalmost that in the de Sitter space-time By using the scale transformation of the stable modelwe construct the BransndashDicke like model We have shown that the BransndashDicke type modeladmits a solution describing an arbitrary expanding evolution of the universe The solutionis stable that is an attractor solution by the construction Therefore even if we started withdifferent initial conditions which are different a little bit with each other the universe willevolve into the stable solution

We should note that the F(R) bigravity models in [22 23] can be rewritten in the scalarndashtensor form in (17) (18) and (19) by using the scale transformation Therefore we can applythe procedures of the stability analysis in this paper to the F(R) bigravity models

When we consider the stability we only consider homogeneous perturbation whichdoes not depend on the spatial coordinates In case of massive gravity however if weconsider inhomogeneous perturbation it has been reported that there could appear ghost

11

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

in inhomogeneous andor anisotropic background [32] and there also appear superluminalmode in general [33] Furthermore it has been shown that the superluminal mode could breakcausality [34] Then we need further investigation by using the inhomogeneous perturbation inorder to show the consistency in the models proposed in this paper The investigation requireshowever highly non-trivial and complicated calculations Therefore we like to reserve thisinhomogeneous perturbation as future works

Acknowledgments

We are grateful to S D Odintsov for useful discussions We are also indebted to S Deser fortelling the problem about the superluminality The work is supported by the JSPS Grant-in-Aidfor Scientific Research (S) 22224003 and (C) 23540296 (SN) and that for Young Scientists(B) 25800136 (KB)

Appendix A The derivation of equations (49) and (50)

In this appendix we derive equations (49) and (50)By using (28) we have

L = K + K

Kminus H

H (A1)

Substituting (28) and (A1) into equations (40)ndash(43) we can eliminate c and L as

ω(η)η2M2g = minus4M2

g (H minus H2) minus 2m2M2effa (τ ) b (τ )

(1 minus K

H

) (A2)

V (η)a(τ )2M2g = M2

g (2H + 4H2) + m2M2eff

(6a (τ )2 minus 5a (τ ) b (τ ) minus a (τ ) b (τ ) K

H

) (A3)

σ (ζ )ζ 2M2f = minus4M2

f K

(H

Hminus K

)minus 2m2M2

eff

(1 minus K

H

)a (τ )3 K

b (τ ) H (A4)

U (ζ )b(τ )2M2f = M2

f

(2HH

K+ 4H2

)+ m2M2

eff

(a (τ )3 H

b (τ ) Kminus 2b (τ )2 + a (τ )3

b (τ )

) (A5)

Furthermore by plugging (A2) into (A4) we find

K minus σ (ζ )ζ 2

4Kminus m2M2

eff

2M2f

(1 minus K

H

)a (τ )3

b (τ ) H= H minus ω(η)η2

4Hminus m2M2

eff

2M2g

(1 minus K

H

)a (τ ) b (τ )

H

(A6)

We also eliminate H from equations (A2) and (A3) and from equations (A4) and (A5) asfollows(

ω (η) η2

2+ V (η) a (t)2

)M2

g = 6M2gH2 + 6m2M2

effa (τ ) (a (τ ) minus b (τ )) (A7)

(H2σ (ζ ) ζ 2

2K2+ U (ζ ) b (t)2

)M2

f = 6M2f H

2 minus 2m2M2eff

(b (τ )2 minus a (t)3

b (t)

) (A8)

By combining (A6) (A7) and (A8) and deleting η and ζ we acquire

0 = 2(K minus H) minus U (ζ )b(τ )2K

2H2+ V (η)a(τ )2

2H+ m2M2

eff

[K

H2M2f

(a(τ )3

b(τ )minus b(τ )2

)

minus 3

HM2g

a(τ )(a(τ )minus b(τ ))+(

1 minus K

H

) (a(τ )3

2M2f b(τ )H

minus a(τ )b(τ )

2M2gH

)] (A9)

12

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

We regard (A3) (A7) (A8) and (A9) as independent equations and study the perturbationfrom the solution as in (46) as in (47) We also choose (48) Thus we obtain

δH =(

minus4H minus m2a2

H

)δH + m2a2

HδK + (

H + 2HH minus 4H3)δη

+(2H + 4H2) minus 6m2a2δ fa + 6m2a2δ fb (A10)

2(H minus H2)δη = 4(HH minus H3)δη + (2H + 4H2 minus 6m2a2)δ fa minus 6HδH + 6m2a2δ fb (A11)

2(H minus H2)δζ = 4(HH minus H3)δζ + (2H + 4H2 + 6m2a2)δ fb

minus 2(H minus H2)

H(δH minus δK) minus 6HδH minus 6m2a2δ fa (A12)

minus H

H2(δH minus δK) =

(H

H+ 2H minus 4H2

)(δη minus δζ ) +

(2

H

H+ 4H

)(δ fa minus δ fb) (A13)

Note that

δV (η) = a(t)minus2(2H + 4HH minus 8H2)δη δU (ζ ) = a(t)minus2(2H + 4HH minus 8H2)δζ

δω(η) = 4(minusH + 2HH)δη δσ (ζ ) = 4(minusH + 2HH)δζ (A14)

By using (A13) we may delete δK in (A10) and (A12) and eventually we find

δH = minus4HδH +[(

H + 2HH minus 4H3) + m2a2

(H

H+ 2H minus 4

H3

H

)]δη

minus m2a2

(H

H+ 2H minus 4

H3

H

)δζ +

[(2H + 4H2

) minus 4m2a2 + 4m2a2 H2

H

]δ fa

+(

4m2a2 minus 4m2a2 H

H

)δ fb (A15)

δζ =(

H

H+ 2H minus 4

H3

H

)δη +

(H + 2H2 + 3m2a2

H minus H2minus 2 minus 4

H2

H

)δ fb minus

(H

H+ 4

H3

H

)δζ

+(

2 + 4H2

Hminus 3m2a2

H minus H2

)δ fa (A16)

Since δK = δ fb equation (A13) can be rewritten as

δ fb = δH +(

HH

H+ 2H2 minus 4

H4

H

)(δη minus δζ ) +

(2H + 4

H3

H

)(δ fa minus δ fb) (A17)

We may examine the stability by using (A11) (A15) (A16) (A17) and the relation

δH = δ fa (A18)

Appendix B Eigenvalue equations for matrices (55) and (58)

In this appendix we present an explicit forms of the eigenvalue equation (51) for the matrices(55) and (58)

13

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

[1] Fierz M and Pauli W 1939 Proc R Soc Lond A 173 211[2] Hinterbichler K 2012 Rev Mod Phys 84 671 (arXiv11053735 [hep-th])[3] Boulware D G and Deser S 1975 Ann Phys 89 193[4] Boulware D G and Deser S 1972 Phys Rev D 6 3368[5] van Dam H and Veltman M J G 1970 Nucl Phys B 22 397

Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

Class Quantum Grav 31 (2014) 075016 K Bamba et al

For the matrix (55) we find

c4 = minus8n2 + 4n + 2

τ

c3 = 2

3τminus2(n+1)[2D0τ

25n2 + n(τ 2 + 4) minus τ 2 + 3n(16n3 + 16n2 + 4n + 5)τ 2n]

c2 = minus2

3τminus4nminus3

[2D2

0n(2n + 1)τ 4 + 2D040n4 + n3(8τ 2 + 44) + n2(10 minus 8τ 2)

+ n(2τ 2 + 5) minus 2τ 2τ 2n+2 + 3n(32n4 + 16n3 + 20n2 + 4n + 3)τ 4n]

c1 = 4

3nτminus4(n+1)[minus2D0minus32n5 + 8n4(4τ 2 minus 13) minus 4n3(2τ 2 + 13)

minus n2(24τ 2 + 29) + n(τ 2 minus 14) minus τ 2 minus 3τ 2n+2 + 4nτ 4(2D0n + D0)2

+ 3(32n4 + 8n3 minus 2n2 minus n minus 1)τ 4n]

c0 = 4

3n2(2n + 1)τminus4nminus5[2D2

0τ4minus8n3 + 8n2(τ 2 minus 2) minus 8nτ 2 minus 3

+ 4D0(4n2 minus 5n + 1)(n minus τ 2)τ 2n+2 + 3(16n3 + 8n2 + 2n + 1)τ 4n] (B1)

and for and (58)

c4 = minus8n2 + 4n + 2

τ

c3 = minus1

3τminus2(n + 1)[D0τ

2minus2n2 minus n(10τ 2 + 7)+ τ 2minus 3τ 2n24n4 + 28n3 minus 2n2 + n(6τ 2 + 5)+ 3τ 2]

c2 = minus1

3τminus4nminus3

[2D2

0(2n + 1)τ 6 + D016n4 + 8n3(10τ 2 + 9)

minus 10n2 + n(34τ 2 + 3)+ 3τ 2τ 2n + 2 + 348n5 minus 72n4 + 4n3(12τ 2 minus 5)

+ 2n2(16τ 2 minus 9)+ 2n(7τ 2 minus 2)+ 5τ 2τ 4n]

c1 = minus1

3τminus4(n + 1)

[2D2

0(2n + 1)τ 46n3 + n2(1 minus 14τ 2)+ n(2 minus 5τ 2) minus 2τ 2minusD0minus48n6 + 16n5(3τ 2 + 22)+ n4(96τ 2 minus 88)+ 8n3(40τ 2 minus 13)

+ 3n2(48τ 2 minus 5)+ n(57τ 2 + 2)+ 10τ 2τ 2n + 2 + 3(2n + 1)

times144n5 minus 8n4(9τ 2 + 1) minus 4n3(5τ 2 minus 7)minus 2n2(3τ 2 + 1)minus 3nτ 2 + 2τ 2τ 4n]

c0 = minus1

3n(2n + 1)τminus6nminus5

[2D3

0τ6minus2n2 + n(2τ 2 minus 1)+ τ 2

+ 2D20minus8n4 + 8n3(τ 2 + 3)+ 2n2(12τ 2 minus 7)+ n(8τ 2 minus 2)+ 5τ 2τ 2n + 4

+ D080n5 minus 16n4(5τ 2 + 13)+ 32n3(5τ 2 + 1) minus 4n2(8τ 2 minus 3)

+ 3n(8τ 2 + 1)+ 9τ 2τ 4n + 2 + 9(8n3 + 4n2 + 2n + 1)(4n2 minus 2nτ 2 minus τ 2)τ 6n]

(B2)

References

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Zakharov V I 1970 JETP Lett 12 312Zakharov V I 1970 Pisrsquoma Zh Eksp Teor Fiz 12 447

[6] Vainshtein A I 1972 Phys Lett B 39 393

14

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