Symmetry analysis and exact solutions of modified Brans-Dicke cosmological equations

Metin Ar�k* and Mikhail B. Sheftel+

Department of Physics, Bogazici University, 34342 Bebek, Istanbul, Turkey(Received 23 July 2008; published 25 September 2008)

We perform a symmetry analysis of modified Brans-Dicke cosmological equations and present exact

solutions. We discuss how the solutions may help to build models of cosmology where, for the early

universe, the expansion is linear and the equation of state just changes the expansion velocity but not the

linearity. For the late universe the expansion is exponential and the effect of the equation of state on the

rate of expansion is just to change the constant Hubble parameter.

DOI: 10.1103/PhysRevD.78.064067 PACS numbers: 04.50.Kd, 04.20.Jb, 98.80.Bp, 98.80.Es

I. INTRODUCTION

The standard model of cosmology [1] has undergoneseveral modifications in the past 20 years. With the dis-covery [2] of cosmic microwave background radiation theearly cosmological models [3–6] developed into the firststandard model which had early radiation dominated andlate dust dominated stages. Observational and theoreticalconsistency has forced modifications on the standardmodel. We now believe that a universe is born in a radiationdominated stage, undergoes an inflationary exponentiallyexpanding stage, becomes radiation dominated again, andthen continues with matter domination, which by now hasevolved into a dark energy dominated, slowly but expo-nentially expanding, stage.

With the discovery of dark energy, which in classicalcosmology is equivalent to a cosmological constant, mod-els where the late cosmological constant is generated by ascalar field, called quintessence, have been considered [7–10]. In its most general form, called extended quintessence,the Lagrangian density consists of an arbitrary function ofthe scalar curvature and the quintessence field, as well as akinetic term for quintessence [11,12]. A special case of thistype of model is the Brans-Dicke-Jordan-Thirry [13–15]theory where the curvature scalar occurs only linearly inthe Lagrangian density. Whether the quintessence field canbe identified with the Brans-Dicke-Jordan-Thirry field isan interesting question [16–19]. In addition to explainingdark matter, Brans-Dicke theory may have other advan-tages. In particular, it has been remarked that Brans-Dicketheory can be imbedded in electroweak theory [20] and itcan explain the cosmic coincidence problem [21].

In standard cosmology the rate of expansion of theuniverse strongly depends on the equation of state of thematter energy that fills it. One immediate question whicharises is whether there is any consistent modification ofEinstein’s equations such that the expansion of the universeis independent of its content. In this paper we would like tostudy such a model. The model consists of a modified

Brans-Dicke-Jordan-Thirry [13–15] model where the signsof the kinetic and potential terms of the ‘‘scalar field’’ arenegative when written on the right-hand sides of theEinstein equations. Such models have been termed quin-tom [22,23], a word derived from quintessence and phan-tom. The term phantom refers to the fact that the equationof state for such a scalar field would violate the positivityof energy. However, one should remark that these terms arepositive just like the geometric term containing the squareof the time derivative of the scale size of the universe whenwritten on the left-hand (marble) side. Thus the Friedmanequation

3

�_a2

a2þ k

a2

�¼ 8�G� (1.1)

is changed into

3

�_a2

a2þ k

a2

�þ 2!

_�2

�2þ . . . ¼ 4!

�2� (1.2)

where � is a new geometric field inspired by the Brans-Dicke-Jordan-Thirry model. The reason we call � a geo-metric field is that its appearance is quite similar to thescale size a of the universe. The terms . . . will be deter-mined by the covariant action. In fact, the analogousequation for a Kaluza-Klein theory with n internal dimen-sions is [24]

3

�_a2

a2þ k

a2

�þ 1

2nðn� 1Þ

_b2

b2þ . . . ¼ 8�G� (1.3)

where b is the scale size of the internal manifold.In Sec. II we present the basic equations of the model as

applied to cosmology. In Sec. III we perform a symmetryanalysis based on dilatational symmetries and show thatsymmetries exist under the conditions that the mass of thescalar field vanishes and/or the curvature of spacelikesections is zero. The solutions are analyzed in Secs. IV,V, and VI.In Sec. VII we conclude with a model of expansion for

the universe where, in the early stages, the mass term of thescalar field can be neglected (m ¼ 0) but the curvatureterm is important. For closed (k ¼ 1) spacelike sections the

*metin.arik@boun.edu.tr+mikhail.sheftel@boun.edu.tr

PHYSICAL REVIEW D 78, 064067 (2008)

1550-7998=2008=78(6)=064067(7) 064067-1 � 2008 The American Physical Society

universe expands linearly. This linear expansion is inde-pendent of the equation of state. However, the rate ofchange of the Newtonian gravitational constant dependson the equation of state. This is in strong contrast tostandard general relativistic cosmology where the rate ofexpansion strongly depends on the equation of state.

Whereas the curvature term is important in the earlystages, we argue that in later stages, after the universe hasexpanded to a large size, it can be neglected. The mass termthen causes a slow exponential expansion which can beidentified with the dark energy phenomenon of the present-day universe. This phenomenon is also independent of theequation of state which only slightly influences the expo-nential rate of expansion.

II. BASIC EQUATIONS OF THE MODEL

The action is the following:

S ¼Z

d4xffiffiffig

p �� 1

8!�2R� 1

2g��@��@��

þ 1

2m2�2 þ LM

�; (2.1)

where � represents the Brans-Dicke scalar field, and !denotes the dimensionless Brans-Dicke parameter taken tobe much larger than 1, !� 1. The scalar field has akinetic term � 1

2 g��@��@�� with the wrong sign, and

the potential of the scalar field contains only a mass termwith the wrong sign 1

2m2�2. LM, on the other hand, is the

matter Lagrangian such that the scalar field � does notcouple with it. R is the Ricci scalar. For simplicity, we alsorestrict our analysis to the Robertson Walker metric toemphasize that � is necessarily spatially homogeneous:

d s2 ¼ dt2 � a2ðtÞ d~x2

½1þ k4~x2�2 ; (2.2)

where k is the curvature parameter with k ¼ �1, 0, 1corresponding to open, flat, closed universes, respectively,and aðtÞ is the scale factor of the universe. After applyingthe variational procedure to the action and assuming � ¼�ðtÞ and energy-momentum tensor of matter and radiationexcluding � to be in the perfect fluid form of T�� ¼diagð�;�p;�p;�pÞ, where � is the energy density andp is the pressure, and also noting that the right-hand side ofthe � equation must be zero in accordance with our pre-vious argument on LM being independent of �, the fieldequations reduce to (dots denote d=dt)

3

4!�2

�_a2

a2þ k

a2

�þ 1

2_�2 þ 1

2m2�2 þ 3

2!

_a

a_�� ¼ �;

(2.3)

�1

4!�2

�2€a

aþ _a2

a2þ k

a2

�� 1

!

_a

a

_�

�� 1

2!€��

þ�1

2� 1

2!

�_�2 � 1

2m2�2 ¼ p; (2.4)

€�þ 3_a

a�þ

�m2 þ 3

2!

�€a

aþ _a2

a2þ k

a2

��� ¼ 0: (2.5)

The following identity shows that the continuity equationcould be considered as a consequence of these three equa-tions:

��d

dtð2:3Þ þ 3

_a

a� ðð2:3Þ þ ð2:4ÞÞ � _�� ð2:5Þ

�

� _�þ 3_a

aðpþ �Þ ¼ 0: (2.6)

Instead, we prefer to consider a more complicated pressure

equation (2.4) as an algebraic consequence of dð2:3Þdt , (2.3)

and (2.5), and the continuity equation in (2.6).We assume the power law for the density

� ¼ Ca�; (2.7)

which implies that the continuity equation (2.6) becomesthe equation of state:

p ¼ �ð�þ 3Þ3

�; (2.8)

so that for dust we have � ¼ �3, p ¼ 0 and for radiation� ¼ �4, p ¼ �=3.Now we define the new unknowns FðaÞ and HðaÞ by the

formulas

F ¼_�

�; H ¼ _a

a) _� ¼ ��H;

€a

a¼ aHH0 þH2;

€�

�¼ aHF0 þ F2;

(2.9)

where a prime denotes the derivative of a function of a, andthe relations _H ¼ H0ðaÞ _a ¼ aHH0, _F ¼ F0ðaÞ _a ¼ aHF0were used. Equations (2.3), (2.4), and (2.5) become theequations for the unknowns HðaÞ and FðaÞ,

H2 þ 2HFþ 2!

3ðF2 þm2Þ þ k

a2¼

�4!

3

��

�2

��4!

3

�Ca�

�2; (2.10)

2

3aHðH0 þ F0Þ þH2 þ 4

3HFþ 2

3ð2�!ÞF2

þ k

3a2þ 2!

3m2 ¼

�� 4!

3

�p

�2; (2.11)

METIN AR�K AND MIKHAIL B. SHEFTEL PHYSICAL REVIEW D 78, 064067 (2008)

064067-2

aH

�1

2H0 þ!

3F0�þH2þ!HFþ!

3ðF2þm2Þþ k

2a2¼ 0:

(2.12)

The more lengthy equation (2.11) is still an algebraic

consequence of dð2:10Þdt , (2.10) and (2.12), and the continuity

equation in (2.6) due to the linear relation

dð2:10Þdt

þ 2F� ð2:10Þ þ 3H � ðð2:10Þ� ð2:11ÞÞ þ 4F� ð2:12Þ ¼ 0;

so we choose (2.10) and (2.12) together with the stateequation (2.8) as independent dynamic equations.However, these equations do not form a closed systemfor determining HðaÞ and FðaÞ because of � on theright-hand side of (2.10). To eliminate � from the �equation (2.10), we differentiate (2.10) with respect totime, use

_a¼aH; _�¼F�; _H¼aHH0ðaÞ; _F¼aHF0ðaÞ;and eliminate � with the aid of (2.10) with the result

ðH2 þHFÞH0 þ�H2 þ 2

3!HF

�F0

¼ �

2aH3 þ ð�� 1Þ

aH2Fþ ð�!� 6Þ

3aHF2 � 2!

3aF3

þ k

a3

�ð�þ 2Þ2

H � F

�þm2!

3að�H � 2FÞ: (2.13)

Solving algebraically Eqs. (2.13) and (2.12) with respect toH0 and F0, we obtain the closed system of two first orderequations in normal form that determines the two unknownfunctions HðaÞ and FðaÞ:

ð2!� 3ÞaH dHda

¼ ð�!þ 6ÞH2 þ 2ð�þ 4Þ!HF

þ 2!

3½ð�þ 6Þ!� 3�F2

þ 2!

3ð�!þ 3Þm2

þ k

a2½ð�þ 2Þ!þ 3�; (2.14)

�ð2!� 3ÞaH dFda

¼ 3

2ð�þ 4ÞH2 þ 3ð2!þ �þ 1ÞHF

þ ½ð�þ 8Þ!� 6�F2

þ ð�þ 2Þ!m2 þ 3kð�þ 4Þ2a2

: (2.15)

Equation (2.10) serves as an initial condition for thisdynamical system.

III. SYMMETRY GROUP ANALYSIS OF BASICDYNAMICAL EQUATIONS

We will use classical Lie group analysis [25] for sym-metries of the system (2.14) and (2.15) as a tool for findingexact solutions of these equations. Since the equations areof first order, it is impossible to find all Lie groups of pointsymmetries that are admitted by this system [a symmetrycondition cannot be split with respect to the derivatives ofunknowns to generate an overdetermined system, whichcan be solved and can yield all the point symmetries,because these derivatives are expressed from (2.14) and(2.15)]. Therefore, we make an ansatz for the form ofsymmetries that could be admitted by these equationsunder certain conditions. We note that Eqs. (2.14) and(2.15) are very close to the ones that admit scaling (ordilatational) symmetries, and an obstacle to this is the massterm and/or curvature term (the one with k) in each equa-tion. Indeed, a scaling symmetry group of transformationsof independent and dependent variables is defined by

~a ¼ ��a; ~F ¼ ��F; ~H ¼ �H: (3.1)

It is generated by the infinitesimal generator

X ¼ �a@

@aþ �F

@

@Fþ H

@

@H: (3.2)

Under transformations (3.1), Eqs. (2.14) and (2.15) become

ð2!� 3ÞaH dHda

¼ ð�!þ 6ÞH2 þ ���2ð�þ 4Þ!HF

þ �2ð��Þ 2!3

½ð�þ 6Þ!� 3�F2

þ ��2 2!

3ð�!þ 3Þm2

þ ��2ð�þÞ ka2

½ð�þ 2Þ!þ 3�;(3.3)

�ð2!�3ÞaHdFda

¼���3

2ð�þ4ÞH2þ3ð2!þ�þ1ÞHF

þ���½ð�þ8Þ!�6�F2

þ��ð�þÞð�þ2Þ!m2

þ��ð2�þ�þÞ3kð�þ4Þ2a2

: (3.4)

The condition of invariance of Eqs. (2.14) and (2.15) underthe Lie group of transformations (3.1) is that the trans-formed Eqs. (3.3) and (3.4) should coincide withEqs. (2.14) and (2.15), that is, all the � dependence shouldvanish.The first obvious condition is

¼ �; (3.5)

because it alone eliminates � in three terms, so that � willbe present only in the terms with k and m. We have to

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consider several cases of constraints on k andm needed forthe vanishing of the remaining dependence on �.

(1) Case 1 (generic).�þ � ¼ 0 and � ¼ 0, so that � ¼ � ¼ ¼ 0 andhence the symmetry generator X ¼ 0 in (3.2). Thereare no symmetries in the generic case.

(2) Case 2.k ¼ 0, arbitrary m. Then � ¼ ¼ 0 and � is arbi-trary, so it can be set to 1. The symmetry (3.2)becomes

X ¼ a@

@a: (3.6)

(3) Case 3.m ¼ 0, arbitrary k. Then �þ � ¼ 0 and, since ¼�, we can set � ¼ ¼ 1 and � ¼ �1 in (3.2), sothat the symmetry becomes

X ¼ H@

@Hþ F

@

@F� a

@

@a: (3.7)

(4) Case 4.k ¼ m ¼ 0. Then � ¼ is the only condition andwe have two inequivalent choices:

(a) � ¼ ¼ 0, � ¼ 1 in (3.2) with the resulting sym-metry

X1 ¼ a@

@a; (3.8)

(b) � ¼ ¼ 1, � ¼ 0, and the symmetry is

X2 ¼ H@

@Hþ F

@

@F: (3.9)

To summarize, we have only one symmetry in cases 2 and3 and two symmetries in case 4. One symmetry is notenough for integrating a system of two first order equa-tions, so in cases 2 and 3 we can only find invariantsolutions [25], which are particular solutions of our equa-tions. In case 4 with two symmetries we can find allsolutions to our equations, that is, integrate our system inquadratures.

IV. CASE OF FLAT SPACELIKE SECTIONS: K ¼ 0

In this case we have only one symmetry (3.6) with thebasis of invariants fH;Fg. The equation expressed only interms of invariants with solutions of the form F ¼ FðHÞcan be obtained by dividing one of Eqs. (2.14) and (2.15)over the other one, with k ¼ 0, with a result of the formdF=dH ¼ GðH;FÞ which is a first order equation with nomore known symmetries form � 0, and hence it cannot beintegrated. Though we cannot find the general solution incase 2, we still can search for invariant solutions with

respect to the symmetry (3.6), which satisfy the conditions

dH

da¼ 0;

dF

da¼ 0 (4.1)

and produce a ‘‘static’’ solution, fH;Fg ¼ constant. TheseH and F are roots of the algebraic equations

ð�!þ 6ÞH2 þ 2ð�þ 4Þ!HFþ 2!

3½ð�þ 6Þ!� 3�F2

þ 2!

3ð�!þ 3Þm2 ¼ 0;

3

2ð�þ 4ÞH2 þ 3ð2!þ �þ 1ÞHFþ ½ð�þ 8Þ!� 6�F2

þ ð�þ 2Þ!m2 ¼ 0: (4.2)

There are four roots of this system of two quadratic equa-tions. Two of them are real:

H ¼ � 2mffiffiffiffi!

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�!ð�2 þ 6�Þ � 12

p ;

F ¼ � m�ffiffiffiffi!

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�!ð�2 þ 6�Þ � 12

p ) F ¼ �

2H;

(4.3)

where the expressions under the square roots in the denom-inators are positive for �6 � �2 < �< �1 � 0 with the

roots of the denominator �1;2 ¼ 3ð�1� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 4=ð3!Þp Þ

and !> 104. The latter range of values of � includes theinteresting values �dust ¼ �3 and �radiation ¼ �4. For these� and!, two other roots of (4.2) are imaginary and will notbe considered,

H ¼ � 2imð!� 1Þ ffiffiffiffi!

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi6!2 � 17!þ 12

p ;

F ¼ � imffiffiffiffi!

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi6!2 � 17!þ 12

p ;

(4.4)

where the roots of the denominator are !1 ¼ 3=2 and!2 ¼ 4=3, so that 6!2 � 17!þ 12> 0.The equations that define H and F,

_a

a¼ H ¼ constant;

_�

�¼ F ¼ �

2H;

are integrated to give time dependence of the solution,

a ¼ a0eHt; � ¼ �0e

ð�=2ÞHt: (4.5)

The solution (4.5) satisfies the initial condition (2.10) ifarbitrary constants of integration �0 and a0 are related bythe formula

�0 ¼ �a�=20

m

ffiffiffiffiC

3

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið�2 þ 6�Þ!þ 12

�ð!� 1Þ þ 1

s; (4.6)

where C is the coefficient of the power law for the density(2.7).

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V. MASSLESS CASE: M ¼ 0

In this case we also have only one symmetry (3.7) withthe basis of invariants f’ ¼ aF; ¼ aHg (indeed, X’ ¼0, X ¼ 0). Substituting F ¼ ’=a and H ¼ =a into(2.14) and (2.15) with m ¼ 0, we obtain equations withthe new unknowns ’ and :

ð2!� 3Þa d da

¼ ½ð�þ 2Þ!þ 3�ð 2 þ kÞ þ 2ð�þ 4Þ!’

þ 2!

3½ð�þ 6Þ!� 3�’2; (5.1)

�ð2!� 3Þa d’da

¼ 3

2ð�þ 4Þð 2 þ kÞ

þ ½4!þ 3ð�þ 2Þ�’ þ ½ð�þ 8Þ!� 6�’2: (5.2)

We can obtain only one equation expressed solely in termsof invariants by dividing one of these equations over an-other with the result of the form d =d’ ¼ Gð’; Þ, butthis equation has no more known symmetries for k � 0 andhence it cannot be integrated. Though we cannot find thegeneral solution in case 3, again we can search for invariantsolutions with respect to the symmetry (3.7) now, whichsatisfy the conditions

d’

da¼ 0;

d

da¼ 0; (5.3)

so that now ’ and are constants independent of a. Withthese conditions, Eqs. (5.1) form an algebraic system:

½ð�þ 2Þ!þ 3�ð 2 þ kÞ þ 2ð�þ 4Þ!’ þ 2!

3½ð�þ 6Þ!� 3�’2 ¼ 0;

3ð�þ 4Þð 2 þ kÞ þ 2½4!þ 3ð�þ 2Þ�’ þ 2½ð�þ 8Þ!� 6�’2 ¼ 0; (5.4)

which again has four roots. Two real roots are

’ ¼ � ð�þ 2Þ ffiffiffiffiffi3k

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�2½ð�2 þ 8�Þ!þ 12!þ 6�p ;

¼ �ffiffiffiffiffi6k

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�2½ð�2 þ 8�Þ!þ 12!þ 6�p ;

(5.5)

which also yield F ¼ ’=a and H ¼ =a. Formulas (5.5)

imply

’ ¼ ð�þ 2Þ2

) F ¼ ð�þ 2Þ2

H: (5.6)

The expressions under the square roots in the denominatorsare positive for the values of � satisfying �6 � �2 < �<

�1 � �2, where the roots of the denominators �1;2 ¼�4� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4� 6=!p

and !> 104. Physically interesting val-ues �dust ¼ �3 and �radiation ¼ �4 again lie in this range,so the roots (5.5) are indeed real. Two other roots are ’ ¼0 ) F ¼ 0 ) � ¼ �0 ¼ constant, ¼ � ffiffiffiffiffiffiffi�kp

. Thus, ifk ¼ 0, H ¼ F ¼ 0; if k ¼ 1, H is imaginary; if k ¼ �1, ¼ �1, H ¼ �1=a and a ¼ a0 � t, which is a trivialsolution.For the first two real roots we have _a ¼ ¼ constant

and so a ¼ tþ a0. Then the relation _�=� ¼ ’=a yieldstime dependence of our solution,

� ¼ �0ð tþ a0Þ’= ¼ �0ð tþ a0Þð�þ2Þ=2;

a ¼ tþ a0;(5.7)

where we have used (5.6). The solution (5.7) satisfies theinitial condition (2.10) if �0 is expressed in terms of otherconstants as follows:

�0 ¼ � 2ffiffiffiffiffiffiffiffiC!

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3ð 2 þ kÞ þ 6’ þ 2!’2

p¼ � 4

ffiffiffiffiffiffiffiffiC!

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið�þ 2Þ½ð�þ 2Þð2!þ 3Þ þ 12� 2 þ 12k

p ;

(5.8)

where C is again the coefficient of the power law for thedensity (2.7).

VI. MASSLESS CASE WITH FLAT SPACELIKESECTIONS: K ¼ M ¼ 0

In this case we have two commuting symmetries, (3.8)and (3.9), which is enough for the complete integration ofour equations in quadratures that will yield their generalsolution. We apply first the symmetry generator X1 ¼a@=@a with the invariants H and F. A solution expressedsolely in terms of invariants has the form F ¼ FðHÞ.Dividing Eq. (2.15) over (2.14) with k ¼ m ¼ 0, we obtaina single first order equation for FðHÞ:

� dF

dH¼ ½ð�þ 8Þ!� 6�F2 þ 3ð2!þ �þ 1ÞHFþ ð3=2Þð�þ 4ÞH2

ð2!=3Þ½ð�þ 6Þ!� 3�F2 þ 2ð�þ 4Þ!HFþ ð�!þ 6ÞH2: (6.1)

This equation still admits one symmetry X2 ¼ F @@FþH @

@H with the invariant G ¼ F=H [and also a, not containedexplicitly in (6.1)], so that a new independent variable is H and a new unknown is G ¼ GðHÞ. In the new variables fH;GgEq. (6.1) becomes the one with separated variables, H and G,

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� dH

H¼ 2

2!½ð�þ 6Þ!� 3�G2 þ 6ð�þ 4Þ!Gþ 3ð�!þ 6Þf2½ð�þ 6Þ!� 3�Gþ 3ð�þ 4Þgð2!G2 þ 6Gþ 3Þ dG; (6.2)

where the cubic polynomial in G in the denominator wasfactorized. Integrating both sides of this equation in H andG and getting rid of logarithms, we obtain explicitly thefunction HðGÞ,

H ¼ H0hf2½ð�þ 6Þ!� 3�Gþ 3ð�þ 4Þg�� ð2!G2 þ 6Gþ 3Þ3i�b

��

2!G2 þ 6Gþ 3

2½ð�þ 6Þ!� 3�Gþ 3ð�þ 4Þ�q

� exp

�ð�þ 3Þltan�1

�2!Gþ 3ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3ð2!� 3Þp ��

; (6.3)

where we have introduced the following constants:

b ¼ ð�þ 6Þ!� 3

ð�þ 6Þ2!� 6ð�þ 5Þ ;

l ¼ 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3ð2!� 3Þp

ð�þ 6Þ2!� 6ð�þ 5Þ ;

q ¼ 3ð�þ 4Þð�þ 6Þ2!� 6ð�þ 5Þ :

(6.4)

Using (6.3), from the definition of G we also have F as afunction of G: F ¼ GHðGÞ.

The dependence GðaÞ can be determined from (2.14)with F ¼ GHðGÞ at k ¼ 0 and m ¼ 0:

da

a¼ 3ð2!� 3Þ

2!½ð�þ 6Þ!� 3�G2 þ 6ð�þ 4Þ!Gþ 3ð�!þ 6Þ� dH

H;

where dH=H is expressed from (6.2). The resulting equa-tion is

da

a¼ �6ð2!� 3ÞdG

f2½ð�þ 6Þ!� 3�Gþ 3ð�þ 4Þgð2!G2 þ 6Gþ 3Þ :(6.5)

Integrating both sides of (6.5) and eliminating logarithms,we obtain an explicit dependence aðGÞ,

a ¼ a0

�2!G2 þ 6Gþ 3

f2½ð�þ 6Þ!� 3�Gþ 3ð�þ 4Þg2b

� exp

�ltan�1

�2!Gþ 3ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3ð2!� 3Þp ��

: (6.6)

Formulas (6.3) and (6.6) together with F ¼ GHðGÞ yield aparametric representation, with a parameter G, of thegeneral solution HðaÞ, FðaÞ to Eqs. (2.14) and (2.15)with k ¼ m ¼ 0.

From _�=� ¼ F ¼ GH we have

d�

�¼ GHdt ¼ G

da

a

¼ �6ð2!� 3ÞGdGf2½ð�þ 6Þ!� 3�Gþ 3ð�þ 4Þgð2!G2 þ 6Gþ 3Þ ;

where we have expressed da=a from (6.5). Integrating bothsides of this equation and eliminating logarithms, we ob-tain the explicit dependence �ðGÞ,

� ¼ �0

�2½ð�þ 6Þ!� 3�Gþ 3ð�þ 4Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2!G2 þ 6Gþ 3p

�q

� exp

��ð�þ 6Þl2

tan�1

�2!Gþ 3ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3ð2!� 3Þp ��

; (6.7)

and because of the known dependence aðGÞ from (6.6), wehave determined �ðaÞ in a parametric representation. Inorder to satisfy the initial condition (2.10), the constants inthe formulas (6.3), (6.6), and (6.7) should be related asfollows:

�0 ¼ �2ffiffiffiffiffiffiffiffiC!

p a�=20

H0

: (6.8)

To determine time dependence of the solution, we use_a=a ¼ H to obtain

t� t0 ¼Z da

aH

and then express da=a from (6.5) and H from (6.3). The

integrand simplifies if we introduce the new parameter g ¼ð2!Gþ 3Þ= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3ð2!� 3Þpwith the resulting integral

t� t0 ¼ � 3ð2!� 3ÞH0

�2

!

�ð�þ3Þb

�Z f½ð�þ 6Þ!� 3�g� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3ð2!� 3Þp g2ð�þ3Þb�1

ðg2 þ 1Þð�þ3Þb

� exp½�ð�þ 3Þltan�1g�dg; (6.9)

which gives us explicitly the functions tðgÞ and hence tðGÞ,and implicitly the time dependence GðtÞ. Then from thefunctions aðGÞ and �ðGÞ, determined by (6.6) and (6.7),respectively, we know the time dependencies aðtÞ and �ðtÞin a parametric representation with the parameter g interms of the quadrature (6.9).For dust, � ¼ �dust ¼ �3 and the integral (6.9) simpli-

fies substantially with the result

t� t0 ¼ � ð2!� 3Þð!� 1ÞH0

ln½3ð!� 1Þg�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3ð2!� 3Þ

p�;

(6.10)

METIN AR�K AND MIKHAIL B. SHEFTEL PHYSICAL REVIEW D 78, 064067 (2008)

064067-6

that can be inverted easily to yield gðtÞ and so an explicitexpression for GðtÞ:

G¼ 1

2!ð!�1Þ� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2!�3

3

sexp

�� ð!�1Þð2!�3ÞH0ðt� t0Þ

��!

�:

(6.11)

For radiation, � ¼ �radiation ¼ �4 and the simplification isnot enough for evaluating the integral in (6.9).

VII. CONCLUSION

We have shown that, in the model of cosmology we areconsidering, for invariant solutions, the equation of statedoes not affect whether the expansion law of the universe islinear or exponential. Thus, for the m ¼ 0 case the expan-sion of the universe, following the invariant solution (5.7),is always linear. What is interesting is that according to(5.5) this solution is nontrivial only for k ¼ 1, i.e. a closeduniverse. It is also interesting to note that in this case theequation of state which determines the coefficient � onlyaffects the velocity of the linear expansion. This case maybe relevant for the early universe where the curvature termcannot be neglected. Equation (5.7) shows that with a0 ¼0, for t m�1

_�

�¼ �þ 2

2

1

t� �þ 2

2m; (7.1)

so that mj�j j _�j, and for early times the mass term canindeed be neglected. After the universe expands to a largesize, the curvature term can be neglected, and in this casethe k ¼ 0 invariant solution becomes important.Equations (4.3) and (4.5) show that the universe expandsexponentially for all equations of state. The equation ofstate is important just for the value of the Hubble constant.We identify this behavior with the dark energy or cosmo-logical constant which is observed today.For the k ¼ 1, m ¼ 0 and the k ¼ 0, m � 0 cases, we

have been able to present only the invariant solutions. Onthe other hand, for the k ¼ 0, m ¼ 0 case the generalsolution has been found. What seems to be relevant forcosmology is to find the k � 0, m � 0 general solutionwhich interpolates between the two invariant solutions forthe early and late universes.

ACKNOWLEDGMENTS

The research of the authors is partly supported by aresearch grant from Bogazici University ScientificResearch Fund, Research Project No. 07B301.

[1] M. Roos, arXiv:0802.2005v1.[2] A. A. Penzias and R.W. Wilson, Astrophys. J. 142, 419

(1965).[3] A. Friedman, Z. Phys. 10, 377 (1922).[4] A. Friedman, Gen. Relativ. Gravit. 31, 1991 (1999).[5] G. Lemaıtre, Ann. Sci. Soc. Brussels A 47, 41 (1927).[6] G. Lemaıtre, Mon. Not. R. Astron. Soc. 91, 483 (1931).[7] B. Ratra and P. J. E. Peebles, Phys. Rev. D 37, 3406

(1988).[8] R. R. Caldwell, R. Dave, and P. J. Steinhardt, Phys. Rev.

Lett. 80, 1582 (1998).[9] A. R. Liddle and R. J. Scherrer, Phys. Rev. D 59, 023509

(1998).[10] P. J. Steinhardt, L. Wang, and I. Zlatev, Phys. Rev. D 59,

123504 (1999).[11] F. Perrotta, C. Baccigalupi, and S. Matarrese, Phys. Rev. D

61, 023507 (1999).[12] T. Rador, Phys. Lett. B 652, 228 (2007).[13] P. Jordan, Ann. Phys. (Leipzig) 436, 219 (1947).[14] Y. R. Thirry, C.R. Hebd. Seances Acad. Sci. 226, 216

(1948).[15] C. Brans and C.H. Dicke, Phys. Rev. 124, 925 (1961).[16] S. Sen and A.A. Sen, Phys. Rev. D 63, 124006 (2001).[17] N. Banerjee and D. Pavon, Classical Quantum Gravity 18,

593 (2001).[18] M. Arik and M.C. Calik, J. Cosmol. Astropart. Phys. 01

(2005) 013.[19] M. Arik and M.C. Calik, Mod. Phys. Lett. A 21, 1241

(2006).[20] M.N. Chernodub and A. J. Niemi, Phys. Rev. D 77,

127902 (2008).[21] S. Carneiro, Int. J. Mod. Phys. D 14, 2201 (2005).[22] B. Feng, X. Wang, and X. Zhang, Phys. Lett. B 607, 35

(2005).[23] J. Sadeghi, M. R. Setare, A. Banijamali, and F. Milani,

Phys. Lett. B 662, 92 (2008).[24] T. Dereli and R.W. Tucker, Phys. Lett. 125B, 133 (1983).[25] P. Olver, Applications of Lie Groups to Differential

Equations (Springer-Verlag, New York, 1986).

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