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Dark Energy From Fifth Dimension H. Alavirad ∗ and N. Riazi † Physics Department and Biruni Observatory, Shiraz University, Shiraz 71454, Iran Observationalevidencefortheexistenceofdarkenergyisstrong. Herewesuggestamodel which is based on a modified gravitationaltheory in 5D and interpret the 5th dimension as a manifestation of dark energy in the 4D observable universe. We also obtain an equation 8 of state parameter which varies with time. Finally, we match our model with observations 0 0 by choosing the free parameters of the model. 2 Keywords: dark energy, extra dimensions, Kaluza-Klein theory. n a J 1 I. INTRODUCTION 2 ] h RecentobservationssuggestthattheUniverseisnotonlyexpanding,butalsoaccelerating [1,2]. t - p The first candidate for dark energy was a cosmological constant. It was originally introduced by e h Einstein in 1917 to achieve a static universe. However, after Hubble observations which suggested [ that the universe is not static, it was abandoned. 1 v Inparticlephysics,thecosmologicalconstantnormallyarisesasanenergydensityofthevacuum. 4 6 The energy scale of the cosmological constant (Λ) should be much larger than that of the present 1 3 Hubble constant H , if it originates from the vacuum energy density. This is the cosmological . 0 1 0 constant problem. 8 0 Inadditiontothecosmologicalconstantproposal,therearealotofalternativerouteswhichhave : v been proposed to explain the accelerated expansion of the Universe. Some of the most important i X are as follows [3]: r a 1-Quintessence models[4]; 2-Scalar field models[5]; 3-Chameleon fields [6]; 4-K-essence [7]; 5-Modified gravity arising out of string theory [8] or generalization of GR [9, 10]; 6-Phantom dark energy [11]; .... [email protected] ∗ † [email protected] 2 The Einstein field equations consist of two parts. The first part contains the geometry and the second part contains the matter. Some of dark energy (DE) models include the cosmological constant and a scalar field which modify the rhs by introducing some extra terms in it. There is another way: modifying the geometry (i.e. the lhs of the Einstein’s equations). The geometrical modifications can arise from quantum effects such as higher curvature corrections to the Einstein- Hilbert (EH) action. In [12] by introducing a quadratic term in R, an inflationary solution in the early universe was obtained. However, it was pointed out in [9, 10] that the late time acceleration can be realized by adding a term containing inverse power of Ricci scalar to the EH action. The structure of this paper is as follows: in section II, we review the modified gravity theories. In section III, we discuss briefly the Kaluza-Klein gravity and the induced matter in this theory. In section IV, we apply the STM formalism in CDTT model. Then in section V, we extract energy density and the pressureof dark energy from STMformalism. Finally, in section VI, we will match our model with observations by choosing the free parameters of the model. II. MODIFIED GRAVITY We first investigate a fairly general way of modifying gravitational theory which is known as f(R) gravity. The formalism starts with the introduction of an action in the form: S = √ gf(R)d4x+ √ g d4x (1) M Z − Z − L where f(R) is an arbitrary function of R such as R +αR2, R + αlnR, R + α etc, and is a R LM matter Lagrangian density. By varying the action (1) with respect to the metric g , one obtains µν the following field equations[10]: 1 1 g f(R) R f′(R) f′(R) g 2f′(R)+ T = 0 (2) µν µν µ ν µν µν 2 − −∇ ∇ − ∇ 2 where µ ,ν=0...3 and a prime shows differentiate with respect to R . By taking the trace of (3) in the absence of matter and constant curvature (▽ R = 0) case, we find: α 1 f(R) Rf′(R) = 0 (3) − 2 In [9] Carroll et.al. considered a f(R) function of the form (CDTT model): µ4 f(R)= R (4) − R where µ is a new parameter with dimension of [time]−1. By means of (3), we obtain the following field equations for CDTT model: 1 µ4 µ4 1 1 g (R ) R (1+ ) µ4( +g 2) + T = 0 (5) 2 µν − R − µν R2 − ∇µ∇ν µν∇ R2 2 µν 3 The constant-curvature vacuum solutions in 4D, for which ▽ R = 0, satisfy R = √3µ2. µ ± Thus one finds the interesting result that the constant curvature vacuum solutions in 4D are not MinkowskispacebutratheraredeSitterorantideSitterspaces. Moreoverinsphericallysymmetric case (the black hole solutions), one obtains a Schwartzchild-de Sitter black hole. By considering a perfect fluid case i.e: TM = (ρ +p )U U +p g (6) µν M M µ ν M µν where Uµ is the fluid rest-frame four velocity, ρ is the energy density, p is the pressure and we M M consider an equation of state of the form p = ωρ , also we take the metric of the flat Robertson- M M Walker (in four dimension) form, i.e. ds2 = dt2+a(t)2dx2. One obtains the following equation: − µ4 3H2 (2HH¨ +15H2H˙ +2H˙2+6H4) = ρ (7) − 12(H˙ +2H2)2 M for the time-time component of the field equation and: 3 µ4 1 1 1 H˙ + (4H˙ +9H2 a2∂ ∂ ( ) 2a2H∂ ( )) = p (8) 2 − 12(H˙ +2H2)2 − 0 0 a2 − 0 a2 −2 M for the space-space component. Equation (7) reduces to the usual Friedman equation by setting µ = 0. In[9]Carrollet.al. obtainedsomesolutionssuchas eternaldeSitter, powerlawacceleration and future singularity in vacuum case. As mentioned in the Introduction, the solutions of this theory contain instabilities [13]. In [14] Dolgove and Kawasaki also find instability in the matter section of the above modified gravity model. In [15] Faraoni suggested that generally a model is stable if it satisfies the condition:f′′ > 0 and is unstable if f′′ < 0. Dicke in [16] discussed the Newtonian limit in singular nonlinear modified gravity models (i.e.f(0) = ) and found that a model has Newtonian limit if it satisfies f′′(R ) = 0 where 0 ∞ R is the solution of (3). Therefore the f(R) model which is introduced in (4), hasn’t Newtonian 0 limit and is unstable. III. KALUZA-KLEIN GRAVITY AND INDUCED MATTER Kaluza unified electromagnetism with gravity in 1921 by applying Einstein’s general theory of relativity to a five rather than four-dimensional spacetime manifold [17]. The Einstein equations, in five dimensions with no five dimensional energy-momentum tensor, are: Gˆ = 0 (9) AB 4 or equivalently: Rˆ = 0 (10) AB where hat denotes a five dimensional quantity and A,B=0...4 and Gˆ Rˆ 1Rˆgˆ is the AB ≡ AB − 2 AB Einstein tensor, Rˆ and Rˆ = gˆ RÂB are the five dimensional Ricci tensor and Ricci scalar and AB AB gˆ is the five dimensional metric. The absence of matter in the five dimensional universe is the AB Einstein’s idea which says that the universe in higher dimensions is empty. Thefive dimensional Ricci tensor and Christoffel symbols are defined in terms of metric exactly as in four dimensions (minimal extension). Then everything depends on one’s choice for the form of the five dimensional metric. Kaluza proposed the following metric: g +κ2φ2A A κφ2A αβ α β α gÂB =   (11) κφ2A φ2 β   wheretheelectromagnetic potentialisscaled byaconstantκinordertogettherightmultiplicative factor in action. There is an important question with the Kaluza’s assumption: where is the fifth dimension? Why do we not observe it? Kaluza proposed the cylindrical condition to overcome this problem which means dropping all derivatives with respect to the fifth coordinate. In1926KleinshowedthatKaluza’scylinderconditionwouldarisenaturallyifthefifthdimension has (1) a circular topology, in which case physical fields would depend on it only periodically, and couldbeFourier-expanded;and(2)asmallenough(”compactified”)scaleinwhichcasetheenergies of all Fourier modes above the ground state could be made so high as to be unobservable. This version of Klein theory is called compactified Kaluza-Klein gravity. If one applies the cylinder condition and use the metric (11) and field equations (9), then one will find that the αβ , α4 and 44-components of the five dimensional field equations (9) reduce − − respectively to the following field equations in four dimensions: κ2φ2 1 G = TEM [▽ (∂ φ) g 2φ] αβ 2 αβ − φ α β − αβ ∂αφ ▽αF = 3 F αβ αβ − φ κ2φ3 2φ= F Fαβ (12) αβ 4 5 where G R Rg /2 is the Einstein tensor in four dimensions, TEM g F Fγδ FγF αβ ≡ αβ− αβ αβ ≡ αβ γδ 4 − α βγ is the electromagnetic energy-momentum tensor, and F ∂ A ∂ A , where α,β = 0...3. If αβ α β β α ≡ − the scalar field φ is constant throughout spacetime, then the first two equations of (12) are just the Einstein and Maxwell equations : G = 8πGφ2TEM αβ αβ ▽αF = 0 (13) αβ where we have identified the scaling parameter κ in terms of the gravitational constant G (in four dimensions) by: κ 4√πG (14) ≡ This is the result originally obtained by Kaluza and Klein, who set φ = 1. The condition φ = constant. is consistent with the third equation of (12) when F Fαβ = 0, which was first pointed αβ out by Jordan [18]. An alternative is to abandon the cylindrical condition [19, 20]. Thereforethe metric dependson the fifth dimension and this dependence allows one to obtain electromagnetic radiation, dust and other forms of cosmological matter. These types of theories are called noncompactified Kaluza- Klein theories. Wesson proposed that the fifth coordinate ψ might be related to the rest mass. Dimensionally x4 = Gm allows us to treat the rest mass m of a particle as a length coordinate, in analogy with c2 x0 = ct. We can say that the four-dimensional matter is a manifestation of the five-dimensional geometry [19]. In noncompactified Kaluza-Klein theories we begin with a metric of the form: g 0 αβ gÂB =   (15) 0 ǫφ2   where we have introduced the factor ǫ in order to allow a timelike or a spacelike signature for the fifth dimension (and ǫ2 = 1). Now by using the five dimensional field equations (9) in vacuum and keeping derivatives with respect to the fifth coordinate x4, the resulting expression for the αβ α4 and 44 parts of the − − − 6 five dimensional Ricci tensor R are : αβ ▽ (∂ φ) ǫ ∂ φ∂ g Rˆ = R β α + ( 4 4 αβ ∂ g αβ αβ − φ 2φ2 φ − 4 αβ gγδ∂ g ∂ g + gγδ∂ g ∂ g 4 γδ 4 αβ) (16) 4 αγ 4 βδ − 2 g44gβγ ∂ gβγ∂ g Rˆ = (∂ g ∂ g ∂ g ∂ g )+ β 4 γα α4 4 βγ 4 44 γ 44 4 αβ 4 − 2 gβγ∂ (∂ g ) ∂ gβγ∂ (g ) gβγ∂ ∂ g 4 β γα α 4 βγ 4 α βγ + (17) 2 − 2 − 2 gβγgδǫ∂ g ∂ g ∂ gβγ∂ g 4 γα β δβ 4 α βγ + + 4 4 ∂ gαβ∂ g gαβ∂ (∂ ) ∂ φgαβ∂ g Rˆ = ǫφ2φ 4 4 αβ 4 αβ + 4 4 αβ (18) 44 − − 2 − 2 2φ gαβgγδ∂ g ∂ g 4 γβ 4 αδ − 4 Then Eq. (16) gives the following expression for the four dimensional Ricci tensor: ▽ (∂ φ) ǫ ∂ φ∂ g β α 4 4 αβ R = [ ∂ (∂ g ) αβ φ − 2φ2 φ − 4 4 αβ gγδ∂ g ∂ g + gγδ∂ g ∂ g 4 γδ 4 αβ] (19) 4 αγ 4 βδ − 2 Theabove equation allows us to interpret the four dimensional matter as a manifestation of the five dimensional geometry. We assume that the Einstein Field equations hold in four dimensions i.e.: 1 8πGT = R Rg (20) αβ αβ αβ − 2 where T is the four-dimensional matter energy momentum tensor. By contracting (19) with g , αβ αβ we obtain the following expression for Ricci scalar: ǫ R = [∂ gαβ∂ g +(gαβ∂ g )2] (21) 4φ4 4 4 αβ 4 αβ inserting (21) and (19) into (20) one finds: ▽ (∂ φ) ǫ ∂ φ∂ g 8πGT = β α [ 4 4 αβ ∂ (∂ g )+gγδ∂ g g µν φ − 2φ2 φ − 4 4 αβ 4 αγ βγ gαγ∂ g ∂ g g 4 γδ 4 αβ + αβ(∂ gγδ∂ g +(gγδ∂ g )2)] (22) 4 4 γδ 4 γδ − 2 4 If we use this expression for T , the four-dimensional Einstein equations G = 8πGT are µν αβ αβ contained in the five-dimensional vacuum ones Gˆ = 0. The matter described by T is a AB µν manifestation of pure geometry in the five dimensional world. There are solutions for different types of metric and energy-momentum tensor, such as the spherically symmetric case [20], the isotropic and homogenous case [21], etc. 7 IV. 5D MODIFIED GRAVITY IN STM FORMALISM It is easy to check that all f(R) gravity theories in the vacuum and constant curvature case are equivalent to the Einstein field equations in the presence of a cosmological constant, so we can use any f(R) model. In the constant curvature and no matter case, Eq.(3) reduces to : 1 R f′(R) g f(R)= 0 (23) µν µν − 2 From (4) we obtain R and by substituting it in f(R) and its derivative, and theb by replacing it 0 in (23), the following relation is obtained: R +β(R )g = 0 (24) µν 0 µν where: f(R ) 0 β(R )= (25) 0 −f′(R ) 0 which can berewritten as Einstein field equation in thepresenceof acosmological constant, where: R 0 Λ= β+ (26) 2 In this section we use the CDTT model and work with a 5D extension of the flat RW metric in the form: ds2 = dt2 a(t)2(dr2+r2dΩ2) R(t)2dψ2 (27) − − wherea(t)isthescalefactorofordinary3DspatialdimensionsandR(t)isthescalefactorofthefifth dimension. In five dimensions, the solution of (3) for CDTT model is (with MP (8πG)−12 = 1): ≡ 7 Rˆ = µ2 (28) ±r3 Here we choose the minus sign. Now we try to find out a(t) and R(t). By replacing (27) and (28) in (23) in five dimensinal and vacuum (STM formalism) and constant curvature case i.e: µ4 1 µ4 H = (1+ )Rˆ (1 )Rˆgˆ = 0 (29) AB Rˆ2 AB − 2 − Rˆ2 AB we obtain the following equations: a¨(t) R¨(t) √21 H0 = 15 5 + µ2 = 0 (30) 0 − a(t) − R(t) 3 8 a¨(t) a˙(t) a˙(t)R˙(t) √21 H1 = H2 = H3 = 5 10( )2 5 + µ2 = 0 (31) 1 2 3 − a(t) − a(t) − a(t)R(t) 3 R¨(t) a˙(t)R˙(t) √21 H4 = 5 15 + µ2 = 0 (32) 4 − R(t) − a(t)R(t) 3 By computing Rˆ for metric (27) and replacing it in (28) we find: a¨(t) R¨(t) a˙(t) a˙(t)R˙(t) 21µ2 6 +2 +6( )2+6 = 0 (33) a(t) R(t) a(t) a(t)R(t) − p3 by solving simultaneously (30),(31),(32),(33) we find a(t) and R(t) : 1 1 µt√51894 3 2µt√51894 r 7µe 15 √51894(e 15 C2 C3) a(t) = − − (34) ± µt√518914 µe 15 3 1 3 1 R(t)= C (e4µt√1553474 C22−C32)41(e2µt√1553474 C2+C3)34 (35) 1 3 1 3 1 2µt√53474 3 4µt√53474 1 (e 15 C2 C3)4(e 15 )8 − In (34 ),(35) we have three free parameters: C ,C ,C . In section IV we will try to determine them 1 2 3 from the available observational parameters. V. DENSITY AND PRESSURE FROM STM We consider a perfect fluid energy momentum tensor for dark energy in four dimensions of the form: Tµ = diag(ρ (t), p (t), p (t), p (t)) (36) ν DE DE DE DE − − − Wesson [22] suggested that the new terms due to fifth dimension which depend on R(t) (and ψ in noncompactified Kaluza-Klein cosmology) in H0 and Ha (where a=1, 2,3), are density and 0 a pressure of matter respectively. The main proposal of our model is to choose them as density and pressure of the dark energy. By using equations (30) and (31), we obtain the following expressions for ρ (t) and p (t): DE DE Hˆ0 = f(a(t))+ρ (t) (37) 0 DE 9 Hâ = g(a(t)) p (t) (38) a DE − and: 10R¨(t) ρ (t)= (39) DE 7 R(t) 10a˙(t)R˙(t) p (t) = (40) DE − 7 a(t)R(t) It is clear that ρ (t) and p (t) vanish for a model with constant R(t). DE DE VI. TYPICAL VALUES FOR THE CONSTANTS To specify C , C and µ we appeal to the observational value of some cosmological param- 2 3 eters. From recent observations, we know that the Hubble constant at present time is about 73 8kms−1Mpc−1 or in SI : ± a˙(t) H(t ) = ⋍ 23.6 3 10−19s−1 (41) 0 a(t)|t0 ± × We also know that: a˙(t) H(t) = (42) a(t) Theothercosmological parameteristhedensityparameterΩ = 8πGρ. Recentobservations suggest 3H02 thattheΩ = Ω +Ω 1andtheΩ 0.7. Anothercosmological parameteristheequation tot DE m DE ≃ ≃ p(t) of state parameter (EoS) w(t) = . Recent observations limit the value of ω between 0.4 ρ(t) DE − and 1.02. Another important cosmological parameter is the transition redshift z , in which the T − cosmic evaluation transfersfromthedecelerated eratotheaccelerated eraorequivalently thevalue of w is getting less than 0.3. Observations suggest that the value of z is between 0.15 to DE T − 0.5 [23]. We obtain these cosmological parameters for our model and by this means we can get some idea about the free parameters of the model. Here, we make five choices and discuss them separately, in order to get an idea about the behavior of the solutions. The constant C can be 1 determined by observations on variation of some parameters during the history of the Universe such as the fine structure constant α [24]. 10 A. Choice 1: C2 = 1,C3 =0, µ=8.5 10−18s−1 − × In this case w (t) and Ω (t) both are constant : DE DE w (t) = 1 Ω (t) = 0.47 (43) ED DE − Thiscasecorrespondstoacosmologicalconstant,butΩ (t)doesnotmatchwiththeobservations. DE In this case H = 0.23 10−17s−1 in SI, which is in line with observations. 0 × B. Choice 2: C2 = 1, C3 = 1, µ=7 10−18s−1 − − × We plot w (t) and Ω (t) for this choice in Fig.1. In this case at z = 0 we have : DE DE ω (0) = 0.55 Ω (0) =0.61 (44) ED DE − Observations suggest that at the present time (z = 0); Ω = Ω + Ω 1 so we obtain 0 DE0 m0 ≃ Ω = 0.39. In this case H = 0.23 10−17s−1 in SI, which is in line with observations. The value m0 0 × −18 −1 FIG. 1: wDE(t) (left) and ΩDE(t) (right) for the case C2 = 1 and C3 = 1,µ=7 10 s − − × of z in this case is 0.2. T C. Choice 3: C2 = 9, C3 = 9, µ=6 10−18s−1 − − × We plot the w (t) and Ω (t) in Fig.2. In this case at z = 0 we have : DE DE w (0) = 0.39 Ω (0) = 0.68 (45) ED DE − In this case H = 0.19 10−17s−1 in SI, which is in agreement with observations. The value of z 0 T × in this case is 0.1.