Quantum induced w = -1 crossing of the quintessence and phantom models
aa r X i v : . [ g r- q c ] J a n Quantum induced ω = − M. Alimohammadi ∗ and L. Sadeghian † Department of Physics, University of Tehran,North Karegar Ave., Tehran, Iran.
Abstract
Considering the single scalar field models of dark energy, i.e. thequintessence and phantom models, it is shown that the quantum effectscan cause the system crosses the ω = − One of the most important aspects of the present universe is its accelerated ex-pansion. Since two independent observations based on redshift-distance relationof type Ia supernovas in 1998 [1, 2], numerous observations [3] consistently in-dicate that our universe is dominated by a perfect fluid with negative pressure,dubbed dark energy, which constitutes two third of the present universe.The first candidate which has been introduced for dark energy is a cos-mological constant Λ of order (10 − eV) , with equation of state parameter ω = p/ρ = −
1. This model suffers from fine tuning and coincidence prob-lems [4]. As an alternative to cosmological constant, the dynamical modelshave been introduced. In these models, the equation of state parameters ω ( z ) isone of the main parameters which is usually used in studying the time variationof dark energy. In accelerating universe, ω satisfies ω < − / ϕ [5], ω is always ω > −
1. In phantom model, which is a scalar fieldtheory with a field σ with negative kinetic energy, ω always satisfies ω < − ω = −
1, the so-called phantom-divide-line, crossing [7]. Thisphenomenon can not be explained by none of these two models, the quintessenceor phantom models. A possible way to overcome this problem is to consider twoscalar fields in the models known as hybrid models. One of these models is themodel consists of one quintessence and one phantom field, the so-called quintommodel [8]. Recently, it has been shown that the ω > − ω < − ∗ [email protected] † [email protected]
1f one considers one scalar field, but with suitable interaction with backgrounddark matter, again this transition can be occurred [10].In present paper, we study the contribution of quantum effects in ω = − ϕ (cid:3) ϕ [12], exist in theLagrangian. But, as we see, it is not the case at the quantum level. This tran-sition can be induced quantum-mechanically, with no need to higher derivativeterms. The quantum effects are described via the account of conformal anomaly,reminding about anomaly-driven inflation [13]. The contribution of this quan-tum effect in preserving the most of the energy conditions of phantom matterhas been discussed in [14] and its influence in moderating the sudden futuresingularity (Big Rip) of phantom model has been studied in [15].The scheme of the paper is as follows. In section 2 we briefly review thequintessence and phantom models and introduce the perturbative method ofstudying the phantom-divide-line crossing of these models. The energy densityand pressure resulting from the conformal anomaly are also quoted. In section3 we apply our method to quintessence and phantom models and show that thesystem, except for very special initial conditions, has a transition from ω < − ω > −
1, or vise versa, resulting from quantum effects. In special free purephantom model, it is shown that this transition always occurs from ω < − ω > − ~ = c = G = 1 throughout the paper. w = − crossing Consider a spatially flat Friedman-Robertson-Walker space-time in co-movingcoordinates ( t, x, y, z )d s = − d t + a ( t )(d x + d y + d z ) (1)where a ( t ) is the scale factor. It is assumed that the universe is filled with (dark)matter and a single scalar field. The evolution equation of matter density ρ m is˙ ρ m + 3 Hγ m ρ m = 0 , (2)in which γ m = 1 + ω m . ω m , the equation of state parameter of matter field,is defined through ω m = p m /ρ m , in which p m is the pressure of matter field. H ( t ) = ˙ a ( t ) /a ( t ) is the Hubble parameter and ”dot” denotes the time derivative.The dark energy consists of the quintessence field ϕ (or phantom field σ ), whichin the case of homogenous field, its energy density ρ D and pressure p D (” D ”denotes the dark energy) are [5, 6] ρ quintessence = 12 ˙ ϕ + V ( ϕ ) , p quintessence = 12 ˙ ϕ − V ( ϕ ) , (3)or ρ phantom = −
12 ˙ σ + V ( σ ) , p phantom = −
12 ˙ σ − V ( σ ) . (4)2he Friedman equations are H = 8 π ρ tot. , (5)and ˙ H = − π ( ρ tot. + p tot. ) . (6)The equation of state parameter ω = p tot . /ρ tot . is found as ω = − −
23 ˙ HH . (7)For quintessence phase ω > −
1, we have ˙
H < ω < − H obeys ˙ H >
0. If ˙ H ( t ) = 0 and H ( t ) has a relative extremum at t = t , thesystem crosses ω = − t = t If we restrict ourselves to t − t ≪ h − , where h = H ( t ) and h − is oforder of the age of universe, H ( t ) can be expanded as H ( t ) = h + h ( t − t ) α + h ( t − t ) α +1 + O (cid:0) ( t − t ) α +2 (cid:1) . (8) α ≥ H ( t ) at t = t and h = α ! H ( α ) ( t ). H ( n ) ( t ) is the n -th derivative of H ( t ) at t = t . The transitionfrom ω > − ω < − α is even positive integerand h >
0. In reverse case, h must be negative. In the case of quintom model,it has been shown that for slowly-varying potentials V ( ϕ, σ ), α = 2 and h > ω > − ω < − T = b ( F + 23 (cid:3) R ) + b ′ G + b ′′ (cid:3) R, (9)where F is the square of 4d Weyl tensor and G is Gauss-Bonnet invariant, givenby: F = 13 R − R ij R ij + R ijkl R ijkl ,G = R − R ij R ij + R ijkl R ijkl . (10)Generally for N scalars, N / spinors, N vector fields, N (= 0 or 1) gravitonsand N HD higher derivative conformal scalars (including phantom), b , b ′ and b ′′ are given by b = N + 6 N / + 12 N + 611 N − N HD π ) ,b ′ = − N + 11 N / + 62 N + 1411 N − N HD π ) , b ′′ = 0 . (11)3sing eq.(9), one can find the contributions due to conformal anomaly to ρ and p as follows [17] ρ A = − a (cid:8) b ′ (6 a H + 12 a H )+( 23 b + b ′′ ) h a ( − H ¨ H − H ˙ H + 3 ˙ H ) + 6 a H i − b + 6 b ′ − b ′′ } , (12)and p A = b ′ (cid:20) H + 8 H ˙ H + 1 a (4 H + 8 ˙ H ) (cid:21) + (cid:18) b + b ′′ (cid:19) [ − H − H ¨ H − H ˙ H − H + 1 a (2 H + 4 ˙ H )] − − b + 6 b ′ − b ′′ a . (13)Now it looks reasonable to solve the Friedman equations with these quantumcorrections and see if there exists any new result in the phantom-divide-line-crossing issue due to this correction. In this section we consider the expansion (8) for H ( t ) and try to solve theequations (2), (5) and (6) with ρ tot. = ρ m + ρ D + ρ A and p tot. = p m + p D + p A .We want to find any consistent solution of these equations with ω = − In the case of quintessence field, one has N = 1 and N / = N = N = N HD =0. So b = − b ′ = 1120(4 π ) . (14)Eqs. (2) and (8) (with α ≥ t ≡
0) result in ρ m ( t ) = ρ m (0)[1 − h γ m t + 92 γ m h t + ... ] . (15)By expanding both sides of eq.(5) near t = 0, one finds h + 2 h h t α + ... = 8 π ρ tot. (0) + ˙ ρ tot. (0) t + 12 ¨ ρ tot. t + ... ] , (16)in which ρ tot. ( t ) = ρ quintessence + ρ m + ρ A = ρ cl. + ρ A . (17)In above equation, ”cl.” denotes ”classical”. Eq.(16) then results in the followingtwo relations: h = 8 π ρ tot. (0) = 8 π (cid:20) ρ cl. (0) + 2 b (cid:18) h + 4 h h δ α, + 2 a (cid:19)(cid:21) , (18)4 = ˙ ρ tot. (0) = ˙ ρ cl. (0) + 2 b (cid:18) h h δ α, + 12 h h δ α, + 12 h h δ α, − h a (cid:19) , (19)in which a is the scale factor at transition time t = 0. The same expansion forthe second Friedman equation (6) results in the following extra relation:0 = δ (0) = ρ cl. (0) + p cl. (0) − b (cid:18) h h δ α, + 6 h δ α, + 6 h δ α, − a (cid:19) , (20)in which δ ( t ) = ρ tot. + p tot. = ρ cl. + p cl. + ρ A + p A . (21)In eqs.(18)-(20), ρ m ( t ) is given by eq.(15) and ρ m + p m = γ m ρ m . ρ quintessence and p quintessence are those in eq.(3), ρ A and p A are given by eqs.(12) and (13)with H ( t ) = h + h t α + ... , b and b ′ from eq.(14), and b ′′ = 0. Note that eqs.(19)and (20) indicate the energy conservation law ˙ ρ tot. + 3 H ( ρ tot. + p tot. ) = 0 at t = 0. Since δ (0) = ρ tot. (0) + p tot. (0) = 0, ˙ ρ tot. must satisfy ˙ ρ tot. (0) = 0.Let us first consider the equation (20). Since ρ cl. + p cl. = ˙ ϕ + γ m ρ m , eq.(20)for α ≥ ϕ (0) + γ m ρ m (0) + 16 b a = 0 , (for α ≥ . (22)The above equation has no solution except ˙ ϕ (0) = 0, ρ m (0) = 0, and a → ∞ ,which is unphysical. So we have only two choices α = 2 and α = 3.For α = 2, three equations (18)-(20) can be used to obtain the coefficients h , h and h . h is found as h = − ¨ ϕ (0) + (d V / d ϕ ) ϕ (0) , (23)which is nothing but the evolution equation¨ ϕ + 3 H ˙ ϕ + d V ( ϕ )d ϕ = 0 (24)at t = 0. h , in terms of h , is found from eq.(18) as following h = 18 bh (cid:26) π h − (cid:20) ρ cl. (0) + 2 bh + 4 ba (cid:21)(cid:27) , (25)in which ρ cl . (0) = ˙ ϕ (0) + V (0) + ρ m (0). Finally, h can be expressed interms of h by using eqs.(19) or (20). The next-leading relations from Friedmanequations, that is the coefficients of t of eq.(5) and t of eq.(6), are:2 h h = 12 ¨ ρ tot. (0) . (26)and 2 h = − π ˙ δ (0) . (27)Each of the above equations can be used to determine the parameter h ofexpansion (8), which are the same if one uses the field equation (24).5t is interesting to note that the quantum correction terms (the third andfourth terms in the right-hand-side of eq.(25)) are much smaller than the clas-sical terms: h ∼ a ≪ h . (28)The important observation is that if we set b →
0, then eq.(20) results in˙ ϕ (0) = 0 and ρ m (0) = 0, from which eq.(27) reduces to h = 0 (note that δ (0) =˙ ϕ (0) + γ m ρ m (0) and ˙ δ (0) = 2 ˙ ϕ (0) ¨ ϕ (0) − h γ m ρ m (0)). So without quantumeffects, there is no non-trivial solution for eqs.(5) and (6) with H ( t ) given byeq.(8), and therefore there is no phantom-divide-line crossing in quintessencemodel in classical level. But by considering the quantum effects, h has non-trivial solution (25) which can be positive or negative, depending on the values h , a , ˙ ϕ (0), V (0) and ρ m (0). So quantum effects can induce the ω = − α = 3. In this case, one again finds the fieldequation (23), which expresses the Hubble parameter H ( t ) at t = 0 in terms of ϕ (0), ˙ ϕ (0) and ¨ ϕ (0). But now the first equation, eq.(18), also gives h in termsof the field ϕ and its derivatives, the matter density ρ m , and the scale factor a ( t ) at t = 0: h o = πb ± s(cid:18) πb (cid:19) − b (cid:18) ρ cl. (0) + 4 ba (cid:19) / . (29)So α = 3 solution exists only if the right-hand-sides of eqs.(23) and (29) areequal, which is a very special choice of initial values. Under these conditions, ofcourse, there is no ω = − α = 2 and the quantum effects induce the ω = − h is large ( see, for example, eq.(25)). Thisis, of course, consistent with our physical intuition about the role of quantumeffects in gravitational phenomena. To study the phantom model one must consider N = N / = N = N = 0 and N HD = 1 in eq.(11). So b ′ = − b = 790(4 π ) . (30)6he procedure of previous subsection can be followed here. In this way we findequations similar to (18)-(20), but now ρ D and p D are those introduced in eq.(4)and ρ A and p A are given by eqs.(12) and (13) with b ′ = − (7 / b and b ′′ = 0.The result is: h = 8 π (cid:20) ρ cl. (0) − b ′ (cid:18) h + 10 h a + 8 h h δ α, + 9 a (cid:19)(cid:21) , (31)0 = ˙ ρ cl. (0) − b ′ (cid:18) h h δ α, + 24 h h δ α, + 24 h h δ α, − h a − h a (cid:19) , (32)and0 = ρ cl. (0) + p cl. (0) + 8 b ′ (cid:18) h h δ α, + 6 h δ α, + 6 h δ α, − h a − a (cid:19) , (33)in which ρ cl. = −
12 ˙ σ + V ( σ ) + ρ m , ρ cl. + p cl. = − ˙ σ + γ m ρ m . (34)For α ≥
3, eq.(31) does not depend on h , so h is determined from two inde-pendent equations. The first one is the equation of motion of phantom field h = − ¨ σ (0) − (d V / d σ ) σ (0) , (35)and second one is eq.(31). These two expressions are equal only if a very specificinitial conditions has been chosen. So the only typical solution, which alwaysexists, is α = 2.In the case α = 2, the equations (31)-(33) can be used to calculate h , h and h . h is specified by eq.(35), and h and h can be expressed in terms of h . Using (31), h is found as: h = 748 b ′ h (cid:26) ρ cl . (0) − (cid:20) π h + 60 b ′ h a + 6 b ′ h + 54 b ′ a (cid:21)(cid:27) . (36)Like the quintessence case, it can be easily shown that the resulting equationshave no non-trivial solution in b ′ → ω = − b ′ = 0, h becomes non-trivial and isdetermined by eq.(36).In special case ρ m (0) = 0 and V = 0, one has ρ cl. = − (1 /
2) ˙ σ (0) andtherefore eq.(36) clearly results in: h < , for free pure phantom model (37)which proves the transition from ω < − ω > − h is determined by relation h = 110368 πb ′ ˙ σ (0) a (cid:8) π ˙ σ (0) a + 63¨ σ (0) ˙ σ (0) a + 112 πb ′ ¨ σ (0) a +2400 πb ′ ¨ σ (0) ˙ σ (0) a + 27216 πb ′ ˙ σ (0) (cid:9) (38)which is a positive quantity. 7herefore for arbitrary matter density and phantom potential, the sign of h and h can be positive or negative, depending on values h , a , ˙ σ (0), V (0),and ρ m (0). But in free ( V = 0) and pure ( ρ m = 0) phantom model, the sign of h and h are uniquely determined by Friedman equations. h is negative and h is positive. Acknowledgement:
This work was partially supported by the ”center of ex-cellence in structure of matter” of the Department of Physics of University ofTehran, and also a research grant from University of Tehran.
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