From Big to Little Rip in modified F(R,G) gravity

0.57 MB·English

by  A. N. Makarenko

From Big to Little Rip in modified F(R,G) gravity
Makarenko Andrey N.∗, Obukhov Valery V.†, Kirnos Ilya V.‡
Department of Theoretical Physics, Tomsk State Pedagogical University, Tomsk, 634041 Russia
We discuss the cosmological reconstruction in modified Gauss-Bonnet (GB) gravity. It
is demonstrated that the modified GB gravity may describe the most interesting features
2 of late-time cosmology. We derive explicit form of effective phantom cosmological models
1
0 ending by the finite-time future singularity (Big Rip) and without singularities in the future
2
(Little Rip).

n
a
J

4 I. INTRODUCTION
2

]
Recent observational data indicate that our Universe is currently in accelerated phase [1]. Since
c
q
- the discovery of the acceleration, a large number of possible mechanisms have been proposed to
r
g
explain the origin of the dark energy. General Relativity in its standard form can not explain the
[

2 acceleratedexpansionwithoutextratermsorcomponents,whichhavebeengatheredunderthename
v
2 of dark energy. There are various proposals to construct an acceptable dark energy model, such
4
7 as: a scalar field with quintessence-like or phantom-like behavior [2], spinor, (non-)abelian vector
4
. theory, cosmological constant, fluid with a complicated equation of state, and higher dimensions
1
0
[3]. As a result, we have a number of competing scenarios which to describe late-time acceleration.
2
1
Working in framework of one of the above proposals, we are forced to introduce the following extra
:
v
i cosmological components: inflaton, a dark component and dark matter.
X
r We concentrate on the modified gravity [4, 5], which represents a classical generalization of
a
general relativity (modifications of the Hilbert-Einstein action by introducing different functions of
the Ricci scalar [5, 7] or Gauss-Bonnet invariant [8, 9]), should consistently describe the early-time
inflation and late-time acceleration, without the introduction of any other dark component.
In the present work we will focus on the Gauss-Bonnet gravity, where the gravitational action
includes functions of the Gauss-Bonnet invariant
G = R2−4R Rµν +R Rµνλσ . (1)
µν µνλσ
∗ Electronic address: [email protected]
† Electronic address: [email protected]
‡ Electronic address: [email protected]
2
It is interesting, that the theory of this type is closely related to higher order string curvature
corrections [10]. The starting action is
(cid:90) √ (cid:20) R 1 (cid:21)
S = d4x −g − ∂ φ∂µφ−V(φ)−ξ (φ)G . (2)
2κ2 2 µ 1
Here G is the Gauss-Bonnet invariant (1) and the scalar field φ is canonical in (2).
One may redefine the scalar field φ by φ = (cid:15)ϕ. The action takes the following form
(cid:90) √ (cid:20) R (cid:15)2 (cid:21)
S = d4x −g − ∂ φ∂µφ−V˜(ϕ)−ξ˜(ϕ)G . (3)
2κ2 2 µ 1
Here
V˜(ϕ) ≡ V((cid:15)ϕ) , ξ˜(ϕ) ≡ ξ ((cid:15)ϕ) . (4)
1 1
If a proper limit (cid:15) → 0 exists, the action (3) reduces to
(cid:90) √ (cid:20) R (cid:21)
S = d4x −g −V˜(ϕ)−ξ˜(ϕ)G . (5)
2κ2 1
Then ϕ is an auxiliary field. By variation of ϕ, we find
0 = V˜(cid:48)(ϕ)−ξ˜(cid:48)(ϕ)G , (6)
1
which may be solved with respect to ϕ as
ϕ = Φ(G) . (7)
Substituting (8) into the action (5), we get for the F(G)-gravity
(cid:90) √ (cid:20) R (cid:21)
S = d4x −g −F(G) , F(G) ≡ V˜ (Φ(G))−ξ˜ (Φ(G))G . (8)
2κ2 1
On the other hand, the action ?? is associated with non-local models [11] Let’s start with the
next action
(cid:90) √ (cid:18) R κ2 (cid:19)
S = d4x −g − G(cid:3)−1G+L . (9)
2κ2 2α m
Here L is the matter Lagrangian. By introducing the scalar field φ, one may rewrite the action
m
(9) in a local form:
(cid:90) (cid:18) (cid:19)
R α
S = d4x − ∂ φ∂µφ+φG+L . (10)
2κ2 2κ2 µ m
3
In fact, from the φ-equation we find φ = −κ2(cid:3)−1G. By substituting this expression into (10), we
α
reobtain (9).
Thus, by studying F(G) - gravity, we are also exploring several other models. Note that in some
cases, F(G) or even F(R,G) gravity maybe transformed to F(R) gravity [12].
The five-year Wilkinson Microwave Anisotropy Probe (WMAP) data [1] give the bounds to the
value of the equation of state (EoS) parameter w , which is the ratio of the pressure of the dark
DE
energy to the energy density of it, in the range of -1.11 < w < -0.86. This could be consistent
DE
if the dark energy is a cosmological constant with w = -1 and therefore our universe seems to
DE
approach asymptotically de Sitter universe.
Although the accelerating expansions seem to be de Sitter type, the possibility that the current
acceleration could be quintessence type, in which w > -1, or phantom type, in which w < -1,
DE DE
is not completely excluded.
It is often assumed that the early universe started from the singular point often called Big Bang.
However, if the current (or future) universe enters the quintessence/phantom stage, it may evolve
to the finite-time future singularity depending on the specific model under consideration and the
value of the effective EoS parameter (Big Rip) [6]. An elegant solution to this problem has been
recently proposed [14] under the form of the so-called Little Rip cosmology which appears to be a
realistic alternative to the ΛCDM model.
II. [R+f(G)] GRAVITY
Consider the following action (see Ref. [9]):
(cid:90) √ (cid:18) 1 (cid:19)
S = d4x −g R+f(G)+L (11)
2κ2 m
where κ2 = 8πG , G being the Newton constant, and the Gauss-Bonnet invariant is defined
N N
by(1). By varying the action (11) over g , are obtained:
µν
(cid:18) (cid:19)
1 1 1
0 = −Rµν + gµνR +Tµν + gµνf(G)−2f RRµν +4f RµRνα−2f RµαβτRν −
2k2 2 2 G G α G αβτ
− 4f RµαβνR +2(∇µ∇νf )R−2gµν(∇2f )R−4(∇ ∇µf )Rνρ
G αβ G G ρ G
− 4(∇ ∇νf )Rµρ+4(∇2f )Rµν +4gµν(∇ ∇ f )Rρσ −4(∇ ∇ f )Rµανβ,
ρ G G ρ σ G α β G
where f = f(cid:48)(G) and f = f(cid:48)(cid:48)(G) ((cid:48) = ∂/∂G).
G GG
4
We consider a spatially-flat universe
3
(cid:88)
ds2 = −dt2+a(t)2 (dxi)2,. (12)
i=1
Here a(t) is the scale factor at cosmological time t. It is easy to obtain the FRW equations
3
0 = − H2+Gf −f(G)−24G˙H3f +ρ ,
κ2 G GG m
1
0 = 8H2f¨ +16H(H˙ +H2)f˙ + (2H˙ +3H2)+f −Gf +p . (13)
G G κ2 G m
Here H is the Hubble rate (H = a˙/a,˙= ∂/∂t), ρ is matter energy density:
m
ρ˙ +3H(1+w)ρ = 0 , (14)
m m
while G, its derivative and R can be defined as functions of the Hubble parameter as
G = 24(H˙H2+H4), R = 6(H˙ +2H2),
G˙ = 24H(H¨H +2H˙2+4H2).
It is easy to see that the second equation in (13) is a differential consequence on the solution.
The first equation (13) can thus be expressed as follows
3
0 = − H2+24(H˙H2+H4)f −f(G)−242H4(H¨H +2H˙2+4H2)f +ρ , (15)
κ2 G GG m
ByselectingcertainoftheHubbleparameterH weobtainadifferentialequationforthefunction
f. This is the method of reconstruction (this method has been implemented in Ref. [13] for f(R)
gravity and Ref. [9] for f(G) gravity).
III. f(R, G) GRAVITY
Let us now consider a more general model for a kind of modified Gauss-Bonnet gravity. This
can be described by the following action
(cid:90) √ (cid:20) 1 (cid:21)
S = d4x −g F(R,G)+L . (16)
2k2 m
Varying over g the gravity field equations are obtained [9],
µν
1
0 = Tµν + gµνF(G)−2F RRµν +4F RµRνρ
2 R,G G ρ
−2F RµρστRν −4F RµρσνR +2(∇µ∇νF )R−2gµν(∇2F )R
G ρστ G ρσ G G
−4(∇ ∇µF )Rνρ−4(∇ ∇νF )Rµρ+4(∇2F )Rµν +4gµν(∇ ∇ F )Rρσ
ρ G ρ G G ρ σ G
−4(∇ ∇ F )Rµρνσ −F Rµν +∇µ∇νF −gµν∇2F . (17)
ρ σ G R R R
5
In the case of a flat FRW Universe, described by the metric (??), the first FRW equation yields
1
0 = (GF −F −24H3F )+3(H˙ +H2)F −3HF +k2ρ . (18)
G Gt R Rt m
2
For simplicity, we consider the following subfamily of functions
F(R,G) = f (G)+f (R) . (19)
1 2
Correspondingly, the Friedmann equation (18) can be split into two equations, as
0 = −24H3G˙f +Gf −f ,
1GG 1G 1
1
0 = −3HR˙f +3(H˙ +H2)f − f +κ2ρ (20)
2RR 2R 2 m
2
IV. LITTLE RIP MODEL
In Ref. [15] is proposed Little Rip model on the basis of a viscous fluid. Using the obtained in
this study form the scale factor, we construct a model of the GB gravity.
Consider the scale factor as [15]
(cid:16) (cid:17)
a(t) = expα eβt−1 (21)
In this case the Hubble parameter has the form
H(t) = αβeβt (22)
Then the derivative of the Hubble parameter is proportional to itself, and
(cid:18) (cid:19)
H
G = 24H3 −α . (23)
β
FRW equation for the R+f(G) theory gravity takes the form
3 9β2H +19βH2+4H3 df βH2 d2f
0 = − H2−f + − +ρ . (24)
κ2 (3β +4H)2 dH 3β+4H dH2 m
Choose the energy density in the form ρ = const we obtain the following solution
m
18β3H +3βH3+3H4+2β2(cid:0)6H2−κ2ρ (cid:1) H3(β+H)C
m 1
f(H) = − + + (25)
2β2κ2 1+β
HC2(cid:16)−βeHβ (cid:0)3β2+2βH +H2(cid:1)+H2(β +H)ExpIntegralEi(cid:104)Hβ(cid:105)(cid:17)
+
2β3(1+β)
6
200
4(cid:180)1018
100
3(cid:180)1018
0.2 0.4 0.6 0.8 1.0
2(cid:180)1018
(cid:45)100
1(cid:180)1018
(cid:45)200
2 4 6 8 10
b
a
FIG. 1: Plot of f(t), a) (t,0,1), b) (t,0,10) (Green line - C = 0, C = 10, red line - C = 0, C = 0, blue
2 1 2 1
line - C =10, C =10, black line - C =10, C =0).
2 1 2 1
Here ExpIntegralEi[z] gives the exponential integral function Ei[z], where Ei(z) =
−(cid:82)∞ e−t(cid:14)tdt.
−z
This solution can be illustrated by the graph (we assume that α = 1, β = 1κ = 1, ρ = 0):
m
Figures show that at the initial time all the solutions have the same behavior (dominated by the
first term in (25)), then begins to play the role of the second term (C (cid:54)= 0), and finally turned the
1
third term (C (cid:54)= 0), this is clearly seen in the figure 1.
2
500
2500 400
2000
300
1500
200
1000
500 100
0.5 1.0 1.5 2.0
50 100 150 200
(cid:45)500
FIG. 3: Plot of f(G), (G,0,200) (Red line
FIG. 2: Plot of f(t), (t,0,2.2) (Red line -
- C = −10, C = 100, blue line - f(G) =
C =1, C =1, blue line - f(t)=4eet). 2 1
2 1
8G1/8eG1/4).
We can consider the dependence of the F(G). Obviously not possible to do this, but for large H,
we can assume that G = 24H4. From (25) one can see that there are 3 terms: the first dominates
at small G, then by 30 < G < 100 included the last term (C (cid:54)= 0 and f(G) ∼ G1/8eG1/4, figure 3),
2
then included the second term (C (cid:54)= 0 and f(G) ∼ G, figure 4a) and then G− > ∞ (figure 4b)
1
7
again dominated by the last term.
15000 150000
10000 100000
5000 50000
2000 4000 6000 8000 10000 20000 40000 60000 80000 100000
a b
FIG. 4: Plot of f(G) a) (G,0,10000), b) (G,0,100000) (Red line - C = −10, C = 100, blue line - f(G) =
2 1
1.8G).
This behavior is due to the fact that the last term consists of two parts with different signs. For
small values of G terms rather different, and with increasing G, they become the same order and
the difference becomes smaller than other terms. When G tends to infinity, the difference between
them begins to grow faster than the other terms.
By selecting different values of the constants C and C can be different depending f from G.
1 2
For example, we construct the graph for the following f(G) values of the constants of integration:
C = 1, −20 < C < 20 (figure 5).
2 1
10 20 30 40 50 60 70
(cid:45)10
(cid:45)20
(cid:45)30
(cid:45)40
(cid:45)50
FIG. 5: Plot of f(G), (G,0,70) (C =1, −20<C <20).
2 1
8
Consider now the F(G,R) - gravity. Let us solve the equations (20)
1
f (H) = (26)
2 (4α(2+α(2+(−5+α)α))β4)
(cid:18) (cid:18) (cid:20) (cid:21)(cid:19)(cid:19)
βeHβH(cid:0)β2+4βH −H2(cid:1)C2+H(cid:0)2β3+2β2H −5βH2+H3(cid:1) 4C1+C2ExpIntegralEi H
β
1
f (H) = (27)
1 2α3(1+α)β4
(cid:18) (cid:18) (cid:20) (cid:21)(cid:19)(cid:19)
−α2βeHβH(cid:0)3β2+2βH +H2(cid:1)C[2]+H3(β +H) 2C[1]+α2C[2]ExpIntegralEi H
β
Using the
1 (cid:16) (cid:112) (cid:17)
H = −3β + 9β2+12R . (28)
12
We obtain the dependence of the function f from R
2
(cid:18) √
(cid:16) (cid:112) (cid:17) −3β+ 9β2+12R (cid:16) (cid:16) (cid:112) (cid:17)(cid:17)
f2(R) = −3β+ 9β2+12R −6βe 12β 2R+3β β −3 9β2+12R C2+
(cid:16) (cid:112) (cid:112) (cid:17)
117β3−69βR+57β2 9β2+12R+R 9β2+12R (29)
(cid:32) (cid:34) (cid:112) (cid:35)(cid:33)(cid:33)
−3β+ 9β2+12R
4C +C ExpIntegralEi
1 2
12β
We can consider the dependence of the f (G). Obviously not possible to do this, but for large
1
H, we can assume that G = 24H4. Then
G1/4 (cid:16) √ √ √ (cid:17)
f1(G) = −6al2βe23/431/4β 12 3β2+4 61/4βG1/4+ 2 G G1/4C2+
(cid:32) (cid:34) (cid:35)(cid:33)
(cid:16) √ (cid:17) G1/4
+ 6 2βG3/4+63/4G 2C +al2C ExpIntegralEi (30)
1 2
23/431/4β
V. POWER LAW SOLUTIONS
Consider a power-law solutions. Choose the Hubble parameter as
α
H(t) = . (31)
t −t
s
9
Then the scale factor is given by
C
a(t) = . (32)
(t −t)α
s
This scale factor can meet the various phases of expansion, which are characterized by different
values of the equation of state (EoS) parameter w . To easily find the value w for (32). The
DE DE
equations FRW have the form:
3 1 (cid:16) (cid:17)
ρ = H2, p = − 2H˙ +3H2 . (33)
κ2 κ2
Comparing with the (13), we can write formally the same equations by removing all unnecessary to
the effective pressure and energy density;:
ρ = Gf −f(G)−24G˙H3f +ρ
eff G GG m
p = 8H2f¨ +16H(H˙ +H2)f˙ +f −Gf +p . (34)
eff G G G m
So
p 12H˙ +3H2
eff
w = = − . (35)
DE ρ 3 H2
eff
For (31) have w = −1/3(2/α+3).
DE
Consider the different values of w . If w = -1 this could be consistent if the dark energy is
DE DE
a cosmological constant and therefore our universe seems to approach to asymptotically de Sitter
universe. This value is not possible for the chosen scale factor. This value is the min in the
permissible region, but the possibility that the current acceleration could be quintessence type, in
whichw >-1, orphantomtype, inwhichw <-1, isnotcompletelyexcluded. Forquintessence
DE DE
type, the value of α is negative. For phantom type, the value of α is positive.
It is easy to write of the equation (15) for this cases as an equation for f(t)
0 = −4(1+α)(cid:0)3α2−κ2ρ (t −t)2(cid:1)+
0 s
+ κ2(t −t)2(cid:0)−4(1+α)f(t)−(t −t)(cid:0)−(6+α)f(cid:48)(t)+(t −t)f(cid:48)(cid:48)(t)(cid:1)(cid:1) (36)
s s s
where ρ = const.
0
6α2(1+α) C
f(t) = ρ − +(t −t)−1−αC + 2 (37)
0 (−1+α)κ2(t −t)2 s 1 (t −t)4
s s
10
We now express t through G
23/431/4(cid:0)α3+α4(cid:1)1/4
t −t = . (38)
s
G1/4
And we obtain the following form of the function f(G)
(cid:113)
(cid:112)
3 α3(1+α)G
f(G) = − (2−1+α)ακ2 +ρ0−i−2α2−34(1+α)314(−1−α)G1+2α (cid:0)α3(1+α)G(cid:1)14(−1−α)C1 (39)
It is seen that if α = 3 the equation is considerably simplified
√
3 G
− √ +ρ . (40)
0
2κ2
Now consider the space filled with dust ρ = ρ0.
m a3
(cid:113)
(cid:112)
32 α3(1+α)G 22+94α33α/4(1+α)G−3α/2(cid:0)α3(1+α)G(cid:1)3α/4p0
f(G) = − + −
(−1+α)ακ2 (4+19α+12α2)C3
−2−43(1+α)314(−1−α)e−iαπG1+2α (cid:0)α3(1+α)G(cid:1)41(−1−α)C1 (41)
It is seen that if α = 3 the equation is considerably simplified
√
3 G 644972544 21/4ρ
0
f(G) = −√ + . (42)
2κ2 169C3G9/4
Consider now the F(R,G) - gravity (20). Equations take the form
0 = 4(1+α)f −(t −t)(cid:0)(6+α)f(cid:48) −(t −t)f(cid:48)(cid:48)(cid:1), (43)
1 s 1 s 1
0 = (2+4α)f −(t −t)(cid:0)(4+α)f(cid:48) −(t −t)f(cid:48)(cid:48)(cid:1). (44)
2 s 2 s 2
The solution of these equations have the form
(t−t )3−αC +C 23/431/4(cid:0)α3G+α4G(cid:1)1/4
s 1 2
f = , t −t = − √ (45)
1 (ts−t)4 s G
f = (t−t )12(cid:16)−3−α−√1+(−10+α)α(cid:17)(cid:16)C +(t−t )√1+(−10+α)αC (cid:17), t −t = √6(cid:112)α√(1+2α). (46)
2 s 1 s 2 s
R
Or finally get
f (G) = C G(1+α)/4, (47)
1 1
f2(R) = R41(cid:16)3+α+√1+(−10+α)α(cid:17)(cid:16)C1+R−21√1+(−10+α)αC2(cid:17). (48)