PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 94 (108) (2013), 187–196 DOI: 10.2298/PIM1308187D

A NEW MODEL OF NONLOCAL MODIFIED GRAVITY

Ivan Dimitrijević, Branko Dragovich, Jelena Grujić, and Zoran Rakić

Abstract. We consider a new modified gravity model with nonlocal term of the form R^{−1} F(□)R. This kind of nonlocality is motivated by investigation of applicability of a few unusual ansätze to obtain some exact cosmological solutions. In particular, we find attractive and useful quadratic ansatz □R = qR^{2} .

1. Introduction

In spite of the great successes of General Relativity (GR) it has not got the status of a complete theory of gravity. To modify GR there are motivations coming from its quantum aspects, string theory, astrophysics and cosmology. For example, cosmological solutions of GR contain Big Bang singularity, and Dark Energy as a cause for accelerated expansion of the Universe. This initial cosmological singularity is an evident signature that GR is not appropriate theory of the Universe at cosmic time t = 0. Also, GR has not been verified at the very large cosmic scale and dark energy has not been discovered in the laboratory experiments. This situation gives rise to research for an adequate modification of GR among numerous possibilities (for a recent review, see [1]).

Recently it has been shown that nonlocal modified gravity with action

(1.1)  S = ∫ d^{4}x\sqrt{−g} ( \frac{R − 2Λ}{16πG} + CRF(□)R) ,

where R is scalar curvature, Λ–cosmological constant, F(□) = ∑ _{n=0}^{∞} f_{n}□^{n} is an analytic function of the d’Alembert–Beltrami operator

□ = \frac{1}{\sqrt{−g}} ∂_{μ} \sqrt{−g} g^{μν} ∂_{ν}, g = det(g_{μν} )

and C is a constant, has nonsingular bounce cosmological solutions, see [2, 3, 4, 5]. To solve equations of motion the ansatz □R = rR+s was used. In [6] we introduced some new ansätze, which gave trivial solutions for the above nonlocal model (1.1). Here we consider some modification of the above action in the nonlocal sector, i.e.,

2010 Mathematics Subject Classification: Primary 83D05, 83F05, 53C21; Secondary 83C10, 83C15.

Partially supported by the Serbian Ministry of Education, Science and Technological Development, Project 174012.

(1.2)  S = ∫ d^{4}x\sqrt{−g} ( \frac{R}{16πG} + R^{−1} F(□)R)

and look for nontrivial cosmological solutions for the new ansätze (see [6]). Note that the cosmological constant Λ in (1.2) is hidden in the term f_{0}, i.e., Λ = −8πGf_{0}. To the best of our knowledge action (1.2) has not been considered so far. However, there are investigations of gravity modified by 1/R term (see, e.g. [7] and references therein), but it is without nonlocality.

2. Equations of motion

By variation of action (1.2) with respect to metric g^{μν} one obtains the equations of motion for g_{μν}

(2.1) R_{μν} V − (∇_{μ}∇_{ν} − g_{μν} □)V − \frac{1}{2}g_{μν} R^{−1} F(□)R

+ ∑_{n=1}^{∞} \frac{f_{n}}{2} ∑_{l=0}^{n−1} (g_{μν} (∂_{α} □^{l}(R^{−1} )∂^{α} □^{n−1−l} R + □^{l}(R^{−1} )□^{n−l} R)

− 2∂_{μ}□^{l}(R^{−1} )∂_{ν}□^{n−1−l} R) = −\frac{G_{μν}}{16πG},

V = F(□)R^{−1} − R^{−2} F(□)R.

The trace of the equation (2.1) is

(2.2) RV + 3□V + ∑_{n=1}^{∞} f_{n} ∑_{l=0}^{n−1} (∂_{α} □^{l}(R^{−1} )∂^{α} □^{n−1−l} R + 2□^{l}(R^{−1} )□^{n−l} R)

− 2R^{−1} F(□)R = \frac{R}{16πG}.

The 00 component of (2.1) is

(2.3) R_{00}V − (∇_{0}∇_{0} − g_{00}□)V − \frac{1}{2}g_{00}R^{−1} F(□)R

+ ∑_{n=1}^{∞} \frac{f_{n}}{2} ∑_{l=0}^{n−1} (g_{00} (∂_{α} □^{l}(R^{−1} )∂^{α} □^{n−1−l} R + □^{l}(R^{−1} )□^{n−l} R)

− 2∂_{0}□^{l}(R^{−1} )∂_{0}□^{n−1−l} R) = −\frac{G_{00}}{16πG}.

We use Friedmann–Lemaître–Robertson–Walker (FLRW) metric ds^{2} = −dt^{2} + a^{2}(t)( \frac{dr^{2}}{1−kr^{2}} + r^{2}dθ^{2} + r^{2} sin^{2} θdϕ^{2}) and investigate all three possibilities for curva-

ture parameter k (0, ±1). In the FLRW metric scalar curvature is R = 6 ( \frac{ä}{a} + \frac{ȧ^{2}}{a^{2}} + \frac{k}{a^{2}} )

and □h(t) = −∂_{t}^{2}h(t) − 3H∂_{t}h(t), where H = \frac{ȧ}{a} is the Hubble parameter. In the sequel we shall use three kinds of ansätze (two of them introduced in [6]) and solve equations of motions (2.2) and (2.3) for cosmological scale factor in the form a(t) = a_{0}|t − t_{0}|^{α}.

3. Quadratic ansatz: □R = qR^{2}

Looking for solutions in the form a(t) = a_{0}|t − t_{0}|^{α} this ansatz becomes

α(2α − 1)(qα(2α − 1) − (α − 1))(t − t_{0})^{−4} (3.1)

+ \frac{αk}{3a_{0}^{2}} (1 − α + 6q(2α − 1))(t − t_{0})^{−2α−2} + \frac{qk^{2}}{a_{0}^{4}}(t − t_{0})^{−4α} = 0.

Equation (3.1) is satisfied for all values of time t in six cases:

(1) k = 0, α = 0, q ∈ ℝ,  (4) k = −1, α = 1, q ≠ 0, a_{0} = 1,
(2) k = 0, α = \frac{1}{2}, q ∈ ℝ,  (5) k ≠ 0, α = 0, q = 0,
(3) k = 0, α ≠ 0, \frac{1}{2}, q = \frac{α−1}{α(2α−1)} , (6) k ≠ 0, α = 1, q = 0.

In cases (1), (2) and (4) we have R = 0 and therefore R^{−1} is not defined. Case (5) yields a solution which does not satisfy the equations of motion. Hence there remain two cases for further consideration.

3.1. Case k = 0, q = \frac{α−1}{α(2α−1)} . For this case, we have the following expressions depending on the parameter α:

q = \frac{α − 1}{α(2α − 1)} ,  R = 6α(2α − 1)(t − t_{0})^{−2} ,
a = a_{0}|t − t_{0}|^{α} ,  H = α(t − t_{0})^{−1} ,
R_{00} = 3α(1 − α)(t − t_{0})^{−2} ,  G_{00} = 3α^{2}(t − t_{0})^{−2} .

We now express □^{n}R and □^{n}R^{−1} in the following way:

□^{n}R = B(n, 1)(t − t_{0})^{−2n−2} , □^{n}R^{−1} = B(n, −1)(t − t_{0})^{2−2n} ,
B(n, 1) = 6α(2α − 1)(−2)^{n}n! ∏_{l=1}^{n} (1 − 3α + 2l), n ⩾ 1,

B(n, −1) = (6α(2α − 1))^{−1} 2^{n} ∏_{l=1}^{n} (2 − l)(−3 − 3α + 2l), n ⩾ 1,
B(0, 1) = 6α(2α − 1), B(0, −1) = B(0, 1)^{−1} .

Note that B(1, −1) = −\frac{3α+1}{3α(2α−1)} = −2(3α + 1)B(0, 1)^{−1} and B(n, −1) = 0 if n ⩾ 2. Also, we obtain

F(□)R = ∑_{n=0}^{∞} f_{n}B(n, 1)(t − t_{0})^{−2n−2} ,
F(□)R^{−1} = f_{0}B(0, −1)(t − t_{0})^{2} + f_{1}B(1, −1).

Substituting these equations into trace and 00 component of the EOM one has

(3.2)  \begin{matrix}r^{−1} ∑_{n=0}^{∞} f_{n}B(n, 1) (−3r + 6(1 − n)(1 − 2n + 3α)) (t − t_{0})^{−2n} \\ + r ∑_{n=0}^{1} f_{n} (rB(n, −1) + 3B(n + 1, −1)) (t − t_{0})^{−2n} \\ + 2r ∑_{n=1}^{∞} f_{n}γ_{n}(t − t_{0})^{−2n} = \frac{r^{2}}{16πG}(t − t_{0})^{−2} , \\ ∑_{n=0}^{∞} f_{n}r^{−1} B(n, 1) ( \frac{r}{2} − A_{n}) (t − t_{0})^{−2n} \\ + ∑_{n=0}^{1} f_{n}rB(n, −1) A_{n} (t − t_{0})^{−2n} + \frac{r}{2} ∑_{n=1}^{∞} f_{n}δ_{n}(t − t_{0})^{−2n} \\ = \frac{−r^{2}}{32πG} \frac{α}{2α − 1} (t − t_{0})^{−2} , \end{matrix}

where r = B(0, 1) and

γ_{n} = ∑_{l=0}^{n−1} B(l, −1)(B(n − l, 1) + 2(1 − l)(n − l)B(n − l − 1, 1)),
δ_{n} = ∑_{l=0}^{n−1} B(l, −1)(−B(n − l, 1) + 4(1 − l)(n − l)B(n − l − 1, 1)),
A_{n} = 6α(1 − n) − r \frac{α − 1}{2(2α − 1)} = \frac{r}{2} \frac{3 − 2n − α}{2α − 1} .

Equations (3.2) can be split into system of pairs of equations with respect to each coefficient f_{n}. In the case n > 1, there are the following pairs:

f_{n}(B(n, 1) (−3r + 6(1 − n)(1 − 2n + 3α)) + 2r^{2}γ_{n}) = 0,

f_{n}( B(n, 1) ( \frac{r}{2} − A_{n}) + \frac{r^{2}}{2} δ_{n}) = 0.

Taking \frac{3α−1}{2} to be a natural number one obtains:

B(n, 1) = 6α(2α − 1)4^{n}n!\frac{(\frac{3}{2}(α − 1))!}{(\frac{3}{2}(α − 1) − n)!} , n < \frac{3α − 1}{2} ,

B(n, 1) = 0, n ⩾ \frac{3α − 1}{2} ,

γ_{n} = 2B(0, −1)B(n − 1, 1)(3nα − 2n^{2} − 3α − 1), n ⩽ \frac{3α − 1}{2} ,

δ_{n} = 2B(0, −1)B(n − 1, 1)(2n^{2} + 3n + 3α − 3αn + 1), n ⩽ \frac{3α − 1}{2} ,

γ_{n} = δ_{n} = 0, n > \frac{3α − 1}{2} .

If n > \frac{3α−1}{2} , then B(n, 1) = γ_{n} = δ_{n} = 0 and hence the system is trivially satisfied for arbitrary values of coefficients f_{n}. On the other hand for 2 ⩽ n ⩽ \frac{3α−1}{2} the system has only the trivial solution f_{n} = 0.

When n = 0, the pair becomes

f_{0}( − 2r + 6(1 + 3α) + 3rB(1, −1)) = 0, f_{0} = 0

and its solution is f_{0} = 0. The remaining case n = 1 reads

f_{1}( − 3r^{−1} B(1, 1) + rB(1, −1) + 2γ_{1}) = \frac{r}{16πG},
f_{1}( A_{1}(rB(1, −1) − r^{−1} B(1, 1)) + \frac{1}{2}(B(1, 1) + rδ_{1}))  = \frac{−r^{2}}{32πG} \frac{α}{2α − 1} ,

and it gives f_{1} = −\frac{3α(2α−1)}{32πG(3α−2)} .

3.2. Case k ≠ 0, α = 1, q = 0. In this case

a = a_{0}|t − t_{0}|, H = (t − t_{0})^{−1} , R = s(t − t_{0})^{−2} ,

s = 6( 1 + \frac{k}{a_{0}^{2}} ) , □R = 0, R_{00} = 0,

□^{n}R^{−1} = D(n, −1)(t − t_{0})^{2−2n} ,

D(0, −1) = s^{−1} , D(1, −1) = −8s^{−1} , D(n, −1) = 0, n ⩾ 2.

Substitution of the above expressions in the trace and 00 component of the EOM yields

3f_{0} + ∑_{n=0}^{1} f_{n}sD(n, −1)(t − t_{0})^{−2n} + 4f_{1}(t − t_{0})^{−2} = \frac{s}{16πG}(t − t_{0})^{−2} ,

− 6f_{0}s^{−1} + \frac{1}{2}f_{0} + 6 ∑_{n=0}^{1} f_{n}D(n, −1)(1 − n)(t − t_{0})^{−2n}

+ 2f_{1}(t − t_{0})^{−2} = −\frac{s}{32πG}(t − t_{0})^{−2} .

This system leads to conditions for f_{0} and f_{1}:

−2f_{0} − 4f_{1}(t − t_{0})^{−2} = \frac{s}{16πG}(t − t_{0})^{−2} ,
\frac{1}{2}f_{0} + 2f_{1}(t − t_{0})^{−2} = −\frac{s}{32πG}(t − t_{0})^{−2} .

The corresponding solution is

f_{0} = 0, f_{1} = \frac{−s}{64πG}, f_{n} ∈ ℝ, n ⩾ 2.

192 DIMITRIJEVIĆ, DRAGOVICH, GRUJIĆ, AND RAKIĆ

4. Ansatz □^{n}R = c_{n}R^{n+1} , n ⩾ 1

Presenting □^{n+1} R in two ways:

□^{n+1} R = □c_{n}R^{n+1} = c_{n}((n + 1)R^{n}□R − n(n + 1)R^{n−1} Ṙ^{2})
= c_{n}(n + 1)(c_{1}R^{n+2} − nR^{n−1} Ṙ^{2}) = c_{n+1} R^{n+2}

it follows Ṙ^{2} = R^{3}, (4.1) c_{n+1} = c_{n}(n + 1)(c_{1} − n), (4.2) where Ṙ^{2} means (Ṙ)^{2}.

The general solution of equation (4.1) is

(4.3)  R = \frac{4}{(t − t_{0})^{2}} , t_{0} ∈ ℝ.

Taking n = 1 in the ansatz yields

(4.4)  □R = c_{1}R^{2}.

The substitution of (4.3) in (4.4) gives H = \frac{2c_{1} +3}{3(t−t_{0} )} . This implies

(4.5)  a(t) = a_{0}|t − t_{0}|^{\frac{2c_{1} +3}{3}} , a_{0} > 0.

Using (4.3) in the equation

R = 6 ( \frac{ä}{a} + \frac{ȧ^{2}}{a^{2}} + \frac{k}{a^{2}} )

gives

(t − t_{0})^{2} ÿ − \frac{4}{3}y = −2k(t − t_{0})^{2}, where y = a^{2}(t).

It can be shown that the general solution of the last equation is

a^{2}(t) = C̃_{1}|t − t_{0}|^{\frac{3+\sqrt{57}}{6}} + C̃_{2}|t − t_{0}|^{\frac{3−\sqrt{57}}{6}} − 3k|t − t_{0}|^{2}, C̃_{1}, C̃_{2} ∈ ℝ. By comparison of the last equation with (4.5) one can conclude:

(1) If c_{1} = 0, then k must be equal to −1. In this case □^{n}R = 0, n ⩾ 1.

(2) If c_{1} ≠ 0, then k must be equal to 0. In this case c_{1} = \frac{1}{8}( − 9 ± \sqrt{57}). 4.1. Case □^{n}R = c_{n}R^{n+1} , c_{1} = 0. From the previous analysis, it follows:

(4.6)  \begin{aligned}k & = −1, & a(t) & = \sqrt{3}|t − t_{0}|, & H(t) & = \frac{1}{t − t_{0}} , \\ R & = \frac{4}{(t − t_{0})^{2}} , & □^{n}R & = 0, n ⩾ 1, & F(□)R & = f_{0}R. \end{aligned}

It can be shown that

(4.7)  □^{n}R^{−1} = (−1)^{n}4^{n−1} ∏_{l=0}^{n−1} (1 − l)(2 − l)(t − t_{0})^{2−2n} .

From (4.7) it follows □^{n}R^{−1} = 0, n > 1. Then
(4.8)  F(□)(R^{−1} ) = f_{0}R^{−1} + f_{1}□R^{−1} .

Substituting (4.6) and (4.8) in the 00 component of the EOM one obtains

\frac{f_{0}}{2} (t − t_{0})^{2} + 2f_{1} + \frac{1}{8πG} = 0

and it follows

(4.9)  f_{0} = 0, f_{1} = \frac{−1}{16πG}, f_{n} ∈ ℝ, n ⩾ 2.

Substituting (4.6) and (4.8) in the trace equation one has

−2f_{0}(t − t_{0})^{2} − 4f_{1} − \frac{1}{4πG} = 0

and it gives the same result (4.9).

4.2. Case □^{n}R = c_{n}R^{n+1} , c_{1} = \frac{1}{8}( − 9 ± \sqrt{57}). In this case:

k = 0, R = \frac{4}{(t − t_{0})^{2}} , H = \frac{2c_{1} + 3}{3(t − t_{0})} , a = a_{0}|t − t_{0}|^{\frac{1}{3} (2c_{1} +3)} , a_{0} > 0,

R_{00} = 3α(1 − α)(t − t_{0})^{−2} , G_{00} = (3α(1 − α) + 2)(t − t_{0})^{−2} , α = \frac{2c_{1} + 3}{3} ,

□^{n}R = 4^{n+1} c_{n}(t − t_{0})^{−2n−2} , c_{0} = 1.

One can show that □^{n}R^{−1} = M(n, −1)(t − t_{0})^{2−2n} , where

M(0, −1) = \frac{1}{4}, M(1, −1) = −(c_{1} + 2), M(n, −1) = 0, n > 1.

Also one obtains

(4.10)  \begin{aligned}F(□)R & = ∑_{n=0}^{∞} 4^{n+1} f_{n}c_{n}(t − t_{0})^{−2n−2} , \\ F(□)R^{−1} & = f_{0}M(0, −1)(t − t_{0})^{2} + f_{1}M(1, −1). \end{aligned}

Substituting (4.10) in the trace equation it becomes

(4.11) − \frac{1}{4πG}(t − t_{0})^{−2} − 2f_{0} − 3 ∑_{n=1}^{∞} 4^{n}f_{n}c_{n}(t − t_{0})^{−2n}

+ ∑_{n=1}^{∞} f_{n}( ∑_{l=0}^{n−1} M(l, −1)4^{n−l+1} ((1 − l)(n − l)c_{n−1−l} + 2c_{n−l} )) (t − t_{0})^{−2n}

+ f_{1}(4M(1, −1) + 3M(2, −1))(t − t_{0})^{−2} = 0.

To satisfy equation (4.11) for all values of time t one obtains:

f_{0} = 0, f_{1}(2c_{1} + 1) = −\frac{1}{16πG},

f_{n}( − 3c_{n} + ∑_{l=0}^{1} M(l, −1)4^{1−l} ((1 − l)(n − l)c_{n−1−l} + 2c_{n−l} )) = 0, n ⩾ 2.

Suppose that f_{n} ≠ 0 for n ⩾ 2, then from the last equation it follows

−3c_{n} + ∑_{l=0}^{1} M(l, −1)4^{1−l} ((1 − l)(n − l)c_{n−1−l} + 2c_{n−l} ) = 0

and it becomes

(4.12)  c_{n−1} (n^{2} − c_{1}n − 2c_{1} − 4) = 0.

Since c_{n−1} ≠ 0, condition (4.12) is satisfied for n = −2 or n = c_{1} + 2. Hence, we conclude that f_{n} = 0 for n ⩾ 2. In such a case the 00 component of the EOM becomes

(4.13) \frac{1}{16πG}(−3α^{2} + 3α + 2)(t − t_{0})^{−2}

+ \frac{1}{2}f_{0}( \frac{3}{2}α^{2} − \frac{9}{2}α + 1) + f_{1}c_{1}(3α^{2} − 3α + 2)(t − t_{0})^{−2}

+ 8f_{1}M(0, −1)(1 − c_{1})(t − t_{0})^{−2} + 3α(3 − α)M(0, −1)f_{0}

− 3α(α − 1)M(1, −1)f_{1}(t − t_{0})^{−2} = 0.

In order to satisfy equation (4.13) for all values of time t it has to be

f_{0} = 0, f_{1}( \frac{4}{3}c_{1}^{3} + \frac{10}{3}c_{1}^{2} + 2c_{1} + 1) = \frac{1}{16πG}( \frac{2}{3}c_{1}^{2} + c_{1} − 1) .

The necessary and sufficient condition for the EOM to have a solution is

c_{1}(8c_{1}^{2} + 18c_{1} + 3) = 0.

Since c_{1} = \frac{1}{8}( − 9 ± \sqrt{57}), the last condition is satisfied.

5. Cubic ansatz: □R = qR^{3}

Recall that we are looking for solutions in the form a(t) = a_{0}|t − t_{0}|^{α}. In the explicit form the ansatz reads

α(α − 1)( 3(2α − 1)(t − t_{0})^{−4} + \frac{k}{a_{0}^{2}} (t − t_{0})^{−2α−2} )

= 18q( α(2α − 1)(t − t_{0})^{−2} + \frac{k}{a_{0}^{2}} (t − t_{0})^{−2α} )^{3} .

It yields the following seven possibilities:

\begin{matrix}(1) k = 0, α = 0, q ∈ ℝ, & (4) k = 0, α = 1, q = 0, \\ (2) k = 0, α = \frac{1}{2}, q ∈ ℝ, & (5) k ≠ 0, α = 0, q = 0, \\ (3) k = −1, α = 1, q ≠ 0, a_{0} = 1, & (6) k ≠ 0, α = 1, q = 0, \end{matrix}

(7) k ≠ 0, α = \frac{1}{2}, q = −a_{0}^{4}/72.

Cases (1), (2) and (3) contain scalar curvature R = 0, and therefore we will not discuss them. Cases (4), (5) and (6) are also obtained from the quadratic ansatz

NONLOCAL MODIFIED GRAVITY 195

and have been discussed earlier. The last case contains:

a(t) = a_{0} \sqrt{|t − t_{0}|} , H(t) = \frac{1}{2(t − t_{0})} ,
R(t) = \frac{6k}{a_{0}^{2}}|t − t_{0}|^{−1} , R_{00} = \frac{3}{4(t − t_{0})^{2}} .

One can derive the following expressions:

(5.1)  \begin{aligned}□^{n}R & = N(n, 1)|t − t_{0}|^{−2n−1} , □^{n}R^{−1} = N(n, −1)|t − t_{0}|^{1−2n} , \\ N(0, 1) & = \frac{6k}{a_{0}^{2}}, N(0, −1) = N(0, 1)^{−1} , \\ N(n, 1) & = N(0, 1)(−1)^{n} ∏_{l=0}^{n−1} (2l + 1)( 2l + \frac{1}{2}) , n ⩾ 1, \\ N(n, −1) & = N(0, 1)^{−1} (−1)^{n} ∏_{l=0}^{n−1} (2l − 1)( 2l − \frac{3}{2}) , n ⩾ 1, \\ F(□)R & = ∑_{n=0}^{∞} f_{n}N(n, 1)|t − t_{0}|^{−2n−1} , \\ F(□)R^{−1} & = ∑_{n=0}^{∞} f_{n}N(n, −1)|t − t_{0}|^{1−2n} . \end{aligned}

Substituting (5.1) in the trace equation we obtain

− 2N(0, 1)^{−1} ∑_{n=0}^{∞} f_{n}N(n, 1)|t − t_{0}|^{−2n}

+ N(0, 1) ∑_{n=0}^{∞} f_{n}(N(n, −1) − N(0, 1)^{−2} N(n, 1))|t − t_{0}|^{−2n}

+ 3 ∑_{n=0}^{∞} f_{n}(N(n, −1) − N(0, 1)^{−2} N(n, 1))|t − t_{0}|^{−1−2n}

+ ∑_{n=1}^{∞} f_{n} ∑_{l=0}^{n−1} N(l, −1)((1 − 2l)(−2n + 2l + 1)N(n − l − 1, 1)

+ 2N(n − l, 1))|t − t_{0}|^{−2n} = \frac{N(0, 1)}{16πG}|t − t_{0}|^{−1} .

This equation implies the following conditions on the coefficient f_{0}:

f_{0} = 0, \frac{N(0, 1)}{16πG} = 0.

Since N(0, 1) ≠ 0, the last equation never holds and therefore there is no solution in this case.

6. Concluding remarks

Using a few new ansätze we have shown that equations of motion for nonlocal gravity model given by action (1.2) yield some bounce cosmological solutions of the form a(t) = a_{0}|t − t_{0}|^{α}. When t → ∞ then R → 0 and these solutions lead to f_{0} = 0 and hence Λ = 0. In particular, the quadratic ansatz □R = qR^{2} is very promising. Note that ansatz □^{n}R = c_{n}R^{n+1} , n ⩾ 1, can be viewed as a special case of ansatz □R = qR^{2}.

It is worth noting that equations of motion (2.2) and (2.3) have the de Sitter solutions a(t) = a_{0} cosh(λt), k = +1 and a(t) = a_{0}e^{λt}, k = 0, when f_{0} = \frac{−3λ^{2}}{8πG} = \frac{−Λ}{8πG} , f_{n} ∈ ℝ, n ⩾ 1.

This investigation can be generalized to some cases with R^{−p} F(□)R^{q} nonlocal term, where p and q are some natural numbers satisfying q − p ⩾ 0. It will be presented elsewhere with discussion of various properties.

References

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5. I. Dimitrijević, B. Dragovich, J. Grujić, Z. Rakić, New cosmological solutions in nonlocal modified gravity, Rom. J. Phys. (2013), to appear, [arXiv:1302.2794 [gr-qc]].

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(I. Dimitrijević, and Z. Rakić) Faculty of Mathematics University of Belgrade Belgrade, Serbia

ivand@matf.bg.ac.rs
zrakic@matf.bg.ac.rs

Institute of Physics University of Belgrade Belgrade, Serbia

dragovich@ipb.ac.rs

Teachers Training Faculty University of Belgrade Belgrade, Serbia

jelenagg@gmail.com