Some Cosmological Solutions of a Nonlocal Modified Gravity

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1 Filomat 9:3 15), DOI 1.98/FIL153619D Published by Faculty of Sciences and athematics, University of Niš, Serbia Available at: Some Cosmological Solutions of a Nonlocal odified Gravity Ivan Dimitrijević a, Branko Dragovich b, Jelena Grujić c, Zoran Rakić a a Faculty of athematics, University of Belgrade, Studentski trg 16, Belgrade, Serbia b Institute of Physics, University of Belgrade, Pregrevica 118, 118 Belgrade, Serbia c Teacher Education Faculty, University of Belgrade, Kraljice Natalije 43, Belgrade, Serbia Abstract. We consider nonlocal modification of the Einstein theory of gravity in framework of the pseudo- Riemannian geometry. The nonlocal term has the form HR)F )GR), where H and G are differentiable functions of the scalar curvature R, and F ) = f n n is an analytic function of the d Alambert operator n=. Using calculus of variations of the action functional, we derived the corresponding equations of motion. The variation of action is induced by variation of the gravitational field, which is the metric tensor µν. Cosmological solutions are found for the case when the Ricci scalar R is constant. 1. Introduction Although very successful, Einstein theory of gravity is not a final theory. There are many its modifications, which are motivated by quantum gravity, string theory, astrophysics and cosmology for a review, see [1]). One of very promising directions of research is nonlocal modified gravity and its applications to cosmology as a review, see [, 3] and [4]). To solve cosmological Big Bang singularity, nonlocal gravity with replacement R R + CRF )R in the Einstein-Hilbert action was proposed in [5]. This nonlocal model is further elaborated in the series of papers [6 1]. In [13] we introduced a new approach to nonlocal gravity given by the action S = d 4 x R 16πG + R 1 F )R ), 1) where the d Alembert operator is = 1 µ µν ν, = det µν ). The nonlocal term R 1 F )R = f + R 1 f n n R contains f which can be connected with the cosmological constant as f = Λ 8πG. This term is also invariant under transformation R CR, where C is a constant, i.e. this nonlocality does not depend on magnitude of the scalar curvature R. 1 athematics Subject Classification. 83D5, 53B1, 53B5; Secondary 53C5, 83F5 Keywords. nonlocal gravity, cosmological solutions, equations of motion, calculus of variations, pseudo-riemannian manifold. Received: 11 November 14; Accepted: 1 December 14 Communicated by Ljubica Velimirović and ića Stanković Work on this paper was supported by inistry of Education, Science and Technological Development of the Republic of Serbia, grant No addresses: ivand@matf.bg.ac.rs Ivan Dimitrijević), dragovich@ipb.ac.rs Branko Dragovich), jelenagg@gmail.com Jelena Grujić), zrakic@matf.bg.ac.rs Zoran Rakić)

2 Ivan Dimitrijevic et al. / Filomat 9:3 15), In this paper we consider n-dimensional pseudo-riemannian manifold with metric µν of signature n, n + ). Our nonlocal gravity model here is larger than 1) and given by the action S = R Λ 16πG + HR)F )GR)) d n x, ) which is a functional of metric gravitational field) µν, where H and G are differentiable functions of the scalar curvature R, and Λ is cosmological constant.. Variation of the action functional Let us introduce the following auxiliary functionals S = R Λ) d n x, S 1 = HR)F )GR) d n x. 3) Then the variations of S and S 1 can be considered separately and the variation of ) can be expressed as δs = 1 16πG δs + δs 1. 4) Note that variations of the metric tensor elements and their first derivatives are zero on the boundary of manifold, i.e. δ µν =, δ λ µν =. Lemma.1. Let be a pseudo-riemannian manifold. Then the following basic relations hold: µν x σ = µα Γ ν σα να Γ µ σα, δ = µν δ µν = µν δ µν, Γ µ µν = x ν ln, δ = 1 µν δ µν, = µ µ = 1 µ µν ν ), δr = R µν δ µν + µν δ µν µ ν δ µν. Lemma.. On the manifold holds µν δr µν d n x =. Proof. Let W ν = µα δγ ν µα + µν δγ α µα. Then it follows 1 x ν W ν ) = Wν x ν + 1 Wν x ν. 5) Using Lemma.1 we get 1 x ν W ν ) = x ν µα δγ ν µα) + x ν µν δγ α µα) + µα δγ ν µα + µν δγ α µα)γ β νβ = µα x ν δγν µα µα δ Γν µα x ν + µν x ν δγα µα + µν δ Γα µα x ν + µα δγ ν µα + µν δγ α µα)γ β νβ. 6)

3 oreover, using again Lemma.1 we obtain Ivan Dimitrijevic et al. / Filomat 9:3 15), x ν W ν ) = αβ Γ µ νβ δγν µα + µβ Γ α νβ δγν µα βν Γ µ νβ δγα µα µβ Γ ν νβ δγα µα µα δ Γν µα x ν + µν δ Γα µα x ν Finally, we have µα Γ β νβ δγν µα + µν Γ β νβ δγα µα µν δr µν = 1 x ν W ν ) and = µν δ Γα µν x α + δ Γα µα x ν + Γβ αµδγ α βν + Γα νβ δγβ µα Γ β νµδγ α βα ) Γα βα δγβ µν = µν δ Γα µν x α + Γα µα x ν + Γβ αµγ α βν ) Γα βα Γβ µν = µν δr µν. 7) µν δr µν d n x = x ν W ν )d n x. 8) Using the Gauss-Stokes theorem one obtains x ν W ν )d n x = W ν dσ ν. 9) Since δ µν = and δ λ µν = at the boundary we have W ν =. Then we have Wν dσ ν =, that completes the proof. Lemma.3. The variation of S is δs = G µν δ µν d n x + Λ µν δ µν d n x, 1) where G µν = R µν 1 R µν is the Einstein tensor. Proof. The variation of S can be found as follows δs = δr Λ) ) d n x = δr ) d n x Λ δ d n x = ) δr + Rδ + Λ µν δ µν d n x δ = µν R µν ) 1 R µν δ µν + Λ µν δ µν) d n x = G µν δ µν d n x + Λ µν δ µν d n x + µν δr µν d n x. 11) Using Lemma., from the last equation we obtain the variation of S. Lemma.4. For any scalar function h we have hδr d n x = hrµν + µν h µ ν h ) δ µν d n x. 1) Proof. Using Lemma.1, for any scalar function h we have hδr d n x = hrµν δ µν + h µν δ µν h µ ν δ µν) d n x. 13)

4 Ivan Dimitrijevic et al. / Filomat 9:3 15), The second and third term in this formula can be transformed in the following way: h µν δ µν ) d n x = h µ ν δ µν d n x = µν h)δ µν d n x, 14) µ ν h δ µν d n x. 15) To prove the first of these equations we use the Stokes theorem and obtain h µν δ µν d n x = = h µν α α δ µν d n x = α h µν ) α δ µν d n x µν α α h δ µν d n x = Here we have used γ µν = and α α = α α = to obtain the last integral. To obtain the second equation we first introduce vector µν h δ µν d n x. 16) N µ = h ν δ µν ν hδ µν. 17) From the above expression follows µ N µ = µ h ν δ µν ν hδ µν ) = µ h ν δ µν + h µ ν δ µν µ ν h δ µν ν h µ δ µν = h µ ν δ µν µ ν h δ µν. 18) Integrating µ N µ yields µn µ d n x = Nµ n µ d, where n µ is the unit normal vector. Since N µ = we have that the last integral is zero, which completes the proof. Lemma.5. Let θ and ψ be scalar functions such that δψ =. Then one has αβ α θ β ψ µν δ µν d n x θδ ψ d n x = 1 + θ δψ d n x + 1 µ θ ν ψδ µν d n x µν θ ψδ µν d n x. 19) Proof. Since θ and ψ are scalar functions such that δψ = we have θδ ψ d n x = θ α δ αβ β ψ) d n x + θδ = α θδ αβ β ψ)) d n x α θ δ αβ β ψ) d n x α αβ β ψ) d n x θ µν ψδ µν d n x. )

5 Ivan Dimitrijevic et al. / Filomat 9:3 15), It is easy to see that αθδ αβ β ψ)) d n x =. From this result it follows θδ ψ d n x = = = 1 + At the end we have that αβ α θ β ψδ ) d n x α θ β ψδ αβ d n x θ µν ψδ µν d n x αβ α θ β ψ µν δ µν d n x µ θ ν ψδ µν d n x β αβ α θ δψ) d n x + β αβ α θ) δψ d n x αβ α θ β δψ d n x + 1 µν θ ψδ µν d n x αβ α θ β ψ µν δ µν d n x θ δψ d n x + 1 θδ ψ d n x = 1 αβ α θ β ψ µν δ µν d n x µ θ ν ψδ µν d n x + θ δψ d n x + 1 µ θ ν ψδ µν d n x µν θ ψδ µν d n x. 1) µν θ ψδ µν d n x. ) Now, after this preliminary work we can get the variation of S 1. Lemma.6. The variation of S 1 is δs 1 = 1 µν HR)F )GR)δ µν d n x + Rµν Φ K µν Φ ) δ µν d n x f n µν α l HR) α n 1 l GR) + l HR) n l GR) ) µ l HR) ν n 1 l GR) ) δ µν d n x, where K µν = µ ν µν, Φ = H R)F )GR) + G R)F )HR) and denotes derivative with respect to R. Proof. The variation of S 1 can be expressed as ) δs 1 = HR)F )GR)δ ) + δhr))f )GR) + HR)δF )GR)) d n x. 3) For the first two integrals in the last equation we have I 1 = HR)F )GR)δ ) d n x = 1 µν HR)F )GR)δ µν d n x, 4) I = δhr))f )GR) d n x = H R)δR F )GR) d n x.

6 Ivan Dimitrijevic et al. / Filomat 9:3 15), Substituting h = H R) F )GR) in equation 1) we obtain I = Rµν H R)F )GR) K µν H R)F )GR) )) δ µν d n x. 5) The third integral can be presented using linear combination of the following integrals J n = HR)δ n GR)) d n x. 6) J is the integral of the same form as I so J = Rµν G R)HR) K µν G R)HR) )) δ µν d n x. 7) For n >, we can find J n using 19). In the first step we take θ = HR) and ψ = n 1 GR) and obtain J n = 1 αβ α HR) β n 1 GR) µν δ µν d n x µ HR) ν n 1 GR)δ µν d n x + HR) δ n 1 GR) d n x + 1 µν HR) n GR)δ µν d n x. 8) In the second step we take θ = HR) and ψ = n GR) and get the third integral in this formula, etc. Using 19) n times one obtains J n = 1 n 1 µν α l HR) α n 1 l GR) + µν l HR) n l GR) µ l HR) ν n 1 l GR) ) δ µν d n x + Rµν G R) n HR) K µν G R) n HR) )) δ µν d n x. 9) Using the equation 1) we obtain the last integral in the above formula. Finally, one can put everything together and obtain δs 1 = I 1 + I + f n J n = 1 µν HR)F )GR)δ µν d n x + n= f n µν α l HR) α n 1 l GR) + l HR) n l GR) ) Rµν Φ K µν Φ ) δ µν d n x µ l HR) ν n 1 l GR) ) δ µν d n x. 3) Theorem.1. The variation of the functional ) is equal to zero iff G µν + Λ µν 1 16πG µνhr)f )GR) + R µν Φ K µν Φ ) f n µν α l HR) α n 1 l GR) µ l HR) ν n 1 l GR) + µν l HR) n l GR) ) =. 31) Proof. Since we have δs = 1 16πG δs + δs 1 the theorem follows from Lemmas.3 and.6.

7 Ivan Dimitrijevic et al. / Filomat 9:3 15), Signature 1, 3) In the physics settings, where functional S represents an action, theorem.1 gives the equations of motion. From this point we assume that manifold is the four-dimensional homogeneous and isotropic one with signature 1, 3). Then the metric has the Friedmann-Lemaître-Robertson-Walker FLRW) form: ) ds = dt + at) dr 1 kr + r dθ + r sin θdϕ. 3) Theorem 3.1. Suppose that manifold has the FLRW metric. Then the expression 31) has two linearly independent equations: 4Λ R HR)F )GR) + RΦ + 3 Φ) 16πG n 1 + f n µ l HR) µ n 1 l GR) + l HR) n l GR) ) =, 33) G + Λ 1 16πG HR)F )GR) + Φ K Φ) f n α l HR) α n 1 l GR) l HR) n 1 l GR) + l HR) n l GR) ) =. 34) Proof. The FLRW metric satisfies R µν = R 4 ä µν and scalar curvature R = 6 a + ) ) ȧ a + k depends only on t, a hence equations 31) for µ ν are trivially satisfied. On the other hand, equations with indices 11, and 33 can be rewritten as µµ R 4 + Λ 8πG HR)F )GR) + R Φ + α l HR) α n 1 l GR) + l HR) n l GR) )) =. f n n 1 Therefore these three equations are linearly dependent and there are only two linearly independent equations. The most convenient choice is the trace and -equation. Corollary 3.1. For HR) = R p and GR) = R q the action ) becomes R Λ S = 16πG + Rp F )R q) d n x, 35) and equations of motion are 1 16πG G µν + Λ µν ) 1 µνr p F )R q + R µν Φ K µν Φ ) f n µν α l R p α n 1 l R q µ l R p ν n 1 l R q + µν l R p n l R q) =, where Φ = pr p 1 F )R q + qr q 1 F )R p. Corollary 3.. For HR) = R p and GR) = R q the equations of motion 36) are equivalent to the following two equations: 1 16πG 4Λ R) Rp F )R q + RΦ + 3 Φ) + f n n 1 36) µ l R p µ n 1 l R q + l R p n l R q) =, 37)

8 Ivan Dimitrijevic et al. / Filomat 9:3 15), πG G + Λ ) 1 R p F )R q + Φ K Φ) f n α l R p α n 1 l R q + l R p n l R q l R p n 1 l R q) =. 38) 4. Constant scalar curvature Theorem 4.1. Let R = = constant. Then, solution of equations of motion 36) has the form R 6k 1. For >, at) = + σe 3 t + τe 3 t, where 9k = R στ, σ, τ R.. For =, at) = kt + σt + τ, where σ + 4kτ =, σ, τ R 3. For <, at) = where k is curvature parameter. Proof. Since R = one has 6k + σ cos 3 t + τ sin 3 t, where 36k = R σ + τ ), σ, τ R, 6 ä a + ȧ) k ) + = R. a a 39) The change of variable bt) = a t) yields second order linear differential equation with constant coefficients 3 b b = 6k. 4) Depending on the sign of we have the following solutions for bt) >, bt) = 6k + σe 3 t + τe 3 t, =, bt) = kt + σt + τ, <, bt) = 6k + σ cos R 3 t + τ sin R 3 t. Putting R = = const into 37) and 38) one obtains the following two equations: ) f R p+q 4Λ p + q = 16πG, Equations 4) will have a solution if and only if f R p+q 1 1 R ) + p + q) 41) = G + Λ 16πG. 4) R p+q ) + Λ )p + q)) =. 43) In the first case we take + 4 = that yields the following conditions on the parameters σ and τ: >, 9k = R στ, =, σ + 4kτ =, <, 36k = R σ + τ ). 44)

9 Ivan Dimitrijevic et al. / Filomat 9:3 15), Solutions given in 41) together with conditions 44) restrict the possibilities for the parameter k. Theorem If > then for k = there is solution with constant Hubble parameter, for k = +1 the ) 1 solution is at) = cosh 1 R ) 3 t + ϕ 1 and for k = 1 it is at) = sinh 1 R 3 t + ϕ, where σ + τ = 6 cosh ϕ and σ τ = 6 sinh ϕ.. If = then for k = the solution is at) = τ = const and for k = 1 the solution is at) = t + σ. ) 1 3. If < then for k = 1 the solution is at) = cos 1 3 t ϕ, where σ = 6 cos ϕ and τ = 6 sin ϕ. Proof. Let >. Set k = then we obtain solution with constant Hubble parameter. Alternatively, if we set k = +1 then there is ϕ such that σ + τ = 6 cosh ϕ and σ τ = 6 sinh ϕ. oreover, we obtain bt) = 1 cosh 1 R 1 3 t + ϕ, at) = cosh 1 At the end, if we set k = 1 one can transform bt) to bt) = 1 sinh 1 R 1 3 t + ϕ, at) = sinh 1 R R 3 t + ϕ. 45) 3 t + ϕ. 46) Let =. If k = then function bt) and consequently at) become constants which leads to a solution at) = τ = const. On the other hand if k then we can write bt) = kt σ k ). 47) If k = +1 then there are no solutions for the scale factor at), because bt). On the other hand, when k = 1 the scale factor becomes at) = t + σ. 48) In the last case, let <. If k = 1 we can find ϕ such that σ = 6 cos ϕ and τ = 6 sin ϕ and rewrite at) and bt) as bt) = 1 cos 1 R 1 3 t ϕ, at) = cos 1 R 3 t ϕ. 49) In the case k = +1 one can transform bt) to bt) = 1 sin 1 3 ), t ϕ which is non positive and hence yields no solutions. Theorem 4.3. If in 43) we take R + p + q)λ ) ) = 5) then: R p+q 1 1. For p + q 1 there is obvious solution =. In particular if p + q = 1 then 5) is satisfied for any if Λ =.. For p + q = there is no solution. Λp + q) 3. For p + q, 1 there is a unique value = that gives a solution. Since p and q are integers the p + q 1 value of in the last equation is always positive, and for k = the solution bt) is a linear combination of exponential functions. Proof of this theorem is evident.

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