Einstein-Yang-Mills AdS Black Brane Solution in Massive Gravity and Viscosity Bound
aa r X i v : . [ h e p - t h ] O c t Einstein-Yang-Mills AdS Black BraneSolution in Massive Gravity and ViscosityBound
Mehdi Sadeghi ∗ Department of Sciences, University of Ayatollah Ozma Borujerdi, Borujerd, Lorestan, Iran
October 23, 2018
Abstract
We introduce the Einstein-Yang-Mills AdS black brane solution in context of mas-sive gravity. The ratio of shear viscosity to entropy density is calculated for thissolution. This value violates the KSS bound if we apply the Dirichlet boundary andregularity on the horizon conditions.PACS numbers: 11.10.Jj, 11.10.Wx, 11.15.Pg, 11.25.Tq
Keywords:
Black Brane, Shear viscosity, Entropy density, Gauge/Gravity duality,Fluid/Gravity duality, Green-Kubo Formula
General theory of relativity introduced by Albert Einstein in 1915 is a theory thatgraviton is massless within it. This theory predicts the gravitational waves whichobserved by advanced LIGO in 2016, but there are some phenomena that GR can-not explain them including the current acceleration of the universe, the cosmologicalconstant problem, dark energy and dark matter. In recent decades, GR had beengeneralized to explain these problems such as massive gravity[1], bimetric gravity[2],scalar-tensor gravity[3] and modified gravity [4]-[7]. On the other hand, the hierarchyproblem and brane-world gravity solutions suggest that graviton isn’t massless.Massive gravity helps us to study the quantum gravity effects and this theory in-cludes some interesting properties: (i) it could explain the accelerated expansion of theuniverse without considering the dark energy, (ii) the graviton behaves like a lattice ex-citation and exhibits a Drude peak in this theory, (iii) current experimental data from ∗ [email protected] he observation of gravitational waves by advanced LIGO requires the graviton mass[8].Massive gravity introduced by Fierz-Pauli [9] suffers from vDVZ (van Dam-Veltman-Zakharov) discontinuity problem. To resolve this problem, it must be consideredin a nonlinear framework according to Vainshtein proposal. This proposal containsBoulware-Deser ghost. Finally de Rham, Gabadadze and Tolley (dRGT) solve thisproblem[1].Gauge/Gravity duality [10]-[13] relates two different types of theories: gravity in( d + 1)-dimension and gauge theory in d -dimension. Perturbation theory is not appli-cable to the strongly coupled gauge theories but gauge/gravity duality opens a windowfor solving these theories by introducing a dictionary which can translate the infor-mation of strongly coupled gauge theory into a weakly gravity theory and vice versa.Gauge theory lives on the boundary of AdS and gravity on the bulk of AdS in thisduality, thus this forced us the background to be Anti-de sitter spacetime. In thelong wavelength regime this duality leads to fluid/gravity duality [14]-[18]. Any fluidis characterized by some transport coefficients. One of these transport coefficients isthe shear viscosity. This duality is a powerful method for the the computation oftransport coefficients in strongly coupled gauge theories in the hydrodynamic limit.There is three ways to calculate this coefficient: pole method, Green-Kubo formulaand membrane paradigm. In section 3 of this paper, shear viscosity is calculated viaGreen-Kubo formula.In the Green-Kubo formula approach, transport coefficients of plasma are related totheir thermal correlators and we use gauge/gravity duality for finding this correlator[19]-[23]. η = lim ω → ω Z dt d~x e iωt (cid:10) [ T xy ( x ) , T xy (0)] (cid:11) = − lim ω → ω ℑ G x xy y ( ω,~ . (1)The ratio of shear viscosity to entropy density is proportional to the inverse squarecoupling of quantum thermal gauge theory. It means the stronger the coupling, theweaker the shear viscosity per entropy density.The study of Quark Gluon Plasma (QGP) arises from the fact that after Big Bang theuniverse was filled with very hot and dense soup, known as QGP, which is stronglycoupled. In laboratory, it is created by head-on collision between heavy ions such asgold or lead nuclei. The lower bound of ηs is related to QGP in all fluids in the naturedue to its strongly coupled charactrestic. The KSS bound supports both string theoryoutcomes [21] and quark-gluon plasma experimental data [24].In this paper we consider massive gravity in the presence of Yang-Mills gauge fieldand introduce the black-brane solution. This model can be considered as a general-ization of the Einstein-Hilbert model for studying the unknown part of the universe,dark matter and dark energy. Finally, we check the Dirichlet boundary and regularityon the horizon conditions for the value of ηs . Einstein-Yang-Mills AdS Black Brane Solu-tion in Massive Gravity
The action of 5-dimensional Einstein-massive gravity with negative cosmological con-stant in the presence of Yang-Mills source is as below, S = Z d x √− g (cid:16) R − − γ ab F ( a ) µν F ( b ) µν + m X i =1 c i U i ( g, f ) (cid:17) , (2)where R is the Ricci scalar,Λ = − l the cosmological constant, l the AdS radius and F ( a ) µν the SO (5 ,
1) Yang-Mills gauge field tensor [25], F ( a ) µν = ∂ µ A ( a ) ν − ∂ ν A ( a ) µ + 12 e C abc A ( b ) µ A ( c ) ν a = 1 , , ..., N, (3)where e is the gauge coupling constant, C ’s are the gauge group structure constants, A ( a ) ν ’s the gauge potentials and γ ab = − Γ ab | det Γ ab | N the metric tensor of the gauge groupin which Γ ab = C cad C dbc and | det Γ ab | > c i ’s are constants and U i are symmetric polynomials of the eigenvalues ofthe 5 × K µν = √ g µα f αν U = [ K ] U = [ K ] − [ K ] U = [ K ] − K ][ K ] + 2[ K ] U = [ K ] − K ][ K ] + 8[ K ][ K ] + 3[ K ] − K ] (4)Variation of the action (2) with respect to the metric tensor g µν and the Yang-Millsfield tensor F ( a ) µν leads to, G µν + Λ g µν = 8 πT µν , (5) F ( a ) µν ; ν = j ( a ) µ , (6)where G µν is the Einstein tensor and the gauge current and the stress-energy tensorcarried by the gauge fields are as follows, T µν = 14 π γ ab ( F ( a ) λµ F ( b ) νλ − F ( a ) λσ F ( b ) λσ g µν ) + m χ µν , (7) j ( a ) ν = 1 e C abc A ( b ) µ F ( c ) µν , (8) χ µν = − c (cid:18) U g µν − K µν (cid:19) − c (cid:18) U g µν − U K µν + 2 K µν (cid:19) − c (cid:18) U g µν − U K µν + 6 U K µν − K µν (cid:19) − c (cid:18) U g µν − U K µν + 12 U K µν − U K µν + 24 K µν (cid:19) . (9) χ µν is the massive term. he invariant scalar F ≡ γ ab F ( a ) λσ F ( b ) λσ for the YM fields is, F Y M = 6 e r . (10)We consider the following metric ansatz for a five-dimensional planar AdS black brane, ds = − r N ( r ) l f ( r ) dt + l dr r f ( r ) + r h ij dx i dx j , (11)A generalized version of f µν was proposed in [26] with the form f µν = diag (0 , , c h ij ),where h ij = l δ ij .The values of U i are calculated as below, U = 3 c r , U = 6 c r , U = 6 c r , U = 0Inserting this ansatz into the action Eq.(2) yields, I = Z d x N ( r ) l ddr " r (cid:18) − f ( r ) − e l r ln r (cid:19) + m l c (cid:18) c r c c r + 2 c c r (cid:19) (12)We can find the equation of motion by variation of N ( r ) [27], ddr " r (cid:18) − f ( r ) − e l r ln r (cid:19) + m l (cid:18) c c r c c r + 2 c c r (cid:19) = 0 f ( r ) is found by solving the following equation, r (cid:18) − f ( r ) − e l r ln r (cid:19) + m l (cid:18) c c r c c r + 2 c c r (cid:19) = b (13)where r and b are integration constants. f ( r ) is found by solving Eq.(13) which yields, f ( r ) = 1 − b r − e l r ln r + m l (cid:18) c c r + c c r + 2 c c r (cid:19) . (14)Event horizon is where f ( r ) = 0 and we can find b by applying this condition, b = r " − e l r ln r + m l (cid:18) c c r + c c r + 2 c c r (cid:19) ≡ r (cid:18) − e l r ln r + ∆ (cid:19) (15)where ∆ is, ∆ ≡ m l (cid:18) c c r + c c r + 2 c c r (cid:19) . (16)By substituting b in f ( r ) we will have, f ( r ) = l r (cid:16) c c m ( r − r )+3 c c m ( r − r )+ c c m ( r − r )+ 3 l ( r − r ) − e ln( rr ) (cid:17) . (17) t is easy to show that N ( r ) is constant by variation of f ( r ) . By applying the speedof light c = 1 in the boundary, N is found as the following,lim r → r b N f ( r ) = 1 → N = 11 − e l r b ln( r b r ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r b = ∞ = 1 (18)Where r b is boundary of AdS.Einstein-Yang Mills solutions are different from Einstein-Maxwell equation in d > T = κ ( r )2 π = 12 π √ g rr ddr √ g tt | r = r = 14 π √ g rr g tt ∂ r g tt | r = r (19)where, κ ( r ) is the surface gravity on the event horizon. In our case temperature is, T = r πl (cid:20) − e l r − m l (cid:18) c c r + 2 c c r + 6 c c r (cid:19)(cid:21) . (20)The entropy can be found by using Hawking-Bekenstein formula, A = Z d x √− g | r = r ,t = cte = r V l S = A G = r V l Gs = SV = 4 πr l (21)where V is the volume of the constant t and r hyper-surface with radius r and in thelast line we used πG = 1 so G = 4 π . The black brane solution is, ds = − f ( r ) l dt + l f ( r ) dr + r l (cid:16) dx + dy + dz (cid:17) (22) f ( r ) = r f ( r ) r is the radial coordinate that put us from bulk to the boundary.From linear response theory we know for calculating the shear viscosity we must perturbthe metric as g µν → g µν + δg xy . As we are looking for zero frequency solution, we takethe metric perturbation to be δg xy = r l φ ( r ) e iωt and set ω = 0. ds = − f ( r ) l dt + l f ( r ) dr + r l dx + dy + dz + 2 φ ( r ) dxdy ! , (23) hen by plugging the perturbed metric in the action and keeping terms up to φ wewill have, S = − Z d x (cid:16) K φ ′ − K φ (cid:17) . (24)where K = r f ( r ) l = √− gg rr = r l (cid:16) c c m ( r − r ) + 3 c c m ( r − r )+ c c m ( r − r ) + 3 l ( r − r ) − e ln( rr ) (cid:17) , (25) K = 12 l (2 c c m r + c c m r ) . (26)Then the equation of motion will be as follows,( K φ ′ ) ′ + K φ = 0 . (27)We are going to solve the equation of motion perturbatively in terms of m and e . Atfirst we consider m = e = 0. Then the EoM is as follows,( K φ ′ ) ′ = 0 . (28)Where K = K ( m = 0). The solution is as follows, φ = C Z drK + C . (29)By using near horizon approximation we will have, φ ( r ) = C + C ( M log( r − r ) + ... ) . (30)Where M is a constant (depending on e , l and r ) and ellipses are regular terms. Thenapplying the boundary conditions (regularity at horizon and φ = 1 at the boundary)gives C = 0 and C = 1 which means that φ ( r ) = 1 is a constant solution. In thiscase, according to equation (16) in [28], ηs = 14 π φ ( r ) = 14 π (31)Now consider m and e to be a small parameters and try to solve Eq.(27). By Putting φ = φ + m φ ( r ) + e φ ( r ) where φ = 1 and expanding EoM in terms of powers of m and e , we will find, m (cid:16) c c l r + 2 c c l r + 2 r ( r − r ) φ ′′ ( r ) + 2(5 r − r ) φ ′ ( r ) (cid:17) = 0 (32) e ddr (cid:16) r ( r − r ) φ ′ ( r ) (cid:17) = 0 (33)Then the solutions are as follows, φ ( r ) = C − r (cid:16) c c l r ArcT an rr + 24 C log r + 3 c c l r log( r − r )++ c c l r log( r − r ) − c c l r log( r + r ) − C log( r − r ) (cid:17) . (34) ( r ) = C + C r (cid:18) log( r − r ) − r (cid:19) (35) φ ( r ) is as follows, φ ( r ) = Φ − m r (cid:16) c c l r ArcT an rr + 24 C log r + 3 c c l r log( r − r )+ c c l r log( r − r ) − c c l r log( r + r ) − C log( r − r ) (cid:17) + e C r (cid:18) log( r − r ) − r (cid:19) (36)where Φ ( r ) = φ + e C + m C .To be regular at the horizon we should get rid of log( r − r ) by choosing C = ( r l c c + 3 c c l r ) − e m r C .By substituting C in the Eq.(36) we have, φ ( r ) = Φ − m r (cid:16) c c l r ArcT an rr + 3 c c l r log( r − r )+ c c l r log( r − r ) − c c l r log( r + r ) − c c l r log( r − r ) − c c l r log( r − r ) + 4 c c l r log r + 12 c c l r log r (cid:17) (37)The second boundary condition is at the boundary, φ ( r = ∞ ) = 1, which gives,Φ = 1 + m l πc r ( c r + i (3 c c + c r )) (38)Thus we have found φ ( r ) everywhere, from horizon to boundary. The solution is asfollows, φ ( r ) = 1 + m c l r (cid:16) (6 c c + c r ) log( r + r ) − c r ArcT an rr ++ 2 c r log( r + r ) − c c + c r ) log r + πc r (cid:17) (39)Now find it at the horizon, φ ( r ) = 1 + m φ ( r ) + e φ ( r ), by taking the limit r → r and putting the result in viscosity formula we will have, ηs = 14 π φ ( r ) = 14 π (cid:0) Φ + m φ ( r ) + e φ ( r ) (cid:1) = 14 π m c l r (cid:16) c c log 2 + c r ( π + 6 log 2) ! . (40)As mentioned before, we applied these 2 conditions:(i) φ is regular at horizon r = r ,(ii) goes like to φ = 1 near the boundary as r → ∞ .By applying the Petrov-like boundary condition the KSS bound will be preserved.The procedure is the same as the second example of [29]. Conclusion
In this paper, applying different boundary conditions gives us different values of ηs , andsubsequently different couplings or theories on the boundary. The Petrov-like bound-ary [29] condition on the hypersurface preserves KSS bound but the Dirichlet boundaryand regularity on the horizon conditions violate KSS bound [28]. We showed that ourresult (40) is in agreement to the literature and it is valid perturbatively in m and e . When the KSS bound violates it means the model behaves effectively like higherderivative gravity theories [27, 30, 31, 32, 33] and when the KSS bound saturates itmeans the model behaves effectively like Einstein-Hilbert gravity. Acknowledgment
Author would like to thank Shahrokh Parvizi for usefuldiscussions and Ahmad Moradpur for useful comment and the referee of EPJC forvaluable comments which helped to improve the manuscript.
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