Einstein–Yang–Mills Ads Black Brane Solution in Massive Gravity and Viscosity Bound

Eur. Phys. J. C (2018) 78:875 https://doi.org/10.1140/epjc/s10052-018-6360-5

Regular Article - Theoretical Physics

Einstein–Yang–Mills AdS black brane solution in massive gravity and viscosity bound

Mehdi Sadeghia Department of Sciences, University of Ayatollah Ozma Borujerdi, Borujerd, Lorestan, Iran

Received: 16 January 2018 / Accepted: 22 October 2018 / Published online: 29 October 2018 © The Author(s) 2018

Abstract Weintroduce the Einstein–Yang–MillsAdS black in d-dimension. Perturbation theory is not applicable to the brane solution in context of massive gravity. The ratio of shear strongly coupled gauge theories but gauge/gravity duality viscosity to entropy density is calculated for this solution. opens a window for solving these theories by introducing a This value violates the KSS bound if we apply the Dirichlet dictionary which can translate the information of strongly boundary and regularity on the horizon conditions. coupled gauge theory into a weakly gravity theory and vice versa. Gauge theory lives on the boundary of AdS5 and gravity on the bulk of AdS5 in this duality, thus this forced us 1 Introduction the background to be Anti-de sitter spacetime. In the long wavelength regime this duality leads to fluid/gravity dual- General theory of relativity introduced by Albert Einstein ity [15–19]. Any fluid is characterized by some transport in 1915 is a theory that graviton is massless within it. This coefficients. One of these transport coefficients is the shear theory predicts the gravitational waves which observed by viscosity. This duality is a powerful method for the the com- advanced LIGO in 2016, but there are some phenomena that putation of transport coefficients in strongly coupled gauge GR cannot explain them including the current acceleration of theories in the hydrodynamic limit. There is three ways to the universe, the cosmological constant problem, dark energy calculate this coefficient: pole method, Green-Kubo formula and dark matter. In recent decades, GR had been generalized and membrane paradigm. In Sect. 3 of this paper, shear vis- to explain these problems such as massive gravity [1], bimet- cosity is calculated via Green–Kubo formula. In the Green– ric gravity [2], scalar-tensor gravity [3] and modified gravity Kubo formula approach, transport coefficients of plasma are [4–7]. On the other hand, the hierarchy problem and brane- related to their thermal correlators and we use gauge/gravity world gravity solutions suggest that graviton is not massless. duality for finding this correlator [20–24]. Massive gravity helps us to study the quantum gravity η = 1 iωt [ x ( ), x ( )] effects and this theory includes some interesting properties: lim dt dxe Ty x Ty 0 ω→0 2ω (i) it could explain the accelerated expansion of the universe =− 1 xx(ω, ). without considering the dark energy, (ii) the graviton behaves lim G yy 0 (1) ω → 0 ω like a lattice excitation and exhibits a Drude peak in this theory, (iii) current experimental data from the observation of The ratio of shear viscosity to entropy density is proportional gravitational waves by advanced LIGO requires the graviton to the inverse square coupling of quantum thermal gauge mass [8]. theory. It means the stronger the coupling, the weaker the Massive gravity introduced by Fierz–Pauli [9]suffers shear viscosity per entropy density. from vDVZ (van Dam–Veltman–Zakharov) discontinuity The study of Quark Gluon Plasma (QGP) arises from the problem. To resolve this problem, it must be considered in a fact that after Big Bang the universe was filled with very hot nonlinear framework according to Vainshtein proposal. This and dense soup, known as QGP,which is strongly coupled. In laboratory, it is created by head-on collision between heavy proposal contains Boulware–Deser ghost. Finally de Rham, η Gabadadze and Tolley (dRGT) solve this problem [1]. ions such as gold or lead nuclei. The lower bound of s is Gauge/gravity duality [10–14] relates two different types related to QGP in all fluids in the nature due to its strongly of theories: gravity in (d + 1)-dimension and gauge theory coupled charactrestic. The KSS bound supports both string theory outcomes [22] and quark-gluon plasma experimental a e-mail: [email protected] data [25]. 123 875 Page 2 of 6 Eur. Phys. J. C (2018) 78 :875 In this paper we consider massive gravity in the presence of 1 (a)λ (b) 1 (a)λσ (b) 2 Tμν = γab Fμ Fνλ − F Fλσ gμν + m χμν, Yang–Mills gauge field and introduce the black-brane solu- 4π 4 tion. This model can be considered as a generalization of the (7)

Einstein–Hilbert model for studying the unknown part of the (a)ν 1 a (b) (c)μν j = C Aμ F , (8) universe, dark matter and dark energy. Finally, we check the e bc Dirichlet boundary and regularity on the horizon conditions c1 η χμν =− (U1gμν − Kμν) for the value of s . 2 c2 2 − (U gμν − 2U Kμν + 2Kμν) 2 2 1 c3 2 Einstein–Yang–Mills AdS black brane solution in − (U3gμν − 3U2Kμν massive gravity 2 2 3 c4 + 6U Kμν − 6Kμν) − (U gμν − 4U Kμν 1 2 4 3 The action of 5-dimensional Einstein-massive gravity with + 12U K2 − 24U K3 + 24K4 ). (9) negative cosmological constant in the presence of Yang– 2 μν 1 μν μν Mills source is as below, χμν is the massive term. √ ( ) 5 (a) (b)μν F ≡ γ (a)λσ b S = d x −g R − 2 − γabFμν F The invariant scalar abF Fλσ for the YM fields is, 4 2 +m ci Ui (g, f ) , (2) 6e2 F = . (10) i=1 YM r 4 − where R is the Ricci scalar, = 6 the cosmological con- l2 We consider the following metric ansatz for a five-dimen- (a) ( , ) stant, l the AdS radius and Fμν the SO 5 1 Yang–Mills sional planar AdS black brane, gauge field tensor [26], r 2 N(r)2 l2dr2 (a) (a) (a) 2 2 2 i j F = ∂μ A − ∂ν A ds =− f (r)dt + + r hijdx dx , (11) μν ν μ l2 r 2 f (r) 1 a (b) (c) + C Aμ Aν a = 1, 2,...,N, (3) 2e bc A generalized version of fμν was proposed in [27] with the 1 fμν = diag( , , c2h ) h = δ form 0 0 0 ij , where ij l2 ij. where e is the gauge coupling constant, C’s are the gauge U (a) The values of i are calculated as below, group structure constants, Aν ’s the gauge potentials and γ =− ab ab 1 the metric tensor of the gauge group in 3c 6c2 6c3 | | N 0 0 0 det ab U1 = , U2 = , U3 = , U4 = 0 = c d | | > r r 2 r 3 which ab CadCbc and det ab 0. U In (2), ci ’s are constants and i are symmetric√ polynomials μ μα Inserting this ansatz into the action Eq.(2) yields, of the eigenvalues of the 5 × 5matrixKν = g fαν 3N(r) d 2e2l2 I = d5x r 4 1 − f (r) − ln r U1 =[K] l5 dr r 4 U =[K]2 −[K2] 3 2 2 2 c1r 2 2 +m l c0 + c0c2r + 2c c3r (12) 3 2 3 0 U3 =[K] − 3[K][K ]+2[K ] 3 4 2 2 3 2 2 4 ( ) U4 =[K] − 6[K ][K] + 8[K ][K]+3[K ] − 6[K ] We can find the equation of motion by variation of N r [28], (4) d 2e2l2 r 4 1 − f (r) − ln r dr r 4 Variation of the action (2) with respect to the metric tensor (a) gμν and the Yang-Mills field tensor Fμν leads to, c c r 3 +m2l2 0 1 + c2c r 2 + 2c3c r = 0 3 0 2 0 3 Gμν + gμν = 8πTμν, (5) f (r) is found by solving the following equation, (a)μν (a)μ F;ν = j , (6) 2e2l2 r 4 1 − f (r) − ln r r 4 where Gμν is the Einstein tensor and the gauge current and the stress-energy tensor carried by the gauge fields are as c c r 3 +m2l2 0 1 + c2c r 2 + 2c3c r = b4 (13) follows, 3 0 2 0 3 123 Eur. Phys. J. C (2018) 78 :875 Page 3 of 6 875 where r0 and b are integration constants. f (r) is found by The entropy can be found by using Hawking–Bekenstein for- solving Eq. (13) which yields, mula, √ 3 3 r0 V3 4 2 2 2 3 = − | = , = = b 2e l c c c c2 2c c3 A d x g r r0 t cte f (r) = − − r +m2l2 0 1 + 0 + 0 . l3 1 4 4 ln 2 3 r r 3r r r 3 A r V3 (14) S = = 0 4G 4l3G π 3 4 S 4 r0 Event horizon is where f (r0) = 0 and we can find b by s = = 3 (21) applying this condition, V3 l 2 2 2 3 where V3 is the volume of the constant t and r hyper-surface 4 4 2e l 2 2 c0c1 c0c2 2c0c3 1 = − + + + with radius r0 and in the last line we used π = 1so b r0 1 4 ln r0 m l 2 3 16 G r0 3r0 r0 r0 1 = π 4G 4 . 2 2 ≡ 4 − 2e l + r0 1 4 ln r0 (15) r0 3 Shear viscosity to entropy density where is, 2 3 The black brane solution is, c c c c2 2c c3 ≡ m2l2 0 1 + 0 + 0 . (16) 3r 2 3 0 r0 r0 f (r) l2 r 2 ds2 =− 1 dt2 + dr2 + (dx2 + dy2 + dz2) l2 f (r) l2 4 ( ) 1 By substituting b in f r we will have, 2 f1(r) = r f (r) (22) l2 f (r) = 6c3c m2(r − r ) + 3c2c m2(r 2 − r 2) 3r 4 0 3 0 0 2 0 r is the radial coordinate that put us from bulk to the boundary. From linear response theory we know for calculating the 2 3 3 3 4 4 +c0c1m (r − r ) + (r − r ) → + 0 l2 0 shear viscosity we must perturb the metric as gμν gμν δ r gxy. As we are looking for zero frequency solution, we take − 2 . 2 6e ln (17) δg = r φ(r)eiωt r0 the metric perturbation to be xy l2 and set ω = 0. It is easy to show that N(r) is constant by variation of f (r). 2 = f1(r) l By applying the speed of light c 1 in the boundary, N is ds2 =− dt2 + dr2 2 ( ) found as the following, l f1 r r 2 + (dx2 + dy2 + dz2 + 2φ(r)dxdy), (23) 2 1 l lim N 2 f (r) = 1 → N 2 = = 1 r→r 2e2l2 rb b 1 − ln( ) r =∞ Then by plugging the perturbed metric in the action and keep- r4 r0 b b ing terms up to φ2 we will have, (18) − 1 5 2 2 S2 = d x(K1φ − K2φ ). (24) where rb is boundary of AdS. 2 Einstein–Yang–Mills solutions are different from Einstein– where Maxwell equation in d > 4. Hawking temperature is defined r 5 f (r) √ r by, K = 1 = −ggrr = 6c3c m2(r − r ) 1 l3 3l3 0 3 0 κ(r0) 1 d √ T = = √ g | = π π tt r r0 + 3c2c m2(r 2 − r 2) 2 2 grr dr 0 2 0 1 2 3 3 3 4 4 2 r = √ ∂ | = + c c m (r − r ) + (r − r ) − e , r gtt r r0 (19) 0 1 0 2 0 6 ln 4π grrgtt l r0 (25) where, κ(r0) is the surface gravity on the event horizon. In 1 our case temperature is, K = (2c2c m2r + c c m2r 2). (26) 2 2l3 0 2 0 1 2 2 2 3 r 2e l c c 2c c2 6c c3 T = 0 4 − − m2l2 0 1 + 0 + 0 . Then the equation of motion will be as follows, π 2 4 2 3 4 l r0 3r0 r0 r0

(20) (K1φ ) + K2φ = 0. (27) 123 875 Page 4 of 6 Eur. Phys. J. C (2018) 78 :875

− ( 4 − 4)) We are going to solve the equation of motion perturbatively 6C1 log r r0 2 2 2 2 in terms of m and e . At ﬁrst we consider m = e = 0. 2 + e C4 ( ( 4 − 4) − ) Then the EoM is as follows, 8 log r r0 4logr (36) 4r0 ( φ ) = . K3 0 0 (28) 2 2 where 0(r) = φ0 + e C3 + m C2. To be regular at the horizon we should get rid of log(r − r ) = ( 2 = ) 0 Where K3 K1 m 0 . The solution is as follows, = 1 ( 3 2 + 2 2 2) − e2 by choosing C1 6 r0 l c1c0 3c0c2l r0 2 4 C4. m r0 dr By substituting C4 in the Eq. (36)wehave, φ0 = C1 + C2. (29) K3 m2 r φ(r) = − 2c c l2r 3 ArcT an By using near horizon approximation we will have, 0 4 0 1 0 24r0 r0 2 2 2 4 4 + 3c c2l r log(r − r ) φ (r) = C + C (M log(r − r ) +···). (30) 0 0 0 0 2 1 0 + 2 3 ( − ) − 2 3 ( + ) c0c1l r0 log r0 r c0c1l r0 log r0 r − 2 3 ( 4 − 4) Where M is a constant (depending on e , l and r0) and ellipses c0c1l r0 log r r0 are regular terms. Then applying the boundary conditions − 2 2 2 ( 4 − 4) + 2 3 3c0c2l r0 log r r0 4c0c1l r0 log r (regularity at horizon and φ = 1 at the boundary) gives C1 = 0 and C = 1 which means that φ (r) = 1 is a constant + 2 2 2 2 0 12c0c2l r0 log r (37) solution. In this case, according to Eq. (16) in [29],

η The second boundary condition is at the boundary, φ(r = 1 2 1 = φ(r0) = (31) ∞) = s 4π 4π 1, which gives,

2 2 Now consider m2 and e2 to be small parameters and try to m l πc0 = 1 + (c r + i(3c c + c r )) (38) φ = φ + 2φ ( ) + 2φ ( ) 0 2 1 0 0 2 1 0 solve Eq. (27). By Putting 0 m 1 r e 2 r 24r0 2 where φ0 = 1 and expanding EoM in terms of powers of m and e2, we will ﬁnd, Thus we have found φ(r) everywhere, from horizon to boundary. The solution is as follows, 2( 2 2 + 2 2 m c0c1l r 2c0c2l r m2c l2 + 2r(r 4 − r 4)φ (r) + 2(5r 4 − r 4)φ (r)) = 0(32)φ( ) = + 0 ( + ) ( 2 + 2) 0 1 0 1 r 1 2 6c0c2 c1r0 log r r0 24r0 2 d ( ( 4 − 4)φ ( )) = e r r r0 2 r 0 (33) r dr − 2c1r0 ArcT an r0 Then the solutions are as follows, + 2c1r0 log(r +r0)−4(3c0c2 +c1r0) log r +πc1r0

φ ( ) = − 1 ( 2 3 r + (39) 1 r C2 4 2c0c1l r0 ArcT an 24C1 log r 24r0 r0 + 2 2 2 ( 4 − 4) φ( ) = + 2φ ( )+ 2φ ( ) 3c0c2l r0 log r0 r Now ﬁnd it at the horizon, r0 1 m 1 r0 e 2 r0 , 2 3 2 3 by taking the limit r → r0 and putting the result in viscosity + c0c1l r log(r0 − r) − c0c1l r log(r0 + r) 0 0 formula we will have, − ( 4 − 4)). 6C1 log r r0 (34) η 1 1 C4 4 4 2 2 2 2 φ (r) = C + (log(r − r ) − 4logr) (35) = φ(r0) = 0 + m φ1(r0) + e φ2(r0) 2 3 8 0 s 4π 4π 4r0 1 m2c l2 = 1 + 0 (12c c log 2 + c r (π + 6log2) . φ( ) π 2 0 2 1 0 r is as follows, 4 24r0 (40) m2 r φ(r) = − (2c c l2r 3 ArcT an + 24C log r 0 4 0 1 0 1 As mentioned before, we applied these 2 conditions: 24r0 r0 2 2 2 4 4 + 3c c2l r log(r − r ) 0 0 0 (i) φ is regular at horizon r = r , + 2 3 ( − ) − 2 3 ( + ) 0 c0c1l r0 log r0 r c0c1l r0 log r0 r (ii) goes like to φ = 1 near the boundary as r →∞. 123 Eur. Phys. J. C (2018) 78 :875 Page 5 of 6 875

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