Second order transport coefficients of nonconformal fluids from compactified Dp-branes
aa r X i v : . [ h e p - t h ] D ec Prepared for submission to JHEP
Second order transport coefficients ofnonconformal fluids from compactifiedDp-branes
Chao Wu and Yanqi Wang
School of Physics and Materials Science, Anhui University, Hefei 230601, China
E-mail: [email protected], [email protected]
Abstract:
All the 7 dynamical second order transport coefficients of the noncon-formal fluids that correspond to Dp-branes with one or more world-volume directionscompactified are derived via fluid/gravity correspondence. The conditions that con-sidered in this paper include D4-brane with 1, 2 or 3 compact directions, D3-branewith 1 or 2 compact directions, as well as D2-brane with 1 direction compactified. Thederived second order transport coefficients satisfy the Haack-Yarom, Romatschke andKleinert-Probst relations.
Keywords:
Holography and quark-gluon plasmas, Gauge-gravity correspondence, D-branes ontents p − q + 2 dimensions 33 The first order calculation 7 Fluid/gravity correspondence [1, 2] is an effective method of calculating the dynamicalsecond order transport coefficients for various kinds of gravity backgrounds. After it wasproposed, this framework has been used in asymptotically Anti de Sitter (AdS) blackhole in various dimensions [3–5], with background scalar field [6] and with backgroundvector charge [7, 8]. – 1 –ranes in string/M theory are the vacuum solutions of 10/11 dimensional super-gravity [9]. The methods used in previous studies on the transport properties of branesare the Minkowski AdS/CFT correspondence [10–16], the stretched horizon formalism[17–19] and the boundary derivative expansion method in Fefferman-Graham coordi-nate [20]. Reference [21] is very important in the development of relativistic hydro-dynamics since it proposes a standard and general formulation of the second orderconstitutive relation for relativistic fluid.Based on these previous works, [22–25] calculate the first and second order trans-port coefficients for D-branes via the fluid/gravity correspondence and have got somenew results. Thereinto, [22, 23] studies the D4-brane with 1 world-volume directioncompact and [25] calculates all the dynamical second order transport coefficients forDp-brane. This paper can be seen as a follow-up work of [25] and we are aiming tocalculate all the dynamical second order transport coefficients for Dp-brane with q ofits world-volume direction(s) compactified, where 1 ≤ q ≤ p −
1. This can be seen asour first motivation. The maximum of q can not be p since we need to leave at least onedimension on the world-volume to let the fluid live in. If we denote a Dp-brane with q direction(s) compact as D(p-q)-brane, such that the cases considered here are D(4-1),D(4-2), D(4-3), D(3-1), D(3-2) and D(2-1)-brane. Actually, the D(4-1) case has beenstudied in [22, 23], thus we will not offer the details for this case but will include itsresults in the final expression for the second order constitutive relation.Our second motivation is to give a final answer to the question proposed in [13]which calculates the first order transport coefficients for Dp-brane. In the conclusionof that paper, in order to make their result compatible with that of [12], the authorsproposed an expression for the radio ζ /η . We quote it here in our convention as ζη = 2( p − + 2 q (5 − p )( p − q )(9 − p ) . (1.1)After discovering the above can cover both their results and that of [12], the authorsof [13] say that they do not know whether this compatibility is just a coincidence ordue to some reason behind the formula. Now we can answer this question: the aboveexpression of ζ /η is for the compactified Dp-brane, or in our notation, the D(p-q)-brane.When setting q = 0, it will be back to the cases in [13] and when setting p = 4 , q = 1,it reproduces the result of [12].The structure of the paper is as follows. In Section 2 we will reduce the 10 di-mensional background of near-extremal black Dp-brane to a p − q + 2 dimensionalEinstein-dilaton theory. In Section 3 we will answer the question proposed in [13] byoffering the result of the first order constitutive relation for D(p-q)-brane. Section 4is a preparation for the second order and then in Section 5 we will solve the differen-– 2 –ial equations for all the second order perturbations. The final result of second orderstress energy tensor and transport coefficients will be given in Section 6. Section 7 is asummary about this paper. p − q + 2 dimensions This section shows how to reduce the 10 dimensional background of Dp-brane to a p − q + 2 dimensional spacetime. Since this work is subsequent to [25], the conventionsare the same as that reference. We will then be brief in offering the details of thedefinitions which are the same as before and will also stress the differences.The action for Dp-brane in 10 dimension contains the bulk term, the Gibbons-Hawking surface term and the counter term. It reads as S = 12 κ Z d x √− G (cid:20) R −
12 ( ∇ ˆ M φ ) − g s − p )! e p − φ F − p (cid:21) − κ Z d x √− H K + 1 κ Z d x √− H − p L e − p − p ) φ . (2.1)The equations of motion (EOM) derived from this action admit the following back-ground solution: ds = (cid:18) rL p (cid:19) (7 − p )28 (cid:16) − f ( r ) dt + δ ij dx i dx j (cid:17) + (cid:18) L p r (cid:19) ( p +1)(7 − p )8 dr f ( r ) + (cid:18) rL p (cid:19) (7 − p )28 δ mn dy m dy n + (cid:18) L p r (cid:19) ( p +1)(7 − p )8 r d Ω − p , (2.2) e φ = (cid:18) rL p (cid:19) ( p − − p )4 , F θ ··· θ − p = g − s Q p √ γ − p , (2.3)where f ( r ) = 1 − r − pH r − p and Q p = (7 − p ) L − pp . Here L − pp = (2 πl s ) − p g s N (7 − p )Ω − p , which relatesthe charge of the Dp-brane and will become the characteristic length of the reducedspacetime.The 10 dimensional line element (2.2) has three parts and represents a spacetimewith structure of M p − q +2 × T q × S − p , of which the coordinate is x ˆ M = ( x M , y m , θ a ).Thereinto, x M = ( x µ , r ) are the coordinates of M p − q +2 with µ = 0 , , · · · , p − q ; y m arethe coordinates of the compactified directions on D-branes that forms a q -dimensionaltorus T q with m = 1 , · · · , q ; and θ a are the coordinates of 8 − p dimensional sphere S − p with a = 1 , · · · , − p . – 3 –e use the ansatz ds = e α A g MN dx M dx N + e α A (cid:0) e β B δ mn dy m dy n + e β B L p d Ω − p (cid:1) (2.4)to perform the dimensional reduction. If we denote the metric for the above ansatz as G ˆ M ˆ N = { e α A g MN , e α A +2 β B δ mn , e α A +2 β B L p γ ab } then the 9D induced boundarymetric is H ˆ M ˆ N = G ˆ M ˆ N − n ˆ M n ˆ N and the corresponding line element is ds = e α A h MN dx M dx N + e α A (cid:0) e β B δ mn dy m dy n + e β B L p d Ω − p (cid:1) , (2.5)where h MN is the boundary metric of M p − q +2 . The unit norm on the 9D boundary is n ˆ M = ∇ ˆ M r q G ˆ N ˆ P ∇ ˆ N r ∇ ˆ P r = ( n M , n m , n a ) = (cid:0) e α A n M , , (cid:1) , (2.6)with n M the unit norm on the boundary of M p − q +2 .As said in [22], the requirement that the reduced action is also in Einstein framegives two relations among those 4 parameters in (2.4). So one may set them as α = − − p + qp − q , α = 1 , β = − − pq , β = 1 . (2.7)The reduced action for the p − q + 2 dimensional theory turns out to be S = 12 κ p − q +2 Z d p − q +2 x √− g (cid:20) R −
12 ( ∂φ ) − − p + q ) p − q ( ∂A ) − (8 − p )(8 − p + q ) q ( ∂B ) + V ( φ, A, B ) (cid:21) − κ p − q +2 Z d p − q +1 x √− hK + 1 κ p − q +2 Z d p − q +1 x √− h − p L p exp (cid:20) − − p + qp − q A + 3 − p − p ) φ (cid:21) , (2.8) V ( φ, A, B ) = (7 − p )(8 − p ) L p exp (cid:18) − p − q A − B (cid:19) − Q p L − p ) p exp (cid:20) p − φ − p − q )(7 − p ) + 16 p − q A − − p ) B (cid:21) . (2.9)The reduced background fields that solves it is ds = (cid:18) rL p (cid:19) − pp − q (cid:0) − f ( r ) dt + d x (cid:1) + (cid:18) rL p (cid:19) ( p − p +9)+ q (7 − p ) p − q dr f ( r ) , (2.10)– 4 – φ = (cid:18) rL p (cid:19) ( p − − p )4 , (2.11) e A = (cid:18) rL p (cid:19) ( p − + q (5 − p )2(8 − p + q ) , e B = (cid:18) rL p (cid:19) − q (5 − p )2(8 − p + q ) . (2.12)For the cases of p = 3, the scalar fields A, B still not vanish. This suggests that theD3-brane will correspond to nonconformal relativistic fluids after compactification. Toput it more generalized, dimensional reduction can make the dual fluid of a conformalbackground nonconformal. This tells us that we can perform a dimensional reductionon AdS black hole backgrounds and calculated the second order transport coefficientsof the dual fluids, which may give a direct check on the proposal in [20].The reduced theory is actually a p − q + 2 dimensional Einstein gravity coupledwith one background scalar—for which one can choose the dilaton. This is because thethree scalars in (2.11) and (2.12) are not independent as can be seen from the following: A = (cid:20) p − − p ) + 2 q (5 − p )(8 − p + q )( p − − p ) (cid:21) φ,B = − q (5 − p )(8 − p + q )( p − − p ) φ. (2.13)The above are not appropriate for p = 3 because of the terms p − A and B are actually inverse proportionalto each other when p = 3 and dilaton vanishes. Anyway, there is always just oneindependent scalar field in the background of the p − q + 2 dimensional reduced gravitytheory.In Eddington-Finkelstein coordinate, the boosted form of (2.10) is ds = − r − pp − q f ( r ) u µ u ν dx µ dx ν + r − pp − q P µν dx µ dx ν − r ( p − p − q (7 − p )2( p − q ) u µ dx µ dr,u µ = (1 , β ) q − β , P µν = η µν + u µ u ν (2.14)where now x = v is the in-going Eddington time. The above represents ideal relativisticfluid living on the boundary of M p − q +2 . The fluid/gravity correspondence [1, 2] statesthat to mimic the real fluid flow, one should first promote the boost parameter u µ and r H to be x µ dependent, then add the (also x µ dependent) metric perturbations, andfinally solve them from EOM in the order of partial derivatives of x µ in some localpatch on the boundary. The global solution of the metric that is dual to arbitraryboundary flow can then be got by stitching together all local solutions. The boundary– 5 –oordinate dependent global metric ansatz can be set as ds = − r − pp − q [ f ( r H ( x ) , r ) − k ( r H ( x ) , u α ( x ) , r )] u µ ( x ) u ν ( x ) dx µ dx ν + 2 r − pp − q P ρµ ( u α ( x )) w ρ ( u α ( x ) , r ) u ν ( x ) dx µ dx ν + r − pp − q [ P µν ( u α ( x )) + α µν ( r H ( x ) , u α ( x ) , r ) + h ( r H ( x ) , u α ( x ) , r ) P µν ( u σ ( x ))] dx µ dx ν − r ( p − p − q (7 − p )2( p − q ) [1 + j ( r H ( x ) , u α ( x ) , r )] u µ ( x ) dx µ dr. (2.15)This metric contains not only the boosted black brane metric but also the artificially setperturbations, which are the tensor perturbation α µν ( r H ( x ) , u α ( x ) , r ), the vector per-turbation w ρ ( u α ( x ) , r ) and the scalar perturbations k ( r H ( x ) , u α ( x ) , r ), h ( r H ( x ) , u α ( x ) , r )and j ( r H ( x ) , u α ( x ) , r ). They depend on the boundary coordinates x µ through theproper velocity u µ and the horizon parameter r H . All the perturbations will be solvedtubewisely [1] in the bulk of the reduced p − q + 2 dimensional nonconformal spacetimeby choosing some special point in x µ directions—which is just x µ = 0, following [1].The tubewise solving procedure of (2.15) is preformed by expanding it in terms ofderivatives of boundary coordinates of u µ and r H . The EOM derived from (2.8) that(2.15) satisfies order by order reads E MN − T MN = 0 , (2.16) ∇ φ − ( p − − p ) L p exp (cid:20) p − φ − p − q )(7 − p ) + 16 p − q A − − p ) B (cid:21) = 0 , (2.17) ∇ A − (7 − p )(8 − p )(8 − p + q ) L p e − p − q A − B + (7 − p ) [( p − q )(7 − p ) + 8]16(8 − p + q ) L p × exp (cid:20) p − φ − p − q )(7 − p ) + 16 p − q A − − p ) B (cid:21) = 0 , (2.18) ∇ B − q (7 − p )(8 − p + q ) L p e − p − q A − B + q (7 − p ) − p + q ) L p × exp (cid:20) p − φ − p − q )(7 − p ) + 16 p − q A − − p ) B (cid:21) = 0 , (2.19)where E MN is the Einstein tensor and the bulk energy momentum tensor is defined as T MN = 12 (cid:18) ∂ M φ∂ N φ − g MN ( ∂φ ) (cid:19) + 8(8 − p + q ) p − q (cid:18) ∂ M A∂ N A − g MN ( ∂A ) (cid:19) + (8 − p )(8 − p + q ) q (cid:18) ∂ M B∂ N B − g MN ( ∂B ) (cid:19) + 12 g MN V. (2.20)– 6 –he EOM of φ, A and B are actually not independent from each other. This is becauseboth A and B are proportional to φ , as can be seen from (2.13). The consequence of thisfact is that the differential equations that derived from (2.17) to (2.19) are the same, ofwhich we will take advantage in solving the scalar perturbations for the compactifiedD3-brane. Since there is no dilaton’s EOM at hand for D3-brane, we will use the EOMof A instead. We need only to solve the compactified Dp-brane with 2 ≤ p ≤
4. Because D5 andD6-brane do not have dual fluids, as has been pointed out in [25] and references therein.D1-brane can not be compactified since it has only one spatial dimension. We need atleast 1 spatial dimension to let the relativistic fluids live in and support their viscousscalar terms.We expand (2.15) to the first order of the derivative of x µ at x µ = 0 and get ds = r − pp − q (cid:20) − (cid:18) f ( r ) − (7 − p ) r − pH r − p δr H − k (1) ( r ) (cid:19) dv + 2 (cid:0) ( f − δβ i − w (1) i ( r ) (cid:1) dvdx i + ( δ ij + α (1) ij ( r ) + h (1) ( r ) δ ij ) dx i dx j (cid:21) + 2 r ( p − p − q (7 − p )2( p − q ) (1 + j (1) ( r )) dvdr − r ( p − p − q (7 − p )2( p − q ) δβ i dx i dr (3.1)All the perturbations depend only on r after derivative expansion, which coincides withthe fact that the fluid within the tube associated with that point is in local equilibrium.The above first order expanded metric depends on x µ linearly through g (1) vv , g (1) vi and g (1) ir , which contains δr H = x µ ∂ µ r H and δβ i = x µ ∂ µ β i . The dependence on r is muchmore complicated than on x µ since the perturbations will turn out to be complicatefunctions of r . The perturbations will be solved in groups of how they transformedunder the action of SO ( p − q ).The tensor perturbation satisfies the traceless tensor part of the Einstein equation E ij − p − q δ ij δ kl E kl − (cid:18) T ij − p − q δ ij δ kl T kl (cid:19) = 0 , (3.2)After we put (3.1) in, it gives ∂ r ( r − p f ( r ) ∂ r α (1) ij ( r )) + (9 − p ) r − p σ ij = 0 (3.3)– 7 –hich does not depend on q . The reason is that compactification of world-volumedirection only affects the trace of the metric, but α ij is the traceless part of the metricperturbation. The solution of the above has been solved in [25] as F ( r ) = 13 r / H (cid:20) √ √ rr H r − r H + ln ( √ r + √ r H ) ( r + √ rr H + r H ) ( r + rr H + r H ) r (cid:21) , (D4-brane) F ( r ) = 12 r H (cid:20) r H r + ln ( r + r H ) ( r + r H ) r (cid:21) , (D3-brane) F ( r ) = 25 r / H " π π √ rr H r − r H − π π √ rr H r − r H − π π √ rr H r + r H − π π √ rr H r + r H + ln ( √ r + √ r H ) ( r + r r H + r r H + rr H + r H ) r . (D2-brane) (3.4)There are two components of the Einstein equation can be used to solve the vectorperturbation: the (0 i ) and ( ri ) components. In the original framework, ( ri ) componentof Einstein equation E ri − T ri = 0 (3.5)is used to solve w (1) i as w (1) i ( r ) = a ( r ) ∂ β i , a ( r ) = − − p ) r − p (3.6)The linear combination of the (0 i ) and ( ri ) components of Einstein equation g r ( E i − T i ) + g rr ( E ri − T ri ) = 0 , (3.7)is used to give a constraint relation between the two first order vector viscous terms as1 r H ∂ i r H = − − p ∂ β i , (3.8)Note that both the first order vector perturbation (3.6) and constraint equation (3.8)are not dependent on q since the vector part is still not affected by compactification.As we have analyzed that compactifying the Dp-brane only affects the trace part ofthe metric, thus will only change the scalar perturbation h . In the scalar part solving– 8 –rocedure, we have (00) , (0 r ) , ( rr ) and ( ii ) (with i summed) components of Einsteinequation, as well as the EOM of φ . But there are only 3 unknown scalar functions, thustwo of the five equations are redundant. The ( ii ) (with i summed) component turnsout to be more complex than the others thus can be omitted. A linear combination ofthe (00) and ( r
0) components of Einstein equation g r ( E − T ) + g rr ( E r − T r ) = 0 (3.9)does not contain any scalar perturbations and it is called the first scalar constraintequation. To put (3.1) into the above one has1 r H ∂ r H = − − p ∂β. (3.10)A linear combination of the ( r
0) and ( rr ) components of Einstein equation (which isalso called the second scalar constraint equatioin) g rr ( E rr − T rr ) + g r ( E r − T r ) = 0 , (3.11)the ( rr ) component itself E rr − T rr = 0 , (3.12)and the EOM of φ (2.17) are chosen to solve the 3 scalar perturbations, whose firstorder differential equations are( r − p k (1) ) ′ + 2(7 − p ) r − p j (1) − (cid:20) ( p − q ) r − p − p − q )9 − p r − pH (cid:21) h ′ (1) − r − p ∂β = 0 , (3.13)2( p − q ) rh ′′ (1) + (7 − p )( p − q ) h ′ (1) − − p ) j ′ (1) = 0 , (3.14)2( r − p k (1) ) ′ + 2 r − p f j ′ (1) + 4(7 − p ) r − p j (1) − ( p − q ) r − p f h ′ (1) − r − p ∂β = 0 . (3.15)Note that there are always a factor of ( p − q ) associated with h (1) in the above 3differential equations, suggesting that the compactification of Dp-brane only affect thescalar perturbation h . The solution for the 3 scalar perturbations are F h = 1 p − q F, F j = − − p r − p − r − p H r − p − r − pH + 5 − p − p ) F,F k = 4(9 − p ) r − p − − p (cid:18) − p + 2 r − pH r − p (cid:19) F. (3.16)where χ = F χ ∂β ( χ = { h, j, k } ). The factor of p − q in the denominator of theexpression of F h shows that the dimensions of spatial directions of the compactifiedDp-brane is now p − q but not p as in [25].– 9 – .2 Constitutive relation on the boundary The constitutive relation of the fluid living on the boundary can be got from takinglarge r limit for the Brown-York tensor of the bulk theory: T µν = 1 κ p − q +2 lim r →∞ (cid:18) rL p (cid:19) (9 − p )( p − q − p − q ) K µν − h µν K − − p L p (cid:18) rL p (cid:19) − ( p − q (5 − p )2( p − q ) h µν , (3.17)The third term in the parenthesis of the above is from the counter term of the action.Note that the difference of the two powers of the factor rL p is always 2 for all Dp-branewith or without compactification. Here K µν = −
12 ( n ρ ∂ ρ h µν + ∂ µ n ρ h ρν + ∂ ν n ρ h ρµ ) , (3.18)and K = h µν K µν . Putting (3.1) into (3.17) with all solved first order perturbations wethen have the first order constitutive relation of the boundary fluid as T µν = 12 κ p − q +2 (cid:20) r − pH L − pp (cid:18) − p u µ u ν + 5 − p P µν (cid:19) − (cid:18) r H L p (cid:19) − p (cid:18) σ µν + 2( p − + 2 q (5 − p )( p − q )(9 − p ) P µν ∂u (cid:19) . (3.19)Then thermal and transport coefficients of first order can be got as ε = 12 κ p − q +2 − p r − pH L − pp , p = 12 κ p − q +2 − p r − pH L − pp ,η = 12 κ p − q +2 (cid:18) r H L p (cid:19) − p , ζ = 12 κ p − q +2 p − + 2 q (5 − p )( p − q )(9 − p ) (cid:18) r H L p (cid:19) − p . (3.20)In the above, only the bulk viscosity depends on the number of compact directionssince it relates with h (1) . From these results, one can see that the ratio ζ /η that weget via fluid/gravity correspondence is exactly what is proposed in [13] as quoted inour (1.1). Thus the expression for the ratio ζ /η given in [13] is indeed for compactifiedDp-brane. In the first order calculation we have two equations that without any perturbation,which are actually the Navier-Stokes equations of the boundary fluid at the first or-der. But from the gravity viewpoint they are the first order scalar (3.10) and vector– 10 –3.8) constraint. They relate the following 5 first order partial derivative terms of thecollective modes of different type: ∂ r H , ∂β, ∂ i r H , ∂ β i , σ ij = ∂ ( i β j ) − p − q δ ij ∂β. (4.1)Thus only three of the above are actually independent. We choose ∂β , ∂ β i and σ ij asthe first order scalar, vector and tensor viscous term, respectively. That’s why the firstorder perturbations can be written as χ (1) = F χ ∂β , w (1) i = a ( r ) ∂ β i and α (1) ij = F ( r ) σ ij .The second order solving procedure follows the same rule. But now we have muchmore viscous terms since its complexity grows nonlinearly. These second order viscousScalars of SO( p − q ) Vectors of SO( p − q ) Tensors of SO( p − q ) s = r H ∂ r H v i = r H ∂ ∂ i r H t ij = r H ∂ i ∂ j r H − p − q δ ij s s = ∂ ∂ i β i v i = ∂ β i t ij = ∂ Ω ij s = r H ∂ i r H v i = ∂ j β i t ij = ∂ σ ij S = ∂ β i ∂ β i v i = ∂ j Ω ij T ij = ∂ β i ∂ β j − p − q δ ij S S = ( ∂ i β i ) v i = ∂ j σ ij T ij = σ k [ i Ω j ] k S = Ω ij Ω ij V i = ∂ β i ∂β T ij = Ω ij ∂β S = σ ij σ ij V i = ∂ β j Ω ij T ij = σ ij ∂β V i = ∂ β j σ ij T ij = Ω ki Ω jk − p − q δ ij S T ij = σ ki σ jk − p − q δ ij S T ij = σ k ( i Ω j ) k Table 1 . The list of SO( p − q ) invariant second order viscous terms for 1 ≤ p − q ≤ T , , do not exist when p − q = 2. At p − q = 1, only s , , , S , , v , , and V exist. terms satisfy 6 identities which can be extract from ∂ µ ∂ ρ T (0) ρν = 0. T (0) µν here is thethermodynamical part of (3.19) with r H and u µ depend on x µ . After we expand T (0) µν and take different components of ∂ µ ∂ ρ T (0) ρν = 0, it gives9 − p r H ∂ r H + ∂ ∂β − − p ( ∂β ) − − p ∂ β i ∂ β i = 0 , (4.2)5 − p r H ∂ i r H + ∂ ∂β − − p ∂ β i ∂ β i − − p − p ( ∂β ) + ∂ i β j ∂ j β i = 0 , (4.3)– 11 – − p r H ∂ ∂ i r H + ∂ β i − − p − p ∂ β i ∂β + ∂ β j ∂ j β i = 0 , (4.4)9 − p r H ∂ ∂ i r H + ∂ i ∂β − − p ∂ β i ∂β − − p ∂ β j ∂ i β j = 0 , (4.5) ∂ Ω ij − − p − p Ω ij ∂β − ∂ k β [ i ∂ j ] β k = 0 , (4.6)5 − p r H ∂ i ∂ j r H + ∂ ∂ ( i β j ) − − p ∂ β i ∂ β j − − p − p ∂ ( i β j ) ∂β + ∂ k β ( i ∂ j ) β k = 0 . (4.7)These are the same as the cases of Dp-brane without compact direction. But afterreexpressed in terms of the spatial viscous tensors in Table 1, they actually depend on q : s + 29 − p s − − p )(5 − p ) S − − p ) S = 0 , (4.8) s + 5 − p s − − p S + ( p − + q (5 − p )( p − q )(9 − p ) S − S + S = 0 , (4.9) v + 25 − p v + 2( p − p + 9 + q (7 − p ))( p − q )(9 − p )(5 − p ) V − − p V + 25 − p V = 0 , (4.10) v + 2( p − q )( v + v )( p − q − − p ) − p − q + 2)( p − q )(9 − p )(5 − p ) V − V + V )(9 − p )(5 − p ) = 0 , (4.11) t − T + p − p + 18 + q (5 − p )( p − q )(9 − p ) T = 0 , (4.12) t + 25 − p t − − p ) T + 2( p − p + 18) + 2 q (5 − p )( p − q )(9 − p )(5 − p ) T − − p T + 25 − p T = 0 . (4.13)The above 6 identities are used in deriving the differential equations for the sec-ond order perturbations as well as the second order Navier-Stokes equations. We willderive the latter in this section from a hydrodynamical viewpoint. They can be gainedby expanding (3.19) to the second order of partial derivatives and putting it into ∂ µ T (0+1) µν = 0, where we add the superscript (0 + 1) to emphasize the difference with T (0) µν . The second order Navier-Stokes equations are1 r ( p − / H ∂ r (1) H = 4( p − + 4 q (5 − p )( p − q )(9 − p ) (7 − p ) S + 4(9 − p )(7 − p ) S , (4.14)1 r ( p − / H ∂ i r (1) H = [4( p − + 4 q (5 − p )] v + 16( p − q ) v ( p − q − − p )(7 − p )(5 − p )+ 2( p − p − p + 77) − q ( p − p + 85)( p − q )(9 − p )(7 − p )(5 − p ) V – 12 – − p )(9 − p )(7 − p )(5 − p ) V − p − p + 77)(9 − p )(7 − p )(5 − p ) V . (4.15)They are more complex than the Dp-brane case, when set q = 0, they are back to theform of Dp-brane. The above second order Navier-Stokes equations can also be gotthrough the constraints of Einstein equation (3.7) and (3.9), which will be specifiedlater when we solve the second order perturbations. The calculation of the second order needs first to expand (2.15) to the second order ofpartial derivatives of r H and β i , and the result is ds = − r − pp − q (cid:20) f − (1 − f ) δβ i δβ i − (7 − p ) r − pH r − p δr H − (7 − p ) r − pH r − p δ r H − (7 − p ) r − pH r − p δr (1) H − (7 − p )(6 − p ) r − pH r − p ( δr H ) − ( F k + δF k ) ∂β − F k ( δ∂β + δβ i ∂ β i ) − a ( r ) δβ i ∂ β i − k (2) ( r ) (cid:21) dv − r − pp − q (cid:20) (1 − f )( δβ i + 12 δ β i ) + a ( ∂ β i + δ∂ β i + δβ j ∂ j β i )+ (7 − p ) r − pH r − p δr H δβ i + F k ∂βδβ i + F δβ j ∂ ( i β j ) + w (2) i ( r ) (cid:21) dvdx i + 2 r ( p − p − q (7 − p )2( p − q ) (cid:20) F j + δF j ) ∂β + F j ( δ∂β + δβ i ∂ β i ) + 12 δβ i δβ i + j (2) ( r ) (cid:21) dvdr + r − pp − q (cid:20) δ ij + (1 − f ) δβ i δβ j + 2 aδβ ( i ∂ | | β j ) + ( F + δF ) ∂ ( i β j ) + F (cid:0) δ∂ ( i β j ) + δβ ( i ∂ | | β j ) (cid:1) + α (2) ij ( r ) + h (2) ( r ) δ ij (cid:21) dx i dx j − r ( p − p − q (7 − p )2( p − q ) (cid:18) δβ i + 12 δ β i + F j ∂βδβ i (cid:19) dx i dr. (4.16)Here we have δ ψ = x µ x ν ∂ µ ∂ ν ψ with ψ either r H or β i , while δ F ( r H ( x ) , r ) = − (5 − p ) F ( r ) + 2 r F ′ ( r )2 r H δr H (4.17)with F standing for any of the F, F j and F k . The above does not change comparedwith the Dp-brane case since F, F j and F k do not change.Each component of the above second order expanded metric has the structure of g MN = r ( ··· ) [ · · · ], the expressions inside the brackets [ · · · ] do not change while termsinside the parenthesis ( · · · ) have changed compared with the Dp-brane case. Thecomponents of the second order expanded metric are now the second order polynomialsof the boundary coordinates x µ through terms like δ ψ or ( δψ ) with ψ being r H or β i .– 13 – Solving the second order perturbations
In the following subsections we will solve the second order perturbations for all thecompactified Dp-branes with p = 2 , , ≤ q ≤ p −
1. The cases can be dividedinto three classes, in terms of p − q , they are p − q = 1 , ,
3. Since p − q = 3 is justthe D4-brane with 1 direction compactified. We refer the reader to [23] for details andwill omit the solving procedure in this work despite the conventions used here is a littledifferent from that of [23]. We will solve the second order tensor perturbations for all compactified Dp-brane inthis subsection. Since we omit the D(4-1)-brane, the cases contained here are only p − q = 2 which include D(4-2) and D(3-1)-brane. Because the spatial viscous tensorscan only be constructed in spacetime with more than 1 spatial dimensions.The differential equation of the tensor perturbation is ddr r − p f ( r ) dα (2) ij dr ! = S ij ( r ) , (5.1)which is an exact second order linear inhomogeneous differential equation. The inho-mogeneous term S ij ( r ) is called the source term in [1]. This equation is singular at r = 0, r = r H and r → ∞ . Its solution can be got by integration twice as α (2) ij ( r ) = Z ∞ r − x − p f dx Z x S ij ( y ) dy, (5.2)We will solve the above equation by specifying p, q with explicit values. Whenputting the second order expanded metric (4.16) into (3.2) for D4-brane with 2 compactdirections, one gets ∂ r ( r f ∂ r α (2) ij ) = (cid:18) r − r F − r F ′ (cid:19) ( t + T ) + (cid:20) r − r F − r F ′ − r F ′′ + 12 r f F ′ + (4 r − F F ′ + r f F F ′′ + 5 r F j − r F F j + 2(4 r − F ′ F j − (5 r − F F ′ j + r f F ′ F ′ j + 2 r f F ′′ F j − r F F k + 4 r F ′ F k − r F F ′ k + r F ′ F ′ k + r F ′′ F k − r F F ′′ k (cid:21) T − (cid:16) r + 5 r F + 4 r F ′ (cid:17) T , (5.3)– 14 –he source term on the right hand side of the above equation has the same structureas D2-brane. The differences lie only in the coefficient functions of each spatial viscoustensors. Using (5.2), one has α (2) ij = (cid:18) − π √ − ln 33 (cid:19) r ( t + T ) + (cid:20) r + (cid:18) − π √ − ln 315 (cid:19) r (cid:21) T + (cid:20) r − (cid:18) π √ (cid:19) r (cid:21) T (5.4)This solution is in a form of an asymptotic series of r , which is got by making a seriesexpansion in between the two sequential integrations for y and x , which has beenspecified in [23]. Expanding the source term at the beginning will lead to wrong resultsfor α (2) ij and h (2) . But this will not affect the integration for w (2) i , j (2) and k (2) . We thinkthe reason is that both the integrations for α (2) ij and h (2) depend on the emblackeningfactor f ( r ) while w (2) i do not. As for j (2) and k (2) , they are calculated by integrationonly once. This exchangeability between the integral of y and asymptotic expansionwill give us great help for the calculations.Though near extremal D3-brane background can be reduced trivially to AdS blackhole metric and the dual fluid is conformal. After compactification, it will correspond tononconformal fluid. The tensor perturbation for D(3-1)-brane satisfies the differentialequation ∂ r ( r f ∂ r α (2) ij ) = (cid:0) r − r F − r F ′ (cid:1) ( t + T ) + (cid:20) r − r F − r F ′ − r F ′′ + (5 r − F F ′ + r f F F ′′ + 12 r f F ′ + 6 r F j − r F F j + 2(5 r − F ′ F j − r − F F ′ j + r f F ′ F ′ j + 2 r f F ′′ F j − r F F k + 5 r F ′ F k − r F F ′ k + r F ′ F ′ k + r F ′′ F k − r F F ′′ k (cid:21) T − (cid:0) r + 6 r F + 4 r F ′ (cid:1) T , (5.5)using (5.2) one gets the solution as α (2) ij = (cid:18) − ln 24 (cid:19) r ( t + T ) + (cid:20) r + (cid:18) − ln 212 (cid:19) r (cid:21) T + (cid:18) r − ln 22 r (cid:19) T . (5.6)Compared with D3-brane case where the coefficient function of t + T is proportionalto that of T , these two coefficient functions here are not proportional to each other,suggesting the background is not conformal now.– 15 – .2 The vector part There are a constraint plus a dynamical equation for the vector part. Both the casesof p − q = 2 and 1 need to be analyzed here. The vector constraint equation can be got by feeding (4.16) into (3.7). The result isjust the spatial component of the Navier-Stokes equation of second order, as one cancheck by substituting specific values for p and q into (4.15).Let’s start with p − q = 2 first. For D(4-2)-brane, the vector constraint equation is ∂ i r (1) H = (cid:20) − r − r f F ′ + 43 (5 r − F j + 103 r F k + 43 r F ′ k (cid:21) v + (cid:20)
43 (5 r − F j + 103 r F k + 43 r F ′ k (cid:21) v + (cid:20) r + 16 r (13 r − F ′ − (15 r + 2) F j − r F k − r F ′ k (cid:21) V + (cid:20) − r − r f F ′ + 23 (5 r − F j + 53 r F k + 23 r F ′ k (cid:21) V + (cid:20) − r − r f F ′ + 23 (5 r − F j + 53 r F k + 23 r F ′ k (cid:21) V . (5.7)After expansion with respect to 1 /r , we have ∂ i r (1) H = 45 v + 3215 v − V − V − V (5.8)By the same token, one has the vector constraint for D(3-1)-brane as ∂ i r (1) H = (cid:20) − r − r f F ′ + (3 r − F j + 32 r F k + 12 r F ′ k (cid:21) v + (cid:20) (3 r − F j + 32 r F k + 12 r F ′ k (cid:21) v + (cid:20) r + 116 r (9 r − F ′ −
34 (5 r + 1) F j − r F k − r F ′ k (cid:21) V + (cid:20) − r − r f F ′ + 12 (3 r − F j + 34 r F k + 14 r F ′ k (cid:21) V + (cid:20) − r − r f F ′ + 12 (3 r − F j + 34 r F k + 14 r F ′ k (cid:21) V . (5.9)After expansion, the above becomes ∂ i r (1) H = 16 v + 23 v + 18 V − V − V . (5.10)– 16 –he above two cases have two spatial directions for the dual fluid. Next we considersituations for p − q = 1, the vector viscous term in this case are only v , , and V .Thus the equations will turn out to be much simple now. In the order of D(4-3), D(3-2)and D(2-1)-brane, the vector constraint equations can be read as ∂ i r (1) H = (cid:20)
23 (5 r − F j + 53 r F k + 23 r F ′ k (cid:21) v + (cid:20) r + 23 r ( r + 2) F ′ −
83 (5 r + 1) F j − r F k − r F ′ k (cid:21) V , (5.11) ∂ i r (1) H = (cid:20)
12 (3 r − F j + 34 r F k + 14 r F ′ k (cid:21) v + (cid:20) r + 14 r ( r + 1) F ′ − (3 r + 1) F j − r F k − r F ′ k (cid:21) V , (5.12) ∂ i r (1) H = (cid:20)
215 (7 r − F j + 715 r F k + 215 r F ′ k (cid:21) v + (cid:20) r + 245 r (3 r + 2) F ′ −
845 (7 r + 3) F j − r F k − r F ′ k (cid:21) V , (5.13)which, after expansion with respect to 1 /r , lead to ∂ i r (1) H = 1615 v − V , D(4-3)-brane (5.14) ∂ i r (1) H = 13 v − V , D(3-2)-brane (5.15) ∂ i r (1) H = 16105 v − V . D(2-1)-brane (5.16)
The differential equation of w (2) i can be got by plugging the second order expandedmetric (4.16) into (3.5): ddr r − p dw (2) i dr ! = S i ( r ) , (5.17)where S i in the right hand side is the source term. This equations is similar like (5.1),the only difference is that the above does not have the emblackening factor in the lefthand side, thus no singularity at r = r H . The above equation can be solved by directintegration as w (2) i = Z ∞ r x − p dx Z ∞ x S i ( y ) dy. (5.18)– 17 –et’s solve the p − q = 2 cases first. For D(4-2)-brane, the differential equation of w (2) i is ∂ r ( r ∂ r w (2) i ) = (cid:16) − r − r F ′ + 5 r F j − r F ′ j (cid:17) v + (cid:16) r F j − r F ′ j (cid:17) v + (cid:18) r − r F ′ − r F ′′ + 254 r F j + 52 r F ′ j − r F ′′ j (cid:19) V + (cid:18) r − r F ′ + 52 r F j − r F ′ j (cid:19) V + (cid:18) − r − r F ′ + 52 r F j − r F ′ j (cid:19) V , (5.19)of which the solution is very simple w (2) i = − r V − r V − r V . (5.20)The vector dynamical equation for the second order perturbation of D(3-1)-brane is ∂ r ( r ∂ r w (2) i ) = (cid:0) − r − r F ′ + 6 r F j − r F ′ j (cid:1) v + (cid:0) r F j − r F ′ j (cid:1) v + (cid:18) r − r F ′ − r F ′′ + 152 r F j + 32 r F ′ j − r F ′′ j (cid:19) V + (cid:18) r − r F ′ + 3 r F j − r F ′ j (cid:19) V + (cid:18) − r − r F ′ + 3 r F j − r F ′ j (cid:19) V , (5.21)with the solution is w (2) i = − r V − r V − r V . (5.22)For the cases of p − q = 1, the differential equation for w (2) i becomes simple again.In the order of D(4-3), D(3-2) and D(2-1)-brane, they are ∂ r ( r ∂ r w (2) i ) = (cid:18) r F j − r F ′ j (cid:19) v + (cid:18) − r F ′ − r F ′′ + 152 r F j + 2 r F ′ j − r F ′′ j (cid:19) V , (5.23) ∂ r ( r ∂ r w (2) i ) = (cid:0) r F j − r F ′ j (cid:1) v + (cid:0) − r F ′ − r F ′′ + 9 r F j + r F ′ j − r F ′′ j (cid:1) V , (5.24) ∂ r ( r ∂ r w (2) i ) = (cid:18) r F j − r F ′ j (cid:19) v + (cid:18) − r F ′ − r F ′′ + 212 r F j + 23 r F ′ j − r F ′′ j (cid:19) V . (5.25)– 18 –he above 3 equations can also be identified by the power of r in the left hand side:this power is 8 − p for D(p-q)-brane. The solutions of the above are w (2) i = − r V , D(4-3)-brane (5.26) w (2) i = − r V , D(3-2)-brane (5.27) w (2) i = − r V , D(2-1)-brane (5.28)
The scalar perturbations plays an important role in the nonconformal background.They are also more difficult to solve than the tensor and vector part because thescalar perturbations are mixed together in the differential equations derived from scalarcomponents of Einstein equation and the EOM of dilaton (or the scalar A ).The differential equation of h (2) for D(p-q)-brane is ddr (cid:18) r − p f ( r ) dh (2) dr (cid:19) = S h ( r ) , (5.29)whose solution can be written as h (2) ( r ) = Z ∞ r − x − p f dx Z x S h ( y ) dy. (5.30)This is very like α (2) ij , which can be seen from (5.1) and (5.2).The differential equations for j and k are actually first order ones thus they can besolved by integrating once as j (2) ( r ) = − Z ∞ r S j ( x ) dx (5.31) k (2) ( r ) = − r − p Z ∞ r S k ( x ) dx. (5.32) r − p is present in from of the integration. This is because k always appear as ( r − p k )in the differential equations. We still first deal with the cases of p − q = 2 in the scalar sector. The first scalarconstraint does not contain any unknown perturbation and it can be calculated byplugging the second order expanded metric into (3.9), which gives for the D(4-2)-branecase as ∂ r (1) H = (cid:18) r + 415 r − F + 215 r f F ′ − r f F j − r F k (cid:19) s + 25 r s – 19 – (cid:18) − r + 415 r − F + 215 r f F ′ − r f F j − r F k (cid:19) S + (cid:18) r + 215 r − F + 1175 r f F ′ + 475 r f F ′′ − r f F j − r f F ′ j − r F k (cid:19) S − r S + (cid:18) r + 415 r + 215 r f F ′ (cid:19) S (5.33)After expanding in powers of 1 /r , one has ∂ r (1) H = 225 S + 415 S + 45 r / (cid:18) s + 12 s − S + 310 S − S + S (cid:19) . (5.34)The expression inside the parenthesis is just the identity (4.9) that satisfied by thesecond order spatial viscous terms thus gives a zero. So the above finally becomes ∂ r (1) H = 225 S + 415 S (5.35)which is actually (4.14) in the case of p = 4, q = 2.(3.9) gives us for D(3-1)-brane the first scalar constraint as ∂ r (1) H = (cid:18) r + 16 r − F + 112 r f F ′ − r f F j − r F k (cid:19) s + 13 r s + (cid:18) − r + 16 r − F + 112 r f F ′ − r f F j − r F k (cid:19) S + (cid:18) r + 112 r − F + 772 r f F ′ + 136 r f F ′′ − r f F j − r f F ′ j − r F k (cid:19) S − r S + (cid:18) r + 16 r + 112 r f F ′ (cid:19) S , (5.36)which gives ∂ r (1) H = 136 S + 16 S + 13 r (cid:18) s + s − S + 16 S − S + S (cid:19) (5.37)Again, terms in the parenthesis sum to zero by (4.9) and one finally has the scalarcomponent of Navier-Stokes equation as ∂ r (1) H = 136 S + 16 S . (5.38)Then we solve the situations of p − q = 1. The first scalar constraint for D(4-3),D(3-2) and D(2-1)-brane can be listed separately as ∂ r (1) H = (cid:18) r + 415 r − F + 215 r f F ′ − r f F j − r F k (cid:19) s + 25 r s – 20 – (cid:18) − r + 415 r − F + 215 r f F ′ − r f F j − r F k (cid:19) S + (cid:18) r + 415 r − F + 1675 r f F ′ + 475 r f F ′′ − r f F j − r f F ′ j − r F k (cid:19) S , (5.39) ∂ r (1) H = (cid:18) r + 16 r − F + 112 r f F ′ − r f F j − r F k (cid:19) s + 13 r s + (cid:18) − r + 16 r − F + 112 r f F ′ − r f F j − r F k (cid:19) S + (cid:18) r + 16 r − F + 536 r f F ′ + 136 r f F ′′ − r f F j − r f F ′ j − r F k (cid:19) S , (5.40) ∂ r (1) H = (cid:18) r + 435 r − F + 235 r f F ′ − r f F j − r F k (cid:19) s + 27 r s + (cid:18) − r + 435 r − F + 235 r f F ′ − r f F j − r F k (cid:19) S + (cid:18) r + 435 r − F + 24245 r f F ′ + 4245 r f F ′′ − r f F j − r f F ′ j − r F k (cid:19) S . (5.41)After expansion in terms of powers of inverse r , the above become ∂ r (1) H = 1675 S + 45 r / (cid:18) s + 12 s − S + 45 S (cid:19) , D(4-3)-brane (5.42) ∂ r (1) H = 19 S + 13 r (cid:18) s + s − S + 23 S (cid:19) , D(3-2)-brane (5.43) ∂ r (1) H = 16245 S + 421 r / (cid:18) s + 32 s − S + 47 S (cid:19) . D(2-1)-brane (5.44)Once more, terms in the parenthesis of the above are all zero in the name of (4.9).Thus one has finally ∂ r (1) H = 1675 S , D(4-3)-brane (5.45) ∂ r (1) H = 19 S , D(3-2)-brane (5.46) ∂ r (1) H = 16245 S . D(2-1)-brane (5.47)– 21 – .3.2 The scalar dynamical equations
To solve the 3 scalar perturbations, we need the second scalar constraint (3.11), the ( rr )component of Einstein equation (3.12) and the EOM of dilaton (2.17) or the scalar A (2.18). Note the compactified D3-brane does not have dilaton field in the background,so we need EOM of A to solve the scalar perturbations. For compactified D4 andD2-brane, one can just use EOM of dilaton.We still tackle the cases of p − q = 2 first. In the case of D(4-2)-brane, after feedingthe equations mentioned above with (4.16), we get5( r k (2) ) ′ + 30 r j (2) − r − h ′ (2) = (cid:16) − r + 5 r F + 2 r F ′ (cid:17) s + (cid:16) r + 5 r F + 2 r F ′ (cid:17) S + (cid:20) − r + r F + 265 r F ′ + 45 r F ′′ −
12 (5 r − F F ′ + 14 r f F ′ − r F j + 45 r F j − r − F ′ F j − r F ′ F k − r F ′ F ′ k + 30 r F j F k + 10 r F j F ′ k (cid:21) S + (cid:18) r + 20 r (cid:19) S − (cid:20) r + (5 r − F F ′ + 12 r f F ′ (cid:21) S , (5.48)2 rh ′′ (2) + 3 h ′ (2) − j ′ (2) = (cid:18) F F ′ + 12 rF F ′′ + 14 rF ′ + rF ′ F ′ j − F j F ′ j (cid:19) S + 2 r S + (cid:18) F F ′ + rF F ′′ + 12 rF ′ (cid:19) S , (5.49)and( r k (2) ) ′ + r f j ′ (2) + 6 r j (2) − r f h ′ (2) = (cid:18) − r + 12 r F (cid:19) s + (cid:18) r + 12 r F (cid:19) S + (cid:20) − r + 110 r F + 15 r F ′ − r f F F ′ − r F j + 9 r F j − r f F ′ F j + 3 r f F j F ′ j − r F ′ F k + 6 r F j F k + r F ′ j F k + 2 r F j F ′ k (cid:21) S + 2 r S − (cid:18) r + 12 r f F F ′ (cid:19) S , (5.50)respectively. Then the perturbations can be solved as h (2) = (cid:18) − π √ − ln 36 (cid:19) r ( s + S ) + (cid:20) r + (cid:18) − π √ − ln 330 (cid:19) r (cid:21) S + 2 r S + (cid:18) r + 23 r (cid:19) S , (5.51) j (2) = − (cid:18) − π √ − ln 36 (cid:19) r ( s + S ) − (cid:18) − π √ − ln 330 (cid:19) r S − r S , (5.52)– 22 – (2) = (cid:20) − (cid:18) − π √ − ln 315 (cid:19) r + 110 r − (cid:18) − π √ − ln 36 (cid:19) r (cid:21) s + (cid:20) r − (cid:18) − π √ − ln 315 (cid:19) r + 110 r − (cid:18) − π √ − ln 36 (cid:19) r (cid:21) S + (cid:20) − (cid:18) − π √ − ln 375 (cid:19) r + 1235 r / + 4725 r − (cid:18) − π √ − ln 330 (cid:19) r (cid:21) S − r S − (cid:18) r − r / − r + 23 r (cid:19) S . (5.53)By the same token as the previous case, one has for D(3-1)-brane the differentialequations derived from the second scalar constraint, the ( rr ) component of Einsteinequation and the EOM of the scalar A as3( r k (2) ) ′ + 24 r j (2) − r − h ′ (2) = (cid:0) − r + 3 r F + r F ′ (cid:1) s + (cid:0) r + 3 r F + r F ′ (cid:1) S + (cid:20) − r + r F + 83 r F ′ + 13 r F ′′ −
12 (3 r − F F ′ + 18 r f F ′ − r F j + 36 r F j − r − F ′ F j − r F ′ F k − r F ′ F ′ k + 24 r F j F k + 6 r F j F ′ k (cid:21) S + (cid:18) r + 6 r (cid:19) S − (cid:20) r + (3 r − F F ′ + 14 r f F ′ (cid:21) S , (5.54) rh ′′ (2) + 2 h ′ (2) − j ′ (2) = (cid:18) F F ′ + 14 rF F ′′ + 18 rF ′ + 12 rF ′ F ′ j − F j F ′ j (cid:19) S + 1 r S + (cid:18) F F ′ + 12 rF F ′′ + 14 rF ′ (cid:19) S , (5.55)( r k (2) ) ′ + r f j ′ (2) + 8 r j (2) − r f h ′ (2) = (cid:18) − r + 12 r F (cid:19) s + (cid:18) r + 12 r F (cid:19) S + (cid:20) − r + 16 r F + 16 r F ′ − r f F F ′ − r F j + 12 r F j + 3 r f F j F ′ j − r f F ′ F j − r F ′ F k + 8 r F j F k + r F ′ j F k + 2 r F j F ′ k (cid:21) S + r S − (cid:18) r + 12 r f F F ′ (cid:19) S . (5.56)The solutions of the above are h (2) = (cid:18) − ln 28 (cid:19) r ( s + S ) + (cid:20) r + (cid:18) − ln 224 (cid:19) r (cid:21) S + 12 r S + (cid:18) r + 14 r (cid:19) S , (5.57)– 23 – (2) = − (cid:18) − ln 28 (cid:19) r ( s + S ) − (cid:18) − ln 224 (cid:19) r S − r S , (5.58) k (2) = (cid:20) − (cid:18) − ln 212 (cid:19) r + 118 r − (cid:18) − ln 28 (cid:19) r (cid:21) s + (cid:20) r − (cid:18) − ln 212 (cid:19) r + 118 r − (cid:18) − ln 28 (cid:19) r (cid:21) S + (cid:20) − (cid:18) − ln 236 (cid:19) r + 115 r + 37108 r + ln 224 r (cid:21) S − r S − (cid:18) r − r − r + 14 r (cid:19) S . (5.59)For the situations of one spatial dimension after compactification, i.e. p − q = 1,the D(4-3)-brane has the differential equations as5( r k (2) ) ′ + 30 r j (2) − (5 r − h ′ (2) = (cid:16) − r + 5 r F + 2 r F ′ (cid:17) s + (cid:16) r + 5 r F + 2 r F ′ (cid:17) S + (cid:20) − r + r F + 265 r F ′ + 45 r F ′′ − (5 r − F F ′ − r F j + 45 r F j − r − F ′ F j − r F ′ F k − r F ′ F ′ k + 30 r F j F k + 10 r F j F ′ k (cid:21) S , (5.60)2 rh ′′ (2) + 3 h ′ (2) − j ′ (2) = (cid:0) F F ′ + 2 rF F ′′ + rF ′ + 2 rF ′ F ′ j − F j F ′ j (cid:1) S , (5.61)( r k (2) ) ′ + r f j ′ (2) + 6 r j (2) − r f h ′ (2) = (cid:18) − r + 12 r F (cid:19) s + (cid:18) r + 12 r F (cid:19) S + (cid:20) − r + 110 r F + 15 r F ′ − r f F F ′ − r F j + 9 r F j + 3 r f F j F ′ j − r f F ′ F j − r F ′ F k + 6 r F j F k + r F ′ j F k + 2 r F j F ′ k (cid:21) S . (5.62)Which can be solved as h (2) = (cid:18) − π √ − ln 33 (cid:19) r ( s + S ) + (cid:20) r + (cid:18) − π √ − ln 315 (cid:19) r (cid:21) S (5.63) j (2) = − (cid:18) − π √ − ln 36 (cid:19) r ( s + S ) − (cid:18) − π √ − ln 330 (cid:19) r S (5.64) k (2) = (cid:20) − (cid:18) − π √ − ln 315 (cid:19) r + 110 r − (cid:18) − π √ − ln 36 (cid:19) r (cid:21) s + (cid:20) r − (cid:18) − π √ − ln 315 (cid:19) r + 110 r − (cid:18) − π √ − ln 36 (cid:19) r (cid:21) S – 24 – (cid:20) − (cid:18) − π √ − ln 375 (cid:19) r + 3235 r / + 11950 r − (cid:18) − π √ − ln 330 (cid:19) r (cid:21) S . (5.65)For the D(3-2)-brane, we have3( r k (2) ) ′ + 24 r j (2) − (3 r − h ′ (2) = (cid:0) − r + 3 r F + r F ′ (cid:1) s + (cid:0) r + 3 r F + r F ′ (cid:1) S + (cid:20) − r + r F + 83 r F ′ + 13 r F ′′ − (3 r − F F ′ − r F j + 36 r F j − r − F ′ F j − r F ′ F k − r F ′ F ′ k + 24 r F j F k + 6 r F j F ′ k (cid:21) S , (5.66) rh ′′ (2) + 2 h ′ (2) − j ′ (2) = (cid:18) F F ′ + rF F ′′ + 12 rF ′ − F j F ′ j + rF ′ F ′ j (cid:19) S , (5.67)( r k (2) ) ′ + r f j ′ (2) + 8 r j (2) − r f h ′ (2) = (cid:18) − r + 12 r F (cid:19) s + (cid:18) r + 12 r F (cid:19) S + (cid:20) − r + 16 r F + 16 r F ′ − r f F F ′ − r F j + 12 r F j + 3 r f F j F ′ j − r f F ′ F j − r F ′ F k + 8 r F j F k + r F ′ j F k + 2 r F j F ′ k (cid:21) S . (5.68)Note the last differential equation is from the EOM of A . The solutions of the aboveare h (2) = (cid:18) − ln 24 (cid:19) r ( s + S ) + (cid:20) r + (cid:18) − ln 212 (cid:19) r (cid:21) S , (5.69) j (2) = − (cid:18) − ln 28 (cid:19) r ( s + S ) − (cid:18) − ln 224 (cid:19) r S , (5.70) k (2) = (cid:20) − (cid:18) − ln 212 (cid:19) r + 118 r − (cid:18) − ln 28 (cid:19) r (cid:21) s + (cid:20) r − (cid:18) − ln 212 (cid:19) r + 118 r − (cid:18) − ln 28 (cid:19) r (cid:21) S + (cid:20) − (cid:18) − ln 236 (cid:19) r + 415 r + 2354 r − (cid:18) − ln 224 (cid:19) r (cid:21) S . (5.71)The differential equations for the second order scalar perturbations of D(2-1)-branecan be got as7( r k (2) ) ′ + 70 r j (2) − (7 r − h ′ (2) = (cid:18) − r + 7 r F + 2 r F ′ (cid:19) s – 25 – (cid:18) r + 7 r F + 2 r F ′ (cid:19) S + (cid:20) − r + 3 r F + 387 r F ′ + 47 r F ′′ − (7 r − F F ′ − r F j + 105 r F j − r − F ′ F j − r F ′ F k − r F ′ F ′ k + 70 r F j F k + 14 r F j F ′ k (cid:21) S , (5.72)2 rh ′′ (2) + 5 h ′ (2) − j ′ (2) = (cid:0) F F ′ + 2 rF F ′′ + rF ′ − F j F ′ j + 2 rF ′ F ′ j (cid:1) S , (5.73)( r k (2) ) ′ + r f j ′ (2) + 10 r j (2) − r f h ′ (2) = (cid:18) − r + 12 r F (cid:19) s + (cid:18) r + 12 r F (cid:19) S + (cid:20) − r + 314 r F + 17 r F ′ − r f F F ′ − r F j + 15 r F j + 3 r f F j F ′ j − r f F ′ F j − r F ′ F k + 10 r F j F k + r F ′ j F k + 2 r F j F ′ k (cid:21) S . (5.74)Solving these perturbations is tough because F ( r ) of D2-brane is more complex thanthe other Dp-brane, as can be seen from (3.4). One can simplify the calculation by firstmaking the series expansion with respect to 1 /r and then do the integral when solvingthe differential equations of j (2) and k (2) . The solutions are h (2) =
415 + π s − √ √ √ − ln 510 ! r ( s + S )+ " r + π s − √ √ √ −
370 ln 5 ! r S , (5.75) j (2) = −
215 + π s − √ √ √ − ln 520 ! r ( s + S ) − π s − √ √ √ − ! r S , (5.76) k (2) = (cid:20) −
435 + 3 π s − √ √ √ − ! r + 128 r −
215 + π s − √ √ √ − ln 520 ! r (cid:21) s + (cid:20) r −
435 + 3 π s − √ √ √ − ! r + 128 r – 26 –
215 + π s − √ √ √ − ln 520 ! r (cid:21) S + (cid:20) − π s − √ √ √ − ! r + 32273 r + 2531764 r − π s − √ √ √ − ! r (cid:21) S . (5.77) In this section, we will offer all the results of the second order constitutive relations forthe D(p-q)-brane with 2 ≤ p ≤ ≤ q ≤ p −
1. We will omit D(4-1)-brane casehere as in the previous sections, for its results can be found in [23]. One will find thefollowing substitutions very helpful to gain the final form of the constituent relations: t ij = ∂ σ ij → h D ∂ µ u ν i , T ij = ∂ β i ∂ β j − p − q δ ij s → Du h µ Du ν i , T ij = σ ij ∂β → σ µν ∂ ρ u ρ , T ij = σ k ( i Ω j ) k → σ ρ h µ Ω ν i ρ (6.1)are for the spatial viscous tensors and s + S = ∂ ∂β + ∂ β i ∂ β i → D∂u, S = ( ∂β ) → ( ∂u ) , S = σ ij → σ µν (6.2)are for the spatial viscous scalar terms.The situation of p − q = 2 includes the D(4-2) and D(3-1)-brane cases. For theD(4-2)-brane, one has T µν = 12 κ ( r H L (cid:18) u µ u ν + 12 P µν (cid:19) − (cid:18) r H L (cid:19) (cid:18) σ µν + 35 ∂ ρ u ρ P µν (cid:19) + r H L (cid:20)(cid:18) − π √ − ln 32 (cid:19) · (cid:18) h D σ µν i + 12 σ µν ∂u (cid:19) + (cid:18)
65 + π √ (cid:19) σ µν ∂u − (cid:18) π √ (cid:19) · σ ρ h µ Ω ν i ρ (cid:21) + r H L P µν (cid:20)(cid:18) − π √ − (cid:19) D ( ∂u )+ (cid:18) − π √ − (cid:19) ( ∂u ) + 310 · σ ρλ (cid:21)) . (6.3)With the second order transport coefficients can be read from the above as ητ π = 12 κ (cid:18) − π √ − ln 32 (cid:19) r H L , ητ ∗ π = 12 κ (cid:18)
65 + π √ (cid:19) r H L , – 27 – = − κ (cid:18) π √ (cid:19) r H L , ζ τ Π = 12 κ (cid:18) − π √ − (cid:19) r H L ,ξ = 12 κ r H L , ξ = 12 κ (cid:18) − π √ − (cid:19) r H L . (6.4)The D(3-1)-brane has the second order stress energy tensor as T µν = 12 κ ( r H L (cid:18) u µ u ν + P µν (cid:19) − (cid:18) r H L (cid:19) (cid:18) σ µν + 13 P µν ∂ ρ u ρ (cid:19) + r H L (cid:20) (cid:18) − ln 22 (cid:19) · (cid:18) h D σ µν i + 12 σ µν ∂u (cid:19) + (cid:18)
13 + ln 26 (cid:19) σ µν ∂u − ln 2 · σ ρ h µ Ω ν i ρ (cid:21) + P µν r H L (cid:20) (cid:18) − ln 26 (cid:19) D ( ∂u ) + (cid:18) − ln 218 (cid:19) ( ∂u ) + 112 · σ ρλ (cid:21)) . (6.5)The corresponding second order transport coefficients are ητ π = 12 κ (cid:18) − ln 22 (cid:19) r H L , ητ ∗ π = 12 κ (cid:18)
13 + ln 26 (cid:19) r H L , λ = − κ ln 2 r H L ,ζ τ Π = 12 κ (cid:18) − ln 26 (cid:19) r H L , ξ = 12 κ r H L , ξ = 12 κ (cid:18) − ln 218 (cid:19) r H L . (6.6)When p − q = 1, one has the D(4-3), D(3-2) and D(2-1)-brane. These cases are likethe D1-brane in [25]. The spatial viscous terms do not contain σ ij or Ω ij since thereis only one spatial direction. Thus the results will be much simpler than the cases of p − q = 2. The second order constituent relation for D(4-3)-brane is T µν = 12 κ ( r H L (cid:18) u µ u ν + 12 P µν (cid:19) − (cid:18) r H L (cid:19) P µν ∂ ρ u ρ + r H L P µν (cid:20)(cid:18) − π √ − (cid:19) D ( ∂u ) + (cid:18) − π √ − (cid:19) ( ∂u ) (cid:21)) . (6.7)There are only two transport coefficients at the second order: ζ τ Π = 12 κ (cid:18) − π √ − (cid:19) r H L , ξ = 12 κ (cid:18) − π √ − (cid:19) r H L (6.8)In the D(3-2)-brane case one has T µν = 12 κ ( r H L (cid:18) u µ u ν + P µν (cid:19) − (cid:18) r H L (cid:19) P µν ∂ ρ u ρ – 28 – P µν r H L (cid:20)(cid:18) − (cid:19) D ( ∂u ) + (cid:18) − (cid:19) ( ∂u ) (cid:21) ) , (6.9)whose second order transport coefficients can be read as ζ τ Π = 12 κ (cid:18) − (cid:19) r H L , ξ = 12 κ (cid:18) − (cid:19) r H L . (6.10)The situation of D(2-1)-brane gives us T µν = 12 κ ( r H L (cid:18) u µ u ν + 32 P µν (cid:19) − (cid:18) r H L (cid:19) P µν ∂u + P µν r H L " π s − √ √ √ − ! D∂u + π s − √ √ √ − ! ( ∂u ) , (6.11)with the second order coefficients are ζ τ Π = 12 κ π s − √ √ √ − ! r H L ,ξ = 12 κ π s − √ √ √ − ! r H L . (6.12)All the results listed in this section can be rewritten in a universal form by introduc-ing the Harmonic number and the second order constitutive relations for the Dp-branewith q directions of their world-volume compactified can be recast in the form as T µν = 12 κ p − q +2 ( r − pH L − pp (cid:18) − p u µ u ν + 5 − p P µν (cid:19) − (cid:18) r H L p (cid:19) − p (cid:18) σ µν + 2( p − + 2 q (5 − p )( p − q )(9 − p ) P µν ∂u (cid:19) + r H L p "(cid:18) − p + 17 − p H − p − p (cid:19) · (cid:18) h D σ µν i + 1 p − q σ µν ∂u (cid:19) + (cid:18) p − + 3 q (5 − p )(5 − p )(9 − p ) − ( p − + q (5 − p )(7 − p )(9 − p ) H − p − p (cid:19) σ µν ∂up − q + 15 − p · σ ρ h µ σ ν i ρ + (cid:18) − − p + 27 − p H − p − p (cid:19) · σ ρ h µ Ω ν i ρ – 29 – P µν r H L p "(cid:18) p − + 2 q (5 − p )( p − q )(5 − p )(9 − p ) + 2( p − + 2 q (5 − p )( p − q )(7 − p )(9 − p ) H − p − p (cid:19) D ( ∂u )+ [2( p − + 2 q (5 − p )][(3 p − p + 18) + 3 q (5 − p )]( p − q ) (5 − p )(9 − p ) + (5 − p )[2( p − + 2 q (5 − p )]( p − q )(7 − p )(9 − p ) H − p − p ! ( ∂u ) + ( p − + q (5 − p )( p − q )(5 − p )(9 − p ) · σ αβ . (6.13)from which one can read all the second order transport coefficients as ητ π = 12 κ p − q +2 (cid:18) − p + 17 − p H − p − p (cid:19) r H L p ,ητ ∗ π = 12 κ p − q +2 (cid:20) p − + 3 q (5 − p )(5 − p )(9 − p ) − ( p − + q (5 − p )(7 − p )(9 − p ) H − p − p (cid:21) r H L p ,λ = 12 κ p − q +2 − p r H L p , λ = 12 κ p − q +2 (cid:18) − − p + 27 − p H − p − p (cid:19) r H L p ,ζ τ Π = 12 κ p − q +2 " p − + 2 q (5 − p )( p − q )(5 − p )(9 − p ) + 2( p − + 2 q (5 − p )( p − q )(7 − p )(9 − p ) H − p − p r H L p ,ξ = 12 κ p − q +2 ( p − + q (5 − p )( p − q )(5 − p )(9 − p ) r H L p ,ξ = 12 κ p − q +2 " [2( p − + 2 q (5 − p )][(3 p − p + 18) + 3 q (5 − p )]( p − q ) (5 − p )(9 − p ) + (5 − p )[2( p − + 2 q (5 − p )]( p − q )(7 − p )(9 − p ) H − p − p r H L p . (6.14)We have added the result of D(4-1)-brane into the above formulas, that’s why theviscous tensor related with λ appears. The definition for the Harmonic number withits special value for the cases of Dp-brane with 1 ≤ p ≤ q = 0 and p = 4 , q = 1. Also note that among the above 7 dynamical second order transportcoefficients, ητ ∗ π , ζ τ Π , ξ , ξ relate with the compactified dimensions of the world-volume q while ητ π , λ , λ do not. The reason is that the fluid flow of the formerfour coefficients are scalar modes, thus should be affected by the change of scalarperturbation. Whereas the latter’s are purely tensor modes.– 30 – − p (cid:16) π − p (cid:17) − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p )5 − p Λ ( p − − p p − p (cid:16) π − p (cid:17) − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p )5 − p Λ ( p − − p η (cid:16) π − p (cid:17) − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p − p Λ ( p − − p ζ p − +2 q (5 − p )( p − q )(9 − p ) (cid:16) π − p (cid:17) − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p − p Λ ( p − − p ητ π (cid:16) − p + − p H − p − p (cid:17) − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p Λ ( p − − p ητ ∗ π h p − +3 q (5 − p )(5 − p )(9 − p ) − ( p − + q (5 − p )(7 − p )(9 − p ) H − p − p i − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p Λ ( p − − p λ − p − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p Λ ( p − − p λ (cid:16) − − p + − p H − p − p (cid:17) − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p Λ ( p − − p ζ τ Π h p − +2 q (5 − p )( p − q )(5 − p )(9 − p ) + p − +2 q (5 − p )( p − q )(7 − p )(9 − p ) H − p − p i − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p Λ ( p − − p ξ p − + q (5 − p )( p − q )(5 − p )(9 − p ) 2 − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p Λ ( p − − p ξ h [2( p − +2 q (5 − p )][(3 p − p +18)+3 q (5 − p )]( p − q ) (5 − p )(9 − p ) + (5 − p )[2( p − +2 q (5 − p )]( p − q )(7 − p )(9 − p ) H − p − p i × − p )5 − p π − p − p (7 − p ) − p − p Γ (cid:0) − p (cid:1) − p λ p − − p p +1 N V q T − p Λ ( p − − p Table 2 . The dual field theory reformulation of the results of compactified Dp-brane. Here V q is the volume of the compact dimensions of Dp-brane. λ and ξ for the compactifiedDp-brane are both zero. One can also reformulate the results of this work in terms of field theory quantities,which can be found in Table 2.As one can check, the results of (6.14) satisfy the Haack-Yarom relation 4 λ + λ =– 31 – ητ π [7], the Romatschke relations [21] τ π = τ Π , ξ = 1 p − q [1 − ( p − q ) c s ] λ (6.15)and also the Kleinert-Probst relations [26] ητ ∗ π = (cid:0) − ( p − q ) c s (cid:1) (4 λ − ητ π ) ,ξ = 2( p − q ) (cid:0) − ( p − q ) c s (cid:1) (cid:2) (1 − p − q ) c s )2 λ + ( p − q ) c s ητ π (cid:3) . (6.16)Since p − q is the spatial dimension of the dual relativistic fluid, if we denote d as thedimension of the relativistic fluid, then we have p − q = d −
1. So the above Romatschkeand Kleinert-Probst relations still have the same form as in [25].
In this paper, we derive all the second order dynamical transport coefficients for Dp-brane with q directions of their world-volume compactified. To be more specific, thesituations considered in this paper include the D4-brane with 2 and 3 directions of itsworld-volume compactified, D3-brane with 1 and 2 directions compactified, as well asD2-brane with 1 direction compactified. This work can be seen as a generalization of[25] which considers only the Dp-brane, thus by setting q = 0 one can reproduce allthe results of [25]. Through [22, 23, 25] and this work, we have finished the calculationfor all the dynamical second order transport coefficients for Dp-brane with or withoutcompactified dimension(s).Through the calculation we can see that the compactification of the world-volume ofDp-brane only affect the scalar perturbation h , which can be seen as the trace part of thetensor perturbation. This causes expressions of ητ ∗ π , ζ τ Π , ξ , ξ depend on the numberof compact dimension q since they relate with the viscous scalars thus are sensitiveto the spatial dimensions that fluids live in. While the results of ητ π , λ , λ do notcontain q since they relate to viscous tensors and are not affected by compactification.Just like Dp-brane, the second order transport coefficients for the compactifiedDp-brane also satisfy the identities like Haack-Yarom, Romatschke and Kleiner-Probstrelations. The dispersion relations of the D(p-q)-brane do not change compared withDp-brane case. Thus one may conclude that the dispersion relations are not affectedby compactification.We know from [25] that near-extremal black D3-brane will lead to the asymptoti-cally AdS black hole after integrating out the unit 5-sphere. Keep on compactifyingone or more world-volume directions of the near-extremal black D3-brane equals to– 32 –ompactify the AdS black hole. From the results of the compactified D3-brane in thiswork, one can see that compactification on AdS black hole can lead to nonconformalresults for the dual fluid’s transport coefficients. This reminds us that we can alsoget nonconformal transport coefficients from compactified AdS black holes in otherdimensions. This calculation may give a direct check on the method of obtaining non-conformal hydrodynamical stress-energy tensor from a conformal one that is proposedin [20]. Acknowledgement
C. Wu would like to thank Yu Lu for discussions. This work is supported by theYoung Scientists Fund of the National Natural Science Foundation of China (GrantNo. 11805002).
A Christoffel symbol and Ricci quantities of the reductionansatz
The reduction ansatz used in this work is ds = e α A g MN dx M dx N + e α A (cid:0) e β B δ mn dy m dy n + e β B L p d Ω − p (cid:1) . (A.1)We separately denote e Γ ˆ M ˆ N ˆ P and Γ MNP as the Christoffel symbols in 10 dimension and p − q +2 dimensional reduced theory. Ω Γ abc is the Christoffel symbol on 8 − p dimensionalunit sphere. The Christoffel symbols of the reduction ansatz can be listed as e Γ MNP = Γ
MNP + α ( δ MN ∂ P A + δ MP ∂ N A − g NP ∇ M A ) , e Γ Mmn = − ( α ∇ M A + β ∇ M B ) e ( − α +2 α ) A +2 β B δ mn , e Γ nMm = ( α ∂ M A + β ∂ M B ) δ nm , e Γ Mab = − ( α ∇ M A + β ∇ M B ) e ( − α +2 α ) A +2 β B γ ab L p , e Γ aMb = ( α ∂ M A + β ∂ M B ) δ ab , e Γ abc = Ω Γ abc . (A.2)The following relations are useful in the calculation: e Γ NMN = Γ
NMN + ( p − q + 2) α ∂ M A, e Γ ˆ NM ˆ N = Γ NMN + [( p − q + 2) α + (8 − p + q ) α ] ∂ M A + [ qβ + (8 − p ) β ] ∂ M B. (A.3)– 33 –ere e Γ ˆ NM ˆ N = e Γ NMN + e Γ nMn + e Γ aMa . The 10 dimensional Ricci tensors are R MN = R MN − [( p − q ) α + (8 − p + q ) α ] ∇ M ∇ N A − α g MN ∇ P ∇ P A − [ qβ + (8 − p ) β ] ∇ M ∇ N B + (cid:2) ( p − q ) α + 2(8 − p + q ) α α − (8 − p + q ) α (cid:3) ∂ M A∂ N A − (cid:2) ( p − q ) α + (8 − p + q ) α α (cid:3) g MN ( ∂A ) + ( α − α )[ qβ + (8 − p ) β ]( ∂ M A∂ N B + ∂ N A∂ M B ) − α [ qβ + (8 − p ) β ] g MN ∂ P A∂ P B − (cid:2) qβ + (8 − p ) β (cid:3) ∂ M B∂ N B ; (A.4) R mn = − (cid:2) α ∇ A + β ∇ B + (cid:0) ( p − q ) α α + (8 − p + q ) α (cid:1) ( ∂A ) + (( p − q ) α β + (8 − p + 2 q ) α β + (8 − p ) α β ) ∂A∂B + (cid:0) qβ + (8 − p ) β β (cid:1) ( ∂B ) (cid:3) e ( − α +2 α ) A +2 β B δ mn ; (A.5) R ab = (7 − p ) γ ab − (cid:2) α ∇ A + β ∇ B + (cid:0) ( p − q ) α α + (8 − p + q ) α (cid:1) ( ∂A ) + (( p − q ) α β + qα β + (16 − p + q ) α β ) ∂A∂B + (cid:0) qβ β + (8 − p ) β (cid:1) ( ∂B ) (cid:3) e ( − α +2 α ) A +2 β B γ ab L p . (A.6)From the above we can get the Ricci scalar as R = (7 − p )(8 − p ) L p e − α A − β B + e − α A (cid:2) R − p − q + 1) α + (8 − p + q ) α ) ∇ A − qβ + (8 − p ) β ) ∇ B − (cid:0) ( p − q )( p − q + 1) α + 2( p − q )(8 − p + q ) α α +(8 − p + q )(9 − p + q ) α (cid:1) ( ∂A ) − q ( q − p ) α β + (8 − p )( p − q ) α β + q (9 − p + q ) α β + (8 − p )(9 − p + q ) α β ) ∂A∂B − (cid:0) q ( q + 1) β + 2 q (8 − p ) β β + (8 − p )(9 − p ) β (cid:1) ( ∂B ) (cid:3) . (A.7) References [1] S. Bhattacharyya, V. E. Hubeny, S. Minwalla, and M. Rangamani,
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