Second order transport from anomalies
aa r X i v : . [ h e p - t h ] D ec Preprint typeset in JHEP style - HYPER VERSION
Second order transport from anomalies
Sayantani Bhattacharyya, a Justin R. David b and Somyadip Thakur b . a Physics Department,Ramkrishna Mission Vivekananda University,Belur Math, Howrah 711202, India. b Centre for High Energy Physics, Indian Institute of Science,C.V. Raman Avenue, Bangalore 560012, India. [email protected], justin,[email protected]
Abstract:
We study parity odd transport at second order in derivative expansionfor a non-conformal charged fluid. We see that there are 27 parity odd transportcoefficients, of which 12 are non-vanishing in equilibrium. We use the equilibriumpartition function method to express 7 of these in terms of the anomaly, shear viscos-ity, charge diffusivity and thermodynamic functions. The remaining 5 are constrainedby 3 relations which also involve the anomaly. We derive Kubo formulae for 2 of thetransport coefficients and show these agree with that derived from the equilibriumpartition function. ontents
1. Introduction 1
2. Anomalous transport from equilibrium partition function 7
3. Kubo formula for the transport coefficients Φ , Φ
1. Introduction
Fluid dynamics is an effective description of near-equilibrium systems. Propertiesof a fluid system are always slowly varying compared to some intrinsic length scale,for example a mean free path. This length scale is determined by the details ofthe underlying microscopic theory. The fundamental variables of fluid dynamics arelocal velocities u µ ( x ), temperature T ( x ), and all the other conserved charges or theirchemical potentials µ a ( x ). The conservation equations for the stress tensor T µν andthe other conserved currents J aµ govern the time evolution of fluid dynamics.The stress tensor and the conserved currents are related to the fluid variables { u µ , T, µ a } via constitutive relations. Since fluid systems are always slowly varying itis appropriate to organize the constitutive relations in terms of a derivative expansionof the fluid variables. At every order in the derivative expansion, the independentterms of the constitutive relation are constructed out of the independent derivativesof the fluid variables. The independent terms in the constitutive relation are mul-tiplied by coefficients which are functions of temperature and chemical potentials.These coefficients are called transport coefficients. In this paper we will study thetransport coefficients that occur in the parity odd sector, at second order in the– 1 –erivative expansion. These are terms constructed out of various derivatives in thefluid variables which are odd under parity. We will consider relativistic fluid systemswith one additional conserved current.It is usually difficult to compute transport coefficients from the microscopic the-ory and they are generically determined from experiments. However parity oddtransport coefficients which occur at the first order in the derivative expansion havebeen related to quantum anomalies of the microscopic theory [1, 2, 3, 4, 5, 6, 7, 8] .Our goal is to see if a similar phenomenon occurs at the second order in the deriva-tive expansion of the constitutive relations. One motivation to consider second orderfluid dynamics is that first order fluid dynamics is known to have problems withcausality and numerical stability. Transport coefficients which occur at second orderprovide important constraints for spectral densities through sum rules [10]. Par-ity odd transport coefficients at second order affect the dispersion relation of chiralmodes [11] . This phenomenon has important experimental consequences like thespatial separation of particles of different chirality. In the relativistic context thisphenomenon was first observed holographically [12] and then understood due to thepresence of a parity odd transport coefficient at second order by [13].Parity odd transport coefficients at second order has been studied earlier in [13]for conformal fluids. They used the principle that parity odd terms which are evenunder time-reversal invariance should not contribute to local entropy production.With this principle they could constrain these transport coefficients and determinesome of them. We will use the method developed in [14, 15] to determine andconstrain the parity odd transport coefficients. We consider non-conformal fluidsin 3 + 1 dimensions which admits one anomalous charge current. This method isbased on the requirement that the fluid equations have to be consistent with theexistence of an equilibrium partition function. Therefore the approach first relies onthe physical requirement of the existence of equilibrium. More precisely: • In a time independent background, that is a space-time metric with a time likeKilling vector and background gauge fields independent of the time direction,any fluid equation will admit a time independent solution.The second assumption is: • The stress tensor and the charge current evaluated on this time independentsolution can be obtained from the partition function by varying it with respectto the background metric and the gauge field.This method is implemented as follows. See [9] for a recent review with a complete list of references. We thank Yashodhan Hatwalne for bringing this reference to our attention. – 2 –. We first classify all the parity odd transport coefficients till the second orderin the derivative expansion of the stress tensor and the charge current usingsymmetries.2. We then evaluate the stress tensor and the charge current on the equilibriumfluid configuration to the second order in the derivative expansion.3. The equilibrium partition function is written to the second order in derivativeexpansion taking all the parity odd terms into consideration. This is also donebased on symmetries4. This stress tensor and the charge current obtained from the equilibrium par-tition function is required to agree with that obtained from the stress tensorevaluated on the equilibrium fluid configuration.From the description of the method it is clear that only transport coefficientswhich do not vanish in the equilibrium fluid configuration will be constrained ordetermined. We will see that in total there are 27 parity odd transport coefficients.Out of these 12 do not vanish in the equilibrium fluid configuration. Among the12, we determine 7 which we label as Φ i , i = 1 , · · · . and show that theyare related to the anomaly. The rest of the 5 are constrained by 3 relations. Theserelations also involve the anomaly. The results are summarized in (1.9) and (1.10).We will then derive Kubo formulae for two of the transport coefficients Φ , Φ andshow that it agrees with the equilibrium partition function method. The remainingtransport coefficients seem to be related to three point functions.The organization of the paper is as follows. In the rest of the introduction wesummarize our main results. In section 2 we implement the method of [14] to relatethe transport coefficients which do not vanish in equilibrium to the anomalies. Insection 3 we use the Kubo formalism to derive two of the transport coefficients. Insection 4 we study the effects of the second order transport coefficients on linearizeddispersion relations. We also verify the relation between the transport coefficientΦ and the anomaly coefficient obtained using the holographic evaluation of thistransport coefficient for the case of N = 4 Yang-Mills. Using holography we alsoshow that for the conformal case of N = 4 Yang-Mills the transport coefficient Φ vanishes. In appendix A we discuss some of the details involving the classificationof the parity odd data at second order in derivatives. In appendix B we show theconsistency of the velocity profile used to derive the Kubo formulae for transportcoefficients. As we mentioned earlier, the aim of this note is to constrain the parity odd sec-ond order transport coefficients of an anomalous charged fluid in the presence of– 3 –ackground electric and magnetic fields. To define the transport coefficients unam-biguously we first must have an unambiguous definition of fluid variables, that is thevelocities, the temperature and the chemical potentials. { u µ , T, µ } . We will work inLandau frame which is defined by the following two conditions for the charged fluid. J µ u µ = − q, T µν u µ = − Eu ν . (1.1)Let us now consider the expansion of the charge current J µ and the stress tensor T µν in terms of the number of space time derivatives. This is given by J µ = J µ (0) + J µ (1) + · · · , T µν = T µν (0) + T µν (1) + · · · , (1.2)where the subscript ( i ) refer to the number of space-time derivatives. The terms J µ ( i ) and T µν ( i ) for i = 0 are all perpendicular to the velocity u µ . The equations ofmotion for the fluid in the presence of external electromagnetic field are given by thefollowing conservation laws ∇ µ T µν = F νµ J µ + c m ∇ µ [ ǫ αβγδ F αβ R µνγδ ] , (1.3) ∇ µ J µ = − C ǫ αβγδ F αβ F γδ + c m ǫ αβγδ R µναβ R νµγδ = CE µ B µ + c m ǫ αβγδ R µναβ R νµγδ . Here E µ = F µν u ν , B µ = ǫ µναβ u ν F αβ and C is the gauge anomaly coefficient and c m is the coefficient of the mixed gauge-gravitational anomaly. Note that the termsproportional to the mixed anomaly are fourth order in derivatives, therefore they donot affect the analysis of the equations of motion to 2nd order in derivatives. How-ever the gravitational anomaly does enter the discussion of the equilibrium partitionfunction at the first order in the derivative expansion [16].We will now state the known results for the form of the stress tensor and thecurrent up to first order in the derivative expansion. For i = 0, the part withno space-time derivative the stress tensor and current is completely determined bythermodynamics. At first order in derivative expansion, that is i = 1, the form ofthe current and stress tensor is explicitly known [14], [1]. This form is consistentwith all physical requirements in the presence of anomaly as well as an externalelectromagnetic field [14], [1]. The final result for the stress tensor and the chargecurrent to first order in derivatives is given by[ T (0) ] µν = ( E + P ) u µ u ν + P G µν , [ T (1) ] µν = − ησ µν − ζ Θ P µν ,J µ (0) = qu µ ,J µ (1) = ∆ V µ + ξ l l µ + ξ B B µ . (1.4)– 4 –ere E is the energy density, P the pressure and q , the charge density and G µν is the background metric. The variables Θ , σ µν , V µ , l µ and B µ are all on-shellindependent terms which are first order in derivatives. These are defined in table 1.The variables η, ζ , ∆ , ξ l and ξ B refer to the first order transport coefficients.Scalars Vectors Pseudo Vectors Tensors(1) (3) (2) (1)Θ = ∇ µ u µ u µ ∇ µ u ν l µ = ǫ µναβ u ν ∂ α u β σ µν = ∇ h µ u ν i P µν ∇ ν ν B µ = ǫ µναβ u ν F αβ V µ = (cid:0) E µ T − P µν ∇ ν ν (cid:1) Table 1:
Data at 1st order in derivative
Throughout this paper, the symbol A h µν i on any tensor A µν denotes the projected,traceless, symmetric part of the tensor. A h µν i = P αµ P βν (cid:18) A αβ + A βα − P γθ A γθ G αβ (cid:19) , (1.5)and P µν is the projector P µν = u µ u ν + G µν . (1.6)The variable ν is related to the chemical potential by ν = µT . (1.7)Let us now proceed to the stress tensor and the charge current at second orderin derivatives which we denote as [ T (2) ] µν and J µ (2) . Purely from symmetry consider-ations, the number of independent transport coefficients upto second order is equalto total number of possible scalars which appear in the trace of the stress tensor,together with the number of possible vectors which appear in the current and thepossible symmetric traceless tensors which appear in the traceless part of the stresstensor. But not all of them are independent, they can be related using the equationsof motion given in (1.3). In Table 2 we have listed all the parity odd and on-shell in-dependent scalars, vectors and tensors containing two space-time derivatives. Someof the details that went into this classification is discussed in appendix A.From the table it can be seen that at second order in derivatives there are 27parity odd transport coefficients which appear in the current and the stress tensor.Therefore the most general parity odd contributions at second order in the stress– 5 –seudo-scalars Pseudo Vectors Pseudo-tensors(6) (9) (12) S = l µ ∂ µ ν V µ (1) = ǫ µναβ u ν B α l β τ (1) µν = ∇ h µ l ν i S = B µ ∂ µ ν V µ (2) = ǫ µναβ u ν ( ∂ α ν )( ∂ α T ) τ (2) µν = ∇ h µ B ν i S = l µ ∂ µ T V µ (3) = Θ l µ τ (3) µν = l h µ ∂ ν i ν S = B µ ∂ µ T V µ (4) = ǫ µνλσ u ν ∇ λ V σ τ (4) µν = B h µ ∂ ν i ν S = l µ V µ V µ (5) = σ µν l ν τ (5) µν = l h µ ∂ ν i T S = B µ V µ V µ (6) = Θ B µ τ (6) µν = B h µ ∂ ν i T V µ (7) = σ µν B ν τ (7) µν = l h µ V ν i V µ (8) = ǫ µναβ u ν ( ∂ α T ) V β τ (8) µν = B h µ V ν i V µ (9) = ǫ µναβ u ν ( ∂ α ν ) V β τ (9) µν = u θ σ h µα ( ∂ β T ) ǫ θαβν i τ (10) µν = u θ σ h µα ( ∂ β ν ) ǫ θαβ ν i τ (11) µν = u θ σ h µα V β ǫ θαβν i τ (12) µν = u θ ∇ β σ h µα ǫ θαβ ν i Table 2:
Parity odd data at 2nd order in derivatives tensor and current can be parametrized as follows.[ T (2) ] µν = X i =1 Φ i τ ( i ) µν + P µν " X i =1 χ i S i ,J µ (2) = X i =1 ∆ i V µ ( i ) . (1.8)Where τ ( i ) µν , S i and V µ ( i ) are defined in Table 2. Our goal is to constrain the transportcoefficients Φ i , χ i and ∆ i , using the existence of an equilibrium partition function.It can be shown among these 27 terms only 12 can be non-zero in a time independentequilibrium fluid configuration. These are the first 4 in the list of scalars S · · · S , thefirst 2 in the list of vectors V µ (1) , V µ (2) and the first 6 in the list of tensors τ (1) µν · · · τ (6) µν .Therefore the analysis using the equilibrium partition function can be used to con-strain the 12 transport coefficients multiplying these non-vanishing terms. These are χ · · · χ and ∆ , ∆ and Φ · · · Φ . The final result of this analysis is the following.Φ = η b , Φ = 2 η b , Φ = η (cid:18) ∂b ∂ν (cid:19) , Φ = 2 η (cid:18) ∂b ∂ν (cid:19) , Φ = η (cid:20) − b T + ∂b ∂T (cid:21) , Φ = 2 η (cid:20) − b T + ∂b ∂T (cid:21) , (1.9)– 6 – = − ∆ b ,T R (cid:20) χ − ζ (cid:18) ∂b ∂T − b T (cid:19)(cid:21) − R (cid:20) χ − ζ (cid:18) ∂b ∂ν − b T (cid:19)(cid:21) = 0 ,T R (cid:20) χ − ζ (cid:18) ∂b ∂T − b T (cid:19)(cid:21) + R (cid:20) χ − ζ (cid:18) ∂b ∂ν (cid:19)(cid:21) = 0 ,R T ∆ + (cid:20) χ − ζ (cid:18) ∂b ∂ν (cid:19)(cid:21) − q ( E + P ) (cid:20) χ − ζ (cid:18) ∂b ∂ν − b T (cid:19)(cid:21) = 0 , (1.10)where b = T E + P (cid:18) Cν − C ν (cid:19) , b = T E + P (cid:18) Cν − C (cid:19) ,R = (cid:18) ∂P∂E (cid:19) q , R = (cid:18) ∂P∂q (cid:19) E . Note that C is the gauge anomaly coefficient and C is related to the coefficient ofthe mixed gauge-gravitational anomaly[16] by C = 8 π c m . (1.11)Therefore we see that the coefficients Φ , · · · Φ and ∆ are determined in terms ofthe anomaly, the shear viscosity η , the charge diffusivity ∆ and the thermodynamicfunctions E, P, T, ν . The rest ∆ , χ , · · · χ are constrained by 3 relations whichinvolve the anomaly and the bulk viscosity ζ .
2. Anomalous transport from equilibrium partition function
In sub-section 2.1 we will briefly outline the general procedure we use to relate parityodd transport at the second order in derivatives to the anomaly. This method hasbeen described and analyzed in detail by [14] and [15]. It relies on the analysis ofthe equilibrium partition function. Thus we will be able to constrain only thosetransport coefficients which do not vanish in the equilibrium configuration. Thissection also will serve to introduce the notation and conventions used in the paper.In sub-section 2.2 we implement this method and derive the relations given in (1.9)and (1.10).
We are interested in a fluid flow on a static background metric and a static externalelectromagnetic field. The most general static metric and gauge field can be written The relation (1.11) is derived in [16] using properties of the partition function on cones. It isan equation that relates coefficients at different orders in derivative expansion. In our analysis weshall simply assume their result. We will subsequently see that this identification is also consistentwith the analysis of [9] which studies the effect of gravitational anomalies on hydrodynamics. – 7 –n the following form. ds = G µν dx µ dx ν = − e σ ( dt + a i dx i ) + g ij dx i dx j , Gauge Field : A µ dx µ , (2.1) σ , a i , g ij , A and A µ are all slowly varying functions of the spatial co-ordinates ( ~x ).Our notations are as follows: • Greek indices run from 1 to 4. • Latin indices run from 1 to 3. • All Greek indices are lowered or raised by the 4 dimensional metric G µν unlessexplicitly mentioned. • All Latin indices are lowered or raised by the 3 dimensional metric g ij unlessexplicitly mentioned. • ¯ ∇ µ is covariant derivative with respect to the metric G µν and ∇ i is the covariantderivative with respect to the metric g ij . • For any tensor A µν the notation A h µν i denotes the traceless symmetric part ofthe tensor, projected in direction perpendicular to the fluid velocity. A h µν i = P αµ P βν (cid:18) A αβ + A βα − P γθ A γθ G αβ (cid:19) , where P µν ≡ u µ u ν + G µν is referred to as the projector.Our basic assumption is that in such a background any fluid equation will admita time independent solution. The stress tensor and the current, evaluated on thistime independent solution, can be generated by varying the partition function of thesystem with respect to the background metric and gauge field. If Z is the partitionfunction at temperature T then the stress tensor and the current evaluated on theequilibrium are given by the following formulae. T | equilibrium = − T e σ √− G (cid:20) δZδσ (cid:21) = T √ g (cid:20) δZδ ¯ T (cid:21) ,T i | equilibrium = T √− G (cid:20) δZδa i − A δZδA i (cid:21) = ¯ T √ g (cid:20) δZδa i − A δZδA i (cid:21) ,T ij | equilibrium = − T √− G g il g jm (cid:20) δZδg lm (cid:21) = − T √ g g il g jm (cid:20) δZδg lm (cid:21) ,J | equilibrium = − T e σ √− G (cid:20) δZδA (cid:21) = − e σ √ g (cid:20) δZδ ¯ ν (cid:21) ,J i | equilibrium = T √− G (cid:20) δZδA i (cid:21) = ¯ T √ g (cid:20) δZδA i (cid:21) , (2.2)– 8 –here ¯ T = T e − σ , ¯ ν = A T , A i = A i − A a i . The strategy we will adopt to constrain the parity odd coefficients which occurat the second order is the following:1. Write down the most general partition function upto a given order in derivativeexpansion and consistent with all the symmetries. It will be a function of a i , σ and A i and their derivatives.2. Vary the partition function Z , to obtain the most general possible expressionfor T µν | equilibrium and J µ | equilibrium .3. Parametrize the most general possible fluid stress tensor and current up to somegiven order in derivative expansion using symmetries. This will give the max-imum number of independent transport coefficients possible constrained onlyby symmetry. We have stated the results of this analysis already in equations(1.4) and (1.8).4. Evaluate the most general fluid stress tensor and current on the equilibriumsolution. The final outcome will contain some of the unknown transport coef-ficients.5. Equate the final outcome of the previous step with the stress tensor and currentwe have already obtained by varying the partition function.6. This will express the transport coefficients which appear in the stress tensorand the current evaluated in equilibrium, in terms of the coefficients appearingin the partition function.7. Eliminating the coefficients which appear in the partition function results inthe desired relations among the transport coefficients.One might wonder that to execute the fourth step, the equilibrium solution forthe velocity, temperature and the other fluid variables needs to be independentlyfound. But as it has been explained in [14] using this method one can perturba-tively determine both the solution and the transport coefficients in terms of the freefunctions appearing in the partition function. In this subsection we shall apply the general procedure described in 2.1 to the par-ticular case of the parity-odd sector of the charged fluid at second order in derivativeexpansion. The first contribution to the parity-odd sector comes at first order inderivative expansion. This has been analyzed in detail in section 3 of [14]. We willnot repeat the first order computation here but we will extensively use their result.– 9 – tress tensor and current from the partition function
Since the transport coefficients we are interested in belong to the parity-odd sector,we will restrict our attention to only the parity odd part of the partition function.The partition function Z (2) at second order in derivatives is a gauge invariant scalarfunctional of the background metric and gauge-fields. Hence we need to list allpossible parity odd scalars that can be constructed from the metric functions andgauge fields that contain two space-derivatives. Note that since all the functions aretime independent no time derivatives occur. There are four such scalars:1. ǫ ijk ∂ i ¯ νf jk ,2. ǫ ijk ∂ i ¯ T f jk ,3. ǫ ijk ∂ i ¯ νF jk ,4. ǫ ijk ∂ i ¯ T F jk ,where ¯ ν = A T , ¯ T = T e − σ , A i = A i − A a i and f jk = ∂ j a k − ∂ k a j , F jk = ∂ j A k − ∂ k A j .Therefore naively, the parity odd second order partition function at two deriva-tive order can have 4 free parameters, but two of them can be related by totalderivatives. Ignoring the total derivative terms the most general second order parti-tion function can be written as Z (2) = Z √ g (cid:2) M ( ¯ T , ¯ ν ) ǫ ijk ∂ i ¯ νF jk + T M ( ¯ T , ¯ ν ) ǫ ijk ∂ i ¯ νf jk (cid:3) . Varying this partition function with respect to the metric and the gauge fields andusing (2.2) we get the following second order correction to the stress tensor andcharge current.[Π (2) ] = T (cid:20)(cid:18) ∂M ∂ ¯ T (cid:19) ǫ ijk ∂ i ¯ νF jk + T (cid:18) ∂M ∂ ¯ T (cid:19) ǫ ijk ∂ i ¯ νf jk (cid:21) , [Π (2) ] i = 2 T ¯ T (cid:18) ∂M ∂ ¯ T − ¯ ν ∂M ∂ ¯ T (cid:19) ǫ ijk ( ∂ j ¯ T )( ∂ k ¯ ν ) , [Π (2) ] ij = 0 , [ j (2) ] = T ¯ T (cid:20)(cid:18) ∂M ∂ ¯ T (cid:19) ǫ ijk ∂ i ¯ T F jk + T (cid:18) ∂M ∂ ¯ T (cid:19) ǫ ijk ∂ i ¯ T f jk (cid:21) , [ j (2) ] i = 2 ¯ T (cid:18) ∂M ∂ ¯ T (cid:19) ǫ ijk ( ∂ j ¯ T )( ∂ k ¯ ν ) . (2.3)It is important to note the third equation in (2.3). The fact that the spatial compo-nents of the equilibrium stress tensor vanish at second order will serve as an importantconstraint in determining the transport coefficients. Though the system is anomalous, all the effects of anomaly i.e. the anomalous transformationproperty of the partition function under the gauge transformation is accounted by the first orderpart Z (1) . Therefore in Z (2) we need to consider only gauge invariant scalars. – 10 – tress tensor from fluid dynamics We have to evaluate the fluid dynamical stress tensor on the equilibrium, that isa time independent solution in the given static background metric and gauge field.This equilibrium solution for the velocity field, temperature or chemical potential interms of the background metric and gauge field can also be expanded in terms ofderivatives. We shall use the following notation. u µ | eq = ¯ u µ + δu µ (1) + δu µ (2) + · · · ,T | eq = ¯ T + δT (1) + δT (2) + · · · ,ν | eq = ¯ ν + δν (1) + δν (2) + · · · , (2.4)where δu µ ( i ) , δT ( i ) and δν ( i ) are i th corrections to the zeroth order equilibrium solutioncontaining i derivatives on the background data. Now we have to substitute (2.4) influid stress tensor and current given in (1.4)) and extract the part that will be parityodd and which involve exactly two derivatives on the background data. This is thepart which have to be equated with (2.3).From the analysis done in [14] we know that ¯ u µ = e − σ { , , , } , ¯ T = T e − σ , ¯ ν = A T , [ δu (1) ] = 0 , δT (1) = 0 , δν (1) = 0 , [ δu (1) ] = − a i [ δu (1) ] i , [ δu (1) ] i = (cid:18) b (cid:19) ¯ l i + b ¯ B i , [ δu (1) ] i = g ij [ δu (1) ] j , (2.5)where F jk ≡ ∂ j A k − ∂ k A j , ¯ l i = − e σ ǫ ijk f jk , ¯ B i = 12 (cid:0) ǫ ijk F jk + A ǫ ijk f jk (cid:1) = 12 ǫ ijk F jk − ¯ T ¯ ν ¯ l i ,b = T E + P (cid:18) Cν − C ν (cid:19) ,b = T E + P (cid:18) Cν − C (cid:19) . (2.6)Here C is the anomaly coefficient and C is related to the mixed anomaly in (1.3)by C = 8 π c m . (2.7) Note that equation (2.5) is valid only if we choose Landau frame. This is the place where thechoice of a fluid frame enters our analysis. We are going to use these equations in all our subsequentcalculation. The relation (2.7) was derived in [16]. Here we simply use it. – 11 –f we also expand E , P and q in terms of a derivative expansion as E | eq = ¯ E + δE (1) + δE (2) + · · · , P | eq = ¯ P + δP (1) + δP (2) + · · · , q | eq = ¯ q + δq (1) + δq (2) + · · · , then from (2.5) it follows that δE (1) = δP (1) = δq (1) = 0 . (2.8)Using the fact that ( u µ u µ = −
1) to all order in derivative expansion we find that[ δu (2) ] ∝ [ δu (1) ] i [ δu (1) ] i = Parity even ∼ . (2.9)Also using the Landau gauge condition on the second order stress tensor and currentone can show that in equilibrium[ T (2) ] = [ T (2) ] i = [ J (2) ] = 0 . (2.10)Note that [ T (2) ] , [ T (2) ] i , [ J (2) ] are the components of the stress tensor and the chargecurrent which are second order in derivatives obtained from the equation (1.8). While[Π (2) ] , [Π (2) ] i , [ j (2) ] refer to the components of the stress tensor and the chargecurrent obtained from the equilibrium partition function using (2.3).Using (2.5), (2.8), (2.9) and (2.10) we get the following form for the second orderstress tensor and current evaluated in equilibrium.[Π (2) ] = δE (2) ¯ u + δ ( − ησ ) + δ ( − ζ Θ P ) , [Π (2) ] i = ( ¯ E + ¯ P )¯ u [ δu (2) ] i + δ ( − ησ i ) + δ ( − ζ Θ P i ) , [ j (2) ] = δq (2) ¯ u + δ (∆ V ) + δ ( ξ l l ) + δ ( ξ B B ) , (2.11)where δ ( − ησ µν ), δ ( ζ Θ P µν ) and δ (∆ V µ ) are the two derivative corrections of ( − ησ µν ),( ζ Θ P µν ) and (∆ V µ ) when evaluated on δu µ (1) , δT (1) and δν (1) .Now if Q (1) µν is a tensor which is first order in the derivative expansion satisfyingthe following two conditions Q (1) µν u µ = 0 at all orders and Q (1) µν | (¯ u µ , ¯ T , ¯ ν ) = 0 , (2.12)then one can show in general that δQ (1)00 = δ [ Q (1) ] i = 0 . (2.13)Similarly if Q (1) µ is a vector which is first order in the derivative expansion satisfying Q (1) µ u µ = 0 at all orders and Q (1) µ | (¯ u µ , ¯ T , ¯ ν ) = 0 , (2.14) δ h u µ Q (1) µν i = ¯ u µ δQ (1) µν + δu µ h Q (1) µν | (¯ u α , ¯ T , ¯ ν ) i = e − σ δQ (1)0 ν . – 12 –hen it follows that δQ (1)0 = 0 . (2.15)This argument allows us to conclude that δ ( − ησ ) = δ ( ζ Θ P ) = δ ( − ησ i ) = δ ( ζ Θ P i ) = δ (∆ V ) = 0 . Similarly δ ( ξ l l µ ) and δ ( ξ B B µ ) are the two derivative corrections of ( ξ l l µ ) and ( ξ B B µ )when evaluated on δu µ (1) , δT (1) and δν (1) . But l µ and B µ are already parity odd.Therefore δ ( ξ l l µ ) and δ ( ξ B B µ ) are going to be parity even and hence we can set themto zero for our purpose. Putting all this together we have the following result forthe components of the stress tensor and the current at the second order in derivativeexpansion [Π (2) ] = δE (2) ¯ u = e σ δE (2) , [Π (2) ] i = ( ¯ E + ¯ P )¯ u [ δu (2) ] i = − e σ ( ¯ E + ¯ P )[ δu (2) ] i , [ j (2) ] = − e σ δq (2) . (2.16)Inverting (2.16) and using (2.3) we can determine the second order piece of the equi-librium solution for velocity, temperature and chemical potential in terms of back-ground data. Now the components [Π (2) ] ij and [ j (2) ] i will give rise to the constraintson the transport coefficients.[Π (2) ] ij = δP g ij − η δσ ij − ζ δ Θ g ij + [ T (2) ] ij , [ j (2) ] i = ¯ q [ δu (2) ] i + ∆ δ [ V i ] + [ J (2) ] i . (2.17)We have to use (2.16) to determine δP and [ δu (2) ] i .The transport coefficients are determined by demanding that (2.17) is satisfied.For convenience let us further split the first equation in (2.17) in two parts, the tracepart which is obtained by contracting the equation with g ij and a traceless part whichis obtained by subtracting the trace part of the equation from the first equation of(2.17) . These are given byTrace part : δP − ζ δ Θ + 13 ( − ηδσ ij + [ T (2) ] ij ) g ij = 0 . Traceless part : − η (cid:18) δσ ij − g ij δσ lm g lm ) (cid:19) + (cid:18) [ T (2) ] ij − g ij T (2) ] lm g lm ) (cid:19) = 0 . (2.18)In (2.18) we have used (2.3) to set [Π (2) ] ij to zero. Therefore the RHS of (2.18)vanishes. δ [ u µ Q (1) µ ] = ¯ u µ δQ (1) µ + δu µ [ Q (1) µ | (¯ u α , ¯ T , ¯ ν ) ] = e − σ δQ (1)0 . Note that here trace is taken with the 3 dimensional metric. – 13 – ransport in the traceless part of the stress tensor
In this subsection we shall analyze the second equation of (2.18).Traceless part : − η (cid:18) δσ ij − g ij δσ lm g lm ) (cid:19) + (cid:18) [ T (2) ] ij − g ij T (2) ] lm g lm ) (cid:19) = 0 . (2.19)We now go through the following steps: • We first evaluate the second order contribution from σ ij . • We then write down the most general form for [ T (2) ] h µν i from usual symmetryanalysis and on shell independence. In general it will contain many unknowntransport transport coefficients to begin with. But if we restrict our attentionto only the parity odd ones, the total number of terms is 18 ( Φ i , ( i = 1 , · · · , χ , ( i = 1 , · · · , • One can see that all the 6 χ i s are not going to appear in the combination thatwe are going to evaluate in (2.19). • So we have to evaluate the rest of the 12 terms (multiplying Φ i s on the equi-librium time-independent solution.For this it is sufficient to substitute only the zeroth order equilibrium solution,since each of these terms already contain two derivatives. • As mentioned in section 1.1, it turns out that 6 of these twelve terms evaluateto zero in equilibrium. So there are only 6 transport coefficients which can beconstrained by this equilibrium analysis. These 6 terms are the following. τ (1) µν = ¯ ∇ h µ l ν i , τ (2) µν = ¯ ∇ h µ B ν i , τ (3) µν = ( ∂ h µ ν ) l ν i , (2.20) τ (4) µν = ( ∂ h µ ν ) B ν i , τ (5) µν = ( ∂ h µ T ) l ν i , τ (6) µν = ( ∂ h µ T ) B ν i , where l µ ≡ ǫ µναβ u ν ∂ α u β and B µ ≡ ǫ µναβ u ν ∂ α A β . Let us now rewrite thetraceless part of the second order fluid stress tensor with these six terms, thisis given by[ T (2) ] µνodd = X a =1 Φ a [ τ ( a ) ] µν + trace part + terms that vanish in equilibrium . (2.21) • From (2.19) we see that each of these six Φ a ’s has to be such that, they cancelthe contribution of δσ µν when all of them are evaluated in equilibrium.– 14 – Evaluating the spatial components of these six terms on the zeroth order equi-librium solution given in (2.5) we obtain[ τ (1) ] ij = ¯ ∇ h i l j i = g il g jm (cid:20) ∇ l ¯ l m + ∇ m ¯ l l − g lm ∇ k ¯ l k ) (cid:21) + O ( ∂ ) , [ τ (2) ] ij = ¯ ∇ h i B j i = g il g jm (cid:20) ∇ l ¯ B m + ∇ m ¯ B l − g lm ∇ k ¯ B k ) (cid:21) + O ( ∂ ) , [ τ (3) ] ij = ( ¯ ∇ h µ ν ) l ν i = g il g jm (cid:20) ( ∇ l ¯ ν )¯ l m + ( ∇ m ¯ ν )¯ l l − g lm ∇ k ¯ ν ¯ l k ) (cid:21) + O ( ∂ ) , [ τ (4) ] ij = ( ¯ ∇ h µ ν ) B ν i = g il g jm (cid:20) ( ∇ l ¯ ν ) ¯ B m + ( ∇ m ¯ ν ) ¯ B l − g lm B k ∇ k ¯ ν ¯) (cid:21) + O ( ∂ ) , [ τ (5) ] ij = ( ¯ ∇ h µ T ) l ν i = g il g jm (cid:20) ( ∇ l ¯ T )¯ l m + ( ∇ m ¯ T )¯ l l − g lm l k ∇ k ¯ T ) (cid:21) + O ( ∂ ) , [ τ (6) ] ij = ( ¯ ∇ h µ T ) B ν i = g il g jm (cid:20) ( ∇ l ¯ T ) ¯ B m + ( ∇ m ¯ T ) ¯ B l − g lm B k ∇ k ¯ T ) (cid:21) + O ( ∂ ) , (2.22)where ¯ l i ≡ − e σ ǫ ijk f jk and ¯ B i ≡ ǫ ijk ( F jk + A f jk ). To evaluate equation (2.22)we have extensively used section 2 of [14]. • The spatial components of the shear tensor gives the following contribution to[Π (2) ] ij − η δσ ij = − η (cid:20) ¯ P iµ ¯ P jν (cid:26) ¯ ∇ µ [ δu (1) ] ν + ¯ ∇ ν [ δu (1) ] µ − ¯ ∇ α [ δu (1) ] α (cid:27) + (cid:0) ¯ P jµ δP iν + ¯ P iν δP jµ (cid:1) (cid:18) ¯ ∇ µ ¯ u ν + ¯ ∇ ν ¯ u µ (cid:19) (cid:21) = − η (cid:20) g il g jm (cid:26) ∇ l [ δu (1) ] m + ∇ m [ δu (1) ] l − g lm ∇ k [ δu (1) ] k − g lm ∂ k σ )[ δu (1) ] k (cid:27) + δu i g jk ( ∂ k σ ) + [ δu (1) ] j g ik ( ∂ k σ )2 (cid:21) , (2.23)where ¯ P iµ = ¯ u i ¯ u µ + G iµ = G iµ , ¯ P jµ δP iν = G jµ ¯ u ν [ δu (1) ] i . The last term in the last line of (2.23) results from this expansion of the pro-jectors. Substituting (2.5) in (2.23) we obtain the following − ηδσ ij = − η (cid:20) b τ (1) ] ij + b [ τ (2) ] ij + 12 (cid:18) ∂b ∂ν (cid:19) [ τ (3) ] ij + (cid:18) ∂b ∂ν (cid:19) [ τ (4) ] ij + 12 (cid:18) − b T + ∂b ∂T (cid:19) [ τ (5) ] ij + (cid:18) − b T + ∂b ∂T (cid:19) [ τ (6) ] ij (cid:21) . (2.24)– 15 –ext we have to substitute (2.23) and (2.24) in (2.19). Now to satisfy (2.19) thecoefficient of each independent expression should vanish. From examining (2.23) and(2.24) it seems that demanding every independent term to vanish results in moreequations than the unknowns which are the transport coefficients. However from thestructure of the equations in (2.19) it is clear that the equation admits the followingunique solution for the transport coefficientsΦ = η b , Φ = 2 η b , Φ = η (cid:18) ∂b ∂ν (cid:19) , Φ = 2 η (cid:18) ∂b ∂ν (cid:19) , Φ = η (cid:20) − b T + ∂b ∂T (cid:21) , Φ = 2 η (cid:20) − b T + ∂b ∂T (cid:21) , (2.25)where b = T E + P (cid:18) Cν − C ν (cid:19) , b = T E + P (cid:18) Cν − C (cid:19) , (2.26)and E, P, T is the energy density, pressure and temperature respectively. ν = µT refersto the chemical potential, and η is the shear viscosity. C is the gauge anomaly coef-ficient, while C is related to the mixed anomaly by (1.11). The transport coefficientΦ gives rise to chiral dispersion relations in the shear mode [13]. Transport in the trace of the stress tensor and current
We will now constrain the transport coefficients that occur in the trace part of thestress tensor and the current using the equilibrium partition function. We will firstrewrite the relevant part of the current and the trace part of the stress tensor up to2nd order in derivative expansion. Using (2.3) and (2.17) we obtain the followingequation for the trace part of the stress tensor and current at second order.Trace part : 0 = δP − ζ δ Θ + 13 ( − ηδσ ij + [ T (2) ] ij ) g ij = δP − ζ δ Θ + " X χ i S i eq = " − ζ δ Θ + (cid:18) ∂P∂E (cid:19) q δE (2) + (cid:18) ∂P∂q (cid:19) E δq (2) + " X χ i S i eq (2.27)Current : [ J (2) ] i = [ j (2) ] i − δ [∆ V i ] − q [ δu (2) ] i ⇒ " X k =1 ∆ k V i ( k ) eq = 2 ¯ T (cid:18) ∂M ∂ ¯ T (cid:19) ǫ ijk ( ∂ j ¯ T )( ∂ k ¯ ν ) − δ [∆ V i ] − q [ δu (2) ] i . (2.28)Both in (2.27) and (2.28) we have used (1.8) to write the fluid stress tensor andthe current in terms of the transport coefficients. Also in (2.27) to evaluate thecombination ( − ηδσ ij + [ T (2) ] ij ) g ij we have used (2.24) and (2.25). One can see thatthe transport coefficients Φ i s drop out.– 16 –sing (2.16) and (2.3) we evaluate δE , δq and [ δu (2) ] i , this results in δE = e − σ Π (2)00 = 2 ¯ T (cid:18) ∂M ∂ ¯ T (cid:19) ( ¯ B i ∂ i ¯ ν ) − T (cid:18) ∂M ∂ ¯ T − ¯ ν ∂M ∂ ¯ T (cid:19) (¯ l i ∂ i ¯ ν ) ,δq = e − σ j (2)0 = 2 (cid:18) ∂M ∂ ¯ T (cid:19) ( ¯ B i ∂ i ¯ T ) − T (cid:18) ∂M ∂ ¯ T − ¯ ν ∂M ∂ ¯ T (cid:19) (¯ l i ∂ i ¯ T ) , [ δu (2) ] i = − (cid:18) e − σ ¯ E + ¯ P (cid:19) [Π (2) ] i = − (cid:18) T E + P (cid:19) (cid:18) ∂M ∂ ¯ T − ¯ ν ∂M ∂ ¯ T (cid:19) ǫ ijk ( ∂ j ¯ T )( ∂ k ¯ ν ) . (2.29)Now we have to compute δ Θ and δ (∆ V i ). δ Θ = ¯ ∇ µ [ δu (1) ] µ = ∇ k [ δu (1) ] k − (cid:18) ∂ k ¯ T ¯ T (cid:19) [ δu (1) ] k = 12 (cid:18) ∂b ∂ ¯ T − b ¯ T (cid:19) (¯ l i ∂ i ¯ T ) + 12 (cid:18) ∂b ∂ ¯ ν − b ¯ T (cid:19) (¯ l i ∂ i ¯ ν )+ (cid:18) ∂b ∂ ¯ T − b ¯ T (cid:19) ( ¯ B i ∂ i ¯ T ) + (cid:18) ∂b ∂ ¯ ν (cid:19) ( ¯ B i ∂ i ¯ ν ) . (2.30)To derive (2.30) we have used the following identities ∇ i ¯ l i = − (cid:18) ∂ i ¯ T ¯ T (cid:19) ¯ l i , ∇ i ¯ B i = − ¯ T ¯ l i ∂ i ¯ ν. (2.31)Similarly δV i is given by the following expression δ { ∆ V i } = ∆ G iµ F µν δu ν = ∆ ǫ ijk [ δu (1) ] j ¯ B k = ∆ (cid:18) b (cid:19) ǫ ijk ¯ l j ¯ B k , (2.32)where F µν = ∂ µ A ν − ∂ ν A µ .Now we have to evaluate the 6 scalars ( S i , i = 1 , · · · ,
6) and the i th compo-nent of the 9 vectors ( V i ( k ) , k = 1 , · · · ,
9) on the equilibrium solution. We needthe equilibrium solution only at zeroth order since all of them already contain twoderivatives. Explicit expressions for all these terms are listed in table 2.As mentioned in section 1.1, only 6 terms, 4 scalars and 2 vectors are non-vanishing on the zeroth order equilibrium solution. Thus the relevant parts of thesecond order current and trace of the stress tensor are given by[ T (2) ] αβ | trace part = P αβ [ χ ( l µ ∂ µ ν ) + χ ( B µ ∂ µ ν ) + χ ( l µ ∂ µ T ) + χ ( B µ ∂ µ T )] ,J µ (2) = ∆ [ ǫ µναβ u ν ( ∂ α ν )( ∂ β T )] + ∆ [ ǫ µναβ u ν B α l β ] . (2.33)– 17 –e then evaluate all these six terms on the zeroth order equilibrium solution.( l µ ∂ µ ν ) = (¯ l i ∂ i ¯ ν ) , ( B µ ∂ µ ν ) = ( ¯ B i ∂ i ¯ ν ) , ( l µ ∂ µ T ) = (¯ l i ∂ i ¯ T ) , ( B µ ∂ µ T ) = ( ¯ B i ∂ i ¯ T ) ,ǫ iναβ u ν ( ∂ α ν )( ∂ β T ) = ǫ ijk ( ∂ j ¯ ν )( ∂ k ¯ T ) ,ǫ iναβ u ν B α l β = ǫ ijk ¯ B j ¯ l k . (2.34)Substituting (2.34) in (2.33) we express the LHS of (2.33) in terms of the back-ground data and the two arbitrary functions M and M of the second order parityodd partition function. Now using (2.27) and (2.3) we can express the 5 transportcoeffcients appearing in (2.33) in terms of M and M . Since one cannot generate aterm of the form ǫ ijk l j B k from the partition function, the 6th transport coefficients∆ will be completely determined by the correction of the first order current. This issimilar to the way the transport coefficients Φ i ’s, which appear in the traceless partof the stress tensor were determined. The end result of this step is∆ = − ∆ b ,χ = − R ¯ T (cid:20) ¯ ν ∂M ∂ ¯ T − ∂M ∂ ¯ T (cid:21) + ζ (cid:18) ∂b ∂ν − b ¯ T (cid:19) ,χ = − R ¯ T (cid:18) ∂M ∂ ¯ T (cid:19) + ζ (cid:18) ∂b ∂ν (cid:19) ,χ = − R ¯ T (cid:20) ¯ ν ∂M ∂ ¯ T − ∂M ∂ ¯ T (cid:21) + ζ (cid:18) ∂b ∂ ¯ T − b ¯ T (cid:19) ,χ = 2 R (cid:18) ∂M ∂ ¯ T (cid:19) + ζ (cid:18) ∂b ∂ ¯ T − b ¯ T (cid:19) , ∆ = 2 ¯ T (cid:18) ∂M ∂ ¯ T (cid:19) + (cid:18) qT E + P (cid:19) (cid:18) ∂M ∂ ¯ T − ¯ ν ∂M ∂ ¯ T (cid:19) , (2.35)where R = (cid:18) ∂P∂E (cid:19) q , R = (cid:18) ∂P∂q (cid:19) E ,b = T E + P (cid:18) Cν − C ν (cid:19) , b = T E + P (cid:18) Cν − C (cid:19) . Eliminating M and M from these expressions we get three relations among the– 18 –emaining 5 second order transport coefficients.∆ = − ∆ b ,T R (cid:20) χ − ζ (cid:18) ∂b ∂T − b T (cid:19)(cid:21) − R (cid:20) χ − ζ (cid:18) ∂b ∂ν − b T (cid:19)(cid:21) = 0 ,T R (cid:20) χ − ζ (cid:18) ∂b ∂T − b T (cid:19)(cid:21) + R (cid:20) χ − ζ (cid:18) ∂b ∂ν (cid:19)(cid:21) = 0 ,R T ∆ + (cid:20) χ − ζ (cid:18) ∂b ∂ν (cid:19)(cid:21) − q ( E + P ) (cid:20) χ − ζ (cid:18) ∂b ∂ν − b T (cid:19)(cid:21) = 0 . (2.36)
3. Kubo formula for the transport coefficients Φ , Φ In this section we derive the relations obtained for the transport coefficients Φ andΦ given in (2.25) using the Kubo formula. We consider the following equilibriumbackground for the metric, gauge field and the velocity g (0) µν = diag( − , , , , A µ = ( ν (0) T , , , , u µ = (1 , , , . (3.1)The chemical potential ν (0) and the temperature T are constants and do not dependon space-time. Since the energy E (0) and the pressure P (0) can be thought of asfunctions of the temperature and the chemical potential, they are also constants inspace-time. Now consider the following non-zero metric perturbations about thisbackground δg tx = h tx , δg tz = h tz , δg yx = h yx , δg yz = h yz . (3.2)The gauge field perturbations are given by δ A µ = (0 , a x , , a z ) . (3.3)The fluid velocity is close to the rest frame and its perturbations are given by δu µ = (0 , v x , , v z ) . (3.4)All perturbations are assumed to have dependence only in the time t and y -direction.In appendix B we will show that the background and the perturbations consideredin equations (3.1) to (3.4) consistently solve the linearized fluid equations withoutthe need for turning on any other perturbations. A simple reason that these per-turbations consistently solve the linearized fluid equations is that they are all in thespin-2 shear sector and therefore they decouple from the rest.To derive Kubo formulae for transport coefficients, we consider the constitutiverelations for the stress tensor and the charge current as one point functions in thepresence of external sources. We then obtain two point functions for the currents bydifferentiating with respect to the metric and the gauge field perturbations. Working– 19 –his out to the linear order in perturbations will result in Kubo formulae for thetransport coefficients Φ , Φ .To proceed we will require the the Christoffel symbols to the linear order inperturbations. The non-vanishing elements at the linear order are given byΓ txy = −
12 ( ∂ y h tx − ∂ t h yx ) , Γ tzy = −
12 ( ∂ y h tz − ∂ t h zy ) , (3.5)Γ xtt = ∂ t h tx , Γ xty = 12 ( ∂ y h tx + ∂ t h yx ) , Γ xyy = ∂ y h xy , Γ ztt = ∂ t h z , Γ zty = 12 ( ∂ y h tz + ∂ t h yz ) , Γ xyy = ∂ y h zy , Γ ytx = 12 ( ∂ t h yx − ∂ y h tx ) , Γ ytz = 12 ( ∂ t h yz − ∂ y h tz ) . Evaluating the inverse metric to the linear order we obtain h tx = h tx , h tz = h tz , h yx = − h yx , h yz = − h yz . (3.6)The covariant components of the velocity are given by u µ = ( − , v x + h xt , , v z + h zt ) . (3.7)We now evaluate various components of the stress tensor to the linear order inthe perturbations. From the list of terms that contribute at 2nd order in derivativesgiven in table 2 we see that in the background we have chosen all the scalars S i vanish. The reason for this is for the background all the thermodynamic functionsare independent of space-time. We now examine the traceless part of the stresstensor. Note that the contributions from τ ( i ) µν for i = 3 , , , , , , , ,
11 vanishsince the thermodynamic functions are independent of space-time. What remainsto be evaluated are the contributions from τ ( i ) µ for i = 1 , i = 12. Let us firstexamine the tx and the ty component of the stress tensor. To the second order inderivatives and to the linear order in the perturbation this is given by T tx = ( E (0) + P (0 ) v x + P (0) h tx , (3.8) T tz = ( E (0) + P (0) ) v z + P (0) h tz ,T ty = 0 . Note that σ tx , σ tz , [ τ ( i ) ] tx and [ τ ( i ) ] tz for i = 1 , ,
12 do not contribute at the linearorder. The reason is that the term ∇ α u β and ∇ α l β is a first order term, therefore onehas to evaluate the projector for these components say P tα P xβ at the zeroth order,which vanishes. – 20 –ets examine the yx component of the stress tensor. We need to evaluate thecontributions from [ τ ( i ) ] yx for i = 1 , ,
12. These are given by [ τ (1) ] yx = 12 ∂ y ( v z + h zt ) , (3.9)[ τ (2) ] yx = 12 ∂ y a z , [ τ (12) ] yx = 12 ∂ y ( ∂ y v z + ∂ t h yz ) . We also need the contribution of the shear tensor to the linear order. This is givenby σ yx = 12 ( ∂ y v x + ∂ t h yx ) . (3.10)Considering all these contributions along with the contribution to the stress tensorto the zeroth order in derivative we obtain T yx = − P (0) h yx − η ( ∂ y v x + ∂ t h yx ) (3.11)+ 12 Φ ∂ y ( v z + h zt ) + 12 Φ ∂ y a z + Φ ∂ y ( ∂ y v z + ∂ t h yz ) . The equations of motion for the x component of the stress tensor to the linear orderin the fields is given by ∂ t T tx + ∂ t h tx T tt + ∂ y T yx = 0 . (3.12)Here T tt is the zeroth order tt component of the stress tensor which is given by T tt = E (0) . (3.13)Fourier transforming these equations and taking the zero frequency limit or taking thetime independent situation we obtain the Ward identities for the one point functionof the stress tensor. lim ω → T yx ( k ) = 0 . (3.14)We can now differentiate these with respect to the background fields h zt , and a z andobtain Kubo formulae for the transport coefficients Φ and Φ respectively. A similarprocedure for the yz component of the stress tensor yields the same result. In evaluating these contributions we take ǫ = √− g . – 21 –o proceed we first eliminate v x and v z using (3.8). This results in the followingequations T yx = − P (0) h yx − η (cid:18) ∂ y T tx − P (0) ∂ y h tx E (0) + P (0 (cid:19) (3.15)+ Φ ∂ y T tz + E (0) ∂ y h tz E (0) + P (0) ! + Φ ∂ y a z + Φ ∂ y T tz − P (0) ∂ y h tz E (0) + P (0 ! . Here we have already used time-independence to drop the time derivatives. Fouriertransforming these equations we obtain T yx ( k ) = − P (0) h yx − η (cid:18) ikT tx − ikP (0) h tx E (0) + P (0 (cid:19) (3.16) − Φ (cid:18) k T tz + E (0) k h tz E (0) + P (0) (cid:19) − Φ k a z − Φ k (cid:18) T tz − P (0) h tz E (0) + P (0) (cid:19) = 0 . The last equality in the above equation implements the Ward identity given in (3.14)and k is the momentum in the y direction. Let us now focus on the expression for T yx a similar analysis can be repeated for T yz . Differentiating the Ward identity for T yx with respect to h zt and a z and setting the other backgrounds to zero we obtainthe following two equations k E (0) + P (0) (cid:2) (Φ h T tz ( k ) T tz ( − k ) i + E (0) ) + Φ ( h T tz ( k ) T tz ( − k ) i − P (0) ) (cid:3) = − ik ηE (0) + P (0) h T tx ( k ) T tz ( − k ) i , (3.17)Φ + Φ E (0) + P (0) k h T tz ( k ) j z ( − k ) i + Φ k = − ik ηE (0) + P (0) h T tz ( k ) j z ( − k ) i . To obtain the first equation we have differentiated with respect to h tz and set allthe other backgrounds to zero. The second equation is obtained by differentiatingthe Ward identity with respect to a z and setting the remaining backgrounds to zero.The equations in (3.17) are sufficient to determine the transport coefficients Φ , Φ .From [9] we have the following results for the various two point functions.lim k → ,ω → h T tx ( k ) T tz ( − k ) i = ik (cid:18) C ν (0) T (0) ) − C ( T (0) ) ν (0) (cid:19) , (3.18)lim k → ,ω → h T tx ( k ) j z ( − k ) i = ik (cid:18) C ν (0) T (0) ) − C ( T (0) ) (cid:19) . – 22 –hese results are given in equations (79)-(81) for a system with 3 chemical potentialsand equations (123)-(125) for a system with a single chemical potential of [9]. Thedefinition of the variables for the two point functions used is given in equation (47).This reference also uses the normalization − C π , c m π , C = 124 , (3.19)for the gauge anomaly and we have rewritten the chemical potential µ in terms ofthe variable ν . From [17] we can read out the following correlatorslim k → ,ω → h T tz ( k ) T tz ( − k ) i = P (0) , lim k → ,ω → h T tz ( k ) j z ( − k ) i = 0 . (3.20)These correlators are mentioned below equation (2.16) of reference [17]. Substitutingthe formulae for the two point functions given in (3.18) and (3.20) into the equationsgiven in (3.17) we obtainΦ = 2 ηE (0) + P (0) (cid:18) C ν (0) T (0) ) − C ( T (0) ) ν (0) (cid:19) , (3.21)Φ = 2 ηE (0) + P (0) (cid:18) C ν (0) T (0) ) − C ( T (0) ) (cid:19) . These expressions agree with that derived using the equilibrium partition functionmethod which are given in (2.25).
4. Chiral shear waves
In this section we examine the effects of the second order parity transport coefficientson linearized dispersion relations about the equilibrium characterized by the followingbackground given in (3.1). Note that the temperature T and the chemical potential ν (0) are independent of space-time and therefore all other thermodynamic variablesare constants in space-time. We consider shear modes , for this we examine velocityperturbations of the form δu µ = (0 , v x , , v z ) . (4.1)These perturbations depend only on time t and the y -direction. We include all theterms in the stress tensor to the 2nd order in derivatives given by T µν = [ T (0) ] µν + [ T (1) ] µν + [ T (2) ] µνodd , (4.2)where [ T (0) ] µν and [ T (1) ] µν are given in (1.4) and [ T (2) ] µνodd is given in (1.8). We nowwrite down the contribution to the stress tensor from these velocity fluctuations We have shown that none of the parity odd transport coefficients modify the sound or thecharge dissipation mode for the equilibrium characterized by (3.1). – 23 –hich are linear order in the fluctuations. It can be seen that the only contributionsfrom 2nd order which arise are from the term involving Φ and Φ . Thus the stresstensor to the linear order in velocity fluctuations is given by δT tx = ( E (0) + P (0) ) v x , δT tz = ( E (0) + P (0) ) v z , (4.3) δT yx = − η∂ y v x + Φ ∂ y v z + Φ ∂ y v z ,δT yz = − η∂ y v z − Φ ∂ y v x − Φ ∂ y v x . We now substitute these values in the equations of motion for the stress tensor givenby ∂ t δT tx + ∂ y T yx = 0 , (4.4) ∂ t δT tz + ∂ y T yz = 0 . Substituting the expressions for the stress tensor from (4.3) into the above equationsand then taking the Fourier transform of these equations result in the following setof coupled equations( − iω ( E (0) + P (0) ) + ηk ) v x − i k + Φ ) v z = 0 , (4.5)( − iω ( E (0) + P (0) ) + ηk ) v z + i k + Φ ) v x = 0 , From these equations we see that the two shear modes split depending on theirchirality. The dispersion relations for these modes are given by ω = − i ηE (0) + P (0) k ∓ i E (0) + P (0) ) (Φ + Φ ) k . (4.6)Thus in the basis we have used to list the second order transport coefficients bothΦ as well as Φ contribute to the splitting of the shear modes. Earlier studies ofthe chiral shear modes were restricted to the conformal transport at second order,therefore the contribution of Φ to the splitting was not noticed. Let us call thecoefficient of k as the chiral dispersion coefficient and define D = 12( E (0) + P (0) ) (Φ + Φ ) . (4.7)Using holography we now show that for the case N = 4 Yang-Mills, Φ = 0.We will also check the relation of Φ to the anomaly coefficient derived in this paperusing the holographic result for this transport coefficient. The holographic dual ofthis system is given by the Reisner Nordstr¨om black hole in AdS . We will use thenotations of [2]. The five dimensional action we consider is given by S = 116 πG Z √− g (cid:18) R + 12 − F AB F AB − κ ǫ LABCD A L F AB F CD (cid:19) . (4.8)– 24 –he equations of motion of the above action are G AB − g AB + 2 (cid:18) F AC F CB + 14 g AB F CD F CD (cid:19) = 0 , (4.9) ∇ B F AB + κǫ ABCDE F BC F DE = 0 , where G AB is the five dimensional Einstein tensor. The Reisner-Nordstr¨om blackbrane solution in Eddington Finkelstein coordinates is given by ds = − u µ dx µ dr − r V ( r, m, q ) u µ u ν dx µ dx ν + r P µν dx µ dx ν , (4.10) A = √ q r u µ dx µ and u µ dx µ = dv, V ( r, m, q ) = 1 − mr + q r . (4.11)Let R be the radius of the outer horizon of the black hole. We then define thequantities M ≡ mR , Q = qR , Q = M − . (4.12)The last equation results from the fact that R is the largest root of V ( r ) = 0. Thethermodynamic quantities of this black hole are given by T = R π (2 − Q ) , µ = 2 √ QR, (4.13)where T is the Hawking temperature and Q the charge density. The energy density,pressure and the shear viscosity in terms of these variables are given by E (0) = 3 M R πG , P (0) = M R πG η = R πG = s π . (4.14)Before we proceed we first relate the anomaly coefficient C to the Chern-Simonscoefficient κ . The boundary current is given by J µ = 116 πG √− g F rµ | r →∞ , (4.15)where r is the radial direction. Here we are using the definition of the current whichis consistent with the Page charge using which the charge density of the black holeis evaluated. There are other definitions of current as discussed in footnote 3 of [2].The bulk equations of motion for the gauge field results in the following conservationlaw for the current. ∂ µ J µ = − κ πG ǫ µνρσ F µν F ρσ . (4.16)We now have to identify the relation between the gauge field used in field theory andthat of the bulk gauge field. Note that the chemical potential value is related to the– 25 –orizon value of the gauge field. Comparing the horizon value of the gauge field in(4.10) and (4.13) we see that the relation between the bulk gauge field and the gaugefield is given by A fieldµ = 4 A bulkµ . (4.17)The relation is important since we have already used this normalization to define thethermodynamics of the boundary gauge theory. The gauge fields in the field theorymust be defined consistent with this thermodynamics. Substituting the relation(4.17) into the conservation law (4.16) we obtain ∂ µ J µ = − κ πG ǫ µνρσ F fieldµν F fieldρσ . (4.18)Now all quantities are defined in the field theory. Comparing with the current con-servation law in (1.3) we obtain C = κ πG . (4.19)The transport coefficient Φ can be identified to the coefficient N in the notationof [2], equation (4.37) see also [18, 3]. Reading out the holographic value of N fromequation (4.38) of [2] we haveΦ πG √ M ( M − R κ. (4.20)Note that the action given in (4.8) captures the situation when the gravitationalanomaly is zero. Using the relations in (4.12), (4.13) and (4.14) it can be seen thatΦ can be written as Φ = µ ηE (0) + P (0) κ πG . (4.21)We can now compare it with the expression derived earlier for this paper for Φ forcharge fluid with an anomaly which is given byΦ = 23 µ ηE (0) + P (0) C. (4.22)Therefore we obtain C = κ πG . (4.23)This is precisely the relation between the Chern-Simons coefficient and C obtaineddirectly using equations of motion in (4.19). This serves as a check for the relationbetween the transport coefficient Φ and the anomaly coefficient derived in this paper.We will now use the holographic value of Φ given in (4.20) to evaluate itscontribution to the chiral dispersion coefficient for the Riesner-Nordstr¨om black hole D = √ Q κ M R + 12( E (0) + P (0) ) Φ . (4.24)– 26 –y the AdS/CFT correspondence the chiral dispersion relation corresponds to thequasi-normal modes seen in the shear channel of the graviton fluctuations [19]. Wetherefore compare this dispersion coefficient with that obtained in [12] by studyingthe quasi-normal modes in the shear sector. They find that the dispersion coefficientis given by D QNM = κ SY ( Q SY ) m R . (4.25)By comparing the action and the background solution given in [12] to that given in(4.8) and (4.10) we obtain the following relations between the variables of [12] andthat used here κ SY = 2 κ , ( Q SY ) = 3 q = 3 Q R . (4.26)Substituting the above relations in (4.25) we see that D QNM = √ Q κ M R . (4.27)Since D QNM must be equal to D evaluated in (4.24) we see that we must have Φ = 0for this system.
5. Conclusions
We have used the equilibrium partition function to obtain expressions for 7 par-ity odd transport coefficients which occur at 2nd order for a non-conformal fluidwith a single conserved charge. These transport coefficients can be expressed interms of the anomaly, shear viscosity, bulk viscosity, charge diffusivity and thermo-dynamic functions. Out of these 2 transport coefficients can also be derived usingthe Kubo formulae. These formulae agree with that obtained by the partition func-tion method. The equilibrium partition function also gives 3 constraints for 5 otherparity odd transport coefficients at this order. The transport coefficient Φ affectschiral dispersion relations [13].As we have mentioned earlier, parity odd coefficients have be examined earlierfor conformal charge transport in [13]. There the principle used was that thesecoefficients should not contribute to entropy production. In general the constraintsobtained by examining the zero entropy production condition will be more than thatobtained from the equilibrium partition function. It will be interesting to carry outthe analysis of [13] to non-conformal fluids and compare with the results obtained inthis paper.Our analysis of these transport properties were motivated by the possibility ofstudying them in holography. While this work was being done we received preprint[18] which evaluates all 2nd order transport coefficients for a charged conformal fluidin the framework of AdS/CFT. It is useful to compare the relations we have obtainedfor the parity odd sector with the expressions of [18].– 27 –inally it will be useful to develop Kubo like expressions for all the parity oddtransport coefficients. From the constitutive relations it seems that the remainingtransport coefficients involves 3 point functions. Determining these relations willprovide an alternate method to obtain the transport coefficients from holography. A. Classification of parity odd data at 2nd order in derivatives
In this appendix we provide some details which led to the classification of the parityodd data at second order in derivatives given in table 2. We consider the followingbasis of vectors to construct the the second order terms: • Parity odd vectors:Vorticity : l µ = ǫ µναβ u ν ∂ α u β Magnetic field : B µ = 12 ǫ µναβ u ν F αβ . (A.1) • Parity even vectors u µ , ∂ µ T, ∂ µ ν, (A.2)Electric field : → V µ = E µ T − P µρ ∂ ρ ν. Among these vectors, the electric field V µ vanishes on the equilibrium fluid configu-ration given in (2.1). We also consider the shear tensor σ µν = ∇ h µ u ν i . (A.3)and the scalar Θ = ∇ µ u µ . (A.4)Note that both σ µν and the scalar Θ vanishes on the equilibrium configuration in(2.1). We use these basic quantities we can construct the various parity odd termsthat occur at second order in derivatives given in table 2. These terms are indepen-dent and cannot be related to each other by equations of motion to first order inderivatives.Let us consider the scalars listed in table 2: One would have thought that scalarsof the form ∇ µ l µ and ∇ µ B µ should be listed. But it can be shown that these scalarscan be related to the ones listed in the table by calculations which lead to thefollowing identities. ∇ µ l µ = (cid:18) qTE + P (cid:19) V µ l µ − (cid:18) ∂ µ TT (cid:19) l µ (A.5) ∇ µ B µ = (cid:18) qTE + P (cid:19) V µ B µ − (cid:18) ∂ µ TT (cid:19) B µ − T l µ ( V µ − ∂ µ ν ) (A.6)– 28 –hese identities can be verified by straight forward calculations.Let us now consider the vectors listed in table 2: Note that the vectors V µ (3) to V µ (9) vanish on the equilibrium configuration (2.1). Again one would have naivelyexpected to list terms such as P µα ( u. ∇ ) B α . We will show now using the equations ofmotion to the zeroth order that this term is related to the vectors already listed intable 2. The manipulations are as follows: P µα ( u. ∇ ) B α = 12 P µα u θ ∇ θ (cid:2) ǫ ανλβ u ν F λβ (cid:3) , = 12 P µα ǫ ανλβ a ν F λβ + 12 ǫ µνλβ u ν u θ ∇ θ F λβ , = − ǫ µνλβ a ν u λ u θ F θβ − ǫ µνλβ u ν u θ ∇ λ F βθ , = ǫ µνλβ a ν u λ E β − ǫ µνλβ u ν ∇ λ E β + ǫ µνλβ u ν ( ∇ λ u θ ) F βθ , = − T ǫ µναβ u ν ∇ α V β + σ µν B ν − Θ (cid:20) B µ − T (cid:18) s ∂ν∂s + q ∂ν∂q (cid:19) l µ (cid:21) − (cid:18) qT E + P (cid:19) ǫ µνλσ u ν V λ ∂ σ ν − ǫ µνλσ u ν l λ B σ , = − T ǫ µναβ u ν ∇ α V β + V µ − V µ T (cid:18) s ∂ν∂s + q ∂ν∂q (cid:19) V µ + (cid:18) qT E + P (cid:19) V µ (9) + 12 V µ (1) . (A.7)Here a ν = ( u. ∇ ) u ν . In the RHS of equation (A.7), all terms except the last onevanishes at equilibrium. Therefore ( u. ∇ ) B µ does not vanish in equilibrium but itcan be related to all the other vectors listed in the table. A similar analysis can bedone for the vector of the form P µα ( u. ∇ ) l α . We have P µα ( u. ∇ ) l α = Rǫ µναβ u ν ∇ α V β + (cid:18) ∂R∂T (cid:19) ǫ µναβ u ν ( ∇ α T ) V β (A.8)+ (cid:18) ∂R∂ν (cid:19) ǫ µναβ u ν ( ∇ α ν ) V β + σ µν l ν + (cid:18) sT ∂ν∂s + qT ∂ν∂q − (cid:19) Θ l µ , where R = qTE + P . (A.9)From equation (A.8) it is clear that P µα ( u. ∇ ) l α vanishes at equilibrium and that it isrelated to the other vectors listed in table 2.In fact using symmetries we can list out all the independent terms appearing atthe second order. We can show that out at two derivatives either involving only thefluid variables or the velocity and derivative of the field strength, only one pseudovector can be constructed if we demand on-shell independence for all the terms. Thismakes it clear that once we have chosen ǫ µναβ u ν ∇ α V β , all other two derivative pseudovectors of the form mentioned earlier must be related to V µ (4) by equations of motion.– 29 –rom this argument it is possible to observe that it is not necessary to list ( u. ∇ ) B µ ,( u. ∇ ) l µ , ∇ µ l µ or ∇ µ B µ as independent data. A similar analysis can be performedfor the Pseudo-tensor. This leads to the the 12 tensors listed in 2. B. Consistency of the fluid profiles at the linear order
In this section we show that the velocity perturbations and the background fieldconfiguration considered in equations (3.1) to (3.4) consistently solve the linearizedfluid equations of motion without the need of the any other fluctuations. The per-turbations depends only on the time, t and the spatial direction, y .We will consider the perturbations to the linear order. We first write down all thecomponents of the stress tensor to 2nd order in derivatives and to linear order in theperturbations. This are given by T tt = E (0) , (B.1) T tx = = ( E (0) + P (0) ) v x + P (0) h tx ,T tz = ( E (0) + P (0) ) v z + P (0) h tz ,T ty = 0 ,T xx = P (0) ,T xy = − P (0) h yx − η ( ∂ y v x + ∂ t h yx )+ 12 Φ ∂ y ( v z + h zt ) + 12 Φ ∂ y a z + 12 Φ ∂ y ( ∂ y v z + ∂ t h yz ) ,T xz = 0 ,T yy = P (0) ,T yz = − P (0) h yz − η ( ∂ y v z + ∂ t h yz ) −
12 Φ ∂ y ( v x + h zt ) −
12 Φ ∂ y a x −
12 Φ ∂ y ( ∂ y v x + ∂ t h y ) ,T zz = P (0) . Now there are 4 equations of motion for the stress tensor. We will show that that the t and y components are trivially satisfied. The x and z components of this equationscan be used to determine the velocity profiles. The t component of the equations ofmotion of the stress tensor is given by ∂ t T tt + ∂ y T yt + Γ µµt T tt + Γ ttt T tt + Γ txx T xx + Γ tyy T yy + Γ tzz T zz = 0 . (B.2)Note that we have used the fact that all Christofell symbols are first order in the fields.We have also used the fact that the only dependence is through time t and y . Now– 30 –xamining the Christofell symbols given in (3.5) we see that there is no contributionto the equation in (B.2) from any term involving the Christofell symbols. Also since T tt = E (0) , the first term also vanishes. The second term in the equation vanishesbecause T yt = 0. Thus this equation is satisfied and imposes no constrains on thevelocity configuration chosen. Now let examine the y component of the equations ofmotion of the stress tensor. We have ∂ t T ty + ∂ y T yy + Γ µµy T yy + Γ ytt T tt + Γ yxx T xx + Γ yyy T yy + Γ yzz T zz = 0 . (B.3)Again, examining each term one can see that this equation is also satisfied. Thusthe t and y equations are satisfied. Thus v x and v y are determined by the x and z component of the equations of motion ∂ t T tx + ∂ t h tx T tt + ∂ y T yx = 0 , (B.4) ∂ t T tz + ∂ t h tz T tt + ∂ y T yz = 0 . To complete the analysis we show that the equations of motion of the chargecurrent are also trivially satisfied. The only non-zero values of the the vorticity tothe linear order is given by l x = ∂ y v z , l z = − ∂ y v x . (B.5)Similarly the non-zero values of the magnetic field to the linear order B x = 12 ∂ y a z , B z = − ∂ y a x . (B.6)One can also see that for the background in (3.1) and fluctuations to the linear orderthe electric field V µ vanishes to the linear order in the fields. It can also be seen thatto the linear order in fields the vectors V µ ( i ) with i = 1 , · · · J t = − q (0) , (B.7) J x = q (0) v x + ξl x + ξ B B x ,J y = 0 ,J z = q (0) v z + ξl z + ξ B B z . Now the current conservation equation to this order reads ∂ t ( √ gJ t ) + ∂ y ( √ gJ y ) = 0 , (B.8)and since √ g does not change to the linear order, this equation is satisfied and doesnot impose any further conditions on the velocities v x and v y . Here we have usedthat the only dependence is through t and y .– 31 –hus the constant background with the linear velocity profiles v x , v y given in(3.1) to (3.4) consistently solve the equations of motion. There are only 2 equationswhich determine the velocity profiles v x , v y . A simple way of stating this is that wehave turned only the shear fluctuations which decouple from the rest of the modes. Acknowledgments
We thank Aninda Sinha for organizing a stimulating workshop on “Non-perturbativegauge theories, holography and all that” which initiated this collaboration. Wethank Nabamita Banarjee, Bobby Ezuthachan, Yashodhan Hatwalne, Sachin Jain,Shiraz Minwalla and Rahul Pandit for useful and stimulating discussions. S.B thanksthe CHEP, IISc and the ICTS-TIFR for hospitality during the early phase of thiswork. The work of J.R.D is partially supported by the Ramanujan fellowship DST-SR/S2/RJN-59/2009.
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