aa r X i v : . [ nu c l - t h ] O c t Relativistic viscous hydrodynamics order by order
Jian-Hua Gao ∗ Shandong Provincial Key Laboratory of Optical Astronomy andSolar-Terrestrial Environment, Institute of Space Sciences,Shandong University, Weihai 264209, China andKey Laboratory of Quark and Lepton Physics (MOE),Central China Normal University, Wuhan 430079, China
Shi Pu
Institute for Theoretical Physics, Goethe University,Max-von-Laue-Straße 1, 60438 Frankfurt am Main, GermanyDepartment of Physics, National Center for Theoretical Sciences,and Leung Center for Cosmology and Particle Astrophysics,National Taiwan University, Taipei 10617, Taiwan andInterdisciplinary Center for Theoretical Study and Department of Modern Physics,University of Science and Technology of China, Hefei 230026, China
Abstract
In this paper, we propose a method of solving the viscous hydrodynamics order by order in aderivative expansion. In such a method, the zero-order solution is just one of the ideal hydro-dynamics. All the other higher order corrections satisfy the same first-order partial differentialequations but with different inhomogeneous terms. We take the Bjorken flow as an example to testthe validity of our method and present how to deal with the problems about the initial conditionand perturbation evolution in our formalism.
PACS numbers: 12.38.Mh, 25.75.-q, 52.27.Ny ∗ Electronic address: [email protected] . INTRODUCTION Relativistic hydrodynamics has been an important and useful theoretical tool in highenergy heavy-ion physics such as at BNL Relativistic Heavy Ion Collider (RHIC) and CERNLarge Hadron Collider (LHC), which have succeeded greatly in describing the collective flowfrom the data of those colliders [1–8]. Hydrodynamics can be considered as an macroscopiceffective field theory of more fundamental microscopic theory such as quantum field theory,in describing the non-equilibrium evolution of a given system. However, it is not trivial tobuild a consistent and causal relativistic hydrodynamics beyond the ideal hydrodynamics.The relativistic first order viscous hydrodynamics was first proposed by Eckart in [9], andLandau and Lifshitz in [10], both of which showed that dissipative fluctuation may propagateat an infinite speed, which is inconsistent with the relativistics. This is the so-called causalityproblem. In addition, the solution is also unstable due to small perturbation of the equilibriain these viscous hydrodynamics [11]. The nonrelativistic causal viscous hydrodynamics wasfirst presented by M¨uller in [12] and was later generalized into the relativistic version by Israeland Stewart in [13]. They remedied the previous viscous hydrodynamics by introducing somesecond order terms in deviations away from equilibrium into the entropy current. Thereforethese formalisms are also referred to as the second order theory of viscous hydrodynamics.For example, the relaxation time for shear viscous tensor, which is one of the well-knownsecond order parameters, describes how long it will take for the system to return to theequilibrium states after a small perturbation via shear viscosity. Therefore, if all thesehydrodynamic parameters satisfy certain constraints, the system will be causal and stable(e.g., for shear viscous tensor [14, 17], bulk viscous pressure [15], and heat conductingflow [16, 17]). Recently, some other authors discussed further the second order viscoushydrodynamics [18–20], especially from the point of view of effective theory, in which all thesecond derivative terms are included.There exist two different methods solving the viscous hydrodynamic equations, eitherexpressing the dissipative contributions in energy-momentum tensor or charge current interms of differentials on primary variables such as fluid 4-velocity u µ and chemical potential,then substituting them into the hydrodynamic equation, or regarding the dissipative quan-tities as independent dynamical quantities which satisfies extra differential equations. Thesame difficulty in both methods is that the contributions from different orders are mixed2ogether, which implies small errors in high orders might also cause big uncertainty in thenumerical simulations after time evolutions. On the other hand, the point of view of effec-tive theory, where higher order terms should always be small corrections to the lower orderduring the whole evolution, might give us some hints to simplify these problem. Besides,from the second order to the third or even higher orders, one has to deal with more andmore complicated differential hydrodynamic equations.In Sec. II of this paper, we will try to present a consistent formalism of solving theviscous hydrodynamic equation order by order in comparison with microscopic theories. Wewill show that the zero order solution is just the one of the ideal hydrodynamics in ourmethod and all the other higher order corrections satisfy the same first-order partial differ-ential equation but with different inhomogeneous source terms. We find that our methodis a recursion process, the next order solution can be obtained only after we get all theprevious order solution. In every order calculation, we only need to deal with the same firstorder differential equations with different inhomogeneous source terms. Such method can bemanipulated to any higher order. In Sec.III, we will discuss how to deal with the problemsabout the initial condition and stability in our formalism. In Sec.IV, we choose the Bjorkenflow as a test to illustrate the validity of our method and how to manipulate the initialcondition and perturbation evolution specifically. Finally, there is the conclusion in Sec V. II. HYDRODYNAMICS ORDER BY ORDER
Since we will present our method mainly theoretically or formally, for simplicity, we willrestrict ourselves to the conformal non-charged fluid. In such a system, the dissipativeterms are constrained greatly due to the conformal symmetry. More general cases can beextended straightforwardly and will be presented elsewhere. Since the fluid is not charged,only energy-momentum conservation is involved, ∂ ν T νµ = 0 , (1)where the energy-momentum tensor T µν is assumed to be able to expand as the primaryhydrodynamic variable, local fluid velocity u µ ( x ) ( u = −
1) and local temperature T ( x ).In the following, we will always work in the Landau frame and adopt the convention of themetric tensor g µν = [ − , +1 , +1 , +1]. In such frame and convention, the energy-momentum3ensor can be generally decomposed into T µν = ( ǫ + P ) u µ u ν + P g µν + Π µν , (2)where ǫ is the energy density, P is the pressure, and Π µν includes all the dissipative termsand satisfies u µ Π µν = 0.Generally, in long wavelength and low frequency limit, if Knudsen number K = ℓ mfp ∂ µ ≪
1, with ℓ mfp the mean free path of particles and ∂ µ the space-time derivatives, the hydro-dynamic is workable [19, 20]. In this case, we can expand all hydrodynamic quantities andequations in the power of the Knudsen number. In the leading order, we will get the idealfluid. In the first order, Π µν will be introduced and can be expanded as the differentialsof the local velocity u µ order by order. In a conformal theory, this dissipative term can begenerally written as [18, 21],Π µν = − ησ µν + π µν , (3) σ µν ≡ ∆ µα ∆ νβ (cid:18) ∂ α u β + ∂ β u α −
23 ∆ αβ ∇ · u (cid:19) , (4)∆ µν ≡ g µν + u µ u ν , ∇ · u ≡ ∆ αβ ∂ α u β , (5)where η is the shear viscosity, π µν is the second-order differential terms and in a conformaltheory, can be generally decomposed into the following form, π µν = ητ Π (cid:20) u α ∂ α σ µν + 13 σ µν ∂ α u α (cid:21) + λ (cid:20) σ µα σ να −
13 ∆ µν σ αβ σ αβ (cid:21) + 12 λ h σ µα Ω να σ ν α Ω µα i + λ (cid:20) Ω µα Ω να −
13 ∆ µν Ω αβ Ω αβ (cid:21) , (6)Ω µν ≡
12 ∆ µα ∆ νβ (cid:0) ∂ α u β − ∂ β u α (cid:1) , (7)where τ Π , λ , , are transport coefficients in the second order theory and Ω µν is the vorticitytensor. The entropy current S µ is defined as [13] S µ = PT u µ − T u ν T νµ − Q µ , (8)where Q µ represents a possible second order correction. In the leading order, S µ = su µ ,with s the entropy density sT = ǫ + P. (9)4ow we will propose our method. Firstly, it is quite natural and straightforward thatwe will treat fluid 4-velocity u µ ( x ) and temperature T ( x ) as the primary variables, energydensity and pressure can be expressed as the function of T ( x ) by the equation of state. Wecan imagine the final solution of u µ ( x ) and T ( x ) can be obtained by the serials expansion as u µ ( x ) = u µ ( x ) + u µ ( x ) + u µ ( x ) + ..., (10) T ( x ) = T ( x ) + T ( x ) + T ( x ) + ..., (11)where the series are expanded in the power of Knudsen number. Then, we assume that allhydrodynamic quantities and equations can be expanded in the power of Knudsen number.For example, it follows that the energy-momentum tensor can expanded as T νµ = T νµ + T νµ + T νµ + ... (12)The zero-order energy-momentum tensor is given by T νµ = ( ǫ + P ) u µ u ν + P g µν (13)where ǫ ≡ ǫ ( T ) and P ≡ P ( T ). It is just the decomposition of the ideal fluid.Secondly, in order to avoid the mixture of different orders, we assume the differentialhydrodynamic equations satisfy the conservation law order by order, i.e. we let ∂ µ T µνi = 0 , ( i = 0 , , , ... ) . (14)It looks very robust and adds more constraints to the hydrodynamic equations, but it isreasonable. From the classical kinetic theory, i.e., the Boltzmann equations, the distributionfunction f can be expanded in power of K , f = f + f + f + ... and obtained order by order.Provided the time reversal symmetry is protected, we can get ∂ µ T µνi = 0 (also see AppendixI). These kinds of methods are widely used in theoretical physics; e.g., for quantum kinetictheory, a similar treatment will give the exact transport coefficients of chiral magnetic andvortical effects [22, 23] or Hall effects [24], and other related hydrodynamics [25–27].)Back to our case, the zero-order approximation u µ and T can be obtained by solving theideal hydrodynamic equation, ∂ ν T νµ = 0 . (15)5s usual, we can decompose them into a component parallel to u µ by contracting Eq.(15)with u µ , ( ǫ + P ) ∇ · u + ǫ ′ ˙ T = 0 , (16)and the other three components orthogonal to u µ by projecting Eq.(15) with ∆ µν ∆ µν (cid:18) ˙ u ν + 1 T ∂ ν T (cid:19) = 0 (17)where ˙ T ≡ u µ ∂ µ u , ǫ ′ ≡ dǫdT | T = T = dǫ dT and ∆ µν ≡ g µν + u µ u ν . The zero-order entropycurrent is given by S µ = P T u µ − T u ν T νµ = ǫ + P T u µ = s u µ . (18)It is well known that it is conserved for the ideal fluid ∂ µ S µ = 0 . (19) A. The first-order equations
Now let us continue to deal with the next-to-leading order, in which the energy-momentum tensor is given by T νµ = ( ǫ ′ + P ′ ) T u µ u ν + P ′ T g µν + ( ǫ + P ) (¯ u µ u ν + u µ ¯ u ν ) − η σ µν (20)where ¯ u µ ≡ ∆ µν u ν and σ µν ≡ ∆ µα ∆ νβ (cid:18) ∂ α u β + ∂ β u α −
23 ∆ αβ ∇ · u (cid:19) . (21)Note that since there are corrections to the temperature T ( x ), the energy density andpressure will also have some corrections, ǫ ( T ) = ǫ + ǫ ′ T + O ( K ) , P ( T ) = P + P ′ T + O ( K ) . (22)It should be noted that we have constrained the normalization condition for u µ as( u + u ) = − , (23)which leads to the following relation: u · u = 1 − q u . (24)6ence only three components of u µ are independent and the component parallel to u µ canbe totally determined by ¯ u µ . In addition, we can notice that u · u only contributes toat least second order, that is why only ¯ u µ is involved in the first-order energy-momentumtensor (20).Since the zero-order energy-momentum tensor has already satisfied the conservation equa-tion, we need the first-order energy-momentum tensor to satisfy the conservation equationindependently ∂ ν T νµ = 0 . (25)The component parallel to u µ reads ∇ · ¯ u + 1 T (cid:18) v s + 1 (cid:19) ¯ u µ ∂ µ T + 1 T v s ˙ T + T (cid:18) T v s (cid:19) ′ ˙ T = 1 ǫ + P u µ C µ , (26)where u µ C µ = η σ µν ∂ ν u µ (27)and the components orthogonal to u µ reads∆ µα ( ˙¯ u α + ¯ u ν ∂ ν u α ) + 1 T ¯ u µ ˙ T + 1 T ∆ µα ∂ α T − T T ∆ µν ∂ ν T = 1 ǫ + P ∆ µα C α (28)where ∆ µα C α = ∆ µα ∂ ν ( η σ να ) . (29)The first-order correction to the entropy current S µ is given by S µ = 1 T [ ǫ ′ T u µ + ( ǫ + P ) ¯ u µ ] . (30)It is easy to show that the divergence of the entropy current is always positive and consistentwith the second thermal law, ∂ µ S µ = η T σ µν σ µν ≥ . (31) B. The second-order equations
We now turn to the second order, in which the energy-momentum tensor is given by T νµ = ( ǫ ′ + P ′ ) T u µ u ν + P ′ T g µν + ( ǫ + P ) (¯ u µ u ν + u µ ¯ u ν )+ 12 T [ P ′′ g µν + ( ǫ ′′ + P ′′ ) u µ u ν ] + ( ǫ + P ) (cid:0) ¯ u u µ u ν + ¯ u µ ¯ u ν (cid:1) + ( ǫ ′ + P ′ ) T (¯ u µ u ν + ¯ u µ u ν ) − η ′ T σ µν − η σ µν + Π νµ , (32)7here ¯ u µ ≡ ∆ µν u ν . For energy density and pressure, we have ǫ = ǫ + ǫ ′ T + ǫ ′ T + ǫ ′′ T ,and P = P + P ′ T + P ′ T + P ′′ T . Just as we did for u µ , we have constrained thenormalization condition for u µ as ( u + u + u ) = − , (33)which results in u · u = q u − q u + 2¯ u · ¯ u + ¯ u . (34)It is easy to show that u · u only contributes to at least third order, which can be droppedoff for the second order T µν . However, we must consider u · u which has been neglected atthe first order T µν . Since both the zero-order and the first order energy-momentum tensorshave already satisfy the conservation equation, we need the second-order energy-momentumtensor to satisfy the conservation equation independently, i.e., ∂ ν T νµ = 0 . (35)The component parallel to u µ reads, ∇ · ¯ u + 1 T (cid:18) v s + 1 (cid:19) ¯ u µ ∂ µ T + 1 T v s ˙ T + T (cid:18) T v s (cid:19) ′ ˙ T = 1 ǫ + P u µ C µ , (36)where u µ C µ = ǫ u µ ∂ µ (cid:20) ( ǫ + P )2 (cid:18) ǫ + P (cid:19) ′′ T − ¯ u (cid:21) + u µ ∂ µ (cid:20) ( ǫ + P )2 (cid:18) P ǫ + P (cid:19) ′′ T − P ¯ u (cid:21) − h ǫ ′ T u µ + ( ǫ + P )¯ u µ i ∂ µ (cid:20) ( ǫ ′ + P ′ )( ǫ + P ) T (cid:21) − (cid:20)(cid:18) ǫ ′ + P ′ ǫ + P η − η ′ (cid:19) T σ µν − η σ µν + π νµ (cid:21) σ µν + ( ǫ + P ) ¯ u ν u µ ∂ ν ¯ u µ , (37)and the components orthogonal to u µ reads∆ µα ( ˙¯ u α + ¯ u ν ∂ ν u α ) + 1 T ¯ u µ ˙ T + 1 T ∆ µα ∂ α T − T T ∆ µν ∂ ν T = 1 ǫ + P ∆ µα C α , (38)8here ∆ µα C α = P ∆ µν ∂ ν (cid:20) − ( ǫ + P )2 (cid:18) ǫ + P (cid:19) ′′ T + ¯ u (cid:21) +∆ µν ∂ ν (cid:20) ( ǫ + P )2 (cid:18) P ǫ + P (cid:19) ′′ T − P ¯ u (cid:21) − ∆ µα ∂ ν (cid:20)(cid:18) η ′ − ǫ ′ + P ′ ǫ + P η (cid:19) σ να + η σ να − π να (cid:21) + T να ∆ µα ∂ ν (cid:20) ( ǫ ′ + P ′ )( ǫ + P ) T (cid:21) + ∆ µα ∂ ν [( ǫ + P ) ¯ u α ¯ u ν ] . (39)The second-order correction to the entropy current is S µ = − T u ν T νµ + 12 ¯ u S µ − T T S µ − T ¯ u ν T νµ + T T (cid:18) ǫ ′ T u µ + P ′ T ¯ u µ (cid:19) . (40)It is straightforward to derive the rate of entropy production for the second order as ∂ µ S µ = 12 T ( η ′ T σ µν + 2 η σ µν − π µν ) σ µν . (41)Generally, they are not positive definite, however they do not violate the second law ofthermodynamics since the third order terms must be small compared to the second orderterm in the domain of applicability of hydrodynamics. Similar possible negative signs andcomments can also be found in [18].Actually, such a recursion process can be generalized to any higher orders without anydifficulty. It is important to note that all the equations have a similar form, i.e., thecomponent parallel to u µ reads ∇ · ¯ u n + 1 T (cid:18) v s + 1 (cid:19) ¯ u µn ∂ µ T + 1 T v s ˙ T n + T n (cid:18) T v s (cid:19) ′ ˙ T = 1 ǫ + P u µ C µn , (42)and the components orthogonal to u µ reads∆ µα ( ˙¯ u αn + ¯ u νn ∂ ν u α ) + 1 T ¯ u nµ ˙ T + 1 T ∆ µα ∂ α T n − T n T ∆ µν ∂ ν T = 1 ǫ + P ∆ µα C αn , (43)where C αn depends only on the u µ , T , ¯ u µm , and T m ≤ m ≤ n − III. INITIAL CONDITIONS AND STABILITY
In order to solve the hydrodynamic equations, we must give some specific initial condi-tions, e.g., u ( t , ~x ) and T ( t , ~x ), where t is the initial time. Generally, we can decompose9hem into u µ ( t , ~x ) = u µ ( t , ~x ) + u µ ( t , ~x ) + u µ ( t , ~x ) + ...,T ( t , ~x ) = T ( t , ~x ) + T ( t , ~x ) + T ( t , ~x ) + ..., (44)in any way as long as they satisfy u µ ( t , ~x ) ≫ u µ ( t , ~x ) ≫ u µ ( t , ~x ) ≫ ..., (45) T ( t , ~x ) ≫ T ( t , ~x ) ≫ T ( t , ~x ) ≫ ..., (46)With different decompositions, the final result should differ only in higher orders. Forsimplicity, we can just set u µ ( t , ~x ) = u µ ( t , ~x ) , u µ ( t , ~x ) = 0 , u µ ( t , ~x ) = 0 , ...T ( t , ~x ) = T ( t , ~x ) , T ( t , ~x ) = 0 , T ( t , ~x ) = 0 , ... (47)With the initial state u µ ( t , ~x ), we can solve the zeroth-order equations (16) and (17) andobtain the solution u µ ( t, ~x ). With this zeroth-order solution, we can calculate the first-orderinhomogeneous term u µ C µ and ∆ µα C α which includes the first derivative of u µ ( t, ~x ) andsolve the first-order equations (42) and (43) under the initial conditions u µ ( t , ~x ) = 0 and T ( t , ~x ) = 0. After getting the first-order solution, we can proceed further to obtain thesecond-order contribution and so on. Hence, to solve the n th-order equations, there is noneed to know the initial value of the derivative of n th-order correction; we only need thederivative of lower order corrections which have been solved already. This should be a goodadvantage in our iterative method compared to other methods. Besides, using the initialconditions (47), we actually rule out all the free modes which will lead to instability from thehomogenous solutions in Eqs.(42) and (43) for n ≥
1. Only the particular solution which isproportional to the inhomogeneous term survives. However, in the numerical simulation, thecomputation error can be inevitable and make the above argument invalid. The interestingthing in our method is that whether the instability arises or not depends only on the zeroth-order solution u µ , as shown in Eq.(42) and Eq.(43) In the next section, we will use Bjorkenflow as a simple example to illustrate how the perturbations evolve.10 V. BJORKEN FLOW
In this section, we will choose the (1+1)-dimensional Bjorken flow [28] as an exampleto illustrate the validity of our formalism. In order to do that, we will use the coordinatesystem, such that τ = √ t − z , η = tanh − zt = 12 ln t + zt − z . (48)Bjorken flow is given by u µ ( τ ) = (cosh η, , , sinh η ) . (49)In the following, we will explicitly solve for the velocity field u µ and the energy density ǫ ortemperature T with the initial conditions u µ ( τ ) = (cosh η, , , sinh η ) , ǫ ( τ ) = 3 P ( τ ) = aT ( τ ) = Cτ / , (50)where C is a constant and we have used ǫ = 3 P = aT with a a constant for the conformalfluid. We will choose the decomposition in Eq.(47), i.e., u µ ( τ ) = u µ ( τ ) , u µ ( τ ) = 0 , u µ ( τ ) = 0 , ...,T ( τ ) = T ( τ ) , T ( τ ) = 0 , T ( τ ) = 0 , ... (51)Hence we have designated full initial configuration to the the zeroth-order equations orideal hydrodynamic equations (16) and (17). From the uniqueness of the solution for thedifferential equations, the solution must be the Bjorken’s solution u µ ( τ ) = (cosh η, , , sinh η ) , ǫ ( τ ) = 3 P ( τ ) = aT ( τ ) = Cτ / . (52)Substituting the Bjorken solutions (52) into the first-order equations (42) and (43) yields ∇ · ¯ u + 3 T ∂ τ T + 1 τ T T = η ǫ τ , (53) ∂ τ ¯ u µ + v µ τ ( v · ¯ u ) − τ ¯ u µ + 1 T ∆ µα ∂ α T = 0 , (54)where v µ = (cid:0) zτ , , , tτ (cid:1) . Given the initial condition T ( τ ) = 0 and ¯ u µ ( τ ) = (0 , , , u µ = (0 , , , , T = ˆ η τ / (cid:20) − (cid:16) τ τ (cid:17) / (cid:21) T , (55)11here ˆ η and κ are both constants and defined as in Ref.[18] η = bT = C ˆ η (cid:16) ǫ C (cid:17) / = C ˆ η τ , κ = (cid:18) Ca (cid:19) / . (56)The energy density of the first order can be given by ǫ = 2ˆ η τ / (cid:20) − (cid:16) τ τ (cid:17) / (cid:21) ǫ . (57)Now substituting the first-order solution (55) into the second-order equations (36) and (38),we can have ∇ · ¯ u + 3 T ∂ τ T + 1 τ T T = 2 (cid:16) ˆ η ˆ τ Π0 − ˆ λ , (cid:17) τ / , (58) ∂ τ ¯ u µ + v µ τ ( v · ¯ u ) − τ ¯ u µ + 1 T ∆ µα ∂ α T = 0 , (59)where τ Π0 = ˆ τ Π0 (cid:16) ǫ C (cid:17) − / , λ , = C ˆ λ , (cid:16) ǫ C (cid:17) / . (60)Again with the initial condition T ( τ ) = 0 and ¯ u µ ( τ ) = (0 , , , u µ = (0 , , , , T = (cid:16) ˆ η ˆ τ Π0 − ˆ λ , (cid:17) τ / (cid:20) − (cid:16) τ τ (cid:17) / (cid:21) T , (61)or the energy density ǫ = 3ˆ η τ / (cid:20) − (cid:16) τ τ (cid:17) / (cid:21) ǫ + 2 (cid:16) ˆ η ˆ τ Π0 − ˆ λ , (cid:17) τ / (cid:20) − (cid:16) τ τ (cid:17) / (cid:21) ǫ . (62)Up to the second-order contribution, the energy density is given by ǫ = Cτ / ( η τ / (cid:20) − (cid:16) τ τ (cid:17) / (cid:21) + 3ˆ η τ / (cid:20) − (cid:16) τ τ (cid:17) / (cid:21) + 2 (cid:16) ˆ η ˆ τ Π0 − ˆ λ , (cid:17) τ / (cid:20) − (cid:16) τ τ (cid:17) / (cid:21) . (63)It is obvious that our expansion method will be valid as long asˆ η τ / ≪ , (cid:12)(cid:12)(cid:12) ˆ η ˆ τ Π0 − ˆ λ , (cid:12)(cid:12)(cid:12) τ / ≪ . (64)12ur result in Eq.(63) is consistent with the result obtained in Ref.[18] once we drop theterms including τ which enters due to the constraint of the initial conditions.Now let us take into account the stability problems in this specific example. We willfollow the method given by Gubser and Yarom in Ref.[29]. In order to do that, we rewritethe same homogeneous differential equations corresponding to Eqs.(53) and (54) or Eqs.(58)and (59) or even higher orders as ∂ τ δ ˆ T + 13 τ ∂ η δu η + 13 ∇ ⊥ · δu ⊥ = 0 , (65) ∂ τ δu η + 23 τ δu η + 1 τ ∂ η δ ˆ T = 0 , (66) ∂ τ δ u ⊥ − τ δ u ⊥ + ∇ ⊥ δ ˆ T = 0 , (67)where δ ˆ T = δT /T , δu η = v · δu and δ u ⊥ = (0 , δu x , δu y , δT and δu µ to denote T , T , ..., T n and ¯ u µ , ¯ u µ , ..., ¯ u nµ respectively. In the momentum space, δ ˆ T = ˆ δ ˆ T e ik η η + i k ⊥ · x ⊥ dk η d k ⊥ , δ ˆ u µ = ˆ δ ˆ U µ e ik η η + i k ⊥ · x ⊥ dk η d k ⊥ . (68)It follows that ∂ τ δ ˆ T + i τ k η δU η + i k ⊥ · δ U ⊥ = 0 , (69) ∂ τ δU η + 23 τ δU η + iτ k η δ ˆ T = 0 , (70) ∂ τ δ U ⊥ µ − τ δ U ⊥ µ + i k ⊥ µ δ ˆ T = 0 . (71)We can decompose δ U ⊥ µ into δ U ⊥ = k ⊥ δW + ˜ k ⊥ δ ˜ W , (72)where ˜ k ⊥ is a constant transverse vector satisfying k ⊥ · ˜ k ⊥ = 0. Then we find δ ˜ W decoupleswith the other functions ∂ τ δ ˆ T + i τ k η δU η + i k ⊥ δW = 0 , (73) ∂ τ δU η + 23 τ δU η + iτ k η δ ˆ T = 0 , (74) ∂ τ δW + iδ ˆ T − τ δW = 0 , (75) ∂ τ δ ˜ W − τ δ ˜ W = 0 . (76)13he solution for δ ˜ W is given by δ ˜ W = δ ˜ W (cid:18) ττ (cid:19) / . (77)We cannot get the analytic solutions for the other functions with the arbitrary k η and k ⊥ .However, we can take two limits k η = 0 and k ⊥ = 0. When k η = 0, we can have δW = C τ / J (cid:18) k ⊥ τ (cid:19) + C τ / N (cid:18) k ⊥ τ (cid:19) , (78) δU η = C τ / , δ ˆ T = i (cid:18) ∂ τ − τ (cid:19) δW, (79)where J and N denote Bessel and Neumann functions respectively and C , C , and C areall integration constants. When k ⊥ = 0, we can have the solution δU η = C (cid:18) τ (cid:19) √ − k η + C (cid:18) τ (cid:19) − √ − k η , (80) δ ˆ T = iτk η (cid:18) ∂ τ + 23 τ (cid:19) δU η , (81) δW = C τ / + iC k η τ − √ − k η + iC k η τ √ − k η , (82)where C , C , and C are also integration constants. From the results of both limits, we cannoticed that the perturbations δU η and δ ˆ T always decay with the proper time τ increasing.The perturbation δ ˜ W increases as τ / with the proper time. The evolution of the perturba-tion δW is more complicated and depends on the specific k η and k ⊥ . However, it is obviousthat there is no exponential increase and the behavior of the perturbation increase must beless than the first order of τ . V. CONCLUSION
In this paper, we have presented a perturbative procedure for solving the viscous hydro-dynamic equation order by order in the framework of an effective theory.For simplicity, we only consider a conformal fluid and more general cases can be straight-forward to be obtained. Firstly, we expand all hydrodynamic quantities and differentialequations in the power of the Knudsen number. Secondly, we assume the conservation equa-tions are satisfied order by order independently. In the leading order, we get the solutions ofan ideal fluid. By solving the differential equations at first and second order, we find these14quations have a uniform expression with different sources. Therefore, we argued that ourmethod can be extended to any orders. We have taken the Bjorken flow as an exampleand found that our method is very powerful and has good advantage to deal with the initialcondition and perturbation evolution. It should be noticed that in our current work welimited ourselves to the theoretical analysis; we postpone the complete numerical analysisand manipulation to a future study.
Acknowledgments
J.H.G. was supported in part by the Major State Basic Research Development Program inChina (Grant No. 2014CB845406), the National Natural Science Foundation of China underthe Grant No. 11105137, 11475104 and CCNU-QLPL Innovation Fund (QLPL2014P01). S.Pwas supported in part by the NSFC under the Grant No. 11205150.
I. ORDER EXPANSION IN KINETIC THEORY
Our method is inspired by microscopic kinetic theory. As a macroscopic effective theory,hydrodynamic equations can be obtained from other microscopic theories. In most of thosemicroscopic theories, the differential equations are expanded in terms of scaling, then aresolved in each order independently. As an example, let us consider the relativistic kinetictheory Boltzmann equations without external fields, dfdt ≡ p µ E p ∂ µ f = C [ f ] , (1)where f is the distribution functions of particles, p µ = ( E p , p ) is the four-momentum ofparticles and C [ f ] is the collision term. We can expand f and C [ f ] in a gradient expansionway, f = f + f + ... , C = C + C , which is equivalent to expanding in powers of K . Forsimplicity, we neglect the higher order terms in the collision term, and simply set C = C . Inthis case, the current and energy-momentum tensor in each order are given by the integrationover momentum, i.e., j µn = ´ d p (2 π ) p µ E p f n and T µνn = ´ d p (2 π ) p µ p ν E p f n , where the lower index n means the n -th order in the gradient expansion. Taking covariant derivatives, we get, ∂ µ j µn = ˆ d p (2 π ) p µ E p ∂ µ f n = ˆ d p (2 π ) C [ f n − ] = 0 ,∂ µ T µνn = ˆ d p (2 π ) p µ p ν E p ∂ µ f n = ˆ d p (2 π ) p ν C [ f n − ] = 0 , (2)15here in the last line, we used the results of the time-reversal symmetry of the collision term,which guarantees total energy-momentum and number conservation. That implies that ineach order the currents and energy-momentum tensor are conserved independently, whichis very similar to our method. [1] U. Heinz and R. Snellings, Annu. Rev. Nucl. Part. Sci. , 123 (2013).[2] C. Gale, S. Jeon and B. Schenke, Int. J. Mod. Phys. A , 1340011 (2013).[3] D. A. Teaney, arXiv:0905.2433 [nucl-th].[4] P. Romatschke and U. Romatschke, Phys. Rev. Lett. , 172301 (2007);[5] H. Song and U. Heinz, Phys. Lett. B658 , 279 (2008); Phys. Rev. C , 064901 (2008); Phys.Rev. C , 024902 (2008);[6] K. Dusling and D. Teaney, Phys. Rev. C , 034905 (2008).[7] D. Molnar and P. Huovinen, J. Phys. G , 104125 (2008).[8] B. Schenke, S. Jeon and C. Gale, Phys. Rev. Lett. , 042301 (2011); Phys. Rev. C ,024901 (2012).[9] C. Eckart, Phys. Rev. , 919 (1940).[10] L. D. Landau and E. M Lifshitz, Fluid Mechanics (Pergamon, London, 1959)[11] W. A. Hiscock and L. Lindblom, Annals Phys. , 466 (1983). W. A. Hiscock and L. Lind-blom, Phys. Rev. D , 725 (1985). W. A. Hiscock and L. Lindblom, Phys. Rev. D , 3723(1987).[12] I. M¨uller, Z. Phys. 198, 329(1967)[13] W. Israel and J. M. Stewart, Annals Phys. (1979) 341.[14] S. Pu, T. Koide and D. H. Rischke, Phys. Rev. D , 114039 (2010)[15] G. S. Denicol, T. Kodama, T. Koide and P. Mota, J. Phys. G , 115102 (2008)[arXiv:0807.3120 [hep-ph]].[16] S. Pu, T. Koide and Q. Wang, AIP Conf. Proc. , 186 (2010).[17] S. Pu, arXiv:1108.5828 [hep-ph].[18] R. Baier, P. Romatschke, D. T. Son, A. O. Starinets and M. A. Stephanov, JHEP , 100(2008)[19] B. Betz, D. Henkel and D. H. Rischke, Prog. Part. Nucl. Phys. , 556 (2009)
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