Relativistic quantum fluid with boost invariance
RRelativistic quantum fluid with boost invariance
D. Rindori, L. Tinti,
2, 3
F. Becattini, and D.H. Rischke
2, 4 Università di Firenze and INFN Sezione di FirenzeVia G. Sansone 1, Sesto Fiorentino, I-50019 Florence, Italy Institut für Theoretische Physik, Johann Wolfgang Goethe-UniversitätMax-von-Laue-Str. 1, D-60438 Frankfurt am Main, Germany Instytut Fizyki, Uniwersytet Jana Kochanowskiego w Kielcachul. Uniwersytecka 7, PL 25-406 Kielce, Poland Helmholtz Research Academy Hesse for FAIR, Campus Riedberg,Max-von-Laue-Str. 12, D-60438 Frankfurt am Main, Germany
We study a relativistic fluid with longitudinal boost invariance in a quantum-statistical frameworkas an example of a solvable non-equilibrium problem. For the free quantum field, we calculate theexact form of the expectation values of the stress-energy tensor and the entropy current. For thestress-energy tensor, we find that a finite value can be obtained only by subtracting the vacuumof the density operator at some fixed proper time τ . As a consequence, the stress-energy tensoracquires non-trivial quantum corrections to the classical free-streaming form. I. INTRODUCTION
Spurred by a successful description of experimental data in high-energy nuclear collisions, relativistic hydrody-namics has recently made major progress, both regarding its theoretical foundations as well as its phenomenologicalapplications. Lately, the quantum-statistical foundations of relativistic hydrodynamics have attracted a great deal ofattention [1–5], in particular to describe quantum phenomena in relativistic fluids such as chirality [6] and polariza-tion [7]. In a quantum-statistical framework, hydrodynamic quantities, such as the stress-energy tensor and conservedcurrents, are the expectation values of the corresponding quantum operators with respect to a suitable statistical (ordensity) operator (cid:98) ρ : T µν = tr( (cid:98) ρ (cid:98) T µν ) ren , (1)where the subscript “ ren ” implies renormalization of the otherwise divergent expectation value.In general, the form of the stress-energy tensor and the currents crucially depends on the density operator. Exactexpressions are known only in a few cases, including the familiar global thermodynamic equilibrium and, as a recentdevelopment, global thermodynamic equilibrium with rotation and acceleration. However, no exact form is known inlocal thermodynamic equilibrium, which is defined by [1, 2, 8, 9]: (cid:98) ρ LE = 1 Z exp (cid:20) − (cid:90) Σ dΣ µ (cid:16) (cid:98) T µν β ν − ζ (cid:98) j µ (cid:17)(cid:21) , (2)where β ( x ) is a four-temperature field [equal to the four-velocity u ( x ) divided by the temperature T ( x ) ], ζ ( x ) isa scalar field [equal to the ratio of the chemical potential µ ( x ) associated with the conserved current (cid:98) j and thetemperature]. The hypersurface Σ is a three-dimensional space-like hypersurface, on which local equilibrium is defined.The calculation of expectation values of operators using Eq. (2) can be performed only in the hydrodynamic limitof slowly varying fields [1]. For the stress-energy tensor, the leading-order term coincides with the familiar perfect-fluid expression. Beyond this approximation, quantum corrections appear, which have been estimated by means of aperturbative expansion only in the global-equilibrium case [10].Recently, S. Akkelin [11, 12] has derived an exact solution of a particular non-equilibrium problem, a free neutralscalar field with the density operator: (cid:98) ρ = 1 Z exp (cid:34) − T ( τ ) (cid:90) Σ( τ ) dΣ µ (cid:98) T µν u ν (cid:35) , (3)with Σ( τ ) being a proper-time τ hyperbola in the future light-cone in two dimensions (see Fig. 1) and u ( x ) thefour-velocity field coinciding with the unit vector perpendicular to Σ . The density operator (3) is invariant underlongitudinal boosts, a symmetry which has been often used to study general features of relativistic hydrodynamicsproblems. Lately, longitudinal boost-invariant solutions have been studied in the context of spin-hydrodynamics[13] and magneto-hydrodynamics [14–16]. This symmetry and the solution found in Ref. [11] also offers a specialopportunity to explore in detail some essential features of quantum relativistic hydrodynamics in a non-equilibrium a r X i v : . [ h e p - t h ] F e b Figure 1: Two-dimensional section of the future light cone. Curves of constant Milne time τ are hyperbolae, whilecurves of constant space-time rapidity η are lines through the origin. The thicker hyperbolae are two-dimensionalsections of the three-dimensional hypersurfaces Σ( τ ) and Σ( τ ) at constant τ and constant τ , respectively.situation and, in particular, to determine the pure quantum corrections to classical hydrodynamics and kinetic equa-tions, including those to the stress-energy tensor and to the entropy current. In other words, this solution provides abenchmark test of a relativistic quantum fluid.In this work, we extend the results of Ref. [11] and study the stress-energy tensor with longitudinal boost-invariantsymmetry. We find that, even for the simplest case of a free scalar field, there are relevant quantum correctionsrelated to its renormalization by subtraction of the vacuum expectation value. Indeed, while the traditional vacuumof the field, expanded in plane waves, the so-called Minkowski vacuum, fails to provide a finite energy density, thesubtraction of the vacuum expectation value with respect to the vacuum of the density operator does. Conversely, forthe entropy current, no significant quantum correction is found.This paper is organized as follows. We will start in Sec. II with a review of the density-operator approach inrelativistic quantum-statistical mechanics with special emphasis on symmetry considerations. In Sec. III we willspecialize to the symmetry of concern for this work, that is boost invariance. As underlying quantum field theory, inSec. IV we will present the field theory of the free neutral scalar field in the future light cone, including a diagonalizationof the density operator. This will put us in the position to calculate the thermal expectation value of the stress-energytensor in Sec. V both in local thermodynamic equilibrium and out of equilibrium. Finally, we will discuss the entropycurrent and entropy production in Sec. VI, before concluding this paper in Sec. VII.In this work, we use natural units (cid:126) = c = k B = 1 . Operators in Hilbert space are denoted with a wide upper hat,e.g., (cid:98) O , while vectors of unit length have a regular hat, that is ˆ n µ . Repeated indices are assumed to be contracted.We adopt the “mostly-minus” convention, so the Minkowski metric is g µν = diag(1 , − , − , − . For the Levi-Civitasymbol we use the convention (cid:15) = 1 . II. LOCAL THERMODYNAMIC EQUILIBRIUM, DENSITY OPERATOR, AND SYMMETRIES
In quantum-statistical mechanics, the local-equilibrium density operator (LEDO) (cid:98) ρ LE , Eq. (2), is obtained bymaximizing the entropy S = − Tr( (cid:98) ρ LE log (cid:98) ρ LE ) under the constraints of fixed energy-momentum and, possibly, chargedensities on a given three-dimensional space-like hypersurface Σ . The hypersurface can be either specified a priori orcan be found in a self-consistent procedure by using the thermodynamic fields themselves [1].The energy-momentum densities on a hypersurface Σ are obtained by contracting the stress-energy tensor with itsnormal unit vector n , so that the constraints read: n µ Tr( (cid:98) ρ LE (cid:98) T µν ) ren = n µ T µν , (4)and likewise for the conserved currents. The densities on the right-hand side of Eq. (4) are meant to be the actual ones,no matter how they are known or defined, and they are supposedly finite. It is crucial to specify that the expectationvalues on the left-hand side must be suitably renormalized because, in general, the expectation value of the operator (cid:98) T µν with a density operator such as in Eq. (2) is divergent. For instance, in free field theory, the renormalizationprocedure is most readily established by subtracting the vacuum expectation value, that is: Tr( (cid:98) ρ (cid:98) T µν ) ren = Tr( (cid:98) ρ (cid:98) T µν ) − (cid:104) | (cid:98) T µν | (cid:105) , (5)which is tantamount to normal-ordering of the creation and annihilation operators because the currents are quadraticin the fields. We will delve into the question of vacuum subtraction in Sec. III A.With the constraints (4), the function to be maximized with respect to (cid:98) ρ LE is: − Tr( (cid:98) ρ LE log (cid:98) ρ LE ) + (cid:90) Σ( τ ) dΣ µ (cid:104)(cid:16) tr( (cid:98) ρ LE (cid:98) T µν ) ren − T µν (cid:17) β ν − ζ (cid:16) tr( (cid:98) ρ LE (cid:98) j µ ) ren − j µ (cid:17)(cid:105) , (6)where the thermodynamic fields β and ζ are Lagrange multipliers introduced to enforce the constraints (4). Thesolution is Eq. (2) and it should be pointed out that it can be kept in that simple form without subtraction of thevacuum expectation value because the latter is not an operator and would appear in the partition function Z as well(in order to make Tr (cid:98) ρ LE = 1 ), hence cancelling out in the ratio. With the energy-momentum densities given by theright-hand side of Eq. (4), the thermodynamic fields β and ζ are determined by solving them with (cid:98) ρ LE given by Eq.(2); there are five equations with five unknowns ( β µ and ζ ), which in general can be solved.Unless β is a Killing field and ζ constant, which characterizes a state of global thermodynamic equilibrium, theoperator (2) is not independent of the hypersurface, hence it cannot be the actual density operator in the Heisenbergrepresentation. In fact, the true density operator is, for a system which supposedly achieves local thermodynamicequilibrium at some time τ , the so-called non-equilibrium density operator (NEDO), which is just Eq. (2) at time τ : (cid:98) ρ = 1 Z exp (cid:34) − (cid:90) Σ( τ ) dΣ µ (cid:16) (cid:98) T µν β ν − ζ (cid:98) j µ (cid:17)(cid:35) . (7)This can be recast by using Gauss’ theorem as [17] (cid:98) ρ = 1 Z exp (cid:34) − (cid:90) Σ( τ ) dΣ µ (cid:16) (cid:98) T µν β ν − (cid:98) j µ ζ (cid:17)(cid:35) = 1 Z exp (cid:34) − (cid:90) Σ( τ ) dΣ µ (cid:16) (cid:98) T µν β ν − (cid:98) j µ ζ (cid:17) + (cid:90) Ω dΩ (cid:16) (cid:98) T µν ∇ µ β ν − (cid:98) j µ ∇ µ ζ (cid:17)(cid:35) . (8)In the exponent on the right-hand side, the first term is just the operator of local equilibrium at time τ , while thesecond term contains dissipative corrections [17].Suppose now that the actual density operator, the NEDO, has some symmetry, meaning that it commutes withsome unitary representation (cid:98) U ( g ) in Hilbert space of a group or a subgroup G of transformations, to be specific ofthe proper orthochronous Poincaré group IO(1,3) ↑ + . We have: (cid:98) U ( g ) (cid:98) ρ (cid:98) U ( g ) − = 1 Z exp (cid:34) − (cid:90) Σ( τ ) dΣ µ ( x ) (cid:16) (cid:98) U ( g ) (cid:98) T µν ( x ) (cid:98) U ( g ) − β ν ( x ) − ζ ( x ) (cid:98) U ( g ) (cid:98) j µ ( x ) (cid:98) U ( g ) − (cid:17)(cid:35) = 1 Z exp (cid:34) − (cid:90) Σ( τ ) dΣ µ ( x ) (cid:16) D ( g − ) µρ D ( g − ) νσ (cid:98) T ρσ ( g ( x )) β ν ( x ) − ζ ( x ) D ( g − ) µρ (cid:98) j ρ ( g ( x )) (cid:17)(cid:35) . Let us now set y = g ( x ) and we obtain, remembering dΣ µ ( x ) = D ( g ) νµ dΣ ν ( y ) , (cid:98) U ( g ) (cid:98) ρ (cid:98) U ( g ) − = 1 Z exp (cid:34) − (cid:90) g (Σ( τ )) dΣ ρ ( y ) (cid:16) (cid:98) T ρσ ( y ) D ( g − ) νσ β ν ( g − ( y )) − ζ ( g − ( y )) (cid:98) j ρ ( y ) (cid:17)(cid:35) . Thus, if the hypersurface is invariant under the transformation g and if: D ( g − ) νσ β ν ( g − ( y )) = β σ ( y ) , ζ ( g − ( y )) = ζ ( y ) , (9)then the operator (cid:98) ρ is invariant under the transformation (cid:98) U ( g ) (cid:98) ρ (cid:98) U ( g ) − . Equations (9) specify the symmetry conditionson the transformations of the thermodynamic fields β and ζ . An invariance of (cid:98) ρ has straightforward consequences forthe expectation values of operators. For instance, for the stress-energy tensor: T µν ( x ) = Tr[ (cid:98) ρ (cid:98) T µν ( x )] = Tr (cid:104)(cid:98) ρ (cid:98) U ( g ) − (cid:98) T µν ( x ) (cid:98) U ( g ) (cid:105) = D ( g ) µρ D ( g ) νσ Tr (cid:104)(cid:98) ρ (cid:98) T µν ( g − ( x )) (cid:105) = D ( g ) µρ D ( g ) νσ T µν ( g − ( x )) . (10)If we consider a one-parameter subgroup of transformations g φ [e.g., a rotation, g φ = exp( − iφ J ) , around some axis],Eqs. (9) and (10) have the consequence that the Lie derivative along the vector field X ( x ) = d g φ ( x ) / d φ of the fieldunder consideration vanishes, that is: L X ( β ) µ = 0 , L X ( T ) µν = 0 . (11)An important question concerns the persistence of the symmetry of the local thermodynamic equilibrium operator,that is whether the implication: (cid:98) ρ = (cid:98) U ( g ) (cid:98) ρ (cid:98) U ( g ) − = ⇒ (cid:98) ρ LE ( τ ) = (cid:98) U ( g ) (cid:98) ρ LE ( τ ) (cid:98) U ( g ) − is true for any τ . Indeed, it can be shown that if the subgroup G transform Σ( τ ) into itself and if the fields β and ζ are also symmetric under G, namely they fulfill Eqs. (9) or (11), this is the case. Indeed, by definition, (cid:98) ρ LE ( τ ) is the solution of maximizing a function which is invariant under any unitary transformation, the entropy, with theconstraint (4). If a particular (cid:98) ρ LE fulfills Eq. (4), so will (cid:98) U ( g ) (cid:98) ρ LE ( τ ) (cid:98) U ( g ) − as it can be readily checked. Therefore,either (cid:98) U ( g ) (cid:98) ρ LE ( τ ) (cid:98) U ( g ) − is a different solution of the constrained maximization problem, or it coincides with (cid:98) ρ LE ( τ ) .In both cases, it is possible to generate one symmetric solution under the subgroup G by using a particular solution (cid:98) ρ (0)LE and summing over all g ’s: (cid:98) ρ LE ( τ ) = 1 M ( G ) (cid:88) g ∈ G (cid:98) U ( g ) (cid:98) ρ (0)LE (cid:98) U ( g ) − . It is then obvious that the sufficient condition for (cid:98) ρ LE ( τ ) , given by Eq. (2), to be symmetric under G is that the fields β and ζ fulfill Eqs. (9) at time τ . This is a crucial point for the purpose of this work. III. RELATIVISTIC QUANTUM FLUID WITH LONGITUDINAL BOOST INVARIANCE
Suppose that the density operator is given by Eq. (7) with Σ( τ ) being the hyperboloid τ = √ t − z = τ inMinkowski space-time and with β µ = 1 T ( τ ) 1 τ ( t, , , z ) = 1 T ( τ ) u µ , (12)where T ( τ ) and ζ ( τ ) are constant on the hypersurface. This vector field is time-like on the hypersurface Σ( τ ) ,hence thermodynamically meaningful.The field β in Eq. (12) and the field ζ fulfill Eq. (9) for any longitudinal boost with hyperbolic angle ξ along the z axis, L z ( ξ ) , and manifestly for translations and rotations in the xy plane. Besides, the hypersurface Σ( τ ) is invariantunder the same transformations. Therefore, the density operator has the symmetry group IO(2) ⊗ SO(1 , , that isthe Euclidean group in the transverse plane times Lorentz transformations in the longitudinal direction. Furthermore,the density operator is also invariant under a space-reflection transformation turning x, y, z into − x, − y, − z .This symmetry group dictates the possible forms of vector and tensor fields, which are most easily found by usingMilne coordinates, ( τ, x, y, η ) , instead of the usual Cartesian ones, ( t, x, y, z ) : t = τ cosh η , z = τ sinh η ,τ = (cid:112) t − z , η = 12 log (cid:18) t + zt − z (cid:19) , whence it turns out that the coordinate basis vectors are: ∂∂τ = 1 τ ( t, , , z ) = (cosh η, , , sinh η ) = u , ∂∂η = τ (sinh η, , , cosh η ) = ( z, , , t ) ≡ τ ˆ η ,∂∂x = ˆ i , ∂∂y = ˆ j , and the metric tensor: d s = d t − d x − d y − d z = d τ − d x − d y − τ d η . The vector fields X ( x ) associated with the symmetry group along which the Lie derivatives vanish can be readilyfound: d T ( a ) x d a = ˆ i , d T ( a ) y d a = ˆ j , (13) d R ( φ ) x d φ = (0 , − y, x, ≡ r ˆ ϕ , d L ( ξ ) x d ξ = ( z, , , t ) = τ ˆ η , where T i are translations in the coordinate directions of the xy plane, r = (cid:112) x + y , R ( ϕ ) is a rotation with angle ϕ in the same plane, and L ( ξ ) is a longitudinal boost with hyperbolic angle ξ . Note that three vector fields are justthe Milne-coordinate basis vectors, which, by construction, have vanishing Lie derivatives among each other, that isvanishing Lie commutators.As has been mentioned, the condition of vanishing Lie derivatives along the vector fields (13) puts strong limitationson the form of the fields in general. For instance, a vector field V ( x ) can be decomposed onto the coordinate basisvectors: V ( x ) = A ( τ ) u + B ( τ )ˆ i + C ( τ )ˆ j + D ( τ ) τ ˆ η , where the coefficients depend on the variable τ only as a consequence of L X ( V ) = 0 , where X is either ˆ i , or ˆ j , or τ ˆ η . Also, by implementing L ˆ ϕ ( V ) = 0 one obtains that both B and C are in fact zero. Furthermore, by reflectioninvariance, the component proportional to ˆ η must be vanishing because a reflection turns η into − η and the vectorfield has just one component: V ( x ) = A ( τ ) u. (14)Similarly, the form a symmetric tensor field like the stress-energy tensor T µν can be obtained by iterated projectionsonto vectors and orthogonal components. The result is: T µν = E ( τ ) u µ u ν + P T ( τ ) (cid:16) ˆ i µ ˆ i ν + ˆ j µ ˆ j ν (cid:17) + P L ( τ )ˆ η µ ˆ η ν . (15)The form (15) is different from the usual perfect fluid form, for which P T = P L . The difference between transverseand longitudinal pressure is owing to the lack of full rotational symmetry and it is to be expected, on general grounds,that this difference is a quantum effect, as already observed for global equilibrium [10].In order to determine the three scalar functions in Eq. (15), we have to calculate the expectation values of operatorswith the density operator (7). The unit four-vector orthogonal to the hyperboloid with fixed τ is u itself, so theoperator (7) becomes: (cid:98) ρ = 1 Z exp (cid:34) − (cid:98) Π( τ ) T ( τ ) (cid:35) , (16)with: (cid:98) Π( τ ) = (cid:90) Σ( τ ) dΣ u µ u ν (cid:98) T µν = τ (cid:90) d x d y d η u µ u ν (cid:98) T µν , where we have used the measure in Milne coordinates. It should be stressed that the operator (cid:98) Π( τ ) is not conservedbecause the divergence of the integrand is not zero: ∂ µ (cid:16) u ν (cid:98) T µν (cid:17) = (cid:98) T µν ∂ µ u ν (cid:54) = 0 , so it depends on τ . We can also write down a general form of the local equilibrium operator (cid:98) ρ LE ( τ ) at any Milnetime τ by taking the hyperboloid τ = const. as local-equilibrium hypersurface, which is invariant under the sametransformations as Σ( τ ) , according to the discussion in Sec. II. Since the field β ( τ ) must fulfill Eqs. (9) and (11), itcan only be of the form (14): β = 1 T ( τ ) u = 1 T ( τ ) (cosh η, , , sinh η ) , thus the constraint (4) becomes, by using Eq. (15): n µ Tr( (cid:98) ρ LE (cid:98) T µν ) ren ≡ n µ T µν LE = u µ T µν LE = E ( τ ) LE u ν = n µ T µν = E ( τ ) u ν . This vector equation comes down to one scalar equation E ( τ ) LE = E ( τ ) with T ( τ ) as unknown to be determinedonce the actual E ( τ ) is determined by using the actual density operator (7). The local thermodynamic equilibriumoperator will be of the same form as Eq. (16), that is: (cid:98) ρ LE ( τ ) = 1 Z exp (cid:34) − (cid:98) Π( τ ) T ( τ ) (cid:35) , (17)with (cid:98) Π( τ ) (cid:54) = (cid:98) Π( τ ) . A. Vacuum effects
A very interesting feature of a relativistic quantum fluid with the four-temperature field (12) is that the spectrum of (cid:98) Π( τ ) , and particularly the lowest-lying eigenvector, the (cid:98) Π vacuum, may depend on τ , as it is clear from Ref. [11]. This τ -dependent vacuum | τ (cid:105) is in general also different from the vacuum of a quantum field theory – even for free fields – inflat space-time obtained by quantizing in Cartesian coordinates, the so-called Minkowski vacuum | M (cid:105) . This is clearlyat variance with familiar equilibrium quantum thermodynamics, where the Hamiltonian operator achieves its minimaleigenvalue in the Minkowski vacuum. The distinction between vacua is very important as to the renormalization ofseveral quantities, including, e.g., the stress-energy tensor. In a free field theory, the renormalization of the expectationvalue of an operator (cid:98) O involves the subtraction of its vacuum expectation value. If more vacua are present, there isan ambiguity as we could define, as usual: (cid:104) (cid:98) O (cid:105) ren ≡ Tr( (cid:98) ρ (cid:98) O ) − (cid:104) M | (cid:98) O | M (cid:105) , (18)[see Eq. (5)] or, in our case: (cid:104) (cid:98) O (cid:105) ren ≡ Tr( (cid:98) ρ (cid:98) O ) − (cid:104) τ | (cid:98) O | τ (cid:105) . (19)Note that the (cid:98) Π vacuum can be subtracted by taking the limit T ( τ ) → of the unrenormalized expression since: lim T ( τ ) → (cid:98) ρ LE ( τ ) = lim T ( τ ) → Z exp[ − (cid:98) Π( τ ) /T ( τ )] = | τ (cid:105)(cid:104) τ | ≡ P τ . For this reason, in general the vacuum | τ (cid:105) will have the same symmetries as the original density operator, but it willbe less symmetric than the supposedly Poincaré-invariant Minkowski vacuum | M (cid:105) .It should be pointed out that the vacuum | τ (cid:105) is τ -dependent, hence a subtraction like in Eq. (19) implies that theexpectation value can get an undesired time dependence. For instance, if we define the renormalized stress-energytensor as: T µν ≡ Tr( (cid:98) ρ (cid:98) T µν ) − (cid:104) τ | (cid:98) T µν | τ (cid:105) = Tr[( (cid:98) ρ − P τ ) (cid:98) T µν ] , then: ∂ µ T µν = ∂ µ Tr[( (cid:98) ρ − P τ ) (cid:98) T µν ] = Tr[( (cid:98) ρ − P τ ) ∂ µ (cid:98) T µν ] + Tr[ − ( ∂ µ P τ ) (cid:98) T µν ] = − Tr (cid:20) u µ ∂ P τ ∂τ (cid:98) T µν (cid:21) (cid:54) = 0 , where we used ∂ µ (cid:98) T µν = 0 and the time independence of the density operator. Therefore, the expectation value T µν would no longer fulfill a conservation equation even though the operator (cid:98) T does.Therefore, in order to have a properly finite, conserved stress-energy tensor for a relativistic quantum fluid, thevacuum must be necessarily fixed, just like the density operator. Of course the Minkowski vacuum | M (cid:105) meets thisrequirement and is seemingly the most obvious choice. However, we will see in Sec. V that the subtraction of thevacuum expectation value of (cid:98) T µν of a free field with respect to | M (cid:105) does not give rise to a finite value, for theparticular symmetry we are dealing with, and an alternative definition is needed. This does not mean that the vacuum | τ (cid:105) is degenerate, but that Poincaré transformations will give rise to non-vanishing componentsof excited states. IV. FREE SCALAR FIELD IN MILNE COORDINATES
As has been mentioned in the Introduction, a closed analytic form of the stress-energy tensor with the four-temperature field (12) exists for the case of free fields, providing the opportunity to determine exact quantum correc-tions to the classical expressions in the non-equilibrium case. The system which is described by the operator (7) and afree scalar field is that of a fluid where interactions effectively cease at the hypersurface Σ( τ ) with temperature T ( τ ) and a four-velocity u = T β , with particles freely streaming thereafter. We thus expect to recover, in the classical limit,the classical kinetic-theory solutions of the free-streaming Boltzmann equation starting from the local thermodynamicequilibrium expressions with proper temperature T ( τ ) and flow velocity u ( τ ) .The calculation of the stress-energy tensor for the massive free scalar field (cid:98) ψ ( x ) requires the solution of the Klein-Gordon equation in Milne coordinates: (cid:20) τ ∂ τ ( τ ∂ τ ) − ∂ x − ∂ y − τ ∂ η + m (cid:21) (cid:98) ψ ( τ, x T , η ) = 0 . This is a well-known problem in the literature [18, 19], which has even raised some discussion. It has been convincinglydemonstrated [19] that, within the future light cone, there is a complete set of solutions of the Klein-Gordon equationin Milne coordinates, which allow an expansion in terms of the familiar plane waves and which do not mix positiveand negative frequencies. These mode functions can be obtained starting from the usual expansion of the scalar field[11] in plane waves. We will recapitulate the salient points of the derivation presented in Ref. [11]. The obtained fullexpansion of the field in Milne coordinates reads: (cid:98) ψ ( τ, x T , η ) = (cid:90) d p T d µ π √ (cid:104) h ( p , τ )e i ( p T · x T + µη ) (cid:98) b p + h ∗ ( p , τ )e − i ( p T · x T + µη ) (cid:98) b † p (cid:105) , (20)where p = ( p T , µ ) to distinguish it from the Cartesian vector p = ( p T , p z ) . Here, (cid:98) b † p and (cid:98) b p are creation andannihilation operators satisfying the usual algebra: [ (cid:98) b p , (cid:98) b † p (cid:48) ] = δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , [ (cid:98) b p , (cid:98) b p (cid:48) ] = 0 = [ (cid:98) b † p , (cid:98) b † p (cid:48) ] . (21)The relation between the operators (cid:98) b † p and the familiar (cid:98) a † ( p ) of the plane-wave expansion reads: (cid:98) a † ( p ) = 1 √ πm T cosh y (cid:90) + ∞−∞ d µ e − iµy (cid:98) b † p , (22)where y is the particle rapidity in longitudinal direction, which can be easily inverted to obtain p z . Since there is nomixing between creation and annihilation operators, the vacuum of the b p operators is the same Minkowski vacuum | M (cid:105) as for the operators a ( p ) , which is a consequence of the fact that the functions h ( p , τ ) can be expressed as a linearcombination of plane waves with just positive frequency [20]. In Eq. (20) µ is the eigenvalue of the boost operator (cid:98) K z , so that: (cid:98) U ( L ( ξ )) (cid:98) b † p (cid:98) U ( L ( ξ )) − = e − iξ (cid:98) K z (cid:98) b † p e iξ (cid:98) K z = e − iξµ (cid:98) b † p , i.e., (cid:98) b † p creates a state with eigenvalue µ . The τ -dependent functions in Eq. (20) are h ( p , τ ) = − i e π µ H (2) iµ ( m T τ ) , h ∗ ( p , τ ) = i e − π µ H (1) iµ ( m T τ ) , (23)where the Hankel functions are [21]: H (2) iµ ( m T τ ) = − iπ e − π µ (cid:90) + ∞−∞ d θ e − im T τ cosh θ + iµθ , H (1) iµ ( m T τ ) = 1 iπ e π µ (cid:90) + ∞−∞ d θ e im T τ cosh θ − iµθ , (24)with m T = (cid:112) p + m being the transverse mass. The integration variable θ in Eq. (24) is related to the Milnecoordinates and rapidity by [11]: θ = y − η . (25)The functions (24) solve the differential equations: (cid:20) τ ∂ τ ( τ ∂ τ ) + m T + µ τ (cid:21) h ( p , τ ) = 0 , which are indeed Bessel’s differential equations. It is also useful to define: ω = m + µ τ . (26)Let us now work out the density operator, particularly the operator (cid:98) Π( τ ) in Eq. (17). In a non-equilibrium situationit is known that the density operator depends on the particular stress-energy tensor operator which is employed,however for the free scalar field we will be using the canonical tensor: (cid:98) T µνC = 12 (cid:16) ∂ µ (cid:98) ψ∂ ν (cid:98) ψ + ∂ ν (cid:98) ψ ∂ µ (cid:98) ψ (cid:17) − g µν (cid:98) L , (cid:98) L = 12 (cid:16) g µν ∂ µ (cid:98) ψ∂ ν (cid:98) ψ − m (cid:98) ψ (cid:17) , where (cid:98) L is the Lagrangian density. Hence: (cid:98) T µνC u µ u ν = 12 (cid:20)(cid:16) ∂ τ (cid:98) ψ (cid:17) + (cid:16) ∂ x (cid:98) ψ (cid:17) + (cid:16) ∂ y (cid:98) ψ (cid:17) + 1 τ (cid:16) ∂ η (cid:98) ψ (cid:17) + m (cid:98) ψ (cid:21) . (27)By using the above equation along with Eq. (20) and taking advantage of the invariance by reflection p → − p of thefunctions h ( p , τ ) , one can obtain the following expression for (cid:98) Π( τ ) : (cid:98) Π( τ ) = τ (cid:90) d x d y d η (cid:98) T µν u µ u ν = (cid:90) d p T d µ ω (cid:104) K (cid:16)(cid:98) b p (cid:98) b † p + (cid:98) b † p (cid:98) b p (cid:17) + Λ (cid:98) b p (cid:98) b − p + Λ ∗ (cid:98) b † p (cid:98) b †− p (cid:105) , (28)where the positive real function K ( p , τ ) and the complex function Λ( p , τ ) are defined as: K ( p , τ ) = πτ ω (cid:16) | ∂ τ h ( p , τ ) | + ω | h ( p , τ ) | (cid:17) , (29) Λ( p , τ ) = πτ ω (cid:110) [ ∂ τ h ( p , τ )] + ω h ( p , τ ) (cid:111) . (30)Note that, with ω and h being invariant under a reflection p → − p , so are K and Λ , and: K ( p , τ ) − | Λ( p , τ ) | = 1 , (31)as K − | Λ | is proportional to the Wronskian of the Hankel functions K ( p , τ ) − | Λ( p , τ ) | = − (cid:16) πm T τ (cid:17) (cid:16) W [H (2) iµ ( m T τ ) , H (1) iµ ( m T τ )] (cid:17) , which is known to be a very simple function [21]: W [H (2) iν ( x ) , H (1) iν ( x )] = H (2) (cid:48) iν ( x )H (1) iν ( x ) − H (1) (cid:48) iν ( x )H (2) iν ( x ) = 4 iπx . (32)The above relation is not accidental but it is related to the invariance of the Klein-Gordon scalar product of the modefunctions [20]. Equation (31) allows to write: K ( p , τ ) = cosh 2Θ( p , τ ) , Λ( p , τ ) = sinh 2Θ( p , τ ) exp[ iχ ( p , τ )] , (33)which is very important to highlight the vacuum effects, as it will become clear later.Due to the terms proportional to Λ and Λ ∗ , (cid:98) Π( τ ) in Eq. (28) is not diagonal in the creation and annihilationoperators. If it were, we could easily calculate the expectation values of products of creation and annihilationoperators, hence of operators quadratic in the field, using standard methods. We thus look for a suitable Bogolyubovtransformation that diagonalizes (cid:98) Π( τ ) , (cid:98) ξ † p ( τ ) = A ( p , τ ) (cid:98) b † p − B ( p , τ ) (cid:98) b − p , (cid:98) ξ p ( τ ) = A ∗ ( p , τ ) (cid:98) b p − B ∗ ( p , τ ) (cid:98) b †− p , (34)where A and B are complex functions to be determined. We require (cid:98) ξ † p and (cid:98) ξ p to fulfill the usual algebra: [ (cid:98) ξ p ( τ ) , (cid:98) ξ † p (cid:48) ( τ )] = δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , [ (cid:98) ξ p ( τ ) , (cid:98) ξ p (cid:48) ( τ )] = 0 = [ (cid:98) ξ † p ( τ ) , (cid:98) ξ † p (cid:48) ( τ )] , (35)so that, by enforcing the commutation relations (21), we find respectively (cid:0) | A ( p , τ ) | − | B ( p , τ ) | (cid:1) δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) = δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , [ A ∗ ( − p , τ ) B ∗ ( p , τ ) − A ∗ ( p , τ ) B ∗ ( − p , τ )] δ ( p T + p (cid:48) T ) δ ( µ + µ (cid:48) ) = 0 , [ A ( p , τ ) B ( − p , τ ) − A ( − p , τ ) B ( p , τ )] δ ( p T + p (cid:48) T ) δ ( µ + µ (cid:48) ) = 0 . The above equation is fulfilled if: A ( p , τ ) = A ( − p , τ ) , B ( p , τ ) = B ( − p , τ ) , | A ( p , τ ) | − | B ( p , τ ) | = 1 , (36)so we can set: A ( p , τ ) = cosh θ ( p , τ ) e iχ A ( p ,τ ) , B ( p , τ ) = sinh θ ( p , τ ) e iχ B ( p ,τ ) . (37)The conditions (36) make it easier to invert Eq. (34): (cid:98) b p = A ( p , τ ) (cid:98) ξ p ( τ ) + B ∗ ( p , τ ) (cid:98) ξ †− p ( τ ) , (cid:98) b † p = A ∗ ( p , τ ) (cid:98) ξ † p ( τ ) + B ( p , τ ) (cid:98) ξ − p ( τ ) . (38)Plugging Eq. (38) into Eq. (28) we obtain (cid:98) Π( τ ) = (cid:90) d p T d µ ω (cid:110)(cid:2) K (cid:0) | A | + | B | (cid:1) + Λ AB ∗ + Λ ∗ A ∗ B (cid:3) (cid:16)(cid:98) ξ p (cid:98) ξ † p + (cid:98) ξ † p (cid:98) ξ p (cid:17) + (cid:0) KAB + Λ A + Λ ∗ B (cid:1) (cid:98) ξ p (cid:98) ξ − p + (cid:16) KA ∗ B ∗ + Λ ∗ A ∗ + Λ B ∗ (cid:17) (cid:98) ξ † p (cid:98) ξ †− p (cid:111) , (39)where we used the invariance of the integral under reflections p (cid:55)→ − p . In order to make (cid:98) Π( τ ) diagonal, the secondline of Eq. (39) must vanish: KAB + Λ A + Λ ∗ B = 0 (the other equation is just the complex conjugate). This can be rewritten by using Eqs. (33) and (37): cosh 2Θ sinh 2 θ e i ( χ A + χ B ) + sinh 2Θ cosh θ e i ( χ +2 χ A ) + sinh 2Θ sinh θ e i (2 χ B − χ ) = 0 , the solution of which is: χ B − χ A = χ , θ = − Θ . We can then set χ A = 0 and find A, B fulfilling the Bogolyubov relations (34) A = cosh Θ , B = − sinh Θ e iχ , (40)whence, by using Eq. (33) K (cid:0) | A | + | B | (cid:1) + Λ AB ∗ + Λ ∗ A ∗ B = cosh − (cid:0) sinh 2Θ e iχ cosh Θ sinh Θ e − iχ (cid:1) = 1 . With these solutions, Eq. (38) becomes: (cid:98) b p = cosh Θ( p , τ ) (cid:98) ξ p ( τ ) − sinh Θ( p , τ )e − iχ (cid:98) ξ †− p ( τ ) , (cid:98) b † p = cosh Θ( p , τ ) (cid:98) ξ † p ( τ ) − sinh Θ( p , τ )e iχ (cid:98) ξ − p ( τ ) , (41)and the operator (39): (cid:98) Π( τ ) = (cid:90) d p T d µ ω (cid:16)(cid:98) ξ p ( τ ) (cid:98) ξ † p ( τ ) + (cid:98) ξ † p ( τ ) (cid:98) ξ p ( τ ) (cid:17) = (cid:90) d p T d µ ω (cid:18)(cid:98) ξ † p ( τ ) (cid:98) ξ p ( τ ) + 12 (cid:19) , (42)where in the last equality we have used the commutation relations (35).0 A. Discusssion
The non-trivial Bogoliubov transformation (41) between different sets of creation and annihilation operators isreminiscent of the Unruh effect [22] and indeed the velocity field u implied in Eq. (12) has non-vanishing acceleration.However, we are facing essentially different physics here; as it has been pointed out, the relation (22) between plane-wave creation operators and the creation operators appearing in the field expansion in curvilinear coordinates doesnot mix creation and annihilation operators. In other words, unlike in the Unruh effect, the observers associated withMilne coordinates (defined by η = x T = const ) count the same particles as the inertial observer.In fact, the Bogolyubov transformation (41) stems from the somewhat unexpected form of the local thermodynamicequilibrium operator (cid:98) Π in Eq. (28) involving quadratic combinations of two annihilation and two creation operators,unlike the Hamiltonian in global-equilibrium thermal field theory. We thus have a concrete situation where the vacuum | τ (cid:105) , which is the lowest-lying eigenvector of (cid:98) Π( τ ) annihilated by all (cid:98) ξ p ( τ ) ’s, (cid:98) ξ p ( τ ) | τ (cid:105) = 0 , is different from the Minkowski vacuum | M (cid:105) , which is annihilated by the (cid:98) b p , as envisioned in Sec. III A. The fullexpression of the vacuum | τ (cid:105) can be obtained from the coefficients in Eq. (41) with known methods [20] and reads: | τ (cid:105) = (cid:89) p | cosh Θ( p , τ ) | / exp (cid:20) −
12 tanh Θ( p , τ )e − iχ ( p ,τ ) (cid:98) b † p (cid:98) b †− p (cid:21) | M (cid:105) . (43)With (cid:98) Π diagonal in Eq. (42), we can readily obtain the expectation values of products of creation and annihilationoperators in local thermodynamic equilibrium. The form (42) is essentially the same as the equilibrium Hamiltonianoperator of the free field with the replacements µ → p z and ω → ε . We thus have: (cid:104) (cid:98) ξ † p ( τ ) (cid:98) ξ p (cid:48) ( τ ) (cid:105) LE = n B ( p , τ ) δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , (cid:104) (cid:98) ξ p ( τ ) (cid:98) ξ † p (cid:48) ( τ ) (cid:105) LE = [ n B ( p , τ ) + 1] δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , (cid:104) (cid:98) ξ p ( τ ) (cid:98) ξ p (cid:48) ( τ ) (cid:105) LE =0 = (cid:104) (cid:98) ξ † p ( τ ) (cid:98) ξ † p (cid:48) ( τ ) (cid:105) LE , (44)where (cid:104)·(cid:105) LE stands for Tr( (cid:98) ρ LE · ) and n B is the Bose-Einstein distribution function: n B ( p , τ ) = 1e ω ( τ ) /T ( τ ) − , (45)with ω ( τ ) given by Eq. (26).It is important to emphasize that Eq. (45) is by no means a density of particles as usually in Minkowski space-time.Equation (45) accounts for the mean number of excitations of the (cid:98) ξ † p ( τ ) operator, which is not the mean number ofexcitations of the Minkowski vacuum as expressed by the (cid:98) a ( p ) ’s or (cid:98) b p ’s. Indeed, the expectation values of the variouscombinations can be found by means of Eq. (38) including the solution (40) and Eq. (44): (cid:104) (cid:98) b p (cid:98) b p (cid:48) (cid:105) LE = −
12 sinh(2Θ) e − iχ (2 n B + 1) δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , (46) (cid:104) (cid:98) b † p (cid:98) b † p (cid:48) (cid:105) LE = −
12 sinh(2Θ) e iχ (2 n B + 1) δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , (cid:104) (cid:98) b p (cid:98) b † p (cid:48) (cid:105) LE = (cid:2) n B cosh(2Θ) + cosh Θ (cid:3) δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , (cid:104) (cid:98) b † p (cid:98) b p (cid:48) (cid:105) LE = (cid:2) n B cosh(2Θ) + sinh Θ (cid:3) δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) . As is clear from Eq. (46), field vacuum effects are encoded in a non-vanishing value of the angle Θ( p , τ ) , which is botha function of the modes and of the Milne time τ and whose value can be determined through the relations (33). Weare now in a position to calculate the expectation values of all operators which are quadratic in the field. V. THE STRESS-ENERGY TENSOR AND ITS RENORMALIZATION
We now come to the main point of this work, namely the determination of the stress-energy tensor. We start bycalculating it in local thermodynamic equilibrium.1
A. Local thermodynamic equilibrium
As the symmetries of (cid:98) ρ LE are the same as (cid:98) ρ (see the discussion in Sec. II) the structure must be the same as in Eq.(15): Tr( (cid:98) ρ LE (cid:98) T µν ) = (cid:104) (cid:98) T µν (cid:105) LE = E ( τ ) LE u µ u ν + P T ( τ ) LE (cid:16) ˆ i µ ˆ i ν + ˆ j µ ˆ j ν (cid:17) + P L ( τ ) LE ˆ η µ ˆ η ν , (47)Hence, by using Eq. (27) with the expansion (20) we obtain: E ( τ ) LE = (cid:104) (cid:98) T µνC (cid:105) LE u µ u ν = (cid:90) d p T d µ d p (cid:48) T d µ (cid:48) π ) × (cid:18)(cid:26) [ ∂ τ h ( p , τ )][ ∂ τ h ( p (cid:48) , τ )] − (cid:18) p x p (cid:48) x + p y p (cid:48) y + 1 τ µµ (cid:48) − m (cid:19) h ( p , τ ) h ( p (cid:48) , τ ) (cid:27) × e i [( p T + p (cid:48) T ) · x T +( µ + µ (cid:48) ) η ] (cid:104) (cid:98) b p (cid:98) b p (cid:48) (cid:105) LE + (cid:26) [ ∂ τ h ( p , τ )][ ∂ τ h ∗ ( p (cid:48) , τ )] + (cid:18) p x p (cid:48) x + p y p (cid:48) y + 1 τ µµ (cid:48) + m (cid:19) h ( p , τ ) h ∗ ( p (cid:48) , τ ) (cid:27) × e i [( p T − p (cid:48) T ) · x T +( µ − µ (cid:48) ) η ] (cid:104) (cid:98) b p (cid:98) b † p (cid:48) (cid:105) LE + (cid:26) [ ∂ τ h ∗ ( p , τ )][ ∂ τ h ( p (cid:48) , τ )] + (cid:18) p x p (cid:48) x + p y p (cid:48) y + 1 τ µµ (cid:48) + m (cid:19) h ∗ ( p , τ ) h ( p (cid:48) , τ ) (cid:27) × e − i [( p T − p (cid:48) T ) · x T +( µ − µ (cid:48) ) η ] (cid:104) (cid:98) b † p (cid:98) b p (cid:48) (cid:105) LE + (cid:26) [ ∂ τ h ∗ ( p , τ )][ ∂ τ h ∗ ( p (cid:48) , τ )] − (cid:18) p x p (cid:48) x + p y p (cid:48) y + 1 τ µµ (cid:48) − m (cid:19) h ∗ ( p , τ ) h ∗ ( p (cid:48) , τ ) (cid:27) × e − i [( p T + p (cid:48) T ) · x T +( µ + µ (cid:48) ) η ] (cid:104) (cid:98) b † p (cid:98) b † p (cid:48) (cid:105) LE (cid:17) . (48)Plugging the relations (46) into Eq. (48) we obtain: E ( τ ) LE = 14(4 π ) (cid:90) d p T d µ (cid:26) − (cid:2) ( ∂ τ h ) + ω h (cid:3) (2 n B + 1) sinh(2Θ) e − iχ + c . c . + (cid:0) | ∂ τ h | + ω | h | (cid:1) (2 n B + 1) cosh(2Θ) (cid:111) , and, by using Eqs. (29), (30), and (33): E ( τ ) LE = 116 π τ (cid:90) d p T d µ ω (cid:2) (2 n B + 1) cosh (2Θ) − sinh (2Θ)(2 n B + 1) (cid:3) = 1(2 π ) τ (cid:90) d p T d µ ω (cid:18) n B + 12 (cid:19) . (49)The longitudinal and transverse pressures can be worked out in a similar fashion: the Wronskian of the Hankelfunction is again recovered and the expressions greatly simplify. One obtains: P T ( τ ) LE = 1(2 π ) τ (cid:90) d p T d µ | p T | (cid:18) n B + 12 (cid:19) , P L ( τ ) LE = 1(2 π ) τ (cid:90) d p T d µ µ τ (cid:18) n B + 12 (cid:19) . (50)Equations (49) and (50) can be written in a compact fashion by introducing the functions: K γ ( p , τ ) = πτ ω (cid:2) | ∂ τ h ( p , τ ) | + γ ( p , τ ) | h ( p , τ ) | (cid:3) , (51) Λ γ ( p , τ ) = πτ ω (cid:8) [ ∂ τ h ( p , τ )] + γ ( p , τ )[ h ( p , τ )] (cid:9) , (52)where γ is defined as γ ( p , τ ) = ω ( p , τ ) = m + µ τ , for E ( τ ) LE , − m ≡ − µ τ − m , for P T ( τ ) LE , − m + µ τ , for P L ( τ ) LE . (53)2Thanks to the Wronskian of the Hankel functions, they satisfy the relation: K γ ( p , τ ) − | Λ γ ( p , τ ) | = γ ( p , τ ) ω ( p , τ ) . With this in mind, and setting Γ γ = {E , P T , P L } , we have for the thermodynamic function of the stress-energy tensor: Γ γ ( τ ) LE = (cid:90) d p T d µ (2 π ) τ ω ( p , τ ) [ K γ ( p , τ ) K ( p , τ ) − Re (Λ γ ( p , τ )Λ ∗ ( p , τ ))] (cid:18) n B ( p , τ ) + 12 (cid:19) , where the combination in square brackets reads K γ ( p , τ ) K ( p , τ ) − Re (Λ γ ( p , τ )Λ ∗ ( p , τ )) = ω ( p , τ ) + γ ( p , τ )2 ω ( p , τ ) , hence Γ γ ( τ ) LE = (cid:90) d p T d µ (2 π ) τ ω ( p , τ ) ω ( p , τ ) + γ ( p , τ )2 (cid:20) n B ( p , τ ) + 12 (cid:21) . (54)The above integrals can be written in a familiar form by changing the integration variable to p z = µ/τ . This implies: ω ( p , τ ) = (cid:114) | p T | + µ τ + m = (cid:113) p x + p y + p z + m = ε , which is just the on-shell energy, and: d p T d µτ = dp x dp y dp z , In turn, the distribution n B ( p , τ ) becomes the energy-dependent Bose-Einstein phase-space distribution n B ( ε, T ( τ )) .Hence, the first term of the energy density (49) as well as the transverse and longitudinal pressures (50) can bewritten as the familiar momentum integrals of the relativistic uncharged Bose gas. Altogether, the unrenormalizedstress-energy tensor in local equilibrium reads: Tr (cid:104)(cid:98) ρ LE (cid:98) T µν ( x ) (cid:105) = (cid:90) d pε p µ p ν (cid:20) β ( x ) · p − (cid:21) , (55)where β is the four-temperature in Eq. (12). Hence, the thermodynamic functions Γ γ are just the familiar functions of T ( τ ) as for the ideal relativistic gas. In particular, the transverse and the longitudinal pressures are in fact identical,namely P T ( τ ) LE = P L ( τ ) LE ≡ P ( τ ) LE . (56) B. Actual stress-energy tensor
The actual (unrenormalized) expectation value of the stress-energy tensor can be calculated by using the densityoperator (16), that is: Tr (cid:104)(cid:98) ρ (cid:98) T µν ( x ) (cid:105) = 1 Z Tr (cid:110) exp[ − (cid:98) Π( τ ) /T ( τ )] (cid:98) T µν ( x ) (cid:111) . Symmetries dictate that its form is given by Eq. (15), so we need to determine the three functions Γ γ . It is readilyfound that the same expression as in Eq. (48) is obtained, with the simple replacement of the local-equilibrium valuesof the quadratic combinations of (cid:98) b p and (cid:98) b † p with their actual expectation values, for instance: (cid:104) (cid:98) b † p (cid:98) b p (cid:105) = 1 Z Tr (cid:110) exp[ − (cid:98) Π( τ ) /T ( τ )] (cid:98) b † p (cid:98) b p (cid:111) . at time τ , i.e., expressing theconstant (cid:98) b p ’s as functions of the operators diagonalizing (cid:98) Π( τ ) instead of (cid:98) Π( τ ) . We thus get the same formulae as Eq.(46), with τ replaced by τ : (cid:104) (cid:98) b p (cid:98) b p (cid:48) (cid:105) = −
12 sinh[2Θ( τ )] e − iχ ( τ ) [2 n B ( τ ) + 1] δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , (57) (cid:104) (cid:98) b † p (cid:98) b † p (cid:48) (cid:105) = −
12 sinh[2Θ( τ )] e iχ ( τ ) [2 n B ( τ ) + 1] δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , (cid:104) (cid:98) b p (cid:98) b † p (cid:48) (cid:105) = (cid:8) n B ( τ ) cosh[2Θ( τ )] + cosh Θ( τ ) (cid:9) δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) , (cid:104) (cid:98) b † p (cid:98) b p (cid:48) (cid:105) = (cid:8) n B ( τ ) cosh[2Θ( τ )] + sinh Θ( τ ) (cid:9) δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) . We note in passing that, as expected, the expectation value of excitations of the Minkowski vacuum, described by (cid:104) (cid:98) b † p (cid:98) b p (cid:105) for each mode, is constant in time, the density operator being fixed and the operators (cid:98) b p being time-independentby construction. The mean number of particles with momentum p can be obtained by using Eq. (22): (cid:104) (cid:98) a † ( p ) (cid:98) a ( p (cid:48) ) (cid:105) = 12 πm T √ cosh y cosh y (cid:48) (cid:90) + ∞−∞ d µ e − iµ ( y − y (cid:48) ) (cid:104) (cid:98) b † p T ,µ (cid:98) b p (cid:48) T ,µ (cid:105) , where y is the rapidity.Now, by taking advantage of the right-hand side of Eq. (48), and by using Eq. (57), as well as Eqs. (29), (30), and(33), it can be shown that: E ( τ ) = 1 Z Tr (cid:110) exp[ − (cid:98) Π( τ ) /T ( τ )] (cid:98) T µν (cid:111) u µ u ν = 1(2 π ) τ (cid:90) d p T d µ ω ( τ ) { K ( τ ) K ( τ ) − Re [Λ( τ )Λ ∗ ( τ )] } (cid:20) n B ( τ ) + 12 (cid:21) . (58)The pressures can be derived likewise and we finally have: Γ γ ( τ ) = 1(2 π ) τ (cid:90) d p T d µ ω ( τ ) { K γ ( τ ) K ( τ ) − Re [Λ γ ( τ )Λ ∗ ( τ )] } (cid:20) n B ( τ ) + 12 (cid:21) . (59)Of course, at the time τ = τ we recover the local thermodynamic equilibrium expression (54), as required byconstruction. However, at later times τ > τ the stress-energy tensor differs from the local equilibrium form. Indeed,since we are dealing with a free field, one expects to find the same expression as for the free-streaming solution ofthe Boltzmann equation in Milne coordinates, see Appendix A. However, there are quantum corrections due to thevacuum subtraction. C. Renormalization and comparison with classical limits
The expressions found include divergent terms, both in the stress-energy tensor in local equilibrium (54) and theactual one (59). As we have seen in Sec. III A, in order to fulfill the continuity equation, the stress-energy tensor shouldbe renormalized by subtracting a vacuum expectation value (VEV) with a constant vacuum: either with respect tothe Minkowskian vacuum | M (cid:105) , like in Eq. (18), or with respect to the vacuum | τ (cid:105) of the operator (cid:98) Π( τ ) , like inEq. (19).The Minkowski VEV of the stress-energy tensor is calculated in Appendix B. For the stress-energy tensor it isfound: E M ≡ (cid:104) M | (cid:98) T µν u µ u ν | M (cid:105) = 1(2 π ) τ (cid:90) d p T d µ ω ( τ ) K ( τ )2 , and the renormalized energy density is then: E ( τ ) ren = 1(2 π ) τ (cid:90) d p T d µ ω ( τ ) (cid:18) { K ( τ ) K ( τ ) − Re [Λ( τ )Λ ∗ ( τ )] } (cid:20) n B ( τ ) + 12 (cid:21) − K ( τ ) (cid:19) . The main drawback of this expression is that it is still divergent. This is most easily seen at τ = τ where: E ( τ ) ren = 1(2 π ) τ (cid:90) d p T d µ ω ( τ ) n B ( τ ) − π ) τ (cid:90) d p T d µ ω ( τ ) [ K ( τ ) − . (60)4While the first term is finite, the second is not due to the behaviour of the K function for large values of its effectiveargument, which is m T τ , at fixed µ [see Eqs. (29) and (23)]. The asymptotic behaviour for large transverse mass m T of the K function is derived in Appendix C and one has, at leading order: K ( τ ) − τ ) − (cid:39) Θ (cid:39) m τ , which makes the integral in Eq. (60) divergent.In conclusion, in order to have a finite stress-energy tensor, we are left with the option to subtract the VEV’s withrespect to | τ (cid:105) , which can be readily done by taking the limit T ( τ ) → in Eq. (59) and subtracting what is left,taking into account that lim T → n B = 0 . We thus have: Γ γ ( τ ) ren = 1(2 π ) τ (cid:90) d p T d µ ω ( τ ) { K γ ( τ ) K ( τ ) − Re [Λ γ ( τ )Λ ∗ ( τ )] } n B ( τ ) , (61)which incorporates the relation between the energy density and the pressures.It is interesting to study the behaviour of the functions (61) at late times τ , which means for large values of mτ (see Appendix C). In this limit, we have Θ( τ ) → , hence K ( τ ) → and Λ( τ ) → , implying that the Minkowskianvacuum is recovered asymptotically. This is also clear from Eq. (43), which shows that | τ (cid:105) → | M (cid:105) . For the energydensity, at late times we have: E ( τ ) ren (cid:39) τ →∞ π ) τ (cid:90) d p T d µ ω ( τ ) K ( τ ) n B ( τ ) (62) = 1(2 π ) τ (cid:90) d p T d µ ω ( τ ) n B ( τ ) + 1(2 π ) τ (cid:90) d p T d µ ω ( τ )[ K ( τ ) − n B ( τ )= 1(2 π ) τ (cid:90) d p T d µ ω ( τ ) n B ( τ ) + 2(2 π ) τ (cid:90) d p T d µ ω ( τ ) sinh Θ( τ ) n B ( τ ) . It can be shown that the first term in Eq. (62) is the classical free-streaming solution in Milne coordinates (seeAppendix A), while the second term is a pure quantum-field correction due to the difference between vacua, since itvanishes only if Θ( τ ) = 0 . Somewhat surprisingly, the quantum correction to energy density does not vanish at latetimes, and it can even be comparable with the classical term if the main argument of Θ( τ ) , that is m T τ is O (1) ,that is for an early decoupling of the system.Similar expressions can be obtained for the pressures. For large times, the leading term of the Λ γ ( τ ) function hasan oscillating behaviour ∼ exp( − im T τ ) [see Appendix C, Eq. (C4)], so the integrals in p T or m T involving Λ γ ( τ ) are expected to decay as τ → ∞ . Therefore, only the first term of Eq. (61) is left and one has: P γ ( τ ) ren (cid:39) τ →∞ π ) τ (cid:90) d p T d µ ω ( τ ) K γ ( τ ) K ( τ ) n B ( τ ) . (63)Also, at late times [see Appendix C, Eq. (C3)]: ω ( τ ) K γ ( τ ) (cid:39) τ →∞ m + γ m T , so Eq. (63) becomes: P γ ( τ ) ren (cid:39) τ →∞ π ) τ (cid:90) d p T d µ m + γ m T K ( τ ) n B ( τ ) (64) = 1(2 π ) τ (cid:90) d p T d µ m + γ m T n B ( τ ) + 2(2 π ) τ (cid:90) d p T d µ m + γ m T sinh Θ( τ ) n B ( τ ) , with γ from Eq. (53). Again, the first term is the leading approximation of the classical free-streaming solution inMilne coordinates for large m T τ and fixed µ , whereas the second term is a pure quantum correction. VI. ENTROPY CURRENT
The need of subtracting the vacuum | τ (cid:105) to obtain a finite value for the stress-energy tensor for the free field has someinteresting connection to the way the entropy and the entropy current of a relativistic fluid in local thermodynamic5equilibrium are calculated. This problem has been approached in the framework of the relativistic density operatorin Ref. [23]. We first observe that the entropy of a relativistic fluid in local equilibrium, S = − tr( (cid:98) ρ LE log (cid:98) ρ LE ) , with (cid:98) ρ LE given by the Eq. (48), is independent of the vacuum subtraction because, as remarked in Sec. II, the densityoperator (48) turns out to be independent of any non-operator term which is subtracted from the stress-energy tensoroperator, as it cancels out in the ratio with the normalizing Z LE .However, it was pointed out in Ref. [23] that, provided that the vacuum is non-degenerate, there is only one goodchoice of the vacuum if one has to make log Z LE extensive, i.e.: log Z LE = (cid:90) Σ dΣ µ φ µ , and this is the vacuum (meant as the eigenvector with minimal eigenvalue) of the operator (cid:98) Π( τ ) , which we havedenoted with | τ (cid:105) . Therefore, the entropy current reads: s µ = φ µ + (cid:104) Tr( (cid:98) ρ LE (cid:98) T µν ) − (cid:104) τ | (cid:98) T µν | τ (cid:105) (cid:105) β ν , (65)with: φ µ = (cid:90) ∞ d λ (cid:110) Tr[ (cid:98) ρ LE ( λ ) (cid:98) T µν ] − (cid:104) τ | (cid:98) T µν | τ (cid:105) (cid:111) , where (cid:98) ρ LE ( λ ) is the operator defined by: (cid:98) ρ LE ( λ ) = 1 Z LE ( λ ) exp (cid:18) − λ (cid:90) Σ dΣ µ (cid:98) T µν β ν (cid:19) . The renormalized value: T µν LE = Tr( (cid:98) ρ LE (cid:98) T µν ) − (cid:104) τ | (cid:98) T µν | τ (cid:105) of the stress-energy tensor in local thermodynamic equilibrium with subtraction of the VEV with respect to | τ (cid:105) canbe found by taking the limit T ( τ ) → , as we have seen in Sec. III A. Hence, for the free scalar field, it is readily foundfrom Eq. (55) that we are left with the classical expression: T µν ( x ) LE = (cid:90) d pε p µ p ν β ( x ) · p − . It is now easy to show that φ µ = P LE β µ , with P LE being the one pressure in Eq. (56), and that the entropy currentcoincides with the classical equilibrium expression: s µ = ( E LE + P LE ) β µ , where E LE and P LE are related by the usual equation of state of a free relativistic gas, without apparent quantumcorrection.We end this section by discussing the entropy-production rate equation established in Refs. [8, 9] [for a derivationsee Ref. [17]], which for ζ = 0 reads: ∇ µ s µ = ( T µν − T µν LE ) ∇ µ β ν . (66)In the above equation it is usually understood that T µν and T µν LE are the renormalized stress-energy tensor expectationvalues, fulfilling the constraint equation (4), and usually obtained by subtracting the Minkowski VEV of both. How-ever, in our case, in order to obtain finite values for the constraint equation (4) and to find an appropriate expressionof the entropy current, we need to subtract different VEV’s, as we have seen. In particular: T µν LE = Tr( (cid:98) ρ LE (cid:98) T µν ) − (cid:104) τ | (cid:98) T µν | τ (cid:105) ,T µν = Tr( (cid:98) ρ (cid:98) T µν ) − (cid:104) τ | (cid:98) T µν | τ (cid:105) . One may thus wonder whether such a difference in the VEV subtraction introduces a new quantum term in theentropy production rate. The answer is again no, provided that In this section, for the sake of simplicity, we assume vanishing chemical potentials, that is ζ = 0 ; the extension of these arguments to anon-vanishing chemical potential is straightforward.
6• the renormalized expectation value T µν is finite;• the renormalized expectation value T µν fulfills the continuity equation;• the renormalized expectation value in local equilibrium T µν LE fulfills the constraint (4).The proof of Eq. (66) [17] can be shown to hold. VII. SUMMARY AND CONCLUSIONS
To summarize, we have studied a relativistic quantum fluid with longitudinal boost invariance, which, for thefree scalar field, is an exactly solvable non-equilibrium problem, further developing and extending the results ofRefs. [11, 12]. By using the non-equilibrium density operator, we have derived an exact solution for the stress-energy tensor and the entropy current for the free scalar field initially in local thermodynamic equilibrium. Themost remarkable feature of the solution is the difference between the vacuum of the density operator and the familiarvacuum of the field in Minkowski space-time. We have found that a finite, renormalized value of the stress-energytensor can be achieved only by subtracting the vacuum of the density operator, and not the vacuum of the field. Withrespect to the known classical free-streaming solution, we have found quantum corrections related to the differencebetween the vacuum of the density operator and the Minkowski vacuum. These corrections are numerically relevantfor an early decoupling of the field, that is if mτ = O (1) , where τ is the hyperbolic time; in this case they surviveat late times and affect the relation between energy density and pressure as compared to the classical free-streamingcase. ACKNOWLEDGMENTS
D.R. would like to express his gratitude to D.H.R. for his kind hospitality and support at the Institut für TheoretischePhysik of Goethe University Frankfurt am Main during the course of this work. The work of L.T. and D.H.R. wassupported by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) through the CRC-TR 211“Strong-interaction matter under extreme conditions”, project number 315477589 TRR 211.
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The collisionless Boltzmann equation in classical relativistic kinetic theory reads: p · ∂f ( x, p ) = 0 , (A1)and its explicit solution in Cartesian coordinates is: f ( x, p ) = f (cid:18) x − t − t ε p , p (cid:19) , (A2)where f ( x , p ) = f ( t , x ; p ) is the initial condition in a generic inertial reference frame, and ε = (cid:112) m + p is the(on-shell) energy.In longitudinal boost-invariant symmetry, the initial condition is given at some Milne time τ rather than a time t in Cartesian coordinates. Nevertheless, there is a very simple solution in this case, too. Since the distributionfunction is a scalar, it must be invariant under the symmetry transformations at stake, that are longitudinal boostsas well as rotations and translations in the transverse plane. Hence, it depends only on the independent scalarsthat may be formed with combinations of space-time and momentum vector which are invariant under the group oftransformations IO(2) ⊗ SO(1 , . These scalars are: τ = (cid:112) t − z , p T = (cid:113) p x + p y , w = zε − tp z . (A3)The last variable can be shown to be equivalent to the covariant component p η of the four-momentum vector in Milnecoordinates. Indeed, there is a fourth invariant scalar: v = tε − zp z = τ (cid:114) m + p + w τ , (A4)but it is redundant because of the on-shell condition (and positivity) of the energy and because t > | z | in the futurelight cone. The reflection invariance (see Sec. II) makes f dependent on the square of w rather than just w . Byutilizing these arguments, Eq. (A1) becomes: vτ ∂∂τ f ( τ, p T , w ) = 0 , (A5)since the contribution in the partial derivatives with respect to w cancels out. The free-streaming solution is thenvery simple, a constant in τ : f ( τ, p T , w ) = f ( τ , p T , w ) ≡ f ( p T , w ) . We are now in a position to calculate the free-streaming solution for the stress-energy tensor from its classicalkinetic definition: T µν = 1(2 π ) (cid:90) d p ε p µ p ν f ⇒ E = u µ u ν T µν = π ) (cid:82) d p ε v τ f , P T = (cid:16) ˆ i µ ˆ i ν + ˆ j µ ˆ j ν (cid:17) T µν = π ) (cid:82) d p ε p f , P L = ˆ η µ ˆ η ν T µν = π ) (cid:82) d p ε w τ f , (A6)and changing the integration variables: w = zε − tp z ⇒ d w = (cid:12)(cid:12)(cid:12) − vε (cid:12)(cid:12)(cid:12) d p z ⇒ d p z ε = d wv , (A7)one obtains: E = 1(2 π ) (cid:90) d p T d wv v τ f ( p T , w ) = 1(2 π ) τ (cid:90) d p T d w (cid:114) m + w τ f ( p T , w ) , (A8)and P T = 1(2 π ) (cid:90) d p T d wv p f ( p T , w ) = 1(2 π ) τ (cid:90) d p T d w (cid:112) m + w /τ p f ( p T , w ) , (A9) P L = 1(2 π ) (cid:90) d p T d wv w τ f ( p T , w ) = 1(2 π ) τ (cid:90) d p T d w (cid:112) m + w /τ w τ f ( p T , w ) . w → µ and with the initialdistribution equal to the local equilibrium Bose-Einstein distribution function f = n . Appendix B: Minkowski vacuum expectation values
In order to calculate the scalars Γ γ ( τ ) M of the stress-energy tensor in the Minkowski vacuum, we take advantage ofit being annihilated by all the (cid:98) b p ’s as it is clear from Eq. (22). Hence, the only product of (cid:98) b p and (cid:98) b † p with non-vanishingexpectation value with respect to | M (cid:105) is ˆ b p ˆ b † p (cid:48) , and using the commutation relations (21): (cid:104) M | (cid:98) b p (cid:98) b † p (cid:48) | M (cid:105) = (cid:104) M | (cid:98) b † p (cid:48) (cid:98) b p | M (cid:105) + (cid:104) M | [ (cid:98) b p , (cid:98) b † p (cid:48) ] | M (cid:105) = δ ( p T − p (cid:48) T ) δ ( µ − µ (cid:48) ) . (B1)We can now replace these VEV’s to obtain Γ γ ( τ ) M in the stress-energy tensor expression contracted with suitablevectors. For instance, for the energy density, we can use Eq. (48) by simply replacing the local equilibrium expectationvalues with those in the Minkowski vacuum and obtain: E ( τ ) M ≡ (cid:104) M | (cid:98) T µν u µ u ν | M (cid:105) = (cid:90) d p T d µ π ) (cid:0) | ∂ τ h | + ω | h | (cid:1) = (cid:90) d p T d µ (2 π ) τ ω K . (B2)Similarly, for the pressures, one finds: Γ γ ( τ ) M = (cid:90) d p T d µ (2 π ) τ ω K γ . (B3) Appendix C: Asymptotics
It is interesting to study the behaviour of the stress-energy tensor and related quantities for late times τ . With h ( τ ) = − i e π µ H (2) iµ ( m T τ ) , (C1)one can make use of the asymptotic expansion for large arguments [21] H (2) ν ( x ) ∼ (cid:114) πx e − i ( x − π ν − π ) (cid:88) n ix ) n Γ( ν + 1 / n ) n !Γ( ν + 1 / − n ) , (C2)which is valid for Re( ν ) > − / and | arg( x ) | < π . Making use of the property z Γ( z ) = Γ( z + 1) , substituting x = m T τ and ν = iµ , and plugging this into Eq. (C1) we get: h ( τ ) ∼ (cid:114) − iπm T τ e − im T τ (cid:88) n im T τ ) n (cid:0) iµ + − n (cid:1) (2 n ) n ! , valid for large m T τ . Similarly, using the exact relation [21] z∂ z H (2) ν ( z ) = ν H (2) ν ( z ) − z H (2) ν +1 ( z ) , along with the expansion (C2), one obtains the expansion for the proper-time derivative ∂ τ h : ∂ τ h ( τ ) ∼ − im T (cid:114) − iπm T τ e − im T τ (cid:34) (cid:88) n> im T τ ) n (cid:32) − iµ (cid:0) iµ + − n (cid:1) (2 n − ( n − (cid:0) iµ + − n (cid:1) (2 n ) n ! (cid:33)(cid:35) . In particular, retaining the terms up to first order (i.e., next-to-leading order) in m T τ we get: h ( τ ) (cid:39) (cid:114) − iπm T τ e − im T τ (cid:20) − i m T τ (cid:18) iµ − (cid:19) (cid:18) iµ + 12 (cid:19)(cid:21) = (cid:114) − iπm T τ e − im T τ (cid:18) i µ m T τ (cid:19) ,∂ τ h ( τ ) (cid:39) − im T (cid:114) − iπm T τ e − im T τ (cid:18) − i − µ m T τ (cid:19) . K γ (cid:39) m + γ m T ω , (C3) Λ γ (cid:39) m T ω e − im T τ (cid:20) − m (cid:18) − i − µ m T τ (cid:19) + γ (cid:18) i µ m T τ (cid:19)(cid:21) . (C4)The rest is of the order of / [ m T ω ( m T τ ) ] for K γ and exp( − im T τ ) / [ m T ω ( m T τ ) ] for Λ γ .Equations (C3) and (C4) are very useful to study the large p T (hence, large m T ) behaviour as well as the long-timebehaviour. For large p T , Eq. (C3) implies that K → , hence to leading order K − is simply zero. However, fromEq. (C4) and the exact relation (33) one can obtain the terms up to second order. Indeed, for γ = ω , in the large m T limit: Λ (cid:39) m T τ e − im T τ , (C5)hence: | Λ | = sinh Θ (cid:39) Θ = 12 m T τ , (C6)and in the limit of large m T K − (cid:39)
12 Θ (cid:39) m τ . (C7)Similarly, the leading expressions at late time τ → ∞ can be derived. By using the asymptotic expansions (C3)and (C4) and expanding ω ( τ ) and γ ( τ ) for large τ one obtains: K γ (cid:39) m + ˜ γ m , (C8) Λ γ (cid:39) m e − im T τ (cid:20) ˜ γ − m + i m (3 − µ ) + ˜ γ (1 + 4 µ )4 m T τ (cid:21) , at first order in /τ , with ˜ γ : ˜ γ = lim τ →∞ γ = m , for E , − m , for P T , − m , for P L . (C9)It is important to note that, except for the energy density and only at the leading order, the Λ γ ’s have a rapidlyoscillating phase that prevents a proper limit in the function domain. However, they converge in the distributiondomain, which is fine since they have to be integrated. In fact the limits: lim τ →∞ sin(2 m T τ ) , lim τ →∞ cos(2 m T τ ) are proportional to Dirac deltas. We make use of the formula for the delta families δ ( x ) = lim (cid:15) → (cid:15) f ( x/(cid:15) ) , for any f function normalized to and set (cid:15) = 1 /τ in the former case and (cid:15) = 1 / √ τ in the latter: lim (cid:15) → π sin( x/(cid:15) ) x = δ ( x ) ⇒ sin(2 m T τ ) τ →∞ −→ πm T δ (2 m T ) , lim (cid:15) → (cid:15) √ π cos (cid:18) x (cid:15) (cid:19) = δ ( x ) ⇒ cos(2 m T τ ) τ →∞ −→ (cid:114) πτ δ ( √ m T ) ..