Rotation in an exact hydro model
RRotation in an exact hydro model
L.P. Csernai, D.J. Wang
Institute of Physics and Technology, University of Bergen, Allegaten 55, 5007 Bergen, Norway
T. Cs¨org˝o
Wigner Research Center for Physics, H-1121 Budapesti 114, POBox 49, Hungary (Dated: October 22, 2018)We study an exact and extended solution of the fluid dynamical model of heavy ion reactions, andestimate the rate of slowing down of the rotation due to the longitudinal and transverse expansionof the system. The initial state parameters of the model are set on the basis of a realistic 3+1Dfluid dynamical calculation at TeV energies, where the rotation is enhanced by the build up of theKelvin Helmholtz Instability in the flow.
PACS numbers: 25.75.-q, 24.70.+s, 47.32.Ef
I. INTRODUCTION
In peripheral heavy ion collisions the system has an-gular momentum. It has been shown in hydrodynamicalcomputations that this leads to a large shear and vor-ticity [1]. Furthermore when the Quark-Gluon Plasma(QGP) is formed with low viscosity [2], interesting newphenomena may occur like rotation [3], or turbulence,which shows up in form of a starting Kelvin-HelmholtzInstability (KHI) [4, 5].Surprisingly, the effects arising from non-vanishing ini-tial angular momentum in fireball hydrodynamics can bestudied with the help of exact and explicit, analytic so-lutions of the equations of hydrodynamics. The pioneersto the application of the hydrodynamical method to highenergy particle and nuclear collisions initially neglectedthe effects from the non-vanishing angular momentum:Belenkij and Landau considered the 1+1 dimensional ex-plosion of a non-expanding region of very high energydensity [6], while Hwa and Bjorken considered [7, 8] aboost-invariant, already asymptotic expansion as the ini-tial condition of 1+1 dimensional expansion governed byrelativistic hydrodynamics.However, if the collision of two protons or two heavyions is not exactly head-on, there is a non-vanishing ini-tial angular momentum present in the initial conditions,which was for a long time neglected. However, ratherrecently, numerical investigations of relativistic hydrody-namics [1–5], as well as exact and explicit analytic so-lutions of relativistic and non-relativistic hydrodynamicswere found for rotating fluids [9, 10]. It took 35 yearsafter the publication of the first exact, non-rotational so-lution in a similar class [11] to find and publish rotatingsolutions of hydrodynamics, a long road with lots of sur-prises and unexpected turns, that were recently brieflysummarized in ref. [10]. In ref. [12] the AdS/CFTholography method is used to study the QGP created inheavy ion collisions. The authors chose two types of ve-locity profiles, which do not favor instability classically,and they use the planar black hole geometry to computethe fluid shear by using parameters predicted in [4]. The discussion shows that for the two clasically stable con-figurations in the holographic model instability developsvery slowly for lower chemical potentials, but turbulencemay still exist for high chemical potentials. Finding ofthese rotating solutions started to shed more light on theeffects of rotation, which plays a very important role [13]in shaping the events in astrophysical, hydrodynamicallyevolving systems, turbulence, or in eddy and vortex for-mation in fluid dynamics. This motivates the subject ofthe present paper, which evaluates numerically some ofthe important characteristics of the fireball (like the sizeof the fireball along the axis of rotation and the size ofthe fireball in the plane perpendicular to the rotation) forquasi-realistic initial conditions and compares it with thecase when the initial angular momentum is negligible, oris neglected.Based on ref. [4] we can extract some basic parametersof the rotation obtained with numerical fluid dynamicalmodel PICR. These parameters are extracted from modelcalculation of a Pb+Pb collisions at √ s NN = 2 .
76 A TeVand at the impact parameter of b = 0 . b max , with highresolution and thus small numerical viscosity. Thus, inthis collision the KHI occurs and enhances rotation atintermediate times, because the turbulent rotation gainsenergy from the original shear flow. The turbulent rota-tions leads to a rotation profile where the rotation of theexternal regions lags behind the rotation of the internalzones. This is a typical growth of the KHI. See Table I.The initial angular momentum of the system is large, L y = − . × (cid:126) in mid-peripheral collision. This isarising from the pre-collision state. In refs. [1, 3, 4] thesame Initial State (IS) model is used, before the PICRfluid dynamical model started. The IS model [14, 15]describes the first 4 fm/c time period after the momentwhen the Lorentz contracted, thin projectile and targetslabs (like in the CGC model) penetrate each other. Thematter is then divided into longitudinally ( z directed)expanding (fire)streaks . The [ x, y ] transverse plane isdivided into surface elements according to the resolu-tion of the PICR model, and each element belongs toone streak. The streak expansion is described in a one- a r X i v : . [ h e p - ph ] J u l t Y ˙ Y θ R ˙ R ω (fm/c) (fm) (c) (Rad) (fm) (c) (c/fm)0. 4.38 0.90 0.000 3.68 0.60 0.01752. 6.18 0.88 0.035 4.87 0.84 0.03504. 7.91 0.84 0.105 6.56 0.97 0.05206. 9.54 0.80 0.209 8.49 0.86 0.07008. 11.09 0.76 0.349 10.21 0.81 0.0350TABLE I. Time dependence of some characteristic parame-ters of the fluid dynamical calculation presented in ref. [4]. R is the average transverse radius ( R ≈ √ XZ ), Y is the lengthof the system in the direction of the axis of the rotation y , θ is the polar angle of the rotation of the interior region ofthe system measured versus the z directed, beam axis, of thereaction plane, [ x, z ] plane, ˙ R, ˙ Y are the speeds of expansionin radial and rotational axis directions, and ω is the angu-lar velocity of the internal region of the matter during thecollision. dimensional classical Yang-Mills field, spanned by thecolor charges at the two ends of the streak. This fieldslows down the expansion due to the large string-ropetension . The IS dynamics satisfies the momentum con-servation Streak by Streak , and this way the total angu-lar momentum of the IS is also exactly conserved. Duringthe IS model the dynamics is one dimensional ( z ), with z directed velocities only, but the expansion is differentin different transverse points.After 4 fm/c of the IS model, local equilibrium isreached and the PICR (3+1)D fluid dynamical modelstarts (with t = 0 fm/c fluid dynamical model time).Then, due to the fluid dynamical development and equili-bration, the x -directed velocity starts to increase and theaverage of the z -directed velocity decreases. This way theangular momentum is exactly conserved in the (3+1)Dfluid dynamical calculation. Thus the x -directed velocityor otherwise the rotation in the horizontal plane starts updelayed. This is a fully realistic model of the initial longi-tudinally transparent non-equilibrium dynamics, and thesubsequent equilibration of rotation on a larger scale [4].As the initial conditions and initial times in the exactmodel and in the PICR (3+1)D model are not identicalwe have matched the time coordinates such that t exact =0 fm/c in the exact model corresponds to: t = 3 fm/c inthe full (3+1)D calculation.If we compare the rotation of the horizontal plane ( x directed velocity) only, then the PICR model and theexact model are becoming similar at t = 6 ( t exact = 3)fm/c and after (see Table I and Table II). If however,we estimate the average of the z directed velocity also,and consider then the average of the x and z directed ve-locities, the two models are becoming similar already at t = 4 ( t exact = 1) fm/c and after. Thus, the applicabil-ity of the exact model with the parameters chosen here,starts approximately after t = 5 fm/c on the timescale ofthe PICR model. The radius, R , parameters are matchedto each other in the two models s0 that in the (3+1)Dmodel at the same time moments, t = 5 and 8 fm/c the radii are 8.49 and 10.21 fm for a sharp matter surface,while in the exact model the corresponding radii are 3.97and 5.36 fm respectively (i.e. about half of the previousvalues) but these represent the width parameter of aninfinite Gaussian matter distribution.The initial part of the (3+1)D model describes theequilibration of the rotational flow from the initial shearflow, which rotation then leads to a maximal, azimuthallyaveraged angular velocity. Then the system expands andthe angular velocity decreases. The exact model, assum-ing uniform rotation can only describe this second phaseof the process. It is important to mention that the KHIfacilitates the equilibration and speed up of the rotationand leads to an earlier and bigger maximal angular ve-locity. In this case the applicability of the exact model ismore extended in time, and spans the range between theequilibration of the rotation and the freeze out. At lowerbeam energies and small impact parameters (i.e. at loverangular momentum) the timespan of the applicability ofexact model should be tested separately.We want to use these fluid dynamical calculations totest a new family of exact rotating solutions [10], whichmay provide more fundamental insight to the interactionbetween the rotation and expansion of the system. Thismodel offers a few possible variations, here we chose theversion to test. We change the axis labeling of ref.[10], so that the axis of the rotation is y while the trans-verse plane of the rotation is the [ x, z ] plane. Thus thevalues extracted from the results of the fluid dynamicalmodel, [4], should take this into account. The initial ra-dius parameter, R , corresponds to the system size in the x or z directions in hydro, and we assume an x, z symme-try in the exact model. The rotation axis is the y axis inhydro and now also in the exact model. The exact modelassumes azimuthal symmetry, so it cannot describe thebeam directed elongation of the system, but this is aris-ing from the initial beam momentum, and we intend todescribe the rotation of the interior part of the reactionplane and the rotation there.In Section II we recapitulate some of the central resultsof ref. [10] for clarity and so that the manuscript be self-contained. New studies will start in Section III. II. FROM THE EULER EQUATION TOSCALING
In ref. [10] it is assumed that the temperature and thedensity have time independent distributions with respectto a scaling variable: s = r x /X + r y /Y + r z /Z . Nowinstead we assume azimuthal symmetry and thus use thecorresponding cylindrical coordinates instead of ( x, y, z ).However, we also use the coordinates in length dimension,( r ρ , r ϕ , r y ), so that r ρ = ρ, r ϕ = r ρ ϕ, r y = y . These represent the so called ”out, side, long” direc-tions. The boundary values of these coordinates are then(
R, S, Y ). The scaling variables can be introduced as s ρ = r ρ /R , s ϕ = r ϕ /S s y = r y /Y , where S is the roll-length on the outside circumference,starting from ϕ = 0 and S = 0 at t , S = Rϕ and ˙ ϕ = ω and this displacement is orthogonal to thelongitudinal and transverse displacements. The internalroll-length is r ϕ = ϕ r ρ , the corresponding velocity is v ϕ = ω r ρ , and so v ϕ = ω r ρ . On the other hand fromthe scaling of r ρ , it follows that r ρ = R s ρ .Nevertheless, in case of these scaling variables the dis-tributions of density and temperature, n ( s ) and T ( t, s )should not depend on s ϕ or r ϕ , just on the radius andthe longitudinal coordinates. Therefore in this work weintroduce another scaling variable: s ≡ s ρ + s y . Our reference frame is then spanned by the directions( r ρ , r ϕ , r y ). In this case due to the azimuthal symmetrythe derivatives, ∂s/∂r ϕ vanish. In this coordinate systemthe volume is V = 2 πR Y .The derivatives, ˙ R ( t ) and ˙ Y ( t ) in this exact modelshould not equal the ones obtained from the fluid dy-namical model, because in the more realistic model thedensity and velocity profiles do not agree with the exactmodel’s assumptions. Also initially in the realistic fluiddynamical model the angular momentum increases in thecentral region due to the developing turbulence, while inthe exact model it is monotonously decreasing due to thescaling expansion.For simplicity we also assume that the Equation ofState (EoS), (cid:15) = (cid:15) ( n, p ), with a constant κ is: (cid:15) = κp and p = nT , (1)where n is the conserved net baryon charge and T is thetemperature.Now we calculate the eq. (15) in ref. [10]. n m ( ∂ t + v · ∇ ) v = −∇ p . (2)For the variables of this equation we have: T = T (cid:18) V V (cid:19) /κ T ( s ) ,n = n V V ν ( s ) ,ν ( s ) = 1 T ( s ) e − (cid:82) s du T ( u ) , (3)and in addition in ref. [10] it is assumed that the tem-perature and the density have time independent distri-butions with respect to the scaling variable s . Thus, for the right hand side of eq. (2): − ∇ p = −∇ nT = − n V V T (cid:18) V V (cid:19) /κ ∇ e − (cid:82) s du T ( u ) = − n V V T (cid:18) V V (cid:19) /κ e − (cid:82) s du T ( u ) ( −
12 ) 1 T ( s ) ∇ s = nmQ/V γ (cid:16) r ρ R e ρ + r y Y e y (cid:17) , (4)where γ = 1 /κ and Q ≡ T V γ m .Using the ρ, ϕ, y coordinates, the rotation would showup as an independent orthogonal term. However, theclosed system has no external torque, and the inter-nal force from the gradient of the pressure is radial,which does not contribute to tangential acceleration. Thechange of the angular velocity arises from the angularmomentum conservation in the closed system as a con-straint, so we do not have to derive additional dynamicalequations to describe the evolution of the rotation.Now for the left hand side of eq. (2), the velocity, v = v ( t, r ρ , r ϕ , r y ), scales as v = v ρ e ρ − v ϕ e ϕ + v y e y = ˙ RR r ρ e ρ − ωr ρ e ϕ + ˙ YY r y e y . (5)Let us first calculate the time derivatives for the compo-nents. (See e.g. [16]): ∂ t v ρ = (cid:34)(cid:32) ¨ RR − ˙ R R (cid:33) − ω (cid:35) r ρ ,∂ t v ϕ = − ω ˙ RR r ρ , ∂ t v y = (cid:34) ¨ YY − ˙ Y Y (cid:35) r y . (6)The other term of the comoving derivative gives:( v · ∇ ) v = ˙ R R r ρ e ρ + ω ˙ RR r ρ e ϕ + ˙ Y Y r z e z . (7)Adding eq. (6) and (7) we get: mn ( ∂ t + v · ∇ ) v ρ = mn (cid:104)(cid:16) ¨ R/R (cid:17) − ω (cid:105) r ρ ,mn ( ∂ t + v · ∇ ) v z = mn (cid:16) ¨ Y /Y (cid:17) r z . (8)Then the equality of the right hand side and left handside of the Euler equation (2) leads to the ordinary dif-ferential equations. Multiplying the two non-vanishingequations with R and Y respectively yields: R ¨ R − W/R = Y ¨ Y = T m (cid:18) V V (cid:19) γ , (9)where W ≡ ω R . Notice that from the angular mo-mentum conservation ω = ω R /R , thus the rotationalterm, R ω in the equation, takes the form W/R .Notice that due to the EoS the pressure is proportionalto the baryon density n , just as the r.h.s. of the Eulerequation, therefore the equation of motion does not de-pend on n or n . III. CONSERVATION LAWS
If we want to calculate the energy of the whole system,then we should actually integrate it for the whole volume, V . Thus, not only the scaling of v = ( v ρ , v ϕ , v z ) but theparticle density distribution, n ( s ) will also play a role.The rotational energy at the surface is E Side ≡ m ˙ S = mR ω , and if we express ω via ω usingthe relation ω = ω R /R , then E Side = W/R , i.e.just as before. The expansion energy at the surfaceis E Out ≡ m ˙ R , and for the longitudinal direction, E Long ≡ m ˙ Y .For the interior we can calculate the radial and longitu-dinal expansion velocities and the corresponding kineticenergies, as well as the kinetic energy of the rotation. Inthe evaluation of the internal and kinetic energies the ra-dial and longitudinal density profiles of the system shouldbe taken into account. We now assume that the temperature profile is flat,and consequently the density profiles are Gaussian andseparable [17]. With this approximation the different in-tegrated energies are calculated. The boundary of thespatial integrals can be set to infinity or to finite values( s ρM , s yM ) as well. To be consistent with the earlierexact model results we integrate now to infinity, the in-tegrals are finite.Adding up the kinetic energies yields E K = 12 mN B (cid:16) α ˙ R + α R ω + β ˙ Y (cid:17) , (10)where α ≡ √ C n I B ( s yM ) I C ( s ρM ) and β ≡ √ C n I A ( s ρM ) I D ( s yM ) , where C n = 1 (cid:14)(cid:2) √ I A (0 . s ρM ) I B (0 . s yM ) (cid:3) . Now we ex-tend the boundaries to infinity, thus α = 2 and β = 1.Here α and β are time independent, because theydepend on the scaling variables only. If we divide thisresult by the conserved baryon charge, N B , we will get E K N B = 12 m (cid:104) α (cid:16) ˙ R + R ω (cid:17) + β ˙ Y (cid:105) . (11)Based on the EoS, (cid:15) = κp = κnT , one can calculatethe compression energy also based on the density profilesof n ( s ) and (cid:15) ( s ) = κ n ( s ) T . Here we made the samesimplifying assumptions on the density profiles as before.Then volume integrated internal energy and net baryoncharge will have the same density profile, normalized to N B : I A ( u ) = 1 − exp( − u ), I B ( u ) = √ π Φ( √ u ), I C ( u ) = 1 − (1 + u ) exp( − u ), I D ( u ) = √ π Φ( √ u ) −√ ue − u , where Φ( u ) = erf( u ) ≡ √ π (cid:82) u exp( − x ) dx [18]. E int = κ (cid:90) pdV = κ (cid:90) nT dV = κN B T ( V /V ) γ C n × V πR Y (cid:90) s yM (cid:90) s ρM ν ( s ) ds ρ ds y √ s y = κN B T ( V /V ) γ = κT (cid:18) V V (cid:19) γ (12)where C n is the normalization constant. IV. REDUCTION TO A SINGLEDIFFERENTIAL EQUATION
Now following the method of ref. [17] , we study thefollowing combination: F = 12 ∂ t (cid:0) α R + β Y (cid:1) = ∂ t (cid:16) α R ˙ R + β Y ˙ Y (cid:17) = α ˙ R + β ˙ Y + α R ¨ R + β Y ¨ Y . (13)Here we used the notation ∂ t = ∂∂t and ∂ t = ∂ ∂t . Now we can replace the last two terms, α R ¨ R, β Y ¨ Y ,by using eqs. (9), i.e. we use the Euler equation (2).Then we obtain: F = α ˙ R + β ˙ Y + α WR + ( α + β ) Q (2 πR Y ) γ , (14)On the other hand from the energy conservation , E tot = E k + E int , we get that E tot N B m = 12 (cid:20) α ˙ R + β ˙ Y + α WR + 2 κQ (2 πR Y ) γ (cid:21) , (15)where we used the EoS and thus the parameter κ nowappears in the expression of the energy.Now, if our EoS is such that κ = ( α + β ) / , (16)then F = 2 E tot / ( N B m ) = const., and in the same typeof calculation as in ref. [17], we can introduce U ( t ) ≡ α R ( t ) + β Y ( t ) , (17)which satisfies ∂ t (cid:0) α R + β Y (cid:1) = ∂ t U ( t ) = 2 F . (18)Thus, the solution of eq. (18), can be parametrized as: U ( t ) = A ( t − t ) + B ( t − t ) + C , (19)where : A = α ˙ R + β ˙ Y + α W/R + ( α + β ) T mB = 2 α R ˙ R + 2 β Y ˙ Y C = α R + β Y . (20) t Y ˙ Y θ R ˙ R ω (fm/c) (fm) (c) (Rad) (fm) (c) (c/fm)0.0 4.000 0.300 0.000 2.500 0.250 0.1501.0 4.349 0.393 0.135 2.852 0.441 0.1152.0 4.776 0.458 0.235 3.360 0.567 0.0833.0 5.258 0.503 0.307 3.970 0.646 0.0594.0 5.777 0.534 0.358 4.642 0.696 0.0445.0 6.322 0.555 0.397 5.356 0.729 0.0336.0 6.886 0.571 0.426 6.096 0.752 0.0257.0 7.462 0.582 0.449 6.856 0.767 0.0208.0 8.049 0.591 0.467 7.629 0.779 0.016TABLE II. Time dependence of characteristic parameters ofthe exact fluid dynamical model [10]. R is the transverse ra-dius, Y is the (rotation axis directed) length of the system,˙ R, ˙ Y are the speed of expansion in transverse and axis direc-tions, θ id the angle of rotation, and ω is the angular velocityof the matter. Let us take one of the Euler equations from eq. (9),¨ Y = QY (2 πR Y ) γ , (21)and express R in terms of U ( t ) which is known basedon the energy conservation: R ( t ) = ( U ( t ) − β Y ) /α . (22) E t o t E k E in t E r o t (MeV/n) t ( f m / c ) FIG. 1. (color online) The time dependence of the kinetic en-ergy of the expansion, E K , of the internal energy, E int , andthe rotational energy, E rot per nucleon in the exact modelwith the initial conditions described above. The kinetic en-ergy of the expansion is increasing, at the cost of the de-creasing internal energy and the slower decreasing rotationalenergy. The rotational energy is decreasing to the half of theinitial one in 2.1 fm/c. This leads to a 2nd order differential equation for Y ( t ):¨ Y = α γ QY [2 πY ( U ( t ) − β Y )] γ = f ( Y, t ) , (23) which can be solved. Then R ( t ) and ˙ R ( t ) are given byeqs. (22) and (14) respectively.In the main steps we followed ref. [17], however itturned out that the modified last step of the methodprovides a more straightforward solution. We show themodel provides an excellent and simple semi-analytic toolto study the effects and consequences of an expandingand rotating system.We used the Runge Kutta [19] method to solve thisdifferential equation. We need the constants, Q and W ,as well as the initial conditions for R and Y .Based on fluid dynamical model calculation results,presented in table I, we chose the parameters: T =250 MeV , m = 939 . ω = 0 .
15 c/fm. For theinternal region we take the initial radius parameters as R = 2 . R = 0 . Y , corresponds tothe out of plane, y direction in the fluid dynamical model(and not to the beam direction!). Due to the eccentricityat finite impact parameters, with an almond shape pro-file, the initial out of plane size is larger than the in planetransverse size, so we chose Y = 4 . Y = 0 . v (c) t ( f m / c ) w = 0 . 1 5 c / f m v Y v R w = 0 c / f m v Y v R FIG. 2. (color online) The time dependence of the velocityof expansion in the transverse radial direction, v R and in thedirection of the axis of the rotation, v Y . the exact solution is able to describe the monotonic ex-pansion, and so the steady decrease of the rotation, westart from a higher initial angular velocity than shown bythe fluid dynamical model, PICR, as the angular veloc-ity was measured versus the horizontal plane where theangular velocity starts from zero.With these initial parameters the exact model yields adynamical development shown in Table II. According toexpectations the radius, R , and the axis directed size, Y ,are increasing, the angular velocity, ω decreases, The to-tal energy is conserved, while the kinetic energy of expan-sion is increasing, and that of the rotation and internalenergy are decreasing. See Fig. 1.The change of the expansion velocity is shown in Fig.2. The expansion velocity is increasing in both directions.While in the axis, y , direction the velocity increases from0.4 c to 0.5 c in 8 fm/c time, the radial expansion in-creases faster, in part due to the centrifugal force fromthe rotation. The radial expansion velocity increases bynear to 10 percent due to the rotation, which is signifi-cant, while the expansion in the direction of the axis ofrotation is hardly changed. In both cases the expansionin the radial direction is large. This is due to the choiceof a small initial radius parameter. This exact, perfectfluid model overestimates the radial expansion velocitydue to the lack of dissipation and the Freeze out happensearlier than 8 fm/c as at this time the size of the systemis already reaching 16 fm (see Fig. 3) larger than theestimates based two particle correlation experiments. Atthe same time, although the rotation is non-relativistic,the Hubble flow terms are large and the Hubble flows arerelativistic. This problem is not new, it is also present inref. [15] and in earlier exact models. The radial (direc-tional Hubble) solutions go smoothly over to a relativisticexact solution of hydrodynamics [20]. As the velocity ofrotation decreases faster asymptotically than the Hubbleflow, the rotation is not expected to influence essentiallythe asymptotic relativistic behaviour of the flow.The more rapid velocity change arises partly from thecentrifugal acceleration of the rotation, but also from thefact that the initially smaller transverse size increasesfaster in the direction of equal sizes in both directions.See Fig. 3. Y R (fm) t ( f m / c )
FIG. 3. (color online) The time dependence of the Radial, R , and the y axis directed size, Y , of the expanding system.As the y directed velocity is initially larger and its change isrelatively smaller the change of the rate of increase is hardlyvisible. V. CONCLUSIONS
In conclusion, the exact model can be well realizedwith parameters extracted from detailed, high resolution,3+1D relativistic fluid dynamical model calculations withthe PICR code. The exact model describes a system withwith one single, uniform ω representing the whole mat-ter at a given moment of time. Depending on the impactparameter, the system size, the beam energy, and thetransport properties, the uniform flow can develop fromthe initial shear flow at different times. It is importantto know when this time is reached in a collision and withwhich parameters. Then this will enable us to concludeabout the material parameters and the equilibration dy-namics. The exact model provides us with a simple andstraightforward tool to give a precise estimate about thetime moment when the rotation equilibrated and the pa-rameters of the matter at that moment.The exact model can be used as a tool, when the ro-tations can be detected at freeze out. Then it providedan estimate of the rate of decrease of angular speed androtational energy due to the expansion in an explosivelyexpanding system. This indicates that the effects of ro-tation can be observable in case of rapid freeze out andhadronization, and the rate if conversion from rotationalenergy to expansion can be studied in detail dependingon the parameters of the model.At the same time these studies also show that the ini-tial rotation is also influencing the rate of decrease ofrotation. Here especially the enhancement of initial rota-tion due to the Kelvin Helmholtz Instability is essential,although this is a special (3+1)D instability, which in it-self cannot be included in the exact rotational model.Still the presence of the KHI is essential to generatethe rotation, and thus the observation of the rotation isstrongly connected to the evolving turbulent instabilityin low viscosity Quark-gluon plasma.Due to the difference of the time evolution betweenthe numerical and the exactly solvable hydrodynamicalmodels at early times one expects that the predictionsof the two models in the sector of penetrating probes(dilepton spectrum, direct photon spectra, elliptic flow,HBT) will be different and these differences later on canbe used to explore the mechanism of equilibration. ACKNOWLEDGEMENTS
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