Department of Physics

Transcription

1 Department of Physics Early time dynamics in heavy ion collisions from AdS/CFT correspondence Anastasios Taliotis based on work done with Yuri Kovchegov arxiv: [hep-ph]

2 The problem: Study the collision of two heavy ions Study the energy density as function of (early) time Show system is initially anisotropic Combine knowledge for late times isotropic estimate isotropization time Work in framework of AdS/CFT

3 Presentation layout I. Discuss key words II. Define problem III. Refer to late time solutions IV. Solve for early times V. Match the two solutions estimate isotropization time VI. Conclusions

4 I. Key words Isotropization Thermalazation Hydrodynamics Experimental data AdS/CFT: Introduction

5 Isotropization: example Consider a fluid T µν p 0 0 Let stress energy tensor: transversely symmetric Define isotropization the ε p p 3 lim p 3 p

6 Thermalization T µν = T µν ( temperature) Isotropization Thermalization Isotropization ûthermalization

7 How phenomena are related: Thermalization => equilibrium =>Ignore viscosity effects => ignore anisotropy=>isotropization In all: Initially anisotropic medium can become isotropic (reach equilibrium). then two possibilities: thermal non thermal

8 Where anisotropy comes from? Assume two heavy ions collision in x3-axis: Then the -axis is a preferred direction x 3

9 Hydrodynamics Why introduce hydrodynamics? Hydrodynamic simulations strongly support that in heavy ions collisions, for τ << 1(early times?? τ << 1 Qs Qs saturation scale?? ~1fm/c??) system behaves hydrodynamically. Consider following figure:

10 Data motivate hydrodynamic behavior Boltzmann distribution (can describe exp. Data at RHIC)

11 Assumptions to be justified by the result: Assume hydrodynamic behavior soon after collision (yellow area ). The mean free path is 1/ 3 small (strongly correlated): λ / V << 1 Consistent with large densities at early times: λ ~ 1 ρσ If medium is thermal, then T=200Mev as estimated from Hydrodynamic simulations data This implies large coupling regime -consistent with large cross section => small mean free path is justified.

12 Necessity for alternative methods to attack the problem pqcd does not give complete description Moreover, cannot reproduce small initialization time for Hydro behavior (where yellow area begins ) What can we do?

13 It has been suggested that isotropization and (possible) thermalization, is a strong coupling effect Argument becomes stronger from the fact that if thermalization occurs, it occurs at τ H ~1/ Temp. g 4 (pertub. calcu.) => early initialization (experimental fact) => large coupling. Motivation for introducing AdS/CFT

14 AdS/CFT: Introduction D3 branes (3+1) dim. Objects Open strings: gauge fields that live on the branes They have end points on branes( Dirichlet B.C.)

15 How does it work? Describe a system of N coincident D3 branes from two different viewpoints: Gauge way Vs Super String way

16 The Gauge way Consider N coincident D3 branes: U(N)=SU(N)xU(1) ~N^2 combinations to couple the branes = # of generators In particular : Conformal N =4 SYM theory Closed strings flow around in empty space (flat space) SU(N)SuYM+ (closed)flat+ Interactions

17 The SuperString way Background (type II.b) : (Closed) flat + (closed) throat + Interactions D 3 z S 5 S 5 - radius R surrounds branes 5 - Horizon (AdS x ) S D3 D 3 z x µ Minkowski

18 Duality: motivations (a) Algebras - gauge theory: (6+4) Poincare + (1+4) conformal- scale - IIb Super String Theory: 15 generators of isometries Fact: both sets of generators have same algebras

19 (b) Decoupled systems Turn off interactions Take low energy limit in this limits => both theories have two decoupled subsystems Gauge th. :SU(N) (closed)flat String th. : (closed)ads (closed)flat SU(N) (closed)ads? This is the conjecture Note: Closed strings contain graviton as massless excitations=> conjecture implies Gauge SU(N) ( gravity )AdS

20 Scales Vs Parameters/convenient limits Type IIB superstr. N =4 Super YMSU( ) N c l / p(10) L Q. gravity & fields Q. strings 1/ Nc l p Clas. fields & part. Clas. Strings L λ ( 10) / L << 1 =Radius of S = `t Hooft = = 5 L l 4 4 / s G ls (Ignore QM/small ) & / L << 1 (Ignore extended objects/small ) N c (10) g 2 YM Large & Large l s / N L c λ g 2 YM = 4π g s 1/λ 2 L / l = N ~1/ G π 4 4 (10) p(10) c g s

21 Practical use of correspondence φ, g, B SUGRA contains, other Form Fields S µν µν 5 Compactify fields on (dim reduction/known from 70 s) Equations of motion ( varying SUGRA Lagranian) - keep only φ, g µν - set everything else to zero/it is a consistent set of D.E. and solutions!!

22 2 1 Remain with Rµν = Λ gµν + µ φ ν φ, Λ = 6 and φ =0 3 2 Λ= 6 it is not input of theory but results from S previous dim. reduction on. 5 Dictionary assigns O SUGRA in some particular manner. In particular: O SU ( Nc ) g µν T µν This is what we use in this paper.

23 II. Define problem We use AdS/CFT to calculate ε (τ ) energy density of the collision of two heavy ions We study the limit of early times We solve Einstein s equations (string side) and match the solutions to Gauge theory side.

24 τ = Kinematics-Area of interest Light cone coordinates: 2x + x _ y = 1 ln 2 1 x = ± ( x0 x3 ) 2 ± Figure shows lines of constant y and τ Hydrodynamics begins early Janik and Peschanski showed ideal hydrodynamics apply in later times in framework of AdS/CFT This presentation deals with early times : τ <<1 x x + _ xt 0 η y τ zx 3

25 Late time solutions τ >>1 This is the most (at x 3 = 0) general translationally invariant stress energy tensor in transverse plane that obeys the symmetry x for two particles + x moving opposite to each other. [Kovchegov/hep-ph/ ] µν µ ν µ ν µ ν µ ν T ( τ ) = A( τ ) u u + B( τ ) u υ + υ u ) + C( τ ) υ υ + D( τ ) η µν x Where, υ ( + x u,,0,0) the 4 th µ ν = ± velocity of the two oppositely τ τ moving particles. Bjorken hydrodynamics:perfect fluid lim p 3 p

26 µ dε ( τ ) ε Conservation : µ T = 0 = dτ Isotropy: p3=p & + + ( T x ) T ( g) L µν µ µ ν = µ = β g => together imply ετ ( )~ 1 τ Tracelessness because N =4 is scale invariant (and conformal) => has β(g)=0.this is implied from conservation of dilatation current: p τ Remarkably: Janik and Pechanski derived this at late times (where hydro holds) using AdS. 3 4/3 Their way to filter out correct solution: absence of singularities for curvature invariant (details follow)

27 Holographic renormalization Use 5 diffeo. to write metric as Here µ,ν take values 0,1 (or +.-),2,3 while g s depend from µ x, z Expansion of gravity solution near the UV boundary (z=0): schematically g 5x5 4x z 2 First term: Minkowski metric Second term : vanishes Third term: Interesting!

28 Solving the problem : Negative cosmological constant. Then,, Einstein's φ equations (with =0) read: x x -y +y ( + _ ) (1) -diffeomorphisms (5) Symmetries -boost invariance (3) => can write metric as : -rotational symmetry in transverse plane(1)

29 Recall: < T µν >= 2 Nc 2 2π g (4) µν ( x) We expand the exponents of the metric in series of z for a(τ,z) (expand b, c similarly): Using T above, we arrive at: =>Solving Einstein's equations, then: b o ( τ ), c ( τ ) Similar hold for o. These give p(τ) & p3(τ) Note: Generally anisotropic N g xz ετ ( ) lim 2π z 00 c 1 + (, ) = 2 za0 4

30 Power low solution Try ε ( τ ) ~ τ Solve Einstein s equations order to order in z Motivation: Guess (if any) convenient scaling variable : - Late times: -Early times: These transformations makes Einstein's equations simpler. 4 u = a0 u = z τ z 4 τ ds 2 ( z, τ ) a ds 2 ( u, τ )

31 III. Late time solutions Janik & Peeschanski obtained (keeping u fixed/varying τ as it becomes large): a u D Du 2 D Du D D 4 4 ( ) = (1 ) ln(1 + ) + (1 + ) ln(1 ), = ( ) > 0 Recall a(u( τ,z)) : τ Similar solutions hold for b,c (different coefficients) J&P: Filtered out choosing = 4 3: demanding absence of curvature singularities: < Same result: Demanding single valued real metric Result: Perfect fluid of Bjorken Hydrodynamics!

32 Branch Cuts begin at: 1 Du 4 0 a u D Du 2 D Du D D 4 4 ( ) = (1 ) ln(1 + ) + (1 + ) ln(1 ), = ( ) > 0 Remember the function(s) a(u) (b(u) and c(u)) need to be exponentiated to obtain the metric. Requiring the coefficients in front of the logarithm(s) to be integers l,m,n: After some algebra one obtains that the only allowed power is = giving the Bjorken hydrodynamics reproducing the result of Janik and Peschanski: 4 3

33 IV. Early time solutions Power low ε ~ τ : Recall scaling Bond. Cond.: #, is known polynomial of obtained from pertubative solutions in z Similar hold for the momentum functions Solving analytically at, obtain: e 4 τ τ + = e a(, u) a( u) a u u u 4 ( ) = #, 0 u = z τ b( u), c( u) τ 0 A: component of metric. Similarly there exist B,C F: hypergeometric function.

34 F s and so A, B, C(u) appear branch cut at u 1=> Metric is not real singled valued. Stress-Energy tensor constrains (J&P):[hep-th/ v2] µ ν Let t µ any time-like vector. It must be Tµνt t 0 (a) Boost invariance (symmetry assumption, x ) + x (b) Rotational symmetry on plane (c) Conservation T µν µ = 0 (d) Tracelessness T µ µ (e) ε τ < (true for early times only) Two choices of achieve single =-2, violates energy conservation valued metric for A, B, C(u) =0, acceptable 1 0 Conclusion: by assumption: ε ~ τ => ε Øconst. Similarly pzø -ε, pø ε =>also const. Note: System is anisotropic at early times Vs isotropic found at late times (J&P)

35 Logarithmic: Motivated by pqcd Consider: B.C.: Found from N g xz ετ ( ) lim 2π z 00 c 1 + (, ) = 2 za0 4 Solve Einstein s equations as function of δ Choose the uu,ux3,x3x3 components of equation above and solve. Remarkably, equations are exactly (non trivial) solvable. Get nice behaved solutions " δ

36 Constrain δ: Plug solutions to for µ=ν=τ and µ=ν=y, that is the remaining components Find that only δ=0 is consistent => As τ a1 a0 then ε ~ a 0 + is constant again ln 1 τ

37 V. Estimate isotropization time Recall power low solution for early time. scaling variable: z u= τ Branch cut of hypergeometric F ( u 2 ) ~ln(1 u ); u = a z τ /3 o solution begins at u=1, so Branch area Late times (J&P). Scaling variable: 4 4 4/3 o Recall singular part of this solution is u = a z τ 4 ln(1 D( ) u ), D( = 4/3) = 1/3 So, branch cut of J&P solution ( 1 4 ~ln(1 ) 3 u ) begins at:

38 Equate => a 0 N 1 ετ ( ) = a 2π τ c 2 0 4/3 Eliminate. Recall => Where we used (J&P) result: Evaluate ε o : Au+Au collision (RHIC) at:

39 Hydrodynamic simulations describe RHIC data accurately. They give: ε 3 = 15 GeV / fm for τ = 0.6 / ε o Use ετ ( ) = 4/3 to obtain τ fm c ε = 38 fm 0 8/3 Use above and N c = 3, get τ iso 0.29 fm / c

40 VI. Conclusions Filter solutions down to demanding real single valued solutions instead of absence of singularities of curvature invariant (J&P). pqcd, suggests logarithmic behavior at early times. AdS/CFT gives constant. Power low gives also constant. Found anisotropic system here at τ<<1 Vs isotropic at τ>>1 (J&P) Estimate isotropization time, that is the transition from anisotropic isotropic medium. Result agrees to at least 50% with RHIC simulations.

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