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Monte Carlo Simulation for Elastic Energy Loss ofHigh-Energy Partons in Quark-Gluon Plasma

J. Auvinen

Department of PhysicsUniversity of Jyväskylä, Finland

April 4th, 2011

J. Auvinen (JYFL, Finland) MC Simulation for Elastic Energy Loss April 4th, 2011 1 / 33

The papers

J. Auvinen, K. J. Eskola and T. Renk,“A Monte-Carlo model for elastic energy loss in a hydrodynamicalbackground,”Phys. Rev. C 82, 024906 (2010) (arXiv:0912.2265 [hep-ph]).J. Auvinen, K. J. Eskola, H. Holopainen and T. Renk,“Elastic energy loss with respect to the reaction plane in aMonte-Carlo model,”Phys. Rev. C 82, 051901 (2010) (arXiv:1008.4657 [hep-ph]).T. Renk, H. Holopainen, J. Auvinen and K. J. Eskola, in progress.


Motivation: Suppression of high-energy hadrons

Nuclear modification factor:

RhAA(b) =

d2NAA→h+X

dpTdy

< TAA(b) > d2σpp→h+X

dpTdy

,

where the nuclear overlap function is defined

TAA(b) =

∫d2sTA(s + b/2)TA(s− b/2),

TA(s) =

∫ ∞−∞

dz nA(√

s2 + z2).

nA(√

s2 + z2) is the Woods-Saxon potential.


Motivation: Suppression of high-energy hadrons

(GeV/c)T

p0 2 4 6 8 10 12 14 16 18 20

AA

R

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6 , 0-10%γdirect , 0-10% 0π

, 0-10% η

, 0-10%φ

, 0-20%ΨJ/, 0-20%ω

= 200 GeVNNsAu+Au, PHENIX preliminary

a)

Measured nuclear modifica-tion factors for direct pho-tons and various mesons byPHENIX detector at RHICa.

aK. Reygers [for thePHENIX Collaboration], J.Phys. G 35, 104045 (2008).

Strong suppression seen for mesons; weak suppression for photons.How well could this effect be reproduced by simulating hard partonslosing energy to a strongly interacting medium?


Scattering rate

Scattering rate for a particle of energy E1, assuming a scattering processij → kl (gg → gg , gq → gq etc.) with scattering amplitude |M|2ij→kl :

Γij→kl =1

2E1

∫d3p2

(2π)32E2f (p2,T )

∫d3p3

(2π)32E3

∫d3p4

(2π)32E4

· (2π)4δ(4)(p1 + p2 − p3 − p4) |M|2ij→kl ,

where f (p2,T ) is Bose-Einstein distribution

fBE (p,T ) =gg

eu·pT − 1

when the thermal particle is a gluon, and Fermi-Dirac -distribution

fFD(p,T ) =gq

eu·pT + 1

when the thermal particle is a quark.


Scattering rate

Γ can be written in terms of particle density n and cross section σij→kl (s):

Γij→kl =

∫d3p2

(2π)3 f (p2,T )(1− cos θ12)σij→kl (s),

≡ n(T )〈(1− cos θ12)σij→kl 〉.

Cross sections have t, u → 0 -singularities which need to be regulated. Theregulator m = smgT = sm

√4παsT is implemented on the limits of the σ

integral:

σij→kl (s) =1

16πs2

∫ −m2

−s+m2dt |M|2ij→kl .

This choice also requires us to have s ≥ 2m2.


Total scattering rate

The total scattering rate for a hard parton i is defined as

Γi =∑j ,(kl)

Γij→kl .

The total scattering rate for a gluon:

Γg = Γgg→gg + Γgg→qq̄ + Γgq→gq + Γgq̄→gq̄

The total scattering rate for a quark qi :

Γqi =Γqig→qig + Γqiqj→qiqj + Γqiqi→qiqi

+ Γqi q̄i→qj q̄j + Γqi q̄i→qi q̄i + Γqi q̄i→gg


Contribution of different processes on gluon scattering rate

0.0 0.2 0.4 0.6 0.8 1.010-5

10-4

10-3

10-2

10-1

1

s = 0.3sm = 1.0E1 = 5.0 GeV

[GeV

]

T [GeV]

Total gluon scattering rate g + g g + g g + g q + q

g + q g + q + g + q g + q

(a)


Contribution of different processes on quark scattering rate

0.0 0.2 0.4 0.6 0.8 1.010-5

10-4

10-3

10-2

10-1

1

Total quark scattering rate qi + g qi + g qi + qi qi + qi

qi + qj qi + qj + qi + qj qi + qj qi + qi qj + qj

qi + qi qi + qi qi + qi g + g

s = 0.3sm = 1.0E1 = 5.0 GeV

[GeV

]

T [GeV]

(b)


The Monte Carlo simulation

The hard parton is propagated in small time steps ∆t ∼ O(10−2) fm.The probability for having (at least) one collision isP(number of collisions ≥ 1) = 1− e−Γi ∆t .Γi = Γi (E1,T ) is calculated in the local rest frame of the fluid →boost E1,∆t to LRF.Finite scattering angles; non-eikonal propagation.After scattering, the parton with higher energy is selected as the hardparton.The simulation ends when the temperature of the medium is lowenough (≈ 160 MeV).


Results from the first paper (Central collisions)

J. Auvinen, K. J. Eskola and T. Renk,“A Monte-Carlo model for elastic energy loss in a hydrodynamicalbackground”,Phys. Rev. C 82, 024906 (2010) (arXiv:0912.2265 [hep-ph]).



Hydrodynamical background 1:Initial conditions from the EKRT model 2.Assuming longitudinal boost-invariance reduces the hydrodynamicalevolution equations into (2+1) dimensions.The azimuthal symmetry of central collisions leaves (1+1)-dimensionalevolution equations to be solved.

1K. J. Eskola, H. Honkanen, H. Niemi, P. V. Ruuskanen and S. S. Rasanen, Phys.Rev. C 72, 044904 (2005).

2K. J. Eskola, K. Kajantie, P. V. Ruuskanen and K. Tuominen, Nucl. Phys. B570(2000) 379-389.



Results: The nuclear modification factor RAA for π0

The simulation is done both with and without plasma effects.The pT distributions are folded with KKP fragmentation functions 3.

4 6 8 10 12 14 16 18 20

0.1

1

10

s = 0.3sm = 1.0

Simulation, hadron level Simulation, parton levelPHENIX 0-10% centrality

RAA(

)

pT [GeV]

PHENIX data fromA. Adare et al. [PHENIXCollaboration], Phys. Rev.Lett. 101, 232301 (2008).

3B. A. Kniehl, G. Kramer and B. Potter, Nucl. Phys. B 582, (2000) 514.J. Auvinen (JYFL, Finland) MC Simulation for Elastic Energy Loss April 4th, 2011 13 / 33

Results from the first paper (Central collisions)Sensitivity to the value of the strong coupling constant:

4 5 6 7 8 9 10 11 12 13 14 15

0.1

1

10sm = 1.0 s = 0.15

s = 0.3 s = 0.6

Par

toni

c R

AA

pT [GeV]

We have also studied the tomography of the punch-through partons andthe degree of non-eikonality.


Results from the second paper (Peripheral collisions)

J. Auvinen, K. J. Eskola, H. Holopainen and T. Renk,“Elastic energy loss with respect to the reaction plane in a Monte-Carlomodel”,Phys. Rev. C 82, 051901 (2010).


Results from the second paper (Peripheral collisions)

Hydrodynamical background 4:The smooth sWN profile 5 is used as an initial state.Assuming longitudinal boost-invariance reduces the hydrodynamicalevolution equations into (2+1) dimensions.Centrality classes defined using the optical Glauber model.

4H. Holopainen, H. Niemi and K. J. Eskola, Phys. Rev. C 83, 034901 (2011).5P. F. Kolb, U. W. Heinz, P. Huovinen, K. J. Eskola and K. Tuominen, Nucl. Phys. A

696, 197 (2001).J. Auvinen (JYFL, Finland) MC Simulation for Elastic Energy Loss April 4th, 2011 16 / 33

Results from the second paper (Peripheral collisions)The π0 nuclear modification as a function of the reaction plane angle∆φRP :

More matter in out-of-plane direction than in-planeRAA varies with reaction plane angle φ?

b

φRP


Results from the second paper (Peripheral collisions)The π0 nuclear modification as a function of the reaction plane angle∆φRP :

5 6 7 8 90.0

0.2

0.4

0.6

0.8

1.0

1.2 0

s = 0.5sm = 1.0

RA

A(p

T)

pT [GeV]

0-10% centrality

5 6 7 8 90.0

0.2

0.4

0.6

0.8

1.0

1.2 040-50% centrality

pT [GeV]5 6 7 8 9

0.0

0.2

0.4

0.6

0.8

1.0

1.2 050-60% centrality

pT [GeV]

MC RP = 0-15° PHENIX RP = 0-15° MC RP = 75-90° PHENIX RP = 75-90° MC average PHENIX average

RAA(φRP) data from S. Afanasiev et al. [PHENIX Collaboration],“High-pT π0

Production with Respect to the Reaction Plane in Au + Au Collisions at√(sNN) = 200 GeV”, Phys. Rev. C 80, 054907 (2009).J. Auvinen (JYFL, Finland) MC Simulation for Elastic Energy Loss April 4th, 2011 18 / 33

Work in Progress: Initial state fluctuations

T. Renk, H. Holopainen, J. Auvinen and K. J. Eskola - coming soon!

Hydrodynamics:

Event-by-event hydro calculations with fluctuating initial state 6:The eBC profile 7 is used as an initial state.Centrality classes defined using the Monte Carlo Glauber.

6H. Holopainen, H. Niemi and K. J. Eskola, Phys. Rev. C 83, 034901 (2011).7P. F. Kolb, U. W. Heinz, P. Huovinen, K. J. Eskola and K. Tuominen, Nucl. Phys. A

696, 197 (2001).J. Auvinen (JYFL, Finland) MC Simulation for Elastic Energy Loss April 4th, 2011 19 / 33

Work in Progress: Initial state fluctuationsRAA at pT = 10 GeV as a function of the angle of outgoing partons withthe reaction plane φ:

The initial state fluctuations average out in central collisions.In non-central collisions, the average over fluctuations gives slightlyless suppression compared to smooth hydro.


Final thoughts

With reasonable parameter values, elastic scattering would seem togive notable contribution to the suppression of high-energy hadrons.However, this kind of fully incoherent model fails to reproduce themeasured dependencies on the reaction plane angle and centrality.The initial state fluctuations average out in central collisions; however,they seem to have small but clear effect on the nuclear modification innon-central case.Next step: Implement also radiative energy loss (inelastic collisions)and coherence effects.Running coupling? Different regularisation method?Heavy quark elastic energy loss (charm).


Thanks for listening!


Extra slides


Results: Tomography (Central collisions)

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

Ninit = 150 000Nfinal = 35 663|yfinal| < 0.35pTfinal > 5 GeVsm = 1.0

s = 0.15

x [fm]

y [fm

]

0.002000.005000.008000.01100.01400.01700.0200

(a)

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x [fm]

y [fm

]

0.002000.005000.008000.01100.01400.01700.0200

Ninit = 150 000Nfinal = 18 891|yfinal| < 0.35pTfinal > 5 GeVsm = 1.0

s = 0.3

(b)

The initial production points of the punch-through partons in (transverse)xy -plane for α = 0.15 (left) and α = 0.3 (right). The coordinates of the

particle are rotated in the transverse plane so that it moves in the directionof x-axis.


Rapidity distributionThe initial rapidity distribution of the punch-through partons. The finalinterval |y | ≤ 0.35 is shown by the dashed lines.

-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

s = 0.3sm = 1.0

1 dN

N d

y

yinitial


Parton starting points and binary collision verticesParton starting points correlated with binary collision vertices vs. partonstarting points sampled from the corresponding nuclear overlap functionTAA(b).


The effect of energy on gluon scattering rate

0 10 20 30 40 5010-5

10-4

10-3

10-2

10-1

1

Total gluon scattering rate g + g g + g g + g q + q

g + q g + q + g + q g + q

s = 0.3sm = 1.0T = 0.5 GeV

[GeV

]

E1 [GeV]

(a)


The effect of energy on quark scattering rate

0 10 20 30 40 5010-5

10-4

10-3

10-2

10-1

1

Total quark scattering rate qi + g qi + g qi + qi qi + qi

qi + qj qi + qj + qi + qj qi + qj qi + qi qj + qj

qi + qi qi + qi qi + qi g + g

s = 0.3sm = 1.0T = 0.5 GeV

[GeV

]

E1 [GeV]

(b)


Initial state

Initial pT distribution of the hard partons produced using CTEQ6L1 8

parametrization for parton distribution functions xi fi (x ,Q2):

dσAA→f +X

dp2Tdy1

=

∫dy2

∑ijk

x1fi/A(x1,Q2)x2fj/A(x2,Q2)dσij→fk

dt.

The initial rapidity interval is chosen to be |yi | ≤ 1.0.

8J. Pumplin, D. R. Stump, J. Huston, H. L. Lai, P. Nadolsky, W. K. Tung, JHEP0207:012(2002), hep-ph/0201195


Initial state

Starting point on xy -plane determined by sampling the nuclear overlapfunction

TAA(b) =

∫d2sTA(s + b/2)TA(s− b/2).

Nuclear density is given by Woods-Saxon distribution

nA(r) = n0(1 + er−RA

d )

with n0 = 0.17 fm−3, RA = 1.12A1/3 − 0.86A−1/3 fm, d = 0.54 fm.The initial rapidity and the starting (proper) time τ0 define the timet0 = τ0 cosh(yi ) and z-coordinate z0 = τ0 sinh(yi ).


The Monte Carlo simulation: Producing the thermal particle

The energy E2 and the collision angle θ12 of the plasma particle aresampled from the expression of Γ as follows.

The scattering rate for any particular process can be written in theform

ΓX ∼∫ ∞

m22E1

dE2 f (E2,T )(H(4E1E2

m2 )− H(2)),

where H(x) is the integral function of σX (s) with the variablex = s

m2 = 2E1m2 E2(1− cos θ12). E2 can be sampled from this expression.

After E2 has been found, the collision angle can be determined bypicking a random number r from the interval [H(2),H(4E1E2

m2 )].Solving now x0 from the equation H(x0)− r = 0 givescos θ12 = 1− x0m2

2E1E2.


The Monte Carlo simulation: Scattering angle

The cross section of the process determines the distribution of scatteringangle θ13.

The scattering angle is determined in the CMS frame of the collision.The method is very similar to one used with finding the collision angle.The integral function is now ΣX (t) =

∫dt |M|2X . Picking a random

number R from the interval [ΣX (−s + m2),ΣX (−m2)] and solving t0from the equation ΣX (t0)− R = 0 gives the scattering angle ascos θ13 = 2t0

s + 1.


Mixed phase (Central collisions)

In the mixed phase, temperature stays constant (T = TC ) whileenergy density ε keeps decreasing.The scattering rate Γ depends on T but not ε → How to implementthe effect of mixed phase?Effective temperature Teff (R, τ) = 30

gQπ2 (ε(R, τ)− B)1/4 with bagconstant B = (239 MeV)4.The simulation ends when the boundary of mixed phase and hadrongas phase is reached (i.e. when T < TC ).
