Analytic hydrodynamics and time evolution in high energy collisions · 2013. 7. 1. · Analytic...

Analytic hydrodynamicsand time evolution in high energy collisions

Mate CsanadEotvos University, Budapest

51st International School of Subnuclear Physics, Erice, Sicily, 2013

July 1, 2013

Mate Csanad, Eotvos University 51st ISSP, Erice, 2013 1 / 12

Time evolution of the sQGPStrongly interacting QGP discovered at RHIC & created at LHC

A hot, expanding, strongly interacting, perfect QG fluid

Hadrons created at the freeze-out

Leptons, photons “shine through”


Perfect hydrodynamics

Applicable on many scales (10−15 m → 1010 m)

Exact, analytic solutions: important to get insight

Non-relativistic hydro: many solutions, applicable, but inconsistent!

Relativistic perfect hydro: formulated by LandauFamous 1+1D solutions: Landau-Khalatnikov, Hwa-BjorkenRevival of interest, many new solutions, mostly 1+1D, few 1+3DAll (up to now) using constant Equation of State (speed of sound)!

Energy-momentum conservation:

∂νTµν = 0 with Tµν = (ε+ p)uµuν − pgµν

Continuity of a conserved charge n or entropy density σ

Equation of State: ε = κp (κ = const. ⇒ c2s = ∂p/∂ε = 1/κ)

EoS determines time evolution!

Temperature defined via number density or entropy densityMate Csanad, Eotvos University 51st ISSP, Erice, 2013 3 / 12

Constant EoS hydro compared to data

Time evolution versus EoS

Take first exact, analytic and truly 3D relativistic solutionCsorgo, Csernai, Hama et al., Heavy Ion Phys. A21, 73 (2004), nucl-th/0306004

Calculate observables for identified soft hadronsCsanad, Vargyas, Eur. Phys. J. A 44, 473 (2010), arXiv:0909.4842

This fixes only the final state, initial state depends on EoS (κ)!Csanad, Nagy, Csorgo, Eur. Phys. J. ST, 19 (2008), arXiv:0710.0327


Constant EoS hydro compared to data

Time evolution via penetrating probesPhotons and leptons are created throughout the evolution

Their distribution reveals information about the EoS!

Calculated direct photon & lepton observables, compare to dataCsanad, Majer, Central Eur. J. Phys. 10 (2012), arXiv:1101.1279

Average EoS: cs = 0.36± 0.02stat ± 0.04syst (i.e. κ = 7.7)This is an average EoS, c.f. PHENIX estimate cs = 0.35

Constant EoS throughout the whole evolution?? Mate Csanad, Eotvos University 51st ISSP, Erice, 2013 5 / 12

Solutions with a QCD EoS

A new type of solution for arbitrary EoS

The first analytic hydro solution for arbitrary EoS (κ(T )):Csanad, Nagy, Lokos, Eur. Phys. J. A 48, 173 (2012), arXiv:1205.5965

σ = σ0τ3

0

τ3,

uµ =xµ

τ,

τ30

τ3= exp

∫ T

T0

(κ(β)

β+

1

κ(β) + 1

dκ(β)

dβ

)dβ

β is the integration variable here, i.e. T

Arbitrary κ(T ) functions may be used, the QCD one as well

Other new solutions with conserved charge, also for κ(p).


Solutions with a QCD EoS

Result for the lattice QCD EoS

lQCD EoS calculated for physical quark masses, continuum limitBorsanyi, Fodor, Katz et al. JHEP 1011, 077 (2010), arXiv:1007.2580

T (τ) from QCD EoS + hydroCsanad, Nagy, Lokos, Eur. Phys. J. A 48, 173 (2012), arXiv:1205.5965

RHIC 200 GeV Au+Au: τini/τfinal ≈ 0.1− 0.2

LHC 2.76 TeV Pb+Pb: thermalize fast, longer evolution: higher Tini


The p+p system

What about p+p?

What initial energy densities are possible at LHC p+p?

RHIC Au+Au: 5-15 GeV/fm3, RHIC p+p: few × 10 MeV/fm3

Lattice QCD, fenomenology:εcrit ≈ 1 GeV/fm3

The Bjorken estimatePhys.Rev.D27,1983, 2000 citations

E = NdE

dy∆y = N

dE

dy

1

2

2d

t= εAd

εBj =N

R2πτ0

dE

dy=〈E 〉

R2πτ0

dN

dy

Bjorken’s solution: no acceleration, flat rapidity (y) distribution

Contradicts measurements: finite dN/dy !

Need to estimate the work needed for accelerationMate Csanad, Eotvos University 51st ISSP, Erice, 2013 8 / 12

The p+p system

Effect of longitudinal acceleration

First explicit accelerating solution (control parameter λ, Bjorken: 1)Nagy, Csorgo, Csanad, Phys.Lett.B663,306 & Phys.Rev.C77,024908

λ controls rapidity distribution width (λ = 1: no acceleration, flat)


The p+p system

Initial energy density in p+p at LHC

Basic observed quantities:Number of particles per unit rapidity: 5.89±0.01 (CMS, ALICE)Average transverse energy: 0.562±0.002 GeV (CMS)Initial system size: 1.081±0.005 fm (TOTEM σ or ALICE HBT)

Initial acceleration calculated from dN/dy shape → 25% correctionεini estimate at average multiplicity (Csanad, Csorgo, in prep.)

εini = 1.14± 0.01(stat)+0.21−0.16(syst) GeV/fm3

Proportional to multiplicity


The p+p system

Is it unprecedented? Consequences?

Bjorken and Landau worked out hydro for pp and pA

,,It is not clear what the produced quanta which carry thisenergy really are: constituent quarks? current quarks?gluons? hadrons? However, this uncertainty should not affectthe estimated energy density.” (Bjorken)

If (high multiplicity) p+p is a supercritical system:

Radial flow (centrality?)γ/π enhancement in high multiplicity events? (M.J.T.)Scaling of azimuthal asymmetry?Scaling of Bose-Einstein correlation radii?Low mass dilepton enhancement?Direct photon enhancement?

High momentum jet suppression: need to divide by length scale!

Proper medium measure: energy loss per unit length?


Summary

Summary

New analytic hydro solutions with arbitrary EoS:Constant EoS solutions work on RHIC & LHC data (h, γ, `±)

Large temperature change: variable EoS more realistic

Found first exact solutions with arbitrary EoS

Estimated T (τ) based on a lattice QCD EoS

High temperatures not inconsistent with LHC p+p?

Supercritical medium in p+p?Estimated εini based on Bjorken’s idea

Supercritical energy density not inconsistent with p+p

Proper medium measure: energy loss per unit length


Thank you for your attention!


Hydrodynamics in heavy ion collisions

The strongly interacting QGP discovered at RHIC

Jet suppression: new phenomenon of missing high energy jetsPHENIX Coll., Phys.Rev.Lett. 88.022301 (2002)

No jet suppression in d+Au: new form of matter in central Au+AuPHENIX Coll., Phys. Rev. Lett. 91, 072303 (2003)

Collective dynamics: it is a liquid!PHENIX Coll., Nucl. Phys. A 757, 184-283 (2005)

Scaling properties: appearance of quark degrees of freedomPHENIX Coll., Phys. Rev. Lett. 98, 162301 (2007)

Energy loss of heavy quarks: nearly perfect liquidPHENIX Coll., Phys. Rev. Lett. 98, 172301 (2007)

Thermal photons: very high initial temperaturePHENIX Coll., Phys. Rev. Lett. 104, 132301 (2010)

Flow harmonics : extremely small kinematic viscosityALICE Coll., Phys.Rev.Lett. 107, 032301 (2011),CMS Coll., Eur. Phys. J. C 72, 2012 (2012),

ATLAS Coll., Phys.Rev. C86, 014907 (2012)



Soft hadron creation in A+A via hydro

Take first exact, analytic and truly 3D relativistic solutionCsorgo, Csernai, Hama et al., Heavy Ion Phys. A21, 73 (2004), nucl-th/0306004

Calculate observables for identified hadronsTransverse momentum distribution N1 (pt)Azimuthal asymmetry v2 (pt)Bose-Einstein correlation radii Rout,side.long (pt)

Compared to data successfully (RHIC shown, LHC done as well)Csanad, Vargyas, Eur. Phys. J. A 44, 473 (2010), arXiv:0909.4842

Data: PHENIX Coll., PRC69034909(2004), PRL91182301(2003), PRL93152302(2004)



Penetrating probes: photons and leptons

Photons and leptons are created throughout the evolution


Compared to PHENIX data (spectra and flow) successfully

Predicted photon HBT radiiCsanad, Majer, Central Eur. J. Phys. 10 (2012), arXiv:1101.1279

Data: PHENIX Collaboration, arXiv:0804.4168 and arxiv:1105.4126

Average EoS: cs = 0.36± 0.02stat ± 0.04syst (i.e. κ = 7.7)

Compatible with soft dilepton data as well



Equations of hydrodynamics

Energy-momentum conservation:

∂νTµν = 0 with Tµν = (ε+ p)uµuν − pgµν

In case of a conserved charge n (chem. pot. 6= 0):

∂µ(nuµ) = 0

If no n: entropy conservation (via ε = p + Tσ and dε = Tdσ):

∂ν(σuν) = 0

Equation of State: ε = κp (κ = const. ⇒ c2s = ∂p/∂ε = 1/κ)

This is a full set of equations for uµ, n and p

Temperature p = nT (if ∃ n), or p = Tσ/(κ+ 1) (if no n)

The same (!) resulting temperature equation, if κ = const.:

T∂µuµ + κuµ∂µT = 0

All κ = const. solutions valid for both {uµ, n, T} and {uµ, σ, T}.Mate Csanad, Eotvos University 51st ISSP, Erice, 2013 17 / 12


Equations of hydrodynamics with a conserved charge

There may be a conserved charge or number n (chem. pot. 6= 0)

Basic eqs: continuity and energy-mom. conservation

∂µ(nuµ) = 0 and ∂νTµν = 0

Energy-momentum tensor in perfect fluid: Tµν = (ε+ p)uµuν − pgµν

Equation of State (if κ = const., c2s = ∂p/∂ε = 1/κ): ε = κp

This is a full set of equations for uµ, n and p

Introduce temperature with p = nT , temperature eq. from here:

T∂µuµ + κuµ∂µT = 0

Write up solutions for uµ, n and T instead!



Equations of hydrodynamics without conserved charges

Energy-momentum conservation is the same

Let us introduce entropy density σ

Fundamental thermodynamical relations without a conserved charge

ε+ p = Tσ ⇒ dε = Tdσ and dp = σdT

The same continuity equation for σ follows from here:

∂ν(σuν) = 0 ,

EoS can be used the same manner here, but different p – T relation!

ε = κp and p = Tσ/(κ+ 1)

If κ = const., we get the same equation on the temperature as with n:

T∂µuµ + κuµ∂µT = 0.

All κ = const. solutions valid for both {uµ, n, T} and {uµ, σ, T}.Mate Csanad, Eotvos University 51st ISSP, Erice, 2013 19 / 12


A known ellipsoidal solution with constant κ

First exact, analytic and truly 3D relativistic solutionCsorgo, Csernai, Hama et al., Heavy Ion Phys. A21, 73 (2004), nucl-th/0306004

uµ =xµ

τ, with τ =

√xµxµ,

n = n0V0

Vν(s), T = T0

(V0

V

)1/κ

ν(s)−1,

ν(s) is an arbitrary function and

s =r 2x

X 2+

r 2y

Y 2+

r 2z

Z 2, V = τ3,

X ,Y ,Z : principal axes of expanding ellipsoid, X (t) = X0t etc.This solution is non-accelerating, ie. obeys uν∂νuµ = 0.Source function is:

S(x , p)d4x = N pµ d3Σµ(x)H(τ)dτ

n(x) exp (pµuµ(x)/T (x))− 1



Direct photon observables compared to data

Photons are created throughout the evolution


The source function of photon creation is assumed as:

S(x , p)d4x = N pµuµ

exp (pµuµ(x)/T (x))− 1d4x

Integrated over energy (i.e. four-momentum): emission ∝ T 4

Analyzed systematic change with T power, based on

rate(A + B → X ) = nAnB 〈vσA+B→X 〉 ∝ T 6

Transverse mom. distribution, elliptic flow and HBT radii calculableCsanad, Majer, Central Eur. J. Phys. 10 (2012), arXiv:1101.1279



A realistic EoS

Take EoS from lQCD: trace anomaly I (T )/T 4 parametrized

Pressure is given as p(T )T 4 =

∫dTT

I (T )T 4

EoS in form of κ(T ) analytically givenCsanad, Nagy, Lokos, arXiv:1205.5965

Problem: d(κT )/dT ≤0 at T = 173− 225 MeV

Recall: conserved charges not compatible with this EoS!



A new κ(T ) type of solution with conserved charge

A new solution with conserved charge n, for arbitrary κ(T ):Csanad, Nagy, Lokos, arXiv:1205.5965

n = n0τ3

0

τ3,

uµ =xµ

τ,

τ30

τ3= exp

∫ T

T0

(1

β

dκ(β)β

dβ

)dβ

Arbitrary κ(T ) functions may be used

For some choices of the κ(T ) function this solution becomes ill-defined

If d(κT )/dT ≤ 0, the last equation cannot be inverted

Same problem as mentioned earlier



A new κ(p) type of solution without conserved charge

A partly implicit new solution:Csanad, Nagy, Lokos, arXiv:1205.5965

σ = σ0τ3

0

τ3,

uµ =xµ

τ,

τ30

τ3=

∫ p

p0

(κ(β)

β+

dκ(β)

dβ

)dβ

κ(β) + 1

This solution may be used if κ is given as a function of pressure p.



Solutions with arbitrary EoS?

Constant EoS may not be realistic (temperature may change rapidly)

If κ(T ), new solutions have to be found

With a conserved charge (and ε = κnT ), the temperature equation is:

T∂µuµ +

[κ+ T

dκ

dT

]uµ∂µT = 0.

Works only if d(κT )/dT 6= 0!

In case of no conserved charges (and ε = κTσ/(κ+ 1)):

T∂µuµ +

[κ+

T

κ+ 1

dκ

dT

]uµ∂µT = 0,

Remarkable: these are not the same, but coincide if κ = const.!

If uµ known, these can be solved for an arbitrary κ(T )!



Lattice QCD EoS

lQCD EoS calculated for physical quark masses, continuum limit

Trace anomaly I (T )/T 4 given, c2s (T ) or κ(T ) calculable

Borsanyi, Fodor, Katz et al. JHEP 1011, 077 (2010), arXiv:1007.2580

Csanad, Nagy, Lokos, arXiv:1205.5965

Problem: d(κT )/dT ≤0 at T = 173− 225 MeV

Recall previous slide: conserved charges not compatible with this EoS!



Input parameters in estimating εini

parameter value stat. syst. eff. on ε

λ 1.073 0.1% 0.4% (from data)c2s 0.1 - -2%+0.2%(if 0.05 < c2

s < 0.5)τf /τ0 2 - -4%+10% (for 1.5–4)τ0 1 fm/c - only underestimates ε

R (from σinel) 1.081 fm 0.5% 1.5%〈mt〉 0.562 GeV/c2 0.5% 3%

dN/dη 5.895 0.2% 3%


Analytic hydrodynamics and time evolution in high energy collisions · 2013. 7. 1. · Analytic...

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