Thermodynamics at =0 on the lattice...Thermodynamics at µ = 0 on the lattice Y. Aoki U. Wuppertal RHIC Physics in the Context of the Standard Model Y. Aoki (U. Wuppertal) Thermodynamics

Thermodynamics at µ = 0 on the lattice

Y. Aoki

U. Wuppertal

RHIC Physics in the Context of the Standard Model

Y. Aoki (U. Wuppertal) Thermodynamics at µ = 0 on the lattice RBRC workshop 1 / 19

Staggered thermodynamics

Equation of State : YA, Z. Fodor, S.D. Katz, K.K. SzabóJHEP 0601:089,2006 [hep-lat/0510084]

Order of the transition : YA, G. Endrodi, Z. Fodor, S.D. Katz, K.K. Szabó

◮ Control most of the systematic errors in staggered thermodynamics◮ We cannot address locality problem of staggered


Staggered thermodynamics

Equation of State : YA, Z. Fodor, S.D. Katz, K.K. SzabóJHEP 0601:089,2006 [hep-lat/0510084]

Order of the transition : YA, G. Endrodi, Z. Fodor, S.D. Katz, K.K. Szabó

◮ Control most of the systematic errors in staggered thermodynamics◮ We cannot address locality problem of staggered


Systematic errors of the EoS

0.0

1.0

2.0

3.0

4.0

5.0

1.0 1.5 2.0 2.5 3.0 3.5 4.0

T/Tc

p/T4

pSB/T4

3 flavour2+1 flavour

2 flavour

Bielefeld 2000.

finite MD step size ∆τ

a dependence is notinvestigated

ma fixed while changing a(T = 1/(Nta))

Solution

Use exact algorithm: RHMC

Do several a with improvedaction

Tune ma to reproduce theT = 0 experimental spectrum:LCP


Systematic errors of the EoS

0.0

1.0

2.0

3.0

4.0

5.0

1.0 1.5 2.0 2.5 3.0 3.5 4.0

T/Tc

p/T4

pSB/T4

3 flavour2+1 flavour

2 flavour

Bielefeld 2000.

finite MD step size ∆τ

a dependence is notinvestigated

ma fixed while changing a(T = 1/(Nta))

Solution

Use exact algorithm: RHMC

Do several a with improvedaction

Tune ma to reproduce theT = 0 experimental spectrum:LCP


Discritization error

Rorational symmetry breaking

Taste symmetry breaking


Improving discritization error

Rotational symmetry breaking◮ Symanzik (tree level) gauge action→ improves gauge sector◮ Single (stout) link action→ does not improve f at T =∞

⋆ Large deviation from continuum SB⋆ a → 0 behavior is very good

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

4 6 8 10 12 14 16

Nτ

fF(0)(Nτ)/fcont, F

(0)STANDARD

NAIKP4

Heller, Karsch, Sturm





0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

4 6 8 10 12 14 16

Nτ

fF(0)(Nτ)/fcont, F

(0)STANDARD

NAIKP4

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

1/Nt2 (∝ a

2)

0

0.5

1

1.5

2

2.5

3

f(N

t) / f

cont

Nt=6 4

Standard (stout, ours)Naik (Asqtad, MILC)p4 (Bielefeld)

Heller, Karsch, SturmY. Aoki (U. Wuppertal) Thermodynamics at µ = 0 on the lattice RBRC workshop 5 / 19




0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

4 6 8 10 12 14 16

Nτ

fF(0)(Nτ)/fcont, F

(0)STANDARD

NAIKP4

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

1/Nt2 (∝ a

2)

0

0.5

1

1.5

2

2.5

3

f(N

t) / f

cont

Nt=6 4






0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

1/Nt2 (∝ a

2)

0

0.5

1

1.5

2

2.5

3

f(N

t) / f

cont

Nt=8 6


0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

1/Nt2 (∝ a

2)

0

0.5

1

1.5

2

2.5

3

f(N

t) / f

cont

Nt=6 4




Taste symmetry breaking,local fluctuation of the gaugefield is responsible for

Reduced by the stout linksmearing(Morningstar & Peardon).

0 0.1 0.2 0.3 0.4 0.5

mG2

0

0.2

0.4

0.6

0.8

1

mπ’

2

Naive

a-1

=1 GeV



Taste symmetry breaking,local fluctuation of the gaugefield is responsible for

Reduced by the stout linksmearing(Morningstar & Peardon).

0 0.1 0.2 0.3 0.4 0.5

mG2

0

0.2

0.4

0.6

0.8

1

mπ’

2

NaiveStout Link

a-1

=1 GeV


LCP: to be always on the physical pointThe line of constant physics

mud(β), ms(β).

Tune ms using Nf = 3 degeneratesimulations and LO ChPT

m2PS

m2V

∣

∣

∣

∣

Nf =3

(m = ms) =m2

ηs

m2φ

=2m2

K −m2π

m2φ

.

obtain ms(β)

mud = ms/25

should be checked in Nf = 2 + 1simulation.

LCP1

3 3.2 3.4 3.6 3.8 4β

0

0.05

0.1

0.15

0.2

ms


How close to the physical and continuum limit ?

0 5 10 15mπ

2/Tc

2

0

5

10

15

(mπ’

2 -mπ2 )/

Tc2

MILC, standard

Bielefeld, p4

This work, stout

Nf=2, Nt=6

Nf=3, Nt=4

MILC, AsqtadNf=2+1, Nt=6

Nf=2+1, Nt=6

∆′

π =m2

π′ −m2π

T 2c

= (m2π′ −m2

π)N2t

(T = 0 masses, measured at βc(Nt ))


EoS procedureintegral along LCP

pT 4

∣

∣

∣

(β,m)

(β0,m0)= −

fT 4

∣

∣

∣

∣

(β,m)

(β0,m0)

= −N4t

∫ (β,m)

(β0,m0)

d(β,mud ,ms)

〈 − Sg/β〉〈ψψud〉〈ψψs〉

.

pT 4

∣

∣

∣

(β,m)

=p

T 4

∣

∣

∣

(β,m)

(β0,m0)(T 6= 0)−

pT 4

∣

∣

∣

(β,m)

(β0,m0)(T = 0).

RHMC algorithm does not require the extrapolation in step size

T=0: no msimud = mphys

ud data.

pT 4

∣

∣

∣

(β,mphysud )

=p

T 4

∣

∣

∣

(β,msimud )− N4

t

∫ mphysud

msimud

dmud 〈ψψud 〉.

Extrapolation only needed for 〈ψψud〉(mphysud )← msim

ud = {3, 5, 7, 9} ×mphysud .



pT 4

∣

∣

∣

(β,m)

(β0,m0)= −

fT 4

∣

∣

∣

∣

(β,m)

(β0,m0)

= −N4t

∫ (β,m)

(β0,m0)

d(β,mud ,ms)

〈−Sg/β〉〈ψψud 〉〈ψψs〉

.

pT 4

∣

∣

∣

(β,m)

=p

T 4

∣

∣

∣

(β,m)

(β0,m0)(T 6= 0)−

pT 4

∣

∣

∣

(β,m)

(β0,m0)(T = 0).



ud data.

pT 4

∣

∣

∣

(β,mphysud )

=p

T 4

∣

∣

∣

(β,msimud )− N4

t

∫ mphysud

msimud

dmud 〈ψψud 〉.


ud = {3, 5, 7, 9} ×mphysud .



pT 4

∣

∣

∣

(β,m)

(β0,m0)= −

fT 4

∣

∣

∣

∣

(β,m)

(β0,m0)

= −N4t

∫ (β,m)

(β0,m0)

d(β,mud ,ms)

〈−Sg/β〉〈ψψud 〉〈ψψs〉

.

pT 4

∣

∣

∣

(β,m)

=p

T 4

∣

∣

∣

(β,m)

(β0,m0)(T 6= 0)−

pT 4

∣

∣

∣

(β,m)

(β0,m0)(T = 0).



ud data.

pT 4

∣

∣

∣

(β,mphysud )

=p

T 4

∣

∣

∣

(β,msimud )− N4

t

∫ mphysud

msimud

dmud 〈ψψud 〉.


ud = {3, 5, 7, 9} ×mphysud .


Simulation Procedure

many β’s (16 pts for Nt = 4, 14 pts for Nt = 6)

given β, msims = mphys

s (β) fixed. → mphysud = mphys

s /25.T 6= 0:

◮ msimud = mphys

ud .◮ Ns = 3Nt .

T = 0:◮ msim

ud = {3, 5, 7, 9} ×mphysud

◮ keeping Lsmπ > 3

finite size effect?◮ less than stat. error for several β’s checked


EoS

1 1.5 2 2.5 3T/Tc

0

2

4

6

8

10

p/T

4

SB limit

Nt=4

Nt=6

Nt=∞Nt=4

Nt=6

1 1.5 2 2.5 3T/Tc

0

5

10

15

20

25

30

ε/T

4

SB limit

Nt=4

Nt=6

Nt=∞

Nt=4

Nt=6

Multiplied with cSB/cNt (cSB = limNt→∞cNt ), Nt = 4(red), 6(blue).


EoS

Multiplied with cSB/cNt (cSB = limNt→∞cNt ), Nt = 4(red), 6(blue).

Tc: inflection point of χI/T 2.


ms −mud phase diagram

Columbia 1990.



Non-lattice approachms = mud →∞: global Z3 symmetry

◮ 3d, Z3 symmetric spin model:◮ low T : symmetric phase: confined◮ high T : broken phase: deconfined◮ 1st order transition

( Yaffe & Svetitsky )

1/m = 0+◮ small perturbation by external field

h ∝ e−βm (Banks & Ukawa)◮ 1st order transition



Non-lattice approach

chiral symmetry:(Pisarski & Wilczek)

◮ ms →∞, mud = 0: Nf = 2:2nd order with O(4)

◮ ms = mud → 0: Nf = 3:1st order

◮ m = 0+: 1st order



Non-lattice approach

chiral symmetry:(Pisarski & Wilczek)

◮ ms →∞, mud = 0: Nf = 2:2nd order with O(4)

◮ ms = mud → 0: Nf = 3:1st order

◮ m = 0+: 1st order



Nt = 4Columbia 1990

Lattice approach



Lattice approachms = mud →∞:

◮ 1st order confirmed by lattice⋆ 1st order (Columbia)⋆ 2nd order (Ape)⋆ 1st order by Finite Size Scaling

(Fukugita et al)



Lattice approachms = mud →∞:

◮ 1st order confirmed by lattice⋆ 1st order (Columbia)⋆ 2nd order (Ape)⋆ 1st order by Finite Size Scaling

(Fukugita et al)



Lattice approachms →∞, mud = 0: Nf = 2:

◮ Wilson fermion confirmed O(4)scaling

◮ Staggered: no O(4), O(2) scalingobserved.1st order ? (Pisa)

◮ possible ∆τ artifact(Philipsen Lattice2005)



Lattice approachms = mud : Nf = 3: end point(Bielefeld-Swansea)

◮ mπ,c = 290 MeV [standard, Nt = 4]◮ mπ,c = 67 MeV [p4, Nt = 4]◮ Large discretization error!→ needs investigation on adependence



Very important to determine theorder at the physical point: all µ 6= 0physics depend on this.

Position of the physical point:open question in the continuum limit.

Very demanding to explore all themass range.

Perhaps doable to determine theorder at the physical point.


Strategy

Exact algorithm (RHMC)

Finite size scalingCorrect quark masses

◮ tune ms and mu so thatratios of mπ, mK , fK takephysical values.

◮ LCP2: mu = ms/27.3, ms →

Continuum limit


Strategy

Exact algorithm (RHMC)

Finite size scalingCorrect quark masses

◮ tune ms and mu so thatratios of mπ, mK , fK takephysical values.

◮ LCP2: mu = ms/27.3, ms →

Continuum limit

LCP2

3 3.2 3.4 3.6 3.8 4β

0

0.05

0.1

0.15

0.2

0.25

0.3

ms

LCP1

LCP2


Finite Size Scaling of chiral susceptibility χud

Ns/Nt = 3− 5(6).No volume dependence found.How about in the continuum limit ?


Finite Size Scaling of chiral susceptibility χud

Ns/Nt = 3− 5(6).No volume dependence found.How about in the continuum limit ?


Continuum limit of the peak hightχ has power divergence, which should be subtracted.

∆χ = χ(T 6= 0)− χ(T = 0).

To make RG invariant quantity, mass squared is multiplied

m2ud∆χ

T 4c /(m

2ud∆χ).


Finite Size Scaling after continuum extrapolation

0 0.01 0.02 0.03 0.04

1/(VTc3)

0

10

20

30

T4 /(

m2 ∆χ

)

1st order

χ2 /dof=87/2

Susceptibility does not diverge as V →∞

→ Crossover


Summary

Stout-link smearing improvement was used to reduce the tasteviolation. Works quite well.

EoS

The equation of state was calculated with Nt = 4 and 6 lattices,using LCP1 (approximation).

For the reliable continuum extrapolation, Nt = 8 simulation isneeded.

Order of the transition

Fine tuned LCP2 was used.

Continuum limit of chiral susceptibility was obtained usingNt = 4, 6, 8, 10.

Finite Size Scaling was applied.

Cross-over was found for the physical point.

Tc ?Y. Aoki (U. Wuppertal) Thermodynamics at µ = 0 on the lattice RBRC workshop 19 / 19

Documents

Thermodynamics at =0 on the lattice...Thermodynamics at µ = 0 on the lattice Y. Aoki U. Wuppertal RHIC Physics in the Context of the Standard Model Y. Aoki (U. Wuppertal) Thermodynamics