Equation of state Pasi Huovinen Uniwersytet Wroc lawski Collective Flows and Hydrodynamics in High Energy Nuclear Collisions Dec 14, 2016, University of Science and Technology of China, Hefei, China The speaker has received funding from the European Union s Horizon 2020 research and innovation programme under the Marie Sk lodowska-curie grant agreement No 665778 via the National Science Center, Poland, under grant Polonez DEC-2015/19/P/ST2/03333
Equation of state in form P = P (ɛ, n) needed to close the system of hydrodynamic equations P. Huovinen @ Hefei School, Dec 14, 2016 1/34
Remark: P = P (ɛ, n) is not a complete equation of state in a thermodynamical sense. A complete equation of state allows to compute all thermodynamic variables. For example, s = s(ɛ, n): ds = 1/T dɛ µ/t dn (1st law of thermod.) 1 T = s ɛ n, P = P (ɛ, n) does not work! µ T = s n ɛ, P ɛ n =? P n ɛ =? P = T s + µn ɛ However, P = P (T, µ) does work! dp = sdt + ndµ s = P T µ, n = P µ T P. Huovinen @ Hefei School, Dec 14, 2016 2/34
Nuclear phase diagram Temperature [MeV] 2 10 12 K ~170 ǫ 0.7 GeV/fm 3 Quark gluon plasma? Tricritical point? Hadrons : baryons p, n,... mesons π, K,... SC phases? 0 Everyday world n 0 = 1.4 10 44 /m 3 ǫ 0.16 GeV/fm 3 n 0 ~5 10 Baryon density 1 3 (n q n q ) P. Huovinen @ Hefei School, Dec 14, 2016 3/34
0th approximation for equation of state at µ B = n B = 0: Hadronic phase: ideal gas of massless (boltzmann) pions, g π = 3 ɛ π = 3g π π 2 T 4 P π = g π π 2 T 4 = 1 3 ɛ π Partonic phase: ideal gas of partons + bag constant of the bag model, B ɛ QGP = 3g QGP π 2 P QGP = g QGP π 2 T 4 + B Number of DOF in partonic phase: 2 quark flavours and gluons g QGP = 40 T 4 B = 1 3 ɛ QGP 4 3 B P. Huovinen @ Hefei School, Dec 14, 2016 4/34
Gibbs criterion: P QGP (T c ) = P π (T c ) T c = First order phase transition: Pressure: ( π 2 g QGP g π )1 4 B 1 4 Energy density: P. Huovinen @ Hefei School, Dec 14, 2016 5/34
QCD equation of state lattice QCD (Karsch & Laermann, hep-lat/0305025): 16 14 ε/t 4 RHIC ε SB /T 4 12 10 8 6 4 2 0 SPS 3 flavour 2 flavour 2+1-flavour T c = (173 +/- 15) MeV ε c ~ 0.7 GeV/fm 3 LHC T [MeV] 100 200 300 400 500 600 - EoS from first principles P. Huovinen @ Hefei School, Dec 14, 2016 6/34
lattice QCD: Budapest-Wuppertal collaboration, arxiv:1309.5258: 5 4 N t =8 10 12 16 4 stout crosscheck (ε-3p)/t 4 3 2 1 continuum limit 0 200 300 400 500 T[MeV] Trace anomaly ɛ 3P T 4 P. Huovinen @ Hefei School, Dec 14, 2016 7/34
lattice QCD: Budapest-Wuppertal collaboration, arxiv:1309.5258: 5 4 N t =8 10 12 16 4 stout crosscheck (ε-3p)/t 4 3 2 1 continuum limit 0 200 300 400 500 T[MeV] obtain pressure via What is P (T 0 )? P T 4 P 0 T 4 0 = T dt ɛ 3P T 0 T 5 P. Huovinen @ Hefei School, Dec 14, 2016 8/34
6 5 Lattice vs. HRG lattice continuum limit SB 4 p/t 4 3 2 1 0 HTL NNLO HRG 200 300 400 500 T[MeV] - Lattice agrees with Hadron Resonance Gas at low T P. Huovinen @ Hefei School, Dec 14, 2016 9/34
Hadron Resonance Gas model Dashen-Ma-Bernstein: Phys. Rev. 187, 345 (1969) EoS of interacting hadron gas well approximated by non-interacting gas of hadrons and resonances P (T, µ) = ±g i (2π) 3 T d 3 p ln (1 ± e i valid when - interactions mediated by resonances - resonances have zero width p 2 +m 2 µ i T Prakash & Venugopalan, NPA546, 718 (1992): experimental phase shifts Gerber & Leutwyler, NPB321, 387 (1989): chiral perturbation theory HRG good approximation at low temperatures lattice should reproduce HRG at T 120 140 MeV and it does ) P. Huovinen @ Hefei School, Dec 14, 2016 10/34
End of evolution I when fluid dynamical description breaks down, so-called freeze-out convert fluid to particles energy conservation iff EoS is the same before and after freeze-out in HRG this is by definition true P. Huovinen @ Hefei School, Dec 14, 2016 11/34
End of evolution II Ratios 1 p/p Λ/ Λ Ξ/ Ξ + Ω / Ω π - /π - + - - - K /K K /π p/π K *0 - /h - φ/h - Λ/h - - Ξ /hω/π *10 p/p - + - - - K /K K /π /π p Ω/h *50-10 -1 10-2 STAR PHENIX PHOBOS BRAHMS s NN =130 GeV Model re-fit with all data T = 176 MeV, = 41 MeV µ b =200 GeV s NN Model prediction for T = 177 MeV, = 29 MeV µ b Particle ratios T 160 170 MeV temperature p T -distibutions T 100 140 MeV temperature Evolution out of chemical equilibrium P. Huovinen @ Hefei School, Dec 14, 2016 12/34
Chemical non-equilibrium Treat number of pions, kaons etc. as conserved quantum numbers below T ch (Bebie et al, Nucl.Phys.B378:95-130,1992) number of pions: thermal pions + everything from decays: ˆn π = n π + 2n ρ + n + entropy per pion, kaon etc. conserved ˆn π s (T, {µ i}) = ˆn π s (T ch, 0) ˆn K s (T, {µ i}) = ˆn K s (T ch, 0). P. Huovinen @ Hefei School, Dec 14, 2016 13/34
Chemical non-equilibrium Treat number of pions, kaons etc. as conserved quantum numbers below T ch (Bebie et al, Nucl.Phys.B378:95-130,1992) P = P (ɛ, n b ) changes very little, but T = T (ɛ, n b ) changes... P. Huovinen @ Hefei School, Dec 14, 2016 14/34
HRG below T 160 170 MeV Parametrize lattice using: 5 Procedure for EoS ɛ 3P T 4 = d 2 T 2 + d 4 T 4 + c 1 T n 1 + c 2 T n 2 4 N t =8 10 12 16 4 stout crosscheck (ε-3p)/t 4 3 2 1 continuum limit 0 200 300 400 500 T[MeV] P. Huovinen @ Hefei School, Dec 14, 2016 15/34
HRG below T 160 170 MeV Parametrize lattice using: Procedure for EoS ɛ 3P T 4 = d 2 T 2 + d 4 T 4 + c 1 T n 1 + c 2 T n 2 Require that: ɛ 3P T 4, T0 d dt ɛ 3P T 4, T0 d 2 ɛ 3P dt 2 T 4 T0 are continuous = T 0, c 1, c 2 fixed χ 2 fit to lattice above T 0 MeV P. Huovinen @ Hefei School, Dec 14, 2016 15/34
Final result, P/T 4 P/T 4 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 CF CE 100 200 300 400 500 T [MeV] P. Huovinen @ Hefei School, Dec 14, 2016 16/34
Nuclear phase diagram Temperature [MeV] 2 10 12 K ~170 ǫ 0.7 GeV/fm 3 Quark gluon plasma? Tricritical point? Hadrons : baryons p, n,... mesons π, K,... SC phases? 0 Everyday world n 0 = 1.4 10 44 /m 3 ǫ 0.16 GeV/fm 3 n 0 ~5 10 Baryon density 1 3 (n q n q ) P. Huovinen @ Hefei School, Dec 14, 2016 17/34
Taylor expansion for pressure where P T 4 = Σ i,j c ij (T ) ( µb T ) i ( µs ) j, T c ij (T ) = 1 i j P i!j! (µ B /T ) i (µ S /T ) j T 4, i.e. moments of baryon number and strangeness fluctuations and correlations an EoS based on lattice calculations of these? But: Only limited set extrapolated to continuum P. Huovinen @ Hefei School, Dec 14, 2016 18/34
Parametrization c ij (T ) = a 1ij ˆT n 1ij + a 2ij ˆT n 2ij + a 3ij ˆT n 3ij + a 4ij ˆT n 4ij + a 5ij ˆT n 5ij + a 6ij ˆT n 6ij + c SB ij, where n kij are integers with 1 < n kij < 42, and ˆT = T T s R, with T s = 0.1 or 0 GeV, and R = 0.05 or 0.15 GeV. P. Huovinen @ Hefei School, Dec 14, 2016 19/34
Constraints: c ij (T sw ) = c HRG ij (T sw ) d dt c ij(t sw ) = d dt chrg ij (T sw ) d 2 dt 2 c ij(t sw ) = d2 dt 2 chrg ij (T sw ) d 3 dt 3 c ij(t sw ) = d3 dt 3 chrg ij (T sw ) at T sw = 155 MeV 3rd derivative to quarantee smooth behaviour of speed of sound: c 2 s d2 dt 2 c ij P. Huovinen @ Hefei School, Dec 14, 2016 20/34
0.2 0.01 c 20 c 40 0.15 0.008 0.006 0.1 p4, N τ =4 N τ =8, HISQ 0.004 0.05 0.002 0 T [MeV] 150 200 250 300 350 400 450 0 T [MeV] 150 200 250 300 350 400 450 P. Huovinen @ Hefei School, Dec 14, 2016 21/34
P/T 4 4.5 4 3.5 3 2.5 P [GeV/fm -3 ] 2 1.5 1 0.5 400 0 350 300 250 T [MeV] 200 150 1000 50 100 250 200 150 µ B [MeV] 300 P. Huovinen @ Hefei School, Dec 14, 2016 22/34
P/T 4 4.5 4 3.5 3 2.5 P [GeV/fm -3 ] 2 1.5 1 0.5 400 0 350 300 250 T [MeV] 200 150 1000 50 100 250 200 150 µ B [MeV] 300 P. Huovinen @ Hefei School, Dec 14, 2016 23/34
Speed of sound c s 2 0.3 0.28 0.26 0.24 0.22 0.2 0.18 0.16 0.14 0.12 140 160 180 200 220 240 260 T [MeV] µ B = 0 s95p-v1 parametrization by P. Petreczky and P.H. P. Huovinen @ Hefei School, Dec 14, 2016 24/34
Speed of sound c s 2 0.3 0.28 0.26 0.24 0.22 0.2 0.18 0.16 0.14 0.12 140 160 180 200 220 240 260 T [MeV] µ B = 0 s/n B = 400 P. Huovinen @ Hefei School, Dec 14, 2016 25/34
Speed of sound c s 2 0.3 0.28 0.26 0.24 0.22 0.2 0.18 0.16 0.14 0.12 140 160 180 200 220 240 260 T [MeV] µ B = 0 s/n B = 400 s/n B = 100 P. Huovinen @ Hefei School, Dec 14, 2016 26/34
Speed of sound c s 2 0.3 0.28 0.26 0.24 0.22 0.2 0.18 0.16 0.14 0.12 140 160 180 200 220 240 260 T [MeV] µ B = 0 s/n B = 400 s/n B = 100 s/n B = 65 P. Huovinen @ Hefei School, Dec 14, 2016 27/34
Transverse expansion and flow Define speed of sound c s : c 2 s = P ɛ s/nb large c s stiff EoS small c s soft EoS For baryon-free matter in rest frame (ɛ + P )Du µ = µ P τ u µ = c2 s s µs The stiffer the EoS, the larger the acceleration P. Huovinen @ Hefei School, Dec 14, 2016 28/34
Blast wave (Siemens and Rasmussen, PRL 42, 880 (1979)) Freeze-out surface a thin cylindrical shell radius r, thickness dr, expansion velocity v r, decoupling time τ fo, boost invariant Cooper-Frye for Boltzmannions dn = g ( dy p T dp T π τ vr γ r p ) ( T γr m ) T fo r m T I 0 K 1 T T P. Huovinen @ Hefei School, Dec 14, 2016 29/34
effect of temperature and flow velocity v r = 0.3 T = 120MeV The larger the temperature, the flatter the spectra The larger the velocity, the flatter the spectra blueshift The heavier the particle, the more sensitive it is to flow (shape and slope) P. Huovinen @ Hefei School, Dec 14, 2016 30/34
EoS vs. T fo hard EoS high T fo soft EoS low T fo dn/dyp T dp T [GeV -2 ] 10000 1000 100 10 1 0.1 0.01 h - p-p NA49 bag model lattice 0 0.5 1 1.5 2 2.5 p T [GeV] T fo 120 MeV (bag), T fo 130 MeV (lattice) P. Huovinen @ Hefei School, Dec 14, 2016 31/34
v 2 and EoS ideal hydro, Au+Au at s NN = 200 GeV s95p: T dec = 140 MeV EoS Q: first order phase transition at T c = 170 MeV, T dec = 125 MeV v 2 (p T ) of protons sensitive to phase transition! P. Huovinen @ Hefei School, Dec 14, 2016 32/34
v 2 and EoS ideal hydro, Pb+Pb at s NN = 18 GeV dn/dyp T dp T [GeV -2 ] 10000 1000 100 10 1 0.1 0.01 h - p-p NA49 bag model lattice 0 0.5 1 1.5 2 2.5 p T [GeV] v 2 0.18 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 bag model lattice π - 0 0.2 0.4 0.6 0.8 1 1.2 1.4 p T [GeV] p T fo 120 MeV (bag) T fo 130 MeV (lattice) protons no longer sensitive to phase transition! P. Huovinen @ Hefei School, Dec 14, 2016 33/34
fit to p T, v n, multiplicities etc. Global analysis Pratt et al., PRL 114, 202301 (2015) Bayesian analysis using emulators P. Huovinen @ Hefei School, Dec 14, 2016 34/34