Van der Waals equation: event-by-event fluctuations, quantum statistics and nuclear matter Volodymyr Vovchenko In collaboration with Dmitry Anchishkin, Mark Gorenstein, and Roman Poberezhnyuk based on: Vovchenko, Anchishkin, Gorenstein, arxiv:50.03785 [nucl-th], J. Phys. A in print and arxiv:504.0363 [nucl-th], Phys. Rev. C in print Transport Meeting FIAS, Frankfurt 0 June 205 / 3
Outline Introduction 2 Van der Waals equation in Grand Canonical Ensemble 3 Particle number fluctuations Scaled variance Non-gaussian fluctuations 4 VDW equation with quantum statistics 5 Applications: nuclear matter as VDW gas of nucleons 6 Summary 2 / 3
Van der Waals equation Van der Waals equation p(t, V, N) = NT V bn an2 V 2 Formulated in 873. Nobel Prize in 90. Is the simplest analytical model of interacting system with st order phase transition and critical point. Motivation: A toy model to study QCD critical point E.-by-e. fluctuations can be used to study QCD phase transition Stephanov, Rajagopal, Shuryak, Phys. Rev. D (999) Ejiri, Redlich, Karsch, Phys. Lett. B (2005) 3 / 3
Van der Waals equation Van der Waals equation Two ingredients: p(t, V, N) = NT V bn an2 V 2 ) Short-range repulsion: particles are hard spheres, b = 4 4πr 3 3 2) Attractive interaction in mean-field approximation Critical point p v = 2 p = 0, v = V /N v 2 p C = a 27b 2, n C = 3b, T C = 8a 27b Reduced variables p = p p C, ñ = n n C, T = T T C 4 / 3
Van der Waals equation Van der Waals equation in reduced variables p( T, ñ) = 8 T ñ 3 ñ 3ñ2 2.0 P.5.0 0.5 0.0 T=. T= T=0.9 T=0.8 T gas mixed phase liquid -0.5 2 3 4 v n Below T C isotherms are corrected by Maxwell s rule of equal areas This results in appearance of mixed phase 5 / 3
VDW equation in GCE VDW equation has a simple and familiar form in canonical ensemble p(t, V, N) = NT V bn an2 V 2 In GCE one needs to define p = p(t, µ) What are the advantages of the GCE formulation? ) Hadronic physics applications: number of hadrons usually not conserved, and GCE formulation is a good starting point to insert VDW interactions in multi-component hadron gas. 2) CE cannot describe particle number fluctuations. E.g., N-fluctuations in a small (V V 0 ) subsystem follow GCE results. 3) GCE formulation is a good starting point to include effects of quantum statistics. 6 / 3
From CE to GCE Variables T, V, N are not the natural variables for the pressure function. Therefore p(t, V, N) does not contain full information about system. One needs instead free energy F (T, V, N). ( ) F V = p(t, V, N) F(T, V, N) = F(T, V 0, N) dv p(t, V, N) V T,N V 0 How to fix integration constant F (T, V 0, N)? Ideal gas at V 0! VDW free energy φ(t ; d, m) = F(T, V, N) = F id (T, V bn, N) a N2 V F id (T, V, N) = NT d 2π 2 0 [ + ln ] V φ(t ; d, m) N k 2 dk exp( k 2 + m 2 /T ) = d m2 T ( m ) 2π 2 K 2 T 7 / 3
From CE to GCE Chemical potential: ( ) F µ = = T ln N T,V (V bn) φ(t ; d, m) N + b NT V bn 2a N V Transcendental equation for n(t, µ) N V n(t, µ) = n id (T, µ ) + b n id (T, µ ), n T µ = µ b b n + 2a n With n(t, µ) one then recovers p(t, µ): p(t, µ) = Energy density: Tn bn an2 = p id (T, µ ) an 2, where n n(t, µ) ε(t, µ) = [ɛ id (T ) an] n Average energy per particle reduced by attractive mean field an, excluded volume has no effect 8 / 3
Excluded-volume model Let a = 0: only repulsive interactions. Then particle density: n(t, µ) = n id (T, µ ) + b n id (T, µ ), n T µ = µ b b n pressure: p(t, µ) = Tn bn = p id[t, µ bp(t, µ)] energy density: ε(t, µ) = ɛ id (T ) n(t, µ) This reproduces the excluded-volume model Rischke, Gorenstein, Stoecker, Greiner, Z. Phys. C5, 485 9 / 3
Scaled variance in VDW equation Scaled variance is an intensive measure of N-fluctuations ω[n] N2 N 2 = T ( ) n N n µ In ideal Boltzmann gas fluctuations are Poissonian and ω id [N] =. ω[n] in VDW gas (pure phases) [ ω[n] = ( bn) 2 2an T Repulsive interactions suppress N-fluctuations while attractive interactions cause enhancement ] T 0 / 3
Scaled variance outside mixed phase region 3 In reduced variables ω[n] = [ 9 (3 ñ) ñ 2 4 T 0.5 0. ] 0.00 2 T gas 2 0 mixed phase liquid 0 0 2 3 n ω[n] at the critical point.00 0.0 0.0 / 3
Scaled variance in metastable phases VDW predicts existence of metastable liquid and gas phases ω[n] = [ 9 (3 ñ) ñ ] 2 4 T 3 0.5 0. 2 0.00 T gas metastable 2 0 unstable metastable liquid 0 0 2 3 n ω[n] at the spinodal instability line, i.e. when p/ v = 0.00 0.0 0.0 2 / 3
Scaled variance in mixed phase region In mixed phase N g + N l = V [ξn g + ( ξ)n l ] In addition to GCE fluctuations in gaseous and liquid phases there are also fluctuations of volume fractions ω[n] = ξ 0ñg 9ñ + (ñ g ñ l ) 2 9ñ [ (3 ñ g ) ñg 2 4 T [.5 ñ g ξ 0 (3 ñ g ) 2 ] + ( ξ 0)ñ l 9ñ ñg 2 ξ 0 4 T + [ (3 ñ l ) 2 ñl 4 T ñ l ] ( ξ 0 )(3 ñ l ) ñ l 2 ( ξ 0 )4 T 2 ] 0.00 T.0 gas 0 2 0.5 liquid 0.5 mixed phase 0.0 0 2 3 n 0..00 0.0 0.0 3 / 3
Scaled variance in whole phase diagram 3 0.5 0. 0.00 2 T 2 gas 0 liquid.00 mixed phase 0 0 2 3 n 0.0 0.0 4 / 3
Non-gaussian fluctuations Higher-order (non-gaussian) fluctuations are expected to be more sensitive to the proximity of the critical point Stephanov, Phys. Rev. Lett. (2009); Karsch, Redlich, Phys. Lett. B (200) Skewness Kurtosis s[n] = Sσ = κ 3 κ 2 κ[n] = κσ 2 = κ 4 κ 2 Cumulants κ i = i (p/t 4 ) (µ/t ) i 5 / 3
Skewness Skewness in VDW gas (pure phases) [ ] [ ] 3bn s[n] = ω 2 [N] ( bn) 3 = ω 2 ñ [N] ( 3ñ)3 6 / 3
Skewness Skewness in VDW gas (pure phases) [ ] [ ] 3bn s[n] = ω 2 [N] ( bn) 3 = ω 2 ñ [N] ( 3ñ)3 3 0 S 40.00 0.00 2 - T gas metastable 0-0 unstable metastable liquid 0 0 2 3 n Skewness is positive (right-tailed) in gaseous phase and negative (left-tailed) in liquid phase 6 / 3.00 0.00 -.00-0.00-40.00
Kurtosis Kurtosis in VDW gas (pure phases) κ[n] = 3s 2 [N] 2ω[N]s[N] 6ω 3 b 2 n 2 [N] ( bn) 4 7 / 3
Kurtosis Kurtosis in VDW gas (pure phases) T κ[n] = 3s 2 [N] 2ω[N]s[N] 6ω 3 b 2 n 2 [N] ( bn) 4 3 2 gas 00 metastable 0-0 -0-00 - unstable 0 00 0 0 2 3 0 n metastable liquid Kurtosis is negative (flat) above critical point (crossover), positive (peaked) elsewhere and very sensitive to the proximity of the critical point 7 / 3 40.00 0.00.00 0.00 -.00-0.00-40.00
VDW equation with quantum statistics Boltzmann statistics does not work well everywhere Problems with Boltzmann statistics can already be seen on an ideal gas level For non-relativistic gas s id Boltz = nid T [m + 52 T µ ] At T 0 [ sboltz id 5 = n 0 2 + 3 ] 2 ln(t /c 0) sboltz id can be negative at high n or at T 0 Quantum statistics needed in such case Requirements for VDW equation with quantum statistics ) Reduce to ideal quantum gas at a = b = 0 2) Reduce to classical VDW when quantum statistics are negligible 3) s 0 and s 0 as T 0 8 / 3
VDW equation with quantum statistics in GCE Ansatz: Take pressure in the following form p(t, µ) = p id (T, µ ) an 2, µ = µ b p a b n 2 + 2an where p id (T, µ ) is pressure of ideal quantum gas. ( ) p n(t, µ) = µ ( ) p s(t, µ) = T T µ = = n id (T, µ ) + b n id (T, µ ) s id (T, µ ) + b n id (T, µ ) ε(t, µ) = Ts + µn p = [ɛ id (T, µ ) an] n This formulation explicitly satisfies requirements -3 Algorithm for GCE ) Solve system of eqs. for p and n at given (T, µ) (there may be multiple solutions) 2) Choose the solution with largest pressure 9 / 3
Quantum VDW: from GCE to CE One can define pressure as a function of CE variables T and n Recall that n id (T, µ ) n(t, µ) = + b n id (T, µ ) ( Therefore µ (n, T ) = µ id n n = d 2π 2 0 n id (T, µ ) = n(t, µ) b n(t, µ) bn, T ) where µ id (n, T ) is solution to ( ) dk k [exp 2 m2 + k 2 µ id (n, T ) + η T ] VDW equation with quantum statistics in CE p = p id[ T, µ id( n )] bn, T a n 2 20 / 3
Nuclear matter as a VDW gas of nucleons Nuclear matter is known to have a liquid-gas phase transition at T 20 MeV and exhibit VDW-like behavior Usually studied analyzing nuclear fragment distribution Theory: Csernai, Kapusta, Phys. Rept. (986) Stoecker, Greiner, Phys. Rept. (986) Serot, Walecka, Adv. Nucl. Phys. (986) Bondorf, Botvina, Ilinov, Mishustin, Sneppen, Phys. Rept. (995) Experiment: Pochodzalla et al., Phys. Rev. Lett. (995) Natowitz et al., Phys. Rev. Lett. (2002) Karnaukhov et al., Phys. Rev. C (2003) Our description: Nuclear matter as a system of nucleons (d = 4, m = 938 MeV) described by VDW equation with Fermi statistics. Pions, resonances and nuclear fragments are neglected 2 / 3
VDW gas of nucleons: zero temperature How to fix a and b? Saturation density and binding energy at T = 0 From E B = 6 MeV and n = n0 = 0.6 fm 3 at T = p = 0 we obtain: a = 329 MeV fm 3 and b = 3.42 fm 3 (r = 0.59 fm) 4 40 (a) (b) 3 30 P (MeV/fm 3 ) 2 0 - E B (MeV) 20 0 0-0 -2 0.00 0.04 0.08 0.2 0.6 0.20 0.24 0.28 n (fm -3 ) -20 0.00 0.04 0.08 0.2 0.6 0.20 0.24 0.28 n (fm -3 ) Mixed phase at T = 0 is rather special: A mix of vacuum (n = 0) and liquid at n = n 0 22 / 3
VDW gas of nucleons: pressure isotherms P (MeV/fm 3 ) 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0. T=2 MeV T=9.7 MeV T=8 MeV T=5 MeV CE pressure p = p id[ T, µ id( n )] bn, T a n 2 0.0 0.00 0.04 0.08 0.2 0.6 0.20 0.24 0.28 n (fm -3 ) (b) P (MeV/fm 3 ) 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0. 0.0 T=8 MeV T=5 MeV T=2 MeV T=9.7 MeV -0. 5 0 5 20 25 30 35 40 45 50 55 60 65 70 v (fm 3 ) Behavior qualitatively same as for Boltzmann case Mixed phase results from Maxwell construction Critical point at T c = 9.7 MeV and nc = 0.07 fm 3 (a) 23 / 3
VDW gas of nucleons: (T, µ) plane T (MeV) 35 30 25 20 5 0 5 Density in (T, µ) plane gas liquid 0 880 890 900 90 920 930 (MeV) n (fm -3 ) Crossover region at µ < µ C = 908 MeV is clearly seen 24 / 3 0.8 0.6 0.3 0. 0.09 0.07 0.04 0.02 0.00
VDW gas of nucleons: (T, µ) plane T (MeV) 35 30 25 20 5 0 5 Boltzmann CP Density in (T, µ) plane gas liquid 0 880 890 900 90 920 930 (MeV) n (fm -3 ) Boltzmann: T C = 28.5 MeV. Fermi statistics important at CP 0.8 0.6 0.3 0. 0.09 0.07 0.04 0.02 0.00 25 / 3
VDW gas of nucleons: scaled variance Scaled variance in quantum VDW: [ ω[n] = ω id (T, µ ) ( bn) 2 2an T ω id(t, µ ) ] T (MeV) 35 30 25 20 5 0 5 0.5 2 gas 0 liquid 0 880 890 900 90 920 930 (MeV) 0. 0.00.00 0.0 0.0 26 / 3
VDW gas of nucleons: skewness s[n] = T (MeV) Skewness in quantum VDW: ω[n] T ω id (T, µ ) [ω id (T, µ )] 2 ( bn) 2 + ω2 [N] µ ω id (T, µ ) 35 30 25 20 5 0 5 0 - gas 0 0 880 890 900 90 920 930-0 (MeV) - liquid S 3bn ( bn) 3 40.00 0.00.00 0.00 -.00-0.00-40.00 27 / 3
VDW gas of nucleons: kurtosis Kurtosis in quantum VDW: ( ) s[n] κ[n] = (s[n]) 2 + T =... µ T T (MeV) 35 30 25 20 5 0 5 - -0 0 0 gas -00 00 00 0 880 890 900 90 920 930 (MeV) 0 liquid Crossover region is clearly characterized by large negative kurtosis This also been suggested for QCD CP in Stephanov, PRL (20) 40.00 0.00.00 0.00 -.00-0.00-40.00 28 / 3
Summary Classical VDW equation is reformulated in GCE as a transcendental equation for particle density. 2 Scaled variance, skewness, and kurtosis of particle number fluctuations are calculated for VDW equation. Fluctuations remain finite both inside and outside the mixed phase region but diverge at the critical point. 3 VDW equation with Fermi statistics is presented and is able to qualitatively describe properties of symmetric nuclear matter. Fermi statistics effects remain quantitatively important near the critical point of nuclear liquid-gas transition. 4 Non-gaussian fluctuations are very sensitive to the proximity of the critical point. Gaseous phase is characterized by positive skewness while liquid phase corresponds to negative skewness. The crossover region is clearly characterized by negative kurtosis in VDW model. Possible tasks: Other fluctuation measures, e.g., strongly intensive quantities Inclusion of VDW interactions in multi-component systems, e.g., nuclear fragments, HRG etc. BEC in interacting VDW gas, e.g., gas of pions Transport coefficients 29 / 3
Thanks for your attention! 30 / 3
Backup slides 30 / 3
Scaled variance in mixed phase region Inside the mixed phase: V g = ξ V, V l = ( ξ) V, F(V, T, N) = F(V g, T, N g ) + F (V l, T, N l ) ω[n] = ξ 0n g n N = N g + N l = V [ξn g + ( ξ)n l ] [ + (n g n l ) 2 V n ( bn g ) 2 2an g T ] + ( ξ 0)n l n [ ξ 2 ξ 2], ξ 0 = n l n n l n g [ ( bn l ) 2 2an l T In addition to GCE fluctuations in gaseous and liquid phases there are also fluctuations of volume fractions [ W (ξ) = C exp ( 2 ) ] F 2T ξ 2 (ξ ξ 0 ) 2 ξ=ξ 0 ξ 2 ξ 2 = T V [ ] n g T ξ 0 ( bn g ) 2 2an2 g n l T + ξ 0 ( ξ 0 )( bn l ) 2 2an2 l ξ 0 ] 3 / 3