K_1/K^* enhancement as a signature of chiral symmetry restoration in heavy ion collisions
Haesom Sung, Sungtae Cho, Juhee Hong, Su Houng Lee, Sanghoon Lim, Taesoo Song
aa r X i v : . [ nu c l - t h ] F e b K /K ∗ enhancement as a signature of chiral symmetry restoration in heavy ioncollisions Hae-Som Sung, ∗ Sungtae Cho, † Juhee Hong, ‡ Su Houng Lee, § Sanghoon Lim, ¶ and Taesoo Song ∗∗ Department of Physics and Institute of Physics and Applied Physics, Yonsei University, Seoul 03722, Korea Division of Science Education, Kangwon National University, Chuncheon 24341, Korea Department of Physics, Pusan National University, Pusan, Korea GSI Helmholtzzentrum f¨ur Schwerionenforschung GmbH, Planckstrasse 1, 64291 Darmstadt, Germany
Based on the fact that the mass difference between the chiral partners is an order parameter ofchiral phase transition and that the chiral order parameter reduces substantially at the chemicalfreeze-out point in ultra-relativistic heavy ion collisions, we argue that the production ratio of K over K ∗ in such collisions should be substantially larger than that predicted in the statisticalhadronization model. We further show that while the enhancement effect might be contaminated bythe relatively larger decrease of K meson than K ∗ meson during the hadronic phase, the signal willbe visible through a systematic study on centrality as the kinetic freeze-out temperature is higherand the hadronic life time shorter in peripheral collisions than in central collisions. I. INTRODUCTION
A recent lattice calculation finds the pseudo-criticaltemperature for QCD chiral crossovers at zero baryonchemical potential to be around 156 . ± . ρ and a is dueto the chiral symmetry breaking as is well represented inthe Weinberg relation [6]. Hence, when hadrons are pro-duced at the chemical freeze-out point in heavy ion colli-sions, the chiral partners will have masses much closer toeach other than their vacuum values. This does not meanthat the masses should vanish at the chemical freeze-outpoint. In fact, a recent study shows that while chiralsymmetry breaking is responsible for the mass differencebetween the ρ and a meson, it only accounts for a smallfraction of the common mass [7].Statistical hadronization model analysis indicates thatthe abundances of hadrons are determined at the chem-ical freeze-out point [8]. This suggests that the produc-tion of chiral partners at the chemical freeze-out pointwill be similar. Unfortunately, a has a large vacuum de-cay width and a large dissociation cross section so that ∗ [email protected] † [email protected] ‡ [email protected] § [email protected] ¶ [email protected] ∗∗ [email protected] many of the a produced at the chemical freeze-out willnot survive during the hadronic phase. Indeed, the mea-sured yields of resonances with large decay widths tendto be smaller than the statistical hadronization modelpredictions.On the other hand, recently it has been emphasizedby one of us that the K and K ∗ are also chiral partnersand that both have vacuum widths that are smaller than100 MeV, making them ideal objects to study the effectsof partial chiral symmetry restoration in a nuclear targetexperiment [9]. In this work, we will show that the en-hancement of the ratio K /K ∗ in heavy ion collisions canbe used as a signature of chiral symmetry restoration inthe early stages of heavy ion collisions. Effects of chiralsymmetry restoration in heavy ion collisions were studiedin K + /π + , (Λ + Σ ) /π − ratios [10]. Here, we will studythe K /K ∗ ratio as their properties are direct order pa-rameters of spontaneous chiral symmetry breaking. II. K , K ∗ COUPLINGS IN AN EFFECTIVELAGRANGIAN
From a symmetry argument alone, one expects K tobe the chiral partner of K ∗ . In fact, arguments basedsolely on chiral symmetry predict that when chiral sym-metry is restored, the vector and axial particles becomedegenerate [11]. For these effects to be observable inheavy ion collisions, the vacuum widths as well as thehadronic dissociation of both vector and axial particlesshould be small so that the signal will not be smearedout during the hadronic phase. Since the vacuum widthsof both the K and K ∗ meson are already smaller than100 MeV, we focus here on their hadronic absorptionsduring the hadronic phase as well as on the centralitydependence of these effects in a heavy ion collision. Still,in order to work out the suppression of the initially pro-duced K mesons during the hadronic phase of heavy ioncollisions, we need to estimate the hadronic cross sectionswith other hadrons.For that purpose, we use the Lagrangian involvingspin-1 and spin-0 mesons following the massive Yang-Mills approach [12]: L V = 18 F π Tr (cid:0) D µ U D µ U † (cid:1) + 18 F π Tr (cid:2) M (cid:0) U + U † − (cid:1)(cid:3) −
12 Tr ( F µνL F µνL + F µνR F µνR )+ m Tr ( A µL A µL + A µR A µR ) , (1)where U = exp (cid:0) i P/ √ f π (cid:1) with the octet of pseu-doscalar mesons P = P a λ a / √ F π = √ f π = 132MeV. The covariant derivative and the field strength ten-sor are given by D µ U = ∂ µ U − i g A µ L U + i g U A µ R ,F L,Rµ ν = ∂ µ A L,Rν − ∂ ν A L,Rµ − ig (cid:2) A L,Rµ , A
L,Rν (cid:3) , (2)where V µ = A Lµ + A Rµ and A µ = A Lµ − A Rµ with thevector and axial-vector mesons V µ = V aµ λ a / √ A µ = A aµ λ a / √ K ∗ Kπ and the K Kρ interaction terms. L (3) = i g (cid:16) V µ P ←→ ∂ µ P (cid:17) + i g F π V µ P A µ − V µ A µ P )+ i g ∂ µ V ν − ∂ ν V µ ) V µ V ν ]+ i g Tr [( ∂ µ V ν − ∂ ν V µ ) A µ A ν − ∂ ν A µ A µ V ν ] . (3)Specifically, L V P P = i g K ∗ µ [( ~τ · ∂ µ ~π ) K − ( ~τ · ~π ) ∂ µ K ]+ i g K [( ~τ · ~ρ µ )] ∂ µ K + H.C., (4) L AV P = i g F π K µ (cid:2) ( ~τ · ~ρ µ ) K − ( ~τ · ~π ) K ∗ µ (cid:3) + H.C., (5)with K ≡ ( K + , K ) T , K ∗ ≡ ( K + , K ∗ ) T and K ≡ ( K +1 , K ) T being the strange pseudoscalar, vector andaxial-vector meson isospin doublets, respectively. ~π and ~ρ , with Pauli matrices ~τ , are the pion and ρ meson isospintriplets.As it stands, the interaction terms for K K ∗ π and K Kρ have a common coupling as given from Eq. (3).This leads to a larger calculated decay width of K K ∗ π compared with that of K Kρ , because the phase spacefor the former two-body decay is larger than that forthe latter. On the other hand, experimentally one findsΓ K → K ∗ π = 25 MeV and Γ K → Kρ = 38 MeV. The reso-lution of this puzzles comes from mixing effects. In thevacuum, the low-lying modes that couple to the vectorcurrent are K ∗ (892) and K ∗ (1410) while those for theaxial vector current are K (1270) and K (1400). Thereis a subtlety in the nature of the two K states: they are assumed to be a mixture of the P and P quark-antiquark pair in the quark model [13]. We thereforeintroduce two mixing angles θ A and θ V so that the chi-ral Lagrangian in Eq. (4) and (5) are constructed forboth ( K ∗ A , K A ) and ( K ∗ B , K B ) that are related to thephysical mesons as follows: K A = cos θ A K (1270) + sin θ A K (1400) ,K B = − sin θ A K (1270) + cos θ A K (1400) ,K ∗ A = cos θ V K ∗ (892) + sin θ V K ∗ (1410) ,K ∗ B = − sin θ V K ∗ (892) + cos θ V K ∗ (1410) . (6)Now, we re-express the sum of the two chiral La-grangians in terms of K and K ∗ mesons and the twoindependent couplings g A , g B so that the SU(2) chiralsymmetry is still preserved. L ′ V P P = i g K ∗ Kπ ¯ K ∗ µ h(cid:16) √ ~τ · ∂ µ ~π (cid:17) K − (cid:16) √ ~τ · ~π (cid:17) ∂ µ K i + H.C., (7) L ′ AV P = i g K Kρ ¯ K µ (cid:16) √ ~τ · ~ρ µ (cid:17) K − i g K K ∗ µ π ¯ K µ (cid:16) √ ~τ · ~π (cid:17) K ∗ µ + H.C., (8)where g K ∗ Kπ = 14 √ g A cos θ V − g B sin θ V ) ,g K ∗ (1410) Kπ = 14 √ g A sin θ V + g B cos θ V ) ,g K Kρ = F π √ (cid:0) g A cos θ A − g B sin θ A (cid:1) ,g K (1400) Kρ = F π √ (cid:0) g A sin θ A + g B cos θ A (cid:1) ,g K K ∗ π = F π √ (cid:0) g A cos θ A cos θ V + g B sin θ A sin θ V (cid:1) ,g K (1400) K ∗ π = F π √ (cid:0) g A sin θ A cos θ V − g B cos θ A sin θ V (cid:1) . (9)Using the above interaction Lagrangians, we determinethe mixing angle θ A ( θ B ) and coupling constant g A ( g B )to best fit the experimental data for the six decay pro-cesses given in Table I. The table also summarizes thebest fit couplings and the experimental data as well asthe calculated decay widths of the six processes. III. THE CROSS SECTION OF THE K MESON
We assume that K mesons are in thermal equilibriumwhen they are produced at the chemical freeze-out. Then K mesons interact with other particles until the kinetic ABC K ∗ (892) Kπ K ∗ (1410) Kπ K (1270) Kρ K (1400) Kρ K (1270) K ∗ (892) π K (1400) K ∗ (892) πg ABC − .
02 (GeV) 0.649 (GeV) − .
95 (GeV)Γ A → BC (experiment) 50.0 (MeV) 15.6 (MeV) 37.8 (MeV) 5.2 (MeV) 14.4 (MeV) 163 (MeV)Γ A → BC (calculated) 48.4 (MeV) 14.8 (MeV) 39.7 (MeV) 14.7 (MeV) 9.42 (MeV) 98.6 (MeV)TABLE I. The coupling constants are calculated with g A =15.5, g B = -10.4, cos θ V =0.6469, sin θ V =0.7626, cos θ A =0.724 andsin θ A = -0.6898. With these parameters we also find four point couplings g K (1270) Kππ =7.67(GeV − ) and g K (1400) Kππ = − . − ). freeze-out point. We consider the hadronic effect on the K meson due to interactions with light mesons such aspions and rho mesons; K + π → K + π , K + π → K ∗ + ρ , K + ρ → K ∗ + π , and K + ρ → K + ρ . In additionto the interaction terms given in Eq. (7) and (8), theLagrangian describing the interaction between the K meson and pions and rho mesons is given as follows: L ′ AP P P = 2 g K Kππ ¯ K µ h ( ~τ · ∂ µ ~π ) ( ~τ · ~π ) K + ( ~τ · ~π ) ∂ µ K i , L ′ V P P = i g ρππ [ ρ µ ( ∂ µ π + π − − π + ∂ µ π − )+ ρ + µ ( ∂ µ π − π − π − ∂ µ π )+ ρ − µ ( ∂ µ π π + − π ∂ µ π + )] + H.C. + i g KKρ h ¯ K (cid:16) √ ~τ · ~ρ µ (cid:17) ∂ µ K − ∂ µ ¯ K (cid:16) √ ~τ · ~ρ µ (cid:17) K i , L ′ AV P = i g F π √ { ρ + µ ( π − a µ − π a − µ )+ ρ − µ ( π a +1 µ − π + a µ ) + ρ µ ( π + a − µ − π − a +1 µ ) } ]+ i g K ∗ Ka ¯ K ( √ ~τ · ~a µ ) K ∗ µ L ′ V V V = √ i g K ∗ K ∗ ρ [ ¯ K ∗ ν ~τ K ∗ µ ( ∂ µ ~ρ ν − ∂ ν ~ρ µ )+ ( ∂ µ ¯ K ∗ ν − ∂ ν ¯ K ∗ µ ) ~τ · ~ρ µ K ∗ ν + ¯ K ∗ µ ~τ · ~ρ ν ( ∂ µ K ∗ ν − ∂ ν K ∗ µ )]+ i g ∂ µ ( ~τ · ~ρ ν ) − ∂ ν ( ~τ · ~ρ µ ))( ~τ · ~ρ µ )( ~τ · ~ρ ν )] , L ′ AAV = i √ g K a K ∗ [ ∂ ν ¯ K µ { ( ~a ν · ~τ ) K ∗ µ − ( ~a µ · ~τ ) K ∗ ν } − ¯ K µ ∂ µ ( ~a ν · ~τ ) K ∗ ν ]+ i √ g K K ρ [ { ¯ K µ ( ~ρ ν · ~τ ) − ¯ K ν ( ~ρ µ · ~τ ) } ∂ ν K µ − i g∂ µ ¯ K ν ( ~ρ ν · ~τ ) K µ ] . (10)The coupling constant g = 8 .
49 is determined by fittingthe experimental decay width of ρ → ππ , from which wefind g ρππ = g ρρρ = g/ √ .
006 and g KKρ = g/ √ . g A and g B , which were determined to give the coupling constantsas given in Table I and additionally g K ∗ K ∗ ρ = 0 . g K K ρ = 0 . g K a K ∗ = 9 and g a ρπ = 3 .
36 (GeV).The amplitude for each scattering of different chargestates can be written as two isospin channels because of isospin conservation. For example, the scattering of πK → Kπ consists of the following charge states : K − π − → K − π − , K − π → K − π , ¯ K π − → K − π , K − π + → K − π + , and ¯ K π → K − π + . These five pro-cesses are described by two independent scattering am-plitudes a / and a / for isospin 3 / / h K − π − | S | K − π − i = a / h K − π | S | K − π i = a / + a / h ¯ K π − | S | K − π i = − √ (cid:0) a / − a / (cid:1) h K − π + | S | K − π + i = a / + a / h ¯ K π | S | K − π + i = √ (cid:0) a / − a / (cid:1) . (11)It should be noted that squaring and adding all fiveamplitudes, one finds that the sum is proportional to2 a / + a / reflecting the degeneracy of the isospin.Using the interaction terms in the Lagrangian dis-cussed above, we evaluated the amplitudes for all dia-grams shown in Fig. 1. These represent the absorp-tion amplitudes of the K meson by π and ρ mesons.First, the amplitudes for K absorption by π mesons arerepresented by the processes πK → πK and πK → ρK ∗ . The amplitude for definite isospin channels ofthe π ( p ) K ( p ) → π ( p ) K ( p ) are given as a / = g K Kππ ǫ µ ( p + 3 p ) µ + 3 M s + 2 M t + M u and a / = − g K Kππ ǫ µ p µ − M t − M u for isospin 1/2 and 3/2,respectively. Here, M s = M (1 − b ) , M t = M (1 − c ) , and M u = M (1 − d ) , where the subscript in the right hand siderepresents the diagrams in Fig. 1. The matrix elementsare given as M s = g K K ∗ π g K ∗ Kπ ǫ µ ( p − p ) ν × − g µν + ( p + p ) µ ( p + p ) ν /ss − m K ∗ , M t = g ρππ g K Kρ ǫ µ − ( p + p ) µ t − m ρ , M u = g K K ∗ π g K ∗ Kπ ǫ µ ( p + p ) ν × − g µν + ( p − p ) µ ( p − p ) ν /uu − m K ∗ . (12)The two independent amplitudes for the π ( p ) K ( p ) → ρ ( p ) K ∗ ( p ) are respectively given as a / = 3 M s +2 M t + (1-a) (1-b) (1-c) (1-d)(2-a) (2-b) (2-c) (2-d) (2-e)(3-a) (3-b) (3-c) (3-d)(4-a) (4-b) (4-c) (4-d) (4-e)FIG. 1. Feynman diagrams for the K meson absorption by π, ρ mesons. (1) K π → Kπ , (2) K π → K ∗ ρ , (3) K ρ → K ∗ π and (4) K ρ → Kρ . M u and a / = −M t − M u . Here, M t = M (2 − b ) + M (2 − d ) and M u = M (2 − c ) + M (2 − e ) , with M s = g K K ∗ π g K ∗ k ∗ ρ ǫ α ǫ ∗ γ ǫ ∗ δ − g αβ + ( p + p ) α ( p + p ) β /ss − m K ∗ [ g γβ (2 p + p ) δ − g δβ ( p + 2 p ) γ − g γδ ( p − p ) β ] , M t = g K K ∗ π g ρππ ǫ µ ǫ ∗ µ ǫ ∗ γ t − m π (2 p − p ) γ + g a ρπ g K a K ∗ ǫ µ ǫ ∗ α ǫ ∗ δ − g αβ + ( p − p ) α ( p − p ) β /tt − m a [ g µβ p δ − g µδ p β + g βδ ( p − p ) µ ] , M u = g K Kρ g K ∗ Kπ ǫ µ ǫ ∗ µ ǫ ∗ δ u − m π (2 p − p ) δ + g K K ∗ π g K K ρ ǫ µ ǫ ∗ γ ǫ ∗ α − g αβ + ( p − p ) α ( p − p ) β /uu − m K [ g µβ ( p − p ) γ − g µγ p β − g γβ ( p − p ) µ ] . (13)Second, the amplitudes for absorption by ρ mesons are ρK → πK ∗ and ρK → Kρ . The isospin channelsof ρ ( p ) K ( p ) → π ( p ) K ∗ ( p ) are a / = 3 M s − M t and a / = M t . Here, M s = M (3 − a ) + M (3 − b ) , and M t = M (3 − c ) + M (3 − f ) are given as follows: M s = g K Kρ g K ∗ Kπ ǫ µ ǫ µ ǫ ∗ λ s − m K (2 p + p ) λ + g K K ρ g K Kπ ǫ µ ǫ ν ǫ ∗ λ − g µ ′ λ + ( p + p ) µ ′ ( p + p ) λ /m K s − m K [ g µ ′ ν ( p + p ) µ − g µµ ′ ( p + p ) ν + g µν p µ ′ ] , M t = g K K ∗ π g ρππ ǫ µ ǫ ν ǫ ∗ ν t − m π ( p − p ) µ − g K a K ∗ g a ρπ ǫ µ ǫ ν ǫ ∗ λ − g ν ′ µ + ( p − p ) µ ( p − p ) ν ′ /m a t − m a [ g λν p ν ′ − g νν ′ p λ − g λν ′ ( p − p ) ν ] . (14)The isospin channels of ρ ( p ) K ( p ) → K ( p ) ρ ( p ) are a / = − M s − M t + 2 M u and a / = 2 M t − M u . Here, M s = M (4 − a ) + M (4 − b ) , M t = M (4 − c ) + M (4 − d ) , and M u = M (4 − e ) are given as follows: M s = g K Kρ g ρKK ǫ µ ǫ µ ǫ ∗ λ s − m K ( p + 2 p ) λ + g K K ρ g K Kρ ǫ µ ǫ ν ǫ ∗ λ − g µ ′ λ + ( p + p ) µ ′ ( p + p ) λ /m K s − m K [ g νµ ′ ( p + p ) µ − g µµ ′ ( p + p ) ν + g µν p µ ′ ] , M t = g K Kρ g ρKK ǫ µ ǫ ν ǫ ∗ ν t − m K ( p − p ) µ + g K K ρ g K Kρ ǫ µ ǫ ν ǫ ∗ λ − g ν ′ µ + ( p − p ) µ ( p − p ) ν ′ /m K t − m K [ g νν ′ ( p − p ) λ − g ν ′ λ ( p − p ) ν + g λν p ν ′ ] , M u = g K Kρ g ρρρ ǫ µ ǫ ν ǫ ∗ λ − g µ ′ ν + ( p − p ) µ ′ ( p − p ) ν /m ρ u − m ρ [ g µµ ′ (2 p − p ) λ − g µ ′ λ ( p − p ) µ − g µλ ( p + p ) µ ′ ] . (15)We keep the convention where particles 1 and 2 standfor initial-state mesons, and particles 3 and 4 stand forfinal-state mesons shown on the left and right sides ofthe diagrams, respectively. The Mandelstam variables s = ( p + p ) , t = ( p − p ) , and u = ( p − p ) havealso been used.To take into account the finite size of the hadrons whencalculating amplitudes, we introduce form factors shownbelow at each interaction vertex for the u, t − channel andthe s − channel, respectively, F u,t ( ~q ) = Λ − m ex Λ + ~q , F s ( ~q ) = Λ + m ex Λ + ω , (16)with m ex being the mass of the exchanging particle ateach diagram. In Eq. (16) ~q is the squared three-momentum transfer at t, u − channels, and ω is the totalenergy of the incoming particles at the s − channel in thecenter of mass frame. On the other hand, we apply thefollowing form factor for the four point contact interac-tion, F c ( ~k ) = (cid:18) Λ Λ + ~k (cid:19) , (17)where ~k is the average of the squared three-momentafrom the form factors for the given channels at each pro-cess. We set the cut off parameter Λ = 1.8 GeV [14]. The cross section after spin averaging is given by σ = 164 πs Z dt M | p cm | F , (18)with M being the squared amplitude of all processesevaluated by averaging and summing over the degenera-cies of the initial and final particles, respectively. p cm inEq. (18) represents for the three momenta of the initialparticles in the center of mass frame. Fig. 2 shows thecross section for the absorption of the K meson by π and ρ mesons.For the hadronic production cross section of the K meson during the hadronic stage in heavy-ion collisions,we will use detailed balance relation in the rate equation. IV. TIME EVOLUTION OF THE K MESONABUNDANCE
We now investigate the time evolution of the K me-son abundance using the cross sections evaluated in theprevious section. We construct an evolution equation,similar to that in Ref. [15], composed of densities andabundances for mesons participating in all the processesshown in Fig. 2: the dissociation processes of K are dueto π and ρ mesons. K π → K π K π → K * ρ K ρ → K ρ K ρ → K (cid:0) π s s GeV ) σ m b ) FIG. 2. The cross sections for the dissociation of the K meson by π and ρ mesons via processes K π → Kπ , K π → K ∗ ρ , K ρ → Kρ and K ρ → K ∗ π . dN K ( τ ) dτ = h σ K π → Kπ v K π i n π (cid:20) − N K + N TK N TK N K (cid:21) + h σ K π → K ∗ ρ v K π i (cid:20) − n π N K + n Tπ N TK n Tρ N TK ∗ n ρ N K ∗ (cid:21) + h σ K ρ → K ∗ π v K ρ i (cid:20) − n ρ N K + n Tρ N TK n Tπ N TK ∗ n π N K ∗ (cid:21) + h σ K ρ → Kρ v K ρ i (cid:20) − n ρ N K + n Tρ N TK n Tρ N TK n ρ N K (cid:21) + h Γ K → Kρ i (cid:20) − N K + N TK n Tρ N TK n ρ N K (cid:21) + h Γ K → K ∗ π i (cid:20) − N K + N TK n Tπ N TK ∗ n π N K ∗ (cid:21) , (19)with n i ( τ ) being the density of the light meson i in thehadronic phase at proper time τ . N j ( τ ) = n j ( τ ) V ( τ ),where V ( τ ) is the volume of the hadronic matter givenin Eq. (27). The ones with the superscript T are thethermal abundance at temperature T , which depends onthe proper time τ as given in Eq. (27). n Ti ( τ ) = g i π Z ∞ p dp exp (cid:16)p p + m i /T ( τ ) (cid:17) − ≈ g i π m i T ( τ ) K (cid:20) m i T ( τ ) (cid:21) , (20)where K is the modified Bessel function of the secondkind. We take n i ( τ ) = n Ti ( τ ) for pions and ρ mesons.We use the following two constraint equations involv-ing the τ dependence for the numbers of K and K ∗ . N K + N N K ∗ + N K = N , (21) N K ∗ + N K × BR N = aT + b, (22)where N is a constant determined at the chemical freeze-out point. Eq. (21) follows from neglecting strangenessannihilation during the hadronic phase. BR=0.16 is thebranching ratio of K decaying into K ∗ so that the ra-tio in Eq. (22) is the ratio between the final K ∗ and K numbers when these mesons freeze-out at tempera-ture T . This equation is motivated by the experimentalobservation that show the approximately linear decrease in the observed ratio between K ∗ and K numbers as afunction of the cube root of multiplicity in heavy ion col-lision [16], which can be understood in a hadronic modelcalculation [15] that finds the ratio to decreases linearlywith the temperature of the hadronic phase. The coeffi-cients a and b in Eq. (22) were determined to reproducethe statistical model prediction for the ratio at T = T c and its linear reduction to 0.158 at T = 90 MeV, whichare the observed values for the ratio and the extractedfreeze-out temperature [22] for the lowest and highestcentrality Pb-Pb collision [16], respectively. a = 1.816GeV − and b = − K is equalto that of K ∗ at the chemical freeze-out point, whereas a = 2.272 GeV − and b = − K and K ∗ follow statistical model values at the chemical freeze-outpoint. Solving Eq. (19) with the two constraints givenin Eq. (21) and Eq. (22), we can determine the τ depen-dence and the final freeze-out numbers of N K , N K ∗ , N K .When substituting in the rate equation, we take thedegeneracy factor g i = (2 S + 1), where S is the spin,as the isospin effect is taken into account in the crosssection which includes the contribution from all chargestates. The production contribution has been taken intoaccount by a detailed balance condition, which requireseach square bracket in Eq. (19) to be zero in thermalequilibrium. K π → K π K π → K (cid:1) ρ K ρ → K ρ K ρ → K (cid:2) π K → K ρ K → K (cid:3) π T ( GeV ) < σ v ab > ( m b ) FIG. 3. Thermally averaged cross sections for the absorptionof a K meson via processes K π → Kπ , K π → K ∗ ρ , K ρ → Kρ , K ρ → K ∗ π , K → Kρ , and K → K ∗ π . In Eq. (19), we have considered the thermally averagedcross section, h σ ab → cd v ab i given below for initial two par-ticles in a two-body process ab → cd , h σ ab → cd v ab i = R d ~p a d ~p b f a ( ~p a ) f b ( ~p b ) σ ab → cd v ab R d ~p a d ~p b f a ( ~p a ) f b ( ~p b )= (cid:2) π T m a m b K ( m a /T ) K ( m b /T ) (cid:3) − × Z ∞√ s d √ s K ( √ s/T ) Z t t dt |M| , (23)with √ s = max( m a + m b , m c + m d ), K and K be-ing the modified Bessel function of the second kind.The integration limits are t ( t ) = ( m + m − m + m √ s ) − ( p cm ∓ p cm ) . Here, we approximate f i to be the Boltz-mann momentum distribution of a particle i , f i ( ~p ) = e − √ ~p + m /T . v ab is the relative velocity of particlesof species a and b , v ab = q ( p a · p b ) − m a m b / ( E a E b ). h Γ K i = Γ K ( m K ) K ( m K /T ) /K ( m K /T ) is the ther-mally averaged decay width of the K meson [15]. V. INITIAL TEMPERATURE AND TIMEEVOLUTION IN HEAVY ION COLLISION
The initial temperature inside the nuclear matter pro-duced from heavy ion collisions is studied with hydrosimulation. We will consider Pb-Pb collision at √ s NN =5 .
02 TeV in different centrality ranges: 0–5 %, 40–50 %and 70–80 %. The initial energy density in the transverseplane is calculated by using a two-dimensional Gaussiandistribution of σ = 0 . sonic model [18]. This tuneis required to match the multiplicity with the measureddata in various centrality ranges [19]. Figure 4 shows the initial temperature distribution at τ = 0 . /c of example events in three centrality rangesfrom the hydrodynamic simulation. As can be seen in thefigure, the fraction of area where initial temperature ishigher than 156 MeV is 84 %, 75 %, and 67 % in the 0–5%, 40–50 %, and 70–80 % centrality ranges, respectively.Furthermore, the fraction of initial energy that comesfrom regions where the temperature is higher than 156MeV is larger than 98% in all three centrality ranges.Therefore we can assume that almost all K and K ∗ areproduced through the QGP phase even in 70–80 % cen-tral collisions.For calculation of the number of K and K ∗ mesonsinside the expanding medium, a more simple approachis used by assuming an isentropic expansion of uniformmatter as follows [20]: ∂ τ ( Aτ h T ττ i ) = − pA, (24) ∂ τ ( Aτ s h γ r i ) = 0 , (25)where T ττ = ( e + p ) u τ − p is the energy momentum tensorin the Milne coordinate system with u τ , e and p being thefluid velocity, energy density and pressure, respectively. s is the entropy density, γ r = 1 / p − v r with v r beingthe radial velocity of fluid cell, A = πR with R being thetransverse radius of the uniform matter and h· · · i denotesaverage over the transverse area. Eq. (25) clearly showsthat the total entropy is conserved during the expansionwith Lorentz contraction of transverse area due to theflow velocity taken into account.Assuming that the radial flow velocity is a linear func-tion of the radial distance from the center, i.e., γ r v r = γ R ˙ R ( r/R ), where ˙ R = ∂R/∂τ and γ R = 1 / p − ˙ R , wethen have [20] h u τ i| η =0 = h γ r i = 1 + γ R ˙ R , h γ r i = 23 γ R ˙ R (cid:0) γ R − (cid:1) . (26)Since the energy density e and pressure p are related bythe equation of state of the matter through its temper-ature T [21], Eqs.(24)-(25) are thus simultaneous equa-tions for T , ˙ R .The results of the numerical calculations can beparametrized in the following form for the hadronicphase: V ( τ ) = π h R + v ( τ − τ c ) + a/ τ − τ c ) i τ c,T ( τ ) = T c − ( T h − T f ) (cid:18) τ − τ h τ f − τ h (cid:19) α , (27)with T h and T f being the hadronization and kineticfreeze-out temperature. We take T h = 156 MeV to be thesame as the cross over temperature T c = T h . Eq. (27)can be thought to describe the system of the hadronic − − − − y [f m ] T [GeV]=5 TeV NN sPb+Pb b= 1.8 fm=399 part N10 − − − − y [f m ] T [GeV]=5 TeV NN sPb+Pb b=10.1 fm=74 part N10 − − − − y [f m ] T [GeV]=5 TeV NN sPb+Pb b=13.9 fm=17 part N FIG. 4. Initial temperature distribution for example eventsin 0–5%, 40–50% and 70–80% centrality ranges of Pb-Pb col-lisions at √ s NN = 5 .
02 TeV. phase of the nuclear matter expanding with transversevelocity v and transverse acceleration a starting from itstransverse size R at the chemical freeze-out time τ c . Thevalues of T f according to centrality are taken from Ref.[22]. The values used in Eq. (27) are summarized inTable II. The schematic hydrodynamics in Eqs. (24) -(26) assumes the expansion of a uniform matter whosetemperature or particle density is the same anywhere inthe local rest frame. In the Lab. frame, however, the TABLE II. Values for the volume and temperature profiles inthe phenomenological model Eq.(27).Centrality T f t c t f R v a α (%) (MeV) (fm/c) (fm/c) (fm) (c) ( c /fm )0 −
5% 90 8.7 28.1 14.9 0.93 0.04 0.83540 −
50% 108 4.9 13 7.8 0.78 0.052 0.970 −
80% 147 2.2 2.9 4.43 0.481 0.161 0.847 particle density increases with the radial distance due toLorentz contraction. We note that Eq. (27) is not theparametrization for the volume in the Lab frame but forthe volume with the Lorentz contraction taken into ac-count, that is, Aτ h γ r i in Eq. (25). VI. THE ABUNDANCE OF K MESONS INHEAVY-ION COLLISIONS
Table III shows the abundance of the strange mesonsat the chemical freeze-out point for different centralityranges. The yield ratio between the K and K ∗ mesonsat the kinetic freeze-out point is also given. N K is thesolution of the rate equation Eq. (19) with the initialnumber equal to N K ∗ = N TK ∗ at T c = 156 MeV in thepresence of chiral symmetry restoration. On the otherhand, N TK is the expected yield without chiral symme-try restoration when the initial number is given by itsthermal equilibrium number at T c . Hence, N TK at thekinetic freeze-out point is the number expected in thestatistical model when hadronic dissociation is taken intoaccount. For a clear signature of chiral symmetry restora-tion at T c , we want N K /N K ∗ at the kinetic freeze-outtemperature to be sufficiently larger than N TK /N K ∗ at T c , which is the standard statistical model prediction forthe ratio. One notes that this is true for 40-50 % and70-80 % centrality ranges. In fact, as can be seen inthe last column of Table III, for 70-80 %, one finds that N K /N K ∗ at the kinetic freeze-out point is 0.905, whichis much larger than N TK /N K ∗ at any time for all central-ities: the time dependencies of N TK and N K ∗ are givenin Fig. 5. Fig. 6 summarizes the yield ratios between K and K ∗ mesons for different centralities with (red dots)and without (black squares) chiral symmetry restorationat T c . As can be seen in the figure, the ratio increasessharply as the centrality decreases (because of the higherkinetic freeze-out temperature and thus the shorter life-time of the hadronic phase at more peripheral collisions[23, 24]) when chiral symmetry restoration at T c is takeninto account. It is possible that at the chemical freeze-out point, the chiral order parameter will partly acquireits vacuum value. Then, the initial ratio of N K /N K ∗ at T c would be slightly less than 1. Nevertheless, thelarge increase in the yield ratio towards peripheral colli-sions would still be visible so that one can identify chiralsymmetry restoration. Hence, a systematic study of the TABLE III. The kaon, K ∗ and K meson abundances underhadronic interaction at chemical and kinetic freeze out tem-peratures. N TK is the number of K meson assuming ther-mal equilibrium while N K is the abundance assuming chiralsymmetry at T c . The last column shows our prediction whenchiral symmetry is restored.Centrality T(MeV) N K N K ∗ N TK N K N TK N K ∗ N K N K ∗ −
5% 156 80.5 37.0 5.62 37.0 0.152 1.0090 0.157 1.37 0.006 0.05740 −
50% 156 12.4 5.72 0.867 5.72 0.152 1.00108 0.149 1.17 0.034 0.26870 −
80% 156 1.80 0.827 0.125 0.827 0.152 1.00147 0.108 0.714 0.137 0.905 N K(cid:4) N T (cid:5)(cid:6)6 8
10 12012345 (cid:7) N (cid:8)(cid:9) N T (cid:10)(cid:11) FIG. 5. The time evolution of the K meson abundance un-der hadronic interaction when the initial number is equal to N K ∗ (circle) or N TK (triangle) at T c for 40-50%(upper graph)and 70-80%(lower graph) centrality ranges. ratio N K /N K ∗ for different centralities in a heavy ioncollision will unambiguously show the evidence of chiralsymmetry restoration at T c . VII. SUMMARY
We studied effects of chiral symmetry restoration atthe chemical freeze-out point in a relativistic heavy ion ● Chiral symmetry restoration ■ Without chiral symmetryStatistical model at T c - % - % - % Centrality0.20.40.60.8 N K / N K * FIG. 6. The yield ratio of K over K ∗ with and withoutchiral symmetry restoration for 70-80%, 40-50%, and 0-5%centrality ranges. K feed down to K ∗ should be added whencomparing to experiment. collision. As the mass differences between the chiral part-ners are order parameters of the chiral symmetry restora-tion, the production of the chiral pairs will be degenerateat the chemical freeze-out point, where the chiral orderparameter is found to be substantially reduced from thevacuum value in lattice calculations. For the effects tobe observable, the vacuum width as well as the hadronicdissociation of both particles should be small so that thesignal will not be smeared out during the hadronic phase.The vacuum widths of both the K and K ∗ meson aresmaller than 100 MeV and thus potential candidates tobe studied. We have performed a systematic study ontheir hadronic absorption during the hadronic phase aswell as on the centrality dependence of these effects in aheavy ion collision. Our findings suggest that while theanomalously large initial ratio between the K and K ∗ meson compared to that of the statistical model predic-tion will most likely be smeared out in a central collision,the signal will be visible in a peripheral heavy ion colli-sion due to the shorter life time of the hadronic phaseand higher freeze-out temperature.The chiral partnership between the K ∗ and K ex-ists between the same charge states when the baryonchemical potential is non-zero [9]. In ultra-relativisticheavy ion collisions, the initial state will have almost zerobaryon chemical potential. Hence, one can compare theproduction of any of the charge states through the decayproducts given below or their charge conjugation. Forthe K meson they are K − → ρ K − ρ − ¯ K π K ∗− π − ¯ K ∗ , ¯ K → ρ + K − ρ ¯ K π + K ∗− π ¯ K ∗ , and for K ∗ K ∗− → ( π K − π − ¯ K , ¯ K ∗ → ( π + K − π ¯ K .
0A systematic study of the production of these states de-pending on the centrality will lead us to identify chiralsymmetry restoration in heavy ion collision.
ACKNOWLEDGMENTS
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