The equations of state of asymmetric nuclear matter

The equations of state (EOS) of asymmetric nuclear matter (ANM) in an extended Nambu-Jona-Lasinio (ENJL) model was investigated by means of examining effective potential in one-loop approximation. Our numerical results show that isospin dependence of saturation density in our model is reasonably strong and critical temperature for liquid-gas phase transition decreases with increasing neutron excess.

JOURNAL OF SCIENCE OF HNUE Mathematical and Physical Sci., 2012, Vol 57, No 7, pp 106-112 This paper is available online at http://stdb.hnue.edu.vn THE EQUATIONS OF STATE OF ASYMMETRIC NUCLEAR MATTER Le Viet Hoa and Le Duc Anh Faculty of Physics, Hanoi National University of Education Abstract The equations of state (EOS) of asymmetric nuclear matter (ANM) in an extended Nambu-Jona-Lasinio (ENJL) model was investigated by means of examining effective potential in one-loop approximation Our numerical results show that isospin dependence of saturation density in our model is reasonably strong and critical temperature for liquid-gas phase transition decreases with increasing neutron excess Keywords: The equations of state, asymmetric nuclear matter, isospin Introduction The success of nuclear physics in satisfactorily explaining low and mediate energy nuclear phenomena leads to a strong belief that nucleons and mesons are appropriate degrees of freedom At present, the relativistic treatment of nuclear many-body systems introduced not long ago by Walecka [1-3] turned out to be a quite successful tool for the study of many nuclear properties: binding energies, effective nucleon mass, equation of state, liquid-gas phase transition, ect Along with the success of the Walecka’s model, a four-nucleon model of nuclear matter [4-6] is introduced which consists of only nucleon degrees of freedom In this article we will consider the isospin dependence of the energy of asymmetric nuclear matter in the extended Nambu-Jona-Lasinio (ENJL) model The main goal of such studies is to probe the properties of nuclear matter in the region between symmetric nuclear matter and pure neutron matter This information is important in understanding the explosion mechanism of supernova and the cooling rate of neutron stars Received October 22, 2012 Accepted November 6, 2012 Physics Subject Classification: 62 44 01 03 Contact Le Viet Hoa, e-mail address: hoalv@hnue.edu.vn 106 The equations of state of asymmetric nuclear matter Content 2.1 The equations of state of nuclear matter Let us consider the nuclear matter given by the Lagrangian density ¯ ∂ˆ − M)ψ + Gs (ψψ) ¯ − Gv (ψγ ¯ µ ψ)2 − Gr (ψτ ¯ γ µ ψ)2 + ψγ ¯ µψ, (2.1) £ = ψ(i 2 where ψ is nucleon field operator, M is the "bare" mass of the nucleon, µ = diag(µp , µn ); µp,n = µB ± µI /2 is chemical potential, τ = σ/2 with σ are the isospin Pauli matrices, γµ are Dirac matrices, and Gs,v,r are coupling constants Bosonizing ¯ σ ˇ = ψψ, ¯ µ ψ, ˇbµ = ψτ ¯ γµ ψ, ω ˇ µ = ψγ leads to ¯ ∂ˆ − M + γ µ)ψ + Gs ψˇ ¯σ ψ − Gv ψγ ¯ µω ¯ µ τ ˇbµ ψ £ = ψ(i ˇ µ ψ − Gr ψγ Gr ˇµˇ Gs Gv µ σ ˇ + ω ˇ ω ˇµ + b bµ − 2 (2.2) In the mean-field approximation σ ˇ = σ, ω ˇ µ = ωδ0µ , ˇbaµ = bδ3a δ0µ (2.3) Inserting (2.3) into (2.2) we obtain that ¯ ∂ˆ − M ∗ + γ µ∗ }ψ − U(σ, ω, b), £M F T = ψ{i (2.4) where M ∗ = M − Gs σ, µ ∗ (2.5) = µ − Gv ω − Gr τ b, [Gs σ − Gv ω − Gr b2 ] U(σ, ω, b) = (2.6) (2.7) The solution M ∗ of Eq (2.5) is the effective mass of the nucleon Starting from (2.4) we arrive at the inverse propagator  (k0 +µ∗p )−M ∗ −σ.k 0  ∗ ∗ σ.k −(k0 +µp)−M 0  S −1 (k; σ, ω, b) =  ∗ ∗  0 (k0 +µn )−M −σ.k 0 σ.k −(k0 +µ∗n )−M ∗     (2.8)  107 Le Viet Hoa and Le Duc Anh Thus det S −1 (k; σ, ω, b) = (k0 + E++ )(k0 − E−− )(k0 + E+− )(k0 − E−+ ), (2.9) in which E∓± = Ek± ∓ µB − Gω ω , Ek± = Ek ± ( Gρ µI − b), 2 Ek = k + M ∗2 (2.10) Based on (2.7) and (2.8) the effective potential is derived: T Ω = U(σ, ω, b) + iTrlnS −1 = [Gs σ − Gv ω − Gr b2 ]− 2 π − + ∞ − k dk ln(1+e−E− /T ) + + ln(1+e−E+ /T )+ln(1+e−E− /T )+ln(1+e−E+ /T ) (2.11) The ground state of nuclear matter is determined by the minimum condition ∂Ω = 0, ∂ω ∂Ω = 0, ∂σ ∂Ω = ∂b (2.12) Inserting (2.11) into (2.12) we obtain the gap equations ∞ π2 ω = π2 b = 2π σ = ∞ k dk + − + k dk (n− p − np ) + (nn − nn ) ≡ ρB ∞ M∗ + − + (n− p + np ) + (nn + nn ) ≡ ρs Ek + − + k dk (n− p − np ) − (nn − nn ) ≡ ρI , (2.13) where − n− p = n− ; + n+ p = n+ ; − n+ n = n+ ; + n− n = n− ; ± E∓ /T n± +1 ∓ = e −1 , The pressure P is defined as P = −Ω|taken at minimum (2.14) Combining Eqs (2.11), (2.13) and (2.14) together we get the following expression for the pressure P = − Gr T Gs Gv ρs + ρB + ρ + 2 I π2 − −E+ /T + ln(1 + e 108 + −E− /T ) + ln(1 + e ∞ − k dk ln(1 + e−E− /T ) + ) + ln(1 + e−E+ /T ) (2.15) The equations of state of asymmetric nuclear matter The energy density is obtained by the Legendre transform of P : ε = Ω(σ, ω, b) + T ς + µB ρB + µI ρI ∞ Gs Gv Gr + − + = ρs + ρB + ρI + k dkEk (n− p + np + nn + nn ) (2.16) 2 π with the entropy density defined by ∞ ∂Ω − − − ς = =− k dk n− p ln np + (1 − np ) ln(1 − np ) ∂T π − − − + nn ln nn + (1 − n− n ) ln(1 − nn ) + + + + + + + + n+ p ln np + (1 − np ) ln(1 − np ) + nn ln nn + (1 − nn ) ln(1 − nn ) (2.17) Let us introduce the isospin asymmetry α: α = (ρn − ρp )/ρB , (2.18) in which ρB = ρn + ρp is the baryon density, and ρn , ρp are the neutron, proton densities, respectively Taking into account (2.5), (2.13), and (2.18) together the Eqs (2.15), (2.16) can be rewritten as T Gv Gr α 2 (M − M ∗ )2 ρB + + + P =− 2Gs π + ln(1 + e− ε= Ek +µ∗ p T ) + ln(1 + e− Ek +µ∗ n T (M − M ∗ )2 Gv Gr α 2 ρB + + + 2Gs π ∞ k dk ln(1 + e Ek −µ∗ p T ) ) + ln(1 + e− ∞ − Ek −µ∗ n T ) (2.19) + − + k dkEk (n− p + np + nn + nn ) (2.20) Eqs (2.19) and (2.20) constitute the equations of state (EOS) governing all thermodynamical processes of nuclear matter 2.2 Numerical study In order to understand the role of isospin degree of freedom in nuclear matter, let us carry out the numerical study First we follow the method developed by Walecka [1] to determine the three parameters Gs , Gv , and Gr for symmetric nuclear matter based on the saturation condition: The saturation mechanism requires that at normal density ρB = ρ0 = 0.17f m−3 the binding energy εbin = −M + ε/ρB attains its minimum value 109 Le Viet Hoa and Le Duc Anh (εbin )0 ≃ −15, 8MeV , in which ε is given by (2.20) It is found that G2s = 13.62f m2 and Gv = 0.75Gs As to fixing Gr let us employ the expansion of nuclear symmetry energy (NSE) around ρ0 Esym L ρB − ρ0 = a4 + ρ0 Ksym ρB − ρ0 + 18 ρ0 , (2.21) with a4 being the bulk symmetry parameter of the Weiszaecker mass formula, experimentally we know a4 = 30 − 35MeV ; L and Ksym related respectively to slope and curvature of NSE at ρ0 L = 3ρ0 Ksym = 9ρ0 ∂Esym ∂ρB , ρB =ρ0 ∂ Esym ∂ρ2B ρB =ρ0 Then Gr is fitted to give a4 ≃ 32MeV Its value is Gr = 0.198Gs Thus, all of the model parameters are fixed, which are in good agreement with those widely expected in the literature [1] Now we are ready to carry out the numerical computation In Figures and we plot the density dependence of Ebin (ρB ; α) at several values of of temperature and isospin asymmetry α From these Figures we deduce that for comparison with the results of the chiral approach of nuclear matter [7] the asymmetric nuclear matter in our model is less stiff and the isospin dependence of saturation density is strong enough Figure The density dependence of binding energy at several values of temperature and isospin asymmetry α = and α = 0.25 110 The equations of state of asymmetric nuclear matter Figure The density dependence of binding energy at several values of temperature and isospin asymmetry α = 0.5 and α = The EOS for several α steps at some fixed temperatures is presented in Figures and As we can see from the these figures, the critical temperature for the liquid-gas phase transition decreases with increasing neutron excess Figure The EOS for several α steps at temperatures T = 0MeV and T = 10MeV Figure The EOS for several α steps at temperatures T = 15MeV and T = 20MeV 111 Le Viet Hoa and Le Duc Anh Conclusion In this article we have investigated the isospin dependence of energy and pressure of the asymmetric nuclear matter on the NJL-type model Based on the effective potential in the one-loop approximation we determined the expression of pressure by the effective potential at the minimum As a result, the free energy has been obtained straightforwardly They constitute the equations of state (EOS) of the asymmetric nuclear matter It was indicated that in the asymmetric nuclear matter, the isospin dependence of saturation density is reasonably strong and the critical temperature for the liquid-gas phase transition decreases with increasing neutron excess This is our major success In order to understand better the phase structure of the asymmetric nuclear matter more detail study would be carried out by means of numerical computation This is left for future study REFERENCES [1] J D Walecka, 1974 Ann Phys 83, 491 [2] B D Serot and J D Walecka, 1997 Phys Lett B87, 172 [3] B D Serot and J D Walecka, 1986 Advances in Nuclear Physics, edited by J W Negele and E Vogt (Plenum Press, New York, ), Vol 16, p [4] Tran Huu Phat, Nguyen Tuan Anh and Le viet Hoa, 2003 Nuclear Physics A722, pp 548c-552c [5] Tran Huu Phat, Nguyen Tuan Anh, Nguyen Van Long and Le Viet Hoa, 2007 Phys Rev C76, 045202 [6] Tran Huu Phat, Le viet Hoa, Nguyen Van Long, Nguyen Tuan Anh and Nguyen Van Thuan, 2011 Communications in Physics Vol 21, Number 2, pp 117- 124 [7] Tran Huu Phat, Nguyen Tuan Anh and Dinh Thanh Tam, 2011 Phase Structure in a Chiral Model of Nuclear Matter, Physical Review C84, 024321 112 .. .The equations of state of asymmetric nuclear matter Content 2.1 The equations of state of nuclear matter Let us consider the nuclear matter given by the Lagrangian density... result, the free energy has been obtained straightforwardly They constitute the equations of state (EOS) of the asymmetric nuclear matter It was indicated that in the asymmetric nuclear matter, the. .. these Figures we deduce that for comparison with the results of the chiral approach of nuclear matter [7] the asymmetric nuclear matter in our model is less stiff and the isospin dependence of