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.

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

1 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

2 Content

2.1 The equations of state of nuclear matter

Let us consider the nuclear matter given by the Lagrangian density

£ = ψ(i ˆ¯ ∂ − M)ψ +Gs

2 ( ¯ψψ)

2− Gv

2 ( ¯ψγ

µψ)2− Gr

2 ( ¯ψ~τγ

µψ)2+ ¯ψγ0µψ, (2.1) 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,γµare Dirac matrices, andGs,v,r are coupling constants

Bosonizing

ˇ

σ = ¯ψψ, ωˇµ= ¯ψγµψ, ~ˇbµ= ¯ψ~τγµψ, leads to

£ = ψ(i ˆ¯ ∂ − M + γ0µ)ψ + Gsψˇ¯σψ − Gvψγ¯ µωˇµψ − Grψγ¯ µ~τ.~ˇbµψ

− Gs

2 σˇ

2+ Gv

2 ωˇ

µωˇµ+Gr

In the mean-field approximation

hˇσi = σ, hˇωµi = ωδ0µ, hˇbaµi = bδ3aδ0µ (2.3) Inserting (2.3) into (2.2) we obtain that

£

M F T = ¯ψ{i ˆ∂ − M∗+ γ0µ∗}ψ − U(σ, ω, b), (2.4) where

µ∗ = µ − Gvω − Grτ3b, (2.6) U(σ, ω, b) = 1

2[Gsσ

2

− Gvω2− Grb2] (2.7) The solutionM∗ of Eq (2.5) is the effective mass of the nucleon

Starting from (2.4) we arrive at the inverse propagator

S−1(k; σ, ω, b) =











(k0+µ∗

~σ.~k −(k0+µ∗

n)−M∗ −~σ.~k

n)−M∗









 (2.8)

det S−1(k; σ, ω, b) = (k0 + E++)(k0− E−−)(k0+ E+−)(k0− E−+), (2.9)

in which

E∓± = Ek±∓ µB − Gωω

, Ek±= Ek± (µI

2 −Gρ

2 b), Ek =

q

~k2+ M∗2 (2.10) Based on (2.7) and (2.8) the effective potential is derived:

Ω = U(σ, ω, b) + iTrlnS−1 = 1

2[Gsσ

2

− Gvω2− Grb2]−T

π2

Z ∞

0

k2dk

 ln(1+e−E−− /T) + ln(1+e−E+−/T)+ln(1+e−E−+/T)+ln(1+e−E+/T)



The ground state of nuclear matter is determined by the minimum condition

∂Ω

∂σ = 0,

∂Ω

∂ω = 0,

∂Ω

Inserting (2.11) into (2.12) we obtain the gap equations

π2

Z ∞

0

k2dkM

∗

Ek

 (n−p + n+p) + (n−n + n+n)

≡ ρs

π2

Z ∞

0

k2dk (n−p − n+p) + (n−n − n+n)

≡ ρB

2π2

Z ∞

0

k2dk (n−p − n+p) − (n−n − n+n)

≡ ρI, (2.13) where

n−p = n−−; n+p = n++; n+n = n−+; n−n = n+−; n±∓=

eE±/T + 1−1

, The pressureP 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 = −G2sρ2s+ Gv

2 ρ

2

B +Gr

2 ρ

2

I + T

π2

Z ∞

0

k2dk

 ln(1 + e−E−/T) + ln(1 + e−E+−/T) + ln(1 + e−E−+/T) + ln(1 + e−E+/T)

 (2.15)

The energy density is obtained by the Legendre transform ofP :

ε = Ω(σ, ω, b) + T ς + µBρB + µIρI

= Gs

2 ρ

2

s+ Gv

2 ρ

2

B+Gr

2 ρ

2

I + 1

π2

Z ∞

0

k2dkEk(n−p + n+p + n−n + n+n) (2.16) with the entropy density defined by

ς = ∂Ω

∂T = − 1

π2

Z ∞

0

k2dk



n−p ln n−p + (1 − n−p) ln(1 − n−p) + n−nln n−n + (1 − n−n) ln(1 − n−n)

+ n+p ln n+p + (1 − n+p) ln(1 − n+p) + n+n ln n+n + (1 − n+n) ln(1 − n+n)

 (2.17) Let us introduce the isospin asymmetryα:

in whichρB = ρn+ ρpis the baryon density, andρn,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

P = −(M − M∗)

2

2Gs +

Gv

2 +

Grα2

8



ρ2B + T

π2

Z ∞

0

k2dk ln(1 + e−Ek−µ

∗ p

T ) + ln(1 + e−Ek+µ

∗ p

T ) + ln(1 + e−Ek+µ

∗ n

T ) + ln(1 + e−Ek−µ

∗ n

T )

ε =(M − M∗)2

2Gs

+Gv

2 +

Grα2

8



ρ2B + 1

π2

Z ∞

0

k2dkEk(n−p + n+p + n−n + n+n) (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 energyεbin= −M + ε/ρBattains its minimum value

(εbin)0 ≃ −15, 8MeV , in which ε is given by (2.20) It is found that G2

s = 13.62f m2 and

Gv = 0.75Gs As to fixingGr let us employ the expansion of nuclear symmetry energy (NSE) aroundρ0

Esym = a4+ L

3



ρB− ρ0

ρ0

 +Ksym 18



ρB− ρ0

ρ0

2

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 atρ0

L = 3ρ0



∂Esym

∂ρB



ρ B =ρ 0 ,

Ksym = 9ρ0



∂2Esym

∂ρ2 B



ρ B =ρ 0

ThenGris fitted to givea4 ≃ 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 1 and 2 we plot the density dependence ofEbin(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 1 The density dependence of binding energy at several values of temperature

and isospin asymmetry α = 0 and α = 0.25

Figure 2 The density dependence of binding energy at several values of temperature

and isospin asymmetry α = 0.5 and α = 1

The EOS for severalα steps at some fixed temperatures is presented in Figures 3 and 4 As we can see from the these figures, the critical temperature for the liquid-gas phase transition decreases with increasing neutron excess

Figure 3 The EOS for several α steps at temperatures T = 0MeV and T = 10MeV

Figure 4 The EOS for several α steps at temperatures T = 15MeV and T = 20MeV

3 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.

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[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 1

[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.