[ k4 1 ft[he 3 B X dk mB gaEg2

where the values of time-like components of meson fields are expressed via [he} by solving Eqs. (5.83) and (5.84).

The free parameters of the model are gaE/ma, g^E/mw, gpE/mp, b, and c. Experimental values of mE are used.

Due to the mean-field character of the model and the charge independence of strong interactions, the case of the full baryon octet is not much more complicated than the minimal case of the nucleon matter. Also, going beyond the baryon octet by including A resonances is not a very difficult task within the RMF model.

5.11. The equation of state of the outer core

The theoretical description of the matter at p < 2p0 is within the reach of the modern nuclear theory. The nuclear Hamiltonian, albeit very complicated, is known reasonably well. The calculation of the ground state of nucleon matter requires big computing resources but can be carried out with a reasonable accuracy.

5.11.1 Calculating the equation of state

Consider the matter composed of nucleons, electrons, and possibly muons. Nucleons form a strongly interacting Fermi liquid, while electrons and muons constitute nearly ideal Fermi gases. The energy per unit volume is

E (n n ■> np ■> ne ■> nii ) = EN(nn, Hp) + Ee(ne) + E^(h^) , (5.89)

where EN is the nucleon contribution. In what follows, we will assume full thermodynamic equilibrium. The pressure and energy-density depend on a single parameter; its best choice is usually the baryon density nb. The equilibrium at given nb corresponds to the minimum of E under the condition of electrical neutrality.

We will derive the equilibrium equations using a general method which can also be used in a multi-component matter containing, e.g., hyperons. Let us employ the method of Lagrange multipliers, particularly suitable for calculating the minimum of a function of many variables under additional constraints. In our case, the variables are the number densities Uj, j = u,p,e,j, and the constraints are fixed baryon density: un + up — ub = 0 , (5.90a)

electrical neutrality: ue + u^ — up = 0 . (5.90b)

Let us introduce the auxiliary function E, defined by

E = E + Ai(Ue + U^ — Up) + X2(Un + Up — Ub) . (5.91)

In this case Xi are Lagrange multipliers to be determined from the unconstrained minimization of E by requiring dE/duj = 0 for all j:

with dE /duj = Jj. Eliminating the Lagrange multipliers from Eqs. (5.92) one gets the relation between the chemical potentials

which expresses the equilibrium with respect to the weak-interaction processes u —> p + e + Ve , p + e —> u + ve , (5.94a)

u —> p + J + V^ , p + j —> u + v^ . (5.94b)

We consider a neutron-star core transparent for neutrinos (which occurs, typically, as soon as T < 109 — 10i0 K). In this case neutrinos do not affect the matter thermodynamics, and we can put jVe = jVe = = jvM = 0.

Equations (5.93) supplemented by the constraints (5.90) form a closed system of equations which determine the equilibrium composition of the upej matter.

Electrons are ultra-relativistic, so that (in physical units, in which the electron Fermi momentum is hpFe)

while muons are mildly relativistic

Muons are present only if /ie > m^c2 = 105.65 MeV; in the opposite case we are dealing with the npe matter.

Once the equilibrium is determined, the pressure is calculated from the first law of thermodynamics (at T = 0):

The derivative is taken at the equilibrium composition.

One may worry about density-dependent particle composition. Generally, the density dependence of particle fractions Xj = nj/nb gives a non-vanishing contribution to the density derivative of the energy per nucleon. Let us treat E as a function of nb, xp, xe, and xM. Then

where derivatives are taken at equilibrium. However, using Eqs. (5.90) and (5.93) one can see that the second term on the right-hand-side of Eq. (5.98) vanishes, i.e., both formulae for P give the same result.

5.11.2 The nuclear symmetry energy and the proton fraction

Many-body calculations of the energy per nucleon in an asymmetric nuclear matter with realistic nucleon-nucleon interactions show that, to a very good approximation, the dependence on the neutron excess 5 = 1 — 2xp is quadratic (see, e.g., Lagaris & Pandharipande 1981c; Wiringa et al. 1988; Akmal et al. 1998):

Here, E0 (nb) refers to the symmetric nuclear matter and S(nb) is the symmetry energy. A very high precision of this formula, even for 5 ~ 1, indicates that the higher-order terms of the expansion in 5 are small.

In this context it is instructive to consider the free-Fermi gas (FFG) model of the nuclear matter, where the energy per baryon is

Here, m = (mn + mp)c2/2 = 938.93 MeV is the mean nucleon mass and eF is the Fermi energy in the symmetric nuclear matter at a given nb,

The small-5 expansion of EFFG reads then

which gives the symmetry energy for the free Fermi gas model in the form

\noJ

It is easy to check that the quadratic approximation, Eq. (5.99), is very precise even at 5 = 1. From Eq. (5.100) applied to a pure neutron matter we obtain Effg(nb, 1) = (3/5) 22/3 ep(nb) w 0.9524 ep(nb), while Eq. (5.102) gives 14eF(nb)/15 w 0.9333 eF(nb) which is only 2% smaller!

The simple form of the dependence of EN on xp enables us to clarify the relation between the symmetry energy and the composition of the npe matter at beta-equilibrium. Using Eq. (5.99) we can easily calculate the difference between the chemical potentials of neutrons and protons,

The beta-equilibrium in the npe matter (where xe = xp) implies, therefore,

Accordingly, the proton fraction at a given nb is determined by the symmetry energy. Under typical conditions, the proton fraction is small, xp ^ 1, and

i.e., the dependence of xp on the symmetry energy is very strong.

The proton fraction at the normal nuclear density nb = n0 is directly determined by the symmetry energy Sb(no) = S0 at the saturation point (§ 5.4). As the experimental value of S0 is Sexp ~ 30 MeV, the proton fraction in the neutron-star matter at the normal nuclear density should be xp(n0) ~ 5%, independently of any specific EOS of dense matter. On the other hand, Eq. (5.106) tells us that the actual value of xp(n0) for a given model of dense matter is very sensitive to the value of S0 of that model. In particular, the free Fermi-gas model yields a very small S0, Eq. (5.103), and gives an unrealistically low value xFFG(n0) w 0.0033.

5.12. Equation of state and composition

In the rectangular insert in Fig. 5.6 we show EOSs of the outer core calculated using five models of nucleon matter. They are briefly described in Table 5.3.

Figure 5.6. Pressure versus baryon number density for several EOSs of the npep, matter in beta equilibrium. Labels are the same as in Table 5.3. The tables of the APR and APRb* EOSs have been kindly provided by J.M. Lattimer.

Figure 5.6. Pressure versus baryon number density for several EOSs of the npep, matter in beta equilibrium. Labels are the same as in Table 5.3. The tables of the APR and APRb* EOSs have been kindly provided by J.M. Lattimer.

Two of them, BBB1 and BBB2, are obtained in the framework of the BBG theory assuming different realistic NN potentials and a model of the NNN interaction. Although the NN potentials are essentially different (the local U14 Urbana potential and the non-local momentum-dependent Paris potentials) they fit equally well NN data and give very similar EOSs of the outer neutron-star core.

The APR and APRb* EOSs are calculated using the variational method. The center-of-mass NN potential is the same (Argonne A18). However, the APRb* EOS includes a repulsive boost NN-interaction and uses the Urbana NNN force UIX* readjusted to experimental data (see § 5.5.3). The repulsive component of the UIX* NNN force is much weaker than that of the UIX NNN force in the APR EOS. However, these differences in the Hamiltonian slightly affect the EOS of the outer neutron-star core. Finally, the SLy EOS is based on a Skyrme-type energy density functional; it is very similar to the APR one.

Was this article helpful?

0 0

Post a comment