Equilibrium theory of the symmetry energy of finite nuclei

Abstract

The nuclear symmetry energy of finite nuclei is calculated first in a nonequilibrium scheme in which the binding energy is a function of the central density parameter as well as the mass number and neutron excess parameter, i.e. E(kc, A, ξ), and then in an equilibrium scheme with the central density parameter given as a function of A and ξ in the form kcE(A, ξ) = kc0(1 + q1ξ−q2ξ2 + q3ξ3−q4ξ4 + … ), where kc0 = k∞(1 + ρ0) f−1 and the qj(A) and ρ0(A) depend on Coulomb and surface effects. In the equilibrium scheme, the symmetry energy coefficients are functions of mass number. A connection is made between the symmetry energy coefficients as calculated in the nonequilibrium (NE) and equilibrium schemes, and we find these coefficients to be, β0(A) = β0NE(kc0), β2(A) = β2NE(kc0)−[Formula: see text], etc. We find that the fourth order coefficient β4(A) is large and negative for all A, and is about −47 MeV in the region A ≈ 125 which agrees reasonably well with the −37 MeV value predicted by the Cameron–Elkin exponential mass formula. No linear term is found in the symmetry energy, but third, fifth, and higher order odd symmetry energy coefficients are found to be present. The alternation of the signs of the symmetry energy coefficients as well as the density expansion coefficients are in accordance with Le Chatelier's principle. As in the case of infinite nuclear matter, we find that the binding energy of nuclei with neutron excess is larger than that calculated assuming constant density, and that a negative isospin compression energy must be added to the constant density calculation of the energy if the correct binding is to be predicted. Finally, the general expression for the symmetry energy coefficients of order j is[Formula: see text]