The relation between the fourth-order symmetry energy $E_{ \rm{ sym,4}}(\rho_0)$ of nuclear matter at saturation density $\rho_0$ and its counterpart $a_{ \rm{ sym,4}}(A)$ of finite nuclei in a semiempirical nuclear mass formula is revisited by considering the high-order isospin-dependent surface tension contribution to the latter. We derive the full expression of $a_{ \rm{ sym,4}}(A)$, which includes explicitly the high-order isospin-dependent surface tension effects, and find that the value of $E_{ \rm{ sym,4}}(\rho_0)$ cannot be extracted from the measured $a_{ \rm{ sym,4}}(A)$ before the high-order surface tension is well constrained. Our results imply that a large $a_{ \rm{ sym,4}}(A)$ value of several MeVs obtained from analyzing nuclear masses can nicely agree with the empirical constraint of $E_{ \rm{ sym,4}}(\rho_0)\lesssim 2$ MeV from mean-field models and does not necessarily lead to a large $E_{ \rm{ sym,4}}(\rho_0)$ value of $\approx 20$ MeV obtained previously without considering the high-order surface tension. Furthermore, we also give the expression for the sixth-order symmetry energy $a_{ \rm{ sym,6}}(A)$ of finite nuclei, which involves more nuclear matter bulk parameters and the higher-order isospin-dependent surface tension.
Categories: nucl-th astro-ph.HE nucl-ex