High-order isospin-dependent surface tension contribution to the fourth-order symmetry energy of finite nuclei

The relation between the fourth-order symmetry energy $E_{\mathrm{s}\mathrm{y}\mathrm{m},4}({\rho }_0)$ of nuclear matter at saturation density ${\rho }_0$ and its counterpart $a_{\mathrm{s}\mathrm{y}\mathrm{m},4}(A)$ 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 $a_{\mathrm{s}\mathrm{y}\mathrm{m},4}(A)$, which includes explicitly the high-order isospin-dependent surface tension effects, and find that the value of $E_{\mathrm{s}\mathrm{y}\mathrm{m},4}({\rho }_0)$ cannot be extracted from the measured $a_{\mathrm{s}\mathrm{y}\mathrm{m},4}(A)$ before the high-order surface tension is well constrained. Our results imply that a large $a_{\mathrm{s}\mathrm{y}\mathrm{m},4}(A)$ value of several MeVs obtained from analyzing nuclear masses can nicely agree with the empirical constraint of $E_{\mathrm{s}\mathrm{y}\mathrm{m},4}({\rho }_0)\lesssim 2$ MeV from mean-field models and does not necessarily lead to a large $E_{\mathrm{s}\mathrm{y}\mathrm{m},4}({\rho }_0)$ 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 $a_{\mathrm{s}\mathrm{y}\mathrm{m},6}(A)$ of finite nuclei, which involves more nuclear matter bulk parameters and the higher-order isospin-dependent surface tension.