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

Bao-Jun<EMAIL_ADDRESS>Quantum Machine Learning Laboratory, Shadow
Creator Inc., Shanghai 201208, China  Rui<EMAIL_ADDRESS>Key
Laboratory of Nuclear Physics and Ion-beam Application (MOE), Institute of
Modern Physics, Fudan University, Shanghai 200433, China  Zhen
<EMAIL_ADDRESS>Sino-French Institute of Nuclear
Engineering and Technology, Sun Yat-Sen University, Zhuhai 519082, China  Lie-
Wen Chen444Corresponding author<EMAIL_ADDRESS>School of Physics and
Astronomy and Shanghai Key Laboratory for Particle Physics and Cosmology,
Shanghai Jiao Tong University, Shanghai 200240, China

###### Abstract

The relation between the fourth-order symmetry energy
$E_{\textrm{sym,4}}(\rho_{0})$ of nuclear matter at saturation density
$\rho_{0}$ and its counterpart $a_{\textrm{sym,4}}(A)$ of finite nuclei in a
semi-empirical nuclear mass formula is revisited by considering the high-order
isospin-dependent surface tension contribution to the latter. We derive the
full expression of $a_{\textrm{sym,4}}(A)$ which includes explicitly the high-
order isospin-dependent surface tension effects, and find that the value of
$E_{\textrm{sym,4}}(\rho_{0})$ cannot be extracted from the measured
$a_{\textrm{sym,4}}(A)$ before the high-order surface tension is well
constrained. Our results imply that a large $a_{\textrm{sym,4}}(A)$ value of
several MeVs obtained from analyzing nuclear masses can nicely agree with the
empirical constraint of $E_{\textrm{sym,4}}(\rho_{0})\lesssim 2$ MeV from
mean-field models and does not necessarily lead to a large
$E_{\textrm{sym,4}}(\rho_{0})$ value of $\sim 20$ MeV obtained previously
without considering the high-order surface tension. Furthermore, we also give
the expression for the sixth-order symmetry energy $a_{\textrm{sym,6}}(A)$ of
finite nuclei, which involves more nuclear matter bulk parameters and the
higher-order isospin-dependent surface tension.

##  I Introduction

The fourth-order symmetry energy of nuclear matter plays an important role in
determining the properties of neutron stars such as the proton fraction and
the core-crust transition density/pressure Zha01 ; Ste06 ; Xu09 ; Cai12 ;
Sei14 ; Gon17 ; PuJ17 , with the former being critical to neutron star cooling
Lat91 ; Yak01 ; Yak04 . Mathematically, the fourth-order symmetry energy of
nuclear matter is defined as $E_{\textrm{sym,4}}(\rho)\equiv
24^{-1}\partial^{4}E(\rho,\delta)/\partial\delta^{4}|_{\delta=0}$ where
$E(\rho,\delta)$ is the equation of state (EOS) of an asymmetric nucleonic
matter (ANM) at density $\rho=\rho_{\textrm{n}}+\rho_{\textrm{p}}$ and isospin
asymmetry
$\delta=(\rho_{\textrm{n}}-\rho_{\textrm{p}})/(\rho_{\textrm{n}}+\rho_{\textrm{p}})$
with $\rho_{\textrm{n}}$($\rho_{\textrm{p}}$) denoting the neutron(proton)
density Dan02 ; ditoro ; LCK08 ; Tesym ; Col14 ; Bal16 ; Oer17 ; Garg18 ;
LiBA18 .

Most theoretical model calculations indicate that the
$E_{\textrm{sym,4}}(\rho)$ is much smaller than its preceding term in the ANM
EOS, namely, the nuclear symmetry energy defined similarly as
$E_{\textrm{sym}}(\rho)\equiv
2^{-1}\partial^{2}E(\rho,\delta)/\partial\delta^{2}|_{\delta=0}$, and this
empirical fact can be illustrated from the free Fermi gas (FFG) model. In
particular, in the relativistic FFG model, one has the ratio $\Psi\equiv
E_{\textrm{sym,4}}(\rho)/E_{\textrm{sym}}(\rho)=108^{-1}\cdot(10\nu^{4}+11\nu^{2}+4)/(\nu^{4}+2\nu^{2}+1)$
with $\nu=k_{\textrm{F}}/M$, where $M$ is the nucleon rest mass and
$k_{\textrm{F}}$ is nucleon Fermi momentum in the symmetric nucleonic matter
(SNM). Consequently, the $\Psi$ takes the value in the range of
$1/27\leq\Psi\leq 5/54$ CaiLi2022 ; CaiLi2022a . For example, the
$E_{\textrm{sym,4}}(\rho_{0})$ in the non-relativistic FFG model at the
saturation density $\rho_{0}\approx 0.16\,\textrm{fm}^{-3}$ is about 0.45 MeV
by taking $M\approx 939\,\textrm{MeV}$, which is much less than the value of
$\sim 13$ MeV for the symmetry energy $E_{\textrm{sym}}(\rho_{0})$ in the non-
relativistic FFG model. When considering nucleon-nucleon interactions,
essentially all the predictions on the fourth-order symmetry energy using
either phenomenological models Cai12 ; Sei14 ; Gon17 ; PuJ17 ; Che09 ; Agr17
or microscopic many-body theories Lee98 ; Bom91 ; Kai15  point to the self-
consistent constraint $E_{\textrm{sym,4}}(\rho_{0})\lesssim 2\,\textrm{MeV}$.

On the other hand, a systematic expansion (i.e., the “leptodermous” expansion)
of the energy per nucleon in a finite nucleus is usually made in terms of the
small quantity $I\equiv(N-Z)/A\approx\delta$ and the length parameter
$A^{-1/3}$, such that
$B(N,Z)/A=\sum_{i,j=0,1,2,\cdots}\overline{B}_{ij}I^{2i}A^{-j/3}$ plus terms
like the Coulomb and pairing contributions Mye69 ; Brack85 , here $N$ and $Z$
are the neutron and proton numbers in a finite nucleus with mass number
$A=N+Z$, respectively. Specifically, the classic
Bethe–Weizs$\ddot{\textrm{a}}$cker mass formula gives Mye69

$\displaystyle B(N,Z)=$ $\displaystyle-
a_{\textrm{v}}A+a_{\textrm{sur}}A^{2/3}+a_{\textrm{cou}}\frac{Z^{2}(1-Z^{-2/3})}{A^{1/3}}$
$\displaystyle+a_{\textrm{a}}I^{2}A+a_{\textrm{p}}\frac{(-1)^{N}+(-1)^{Z}}{A^{2/3}},$
(1)

here $-a_{\textrm{v}}$ is the volume energy coefficient, $a_{\textrm{sur}}$ is
the surface energy coefficient, $a_{\textrm{cou}}$ characterizes the Coulomb
interaction between protons, $a_{\textrm{a}}$ is the so-called symmetry energy
coefficient of finite nuclei and $a_{\textrm{p}}$ is the paring energy
coefficient. By analyzing nuclear masses Mye69 , a few coefficients appeared
in Eq. (1) are found to be about $a_{\textrm{v}}\approx
15.7\,\textrm{MeV},a_{\textrm{sur}}\approx
18.6\,\textrm{MeV},a_{\textrm{cou}}\approx 0.7\,\textrm{MeV}$ Mye69 , see also
Ref. Mye96 . By considering the isospin correction to the coefficient of
nuclear surface tension [see formula (3)], the symmetry energy
$a_{\textrm{a}}$ appeared in (1) could be generalized to be Dan03

$a_{\textrm{a}}\to
a_{\textrm{sym}}(A)=\frac{\alpha}{1+(\alpha/\beta)A^{-1/3}},$ (2)

which depends only on the mass number $A$. Here
$\alpha=E_{\textrm{sym}}(\rho_{0})$ and $\beta$ is the so-called surface
symmetry energy, see relevant discussion for details on the $\beta$ in section
II.

Generally, even higher order contributions could be considered in the semi-
empirical Bethe–Weizs$\ddot{\textrm{a}}$cker mass formula in the form of
$a_{\textrm{a}}I^{2}A\to(a_{\textrm{a}}I^{2}+a_{\textrm{a,4}}I^{4}+\cdots)A$,
here $a_{\textrm{a,4}}$ is the fourth-order symmetry energy coefficient of
finite nuclei, which is the main topic of the present work. The
$a_{\textrm{a,4}}$ coefficient was recently found to have a sizable value by
analyzing the double difference of the “experimental” symmetry energy
extracted from nuclear masses AME2012 . Specifically, $a_{\textrm{a,4}}\approx
3.28\pm 0.50$ MeV was extracted in Ref. Jia14  and $a_{\textrm{a,4}}\approx
8.33\pm 1.21$ MeV in Ref. Tia16 . Moreover, in Ref. Jia15 , the fourth-order
symmetry energy of finite nuclei was investigated by fitting nuclear mass data
via the nuclear mass formula with two different forms of the Winger energy,
and the obtained constraint on the $a_{\textrm{a,4}}$ is about $3.91\pm
0.10\,\textrm{MeV}$.

Several questions naturally emerge: How does the sizable $a_{\textrm{a,4}}$
give implications on the $E_{\textrm{sym,4}}(\rho_{0})$? Is the fourth-order
symmetry energy of nuclear matter also large and inconsistent with the
empirical constraint that $E_{\textrm{sym,4}}(\rho_{0})\lesssim
2\,\textrm{MeV}$? In this work, the full formula for the four-order symmetry
energy of finite nuclei is derived, and the conclusion is that although the
$a_{\textrm{a,4}}$ obtained by fitting nuclear masses is sizable, the
corresponding $E_{\textrm{sym,4}}(\rho_{0})$ could still be relatively small
to be consistent with the empirical constraints due to the high-order isospin-
dependent surface tension effects.

The paper is organized as follows: Section II gives a brief review on the
current status on the symmetry energy and the fourth-order symmetry energy of
finite nuclei, starting from the general discussion on Eq. (2). In section
III, the formula for the fourth-order symmetry energy of finite nuclei is
given together with numerical demonstrations, and the emphasis is on the
surface contribution. Section IV is devoted to the analytical expression for
the sixth order symmetry energy of finite nuclei. Section V gives the summary
of the present work.

##  II Symmetry Energies of Finite Nuclei: Relevant Status Review

We start our discussions by reviewing the status of the symmetry energy of
finite nuclei. For a finite nucleus with $N$ neutrons and $Z$ protons, the
total difference $\Delta=N-Z$ could be decomposed into two parts as
$\Delta_{\textrm{v}}=N_{\textrm{v}}-Z_{\textrm{v}}$ and
$\Delta_{\textrm{s}}=N_{\textrm{s}}-Z_{\textrm{s}}$, representing the isospin
difference in the bulk of the nucleus and that distributed on its surface.
Naturally one has $\Delta=\Delta_{\textrm{v}}+\Delta_{\textrm{s}}$. As more
isospin asymmetry moves to the surface, physically the nucleus becomes looser,
indicating that the surface tension of the nucleus becomes smaller compared
with the one in which no isospin asymmetry is distributed on the surface.
Moreover, either neutrons or protons are distributed more on the nucleus
surface, the surface tension $\sigma=a_{\textrm{sur}}/(4\pi r_{0}^{2})$, where
$r_{0}$ is the nuclear matter radius parameter defined by
$4\pi\rho_{0}r_{0}^{3}/3=1$, always decreases. Consequently, the surface
tension with some isospin asymmetry distributed on the nucleus surface could
be written as Dan03

$\sigma=\sigma_{0}-\gamma\mu_{\textrm{a}}^{2}=\sigma_{0}\left[1-(\gamma/\sigma_{0})\mu_{\textrm{a}}^{2}\right],$
(3)

where $\mu_{\textrm{a}}=(\mu_{\textrm{n}}-\mu_{\textrm{p}})/2$ is the chemical
potential difference between neutrons and protons, and $\gamma$ is a
parameter. The leading non-trivial contribution starting at order
$\mu_{\textrm{a}}^{2}$ reflects the aforementioned symmetry between neutrons
and protons.

Based on the relation (3), Ref. Dan03  derived a closed expression for the
symmetry energy of finite nuclei incorporating the surface effects, see
formula (2) where $\alpha\equiv a_{\textrm{a}}^{\textrm{v}}$ is the
coefficient in front of $(N_{\textrm{v}}-Z_{\textrm{v}})^{2}/A$, and
$\beta\equiv 1/16\pi r_{0}^{2}\gamma\equiv a_{\textrm{a}}^{\textrm{s}}$. The
$a_{\textrm{a}}^{\textrm{v}}$ and $a_{\textrm{a}}^{\textrm{s}}$ are the volume
(bulk) symmetry energy and the surface symmetry energy Dan03 ; Dan09 ,
respectively. In the limit of large $A$, the isospin asymmetry gets primarily
stored within the bulk and the coefficient $a_{\textrm{sym}}(A)$ tends towards
$a_{\textrm{a}}^{\textrm{v}}$, i.e., we also have
$a_{\textrm{a}}^{\textrm{v}}=E_{\textrm{sym}}(\rho_{0})$. On the other hand,
in the limit of small mass number $A$, the storage of the isospin asymmetry
gets shifted to the surface and the ratio $a_{\textrm{sym}}(A)/A$ scales as
$a_{\textrm{a}}^{\textrm{s}}/A^{2/3}$.

In addition, further established is the relation between the coefficient
$\beta$ and the ones appeared in the nuclear droplet model Mye69 , i.e.,
$\beta={4Q}/{9}=({4H}/{9})/(1-{2P}/{3J})$, where $J\equiv
E_{\textrm{sym}}(\rho_{0})=\alpha$, $Q$ is the neutron skin stiffness
coefficient Mye96  and the individual constants $H,P$ and $G$, found equal to
$G=3JP/2Q$, describe the dependence of the surface energy on the bulk isospin
asymmetry and on normalized size of the neutron skin.

The uncertainties on the symmetry energy of finite nuclei are mainly due to
those on the surface symmetry energy coefficient $\beta$, e.g., Ref. Dan03
constrained the ranges for the parameters $\alpha$ and $\beta$ as 27
MeV$\lesssim\alpha\lesssim$31 MeV and $11\,\textrm{MeV}\lesssim\beta\lesssim
14\,\textrm{MeV}$ and their ratio as $2.0\lesssim\alpha/\beta\lesssim 2.8$
Dan03 . Similarly, an earlier analysis via the Thomas–Fermi model gives $Q$
about 35.4 MeV, leading to the $\beta$ about 15.7 MeV, and even more earlier
the droplet model gave $Q\approx 16\,\textrm{MeV}$ and $\beta\approx
7.1\,\textrm{MeV}$ Mye69 . Recently based on the analysis on the isobaric
analog states (IAS), the volume symmetry energy coefficient and the surface
symmetry energy coefficient are found to be about
$a_{\textrm{a}}^{\textrm{v}}=\alpha\approx 35.3\,\textrm{MeV}$ and
$a_{\textrm{a}}^{\textrm{s}}=\beta\approx 9.7\,\textrm{MeV}$, respectively
Dan14 , and consequently $\alpha/\beta\approx 3.6$. On the other hand, when
combining the IAS analysis and the neutron skin thickness ($\Delta
r_{\textrm{np}}$) constraint, the relation between the coefficients
$a_{\textrm{a}}^{\textrm{v}}$ and $a_{\textrm{a}}^{\textrm{s}}$ is found to be
slightly changed. Specifically, the $a_{\textrm{a}}^{\textrm{v}}$ and the
$a_{\textrm{a}}^{\textrm{s}}$ obtained in the half-infinite calculation are
found to be about 30.2-33.7 MeV and 14.8-18.5 MeV, leading to the ratio
$\alpha/\beta\approx 1.92\pm 0.24$, while the best values from the combined
analysis from IAS and the $\Delta r_{\textrm{np}}$ are about 33.2 MeV and 10.7
MeV, respectively Dan14 , and thus $\alpha/\beta\approx 3.1$.

We have no attempt to review all the relevant investigations on the
coefficients $a_{\textrm{a}}^{\textrm{v}}$ and $a_{\textrm{a}}^{\textrm{s}}$.
The main point we would like to stress here is that these uncertainties may
further induce essential uncertainties on the fourth-order symmetry energy of
finite nuclei through the ratio $\alpha/\beta$.

It should be pointed out that if the nuclear surface tension is truncated as
in Eq. (3), there would be no higher order symmetry energies from the surface
contribution in finite nuclei. However, it is natural and physical that the
nuclear surface tension should include higher order contributions, i.e.,

$\sigma\approx\sigma_{0}-\gamma\mu_{\textrm{a}}^{2}-\gamma^{\prime}\mu_{\textrm{a}}^{4}-\gamma^{\prime\prime}\mu_{\textrm{a}}^{6}+\cdots,$
(4)

where $\gamma^{\prime}$ and $\gamma^{\prime\prime}$, etc., are effective
parameters. Under this situation, one can obtain higher order terms in the
binding energy per nucleon in finite nuclei, which is the starting point for
the calculations of the present work.

Recently, the bulk term $a_{\textrm{sym,4}}^{\textrm{v}}(A)$ of the fourth-
order symmetry energy $a_{\textrm{sym,4}}(A)$ of finite nuclei was derived
using a variational method in Ref. WangRui17  based on the lowest-order
isospin truncation of Eq. (3) for the surface tension, and the analytic
expression is obtained as

$a_{\textrm{sym,4}}^{\textrm{v}}(A)=\frac{1}{(1+\alpha/\beta
A^{1/3})^{4}}\left(E_{\textrm{sym},4}(\rho_{0})-\frac{L^{2}}{2K_{0}}\right).$
(5)

Based on this formula, Ref. WangRui17  claimed that the
$E_{\textrm{sym,4}}(\rho_{0})$ should be as large as about $20\,\textrm{MeV}$
if $a_{\textrm{a,4}}\approx 3.28\,\textrm{MeV}$ is used Jia14 . Such a large
value of $E_{\textrm{sym,4}}(\rho_{0})\sim 20$ MeV is partially due to the
term $L^{2}/2K_{0}$ which is about 4.2 MeV if $L\approx 45\,\textrm{MeV}$
LiBA13 ; Zha13 ; Oer17  and $K_{0}\approx 240\,\textrm{MeV}$ You99 ; Shl06 ;
Che12 ; Garg18  are adopted, and this confusing value is significantly larger
than the empirical constraint of $E_{\textrm{sym,4}}(\rho_{0})\lesssim
2\,\textrm{MeV}$. In the present work, we show that a high-order surface
contribution will appear in the fourth-order symmetry energy of finite nuclei
due to the high-order isospin dependent surface tension ($\gamma^{\prime}$) as
in Eq. (4), and thus a large $a_{\textrm{a,4}}$ value of several MeVs obtained
from analyzing nuclear masses does not have to lead to a large
$E_{\textrm{sym,4}}(\rho_{0})$ value of $\sim 20$ MeV obtained without
considering the high-order isospin dependent surface tension.

##  III Fourth-order Symmetry Energy of Finite Nuclei: Full Expression

Generally, one can obtain the following equations to determine the bulk and
surface isospin asymmetries as well as the chemical potential difference in
finite nuclei Dan03 ,

$\Delta_{\textrm{v}}+\Delta_{\textrm{s}}=\Delta,~{}~{}2\alpha\Delta_{\textrm{v}}/A-\mu_{\textrm{a}}=0,~{}~{}\Delta_{\textrm{s}}/S+\mathrm{d}\sigma/\mathrm{d}\mu_{\textrm{a}}=0,$
(6)

here $S=4\pi r_{0}^{2}A^{2/3}$ is the surface area of the nucleus via the
relation $R=r_{0}A^{1/3}$. It should be pointed out that the equations shown
in (6) are not complete in the sense that as higher order terms like the
$\gamma^{\prime}\mu_{\textrm{a}}^{4}$ would induce a high order symmetry
energy in finite nuclei related to the surface properties, one needs to add
the bulk term $4\alpha_{4}(\Delta_{\textrm{v}}/A)^{3}\equiv
4E_{\textrm{sym,4}}(\rho_{0})(\Delta_{\textrm{v}}/A)^{3}$ originated from the
fourth-order symmetry energy to the second equation and to be solved self-
consistently. However, since this part was already obtained in Ref. WangRui17
and for the purpose of the present work, we will not give the detailed
derivations here and only focus on the surface contribution.

###  III.1 Effective Symmetry Energies for Finite Nuclei

By introducing the function $f=\sigma/\sigma_{0}$, one obtains in the
situation $f=1-\theta\mu_{\textrm{a}}^{2}\equiv 1+y$ (with
$\theta\equiv\gamma/\sigma_{0}$ and $y\equiv-\theta\mu_{\textrm{a}}^{2}$) the
following expressions for $\Delta_{\textrm{v}},\Delta_{\textrm{s}}$ and
$\mu_{\textrm{a}}$,

$\frac{\Delta_{\textrm{v}}}{\Delta}=\frac{1}{1+\phi},~{}~{}\frac{\Delta_{\textrm{s}}}{\Delta}=\frac{\phi}{1+\phi},~{}~{}\mu_{\textrm{a}}=\frac{2\alpha}{A}\frac{\Delta}{1+\phi},$
(7)

where $\phi=\alpha/\beta A^{1/3}$. The effective symmetry energy (appeared in
the mass formula in the form of $a_{\textrm{a}}^{\textrm{eff}}(N,Z)I^{2}A$) in
finite nuclei could be obtained as,

$a_{\textrm{a}}^{\textrm{eff}}(N,Z)=\left.\left[\frac{\alpha\Delta_{\textrm{v}}^{2}}{A}+\mu_{\textrm{a}}\Delta_{\textrm{s}}+\sigma_{0}S(f-1)\right]\right/AI^{2},$
(8)

which includes the effects from higher order symmetry energies as
$a_{\textrm{a}}^{\textrm{eff}}(N,Z)\approx
a_{\textrm{sym}}(A)+a^{\textrm{s}}_{\textrm{sym,4}}(A)I^{2}+\cdots$. In
particular, for $f=1-\theta\mu_{\textrm{a}}^{2}$ the effective symmetry energy
$a_{\textrm{a}}^{\textrm{eff}}(N,Z)$ reduces to
$a_{\textrm{sym}}(A)=\alpha/(1+\alpha/\beta A^{1/3})$. Moreover, in this
simple model there is only one effective parameter $\theta$, and it is
determined uniquely by the surface symmetry energy coefficient $\beta$ and the
surface tension $\sigma_{0}$. Similarly, the effective fourth-order symmetry
energy of finite nuclei is defined as

$a_{\textrm{a,4}}^{\textrm{eff}}(N,Z)=[{a_{\textrm{a}}^{\textrm{eff}}(N,Z)-a_{\textrm{sym}}(A)}]/{I^{2}},$
(9)

which contains even higher order contributions, e.g., the sixth-order symmetry
energy.

As long as the function $f$ is depending on the combination
$y=-\theta\mu_{\textrm{a}}^{2}$, i.e., $f=f(y)$, it could be proved
straightforwardly that the bulk and the surface isospin asymmetries are given,
by generalizing the first two relations of (7), as

$\displaystyle\frac{\Delta_{\textrm{v}}}{\Delta}=$
$\displaystyle\left(1+\phi\frac{\mathrm{d}f}{\mathrm{d}y}\right)^{-1},~{}~{}\frac{\Delta_{\textrm{s}}}{\Delta}=\phi\frac{\mathrm{d}f}{\mathrm{d}y}\left(1+\phi\frac{\mathrm{d}f}{\mathrm{d}y}\right)^{-1}.$
(10)

In this situation, the effective symmetry energy of finite nuclei can be
obtained as

$\displaystyle a_{\textrm{a}}^{\textrm{eff}}(N,Z)=$
$\displaystyle\frac{\alpha(1+2\phi{\mathrm{d}f}/{\mathrm{d}y})}{(1+\phi{\mathrm{d}f}/{\mathrm{d}y})^{2}}+\frac{4\pi
r_{0}^{2}\sigma_{0}}{I^{2}A^{1/3}}\left[f(y)-1\right],$ (11)

and the $\mu_{\textrm{a}}$ should be self-consistently obtained by solving the
three equations of (6). As $A\to\infty$ and $N-Z\to\infty$ but the ratio $I$
is fixed, the chemical potential difference $\mu_{\textrm{a}}$ will also be
fixed for a given $I$. In particular, all effective models for $f$ tend to be
the same in the large-$A$ limit and
$\mu_{\textrm{a}}\approx\mu_{\textrm{a}}^{\infty}\equiv 2\alpha I$, or
equivalently $2\mu_{\textrm{a}}=\mu_{\textrm{n}}-\mu_{\textrm{p}}\approx
4\alpha\delta$, indicating that the $a_{\textrm{a,4}}^{\textrm{eff}}(N,Z)$
approaches to zero in this limit. It should be remembered that only the
surface contribution to the fourth-order symmetry energy is studied here, see
the comments given at the beginning of this section. If $f=1+y$ is adopted,
then Eq. (11) naturally reduces to $\alpha/(1+\phi)$.

###  III.2 Surface Fourth-order Symmetry Energy

Formula (11) itself could be used to derive the fourth-order symmetry energy
of finite nuclei. We start from formula (11) by assuming the function $f$
takes the form $f\approx 1+y+\kappa y^{2}$ where
$\kappa=-\gamma^{\prime}\sigma_{0}/\gamma^{2}$ is an effective parameter
characterizing the fourth-order contribution to the isospin splitting of the
$f=\sigma/\sigma_{0}$ [see Eq. (4)]. Consequently one has
$f^{\prime}\equiv\mathrm{d}f/\mathrm{d}y=1+2y\kappa$, and the first term in
Eq. (11) becomes

$\displaystyle\frac{\alpha(1+2\phi f^{\prime})}{(1+\phi
f^{\prime})^{2}}\approx\frac{\alpha(1+2\phi)}{(1+\phi)^{2}}-\frac{4\phi^{2}y\alpha}{(1+\phi)^{3}}\kappa,$
(12)

to order $\kappa$. Similarly the second term in Eq. (11) is expanded as ${4\pi
r_{0}^{2}\sigma_{0}(y+y^{2}\kappa)}/{I^{2}A^{1/3}}$, where the first term here
gives

$\frac{4\pi
r_{0}^{2}\sigma_{0}y}{I^{2}A^{1/3}}\approx-\frac{\alpha\phi}{(1+\phi)^{2}}-\frac{16\alpha^{3}\phi^{2}\theta
I^{2}\kappa}{(1+\phi)^{5}}.$ (13)

Summing the first term of Eq. (12) and the first term of Eq. (13) gives the
familiar formula for symmetry energy of finite nuclei, i.e.,
$a_{\textrm{sym}}(A)=\alpha/(1+\phi)$.

Since we assume $\kappa$ is small, the $y$ in $4\pi
r_{0}^{2}\sigma_{0}y^{2}\kappa/I^{2}A^{1/3}$ could be approximated as
$y\approx-{4\theta\alpha^{2}I^{2}}/{(1+\phi)^{2}}$ (recalling that
$\Delta/A=I$), and moreover $\mu_{\textrm{a}}\approx{2\alpha I}/({1+\phi})$.
One then obtains

$\displaystyle-\frac{4\alpha\phi^{2}y}{(1+\phi)^{3}}\approx$
$\displaystyle\frac{16\alpha^{3}\phi^{2}\theta}{(1+\phi)^{5}}I^{2},~{}~{}\frac{4\pi
r_{0}^{2}\sigma_{0}y^{2}}{I^{2}A^{1/3}}\approx\frac{4\alpha^{3}\phi\theta}{(1+\phi)^{4}}I^{2},$
(14)

where the relation ${4\pi
r_{0}^{2}\sigma_{0}y}/{I^{2}A^{1/3}}\approx-{\alpha\phi}/{(1+\phi)^{2}}$ is
used for obtaining ${4\pi r_{0}^{2}\sigma_{0}y^{2}}/{I^{2}A^{1/3}}$ at this
order.

Combining all these terms gives the effective symmetry energy of finite nuclei
to order $I^{2}\kappa$ as

$a_{\textrm{a}}^{\textrm{eff}}(N,Z)\approx\frac{\alpha}{1+\phi}+\frac{4\alpha^{3}\phi\theta\kappa
I^{2}}{(1+\phi)^{4}},$ (15)

the coefficient in front of $I^{2}$ of the second term gives the (surface)
fourth-order symmetry energy of finite nuclei, i.e.,

$a^{\textrm{s}}_{\textrm{sym,4}}(A)=\frac{4\alpha^{3}\theta\kappa\phi}{(1+\phi)^{4}}=\left.\frac{4\alpha^{4}\theta\kappa}{\beta
A^{1/3}}\right/\left(1+\frac{\alpha}{\beta A^{1/3}}\right)^{4}.$ (16)

By combining Eq. (16) with the bulk contribution Eq. (5) derived in Ref.
WangRui17 , we finally obtain the total fourth-order symmetry energy of finite
nuclei as

$\displaystyle a_{\textrm{sym,4}}(A)=$
$\displaystyle\left(1+\frac{E_{\textrm{sym}}(\rho_{0})}{\beta
A^{1/3}}\right)^{-4}$
$\displaystyle\times\left(E_{\textrm{sym,4}}(\rho_{0})-\frac{L^{2}}{2K_{0}}+\frac{4\theta\kappa
E_{\textrm{sym}}^{4}(\rho_{0})}{\beta A^{1/3}}\right).$ (17)

In Fig. 1, the $A$-dependence of the symmetry energy $a_{\textrm{sym}}(A)$
(blue dash-dot line) and the surface fourth-order symmetry energy
$a^{\textrm{s}}_{\textrm{sym,4}}(A)$ (magenta line) of finite nuclei is shown
by adopting $\kappa=0.5$, and moreover $\alpha\approx
30\,\textrm{MeV},\beta\approx 15\,\textrm{MeV},r_{0}\approx 1.12\,\textrm{fm}$
Dan03 , and $\sigma_{0}\approx 0.8\,\textrm{MeV}/\textrm{fm}^{2}$ are used for
illustration. It is found that the fourth-order symmetry energy (due to the
surface) has a weak dependence on the mass number $A$ within the given range.
In addition, we have $a_{\textrm{sym}}(A)\to\beta
A^{1/3},a^{\textrm{s}}_{\textrm{sym},4}(A)\to 4\kappa\theta\beta^{3}A$, and
$a^{\textrm{s}}_{\textrm{sym},4}(A)/a_{\textrm{sym}}(A)\to
4\kappa\theta\beta^{2}A^{2/3}$ as $A\to 0$, and all are independent of the
bulk term $\alpha$ and approach to zero as $A\to 0$. For heavy nuclei the
$\phi=\alpha/\beta A^{1/3}$ is generally smaller than unity since the mass
number $A$ is large, leading to the approximation
$a^{\textrm{s}}_{\textrm{sym,4}}(A)\approx
4\alpha^{3}\theta\kappa\phi={\alpha^{3}A^{2/3}\kappa\phi}/{S\beta\sigma_{0}}$.
The interesting feature of this approximated fourth-order symmetry energy is
that it is linearly proportional to the factor $\phi$ which approaches to zero
in infinite matter limit, i.e.,
$\lim_{A\to\infty}a^{\textrm{s}}_{\textrm{sym,4}}(A)=0\,\textrm{MeV}$. There
is no surprise that $a^{\textrm{s}}_{\textrm{sym},4}(A)$ approaches to zero as
$A\to\infty$ since it only reflects the surface part of the fourth-order
symmetry energy of finite nuclei. Physically the surface disappears as the
mass number $A$ approaches to infinity, i.e., the related fourth-order
symmetry energy becomes zero at this limit, see the inset of Fig. 1. However,
it does not mean that the fourth-order symmetry energy for infinite matter
should be zero, as in the above calculations the relevant term related to
$E_{\textrm{sym,4}}(\rho_{0})$ is not included in the second equation of (6),
i.e., the fourth-order symmetry energy of finite nuclei obtained here is still
characterized by the symmetry energy coefficients $\alpha$ and $\beta$ instead
of the $E_{\textrm{sym,4}}(\rho_{0})$, see Ref. WangRui17  for the relevant
discussions. Nonetheless, (16) is enough for our purpose.

Figure 1: The symmetry energy $a_{\textrm{sym}}(A)$ and the surface fourth-
order symmetry energy $a^{\textrm{s}}_{\textrm{sym,4}}(A)$ of finite nuclei as
functions of nuclear mass number $A$ . Here $\alpha/\beta=2$ and $\kappa=0.5$
are adopted for illustration.

Moreover, the value of $A$ corresponding to the maximum of
$a^{\textrm{s}}_{\textrm{sym},4}(A)$ could be found via $\partial
a^{\textrm{s}}_{\textrm{sym},4}(A)/\partial A=0$, and this gives
$A_{\max}=27\alpha^{3}/\beta^{3}\approx 216$. Consequently,
$a^{\textrm{s}}_{\textrm{sym},4}(A_{\max})=27\alpha^{3}\theta\kappa/64\approx
7.5\,\textrm{MeV}$, see the black dashed line of the inset in Fig. 1. We find
that the empirical ratio $\alpha/\beta$ (near 2-3) coincidentally predicts
that the fourth-order symmetry energy (due to the surface contribution)
maximizes near the ${}^{208}\textrm{Pb}$. The infinite matter (with
$A\to\infty$) and the finite nuclei (with $A$ being around 208) is
fundamentally different from this perspective, and this explains the confusion
given in Ref. WangRui17 .

The relation $2\alpha\Delta_{\textrm{v}}/A=\mu_{\textrm{a}}$ could itself be
solved perturbatively order by order. Under the assumption $f=1+y+\kappa
y^{2}$, one obtains the equation for determining the chemical potential using
the expression for $\Delta_{\textrm{v}}/\Delta$ (see the relations (10)),

$\mu_{\textrm{a}}+\phi\mu_{\textrm{a}}-2\kappa\phi\theta\mu_{\textrm{a}}^{3}=2\alpha
I.$ (18)

In the infinite matter limit the $\phi\to 0$ and the equation gives
$\mu_{\textrm{a}}=\mu_{\textrm{a}}^{\infty}\equiv 2\alpha I$. By treating both
the $\phi$ and $\kappa$ perturbatively on the same order, one writes down the
$\mu_{\textrm{a}}$ to the second order as
$\mu_{\textrm{a}}\approx\mu_{\textrm{a}}^{\infty}(1+\varphi_{1}\phi+\varphi_{2}\kappa+\varphi_{3}\phi\kappa+\varphi_{4}\phi^{2}+\varphi_{5}\kappa^{2})$,
with $\varphi_{1}\sim\varphi_{5}$ five coefficients to be determined. Since
there are already two small quantities $\kappa\phi$ in front of
$\mu_{\textrm{a}}^{3}$ one could safely approximate the cubic of the
$\mu_{\textrm{a}}^{3}$ as $\mu_{\textrm{a}}^{\infty,3}$ to second order. Eq.
(18) becomes

$\displaystyle\mu_{\textrm{a}}^{\infty}\left(1+\varphi_{1}\phi+\varphi_{2}\kappa+\varphi_{3}\phi\kappa+\varphi_{4}\phi^{2}+\varphi_{5}\kappa^{2}\right)$
$\displaystyle+\mu_{\textrm{a}}^{\infty}\phi\left(1+\varphi_{1}\phi+\varphi_{2}\kappa\right)-2\kappa\phi\theta\mu_{\textrm{a}}^{\infty,3}=2\alpha
I,$ (19)

which could be calculated order by order: 1) at 0th order: one has the result
$\mu_{\textrm{a}}^{\infty}\approx 2\alpha I$, or
$\mu_{\textrm{n}}^{\infty}-\mu_{\textrm{p}}^{\infty}\approx 4\alpha I$; 2) at
the 1st order: we have
$\mu_{\textrm{a}}^{\infty}[(\varphi_{1}+1)\phi+\varphi_{2}\kappa]=0$,
consequently $\varphi_{1}=-1,\varphi_{2}=0$; and 3) at the 2nd order: one has
the equation
$\mu_{\textrm{a}}^{\infty}[\varphi_{3}\phi\kappa+\varphi_{4}\phi^{2}+\varphi_{5}\kappa^{2}+\varphi_{1}\phi^{2}+\varphi_{2}\phi\kappa-2\kappa\phi\theta\mu_{\textrm{a}}^{\infty,2}]=0$,
solving it gives $\varphi_{4}=1,\varphi_{5}=0$, and
$\varphi_{3}=2\theta\mu_{\textrm{a}}^{\infty,2}$.

By combining the above results one obtains the chemical potential difference
between neutrons and the protons in finite nuclei to the second order as

$\mu_{\textrm{a}}\approx\mu_{\textrm{a}}^{\infty}\left(1-\phi+\phi^{2}+2\kappa\phi\theta\mu_{\textrm{a}}^{\infty,2}\right).$
(20)

The relation (20) could be written in the slight different form,
$\mu_{\textrm{n}}-\mu_{\textrm{p}}\approx 4\alpha
I(1-\phi+\phi^{2})+4\kappa\phi\theta\mu_{\textrm{a}}^{\infty,3}=4\alpha
I(1-\phi+\phi^{2})+32\alpha^{3}\kappa\phi\theta I^{3}$, via the identity
$2\mu_{\textrm{a}}=\mu_{\textrm{n}}-\mu_{\textrm{p}}$. Moreover, near the
infinite matter limit we have $a^{\textrm{s}}_{\textrm{sym},4}(A)\approx
4\alpha^{3}\theta\kappa\phi$ (the surface contribution), one finally obtains
$\mu_{\textrm{n}}-\mu_{\textrm{p}}\approx
4I\alpha(1-\phi+\phi^{2})+8I^{3}a^{\textrm{s}}_{\textrm{sym},4}(A)$. It is
analogous to the relation $\mu_{\textrm{n}}-\mu_{\textrm{p}}\approx 4\delta
E_{\textrm{sym}}(\rho)+8\delta^{3}E_{\textrm{sym,4}}(\rho)$ often used in
determining the proton fraction $x_{\textrm{p}}=(1-\delta)/2$ in neutron stars
Cai12 ; Gon17 ; PuJ17 . In fact a more similar relation could be found for
neutron and proton chemical potential difference, i.e.,
$\mu_{\textrm{n}}-\mu_{\textrm{p}}\approx
4Ia_{\textrm{sym}}(A)[1+2I^{2}a^{\textrm{s}}_{\textrm{sym},4}(A)/a_{\textrm{sym}}(A)]$
in finite nuclei, by recalling that $1-\phi+\phi^{2}\approx 1/(1+\phi)$ and
$\alpha(1-\phi+\phi^{2})\approx a_{\textrm{sym}}(A)$.

###  III.3 Implications of the Smallness of $E_{\textmd{sym},4}(\rho_{0})$

Since the derivation on the term $4\theta\kappa
E_{\textrm{sym}}^{4}(\rho_{0})/\beta A^{1/3}$ is independent of the bulk
contribution $E_{\textrm{sym,4}}(\rho_{0})-L^{2}/2K_{0}$, the two terms are
additive. Here the $\theta\kappa$-term in Eq. (III.2) approaches to zero as
$A\to\infty$ since the surface disappears for infinite matter. While on the
other hand, the bulk term approaches to a constant (independent of $A$), i.e.,
$E_{\textrm{sym,4}}(\rho_{0})-L^{2}/2K_{0}$ as $A\to\infty$. It is now clear
that one could still have a $E_{\textrm{sym,4}}(\rho_{0})\lesssim
2\,\textrm{MeV}$ to be consistent with microscopic calculations, irrespective
of the value of $a^{\textrm{s}}_{\textrm{sym},4}(A)$ since the surface
contribution only affects the finite nuclei. Solving Eq. (III.2) for
$E_{\textrm{sym,4}}(\rho_{0})$ gives the expression for
$E_{\textrm{sym},4}(\rho_{0})$, from which it is clearly demonstrated a large
value of $a_{\textrm{a,4}}$ (as obtained from nucleus mass formula fitting)
should not necessarily lead to a large $E_{\textrm{sym,4}}(\rho_{0})$, and the
balance strongly depends on the higher order coefficient $\kappa$, which has
very little influence on the nuclear structure quantities like the surface
tension of certain typical finite nuclei.

Figure 2: The correlation between $\alpha/\beta$ and $L/\alpha$ (left) and
between $\beta$ and $L$ (right) via the smallness of
$E_{\textrm{sym},4}(\rho_{0})$. See the text for details.

If one accepts the fact that the $E_{\textrm{sym},4}(\rho_{0})$ is empirically
smaller than about 2 MeV Cai12 ; Gon17 ; PuJ17 , then Eq. (III.2) gives
certain correlations among quantities with sizable magnitude. In this sense,
the resulted correlations are expected to be intrinsic like those obtained
from the unbound nature of a pure neutron matter Cai21 . As an example, we
study the correlation between the surface symmetry energy coefficient $\beta$
and the slope parameter $L$, by uniformly sampling within the empirical ranges
for the symmetry energy as $\alpha=E_{\textrm{sym}}(\rho_{0})\approx 28\sim
36\,\textrm{MeV}$ Chen17 ; LiBA21 , the ratio $\alpha/\beta\approx 1\sim 4$,
the nucleon surface tension $\sigma_{0}\approx 0.6\sim
1.0\,\textrm{MeV}/\textrm{fm}^{2}$ in SNM, the coefficient $\kappa\approx
0\sim 1.0$, the slope parameter $L\approx 30\sim 120\,\textrm{MeV}$ of the
symmetry energy, the incompressibility coefficient $K_{0}\approx 220\sim
260\,\textrm{MeV}$ of SNM, the fourth-order symmetry energy
$a_{\textrm{a},4}\approx 2.78\sim 3.78\,\textrm{MeV}$ Jia14  extracted from
nuclear mass data, $E_{\textrm{sym},4}(\rho_{0})\approx 0\sim
2\,\textrm{MeV}$, $a_{\textrm{cou}}\approx 0.6\sim 0.8\,\textrm{MeV}$, and
$r_{0}\approx 1.12\,\textrm{fm}$. The model $f$ is simply truncated as
$f\approx 1+y+\kappa y^{2}$. The results calculated from the samples using Eq.
(III.2) are then shown as open circles in Fig. 2. For comparison, the results
from various Skyrme interactions reported in Ref. Dan09  are present as open
diamonds. Specifically, the correlation of the ratio $\alpha/\beta$ on the
ratio $L/\alpha$ is shown in the left panel of Fig. 2, which demonstrates well
linear dependence. This is consistent with the findings using the Skyrme
interactions Dan09 . Similarly, the correlation between $\beta$ and $L$ is
shown in the right panel of Fig. 2.

Furthermore, if a macroscopic formula for the neutron skin thickness $\Delta
r_{\textrm{np}}$ for heavy nuclei is adopted, then one can similarly
investigate the correlations between the $\Delta r_{\textrm{np}}$ and the
characteristic parameters like the surface symmetry energy $\beta$, the slope
parameter $L$ of the symmetry energy, the $\kappa$ parameter, and so on. In
particular, if a strong correlation between $\kappa$ and $\Delta
r_{\textrm{np}}$ could be established, then one can use the latter from
experiments like the parity-violating electron scattering experiments (PREX,
CREX) Abr12 ; Hor12 ; Adh21 ; Adh22 ; Rei22 ; Yuk22 ; Zha22 ; HuB22  to
effectively constrain the $\kappa$. However, before the lower order parameters
like the surface symmetry energy $\beta$ are well constrained, it is hard to
investigate the real effects of $\kappa$, since the $\beta$ is directly
related to the $\theta$ parameter as $\theta\sim\beta^{-1}$. Nonetheless the
neutron skin thickness for neutron-rich nuclei may provide a promising probe
to detect the $\kappa$ parameter, and these are left for future studies.

##  IV Expression for the Sixth-order Symmetry Energy of Finite Nuclei

If one considers even higher order isospin dependent terms in the nuclear
surface tension coefficient as $\sigma/\sigma_{0}\approx 1+y+\kappa
y^{2}+sy^{3}$ with $s$ being the effective parameter beyond $\kappa$, one can
directly obtain the sixth-order symmetry energy $a_{\textrm{sym},6}(A)$ of
finite nuclei as a function only of $A$ including both the volume and surface
contributions as

$\displaystyle a_{\textrm{sym,6}}(A)=$
$\displaystyle\left(1+\frac{E_{\textrm{sym}}(\rho_{0})}{\beta
A^{1/3}}\right)^{-6}\Bigg{[}E_{\textrm{sym,6}}(\rho_{0})-L_{\textrm{sym,4}}\left(\frac{L}{K_{0}}\right)$
$\displaystyle\hskip
28.45274pt+\frac{K_{\textrm{sym}}}{2}\left(\frac{L}{K_{0}}\right)^{2}-\frac{J_{0}}{6}\left(\frac{L}{K_{0}}\right)^{3}$
$\displaystyle+\frac{16\theta^{2}E_{\textrm{sym}}^{6}(\rho_{0})}{\beta
A^{1/3}}\left(\frac{4\kappa^{2}}{1+\beta
A^{1/3}/E_{\textrm{sym}}(\rho_{0})}-s\right)\Bigg{]},$ (21)

here $K_{\textrm{sym}}\equiv
9\rho_{0}^{2}\mathrm{d}^{2}E_{\textrm{sym}}(\rho)/\mathrm{d}\rho^{2}|_{\rho=\rho_{0}}$
is the curvature coefficient (see, e.g., Ref. ZhouY19 ) of the symmetry
energy, $J_{0}\equiv
27\rho_{0}^{3}\mathrm{d}^{3}E_{0}(\rho)/\mathrm{d}\rho^{3}|_{\rho=\rho_{0}}$
is the skewness coefficient (see, e.g., Ref. Cai17 ) of the EOS $E_{0}(\rho)$
of SNM, and $L_{\textrm{sym,4}}\equiv
3\rho_{0}\mathrm{d}E_{\textrm{sym},4}(\rho)/\mathrm{d}\rho|_{\rho=\rho_{0}}$
is the slope parameter of the fourth-order symmetry energy
$E_{\textrm{sym,4}}(\rho)$ (see, e.g., Ref. Che09 ). The surface contribution
(the last line) in Eq. (IV) comes from two parts: the higher-order term
originated from the lower-order coefficient $\kappa$ (proportional to
$\kappa^{2}$) and the higher-order term from the coefficient $s$. Naturally
the surface term of $a_{\textrm{sym},6}(A)$ approaches to zero as
$A\to\infty$.

Similar as the investigations given in the last section, it is very difficult
to extract the value of the sixth-order symmetry energy
$E_{\textrm{sym,6}}(\rho_{0})\equiv
720^{-1}\partial^{6}E(\rho_{0},\delta)/\partial\delta^{6}|_{\delta=0}$ for
infinite matter from this expression even if the $a_{\textrm{a,6}}$ appeared
in the mass formula via the term $a_{\textrm{a,6}}I^{6}$ is constrained, since
one could adjust the coefficient $s$ to make $E_{\textrm{sym,6}}(\rho_{0})$
small or large without essentially affecting the properties of finite nuclei
like the nuclear surface tension. For example, without including the surface
term here a “small” $a_{\textrm{a,6}}\approx 1\,\textrm{MeV}$ (which can
hardly be “probed” since the tiny factor $I^{6}$ associated with it) may
induce an $E_{\textrm{sym,6}}(\rho_{0})$ about 7-8 MeV, which is likely to be
conflict with the empirical constraint on the EOS of ANM. In this sense, the
surface contribution to $a_{\textrm{sym},6}(A)$ is fundamental. In addition,
by requiring, e.g., $E_{\textrm{sym},6}(\rho_{0})\lesssim 1\,\textrm{MeV}$ and
$a_{\textrm{sym},6}(A)\lesssim 1\,\textrm{MeV}$, Eq. (IV) then provides a link
relating several characteristics with sizable magnitude, which could be used
to establish certain correlations among them. These are left for future
studies.

##  V Summary and conclusions

We have shown that for the fourth-order symmetry energy of finite nuclei, it
could be naturally decomposed into two terms characterizing the bulk and the
surfaces contributions, respectively. The surface contribution $4\theta\kappa
E_{\textrm{sym}}^{4}(\rho_{0})/\beta
A^{1/3}/(1+E_{\textrm{sym}}(\rho_{0})/\beta A^{1/3})^{4}\sim 4\theta\kappa
E_{\textrm{sym}}^{4}(\rho_{0})/\beta A^{1/3}$ characterized by the product of
$\theta\kappa$, to the fourth-order symmetry energy of finite nuclei is
obtained by introducing the next leading order contribution to the isospin
dependence of the nuclear surface tension coefficient through
$\sigma/\sigma_{0}\approx
1-\theta\mu_{\textrm{a}}^{2}+\kappa\theta^{2}\mu_{\textrm{a}}^{4}$, where
$\kappa$ is an effective parameter characterizing the high order effects.
Although the $\kappa$ parameter may induce a sizable $4\theta\kappa
E_{\textrm{sym}}^{4}(\rho_{0})/\beta A^{1/3}$, it has very little impacts on
the nuclear structure quantities interested such as the surface tension itself
(since $y=-\theta\mu_{\textrm{a}}^{2}$ is generally small than unity). Since
this term is independent of the $E_{\textrm{sym,4}}(\rho_{0})$, it
characterizes the coupling between the symmetry energy
$E_{\textrm{sym}}(\rho_{0})$ and the high order coefficient $\kappa$, i.e., it
is induced by some high-order isospin dependent surface tension effects.
Although both the surface contribution and the bulk term to the fourth-order
symmetry energy of finite nuclei could be large, they contribute little to the
nuclear structure quantities, since generally one has $I^{4}\lesssim 0.003$
for finite nuclei. This means that the fourth-order symmetry energy of finite
nuclei is usually hard to be “probed” and special observables are needed (see,
e.g., Refs. Jia14 ; Jia15 ; Tia16 ).

On the other hand, all the microscopic many-body theories and phenomenological
model predictions give the consistent constraint that
$E_{\textrm{sym,4}}(\rho_{0})\lesssim 2$ MeV, indicating the
$E_{\textrm{sym,4}}(\rho_{0})$ could not be large although its counterpart of
finite nuclei could be sizable. One needs to consider the total fourth-order
symmetry energy of finite nuclei composed both of the bulk and the surface
contributions. In this sense the surface contribution characterized by
$\kappa$ to the fourth-order symmetry energy of finite nuclei is fundamental,
i.e., it is essential for explaining a reasonable
$E_{\textrm{sym,4}}(\rho_{0})\lesssim 2\,\textrm{MeV}$ and a sizable
$a_{\textrm{sym,4}}(A)$ simultaneously. Essentially, it is hard to constrain
the parameter $\kappa$ via finite-nucleus information. In the future, unless
certain quantities/processes determining the coefficient $\kappa$ to some
narrow range are available, it seems that one can hardly constrain the fourth-
order symmetry energy $E_{\textrm{sym,4}}(\rho_{0})$ of nuclear matter from
the nuclear mass formula fitting on $a_{\textrm{sym,4}}(A)$, since a finite
nucleus has a surface while the infinite matter does not.

Finally, we also present the expression for the sixth-order symmetry energy of
finite nuclei, which is related to more nuclear matter bulk parameters and the
higher-order isospin-dependent surface tension. We would like to mention that
although the higher-order symmetry energies of finite nuclei are difficult to
measure in terrestrial nuclei, they could be potentially useful for
understanding the properties of neutron star crust or supernova explosion
where extremely neutron-rich clusters may exist.

## Acknowledgments

This work was supported in part by the National Natural Science Foundation of
China under Grants No. 11905302 and No. 11625521, National SKA Program of
China No. 2020SKA0120300, and the Fundamental Research Funds for the Central
Universities, Sun Yat-Sen University (No. 22qntd1801).

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