Gibbs-Tolman approach to the curved interface effects in asymmetric nuclei
aa r X i v : . [ nu c l - t h ] J un Gibbs–Tolman approach to the curved interface effects in asymmetric nuclei
V.M. Kolomietz and A.I. Sanzhur
Institute for Nuclear Research, 03680 Kiev, Ukraine
We redefine the surface tension coefficient and the symmetry energy for an asymmetric nuclearFermi-liquid drop with a finite diffuse layer. Considering two-component charged Fermi-liquid dropand following Gibbs-Tolman concept, we introduce the equimolar radius R e of sharp surface dropletat which the surface tension is applied and the radius of tension surface R s (Laplace radius) whichprovides the minimum of the surface tension coefficient σ . We have shown that the nuclear Tolmanlength ξ is negative and the modulus of ξ growth quadratically with asymmetry parameter X =( N − Z ) / ( N + Z ). PACS numbers: 24.10.Cn, 68.03.Cd, 21.65.Ef, 21.10.Dr
I. INTRODUCTION
The nucleus is a two component, charged system witha finite diffuse layer. This fact specifies a number ofvarious peculiarities of the nuclear surface and symme-try energies: dependency on the density profile function,non-zero contribution to the surface symmetry energy,connection to the nuclear incompressibility, etc. The ad-ditional refinements appear due to the quantum effectsarising from the smallness of nucleus. In particular, thecurved interface creates the curvature correction to thesurface energy E S and the surface part of symmetry en-ergy E sym of order A / and can play the appreciablerole in small nuclei as well as in neck region of fissionablenuclei.The presence of the finite diffuse layer in nuclei createsthe problem of the correct definition of the radius andthe surface of tension for a small drop with a diffuse in-terface. Two different radii have to be introduced in thiscase [1, 2]: the equimolar radius R e , which gives the ac-tual size of the corresponding sharp-surface droplet, andthe radius of tension R s , which derives, in particular, thecapillary pressure. Bellow we will address this problemto the case of two-component nuclear drop. In general,the presence of the curved interface affects both the bulkand the surface properties. The curvature correction isusually negligible in heavy nuclei. However, this correc-tion can be important in some nuclear processes. Forexample the yield of fragments at the nuclear multifrag-mentation or the probability of clasterization of nucleifrom the freeze-out volume in heavy ion collisions [3]. Inboth above mentioned processes, small nuclei necessar-ily occur and the exponential dependence of the yield onthe surface tension [4] should cause a sensitivity of bothprocesses to the curvature correction. Moreover the de-pendency of the curvature interface effects on the isotopicasymmetry of small fragments can significantly enhance(or suppress) the yields of neutron rich isotopes.In the present paper, we analyze of the interface ef-fects in an asymmetric nuclear Fermi-liquid drop witha finite diffuse layer. We follow the ideology of the ex-tended Thomas-Fermi approximation (ETFA) with effec-tive Skyrme-like forces combining the ETFA and the di- rect variational method with respect to the nucleon den-sities, see Ref. [6]. The proton and neutron densities ρ p ( r ) and ρ n ( r ) are generated by the diffuse-layer profilefunctions which are eliminated by the requirement thatthe energy of the nucleus should be stationary with re-spect to variations of these profiles. In order to formulateproper definition for the drop radius, we use the conceptof the dividing surface, originally introduced by Gibbs[1]. Following the Gibbs method, which is applied to thecase of two component system, we introduce the super-ficial (surface) density as the difference (per unit area ofdividing surface) between actual number of particles A and the number of bulk, A V , and neutron excess, A − , V ,particles which a drop would contain if the particle den-sities were uniform.The plan of the paper is the following. In Sect. IIwe discuss the Gibbs’s derivation of equimolar radius inthe case of two-component system with diffuse layer. Wethen derive in Sect. III the surface energy and the surfacecontribution to symmetry energy. The relation of theleptodermous A − / -expansions for finite nuclei to thenuclear matter equation of state is discussed in Sect. IV.Our conclusions are given in Sect. V. II. DIVIDING SURFACE AND EQUIMOLARRADIUS IN ASYMMETRIC NUCLEI
We consider first the spherical nucleus at zero temper-ature, having the mass number A = N + Z , the neu-tron excess A − = N − Z and the asymmetry parameter X = A − /A . The total binding energy of nucleus is E .An actual nucleus has the finite diffuse layer of particledensity distribution. Thereby, the nuclear size is badlyspecified. In order to formulate proper definition for thenuclear radius, we will use the concept of dividing surfaceof radius R , originally introduced by Gibbs [1]. FollowingRefs. [1, 5], we introduce the formal dividing surface ofradius R , the corresponding volume V = 4 πR / S = 4 πR . Note that the dividing surfaceis arbitrary but it should be located within the nucleardiffuse layer.The energy of a nucleus E , as well as the mass number A and the neutron excess A − , are spitted into the volumeand surface parts, E = E V + E S + E C , (1) A = A V + A S , A − = A − , V + A − , S . (2)Here the Coulomb energy E C is fixed and does not de-pend on the dividing radius R . The bulk energy E V andthe surface energies E S can be written as [4, 5] E V = ( − P V + λ̺ V + λ − ̺ − , V ) V (3)and E S = ( σ + λ̺ S + λ − ̺ − , S ) S . (4)Here P V is the bulk pressure P V = − ∂E V ∂ V (cid:12)(cid:12)(cid:12)(cid:12) A V , (5) σ is the surface tension and ̺ V = A V / V and ̺ − , V = A − , V / V are, respectively, the total (isoscalar) and theneutron excess (isovector) volume densities, ̺ S = A S / S and ̺ − , S = A − , S / S are the corresponding surface den-sities. We have used the isoscalar λ = ( λ n + λ p ) / λ − = ( λ n − λ p ) / λ n and λ p are the chemical potentials of neutron andproton, respectively. The Coulomb energy E C must beexcluded from the chemical potentials λ and λ − becauseof Eqs. (1), (3) and (4). Namely, λ n = ∂E∂N (cid:12)(cid:12)(cid:12)(cid:12) Z , λ p = ∂E∂Z (cid:12)(cid:12)(cid:12)(cid:12) N − λ C , (6)where λ C = ∂E C ∂Z (cid:12)(cid:12)(cid:12)(cid:12) N . Generally, the realistic (experimental) chemical poten-tials λ tot ,n and λ tot ,p contain the contributions of the vol-ume, λ vol , surface, λ surf , symmetry, λ sym , and Coulomb, λ C , parts λ tot ,n = ∂E∂N (cid:12)(cid:12)(cid:12)(cid:12) Z = λ vol + λ surf + λ sym ,λ tot ,p = ∂E∂Z (cid:12)(cid:12)(cid:12)(cid:12) N = λ vol + λ surf − λ sym + λ C , (7)where λ sym = 2 b sym X and b sym is the symmetry energy. The knowledge of thechemical potentials λ tot ,n and λ tot ,p allows us to evalu-ate the Coulomb shift λ C . On the β -stability line, thefollowing condition should be satisfied λ tot ,n − λ tot ,p | X = X ∗ ( A ) = 0 , (8) and Eq. (7) provides the relation λ C = 4 b sym X ∗ . (9)Here X ∗ = X ∗ ( A ) indicates the β -stability line.Notation E V stands for the nuclear matter energy ofthe uniform densities ̺ V , ̺ − , V within the volume V . Thestate of the nuclear matter inside the specified volume V is chosen to have the chemical potentials µ and µ − equalto that of the actual droplet. In more detail, from theequation of state for the nuclear matter one has chemi-cal potentials µ ( ρ, ρ − ) and µ − ( ρ, ρ − ) as functions of theisoscalar, ρ , and isovector, ρ − , densities. Then, the fol-lowing conditions should be fulfilled: µ ( ρ = ̺ V , ρ − = ̺ − , V ) = λ ,µ − ( ρ = ̺ V , ρ − = ̺ − , V ) = λ − (10)to derive the specific values of densities ̺ V and ̺ − , V .The surface part of the energy E S as well as the surfaceparticle number A S and the surface neutron excess A − , S are considered as the excess quantities responsible for“edge” effects with respect to the corresponding volumequantities. Using Eqs. (1) – (4) one obtains σ = E − λA − λ − A − S + P V VS − E C S = Ω − Ω V S . (11)Here the grand potential Ω = E − λA − λ − A − − E C andits volume part Ω V = − P V V = E V − λA V − λ − A − , V wereintroduced. From Eq. (11) one can see how the value ofthe surface tension depends on the choice of the dividingradius R , σ [ R ] = Ω4 πR + 13 P V R . (12)Taking the derivative from Eq. (12) with respect to theformal dividing radius R and using the fact that observ-ables E , λ , λ − and P should not depend on the choiceof the dividing radius, one can rewrite Eq. (12) as P V = 2 σ [ R ] R + ∂∂R σ [ R ] , (13)which is the generalized Laplace equation. The formalvalues of surface densities ̺ , S and ̺ − , S can be foundfrom (2) as ̺ S [ R ] = A πR − ̺ V R ,̺ − , S [ R ] = A − πR − ̺ − , V R . (14)In Eqs. (12) – (14) square brackets denote a formal de-pendence on the dividing radius R which is still arbitraryand may not correspond to the actual physical size ofthe nucleus. To derive the physical size quantity an ad-ditional condition should be imposed on the location ofdividing surface. In general, the surface energy E S forthe arbitrary dividing surface includes the contributionsfrom the surface tension σ and from the binding energyof particles within the surface layer. The latter contri-bution can be excluded for the special choice of dividing(equimolar) radius R = R e which satisfy the condition( ̺ S λ + ̺ − , S λ − ) R = R e = 0 . (15)Here we use the notation R e by the analogy with theequimolar dividing surface for the case of the one-component liquid [3, 5]. For the dividing radius definedby Eq. (15) the surface energy reads E S = σ e S e , (16)where σ e ≡ σ ( R e ) and S e = 4 πR e . Using Eqs. (14), (15),the corresponding volume V e = 4 πR e / V e = λA + λ − A − λ̺ V + λ − ̺ − , V . (17)As seen from Eqs. (10), (17), the droplet radius R e is de-termined by the equation of state for the nuclear matterthrough the values of the droplet chemical potentials λ and λ − .The surface tension σ [ R ] depends on the location ofthe dividing surface. Function σ [ R ] has a minimum atcertain radius R = R s (radius of the surface of tension[5]) which usually does not coincide with the equimo-lar radius R e . The radius R s (Laplace radius) denotesthe location within the interface. Note that for R = R s the capillary pressure of Eq. (13) satisfies the classicalLaplace relation P V = 2 σ [ R ] R (cid:12)(cid:12)(cid:12)(cid:12) R = R s . (18)The dependence of the surface tension σ [ R ] of Eq. (12)on the location of the dividing surface for the nuclei Snand
Pb is shown in Fig. 1.Following Gibbs and Tolman [1, 2], we will assume thatthe physical (measurable) value of the surface tension isthat taken at the equimolar dividing surface. We assume,see also Ref. [5], that the surface tension σ ≡ σ ( R e ) ap-proaches the planar limit σ ∞ as σ ( R e ) = σ ∞ (cid:18) − ξR e + O ( R − e ) (cid:19) , (19)where ξ is the Tolman’s length [2]. Note that the expres-sion (19) can be considered as a particular case of ex-pansion of any observable W in a finite saturated Fermi-system over the dimensionless small parameter r /R e ,where r = (4 πρ / − / and ρ is the bulk particle den-sity. Namely, W = W ∞ + W r R e + W (cid:18) r R e (cid:19) + . . . . (20) ( M e V / f m ) R (fm)SkM Sn Pb R s R s FIG. 1. Surface tension σ as a function of the dividingradius R for nuclei Sn and
Pb. The calculation wasperformed using energy E from Eq. (22) and the SkM force.The Laplace radius R s denotes the dividing radius where σ approaches the minimum value, i.e., the Laplace condition ofEq. (18) is satisfied. Taking Eq. (13) for R = R s and comparing with anal-ogous one for R = R e , one can establish the followingimportant relation (see Eq. (A9) in Appendix A) ξ = lim A →∞ ( R e − R s ) + O ( X ) . (21)This result leads to the conclusion that to obtain thenon-zero value of Tolman length ξ , and, consequently,the curvature correction ∆ σ curv = 0 for a curved surface,the nucleus must have a finite diffuse surface layer. III. MICROSCOPIC CONSIDERATION
We will perform the numerical calculations usingSkyrme type of the effective nucleon-nucleon interac-tion. The energy and the chemical potential for ac-tual droplets can be calculated using a direct variationalmethod within the extended Thomas-Fermi approxima-tion [6]. The energy E of the nucleus is given by thefollowing functional E = Z d r { ǫ kin [ ρ n , ρ p ; ∇ ρ n , ∇ ρ p ]+ ǫ Sk [ ρ n , ρ p ; ∇ ρ n , ∇ ρ p ] + ǫ C [ ρ p ] } , (22)where ǫ kin [ ρ n , ρ p ; ∇ ρ n , ∇ ρ p ] is the kinetic energy den-sity, ǫ Sk [ ρ n , ρ p ; ∇ ρ n , ∇ ρ p ] is the potential energy densityof Skyrme nucleon-nucleon interaction and ǫ C [ ρ p ] is theCoulomb energy density. The equilibrium condition canbe written as a Lagrange variational problem. Namely, δ ( E − λ tot ,n N − λ tot ,p Z ) = 0 , (23)where the variation with respect to all possible smallchanges of ρ n and ρ p is assumed.Using the trial profile function for the neutron ρ n ( r )and proton ρ p ( r ) densities and performing the directvariational procedure, we can evaluate the equilibriumparticle densities ρ ( r ) = ρ n ( r ) + ρ p ( r ) and ρ − ( r ) = ρ n ( r ) − ρ p ( r ), the total energy per particle E/A and thechemical potentials λ tot ,n and λ tot ,p for a fixed asym-metry parameter X , see Ref. [6] for details. We will alsoconsider the asymmetric nuclear matter where the energy E ∞ is given by E ∞ = Z d r { ǫ kin [ ρ n , ρ p ] + ǫ Sk [ ρ n , ρ p ] } . (24)Here, the kinetic energy density ǫ kin [ ρ n , ρ p ] and the po-tential energy density ǫ Sk [ ρ n , ρ p ] do not include the termswhich depend on the gradients of nucleon density provid-ing the bulk particle density ρ = const. Note also thatthe Coulomb energy density ǫ C [ ρ p ] does not contributeto the energy E ∞ . We will derive the volume (bulk) partof energy E V as E V = E ∞ and ̺ V = ρ . (25)Using the energy E V from Eq. (25), the above obtainedvalues of the chemical potentials λ n and λ p and the re-lations ∂E V ∂A (cid:12)(cid:12)(cid:12)(cid:12) V ,A − = λ, ∂E V ∂A − (cid:12)(cid:12)(cid:12)(cid:12) V ,A = λ − , (26)we will evaluate the equilibrium bulk densities ̺ V = ρ and ̺ − , V = ρ − , .The nuclear beta-stability requires the fulfillment ofthe condition (8). In Fig. 2 we compare the resultsfor the beta-stability line Z = Z ∗ ( N ) obtained fromEqs. (1), (7) and (8) with the experimental data (soliddots). One can see that the solid line gives the accept-able description for the experimental data. Note thatthe bulk neutron-proton ratio obtained within the Gibbs-Tolman method might slightly differ from that of an ac-tual drop. The dashed line in Fig. 2 represents function Z V ( N V ) which corresponds to Z ∗ ( N ), where the numberof protons Z V and neutrons N V are taken for the nu-clear matter within the equimolar volume (17). We cansee that for nuclei along the beta-stability line one has X V = ( N V − Z V ) / ( N V + Z V ) < X ∗ . That is because thepart of nucleons (mainly neutrons) are located near thenuclear surface and do not contribute to the volume ratio N V /Z V .For arbitrary dividing radius R and fixed asymme-try parameter X we evaluate then the volume, A V =4 π̺ V R / A − , V = 4 π̺ − , V R /
3, the surface, A S =4 π̺ S R and A − , S = 4 π̺ − , S R , particle numbers andthe volume part of equilibrium energy E V . All evaluatedvalues of E V [ R ] , the bulk densities ̺ V and ̺ − , V and thesurface particle densities ̺ S [ R ] and ̺ − , S [ R ] depend onthe radius R of dividing surface and asymmetry param-eter X . The actual physical radius R e of the droplet Z N SkM
FIG. 2. Solid curve is the line of beta stability Z = Z ∗ ( N )obtained from Eqs. (1), (22), (7) and (8) for Skyrme forceSkM and dots are the experimental data. The dashed lineshows the ratio of neutrons and protons within the equimo-lar volume V e of asymmetric nuclear matter obtained by theGibbs-Tolman method for nuclei with X = X ∗ . -20-1001020 ( M e V f m - ) R (fm) R e SkM, Pb ̺ S λ + ̺ − , S λ − FIG. 3. Specific surface particle density ̺ S λ + ̺ − , S λ − versusdividing radius R for Pb. The calculation was performedusing the SkM force. R e denotes the equimolar radius where ̺ S λ + ̺ − , S λ − becomes zero. can be derived by the condition (15), i.e., by the require-ments that the contribution to E S from the bulk bind-ing energy (term ∼ ( ̺ S λ + ̺ − , S λ − ) in Eq. (4)) shouldbe excluded from the surface energy E S . In Fig. 3 werepresent the calculation of the specific surface particledensity ̺ S λ + ̺ − , S λ − as a function of the radius R ofdividing surface. Equimolar dividing radius R e in Fig. 3defines the physical size of the sharp surface droplet andthe surface at which the surface tension is applied, i.e.,the equimolar surface where Eq. (16) is fulfilled. The R e ( X ) / A / X SkM, A = 208 FIG. 4. Dependence of the equimolar dividing radius R e onthe asymmetry parameter X for nuclei with A = 208. Thecalculation was performed for Skyrme force SkM. dependence of the equimolar dividing radius R e on theasymmetry parameter X is shown in Fig. 4.Note that the value of equimolar radius R e , which isderived by Eq. (17), is not considerably affected by theCoulomb interaction. We have also evaluated the valuesof R e neglecting the Coulomb term in Eq. (22), i.e., as-suming E C = λ C = 0. The difference as compared withdata presented in Fig. 4 does not exceed 0.5%. Omit-ting the Coulomb energy contribution to the total en-ergy E of Eq. (22) and evaluating the bulk energy E V of Eq. (25), one can obtain the surface part of energy E S = E − E V and the surface tension coefficient σ [ R e ](11) on the equimolar dividing surface for nuclei withdifferent mass number A ∼ R e and asymmetry parame-ter X . The dependence of the surface tension coefficient σ [ R e ] on the doubled inverse equimolar radius 2 /R e (seeEq. (19)) is shown in Fig. 5.The surface tension σ [ R e , X ] approaches the planarlimit σ ∞ ( X ) in the limit of zero curvature 2 /R e →
0. Asseen from Fig. 5, the planar limit σ ∞ ( X ) depends on theasymmetry parameter. This dependence reflects the factthat the symmetry energy b in mass formula containsboth the volume b V and surface b S contributions, seeRefs. [7, 8] b ( A ) = b V + b S A − / . (27)In Fig. 6 we show the X -dependence of the surface ten-sion σ ∞ ( X ). This dependence can be approximated by σ ∞ ( X ) = σ + σ − X . (28)The dependence of parameters σ and σ − on the Skyrmeforce parametrization is shown in Table I.The isovector term σ − in the surface tension (28) isrelated to the surface contribution b S in Eq. (27) to the e ( X )( M e V / f m ) / R e ( X ) (fm -1 )SkM X = 0.00.10.2 FIG. 5. The dependence of the surface tension coefficient σ [ R e , X ] on the equimolar radius R e for different values ofthe asymmetry parameter X . The calculation was performedfor Skyrme force SkM. ( X ) / ( ) X SkM
FIG. 6. Dependence of the planar surface tension σ ∞ ( X ) onthe asymmetry parameter X . The calculation was performedfor Skyrme force SkM. symmetry energy as b S ≈ πr σ − , (29)see Appendix A, Eq. (A5). The numerical calculation[8] of the volume symmetry energy gives for SkM force b V =26.5 MeV. Using Eq. (29), we evaluate the surface-to-volume ratio r S/V = | b S /b V | = 1 . ÷ .
47 for Skyrmeforce parametrizations from Table I. Note that in theprevious theoretical calculations, the value of surface-to-volume ratio r S/V varies strongly within the interval1 . ≤ r S/V ≤ .
8, see Refs. [7–9].The slope of curves σ [ R e ] in Fig. 5 gives the Tolman ( X ) / ( ) X SkM
FIG. 7. Dependency of the Tolman length ξ on the asymme-try parameter X . The calculation was performed for Skyrmeforce SkM. length ξ , see Eq. (19). The value of the Tolman length ξ depends significantly on the asymmetry parameter X .In Fig. 7 we show such kind of dependence obtained fromresults of Fig. 5.As seen from Fig. 7, one can expect the enhancementof the curvature effects in neutron rich nuclei. The X -dependence of Tolman length ξ can be approximated as ξ ( X ) = ξ + ξ − X . (30)Both parameters ξ and ξ − as well as the surface tensionparameter σ − are rather sensitive to the Skyrme forceparametrization, see Table I. IV. NUCLEAR MATTER EQUATION OFSTATE AND ( A − / , X )-EXPANSIONS FORFINITE NUCLEI Bellow we will consider the relation of the nuclearmacroscopic characteristics (surface and symmetry ener-gies, Tolman length, incompressibility, etc.) to the bulkproperties of nuclear matter. Assuming a small devia-tions from the equilibrium, the equation of state (EOS)for an asymmetric nuclear matter can be written in theform expansion around the saturation point. One has forthe energy per particle (at zero temperature) E ( ǫ, x ) = E ∞ A = µ ∞ + K ∞ ǫ + b ∞ x + . . . , (31)where ǫ = ρ − ρ ∞ ρ ∞ , x = ρ − ρ , ρ = ρ n + ρ p , ρ − = ρ n − ρ p ,ρ ∞ is the matter saturation (equilibrium) density, µ ∞ isthe chemical potential, K ∞ is the nuclear matter incom-pressibility and b ∞ is the symmetry energy coefficient (all values are taken at the saturation point ǫ = 0 and x = 0). Coefficients of expansion (31) are determinedthrough the derivatives of the energy per particle E ( ǫ, x )at the saturation point: µ ∞ = E ∞ A (cid:12)(cid:12)(cid:12)(cid:12) ρ = ρ ∞ , x =0 ≡ E (0 , ,K ∞ = 9 ρ ∂ E ∞ /A∂ρ (cid:12)(cid:12)(cid:12)(cid:12) ρ = ρ ∞ , x =0 ≡ ρ ∞ E (2 , , (32) b ∞ = 12 ∂ E ∞ /A∂x (cid:12)(cid:12)(cid:12)(cid:12) ρ = ρ ∞ , x =0 ≡ E (0 , . (33)We use the short notation E ( n,m ) ≡ ∂ n + m E ∂ǫ n ∂x m (cid:12)(cid:12)(cid:12)(cid:12) ǫ =0 , x =0 . Some coefficients E ( n,m ) are vanishing. From the condi-tion of minimum of E ( ǫ, x ) at the saturation point onehas E (1 , = E (0 , = 0. Odd derivatives with respectto x , i.e., E ( n,m ) for odd m , also vanish because of thecharge symmetry of nuclear forces.Using E ( ǫ, x ), one can also evaluate chemical potentials µ , µ − and pressure P of the nuclear matter beyond thesaturation point. Namely, µ ( ǫ, x ) = ∂E ∞ ∂A (cid:12)(cid:12)(cid:12)(cid:12) A − ,V = ∂∂ǫ (1 + ǫ ) E − x ∂ E ∂x ,µ − ( ǫ, x ) = ∂E ∞ ∂A − (cid:12)(cid:12)(cid:12)(cid:12) A,V = ∂ E ∂x , (34) P ( ǫ, x ) = − ∂E ∞ ∂V (cid:12)(cid:12)(cid:12)(cid:12) A,A − = ρ ∞ (1 + ǫ ) ∂ E ∂ǫ . (35)Similarly to Eq. (31), in a finite uncharged system theenergy per particle E/A (we use A = N + Z , A − = N − Z , X = A − /A ) of the finite droplet is usually presented as( A − / , X )-expansion around infinite matter using theleptodermous approximation E ≡ E ( X, A − / ) = a V + X b V + A − / ( a S + X b S + a c A − / + X b c A − / ) (36)= a V + a S A − / + a c A − / + X ( b V + b S A − / + b c A − / ) (37)where a V , a S and a c are, respectively, the volume, sur-face and curvature energy coefficients, b V , b S and b c are, TABLE I. Nuclear bulk parameters for different Skyrmeforces. SkM SkM* SLy230b T6 µ ∞ (MeV) -15.77 -15.77 -15.97 -15.96 ρ ∞ (fm − ) 0.1603 0.1603 0.1595 0.1609 K ∞ (MeV) 216.6 216.6 229.9 235.9 K (MeV) 913.5 913.5 1016. 1032. K sym (MeV) -148.8 -155.9 -119.7 -211.5 b ∞ (MeV) 30.75 30.03 32.01 29.97 L ∞ (MeV) 49.34 45.78 45.97 30.86 σ (MeV · fm − ) 0.9176 0.9601 1.006 1.021 ξ (fm) -0.3565 -0.3703 -0.3677 -0.3593 σ − (MeV · fm − ) -3.118 -3.094 -3.131 -2.413 ξ − (fm) -5.373 -5.163 -4.590 -2.944 respectively, the volume, surface and curvature symme-try coefficients. The nuclear chemical potentials λ and λ − are derived as λ ( X, A − / ) = E/A − ∂ E/A∂A − / − X ∂ E/A∂X ,λ − ( X, A − / ) = ∂ E/A∂X . (38)Following Gibbs-Tolman method, one can derive the ac-tual nuclear matter densities ρ and ρ − from the condi-tions µ ( ǫ, x ) = λ ( X, A − / ) ,µ − ( ǫ, x ) = λ − ( X, A − / ) . (39)Using Eq. (39), one can establish the relation of themacroscopic energy coefficients in the liquid drop modelexpansion Eq. (36) to the nuclear matter parameters inEOS (31), see Eqs. (A4) – (A9) of Appendix A. The re-sults of numerical calculations of relevant quantities arerepresented in Tables I and II.The value of the Tolman length ξ can be related tothe nuclear matter incompressibility K ∞ and the surfacetension coefficient σ [10]. Let us consider the expansionlike (20) around the equilibrium state of the symmetricnuclear matter for the bulk density and the chemical po-tential: ̺ V = ρ ∞ + ρ r R e + ρ (cid:18) r R e (cid:19) + . . . ,λ = λ ∞ + λ r R e + λ (cid:18) r R e (cid:19) + . . . , (40) TABLE II. Mass formula coefficients for finite nuclei.SkM SkM* SLy230b T6 a V (MeV) -15.8 -15.8 -16.0 -16.0 a S (MeV) 15.0 15.7 16.5 16.7 a c (MeV) 7.30 7.92 8.26 8.16 b V (MeV) 30.8 30.0 32.0 30.0 b S (MeV) -44.2 -44.1 -44.9 -35.1 b c (MeV) 35.7 35.1 28.6 17.3 r S/V = | b S /b V | where λ ∞ ≡ µ ∞ is the equilibrium chemical potential forthe infinite nuclear matter. We will apply the Gibbs –Duhem relation dP V = ̺ V dλ . (41)Using the generalized Laplace equation (13) andEqs. (19) and (40), we rewrite Eq. (41) as d (cid:18) σ ∞ R e − σ ∞ ξR e + . . . (cid:19) = (cid:18) ρ ∞ + ρ r R e + ρ r R e + . . . (cid:19) × d (cid:18) λ ∞ + λ r R e + λ r R e + . . . (cid:19) . (42)Nuclear incompressibility K ∞ in terms of expansion (40)reads K ∞ = 9 ∂P V ∂̺ V (cid:12)(cid:12)(cid:12)(cid:12) ̺ V = ρ ∞ = 9 ̺ V ∂λ∂̺ V (cid:12)(cid:12)(cid:12)(cid:12) ̺ V = ρ ∞ = 9 ρ ∞ λ ρ . (43)Equating in (42) the terms of the same order in curva-ture R − e and taking the incompressibility definition fromEq. (43), one obtains the following relations ρ = 18 σ ∞ K ∞ r , λ = 2 σ ∞ ρ ∞ r (44)and ξ = − σ ∞ K ∞ ρ ∞ − λ λ r . (45)Equation (45) gives an idea how the Tolman length ξ depends on the incompressibility K ∞ and the surfacetension coefficient σ . In particular, if the second order ∼ R − e correction in the chemical potential λ of Eq. (40)is negligible, namely, λ = λ ∞ + λ r R e , we obtain from Eq. (45) the following important relation ξ ≈ − σ ∞ K ∞ ρ ∞ . (46)That means that the Tolman length disappears in thecase of incompressible Fermi liquid with K ∞ → ∞ . Wenote also the relation of the surface tension coefficient σ to the incompressibility K ∞ and the diffuseness param-eter a of the nuclear surface layer [11] σ ∞ ≈ K ∞ ρ ∞ a . (47)Comparing Eqs. (46) and (47) we conclude that ξ ≈ − a/ . This result leads to the conclusions that the nuclear Tol-man length is negative and the non-zero value of ξ re-quires the finite diffuse layer. V. CONCLUSIONS
Considering a small two-component, charged dropletwith a finite diffuse layer, we have introduced a formaldividing surface of radius R which splits the droplet ontovolume and surface parts. The corresponding splittingwas also done for the binding energy E . Assuming thatthe dividing surface is located close to the interface, weare then able to derive the surface energy E S . In general,the surface energy E S includes the contributions fromthe surface tension σ and from the binding energy of A S particles located within the surface layer. The equimolarsurface and thereby the actual physical size of the dropletare derived by the condition ̺ S λ + ̺ − , S λ − = 0 whichmeans that the latter contribution is excluded from thesurface energy providing E S ∝ σ .In a small nucleus, the diffuse layer and the curvedinterface affect the surface properties significantly. Inagreement with Gibbs-Tolman concept [1, 2], two differ-ent radii have to be introduced in this case. The firstradius, R s , is the surface tension radius (Laplace radius)which provides the minimum of the surface tension coeffi-cient σ and the fulfillment of the Laplace relation (18) forcapillary pressure. The another one, R e , is the equimo-lar radius which corresponds to the equimolar dividingsurface due to the condition (15) and defines the physi-cal size of the sharp surface droplet, i.e., the surface atwhich the surface tension is applied. The difference oftwo radii R e − R s in an asymptotic limit of large sys-tem A → ∞ derives the Tolman length ξ . That meansthe presence of curved surface is not sufficient for thepresence of the curvature correction in the surface ten-sion. The finite diffuse layer in the particle distributionis also required. We point out that the Gibbs-Tolmantheory allows to treat a liquid drop within thermody-namics with minimum assumptions. Once the bindingenergy and chemical potential of the nucleus are knownits equimolar radius, radius of tension and surface energy can be evaluated using the equation of state for the in-finite nuclear matter. For a symmetric liquid the valueof Tolman length is about of half of the diffuseness pa-rameter a for the nuclear surface layer. We have alsoestablished the relation of the macroscopic energy coeffi-cients in the liquid drop model expansion Eq. (36) to thenuclear matter parameters.The sign and the magnitude of the Tolman length ξ depend on the interparticle interaction. We have shownthat the Tolman length is negative for a nuclear Fermiliquid drop. As a consequence, the curvature correctionto the surface tension leads to the hindrance of the yieldof light fragments at the nuclear multifragmentation inheavy ion collisions. We have also shown that the Tolmanlength is sensitive to the neutron excess and its absolutevalue growth significantly with growing asymmetry pa-rameter X . Appendix A: Relation of nuclear matter EOS to thecharacteristics of finite nuclei
We will start from the nuclear matter EOS given byEq. (31) and take into consideration the relations (32)and (33) and the following higher order coefficients K = 6 K ∞ + 27 ρ ∂ E ∞ /A∂ρ (cid:12)(cid:12)(cid:12)(cid:12) ρ = ρ ∞ , x =0 ,L ∞ = 32 ρ ∂ E ∞ /A∂ρ∂x (cid:12)(cid:12)(cid:12)(cid:12) ρ = ρ ∞ , x =0 , (A1) K sym = 92 ρ ∂ E ∞ /A∂ρ ∂x (cid:12)(cid:12)(cid:12)(cid:12) ρ = ρ ∞ , x =0 , (A2)for the expansion (31). Here K is the bulk anharmonic-ity coefficient, L ∞ is the density-symmetry coefficient(symmetry energy slope parameter), K sym is the sym-metry energy curvature parameter. Using (19), we writealso σ ≈ σ ∞ (1 − ξ/R e ) ,σ ∞ ≈ σ + σ − X , ξ ≈ ξ + ξ − X (A3)and a V = µ ∞ , b V = b ∞ . (A4)Using the conditions (39) for the chemical potentialsand both relations (38) and (34), we obtain ρ − ρ ∞ ρ ∞ ≈ A − / a S K ∞ + X (cid:20) − L ∞ K ∞ + A − / (cid:26) b S − a S L ∞ /K ∞ ) K ∞ (cid:18) − L ∞ b ∞ (cid:19) − a S K ∞ (cid:20) L ∞ (cid:18) − K K ∞ (cid:19) + K sym (cid:21)(cid:27)(cid:21) and a S = 4 πr σ , b S = 4 πr (cid:18) σ − + 2 L ∞ K ∞ σ (cid:19) , a c = − πr σ (cid:18) ξ + 3 σ K ∞ ρ ∞ (cid:19) , (A5) b c = − πr σ (cid:26) ξ − + (cid:18) L ∞ K ∞ + σ − σ (cid:19) ξ + 3 σ K ∞ ρ ∞ (cid:20) L ∞ K ∞ (cid:18) K K ∞ (cid:19) − K sym K ∞ (cid:21) + 3 σ − K ∞ ρ ∞ (cid:18) K ∞ σ − b ∞ σ (cid:19)(cid:27) . (A6)Here we have assumed A − / ≪
1. The equimolar, R e , and Laplace, R s , radii defined by Eqs. (17) and (18) read R e ≈ r A / (cid:20) − A − / πr σ K ∞ + X (cid:20) L ∞ K ∞ − A − / (cid:26) πr σ − K ∞ (cid:18) − L ∞ b ∞ + K ∞ µ ∞ (cid:19) + 8 πr σ K ∞ (cid:20) L ∞ K ∞ (cid:18) K K ∞ (cid:19) − K sym K ∞ (cid:21)(cid:27)(cid:21)(cid:21) , (A7) R s ≈ r A / (cid:20) − A − / (cid:18) ξ r + 8 πr σ K ∞ (cid:19) + X (cid:20) L ∞ K ∞ − A − / (cid:26) ξ − r + 8 πr σ − K ∞ (cid:18) K ∞ b ∞ σ − σ (cid:19) + 8 πr σ K ∞ (cid:20) L ∞ K ∞ (cid:18) K K ∞ (cid:19) − K sym K ∞ (cid:21)(cid:27)(cid:21)(cid:21) . (A8)Using the derivations of R e and R s , one obtains R e − R s ≈ ξ + (cid:20) ξ − + 3 σ − b ∞ ρ ∞ (cid:18) σ − σ + 2 L ∞ K ∞ − b ∞ µ ∞ (cid:19)(cid:21) X = ξ + (cid:20) σ − b ∞ ρ ∞ (cid:18) σ − σ + 2 L ∞ K ∞ − b ∞ µ ∞ (cid:19)(cid:21) X . (A9)To describe separately the neutron and proton density distributions we introduce the neutron radius, R n , and theproton radius, R p , as the dividing radii with zero value for the corresponding surface densities ̺ n, S = ( ̺ S + ̺ − , S ) / ̺ p, S = ( ̺ S − ̺ − , S ) / ̺ n, S | R = R n = 0 , ̺ p, S | R = R p = 0 . The value of neutron skin r np = R n − R p is then written as r np = R n − R p ≈ X (cid:20) − σ − b ∞ ρ ∞ + A − / (cid:26) πr σ b ∞ (cid:18) ξ − + ξ σ − σ (cid:19) + 4 πr σ − (cid:20) σ − b ∞ ρ ∞ + 4 σ b ∞ ρ ∞ (cid:18) L ∞ K ∞ + 3 b ∞ K ∞ (cid:19)(cid:21)(cid:27)(cid:21) . (A10) [1] J.W. Gibbs, Influence of Surfaces of Discontinuity uponthe Equilibrium of Heterogeneous Masses. – Theory ofCapillarity.: in The Collected Works, Vol. I (Longmans,Green and Co., New York, 1928), p. 219.[2] R.C. Tolman, J. Chem. Phys. , 118, 333 (1949).[3] V.M. Kolomietz, S.V. Lukyanov and A.I. Sanzhur, Phys.Rev. C 86 (2012) 024304 [4] L.D. Landau and E.M. Lifshitz, Statistical Physics (Perg-amon Press, Oxford, 1958).[5] J.S. Rowlinson and B. Widom,
Molecular Theory of Cap-illarity (Clarendon Press, Oxford, 1982).[6] V.M. Kolomietz and A.I. Sanzhur, Eur. Phys. J. , 345(2008).[7] P. Danielewicz, Nucl. Phys. A727 (2003) 233. [8] V.M. Kolomietz and A.I. Sanzhur, Phys. Rev. C (2010) 024324.[9] W. Satu la, R.A. Wyss and M. Rafalski, Phys. Rev. C , 011301(R) (2006). [10] E.M. Blokhuis, J. Kuipers, J. Chem. Phys.124