Density dependence of the nuclear symmetry energy estimated from neutron skin thickness in finite nuclei
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1 Density dependence of the nuclear symmetry energy estimated from neutron skin thickness in finite nuclei X. Viñas a M. Centelles a M. Warda a,b X. Roca-Maza a,c a Departament d Estructura i Constituents de la Matèria and Institut de Ciències del Cosmos, Universitat de Barcelona, Barcelona, Spain b Katedra Fizyki Teoretycznej, Uniwersytet Marii Curie-Sklodowskiej, Poland c INFN sezione di Milano. Via Celoria 16, Milano, Italy M. Centelles, X. Roca-Maza, X. Viñas and M. Warda, Phys. Rev. Lett (2009) M. Warda, X. Viñas, X. Roca-Maza and M. Centelles, Phys. Rev. C (2009) M. Warda, X. Viñas, X. Roca-Maza and M. Centelles, Phys. Rev. C (2010)
2 Why is important the nuclear symmetry energy? The nuclear symmetry energy is a fundamental quantity in Nuclear Physics and Astrophysics because it governs, at the same time, important properties of very small entities like the atomic nucleus ( R m ) and very large objects as neutron stars ( R 10 4 m ) Nuclear Physics: Neutron skin thickness in finite nuclei, stable nuclei, Heavy-Ion collisions, Giant Resonances... High-Energy Physics: Test of the Standard Model through atomic parity non-conservation observables. Astrophysics: Supernova explosion, Neutron emission and cooling of protoneutron stars, Mass-Radius relations in neutron stars, Composition of the crust of neutron stars...
3 Equation of State in asymmetric matter e(ρ, δ) = e(ρ, 0) + c sym (ρ)δ 2 + O(δ 4 ) ( δ = ρ ) n ρ p ρ Around the saturation density we can write e(ρ, 0) a v K v ɛ 2 and c sym (ρ) J Lɛ+ 1 ( 2 K symɛ 2 ɛ = ρ ) 0 ρ 3ρ 0 ρ fm 3, a v 16MeV, K v 230MeV, J 32MeV However, the values of L = 3ρ c sym (ρ)/ ρ ρ0 and K sym = 9ρ 2 2 c sym (ρ)/ ρ 2 ρ0 which govern the density dependence of c sym near ρ 0 are less certain and predictions vary largely among nuclear theories.
4 Experimental constraints Recent reseach in heavy-ion collisions at intermediate energy is consistent with c sym (ρ) = c sym (ρ 0 ).(ρ/ρ 0 ) γ at ρ < ρ 0. Isospin difussion γ = (L = 88 ± 25 MeV). Isoscaling γ = 0.69 (L 65 MeV) Inferred from nucleon emision ratios γ = 0.5(L 55 MeV). The GDR of 208 Pb analyzed with Skyrme forces suggests a constraint c sym (0.1 fm 3 )= MeV (γ ). The study of the PDR in 68 Ni and 132 Sn predicts L=49-80 MeV. The Thomas-Fermi model of Myers and Swiatecki fitted very precisely to binding energies of 1654 nuclei predicts an EOS that yields γ = 0.51 NEUTRON SKIN THICKNESS?
5 Neutron skin thickness
6 What is experimentally know about neutron skin thickness in nuclei? The neutron skin thickness is defined as S = r 2 n 1/2 r 2 p 1/2, where r 2 n 1/2 and r 2 p 1/2 are the rms radii of the neutron and proton distributions respectively. r 2 p 1/2 is known very accurately from elastic electron scattering measurements (e.g. r ch ( 208 Pb) = ± fm [Angeli (2004)]). r 2 n 1/2 has been obtained with hadronic probes such as: (a) Proton-nucleus elastic scattering (5.522fm < r n ( 208 Pb) < fm [Clark (2003)]). (b) Inelastic scattering excitation of the giant dipole and spin-dipole resonances (r n ( 208 Pb) = 5.67 ± 0.07 fm [Krasznahorkay (1990)]). (c) Antiprotonic atoms: Data from antiprotonic X rays and radiochemical analysis of the yields after the antiproton annihilation (r n ( 208 Pb) = 5.66 ± 0.02 fm) [Trzcińska (2001)].
7 0.2 S (fm) I S = (0.9 ± 0.15)I + ( 0.03 ± 0.02) fm A. Trzcińska et al, Phys. Rev. Lett. 87, (2001) CAN S OF 26 STABLE NUCLEI, FROM 40 Ca TO 238 U, ESTIMATED USING ANTIPROTONIC ATOMS DATA HELP IN CONSTRAINING THE SLOPE AND CURVATURE OF c sym?
8 Symmetry energy and neutron skin thickness in the Liquid Drop Model Symmetry Energy where a sym (A) = J, x A = 9J 1 + x A 4Q A 1/3 E sym (A) = a sym (A)(I + x A I C ) 2 A I = (N Z)/A, I C = e 2 Z/(20JR), R = r 0 A 1/3. Neutron skin thickness S = [ 3/5 t e 2 Z/(70J) + 5 ] 2R (b2 n bp) 2 where t = 3r 0 2 J/Q 1 + x A (I I C )
9 Neutron skin thickness t = 2r 0 3J [J a sym(a)] A 1/3 (I I C )
10 Table: Value of a sym (A) and density ρ that fulfils c sym (ρ) = a sym (A) for A = 208, 116 and 40 in MF models. J and a sym are in MeV and ρ is in fm 3. A = 208 A = 116 A = 40 Model J a sym ρ a sym ρ a sym ρ NL NL-SH FSUGold TF-MS SLy SkX SkM* SIII SGII
11 The c sym (ρ)-a sym (A) correlation There is a genuine relation between the symmetry energy coefficients of the EOS and of nuclei: c sym (ρ) equals a sym (A) of heavy nuclei like 208 Pb at a density ρ = 0.1 ± 0.01 fm 3. A similar situation occurs down to medium mass numbers, at lower densities. We find that this density can be very well simulated by ρ ρ A = ρ 0 ρ 0 /(1 + ca 1/3 ), where c is fixed by the condition ρ 208 = 0.1 fm 3. Using the equality c sym (ρ) = a sym (A) and the LDM, the neutron skin thickness can be finally written as: 3 2r 0 L ( t = 1 ɛ K sym )ɛa 1/3( ) I I C 5 3 J 2L
12 Neutron skin thickness t = 3 2r 0 L ( 1 ɛ K sym )ɛa 1/3( ) I I C 5 3 J 2L
13 Fitting procedure and results We optimize 3 2r 0 L ( t = 1 ɛ K sym )ɛa 1/3( ) I I C 5 3 J 2L using c sym = 31.6( ρ ρ 0 ) γ MeV, ɛ = 1 3(1 + ca 1/3 ), ρ 0 = 0.16fm 3 and taking as experimental baseline the neutron skins measured in 26 antiprotonic atoms. We predict (b n b p ): L = 75 ± 25 MeV
14 S (fm) experiment linear average of experiment Our fit Fe Ca Ni Fe Cd 28 Ni 56 Fe Co Zr Sn Ni Sn Te Te Cd 50 Sn Te 90 Th Te Te Zr Ca Sn I = (N Ζ) / Α Bi Pb S = (0.9 ± 0.15)I + ( 0.03 ± 0.02) fm A. Trzcińska et al, Phys. Rev. Lett. 87, (2001) U
15 Influence of the surface width (b n b p ) S = [ 3/5 t e 2 Z/(70J) + 5 ] 2R (b2 n bp 2 ) b n and b p are obtained at the ETF level.
16 Surface contribution to the neutron skin thickness R (b2 n bp) 2 = σ sw I = (0.3 J Q + c)i
17 Fit and results c = 0.07 c = EXP 0.9 I R np (fm) I J Q =
18 Neutron skin thickness R np (fm) in 208 Pb NL3Λ ν1 NL3Λ ν2 TM1 NL3 NL-SH FSUGold NL3Λ ν3 NL1 SkM* SkX SLy4 SkP T6 SGII D1S SIII 0.1 SVI J / Q NL3Λ ν1 NL3Λ ν2 FSUGold NL3Λ ν3 NL3 TM1 SkM* SkX SLy4 SkP T6 SGII D1S SIII SVI NL1 NL-SH L (MeV) NL3Λ ν1 SVI TM1 NL3 NL-SH NL3Λ ν2 FSUGold SLy4 NL3Λ ν3 SkX SkM* SGII SkP T6 D1S SIII NL J / Q L (MeV) L = MeV
19 Constraints on the slope of the symmetry energy
20 Summary and Conclusions We have described a generic relation between the symmetry energy in finite nuclei and in nuclear matter at subsaturation. We take advantage of this relation to explore constraints on c sym (ρ) from neutron skins measured in antiprotonic atoms. These constraints points towards a soft symmetry energy. We discuss the L values constrained by neutron skins in comparison with most recent observations from reactions and giant resonances. We learn that in spite of present error bars in the data of antiprotonic atoms, the size of the final uncertainties in L is comparable to the other analyses. The generic relation between the symmetry energy in finite nuclei and in nuclear matter at subsaturation plausibly encompasses other prime correlations of nuclear observables with the density content of the symmetry energy as e.g. the constrains of c sym (0.1) from the GDR of 208 Pb (L. Trippa et al. Phys. Rev. C77, (R) (2008)).
21 Thank you for your attention
22 Extra material
23 Some technical details The surface stiffness coeficient Q and the surface widths b n and b p are obtained from self-consistent calculations of the neutron and proton density profiles in asymmetric semi-infinite nuclear matter. To this end one has to minimize the total energy per unit area with the constraint of conservation of the number of protons and neutrons with respect to arbitrary variations of the densities. E const S = [ ε(z) µn ρ n (z) µ p ρ p (z) ] dz, where ε(z) is the nuclear energy density functional. In the non-relativistic framework the densities ρ n and ρ p obey the coupled local Euler-Lagrange equations: δε(z) δρ n µ n = 0, δε(z) δρ p µ p = 0. The relative neutron excess δ = (ρ n ρ p )/(ρ n + ρ p ) is a function of the z-coordinate. When z, the densities ρ n and ρ p approach the values of asymmetric uniform nuclear matter in equilibrium with a bulk neutron excess δ 0.
24 From the calculated density profiles one computes: z oq = zρ q(z)dz ρ q(z)dz, From the relation b 2 q = (z z 0q) 2 ρ q(z)dz. ρ q(z)dz t = z 0n z 0p = 3r 0 2 J Q δ 0, one can evaluate Q from the slope of t at δ 0 = 0. The distance t and the surface widths b n and b p in finite nuclei with neutron excess I = (N Z)/A are obtained using δ 0 given by: I + 3 c 1 Z 2 8 Q δ 0 = A 5/ J 4 Q A 1/3
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