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<strong>Theory</strong> <strong>of</strong> nuclear matter <strong>of</strong> neutron stars<strong>and</strong> core collapsing supernovaeA Dissertation PresentedbyYeunhwan LimtoThe <strong>Graduate</strong> Schoolin Partial Fulfillment <strong>of</strong> theRequirements<strong>for</strong> the Degree <strong>of</strong>Doctor <strong>of</strong> Philosophyin<strong>Physics</strong>Stony Brook UniversityJune 2012


Stony Brook UniversityThe <strong>Graduate</strong> SchoolYeunhwan LimWe, the dissertation committee <strong>for</strong> the above c<strong>and</strong>idate <strong>for</strong> the Doctor <strong>of</strong> Philosophy degree,hereby recommend acceptance <strong>of</strong> this dissertation.James M. Lattimer – Dissertation AdvisorPr<strong>of</strong>essor, Department <strong>of</strong> <strong>Physics</strong> <strong>and</strong> AstronomyMichael Zingale – Chairperson <strong>of</strong> DefenseAssociate Pr<strong>of</strong>essor, Department <strong>of</strong> <strong>Physics</strong> <strong>and</strong> AstronomyAlan C. CalderAssistant Pr<strong>of</strong>essor, Department <strong>of</strong> <strong>Physics</strong> <strong>and</strong> AstronomyDominik SchnebleAssociate Pr<strong>of</strong>essor, Department <strong>of</strong> <strong>Physics</strong> <strong>and</strong> AstronomyDaniel M. DavisPr<strong>of</strong>essor, Department <strong>of</strong> GeosciencesDepartment <strong>of</strong> Geosciences, SUNY Stony BrookThis dissertation is accepted by the <strong>Graduate</strong> School.Lawrence Martinii


Abstract <strong>of</strong> the Dissertation<strong>Theory</strong> <strong>of</strong> nuclear matter <strong>of</strong> neutron stars <strong>and</strong> corecollapsing supernovaebyYeunhwan LimDoctor <strong>of</strong> Philosophyin<strong>Physics</strong>Stony Brook University2012<strong>Nuclear</strong> astrophysics is essential to microphysics <strong>for</strong> the complex hydrodynamicssimulation <strong>of</strong> numerical supernovae explosions <strong>and</strong> neutron star merger calculations.Because many aspects <strong>of</strong> equation <strong>of</strong> state (hereafter, EOS) includingsymmetry <strong>and</strong> thermal properties are uncertain <strong>and</strong> not well constrained by experiments,it is important to develop EOS with easily adjustable parameters.The purpose <strong>of</strong> this thesis is to develop the nuclear matter theory <strong>and</strong> an EOScode <strong>for</strong> hot dense matter. This thesis has two major parts. In the first part, wedevelop a Finite-Range Thomas Fermi (hereafter, FRTF) model <strong>for</strong> supernovae<strong>and</strong> neutron star matter based on the nuclear model <strong>of</strong> Seyler <strong>and</strong> Blanchard,<strong>and</strong> Myers <strong>and</strong> Swiatecki. The nuclear model is extended to finite temperature<strong>and</strong> a Wigner-Seitz geometry to model dense matter. We also extend the modelto include additional density dependent interactions to better fit known nuclearincompressibilities, pure neutron matter, <strong>and</strong> the nuclear optical potential.Using our model, we evaluate nuclear surface properties using a semi-infiniteinterface. The coexistence curve <strong>of</strong> nuclear matter <strong>for</strong> two-phase equilibriumis calculated. Furthermore we calculate energy, radii, <strong>and</strong> surface thickness <strong>of</strong>closed shell nuclei inwhich thespin-orbit interactions canbeneglected. To getanoptimized parameter set <strong>for</strong> FRTF, we explore the allowed ranges <strong>of</strong> symmetryiii


energy <strong>and</strong> the density derivative <strong>of</strong> symmetry energy. We summarize recent experimentalresults, astrophysical inference, <strong>and</strong> theoretical pure neutron mattercalculations. The correlation between symmetry energy <strong>and</strong> the surface symmetryenergy in liquid droplet model is also obtained. The beta equilibrium matteris used to model the neutron star crust.The second part <strong>of</strong> the thesis is devoted to construction <strong>of</strong> a code to computethe nuclear EOS <strong>for</strong> hot dense matter that would be distributed to astrophysicscommunity. With this code, users will be able to generate tables with adjustableparameters describing the symmetry, incompressibility, <strong>and</strong> thermal properties<strong>of</strong> nuclear matter. We use the liquid droplet approach to generate thermodynamicallyconsistent nuclear EOS. table. Compared to previous attempts, weinclude neutron skin, Coulumb diffusion, <strong>and</strong> Coulomb exchange. In addition,we compute the surface tension as a function <strong>of</strong> proton fraction <strong>and</strong> temperatureconsistently with the bulk energy. For comparison, we generate an EOS tableusing the SLy4 non-relativistic Skyrme <strong>for</strong>ce model. For both FRTF <strong>and</strong> SLy4,more than 10 % <strong>of</strong> entries <strong>of</strong> EOS tables consists <strong>of</strong> nuclei, alpha particles, <strong>and</strong>nucleons.iv


To Jaenyeong, Jonghyun, <strong>and</strong> Jongbum


ContentsList <strong>of</strong> FiguresList <strong>of</strong> Mathematical <strong>and</strong> Physical SymbolsList <strong>of</strong> TablesAcknowledgementsixxixvxvi1 Introduction 11.1 Organization <strong>of</strong> Chapters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 The Finite-Range Force Model 62.1 Schematic <strong>Nuclear</strong> <strong>for</strong>ce model . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Realistic nuclear <strong>for</strong>ce model . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2.1 Non-Relativistic Potential Model . . . . . . . . . . . . . . . . . . . . 72.2.2 Relativistic Mean Field Model . . . . . . . . . . . . . . . . . . . . . . 92.3 Finite-Range Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3.1 The Interaction Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3.2 Thermodynamic Quantities . . . . . . . . . . . . . . . . . . . . . . . 112.3.3 Properties <strong>of</strong> Uni<strong>for</strong>m <strong>Matter</strong> . . . . . . . . . . . . . . . . . . . . . . 142.3.4 The Zero-Temperature Limit . . . . . . . . . . . . . . . . . . . . . . . 142.3.5 Uni<strong>for</strong>m <strong>Matter</strong> at Zero Temperature . . . . . . . . . . . . . . . . . . 162.4 Two-Phase Equilibrium <strong>of</strong> Bulk <strong>Matter</strong> . . . . . . . . . . . . . . . . . . . . . 192.4.1 Zero-Temperature Two-Phase Equilibrium . . . . . . . . . . . . . . . 202.4.2 Finite-Temperature Symmetric <strong>Matter</strong> . . . . . . . . . . . . . . . . . 212.4.3 Finite-Temperature Asymmetric <strong>Matter</strong> . . . . . . . . . . . . . . . . 222.5 The <strong>Nuclear</strong> Surface in the Semi-Infinite Approximation . . . . . . . . . . . 232.5.1 The Euler Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.6 Isolated Nuclei <strong>and</strong> Nuclei in Dense <strong>Matter</strong> . . . . . . . . . . . . . . . . . . . 322.6.1 Isolated Nuclei . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.6.2 Dense <strong>Matter</strong> <strong>and</strong> the Wigner-Seitz Approximation . . . . . . . . . . 352.7 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 Modified model 393.1 Original Modification to Myers & Swiatecki Model . . . . . . . . . . . . . . . 393.2 Alternate Modification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42vi


3.3 Optimized Parameter Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.4 <strong>Nuclear</strong> Surface Tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494 Gaussian Type Finite Range model 524.1 <strong>Nuclear</strong> Density Functional <strong>Theory</strong> . . . . . . . . . . . . . . . . . . . . . . . 524.1.1 Energy density functional . . . . . . . . . . . . . . . . . . . . . . . . 524.1.2 Effective mass, Potential, <strong>and</strong> Thermodynamic properties . . . . . . . 544.2 Parameters <strong>for</strong> the Gaussian nuclear density functional . . . . . . . . . . . . 554.2.1 Determination <strong>of</strong> the 1+ǫ power . . . . . . . . . . . . . . . . . . . . 584.2.2 The effective range <strong>of</strong> the nuclear <strong>for</strong>ce : r 0 . . . . . . . . . . . . . . . 584.3 <strong>Nuclear</strong> matter <strong>and</strong> nuclei . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.3.1 Specific heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.3.2 Nuclei at T = 0 MeV . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.3.3 Heavy nuclei in the neutron star crust . . . . . . . . . . . . . . . . . 614.4 Phase transition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.4.1 Uni<strong>for</strong>m matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.4.2 Quark matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.5 Astrophysical application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.5.1 Mass-radius relation <strong>of</strong> a cold neutron star . . . . . . . . . . . . . . . 694.5.2 Moment <strong>of</strong> inertia <strong>of</strong> the neutron star . . . . . . . . . . . . . . . . . . 694.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715 <strong>Nuclear</strong> Energy Density Functional <strong>and</strong> <strong>Neutron</strong> <strong>Stars</strong> 725.1 <strong>Nuclear</strong> energy density functional . . . . . . . . . . . . . . . . . . . . . . . . 725.2 Binding energy <strong>of</strong> single nucleus . . . . . . . . . . . . . . . . . . . . . . . . . 745.3 <strong>Neutron</strong> Star Crust . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.3.1 <strong>Neutron</strong> drip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.3.2 Phase transition to uni<strong>for</strong>m nuclear matter . . . . . . . . . . . . . . . 765.3.3 Statistical <strong>for</strong>mulae <strong>for</strong> heavy nuclei . . . . . . . . . . . . . . . . . . . 765.4 Mass-Radius <strong>of</strong> a cold neutron star . . . . . . . . . . . . . . . . . . . . . . . 785.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806 <strong>Nuclear</strong> Equation <strong>of</strong> State <strong>for</strong> Hot Dense <strong>Matter</strong> 816.1 Construction <strong>of</strong> EOS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.2 Liquid Droplet Model as a numerical technique . . . . . . . . . . . . . . . . 826.3 Choice <strong>of</strong> nuclear <strong>for</strong>ce model . . . . . . . . . . . . . . . . . . . . . . . . . . 866.3.1 Pure neutron matter . . . . . . . . . . . . . . . . . . . . . . . . . . . 876.3.2 Maximum mass <strong>of</strong> a cold neutron star . . . . . . . . . . . . . . . . . 876.3.3 Allowed region <strong>of</strong> mass <strong>and</strong> radii <strong>of</strong> neutron stars . . . . . . . . . . . 876.4 Free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 886.4.1 Bulk <strong>Matter</strong> inside <strong>and</strong> outside nuclei . . . . . . . . . . . . . . . . . . 896.4.2 Coulomb energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.4.3 <strong>Nuclear</strong> surface energy . . . . . . . . . . . . . . . . . . . . . . . . . . 906.4.4 <strong>Nuclear</strong> translational energy . . . . . . . . . . . . . . . . . . . . . . . 946.4.5 Alpha particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94vii


6.5 The Combined model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 956.5.1 Equilibrium conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 956.5.2 Determination <strong>of</strong> Coulomb surface shape parameter D . . . . . . . . 996.5.3 Solving the equilibrium equations . . . . . . . . . . . . . . . . . . . . 996.6 Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1006.6.1 Energy density <strong>and</strong> pressure . . . . . . . . . . . . . . . . . . . . . . . 1016.6.2 Phase Boundaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016.6.3 Atomic number in the heavy nuclei . . . . . . . . . . . . . . . . . . . 1016.7 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1027 Conclusions 104A Finite Range Integration 106A.1 Taylor expansion integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 107A.1.1 1D plane parallel nuclear matter . . . . . . . . . . . . . . . . . . . . . 107A.1.2 3D radial symmetric nuclear matter . . . . . . . . . . . . . . . . . . . 110B JEL thermodynamic integration 116C Phase coexistence 121D Numerical Techniques in Heavy Nuclei in Dense <strong>Matter</strong> 124D.1 Non-uni<strong>for</strong>m electron density approximation . . . . . . . . . . . . . . . . . . 125D.2 Uni<strong>for</strong>m electron density approximation . . . . . . . . . . . . . . . . . . . . 125E <strong>Nuclear</strong> Quantities in Non-relativistic Models 127E.1 Mathematical relations between nuclear parameters . . . . . . . . . . . . . . 127Bibliography 130viii


List <strong>of</strong> Figures1.1 Mass <strong>and</strong> radius relation <strong>of</strong> cold neutron stars . . . . . . . . . . . . . . . . . 21.2 Mass <strong>and</strong> radius relation <strong>of</strong> cold neutron stars in EOS . . . . . . . . . . . . . 42.1 The energy per baryon E <strong>and</strong> pressure p . . . . . . . . . . . . . . . . . . . . 162.2 Bulk equilibrium <strong>of</strong> T = 0 asymmetric nuclear matter . . . . . . . . . . . . . 212.3 Pressure isotherms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4 Coexistence curves <strong>for</strong> asymmetric matter . . . . . . . . . . . . . . . . . . . 232.5 Critical temperature as a function <strong>of</strong> proton fraction . . . . . . . . . . . . . . 242.6 The surface density pr<strong>of</strong>ile as a function <strong>of</strong> position . . . . . . . . . . . . . . 252.7 Surface tension <strong>and</strong> 90-10 surface thickness . . . . . . . . . . . . . . . . . . . 262.8 Surface neutron <strong>and</strong> proton density pr<strong>of</strong>iles . . . . . . . . . . . . . . . . . . 282.9 Surface tension <strong>and</strong> neutron skin thickness . . . . . . . . . . . . . . . . . . . 292.10 The surface tension <strong>of</strong> symmetric matter as a function <strong>of</strong> T . . . . . . . . . . 312.11 The surface tension <strong>of</strong> asymmetric matter as a function <strong>of</strong> T . . . . . . . . . 322.12 Contour plots <strong>of</strong> surface tesion <strong>and</strong> R n −R p . . . . . . . . . . . . . . . . . . 332.13 <strong>Neutron</strong> <strong>and</strong> proton density pr<strong>of</strong>iles <strong>for</strong> the nuclei . . . . . . . . . . . . . . . 342.14 <strong>Neutron</strong> <strong>and</strong> proton density pr<strong>of</strong>iles <strong>for</strong> nuclei in dense matter . . . . . . . . 362.15 <strong>Nuclear</strong> properties in dense matter . . . . . . . . . . . . . . . . . . . . . . . 373.1 S v <strong>and</strong> L contour plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.2 Confidence interval <strong>of</strong> S v <strong>and</strong> L . . . . . . . . . . . . . . . . . . . . . . . . . 473.3 Experimentally allowed region <strong>of</strong> S v <strong>and</strong> L . . . . . . . . . . . . . . . . . . . 483.4 Numerical result <strong>and</strong> analytic fitting <strong>of</strong> the critical temperature . . . . . . . 503.5 Surface tension at T = 0 MeV as a given proton fraction . . . . . . . . . . . 514.1 Energy per baryon using Gaussian model . . . . . . . . . . . . . . . . . . . . 584.2 Semi-infinite nuclear matter surface properities . . . . . . . . . . . . . . . . . 594.3 Specific heat <strong>of</strong> uni<strong>for</strong>m nuclear matter . . . . . . . . . . . . . . . . . . . . . 604.4 Density pr<strong>of</strong>iles <strong>of</strong> the close shell nuclei . . . . . . . . . . . . . . . . . . . . . 614.5 Heavy nuclei in dense matter . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.6 Effective mass in the Wigner-Seitz cell . . . . . . . . . . . . . . . . . . . . . 634.7 Phase transition density in neutron star crust. . . . . . . . . . . . . . . . . . 664.8 <strong>Nuclear</strong> <strong>and</strong> quark matter coexistence . . . . . . . . . . . . . . . . . . . . . . 674.9 Mass <strong>and</strong> radius <strong>of</strong> Hybrid <strong>and</strong> <strong>Neutron</strong> stars . . . . . . . . . . . . . . . . . 684.10 Mass-radius <strong>of</strong> cold neutron stars . . . . . . . . . . . . . . . . . . . . . . . . 694.11 Moment <strong>of</strong> inertia <strong>of</strong> a cold neutron star . . . . . . . . . . . . . . . . . . . . 71ix


5.1 Proton density pr<strong>of</strong>ile <strong>of</strong> 208 Pb. . . . . . . . . . . . . . . . . . . . . . . . . . 745.2 The number <strong>of</strong> protons in neutron star crust . . . . . . . . . . . . . . . . . . 795.3 Mass <strong>and</strong> radius relation from EDF . . . . . . . . . . . . . . . . . . . . . . . 806.1 Contour plot <strong>of</strong> S s /S v <strong>and</strong> S v in incompressible liquid droplet model . . . . . 856.2 Contour plot <strong>of</strong> S s /S v <strong>and</strong> S v in compressible liquid droplet model . . . . . . 866.3 Pure neutron matter pressure <strong>of</strong> Skyrme <strong>for</strong>ce model . . . . . . . . . . . . . 876.4 Maximum mass <strong>of</strong> neutron star <strong>of</strong> Skyrme <strong>for</strong>ce model . . . . . . . . . . . . 886.5 Mass <strong>and</strong> radius <strong>of</strong> neutron stars from Skyrme <strong>for</strong>ce model . . . . . . . . . . 896.6 <strong>Nuclear</strong> <strong>for</strong>ce model c<strong>and</strong>idates <strong>for</strong> EOS table . . . . . . . . . . . . . . . . . 906.7 Particles in the Wigner Seitz cell . . . . . . . . . . . . . . . . . . . . . . . . 916.8 D as a function <strong>of</strong> u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1006.9 Phase boundaries when Y p = 0.5 . . . . . . . . . . . . . . . . . . . . . . . . . 1026.10 Atomic number in dense matter, Y e = 0.45, T = 0.726 MeV . . . . . . . . . . 103A.1 The error between analytic <strong>and</strong> numerical calculation . . . . . . . . . . . . . 110A.2 56 Fe binding energy calculation <strong>and</strong> relative error in log scale . . . . . . . . . 115x


Symbols<strong>Nuclear</strong> <strong>Physics</strong> SymbolE energy density (MeV/fm 3 )p pressure (MeV/fm 3 )BKK ′binding energy per baryon at saturation density (MeV)nuclear incompressibility (MeV)density derivative <strong>of</strong> nuclear incompressibility (MeV)ρ baryon number density (fm −3 )ρ 0 nuclear saturation density (fm −3 )Ttemperature (MeV)µ n (µ p ) neutron (proton) chemical potential (MeV)τ n (τ p ) neutron (proton) momentum density (1/fm 5 )H B ,H g ,H C ,H JHamiltonian density contribution from bulk, gradient, Coulomb, <strong>and</strong> spin-orbitcoupling respectivelyt 0 ,t 1 ,t 2 ,t 3 ,x 0 ,x 1 ,x 2 ,x 3coefficients in the Skyrme <strong>for</strong>ce modelQ nn ,Q pp ,Q npdensity gradient interaction coefficientsJ, J n , J p spin-orbit interactionψσnuclear wave functionPauli spin matrixg ω ,g ρ ,g σ coupling constant in the relativistic mean field modelA µphoton fieldxi


ρ,ω µ ,σ ρ, ω, σ field respectivelym ρ ,m ω ,m σmass <strong>of</strong> ρ, ω, <strong>and</strong> σ boson respectivelyB µν ,F µνP oT oWar 12 ,p 12C L,Utensor fieldzero temperature Fermi-momentum (MeV/c)zero temperature kinetic energy (MeV)total interaction energy (MeV)nuclear diffuseness parameterrelative distance <strong>and</strong> relative momentuminteraction energy density functionalα L,U ,β L,U ,σ L,U ,η L,Ucoefficient in the finite range Thomas Fermi model〈g〉Gaussian finite range integralf(r 12 ) Yukawa-like distance function¯ρ, ˆρI 1 ,I 2 ,I 4˜gaverage densitymomentum integralsfinite range integralF 1/2 ,F 3/2 Fermi integralm ∗ tV teffective mass <strong>for</strong> type t nucleonnuclear potential <strong>for</strong> type t nucleonU n (r,p) optical potentialΨ tdegeneracy parameter <strong>for</strong> type t nucleonp pressure (MeV/fm 3 )f tFermi-Dirac distribution function <strong>for</strong> type t nucleonS entropy density (k B /fm 3 )S vsymmetry energy (MeV)L(= 3S ′ v) density derivative <strong>of</strong> symmetry energy (MeV)ω surface tension <strong>of</strong> nuclear matter (MeV/fm 2 )xii


ω δt 90−10QtT cTaylor expansion coefficient in surface tensionsurface thickness <strong>of</strong> 90% -10% nuclear matter densitygradient term in finite range Thomas Fermi modelneutron skin thickness (fm)critical temperature (MeV)p N pure neutron matter pressure (MeV/fm 3 )V 1L,U ,V 2L,U ,V 3L,U ,v 1L,U ,v 2L,u ,v 3L,U ,t 3coefficient <strong>for</strong> Gaussian type nuclear <strong>for</strong>ce modelk fFermi wave numberD nn ,D np ,D ppgradient terms in thermodynamic instabilityνIρ in,outiE sE diffE exE shellS sN s∆metric function in general relativitymomentum <strong>of</strong> inertia <strong>of</strong> rotating neutron stardensity <strong>of</strong> in <strong>and</strong> out <strong>of</strong> nucleus in parameterized density pr<strong>of</strong>ilesurface energy in liquid droplet modelCoulomb diffuseness coefficient in liquid droplet modelCoulomb exchange coefficient in liquid droplet modelnuclear shell effect energy in liquid droplet modelsurface symmetry energynumber <strong>of</strong> neutrons on the surface <strong>of</strong> nuclei in liquid droplet modelnuclear energy gap (MeV)f N ,f o ,f α free energy per baryon contribution from heavy nuclei, outside nucleons, alphaparticlesf bulk ,f Coul ,f surf ,f transfree energy per baryon from bulk, Coulomb interaction, surface, <strong>and</strong> translationrepectivelyν tr NY pskin density <strong>of</strong> type t nucleon in liquid droplet modelradius <strong>of</strong> heavy nuclei in liquid droplet modeltotal proton fraction in liquid droplet modelxiii


x i,oρ αv αB αuproton fraction <strong>of</strong> heavy nuclei, outside nucleonsalpha particle densityvolume <strong>of</strong> alpha particlealpha particle binding energyvolume fraction <strong>of</strong> heavy nuclei to Wigner-Seitz cell in liquid droplet modelh(x,T) surface tension function in liquid droplet modelˆµ µ n −µ ps(u),c(u),D(u),D(u)geometric functions in liquid droplet modelxiv


List <strong>of</strong> Tables1.1 Range <strong>of</strong> Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1 Truncated model parameter set . . . . . . . . . . . . . . . . . . . . . . . . . 193.1 The critical temperature analytic fitting function . . . . . . . . . . . . . . . 493.2 Surface tension analytic fitting function <strong>for</strong> finite range model . . . . . . . . 504.1 Interaction parameter set in Gaussian model . . . . . . . . . . . . . . . . . . 584.2 Numerical <strong>and</strong> experimental result <strong>of</strong> Closed shell nuclei . . . . . . . . . . . 624.3 <strong>Nuclear</strong> properties in the neutron star crust . . . . . . . . . . . . . . . . . . 625.1 Parameters in EKO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735.2 Parameters in SPM & SLy4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 745.3 <strong>Nuclear</strong> properties <strong>of</strong> single nucleus . . . . . . . . . . . . . . . . . . . . . . . 755.4 <strong>Neutron</strong> drip densities <strong>for</strong> all 3 models . . . . . . . . . . . . . . . . . . . . . 765.5 Phase transition points to uni<strong>for</strong>m nuclear matter . . . . . . . . . . . . . . . 765.6 Fitting <strong>for</strong>mulae <strong>for</strong> Wigner-Seitz cell size after the neutron drip . . . . . . . 775.7 Fitting <strong>for</strong>mulae <strong>for</strong> number <strong>of</strong> neutrons after the neutron drip . . . . . . . . 775.8 Fitting <strong>for</strong>mulae <strong>for</strong> number <strong>of</strong> protons after the neutron drip . . . . . . . . 775.9 Fitting <strong>for</strong>mulae <strong>for</strong> be<strong>for</strong>e the neutron drip . . . . . . . . . . . . . . . . . . 786.1 Surface tension analytic fitting function <strong>for</strong> SLy4 . . . . . . . . . . . . . . . 936.2 The critical temperature analytic fitting function <strong>for</strong> SLy4 . . . . . . . . . . 936.3 The chemical potential fitting function . . . . . . . . . . . . . . . . . . . . . 94B.1 Non-relativistic fermion coefficients . . . . . . . . . . . . . . . . . . . . . . . 117B.2 Fermion coefficients p mn <strong>for</strong> M = N = 3; a=0.433 . . . . . . . . . . . . . . . 118E.1 <strong>Nuclear</strong> Properties in Skyrme <strong>for</strong>ce models . . . . . . . . . . . . . . . . . . . 129xv


AcknowledgementsI deeply appreciate my advisor, Pr<strong>of</strong>.James M. Lattimer, that he taught me nuclear astrophysics<strong>and</strong> how I could simplify the complex problems. His door was always open <strong>and</strong> Icould get invaluable advice.I would also like to thank Pr<strong>of</strong>.Alan Calder who always listened to my questions <strong>and</strong> helpedme to get answers. I appreciate Pr<strong>of</strong>.Mike Zingale <strong>for</strong> serving as my committee chairperson<strong>and</strong> trying to help me when I had important things. I also want to thank Pr<strong>of</strong>.DominikSchneble <strong>and</strong> Pr<strong>of</strong>.Daniel M. Davis <strong>for</strong> serving on my thesis committee.My nuclear astrophysics group friends (Dr.Chris Malone, Dr.Aaron Jackson, Brendan,Adam, Rahul, ...) were always there <strong>for</strong> me. I can’t express my appreciation enough.Last but not least, I would like to thank my family, Jaenyeong, Jonghyun, <strong>and</strong> Jongbum.They always made me happy <strong>and</strong> encouraged me to finish my Ph.D.


Chapter 1Introduction<strong>Stars</strong> give us light by burning hydrogen fuel. The simple Einstein’s equation, E = mc 2 isbehind sun’s energy. Hydrogen turns into helium <strong>and</strong> helium burns into carbon, nitrogen,<strong>and</strong> oxygen. Finally, the neutron rich elements iron is <strong>for</strong>med. After an iron core <strong>for</strong>ms, astar begins to collapse since it cannot support itself against gravitational collapse. Duringthis process, which as known as a Type II supernova explosion, a neutron star is <strong>for</strong>med ifthe mass <strong>of</strong> main sequence star is about 8 ∼ 20 M ⊙ . <strong>Stars</strong> with greater than 20 M ⊙ arethought to <strong>for</strong>m black holes. An enormous amount <strong>of</strong> energy (∼ 10 53 erg) is released fromsupernova explosion. The source <strong>of</strong> the energy is gravitational binding energy <strong>and</strong> the most<strong>of</strong> the energy is carried away by neutrinos. Right after core collapse, the initial temperature<strong>of</strong> neutron star is believed to be 10 11 K [38]. <strong>Neutron</strong> stars cool down quickly by neutrinoemission. After the temperature drops below the Fermi temperature, neutron star is cold.The existence <strong>of</strong> neutron stars was first suggested by Walter Baade <strong>and</strong> Fritz Zwicky in 1934[38] only a year after the discovery <strong>of</strong> neutrons by James Chadwick. L<strong>and</strong>au, Bohr, <strong>and</strong>Rosenfeld discussed the idea <strong>of</strong> one giant nuclei in the core <strong>of</strong> a star be<strong>for</strong>e the discovery <strong>of</strong>neutron [67] but their paper was not published.A neutron star has mostly neutrons (∼90%), protons <strong>and</strong> electrons. If the chemical potential<strong>of</strong> electrons is greater than the rest mass <strong>of</strong> kaon, there might be kaon condensation sothe kaon replaces the role <strong>of</strong> electrons <strong>for</strong> charge neutrality. As the density increases, thechemical potential <strong>of</strong> neutrons <strong>and</strong> protons can be greater than the rest mass <strong>of</strong> hyperons,then hyperons (Σ +,−,0 , Λ, Ξ − ) might appear.A neutron star is divided into an inner core, outer core, inner crust <strong>and</strong> outer crust. Theouter crust consists <strong>of</strong> lattice nuclei with free gas <strong>of</strong> electrons. As density increase, the chemicalpotential <strong>of</strong> neutrons becomes greater than zero <strong>and</strong> neutrons drip out <strong>of</strong> heavy nuclei.Thus, the inner crust has a free gas <strong>of</strong> neutrons with lattice nuclei. Between the boundary<strong>of</strong> inner crust <strong>and</strong> outer core, heavy nuclei become exotic. Because <strong>of</strong> high pressure, thespherical nuclei become oblate to minimize the energy, <strong>and</strong> as the density increases more,oblate nuclei merge together to become a cylindrical phase, a cylindrical phase becomes aslab phase, a slab phase turns into a cylindrical hole, <strong>and</strong> a cylindrical hole becomes a sphericalhole, <strong>and</strong> finally a spherical hole can become uni<strong>for</strong>m nuclear matter. The outer corethus does not have any nuclei structure. The inner core, which is more dense than the outercore, is believed to have exotic nuclear matter. Some nuclear physicists argue that theremight be quark matter in the core <strong>of</strong> neutron stars. Many <strong>of</strong> them use the MIT bag model1


to explain the existence <strong>of</strong> quark matter but it’s still an open question. MIT bag modeldescribes the quarks are freely moving in a bag. If the distance between quarks is increases,the <strong>for</strong>ce increases so strong that they cannot get out <strong>of</strong> the bag. To maintain the bag,negative pressure is introduced, which is known as the bag constant (B). Since we cannot doany experiment to reveal the existence <strong>of</strong> quark matter at such high densities on the earthyet, the theoretical prediction <strong>of</strong> quark matter depends on the parameters, especially thebag constant in quark matter.One thing is <strong>for</strong> sure is that the densities <strong>of</strong> neutron stars are extremely high. It is believedthat the central density <strong>of</strong> the core in neutron stars is 4ρ 0 ∼ 10ρ 0 † , where a ρ 0 is the nuclearsaturation density, 3×10 14 g/cm 3 .The typical radius <strong>of</strong> neutron star is around 10km. The mass <strong>of</strong> neutron is about 1.2M ⊙ to2.0 M ⊙ . Since neutron stars are so compact, the surface gravity <strong>of</strong> neutron stars is extremelyhigh, about 2×10 11 times earth gravity. In these conditions, general relativity is essential t<strong>of</strong>ind out the mass-radius relation <strong>of</strong> neutron stars. The Tolman-Oppenhimer-Volkov (TOV)equation solves the internal structure <strong>of</strong> spherically symmetric neutron stars. Thus, thenuclear physics <strong>and</strong> the TOV equation play together to give mass-relation <strong>of</strong> neutron stars.The nuclear physics beyond ρ 0 is not known well <strong>and</strong> the central density is much higher thanρ 0 , various nuclear <strong>for</strong>ce models give different mass-radius relation <strong>of</strong> neutron stars. Fig.Figure 1.1: Mass <strong>and</strong> radius relation <strong>of</strong> cold neutron stars [19]. Each nuclear <strong>for</strong>ce modelshows different mass-radius relation <strong>of</strong> neutron stars.1.1 shows the mass-radius relation <strong>of</strong> cold neutron stars using various nuclear <strong>for</strong>ce models.† ρ 0 ≃ 0.16fm −3 . ρ 0 denotes nuclear saturation density which is central density <strong>of</strong> heavy nuclei. But, ingeneral, it is a nuclear parameter to be determined from st<strong>and</strong>ard nuclear matter properties.2


Since the mass <strong>of</strong> neutron star is dominated by its core <strong>and</strong> the density <strong>of</strong> the core is muchbeyond the nuclear saturation density, the nuclear <strong>for</strong>ce models should be investigated morethan the current level <strong>of</strong> underst<strong>and</strong>ing.In contrast to neutron stars, supernovae explosions <strong>and</strong> neutron star mergers result in extremelyhigh temperature. Thus, we need to know thermodynamic properties <strong>of</strong> nuclearmatter <strong>for</strong> a wide range <strong>of</strong> densities (10 −9 ∼ 1.6/fm 3 ) , temperature (0 ∼ 30 MeV), <strong>and</strong>proton compositions (0 ∼ 0.56). This thesis concerns nuclear physics under such extremeconditions, which are far beyond those achievable by nuclear physics experiments on earth.Hence, a theoretical extrapolation is needed <strong>for</strong> high density, high temperature, <strong>and</strong> low protonfraction. For zero temperature, the extrapolation <strong>of</strong> nuclear properties at high densitydensity ( 10ρ 0 ) can be checked by the mass-radius relation <strong>of</strong> neutron stars [53]. Un<strong>for</strong>tunately,the extrapolation to high temperature cannot be checked with anything fromexperiment or astrophysical observations at this time.The nuclear thermodynamic in<strong>for</strong>mation is called ‘<strong>Nuclear</strong> Equation Of State’ (EOS) <strong>and</strong>provided as a tabulated <strong>for</strong>m because <strong>of</strong> memory constraints. The table should contain freeenergy density, pressure, entropy as a function <strong>of</strong> baryon number density, proton fraction,<strong>and</strong> temperature.For the simulations <strong>of</strong> supernovae explosions, only a few EOS tables available now. The mostfamous one is Lattimer & Swesty [22] (LS) EOS in which they combined non-relativistic potentialmodel with liquid droplet approach. In their EOS they considered phase transitionsfrom three dimensional nuclei to three dimensional bubble. It is difficult in their code toarbitrarily vary nuclear parameters.H. Shen et al. [59] (STOS) built a table using relativistic mean field model (RMF) <strong>and</strong>Thomas Fermi approximation. To per<strong>for</strong>mthe Thomas Fermi approximation, they employedthe parametrized density pr<strong>of</strong>ile method, in which density pr<strong>of</strong>ile follows a mathematicalpolynomial (see chapter 5) so to avoid the numerical difficulty <strong>of</strong> differential equations. Anew version [60] is available now <strong>and</strong> it contains hyperon interactions.G. Shen et al. [61] (SHT) provided a few version <strong>of</strong> tables using RMF parameter sets. Theyemployed the Hartree approximation to find the nuclear density pr<strong>of</strong>ile. It is difficult <strong>and</strong>very time consuming to develop another table using their code. Table 1.1 show the range <strong>of</strong>Table 1.1: Range <strong>of</strong> TablesLS 220 STOS SHTρ(fm −3 ) 10 −6 ∼ 1 (121) 7.58×10 −11 ∼ 6.022 (110) 10 −8 ∼ 1.496 (328)Y p 0.01 ∼ 0.5 (50) 0 ∼ 0.65 (66) 0 ∼ 0.56 (57)T (MeV) 0.3 ∼ 30 (50) 0.1 ∼ 398.1 (90) 0 ∼ 75.0 (109)the independent variables <strong>and</strong> the number <strong>of</strong> grid points (number in the parenthesis). STOStables deals wide range <strong>of</strong> densities <strong>and</strong> temperature. The number <strong>of</strong> grid points, however,is small so that the interpolations from that table might not be suitable <strong>for</strong> the simulation,which needs high thermodynamic consistencies. That is, except <strong>for</strong> the LS EOS, the pressure<strong>and</strong> entropy densities from STOS <strong>and</strong> SHT are obtained from the numerical derivatives(P = −ρ 2∆F,S = ∆ρ −∆F ). But this numerical <strong>for</strong>malism might not give enough accuracy <strong>for</strong>∆T3


the thermodynamic identity, P = µ n ρ n +µ p ρ p −F if the grid spacing is relatively large. Onthe other h<strong>and</strong>, the LS EOS, which used the liquid droplet approach, provides analytic solutions<strong>for</strong> thermodynamic variables. The liquid droplet approach enables us to get algebraicsolutions <strong>for</strong> all thermodynamic derivatives. All three approaches used Wigner-Seitz cellmethod in which one heavy nucleus presents at the center <strong>of</strong> the cell <strong>and</strong> electrons, nucleons,<strong>and</strong> alpha particle exist around the heavy nucleus.Recently, researchers are trying to build EOS using multi-component nuclei. That is, <strong>for</strong>given independent variables (ρ,Y p ,T), several types <strong>of</strong> nuclei are assumed present. Since theelectron capture or neutrino weak process are depends on average mass number (〈A〉), it isassumed that this multi-component nuclei method would give better EOS than the singleheavy nucleus method. Hempel et al. [63, 64] used relativistic mean field models (TM1,TMA, FSU Gold) <strong>and</strong> nuclear statistical equilibrium method to add the alpha, deutron, <strong>and</strong>triton. Blinnikov et al. [65] used Saha equation to find the fraction <strong>of</strong> multi-componentnuclei <strong>and</strong> nuclear mass <strong>for</strong>mula. Furusawa et al. [66] also used Saha equation <strong>and</strong> relativisticmean filed model. They also considered phase transition using geometric function.Except <strong>for</strong> Hempel et al., these EOS tables are under construction. Hempel et al. EOSprovides the fraction <strong>of</strong> alpha, deutron, <strong>and</strong> triton fraction ; however, their fraction are sosmall that their presence does not effect 〈A〉 very much. Furthermore, even though the idea<strong>of</strong> multi-component nuclei may provide better in<strong>for</strong>mation <strong>for</strong> weak interaction, their choice<strong>of</strong> nuclear <strong>for</strong>ce model might not satisfy observations <strong>of</strong> the mass-radius relation <strong>of</strong> neutronstars. The recent discovery <strong>of</strong> a 1.97M ⊙ neutron star (PSR J16142230) <strong>and</strong> the Steiner et al.Figure 1.2: Mass <strong>and</strong> radius relation <strong>of</strong> cold neutron stars in EOS table. [19]. Only LS220can satisfy 1.97M ⊙ neutron star <strong>and</strong> the mass-radius criteria.[53] mass-radius criteria exclude most <strong>of</strong> the tabulated EOSs. Fig. 1.2 shows the mass-radiusrelationinwhich we canfindfromEOStable. Since the Shenet al. use TM1 parameter, theygives the same mass-radius relation. In general, RMFs give larger radii <strong>for</strong> given neutronstar’s mass than the non-relativistic potential models. FSU Gold was invented to reduce theradius <strong>for</strong> a given mass <strong>of</strong> neutron star, but it cannot reproduce 1.97M ⊙ neutron star. As4


shown in Fig. 1.2, LS220 is the only one EOS which satisfy both astrophysical phenomena.Our research starts from inadequacy <strong>of</strong> current EOS to model allowable variation in uncertainnuclear physics. For this end, we investigate the nuclear <strong>for</strong>ce model (Finite Rangemodel) <strong>and</strong> extract the best parameter set which explains nuclear phenomena such as energies<strong>and</strong> radii, as well as pure neutron matter <strong>and</strong> astrophysical data. With the improvednuclear physics model, we build the new EOS tables which can be used <strong>for</strong> proto-neutronstar, supernovae explosions, <strong>and</strong> binary mergers which involves neutron star.We extend the Lattimer-Swesty [22] approach to produce user friendly code to incorporateadditional physics <strong>and</strong> adjustable parameters. User will be able to generate tables with onlya few cpu hours to compute. This code development (Chapter 6) is contained in the secondpart <strong>of</strong> the thesis.In order to calibrate this liquid droplet <strong>for</strong>malism, we employ the FRTF model which permitsthe computation <strong>of</strong> nuclei immersed in dense matter. This is a more sophisticatedapproach but still too computationally intensive to generate a complete table. In addition,it, as well as the tables <strong>of</strong> H.Shen <strong>and</strong> G.Shen, predicts complex behaviors <strong>of</strong> matter nearthe transition between nuclei <strong>and</strong> uni<strong>for</strong>m matter. This complex behavior results from verysmall free energy difference between configurations with different mass numbers. In reality,matter near the transition will be smoothed because <strong>of</strong> the thermal fluctuation. The liquiddroplet approach is more suitable in this region. However, it is necessary to calibrate toliquid droplet model so that it successfully models at low densities where laboratory datais available. Another great advantage <strong>of</strong> making tables using the liquid droplet approachis that it allows the analytic prediction <strong>of</strong> thermodynamically consistent derivatives. Thetable would have the wide range <strong>of</strong> variables such as ρ : 10 −10 ∼ 1.6 fm −3 , T : 0 ∼ 60MeV,Y p : 0 ∼ 0.6 with more grid points than other tables.1.1 Organization <strong>of</strong> ChaptersThis work focuses on developing a Finite Range <strong>for</strong>ce model <strong>and</strong> its application to neutronstar <strong>and</strong> making nuclear equation <strong>of</strong> state (EOS) table. Chapter 2 explains the FiniteRange Thomas Fermi model <strong>of</strong> Yukawa type using a truncated model. Chapter 3 improvesthe truncated model in Chapter 2. Chapter 4 investigates another type <strong>of</strong> Finite Rangemodel - the Gaussian type. Chapter 5 illustrate nuclear physics <strong>and</strong> neutron stars. Finally,Chapter 6 is devoted to making nuclear EOS table. The appendix describes some numericalapproximations that we developed.5


Chapter 2The Finite-Range Force Model† <strong>Nuclear</strong> physics plays an important role in the underst<strong>and</strong>ing <strong>of</strong> astrophysical phenomenasuch as neutron stars <strong>and</strong> supernovae explosions. As described in the introduction, it isreally important to choose the good nuclear <strong>for</strong>ce model to make E.O.S. tables. There are alot <strong>of</strong> kind <strong>of</strong> nuclear <strong>for</strong>ce models, <strong>and</strong> different types <strong>of</strong> nuclear physics model can describethe same or different nuclear phenomena. However, the number <strong>of</strong> models can be used tomake E.O.S. table is limited. We explain some <strong>of</strong> nuclear <strong>for</strong>ce models which is related withmaking E.O.S. table briefly. Our finite-range <strong>for</strong>ce model will then be described after thebrief explanation.2.1 Schematic <strong>Nuclear</strong> <strong>for</strong>ce modelNear the nuclear saturation density at small temperatures, the energy density <strong>of</strong> the uni<strong>for</strong>mnuclear matter can be approximated by [56],[E(ρ,T,x) = ρ −B + K (1− ρ )ρ+S v (1−2x) 2 +a18 ρ o ρ o( ) 2/3 ]ρoT 2ρ(2.1)where B ≃ −16 MeV is the binding energy per baryon, x is the proton fraction, S v is thesymmetry energy, K(≃ 230MeV) is the nuclear incompressibility, <strong>and</strong> a(≃ 1/15MeV −1 ) isthe nuclear level density parameter. Mathematically, this is just a Taylor expansion <strong>for</strong> ρ,x, <strong>and</strong> T at ρ = ρ o , x = 1/2, <strong>and</strong> T = 0 MeV.From thermodynamic derivatives, we can find the pressure, chemical potentials, <strong>and</strong> entropy† This chapter is based on Y. Lim <strong>and</strong> J.M. Lattimer’s work. This part will be submitted to the journal<strong>Nuclear</strong> <strong>Physics</strong> A.6


density:p = ρ2ρ o[ K9µ n = −B + K 18( ρρ o−1)+S v (1−2x) 2 ](1− ρ ρ o)ρˆµ = 4S v (1−2x),ρ o+2S vρρ o(1−2x)− a 3+ 2a ( ) 2/33 ρ ρoT 2 ,ρ( ) 2/3 ρoT 2 ,ρ(2.2)S = 2aρ(ρoρ) 2/3T .These schematic model quantities give reasonable features <strong>of</strong> nuclear matter, such as thepressure vanishes both at ρ o <strong>and</strong> zero density, <strong>and</strong> becomes negative between them. Theproton <strong>and</strong> neutron chemical potentials approach negative infinity in the limit <strong>of</strong> low density.This schematic model can be used <strong>for</strong> a test problem to generate coexistence curve <strong>of</strong> dense<strong>and</strong> dilute matter <strong>and</strong> to find the critical temperature <strong>of</strong> coexistence. This model givesqualitative numbers so we can estimate the range <strong>of</strong> validity <strong>of</strong> realistic models.2.2 Realistic nuclear <strong>for</strong>ce modelEven though the schematic nuclear model can give a good estimation <strong>of</strong> nuclear matter, itcan not be used in the low density (ρ < 0.01fm −3 ) <strong>and</strong> high temperature (T > 5 MeV)region. Instead we need to rely on more sophisticated models which explain both the high<strong>and</strong> low density regions, namely the widely used non-relativistic potential <strong>and</strong> relativisticmean field models <strong>of</strong> the nuclear <strong>for</strong>ce.As pointed out in Steiner et al. [32], the total Hamiltonian density is the sum <strong>of</strong> thekinetic energy density <strong>and</strong> local density dependent interaction energy density. These localproperties simplify the numerical calculation <strong>of</strong> nuclei <strong>and</strong> nuclear matter since we can applythe variational principle without any mathematical or numerical difficulties.Both potential <strong>and</strong> mean field models explain the properties <strong>of</strong> a single nucleus very well.However, we need to be careful when applying those models in high density regions (seesection 6.3).2.2.1 Non-Relativistic Potential ModelThe non-relativistic potential model, which is <strong>of</strong>ten called the ‘Skyrme <strong>for</strong>ce model’, is adensity <strong>and</strong> momentum density functional in a mathematical sense. In this model, theHamiltonian density is composed <strong>of</strong> bulk, gradient, Coulomb energy, <strong>and</strong> spin-orbit couplingcomponents,H = H B +H g +H C +H J . (2.3)7


The bulk part is divided by kinetic energy density, two-body interaction, <strong>and</strong> many bodyinteractions <strong>of</strong> nucleons [32],H B = 2τ n + 2τ p2m n 2m p[t1+ρ(τ n +τ p ) 1+4(x )1+ t (21+ x ) ]22 4 2[t2( 1+(τ n ρ n +τ p ρ p ) 2)4 2 +x − t (1 11) ]4 2 +x+ t [ (01+ x )0 1) ]ρ −( 22 2 2 +x 0 (ρ 2 n +ρ 2 p)+ t [ (31+ x )]3 1ρ −( 212 2 2 +x 3)(ρ 2 n +ρ2 p ) ρ ǫ ,(2.4)where x 0 ,...,t 3 , <strong>and</strong> ǫ are parameters specifying the interaction strengths, which are determinedfrom nuclear data fitting.The density gradient term, which is responsible <strong>for</strong> the surface tension <strong>of</strong> a single nucleus,is given by [32]H g = 1 ][Q nn (∇ρ n ) 2 +2Q np ∇ρ n ∇ρ p +Q pp (∇ρ p ) 2 , (2.5)2where the Q parameters are given byThe Coulomb energy interaction isQ nn = Q pp = 3 [ ]t 1 (1−x 1 )−t 2 (1+x 2 ) ,16Q np = Q pn = 1 ( [3t 1 1+ x (1)−t 2 1+ x )]2.8 2 2H C (r) = e2 ρ p (r)2∫(2.6)( )d 3 r ′ ρ p (r ′ 1/3)|r−r ′ | − 3e2 3ρ p (r ′ ) 4/3 , (2.7)4 πwhere the first term is a classical Coulomb energy interaction, <strong>and</strong> the second term is anexchange Coulomb interaction.The spin-orbit interaction is given byH J =− W 02 (ρ n∇·J n +ρ p ∇·J p +ρ∇·J)+ t 116 (J2 n +J2 p −x 1J 2 )− t 216 (J2 n +J2 p +x 2J 2 ),(2.8)where J n = ∑ n,i ψ† i σ ×∇ψ i, J = J n +J p . There are more than 100 parameter set in thisSkyrme <strong>for</strong>ce model. Combined with the Hartree-Fock † approximation, this Skyrme <strong>for</strong>ce† In the Hartree-Fock or Hartree Approximation, the trial wave function is assumed as initial guess <strong>and</strong>keep iterating until E N+1 − E N < δ, where δ is a tolerance <strong>for</strong> energy difference. This is a mathematical8


nuclear energy density functional is quite successful in demonstrating the binding energy,root mean square radius, <strong>and</strong> charged radius <strong>of</strong> a single nucleus. This <strong>for</strong>ce model, however,should be chosen carefully when studying high density ( ρ 0 ) region (see section 6.3).2.2.2 Relativistic Mean Field ModelAt high density, <strong>for</strong> example n = ρ 0 , the Fermi momentum at zero temperature is given asp f c = c(3π 2 ρ 0 ) 1/3 ∼ 330MeV. (2.9)There<strong>for</strong>e, the expansion <strong>of</strong> the total energy √ m 2 c 4 +p 2 c 2 to mc 2 + p2is not valid anymore,2m<strong>and</strong> the relativistic effects need to be considered.The relativistic mean field model explains the origin <strong>of</strong> nuclear <strong>for</strong>ce from ρ, ω, <strong>and</strong> σmeson exchanges. In the mean field model, the Lagrangian <strong>for</strong>malism satisfies relativity <strong>and</strong>causality, <strong>and</strong> is given by [32]L =¯Ψ[i∂/−g ω ρ/·τ −M +g σ σ − 1 ]2 (1+τ)A/ Ψ+ 1 2 (∂ µσ) 2−V(σ)− 1 4 f µνf µν + 1 2 m2 ω ωµ ω µ − 1 4 B µνB µν + 1 2 m2 ρ ρµ ρ µ − 1 4 F µνF µν (2.10)+ ζ 24 g4 ω(ω µ ω µ ) 2 + ξ24 g4 ρ(ρ µ ·ρ µ ) 2 +gρf(σ,ω 2 µ ω µ )ρ µ ·ρ µwhere V(σ) <strong>and</strong> f(σ,ω µ ω µ ) areV(σ) = 1 2 m σσ 2 + κ 6 (g σσ) 3 + λ 6 (g σσ) 4 ,f(σ,ω µ ω µ ) =6∑a i σ i +i=13∑(ω µ ω µ ) j .j=1(2.11)Here, A µ is photon field, f µν = ∂ µ ω ν −∂ ν ω µ , B µν = ∂ µ ρ ν −∂ ν ρ µ , <strong>and</strong> F µν = ∂ µ A ν −∂ ν A µ .Like a non-relativistic potential model, the RMF is useful to study single nucleus’ properties.It seems that the RMF should be used to study nuclear physics at high density region since itisrelativistic <strong>and</strong>thehyperon’s appearancecanbecontrolledbycouplingconstant. However,it does not show the proper mass <strong>and</strong> radius relation <strong>of</strong> neutron stars. There<strong>for</strong>e, we are notfocused on this RMF <strong>for</strong>malism anymore.2.3 Finite-Range Model <strong>and</strong> its Extension to FiniteTemperatureThe interaction between nucleons is short ranged, on the order <strong>of</strong> 1 ∼ 2 fm. To take intoaccount thisshortrangedinteraction, weneedtoaddthedensity dependent interaction<strong>for</strong>m.The π meson exchange model was successful in accounting <strong>for</strong> the nuclear interactions, <strong>and</strong>analogue <strong>of</strong> ‘Cauchy criterion’9


we can get the e−r/r 0interaction <strong>for</strong>m from the π exchange model. This idea was enlargedrby several authors [24, 30]. We re-examine this idea <strong>and</strong> add new interaction terms to reflectthe recent nuclear experiments <strong>and</strong> to explain the mass <strong>and</strong> radius <strong>of</strong> cold neutron stars.2.3.1 The Interaction EnergyWe begin with the truncated version <strong>of</strong> the Myers & Swiatecki [26] finite-range model,which is an extended version <strong>of</strong> the Seyler & Blanchard [30, 31] finite-range model. We willgeneralize this <strong>for</strong>ce to finite temperatures. Initially, we ignore the Coulomb contributions tothe total energy. The symmetric matter saturation density is ρ o , <strong>and</strong> the zero-temperatureFermi momentum P o is (P o /) 3 = 3π 2 ρ o /2. The zero-temperature kinetic energy is T o =P 2 o /2m b where m b is the baryon mass. We will assume the interaction energy isW = − 1 h 3 ∫d 3 r 1∫d 3 r 2 f(r 12 /a) ∑ t[∫∫C L f t1 f t2 d 3 p t1 d 3 p t2 +C U f t1 f t ′ 2d 3 p t1 d 3 p t ′ 2](2.12)where f ti is the Fermi occupation function <strong>for</strong> particle ti † . The position index i = 1 or 2 <strong>and</strong>the isospin index t = n or p. The notation is such that if t = n then t ′ = p <strong>and</strong> vice versa.p ti is the momentum <strong>of</strong> the nucleon at position i <strong>of</strong> species t. The finite-range function istaken to be Yukawa-like,f(r 12 /a) =14πr 12 a 2e−r 12/a , r 12 = |r 1 −r 2 |. (2.13)The finite-range function f is normalized so that ∫ d 3 r 2 f(r 12 /a) = 1 <strong>for</strong> all r 1 . The rangeparameter a will be determined by fitting matter surface pr<strong>of</strong>iles or surface energies tolaboratory values. The quantities C L,U are momentum- <strong>and</strong> density-dependent interactionfunctionals <strong>for</strong> like (L) <strong>and</strong> unlike (U) pairs <strong>of</strong> particles. Myers & Swiatecki chose them,with some changes in notation, to beC L,U (p 1 ,p 2 , ¯ρ) = h34 T 3oρ o (4πPo3) 2 [α L,U −β L,U(p12P o) 2−σ L,U( 2¯ρρ o) 2/3 ](2.14)where α L ,α U ,β L ,β U ,σ L ,σ U are constants. These constants are to be determined by fittingsaturation properties <strong>of</strong> cold nuclear matter <strong>and</strong> other considerations. In Eq. (2.14) , p 12 =|p 1 −p 2 |. The quantity ¯ρ(r) is the mean density, which is chosen to be¯ρ 2/3 = (ρ 2/31 +ρ 2/32 )/2 (2.15)where ρ 1 <strong>and</strong> ρ 2 are the baryon densities at positions r 1 <strong>and</strong> r 2 , respectively.† t indicates proton or neutron <strong>and</strong> i represents position <strong>of</strong> the particle.10


2.3.2 Thermodynamic QuantitiesThe total energy is the sum <strong>of</strong> the free particle kinetic energy <strong>and</strong> the interaction energy,which can be expressed as∫E = d 3 2r 1 (τ n +τ p )− T ∫ ∫oρ od 3 r 1 d 3 r 2 f(r 12 /a)×2m b 4∑[α L I 1 (ρ t1 ,ρ t2 )−β L I 2 (ρ t1 ,τ t1 ,ρ t2 ,τ t2 )−σ L I 4 (ρ t1 ,ρ t2 )+ (2.16)t]α U I 1 (ρ t1 ,ρ t ′ 2)−β U I 2 (ρ t1 ,τ t1 ,ρ t ′ 2,τ t ′ 2)−σ U I 4 (ρ t1 ,ρ t ′ 2) ,where the double momentum integrals I 1 , I 2 <strong>and</strong> I 4 are analytic <strong>and</strong> are expressed as( ) 2 ∫ 3 ∞I 1 (ρ 1 ,ρ 2 ) =4πPo3 0( ) 2 ∫ 3 ∞I 2 (ρ 1 ,τ 1 ,ρ 2 ,τ 2 ) =4πPo3 0( ) 2 2= (τ 1 ρ 2 +ρ 1 τ 2 ),ρ o P oI 4 (ρ 1 ,ρ 2 ) = 1 2= 1 2( 3∫ ∞( ) 2 2f 1 d 3 p 1 f 2 d 3 p 2 = ρ 1 ρ 2 ,ρ o0f 1 d 3 p 1∫ ∞0f 2 d 3 p 2(p12P o) 2( ) 2 2/3( ) ∫) 2 ρ 2/31 +ρ 2/3∞2ρ o4πPo3( ) 8/3 2(ρ 1 ρ 2 ρ 2/31 +ρ 2/32ρ o).0∫ ∞f 1 d 3 p 1 f 2 d 3 p 20(2.17)IntheThomas-Fermi approximation, thenumber <strong>and</strong>kinetic densities have their usual <strong>for</strong>msin finite temperature matter:ρ t = 1 ∫ ∞4π 3 3where the Fermi occupation function is0f t =f t d 3 p; τ t = 14π 3 5 ∫ ∞[1+exp0f t p 2 d 3 p, (2.18)( )] −1 ǫt −µ t. (2.19)The single particle energies ǫ t , which are functions <strong>of</strong> the momentum p t <strong>and</strong> the nucleondensities, are evaluated below. The nucleon chemical potentials are µ t .The total energy, beginning with Eq. (2.38) <strong>and</strong> using the results in Eqs. (2.17) <strong>and</strong>T11


(2.18), becomes∫E =[{∑d 3 2r 1 τ t − T ∫od 3 r 2 f(r 12 /a)×2m b ρ otα L ρ t1 ρ t2 +α U ρ t1 ρ t ′ 2 −β L ′ (ρ t1τ t2 +ρ t2 τ t1 )−β U ′ (ρ t1τ t ′ 2 +ρ t ′ 2τ t1 ) (2.20)( ) ( ) ]}−σ L ′ ρ t1ρ t2 ρ 2/3t1 +ρ 2/3t2 −σ U ′ ρ t1ρ t ′ 2 ρ 2/3t1 +ρ 2/3t ′ 2where we introduce the new constantsβ ′ L,U = β L,U( 23π 2 ρ o) 2/3,σ ′ L,U = σ L,U2( 2ρ o) 2/3.The energy density is then the integr<strong>and</strong> <strong>of</strong> the d 3 r 1 integration in Eq. (2.20):E(r) = ∑ { [ 2τ t − T oρ t (α L˜ρ t +α U˜ρ t ′)−β L ′ 2mt b ρ (τ t˜ρ t +ρ t˜τ t )o( ) ( ) ]}−β U ′ (ρ t˜τ t ′ +τ t˜ρ t ′)−σ L ′ ρ t ˜ρ t ρ 2/3 5/3t + ˜ρ t −σ U ′ ρ t ρ 2/3 5/3t ˜ρ t ′ + ˜ρt, ′(2.21)where we define the finite-range integral as∫˜g(r) ≡ d 3 r ′ f(|r−r ′ |/a)g(r ′ ) (2.22)where the notation changes from r 1 → r <strong>and</strong> r 2 → r ′ . Using the definitions <strong>for</strong> the effectivemasses m ∗ t <strong>and</strong> the potential V t in∫δE = d 3 rδE = ∑ ∫ [d 3 r V t δρ t + 2δτ2m ∗ t], (2.23)ttwe find, after noting that r <strong>and</strong> r ′ can be reversed within double integrals, 22m ∗ t= 22m b+ 2T oρ o(β ′ L˜ρ t +β ′ U˜ρ t ′),V t =− 2T [oρ oα L˜ρ t +α U˜ρ t ′ −β L˜τ ′ t −β U˜τ ′ t ′( ) 53 ρ2/3 t ˜ρ t +−σ ′ L˜ρ5/3t−σ ′ U( 53 ρ2/3 5/3t ˜ρ t ′ + ˜ρt)]. ′(2.24)In addition, we note that at zero temperature, the total potential felt by a neutron at the12


point r <strong>and</strong> with momentum p isU n (r,p) = V n (r)+ 2T oρ o[β L˜ρ n +β U˜ρ p] ( pP o) 2. (2.25)With these explicit relations, one can now employ the methodology <strong>of</strong> Lattimer & Ravenhall[21] to determine thermodynamic quantities. The single-particle energy isǫ t = p22m ∗ tIt follows from Eq. (2.18) that the densities <strong>and</strong> kinetic densities are+V t . (2.26)ρ t = 12π 2 ( 2m∗t T 2 ) 3/2F 1/2 (Ψ t ), τ t = 12π 2 ( 2m∗t T 2 ) 5/2F 3/2 (Ψ t ), (2.27)where Ψ t = (µ t − V t )/T is the degeneracy parameter. The inversion <strong>of</strong> the first <strong>of</strong> theseyields the chemical potential:[ ( ) ]µ t = V t +TF −11/22π 2 2 3/2ρ t . (2.28)2m ∗ tTThe entropy density S t is given by L<strong>and</strong>au’s quasi-particle <strong>for</strong>mula [22]S t = − 2 h 3 ∫[]d 3 p f t lnf t +(1−f t )ln(1−f t ) = 526m ∗ tT τ t − µ t −V tρ t . (2.29)TThe pressure follows from the thermodynamic identityp = ∑ t= ∑ t−β ′ U(µ t ρ t +TS t )−E[ { 2τ t − T o3m b ρ o(ρ t τ˜t ′ + 7 3 ρ˜t ′τ tρ t (α L˜ρ t +α U ρ˜t ′)−β L(ρ ′ t˜τ t + 3˜ρ 7 )tτ t)−σ L ′ ρ t( 73 ρ2/3 t ˜ρ t +) (5/3 7 ˜ρ t −σ U ′ ρ t3 ρ2/3tρ˜t ′ +˜ρ5/3t ′ ) }] .(2.30)The pressure relation can be simplified, upon eliminating the terms involving[ 2 τ t3m ∗ tp = ∑ t˜ρ5/3t , to+ ρ t2 (µ t −TΨ t )+ T oρ o([β ′ L˜ρ t +β ′ U˜ρ t ′]τ t + 2 3 ρ5/3 t [σ ′ L˜ρ t +σ ′ U˜ρ t ′] ) ] . (2.31)13


2.3.3 Properties <strong>of</strong> Uni<strong>for</strong>m <strong>Matter</strong>For the case <strong>of</strong> uni<strong>for</strong>m matter, ˜g = g, the energy density becomesE = ∑ { 2τ2m ∗ t − T () }oρ t α L ρ t +α U ρ t ′ −2σ L ′t t ρ ρ5/3 t −2σ U ′ ρ5/3 t ′o= ∑ { 2τ t − T [ ()] }oρ t α L ρ t +α U ρ t ′ −2 β L ′ 2mt b ρ τ t +β U ′ τ t ′ +ρ2/3 t [σ L ′ ρ t +σ U ′ ρ t ′] .o(2.32)Here, the effective masses are 22m ∗ t= 2+ 2T o(β L ′ 2m b ρ ρ t +β U ′ ρ t ′). (2.33)oNote that the energy density reduces to a Skyrme-like model. The potentials becomeV t = − 2T [oα L ρ t +α U ρ t ′ −β L ′ ρ τ t −β U ′ τ t ′ − 8 ]o 3 σ′ L ρ5/3 t −σ U ′ ρ t ′(5 3 ρ2/3 t +ρ 2/3t) (2.34)′The pressure can then be written asp = ∑ { 2 τ t3m ∗ t t= ∑ { 2 τ t− T [oρ t α L ρ t +α U ρ t ′ − 10 (3mt b ρ o 3− T [oρ t α L ρ t +α U ρ t ′ −2(β L ′ ρ τ t +β U ′ τ 10 (t ′)−o 3 ρ2/3 t σ′L ρ t +σ U ′ ρ ) ]}t ′β ′ Lτ t +β ′ Uτ t ′ +ρ 2/3t[ ] )]}σ′L ρ t +σ Uρ ′ t ′ .(2.35)The chemical potentials<strong>and</strong>entropy density aregiven by Eqs. (2.28) <strong>and</strong>(2.29), respectively.2.3.4 The Zero-Temperature LimitThe zero-temperature limit is needed in order to establish the parameters <strong>of</strong> the <strong>for</strong>ce fromexperimental constraints. In the zero-temperature limit, one finds from limiting expressions<strong>for</strong> the Fermi integrals:ρ t = 1 π 2 ( 2m∗t (µ t −V t ) 2 ) 3/2,τ t = 1 π 2 ( 2m∗t (µ t −V t ) 2 ) 5/2= 3 5 (3π2 ) 2/3 ρ 5/3t .(2.36)Furthermore, <strong>for</strong> the case <strong>of</strong> symmetric nuclear matter (ρ n = ρ p = ρ o /2), one obtains∣ ∣ 2τ n =2m b∣∣∣ρ=ρo,x=1/22τ p =2m b∣∣∣ρ=ρo,x=1/2310 T oρ o . (2.37)14


Thus, in the case <strong>of</strong> zero temperature, the energy density iswhere∑{ ( 3 2ρtE =T o ρ t5t( ) 2/3 (1 2β L′′2 ρ o) 2/3− 1 [α L˜ρ t +α U ρ˜t ′−ρ o ρ o[ρ 2/3t ˜ρ t +] [5/3 ˜ρ t +β U′′ρ 2/3tρ˜t ′ +˜ρ5/3t ′ ])]} (2.38)(β (L,U) ′′ = ρo) [ 2/3 6 ( ) ]3π2 2/3 β (L,U) ′ 2 5+2σ′ (L,U) = 6 5 β (L,U) +σ (L,U) . (2.39)The potentials at zero temperature becomeV t = − 2T o[α L˜ρ t +α U˜ρ t ′ − 1 ( ) 2/3 ( 2ρ o 2 ρ o)β L′′ ˜ρ 5/3t +β U′′ ˜ρ 5/3t ′− 5 (3 ρ2/3 t σ′L˜ρ t +σ U˜ρ ) ] ′t ′ (2.40)The expression <strong>for</strong> the effective mass <strong>of</strong> Eq. (2.33) is still valid, so the Euler equations atzero temperature, using Eq. (2.36), take the <strong>for</strong>mµ t = 22m ∗ t[( )( ) 2/33π 2 2/3 2ρtρ t +V t = T o − 2T o×ρ o ρ oα L˜ρ t +α U˜ρ t ′ − 1 ( ) 2/3 [ 2 (˜ρ5/3 β L′′2 ρ ot + 5 )3 ρ2/3 t ˜ρ t +β U′′(˜ρ5/3 t ′ + 5 ρ 2/3t ˜ρ t ′)] ] .(2.41)The pressure at zero temperature, using the notation u t = ρ t /ρ o , becomes[∑ 2p =ρ o T o u t5 22/3 u 2/3t −α L ũ t −α U ũ t ′t[+2 −1/3 β L ′′ (ũ5/3 t + 7 3 u2/3 t ũ t )+β U ′′ (ũ5/3 t+ 7 ′ 3 u2/3 t ũ t ′)] ] .(2.42)˜ρ5/3Once again eliminating terms involving t , the pressure relation can be simplified to∑[ (µt βp = ρ o u t2 +(2u t) 2/3 ′′LT o3 ũt + β′′ U3 ũt ′ − 1 )]. (2.43)10t15


2.3.5 Uni<strong>for</strong>m <strong>Matter</strong> at Zero Temperature <strong>and</strong> the SaturationConstraintsNext, we consider uni<strong>for</strong>m matter at zero temperature. Using the notation u = (ρ n +ρ p )/ρ o<strong>and</strong> x = ρ p /(ρ n +ρ p ), the bulk energy density at zero temperature is{ 3[ ])E = T o ρ o5 22/3 u 5/3 (1−x) 5/3 +x 5/3 −u(α 2 L [(1−x) 2 +x 2 ]+2α U (1−x)x+2 2/3 u 8/3 (β L′′[(1−x) 8/3 +x 8/3 ][ ]) }+β Ux(1−x)′′ (1−x) 2/3 +x 2/3 .(2.44)Figure 2.1: The energy per baryon E <strong>and</strong> pressure p <strong>for</strong> zero-temperature matter as afunction <strong>of</strong> composition Y p . The assumed saturation constraints are: ρ o = 0.16545 fm −3 ,E o = −16.533 MeV, p o = 0, S v = 31.63 MeV <strong>and</strong> S ′ v = 17.93 MeV.Fig. 2.1 shows the energy per nucleon E = E/ρ <strong>and</strong> the pressure p <strong>of</strong> uni<strong>for</strong>m matter as afunction <strong>of</strong> density <strong>and</strong> composition Y p <strong>for</strong> the case <strong>of</strong> zero-temperature matter. Obviously,there are substantial regions in which the incompressibility, (∂P/∂n) T,Yp , is negative <strong>and</strong> thematter is hydrodynamically unstable. It is straight<strong>for</strong>ward to show that the free energy canbe lowered if matter spontaneously divides into two phases <strong>of</strong> differing densities with thesame temperature. For asymmetric matter, each phase has differing compositions as well.These phases are in bulk equilibrium, which is described in the next section.16


For st<strong>and</strong>ard nuclear matter, u = 1 <strong>and</strong> x = 1/2. Letting α = α L + α U <strong>and</strong> β ′′ =β L ′′ +β′′ U = (6/5)β+σ, using β = β L+β U <strong>and</strong> σ = σ L +σ U , <strong>and</strong> defining the binding energyas B = −E o /ρ o , one finds the energy per particle to be[E o 3= −B = T oρ o 5 − α ]2 + β′′, (2.45)2while the pressure becomesp o = 0 = ρ o T o[ 25 − α 2 + 5 6 β′′ ], (2.46)<strong>and</strong> the incompressibility parameter reduces toK o = 9 dP∣ = T o [6−9α+20β ′′ ]. (2.47)dρ∣ρ=ρo,x=1/2These can be combined asK o = 9 5 T o +15B ≃ 316 MeV, (2.48)<strong>and</strong>α = 9 5 + 5BT o≃ 3.995,β ′′ = 3 5 + 3BT o≃ 1.917, (2.49)using T o ≃ 37.68 MeV <strong>and</strong> B ≃ 16.54 MeV. (For the purposes <strong>of</strong> illustration, we use theparameter set established by Myers & Swiatecki [26].) The bulk symmetry parameter isS v = 1 8d 2 (E/ρ)dx 2∣∣∣∣∣ρ=ρo,x=1/2= T o[ 13 + α U −α L2+ 5 9 (2β′′ L −β′′ U ) ]. (2.50)The derivatives <strong>of</strong> the symmetry energy <strong>and</strong> the incompressibilities at saturation areS ′ v = ρ o8∣ [d 3 (E/ρ) 2= Tdρdx∣∣∣∣ρ=ρo,x=1/22 o9 + α U −α L+ 252 27 (2β′′ L −β′′ U], ) (2.51)K ′ o =ρ odKdρ∣∣ρ=ρo,x=1/2= K o +9ρ 2 o( d 3 Edρ 3 )ρ=ρ o,x=1/2T o[4−9α+ 1003 β′′ ]= 395 T o +55B ≃ 1204 MeV.In order to specify given values <strong>of</strong> S v <strong>and</strong> S ′ v, one can manipulate Eqs. (2.50) <strong>and</strong> (2.51):=(2.52)α L = 7 5 + 5B +3S′ v −5S v2T o, α U = 2 5 + 5B +5S v −3S ′ v2T o,β ′′L = 3 10 + 10B +9S′ v −9S v10T o, β ′′U = 3 10 + 20B +9S v −9S ′ v10T o.(2.53)17


For values <strong>of</strong> S v = 31.63 MeV <strong>and</strong> S v ′ = 17.93 MeV established by Myers & Swiatecki <strong>for</strong>the truncated model, one finds α L ≃ 1.112, α U ≃ 2.882, β L ′′ ≃ 0.417, β′′ U ≃ 1.505. Notethat <strong>for</strong> zero temperature uni<strong>for</strong>m matter, E,P <strong>and</strong> µ t depend only on α l ,α U ,β L ′′ <strong>and</strong> β′′ U .To constrain other the parameters one has to consider the properties <strong>of</strong> non-uni<strong>for</strong>m matter<strong>and</strong> finite-temperature matter.The specific heat <strong>of</strong> degenerate nucleonic matter depends on the nucleon effective masses.The symmetric matter effective mass at saturation density is 22m ∗ = 22m b+T o (β ′ L +β ′ U),m ∗= 1m b 1+β . (2.54)A reasonable value <strong>for</strong> m ∗ /m b is 0.724 which implies β ≃ 0.381.A related consideration is the optical model potential. The single particle potentials inst<strong>and</strong>ard uni<strong>for</strong>m nuclear matter at zero temperature are obtained from Eq. (2.40). OneobtainsV n = V p = −T o[α− 3 5 β − 4 3 σ ]= −B −T o (1+β), (2.55)where α <strong>and</strong> σ can be eliminated by using the saturation conditions Eqs. (2.45) <strong>and</strong> (2.46).Thus, the potential felt by a neutron with momentum p in st<strong>and</strong>ard nuclear matter is givenby combining Eqs. (2.25) <strong>and</strong> (2.55):( p) ] 2U n (p) = −B +T o[−1−β +β . (2.56)P oSetting E n = U n +T o (p/P o ) 2 to be the total energy <strong>of</strong> the neutron, one can writeU n (p) = − B1+β −T o + β1+β E n(p). (2.57)This results in a linear relation between U n <strong>and</strong> E n with a slope <strong>of</strong> β/(1+β) = 1−m ∗ /m b .Thevaluem ∗ /m b = 0.724there<strong>for</strong>eimpliesaslope<strong>of</strong>0.276, aresult supportedbyexperiment[24] . In addition, the value <strong>of</strong> the potential <strong>for</strong> a zero-momentum neutron is −(T o (1+β)+B) ≃ −68.59 MeV which is also supported by experiment. For β ≃ 0.381, one obtainsσ = β ′′ −(6/5)β ≃ 1.460.The truncated model predicts the neutron <strong>and</strong> proton effective masses in pure neutronmatter to bem ∗ n0m b= 11+β L,m ∗ p0m b=11+β U, (2.58)but these cannot be established without further input. In the original Myers & Swiateckimodel [26], the β <strong>and</strong> σ parameters are not independent, but are related byσ L −σ U = (β L −β U ) σ β = (β′′ L −β ′′U) σ β ′′ ≃ −0.833, (2.59)using the above values. While it is not necessary to constrain this parameter this way, inthe absence <strong>of</strong> additional observables <strong>for</strong> fitting we follow the method <strong>of</strong> Myers & Swiatecki.18


a α L α U β L β U σ0.5882 1.1124 2.8823 0.08182 0.2991 1.4597ρ 0 B m ∗ /m b ω 0 S v S ′ v0.16545 16.539 0.7241 1.250 31.633 17.931T o K o K ′ o t 90−10,o ω δ t o T co37.6794 315.91 1203.54 1.910 3.640 1.391 17.4Table 2.1: Truncated parameter set. The first row contains parameters <strong>of</strong> the <strong>for</strong>ce model.Thesecondrowcontainstheassumedsaturationproperties<strong>of</strong>matter. Thethirdrowcontainsimplied bulk matter <strong>and</strong> surface properties. Units are expressed in MeV <strong>and</strong> fm.We there<strong>for</strong>e haveσ L,U = σ 2or σ L ≃ 0.313 <strong>and</strong> σ U ≃ 1.146. Also,β L ≃ 0.0818 <strong>and</strong> β U ≃ 0.299. These lead to) (1+ β′′ L,U −β′′ U,L, (2.60)β ′′β L,U = 5 6 (β′′ L,U −σ L,U ), (2.61)m ∗ n0m b≃ 0.924,m ∗ p0m b≃ 0.770, (2.62)The parameters <strong>and</strong> the implied physical properties <strong>of</strong> matter <strong>and</strong> interfaces at saturationare contained in Table 2.1In summary, the binding energy, saturation density, symmetry energy, symmetry energyderivative<strong>and</strong>effectivemasscanconstrain5<strong>of</strong>the6independent parameters<strong>of</strong>thetruncatedmodel, namely α (L,U) ,β (L,U) ,σ <strong>and</strong> a. The parameter a can be determined by the symmetricmatter surface energy or the surface diffuseness. In the truncated model, these choicesdetermine the incompressibility parameter <strong>and</strong> its derivative as well as the surface symmetryenergy parameter. Adjusting the incompressibility, its derivative, <strong>and</strong> surface symmetryenergy parameters requires additional density-dependent terms in the interaction energy.This will be explored in the succeeding sections.2.4 Two-Phase Equilibrium <strong>of</strong> Bulk <strong>Matter</strong>In the general case <strong>of</strong> asymmetric matter <strong>and</strong> finite temperature, <strong>for</strong> two phases to be in bulkequilibrium, one minimizes the total free energy density <strong>for</strong> a given temperature, density <strong>and</strong>average proton fraction. We assume the electrons remain uni<strong>for</strong>mly distributed so they neednot be considered <strong>for</strong> this minimization. We also ignore the fact that protons are charged,an omission that will be included when we consider finite nuclei. We label the denser phase19


as inside or I <strong>and</strong> the other phase as outside or II. The total free energy density iswhere the average density ρ <strong>and</strong> proton fraction x satisfyF = vρ I F I +(1−v)ρ II F II (2.63)ρ = vρ I +(1−v)ρ II , ρx = vρ I x I +(1−v)ρ II x II , (2.64)where v is the volume fraction occupied by the dense phase. Minimizing F <strong>for</strong> a given ρ <strong>and</strong>x results in three bulk equilibrium conditions:p I = p II ≡ p o , µ nI = µ nII ≡ µ n , µ pI = µ pII ≡ µ p (2.65)Note that these conditions do not involve ρ,x or v. For a given value <strong>of</strong> x I = ρ pI /ρ I , the bulkequilibrium conditions there<strong>for</strong>e determine ρ nI ,ρ nII <strong>and</strong> ρ pII . Phase equilibrium is possible<strong>for</strong> a given values <strong>of</strong> ρ, T <strong>and</strong> x if the resulting value <strong>of</strong> v ∈ (0,1).The phase equilibrium <strong>of</strong> dense matter using a Skyrme interaction at finite temperaturewas studied by Küpper, Wegmann & Hilf [14] <strong>for</strong> symmetric matter, <strong>and</strong> by Lattimer &Ravenhall [21], Buchler, J. R. & Barranco [1], Barranco & Buchler [6] , <strong>and</strong> by Lamb,Lattimer, Pethick & Ravenhall [16] <strong>for</strong> the general case. The phase equilibrium <strong>of</strong> densematter using a relativistic field-theoretical interaction has been studied by Glendenning,Csernai & Kapusta [9] <strong>for</strong> symmetric matter <strong>and</strong> Müller <strong>and</strong> Serot [23] <strong>for</strong> asymmetricmatter. Phase equilibrium with the finite-range <strong>for</strong>ce has heret<strong>of</strong>ore not been previouslystudied <strong>for</strong> the general case <strong>of</strong> asymmetric hot matter.2.4.1 Zero-Temperature Two-Phase EquilibriumIn the case <strong>of</strong> zero-temperature, µ pI ≃ −16 MeV < 0 <strong>and</strong> µ nI ≃ −16 MeV < 0 <strong>for</strong> x I ∼ 0.5,so ρ pII = ρ nII = 0. Two-phase equilibrium at zero temperature is then determined byignoring the µ nI = µ nII <strong>and</strong> µ pI = µ pII conditions. However, <strong>for</strong> x I 0.34, µ nI > 0 soρ nII > 0 <strong>and</strong> neutron drip occurs. In this regime, equilibrium is determined by ignoring theµ pI = µ pII condition. Note from Eq. (2.41) that as long as ρ pII = 0, the proton chemicalpotential in the uni<strong>for</strong>m light phase is determined by the local neutron density, i.e.,( β′′ (U 2ρnII) ) 5/3 ρ nIIµ pII (ρ pII = 0) ≡ µ c = T o −2αU . (2.66)2 ρ 0 ρ 0This critical chemical potential (µ c ) represents the threshold <strong>for</strong> proton drip † . When µ pI ≥µ c , it becomes energetically favorable <strong>for</strong> protons to drip. This occurs near x I ≃ 0.09. Thebehavior <strong>of</strong> the chemical potentials, densities <strong>and</strong> pressure in bulk equilibrium are shown inFig. 2.2 as a function <strong>of</strong> the proton fraction in the dense phase.Note that <strong>for</strong> very small values <strong>of</strong> x I , the densities in the two uni<strong>for</strong>m phases approacheach other. For x I < 0.035, in this model, bulk equilibrium becomes impossible <strong>and</strong> matterexists in a single uni<strong>for</strong>m phase.† At zero temperature, proton drip does not happen in neutron star crust but free protons exist in case<strong>of</strong> uni<strong>for</strong>m nuclear matter if the phase transition happens in the outer core <strong>of</strong> neutron stars.20


Figure 2.2: Bulk equilibrium <strong>of</strong> T = 0 asymmetric nuclear matter. The solid curves u I =ρ I /ρ 0 <strong>and</strong> u II = ρ II /ρ 0 show the densities at T = 0 <strong>for</strong> two phases in bulk equilibrium <strong>for</strong>various values <strong>of</strong> x I , the proton fraction in the dense phase. The dashed curves show µ n <strong>and</strong>µ p ; neutron drip † occurs when µ n > 0, or x I ≃ 0.34 <strong>for</strong> the truncated finite-range model.Protondrip occurs when µ p > µ c , shown by the dashed-dot curve. Another solid curve showsthe value <strong>of</strong> x II , the proton fraction in the low-density phase. A dashed curve shows thepressure in bulk equilibrium. All units are in MeV <strong>and</strong> fm.2.4.2 Finite-Temperature Symmetric <strong>Matter</strong>, the Critical Temperature,<strong>and</strong> the Coexistence CurvePressure isotherms are shown in Fig. 2.3 <strong>for</strong> the case <strong>of</strong> symmetric matter. For each temperaturedisplayed, bold dots represent the two densities u i <strong>and</strong> u o , found from the solution <strong>of</strong>Eq. (2.65), thatcoexist inbulkequilibrium. Bothphaseshave equal protonfractions. <strong>Matter</strong>with an average density u that satisfies u o < u < u i lies within the coexistence region; thepressure <strong>of</strong> this matter is constant as the density is varied as indicated by the dotted linesconnecting the bold dots. Coexistence is only possible up to the critical point, indicated bythe open circle. It is clear from this figure that the critical point is defined by∂p∣ = ∂2 p∣ ∣∣T= 0, (2.67)∂ρ T ∂ρ 2which is located at T co ≃ 17.3 MeV <strong>and</strong> u co ≃ 0.350. The critical pressure is p co ≃ 0.315MeV fm −3 <strong>for</strong> the truncated interaction. The coexistence region in the temperature-densityplane is illustrated in the upper panel <strong>of</strong> Fig. 2.3. Dotted lines connect the two panels byindicating the two densities that coexist at each temperature or pressure below the criticalvalues.21


Figure2.3: Pressure isothermsareshowninthelower panel. Bolddotsindicatetheboundarydensities <strong>of</strong> the two-phase coexistence region. The upper panel shows the coexistence regionin a density-temperature plot. The critical point is illustrated in each panel with an opencircle.2.4.3 Finite-Temperature Asymmetric <strong>Matter</strong> <strong>and</strong>Coexistence CurvesAt finite temperature <strong>and</strong> arbitrary average proton fraction, the phase coexistence is determinedby the conditions Eq. (2.65). The coexistence curves are the boundaries <strong>of</strong> thetwo-phase region determined by the conditions v = 0 or v = 1. In contrast to the symmetricmatter case, the densities in each phase vary with the filling factor v at fixed T <strong>and</strong> Y p . Thecoexistence curves are shown in Fig. 2.4. Along the phase boundary, the density <strong>of</strong> one phaseis shown by the solid (coexistence) curve <strong>and</strong> the density <strong>of</strong> the other phase is shown by thecorresponding dotted curve. Along the coexistence curve <strong>for</strong> each Y p there is a critical pointwhere the two phases have equal densities, ρ i = ρ o . As shown in Lattimer & Ravenhall [21],the critical point <strong>for</strong> asymmetric matter is determined fromdµ ndx∣ = d2 µ ∣∣∣∣p,Tn= 0. (2.68)∣ dx 2p,TThe critical temperature as a function <strong>of</strong> dense-phase proton fraction x I is shown in Fig. 2.5.22


Figure 2.4: Coexistence curves <strong>for</strong> asymmetric matter <strong>for</strong> the truncated model at finitetemperature. Solid curves show the boundary <strong>of</strong> the two-phase region <strong>for</strong> dense phase protonfractions ranging from x I = 0.5 (upper-most curve) to x I = 0.02 (lower-most curve) inincrements <strong>of</strong> 0.04. At each density along a solid curve, matter with the indicated x I is inbulk equilibrium with a less dense phase (which has a different proton fraction x II ) whosedensities are indicated by the dotted curves. The densities <strong>of</strong> the two phases are equal atthe critical points, indicated by solid dots.2.5 The <strong>Nuclear</strong> Surface in the Semi-Infinite ApproximationA useful approximation <strong>for</strong> the nuclear surface is to treat the interface between two phasesin bulk equilibrium in the limit that the curvature vanishes <strong>and</strong> Coulomb interactions areneglected. The surface is thus treated in a one-dimensional semi-infinite planar geometry,<strong>and</strong> the density pr<strong>of</strong>iles in the surface region are found by minimization <strong>of</strong> the total freeenergy.2.5.1 The Euler EquationsThe equilibrium matter distributions, ρ n (r),ρ p (r), <strong>for</strong> a given temperature T are obtainedby dem<strong>and</strong>ing that the total free energy F is stationary with respect to variations δρ n <strong>and</strong>δρ p subject to the constraints that the total neutron <strong>and</strong> proton numbers,∫ ∫N = d 3 rρ n , Z = d 3 rρ p , (2.69)23


Figure 2.5: Critical temperature as a function <strong>of</strong> dense-phase proton fraction <strong>for</strong> the truncatedmodel.remain fixed. Note that, <strong>for</strong>mally, F, N <strong>and</strong> Z diverge in the semi-infinite case since theintegrals extend to ±∞. This divergence is irrelevant, however, as is discussed below.Free energy stationarity is equivalent to dem<strong>and</strong>ing that the neutron <strong>and</strong> proton chemicalpotentials, µ t = ∂E/∂ρ t , given by Eq. (2.28), are spatially constant. For a specifictemperature <strong>and</strong> proton fraction in the dense phase far from the surface, x I , the chemicalpotentials are there<strong>for</strong>e equal to their corresponding values <strong>for</strong> bulk equilibrium, as inEq. (2.65). Thus, <strong>for</strong> the appropriate values <strong>for</strong> µ n <strong>and</strong> µ p , Eq. (2.28) becomes the Eulerequations which determine ρ n <strong>and</strong> ρ p at every point.Thesurfacethermodynamicpotential<strong>for</strong>asemi-infiniteinterfaceisthedifferencebetweenthe total thermodynamic potentials <strong>of</strong> the semi-infinite system in which two uni<strong>for</strong>m phasesin bulk equilibrium exist with a discontinuous density jump at the interface <strong>and</strong> the systeminwhich thedensity variesaccording t<strong>of</strong>reeenergystationarity. Since µ n <strong>and</strong>µ p arespatiallyconstant, this is equivalent to Ravenhall, Pethick & Lattimer [29] <strong>and</strong> Kolehmainen, Prakash& Lattimer [13]ω =∫ ∞−∞[E −T(S n +S p )−µ n ρ n −µ p ρ p +p o ]dz = −∫ ∞−∞[p(z)−p o ]dz. (2.70)The last equality is not generally true <strong>for</strong> a non-relativistic potential <strong>and</strong> relativistic fieldtheoreticalnuclear <strong>for</strong>ces, but is true <strong>for</strong> the finite-range nucleon interaction. In general, athermodynamic potential is a function <strong>of</strong> temperature <strong>and</strong> chemical potentials, but in thesemi-infinite case one can write ω = ω(T,x I ) since µ n <strong>and</strong> µ p are themselves functions only<strong>of</strong> T <strong>and</strong> x I in bulk equilibrium. In other words, the values <strong>of</strong> the average density ρ <strong>and</strong>24


proton fraction x are irrelevant <strong>for</strong> this calculation.In practice, the integrals in Eqs. (2.69) <strong>and</strong> (2.70) are taken to extend from z II to z Iwhich are located far to the right <strong>and</strong> left <strong>of</strong> the interface, respectively. The integr<strong>and</strong> <strong>of</strong>Eq. (2.70) must vanish far from the interface, according to Eq. (2.65), so these cut<strong>of</strong>fs donot affect the value <strong>of</strong> ω.Surface Pr<strong>of</strong>iles <strong>and</strong> Tension <strong>for</strong> Zero-Temperature Symmetric <strong>Matter</strong>Figure 2.6: The density as a function <strong>of</strong> position <strong>for</strong> zero temperature symmetric matter isshown by the solid curve. The zero point <strong>of</strong> the z−axis is arbitrary. The function ũ is shownby the dashed curve.Inthecase<strong>of</strong>zerotemperaturesymmetric matter, theenergydensity <strong>and</strong>Euler equationsare[ 3E = T o ρ o5 u5/3 − α ]β′′uũ+2 4 u(ũ5/3 +u 2/3 ũ) , (2.71)[µ = −B = T o u 2/3 −αũ+ β′′2 (ũ5/3 + 5 ]3 u2/3 ũ) . (2.72)The density as a function <strong>of</strong> position is shown in Fig. 2.6. The integr<strong>and</strong> <strong>of</strong> Eq. (2.70) is[p o −p =T o ρ o − 2 5 u5/3 + α β′′uũ−2 4 u(ũ5/3 + 7 ]3 u2/3 ũ)= ρ [ou To]2 5 u2/3 (1−ũ)+B(1−u 2/3 ũ) ,(2.73)where we used Eq. (2.72) to eliminate ũ5/3 <strong>and</strong> α, <strong>and</strong> Eq. (2.49) to eliminate β ′′ . The25


pressure, <strong>and</strong> also the integr<strong>and</strong>, must vanish at each extreme boundary, so that at oneextremity, u → 1 <strong>and</strong> at the other u → 0, as is shown in Fig. 2.6. Obviously, in the limitz → ∞, one has u = ũ = 1 since du/dz → 0. From Eq. (2.72) one has µ = −B =T o (1−α+(4/3)β ′′ ), which also follows from Eq. (2.49). At the point where u = 0, however,du/dx ≠ 0 so ũ ≠ 0. In this case one hasµ = −B = T o()−αũ+ β′′ũ25/3(2.74)<strong>for</strong> which one finds ũ ≃ 0.126. The surface pr<strong>of</strong>ile is shown in Fig. 2.6, found by solving Eq.(2.72) at each point in space. The function ũ also is displayed in Fig. 2.6. It is obvious thatsince ũ <strong>and</strong> u are both less than unity, the integr<strong>and</strong> Eq. (2.73) is always greater than orequal to zero.The surface thermodynamic potential or surface tension depends only on the parametersα, β ′′ <strong>and</strong> a, so that given values <strong>for</strong> B <strong>and</strong>ρ o , the surface tension depends only uponain theabsence <strong>of</strong> additional terms in the energy density. Myers & Swiatecki (1969) have noted thatthe surface tension at zero temperature <strong>for</strong> symmetric matter depends almost linearly on a<strong>and</strong> this is illustrated in Fig. 2.7. For the truncated model parameters, including a = 0.5882fm, one obtains the physically reasonable value <strong>for</strong> the zero-temperature symmetric mattersurface tension, ω(x = 1/2,T = 0) ≡ ω o ≃ 1.25 MeV fm −2 . However, the calculated surfacethickness is too small by about 20%, as will be shown below.Figure 2.7: Surface tension (solid curve) <strong>and</strong> 90-10 surface thickness (dashed curve) <strong>for</strong>symmetric matter at zero temperature in the truncated finite-range Thomas Fermi model asfunctions <strong>of</strong> the diffuseness parameter a. The slopes <strong>of</strong> each are nearly linear.It is instructive to compare these results <strong>for</strong> the surface with those established from a26


potential model analogue <strong>for</strong> the Hamiltonian density <strong>of</strong> cold symmetric matter in which thefree energy density is assumed to be composed <strong>of</strong> a bulk matter part F B (ρ) <strong>and</strong> a gradientterm:F = F B + Q 2 ρ′2 . (2.75)In the above, Q is a constant <strong>and</strong> ρ ′ = dρ/dz. Minimizing the total free energy <strong>for</strong> a fixednumber <strong>of</strong> particles introduces the Lagrangian parameter µ o , which is equivalent to thechemical potential −B <strong>for</strong> symmetric matter,F B −µ o ρ = Q 2 ρ′2 . (2.76)There<strong>for</strong>e the density gradient can be determined as a function <strong>of</strong> density:The 90-10 surface thickness, t 90−10 is defined as√ 2 √ρ ′ = − FB −µ o ρ. (2.77)Qt 90−10 = z(ρ = 0.9ρ i )−z(ρ = 0.1ρ i ) (2.78)in the case that ρ o = 0. For F B we substitute the free energy density <strong>of</strong> the truncated model<strong>for</strong> uni<strong>for</strong>m symmetric matter at zero temperature, <strong>for</strong> which ρ i = ρ o :F B −µ o ρT o ρ o) [ 2 B) ( 3B≡ f B = u(1−u(1+2u 1/3 1/3 + + 3 ( )u 2/3 1+ 1 ) ]T o T o 5 2 u1/3 , (2.79)which defines the dimensionless function f B (u). This leads to∫ 0.9ρo√t 90−10,o = dρ/ρ ′ Qρo= I t , (2.80)0.1ρ oT owhereI t =∫ 0.90.1du√ 2fB≃ 3.042; (2.81)the numerical value is obtained using the truncated model with B = 16.539 MeV <strong>and</strong>T o = 37.679 MeV. The symmetric matter surface tension iswhereω o =∫ ∞−∞[F −µ o ρ]dz = 2I ω =∫ 10∫ ∞−∞[F B −µ o ρ]dz = √ QT o ρ 3 o I ω, (2.82)√2fB du ≃ 0.2612, (2.83)wherethenumerical valueisobtainedwiththesameassumptions aspreviously. Thequantity27


Q can be eliminated by combining Eqs. (2.80) <strong>and</strong> (2.82),ω o = ρ o T o t 90−10,o I ω /I t ≃ 0.5353 t 90−10,o MeV fm −3 = 1.231 MeV fm −2 , (2.84)where the truncated model <strong>and</strong> a realistic value <strong>of</strong> the surface thickness(t 90−10,o = 2.30 fm) is used. This value <strong>of</strong> ω o agrees well with that obtained from thenumerical results <strong>for</strong> the truncated model (1.24 MeV fm −2 ). However, as noted be<strong>for</strong>e, thecalculated value <strong>of</strong> t 90−10,o is 20% smaller than the realistic value. This discrepancy canbe traced to the unrealistically large value <strong>of</strong> the incompressibility parameter K o in thetruncated model.Surface Pr<strong>of</strong>iles <strong>and</strong> Tension <strong>for</strong> Zero-Temperature Asymmetric <strong>Matter</strong>Figure2.8: <strong>Neutron</strong>(solidline) <strong>and</strong>proton(dashedline) density pr<strong>of</strong>iles <strong>for</strong>thecase x I = 0.4<strong>and</strong> T = 0 <strong>for</strong> a semi-infinite interface <strong>for</strong> the truncated interaction are indicated in the mainpanel in logarithmic units. The inset shows normalized pr<strong>of</strong>iles u n = ρ n /ρ o ,u p = ρ p /ρ o assolid curves on a linear scale, <strong>and</strong> the quantities ũ n = ˜ρ n /ρ o <strong>and</strong> ũ p = ˜ρ p /ρ o are displayed bydashed curves. The dotted lines indicate the positions <strong>of</strong> the squared-<strong>of</strong>f neutron <strong>and</strong> protonradii, R n <strong>and</strong> R p respectively.The density pr<strong>of</strong>iles <strong>for</strong> neutrons <strong>and</strong> protons <strong>for</strong> cold asymmetric matter are displayedin Fig. 2.8 <strong>for</strong> the case x I = 0.4. Both the densities ρ t <strong>and</strong> the quantities ˜ρ t are shown. Thesquared-<strong>of</strong>f radii R n <strong>and</strong> R p are also shown; these are defined by∫ zIIz Iρ t dz = ρ tI (R t −z I )+ρ tII (z II −R t ) (2.85)28


with the convention that z II is the boundary in the dilute phase <strong>and</strong> z I is the boundary inthe dense phase. Clearly we must take z I > 0, <strong>and</strong> in the calculation shownin Fig. 2.8 these were respectively chosen to be -8 fm <strong>and</strong> +8 fm.Figure 2.9: Surface tension (left panel) <strong>and</strong> neutron skin thickness (right panel) as functions<strong>of</strong> dense phase proton fraction <strong>for</strong> zero-temperature semi-infinite surfaces are shown as solidcurves. In each case, dashed lines represent analytic estimates based onasimplified potentialmodel as described in the text.The surface thermodynamic potential <strong>for</strong> asymmetric matter at zero temperature isshown in Fig. 2.9. Analytic estimates <strong>of</strong> the symmetric matter surface thermodynamicpotential ω o <strong>and</strong> its dependence on proton fraction can be obtained from the potential approachdiscussedinsection 2.5.1. Inparticular, specification<strong>of</strong>thesymmetricmattersurfacethickness parameter t 90−10,o , determines ω o via Eq. (2.84). The leading order dependence <strong>of</strong>the surface tension on proton fraction is quadratic with a coefficient ω δ such thatω ≃ ω o −ω δ δ 2 , δ ≡ 1−2x I . (2.86)In the potential approach, it can be shown in Steiner et al.[32] that the parameter ω δ is, tolowest order, √ ∫ Qρo ( Sv)ω δ = S v ρ −1 (F B −µ o ρ) −1/2 dρ, (2.87)2 E sym029


where E sym is the density dependent symmetry energy. In the truncated model E sym is(S v −E sym = S v 1−u)×1/3[× 1+u 1/3 +u 2/3 − 3(S v −S v ′)( )u 1/3 1+u 1/3 − T (o1−u)(1+ 1/3 1 ) ] (2.88)2S v 3S v 2 u1/3 .Eliminating Q using Eq. (2.82) <strong>and</strong> expressing the free energy density in terms <strong>of</strong> the truncatedmodel energy density, one findswhereI δ =∫ 10ω δ = ω oS vT oI δI ω, (2.89)( )u Sv√ −1 du ≃ 1.024. (2.90)2fB E symThe numerical value is obtained <strong>for</strong> the truncated model with the previously determinedparameters <strong>and</strong> leads to the ratio ω δ /ω o ≃ 3.293 <strong>and</strong> ω δ ≃ 4.054 MeV fm −2 . It is also easyto show from Eq. (2.85) that the neutron skin thickness t can be written, to lowest orderwhen ρ II = 0, ast = R n −R p =∫ ∞−∞[ρn− ρ ]pdz ≃ρ nI ρ pI2ω δ δS v ρ o (1−δ 2 ) . (2.91)From this, we infer a normalized neutron skin thickness t o = t(1−δ 2 )/δ which is predicted tobe insensitive to asymmetry. The analytic approximations <strong>for</strong> surface properties using Eqs.(2.86), (2.89) <strong>and</strong> Eq. (2.91) are displayed in Fig. 2.9. The agreement between the analytic<strong>and</strong> numerical results is excellent. As an example, the difference between the numerical <strong>and</strong>analytic values <strong>for</strong> t o is about 2·10 −5 fm <strong>for</strong> small asymmetries. In addition, the dependence<strong>of</strong> t on asymmetry is in good agreement with the analytical prediction <strong>of</strong> Eq. (2.91).Surface Tension <strong>for</strong> Finite-Temperature <strong>Matter</strong>For the truncated model, the surface tension (thermodynamic potential density, ω) <strong>and</strong>surface thickness parameter (t 90−10 ) <strong>for</strong> finite-temperature symmetric matter are shown inFig. 2.10. The truncated parameter set was employed. Both quantities vary as T 2 <strong>for</strong> smalltemperatures. The surface tension <strong>for</strong> symmetric matter can be well approximated <strong>for</strong> alltemperatures by the expressionω(x = 1/2,T) ≃ ω o[1−(T/T c ) 2 ] 5/3,(2.92)where T c ≃ 17.4 MeV is the critical temperature <strong>for</strong> two-phase coexistence <strong>of</strong> symmetricmatter <strong>and</strong> ω o = ω(x = 1/2,T = 0). This expression indicates the effective level density30


Figure 2.10: The surface tension <strong>of</strong> symmetric matter as a function <strong>of</strong> temperature is shownby the solid curve <strong>for</strong> the truncated model. The three accompanying dashed/dotted curves,labelled by the parameter p, show the temperature dependence ω o [1 − (T/T co ) 2 ] p whereT co ≃ 17.4 MeV is the critical temperature <strong>for</strong> two-phase coexistence <strong>of</strong> symmetric matter.In addition, the surface thickness parameter t 90−10 , scaled by a factor 10, is displayed. AtT co , it tends to infinity.parameter <strong>for</strong> nuclei has the approximate volume <strong>and</strong> surface contributions(a eff = 2m∗ π) 1/3A+ ( 4π) 1/3 10ω o 2 3ρ o 3ρ 2 o Tco2≃ 0.059A + 0.226A 2/3 MeV −1 .A 2/3(2.93)Note that <strong>for</strong> A ≈ 64 the surface <strong>and</strong> volume contributions to the nuclear specific heat areapproximately equal.For asymmetric matter, the temperature dependence <strong>of</strong> the surface tension is similar tothat <strong>of</strong> symmetric matter. As shown in Fig. 2.11, the temperature dependence can be wellapproximatedby the relation Eq. (2.92) as long as T c is reduced to the value (approximately16.1 MeV) appropriate <strong>for</strong> x II = 0.3. This figure also shows the surface thickness parametert 90−10 <strong>and</strong> the neutron skin thickness R n −R p .Fig. 2.12 shows the surface tension <strong>and</strong> the neutron skin thickness as functions <strong>of</strong> x II<strong>and</strong> T <strong>for</strong> the truncated model.31


Figure 2.11: The same as Fig. 2.10 except <strong>for</strong> an asymmetric case with a dense phase protonfraction x I = 0.3. In this case, the critical temperature is T c ≃ 16.1 MeV. The dashed lineshows the neutron skin thickness R n −R p in fm.2.6 Isolated Nuclei <strong>and</strong> Nuclei in Dense <strong>Matter</strong>For the computation <strong>of</strong> finite nuclei, spherical symmetry <strong>and</strong> zero temperature are assumedin this model. These restrictions will be relaxed in a future publication. The main point<strong>of</strong> this model is to demonstrate the feasibility <strong>of</strong> efficiently computing the thermodynamicproperties <strong>of</strong> nuclei in a consistent Thomas-Fermi approximation without the simplificationsassociated with the liquid droplet model. In contrast to the case <strong>of</strong> a semi-infinite interface,it is preferable in the finite nucleus case to choose values <strong>for</strong> N <strong>and</strong> Z, together with thetemperature. The additional two constraints in Eq. (2.65) then determine the chemicalpotentials <strong>of</strong> the isolated nucleus, i.e., a nucleus in a zero density environment.2.6.1 Isolated NucleiThe total energy <strong>of</strong> a nucleus must include the electrostatic Coulomb energy∫ ∫E Coul = e2d 3 r d 3 r ′[ρ p(r)−ρ e (r)][ρ p (r ′ )−ρ e (r ′ )], (2.94)2|r −r ′ |VVwhere the spatial integrals are over the entire volume. In the case <strong>of</strong> an isolated nucleus,ρ e = 0, <strong>and</strong> the integr<strong>and</strong> is finite only within the nucleus. The contribution <strong>of</strong> the Coulombenergy to the proton chemical potential can be found from a variation <strong>of</strong> E Coul with respect32


Figure 2.12: Contour plots <strong>of</strong> surface tesion <strong>and</strong> R n −R pFinite temperature asymmetric matter: (Left panel) A contour plot <strong>of</strong> the surface tension(ω, MeV-fm −2 ); (Right panel) A contour plot <strong>of</strong> the neutron skin thickness (R n −R p , fm).to the proton density: ∆µ p = δE Coul /δρ p . One obtainsδE Coul = e22∫V∫d 3 rVd 3 r ′|r−r ′ | (δρ p(r)[ρ p (r ′ )−ρ e (r ′ )]+δρ p (r ′ )[ρ p (r)−ρ e (r)]). (2.95)One can reverse the variables r <strong>and</strong> r ′ in the second term <strong>of</strong> the double integral, yielding∫∆µ p (r) =e 2 d 3 r ′V |r −r ′ | [ρ p(r ′ )−ρ e (r ′ )][ ∫ 1 r∫ Rc] (2.96)=4πe 2 r ′2 [ρ p (r ′ )−ρ e (r ′ )]dr ′ + r ′ [ρ p (r ′ )−ρ e (r ′ )]dr ′ ,r0where we assumed spherical symmetry in the second line <strong>and</strong> R c is taken to have a valuemuch larger than that <strong>of</strong> the nucleus. In the limit r → 0, the first integral vanishes. In thelimit r → R c , one has ∆µ p (R c ) = 0 because <strong>of</strong> charge neutrality. The quantity ∆µ p (r) mustbe added to the nucleonic part <strong>of</strong> µ p (r).The total Coulomb energy can be writtenE Coul = 1 ∫[ρ p (r)−ρ e (r)]∆µ p (r)d 3 r, (2.97)2Vr33


Figure 2.13: <strong>Neutron</strong> <strong>and</strong> proton density pr<strong>of</strong>iles <strong>for</strong> the nuclei 40 Ca, 56 Fe <strong>and</strong> 208 Pb in thetruncated model are shown as solid curves. The densities u t are scaled by the saturationdensity ρ s . Dashed curves show the corresponding functions ũ t . The chemical potentials,the charge radius R p , the neutron skin thickness δR = R n −R p , <strong>and</strong> the ground state energyE are included in each subfigure. The vertical lines show R n <strong>and</strong> R p .so that the Coulomb contribution to the energy density is justE Coul = 1 2 (ρ p −ρ e )∆µ p . (2.98)Density pr<strong>of</strong>iles <strong>for</strong> the isolated nuclei 40 Ca, 56 Fe <strong>and</strong> <strong>for</strong> 238 U are displayed in Fig. 2.13<strong>for</strong> the truncated interaction model <strong>and</strong> assuming T = 0. The specific ground state energies,charge radii <strong>and</strong> neutron skin thicknesses <strong>for</strong> these nuclei are included in the figure. Thecharge radius is R p <strong>and</strong> the skin thickness is R n −R p <strong>and</strong> are defined byR p =(∫ )ρp r 2 d 3 r 1/2∫ , Rρp d 3 n =r(∫ )ρn r 2 d 3 r 1/2∫ . (2.99)ρn d 3 rThe ground state energy is measured relative to A free neutrons, <strong>and</strong> hence includes theenergy −(m n −m p )c 2 (Z/A) from the neutron-proton mass difference.34


2.6.2 Dense <strong>Matter</strong> <strong>and</strong> the Wigner-Seitz ApproximationIn the case <strong>of</strong> nuclei at finite density, such as those in a neutron star crust, it is convenientto employ the Wigner-Seitz approximation in which a cell <strong>of</strong> radius R c contains a singlenucleus <strong>and</strong> sufficient electrons to neutralize the cell. In this paper, we will assume thatthe Wigner-Seitz cell has a spherical geometry, but this assumption can be relaxed. Inthe case <strong>of</strong> finite density, there<strong>for</strong>e, the density <strong>of</strong> electrons is finite <strong>and</strong> their distributionmust be considered in addition to those <strong>of</strong> the nucleons. The average density <strong>of</strong> the cell isρ = 3(N + Z)/(4πR 3 c). In a neutron star crust, it is logical to specify the density ρ, <strong>and</strong>optimize the beta-equilibrium energy per particle with respect to the density pr<strong>of</strong>iles <strong>and</strong> N<strong>and</strong> Z, which determines R c , µ n <strong>and</strong> µ p . The two additional equations needed, in additionto the Euler equations Eq. (2.28), are equivalent to the beta equilibrium condition<strong>and</strong> the energy minimizationµ n −µ p +(m n −m p )c 2 = µ e (2.100)( ∂E∂R c)β= 0 (2.101)where beta equilibrium is assumed.In a neutron star crust, where the density ρ >> 10 6 g cm −3 <strong>and</strong> T < 1 MeV, theelectrons are relativistic <strong>and</strong> degenerate. In this case, the electron density, which is modifiedby the Coulomb potential, is given in terms <strong>of</strong> the electron chemical potential µ e at the cellboundary byρ e (r) = 1 ( ) 3 µe +∆µ p (r). (2.102)3π 2 cEqs. (2.96) <strong>and</strong> (2.102) must be solved by iteration. In beta equilibrium, one has µ e =µ n (R c )−µ p (R c )+(m n −m p )c 2 since ∆µ p (R c ) = 0.Nuclei <strong>for</strong> several different locations in a neutron star crust at T = 0 are displayed in Fig.2.14. For densities ρ ≥ 0.00028 fm −3 the neutron chemical potential is positive <strong>and</strong> drippedneutrons appear. In this case, the number <strong>of</strong> neutrons in the nucleus is estimated fromN nuc = ρ n0N −4πρ nRc R 3 c/3ρ n0 −ρ nRc, (2.103)where ρ n0 is the neutron density at the nucleus center <strong>and</strong> ρ nRc is the neutron density at thecell boundary. Note that when neutron drip does not occur, N nuc is not exactly equal to N.In addition, in the case <strong>of</strong> dripped neutrons, the mean-square neutron radius is redefined tobe(∫ ) (ρn −ρ nRc )r 2 d 3 r 1/2R n = ∫ . (2.104)(ρn −ρ nRc )d 3 rThe trend is <strong>for</strong> proton number Z to decrease with increasing density, as is shown inFig. 2.15. The energy minimum is very sensitive to the neutron <strong>and</strong> proton numbers which35


Figure 2.14: <strong>Neutron</strong> <strong>and</strong> proton density pr<strong>of</strong>iles <strong>for</strong> nuclei in the cold crust <strong>of</strong> a neutronstar. The optimum nucleus <strong>for</strong> each density ρ (fm −3 ) is shown, together with the pressure p(MeV fm −3 ), the total internal energy per baryon E (MeV), the charge radius R p (fm), theneutron skin thickness R n −R p , <strong>and</strong> the chemical potentials µ n ,µ p <strong>and</strong> µ e (MeV). N,Z <strong>and</strong>N nuc are defined in the text. R c is the Wigner-Seitz cell size <strong>and</strong> E β is the energy <strong>of</strong> uni<strong>for</strong>mnuclear matter in beta equilibrium.36


Figure 2.15: Proton number per unit Wigner-Seitz cell <strong>for</strong> dense matter <strong>for</strong> the truncatedmodel is shown by the solid curve. The energy advantage (MeV/baryon, multiplied by 100)relative to uni<strong>for</strong>m matter in beta equilibrium is shown by the dashed curve. The fillingfactor u, defined in Eq. (2.105) <strong>and</strong> multiplied by 100, is shown by the dotted curve.accounts <strong>for</strong> the wiggles shown in this figure. This figure also shows the filling factoru = ρ−ρ 0ρ 0 −ρ Rc, (2.105)which increases with density, <strong>and</strong> the energy advantage <strong>of</strong> nuclei over that <strong>of</strong> uni<strong>for</strong>m nuclearmatter (both considered in beta equilibrium), ∆E = E −E β , which decreases with density.∆E decreases to zero be<strong>for</strong>e the filling factor approaches the value 1/2 <strong>for</strong> which mattermight expect to be turned inside out (Lamb et al.[15]).In general, the Wigner-Seitz cell size R c decreases with increasing density <strong>and</strong> the nuclearfilling factor in the absence <strong>of</strong> dripped nucleons, (R/R c ) 3 , where R ≈ (R n + R p )/2,there<strong>for</strong>e increases. However, the nuclear mass fraction decreases with density because <strong>of</strong>dripped neutrons <strong>and</strong>, as a result, the filling factor u increases less rapidly than it would otherwise.The Coulomb energy decreases with density, <strong>and</strong> eventually nuclei become unstableto de<strong>for</strong>mations. In the simple analysis <strong>of</strong> Ravenhall et al.[58], significant de<strong>for</strong>mations fromsphericity occur when the filling factor is larger than about 1/8. From the results displayedhere, this occurs when ρ ≃ 0.06 fm −3 . At this point, the energy difference between nuclei<strong>and</strong> uni<strong>for</strong>m matter is small, being only about 0.13 MeV per baryon, <strong>and</strong> the energy differenceeven disappears by densities near ρ = 0.09 fm −3 . Thus the density domain in whichsignificantly de<strong>for</strong>med nuclei appear is probably relatively small in this case. Lamb et al.[16]demonstrated that the nuclei will tend to turn “inside-out” when a filling factor larger than37


1/2 obtains, but this appears to be unlikely to ever occur <strong>for</strong> this model. To explore thesequestions further requires one to consider non-spherical geometries <strong>for</strong> nuclei.2.7 ConclusionsThis chapter demonstrates the feasibility <strong>of</strong> employing a finite-range nuclear interaction toefficiently compute nuclear properties in the Thomas-Fermi approximation, <strong>and</strong> it extendsthese calculations to the computation <strong>of</strong> the properties <strong>of</strong> hot, dense matter using unitWigner-Seitzcells. Themethodisefficientenoughtoallowthecomputation<strong>of</strong>fulltablesthatare densely-enough spaced in density, temperature, <strong>and</strong> electron fraction <strong>for</strong> astrophysicalapplications. Be<strong>for</strong>e such a table can be generated, however, there are a few details <strong>of</strong> themodel that may have to be modified.The model described in this model does not adequately describe realistic nuclear incompressibilitiesor the properties <strong>of</strong> nuclear surfaces. However, these items can be straight<strong>for</strong>wardlycorrected by the addition <strong>of</strong> density-dependent terms to the nuclear interactions. Inaddition, better fits to nuclear effective masses <strong>and</strong> the optical model potential can be accommodatedby additional momentum-dependent terms. Future work will explore possiblemodifications <strong>and</strong> will undertake a detailed comparison to the properties <strong>of</strong> nuclei.In addition, Danielewicz [7] pointed out anapparent inconsistency with the liquid droplettreatment developed by Myers & Swiatecki [24] <strong>and</strong> by Lattimer et al.[17]. with respect toits incorporation <strong>of</strong> the symmetry <strong>and</strong> Coulomb properties <strong>of</strong> nuclei. Danielewicz proposedan alternate <strong>for</strong>mulation <strong>of</strong> the liquid droplet model. However, Steiner et al.[32] pointedout discrepancies between the predictions <strong>of</strong> the two approaches concerning the symmetryproperties <strong>of</strong> extremely neutron-rich nuclei <strong>and</strong> neutron skin thickness. The present modelshould be immune to any discrepancy involved with the liquid droplet model <strong>and</strong>, furthermore,should provide a suitable plat<strong>for</strong>m with which to resolve this controversy. This pointwill be addressed in a future publication.Followingtheresolution<strong>of</strong>thesepoints, weintendtogenerateaseries<strong>of</strong>three-dimensionaltables suitable <strong>for</strong> astrophysical simulations, <strong>and</strong> we will compare their properties to theLattimer-Swesty [22] <strong>and</strong> Shen et al.[61] equations <strong>of</strong> state.38


Chapter 3Modified model† The truncated model is quite successful in representing isolated nuclei <strong>and</strong> dense matter.However, the truncated model has a too large nuclear incompressibility <strong>and</strong> nuclear surfacetension. We added new density dependent interactions so that we can resolve them.3.1 Original Modification to Myers & Swiatecki ModelWeconsidered analternativedensity dependent interactiontermwhichhasanenergydensitycontribution that is proportional to ρ 1+ǫ :C ηL = − h34 T oρ o( 34πP 3 oC ηU = − h34 T oρ o( 34πP 3 o) 2 ( ) ǫ ( ) 2 ρǫη t1 +ρ ǫ t2Lρ o 2) 2 ( ) ǫ ( ) (3.1)2 ρǫη t1 +ρ ǫ t ′ 2Uρ o 2where η L , η U , <strong>and</strong> ǫ are parameters. The total energy is altered by the amount∆E = 1 ( ) ǫ ∫T o 2d 3 r 1 d 3 r 2 f(r 12 /a)×2ρ o ρ o∑(3.2)[η L ρ t1 ρ t2 (ρ ǫ t1 +ρ ǫ t2)η U ρ t1 ρ t ′ 2(ρ ǫ t1 +ρ ǫ t2 ′).The contribution to the potential is∆V t = T oρ o( 2ρ o) ǫ(η L[(1+ǫ)ρ ǫ t˜ρ t +while the energy density contribution is∆E = 1 T o2ρ o( 2ρ o) ǫ ∑ttρ t[η L(ρ ǫ t˜ρ t +[ ])1+ǫ ˜ρ t]+η U (1+ǫ)ρ ǫ 1+ǫt˜ρ t ′ + ˜ρt, (3.3)′( )]1+ǫ ˜ρ t)+η U ρ ǫ t˜ρ 1+ǫt ′ + ˜ρt. (3.4)′† This chapter is an improved version <strong>of</strong> the truncated model in Chapter 2. This work will be submittedto the journal soon.39


In the case <strong>of</strong> uni<strong>for</strong>m matter, the additional contribution to the energy density is∆E = T ( ) ǫo 2 ∑ρ o ρ ot[ ]ρ 1+ǫt ηL ρ t +η U ρ ′ t . (3.5)For st<strong>and</strong>ard nuclear matter (u = 1, x = 1/2, T=0 MeV), one finds, after defining η =η L +η U ,[ǫ o 3= −B = T oρ o 5 − α 2 + β′′2 + η ],2[ 2p o = 0 = ρ o T o5 − α 2 + 5 6 β′′ + η ]2 (1+ǫ) ,K o = T o[6−9α+20β ′′ + 9η 2 (1+ǫ)(2+ǫ) ],[K o ′ = T o 4−9α+ 1003 β′′ + 9η ]2 (1+ǫ)2 (2+ǫ) ,(3.6)S v = T o[ 13 + α U −α L2S ′ v = T o[ 29 + α U −α L+ 5 9 (2β′′ L −β′′ U )+ 1+ǫ4 ([2+ǫ]η L −2[2−ǫ]η U )2+ 2527 (2β′′ L −β′′ U )],+ (1+ǫ)2 ([2+ǫ]η L −2[2−ǫ]η U )4Practical considerations † limit the value <strong>of</strong> ǫ to the range 0 < ǫ < 2/3. For a given value <strong>of</strong>K o , a solution <strong>of</strong> Eq. (3.6) yields the consequence that K o ′ is nearly independent <strong>of</strong> ǫ.].† To determine the range <strong>of</strong> ǫ, we check the pressure behavior <strong>of</strong> pure neutron matter.40


We choose the value ǫ = 1/3. One may show in this case thatα = 9 5 + 5BT o+ 1 ǫβ ′′ =( 35 + 5B − K )o= 18T o 3T o 5 + 20B − K o,T o T o(1− 3ǫ ) −1 [Ko− 3 2 2T o 10 − 9B ( 9B−ǫ − 9 )]2T o 2T o 10= K o− 6 T o 5 − 12B ,T oη = 1 (1− 3ǫ ) −1 ( 3ǫ 2 5 + 5B − K ) (o 3= 6T o 3T o 5 + 5B − K )o,T o 3T oK o ′ = 143 K o −15B − 3 (5 T o +ǫ K o −15B − 9 )5 T o(3.7)= 5K o −20B − 6 5 T o.There<strong>for</strong>e, to reduce K o from the truncated model value <strong>of</strong> 316 MeV to the experimentalvalue, e.g. 230 MeV, requires a value η ≃ 172/T o ≃ 4.3, <strong>and</strong> α <strong>and</strong> β ′′ are respectivelychanged by +η/2 <strong>and</strong> −η/2. In this case, K ′ o ≃ 780 MeV.For ǫ = 1/3, we findS v = T o[ 13 + α 2 −α L + 5 3 β′′ L − 5 9 β′′ + 4 3 η L − 5 9 η ],[ 2S v ′ = T o9 + α 2 −α L + 259 β′′ L − 2527 β′′ + 16 9 η L − 20 ]27 η .(3.8)Inpractice, <strong>for</strong>K o ≃ 230MeV, β ′′ ≃ 0<strong>and</strong>bothα ≃ 6<strong>and</strong>η ≃ 4arepositive. Symmetricmatter there<strong>for</strong>e has positive pressure <strong>for</strong> u > 1. Pure neutron matter, on the other h<strong>and</strong>,has pressure[ 2p N = T o ρ o u5 (2u)2/3 −α L u+ 5 3 22/3 β Lu ′′ 5/3 + 4 ]3 21/3 η L u 4/3 . (3.9)At ρ o , p N has a value about half <strong>of</strong> the non-interacting Fermi pressure, Combining this resultwith reasonable values S v = 30 MeV <strong>and</strong> S v ′ = 15 MeV, one finds α L ≃ 7.2, β L ′′ ≃ −3.4 <strong>and</strong>η L ≃ 9.4. In the case S v ′ = 30 MeV, one finds α L ≃ 13.3, β L ′′ ≃ −6.3 <strong>and</strong> η L ≃ 17.5. In bothcases, p N turns negative at low densities u < 2.5 because β L ′′ < 0.The problem can be more clearly observed if one <strong>for</strong>ms the combination Q = e N −e s −S v ,where e N is the neutron energy per baryon <strong>and</strong> e s = E/ρ is the symmetric matter energy perbaryon. Nominally, the magnitude <strong>of</strong> Q evaluated at the saturation density, Q 1 ≡ Q(u = 1),41


is expected to be small, |Q 1 | ≤ 1 MeV. In our case,[ ( 3Q = T o5 u2/3 2 2/3 − 14 ) [ β′′+(29 18 + 2/3 − 5 ) ]β L′′ u 5/33[ ( η+18 + 2 1/3 − 4 ] (3.10))η L]u 1+ǫ3Thus, <strong>for</strong> S v = 30 MeV <strong>and</strong> S v ′ = 15 MeV, Q 1 ≃ −6.5 MeV. There<strong>for</strong>e, in general, thenon-quadratic terms involving β L ′′ <strong>and</strong> η L lead to neutron matter energy <strong>and</strong> pressure thatare not well-behaved at higher densities in comparison to symmetric energies <strong>and</strong> pressures.Another possibility is to require that Q 1 ≈ 0, specificallyβ(2 ′′18 + 2/3 − 5 )β L ′′ + η (3 18 + 2 1/3 − 4 )η L = 0. (3.11)3Using Eqs.(3.8) <strong>and</strong> (3.11), we can evaluate α L = −4.0, β L ′′ = −1.4, <strong>and</strong> η L = 4.6 usingǫ = 1/3, S v = 30 MeV, <strong>and</strong> S v ′ = 15 MeV. Once again, the negative value <strong>of</strong> β′′ L rendersneutron matter unstable at high densities.3.2 Alternate ModificationMost Skyrme <strong>for</strong>ces explicitly incorporate the η terms with a quadratic x dependence sothat they, unlike the α terms, vanish in Eq. (3.10), <strong>and</strong> the surviving β L ′′ term can be madesmall. It will be necessary <strong>for</strong> us to <strong>for</strong>mulate a <strong>for</strong>ce with similar properties. One way todo this is to make the extra terms in C L,U functionals <strong>of</strong> ρ 1 only (or ρ 2 ). For example,C ηL = − h34 T oρ o( 34πP 3 oC ηU = − h34 T oρ o( 34πP 3 o) 2 ( ) ǫ ( ) 2 ρǫη 1 +ρ ǫ 2Lρ o 2) 2 ( ) ǫ ( ) (3.12)2 ρǫη 1 +ρ ǫ 2Uρ o 2Then,We find∆E = 1 ∫T o2ρ o= T oρ o∫d 3 r 1 d 3 r 2 f(r 12 /a)×[(ρ1) ǫ+ρ o( ) ǫ ρ ∑d 3 rρ o(ρ2∆E = T ( ) ǫo ρ ∑ρ o ρ otρ o) ǫ ] ∑tρ t[η L˜ρ t +η U ρ˜t ′tρ t1[η L ρ t2 +η U ρ t ′ 2].](3.13)]ρ t[η L˜ρ t +η U ρ˜t ′ , (3.14)42


where∆V t = T oρ o[( ρρ o) ǫ[ ǫρIn uni<strong>for</strong>m matter, one finds∑t] ]ρ t (η L˜ρ t +η U ρ˜t ′)+η L˜ρ t +η U ρ˜t ′ +η Lˆρ t +η Uˆρ t ′ , (3.15)∫ˆρ t (r 1 ) =∆E = T ( )o ρ ǫ[ ]η L (ρ 2 nρ o ρ +ρ2 p )+2η Uρ n ρ p ,o∆V t = T ( ) ǫ [ (o ρ ρ 2 )nǫ η +ρ2 p ρ n ρ pL +2η Uρ o ρ o ρ ρ∆p = T oρ o( ρρ o) ǫ(1+ǫ)( ) ǫd 3 ρ2r 2 f(r 12 /a) ρ t2 . (3.16)ρ o][η L (ρ 2 n +ρ2 p )+2η Uρ n ρ p .+2η L ρ t +2η u ρ t ′],(3.17)In terms <strong>of</strong> the proton fraction,∆E( [)= T o u 1+ǫ η L x 2 +(1−x)]+2η 2 U x(1−x) ,ρ o∆p = T o ρ o u 2+ǫ (η L[x 2 +(1−x) 2 ]+2η U x(1−x)For st<strong>and</strong>ard nuclear matter, we now obtain[ǫ o 3= −B = T oρ o 5 − α 2 + β′′2 + η ]2p o = 0 = ρ o T o[ 25 − α 2 + 5 6 β′′ + η 2 (1+ǫ) ]).(3.18)K o = T o[6−9α+20β ′′ + 9η 2 (1+ǫ)(2+ǫ) ][K o ′ = T o 4−9α+ 1003 β′′ + 9η ]2 (1+ǫ)2 (2+ǫ)[ 1S v = T o3 + α 2 − 5 9 β′′ − η 2 −α L + 5 ]3 β′′ L +η L ,[ 2S v ′ = T o9 + α 2 − 2527 β′′ −(1+ǫ) η 2 −α L + 25 ]9 β′′ L +(1+ǫ)η L(3.19)43


Solving, we haveα = 9 5 + 5B + 1 ( 3T o ǫ 5 + 5B − K )o= 18 T o 3T o 5 + 20B − K o,T o T oβ ′′ = 5K/T o −3−9ǫ−45(1+ǫ)B/T o10−15ǫη = 18−10K/T o +150B/T o15ǫ(2−3ǫ)= K oT o− 6 5 − 12BT o,= 18 5 + 30BT o− 2K T o,K o ′ = 143 K o −15B − 3 (5 T o +ǫ K o −15B − 9 )5 T o(3.20)= 5K o −20B − 6 5 T o.The last expressions are obtained with ǫ = 1/3. We also find, <strong>for</strong> ǫ = 1/3,[ 1S v = T o3 + α 2 −α L + 5 3 β′′ L − 5 9 β′′ +η L − η ],2[ 2S v ′ = T o9 + α 2 −α L + 259 β′′ L − 2527 β′′ + 4 3 η L − 2 ]3 η .Pure neutron matter has pressure[ 2p N = T o ρ o u5 (2u)2/3 −α L u+ 5 ]3 22/3 β Lu ′′ 5/3 +(1+ǫ)η L u 4/3(3.21)(3.22)(3.23)At ρ o , p N has a value about half <strong>of</strong> the non-interacting Fermi pressure. Combining this resultwith reasonable values S v = 30 MeV <strong>and</strong> S v ′ = 15 MeV, <strong>and</strong> assuming ǫ = 1/3, one findsα L ≃ −4.4, β L ′′ ≃ 3.6, <strong>and</strong> η L ≃ −10.7. In the case S v ′ = 30 MeV, one finds α L ≃ −8.4,β L ′′ ≃ 6.7, <strong>and</strong> η L ≃ −19.9. In both cases, since β L ′′ > 0, the neutron matter pressure willcontinuously rise with u.However, we now observe that[ ( 3Q = T o5 u2/3 2 2/3 − 14 ) [ β′′+(29 18 + 2/3 − 5 ) ]β L]u ′′ 5/3 , (3.24)3evaluated at the saturation density <strong>and</strong> using ǫ = 1/3, S v = 30 MeV <strong>and</strong> S v ′ = 15 MeV, isQ 1 ≃ −10.3 MeV. Obviously, the neutron matter properties <strong>and</strong> the symmetry propertiescannot be separately adjusted.Most Skyrme <strong>for</strong>ces have the properties that both Q 1 <strong>and</strong> Q ′ 1 ≡ (dQ(u)/dlnu) 1 = [p n −p s −S v ′ ] have small magnitudes: if they don’t, their neutron matter has problems. Setting44


Q 1 to a small number also results in small values <strong>for</strong> Q ′ 1. since <strong>for</strong> ǫ = 1/3, the value <strong>of</strong> β ′′is nearly zero, setting β L ′′ = 0 will ensure that Q 1 is small. The coupled set <strong>of</strong> equations todetermine α L <strong>and</strong> η L in this case areThese result inS vT o= 1 3 + α 2 − η 2 −α L +η L ,S ′ vT o= 2 9 + α 2 − 2 3 η −α L + 4 3 η L.α L = 2 3 + α 2 − η 2 − 4S v −3S ′ vT o≃ 1.6,η L = 1 3 + η 2 −3S v −S ′ vT o≃ 1.1.(3.25)(3.26)The neutron matter energy <strong>and</strong> pressure are now implicitly closely coupled to S v <strong>and</strong> S ′ vsuch that e N,1 ≃ S v −B <strong>and</strong> p N,1 /ρ o ≃ S ′ v . For Hebler & Schwen [63] estimates <strong>of</strong> e N,1 ≃ 14MeV <strong>and</strong> p N,1 /ρ o ≃ 2 MeV, we find S v ≃ 30 MeV <strong>and</strong> S ′ v ≃ 10 MeV. In this case, α L ≃ 1.2<strong>and</strong> η L ≃ 0.7.Alternatively, one could use a value <strong>of</strong> ǫ < 1/3, resulting in β ′′ > 0. To ensure Q 1 ≡ 0,a positive value <strong>for</strong> β L ′′ is also obtained. This results in different values <strong>for</strong> α L <strong>and</strong> η L <strong>and</strong> adifferent relation between S v , S v ′ <strong>and</strong> e N,1, p N,1 .Thus, it would be interesting to fit nuclei to determine the correlation between S v <strong>and</strong> S v ′ <strong>for</strong>a couple different values <strong>of</strong> ǫ, such as 1/3 or 1/6. One could then also add the correlations<strong>for</strong> S v <strong>and</strong> S v ′ obtained from the relations reduced from theoretical studies <strong>of</strong> neutron matterenergy <strong>and</strong> pressure at the saturation density.3.3 Optimized Parameter SetAddingnewdensitydependentinteractionssolvesnuclearincompressibility, opticalpotential,<strong>and</strong> pure neutron matter properties. To find an optimized parameter set, α L,U , β L,U , <strong>and</strong>,η L,U , we may compare the experimental binding energy <strong>of</strong> single nucleus with theoreticalcalculation from the modified model. For this set, we fix β ′′ = 0, which gives proper nuclearincompressibility (K ≃ 235 MeV) <strong>and</strong> varies S v <strong>and</strong> S v ′ (= L/3). Since the experimentalerror in the binding energy is extremely small, so that it is meaningless if we define χ 2 inconventional way, instead we define the χ 2 asχ 2 = 1 N∑ ( ) 2B ex,i −B th,i(3.27)σwhere N(= 2336) is the total number <strong>of</strong> nuclei in the calculation <strong>and</strong> σ is a mean error(arbitrary). The B ex,i is the data [45] <strong>and</strong> B th,i is the numerical calculation.The χσ has a unit <strong>of</strong> MeV so we can estimate the mean difference between B ex <strong>and</strong> B th ineach calculation. Even though our code is fast to calculate single nucleus’s properties, thenumber <strong>of</strong> calculation <strong>for</strong> this contour plot is 2336×41×41 ∼ 4·10 6 so we wrote a simple45


Figure 3.1: Solid line shows the contour plot <strong>of</strong> S v <strong>and</strong> L from FRTF II model. There is astrong correlation between S v <strong>and</strong> L. The dotted ellipse at the center represent statisticalcalculation from Eq. (3.28) (3.29), which makes χ 2 = 8.54.parallel code <strong>for</strong> it. Fig. 3.1 shows the contour plot <strong>of</strong> S v <strong>and</strong> L using FRTF II. The dottedellipse at the center is a statistical calculation from Eq. (3.28) <strong>and</strong> (3.29), which makesχ 2 = 8.5 to compare with the contour plot from numerical calculation. This contour plotimplies strong correlation between S v <strong>and</strong> L. This strong correlation also implies the strongcorrelation between S s <strong>and</strong> S v in liquid droplet model when we extract the in<strong>for</strong>mation <strong>of</strong>S s from semi-infinite nuclear matter using FRTF model.The χ 2 σ 2 = 7.54 MeV 2 gives mean error √ 7.54/100 ∼ 0.027 MeV per nucleon. This χ 2 σ 2can be reduced upto 4.72 MeV 2 if we add semi-empirical <strong>for</strong>mula <strong>for</strong> pairing 12/ √ A <strong>and</strong>shell effects. We used the Taylor expansion <strong>for</strong> χ 2 σ 2 to find σ Sv <strong>and</strong> σ L which areσ 2[ ]2∑χ 2 (p)−χ 2 0 = δp ij M ij δp ij (3.28)i,j=146


where p = (S v ,L) <strong>and</strong> M ij = 1 ∂ 2 p i∂ pj | p . Then the definition <strong>of</strong> σ Sv , σ L , <strong>and</strong> R Sv,L are givenbyσS 2 2∂L 2v=χ2∂S 2 vχ 2 ∂L 2χ2 −(∂ Sv ∂ L χ 2 ) 2σ2σL 2 2∂S 2 =vχ 2∂S 2 vχ 2 ∂L 2χ2 −(∂ Sv ∂ L χ 2 ) 2σ2 (3.29)R Sv,L = − ∂ S v∂ L χ 2√∂S 2 vχ 2 ∂L 2χ2 Fig. 3.2 shows the 95% confidence interval <strong>of</strong> S v <strong>and</strong> L from FRTF II <strong>and</strong> Kortelainen etal. [46] Our result has a simliar mean points <strong>of</strong> S v <strong>and</strong> L with Kortelainen et al. but strongcorrelation since Kortelainen et al. chose several parameters such as ρ s , R np , K at the sametime. The S v <strong>and</strong> L are on the same line <strong>and</strong> very close to each other.140120confidence interval, Kortelainen et alMean <strong>of</strong> S v <strong>and</strong> L <strong>of</strong> Kotelainen et alconfidence interval, FRTF N2, exchange potMean <strong>of</strong> S v <strong>and</strong> L <strong>of</strong> FRTF N2, exchange pot10080L (MeV)6040200-20-4024 26 28 30 32 34 36 38S v (MeV)Figure 3.2: This figure shows 95% confidence interval from FRTF II <strong>and</strong> Kortelainen et al.[46]. The red (blue) ellipse is the result from Kortelainen et al. (FRTF II). Our result has thesimilar slope with Kortelaine et al. but a stronger correlation. Kortelainen et al. assumedσ = 2 MeV, <strong>for</strong> FRTF II we used σ = 1.6 MeV.Recently Lattimer <strong>and</strong> Lim [47] summarized the allowed region <strong>of</strong> S v <strong>and</strong> L <strong>and</strong> comparedthose result with liquid droplet model. Our optimized values <strong>of</strong> S v <strong>and</strong> L are also inside theoverlapped region <strong>of</strong> S v <strong>and</strong> L from several experimentally allowed region.Fig. 3.3 shows S v <strong>and</strong> L allowed region. The blue sky region is the result from Sn neutronskin [48]. GDR comes from giant dipole resonance [49]. There is a restriction from 208 Pbdipole polarizability [52]. HIC represents the result from heavy ion collision [51]. <strong>Nuclear</strong>masses is the result from Korelainen et al. [46]. The hatched region comes from neutron47


Figure 3.3: Several experimental result can restrict the S v <strong>and</strong> L. Our optimized S v <strong>and</strong> Lare in the white overlapped region in the middle <strong>of</strong> the figure [47].star observations [53]. There are also pure neutron matter calculations with purple polygons[54, 55]. This allowed region <strong>of</strong> S v <strong>and</strong> L will give important constraints to develop newnuclear <strong>for</strong>ce model in the future.48


3.4 <strong>Nuclear</strong> Surface TensionCompared with the truncated model, the modified model has a lower critical temperature<strong>and</strong> surface tension. These properties will be used to construct E.O.S table using liquiddroplet approach. In the liquid droplet approach (chapter 6), the surface tension as a function<strong>of</strong> T <strong>and</strong> proton fraction x is required.The critical temperature <strong>for</strong> asymmetric nuclear matter occurs when the dilute <strong>and</strong> densematter have the same density <strong>and</strong> proton fraction. In general, the proton fractions in bothdense <strong>and</strong> dilute matter is different each other except <strong>for</strong> symmetric nuclear matter. Thesurface tension can be obtained only numerically using Eq. (2.70).The surface tension <strong>for</strong> nuclear matter at finite temperature <strong>and</strong> given proton fraction canbe approximated [17] by[ ( ) 2 ] p Tω(x i ,T) = ω 0 (x i ) 1− . (3.30)T c (x i )For both the truncated model <strong>and</strong> the new model, p = 5/3 explains the numerical behaviorsvery well.The critical temperature at which both the dilute <strong>and</strong> dense phase have the same density<strong>and</strong> proton fraction, is approximated byT c (x i ) = T c (x i = 0.5)(1+a(1−2x i ) 2 +b(1−2x i ) 4 +c(1−2x i ) 6 ). (3.31)The a, b, <strong>and</strong> c can be obtained by χ 2 minimization between the numerical result <strong>and</strong> fittingTable 3.1: The critical temperature analytic fitting functionModel T c,0 a b cTruncated 17.354 -0.5006 0.66126 -1.33316New 14.626 -0.4296 0.34216 -1.00298function. Fig. 3.4 shows the plot <strong>of</strong> critical temperature <strong>of</strong> numerical <strong>and</strong> fitting function.This fitting function is used to calculate finite temperature surface tension in Eq. (3.30).The surface tension at T = 0 MeV <strong>and</strong> Y p = 0.5 is also a function <strong>of</strong> proton factions. ω 0 canbe approximated with fractional function as in LS model,2·2 α +qω 0 (x i ) = ω 0 (x i = 0.5)x −α(3.32)i +q +(1−x i ) −αThe parameter α is left free so that we can choose from 2 to 4.0 <strong>and</strong> q is obtained from χ 2minimization. For both models, the best α’s are 3.6 <strong>and</strong> 3.7.49


181614FRTF II Critical TFRTF II fitting functionFRTF I Critical TFRTF I fitting function12T c (MeV)10864200.50.40.3x i0.20.10Figure3.4: Thecritical temperatureasagivenprotonfraction:Numerical result<strong>and</strong>analyticfitting function.Table 3.2: Surface tension analytic fitting function <strong>for</strong> finite range modelModel ω 0 (x i = 0.5) α qTruncated 1.25728 3.6 39.0552New 1.02824 3.7 56.047950


1.41.2FRTF II ω 0 , T=0 MeVFRTF II fitting functionFRTF I ω 0 , T=0 MeVFRTF I fitting function1ω 0 (MeV/fm 2 )0.80.60.40.200.50.40.3x i0.20.10Figure 3.5: Surface tension at T = 0 MeV as a given proton fraction: Numerical result <strong>and</strong>analytic fitting function.51


Chapter 4Gaussian Type Finite Range model† The finite range <strong>for</strong>ce can be managed with Gaussian distance (f(r) ∼ e −(r/r 0) 2 ) dependence.This distance dependence does not represent the repulsion when the distance betweennucleons so close (< 0.1fm) but it can be justified since the total interaction energy is obtainedfrom phase space integration. It also give a better representation <strong>of</strong> radii <strong>of</strong> nuclei<strong>and</strong> neutron skin than the Yukawa type finite range model.4.1 <strong>Nuclear</strong> Density Functional <strong>Theory</strong>4.1.1 Energy density functionalThe energy <strong>of</strong> nuclear matter is given byE = T kin +E FR +E ZR +E C +E S,L (4.1)where T kin , E FR , E ZR , E C , <strong>and</strong> E S,L are the kinetic energy, nuclear finite-range interaction,zero-range interaction, Coulomb interaction, <strong>and</strong> spin-orbit coupling respectively. Thekinetic energy contribution from nucleons is simply obtained by∫T kin =d 3 r ∑ t 22m τ t, (4.2)where t is the type <strong>of</strong> nucleons.In this Gaussian nuclear density functional (GNDF)theory, the number <strong>and</strong> kinetic densitiesareρ t = 1 ∫ ∞f4π 3 3 t d 3 p; τ t = 1 ∫ ∞f4π 3 5 t p 2 d 3 p, (4.3)where f t is the Fermi-Dirac density function,00f t =1. (4.4)1+e(ǫt−µt)/T† This chapter is based on Y. Lim’s work, arXiv:1101.1194. This work will be modified <strong>and</strong> resubmitted.52


For the finite-range term, we use a Gaussian phenomenological model <strong>for</strong> the nuclear potential,E FR = ∑ ∫1[]d 3 rπ 3/2 r 3 1 d 3 r 2 e −r2 12 /r2 0V 1L ρ t (r 1 )ρ t (r 2 )+V 1U ρ t (r 1 )ρ t ′(r 2 )t 0+ ∑ ∫1[]d 3 rπ 3/2 r 3 1 d 3 r 2 e −r2 12 /r2 0V 2L ρ 1+ǫt (r 1 )ρ 1+ǫt (r 2 )+V 2U ρ 1+ǫt (r 1 )ρ 1+ǫt(r ′ 2 )t 0+ ∑ ∫(4.5)1d 3 rπ 3/2 r 3 1 d 3 r 2 e −r2 12 /r2 0 ×t 0[∫∫]d 3 p t1 d 3 p t2 f t1 f t2 V 3L p 2 12 + d 3 p t1 d 3 p t ′ 2f t1 f t ′ 2V 3U p 2 12 ,where p 12 = |p 1 −p 2 |, r 12 = |r 1 −r 2 |, r 0 is the length <strong>of</strong> interaction, <strong>and</strong> V 1L , V 1U , V 2L , V 2U ,V 3L , <strong>and</strong> V 3U are interaction parameters to be determined. The last term is added to explainthe effective mass <strong>of</strong> nucleons in dense matter.The zero-range term in the nuclear <strong>for</strong>ce can be regarded as the energy contribution fromthree-body nuclear <strong>for</strong>ces. The three-body <strong>for</strong>ce is quite important if the baryon densityincreases beyond two or three times the saturation density. One possible <strong>for</strong>m <strong>of</strong> the threebody<strong>for</strong>ce is [34]E ZR = 1 ∫4 t 3 d 3 rρ n (r)ρ p (r)ρ(r), (4.6)where t 3 istheinteractionstrength<strong>for</strong>athree-body<strong>for</strong>ce, ρ n (ρ p )isneutron(proton)density,<strong>and</strong> the ρ is total density.The energy functional <strong>for</strong> the Coulomb interaction has an exchange term which is absent inclassical physics,E C = E pp∫ ∫= e22C +Eex Cd 3 r 1 d 3 r 2ρ p (r 1 )ρ p (r 2 )r 12− 34π (3π2 ) 1/3 e 2 ∫d 3 rρ 4/3p (r).(4.7)In bulk nuclear matter, the spin-orbit <strong>and</strong> Coulomb interactions constitute a small portion<strong>of</strong> the total energy, so we neglect these two terms. Then the bulk density functional wouldbeE B = T kin +E FR +E ZR∫(4.8)= d 3 rE B (r),53


where E B is the energy density <strong>for</strong> bulk matter. Using Eq.(4.2), (4.5), <strong>and</strong> (4.6), we findE B = 2 τ n2m + 2 τ p2m+V 1L (ρ n 〈ρ n 〉+ρ p 〈ρ p 〉)+V 1U (ρ n 〈ρ p 〉+ρ p 〈ρ n 〉)+V 2L (ρ 1+ǫn 〈ρ1+ǫ n 〉+ρ1+ǫ p 〈ρ 1+ǫp 〉)+V 2U (ρ 1+ǫn 〈ρ1+ǫ p 〉+ρ 1+ǫp 〈ρ 1+ǫn 〉) (4.9)+V 3L (ρ n 〈τ n 〉+τ n 〈ρ n 〉+ρ p 〈τ p 〉+τ p 〈ρ p 〉)+V 3U (ρ n 〈τ p 〉+τ n 〈ρ p 〉+ρ p 〈τ n 〉+τ p 〈ρ n 〉)+ 1 4 t 3ρρ n ρ pwhere we defined the Gaussian-type integral using ‘〈...〉’:〈u(r 1 )〉 = 1 ∫d 3 rπ 3/2 r03 2 e −r2 12 /r2 0 u(r2 ). (4.10)4.1.2 Effective mass, Potential, <strong>and</strong> Thermodynamic propertiesThe effective mass <strong>of</strong> the nucleons at the nuclear saturation density is about 0.7m B . Somenuclear density functionals use the effective mass m ∗ = m B . However, we introduce the momentumdependentinteractionwhich describes theeffective mass<strong>of</strong>nucleons. Thefunctionalderivative δE B gives us the effective masses <strong>and</strong> potentials in the nuclear density functional,∫)δE B = d 3 r(V n δρ n + 2δτ2m ∗ n +V p δρ p + 2δτn 2m ∗ p . (4.11)pNow we get the effective mass <strong>for</strong> neutrons <strong>and</strong> protons,m ∗ t =m1+4m(V 3L 〈ρ t 〉+V 3U 〈ρ t ′〉)/ 2 (4.12)<strong>and</strong> the potentials[V t =2 V 1L 〈ρ t 〉+V 1U 〈ρ t ′〉+(1+ǫ)ρ ǫ t (V 2L〈ρ 1+ǫt 〉+V 2U 〈ρ 1+ǫt〉) ′]+V 3L 〈τ t 〉+V 3U 〈τ t ′〉 + 1 4 t 3(2ρ t +ρ t ′)ρ t ′ .(4.13)The thermodynamic properties are extremely important to properly describe the hot densematter. The degeneracy parameter, which is in the Fermi-Dirac distribution function is thekey, to open the thermodynamic properties. Using the Fermi-integral, we get the baryonnumber density <strong>and</strong> kinetic density with degeneracy parameter φ t = (µ t −V t )/T,ρ t = 12π 2 ( 2m∗t T 2 ) 3/2F 1/2 (φ t ), τ t = 12π 2 ( 2m∗t T 2 ) 5/2F 3/2 (φ t ). (4.14)54


L<strong>and</strong>au’s quasi-particle <strong>for</strong>mula gives us the entropy density S t , which tell us how to findthe pressure in this density functional,S t = − 2 ∫d 3 p [ f 3 t lnf t +(1−f t )ln(1−f t ) ] = 526m ∗ tT τ t − µ t −V tρ t . (4.15)TFrom the thermodynamic identity <strong>and</strong> entropy density given above, we can get pressure,p = µ n ρ n +µ p ρ p +TS n +TS p −E= ∑ ( 52τ6m ∗ t +V t ρ t)−Et t= 2 τ n3m + 2 τ p3m +V 1L(ρ n 〈ρ n 〉+ρ p 〈ρ p 〉)+V 1U (ρ n 〈ρ p 〉+ρ p 〈ρ n 〉)+V 2L (1+2ǫ)(ρ 1+ǫn 〈ρ 1+ǫn 〉+ρ 1+ǫp 〈ρ 1+ǫp 〉)+V 2U (1+2ǫ)(ρ 1+ǫn 〈ρ 1+ǫp 〉+ρ 1+ǫp 〈ρ 1+ǫn 〉)+V 3L (ρ n 〈τ n 〉+ 7 3 τ n〈ρ n 〉+ρ p 〈τ p 〉+ 7 3 τ p〈ρ p 〉)(4.16)+V 3U (ρ n 〈τ p 〉+ 7 3 τ n〈ρ p 〉+ρ p 〈τ n 〉+ 7 3 τ p〈ρ n 〉)+ 1 2 t 3ρρ n ρ p .At zero temperature non uni<strong>for</strong>m matter, the chemical potential <strong>of</strong> the proton <strong>and</strong> neutronare given by,µ t = 2(3π 2 ρ2m ∗ t ) 2/3 +V tt= 22m (3π2 ρ t ) 2/3 +2(V 3L 〈ρ t 〉+V 3U 〈ρ t ′〉)(3π 2 ρ t ) 2/3 (4.17)[]+2 V 1L 〈ρ t 〉+V 1U 〈ρ t ′〉+(1+ǫ)ρ ǫ t (V 2L〈ρ 1+ǫt 〉+V 2U 〈ρ 1+ǫt〉)+V ′ 3L 〈τ t 〉+V 3U 〈τ t ′〉+ 1 4 t 3(2ρ t +ρ t ′)ρ t ′.4.2 Parameters <strong>for</strong> the Gaussian nuclear density functionalEvery nuclear model should reproduce five nuclear matter properties: binging energy, pressure,nuclear incompressibility, symmetry energy <strong>and</strong> effective mass, m ∗ . We use the saturationproperties <strong>of</strong> nuclear matter to determine the parameters <strong>of</strong> the density functional.For zero-temperature uni<strong>for</strong>m nuclear matter, we have the energy density as a function <strong>of</strong>55


u = ρ/ρ 0 <strong>and</strong> x = ρ p /ρ,E B= 3 T 0 ρ 0 5 22/3 u 5/3[ (1−x) 5/3 +x 5/3] +u 2[ v 1L (x 2 +(1−x) 2 )+2v 1U x(1−x) ]+2 1+2ǫ u 2+2ǫ[ v 2L (x 2+2ǫ +(1−x) 2+2ǫ )+2v 2U x 1+ǫ (1−x) 1+ǫ] (4.18)+2 2/3 u 8/3[ v 3L (x 8/3 +(1−x) 8/3 )+v 3U (x(1−x) 5/3 +x 5/3 (1−x)) ]+ 1 4 t′ 3u 3 x(1−x),where we define the parametersT 0 = 22m (3π2 ρ 0 /2) 3/2 , v 1L,U = ρ 0V 1L,U , v 2L,U = 1 ( ρ0) 1+2ǫV2L,UT 0 T 0 2v 3L,U = 4·35/3 π 4/35T 0( ρ0) 5/3V3L,U, t ′ 32= ρ2 0t 3 .T 0(4.19)Now we assume that the momentum-dependent interaction is blind to the type <strong>of</strong> nucleon,so V 3L = V 3U (v 3 = v 3L + v 3U ). The binding energy <strong>of</strong> symmetric nuclear matter (u = 1,x = 1/2) is then given by[E 0 3= −B 0 = T 0ρ 0 5 + v 1L +v 1U2+v 2L +v 2U + v ]32 + t′ 3, (4.20)16where B 0 = 16MeV is the binding energy per baryon at the nuclear saturation density.The pressure at the saturation density vanishes, which means that the energy per baryonhas its minimum at the saturation density,[ 2p 0 = ρ 0 T 05 + v 1L +v 1U+(1+2ǫ)(v 2L +v 2U )+ 5 26 v 3 + 1 ]8 t′ 3 = 0. (4.21)The incompressibility parameter at the saturation density is given byK 0 = 9 dp∣dρ∣ρ=ρ0[= T 0 6+9(v1L +v 1U )+9(1+2ǫ)(2+2ǫ)(v 2L +v 2U )+20v 3 + 27 ]8 t′ 3= 265 MeV.(4.22)56


The symmetry energy in nuclear matter is defined asS v = 1 8d 2 (E/ρ)dx 2∣∣∣∣ρ=ρ0,x=1/2= T 0[ 13 + v 1L −v 1U2= 28 MeV.+(1+ǫ)((1+2ǫ)v 2L −v 2U )+ 5 18 v 3 − 1 16 t′ 3](4.23)Another parameter, which is related to symmetry energy, is given byL = 3ρ 08d 3 (E/ρ)dρdx 2 ∣∣∣∣ρ=ρ0,x=1/2[ 2= T 03 + 3 2 (v 1L −v 1U )+3(1+2ǫ)(1+ǫ)((1+2ǫ)v 2L −v 2U )+ 2518 v 3 − 3 ]8 t′ 3= 54 MeV.(4.24)We choose the effective mass at the saturation density as 0.78m b <strong>and</strong> use this number in theEq. (4.12),m ∗ m=1+2mρ 0 V 3 / = 0.78m b. (4.25)2Thus we can easily recover v 3 from Eq. (4.19). From Eqs. (4.20), (4.21), <strong>and</strong> (4.22), we canhave v 1 = v 1L +v 2L , v 2 = v 2L +v 2U <strong>and</strong> t ′ 3 :v 1 = 5K 0/T 0 +5v 3 (1−3ǫ)−72ǫ−90B 0 /T 0 (1+2ǫ)−1245ǫv 2 = 12+90B 0/T 0 −5K 0 /T 0 −5v 390ǫ(1−2ǫ)t ′ 3 = −8v 1 −16v 2 −8v 3 − 16B 0T 0− 485 . (4.26)Then we can manipulate Eq. (4.23), <strong>and</strong> (4.24) to get v 1L <strong>and</strong> v 2L ,v 1L = 1 2 v 1 + 1 [ 5(1−3ǫ)v 3 + 2ǫ−12ǫ 27 16 t′ 3 + (1+2ǫ)S vT 0v 2L =12(1+ǫ) v 2 −[1 54ǫ(1+ǫ) 2 27 v 3 − t′ 316 + S v− L − 1 T 0 3T 0 9− L − 1+6ǫ3T 0 9],](4.27)<strong>and</strong> we can have v 1U = v 1 −v 1L <strong>and</strong> v 2U = v 2 −v 2L .57


Table 4.1: Interaction parameters when ǫ = 1/6, K = 265 MeV, S v = 28MeV, L = 54MeV.v 1L v 1U v 2L v 2U v 3L,U t ′ 3 r 0 (fm)-1.766 -3.472 0.410 0.931 0.169 1.177 1.2054.2.1 Determination <strong>of</strong> the 1+ǫ powerWe added in Eq. (4.5) the auxiliary density interaction with the 1 + ǫ power. We mightregard 1+ǫ as the many-body effect—<strong>for</strong> example, a three-body <strong>for</strong>ce if ǫ > 1 . It is known,2however, that interactions among more than three-bodies are unimportant in dense matter[34]. Thus we can restrict the ǫ to be less than 1 . As ǫ changes, the t 2 3 parameter changessign, which means that the three-body <strong>for</strong>ce can be attractive or repulsive. In the generalSkyrme model with the three-body <strong>for</strong>ce, the t 3 parameter is positive. We choose ǫ = 1/6 sothat the interaction has the <strong>for</strong>m <strong>of</strong> ρ 7/6t 1ρ 7/6t 2. In zero-temperature, uni<strong>for</strong>m matter, we haveu 7/3 terms in the energy density. From Eq. (4.18), the energy density has u 5/3 , u 2 , u 7/3 , u 8/3 ,<strong>and</strong> u 3 terms if we have ǫ = 1/6 so we can use a statistical approach in uni<strong>for</strong>m matter. Fig4.1 shows the energy per baryon from GNDF <strong>and</strong> APR [35] EOS. We can see that as thedensity increases, the pressure from the two models agrees very well.Figure 4.1: The solid line represents the energy per baryon (uni<strong>for</strong>m matter) using GNDF.The upper (lower) curve represents the energy perbaryon <strong>of</strong>pure neutronmatter (symmetricnuclear matter). The energy per baryon (dotted line) from the APR [35] EOS is added <strong>for</strong>comparison.4.2.2 The effective range <strong>of</strong> the nuclear <strong>for</strong>ce : r 0In the Gaussian-interaction model, we can see that the effective range <strong>of</strong> the <strong>for</strong>ce is givenby r 0 , which is approximately ∼ 1 to 2 fm. We do not have an analytic <strong>for</strong>m <strong>of</strong> r 0 , so we58


Figure 4.2: The left figure shows the quantity p 0 −p(z) at the semi-infinite nuclear surfacewhen r 0 = 1.205 fm <strong>and</strong> Y e = 0.5. The surface tension from this configuration is ω = 1.250MeVfm −2 . Therightfigureshowsthesurfacetension(solidline)<strong>and</strong>t 90−10 thickness (dashedline) as a function <strong>of</strong> r 0 . When r 0 = 1.205, t 90−10 = 2.412 fm. The surface tension <strong>and</strong> t 90−10thickness are both linear functions <strong>of</strong> r 0 .need to rely on the numerical solution <strong>of</strong> the surface tension <strong>of</strong> semi-infinite nuclear matter:ω =∫ ∞−∞∫[ ] ∞[ ]E −TSn −TS p −µ n ρ n −µ p ρ p +p 0 dz = − p(z)−p0 dz, (4.28)where p 0 is the pressure at z = −∞ or z = +∞. In one-dimensional, semi-infinite nuclearmatter, we assume that the nuclear density depends only on the z-axis. The Gaussianintegral then becomes1π 3/2 r 3 0∫d 3 ru(r) = 1π 1/2 r 0∫ +∞−∞−∞dzu(z). (4.29)Experimental values <strong>for</strong> the surface tension <strong>and</strong> surface thickness are ω = 1.250 MeV fm −2<strong>and</strong> t 90−10 = 2.3 fm. Fig. 4.2 shows the numerical calculation, which says that r 0 = 1.205fm from the surface tension <strong>and</strong> r 0 = 1.149 from t 90−10 thickness. There is a 5% discrepancybetween the two results. Table 4.1 shows the interaction parameters which we use in thispaper when K = 265 MeV, S v = 28 MeV, L = 54 MeV, <strong>and</strong> ǫ = 1/6. The simple densitydependent interactions (v 1L,U ) are attractive. On the other h<strong>and</strong>, the auxiliary densitydependent interactions (v 2L,U ), momentum dependent interactions (v 3L,U ) three-body <strong>for</strong>ce(t 3 ) are repulsive in our model.59


4.3 <strong>Nuclear</strong> matter <strong>and</strong> nuclei4.3.1 Specific heatThe specific heat <strong>of</strong> uni<strong>for</strong>m nuclear matter can be obtained byC V = T ∂S∂T ∣ = ∂Eρ∂T ∣ . (4.30)ρFor a non-interacting Fermi gas, the specific heat increases linearly with temperature. Whenthe temperature is low enough, we expect that the specific heat <strong>of</strong> the nuclear matter tendsto behave like a free Fermi gas. The specific heat <strong>for</strong>mula <strong>for</strong> the degenerate gas is given by[36]C V = 1 3 m∗ k F k 2 B T , (4.31)where m ∗ is the effective mass <strong>of</strong> a nucleon, k F is the Fermi momentum, <strong>and</strong> k B is theBoltzman constant.However, as the temperature increases, non-linear behavior <strong>of</strong> the specific heat becomesapparent so that the degenerate gas <strong>for</strong>mula is no longer valid, due to the excitation <strong>of</strong>nucleons deep inside the Fermi surface.Figure 4.3: This figure shows the specific heat per nucleon <strong>of</strong> uni<strong>for</strong>m matter <strong>for</strong> differentdensities. If the temperature is low enough, the specific heat behaves linearly with temperature.To calculate the specific heat <strong>of</strong> uni<strong>for</strong>m nuclear matter, we use the Johns, Ellis, <strong>and</strong> Lattimer(JEL) method [12], which enables us to get the pressure, energy density <strong>and</strong> entropydensity <strong>for</strong> a given degeneracy parameter. Fig. 4.3 shows the specific heat per nucleon <strong>of</strong>uni<strong>for</strong>m nuclear matter. It shows the linear relation between the specfic heat <strong>and</strong> temperatureat low temperature region as in Eq. (4.31).60


Figure 4.4: This figure shows the basic properties <strong>of</strong> the closed-shell nuclei which can beobtained from the GNDF. The solid (dotted) line indicates the neutron (proton) density asa function <strong>of</strong> radius. As the number <strong>of</strong> nucelons in the nuclei increase, the neutron skinthickness (R n −R p ) increases.A detailed calculation <strong>of</strong> the specific heat at sub-nuclear density in the neutron star needs totake into account the beta-equilibrium condition <strong>and</strong> heavy nuclei with a neutron gas. Thespecific heat plays an important role in the cooling process <strong>of</strong> neutron stars. In the neutronstar crust, there are heavy nuclei <strong>and</strong> a free-neutron gas. The effective masses <strong>of</strong> protons<strong>and</strong> neutrons are different from the center <strong>of</strong> the heavy nuclei <strong>and</strong> dilute neutron gas, sowe cannot use Eq. (4.31). In this case the specific heat at the neutron star crust can becalculated numerically by changing temperature <strong>and</strong> comparing the total energy change.4.3.2 Nuclei at T = 0 MeVWe can use the GNDF theory <strong>and</strong> the Lagrange multiplier method to find the radius <strong>and</strong>binding energy per nucleon <strong>for</strong> a single nucleus using the Winger-Seitz cell method. In theLagrange multiplier method, the chemical potentials <strong>of</strong> protons <strong>and</strong> neutrons are constantin the Wigner-Seitz cell to minimize the total free energy. Fig. 4.4 shows the radius <strong>and</strong>binding energy <strong>of</strong> the closed-shell nuclei obtained using this method. These results agree wellwith the experiment [32]. 40 Ca has a larger charge radius than the neutron radius because<strong>of</strong> the Coulomb repulsion between protons. The solid (dotted) line denotes the neutron(proton) density. As the atomic number increases, the central density <strong>of</strong> neutrons increases;on the other h<strong>and</strong> the central density <strong>of</strong> protons decreases. The difference between charge<strong>and</strong> neutron radii increases <strong>and</strong> the neutron skin becomes thicker as the atomic numberincreases. In 208 Pb nuclei, the central density <strong>of</strong> protons is lower than the proton density <strong>of</strong>the outer part <strong>of</strong> nuclei (r = 4−5 fm) because <strong>of</strong> proton Coulomb repulsion.Table 4.2 shows the proton <strong>and</strong> neutron radii <strong>and</strong> binding energy per baryon <strong>of</strong> closedshell nuclei from various nuclear models. The calculation from GNDF theory agrees wellwith experimental results.4.3.3 Heavy nuclei in the neutron star crustIntheneutronstarcrust, heavynucleiare<strong>for</strong>medwithafree-neutrongas. Theseheavynucleiaresuspected to <strong>for</strong>mabodycentered cubic (BCC) structure. Inthestaticequilibrium state,we calculate the density pr<strong>of</strong>ile <strong>of</strong> heavy nuclei with a neutron gas using the Wigner-Seitz61


Table 4.2: Comparsion <strong>of</strong> the results from Steiner (Potential & Field Theoretical) et al.[32], FRTF I(truncated), FRTF II(modified) <strong>and</strong> the GNDFNucleus Property Experiment Potential FT FRTF I FRTF II GNDF208 Pb r ch (fm) 5.50 5.41 5.41 5.38 5.45 5.44BE/A(MeV) 7.87 7.87 7.77 8.01 8.17 8.02δR(fm) 0.12 ± 0.05 0.19 0.20 0.15 0.13 0.140.20 ± 0.0490 Zr r ch (fm) 4.27 4.18 4.17 4.10 4.15 4.17BE/A(MeV) 8.71 8.88 8.65 8.77 9.00 8.72δR(fm) 0.09 ± 0.07 0.075 0.093 0.064 0.054 0.05740 Ca r ch (fm) 3.48 3.40 3.34 3.22 3.26 3.31BE/A(MeV) 8.45 8.89 8.61 8.47 8.77 8.33δR(fm) -0.06 ± 0.05 -0.044 -0.046 -0.036 -0.039 -0.042-0.05 ± 0.04Cell method. The plot on the left side <strong>of</strong> Fig. 4.5 shows the proton (dotted line) <strong>and</strong> neutrondensity (solid line) pr<strong>of</strong>iles from the center (r=0 fm) <strong>of</strong> heavy nuclei when ρ = 0.01fm −3 .There are dripped neutrons outside <strong>of</strong> the heavy nuclei. The cell size (Rc), which is a roughestimate <strong>of</strong> the distance between neighboring heavy nuclei, is determined by nuclear density<strong>and</strong> beta equilibrium conditions (µ n = µ p + µ e ). There is a wave function overlap at theboundary <strong>of</strong> the Wigner-Seitz cell. The actual distance between heavy nuclei is (8π/3) 1/3 Rc.The right side <strong>of</strong> Fig. 4.5 shows the binding energy per baryon as a function <strong>of</strong> Wigner-Seitzcell size. As the density decreases, the cell size increases <strong>and</strong>the energy per baryon convergesto −8.0 MeV.Table 4.3: <strong>Nuclear</strong> properties in the neutron star crustρ (fm 3 ) p (MeV/fm 3 ) ǫ (MeV/fm 3 ) N nuc Z Rc (fm)5.623×10 −2 0.181 53.06 271.6 89.51 25.125.012×10 −2 0.147 47.27 218.0 32.18 18.703.981×10 −2 9.332×10 −2 37.52 137.8 21.69 18.062.512×10 −2 3.965×10 −2 23.65 145.3 27.93 22.771.585×10 −2 1.976×10 −2 14.91 150.7 32.07 26.361×10 −2 1.081×10 −2 9.405 147.7 34.29 29.151×10 −3 7.988×10 −4 9.383×10 −1 116.3 38.50 40.711×10 −4 6.945×10 −5 9.352×10 −2 79.23 38.31 65.471×10 −5 1.480×10 −6 9.326×10 −3 58.02 36.98 131.41×10 −6 2.362×10 −9 9.311×10 −4 47.90 35.14 270.61×10 −7 1.714×10 −9 9.303×10 −5 43.37 34.10 569.7Table 4.3 shows the thermodynamic properties <strong>and</strong> physical dimensions <strong>of</strong> nuclei in theneutron star crust. N nuc <strong>and</strong> Z are the number <strong>of</strong> neutrons <strong>and</strong> protons <strong>of</strong> heavy nuclei inthe Wigner-Seitz Cell respectively. The atomic number <strong>of</strong> heavy nuclei remains Z ∼ 35 <strong>for</strong>62


Figure 4.5: In a neutron star, heavy nuclei exist. The left figure shows the density pr<strong>of</strong>ile<strong>of</strong> proton (dotted line) <strong>and</strong> neutron (solid line). r = 0 fm means the center <strong>of</strong> heavy nuclei.Outside the heavy nuclei, there are dripped neutrons. As density decreases, the cell sizeincreases <strong>and</strong> the energy per baryon converges to −8.0 MeV.a large range <strong>of</strong> densities be<strong>for</strong>e the phase transition to uni<strong>for</strong>m matter. This means thatthe proton fraction decreases as the density increases. For a narrow range <strong>of</strong> densities, theatomic number suddenly increases <strong>and</strong> the heavy nuclei merge with free neutrons to <strong>for</strong>muni<strong>for</strong>m nuclear matter.Figure4.6: Effective mass <strong>of</strong> nucleons inthe Wigner-Seitz cell as a function <strong>of</strong> radial distancefrom the center <strong>of</strong> heavy nuclei. Since the nuclear interaction is weak at the boundary <strong>of</strong> thecell, the effective mass <strong>of</strong> nucleon <strong>and</strong> the pure mass <strong>of</strong> a nucleon become equal.Fig. 4.6 shows the effective mass <strong>of</strong> nucleons in the Wigner-Seitz cell. The effective mass<strong>of</strong> nucleons in the Wigner-Seitz cell is given by Eq. (4.12) <strong>and</strong> Eq. (4.19) combining to give,m ∗ t =1+ 5 3m( ) . (4.32)˜ρv t ρ˜3L ρ 0+v t ′3U ρ 063


Since we assume that the momentum interaction isblind with respect to isospin, the effectivemass is identical <strong>for</strong> different isospin nucleons. The effective mass to pure mass ratio <strong>of</strong>nucleons is 0.78 at the center <strong>of</strong> the heavy nuclei <strong>and</strong> becomes 1 at the outer region <strong>of</strong> theWigner-Seitz cells since the density <strong>of</strong> nuclear matter is low at the outside <strong>of</strong> the heavynuclei, the interaction energy <strong>of</strong> nuclear matter is weak.4.4 Phase transitionIn the neutron star, we can see two types <strong>of</strong> phase transitions: one is the phase transitionfrom nuclei with a neutron gas to uni<strong>for</strong>m matter, <strong>and</strong> the other is the phase transitionfrom uni<strong>for</strong>m nuclear matter to quark matter. During the first phase transition, we can seethe nuclear pasta phase [41] † . That is, spherical nuclei become ellipsoidal, then cylindrical,<strong>and</strong> finally slab phase be<strong>for</strong>e nuclear matter becomes uni<strong>for</strong>m matter. However, the energydifference is quite small, so that the effects on the large scale physics are negligible. On theother h<strong>and</strong>, the second phase transition is quite dramatic. The energy <strong>and</strong> pressure changesignificantly from nuclear matter to quark matter.4.4.1 Uni<strong>for</strong>m matterTo check the phase transition points fromheavy nuclei with a neutron gas to uni<strong>for</strong>m nuclearmatter, we can simply compare the energy per baryon <strong>of</strong> uni<strong>for</strong>m nuclear matter with theenergy per baryon <strong>of</strong> nuclei with a neutron gas since the nuclear matter exists in the lowestenergy states. The energy per baryon in uni<strong>for</strong>m matter can be easily obtained by changingthe ‘〈...〉’ integrals to non integral <strong>for</strong>m from Eq. (4.11) since the Gaussian integrations inuni<strong>for</strong>m matter become unity. Typically there is a phase transition around 0.5ρ 0 .We know that in the outer crust <strong>of</strong> the neutron star the nuclei have a BCC structure.If we assume that the pasta phase exists in the low-density region, we may use the densityperturbation to see the phase transition from nuclei with a neutron gas to uni<strong>for</strong>m nuclearmatter. We use the wave number perturbation to see the energy exchange, which has contributionsfrom the volume effects, gradient effects, <strong>and</strong> Coulomb energy can be approximated[20],v(q) ≃ v 0 +βq 2 + 4πe2 , (4.33)q 2 +kTF2where q is the sinusoidal variation <strong>of</strong> the wave number in the spatially periodic densityperturbation. The volume term is given byv 0 = ∂µ p∂ρ p− (∂µ p/∂ρ n ) 2(∂µ n /∂ρ n ) . (4.34)† Around 1/2ρ 0 , spherical nuclei is de<strong>for</strong>med to be oblate nuclei, cylindrical phase, slab, cylindrical hole,<strong>and</strong> bubble phase to minimize the total free energy density. Since its shape resembles Italian pasta, it iscalled ‘nuclear pasta’ phase.64


The energy exchange from the gradient has the <strong>for</strong>mβ = D pp +2D np ξ +D nn ξ 2 , ξ = − ∂µ p/∂ρ n∂µ n /∂ρ n(4.35)where the coefficients <strong>of</strong> the gradient terms are given by D pp = D np = D nn = 132MeV·fm 5[18]. The k TF in the Coulomb interaction represents the inverse Thomas-Fermi screeninglength <strong>of</strong> the electrons. When we see the change in the sign <strong>of</strong> v, the uni<strong>for</strong>m matter phaseis more stable than the periodic structure <strong>of</strong> the nuclei. The v has a minimum atwhen q 2 min = (4πe2 /β) 1/2 −k 2 TF .v min = v 0 +2(4πe 2 β) 1/2 −βk 2 TF , (4.36)Another way to see the phase transition is to use the thermodynamic instability. Thethermodynamic stability condition can be described using the inequalities [18, 19],( ) ∂P− >0,∂vµ( (4.37)∂µ− >0.∂q c)vwhere P = P b + P e is the total pressure from electrons <strong>and</strong> baryons <strong>and</strong> µ = µ n − µ p isthe difference between the neutron <strong>and</strong> proton chemical potentials, which is the electronchemical potential in beta-stable matter. q c is defined as q c = x p −ρ e /ρ. Mathematically,the inequalities in Eq. (4.37) show that the energy per baryon is convex. Eq. (4.37) can beverified to be [18, 19]( ) [ ∂P− = ρ 2 2ρ ∂E(ρ,x (p)+ρ 2∂2 E(ρ,x p ) ∂ 2 E(ρ,x p )− ρ∂vµ∂ρ ∂ρ 2 ∂ρ∂x p( ∂µ− =∂q c)v( ∂ 2 E(ρ,x p )∂x 2 p) 2/ ] ∂ 2 E(ρ,x p )> 0,∂x 2 p) −1+ µ2 eπ 2 3 ρ > 0. (4.38)The second <strong>of</strong> Eq. (4.38) always holds, so the first will determine the phase transition in theneutron star crust. Xu et al.[18], use a simple equation to determine the instability usingthe thermodynamic relation,2∂E∂ 2 Eρ ∂ρ ∂x 2 p+ ∂2 E∂ρ 2 ∂ 2 E∂x 2 p−( ) ∂ 2 2E= ∂µ ( ) 2n ∂µ p ∂µn− . (4.39)∂ρ∂x p ∂ρ n ∂ρ p ∂ρ pEq. (4.39) is equivalent to the volume part <strong>of</strong> the thermodynamic perturbation Eq. (4.34)method except that there is a ∂µ n /∂ρ n difference. Comparing the two methods (pertubation<strong>and</strong> thermodynamic instability) shows the effects <strong>of</strong> the gradient <strong>and</strong> Coulomb terms in theperturbation method on the transition densities. Fig. 4.7 shows transition densities using65


the perturbation method <strong>and</strong> thermodynamic instability. The perturbation method has alower transition density (0.355ρ 0 ) than thermodynamic instability method (0.406ρ 0 ).Figure 4.7: We can see the transition density from nuclei with a neutron gas to nuclearmatter. The solid line denotes the curve from the perturbation method. The dotted linecorresponds to the thermodynamic instability method, equivalent to v 0 in the perturbationmethod. The solid linehas a transitiondensity <strong>of</strong>0.355ρ 0 <strong>and</strong>the dottedline hasatransitiondensity <strong>of</strong> 0.406ρ 0 .4.4.2 Quark matterInthis model, we do not consider the appearance<strong>of</strong> hyperons [37] since it is not clear how thehyperons <strong>and</strong> nucleons interact. Thus we simply consider the phase transition from uni<strong>for</strong>mmatter to quark matter. We use the MIT bag model [37] <strong>for</strong> the quark matter equation <strong>of</strong>state. At T=0 MeV, the pressure <strong>and</strong> energy density are given by [37]p = −B + ∑ [1µ4π 2 (c) 3 f (µ 2 f −m 2 fc 4 ) 1/2 (µ 2 f − 5 2 m2 fc 4 )f+ 3 (µf +(µ 22 m4 f c8 fln−m2 f c4 ) 1/2 )]m f c 2ǫ = B + ∑ f[3µ4π 2 (c) 3 f (µ 2 f −m2 f c4 ) 1/2 (µ 2 f − 5 2 m2 f c4 )− 1 (µf +(µ 22 m4 fc 8 fln−m2 f c4 ) 1/2 )],m f c 2(4.40)66


Figure 4.8: The left panel shows the energy density <strong>of</strong> nuclear matter (dotted line), quarkmatter (dashed) <strong>and</strong> mixed phase (solid). The right panel shows the pressure <strong>of</strong> nuclearmatter (dotted lined), quark matter (dashed) <strong>and</strong> mixed phase (solid).where the density <strong>for</strong> each quark flavor is given byρ f = (µ2 f −m2 f )3/2π 2 (c) 3 . (4.41)For the pure-quark phase we use m u = m d = 0, m s = 150 MeV <strong>and</strong> B = 100 MeV fm −3† . Inthe mixed phase <strong>of</strong> uni<strong>for</strong>m nuclear matter <strong>and</strong> quark matter, we apply Gibb’s conditionsto minimize the free-energy density with two constraints, which are related to total numberdensity <strong>and</strong> charge neutrality,ρ b = χρ N +(1−χ)ρ QQ = χQ N +(1−χ)Q Q = 0,(4.42)where χ is the volume fraction <strong>of</strong> the uni<strong>for</strong>m nuclear matter in the mixed phase <strong>and</strong> thesubscript N (Q) represents nuclear (quark) matter. In the mixed phase, the total charge isglobally neutral in contrast to pure nuclear matter <strong>and</strong> quark matter. From minimizing thefree energy, we havep N = p Qµ n = µ u +2µ d(4.43)µ p = 2µ u +µ d ,then we have the energy density <strong>of</strong> the mixed phaseǫ = χǫ N +(1−χ)ǫ Q . (4.44)† Here B is the bag constant <strong>for</strong> the MIT bag model not the binding energyper baryon<strong>for</strong> nuclearmatter.67


Figure 4.9: The left panel shows the mass-radius relation <strong>of</strong> the neutron stars (dotted line)from the GNDF <strong>and</strong> the hybrid stars (solid line). The right panel shows the number densitypr<strong>of</strong>ile <strong>of</strong> a hybrid star which have M= 1.44M ⊙ . At r ∼ 5.0 km, the phase transition tonuclear matter takes place, <strong>and</strong> protons <strong>and</strong> neutrons exit. When r=8.2 km, the phasetransition to nuclear matter is completed <strong>and</strong> there is no more quark matter.As in the case <strong>of</strong> uni<strong>for</strong>m matter, we assume beta-stable matter so that the chemical potentials<strong>of</strong> the nuclear matter <strong>and</strong> quark matter have the relationµ n = µ p +µ eµ d = µ s = µ u +µ e .(4.45)Fig. 4.8 shows the energy density <strong>and</strong> pressure as a function <strong>of</strong> number density. The dotted(dashed) line represents nuclear (quark) matter. The solid line denotes the mixed phase.The phase transition begins when the baryon density becomes 1.386ρ 0 <strong>and</strong> all nucleons turninto quark matter when the baryon density becomes 5.236ρ 0 .If there is a phase transition in the core <strong>of</strong> a cold neutron star, the mass <strong>and</strong> radius arequite different from the case <strong>of</strong> a pure-nuclear-matter neutron star. The left panel <strong>of</strong> Fig.4.9 shows the mass-radius relation <strong>of</strong> the neutron stars (dotted line) <strong>and</strong> hybrid stars (solidline). The right panel shows the number density pr<strong>of</strong>iles <strong>of</strong> a quark matter <strong>and</strong> nuclearmatter <strong>of</strong> 1.44M ⊙ hybrid star. The maximum mass <strong>of</strong> a cold neutron star with mixed phase(hybrid star) is1.441M ⊙ <strong>and</strong>the central density <strong>of</strong> theneutron star is 9.585ρ 0 . The mass <strong>and</strong>radius curve with the mixed phase indicates where the mixed phase exists. As the distancefrom the center <strong>of</strong> the hybrid star increases, the phase transition to nuclear matter takesplace so that neutrons <strong>and</strong> protons appear (r ∼ 5.0km). Far from the center, the phasetransition is completed (r = 8.2km); pure nuclear matter exists only <strong>for</strong> larger radii. Theexistence <strong>of</strong> quark matter in the hybrid star can be explained by angular momentum loss<strong>of</strong> the proto-neutron star. That is, a fast-rotating neutron stars loses angular momentumbecause <strong>of</strong> magnetic dipole radiation; the central density <strong>of</strong> the neutron star increases dueto the decrease in centripetal <strong>for</strong>ce, then quark matter appears. Since quark matter has alower energy density than pure nuclear matter, we might expect heating <strong>of</strong> the neutron starfrom latent heat from quark matter.68


4.5 Astrophysical application4.5.1 Mass-radius relation <strong>of</strong> a cold neutron starWe know that the typical radius <strong>of</strong> a neutron star is ∼ 10km <strong>and</strong> the mass is ∼ 1.4M ⊙ . Inthis system, the degeneracy pressure <strong>of</strong> the neutrons provides support against gravitationalcollapse. We can apply our model to calculate the mass <strong>and</strong> radius <strong>of</strong> neutron stars <strong>for</strong>a given central density. We use the Tolman-Oppenheimer-Volkov (TOV) equations whichdescribe general relativistic hydrostatic equilibrium:dpdr = p/c 2 )(ǫ+p)−G(M(r)+4πr3 r(r−2GM(r)/c 2 )c 2dMdr = 4π ǫc 2r2 .(4.46)Fig. 4.10 shows the mass-radius relation <strong>for</strong> a cold neutron star. In the GNDF model, themaximum mass <strong>of</strong> a cold neutron star is 2.163M ⊙ , <strong>and</strong> the corresponding radius is 10.673km.The maximum mass from the GNDF model is in between the FRTF truncated model (FRTFI) <strong>and</strong> the modified model <strong>of</strong> the FRTF (FRTF II)Figure 4.10: Mass-radius relation <strong>for</strong> a cold neutron star. The mass <strong>of</strong> a cold neutron starfrom the Gaussian density functional model has a maximum mass <strong>of</strong> (2.163M ⊙ ) when thecentral density is 6.74ρ 0 .4.5.2 Moment <strong>of</strong> inertia <strong>of</strong> the neutron starIn the slow-motion approximation, the moment <strong>of</strong> inertia is given by [39]I = 8π 3∫ R0r 4 (ρ+p)e (λ−ν)/2 ωdr, (4.47)69


where λ = −ln(1 − 2m/r) <strong>and</strong> ν are the metric coefficients <strong>and</strong> ω is the rotational dragfunction. In terms <strong>of</strong> the function j = e −(λ+ν)/2 , the rotational drag satisfies(dr 4 j dω )= −4r 3 ω djdr dr dr , (4.48)with the boundary conditionsω R = 1− 2IR 3 ,There<strong>for</strong>e, the moment <strong>of</strong> inertia can be written asI = − 2 3∫ R0r 3 ω djdr dr = 1 ∫ R(d r 4 ω dj )= R46 0 dr 6( ) dω= 0. (4.49)dr0dωdr∣ . (4.50)RWenotethatthesecond-orderdifferentialequationthatω satisfies, Eq. (4.48), canbeinsteadwritten as a first order differential equation in terms <strong>of</strong> the function φ = dlnω/dlnr,wheredφdr = −φ r (φ+3)−(4+φ)dlnj dr , (4.51)dlnjdr= − 4πr2 (ρ+p), (4.52)r −2mwith the boundary condition φ(0) = 0. The moment <strong>of</strong> inertia becomesusing the boundary condition <strong>for</strong> ω. This simplifies toI = R36 φ Rω R = φ R6 (R3 −2I), (4.53)I = R3 φ R6+2φ R. (4.54)Lattimer <strong>and</strong> Schutz proposed an empirical approximation <strong>for</strong> the moment <strong>of</strong> inertia [43],[I ≃ (0.237±0.008)MR 2 1+4.2 Mkm ( Mkm) ]M ⊙ R +90 . (4.55)M ⊙ RFig. 4.11 shows the moment <strong>of</strong> inertia <strong>of</strong> a cold neutron star. The color b<strong>and</strong> representsupper <strong>and</strong> lower boundaries <strong>of</strong> the empirical approximation. FRTF I, II <strong>and</strong> GNDF agreequite well with this empirical approximation.70


0.550.5GaussianFRTF IIFRTF I0.450.4I/MR 20.350.30.250.20 0.05 0.1 0.15 0.2 0.25M/R (M sun /km)Figure 4.11: Moment <strong>of</strong> inertia <strong>of</strong> a cold neutron star. The color b<strong>and</strong> represents the upper<strong>and</strong> lower boundaries <strong>of</strong> the empirical approximation eq (4.55). Three different models showdifferent curves; however, they represent the empirical results quite well.4.6 ConclusionsIn the GNDF model, the total energy consists <strong>of</strong> the kinetic energy, the finite-range effect,the zero-range effect, <strong>and</strong> the Coulomb energy. Using the Lagrange multiplier method, wefind the potential energy, pressure, chemical potential <strong>and</strong> thermodynamic properties. Theinteraction parameters are obtained from the properties <strong>of</strong> infinite nuclear matter. We areable to obtain the charge radius <strong>and</strong> binding energy per baryon <strong>of</strong> the closed-shell nuclei.The Wigner-Seitz cell size increases as the density decreases <strong>and</strong> the binding energy perbaryon approaches -8.0MeV. The effective mass becomes 0.78m B at the center <strong>of</strong> the heavynuclei <strong>and</strong> becomes m outside <strong>of</strong> heavy nuclei. We are also able to find the pressure <strong>of</strong>uni<strong>for</strong>m symmetric nuclear matter <strong>and</strong> neutron matter. For finite temperature, we can seethe specific heat <strong>of</strong> nuclear matter follows the general trend <strong>of</strong> the free fermions. Thusthe GNDF model is a good nuclear matter model to study <strong>for</strong> both low <strong>and</strong> high nucleardensities. We can improve the current model if we have more exact experimental results<strong>and</strong> we add additional interaction terms to explain the experimental results. The phasetransition was studied using the GNDF model. The density perturbation suggested that thephase transition from non-uni<strong>for</strong>m nuclear matter to uni<strong>for</strong>m matter takes place at densitiesless than0.5ρ 0 . When we take into account the phase transition fromuni<strong>for</strong>mnuclear matterto quark matter, we see that there is a drastic change in the maximum mass <strong>of</strong> a neutronstar, since the pressure <strong>and</strong> energy density <strong>of</strong> quark matter are significantly different fromnuclear matter. The maximum mass <strong>of</strong> the hybrid star is less than 1.5M ⊙ . Since the bagconstant is fixed in this study, we need to investigate the mass <strong>of</strong> hybrid star by varying thebag constant.71


Chapter 5<strong>Nuclear</strong> Energy Density Functional<strong>and</strong> <strong>Neutron</strong> <strong>Stars</strong>In this chapter, we study single nucleus, heavy nuclei in dense matter <strong>and</strong> mass-radius <strong>of</strong>neutron stars using an energy density functional with parametrized density pr<strong>of</strong>ile. Threedifferent models are used <strong>and</strong> each model shows similar nuclear properties in the subnucleardensity regions (ρ < 0.07 fm −3 ). Analytic <strong>for</strong>mulae <strong>for</strong> the Wigner-Seitz cell size, number <strong>of</strong>neutrons, <strong>and</strong> number <strong>of</strong> protons in the cell is given using least χ 2 fitting between numericalcalculation <strong>and</strong> fitting function. The maximum mass <strong>of</strong> a neutron star is investigated,revealing differences among the models.5.1 <strong>Nuclear</strong> energy density functionalIn the nuclear energy density functional, the interaction energy comes from the contact <strong>for</strong>ceso the Hamiltonian density is a functional <strong>of</strong> number density <strong>and</strong> momentum density. Forcomparison with a finite range <strong>for</strong>ce, we used three energy density functional models. Thefirst model is from Ref. [41] (EKO), the second one is the Skyme-type Potential Model(SPM) [32], <strong>and</strong> the last one is the Skyrme-Lyon <strong>for</strong>ce (SLy4).In EKO, the Hamiltonian density is composed <strong>of</strong> homogeneous <strong>and</strong> inhomogeneous terms,The homogeneous term has the <strong>for</strong>mH = H B (ρ n ,ρ p )+H g (ρ n ,ρ p ,∇ρ n ,∇ρ p )+H c . (5.1)H 0 = 22m nτ n + 22m pτ p +(1−δ 2 )v s (ρ)+δ 2 v n (ρ), (5.2)where ρ = ρ n +ρ p <strong>and</strong> δ = (ρ n −ρ p )/ρ. The quantities v s (ρ) <strong>and</strong> v n (ρ) are potential energydensity, which represents the potential contribution from symmetric nuclear matter <strong>and</strong> pure72


neutron matter respectively. Those contribution are given byv s (ρ) = a 1 ρ 2 + a 2ρ 31+a 3 ρv n (ρ) = b 1 ρ 2 + b (5.3)2ρ 31+b 3 ρ .The inhomogeneity (gradient) term H g which is responsible <strong>for</strong> the surface tension <strong>of</strong> nucleiis being used asH g = F 0 |∇ρ| 2 . (5.4)The Coulomb energy density is given by∫E c = e2d 3 rd 3 r ′ 12 |r−r ′ | (ρ p(r)−ρ e )(ρ p (r ′ )−ρ e ), (5.5)In SPM <strong>and</strong> SLy4, the Hamiltonian density is given in Eq. (2.3, 2.4,2.5 2.6,2.7)For all three models, we use a parametrized density pr<strong>of</strong>ile (PDP) † <strong>for</strong> protons <strong>and</strong> neutronsbe<strong>for</strong>e a phase transition to uni<strong>for</strong>m nuclear matter takes place.In PDP, we assume that nuclear density follows [41],⎧ [ ( ) ]⎪⎨ti 3(ρ ini −ρ out ri ) 1−ρ i (r) =R i+ρ outi , r < R i(5.6)⎪⎩ ρ outi , r ≥ R i .The momentum density is simplified toin the Thomas-Fermi approximation.τ n,p = 3 5 (3π2 ) 2/3 (ρ n,p ) 5/3 (5.7)Table 5.1: Parameters in EKOa 1 a 2 a 3 b 1 b 2 b 3 F 0−458.384 2072.775 3.1668 −227.049 1058.942 2.608 61.917Table 5.1, 5.2 show the parameters <strong>of</strong> models which we used in this study.To find the parameters in EKO, we used st<strong>and</strong>ard nuclear matter properties such as E/A =−16 MeV, K = 235 MeV, p(ρ = ρ 0 ,x = 1/2) = 0 MeV/fm 3 , S v = 33.21 MeV, L = 68 MeV.For b 3 , we used least χ 2 fitting to FPS pure neutron matters. F 0 is also found from least χ 2fitting <strong>of</strong> 2336 nuclei binding energy data.† The parametrized density pr<strong>of</strong>ile is not the actual solution <strong>of</strong> the Thomas Fermi approximation. ThePDP is used to avoid the numerical difficulties <strong>of</strong> ∇ρ.73


Table 5.2: Parameters in SPM & SLy4Parameters SPM SLy4t 0 -2719.7 -2488.91t 1 417.64 486.82t 2 -66.687 -545.39t 3 15042 13777.0ǫ 0.14416 1/6x0 0.16154 0.834x1 -0.047986 -0.344x2 0.027170 -1.000x3 0.13611 1.3545.2 Binding energy <strong>of</strong> single nucleusWe calculate the binding energy <strong>of</strong> a single nucleus such as 40 Ca, 90 Zr, <strong>and</strong> 208 Pb. Thesenuclei have closed shells so we can neglect shell effects.In a single nucleus, ρ outn,p = 0 since there are no dripped nucleons.0.10.08Charge density <strong>of</strong> 208 PbEDFPotentialSLy4ρ n , ρ p (fm -3 )0.060.040.0200 2 4 6 8 10r (fm)Figure 5.1: Proton density pr<strong>of</strong>ile <strong>of</strong> 208 Pb.Table 5.3 shows the properties <strong>of</strong> a single nucleus from both experimental results <strong>and</strong> amodel calculation. In table 5.3, r ch is the charge radius, BE/A is the binding energy perbaryon, <strong>and</strong> δR is defined as r n − r p , where r n,p is the root-mean-square radius <strong>of</strong> neutron<strong>and</strong> proton. The density pr<strong>of</strong>iles <strong>for</strong> the single nucleus are given in Fig 5.1. Each modelagrees well with the experimental result. The PDP shows the maximum baryon density atthe center <strong>of</strong> the nucleus because <strong>of</strong> its mathematical expression. However, many <strong>of</strong> heavynuclei with large number <strong>of</strong> protons (e.g. 208 Pb) have maximum baryon density <strong>of</strong>f the centerbecause <strong>of</strong> Coulomb repulsion. Thus PDP cannot show the exact density pr<strong>of</strong>ile <strong>of</strong> a single74


Table 5.3: <strong>Nuclear</strong> properties <strong>of</strong> single nucleusNucleus Exp EKO SPM Sly440 Ca r ch (fm) 3.48 3.36 3.37 3.42BE/A (MeV) 8.45 8.51 8.51 8.14δR −0.06±0.05, −0.05±0.04 −0.03 −0.04 −0.0490 Zr r ch (fm) 4.27 4.23 4.23 4.28BE/A (MeV) 8.71 8.74 8.68 8.42δR 0.09±0.07 0.05 0.06 0.05208 Pb r ch (fm) 5.50 5.51 5.50 5.55BE/A (MeV) 7.87 7.85 7.74 7.56δR (fm) 0.12±0.05, 0.20±0.04 0.11 0.14 0.12nucleus. Except <strong>for</strong> the density pr<strong>of</strong>ile, PDP can be used as an approximated density pr<strong>of</strong>ileto find the charge radius, binding energy per baryon, <strong>and</strong> δR.5.3 <strong>Neutron</strong> Star CrustIn a neutron star, as the density increases, the pressure from degenerate electrons reaches apoint where the electrons can no longer support the gravitational collapse <strong>and</strong> the neutronsreplace the role <strong>of</strong> electrons soon after the neutron drip. The neutron drip divides the outercrust <strong>and</strong> inner crust <strong>of</strong> the neutron star. There is a phase transition around 0.5 ρ 0 [41]between inner crust <strong>and</strong> liquid core since the nuclear system favors the lowest energy state.Even though there might be sequential phase transition from spherical nuclei with a free <strong>of</strong>neutrongas, cylindrical phase, slab, cylindrical hole, bubble touni<strong>for</strong>mmatter, thedifferencein the energy density <strong>and</strong> press is negligible. Thus we consider the phase transition fromspherical nuclei phase to uni<strong>for</strong>m nuclear matter. We can simply compare the energy density<strong>of</strong> uni<strong>for</strong>m nuclear matter with heavy nuclei with a free <strong>of</strong> neutron gas.5.3.1 <strong>Neutron</strong> dripThe neutron drips from heavy nuclei as their density increases. The drip point can besearched from a lower baryon density region. As the density increases, the neutron chemicalpotential increases, becomes positive, <strong>and</strong> neutrons start to drip. In case <strong>of</strong> protons, thechemical potential remains negative so there are never any dripped protons at T=0 MeV.Table 5.4 shows the neutron drip density <strong>for</strong> different models. In case <strong>of</strong> EKO, the neutronsstart to drip when ρ = 2.26×10 −4 fm −3 .Be<strong>for</strong>e the neutron drip, the total pressure comes from only electron contribution <strong>and</strong>soon after the neutron drip, the pressure from dripped neutrons overwhelm the electroncontribution.75


Table 5.4: <strong>Neutron</strong> drip densities <strong>for</strong> all 3 modelsEKO SPM SLy4ρ (fm −3 ) 2.2594×10 −4 2.4491×10 −4 2.0749×10 −45.3.2 Phase transition to uni<strong>for</strong>m nuclear matterTo see the phase transition from lattice nuclei with a gas <strong>of</strong> free neutrons, we compare theenergy per baryon <strong>for</strong> each model. Since it is believed that the energy differences arise betweenvarious pasta phases, we simply compare the energy per baryon from lattice nucleiwith neutron gas with one from uni<strong>for</strong>m nuclear matter. We may use the Maxwell constructionto find the phase transition points between two phases. The total energy density <strong>of</strong> twophases in Ref. [22] can be written asE = vE h +(1−v)E l , (5.8)where ρ = vρ h + (1 − v)ρ l . The high <strong>and</strong> low density <strong>of</strong> the boundary can be found fromsolving p h = p l <strong>and</strong> µ nh = µ nl at the same time. A numerical calculation does not yield anexact intersection <strong>for</strong> the solution.Table 5.5: Phase transition points to uni<strong>for</strong>m nuclear matterEKO Potential SLy4ρ (fm −3 ) 0.0759 0.0670 0.07425.3.3 Statistical <strong>for</strong>mulae <strong>for</strong> heavy nuclei in the neutron starcrustWe compare the model calculation with a statistical <strong>for</strong>mula <strong>for</strong> number <strong>of</strong> protons <strong>and</strong>neutrons in heavy nuclei <strong>and</strong> Wigner-Seitz cell size. Fitting <strong>for</strong>mulae <strong>for</strong> the number <strong>of</strong>protons <strong>and</strong> neutrons can be used in neutron star cooling process. That is, neutrinos areemitted through e−Z bremsstahlung [62] in the crust,ǫ e−Z = 2.1×10 20Z2A( ρρ 0)T 6 9 erg/cm 3 /s, (5.9)where T 9 = T/10 9 K. For this, we present two <strong>for</strong>mulae be<strong>for</strong>e <strong>and</strong> after the neutron drip<strong>for</strong> each curve (Z nuc , N nuc , <strong>and</strong> R c vs ρ) since the derivative <strong>of</strong> each curve w.r.t ρ B is notcontinuous. After the neutron drip, we suggest different types <strong>of</strong> <strong>for</strong>mulae † <strong>for</strong> each curve† Fitting functions in this study are introduced to reflect the numerical result correctly. Since the number<strong>of</strong> protons increases suddenly be<strong>for</strong>e the uni<strong>for</strong>m matter phase transition, c 6 is added.76


which aref Rc = a 0 +a 1 x+a 2 x 2 +a 3 x 3 +a 4 x 4 +a 5 x 5f Nnuc = b 0 +b 1 z +b 2 z 2 +b 3 z 3 +b 4 z 4 +b 5 x 5f Znuc = c 0 +c 1 z +c 2 z 2 +c 3 z 3 +c 4 z 4 +c 5 z 5 + c 6√ z.(5.10)where x = ln(ρfm 3 ) <strong>and</strong> z = 100ρ fm 3 .All a’s, b’s, <strong>and</strong> c’s parameters can be found from the least χ 2 square fitting, in which χ 2 isgiven byχ 2 = 1 N∑(f i −f model ) 2 . (5.11)NiTable 5.6: Fitting <strong>for</strong>mulae <strong>for</strong> Wigner-Seitz cell size after the neutron dripa 0 a 1 a 2 a 3 a 4 a 5EKO −30.812 −30.344 −8.004 −1.428 −0.131 −4.561×10 −3SPM -39.193 -30.239 -5.414 0.631 −4.313×10 −3 −1.334×10 −3SLy4 20.120 13.383 6.504 0.918 5.748×10 −2 1.367×10 −3Table 5.7: Fitting <strong>for</strong>mulae <strong>for</strong> number <strong>of</strong> neutrons in the heavy nuclei after the neutrondripb 0 b 1 b 2 b 3 b 4 b 5EKO 77.737 93.069 -52.346 20.942 -3.740 0.251SPM 76.989 75.456 -49.337 21.835 -4.239 0.311SLy4 87.680 86.728 -12.821 3.396 -0.560 6.366×10 −2Table 5.8: Statistical <strong>for</strong>mulae <strong>for</strong> number <strong>of</strong> protons in the heavy nuclei after the neutrondripc EKO SPM SLy4c 0 36.947 34.619 40.389c 1 1.877 −2.306 5.447c 2 −2.921 5.854×10 −2 −4.265c 3 0.942 -4.286×10 −2 1.362c 4 −0.143 5.375×10 −3 −0.214c 5 8.195×10 −3 - 1.301 ×10 −2c 6 −0.171 −0.131 −0.20377


Be<strong>for</strong>e the neutron drip, another type <strong>of</strong> <strong>for</strong>mula fits the numerical calculations since thebehavior <strong>of</strong> Winger-Seitz cell size, number <strong>of</strong> neutrons, <strong>and</strong> number <strong>of</strong> protons in the cell isrelatively simple.f Rc = exp(a 0 +a 1 x+a 2 x 2 )where x = ln(ρfm 3 ) <strong>and</strong> x 0 = ln(10 −14 ).f Nnuc = b 0 +b 1 (x−x 0 ) 7 +b 2 (x−x 0 ) 9f Znuc = c 0 +c 1 (x−x 0 ) 7 +c 2 (x−x 0 ) 9 .(5.12)Table 5.9: Fitting <strong>for</strong>mulae be<strong>for</strong>e the neutron dripEKO SPM SLy4a 0 1.2571 1.2489 1.2909a 1 −0.3028 −0.3027 −0.3020a 2 6.2495×10 −4 6.2487×10 −4 6.3943×10 −4b 0 34.1418 33.2440 36.8709b 1 3.2976×10 −9 4.1044 ×10 −9 3.6671 ×10 −9b 2 1.1095 ×10 −11 9.3710 ×10 −12 1.2652×10 −11c 0 27.8058 26.9215 29.7386c 1 2.9168×10 −9 3.2463 ×10 −9 3.1409 ×10 −9c 2 −1.8428×10 −12 −2.8772×10 −12 −1.6480×10 −12Fig. 5.2 shows the atomic number in neutron star crust after the neutron drip happens.The fitting function works very well <strong>for</strong> EKO, SPM, <strong>and</strong> SLy4. The full Thomas Fermiapproximation is added in case <strong>of</strong> truncated FRTF model. In the general case, a full ThomasFermi numerical calculation has such a wiggle because <strong>of</strong> multiple minima ‡ .5.4 Mass-Radius <strong>of</strong> a cold neutron starThe uni<strong>for</strong>m nuclear matter exists in the core <strong>of</strong> a neutron star as a result <strong>of</strong> a nuclear phasetransition. The matter in the core is in the beta equilibrium state to minimize the totalenergy density. Mathematically, the beta equilibrium state equation can be obtained froma Lagrange multiplier method, which saysµ n = µ p +µ e (5.13)where µ n,p,e are chemical potentials <strong>of</strong> the neutron, proton, <strong>and</strong> electron respectively.<strong>Nuclear</strong> matter beyond the nuclear saturation densities exists in the core <strong>of</strong> neutron stars.This means that themass-radius relation<strong>of</strong> neutronstars canbeagoodbarometer <strong>of</strong> nuclearmodels. TOV equation (Eq. 4.46) is used to construct the neutron star’s mass <strong>and</strong> radius.‡ Since the Wigner-Seitz cell size is unknown <strong>and</strong> to be found numerically, R c does not smooth near thephase transition density.78


parametrized[ht]504540EDF, Z nucFitting EDFPOT, Z nucFitting POTSly4, Z nucFitting Sly4Truncated Model, Z nucFitting Truncated Model, Z nuc35Z nuc302520150.001 0.01ρ (fm -3 )Figure 5.2: This figure shows the number <strong>of</strong> protons in neutron star crust from numericalcalculation <strong>and</strong> fitting function. Truncated model calculation is added <strong>for</strong> comparison.79


2EKOPotentialSLy41.5M ( M O•)10.507 8 9 10 11 12 13 14 15R ( km )Figure 5.3: Mass <strong>and</strong> radius relation <strong>of</strong> cold neutron star from energy density functionalThe simple EKO does not satisfy the 2.0M ⊙ neutron star. Other models show that themaximum mass <strong>of</strong> neutron stars is greater than 2.0M ⊙ . <strong>Nuclear</strong> matter at high densitiescannot be described by simple mathematical density functional but we need to consider morecomplex nuclear <strong>for</strong>ce arising from many body effects or QCD (quantum chromodynamics)effects.5.5 ConclusionPDP can be used to study a single nucleus <strong>and</strong> nuclei in dense matter. Even though PDPis not a true solution <strong>of</strong> density pr<strong>of</strong>iles, it can give enough accuracy in the total energy <strong>and</strong>continuouspropertiesinWigner-Seitzcellcalculationlikeliquiddropletapproach. There<strong>for</strong>e,PDP is an alternative method to construct a nuclear equation <strong>of</strong> state <strong>for</strong> hot dense matter.Eventhoughallthreemodelsshowsimilarproperties<strong>of</strong>nuclearmattersuchasbindingenergy<strong>of</strong> single nucleus, phase transition points, proton fractions, <strong>and</strong> Wigner-Seitz cell size, eachmodel shows the different maximum mass <strong>of</strong> neutron star. For example, the simple EKOmodel cannot describe 2.0 M ⊙ neutron star. That means that the potential energy in EKOshould be replaced with a more realistic <strong>for</strong>ce model.The simple division <strong>of</strong> symmetric potential <strong>and</strong> pure neutron potential does not work, suchthat the δ term in the potential should have more than δ 2 interaction. This also means thatthe extrapolation beyond the nuclear saturation density is still an open problem which needsto be checked by mass <strong>and</strong> radius relation <strong>of</strong> neutron stars.80


Chapter 6<strong>Nuclear</strong> Equation <strong>of</strong> State <strong>for</strong> HotDense <strong>Matter</strong>† There aremany dem<strong>and</strong>s <strong>for</strong> a nuclear equation <strong>of</strong> state <strong>for</strong> hot dense matter. Core collapsingsupernova explosions, neutron star binary mergers, <strong>and</strong> proto-neutron star simulationneed a highly thermodynamically consistent EOS table. The recent development <strong>of</strong> technologyenables us to develop full 3D simulations <strong>of</strong> the above phenomenon. However, only afew EOS tables are available. Furthermore, except <strong>for</strong> LS220 [22], no EOS tables can satisfythe mass <strong>and</strong> radius relation <strong>of</strong> neutron stars <strong>and</strong> nuclear experiments at the same time.Our EOS follows the methodology <strong>of</strong> LPRL [17] <strong>and</strong> Lattimer & Swesty [22]. We employFRTF <strong>and</strong> SLy4 as a nuclear <strong>for</strong>ce model <strong>and</strong> the Liquid Droplet approach as a numericaltechnique.6.1 Construction <strong>of</strong> EOSToconstructtheEOStable, weneedtwobasictools. Oneisthenuclear <strong>for</strong>cemodel. Anotheris a numerical technique.For a nuclear <strong>for</strong>ce model, we can use (see chapter 1, 2, & 3) the following,• Non-relativistic potential model• Relativistic mean field model• Finite-Range Thomas Fermi model.As a numerical technique, we might choose,• Liquid Droplet Approach• Thomas Fermi Approximation• Hartree (Fock) Approximation.† This chapter is based on J.M. Lattimer <strong>and</strong> Y. Lim’s work. We are checking the thermodynamicconsistencies <strong>and</strong> will submit our work <strong>and</strong> provide the new EOS table soon.81


These two tools can be combined to construct a <strong>Nuclear</strong> EOS Publicly available EOS tablesare• LS EOS [22]- Combines Skyrme type <strong>for</strong>ce model with Liquid droplet model,- Disregard consider neutron skin,- Consider phase transion using geometric function,- LS 220 is the best until now,• STOS [59]- Uses RMF <strong>and</strong> semi-Thomas Fermi approximation (parametrized density pr<strong>of</strong>ile.)- Awkward grid spacing,- New version available (2011), added hyperon interactions,• SHT [61]- Uses RMF <strong>and</strong> Hartree Approximation,- Larger radius <strong>for</strong> given mass <strong>of</strong> neutron stars,For constructing our EOS table, we reconsider the LPRL numerical technique which includesneutron skin. As an improvement, we add Coulomb diffusion <strong>and</strong> Coulomb exchange energy.The liquid droplet approach is the only method which can guarantee thermodynamicconsistency. We can obtain analytic derivatives <strong>for</strong> ∂p/∂ρ, ∂p/∂T, ∂p/∂Y p , etc.The other approaches (Thomas Fermi, Hartree-Fock approximation) cannot give enoughaccuracy in thermodynamics because they have convergence issues. The thermodynamicderivatives can be obtained through the numerical derivatives in TF or HF. This can alsoproduce a thermodynamic inconsistency.The numerical time needed to construct the whole table can also be a problem. In SHT [61],it took 6,000 CPU-days to calculate 107,000 grid points. If we use the LDM approach, ittakes less than 10 minutes to calculate 121 × 50 ∼ 302,500 grid points. That means thatwe can easily generate another table using a different nuclear <strong>for</strong>ce model. We also considerphase transitions which occur in a neutron star’s inner crust, as was done in the LS models.As a nuclear <strong>for</strong>ce model, we use FRTF models <strong>and</strong> SLy4. SLy4 is believed to be themost accurate Skyreme-type nuclear <strong>for</strong>ce model. We choose it <strong>for</strong> comparison to check ourFRTF model parameters.6.2 Liquid Droplet Model as a numerical techniqueFirst, we demonstrate that the liquid droplet model adequately describes nuclei. The liquiddroplet model provides us with the energy per baryon <strong>of</strong> a single nucleus <strong>and</strong> nuclei in densematter. The idea behindit is that theenergy contribution <strong>of</strong> nuclei comes frombulk, surface,82


Coulomb, <strong>and</strong>, symmetry part [24], in addition, there is a pairing effect such that [32]E(A,Z) =−BA+E s A 2/3 +S v A (1−2Z/A)2 Z 2+E1+S s A −1/3 C/S v A+E diffZ 2A +E exZ 4/3A +a ∆ √A.(6.1)In the above equation, B is the binding energy per baryon in the bulk nuclear matter, E s isthe surface symmetry energy, E C ,E diff , <strong>and</strong> E ex are energy coefficients in classical, diffusion,<strong>and</strong> exchange Coulomb energy. The coefficient a is 1 <strong>for</strong> odd-odd nuclei, 0 <strong>for</strong> odd-evennuclei, <strong>and</strong> -1 <strong>for</strong> even-even nuclei. The paring constant (∆) is kept constant at 12 MeV.This liquid droplet model can be extended by adding the nuclear shell effect. The empirical<strong>for</strong>mula <strong>for</strong> this shell effect is given by [50]E shell = 1 2 a m(N v +P v )+ 1 4 b m(N v +P v ) 2 (6.2)where N v <strong>and</strong> P v is the minimum difference <strong>for</strong> neutrons <strong>and</strong> protons between magic number,2, 8, 20, 28, 50, 82, 126, <strong>and</strong>, 184. For example, A = 50,Z = 23,N = 27, thenZ v = |23−20| = 3, N v = |27−28| = 1.The general way to minimize the total energy at T = 0 MeV † <strong>for</strong> a given Z <strong>and</strong> A, weadopt the µ n approach, in which only neutron skin exists on the surface <strong>of</strong> nuclei. µ αmethod exists [32], however, it gives a different slope (S s /S v vs S v ) from the one µ n method.We confirmed the slope from µ n is closed to the one from FRTF model.For a given nuclear model, the total energy (excluding rest mass energy) is given byE = f B (A−N s )+4πR 2 σ(µ n )+µ n N s +E C (R), (6.3)where f B is the binding energy per baryon <strong>for</strong> a specific nuclear model <strong>and</strong> N s is the totalnumber <strong>of</strong> neutrons on the surface. We include in E C the Coulomb, diffusive surface, <strong>and</strong>exchange energies. We assume that f B = f B (ρ,x).With two constraints,Zx =A−N s(6.4)A−N s = 4πR33 ρwe can find the minimum <strong>of</strong> total binding energy using the Lagrange multiplier method.E(µ n ,N s ,R,ρ,x,λ 1 ,λ 2 ) =f B (ρ,x)(A−N s )+4πR 2 σ(µ n )+µ n N s +E c (R)+λ 1 (A−N s − 4πR33 ρ)+λ 2(Z −x(A−N s )).(6.5)† If T ≠ 0 MeV, we need to minimize the total free energy or free energy density because <strong>of</strong> entropy.83


The derivative <strong>of</strong> E with respect to the unknowns gives,∂E= 0 : 4πR 2 ∂σ +N s = 0∂µ n ∂µ n(6.6a)∂E= 0 : −f B +µ n −λ 1 +xλ 2 = 0∂N s(6.6b)∂E∂R = 0 : 8πRσ+ ∂E C∂R −4πR2 ρλ 1 = 0 (6.6c)∂E∂ρ = 0 : ∂f B∂ρ (A−N 4πs)−λ 13 R3 = 0 (6.6d)∂E∂x = 0 : ∂f B∂x (A−N s)−λ 2 (A−N s ) = 0 (6.6e)∂E= 0 : A−N s − 4π ∂λ 1 3 R3 ρ = 0(6.6f)∂E= 0 : Z −x(A−N s ) = 0.∂λ 2(6.6g)We will correct E(A,Z) by the shell <strong>and</strong> pairing energies so these do not need to be includedin the above. From eq. (6.6e) we havefrom eq. (6.6d) <strong>and</strong> (6.6f), we getUsing λ 1 , λ 2 , <strong>and</strong> eq. (6.6b), we obtainλ 2 = ∂f B∂x , (6.7)λ 1 = ρ ∂f B∂ρ . (6.8)µ n = f B +ρ ∂f B∂ρ −x∂f B∂x . (6.9)If we assume that σ depends on µ n through x not ρ, <strong>and</strong> σ = σ 0 −(1−2x) 2 σ δ , we have fourequations to solve, which are4πR 2 14(1−2x)∂µ n /∂x +N s = 0 (6.10a)8πR 2 σ + ∂E c∂R −4πR2 ρ 2∂f B∂ρ = 0A−N s − 4π 3 R3 ρ = 0Z −x(A−N s ) = 0.(6.10b)(6.10c)(6.10d)84


We can eliminate N s , leading to three equations to solve,16πR 2 (1−2x)σ δ + ∂µ n∂x(A− 4π )3 R3 ρ8πR 2 (σ 0 −(1−2x) 2 σ δ )+ ∂E c∂R −4πR2 ρ 2∂f B∂ρ = 0Z − 4π 3 R3 ρx = 0.= 0 (6.11a)(6.11b)(6.11c)In the case <strong>of</strong> the incompressible model where ρ in the nuclei is fixed <strong>for</strong> all nuclei, we areleft with one equation to solve (eq. 6.11a) :( ) 2/3 ( ) 2/3 3 Z16π (1−2x)σ δ + ∂µ (nA− Z )= 0, (6.12)4πρ 0 x ∂x xwith µ n = f B −x ∂f B. In case <strong>of</strong> the simplest nuclear model,∂xf B = −B +S v (1−2x) 2 + K 18 (1−u)2 . (6.13)Fig. ?? <strong>and</strong> 6.2 shows the contour plot <strong>of</strong> χ 2 <strong>for</strong> S s /S v <strong>and</strong> S v space. The χ 2 from totalFigure 6.1: This is a contour plot <strong>of</strong> χ 2 when ρ 0 = 0.155/fm 3 from incompressible model.85


Figure 6.2: This is a contour plot <strong>of</strong> χ 2 from compressible <strong>and</strong> incompressible model.binding energy <strong>of</strong> nuclei is not much different in both cases but the least χ 2 points withrespect to S v <strong>and</strong> S s /S v are different. The σ Sv <strong>and</strong> σ Ss/S v † from the compressible liquiddroplet model are twice larger than the ones from the incompressible liquid droplet model,which result in the larger contour plot when χ 2 = 3. The correlations between S v <strong>and</strong> S s /S vfrom two models are almost same so the slopes from both models are parallel.χ 2 from above two liquid droplet models are less than the one (χ 2 = 4.75) fromFRTF model.This justifies the use <strong>of</strong> the liquid droplet approach as a numerical technique to constructEOS tables.6.3 Choice <strong>of</strong> nuclear <strong>for</strong>ce modelThe nuclear <strong>for</strong>ce model to make the EOS table should represent both nuclear experimentson Earth <strong>and</strong> neutron star’s mass <strong>and</strong> radius relation. Up to now, only LS220 [22] fits both<strong>of</strong> constraints. To a choose nuclear <strong>for</strong>ce model, we need to take several tests. We showthe result from non-relativistic potential (Skyrme <strong>for</strong>ce) model. There are more than 100models <strong>and</strong> our tests were done on 62 models. As a model <strong>of</strong> single nucleus properties,they are all good enough to represent binding energy <strong>and</strong> root-mean-square radius when weuse Hartree-Fock code. However, most <strong>of</strong> the model did not pass the simple test such as† We adopted arbitrary σ 0 as 1.0 MeV in this case.86


pressure from pure neutron matter, maximum mass <strong>of</strong> a cold neutron star, <strong>and</strong> Steiner etal.[53] allowed area <strong>of</strong> mass <strong>and</strong> radius.6.3.1 Pure neutron matterThe pressure in the pure neutron matter should increase as density increases since neutronalone cannot be bound at all. However, some <strong>of</strong> nuclear <strong>for</strong>ce models cannot represent thissimple behaviour. Fig. 6.3 shows pure neutron matter pressure from some <strong>of</strong> Skyrme <strong>for</strong>ceP (MeV/fm 3 ), x=020151050-5EEsFitAFitBFitKFitKsFitLSISkNF1SkNF2Sk‘SVISVIIZZsZss-10-15-200 0.1 0.2 0.3 0.4 0.5 0.6ρ (fm -3 )Figure 6.3: This figure shows the Skyrme <strong>for</strong>ce models which cannot make positive pressure<strong>of</strong> pure neutron matter.models. These models on the list cannot explain positivity <strong>and</strong> monotonic slope <strong>of</strong> pressure.So, they should be excluded from the c<strong>and</strong>idate <strong>of</strong> EOS nuclear <strong>for</strong>ce.6.3.2 Maximum mass <strong>of</strong> a cold neutron starRecently, PSR J16142230 [57] was found to have a mass <strong>of</strong> 1.97M ⊙ . Demorest et al.[57] analyzedtime delay data <strong>of</strong> the pulsar’s emission as it passed behind its white dwarf companion(Shapiro delay). This mass now becomes the minimum <strong>of</strong> maximum mass <strong>of</strong> neutron stars.Also this mass can be used to rule out the nuclear <strong>for</strong>ce model. Fig 6.4 shows the maximummass <strong>of</strong> neutron star <strong>of</strong> some Skyrme <strong>for</strong>ce model. These models also cannot be used <strong>for</strong>suitable nuclear <strong>for</strong>ce model at high density.6.3.3 Allowed region <strong>of</strong> mass <strong>and</strong> radii <strong>of</strong> neutron starsSteiner et al.[53] analyzed the x-ray data <strong>and</strong> used atmosphere models <strong>of</strong> neutron stars t<strong>of</strong>ind the allowed area <strong>of</strong> mass <strong>and</strong> radii <strong>of</strong> neutron stars. From these regions, we can choose87


2.52SGIISkkT8SkMSkT1SkT1sSkT2SkT3SkT3sSkT7SkT8SkT9M=1.95M ( M O•)1.510.507 8 9 10 11 12 13 14 15R ( km )Figure 6.4: This figure shows the Skyrme <strong>for</strong>ce models which cannot produce a 1.97M ⊙neutron star.final c<strong>and</strong>idates <strong>for</strong> the nuclear <strong>for</strong>ce model <strong>of</strong> EOS table. Fig.6.5 shows the Skyrme <strong>for</strong>cemodels which have too large radii <strong>for</strong> a given mass <strong>of</strong> neutron stars. Mathematically, thesehappen because <strong>of</strong> large L so it should be emphasized that L is another important nuclearconstant which should be extracted from nuclear experiments or astrophysical observations.Of all three tests, we can see that SLy0, ... , SLy10, <strong>and</strong> our FRTF I, II, Gaussian modelsare good enough to be used as a nuclear <strong>for</strong>ce model. Fig.6.6 shows the mass <strong>and</strong> radiusrelation <strong>of</strong> Skyrme <strong>for</strong>ce models <strong>and</strong> our finite range <strong>for</strong>ce models, which are suitable <strong>for</strong>make an EOS table.We did not take a test with any relativistic mean field models † , since they usually have largeL so they cannot make allowed regions <strong>of</strong> mass <strong>and</strong> radius in neutron stars.To summarize (Appendix E), the nuclear <strong>for</strong>ce models which pass the 3 tests would haveS v ≃ 32 MeV, L ≃ 46 MeV, K ≃ 230 MeV, <strong>and</strong> ρ o ≃ 0.16fm −3 .6.4 Free energyThe fundamental idea <strong>of</strong> making an EOS table is to minimize free energy or free energydensity <strong>for</strong> a given ρ, T, <strong>and</strong> Y p . In the Liquid Droplet approach, the free energy in theWigner-Seitz cell consists <strong>of</strong> the contributions from finite nuclei <strong>and</strong> nucleons outside finitenuclei. Fig. 6.7 shows the schematic picture inside the Wigner-Seitz cell. The heavy nuclei isin phase equilibrium with nucleons <strong>and</strong> alpha particles outside. The free energy per baryoncontribution from heavy nuclei can be divided into,f N = f bulk +f Coul +f surf +f trans (6.14)† The R 1.4M⊙ ∝ L 1/4 so RMF cannot make the allowed region.88


2.52GSRsSGISIVSkaSkI1SkI2SkI3SkI5Steiner et alPSR J1614-2230M ( M O•)1.510.507 8 9 10 11 12 13 14 15R ( km )Figure 6.5: The Skyrme <strong>for</strong>ce models on the list have too large radii <strong>for</strong> given mass.where f mbulk is the free energy contribution from inside <strong>of</strong> heavy nuclei, f Coul is the Coulombenergy contribution, f surf is the one from surface, <strong>and</strong> f trans comes from translational (orKinetic) parametrizedenergy <strong>of</strong> heavy nuclei. We assume that heavy nuclei have uni<strong>for</strong>mdensity from the center to the surface so we have the contribution from bulk <strong>and</strong> surface.6.4.1 Bulk <strong>Matter</strong> inside <strong>and</strong> outside nucleiThe bulk free energy per baryon, f bulk (ρ i ,x i ,T) ≡ f bulk /ρ i , is exactly that described in bulkmatter in the finite range model (or SLy4). The subscript i(o) refers to the bulk nuclearmatter inside (outside) nuclei. The use <strong>of</strong> the same functional <strong>for</strong>m <strong>for</strong> bulk matter bothinside the nuclei <strong>and</strong> the ‘dripped’ bulk matter outside the nuclei is necessary <strong>for</strong> a consistenttreatment. In general phase transition (which is <strong>of</strong>ten called, pasta phase) happen when theoverall density (ρ) approach 0.5ρ 0 to minize total energy.6.4.2 Coulomb energyThe coulomb free energy is the electrostatic energy needed to assemble the static charge configuration.As an illustration, one could assume the positive charge <strong>of</strong> protons is distributeduni<strong>for</strong>mly within a spherical nuclear volume (radius r N ), <strong>and</strong> this charge is neutralized byan equal, but opposite, charge distributed within radius (r c ) <strong>of</strong> a neutral (Wigner-Seitz) cell.In this case, the number density <strong>of</strong> nuclei would be n N = 3/(4πrc 3 ), <strong>and</strong> the atomic numberwould be Z = 4πrN 3 ρ ix i /3. As shown by Baym et al.[2], the Coulomb energy density <strong>of</strong>spherical nuclei is then given byf Coul = 3 (Z 2 e 2 ρ N1− 3 5 r N 2 u1/3 + 1 )2 u ≡ 4π 5 (r Nρ i x i e) 2 D(u) (6.15)89


M ( M O•)2.521.51M * = 0.811M * = 0.9M * = 1.0SkI4SkI6SkM‘SLy0SLy1SLy2SLy3SLy4SLy5SLy6SLy7SLy8SLy9SLy10Steiner et alPSR J1614-2230M ( M O•)21.51GNDFTruncatedFRTR N1FRTF N2Steiner et alPSR J1614-22300.50.507 8 9 10 11 12 13 14 15R ( km )09 10 11 12 13 14 15 16R ( km )Figure 6.6: The left panel shows mass <strong>and</strong> radius relation <strong>of</strong> the Skyrme <strong>for</strong>ce models whichpasses all tests. The right figure shows the one from our <strong>for</strong>ce models.where u = (r N /r c ) 3 is the fraction <strong>of</strong> space occupied by the nuclei. This relation defines thefunction D(u).As shown by Ravenhall et al.[58] in many situations it may be energetically favorable <strong>for</strong>nuclei to de<strong>for</strong>m from spheres into cylinders or plates, or even turn ‘inside out’ to <strong>for</strong>m relativelyvacuous bubbles surrounded by spherical nuclei <strong>and</strong> could also de<strong>for</strong>m into spheroidalshapes. In an attempt to account <strong>for</strong> varying shapes that would minimize the systematicfree energy, Lattimer & Swesty [22] defined a Coulomb shape function c(u),f Coul = 4π 5 (r Nρ i x i e) 2 c(u). (6.16)In the low density limit in which nuclei are expected to be spherical, c(u) = uD(u). In thelimit in which nuclei are inside-out, but the resulting bubbles are spherical,c = (1−u)D(1−u).6.4.3 <strong>Nuclear</strong> surface energyIn the liquid droplet model, the surface energy is not calculated dynamically, but is insteadparametrized as a leptodermous expansion in the curvature R −1 where R is the nuclear radius.In this work, we keep only the lowest order term, which is the surface tension <strong>of</strong> asemi-infinite, plane interface between two nuclear phases in mechanical <strong>and</strong> chemical equilibrium.The dense phase is characterized by the densities ρ ni ,ρ pi corresponding to those insidethe nucleus, while the light phase, characterized by the densities ρ no ,ρ po , corresponds to thenuclear vapor outside nuclei. The neutron <strong>and</strong> proton chemical potentials <strong>for</strong> semi-infinitematter must be equal in both phases, <strong>and</strong> they are denoted by µ n <strong>and</strong> µ p respectively. At agiven temperature, the phase equilibrium is determined by a single quantity, which can betaken to be the proton fraction x i in the dense phase, or by the neutron chemical potentialµ n , or by the neutron-proton chemical potential difference µ n − µ p . The surface tension isactually a thermodynamic potential, <strong>and</strong> as a consequence it is <strong>for</strong>mally a function <strong>of</strong> theneutron <strong>and</strong> proton chemical potentials, as well as <strong>of</strong> the temperature. If finite-size effects90


HNsn, p, α,eFigure 6.7: In the Wigner Seitz cell, there are one heavy nuclei <strong>and</strong> nucleons, electrons, <strong>and</strong>alpha particles outside. In this model, we have neutron skin on the surface <strong>of</strong> heavy nuclei.such as the Coulomb <strong>for</strong>ces can be ignored, the phase equilibrium dictates that the chemicalpotentials follow a unique trajectory determined only by the temperature <strong>and</strong> proton fractionin the dense phase. The work <strong>of</strong> Lattimer et al.[17] assumed that the surface tensioncould be there<strong>for</strong>e treated directly as a function <strong>of</strong> x i <strong>and</strong> T.However, Coulomb <strong>and</strong> other finite-size effects are not negligible, <strong>and</strong> the surface tensionis modified by them. The approach we initially employ here is to treat the surface tension asa function <strong>of</strong> the neutron chemical potential <strong>of</strong> the surface (<strong>and</strong> temperature), <strong>and</strong> in turn totreat the neutron chemical potential <strong>of</strong> the surface as a function <strong>of</strong> the dense proton fraction(<strong>and</strong> the temperature).As in chapter 2, the surface tension σ is obtained by minimization <strong>of</strong> the total free energyper unit area, <strong>and</strong> results inσ =∫ ∞−∞(F −µ n ρ n −µ p ρ p −p o )dz (6.17)where F = H −TS B <strong>and</strong> p o is the pressure <strong>of</strong> the bulk matter (if any) in the dilute phasein the limit z → ∞.We employ a simplified description following Lattimer et al.[17], but extend it to containthree adjustable parameters: the surface tension <strong>of</strong> zero-temperature symmetric matter, thezero-temperature surface symmetry parameter, <strong>and</strong> a parameter describing the entropy (or91


specific heat) <strong>of</strong> the surface <strong>of</strong> symmetric mater. This description is given below.What appears in the liquid droplet model, however, is the surface free energy per unitarea σ+µ n ν n +µ p ν p , where ν t represent the areal density <strong>of</strong> particles in the surface. It willbe found that these are determined byν t = − ∂σ∂µ t∣ ∣∣T, (6.18)as is expected from thermodynamics. Since our liquid droplet model takes the referenceinterface to be located at the proton surface, this model explicitly considers a skin madeonly <strong>of</strong> neutrons. There are no ‘surface protons’, <strong>and</strong> ν p = 0 . This is consistent withthe statement above that, at a given temperature, the quantity σ is a function <strong>of</strong> a singlequantity, µ n . There<strong>for</strong>e, the surface free energy per area is simply σ(µ n ,T)+µ n ν n . In whatfollows, the neutron chemical potential <strong>for</strong> the surface is denoted by µ s , but it will be shownthat free energy minimization requires µ s = µ no .The surface free energy density <strong>for</strong> spherical nuclei is thusf surf = 4πr 2 Nρ n (σ(µ s ,T)+µ s ν n ) = 3ur N(σ(µ s ,T)+µ s ν n ). (6.19)Once again, following the discussion <strong>of</strong> Lattimer & Swesty [22], the consequences <strong>of</strong>nuclear de<strong>for</strong>mation or shape change are taken into account by the surface shape factor s(u),so thatf surf = 3s(u)r N(σ(µ s ,T)+µ s ν n ). (6.20)Obviously, s(u) = u <strong>for</strong> a sphere. As is the case <strong>for</strong> c(u), the precise <strong>for</strong>m <strong>for</strong> s(u) will berequired.In practice, the results <strong>of</strong> plane-parallel surface energy calculations are most easily renderedin terms <strong>of</strong> the functions σ(x,T) <strong>and</strong> µ s (x,T), where x is the reference asymptoticproton fraction <strong>of</strong> the dense phase. Note that x is equivalent to x i <strong>for</strong> a plane-parallelsurface, but is not necessarily equivalent to x i <strong>for</strong> a finite liquid droplet nucleus. The quantityx is simply a convenient intermediate variable used to simplify the construction <strong>of</strong> σ(µ s ).As in section 3.4, we write the surface tension as a two-parameter function times atemperature dependence:( )2·2 α +qσ(x,T) = σ o h(x,T). (6.21)x −α +q +(1−x) −αThe quantity σ o is the zero-temperature, symmetric matter, surface tension. In case <strong>of</strong>Skyrme <strong>for</strong>ce model, the numerical calculation <strong>of</strong> surface tension is not stable since it is adifferential equation rather than an integral equation. The numerical method is describedin Ravenhall et al. [29].92


Table 6.1: Surface tension analytic fitting function <strong>for</strong> SLy4Model σ 0 (x i = 0.5) α qSLy4 1.13754 3.4 29.0494The factor h(x,T) contains the explicit temperature dependence <strong>of</strong> the surface free energy,<strong>and</strong> can approximated by [29][ ( )] p Th(x,T) = 1−(6.22)T c (x)where T c (x) is the maximum temperature <strong>for</strong> which nuclei can exist <strong>for</strong> a given value <strong>of</strong> x.For the truncate model <strong>and</strong> the modified model , p = 5/3 <strong>and</strong> T c (x) is given section 3.4. ForSLy4, we use p = 2 [29]. T c (x) <strong>for</strong> SLy4 is given by the same <strong>for</strong>mula, e.g. Eq. (3.31), butdifferent coefficients.Table 6.2: The critical temperature analytic fitting function <strong>for</strong> SLy4Model T c,0 a b cSLy4 14.502 -0.3649 -0.02025 -0.34368The neutron chemical chemical potential µ s , <strong>for</strong> a given value <strong>of</strong> x i <strong>and</strong> T, is that whichapplies to bulk phase equilibrium. For proton fractions larger than approximately 0.3 atzero temperature, the phase outside the nucleus is a vacuum with p o = 0. If the free energyis exp<strong>and</strong>ed about ρ = ρ 0 <strong>and</strong> T=0, the condition p = 0 results inµ s (x,T) = f bulk −x ∂f bulk∂x≃ −B −T215MeV +S v(1−4x 2 ). (6.23)This <strong>for</strong>mula works at low temperature <strong>and</strong> it cannot correctly represent neutron chemicalpotential on the surface. So, we may use a fitting function <strong>for</strong> neutron chemical potential<strong>of</strong> dense matter <strong>for</strong> given proton fraction <strong>and</strong> temperature. With appropriate units <strong>of</strong>coefficients,µ s (x,T) = −B +α(1−2x)+β(1−2x) 2 +γT 2 +δ(1−2x)T 2 (6.24)This fitting function is obtained from the numerical calculation <strong>of</strong> phase equilibrium. Table6.4.3 shows the coefficients <strong>for</strong> the fitting function. There<strong>for</strong>e, given that one <strong>of</strong> the equilibriumconditions will be µ s = µ no , the procedure is to determine the quantity x from µ no <strong>and</strong>T by solving a quadratic equation.The surface neutron density is then found from Eq. (6.18), orν n = − ∂σ(µ s,T)= − ∂σ(x,T)∂µ s ∂x93( ) −1 µs (x,T). (6.25)∂x


Table 6.3: The chemical potential fitting functionModel B α β γ δTruncated 16.539 62.365 -28.900 -0.0455 -0.0414Modified 16.000 79.627 -49.128 -0.0595 -0.1017SLy4 15.987 63.963 -34.546 -0.0469 -0.0325All surface thermodynamic quantities can then be evaluated. We again emphasize that <strong>for</strong>surface quantities, x ≠ x i but close each other.6.4.4 <strong>Nuclear</strong> translational energyThenucleiessentially<strong>for</strong>manon-degenerate, non-relativisticBoltzmanngas. Thefreeenergydensity <strong>of</strong> such a gas, ignoring any internal degrees <strong>of</strong> freedom, isf trans = ρ N (µ H −T) = uρ [ ( ) ]iA T uρiln −1 . (6.26)ρ Q A 5/2Here ρ Q = (mT/2π 2 ) 3/2 , where m is the nucleon mass, <strong>and</strong> µ H represents a nuclearchemical potential. As in Lattimer et al.[17], it is assumed that the translational energydiminishes with temperature <strong>and</strong> disappears at T c with the same behavior as the surfaceenergy. This is en<strong>for</strong>ced with the function h(x i ,T). Following Lattimer & Swesty [22],the simplifying assumption is made to replace A in Eq. (6.26) with a constant A o = 60which is an approximation <strong>of</strong> mass number <strong>of</strong> heavy nuclei in neutron star crust. Thisapproximation simplifies the algebra surrounding the equilibrium conditions. In addition,achieving a consistency in the limit u → 1, where an inside-out phase may replace ordinarynuclei, u in Eq. (6.26) is replaced by u(1−u). These approximations are justifiable, giventhat the translational energy is a relatively unimportant component. To summarize, thetranslational free energy density <strong>and</strong> nuclear chemical potential are taken to bef trans = u(1−u)ρ iA oh(x i ,T)(µ H −T); µ H = T ln6.4.5 Alpha particles(u(1−u)ρ iρ Q A 5/2o). (6.27)To represent the thermodynamics <strong>of</strong> an ensemble <strong>of</strong> light nuclei, a gas <strong>of</strong> non-interacting,but finite volume, alpha particles is assumed. The alpha particles have substantial thermodynamiccontributions in regions limited to those near the nuclear dissociation curves. Thealpha particles <strong>for</strong> those temperatures <strong>and</strong> densities may be treated as a non-interactingBoltzmann gas. Alpha particles are bound relative to free neutrons by an energy B α = 28.3MeV. Alpha particles also occupy a fairly large volume <strong>of</strong> space, v α = 24 fm −3 per alphaparticle, <strong>and</strong> the total volume they occupy must be considered. With these stipulations, the94


free energy density <strong>for</strong> the alpha particles isF α = (1−u)ρ α (µ α −T −B α ), (6.28)where ρ α <strong>and</strong> µ α are the number density <strong>and</strong> chemical potential <strong>of</strong> alpha particles, respectively.It is assumed that the fractional volume u is already occupied by nuclei <strong>and</strong> is notavailable <strong>for</strong> alpha particles to occupy. µ α is related to ρ α by( )ραµ α = T ln . (6.29)8ρ Q6.5 The Combined modelThe free energy <strong>of</strong> each component <strong>of</strong> dense matter has been developed in the previoussections. In dense matter, these components are present in varying concentrations, <strong>and</strong> onehas to allow <strong>for</strong> the volume occupied by nuclei <strong>and</strong> by alpha particles. In the most generalcase, in which the four components H,n,p,α exists, the baryon conservation equation isThe charge conservation equation isρ = uρ i +2s(u) ν nr N+(1−u)[4ρ α +(1−ρ α v α )ρ o ]. (6.30)ρY p = ux i ρ i +(1−u)[2ρ α +(1−ρ α v α )x o ρ o ]. (6.31)The total free energy density <strong>for</strong> the entire system becomes:F =uρ i f bulk (ρ i ,x i ,T)+ 3s(u)r N(σ(µ s ,T)+µ s (x,T)ν n )+ 4π 5 (r Nρ i x i e) 2 c(u)+f trans (u,ρ i ,x i ,T)+F α (ρ α ,u,T)+(1−u)(1−ρ α v α )ρ o f bulk (ρ o ,x o ,T).(6.32)We neglect the free energy contribution from leptons <strong>and</strong> photons since we can treat themseparately in nuclear equilibrium.6.5.1 Equilibrium conditionsThe total free energy density is a function <strong>of</strong> the independent variables (ρ,Y p ,T), but also afunction <strong>of</strong> the 9 independent variables (ρ i ,x i ,r N ,µ s ,ν n ,u,ρ o ,x o ,ρ α ). Note that ρ o <strong>and</strong> x ocan be eliminated by the baryon <strong>and</strong> charge conservation equations:[]1 ρ−uρ i −3sν n /r Nρ o =−4ρ α1−ρ α v α 1−u(6.33)ρY p −ux i ρ i −2ρ α (1−u)x o =ρ−uρ i −3sν n /r N −4ρ α (1−u) .The specification <strong>of</strong> nuclear statistical equilibrium <strong>for</strong> the system dem<strong>and</strong>s that F be minimizedwith respect to each <strong>of</strong> the 7 remaining dependent variables. It is convenient to use95


the seven (ρ i ,x i ,r N ,x,ν n /r N ,u,ρ α ) variables set.The derivative with respect to ρ o <strong>and</strong> x o can be resolved with∂ρ o u= −∂ρ i (1−u)(1−ρ α v α ) ,∂ρ o∂x i= 0,∂x o u(x o −x i )= −∂ρ i ρ o (1−u)(1−ρ α v α ) ,∂x o uρ i= −∂x i ρ o (1−u)(1−ρ α v α ) ,∂ρ o∂(ν n /r N ) = − 3s(u)ρ o (1−u)(1−ρ α v α ) ,∂x o∂(ν n /r N ) =∂ρ o∂u = −ρ i −3s ′ (ν n /r N )+4ρ α +ρ o (1−ρ α v α ),(1−u)(1−ρ α v α )∂x o∂u = ρ i(x o −x i )+2ρ α (1−2x o )+3s ′ x o ν n /r Nρ o (1−u)(1−ρ α v α )∂ρ o∂ρ α= ρ ov α −41−ρ α v α,3s(u)x oρ o (1−u)(1−ρ α v α ) ,(6.34),ρ o (1−ρ α v α ) .∂x o= 4x o −2)∂ρ α96


Here s ′ = ∂s(u)/∂u.The minimizations are :∂F∂ρ i= 0 = u(µ ni −x iˆµ i −µ no +x i ˆµ o + h iA o(1−u)µ H)+ 8π 5 (r Nx i e) 2 ρ i c(u), (6.35a)∂F∂x i= 0 = uρ i (ˆµ o − ˆµ i )+ 8π 5 (r Nρ i e) 2 x i c(u)+h ′ ρ iiA ou(1−u)(µ H −T),(6.35b)∂F= 0 = − 3s(u)σ∂r Nr 2 N∂F∂x = 0 = 3s(u)σr N+ 8π 5 (ρ ix i e) 2 r N c(u), (6.35c)( ∂σ∂x +ν n)∂µ s,, (6.35d)∂x∂F∂(ν n /r N ) = 0 = 3s(u)(µ s −µ no ),∂F[ ]∂u = 0 = ρ i(f buli (ρ i ,x i ,T)−µ no +x i ˆµ o )+ 3s′ σ +ν n (µ s −µ no )r N+ 4π 5 (ρ ix i r N e) 2 c ′ +(1−2u) h iρ iA oµ H +p o +ρ α T ,∂F∂ρ α= 0 = (1−u)(µ α −B α −4µ no +2ˆµ o +P o v α ).(6.35e)(6.35f)(6.35g)Here ˆµ = µ n −µ p , h i = h(x i ), h ′ i = ∂h(x i)/∂x i , <strong>and</strong> c ′ = ∂c(u)/∂u.Eq. (6.35c) represents the <strong>Nuclear</strong> virial theorem, that the surface energy equals twicethe Coulomb energy. Solving <strong>for</strong> r N yieldsr 3 N =which can be written more succinctly asr N = 9σ ( s2β c )1/3 , β ≡15s(u)σ8π(ρ i x i e) 2 c(u) , (6.36)( ) 1/3 243π(ρ i x i σe) 2/3 . (6.37)There<strong>for</strong>e r N can be eliminated from the equilibrium equations. It should be noted thatbecause <strong>of</strong> the <strong>Nuclear</strong> virial theorem, the total <strong>of</strong> the surface <strong>and</strong> Coulomb energy densities597


is(f surf +f Coul = β(cs 2 ) 1/3 1+ 2µ ) (sν n≡ D(u) 1+ 2µ )sν n, (6.38)3σ 3σwhere D(u) is defined [22]. The estimation <strong>of</strong> D(u) is described below.Eq. (6.35e) proves the earlier statement that the neutron chemical potential <strong>of</strong> the surfaceµ s should be equal to µ no . Eq. (6.35d) also proves the earlier definition <strong>of</strong> the surfacedensity <strong>of</strong> neutrons ν n . Eq. (6.35g) determines the alpha particle density, reiterating thatthe alpha particles are in nuclear statistical equilibrium. The last term <strong>of</strong> Eq. (6.35g) representsaneffective repulsive interaction <strong>for</strong>thealpha particles because <strong>of</strong>theexcluded volume.The remaining three equilibrium equations can be manipulated into the expected resultsconcerning equilibrium among chemical potentials <strong>and</strong> pressure. Together with theconservation equations, the seven equations to be solved simultaneously areA 1 = 0 = p i −p o −ρ α T −β(D ′ − 2D )3u+ uρ ih iA oµ H ,(6.39a)A 2 = 0 = µ ni −µ no + 1−u [ ]h i µ H −h ′ iA i(µ H −T) ,o(6.39b)A 3 = 0 = µ pi −µ po + 2βD3ρ i x i u + 1−u []h i µ H +(1−x i )h ′ iA i(µ H −T) ,o(6.39c)A 4 = 0 = ux i ρ i −ρY p +(1−u)(ρ po +r p ρ α ),(6.39d)A 5 = 0 = uρ i −ρ+ 2β]3σ Dν n +(1−u)[ρ no +ρ po +(r n +r p )ρ α , (6.39e)µ s = µ no , (6.39f)µ α = B α +2µ no +2µ po −p o v α . (6.39g)We defined the useful abbreviationsr n = 2−ρ no v α , r p = 2−ρ po v α . (6.40)Note that the functions c <strong>and</strong> s could be completely replaced in Eq. (6.39) in favor <strong>of</strong> D because3D ′ = uc ′ /c+2us ′ /s. The nuclear equilibrium is determined only by the combinationcs 2 , or equivalently, by D.Eq. (6.39g) illustrates that µ α , <strong>and</strong> hence ρ α , is a function <strong>of</strong> ρ no <strong>and</strong> ρ po alone. Inaddition, σ <strong>and</strong> ν n , which are functions <strong>of</strong> x, are also determinable from ρ no <strong>and</strong> ρ po , because<strong>of</strong> Eq. (6.39g). There<strong>for</strong>e, it is necessary to simultaneously solve the 5 equations A i = 0 <strong>and</strong>the two supplementary constraints. Note that it would have been inconsistent to evaluate σ<strong>and</strong> ν n directly as functions <strong>of</strong> x i , which seems the more straight<strong>for</strong>ward procedure, becausethis does not ensure the full minimization <strong>of</strong> F.98


6.5.2 Determination <strong>of</strong> Coulomb surface shape parameter DFollowing the work <strong>of</strong> Ravenhall et al.[58], one can identify the dependence <strong>of</strong> surface <strong>and</strong>Coulomb energies on the dimensionality d <strong>of</strong> the geometry. For d = 3, one has spheres <strong>of</strong>radius r N , For d = 2, one has cylinders <strong>of</strong> radius r N . For d=1, one has plates <strong>of</strong> thicknessr N . The surface <strong>and</strong> Coulomb energy densities <strong>of</strong> an isolated configuration with dimensiond are thenf surf = udσr N, f Coul = 4π 5 (ρ ix o r N e) 2 D d (u) (6.41)whereIn the case d = 2, the limit <strong>of</strong> D d becomesD d (u) = 5 [1− d d 2 −4 2 u1−2/d + d−2 ]2 u . (6.42)D 2 (u) = 5 [u−1−ln(u)]. (6.43)8Thus, one has s(u) = ud/3 <strong>and</strong> c(u) = uD d (u). In the event <strong>of</strong> inside-out matter, one simplyreplaces u by 1−u in these equations.Both s <strong>and</strong> c depend upon d. The minimization <strong>of</strong> F with respect to r N allows theelimination <strong>of</strong> r N <strong>and</strong> the identification f surf +f Coul = βD implies that( ) d 2 1/3D d (u)D = u . (6.44)9The <strong>for</strong>mulas Eq. (6.41) are valid <strong>for</strong> integer values <strong>of</strong> d , but gradual de<strong>for</strong>mations betweenthe integer states could be modeled by allowing d to vary [58]. The minimization <strong>of</strong> f surf +f Coul = βD with respect to d then implies that d is determined from minimization <strong>of</strong> D,or equivalently d 2 D d , at each value <strong>of</strong> u, subject to a maximum <strong>of</strong> 3 <strong>and</strong> a minimum <strong>of</strong> 1<strong>for</strong> d. Fig. 6.8 shows the geometric factor D. D does not depends on density so we canfind a fitting function. Lattimer & Swesty [22] showed that a suitable approximation to thisminimization, but also allowing <strong>for</strong> the possibility <strong>of</strong> inside-out matter, is obtained by usingD(u) = u(1−u) (1−u)D1/3 3 (u)+uD 1/33 (1−u)u 2 +(1−u) 2 +0.6u 2 (1−u) . (6.45)26.5.3 Solving the equilibrium equationsThere are seven equilibrium equations, including the two conservation equations <strong>for</strong> baryonnumber <strong>and</strong> charge, <strong>for</strong> seven variables, which can be taken to be ρ i ,x i ,u,ρ no ,ρ po ,ρ α <strong>and</strong> x.Lattimer & Swesty [22] chose to reduce an analogous system by eliminating two variableswith the conservation constraints. Even though we can reduce the number <strong>of</strong> equations tosolve, we may probably meet numerical difficulties since, <strong>for</strong> example, the density <strong>of</strong> outsidenucleons can be negative <strong>and</strong> code breaks. Thus, we do not eliminate the number <strong>of</strong> unknownsfrom the conservation properties but leave them to solve. However, as seen above,99


0.20.180.160.140.12d=1d=2.0d=3d=1,inside outd=2.0,inside outd=3.0,inside outMinimumD(u)0.10.080.060.040.0200 0.2 0.4 0.6 0.8 1Figure 6.8: D as a function <strong>of</strong> u <strong>for</strong> given dimension(d = 1,2,3). The minimum can beobtained the continuous change <strong>of</strong> d.uthere are two trivial equilibrium conditions so that the variables y i = (ρ α ,x) can be substitudedin terms <strong>of</strong> ρ no <strong>and</strong> ρ po .The most convenient <strong>for</strong>m <strong>for</strong> the five dependent variables isz i = (ρ i ,lnρ no ,lnρ po ,x i ,lnu). The ‘log’ variables were introduced to prevent the variablesfrom becoming negative. It also h<strong>and</strong>les the accuracy <strong>of</strong> the small number.Solutions are obtained via Newton-Raphson iteration, in which successive changes ∆z j inthe independent variables z j are found from the matrix equation (summing over repeatedindices),∆z j = −(B ij ) −1 A i . (6.46)Here A i is the vector <strong>of</strong> equations, defined in Eq. (6.39), to be zeroed, <strong>and</strong>where λ = (ρ,Y p ,T).B ij ≡ ∂A i∂z j∣∣∣y,λ, (6.47)6.6 ResultAs mentioned earlier in this chapter, our code is fast enough to generate an EOS table so wecan easily compare EOS tables with different <strong>for</strong>ce models. The liquid droplet approach also100


gives analytic derivatives <strong>of</strong> thermodynamic quantities. This means that we can guaranteethe thermodynamic consistency compared with the Thomas-Fermi or Hartree-Fock approximation,in which thermodynamic derivatives can be obtained numerically. For these reasons,the liquid droplet approach is the most favorable method to make EOS tables.For each point, we allow the independent variables change with small amounts <strong>of</strong> them sincewe use the previous solution <strong>of</strong> the variable set. By doing this, we can avoid code breaking<strong>and</strong> speed up the convergence. In the subroutine eos(ρ,T,Y p ) which consists <strong>of</strong> two subroutine,Newton-Rapshon code is to seek nucleons, alpha particle, <strong>and</strong> heavy nuclei (npaH)solution at first, if it does not converge, try to find nucleons <strong>and</strong> alpha particle (npa) solution,if it does not exist, nucleons (np) solution is to be found.Now we present some <strong>of</strong> the result from our table † .6.6.1 Energy density <strong>and</strong> pressureAt zero temperature, the energy density <strong>and</strong>pressure is a function <strong>of</strong> density <strong>for</strong> bothFRTFs<strong>and</strong> SLy4 since the Fermi-Dirac distribution function is frozen to 1 so we have a momentumdensity as a function <strong>of</strong> density, Eq (2.18, 2.36). For non-zero temperature, we need tocalculate degeneracy parameters using interpolation or JEL scheme [12], then we can findenergy density <strong>and</strong> pressure <strong>for</strong> given ρ, T, <strong>and</strong> Y p .6.6.2 Phase BoundariesFor a given density <strong>and</strong> proton fraction, the heavy nucleus dissolves into nucleons. This phenomenoncorresponds to phase equilibrium <strong>and</strong> critical temperature. Simply, if the criticaltemperature <strong>of</strong> bulk equilibrium is high, the nucleus can exist in relatively high temperature.Fig. 6.9 shows the phase boundaries <strong>of</strong> hot dense matter <strong>for</strong> a given proton fraction (Y p ). Ingeneral in the region below the curve, there exist nucleons, alpha particle, <strong>and</strong> heavy nuclei(npaH). On the upper left region, nucleons exist with alpha particle. On the right region <strong>of</strong>the curve, only nucleons exist. FRTFs <strong>and</strong> SLy4 have the similar phase boundaries. In SLy4,nuclei exist larger range <strong>of</strong> density than FRTFs. Nuclei can exist at higher temperature inTruncated model because <strong>of</strong> higher critical temperature.6.6.3 Atomic number in the heavy nucleiThe most stable nucleus on the earth is 56 Fe which has Z = 26. However, in the densematter or neutron star crust, there are lots<strong>of</strong> nucleons <strong>and</strong> leptons outside, so the interactionamong them may change the configuration <strong>of</strong> the most stable nuclei. Fig. 6.10 shows theatomic number as a function <strong>of</strong> density at given proton fraction (Y e = 0.45) <strong>and</strong> temperature(T = 0.726 MeV). The phase transition in SLy4 has completed <strong>for</strong> large domain <strong>of</strong> density sothe atomic number increase abruptly <strong>for</strong> narrow range <strong>of</strong> density. SLy4 always have a largenumber <strong>of</strong> protons in heavy nuclei <strong>for</strong> a given density, proton fraction, <strong>and</strong> temperature. Incase <strong>of</strong> FRTF II, the atomic number is the smallest because <strong>of</strong> low surface tension.† The result presented in this thesis is from beta version. We’ll post them on the website soon afterchecking thermodynamic consistency.101


141210Phase diagram, Y p =0.45SLy4FRTF ModifiedFRTF TruncatedT (MeV)864201e-06 1e-05 0.0001 0.001 0.01 0.1 1ρ (fm -3 )Figure 6.9: This figure shows the phase boundaries <strong>of</strong> dense matter when Y p = 0.45.6.7 ConclusionsThe liquid droplet approach is the most promising method to generate thermodynamicallyconsistent nuclear EOS table. The nuclear <strong>for</strong>ce model should be tested be<strong>for</strong>e making anEOS table so that the table represents both nuclear experiments on Earth <strong>and</strong> astrophysicalobservations.To make the EOS consistent with the nuclear physics aspect, we have to use one nuclear<strong>for</strong>ce model to calculate the energy contribution from heavy nuclei, nucleons outside, <strong>and</strong>surface tension. In the thermodynamical sense, the derivatives in liquid droplet approachare written analytically <strong>and</strong> can be compared with numerical derivatives.We treat electrons <strong>and</strong> photons separately since they interact weakly. But we add themto calculate total energy, pressure, entropy <strong>and</strong> their derivative with respect to temperature,density, <strong>and</strong> proton composition.For finite temperature, we have to calculate the Fermi integral (F 1/2 , F 3/2 ) to obtain numberdensity <strong>and</strong> momentum density. The interpolation from the table cannot give enough accuracyat low temperature (T ≤ 1 MeV). JEL is a successful scheme to give enough accuracy<strong>for</strong> all domains (relativistic vs. non-relativistic, degenerate <strong>and</strong> non-degenerate).Phase transitions around half <strong>of</strong> nuclear saturation density (1/2ρ 0 ) can be achieved when weemploy the geometric function D <strong>and</strong> changing the dimension d continuously.We should be able to combine the liquid droplet approach with a relativistic mean fieldmodel in the near future if we h<strong>and</strong>le the large L with acceptable values so new parameters102


200180160Atomic number, Y p =0.45, T=0.726MeVSLy4FRTF ModifiedFRTF Truncated140120Z1008060402001e-06 1e-05 0.0001 0.001 0.01 0.1 1ρ (fm -3 )Figure 6.10: The atomic number increases as density increases in general. Around phasetransition region, the atomic number drops <strong>and</strong> increases again.in RMF can represent the allowed regions <strong>of</strong> mass <strong>and</strong> radius <strong>of</strong> neutron stars.103


Chapter 7ConclusionsThe nuclear equation <strong>of</strong> state is needed to simulate supernovae explosions, proto-neutronstars, <strong>and</strong> compact binary mergers involving neutron stars. For these simulations, we needthermodynamic in<strong>for</strong>mation <strong>for</strong> given baryon density, proton fraction, <strong>and</strong> temperature.Since the EOS should cover wide range <strong>of</strong> the above variables, we need to be careful whenwe choose a nuclear <strong>for</strong>ce model to make the EOS table. A good nuclear <strong>for</strong>ce modelshould represent both low <strong>and</strong> high densities nuclear phenomena as well as pure neutron <strong>and</strong>astrophysical matter.First <strong>of</strong> all, we study nuclear physics using Finite Range <strong>for</strong>ce model. In the Finite Range<strong>for</strong>ce model, the nuclear interaction energy at density each point has the contribution fromevery spacial point with weight factor (e −r/a ). The first model that we investigated is thetruncated model (Chapter 2). To find the parameters in the truncated model, we use thest<strong>and</strong>ard nuclear matter properties <strong>of</strong> symmetric nuclear matter , such as E/A = −B ≃ −16MeV, P = 0 MeV/fm 3 , S v ≃ 32 MeV, L ≃ 60MeV. a, which is a nuclear diffusenessparameter can be obtained from semi-infinite nuclear matter calculation. This truncatedmodel is improved by adding new density dependent interactions to fit optical potential,nuclear incompressibility, <strong>and</strong> pure neutron matter. To get optimized parameter set <strong>for</strong> themodified model, we compare the results <strong>of</strong> total binding energy <strong>of</strong> single nucleus from themodified model with experimental values. The recent constraints <strong>of</strong> S v <strong>and</strong> L [47] confirmthat our parameter set <strong>for</strong> the modified model is a good choice.Boththe truncated <strong>and</strong>modified models model areused tocalculate energy per baryon<strong>of</strong>single nucleus <strong>and</strong> heavy nuclei in dense matter. Since the finite range model gives integralequations instead <strong>of</strong> differential equations, the boundary condition has no difficulty <strong>and</strong>the numerical calculation in the unit (Wigner-Seitz) cell is much easier to per<strong>for</strong>m th<strong>and</strong>ifferential equations. Thomas-Fermi approximation is employed to find the nuclear densitypr<strong>of</strong>ile. In TF, we are seeking the local density at each point instead <strong>of</strong> finding the full wavefunctions. From the density pr<strong>of</strong>ile, we find the plane wave so that we are able to calculatemomentum density to calculate Hamiltonian density or kinetic energy. At zero temperature,the relation between the number density <strong>and</strong> momentum density is simple since the Fermi-Diracdistributionfunctionisfrozento1. Atfinitetemperature, however, wehavetocalculatethe Fermi integral which is not an analytic function. For example, from the number density,we aresupposed t<strong>of</strong>ind thedegeneracy parameter (φ)fromFermi integral, F 3/2 (φ)<strong>and</strong>to getthe momentum density from Fermi integral, F 5/2 (φ). JEL scheme provides polynomial fits104


<strong>for</strong> Fermi integral. This fitting function has remarkably high accuracies to calculate Fermiintegral. We employ JEL scheme to calculate nuclear <strong>for</strong>ce at finite temperature.To find the density pr<strong>of</strong>ile <strong>of</strong> nuclei, we use Euler-Lagrange equations to minimize totalbinding energy, <strong>and</strong> it gives that µ n,i <strong>and</strong> µ p,i are constant in the cell if we set the densityat each grid point as unknown. The multi-dimensional Newton-Raphson method is usedto solve the equations. If the initial guess is good enough, the number <strong>of</strong> iterations to getsolutions is only two or three so the code is fast enough to calculate binding energy <strong>of</strong> allnuclei present on earth.The bulk equilibrium calculation is an example <strong>of</strong> using JEL scheme. Bulk (dense-dilute)matter equilibrium is the simplified case <strong>of</strong> heavy nuclei in dense matter. That is, if bulkequilibrium is possible, that suggests there might be a phase <strong>of</strong> heavy nuclei in dense matter.If bulk equilibrium does not exist, there would be uni<strong>for</strong>m nuclear matter phase <strong>for</strong> givendensity, proton fraction, <strong>and</strong> temperature. The conditions <strong>for</strong> bulk equilibrium are made byminimizing the total free energy density, which are P I = P II , µ nI = µ nII , <strong>and</strong> µ pI = µ pII .I(II) represent dense (dilute) phase. From phase equilibrium calculations, we can get thecriticaltemperatureinwhichbothphaseshavethesamedensity <strong>and</strong>protonfraction. Beyondthe critical temperature, phase equilibrium does not exist any more. Semi-infinite nuclearmatter density pr<strong>of</strong>ile is obtained from the finite range <strong>for</strong>ce models. Since the curvatureeffect <strong>of</strong> finite nuclei is small, the semi-infinite nuclear matter calculation is used to find thesurfacetension<strong>for</strong>nuclei. Withthecriticaltemperaturefromphaseequilibrium, semi-infinitecalculation gives analytic fitting function <strong>for</strong> surface tension <strong>for</strong>mula (ω = ω(x,T)).The importance <strong>of</strong> choosing <strong>of</strong> nuclear <strong>for</strong>ce model to make EOS table can also be seenwhen we study neutron star crust. The nuclear potentials have a lot <strong>of</strong> different <strong>for</strong>m. Thesedifference does not effect the pressure, energy density, <strong>and</strong> atomic number in neutron starcrust. If we extend the potentials, however, to high density, the difference in the mass-radiusrelation <strong>of</strong> neutron stars is apparent.Using the finite-range model with liquid droplet approach, we build EOS table <strong>for</strong> astrophysicalsimulations. Thomas-Fermi <strong>and</strong> Hartree (Fock) approximation gives only numericalquantities <strong>of</strong> thermodynamic quantities. On the other h<strong>and</strong>, liquid droplet approach givesanalytic thermodynamic quantities, that means thermodynamic consistency. Compared toLS EOS <strong>and</strong> LPRL [17], our LDM approach contains neutron skins <strong>and</strong> surface diffusenessto improve the liquid droplet <strong>for</strong>malism. The LDM approach is fast to build the full EOStable so that we can manipulate different nuclear <strong>for</strong>ce model without difficulty <strong>and</strong> we canmake an EOS table with a large number <strong>of</strong> grid points. The atomic number at the lowerdensity region (≃ 0.01/fm 3 ) remains around 30 <strong>and</strong> has experienced abrupt changes betweenthe phase transition regions (0.5ρ 0 ). This is a general feature in both the non-relativisticmean field model <strong>and</strong> relativistic mean field model. In the phase transition region, theatomic number from RMF models is an order <strong>of</strong> magnitude greater than the one from thenon-relativistic potential models. Between SLy4 <strong>and</strong> finite range <strong>for</strong>ce model, the phasetransition to uni<strong>for</strong>m matter happens later in SLy4. Since the truncated model has thehigher critical temperature than SLy4 <strong>and</strong> the modified model, nuclei can exist in highertemperature in the truncated model.It is necessary to compare the EOS tables using the same nuclear <strong>for</strong>ce model but withdifferent numerical techniques (LDM, TF, HF) to see how similar the each EOS table wouldbe. This work will be begun in the near future.105


Appendix AFinite Range IntegrationIn the Yukawa type finite range <strong>for</strong>ce mode, one has∫ ∫˜g(r 1 ) = d 3 r 2 f(r 12 /a)g(r 2 ) = d 3 1r 12/a 2 g(r4πr 12 a 2e−r 2 ).(A.1)This integration is not converged easily. The integration which involves∫f(x)e −x dx has ∼ 1/N 2 convergence properties, where N is number <strong>of</strong> grid points. Mathematically,we can increase N as much as possible, but the number <strong>of</strong> times to calculate theintegration increases as ∼ N 2 <strong>and</strong> the current ability <strong>of</strong> cpu may have numerical noise whenwe increase the number <strong>of</strong> grid points.One way to solve the convergence issues is to use mathematical induction, or, geometricsequence. That is, the the error between ˜g N <strong>and</strong> ˜g 2N decreases asThus we may have˜g 2N − ˜g N = ∆, ˜g 4N − ˜g 2N = 1 4 ∆, ˜g 8N − ˜g 4N = 1 ∆,... (A.2)16˜g ∞ = ˜g N +∆+ 1 4 ∆+ 116 ∆+··· = ˜g N + 4 3 ∆ = 4 3˜g 2N − 1 3˜g N . (A.3)This induction shows enough accuracies when N = 50 <strong>and</strong> N = 100 <strong>for</strong> given Wigner seitzcell(12fm). However, we need to calculate the integration twice to complete the convergence.Now, we show the more efficient method to calculate this Yukawa type integration.106


A.1 Taylor expansion integrationTheintegr<strong>and</strong>intheYukawaintegrationisexp<strong>and</strong>totakeintoaccount thevariationbetweenr <strong>and</strong> r +∆. For example, <strong>for</strong> given interval (r i ,r i +δ),f(r) =f(r i )+f ′ (r i )(r −r i )+ 1 2 f′′ (r i )(r −r i ) 2 + 1 3! f(3) (r i )(r −r i ) 3 +···=f i + f i+1 −f i−1(r −r i )+ f i+1 −2f i +f i−1(r −r2∆ 2∆ 2 i ) 2 +··· .(A.4)By doing this, we can eliminate the contamination in the integral between r i <strong>and</strong> r i+1 whenwe use simple trapezoid rule ( ∫ i+1if(x)dx = 1 2 ∆(f i +f i+1 ) ).A.1.1 1D plane parallel nuclear matterThe study <strong>of</strong> 1D parallel (semi-infinite) nuclear matter sheds light on how the surface tensionchanges <strong>for</strong> a given temperature <strong>and</strong> proton fraction. In the semi-infinite nuclear matter,without loss <strong>of</strong> generality, we can say g(r) = g(x,y,z) = g(z). Then the original finite rangeintegration becomes˜g(z o ) =∫ ∞−∞∫ ∞ ∫ ∞dx dy−∞ −∞√e − x 2 +y 2 +(z−z o) 2 /adz4πa 2√ x 2 +y 2 +(z −z o ) g(z).2(A.5)Changing variables, x 2 +y 2 = ρ 2 , ∫ dx ∫ dy = 2π ∫ ρdρ give˜g(z o ) = 1 2= 12a= 12a∫ ∞−∞∫ ∞∫ ∞dzg(z)−∞∫ ∞dzg(z)e −|z−zo|/a−∞0e − √ρ 2 +(z−z o) 2 /aa 2√ ρ 2 +(z −z o ) 2 g(z)dzg(z)(−)e − √ρ 2 +(z−z o) 2 /a | 0 ∞(A.6)With the above general equation in 1D plane parallel case, we apply Taylor expansion <strong>for</strong>smooth distance dependent function u.ũ(z) = 1 [∫ z∫ ∞]u(z ′ )e (z′ −z)/a dz ′ + u(z ′ )e (z−z′)/a dz ′ ≡ u − (z)+u + (z) (A.7)2a −∞z107


This equation can be calculated by iteration scheme.u − i+1 = 12a= 12a∫ zi+1−∞∫ zi−∞=e −∆/a u − i + 12au(z ′ )e (z′ −z i+1 )/a dz ′u(z ′ )e (z′ −z i )/a e −∆/a dz ′ + 12a∫ zi+1z iu(z)e (z−z i+1)/a dz.∫ zi+1z iu(z ′ )e (z′ −z i+1 )/a dz ′(A.8)For the integration at the end, we exp<strong>and</strong>then we havewheref =e −∆/aI0 − = 12aI1 − = 12au(z) = u i + u i+1 −u i∆ (z −z i)+O(∆ 2 ) (A.9)∫ ∆0∫ ∆0u − i+1 = fu− i +I − 0 u i +I − 1 u i+1 ,(1− z )e z/a e −∆/a dz = a−(a+∆)e−∆/a∆2∆z∆ ez/a e −∆/a dz = ∆−a+ae−∆/a .2∆In the same manner, we have an equation <strong>for</strong> u + i ,u + i = 12a= 12a= 12a∫ ∞z iu(z ′ )e (z i−z ′ )/a dz ′∫ zi+1∫ ∞(A.10)(A.11)u(z ′ )e (z i−z ′)/a dz ′ + 1 u(z ′ )e (z i+1−z ′)/a e −∆/a dz ′ (A.12)z i2a z i+1u(z)e (zi−z)/a dz +e −∆/a u + i+1 .z i∫ zi+1With the first order Taylor expansion be<strong>for</strong>e, we havewhereu + i = fu + i+1 +I+ 0 u i +I + 1 u i+1 ,I 0 + = 12aI 1 + = 12aAs a second order expansion, we have∫ ∆0∫ ∆0(1− z )e −z/a dz ′ = I1− ∆z∆ e−z/a dz ′ = I − 0 .(A.13)(A.14)u(z) = u i + u i+1 −u i−1(z −z i )+ u i+1 −2u i +u i−1(z −z2∆ 2∆ 2 i ) 2 +O(∆ 3 ) (A.15)108


thenwhere0u − i+1 = fu− i +J − 0 u i−1 +J − 1 u i +J − 2 u i+1J0 − = 1 ∫ ∆ (2a e−∆/a − z )0 2∆ + z2e z/a dz = 2a2 −a∆−a(2a+∆)e −∆/a2∆ 2 4∆ 2J1 − = 1 ∫ ∆ ) (1− z22a e−∆/a e z/a dz = −2a2 +2a∆+(2a 2 −∆ 2 )e −∆/a0 ∆ 2 2∆ 2J2 − = 1 ∫ ∆ ( z)2a e−∆/a 2∆ + z2e z/a dz = 2a2 −3a∆+2∆ 2 −a(2a−∆)e −∆/a.2∆ 2 4∆ 2(A.16)(A.17)For u + i , u + i = fu + i+1 +J+ 0 u i−1 +J + 1 u i +J − 2 u i+1, (A.18)whereJ 0 + = 12aJ 1 + = 12aJ2 − = 12a∫ ∆0∫ ∆0∫ ∆0(− z )2∆ + z2e −z/a dz = 2a2 −a∆−a(2a+∆)e −∆/a= J −2∆ 2 4∆ 2 0) (1− z2e −z/a dz = ∆2 −2a 2+ a(a+∆)e−∆/a∆ 2 2∆ 2 ∆ 2( z)2∆ + z2e −z/a dz = 2a2 +a∆−(2∆ 2 +3a∆+2∆ 2 )e −∆/a.2∆ 2 4∆ 2(A.19)For this second order expansion, we have to be careful <strong>for</strong> u − 1 <strong>and</strong> u+ N−1, which will beu − 1 = fu − 0 +I −10 u 0 +I −11 u 1 u + N−1 = fuN 0 +I + 0 u N−1 +I + 1 u N (A.20)because there will be no second order derivatives at i = 1 <strong>and</strong> i = N −1.For the test <strong>of</strong> accuracy <strong>of</strong> this tilde equation, we now compare analytic solution <strong>of</strong> u(z) =11+e z/a where a is chosen as 1 fm <strong>for</strong> the test problem.ũ(z) = 1 (1+ z )2 a ez/a − 1 2 (ez/a −e −z/a )ln(1+z/a).(A.21)Fig A.1.1 shows the error between the analytic solution <strong>and</strong> the numerical calculation.This confirms that the second order expansion gives enough accuracy <strong>for</strong> integration technique.109


0.020.015N=40, SimpsonN=40, 1stN=40, 2nd0.020.015N=80, SimpsonN=80, 1stN=80, 2ndError0.01Error0.010.0050.0050-6 -4 -2 0 2 4 6z0-6 -4 -2 0 2 4 6z0.020.015N=40, SimpsonN=40, 1stN=40, 2nd0.020.015N=80, SimpsonN=80, 1stN=80, 2ndError0.01Error0.010.0050.0050-6 -4 -2 0 2 4 6z0-6 -4 -2 0 2 4 6Figure A.1: The error between analytic solution <strong>and</strong> numerical calculation. The red (green,blue) one represent the absolute error between analytic solution <strong>and</strong> trapezoidal (1st, 2ndTaylor expansion) numerical integration. N is the number <strong>of</strong> zone in the given domain (-6,6)The error decrease as N increases. The error from 2nd order expansion is always smallerthan any others.zA.1.2 3D radial symmetric nuclear matterIn the symmetric nuclei or nuclei with neutron gas matter, we need to calculate∫˜g(r 1 ) = d 3 r 2 f(r 12 /a)g(r 2 )= 12ar 1e −r 1/a∫ r1+ 12ar 1(e r 1/a −e −r 1/a+ 12ar 1(e r 1/a −e −r 1/a0)g(r)r(e r/a −e −r/a dr) ∫ Rg(r)re −r/a drr 1) ∫ ∞g(r)re −r/a dr.R(A.22)The last integration represents the effect from out side nucleons. We now address the coefficientfrom the second order Taylor expansion method.110


We separate the two integrals as in the 1D semi-infinite nuclear matter,˜g(r) = ˜g (−) (r)+ ˜g (+) (r)+ 1 2r(e r/a −e −r/a )e −R/a (a+R)g(R).(A.23)There is no difference in the finite range integration between finite nuclei <strong>and</strong> nuclei in densematter except the final tail term.The discrete integration <strong>for</strong> ˜g (−) (r i+1 ) has the <strong>for</strong>m <strong>of</strong>˜g (−) (r i+1 ) = 12ar i+1e −r i+1/awhere f (−)ii∑k=0∫ rk+1r kg(r)r(e r/a −e −r/a )dr= f (−)i ˜g (−) 1(r i )+2a(r i +∆) e−(r i+∆)/ais given byf (−)i∫ ri +∆r i= r ir i +∆ e−∆/a .The right h<strong>and</strong> side integration can be extended byg(r)r(e r/a −e −r/a )dr,(A.24)(A.25)g(r) = g i + g i+1 −g i−1(r−r i )+ g i+1 −2g i +g i−1(r −r2∆ 2∆ 2 i ) 2 . (A.26)We use the g i ′ = g i+1−g i−1, since this expansion has higher order (O(∆ 3 )) error than the2∆<strong>for</strong>ward (f i ′ = (f i+1 −f i )/∆) <strong>and</strong> backward (f i ′ = (f i −f i−1 )/∆) error (O(∆ 2 )).We can find coefficients which correspond to each g i−1 , g i , <strong>and</strong> g i+1 through the integrationso that˜g (−)i+1 = f(−) i˜g (−)i +w 1i g i−1 +w 2i g i +w 3i g i+1 −w 4i g i−1 −w 5i g i −w 6i g i+1(A.27)= f (−)i ˜g (−)i +(w 1i −w 4i )g i−1 +(w 2i −w 5i )g i +(w 3i −w 6i )g i+1111


wherew 1i =w 2i =w 3i =w 4i =w 5i =w 6i =∫1ri +∆[2a(r i +∆) e−(r i+∆)/a− (r −r i)r i2∆ + (r −r ]i) 2re r/a dr,2∆ 2∫1ri +∆[2a(r i +∆) e−(r i+∆)/a1− (r −r ]i) 2re r/a dr,r i∆ 2∫1ri +∆[ 2a(r i +∆) e−(r i+∆)/a (r −ri )r i2∆ + (r −r ]i) 2re r/a dr,2∆ 2∫1ri +∆[2a(r i +∆) e−(r i+∆)/a− (r −r i)r i2∆ + (r −r ]i) 2re −r/a dr,2∆ 2∫1ri +∆[2a(r i +∆) e−(r i+∆)/a1− (r −r ]i) 2re −r/a dr,r i∆ 2∫1ri +∆[ 2a(r i +∆) e−(r i+∆)/a (r −ri )r i2∆ + (r −r ]i) 2re −r/a dr.2∆ 2(A.28)112


Each w i can be obtained[ae −∆/aw 1i = 6a 2 +2a∆−2ar4∆ 2 i −∆r i(∆+r i )w 2i =w 3i =−e ∆/a (6a 2 +∆(∆+r i )−2a(2∆+r i )) ] ,[e −∆/a−6a 3 +a∆ 2 +2a 2 r2∆ 2 i −∆ 2 r i(∆+r i )+2ae ∆/a (3a 2 +∆(∆+r i )−a(3∆+r i )) ] ,[e −∆/a6a 3 +a∆r4∆ 2 i −2a 2 (∆+r i )(∆+r i )−e ∆/a (6a 3 −2∆ 2 (∆+r i )−2a 2 (4∆+r i )+a∆(5∆+3r i )) ] ,[w 4i = ae−2(r i+∆)/a−6a 2 −∆(∆+r4∆ 2 i )−2a(2∆+r i )(∆+r i ))+e(6a ]∆/a 2 −∆r i +2a(−∆+r i )w 5i = e−2(r i+∆)/a)[e(−6a ∆/a 3 +a∆ 2 −2a 2 r2∆ 2 i +∆ 2 r i(∆+r i )(+2a 3a 2 +∆(∆+r i )+a(3∆+r i )) ] ,w 6i = e−2(r i+∆)/a ( )[ae ∆/a 6a 2 +∆r4∆ 2 i +2a(∆+r i )(∆+r i )(− 6a 3 +2∆ 2 (∆+r i )+2a 2 (4∆+r i )+a∆(5∆+3r i )) ] .In the same manner, we have˜g (+)i = 12ar i(e r i/a −e −r i/a)N−1∑k=i∫ rk+1r kg(r)re −r/a dr= f (+)i ˜g i+1 + 1 ( ) ∫ r i +∆e ri/a −e −r i/a2ar ir ig(r)re −r/a dr(A.29)(A.30)= f (+)i ˜g i+1 +w 7i g i−1 +w 8i g i +w 9i g i+1 ,113


wherei = r i +∆ e ri/a −e −r i/ae (r i+∆)/a−e −(r i+∆)/af (+)r iw 7i = acosh(r i/a)e −(r i+∆)/a2∆ 2 r i(×[−6a 2 −∆(∆+r i )−2a(2∆+r i )+e ∆/a 6a 2 −∆r i +2a(−∆+r i )) ] ,w 8i = cosh(r i/a)e −(r i+∆)/a∆ 2 r i)×[e(−6a ∆/a 3 +a∆ 2 −2a 2 r i +∆ 2 r i(+2a 3a 2 +∆(∆+r i )+a(3∆+r i )) ] ,(A.31)w 9i = cosh(r i/a)e −(r i+∆)/a2∆ 2 r i( )×[ae ∆/a 6a 2 +∆r i +2a(∆+r i )−(6a 3 +2∆ 2 (∆+r i )+2a 2 (4∆+r i )+a∆(5∆+3r i )) ] .The initial points <strong>for</strong> g (−)0 <strong>and</strong> g (+)Nare given byg (−)0 = 0, g (+)N = 0.(A.32)Since g (−)1 <strong>and</strong> g (+)0 can’t have 3 points <strong>for</strong> numerical derivative, we use the first derivativesog (−)1 = 12a∆ e−∆ ∫ ∆0g(r)r ( e r/a −e −r/a) dr[g 0 + g ]1 −g 0∆ r= 1 ∫ ∆2a∆ e−∆ r ( e r/a −e −r/a) dr0]ae[2a−2acosh(∆/a)+∆sinh(∆/a)−∆/a=g∆ 2 0( )e[(2a ]−∆/a 2 +∆ 2 )cosh(∆/a)−2a a+∆sinh(∆/a)+g∆ 2 1 .˜g (+)0 = f (+)0 ˜g (+)1 + 12ar 0(e r 0/a −e −r 0/a) ∫ ∆g(r)re −r/a dr,0(A.33)(A.34)114


By exp<strong>and</strong>ing e r 0/a −e −r 0/a in terms <strong>of</strong> r 0 , we have<strong>and</strong> the integration becomesf (+)0 = 2∆ a1e ∆/a −e −∆/a ,(A.35)˜g (+)0 = f (+)0 ˜g (+)1 + 1 a 2 ∫ ∆0g(r)re −r/a dr= f (+)0 ˜g (+)1 + 1 ∫ ∆[ga 2 0 + g ]1 −g 0∆ r re −r/a dr0[]e −∆/a 2a+∆+(−2a+∆)e ∆/a= f (+)0 ˜g (+)1 +∆+ 2a2 −(2a 2 +2a∆+∆ 2 )e −∆/ag 1 .a∆g 0(A.36)Fig A.1.2 shows the result <strong>of</strong> 56 Fe binding energy calculation as a function <strong>of</strong> number <strong>of</strong>grids. The nuclear <strong>for</strong>ce used is FRTF II. The energy converges quickly if we use 2nd orderTaylor expansion. The right panel shows the log scale error <strong>and</strong> the error <strong>of</strong> 2nd orderTaylor expansion is less than 0.1% even in the 40 zones. These method is also used to find-8.6-8.756 FeSimpsonThe first order TaylorThe second order TaylorEnergy converged-1-1.556 FeSimpsonThe first order TaylorThe second order Taylor-8.8-256 Fe E/A (MeV)-8.9-9-9.1Log 10 (1-E cal /E conv )-2.5-3-3.5-4-9.2-4.5-9.3-5-9.440 60 80 100 120 140 160 180 200n (#<strong>of</strong> zone)-5.540 60 80 100 120 140 160 180 200n (#<strong>of</strong> zone)Figure A.2: 56 Fe binding energy calculation <strong>and</strong> relative error in log scale. The Simpsonmethod doesn’t give the enough accuracy even in N=200.the properties <strong>of</strong> heavy nuclei in the dense matter.115


Appendix BJEL thermodynamic integrationTo compute the properties <strong>of</strong> finite-temperature matter, it is necessary to calculate the Fermiintegrals F 1/2 <strong>and</strong> F 3/2 . Directly calculating these integrals by usual integral methods is notadvisable <strong>for</strong>applicationslike hydrodynamics, which require ahighdegree<strong>of</strong>thermodynamicconsistency. The approach we adopt is due to Johns, Ellis & Lattimer[12] (hereafter referredtoasJEL),whichisamodification<strong>of</strong>alessaccuratescheme originatedbyEggleton, Faulkner& Flannery[8]. These approaches involve polynomial interpolations <strong>for</strong> arbitrary degeneracy<strong>and</strong> relativity. However, in the finite-range model, nucleons are treated non-relativistically,so the interpolation method can be considerably simplified from the general case.The degeneracy parameter Ψ, <strong>and</strong> an associated variable f, are defined in terms <strong>of</strong> thechemical potential <strong>and</strong> temperature assuch thatΨ = µ−VT= 2√f ′ = dfdΨ =1+ f a +ln √1+f/a−1√1+f/a+1,f√1+f/a.(B.1)(B.2)In the above, a is a fitting parameter <strong>and</strong> is given in table B. The Fermi integrals areexpressed as polynomials in f with additional parameters p m , where m = 0...M, that arefixed by the requirements <strong>of</strong> yielding exact results <strong>for</strong> the pressure, energy <strong>and</strong> entropy inthe extremely degenerate <strong>and</strong> non-degenerate limits or by fitting intermediate results. Using116


M 3 2a 0.433 1p 0 (e 2 /a) √ π/32 = 5.34689 (e 2 /a) √ π/32 = 2.31520p 1 16.8441 a −1/4 [π 2 −8+56/(5a)]/3 = 44.35653p 2 a −1/4 [π 2 −8+88/(5a)]/3 = 17.4708 32a −5/4 /15 = 2.13333p 3 32a −5/4 /15 = 6.07364 —the results <strong>of</strong> JEL, one findsTable B.1: Non-relativistic fermion coefficientsF 3/2 (Ψ) = 3f(1+f)1/4−M2 3/2 M∑F 1/2 (Ψ) = f′ (1+f) 1/4−M√2F −1/2 (Ψ) =− f′f +a F 1/2(Ψ)m=0∑ Mm=0+ √ 2 f(1+f)1/4−M1+f/ap m f m ,p m f m [1+m−(M − 1 4 ) f1+fM∑p m f[(1+m) m 2 −m=0−(M − 1 4 ) f1+f (3+2m−(M + 3 4 ) f1+f ) ].],(B.3)The parameters are given in Table B <strong>for</strong> two cases, M = 2 <strong>and</strong> 3. With the 1 free parameter<strong>of</strong> the M = 2 scheme (ı.e., a), accuracy is better than 3%. With the 2 free parameters <strong>of</strong>the M = 3 scheme, i.e., a <strong>and</strong> p 1 , accuracy improves about 100-fold.For EOStable, we have a large range<strong>of</strong> temperature <strong>and</strong>density, so the electron can havefour different regimes such as, non-relativistic non-degenerate, non-relativistic degenerate,relativistic non-degenerate, <strong>and</strong>relativistic degenerate. We treat electrons asnon-interactingfermions except <strong>for</strong> the Coulomb interaction. In this case, we add another parameter g toaccount <strong>for</strong> relativistic effects. Then the pressure from electrons can be written, aswherep = m (e me) 3 fg 5/3 ∑pmn f m g n ,π 2 (1+f) M (1+g) N−3/2g = T m e√1+f ,(B.4)(B.5)<strong>and</strong> the electron’s degeneracy parameter is given eq. (B.1). The coefficient p mn is providedin Ref.[12].Among the thermodynamic quantities in EOS table, we may calculate ∂p/∂n, ∂p/∂T, <strong>and</strong>117


p mn n = 0 n = 1 n = 2 n = 3m = 0 5.34689 18.0517 21.3422 8.53240m = 1 16.8441 55.7051 63.6901 24.6213m = 2 17.4708 56.3902 62.1319 23.2602m = 3 6.07364 18.9992 20.0285 7.11153Table B.2: Fermion coefficients p mn <strong>for</strong> M = N = 3; a=0.433∂s/∂T. In JEL scheme, we treat f,g variables independent.wheredp = ∂p ∂pdf +∂f ∂g dgdn = ∂n ∂ndf +∂f ∂g dgdt = ∂t ∂tdf +∂f ∂g dg,t = Tm e c 2 , g = t(1+f)1/2 .The ∂p/∂n can be obtained when we make dt = 0, then(B.6)(B.7)∂p∂p∂n = ∂n+ ∂p ∂g∂f ∂g ∂f+ ∂n ∂g∂f ∂g ∂f, where∂t∂g∂f = − ∂f∂t∂g(B.8)In this manner, we can show that∂p∂p∂t =+ ∂p ∂g∂f ∂g ∂f∂t+ ∂t ∂g∂f ∂g ∂f, where∂n∂g∂f = − ∂f. (B.9)∂n∂gFor ∂S/∂T, we have∂S∂T T[ = g ] g ∂ 2 pT ∂g −ψ∂n . (B.10)2 ∂gWe can find the derivative <strong>of</strong> ∂p/∂f <strong>and</strong> ∂p/∂g from eq. (B.4)∂p∂f = mc 2 n c g 5/2(1+f) M+2 (1+g) N−3/2 ∑pmn f m g n[ 1+m+f(m−M) ]∂p∂g = mc 2 n c fg 3/2(1+f) M+1 (1+g) N−1/2 ∑pmn f m g n [n+ 5 2 +g(4+n−N) ],(B.11)(where n c = 1 mc) 3.π 2 Wealso need to findthederivatives <strong>of</strong>∂ 2 p/∂f 2 , ∂ 2 p/∂f∂g, <strong>and</strong>∂ 2 p/∂g 2 . Inthis case, instead118


<strong>of</strong> find exact polynomial expression, we might use recursion relation, that is,∂ 2 p M +2∂p=−∂f2 1+f ∂f + mc 2 n c g 5/2(1+f) M+2 (1+g) N−3/2×∑[ (1+m)mpmn f m g n +m(m−M)+m−Mf(∂p 2 1∂f∂g = f − M +1 ) ∂p1+f ∂g + mc 2 n c g 3/2(1+f) M+1 (1+g) N−1/2×∑pmn f m g n m[n+ 5 ]2 +g(4+n−N) ,],(B.12)(∂ 2 p 3∂g = 2 2g − N − 1 21+g) ∂p∂g + mc 2 n c fg 3/2(1+f) M+1 (1+g) N−1/2×∑[ n(n+5pmn f m g n ) 2+n(4+n−N)+(4+n−N)g].From thermodynamic quantities, we can getn = 1 ( ) ∂p= 1 (∂f ∂pT ∂ψTT ∂ψ ∂f + ∂g )∂p∂f ∂g= 1 ( 1 f √ )1+f ∂pmc 2 g 1+f/a ∂f + f2 √ 1+f √ ∂p1+f/a∂g( ) √ ∂p 1+f ∂pns = −nψ =∂T mc 2 ∂g −nψ.ψThe derivatives <strong>of</strong> the above items w.r.t f <strong>and</strong> g are given by( )∂n 1∂f = f − 1n+ 1 f2a(1+f/a) mc 2∂n∂g = 1mc 2∂(ns)∂f∂(ns)∂g( ∂p∂f +2(1+f)∂2 pf√1+f/a(−= 1 √1 ∂p 1+f√2mc 2 1+f ∂g + mc√ 21+f ∂ 2 p=mc 2 ∂g − ∂n2 ∂g ψ.2g √ 1+f √ 1+f/a ×∂f − 1 )g ∂p2 21+f∂g +g ∂2 p,∂f∂g√ √ 1+f ∂p 1+fg 2 ∂f + g)∂ 2 p∂f∂g + 12 √ ∂ 2 p,1+f ∂g 2√∂ 2 p 1+f/a∂f∂g −ψ∂n ∂f −n ,f(B.13)(B.14)Finally we can get thermodynamic derivatives using the above <strong>for</strong>mulae <strong>and</strong> constraints119


with differentials,( ) ∂p∂nT( ) ∂p∂Tn( ) ∂(ns)∂Tn( ∂p=∂f + g2(1+f)√ ( 1+f ∂p=−mc 2√ ( 1+f ∂(ns)=−mc 2 ∂f∂p∂g∂f − ∂p∂g)( ∂n∂f +∂n∂f( ∂n∂g− ∂(ns)∂g∂n∂f) −1∂n∂gg2(1+f)) )( −1 g2(1+f) + ∂n∂f( ∂n) )( −1∂gg2(1+f) + ∂n∂f( ∂n) ) −1 −1∂g( ∂n) ) −1 −1∂g(B.15)120


Appendix CPhase coexistenceBulk equilibrium <strong>of</strong> dense <strong>and</strong> dilute matter is the simplified case <strong>of</strong> heavy nuclei <strong>and</strong> nucleongas in neutron stars’ inner crust. This bulk equilibrium can be made by the free energyminimization process.For given density, we have total free energy densityF =uF I (ρ I )+(1−u)F II (ρ II ),ρ =uρ I +(1−u)ρ II(C.1)where u is the volume fraction <strong>of</strong> phase I <strong>and</strong> F I <strong>and</strong> F II are free energy density <strong>of</strong> eachphase. The free energy density should be minimized with respect to u, ρ I (ρ II ). Then wehave∂F=u ∂F I+(1−u) ∂F II ∂ρ II= u(µ I −µ II ) = 0,∂ρ I ∂ρ I ∂ρ II ∂ρ I∂F∂u =F I −F II +(1−u) ∂F I ∂ρ II∂ρ I ∂u=F I −F II +(1−u) ∂F II∂ρ IIρ II −ρ I(1−u) = 0(C.2)From eq (C.2), we have µ I = µ II <strong>and</strong> p I = p II . In case <strong>of</strong> neutron <strong>and</strong> proton matter, thiscan be generalized asµ nI = µ nII , µ pI = µ pI , p I = p II . (C.3)To find the mixed phase <strong>of</strong> quark <strong>and</strong> nuclear matter, we use a similar method in nuclearmatter cases.F = F N u+F Q (1−u)+ 3 4 c(3π2 ρ N x N ) 1/3 ρ N x N u+ 3 4 c(3π2 ρ Q x Q ) 1/3 ρ Q x Q (1−u), (C.4)where F is the free energy density, u is the volume fraction <strong>of</strong> the nuclear matter, ρ N is thenuclear matter density, ρ Q is the quark matter density, x N is electron fraction <strong>of</strong> nuclear121


matter, <strong>and</strong> x Q is electron fraction <strong>of</strong> quark matter. There are two constrains which areρ B = ρ N u+ρ Q (1−u)ρ B Y e = ρ N x N u+ρ Q x Q (1−u).(C.5)We set G(ρ N ,ρ Q ,x N ,x Q ,u)G(ρ N ,ρ Q ,x N ,x Q ,u) =F −α(ρ B −ρ N u+ρ Q (1−u))−β(ρ B Y e −ρ N x N u+ρ Q x Q (1−u)),<strong>and</strong> find the minimum <strong>of</strong> F using Lagrangian multiplier method.∂G/∂x N givesβ = −µ e,N(C.6)(C.7)From ∂G/∂x Q = 0, we haveµ e,N = µ e,Q . (C.8)With ρ N <strong>and</strong> ρ Q , we have∂G∂ρ N= u+c(3π 2 ρ N x N ) 1/3 x N u+αu+βx N u = 0,(C.9)thusFurthermore, we get∂F N∂ρ N= ∂F Q∂ρ Q= −α.(C.10)∂F N∂ρ N= ∂ρ n∂ρ N∂F N∂ρ n+ ∂ρ p∂ρ N∂F N∂ρ p= (1−x N )µ n +x N µ p(C.11)<strong>and</strong>∂F Q∂ρ Q= ∂ρ u∂ρ Q∂F Q∂ρ u+ ∂ρ d∂ρ Q∂F Q∂ρ d+ ∂ρ s∂ρ Q∂F Q∂ρ s= (1+x Q )µ u + 1 2 (2−x Q)µ d + 1 2 (2−x Q)µ s(C.12)= µ u +µ d +µ s +x Q (µ u − 1 2 µ d − 1 2 µ s)= µ u +2µ d +x Q (µ u −µ d )where we have used massless quark limits. There<strong>for</strong>e(1−x N )µ n +x N µ p = µ u +2µ d +x Q (µ u −µ d ).(C.13)The partial derivative <strong>of</strong> G w.r.t u simply gives pressure equations which areF N −ρ N∂F N∂ρ N= p N = F Q −ρ Q∂F Q∂ρ Q= p Q .(C.14)122


To summarize, we need to solve 5 equations, which arewith 5 unknowns,(1−x N )µ n +x N µ p = µ u +2µ d +x Q (µ u −µ d ),µ e,N = µ e,Q ,p N = p Q ,ρ B = ρ N u+ρ Q (1−u),ρ B Y e = ρ N x N u+ρ Q x Q (1−u),ρ N ,ρ Q ,x N ,x Q ,u.(C.15)(C.16)The above equations <strong>and</strong> unknown can be solved numerically using typical Newton-Raphsonmethod.123


Appendix DNumerical Techniques in HeavyNuclei in Dense <strong>Matter</strong>The heavy nuclei in dense matter exist in the narrow range <strong>of</strong> densities (2.0 × 10 −4 ∼8.0×10 −2 /fm 3 ). However, this is the most difficult part <strong>of</strong> calculations in FRTF model <strong>and</strong><strong>Nuclear</strong> E.O.S. table. We now explain the numerical techniques in FRTF model.The total energy in the cell can be obtained from∫E tot = E np +E e +E C d 3 r,(D.1)where E n p is a nuclear energy density <strong>and</strong>E c =m 4 [ex8π 2 (c) 3 e (2x 2 e +1)√ x 2 e +1−ln(x e + √ ]1+x 2 e )E C = 1 2 (ρ p −ρ e )∆µ p (r).(D.2a)(D.2b)Here,<strong>and</strong>∆µ p = 4πe 2 [ 1r∫ r0x e = cm e(3π 2 ρ e ) 1/3 ,r ′2[ ρ p (r ′ )−ρ e (r ′ ) ] dr ′ +∫ Rcr(D.3)r ′[ ρ p (r ′ )−ρ e (r ′ ) ] dr ′ ]. (D.4)If we assume that the inner cell or numerical boundary (R b ) <strong>for</strong> computational purpose issmaller than the actual Wigner-Seitz cell <strong>and</strong> the nuclear <strong>and</strong> electron densities are constantoutside the inner cell, then we have (r < R b < R c )∫ r[ 1∆µ p (r) =4πe 2 r ′2[ ρ p (r ′ )−ρ e (r ′ ) ] dr ′ +r 0−2πe 2 ρ e,o (Rc 2 −Rb).2∫ Rbrr ′[ ρ p (r ′ )−ρ e (r ′ ) ] dr ′ ](D.5)124


D.1 Non-uni<strong>for</strong>m electron density approximationTotal energy contribution from the outer cell † is given byE tot,o = 4π 3 (R3 c −Rb)(E 3 np,o +E e,o )− 1 ∫ρ e,o ∆µ p (r)d 3 r,2 R b ,R cFor R b < r < R c ,∆µ p (r) = e2r (z −z′ e )− 2π 3 e2 ρ e,o(3R 2 c −2R3 br −r2 )where z e ′ is the total number <strong>of</strong> electrons in the inner cell. Then we can have(D.6)(D.7)E C,o = −πe 2 (z −z e ′ )ρ e,o(Rc 2 8π2−R2 b )+15 ρ2 e,o (2R5 c −5R2 c R3 b +3R5 b ) (D.8)For energy minimization,∫ ( ∫F = E tot −λ 1 (ρ n +ρ p )d 3 r − 4π ) ( ∫ ∫3 R3 cρ −λ 2 ρ p d 3 r −)ρ e d 3 rI chose unknowns with ρ n , ρ p , ρ e , <strong>and</strong> λ 1,2 , then we have equations to solve(D.9)0 = F 0,N : µ n (ρ n ,ρ p )−µ n,0 = 00 = F N+1,2N+1 : µ p (ρ n ,ρ p )+∆µ p (ρ p ,ρ e )−µ n,0 +µ e,0 = 00 = F 2N+2,3N+2 : µ e (ρ p ,ρ e )−∆µ p (ρ p ,ρ e )−µ e,0 = 0∫0 = F 3N+3 : (ρ n +ρ p )d 3 r − 4 3 πR3 c ρ = 0∫ ∫0 = F 3N+4 : ρ p d 3 r − ρ e d 3 r = 0(D.10)D.2 Uni<strong>for</strong>m electron density approximationIn case <strong>of</strong> constant electron density approximation,∆µ p (r) =4πe 2 [ 1r∫ r0r ′2 ρ p (r ′ )dr ′ +− 2 3 πe2 ρ e (3R 2 c −r2 ).∫ Rbrr ′ ρ p (r ′ )dr ′ ](D.11)† We divide the Wigner-Seitz cell by inner cell <strong>and</strong> outer cell. In the inner cell, the nuclear density pr<strong>of</strong>ilevaries. On the other h<strong>and</strong>, the nuclear density pr<strong>of</strong>ile is uni<strong>for</strong>m in the outer cell. The radius <strong>for</strong> inner cellis set to 15 fm <strong>for</strong> the numerical calculations.125


For R b < r < R c , we have∆µ p (r) = ze2r − 2 3 πe2 ρ e (3R 2 c −r2 ).The energy contribution from outside the inner cell is given by(D.12)E C,o = −πzρ e e 2 (Rc 2 4π2−R2 b )+15 ρ2 e (4R5 c −5R2 c R3 b +R5 b ). (D.13)For energy minimization, we have∫ ( ∫F = E tot −λ 1 (ρ n +ρ p )d 3 r − 4π (3 R3 cρ)−λ ∫ 2 ρ p d 3 r − 4π )3 R3 cρY eI chose unknowns with ρ n , ρ p , Y e , µ n,0 , <strong>and</strong> µ p,0 then we have equations to solve(D.14)0 = F 0,N : µ n (ρ n ,ρ p )−µ n,0 = 00 = F N+1,2N+1 : µ p (ρ n ,ρ p )−µ p,0 = 0∫0 = F 2N+2 : (ρ n +ρ p )d 3 r − 4 3 πR3 c ρ = 0∫0 = F 2N+3 : ρ p d 3 r − 4 3 πR3 cρY e = 0(D.15)0 = F 2N+4 : µ n,0 −µ p,0 −µ e (Y e ) = 0In the lower density ( ρ < 0.01/fm 3 ) region, two methods give almost same result. Asdensity increases, the uni<strong>for</strong>m electron density approximation is unstable numerically inFRTF model.126


Appendix E<strong>Nuclear</strong> Quantities in Non-relativisticModelsWe present nuclear properties <strong>of</strong> non-relativistic potential (Skyrme <strong>for</strong>ce) models. Theseproperties can be used to restrict nuclear parameters when we make new <strong>for</strong>ce models.E.1 Mathematical relation between nuclearparameters <strong>and</strong> quantitiesOnce we are given nuclear parameters (x 0 ,...,x 3 , t 0 ,...,t 3 ) in Skyrme <strong>for</strong>ce model, we canextract nuclear quantities (B, K, M ∗ , S v , L, ρ o ) from its mathematical expression <strong>of</strong> bulkmatter Hamiltonian density[32].In the st<strong>and</strong>ard Skyrme <strong>for</strong>ce models,wherep(ρ o )ρ oC = 3210M= 0 = 2 3 Cρ2/3 o(1+ 5 )2 βρ o+ 3 8 t 0ρ o + t 316 (ǫ+1)ρǫ+1 o , (E.1)( ) 3π2 2/3; β = M [ ] 12 2 2 4 (3t 1 +5t 2 )+t 2 x 2 . (E.2)From the pressure expression, we can get ρ o using the Newton-Raphson method, with whichwe can get the binding energy per baryon, effective mass, <strong>and</strong> nuclear incompressibility inst<strong>and</strong>ard nuclear matter,B = − E A = −Cρ2/3 o (1+βρ o )− 3t 08 ρ o − t 316 ρ1+ǫ o ,M ∗ = M1+βρ o,K = 9ρ 2∂2 E/A∂ρ 2 ∣∣∣ρ=ρo= −2Cρ 2/3o +10Cρ 5/3o + 9t 316 ǫ(ǫ+1)ρǫ+1 o .(E.3)127


Other important quantities, S v <strong>and</strong> L can be obtainedS v = ρ2 o ∂ 2 E/A= 5 8 ∂x 2 9 Cρ2 o + 10CM [ (t21+ 5 3 2 6 4 x 2)− 1 ]8 t 1x 1− t (3 13)24 2 +x ρ 1+ǫo − t (o 1o)4 2 +x ρ o ,L = 3ρ o∂ 3 E/A∣=8 ∂ρ∂x∣∣ρ=ρo,x=1/2102 27 Cρ2 o + 50CM [ (t21+ 5 9 2 6 4 x 2− t ( )3 124 (1+ǫ) 2 +x 3 ρ 1+ǫo − t (o 1o)4 2 +x ρ o ,ρ 5/3o)− 1 8 t 1x 1]ρ 5/3o(E.4)In this way, we can obtain nuclear quantities in st<strong>and</strong>ard Skyrme <strong>for</strong>ce model.As an inverse process to find x 0 ,...,t 3 in Skyrme <strong>for</strong>ce model, we can use E/A, p(ρ o ,x =1/2) = 0, M∗, K, S v , L, <strong>and</strong> ρ o . Since we only have 7 quantities <strong>for</strong> 8 parameters, we mayuse the pure neutron matter effective mass M ∗ n or the energy difference Q 1 = e N −e S −S vas described in section (3.2).The maximum mass <strong>of</strong> the cold neutron star can be obtained using T.O.V equation <strong>and</strong>varying the central density <strong>of</strong>neutron stars. To get a moreprecise maximum mass <strong>of</strong>neutronstars, we may use numerical Newton Rapshon with numerical derivative <strong>of</strong>f = M 2 −M 1ρ c2 −ρ c1(E.5)<strong>and</strong> find the solution which makes f = 0.128


Table E.1: Non-relativistic Skyrme <strong>for</strong>ce model <strong>and</strong> maximum mass <strong>of</strong> cold neutron starModel B (MeV) S v (MeV) L (MeV) K (MeV) M ∗ /M ρ 0 (fm −3 ) M/M ⊙Gs -15.6023 31.3749 94.0305 237.3811 0.7838 0.1576 2.192M ∗ (0.811) -15.6992 30.7657 79.3938 237.1923 0.8110 0.1609 2.114M ∗ (0.9) -15.5659 27.7811 69.9636 240.2086 0.8957 0.1610 2.114M ∗ (1.0) -15.4172 24.8108 59.5481 240.2216 1.0000 0.1611 2.087Rs -15.6019 30.5851 85.7058 237.5088 0.7826 0.1578 2.177SGI -15.9073 28.3271 63.8631 261.9091 0.6078 0.1545 2.287SGII -15.6085 26.8270 37.6101 214.7885 0.7855 0.1584 1.663SIV -15.9733 31.2164 63.4859 324.7265 0.4707 0.1510 2.391Ska -16.0055 32.9103 74.6226 263.3100 0.6082 0.1554 2.244SkI1 -15.9660 37.5327 161.0952 242.8726 0.6929 0.1604 2.242SkI2 -15.7895 33.3788 104.3625 241.0820 0.6847 0.1576 2.255SkI3 -15.9944 34.8366 100.5490 258.3483 0.5771 0.1578 2.339SkI4 -15.9573 29.5010 60.3980 248.0753 0.6492 0.1602 2.236SkI5 -15.8603 36.6487 129.3675 255.9506 0.5785 0.1558 2.349SkI6 -15.9310 30.0894 59.7002 248.7457 0.6396 0.1591 2.245SkkT8 -15.9571 29.9178 33.6999 235.8483 0.8326 0.1607 1.691SkM -15.7851 30.7446 49.3262 216.7496 0.7884 0.1604 1.680SkM ′ -15.5749 29.8860 70.3127 231.0207 0.6533 0.1571 2.148SkM ∗ -15.7851 30.0312 45.7590 216.7496 0.7884 0.1604 1.618SkT1 -15.9935 32.0171 56.1694 236.3034 1.0000 0.1611 1.849SkT1s -15.9921 32.0154 56.0964 236.1374 1.0000 0.1603 1.8517SkT2 -15.9581 31.9977 56.1523 235.8796 1.0000 0.1611 1.848SkT3 -15.9592 31.4965 55.3027 235.8903 1.0000 0.1611 1.854SkT3s -15.9924 31.6811 55.8395 236.1412 1.0000 0.1603 1.860SkT4 -15.9694 35.4579 94.1479 235.6446 1.0000 0.1591 2.128SkT5 -16.0120 37.0107 98.5508 201.8260 1.0000 0.1641 1.920SkT7 -15.9519 29.5138 31.0932 235.7859 0.8325 0.1607 1.431SkT8 -15.9571 29.9178 33.6999 235.8483 0.8326 0.1607 1.698SkT9 -15.8968 29.7525 33.7134 235.0521 0.8329 0.1604 1.704SLy0 -15.9860 31.9800 47.1008 229.8049 0.6977 0.1604 2.067SLy1 -15.9997 31.9905 47.0548 229.9593 0.6976 0.1604 2.067SLy2 -16.0003 32.0018 47.4436 230.0627 0.6973 0.1606 2.068SLy3 -15.9850 31.9894 45.2930 230.0423 0.6961 0.1605 2.057SLy4 -15.9868 31.9986 45.9305 230.0602 0.6944 0.1596 2.070SLy5 -15.9991 32.0080 48.1339 230.0656 0.6973 0.1606 2.070SLy6 -15.9342 31.9555 47.4365 230.0047 0.6897 0.1590 2.090SLy7 -15.9145 31.9861 46.9306 229.8984 0.6878 0.1584 2.090SLy8 -15.9828 31.9934 47.1634 230.0383 0.6958 0.1604 2.070SLy9 -15.8081 31.9814 54.8507 229.9855 0.6655 0.1512 2.200SLy10 -15.9159 31.9775 38.7279 229.8270 0.6832 0.1556 2.056SV -16.0615 32.8243 96.1003 305.8488 0.3827 0.1551 2.510129


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