Equation of State of Dense Matter

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1 Department of Physics & Astronomy 449 ESS Bldg. Stony Brook University April 25, 2017 Nuclear Astrophysics

2 Zero Temperature Nuclear Matter

3 Expansions Cold, bulk, symmetric nuclear matter saturates at the nuclear saturation density n s 0.16 fm 3, equivalent to ρ s = m nn s g cm 3. At n s, the energy per particle E B is minimized, so the pressure is zero. The binding energy at saturation is B = E B (n = n s, Y p = 0.5, T = 0) 16.0 MeV. One may expand E in each of the three dimensions (n, Y p, T ): «2 E B (n, 0.5, 0) B + Ks n 1 + K «3 s n n s 162 n s E B (n s, Y p, 0) B + S v (1 2Y p) 2 + S 40 (1 2Y p) 4 + E B (n s, 0.5, T ) B + at 2 + The expansion parameters are K s 240 ± 20 MeV: Incompressibility parameter K s 200 ± 200 MeV: Skewness parameter S v 31 ± 3 MeV: Bulk symmetry energy parameter S 40 0 MeV: Bulk fourth order symmetry energy parameter a (m nπ 2 /2 2 )(1.5π 2 n s) 2/ ± MeV 1 : Level density parameter

4 The Nuclear Symmetry Energy Defined as the difference between the energies of cold pure neutron matter (Y p = x = 0) and symmetric (x = 1/2) nuclear matter. S(n) = E B (n, x = 0, T = 0) E B (n, x = 1/2, T = 0) Near n s and x = 1/2, E B (n, x) = E B (n, 1/2)+(1 2x) 2 S 2 (n)+ S 2 (n) = S v + L n n s + K ( ) 2 sym n ns + 3 n s 18 n s S v 31 ± 3 MeV, L 60 ± 40 MeV, K sym 150 ± 100 MeV. C. Fuchs, H.H. Wolter, EPJA 30(2006) 5 symmetry energy Connections to neutron matter: E B (n s, 0) S v + E B (n s, 1/2) = S v B, p(n s, 0) = Ln s /3 Neutron star matter (in beta equilibrium): [ (E B + E e ) = 0, p(n s, x β ) Ln s x 3 1 ( 4Sv c ) ] 3 4 3S v /L 3π 2 n s

5 The Nuclear Symmetry Energy Because of uncertainties in the nuclear symmetry energy parameters, the nuclear symmetry energy is a major uncertainty in the modeling of dense matter. Constraints on the symmetry energy come from nuclear observables, such as binding energies, giant dipole resonances, neutron skin thicknesses, isobaric analog states, etc. In addition, due to the approximate disappearance of higher-than-quadratic terms in the symmetry energy expansion in powers of 1 2x, the properties of pure neutron matter can inform about the symmetry parameters. Finally, due to the correlation between the neutron star matter pressure near n s and neutron star radii, astrophysical observations can also be used to estimate symmetry energy parameters.

6 The Nuclear Mass Formula or Liquid Drop Model Bethe-von Weizsäcker formula (neglecting pairing and shell effects) E nuc (A, Z) = a v A + a s A 2/3 + a C Z 2 /A 1/3 + S v AI 2. Myers & Swiatecki introduced the surface asymmetry term: E nuc (A, Z) = a v A + a s A 2/3 + a C Z 2 /A 1/3 + I 2 ( S v A S s A 2/3). a v = B, a s 18 MeV, a C 0.75 MeV, S s 45 MeV, I = (N Z)/A Determining the Optimum Nucleus: E nuc /A A E nuc /A I = a s 3 A 4/3 + a C 6 A 1/3 (1 I ) 2 S s 3 I 2 A 4/3 = 0 = a C 2 A2/3 (1 I ) + 2S v I 2S s A 1/3 I A = 2(a s + S s I 2 ) a C (1 I ) 2 48 (1 + 2I + ) 60 a C I = S v A 2/3 4S s /A + a C

7 Isolated Nuclei The binding energy curve is heavily skewed. Certain closed-shell nuclei (He, C, O, Pb) have much larger binding than the average. The optimum value of I increases with mass number A. This trend represents the Valley of Beta Stability.

8 Nuclei at Higher Densities At the end of stellar evolution, when an iron core forms, the central stellar density is about ρ 10 7 g cm 3, implying a filling factor u = ρ/ρ s The intranuclear spacing is about 2u 1/3 600 nuclear radii. Electron screening reduces the nuclear Coulomb energy. Approximating electrons as uniformly distributed, even within nuclei, the nuclear Coulomb energy is: E Coul = 3 Z 2 e 2 ( R 2 u1/3 + u ) 2 The reduction factor is about 0.5% for ρ 10 7 g cm 3. This effect increases the nuclear mass, which is proportional to E 1 Coul, as the average density increases. The optimum I also increases with density due to beta equilibrium: (E nuc /A + E e /A) x = µ n + µ p + µ e = 0.

9 Chemical Potentials and Neutron Drip Chemical potentials are equivalent to nucleon separation energies: ( ) ( ) (Enuc /A) (Enuc /A) µ n =, µ p =, N Z Z µ e = (E e/a) = c(3π 2 n s ux) 1/3, Y e = x, Y ( e ) (Enuc /A) µ n µ p = x At sufficiently high density, about ρ = (3.5 4) g cm 3, as x becomes smaller and A becomes larger, µ n becomes positive. Neutrons thus drip out of nuclei., A N

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11 Nuclear Droplet Model Myers & Swiatecki droplet extension: consider the variation of the neutron/proton asymmetry within the nuclear surface. E nuc = ( B + S v δ 2 )(A N s ) + (a s S s δ 2 )A 2/3 + a C Z 2 A 1/3 + µ n N s. N s is the number of excess neutrons associated with the surface, δ = 1 2x = (A N s 2Z)/(A N s ) is the asymmetry of the nuclear bulk fluid, and µ n = B + S v δ(2 δ) is the neutron chemical potential. Surface tension is the surface thermodynamic potential; adding µ n N s gives the total surface energy. Optimizing E nuc with respect to N s yields N s = S s S v δ 1 δ = A I δ ( 1 δ, δ = I 1 + S ) 1 s, S v A 1/3 E nuc (A, Z) = BA + a s A 2/3 + a C Z 2 /A 1/3 + S v AI 2 ( 1 + S s S v A 1/3 ) 1.

12 Schematic Bulk Free Energy Density F B (n, x, T ); n: number density; x: proton fraction; T : temperature n s 0.16 ± 0.01 fm 3 : nuclear saturation density B 16 ± 1 MeV: saturation binding energy K s 200 ± 200 MeV: K s 240 ± 20 MeV: incompressibility skewness S v 31 ± 3 MeV: bulk symmetry energy K sym 150 ± 100 MeV: L 60 ± 40 MeV: symmetry stiffness symmetry incompressibility a ± MeV 1 : bulk level density parameter [ F B = n B + K ( s 1 n ) ] 2 n ( + S v (1 2x) 2 ns ) 2/3 a T 2 18 n s n s n P B = n 2 (F [ ( ) ] B/n) = n2 Ks n 1 + S v (1 2x) 2 + 2an ( ns ) 2/3T 2 n n s 9 n s 3 n µ n,b = F B n x F B n x = B + K ) ) s (1 (1 nns 3 nns n + 2S v (1 2x) a ( ns ) 2/3T 2 18 n s 3 n ˆµ B = 1 F B n x = µ n n µ p = 4S v (1 2x) n s s B = 1 F ( B n T = 2a ns ) 2/3 T ; εb = F B + nts B n

13 Phase Coexistence Negative pressure region: matter is unstable to separating into two phases of different densities (and possibly proton fractions). Physically, this represents coexistence of nuclei and vapor. Neglecting finite-size effects, bulk coexistence approximates the EOS at subnuclear densities. Free Energy Minimization With Two Phases F = ε nts = uf B,I + (1 u)f B,II, n = un I + (1 u)n II, ny e = ux I n I + (1 u)x II n II. n II = n un I 1 u, x II = ny e un I x I n un I F = u F B,I + (1 u) F ( ) B,II u = u(µ n,b,i µ n,b,ii ) = 0 n I n I n II 1 u F = u F B,I + (1 u) F ( ) B,II uni = un I (ˆµ B,I ˆµ B,II ) = 0 x I x I x [ II n ( un I ) F u = F FB,II nii n I B,I F B,II + (1 u) + F ( )] B,II uni = 0 n II 1 u x II n un I = µ n,b,i = µ n,b,ii, µ p,b,i = µ p,b,ii, P B,I = P B,II

14 Critical Point (Y e = 0.5) ( ) PB = n T ( 2 ) P B n 2 = 0 T n c = 5 12 n s ( ) 1/3 ( 5 5Ks T c = 12 32a ( ) 1/3 ( 12 5Ks a s c = 5 8 ) 1/2 ) 1/2

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16 Macroscopic Hydrodynamic Nuclear Model H = H B (n, α)+h C (n, α)+q(n) ( ) 2 dn, H B (n, α) = H B (n, 0)+v sym(n)α 2 dr n = n n + n p, α = n n n p, v sym = S 2 (n)/n, H C = n p V C /2 r V C (r) = e2 n p (r )d 3 r e 2 ( + r r n p(r )d 3 r Ze2 3 R 2 r 2 ) 2R 2 0 Optimize total energy with respect to n and α subject to fixed A = ρd 3 r and N Z = αd 3 r: µ = [H B + H C ] n δ δ [H µn] = 0, δn = 2 d dr r [ Q dn ] + Q dr n [H µα] = 0. δα ( ) 2 dn, µ = [H B + H C ] dr α Multiply first by dn/dr and second by dα/dr, add, then integrate: ( ) 2 dn Q(n) = H B + H C µn µα dr

17 Surface Energy Consider a symmetric, chargeless nucleus. The total energy, the sum of volume and surface energies, is Hd 3 r = µa + [H µn] d 3 r µa + 4πR 2 [H µn] dx We made a leptodermous expansion above to isolate the surface. [ ( ) ] 2 dn E vol,o + E surf,o = µa + 4πR 2 H B (n, 0) µn + Q(n) dx dx 0 Assume H B (n, 0) = nµ + (nk s /18)(1 u) 2, u = n/n s, Q(n) = Q/n, µ = B: du dz = u(1 u), z = x 18Q a, a = 1, u = K s 1 + e z y [ A = 4πn s a 3 z i ( ) ] i+1 2 dz y i(i + 1) π F 2 (y), F i (y) = ez y i y This fixes the parameter y R/a. The symmetric matter surface tension is σ o = E ( ) 2 surf,o dn 4πR 2 = 2Q dz 0 dz n = 2Q dn dn dz n = Qn s a n s

18 Parameter Determination 0.9 t = a 0.1 dz du = 4a ln 3 = 2.3 fm, du a fm Q = K s 18 σ o = ( t ln 3 ) MeV fm 2, K s = 240 MeV [H µn]dx = 2Q ( ) 2 dn dx 0 dx n = 2Q dn dn n s dr n = Qn s 1.17 MeV fm 2 a ( ) 1/3 3 a s = 4πro 2 σ o 19.2 MeV, r o = fm 4πn s

19 The Isovector Density α H i = [H µα] α = [H B(n, α) + H C (n, α)] α µ = 0 α = µ + H C / α = µ + V C /2 2v sym 2v sym µ + VC /2 N Z = IA = d 3 r µ 2v sym 2 H 0 + G 4 ( r ) i d 3 r, G = V C d 3 r = Ze2 R v sym v sym 2R (3H 0 H 2 ) α = N Z (G V C H 0 )/4 v sym H 0 α o = n ) s (N Z + Ze2 S v H 0 8R H 2, v sym (n s ) = S v n s The polarization results in an increase in asymmetry near the center, α o.

20 Nuclear Structure The total symmetry and Coulomb energy: E sym + E C = v sym α 2 d 3 r + 1 (n α)v C d 3 r 4 (N Z)2 (N Z)G = + 1 nv C d 3 r H 0 4H 0 4 (N Z)2 = + 3 Z 2 e 2 ( H 0 5 R + Ze2 Hs (N Z) 3 ) 8R H 0 5 Dipole polarizability α D : ( δ δα Hd 3 r ɛ α d is the function α(r) which solves this. ) zαd 3 r = 0 α D = 1 2ɛ α d = ɛz + V C zα d d 3 r = 1 4ɛ 2v sym z ɛz + V C d 3 r = R2 H 2 v sym 12

21 Neutron Skin Thickness 4π 3 ( R 3 n R 3 p) = ( nn n no n p = n sa 2n no n po n po ( I α o ) d 3 r = n s ) n s 2n no n po ( α n α ) o d 3 r n s R n R p 2 I α o /n s R 3 1 αo/n 2 s 2 = 2 [ ( I 1 A ) ] Ze2 H 2 3 S v H 0 8RS v H 0 Polarization decreases neutron skin thickness. Symmetric nuclei have a proton skin. rn,p 2 = 1 n n,p r 2 d 3 1 r = (n ± α)r 2 d 3 r (N, Z) 2(N, Z) ])] r np R r n r p R = R2 1 ± I = 3 5 [ 3 5 ± [ I ( I H 2 Ze2 H 0 8RA [ H 4 H2 2 ( ) 5H2 1 5Ze2 3H 0 24RA H 0 ( H 4 H2 2 H 0 )] 1 1 I 2

22 Solutions for Arbitrary v sym Assume v sym can be expanded as S v n s v sym (u) = J b j u j, 0 < u < 1 j=1 If the density to lowest order retains the Fermi shape u = (1 + e (r R)/a ) 1 obtained in the absence of symmetry and Coulomb effects, we find ( r ) i d 3 r H i = = 4πn sa 3+i R v sym S v R i [F i+2 (y) (2 + i)t F i+1 (y) + ] where T = b b b b b 6 +. J j=1 b j = 1 ensures that v sym (u = 1) = S v /n s. Expanding the Fermi integrals to lowest order gives H i 3 [ ] A (i + 3)T a 1 +. i + 3 r o A 1/3 S v

23 Macroscopic Quantities Compare the total symmetry energy in the droplet model with that in the hydrodynamic model: ( E sym = S v AI S ) 1 s = A2 I 2 ( = S S v A 1/3 v AI 2 1 3T a ) 1. H 0 r o A 1/3 It is thus apparent that S s /S v 3T a/r o. It now follows that r np H i i A S v ( 1 + (i + 3)S ) s 3S v A 1/3 ( α D = R2 H 2 AR ) S s 12 20S v 3 S v A 1/3 ( 3 2r o 5(1 I S ) 1 [ s I S s 3Ze2 ) 3 S v A 1/3 S v 140r o S v ( S s S v A 1/3 )].

24 Connections Among the Symmetry Parameters The symmetry parameters S v, L and K sym are defined by the expansion S 2(u) S v + L Ksym (u 1) (u 1)2 +. Fitting to nuclear binding energies has established a significant correlation exists between K sym and L: K sym 3L K s Comparing Taylor expansions of us v /S 2(u) and P 4 j=1 b iu i around u = 1: " b 2 = L «# 2 2L Ks L 3 Ks S v 3S v 3S v 3S v 9S v " b 3 = L 5L + Ks 3S v 6S v " b 4 = L L Ks 3S v 3S v " T = L L 3S v 36S v L + 3 3S v 3S v L 9S v 3S v Ks 27S v 1 3 «# «# 2 L 3S v Ks 18S v «# 2 3 Ks S v We used L/S v = 2 and K s/s v = 8. To leading order, one finds S s/s v 1.35

25 A Nucleus in Dense Matter Bulk equilibrium is the 0th order approximation. Now include finite-size effects: Coulomb, surface and translation. Coulomb energy of uniformly dense sphere: f C = 3 Z 2 e 2 5 R. Coulomb energy of uniformly dense sphere within a larger neutral sphere (Wigner-Seitz cell): [ 3 Z 2 e R + 1 ( ) ] 3 R. 5 R 2 R c 2 Diffuseness correction to uniform sphere (d 0.55 fm): [ π2 Z 2 e 2 dd 2 ( ) ] 3 R 2R 3 1 Exchange correction (effects of antisymmetrized wave functions): ( ) 3e2 3Z 2 2/3 ] [1 RRc 4R 2π R c R c

26 Translational Free Energy Nuclei treated as a non-degenerate gas: [ ( ) ] n f T = T ln 1 n Q VA 3/2 n Q = ( ) 3/2 mb T 2π 2 = µ T T A simplification of the translational free energy density to eliminate dependence on R and incorporate pasta effects: [ ( ) ] F T = f T V = 3uf T 4πR 3 = u(1 u)tn I u(1 u)n I ln 1 A o n Q A 5/2 o u = 4πR3 3V, A o 60

27 Surface Energy According to the droplet model we form the sum of the surface tension and surface neutron terms: f s = 4πR 2 (σ + µ s ν s ) σ is the surface tension, to be considered a function of the surface neutron chemical potential, µ s and T. Both σ and µ s can be expressed as functions of the proton fraction of phase I, x I, and T, and can be found from semi-infinite surface calculations, e.g., σ(t = 0) = σ 0 σ δ (1 2x I ) 2 ν s = N s /(4πR 2 ) is the areal density of neutrons connected with the surface F s = f s V = 3u R (σ + ν sµ s )

28 Inclusion of Pasta Phases F s = 3u R (σ + ν sµ s ) β S 3u R F C0 = β C R 2 u β C R 2 D 3 (u) [ u1/3 + u ] 2 With arbitrary dimensionality d: du F s = β S R β Ss(u) [ F C0 = β C R 2 5u d 2 1 d 4 2 u1 2/d + d 2 ] u 2 β C R 2 c(u) d = 2 : c(u) = (5u/8)(u 1 ln u)

29 Model Free Energy Density F = uf I (n I, x I, T ) + (1 u)f II (n II, x II, T )+ + F s (ν s, µ s, R, u) + F C (n I, x I, R, u) + F T (n I, u, T ) F s = s(u) R (σ(µ s, T ) + µ s ν s ) F C = 4π 5 c(u)(n I x I er) 2 [ 2π 3 u(n I x I e) 2 (πd d ) 2 (1 u) + 9 8π [ ( ) ] F T = un I A o ln n I u n Q A 5/2 o 1 n = un I + (1 u)n II + s(u) R ν s ny e = un I x I + (1 u)n II x II ( ) 1/3 3 ( 1 u 1/3)] π

30 Equilibrium Conditions n II = n un I s(u)ν s /R 1 u ny e un I x I x II = n un I s(u)ν s /R F / n I = 0 µ ni = µ nii + (1 u)µ T /A o F / x I = 0 ˆµ I = ˆµ II 2βD/(3n I x I u) F / (ν s /R) = 0 µ s = µ nii F / µ s = 0 ν s = σ/ µ s F / R = 0 R = (9σ/2β)[s(u)/c(u)] 1/3 F / u = 0 P I = P II + β[d 2D/(3u)] un I µ T /A o ( ) 1/3 243π β = (n I x I σe) 2/3, 5 D(u) = ( s 2 (u)c(u) ) 1/3 D(u) u(1 u) (F s + F C0 ) d = β D d = 0 (1 u)d1/3 3 (u) + ud 1/3 3 (1 u) u 2 + (1 u) u 2 (1 u) 2

31 Nuclei and α Abundances in Neutron Stars

32 Electron Fractions in Neutron Stars

33 Nuclear Details in Neutron Stars

34 Nuclei and α Abundances in Supernova Matter

35 Nuclear Details in Supernova Matter

36 Nuclei in Dense Matter