Nuclear & Particle Physics of Compact Stars

Transcription

1 Nuclear & Particle Physics of Compact Stars Madappa Prakash Ohio University, Athens, OH National Nuclear Physics Summer School July 24-28, 2006, Bloomington, Indiana 1/30

2 How Neutron Stars are Formed Lattimer & Prakash, Science 304, 536 (2004). 2/30

3 Equations of Stellar Structure-I In hydrostatic equilibrium, the structure of a spherically symmetric neutron star from the Tolman-Oppenheimer-Volkov (TOV) equations: dm(r) dr dp (r) dr = 4πr 2 ɛ(r) = GM(r)ɛ(r) c 2 r 2 [ 1 + P (r) ɛ(r) ] [ 1 + 4πr3 P (r) M(r)c ] 2 [ 1 2GM(r) c 2 r ] G := Gravitational constant P := Pressure ɛ := Energy density M(r) := Enclosed gravitational mass R s = 2GM/c 2 := Schwarzschild radius 3/30

4 Equations of Stellar Structure-II The gravitational and baryon masses of the star: M G c 2 = R M A c 2 = m A R m A := Baryonic mass n(r) := Baryon number density 0 dr 4πr 2 ɛ(r) 0 dr 4πr 2 n(r) [ 1 2GM(r) c 2 r The binding energy of the star B.E. = (M A M G )c 2. ] 1/2 To determine star structure : Specify equation of state, P = P (ɛ) Choose a central pressure P c = P (ɛ c ) at r = 0 Integrate the 2 DE s out to surface r = R, where P (r = R) = 0. 4/30

5 Nucleonic Equation of State Energy (E) & Pressure (P ) vs. scaled density (u = n/n 0 ). Nuclear matter equilibrium density n 0 = 0.16 fm 3. Proton fraction x = n p /(n p + n n ). Nuclear matter : x = 1/2. Neutron matter : x = 0. Stellar matter in β equilibrium : x = x. 5/30

6 Results of Star Structure Stellar properties for soft & stiff (by comparison) EOS s. Causal limit : P = ɛ. M g : Gravitational mass R : Radius BE : Binding energy n b : Central density I : Moment of inertia φ : Surface red shift, e φ/c2 = (1 2GM/c 2 R) 1/2. 6/30

7 Constraints on the EOS-I R > R s = 2GM/c 2 M/M R/R s ; R s = 2GM /c km. P c < R > (9/8)R s M/M (8/9)R/R s. Sound speed c s : c s = (dp/dɛ) 1/2 c R > 1.39R s M/M R/(1.39R S ). If P = ɛ above n t 2n 0, R > 1.52R s M/M R/(1.52R s ). 7/30

8 Constraints on the EOS-II M max M obs ; In PSR , M obs = 1.44 M. In PSR , P K = 1.56 ms : Ω K ( M max M ) 1/2 ( Rmax 10 km ) 3/2 s 1 Mom. of Inertia I : ( I max = ) g cm 2 M max ( Rmax ) 2 M 10 km f(m max, R max ) In SN 1987A B.E. (1 2) ergs. 8/30

9 9/30

10 Composition of Dense Stellar Matter Crustal Surface : electrons, nuclei, dripped neutrons, set in a lattice new phases with lasagna, sphagetti, like structures Liquid (Solid?) Core : n, p,, leptons: e ±, µ ±, ν es, ν µs Λ, Σ, Ξ, K, π, condensates u, d, s, quarks Constraints : 1. n b = n n + n p + n Λ + : baryon # conservation 2. n p + n Σ + + = n e + n µ : charge neutrality 3. µ i = b i µ n q i µ l : energy conservation µ Λ = µ Σ 0 = µ Ξ 0 = µ n µ Σ = µ Ξ = µ n + µ e µ p = µ Σ + = µ n µ e µ K = µ e = µ µ = µ n µ p µ d = µ u + µ e = µ s = (µ n + µ e )/3 µ u = (µ n 2µ e )/3 10/30

11 Nuclear Matter-I Consider equal numbers neutrons (N) and protons (Z) in a large volume V at zero temperature (T = 0). Let n = (N + Z)/V = n n + n p denote the neutron plus proton number densities; n = 2k 3 F /(3π2 ), where k F is the Fermi momentum. Given the energy density ɛ(n) inclusive of the rest mass density mn, denote the energy per particle by E/A = ɛ/n, where A = N + Z. Pressure: From thermodynamics, we have P = E V = n 2 d(ɛ/n) dn = de d(a/n) = n dɛ dn ɛ = nµ ɛ, where µ = dɛ/dn is the chemical potential inclusive of the rest mass m. At the equilibrium density n 0, where P (n 0 ) = 0, µ = ɛ/n = E/A. 11/30

12 Incompressibility: Nuclear Matter-II The compressibility χ is usually defined by χ = 1 V V P = 1 n ( dp dn ) 1 However, in nuclear physics applications, the incompressibility factor K(n) = = 9 d dn 9 dp dn [ n 2 d(e/a) ] dn = 9 n d2 ɛ dn, or [ 2 = 9 n 2 d2 (E/A) + 2n d(e/a) ] dn 2 dn is used. At the equilibrium density n 0, the compression modulus K(n 0 ) = 9n 2 d 2 (E/A) d 2 (E/A) 0 dn 2 = kf 02 n0 dkf 2. k 0 F Above, k 0 F = (3π2 n 0 /2) 1/3 denotes the equilibrium Fermi momentum. 12/30

13 Nuclear Matter-III Adiabatic sound speed: The propagation of small scale density fluctuations occurs at the sound speed obtained from the relation ( cs c ) 2 dp = dɛ = 1 µ = dp/dn dɛ/dn dp dn = d ln µ d ln n. Alternative relations for the sound speed squared are ( cs c ) 2 = K 9µ = Γ P P + ɛ, where Γ = d ln P/d ln ɛ is the adiabatic index. It is desirable to require that the sound speed does not exceed that of light. 13/30

14 Schematic Nuclear Matter EOS-I ɛ(n) = n m := nucleon mass, [ m k 2 F 2m + A 2 u + u = n/n 0 := density compression ratio, B ] σ + 1 uσ Second term := mean kinetic energy of isospin symmetric matter, Last two terms := parametrize contributions from density dependent potential interactions. Such terms arise, for example, when local contact interactions are used to model the nuclear forces. The coefficients A, B and σ can be determined by using the empirically determined properties of nuclear matter at the equilibrium density., 14/30

15 Schematic Nuclear Matter EOS-II Using the relations for the state variables, and recalling that the total baryon density n = 2k 3 F /3π2, we get E A m = P = n 0 K 9 = d(p/n 0) du ɛ n m = E0 F u 2/3 + A 2 u + B σ + 1 uσ u 2 d(e/a) du = 2 3 E0 F u 5/3 + A 2 u2 + Bσ σ + 1 uσ+1 = 10 9 E0 F u 2/3 + Au + Bσ u σ. Above, E 0 F = (3/5)( k0 F )2 /2m is the mean kinetic energy per particle at equilibrium, with k 0 F denoting the Fermi momentum at equilibrium. 15/30

16 Schematic Nuclear Matter EOS-III Setting u = n/n 0 = 1 for which P/n 0 = 0, we find σ = B = A = K EF 0 9 [ 1 3 E0 F ( E m)] A ( ) [ ( )] σ E σ 1 3 E0 F A m [( ) E A m 5 ] 3 E0 F B Choosing E/A m = 16 MeV and n 0 = 0.16 fm 3 leads to: K 0 A B σ (MeV) (MeV) (MeV) /30

17 Limitations to watch out for Since σ > 1 for both choices of K 0, the energy density varies more rapidly than n 2 which leads to an acausal behavior in the EOS. Also, input values of K 0 below a certain limit are not allowed. An adequate parametrization that reproduces the nuclear matter energy density of realistic microscopic calculations is provided by ɛ nm = mn 0 u E0 F n 0 u 5/3 + V (u), where the potential contribution V (u) is given by V (u) = 1 2 An 0u 2 + Bn 0u σ B u σ 1 + u d 3 k C i 4 (2π) g(k, Λ i)θ(k 3 F k). i=1,2 The function g(k, Λ i ) is suitably chosen to simulate finite range effects. 17/30

18 18/30

19 Neutron-rich Matter-I α = (n n n p )/n := excess neutron fraction n = n n + n p := total baryon density x = n p /n = (1 α)/2 := proton fraction The neutron and proton densities are then n n = (1 + α) 2 n = (1 x) n & n p = (1 α) 2 n = x n. For nuclear matter, α = 0 (x = 0.5), whereas, for pure neutron matter, α = 1 (x = 0). Write the energy per particle (by simplifying E/A to E) as E(n, α) = E(n, α = 0) + E kin (n, α) + E pot (n, α), or 1st term := energy of symmetric nuclear matter 2nd & 3rd terms := isospin asymmetric parts of kinetic and interaction terms in the many body hamiltonian 19/30

20 In a non relativistic description, Neutron-rich Matter-II ɛ kin (n, α) = 3 2 [ ] (3π 2 n n ) 2/3 n n + (3π 2 n p ) 2/3 n p 5 2m = n E F 1 [ (1 + α) 5/3 + (1 α) 5/3]. 2 E F = (3/5)( 2 /2m)(3π 2 n/2) 2/3 := mean K.E. of nuclear matter. E kin (n, α) = E kin (n, α) E kin (n, α = 0) = = 1 ( ) 3 E F α α Quadratic term above offers a useful approximation ; From experiments, bulk symmetry energy 30 MeV ; Contribution from K.E. amounts to EF 0 /3 (12 13) MeV ; Interactions contribute more to the total bulk symmetry energy. 20/30

21 Neutron-rich Matter-III E(n, x) = E(n, 1/2) + S 2 (n) (1 2x) 2 + S 4 (n) (1 2x) 4 +. S 2 (n), S 4 (n), from microscopic calculations. Chemical Potentials : Utilizing E = ɛ/n, n = n n + n p, x = n p /n, and u = n/n 0, µ n = ɛ n n = E + u E np u x E x x, n µ p = ɛ n p = µ n + E nn x, n µ = µ n µ p = E x n = 4(1 2x) [ S 2 (n) + 2S 4 (n) (1 2x) 2 + ]. µ determines the composition of charge neutral neutron star matter. µ governed by the density dependence of the symmmetry energy. 21/30

22 22/30

23 Charge neutral neutron-rich matter-i Old neutron stars are in equilibrium w.r.t. weak interactions. The ground state consists of strongly interacting hadrons and weakly interacting leptons generated in β-decays and inverse β-decays. Local charge neutrality and chemical equilibrium prevail. First consider matter with neutrons, protons and electrons involved in n (or + n) p (or + n) + e + ν e, p (or + n) + e n (or + n) + ν e In cold catalyzed neutron star matter, neutrinos leave the system leading to µ = µ n µ p = µ e. (Later we will also consider the case in which neutrinos are trapped in matter, which leads instead to µ n µ p = µ e µ νe.) 23/30

24 Charge neutral neutron-rich matter-ii In beta equilibrium, one has x [E b(n, x) + E e (x)] = 0. Charge neutrality implies that n e = n p = nx, or, k Fe = k Fp. Combining these results, x(n) is determined from 4(1 2x) [ S 2 (n) + 2S 4 (n) (1 2x) 2 + ] = c ( 3π 2 nx ) 1/3. When S 4 (n) << S 2 (n), x is obtained from β x (1 2 x) 3 = 0, where β = 3π 2 n ( c/4s 2 ) 3. Analytic solution ugly! For u 1, x << 1, and to a good approximation x (β + 6) 1. Notice the high sensitivity to S 2 (n), which favors the addition of protons to matter. 24/30

25 Charge neutral neutron-rich matter-iii Muons in matter : When E Fe m µ c MeV, electrons to convert to muons through e µ + ν µ + ν e. Chemical equilibrium implies µ µ = µ e. At threshold, µ µ = m µ c MeV. As the proton fraction at nuclear density is small, 4S 2 (u)/m µ c 2 1. Using S 2 (u = 1) 30 MeV, threshold density is n 0 = 0.16 fm 3. Above threshold, µ µ = k 2 F µ + m 2 µc 4 = ( c) 2 (3π 2 nx µ ) 2/3 + m 2 µc 4. x µ = n µ /n b := muon fraction in matter. The new charge neutrality condition is n e + n µ = n p. Muons make x e = n e /n b to be lower than its value without muons. 25/30

26 Charge neutral neutron-rich matter-iv Total energy density & pressure : ɛ tot = ɛ b + l=e,µ ɛ l & P tot = P b + l=e,µ P l ɛ b,l and P b,l := energy density and pressure of baryons (leptons). ɛ l = 2 d 3 k k (2π) 2 + m 2 3 l & P l = d 3 k (2π) 3 k 2 k2 + m 2 l { } 3 ɛ b = mn 0 u + 5 E0 F n 0 u 5/3 + V (u) + n 0 (1 2x) 2 us(u), { ( 2 P b = 5 E0 F n 0 u 5/3 + u dv )} du V + n 0 (1 2x) 2 u 2 ds du. As α em 1/137, free gas expressions for leptons are satisfactory. 26/30

27 Charge neutral neutron-rich matter-v STATE VARIABLES AT NUCLEAR DENSITY Quantity Nuclear Stellar matter matter x ɛ b /n m ɛ e /n P b P e µ n m µ p m µ e = µ n µ p Energies in MeV and pressure in MeV fm 3. The numerical estimates are based on an assumed symmetry energy S 2 (u) = 13u 2/3 + 17u, where u = n/n b. 27/30

28 Nucleonic Equation of State Energy (E) & Pressure (P ) vs. scaled density (u = n/n 0 ). Nuclear matter equilibrium density n 0 = 0.16 fm 3. Proton fraction x = n p /(n p + n n ). Nuclear matter : x = 1/2. Neutron matter : x = 0. Stellar matter in β equilibrium : x = x. 28/30

29 Mass Radius Relationship Lattimer & Prakash, Science 304, 536 (2004). 29/30

30 30/30