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The Equation of State for Neutron Stars Laura Tolós Outline Neutron Star (I)

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The Equation of State for Neutron Stars Laura Tolós Outline Neutron Star (I) The Equation of State for Neutron Stars Laura Tolós Outline Neutron Star (I) • first observations by the Chinese in 1054 A.D. and prediction by Landau after discovery of neutron by Chadwick in 1932 • produced in core collapse supernova explosions • usually refer to compact objects with M≈1-2 M¤ and R≈10-12 Km • extreme densities up to 5-10 times nuclear density ρ0 -3 14 3 (ρ0=0.16 fm => n0=310 g/cm ) -30 3 nUniverse ~ 10 g/cm 3 nSun ~ 1.4 g/cm 3 nEarth ~ 5.5 g/cm Neutron Star (II) • usually observed as pulsars with rotational periods from milliseconds to seconds • magnetic field : B ~ 10 8..16 G • temperature: T ~ 10 6…11 K (1 eV=104 K) Pulsars are magnetized rotating neutron stars emitting a highly focused beam of electromagnetic radiation oriented long the magnetic axis. The misalignment between the magnetic axis and the spin axis leads to a lighthouse effect: from Earth we see pulses Since 1967, ∼ 2500 pulsars have been discovered. http://www.atnf.csiro.au/research/pulsar/psrcat/ Most of them have been detected as radio pulsars, but also some observed in X-rays Their period P ranges from and an increasingly large number detected in 1.396 ms for PSRJ1748−2446ad gamma rays. up to 8.5 s for PSR J2144−3933. Magnetic fields Bulk properties: Masses • > 2000 pulsars known • best determined masses: Hulse-Taylor pulsar M=1.4414 ± 0.0002 M¤ Hulse-Taylor Nobel Prize 94 figures by Weiseberg • PSR J1614-22301 M=(1.97 ± 0.04) M¤; PSR J0348+04322 M=(2.01 ± 0.04) M¤ 1Demorest et al ’10; 2Antoniadis et al ‘13 Lattimer ‘16 Bulk properties: Fortin et al ’15: Ø RP-MSP: Bodganov‘13 Radius Ø BNS-1: Nattila et al ‘16 Ø BNS-2: Guver& Ozel ‘13 analysis of X-ray spectra from Ø QXT-1: Guillot & Rutledge ‘14 neutron star (NS) atmosphere: Ø BNS+QXT: Steiner et al ’13 • RP-MSP: X-ray emission from radio millisecond pulsars • BNS: X-burst from accreting NSs • QXT: quiescent thermal emission of accreting NSs theory + pulsar observations: R1.4M~11-13 Km Lattimer and Prakash ’16 Some conclusions: ü marginally consistent analyses, favored R < 13 Km (?) ü future X-ray telescopes (NICER, eXTP) with precision for M-R of ~ 5% ü GW signals from NS mergers with precision for R of ~1 km Bauswein and Janka ’12; Lackey and Wade ‘15 adapted from Fortin’s talk @ NewCompstar Annual Meeting ‘16 Internal structure and composition: the Core A. Watts et al. ‘15 • Atmosphere few tens of cm, ρ ≤104 g/cm3 made of atoms • Outer crust or envelope few hundred m’s, ρ=104-41011 g/cm3 made of free e- and lattice of nuclei • Inner crust 1-2 km, ρ=41011-1014 g/cm3 made of free e-, neutrons and neutron-rich atomic nuclei - ~ρ0/2: uniform fluid of n,p,e • Outer core - ρ0/2-2ρ0 is a soup of n,e ,μ and 3 possible neutron P2 superfluid or 1 proton S0 superconductor • Inner core (?) 2-10 ρ0 with unknown interior made of hadronic, exotic or deconfined matter The Core of a Neutron Star Fridolin Weber Strange Quark Matter Hypothesis working hypothesis: strange quark matter constitutes the ground 56 state of strong interaction rather than Fe Cristina Manuel’s lecture Pauli principle Bodmer,Witten & Ter az awa Equation of State of dense matter: free Fermi gas The Equation of State (EoS) is the relation between the mass/energy density and pressure: P(n), P(E) Extreme Conditions in the stellar interior! * Atoms are ionized * Particles can be degenerate and ultrarelativistic * Pressure due to radiation can be significant As a first approximation, let’s consider the simplest thermodynamical system, the ideal gas. The ideal gas consists of a large number of particles occupying quantum states whose energy is not affected by the interaction between the particles. Depending on the conditions of the stellar interior the gas can be: Classical or Quantum (degenerate) and the energies of the particles can be: Non-relativistic or Relativistic Ideal Gas Classical vs quantum (degenerate) 1 f(ep)= e µ p− Distribution of the particles e kT 1 2 2 4 2 2 ± in the quantum states, f(ep) ep = m c + p c +1 ➔ Fermi-Dirac distribution - 1 ➔ Bose-Einstein distribution A gas is classical when the average occupation of any quantum state is small e µ 1 p− kT μ decreases or T increases f(ep)= e µ e− 1 p− e kT 1 ' ⌧ ± Relativistic vs non relativistic p2 p Non-relativistic: e = mc2 + v = p 2m p m Ultra-relativistic: ep = pc vp = c General formulae Number of (wave vector) states V 2 gs 4⇡k dk with magnitude between k and k+dk (2⇡)3 _ Broglie relation (p= h/λ=hk) , V g(p)dp = g 4⇡p2dp number of momentum states s h3 between p and p+dp: degeneracy factor 1 Internal energy: E = epf(ep)g(p)dp Z0 1 Number of particles: N = f(ep)g(p)dp Z0 Pressure: dE=T dS- P dV + µ dN 2 dep pc dep dep dp = = vp (definition) = dp ep dV dp dV 1/3 dp p p V − = / ! dV −3V 1 1 N ➔ P = pv f(e )g(p)dp = pv 3V p p 3V h pi Z0 1 1 N P = pv f(e )g(p)dp = pv 3V p p 3V h pi Z0 Two kinematical limits: p2 p Non-relativistic: e = mc2 + v = p 2m p m 2N p2 2 P = = of kinetic energy density 3V h2mi 3 Ultra-relativistic: ep = pc vp = c N 1 P = pc = of kinetic energy density 3V h i 3 Pressure is “smaller” in ultrarelativistic gases!! Up to here, the expressions found are applicable to an ideal gas in its most general form: for a dilute classical gas or for a dense gas where quantum effects are important e µ 1 p− kT Ideal Classical (Diluted) Gas f(ep)= e µ e− 1 p− e kT 1 ' ⌧ ± kT µ 1 ep V kT kT 2 P = e e− gs 3 4⇡p dp V 0 h N (integrating by parts)Z P = kT = ⇢kT V e µ 1 p V 2 N =ekT e− kT g 4⇡p dp s h3 Z0 p2 3 non-relativistic: = kT Comparing h2mi 2 (average kinetic energy expressions for P ultra-relativistic: pc =3kT per particle) h i A more intuitive way of determining if a gas is classical (diluted) is by checking whether the interparticle separation is much larger that the typical de Broglie wave length λ=h/p, which is determined by h h mkT 3/2 non-relativistic: λ = ⇢ p ⇠ p ! ⌧ h2 mkT ✓ ◆ h hc kT 3 ultra-relativistic: λ = ⇢ p ⇠ kT ! ⌧ hc ✓ ◆ Gases formed by particles with mass, such as electrons and ions, can behave classically or quantum mechanically, depending on their density Free Fermi Gas When the concentration of fermions becomes large, the interparticle separation may be comparable with the de Broglie wavelength. In particular, the high density requirement for a non-relativistic quantum gas mkT 3/2 ⇢ h2 ✓ ◆ h2⇢2/3 can be viewed as a low temperature requirement kT ⌧ m Therefore a quantum gas is a cold gas but the coldness is set by the density. Temperatures of 10 6 K can be low if the densities are very high! Let’s consider then a cold gas of fermions (electrons or neutrons at T=0) The ground state is referred to as the free Fermi sea The fermions have fallen into quantum states with the lowest possible energy ➔ Pauli principle! They are distributed so that each quantum state is occupied up to a certain energy (Fermi level, eF) and quantum states above this level are empty. This distribution is the T=0 limit of the Fermi-Dirac distribution (setting μ=eF) 1 1 n(k) f(ek) eF kF eF (the energy of the most energetic particle) is called the Fermi energy and kF is the Fermi wave-number (pF = ħ kF is the Fermi momentum): eF=e(pF) Relation between the Fermi momentum and the density: V V 4 N = g ✓(k ~k )=g d3k✓(k ~k )=g ⇡k3 s F −| | s (2⇡)3 F −| | s (2⇡)3 3 F X~k Z 3 2 1/3 1/3 N kF 6⇡ ⇢ 2⇡ gs λ = =2⇡ = ⇢ = gs 2 kF = F 2 ! V 6⇡ g ! kF 6⇡ ⇢ ✓ s ◆ ✓ ◆ Note that de Broglie wave-length of the most energetic particle is comparable with � -1/3, that is of the order of the average distance between fermions Equation of state for a non-relativistic free Fermi gas 2 Non relativistic case: pF c = ħ kF c << m c 2 1/3 3 6⇡ ⇢ mc gs 3 mc kF = ⇢ 2 8⇡ The Compton wavelength gs ⌧ ~ ! ⌧ 6⇡ h ✓ ◆ ⇣ ⌘ of a particle is equivalent h (neutrons) ~ 10-16 m to the wavelength of a photon whose energy is (electrons) ~10-12 m mc the same as the rest- mass energy of the neutrons are still not relativistic particle when electrons are! Internal energy: p2 e = mc2 + p 2m k F ~2k2 V V 4 ~2k2 3 E = mc2 + g 4⇡k2dk = Nmc2 + g ⇡k3 F 2m s (2⇡)3 s (2⇡)3 3 F 2m 5 Z0 ✓ ◆ 3 ~2k2 = N mc2 + F 5 2m ✓ ◆ 2/3 3 ~2 6⇡2 E = Ne(⇢) with e(⇢)=mc2 + ⇢2/3 5 2m g ✓ s ◆ From internal energy, we can obtain the pressure and chemical potential so we finally obtain for internal energy and pressure For ultrarelativistic free Fermi gas, the EoS becomes less stiff (do it!) Baryonic matter in the core A Fermi gas model for only neutrons inside neutron stars is unrealistic: 1.
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