Theory of Nuclear Matter for Neutron Stars and ... - Graduate Physics

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2.3.5 Uniform Matter at Zero Temperature and the SaturationConstraintsNext, we consider uniform matter at zero temperature. Using the notation u = (ρ n +ρ p )/ρ oand 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 and pressure p for zero-temperature matter as afunction of 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 and S ′ v = 17.93 MeV.Fig. 2.1 shows the energy per nucleon E = E/ρ and the pressure p of uniform matter as afunction of density and composition Y p for the case of zero-temperature matter. Obviously,there are substantial regions in which the incompressibility, (∂P/∂n) T,Yp , is negative and thematter is hydrodynamically unstable. It is straightforward to show that the free energy canbe lowered if matter spontaneously divides into two phases of 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 standard nuclear matter, u = 1 and x = 1/2. Letting α = α L + α U and β ′′ =β L ′′ +β′′ U = (6/5)β+σ, using β = β L+β U and σ = σ L +σ U , and 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)and 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)andα = 9 5 + 5BT o≃ 3.995,β ′′ = 3 5 + 3BT o≃ 1.917, (2.49)using T o ≃ 37.68 MeV and B ≃ 16.54 MeV. (For the purposes of 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 of the symmetry energy and 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 of S v and S ′ v, one can manipulate Eqs. (2.50) and (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
Page 1 and 2: Theory of nuclear matter of neutron
Page 3 and 4: Abstract of the DissertationTheory
Page 5 and 6: To Jaenyeong, Jonghyun, and Jongbum
Page 7 and 8: 3.3 Optimized Parameter Set . . . .
Page 9 and 10: List of Figures1.1 Mass and radius
Page 11 and 12: SymbolsNuclear Physics SymbolE ener
Page 13 and 14: ω δt 90−10QtT cTaylor expansion
Page 15 and 16: List of Tables1.1 Range of Tables .
Page 17 and 18: Chapter 1IntroductionStars give us
Page 19 and 20: Since the mass of neutron star is d
Page 21 and 22: shown in Fig. 1.2, LS220 is the onl
Page 23 and 24: density:p = ρ2ρ o[ K9µ n = −B
Page 25 and 26: nuclear energy density functional i
Page 27 and 28: 2.3.2 Thermodynamic QuantitiesThe t
Page 29 and 30: point r and with momentum p isU n (
Page 31: Thus, in the case of zero temperatu
Page 35 and 36: a α L α U β L β U σ0.5882 1.11
Page 37 and 38: Figure 2.2: Bulk equilibrium of T =
Page 39 and 40: Figure 2.4: Coexistence curves for
Page 41 and 42: proton fraction x are irrelevant fo
Page 43 and 44: potential model analogue for the Ha
Page 45 and 46: with the convention that z II is th
Page 47 and 48: Figure 2.10: The surface tension of
Page 49 and 50: Figure 2.12: Contour plots of surfa
Page 51 and 52: 2.6.2 Dense Matter and the Wigner-S
Page 53 and 54: Figure 2.15: Proton number per unit
Page 55 and 56: Chapter 3Modified model† The trun
Page 57 and 58: We choose the value ǫ = 1/3. One m
Page 59 and 60: where∆V t = T oρ o[( ρρ o) ǫ[
Page 61 and 62: Q 1 to a small number also results
Page 63 and 64: where p = (S v ,L) and M ij = 1 ∂
Page 65 and 66: 3.4 Nuclear Surface TensionCompared
Page 67 and 68: 1.41.2FRTF II ω 0 , T=0 MeVFRTF II
Page 69 and 70: For the finite-range term, we use a
Page 71 and 72: Landau’s quasi-particle formula g
Page 73 and 74: The symmetry energy in nuclear matt
Page 75 and 76: Figure 4.2: The left figure shows t
Page 77 and 78: Figure 4.4: This figure shows the b
Page 79 and 80: Figure 4.5: In a neutron star, heav
Page 81 and 82: The energy exchange from the gradie
Page 83 and 84:
Figure 4.8: The left panel shows th
Page 85 and 86:
4.5 Astrophysical application4.5.1
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0.550.5GaussianFRTF IIFRTF I0.450.4
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neutron matter respectively. Those
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Table 5.3: Nuclear properties of si
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which aref Rc = a 0 +a 1 x+a 2 x 2
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parametrized[ht]504540EDF, Z nucFit
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Chapter 6Nuclear Equation of State
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Coulomb, and, symmetry part [24], i
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We can eliminate N s , leading to t
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pressure from pure neutron matter,
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2.52GSRsSGISIVSkaSkI1SkI2SkI3SkI5St
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HNsn, p, α,eFigure 6.7: In the Wig
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Table 6.1: Surface tension analytic
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free energy density for the alpha p
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Here s ′ = ∂s(u)/∂u.The minim
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6.5.2 Determination of Coulomb surf
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gives analytic derivatives of therm
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200180160Atomic number, Y p =0.45,
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for Fermi integral. This fitting fu
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A.1 Taylor expansion integrationThe
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thenwhere0u − i+1 = fu− i +J
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We separate the two integrals as in
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Each w i can be obtained[ae −∆/
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By expanding e r 0/a −e −r 0/a
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M 3 2a 0.433 1p 0 (e 2 /a) √ π/3
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of find exact polynomial expression
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Appendix CPhase coexistenceBulk equ
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To summarize, we need to solve 5 eq
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D.1 Non-uniform electron density ap
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Appendix ENuclear Quantities in Non
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Table E.1: Non-relativistic Skyrme
Page 147 and 148:
[20] C.J. Pethick and D.G. Ravenhal
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Theory of Nuclear Matter for Neutron Stars and ... - Graduate Physics

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