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The End States of Low Mass Stars

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The End States of Low Mass Stars the end states of low mass stars low mass stellar evolution low mass stellar evolution nuclear energy release the white dwarf stage Sirius, the brightest star Sirius, the brightest star stellar corpses: white dwarfs Surface of the Sun Earth Sirius B white dwarfs? “I do not see how a star which has once got into this highly compressed condition is ever going to get out of it … It would seem that the star will be in the awkward predicament when its supply of subatomic energy fails … a star will need energy to cool. ” Arthur Eddington “The stellar material will have radiated so much that it has less energy than the same matter in normal atoms expanded at the absolute zero. If part of it were removed from the star and the pressure taken off, what could it do? ” R. H . Fowler white dwarf stuff Measuring Cup of White Dwarf Stuff The potential energy per particle varies as E Gm2 4π 1/3 ! = grav = p N = Gm2N 2/3n1/3. g N R 3 p General relativistic effects become important when R = E /E 1. gm grav mass ∼ Thermal and Degeneracy Energy: The behavior of the system depends on the origin of the momentum distribution of the particles. The familiar situation is the one in which short-range interactions between particles effectively exchange the energy so as to randomize the momentum distribution. When such a system is in steady state, we can assume that the local thermodynamical equilibrium, characterized by a temperature T , exists in the system: ! k T ≈ B 2 1/2 1/2 2 2k T k T (2mkBT ) if mc kBT B B # p mc 2 + 2 ≈ mc mc ≈ 2 ! " # $ kBT/c if mc kBT $ In this case, the momentum and the kinetic energy of the particles vanish when T 0. The structures: white dwarfs → mean energy of a system of electrons will not vanish at zero temperature because electrons obey the Pauli exclusion principle, which requires that the number of electrons that can When the nuclear fuel in the star is exhausted, the gravitational force will start occupy any quantum state be two. Because the uncertainty principle requires ∆x∆p h, contracting the matter again and the density will be sufficiently high so that x ≥ we can asstheo cquantumiate the degeneracynumber of pressurequantum willstates dominatewith overa pha thermalse space pressure.volume: The potential energy per particle varies as E p Gm= 2 (3π2n4π)11//33. ! = grav =F p!N = Gm2N 2/3n1/3. g N R 3 p TheGenequanrtital yrelativ!F setsistictheeffecquats bnectumomemecimphanicalortant wscalehen Rof the= energyE /E 1. gm grav mass ∼ 2 2 pF ! 2 2/3 2 Thermal and Degeneracy Energy: The beha=vior of the(3πsystemn) depifendsmc on the!F origin of 2 2 2 4 2 2m 2m # !F = pF c + m c mc − ≈ ) * 2 1/3 2 the momentum distribution of the partpicFles.c =The(!c)(3familiaπ n)r situationif mcis the one!F in which ( $ Quashortntum-mec-rangehanicainteractionsl effectsbetwillweenbparte dominaicles effnectivt if !ely exckhangT e(degenerathe energyte).so asElectronto randomizes have F # B the momentum distribution. When such a system is in steady state, we can assume that the ! m c2 for F ∼ e local thermodynamical equilibrium, characterized3 by a temperature T , exists in the system: ! − 31 3 n = 10 cm− mec ≈ " #! kBT ≈ 7 3 ρ nm 10 g cm− . ∼ p ≈ 2 1/2 1/2 2 2k T k T (2mkBT ) if mc kBT B B # p mc 2 + 2 ≈ mc mc ≈ 2 ! " # $ kBT/c if mc kBT $ In this case, the momentum and the kinetic energy of the particles vanish when T 0. The → mean energy of a system of electrons will not vanish at zero temperature because electrons obey the Pauli exclusion principle, which requires that the number of electrons that can occupy any quantum state be two. Because the uncertainty principle requires ∆x∆p h, x ≥ we can associate the number of quantum states with a phase space volume: 2 1/3 pF = !(3π n) . The quantity !F sets the quantum mechanical scale of the energy 2 2 pF ! 2 2/3 2 = (3π n) if mc !F 2 2 2 4 2 2m 2m # !F pF c + m c mc − ≈ ) * 2 1/3 2 pF c = (!c)(3π n) if mc !F ( $ Quantum-mechanical effects will be dominant if ! k T (degenerate). Electrons have F # B ! m c2 for F ∼ e 3 ! − 31 3 n = 10 cm− m c ≈ " e # 7 3 ρ nm 10 g cm− . ∼ p ≈ The potential energy per particle varies as E Gm2 4π 1/3 ! = grav = p N = Gm2N 2/3n1/3. g N R 3 p 1 General relativistic effects become implortan= (ntσ)w−hen Rgm = Egrav/Emass 1. ∼ Thermal and Degeneracy Energy: The behavior of the system depends on the origin of 2 the momentum distribution of the partNcolicl les.(RThe/l) familiar situation is the one in which ≈ short-range interactions between particles effectively exchange the energy so as to randomize t = (lN /c) (R/c)(R/l) the momentum distribution. Whenesc suchcolla system≈ is in steady state, we can assume that the local thermodynamical equilibrium, characterized by a temperature T , exists in the system: E (aT 4)R3 γ ≈ ! k T ≈ B Eγ 3 4 2 4 L = 2 1/aR2 T l/R = aRT1/l2 2 2k T k Tt ≈ (2mkBT ) if mc kBT B B esc # p mc 2 + 2 ≈ mc mc ≈ 2 ! " # $ kBT/c if mc kBT 4 4 3 $ aRT G 5 2 3 34 1 M − In this case, the momenL =tum and the2 mkineticpmeM energ10y ofergthes particles.vanish when T 0. The structures:σT n ≈ α white≈ dwarfsM ! ∗ " → mean energy of a system of electrons will not vanish at zero temperature because electrons GMmp obey the Pauli exclusion principle, whickBTh=requires that the number of electrons that can R When the nuclear fuel in the star is exhausted, the gravitational force will start occupy any quantum state be two. Because the uncertainty principle requires ∆x∆p h, contracting the matter again and the density will be sufficiently high so that x ≥ L R2T 4 we can asstheo cquantumiate the degeneracynumber of pressurequantum will∝states dominateS with overa pha thermalse space pressure.volume: The potential energy per particle varies as Gm2 2 11//33 Egrav pF =1/p4!(3π1/2n4π) 1./12 2 2/3 1/3 !g = T = L RN−= L Gm N n . N S ∝ R ∝3 p TheGenequanrtital yrelativ!F setsistictheeffecquats bnectumomemecimphanicalortant wscalehen Rof the= energyE /E 1. gm grav mass ∼ 2 2 MpF R ! 2 2/3 2 Thermal and Degeneracy Energy: The beha∝=vior of the(3πsystemn) depifendsmc on the!F origin of 2 2 2 4 2 2m 2m # !F = pF c + m c mc − ≈ ) * 2 1/3 2 the momentum distribution of the partpicFles.c =The(!c)(3familiaπ n)r situationif mcis the one!F in which ( #M 1 $ tstar = Quashortntum-mec-rangehanicainteractionsl effectsbetwillweenbparte dominaicles Leffnectivt∝ifM!ely2 exckhangT e(degenerathe energyte).so asElectronto randomizes have When particles are not relativistic, the conditionF can# beB satisfied at equality if the mo2 mentum distribution. When such a system is in steady state, we can assume that the !F mec for 2 ∼ 1/3 2 Gmpme 2/3 local thermodynamical equilibrn ium,= c2hara2/3 cterized3 2 by aNtemperature T , exists in the system: (3π! ) − ! 31 3 n = ! 10 "cm− mec ≈ " #! kBT ≈ 7 3 ρ nm 10 g cm− . ∼ p ≈ 2 1/2 1/2 2 2k T k T (2mkBT ) if mc kBT B B # p mc 2 + 2 ≈ mc mc ≈ 2 ! " # $ kBT/c if mc kBT $ In this case, the momentum and the kinetic energy of the particles vanish when T 0. The → mean energy of a system of electrons will not vanish at zero temperature because electrons obey the Pauli exclusion principle, which requires that the number of electrons that can occupy any quantum state be two. Because the uncertainty principle requires ∆x∆p h, x ≥ we can associate the number of quantum states with a phase space volume: 2 1/3 pF = !(3π n) . The quantity !F sets the quantum mechanical scale of the energy 2 2 pF ! 2 2/3 2 = (3π n) if mc !F 2 2 2 4 2 2m 2m # !F pF c + m c mc − ≈ ) * 2 1/3 2 pF c = (!c)(3π n) if mc !F ( $ Quantum-mechanical effects will be dominant if ! k T (degenerate). Electrons have F # B ! m c2 for F ∼ e 3 ! − 31 3 n = 10 cm− m c ≈ " e # 7 3 ρ nm 10 g cm− . ∼ p ≈ 1 l = (nσ)− 2 Ncoll (R/l) 1 ≈ l = (nσ)− 1 l = (nσ)− tesc = (lNcoll/c) (R/c)(R/l) N (R/l)2 ≈ coll ≈ N (R/l)2 coll ≈ 4 3 t = (lN /c) (R/c)(R/l)Eγ (aT )R esc coll ≈ ≈ tesc = (lNcoll/c) (R/c)(R/l) ≈ 4 3 Eγ 3 4 2 4 Eγ (aT )LR = aR T l/R = aRT l ≈ tesc ≈ 4 3 Eγ (aT )R ≈ Eγ 3 4 2 4 L = aR T l/R4 = aR4T l 3 tesc ≈ aRT G 5 2 3 34 1 M Eγ − 3 4 2 L4= 2 mpmeM 10 erg s . L = aR T l/R = aRT l σT n ≈ α ≈ M tesc ≈ 4 4 3 ! ∗ " aRT G 5 2 3 34 1 M − L = 2 mpmeM 10 erg s . σT n ≈ α ≈ 3 M 4 4 ∗ aRT G 5 2 3 34 1 M ! "GMmp − L = 2 mpmeM 10 erg s . kBT = σT n ≈ α ≈ M R ! GM∗ " m k T = p B R GMmp 2 4 kBT = L R T R S L R2T 4 ∝ ∝ S 2 4 L R TS 1/4 1/2 1/12 ∝ 1/4 1/2 1T/12S L R− L TS L R− L ∝ ∝ ∝ ∝ 1/4 1/2 1/12 T L R− L S ∝ ∝ M R M R ∝ ∝ M R ∝ #M 1 tstar = #M 1 L ∝ M 2 tstar = #M 1 L ∝ M 2 tstar = L ∝ M 2 2 structures:2 Gm me white dwarfs n1/3 = p N 2/3 2 2 2/3 2 2 Gm me 2 (3π ) !1/3 p 2/3 2 Gm me ! n =" N n1/3 = p N 2/3 (3π2)2/3 !2 (3π2)2/3 !2 ! " ! " n = (3N/4πR3) 3 3 n = (3N/4πR ) With n = (3 N / 4 π R ) and N = ( M /m p ) , this reduces to the following mass-radius1/3 1 relation:1/3 3 1/3 RM NαG−=λ(eMm/mp p) 8.7 10− R m N = (M/mp) ≈ ≈ × # # 1/3 1 1/3 3 1/3 RM αG− λemp 8.7 10− R m 1/3 1 1/3 3 1/3 ≈ ≈ × # R#M αG− λemp 8.7 10− R m ≈ ≈ × # # 1 l = (nσ)− 2 Ncoll (R/l) 1 ≈ l = (nσ)− 1 l = (nσ)− tesc = (lNcoll/c) (R/c)(R/l) N (R/l)2 ≈ coll ≈ N (R/l)2 coll ≈ 4 3 t = (lN /c) (R/c)(R/l)Eγ (aT )R esc coll ≈ ≈ tesc = (lNcoll/c) (R/c)(R/l) ≈ 4 3 Eγ 3 4 2 4 Eγ (aT )LR = aR T l/R = aRT l ≈ tesc ≈ 4 3 Eγ (aT )R ≈ Eγ 3 4 2 4 L = aR T l/R4 = aR4T l 3 tesc ≈ aRT G 5 2 3 34 1 M Eγ − 3 4 2 L4= 2 mpmeM 10 erg s .
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