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DEPARTMENT OF PHYSICS AND ASTRONOMY LIFECYCLES OF STARS Option 2601 M.R. Burleigh 2601/Unit 4 Stellar Physics Unit 1 - Observational properties of stars Unit 2 - Stellar Spectra Unit 3 - The Sun Unit 4 - Stellar Structure Unit 5 - Stellar Evolution Unit 6 - Stars of particular interest M.R. Burleigh 2601/Unit 4 DEPARTMENT OF PHYSICS AND ASTRONOMY Unit 4 Stellar Structure M.R. Burleigh 2601/Unit 4 Starbirth M.R. Burleigh 2601/Unit 4 Young Stars M.R. Burleigh 2601/Unit 4 Globular Clusters M.R. Burleigh 2601/Unit 4 Star Death M.R. Burleigh 2601/Unit 4 Star Death M.R. Burleigh 2601/Unit 4 Star Death M.R. Burleigh 2601/Unit 4 Star Death M.R. Burleigh 2601/Unit 4 Stellar Structure Hydrostatic equilibrium Equations of state Energy transport (not derived) Energy sources Stellar models Mass-Luminosity relation Eddington Limit M.R. Burleigh 2601/Unit 4 Hydrostatic Equilibrium r=R (r) r P + dP M.R. Burleigh 2601/Unit 4 P Centre (r = 0) dr Hydrostatic Equilibrium Equation of hydrostatic equilibrium: However: M dP Gmr r 2 dr r dm 2 4r r dr r R 0 dm (1) (2) R 4 0 Only need to know (r) to determine mass of star radius R M.R. Burleigh 2601/Unit 4 r r dr 2 E.g. Sun’s central pressure G=6.67x10-11 Nm2Kg-2, M=1.989x1030kg, R=6.96x108m (ave) =3M /4R3=1410kgm-3 Surface pressure = 0 Let r = dr = R and M(r)= M Pc~G M (ave) / R = 2.7x1014Nm-2 M.R. Burleigh 2601/Unit 4 Equations of State Assume material is a perfect gas Obeys perfect gas law: Pr nr kT r Number density of particles M.R. Burleigh 2601/Unit 4 (3) Boltzmann’s constant (1.381 10-23JK-1) Equations of State n(r) is dependant on density and composition: mH = 1.67 10-27 kg = mass of a hydrogen atom = mean molecular weight nr r r mH 1 1 3 1 2 2 X 4 Y 2 Z Mass fractions of: H He Metals (all other heavier elements) r kT r Pr r mH M.R. Burleigh 2601/Unit 4 Radiation Pressure In massive stars, radiation pressure also contributes to the total pressure: a T r Prad r 3 4 a = 7.564 10-14Jm-3K-4 = radiation constant ( = ¼ ac) M.R. Burleigh 2601/Unit 4 E.g. Sun’s central temperature Use Pc and estimates, assume ~ ½ Then Tc ~ PcmH/(ave) k ~ 1.2x 107K Gas dissociated into ions & electrons but overall electrically neutral… a plasma M.R. Burleigh 2601/Unit 4 Energy Transport T(r) depends on how energy is transported from interior surface Three processes… 1.Conduction – collision of hot energetic atoms with cooler… poor in gases 2.Convection – mass motions of fluids, need steep temp. gradient… happens in some regions of most stars M.R. Burleigh 2601/Unit 4 Energy Transport Three processes… 1.Conduction 2.Convection 3.Radiation – high energy photons flow outward losing energy by scattering and absorption… opacity sources at high T are i) electron scattering and ii) photoionization M.R. Burleigh 2601/Unit 4 Radiative Transport Equation 64r T r dT L r 3 r r dr 2 3 (r) (opacity) depends only upon N(r), T(r) and (r) L(r) at the surface is the star’s bolometric luminosity M.R. Burleigh 2601/Unit 4 (4) For the Sun L ~ 9.5x1029/ Joules s-1 However, we do not know very well Ranges from 10-3 << << 107 Therefore… – 1022 << L << 1032 Joules s-1 Measured value is 3.9x1026 implies – ~ 2.4x103 M.R. Burleigh 2601/Unit 4 The Virial Theorem Considers total energy in a star Gravitational contraction Gravitational potential energy kinetic energy Kinetic energy in bulk Heat M.R. Burleigh 2601/Unit 4 The Virial Theorem Take the equation of hydrostatic equilibrium: dP Gmr r dr r2 and dm 4r 2 r dr dP mG mG 3 dP 4 r dm dm r 4r 4 But: d dP dr 4r 3 P 4r 3 4r 2 .3P dm dm dm d dr mG 3 2 4r P 4r .3P dm dm r M.R. Burleigh 2601/Unit 4 Integrate over the whole star: 4r 3 M P 0M 0 3P M Gm dm 0 dm r Zero at both limits Twice the thermal (P(m) = 0 marks the (kinetic) energy boundary of the star) -2U P, and r are functions of m Gravitational binding energy 2U 2U 0 Total energy of a star: E U E U 0 M.R. Burleigh 2601/Unit 4 Gravitational Contraction Gravitational contraction But, using Virial theorem: 1 U 2 ½ excess must be lost by radiation 1) Star gets hotter 2) Energy is radiated to space 3) Total energy of the star decreases (becomes more –ve more tightly bound) M.R. Burleigh 2601/Unit 4 Stellar Thermonuclear Reactions Light elements “burn” to form heavier elements Stellar cores have high enough T and for nuclear fusion Work (after 1938) by Hans Bethe and Fred Hoyle M.R. Burleigh 2601/Unit 4 Stellar Thermonuclear Reactions Energy release can be calculated from E=mc2 – e.g. 4 x 11H atoms 1 x 42He atom 4 x 1.6729x10-27kg = 6.6916x10-27kg 1 x 6.6443x10-27kg E = 4.26x10-12J M.R. Burleigh 2601/Unit 4 Stellar Thermonuclear Reactions In the Sun ~10% of its volume is at the T and required for fusion Total energy available is… – Energy per reaction x mass/mass in each reaction Etot = 4.26x10-12 x 2x1029/6.6916x10-27 = 1.27x1044J L = 3.9x1026 t ~ 3.3x1017s ~ 1010yrs M.R. Burleigh 2601/Unit 4 Stellar Thermonuclear Reactions Proton – proton chain (PPI, T < 2 107K) 1 1H 11 H 12 H e 1.44MeV 2 1H 11 H 32 He 5.49MeV 3 3 He 2 2 CNO cycle He 42 He 11 H 11 H 12 6 C 11 H 13 7 N 13 7 N 13 C e 6 13 6 C 11 N 14 7 N 14 7 N 11 H 15 8 O 15 8 O 15 N M.R. Burleigh 2601/Unit 4 7 15 N e 7 4 11 H 12 C 6 2 He 12.9MeV The PPI Chain M.R. Burleigh 2601/Unit 4 PPI Chain a c 1 1H 11 H 12 H e 2 1H 11 H 32 He 3 3 He 2 2 He 42 He 11 H 11 H M.R. Burleigh 2601/Unit 4 b The CNO Cycle M.R. Burleigh 2601/Unit 4 CNO Cycle 12 6 C H N 1 1 13 7 13 7 N 13 C e 6 13 6 C 11 H 14 7 N 14 7 N 11 H 15 8 O 15 8 15 7 O 15 N e 7 4 N 11 H 12 C 6 2 He M.R. Burleigh 2601/Unit 4 Triple Alpha High level reactions ~108K 4 2 He He Be 8 4 Be 24He126C 4 2 8 4 Addition of further alphas 22 24 168O,10 Ne,12 Mg upto Fe M.R. Burleigh 2601/Unit 4 Stellar Models: Equations Hydrostatic equilibrium: Mass equation: dP Gmr r (1) 2 dr r dm 4r 2 r (2) dr Equation of state: Pr nr kT r (3) n = number density of particles nr r r mH = mean molecular weight M.R. Burleigh 2601/Unit 4 Radiation pressure: a 4 Prad r T r 3 a = radiation constant = 7.564 10-14Jm-3K-4 = ¼ ac Radiative transport: 64r 2T 3 r dT Lr 3 r r dr = opacity Energy generation: dL 2 4r r r (5) dr ε = rate of energy production (Js-1kg-1) M.R. Burleigh 2601/Unit 4 (4) Boundary Conditions Need to apply boundary conditions to the equations to use them, i.e. fix/know values at certain values of r (centre or surface) e.g. r=0 M(r) = 0 L(r) = 0 r=R M(r) = M L(r) = L And (r), P(r) 0 M.R. Burleigh 2601/Unit 4 T(r) = Teff From (1) write dP P and dr r Then: P = PS – PC = 0 - PC Surface and Centre r = R M PC R For a perfect gas P T M.R. Burleigh 2601/Unit 4 M TC R M TC R From (4) 3 4 T T RT L R 2 C C C R Also: M 3 R Substitute M TC R 4 M R 4TC4 L M 4 R 3 M R L M Observed relationship is L M3.3 ( is dependant on T and ) M.R. Burleigh 2601/Unit 4 Eddington Limit Hydrostatic equilibrium assumes no net outward motion of material from the star, but the outward flow of radiation imparts a force on the material L Momentum of radiation = 4r 2c Force = T L 2 4r c T = cross-section of electron-photon scattering = 6.7 10-29m2 This is opposed by gravitational force = M.R. Burleigh 2601/Unit 4 mH MG 2 r The forces are equal at the Eddington limit mH MG T L 4cmH GM L 2 2 T r 4r c M LE 1.3 10 M Sun 38 erg s-1 So if L > LE material is expelled M max 20 M.R. Burleigh 2601/Unit 4 M Sun 2 Stellar Structure Hydrostatic equilibrium Equations of state Energy transport (not derived) Energy sources Stellar models Mass-Luminosity relation Eddington Limit M.R. Burleigh 2601/Unit 4