dr + Gm dr Gm2 2πr 5 = Gm2 2πr 5 < 0 (2) Q(r 0) P c, Q(r R) GM 2 /8πR 4. M and R are total mass and radius. Central pressure P c (Milne inequality):

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1 1 Stellar Structure Hydrostatic Equilibrium Spherically symmetric Newtonian equation of hydrostatics: dp/dr = Gmρ/r 2, dm/dr = 4πρr 2. 1) mr) is mass enclosed within radius r. Conditions at stellar centers Q = P + Gm 2 /8πr 4 : dq dr = dp dr + Gm 4πr 4 dm dr Gm2 2πr 5 = Gm2 2πr 5 < 0 2) Qr 0) P c, Qr R) GM 2 /8πR 4. M and R are total mass and radius. Central pressure P c Milne inequality): ) P c > GM2 M 2 ) 4 R 8πR 4 = dynes cm 2. 3) R Average density is ρ = 3M 4πR M M M Estimate of T c from perfect gas law: T c P cµ > M ρn o ) ) 3 R g cm 3. 4) R M ) R R ) K. 5) µ is mean molecular weight. T c too low by factor of 7. Better Estimate: ρ = ρ c [1 r/r) 2], M = 8π/15)ρ c R 3. P = P c 4π [ 3 Gρ2 cr 2 1 r ) 2 2 r ) 4 1 r ) ] ) 2 R 5 R 10 R

2 2 PR) = 0: ) P c = 15GM2 M 2 ) 4 R = dynes cm 2, 7) 16πR4 M R ] r 2 2 [ P = P c [1 1 R) 1 ] r 2 = 2 R) P ) c ρ 2 ) 1 + ρρc. 2 ρ c 8) The central density is ρ c = 15M ) ) 5 M 3 R = 3.6 g cm 8πR3 2 ρ 3, 9) M R and the central temperature becomes T c P ) ) cµ M R K. 10) ρ c N o M R Mean molecular weight: Perfect ionized gas k B = 1) P = T 1 + Z i ) n i ρn o T/µ NT, 11) i Z i is charge of ith isotope. Abundance by mass of H, He and everything else denoted by X, Y, and Z = i>he n ia i /ρn o ). Assuming 1 + Z i A i /2 for i >He: µ = 2X Y + 1 n i 1 + Z i ) 4 ρn o 2 + 6X + Y = X Z. i>he 12) Solar gas X = 0.75, Y = 0.22, Z = 0.03) has µ 0.6. Number of electrons per baryon for completely ionized gas Z i>he A i /2: Y e = X + Y 2 + i>he n i Z i X + Y ρn o 2 + Z 2 = X). 13) 2

3 3 The Virial Theorem Position, momentum, mass of ith particle: r i, p i, m i. Newton s Law F i = p i with p i = m i ri : d pi r dt i = pi r i + p i r ) i = d dt Moment of inertia: I = m i r i 2. Static situation: d 2 I/dt 2 = 0. Non-relativistic gas: 2 m i r i = p i r i = 2K. Total kinetic energy: pairs pairs mi ri r i = 1 2d 2 I d 2 t, 14) K = 1 pi 2 r i = 1 pi r 2 i = 1 Fi r 2 i = 1/2)Ω. 15) Sum is virial of Clausius. For perfect gas, only gravitational forces contribute, since forces involved in collisions cancel. F G i r i = F ij G r ) Gm i m j i r j = Ω. 16) r ij Ω is gravitational potential energy, r ij = r i r j. Perfect gas with constant ratio of specific heats, γ = c p /c v : K = 3/2)NT, U = γ 1) 1 NT, E = U+Ω = U 2K. U is internal energy, E is total energy. E = U + Ω = U 4 3γ) = Ω 3γ 4 3 γ 1). 17) For γ = 4/3, E = 0. γ < 4/3, E > 0, configuration unstable. γ > 4/3, E < 0, configuration stable and bound by energy E. Application: contraction of self-gravitating mass Ω < 0. If γ > 4/3, E < 0, so energy is radiated. However, U > 0, so star grows hotter.

4 4 Relativistic gas: p i r i = c p i = K = Ω. Another derivation: where V = 4πr 3 /3. Its integral is V dp = 1 Gm 3 r dm = 1 dω, 18) 3 V r)dp = PV R 0 P r)dv = 1 3 Ω 19) from Eq. 18). Thus Ω = 3 Pr)dV. Non-relativistic case: P = 2ǫ/3, Ω = 2K, E = Ω + K = Ω/2. Relativistic case similar to non-relativistic case with γ = 4/3: P = ǫ/3, Ω = K, E = 0. The critical nature of γ = 4/3 is important in stellar evolution. Regions of a star which, through ionization or pair production, maintain γ < 4/3 will be unstable, and will lead to instabilities or oscillations. Entire stars can become unstable if the average adiabatic index drops close to 4/3, and this actually sets an upper limit to the masses of stars. As we will see, the proportion of pressure contributed by radiation is a steeply increasing function of mass, and radiation has an effective γ of 4/3. We now turn our attention to obtaining more accurate estimates of the conditions inside stars.

5 5 Polytropic Equations of State The polytropic equation of state, common in nature, satisfies P = Kρ n+1)/n Kρ γ, 20) n is the polytropic index and γ is the polytropic exponent. 1) Non-degenerate gas nuclei + electrons) and radiation pressure. If β = P gas /P total is fixed throughout a star P = N o µβ T = [ ] 1/3 3No 1 β) ρ 4/3 21a) µβa ] 1/3 ρ 1/3 21b) [ 3No 1 β) µβa Here µ and a are the mean molecular weight of the gas and the radiation constant, respectively. Thus n = 3. 2) A star in convective equilibrium. Entropy is constant. If radiation pressure is ignored, then n = 3/2: s = 5 2 ln P = h2 2m h 2 2mT ) 5/3 ρno exp µ ) 3/2 ρn o /µ 2 3 s 5 ) 3 = constant 22a) = Kρ 5/3. 22b) 3) An isothermal, non-degenerate perfect gas, with pairs, radiation, and electrostatic interactions negligible: n =. Could apply to a dense molecular cloud core in initial collapse and star formation. 4) An incompressible fluid: n = 0. This case can be roughly applicable to neutron stars. 5) Non-relativistic degenerate fermions: n = 3/2. Lowdensity white dwarfs, cores of evolved stars.

6 6 6) Relativistic degenerate fermions: n = 3. High-density white dwarfs. 7) Cold matter at very low densities, below 1 g cm 3, with Coulomb interactions resulting in a pressure-density law of the form P ρ 10/3, i.e., n = 3/7. Don t confuse polytropic with adiabatic indices. A polytropic change has c = dq/dt is constant, where dq = TdS. An adiabatic change is a specific case: c = 0. γ = ln P ) ln V = γc v c cp c p c c v ), where the adiabatic exponent γ = ln P/ ln V ) s. If γ = c p /c v, as for a perfect gas, γ = c c p )/c c v ). In the adiabatic case, c = 0 and γ = γ regardless of γ s value. Polytropes Self-gravitating fluid with a polytropic equation of state is a polytrope, with Gm r) dm r) Ω = r = 3 GM 2 5 n R = 3 For a perfect gas with constant specific heats, E = 3γ 4 γ 1 For the adiabatic case n = 1/γ 1), PdV. 23) 1 GM 2 5 n R. 24) E = n 3 GM 2 5 n R. 25) For a mixture of a perfect gas and radiation, [ ] β U = β) PdV = β 4 3γ PdV Ω. 26) γ 1 γ 1

7 7 For β = constant, Eq. 26) gives β times the result found in Eq. 24). A bound star has E < 0 and γ > 4/3. If γ = 5/3, E = 3β/7)GM 2 /R). A nested polytrope has P = Kρ 1+1/n ; ǫ = np ρ < ρ t P = Kρ 1/n 1/n 1 t ρ 1+1/n 1 ; ǫ = n 1 P + n n 1 ) P t ρ > ρ t. ρ t and P t are the transition density and pressure between indices n and n 1. ǫ is the energy density E = n 3 5 n [ Gm 2 R GM2 t R t + 3P t [ Mt ρ t 4π 3 R3 t ] + n 1 3GMt 2 5 n 1 Rt 2 ] [ n 1 5 n n n 1 ]. 27) M t and R t are mass and radius interior to transition point. When n 1 0) and n 3, E = 3 5GM 2 t R t. 28) This could apply to a proto-neutron star with relativistic electron gas up to ρ t, and relatively stiff matter beyond. The energy depends on inner core size only. Structure of polytropes and Lane-Emden equation: r = Aξ, θ = ρ ρ c ) 1/n, A = 1 ξ 2 d dξ ξ 2dθ ) dξ [ n + 1) Kρ 1/n 1 c / 4πG)] 1/2. 29) = θ n. 30) Boundary conditions for are θ=1 and θ = dθ/dξ = 0 at ξ=0. The radius is found from ξ 1 where θξ) = 0.

8 8 n γ θ ξ 1 ξ1 2θ 1 ξ 1/3θ 1 [4πn + 1)θ ξ 2 / /8π 1 2 sinξ)/ξ π π π 2 /3 π/8 3/2 5/ / / / / /5 1/ 1 + ξ 2 /3 3 Analytic solutions exist in the following cases: θ = 1 ξ 2 /6; ξ 1 = 6 n = 0, γ = ; 31a) θ = sinξ/ξ; ξ 1 = π n = 1, γ = 2; 31b) θ = 1/ 1 + ξ 2 /3; ξ 1 = n = 5, γ = c) Radius : R = Aξ 1 32a) Mass : M = 4πA 3 ρ c ξ1 2 θ 1 32b) Density ratio : ρ/ρ c = 3θ 1 /ξ 1 32c) [ Central pressure : P c = GM/ 4π n + 1)θ d) ) n 1 R K = G ) 1/n M 3 n 4π. 33) n + 1 ξ 2 1 θ 1 For n, we have the isothermal Lane-Emden equation: ) 1 d ξ 2 ξ 2dφ = e φ = ρ. 34) dξ dξ ρ c ρ = K/2πGr 2 ; R = ; m = 2Kr/G n =, γ = 1. 35) ξ 1 1 ] 1

9 9 For n = 3 mass does not depend on central density, but only on equation of state. For a relativistic degenerate electron gas, which implies a mass M ch = 4π P = hc 4 3π 2) 1/3 nye ) 4/3, 36) ) K 3/ = 5.76Ye 2 M. 37) G This is the famous Chandrasekhar mass, the limiting mass of a white dwarf as ρ c. A degenerate mass larger than M ch cannot be stabilized by electron pressure alone. For T 0, the pressure has a small thermal component P th = T 2 ) 2/3 3π 2 ρy e. 38) 8 hc For a massive stellar core just prior to collapse, T 0.7 MeV and ρ g cm 3, and the P th /P 0.12, and the effective M ch is 1.12) 3/2 = 1.19 times larger. The negative Coulomb lattice pressure, which is about 4% of the total, lowers this. At densities in excess of 10 6 g cm 3, electron capture decreases Y e. For a 56 Fe white dwarf, the zero-temperature Chandrasekhar mass is only 1.17 M. As the cores of massive stars evolve, there is a general tendency for core convergence to occur, i.e., the evolved cores of all massive stars, regardless of mass, tend to be nearly M ch. We see that this is a result of the general requirement for stability. In fact, there is a slight trend for more massive stars to have larger cores, but this can be traced to the higher entropies in these stars recall that Eq. 5) predicts that T M/R) and their larger effective Chandrasekhar masses.

10 10 Standard Model Stars The Main Sequence Those stars converting H to He. Standard model assumes β = constant. ) 4/3 No 3 K = µβ a [ ] 1 β 1/6 R = µ 4 β 4 ρ 2 R c 1 β M = 18 µ 2 β 2 M ρ c / ρ = 54.2 ) 1/3 1 β). 39) 40) T c = 1.96 βµmr M R 107 K. For n = 3 polytrope, mass is independent of ρ c and for given composition µ, is parametrized by β. For Sun, with M = 1M and µ 0.6: µ 2) 2 β 1 = ; ρ c = = 76.7 g cm 3, 41) T c = K. But µ c > µ since some H He has occurred. Luminosity will depend upon nuclear energy generation ǫ and transport opacity κ). For T > ρ 1/3.5 electron scattering dominates: κ 0.4Y e cm 2 g 1. 42) Where T > 10 4 K, κ dominated by bound-bound and boundfree processes: κ ZY e ρt 3.5 cm 2 g 1. 43)

11 11 For Z < 10 4 have free-free opacity: κ Z)Y e ρt 3.5 cm 2 g 1. The dependence κ ρt 3.5 is known as Kramer s opacity. For T < 10 4 K, matter barely ionized: Energy Transport: κ Z/.02)ρT 10 cm 2 g 1. 44) L r) = 4πr 2 4ac 3κ R ρ T 3dT dr. 45) Lr) is luminosity, κ R is the Rosseland mean opacity, averaged over frequencies. κρ) 1 is photon mean free path, dact 4 /4)/dr is radiation energy density gradient. Multiplied together gives energy flux, and multiplied by area 4πr 2 gives net energy flow. Use hydrostatic equilibrium: dp r dp = κl r) 4πcGm r). 46) Luminosity function ηr) = Lr)M/mr)L, with M and L totals. η is sharply peaked at origin. L d [1 β)p] = κ r)η r) dp. 4πcGM 47) L = 4πcGM 1 β c ) κη 48) κη is a pressure average. ssm: κη = cons.) With Eq. 43) L ssm = 4π) 3 4ac 3κ o η c ) µc βg 7.5 M 5.5 4N o ξ1 ξ 2 1 θ 1 ) 4.5 R ) µ cβ c ) 7.5 η c ZY e,c M M ) 5.5 ) 0.5 R L. 49) R

12 12 With Eq. 42), appropriate instead for more massive stars, 1 L 97.5 βµ c ) 4 M η c Y e,c M ) 3 L. 50) For Sun: µ c = 0.73, Y e,c = 0.75 i.e., X 0.5 and Y 0.5), L 2.8/η c ) L. Alternatively, use proton-proton rate T 6 = T/10 6 ): η = ρx 2 T 2/3 6 exp 33.8T 1/3 6 ) M L. 51) With ρ c = 76.7 g cm 3, T c,6 = from ssm η c =4.05. Try using knowledge of polytropic structure. Assume ideal gas and ǫ = ǫ c ρ ρ c ) λ T T c ) ν. 52) For polytrope, ρ = ρ c θ n and T = T c θ: L = 4πA 3 ρ c ǫ c ξ 2 θ 2nλ+ν dξ 4πA 3 ρ c ǫ c 27π 2 2nλ + ν) 3/2, since 2nλ + ν >> 1. With θ exp ξ 2 /6), n = 3, 53) η c = ǫ c M L = ξ 2 θ ) π 2nλ + ν)3/2 = λ + ν) 3/2, 54) P-p cycle has λ = 1, ν 4, so η c 9.8. CNO cycle has λ = 1, ν 20, so η c 41.

13 13 Scaling Relations for Standard Solar Model For Kramer s opacity, κ Z1 + X)ρT 3.5. For electron scattering, κ 1 + X). Suggests with u = 0, 1. Similarly κ 1 + X)Z u ρ n T s, 55) ǫ X 2 m Z m ρ λ T ν, 56) with m = 01) for p-p CNO) cycle. Using L RT 4 /κρ M ǫ, T µβm/r, ρ M/R 3, 57) we find T eff L M α M µβ) α µ X α X 1 + X) α 1 Z α Z, R M β M µβ) βµ X β X 1 + X) β 1 Z β Z, L/R 2) 1/4 M γ M µβ) γ µ X γ X 1 + X) γ 1 Z γ Z, L T δ T eff µβ)δµ X δ X 1 + X) δ 1 Z δ Z. 58) With i = M, µ, X, 1 + X), Z, T), γ i = α i /4 β i /2 δ i = 2α M β i α i β M )/α M 2β M ), D = ν s + 3n + λ): i M µ T α i D ν3 + 2n) + 9λ + 3n + s2λ 1) 7ν + 3λ4 + s) 0 β i D λ + ν + n s 2 ν 4 s 2D i X 1 Z α i D m3n s) u3λ + s) s + 3λ) 3n s)2 m) β i D u + m 1 2 m

14 14 α M α µ α X α 1 α Z α T low 71/13 101/13-2/13-14/13-16/13 0 high β M β µ β X β 1 β Z β T low 1/13-7/13 4/13 2/13 2/13 2 high 19/23 16/23 1/23 1/23 1/23 2 γ M γ µ γ X γ 1 γ Z γ T low 69/52 87/52-3/26-9/26-7/13-2 high 31/92 15/23-1/46-25/92-1/46-2 δ M δ µ δ X δ 1 δ Z δ T low 0-4/3 44/69 8/23 68/69 284/69 high 0-56/31 6/31 44/31 6/31 276/31 Values refer to low-mass ν = 4, λ = 1, m = 0, u = 1, n = 1, s = 3.5) or high-mass ν = 20, λ = 1, m = 1, u = 0, n = 0, s = 0) M-S stars. 1) As H consumed, X decreases and µ increases µ 4/ 5X + 3) and β is nearly constant. So L increases and T eff increases; stars evolve up the main sequence. This explains why in globular clusters the M-S turnoff luminosity >> L even though M M. Also, the early Sun was less luminous, and cooler, than present. If initial present) X = ), L today L initial 1.4, T eff,today T eff,initial 1.11, T c,today T c,initial 1.09, R today R initial 0.96.

15 15 2) Stars on the p-p cycle ν = 4) have R nearly independent of M: R M 1/13 for Kramer s opacity. For stars on the CNO cycle, however, R M 11/15 for Kramer s opacity and R M 19/23 for electron scattering opacity. 3) Population II stars are characterized by low metal compositions, Z < For a given T eff, L Z δ Z dominates the composition dependence. The Population II M-S is shifted to lower L than the Population I M-S. Also, for a given M, T eff Z γ Z implies a shift of the M-S to higher T eff. 4) For a given M, L Z α Z, which is larger for Population II than for Population I stars. Stellar lifetimes τ M/L are nearly Z since κ Z1+X). Thus, lifetimes of Population II stars are substantially less than Population I for a given mass. This is observed in H-R diagrams of globular clusters.

16 16 1. Idealized Stars Radiative Zero Solution Besides the standard model, we could consider an idealized star in which the energy generation is uniform throughout, i.e., ηr) = 1, and the equation of state is that of an ideal gas alone. If such a star is in radiative equilibrium, we can write d lnt d lnp = 3LκP 16πacGMT 4 = 3κ ol 16πacGM µ where the opacity is assumed to scale as N o ) m P m+1 1.1) T 4+t+m κ = κ o ρ n T s. 1.2) The radiative zero solution is obtained if d lnt/d lnp is constant. Eq. 1.1) then implies that d lnt/d lnp = n + 1)/n + s + 4) and P T 4+s+n)/n+1) ; P ρ 4+s+n)/s+3). 1.3) For a Kramer s opacity law n = 1, s = 3.5) we find the effective polytropic index to be 3.25, and, from Eq. 1.1), L = 4π) 3 4ac ) 4µG 7.5 M 5.5 ) 0.5 ξ1 3κ o 17N o ξ 2 1 θ 1 ) 4.5 R ) µ 7.5 ) 1.4) Ye,c 0.8η c L ssm. µ c Y e Note that ξ 1 and θ 1 are evaluated for the n = 3.25 polytrope, and not the n = 3 polytrope as for the standard model). Had we used the Thomsen opacity n = s = 0) instead, we would have just found Eq. 50) with β c =1.

17 17 Completely Convective Stars To conclude this section, we now consider the idealized completely convective star. This case is especially relevant to the pre-main sequence phase of stellar evolution. For a perfect gas, an n = 3/2 polytrope must result for constant entropy. We immediately find ρ c / ρ = 6, T c = 1.2 µmr M R 107 K, P c = 8.7 M2 R 4 M 2 R4erg cm 3. It is also clear that, dimensionally cf. Eq. 32)) M K 3/2 ρ 1/2 c R K 1/2 ρ 1/6 E = ) c KM 1/3 e 2s/3 for fixed M) 1.6) GM 2 R M7/3 K 1 e 2s/3 for fixed M) where K is given by Eq. 22). In the last two equations, s is the entropy per baryon, not the temperature dependence of the energy generation rate. It is straightforward to show that both the heat flux and luminosity vary as the 3/2 power of the difference of the actual temperature gradient from the purely adiabatic one e.g., p. 257). Typically, near the outside of a star, this difference is only 10 6 of the temperature gradient itself. Therefore it is impossible to determine the luminosity from the transport equation as we did in the radiative case. But because radiation eventually escapes, the transport must become radiative just below the surface. Using the photospheric

18 18 condition for the optical depth τ = κρdr 2/3 one may determine the surface temperature and hence the luminosity. Hydrostatic equilibrium can be rewritten as dp/dτ = g/κ 1.7) where g = GM/r 2 GM/R 2, the surface gravity, is nearly constant throughout the thin surface region. As a zeroth approximation, we may write g p P p 2 = 2 3κ p 3 GMR 2 p κ 1 o ρ n p T s p 1.8) where the subscript p indicates photospheric values. Only if κ varies rapidly in the surface region will this result be inaccurate. Combining Eqs. 1.8) and 33), using values for a n = 3/2 polytrope, and employing the perfect gas law, we can find the photospheric temperature: ) 1/Q 3n+5 K 3+3n T p = ) 2 2GM µ 3κ o Rp 2 N o M 3+n R 3n 1 p ) 1/Q, 1.9) where Q = 5 + 3n 2s. The luminosity follows immediately from L = 4πRpσT 2 p: 4 [ L =σ 4π) n 5 3κo 5 [ =σ 4π) 13+11n 2s 5 3κ o ) 4 ) 10+6n 5No ξ 5 2Gµ 1 θ 1 ) 8 2µG ] ) 2+2n T 6+18n 4s 1/3n 1) p M 6+2n ) 20+12n M 12+4n R 6+18n 4s p ξ 5 θ 1) 4+4n ] 1/Q. 5N o 1.10) This relation shows the tremendous sensitivity of the luminosity to the photospheric temperature: typically in the low density surface regions, s 10.

19 19 In general, a star is convective if its luminosity is large enough to force a superadiabatic temperature gradient. Thus, there must exist a minimum luminosity below which a star cannot be completely convective. A star in convective equilibrium has d lnt/d lnp = 2/5n = 3/2), so from Eq. 33), P c = ) No T 5/2 c K 3/2 ; T c = 2 µ 5ξ 1 θ 1 µgm N o R. 1.11) On the other hand, a star in radiative equilibrium, from Eq. 1.4), satisfies, at the center, d lnt d lnp = 3 κ c P c Lη c 16πacG Tc 4M. 1.12) If the logarithmic temperature gradient at the center falls below 2/5, the star will cease to be completely convective. Therefore from the previous two equations, we find L min = 32πacGM Tc 4 15η c κ c P c = 4ac ) 4π) n+2 2Gµ 4+s ξ1 s 3n M s n+3 3η c κ o 5N o ξ 2 1 θ 1 ) 2+s n R s 3n = 271 µ7.5 M/M ) 5.5 η c R/R ).5 L 1.13) where the last equality holds for Kramer s opacity. This can be compared with the luminosity from the standard solar model for 1 M and 1 R ), which behaves on the physical variables

20 20 in a similar way: ) ) µ 7.5 ) Ye,ssm L min /L ssm = 5β µ ssm Y e 4.5 ξ2 1,3 θ ) 1,3 ξ1,3/2 R 0.5 ssm ηssm. ξ 1,3 R η c ξ 2 1,3/2 θ 1,3/2 The numerical coefficient is equal to We expect that η ssm /η c 2, and µ/µ ssm 0.82, so with R 3R we find that L min 1.6L ssm for a solar-type star. This is larger than the actual minimum luminosity reached along the Hayashi track, but the star overshoots this minimum luminosity as it gradually becomes more and more radiative. We will further explore pre-main sequence stars in the next chapter.