1 Stellar Structure Hydrostatic Equilibrium Spherically symmetric Newtonian equation of hydrostatics: dP/dr = Gmρ/r2, dm/dr =4πρr2. (1) − m(r) is mass enclosed within radius r. Conditions at stellar centers Q = P + Gm2/8πr4: dQ dP Gm dm Gm2 Gm2 = + = < 0 (2) dr dr 4πr4 dr − 2πr5 −2πr5 2 4 Q (r 0) Pc, Q (r R) GM /8πR . → → → → M and R are total mass and radius. Central pressure Pc (Milne inequality): 2 2 4 GM 14 M R 2 Pc > =4 10 ⊙ dynes cm . (3) 8πR4 × M R − ⊙ Average density is 3 3M M R 3 ρ¯ = 1.4 ⊙ g cm . (4) 4πR3 ≃ M R − ⊙ Estimate of Tc from perfect gas law: Pcµ 6 M R Tc > 2.1 10 ⊙ K. (5) ≃ ρN¯ o × M R ⊙ µ is mean molecular weight. Tc too low by factor of 7. Better Estimate: 2 3 ρ = ρc 1 (r/R) , M = (8π/15) ρcR . − h i 2 4 6 4π 2 2 1 r 2 r 1 r P = Pc Gρ R + . (6) − 3 c 2 R − 5 R 10 R 2 P (R)=0: 2 2 4 15GM 15 M R 2 Pc = =3.5 10 ⊙ dynes cm , (7) 16πR4 × M R − ⊙ 2 2 r 2 1 r 2 Pc ρ ρ P = Pc 1 1 = 1+ . − R − 2 R 2 ρc ρc (8) The central density is 3 15M 5 M R 3 ρc = = ρ¯ 3.6 ⊙ g cm , (9) 8πR3 2 ≃ M R − ⊙ and the central temperature becomes Pcµ 6 M R Tc 7.0 10 ⊙ K. (10) ≃ ρcNo ≃ × M R ⊙ Mean molecular weight: Perfect ionized gas (kB = 1) P = T (1 + Z ) n ρNoT/µ NT, (11) i i ≡ ≡ Xi Zi is charge of ith isotope. Abundance by mass of H, He and everything else denoted by X, Y , and Z = i>He niAi/(ρNo). Assuming 1 + Zi Ai/2 for i>He: ≃ 1 P 3 n (1 + Z ) − 4 4 µ = 2X + Y + i i = . 4 ρNo ≃ 2+6X + Y 3+5X Z i>HeX − (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 A /2: i>He ≃ i Y niZi Y Z 1 Ye = X + + X + + = (1 + X) . (13) 2 ρNo ≃ 2 2 2 i>HeX 3 The Virial Theorem Position, momentum, mass of ith particle: ~ri, ~pi, mi. ~ ˙ ˙ Newton’s Law Fi = ~pi with ~pi = mi~ri: d d 1 d2I ~p ~r = ~p˙ ~r + ~p ~r˙ = m ~r˙ ~r = , dt i · i i · i i · i dt i i · i 2 d2t X X X (14) 2 Moment of inertia: I = mi~ri . Static situation: d2I/dt2 = 0. P ˙2 ˙ Non-relativistic gas: mi~ri = ~pi ~ri =2K. Total kinetic energy: · P P 1 1 1 K = ~p ~r˙ = ~p˙ ~r = F~ ~r = (1/2)Ω. 2 i · i −2 i · i −2 i · i − X X X (15) Sum is virial of Clausius. For perfect gas, only gravitational forces contribute, since forces involved in collisions cancel. Gmimj F~ G ~r = F~ G ~r ~r = Ω. (16) i · i ij · i − j − r ≡ pairs pairs ij X X X Ω is gravitational potential energy, r = ~r ~r . ij | i − j| Perfect gas with constant ratio of specific heats, γ = cp/cv: 1 K = (3/2) NT, U =(γ 1)− NT, E = U+Ω = U 2K. − − U is internal energy, E is total energy. 3γ 4 E = U +Ω= U (4 3γ)=Ω − . (17) − 3(γ 1) − 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 Relativistic gas: ~p ~r˙ = c ~p = K = Ω. i · i i − Another derivation: P P 1 Gm 1 V dP = dm = dΩ, (18) −3 r −3 where V =4πr3/3. Its integral is R 1 V (r) dP = PV P (r) dV = Ω (19) 0 − −3 Z Z from Eq. (18). Thus Ω = 3 P (r)dV . − Non-relativistic case: R 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 evo- lution. 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 unsta- ble 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 effec- tive γ of 4/3. We now turn our attention to obtaining more accurate estimates of the conditions inside stars. 5 Polytropic Equations of State The polytropic equation of state, common in nature, satisfies P = Kρ(n+1)/n Kργ′, (20) ≡ n is the polytropic index and γ′ is the polytropic exponent. 1) Non-degenerate gas (nuclei + electrons) and radiation pressure. If β = Pgas/Ptotal is fixed throughout a star N 3N 1/3 P = o o (1 β) ρ4/3 (21a) µβ µβa − 3N 1/3 T = o (1 β) ρ1/3 (21b) 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: 3/2 5 ¯h2 s = ln ρNo/µ =constant (22a) 2 − 2mT ! ¯h2 ρN 5/3 2 5 P = o exp s = Kρ5/3. (22b) 2m µ 3 − 3 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 collap∞se 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. Low- density white dwarfs, cores of evolved stars. 6 6) Relativistic degenerate fermions: n = 3. High-density white dwarfs. 3 7) Cold matter at very low densities, below 1 g cm− , with Coulomb interactions resulting in a pressure-density law of the form P ρ10/3, i.e., n =3/7. Don’t∝ confuse polytropic with adiabatic indices. A poly- tropic change has c = dQ/dT is constant, where dQ = TdS. An adiabatic change is a specific case: c = 0. ∂ ln P cv c cp γ′ = = γ − , ∂ ln V cp (c cv) − where the adiabatic exponent γ = (∂ ln P/∂ ln V ) . If γ = |s cp/cv, as for a perfect gas, γ = (c cp)/(c cv). In the ′ − − adiabatic case, c = 0 and γ′ = γ regardless of γ’s value. Polytropes Self-gravitating fluid with a polytropic equation of state isa polytrope, with Gm (r) dm (r) 3 GM 2 Ω= = = 3 PdV. (23) − r −5 n R − Z − Z For a perfect gas with constant specific heats, 3γ 4 1 GM 2 E = − . (24) − γ 1 5 n R − − For the adiabatic case n =1/(γ 1), − n 3 GM 2 E = − . (25) 5 n R − For a mixture of a perfect gas and radiation, β 4 3γ U = +3(1 β) PdV = β − PdV Ω. (26) γ 1 − γ 1 − Z − Z − 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 1+1/n P = Kρ ; ǫ = nP ρ < ρt 1/n 1/n1 1+1/n P = Kρ − ρ 1; ǫ = n P +(n n ) Pt ρ > ρt. t 1 − 1 ρt and Pt are the transition density and pressure between in- dices n and n1. ǫ is the energy density n 3 Gm2 GM 2 n 3 GM 2 E = − t + 1 − t 5 n R − R 5 n R2 − " t # − 1 t (27) Mt 4π 3 n 1 n1 1 +3Pt Rt − − . ρt − 3 5 n − 5 n − − 1 Mt and Rt are mass and radius interior to transition point. When (n 0) and n 3, 1 ≃ ≃ 3 GM 2 E = t . (28) −5 Rt This could apply to a proto-neutron star with relativistic elec- tron gas up to ρt, and relatively stiff matter beyond. The energy depends on inner core size only. Structure of polytropes and Lane-Emden equation: 1/n ρ 1/n 1 1/2 r = Aξ, θ = , A = (n + 1) Kρc − / (4πG) . ρc h i(29) 1 d dθ ξ2 = θn. (30) ξ2 dξ dξ − Boundary conditions for are θ=1 and θ′ = dθ/dξ =0 at ξ=0. The radius is found from ξ1 where θ(ξ)=0. 8 n γ θ ξ ξ2θ ξ /3θ [4π(n + 1)θ 2] 1 ′ 1 − 1 1′ − 1 1′ 1′ − 0 1 ξ2/6 √6 2√6 1 3/8π ∞ − 1 2 sin(ξ)/ξ π π π2/3 π/8 3/2 5/3 3.654 2.714 5.992 0.7704 2 3/2 4.353 2.411 11.40 1.638 3 4/3 6.897 2.018 54.19 11.05 3.25 17/13 8.019 1.950 88.15 20.36 4 5/4 14.97 1.797 622.3 247.5 5 6/5 1/ 1+ ξ2/3 √3 ∞ ∞ ∞ Analytic solutionsp exist in the following cases: θ =1 ξ2/6; ξ = √6 n =0, γ = ; (31a) − 1 ′ ∞ θ = sin ξ/ξ; ξ1 = π n =1, γ′ =2; (31b) 6 θ =1/ 1+ ξ2/3; ξ = n =5, γ = .