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 = . (31c) 1 ∞ ′ 5 q Radius : R = Aξ1 (32a) 3 2 Mass : M = 4πA ρcξ θ (32b) − 1 1′ Density ratio :ρ/ρ ¯ c = 3θ /ξ (32c) − 1′ 1 2 4 Central pressure : Pc = GM/ 4π (n + 1) θ1′ R (32d) h 1/n i n 1 3 n G M − R − K = 4π 2 . (33) n +1  ξ θ ξ1  − 1 1′ !   For n , we have the isothermal Lane-Emden equation: → ∞ 1 d dφ ρ ξ2 = e φ = . (34) ξ2 dξ dξ − ρ   c ρ = K/2πGr2; R = ; m =2Kr/G n = , γ =1. (35) ∞ ∞ ′ 9

For n = 3 mass does not depend on central density, but only on equation of state. For a relativistic degenerate electron gas, ¯hc 1/3 P = 3π2 (nY )4/3 , (36) 4 e   which implies a mass

3/2 K 2 Mch = 4π 2.018 = 5.76Ye M . (37) − G × ⊙   This is the famous Chandrasekhar mass, the limiting mass of a white dwarf as ρc . A degenerate mass larger than Mch cannot be stabilized→ by ∞ electron pressure alone. For T = 0, the pressure has a small thermal component 6

T 2 2/3 P = 3π2ρY . (38) th 8¯hc e   For a massive stellar core just prior to collapse, T 0.7 MeV 9 3 ≃ and ρ 6 10 g cm− , and the Pth/P 0.12, and the ≃ × 3/2 ≃ effective Mch is (1.12) = 1.19 times larger. The negative Coulomb lattice pressure, which is about 4% of the total, lowers 6 3 this. At densities in excess of 10 g cm− , electron capture 56 decreases Ye. For a Fe white dwarf, the zero-temperature Chandrasekhar mass is only 1.17 M . As the cores of massive stars evolve,⊙ there is a general ten- dency for “core convergence” to occur, i.e., the evolved cores of all massive stars, regardless of mass, tend to be nearly Mch. We see that this is a result of the general requirement for sta- bility. 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

Standard Model Stars – The Main Sequence Those stars converting H to He. Standard model assumes β = constant. N 4/3 3 1/3 K = o (1 β) . (39) µβ a −     1 β 1/6 R = 11.18 − R µ4β4ρ2 ⊙  c  √1 β M = 18 − M µ2β2 ⊙ (40) ρc/ρ¯ = 54.2 βµMR 7 Tc =1.96 ⊙ 10 K. M R × ⊙ For n = 3 polytrope, mass is independent of ρc and for given composition µ, is parametrized by β. For Sun, with M = 1M and µ 0.6: ⊙ ≃ 2 2 3 µ 11.18 3 β 1 =0.9996; ρc = = 76.7 g cm , (41) ≃ − 18 18 −   7 Tc =1.307 10 K. × But µc > µ¯ since some H He has occurred. Luminosity will depend→ upon nuclear energy generationǫ ˙ and transport (opacity κ). For T > 8 106ρ1/3.5 electron scattering dominates: · 2 1 κ 0.4Ye cm g . (42) ≃ − Where T > 104 K, κ dominated by bound-bound and bound- free processes: 25 3.5 2 1 κ 2.5 10 ZYeρT cm g . (43) ≃ · − − 11

4 For Z < 10− have free-free opacity: 22 3.5 2 1 κ 8 10 (1 Z) YeρT cm g . ≃ · − − − 3.5 The dependence κ ρT − is known as Kramer’s opacity. For T < 104K, matter∝ barely ionized: κ 10 32 (Z/.02) ρT 10 cm2 g 1. (44) ≃ − − Energy Transport: 4ac dT L (r)= 4πr2 T 3 . (45) − 3κRρ dr L(r) is luminosity, κR is the “Rosseland mean” opacity, av- 1 eraged over frequencies. (κρ)− is photon mean free path, d(acT 4/4)/dr is radiation energy density gradient. Multiplied together gives energy flux, and multiplied by area 4πr2 gives net energy flow. Use hydrostatic equilibrium: dP κL (r) r = . (46) dP 4πcGm (r) Luminosity function η(r)= L(r)M/m(r)L, with M and L to- tals. η is sharply peaked at origin. L d [(1 β) P ]= κ (r) η (r) dP. (47) − 4πcGM 4πcGM L = (1 βc) (48) κη¯ − κη¯ is a pressure average. (ssm: κη = cons.) With Eq. (43) 7.5 5.5 0.5 3 4ac µcβG M ξ1 Lssm = (4π) 4.5 3κoηc 4No ξ2θ R   − 1 1′   7.5 5.5 0.5 (µcβc) M R
.667 ⊙ L . (49) ≃ ηcZYe,c M R ⊙  ⊙   12

With Eq. (42), appropriate instead for more massive stars,

3 1 4 M L 97.5 (βµc) L . (50) ≃ ηcYe,c M ⊙  ⊙ For Sun: µc =0.73, Ye,c =0.75 (i.e., X 0.5 and Y 0.5), ≃ ≃

L (2.8/ηc) L . ≃ ⊙ 6 Alternatively, use proton-proton rate (T6 = T/10 ):

6 2 2/3 1/3 M η =2.0 10 ρX T − exp 33.8T − ⊙. (51) × 6 − 6 L   ⊙ 3 With ρc = 76.7 g cm− ,Tc,6 = 13.07 from ssm ηc=4.05. Try using knowledge of polytropic structure. Assume ideal gas and

ρ λ T ν ǫ˙ =ǫ ˙ . (52) c ρ T  c  c n For polytrope, ρ = ρcθ and T = Tcθ:

3 2 2nλ+ν 3 27π 3/2 L =4πA ρcǫ˙c ξ θ dξ 4πA ρcǫ˙c (2nλ + ν)− , ≃ r 2 Z (53) since 2nλ + ν >> 1. With θ exp( ξ2/6), n = 3, ≃ −

M 2 2 3/2 3/2 ηc =ǫ ˙c = ξ θ′ (2nλ + ν) =0.31(6λ + ν) , L − 1 27π   r (54) P-p cycle has λ =1,ν 4, so ηc 9.8. ≃ ≃ CNO cycle has λ =1,ν 20, so ηc 41. ≃ ≃ 13

Scaling Relations for Standard Solar Model 3.5 For Kramer’s opacity, κ Z(1 + X)ρT − . For electron scattering, κ (1 + X). Suggests∝ ∝ κ (1 + X) ZuρnT s, (55) ∝ − with u =0, 1. Similarly

ǫ˙ X2 mZmρλT ν, (56) ∝ − with m = 0(1) for p-p (CNO) cycle. Using

L RT 4/κρ Mǫ,˙ T µβM/R, ρ M/R3, (57) ∝ ∝ ∝ ∝ we find L M αM (µβ)αµ XαX (1 + X)α1 ZαZ , ∝ R M βM (µβ)βµ XβX (1 + X)β1 ZβZ , ∝ 1/4 T L/R2 M γM (µβ)γµ XγX (1 + X)γ1 ZγZ , eff ∝ ∝   L T δT (µβ)δµ XδX (1 + X)δ1 ZδZ . ∝ eff (58) With i =(M,µ,X, (1 + X),Z,T ), γ = α /4 β /2 i i − i δ = 2(α β α β )/(α 2β ), D = ν s + 3(n + λ): i M i − i M M − M − i M µ T → α D ν(3+2n)+9λ +3n + s(2λ 1) 7ν +3λ(4 + s) 0 i − β D λ + ν + n s 2 ν 4 s 2D i − − − − i X 1 Z → α D m(3n s) u(3λ + s) (s +3λ) (3n s)(2 m) i − − − − − β D u + m 1 2 m i − 14

αM αµ αX α1 αZ αT low 71/13 101/13 -2/13 -14/13 -16/13 0 high 3 4 0 -1 0 0

βM βµ βX β1 βZ βT low 1/13 -7/13 4/13 2/13 2/13 2 high 19/23 16/23 1/23 1/23 1/23 2

γM γµ γX γ1 γZ γT low 69/52 87/52 -3/26 -9/26 -7/13 -2 high 31/92 15/23 -1/46 -25/92 -1/46 -2

δM δµ δX δ1 δZ δT 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 Teff increases; stars evolve up the main sequence. This explains why in glob- ular 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 =0.75(0.7),

Ltoday Teff,today 1.4, 1.11, Linitial ≃ Teff,initial ≃ Tc,today Rtoday 1.09, 0.96. Tc,initial ≃ Rinitial ≃ 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 composi- δ tions, Z < 0.001. For a given Teff ,L Z 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 ZγZ implies a shift of the M-S to higher T . eff ∝ eff 4) For a given M, L ZαZ , which is larger for Population II than for Population∝ I stars. Stellar lifetimes τ M/L are nearly Z since κ Z(1+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

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., η(r) = 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 ln T 3LκP 3κ L µ m P m+1 = = o (1.1) d ln P 16πacGMT 4 16πacGM N T 4+t+m  o where the opacity is assumed to scale as n s κ = κoρ T − . (1.2) The radiative zero solution is obtained if d ln T/d ln P is con- stant. Eq. (1.1) then implies that d ln T/d ln P =(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), 7.5 5.5 0.5 3 4ac 4µG M ξ1 L = (4π) 4.5 3κo 17No ξ2θ R   1 1′   (1.4) 7.5 − µ Ye,c
0.8ηc Lssm. ≃ µ Y  c  e  (Note that ξ1 and θ1′ 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 βc=1. 17

Completely Convective Stars To conclude this section, we now consider the idealized com- pletely 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 im- mediately find

ρc/ρ¯ =6, µMR 7 Tc =1.2 ⊙ 10 K, M R × (1.5) 2⊙ 4 M R 3 Pc =8.7 ⊙erg cm− . M 2 R4 ⊙ It is also clear that, dimensionally (cf . Eq. (32)) 1/2 M K3/2ρ ∝ c 1/2 1/6 1/3 2s/3 R K ρ− KM e (for fixed M) (1.6) ∝ c ∝ − ∝ 3 GM 2 E = M 7/3K 1 e 2s/3 (for fixed M) 14 R ∝ − ∝ − 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 ac- tual temperature gradient from the purely adiabatic one (e.g., Ref. @Ref.Clayton@, p. 257). Typically, near the outside of 6 a star, this difference is only 10− 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 be- come radiative just below the surface. Using the photospheric 18

condition for the optical depth τ = κρdr 2/3 one may determine the surface temperature and hence≃ the luminosity. Hydrostatic equilibrium can be rewrittenR as dP/dτ = g/κ (1.7) − where g = GM/r2 GM/R2, the surface gravity, is nearly constant throughout≃ the thin surface region. As a zeroth ap- proximation, we may write 2 gp 2 2 1 n s Pp = GMRp− κo− ρp− Tp (1.8) ≃ 3 κp 3 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: 2 1/Q 3n+5 1/Q 2GM µ 3+3n 3+n 3n 1 Tp = 2 K M Rp − ,  3κoRp! No  ∝       (1.9) where Q =5+3n 2s. The luminosity follows immediately 2 4− from L =4πRpσTp : 1/(3n 1) 4 10+6n 2+2n 6+18n 4s − n 5 3κo 5No 5 Tp − L =σ (4π)− − ξ1θ1′ 6+2n " 5 2Gµ − M #       8 20+12n 12+4n 6+18n 4s 1/Q 13+11n 2s 5 2µG M Rp − =σ (4π) − . 3κo 5No 5 4+4n "     ξ θ1′ # − (1.10) This relation shows the tremendous sensitivity of the
luminos- ity to the photospheric temperature: typically in the low den- sity surface regions, s 10. ≈− 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 can- not be completely convective. A star in convective equilibrium has d ln T/d ln P =2/5(n =3/2), so from Eq. (33),

5/2 NoTc 3/2 2 µGM Pc = K ; Tc = − . (1.11) µ − 5ξ θ N R   1 1′ o

On the other hand, a star in radiative equilibrium, from Eq. (1.4), satisfies, at the center,

d ln T 3 κcPcLηc = 4 . (1.12) d ln P 16πacG Tc M

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

4 32πacGM Tc Lmin = 15ηc κcPc 4+s s 3n s n+3 4ac n+2 2Gµ ξ1− M − = (4π) 2+s n s 3n 3ηcκo 5No ξ2θ − R −   − 1 1′ 7.5 5.5 µ (M/M )
= 271 ⊙.5 L ηc (R/R ) ⊙ ⊙ (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 in a similar way: 7.5 7.5 8 µ Ye,ssm Lmin/Lssm = 5β µssm Ye ×    4.5   2 0.5 ξ1,3θ1′ ,3 ξ1,3/2Rssm η ssm. × ξ2 θ  ξ R η 1,3/2 1′ ,3/2  1,3  c The numerical coefficient is equal to 6.518. We expect that ηssm/ηc 2, and µ/µssm 0.82, so with R 3R we find that ≈ ≃ ≃ ⊙ Lmin 1.6Lssm 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.