1
Stellar Structure
Hydrostatic Equilibrium
Spherically symmetric Newtonian equation of hydrostatics:
dP/dr = −Gmρ/r 2,
dm/dr = 4πρr 2.
(1)
m(r) is mass enclosed within radius r.
Conditions at stellar centers Q = P + Gm2/8πr 4:
dQ dP
Gm dm Gm2
Gm2
=
+
−
=−
<0
dr
dr
4πr 4 dr
2πr 5
2πr 5
Q (r → 0) → Pc,
(2)
Q (r → R) → GM 2/8πR4.
M and R are total mass and radius.
Central pressure Pc (Milne inequality):
4
2 2
M
R⊙
GM
14
dynes cm−2.
=
4
×
10
Pc >
4
M⊙
R
8πR
(3)
Average density is
3M
M
R⊙ 3
ρ̄ =
≃ 1.4
g cm−3.
3
M⊙
R
4πR
Estimate of Tc from perfect gas law:
M
R
Pcµ
⊙
Tc ≃
> 2.1 × 106
K.
ρ̄No
M⊙
R
µ is mean molecular weight. Tc too low by factor of 7.
Better Estimate:
i
h
2
ρ = ρc 1 − (r/R) ,
M = (8π/15) ρcR3.
1 r 6
4π 2 2 1 r 2 2 r 4
.
−
+
P = Pc − Gρc R
3
2 R
5 R
10 R
(4)
(5)
(6)
2
P (R) = 0:
15GM 2
2 4
M
R⊙
15
=
3.5
×
10
dynes cm−2, (7)
4
M⊙
R
16πR
2 r 2 2
2
ρ
1 r
Pc ρ
P = Pc 1 −
1+
.
1−
=
R
2 R
2 ρc
ρc
(8)
The central density is
5
M
R⊙ 3
15M
= ρ̄ ≃ 3.6
g cm−3,
(9)
ρc =
3
2
M⊙
R
8πR
and the central temperature becomes
M
R
Pcµ
⊙
K.
(10)
≃ 7.0 × 106
Tc ≃
ρcNo
M⊙
R
Mean molecular weight:
Perfect ionized gas (kB = 1)
X
P =T
(1 + Zi) ni ≡ ρNoT /µ ≡ N T,
(11)
Pc =
i
Zi is charge of ith isotope. Abundance byP
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
X ni (1 + Zi)
3
4
4

 ≃
µ = 2X + Y +
=
.
4
ρNo
2 + 6X + Y
3 + 5X − 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
Zi>He ≃ Ai/2:
X ni Zi
Z
1
Y
Y
≃ X + + = (1 + X) . (13)
Ye = X + +
2
ρNo
2
2
2
i>He
3
The Virial Theorem
Position, momentum, mass of ith particle: ~ri, p~i , mi .
Newton’s Law F~i = p~˙ i with p~i = mi~r˙ i:
2I
X
X
d X
1
d
d
p~i · ~ri =
mi~r˙ i · ~ri =
p~˙ i · ~ri + p~i · ~r˙ i =
,
dt
dt
2 d2 t
(14)
P
Moment of inertia: I =
mi~ri2.
Static situation: d2I/dt2 = 0.
P ˙2 P
Non-relativistic gas:
mi~ri =
p~i · ~r˙ i = 2K.
Total kinetic energy:
1X
1X˙
1X ~
˙
K=
p~i · ~ri = −
p~i · ~ri = −
Fi · ~ri = − (1/2) Ω.
2
2
2
(15)
Sum is virial of Clausius. For perfect gas, only gravitational
forces contribute, since forces involved in collisions cancel.
X Gmimj
X
X
G
G
~
~
Fij · ~ri − ~rj = −
Fi · ~ri =
≡ Ω. (16)
rij
pairs
pairs
Ω is gravitational potential energy, rij = |~ri − ~rj |.
Perfect gas with constant ratio of specific heats, γ = cp/cv :
K = (3/2) N T,
U = (γ − 1)−1 N T,
U is internal energy, E is total energy.
E = U + Ω = U (4 − 3γ) = Ω
E = U +Ω = U −2K.
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
P
˙
Relativistic gas:
p~i · ~ri = c p~i = K = −Ω.
Another derivation:
P
V dP = −
1 Gm
1
dm = − dΩ,
3 r
3
(18)
where V = 4πr 3/3. Its integral is
Z
V (r) dP = P V
R
0
−
Z
1
P (r) dV = − Ω
3
(19)
R
from Eq. (18). Thus Ω = −3 P (r)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
Polytropic Equations of State
The polytropic equation of state, common in nature, satisfies
′
(n+1)/n
γ
P = Kρ
≡ Kρ ,
(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
1/3
No 3No
P =
(1 − β)
ρ4/3
(21a)
µβ µβa
1/3
3No
T =
(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
2
5
h̄
s = − ln 
ρNo/µ = constant (22a)
2
2mT
2
5
h̄2 ρNo 5/3
exp
s−
= Kρ5/3.
(22b)
P =
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 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) 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 ∝ ρ10/3, i.e., n = 3/7.
Don’t confuse polytropic with adiabatic indices. A polytropic change has c = dQ/dT is constant, where dQ = T dS.
An adiabatic change is a specific case: c = 0.
c
−
c
c
∂ ln P
p
v
γ′ =
=γ
,
∂ ln V
cp (c − cv )
where the adiabatic exponent γ = (∂ ln P/∂ ln V )|s. If γ =
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 is a
polytrope, with
Z
Z
3 GM 2
Gm (r) dm (r)
=−
= −3 P dV . (23)
Ω=−
r
5−n R
For a perfect gas with constant specific heats,
3γ − 4 1 GM 2
E=−
.
γ−1 5−n R
For the adiabatic case n = 1/(γ − 1),
(24)
n − 3 GM 2
E=
.
(25)
5−n R
For a mixture of a perfect gas and radiation,
Z Z
β
4 − 3γ
U=
+ 3 (1 − β) P dV = β
P dV − Ω. (26)
γ−1
γ−1
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 = Kρ1+1/n;
1/n−1/n1 1+1/n1
ρ
;
P = Kρt
ǫ = nP
ρ < ρt
ǫ = n1P + (n − n1) Pt
ρ > ρt.
ρt and Pt are the transition density and pressure between indices n and n1. ǫ is the energy density
"
#
2
2
GMt
n − 3 Gm
n1 − 3 GMt2
E=
−
+
5−n
R
Rt
5 − n1 Rt2
(27)
Mt 4π 3 n − 1 n1 − 1
+ 3Pt
−
.
− Rt
ρt
3
5 − n 5 − n1
Mt and Rt are mass and radius interior to transition point.
When (n1 ≃ 0) and n ≃ 3,
3 GMt2
E=−
.
(28)
5 Rt
This could apply to a proto-neutron star with relativistic electron gas up to ρt, and relatively stiff matter beyond. The
energy depends on inner core size only.
Structure of polytropes and Lane-Emden equation:
r = Aξ,
θ=
ρ 1/n
,
ρc
i1/2
1/n−1
A = (n + 1) Kρc
/ (4πG)
.
h
(29)
1 d
2 dθ = −θ n .
ξ
(30)
2
dξ
dξ
ξ
Boundary conditions for are θ=1 and θ′ = dθ/dξ = 0 at ξ=0.
The radius is found from ξ 1 where θ(ξ) = 0.
8
n
γ′
θ
ξ1
√
0
∞
1 − ξ 2/6
6
1
2
sin(ξ)/ξ
π
3/2 5/3
3.654
2
3/2
4.353
3
4/3
6.897
3.25 17/13
8.019
4
5/4
14.97
p
5
6/5 1/ 1 + ξ 2/3 ∞
−ξ12θ1′ −ξ1/3θ1′ [4π(n + 1)θ1′2]−1
√
2 6
1
3/8π
π
π 2/3
π/8
2.714
5.992
0.7704
2.411
11.40
1.638
2.018
54.19
11.05
1.950
88.15
20.36
1.797
622.3
247.5
√
3
∞
∞
Analytic solutions exist in the following cases:
√
θ = 1 − ξ 2/6; ξ1 = 6 n = 0, γ ′ = ∞;
θ = sin ξ/ξ; ξ1 = π
n = 1, γ ′ = 2;
q
6
θ = 1/ 1 + ξ 2/3; ξ1 = ∞ n = 5, γ ′ = .
5
Radius : R = Aξ1
Mass : M = −4πA3ρcξ12θ1′
Density ratio : ρ̄/ρc = −3θ1′ /ξ1
h
i
′2 4
Central pressure : Pc = GM/ 4π (n + 1) θ1 R
K=

G 
4π
n+1
M
−ξ12θ1′
!n−1 
3−n 1/n
R

.
ξ1
(31a)
(31b)
(31c)
(32a)
(32b)
(32c)
(32d)
(33)
For n → ∞, we have the isothermal Lane-Emden equation:
1 d
dφ
2
−φ = ρ .
(34)
ξ
=
e
dξ
ρc
ξ 2 dξ
ρ = K/2πGr 2; 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,
h̄c 21/3
P =
3π
(nYe )4/3 ,
(36)
4
which implies a mass
Mch = −4π
K 3/2
× 2.018 = 5.76Ye2M⊙.
G
(37)
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 6= 0, the
pressure has a small thermal component
2/3
T2 2
Pth =
3π ρYe
.
8h̄c
(38)
For a massive stellar core just prior to collapse, T ≃ 0.7 MeV
and ρ ≃ 6 × 109 g cm−3, and the Pth /P ≃ 0.12, and the
effective Mch 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 106 g cm−3, electron capture
decreases Ye. For a 56Fe 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 Mch.
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
Standard Model Stars – The Main Sequence
Those stars converting H to He. Standard model assumes
β = constant.
1/3
4/3 No
3
.
(39)
(1 − β)
K=
µβ
a
1 − β 1/6
R = 11.18 4 4 2
R⊙
µ β ρc
√
1−β
M = 18 2 2 M⊙
(40)
µ β
ρc/ρ̄ = 54.2
βµM R⊙
× 107K.
Tc = 1.96
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
µ
11.183
= 76.7 g cm−3, (41)
= 0.9996; ρc =
β ≃ 1−
18
18
Tc = 1.307 × 107 K.
But µc > µ̄ since some H→He has occurred.
Luminosity will depend upon nuclear energy generation ǫ̇
and transport (opacity κ).
For T > 8 · 106ρ1/3.5 electron scattering dominates:
κ ≃ 0.4Ye cm2 g−1.
(42)
κ ≃ 2.5 · 1025ZYeρT −3.5 cm2 g−1.
(43)
Where T > 104 K, κ dominated by bound-bound and boundfree processes:
11
For Z < 10−4 have free-free opacity:
κ ≃ 8 · 1022 (1 − Z) Ye ρT −3.5 cm2 g−1.
The dependence κ ∝ ρT −3.5 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 3 dT
T
.
(45)
3κRρ dr
L(r) is luminosity, κR is the “Rosseland mean” opacity, averaged over frequencies. (κρ)−1 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:
L (r) = −4πr 2
dPr
κL (r)
=
.
dP
4πcGm (r)
(46)
Luminosity function η(r) = L(r)M/m(r)L, with M and L totals. η is sharply peaked at origin.
L
d [(1 − β) P ] =
κ (r) η (r) dP.
(47)
4πcGM
4πcGM
(1 − βc)
(48)
L=
κη
¯
κη
¯ is a pressure average. (ssm: κη = cons.) With Eq. (43)
7.5
0.5
5.5
µ
βG
4ac
M
ξ1
c
Lssm = (4π)3
4.5
3κoηc 4No
R
−ξ12θ1′
(µcβc)7.5 M 5.5 R⊙ 0.5
L⊙ .
(49)
≃ .667
ηcZYe,c M⊙
R
12
With Eq. (42), appropriate instead for more massive stars,
3
1
M
L ≃ 97.5
(βµc)4
L⊙ .
ηcYe,c
M⊙
(50)
For Sun: µc = 0.73, Ye,c = 0.75 (i.e., X ≃ 0.5 and Y ≃ 0.5),
L ≃ (2.8/ηc) L⊙.
Alternatively, use proton-proton rate (T6 = T /106):
−1/3 M⊙
6
2 −2/3
η = 2.0 × 10 ρX T6
exp −33.8T6
.
L⊙
(51)
With ρc = 76.7 g cm−3, Tc,6 = 13.07 from ssm η c=4.05. Try
using knowledge of polytropic structure. Assume ideal gas and
ǫ̇ = ǫ̇c
ρ
ρc
λ T
Tc
ν
.
(52)
For polytrope, ρ = ρcθn and T = Tcθ:
r
27π
(2nλ + ν)−3/2 ,
2
(53)
2
since 2nλ + ν >> 1. With θ ≃ exp(−ξ /6), n = 3,
L = 4πA3ρcǫ̇c
Z
ξ 2θ2nλ+ν dξ ≃ 4πA3ρcǫ̇c
r
M
2
ηc = ǫ̇c
= − ξ 2θ′
(2nλ + ν)3/2 = 0.31 (6λ + ν)3/2 ,
L
27π
1
(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
For Kramer’s opacity, κ ∝ Z(1 + X)ρT −3.5. For electron
scattering, κ ∝ (1 + X). Suggests
κ ∝ (1 + X) Z u ρnT −s,
(55)
with u = 0, 1. Similarly
ǫ̇ ∝ X 2−mZ mρλT ν ,
(56)
with m = 0(1) for p-p (CNO) cycle. Using
L ∝ RT 4/κρ ∝ M ǫ̇,
ρ ∝ M/R3, (57)
T ∝ µβM/R,
we find
L ∝ M αM (µβ)αµ X αX (1 + X)α1 Z αZ ,
R ∝ M βM (µβ)βµ X βX (1 + X)β1 Z βZ ,
1/4
2
Tef f ∝ L/R
∝ M γM (µβ)γµ X γX (1 + X)γ1 Z γZ ,
δ
L ∝ TefTf (µβ)δµ X δX (1 + X)δ1 Z δZ .
(58)
With i = (M, µ, X, (1 + X), Z, T ), γi = αi /4 − βi /2
δi = 2(αM βi − αiβM )/(αM − 2βM ), D = ν − s + 3(n + λ):
i→
M
µ
i→
X
1
Z
m(3n − s) − u(3λ + s)
u+m
−(s + 3λ)
1
(3n − s)(2 − m)
2−m
αiD ν(3 + 2n) + 9λ + 3n + s(2λ − 1) 7ν + 3λ(4 + s)
βiD
λ+ν +n−s−2
ν −4−s
αi D
βiD
T
0
2D
14
αM
αµ
αX
α1
αZ
low 71/13 101/13 -2/13 -14/13 -16/13
high
3
4
0
-1
0
αT
0
0
βM
βµ
low 1/13 -7/13
high 19/23 16/23
βX
4/13
1/23
βZ
2/13
1/23
βT
2
2
γM
γµ
low 69/52 87/52
high 31/92 15/23
γX
γ1
γZ
-3/26 -9/26 -7/13
-1/46 -25/92 -1/46
γT
-2
-2
low
high
δM
0
0
β1
2/13
1/23
δµ
δX
δ1
-4/3 44/69 8/23
-56/31 6/31 44/31
δZ
δT
68/69 284/69
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 Tef f 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 = 0.75(0.7),
Ltoday
≃ 1.4,
Linitial
Tc,today
≃ 1.09,
Tc,initial
Tef f,today
≃ 1.11,
Tef f,initial
Rtoday
≃ 0.96.
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 compositions, Z < 0.001. For a given Tef f , L ∝ Z δZ 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 ,
Tef f ∝ Z γZ implies a shift of the M-S to higher Tef f .
4) For a given M , L ∝ Z αZ , 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
m m+1
P
3κoL
d ln T
3LκP
µ
=
(1.1)
=
d ln P
16πacGM T 4 16πacGM No
T 4+t+m
where the opacity is assumed to scale as
κ = κoρnT −s.
(1.2)
The radiative zero solution is obtained if d ln T /d ln P is constant. 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
0.5
5.5
4ac 4µG
M
ξ1
L = (4π)3
4.5 R
3κo 17No
−ξ12θ1′
(1.4)
7.5 Ye,c
µ
Lssm.
≃ 0.8ηc
µc
Ye
(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 β c=1.
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
ρc/ρ̄ = 6,
µM R⊙
× 107K,
M⊙R
4
M 2 R⊙
Pc = 8.7 2 4 erg cm−3.
M⊙R
Tc = 1.2
(1.5)
It is also clear that, dimensionally (cf . Eq. (32))
1/2
M ∝ K 3/2ρc
−1/6
R ∝ K 1/2ρc
∝ KM −1/3 ∝ e2s/3 (for fixed M ) (1.6)
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 actual temperature gradient from the purely adiabatic one (e.g.,
Ref. @Ref.Clayton@, 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
R
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/R2, the surface gravity, is nearly
constant throughout the thin surface region. As a zeroth approximation, we may write
2 gp
2
−n s
Pp ≃
= GM Rp−2κ−1
(1.8)
o ρp Tp
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:
1/Q

!2 3n+5
1/Q
µ
2GM
3+3n
3+n
3n−1

K
∝ M
Rp
,
Tp = 
2
No
3κoRp
(1.9)
where Q = 5 + 3n − 2s. The luminosity follows immediately
from L = 4πRp2σTp4:
#1/(3n−1)
"
4 10+6n 6+18n−4s
2+2n Tp
5No
3κo
−ξ15θ1′
L =σ (4π)−n−5
5
2Gµ
M 6+2n
"
8 20+12n 12+4n 6+18n−4s #1/Q
M
Rp
5
2µG
13+11n−2s
=σ (4π)
.
4+4n
′
5
3κo
5No
−ξ θ1
(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
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 ln T /d ln P = 2/5(n = 3/2), so from Eq. (33),
Pc =
NoTc 5/2 −3/2
K
;
µ
Tc =
−2 µGM
.
5ξ1θ1′ NoR
(1.11)
On the other hand, a star in radiative equilibrium, from Eq. (1.4),
satisfies, at the center,
d ln T
3
κcPcLηc
.
=
d ln P
16πacG Tc4M
(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
32πacGM Tc4
Lmin =
15ηc κcPc
4+s
ξ1s−3n
M s−n+3
4ac
2Gµ
n+2
=
(4π)
2+s−n s−3n
2
′
3ηcκo
5No
R
−ξ1 θ1
µ7.5 (M/M⊙)5.5
= 271
L⊙
.5
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 Ye,ssm
µ
8 7.5
×
Lmin/Lssm =
5β
µssm
Ye

4.5
2
′
ξ1,3θ1,3
ξ1,3/2Rssm 0.5 ηssm

× 2
.
′
ξ1,3R
ηc
ξ1,3/2θ1,3/2
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.