stellar structure & evolution
TRANSCRIPT
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Stellar Structure & Evolution
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Lecture 2: The Equations of Stellar
Evolution
Literature: MWW chapters 1–3
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Spherical stars
r RM m R
r m rrMR
==
=
=
( )
( ) ( )ρπ
ρπ
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3
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m(r) = 4π r20
r
∫ ρ (r) dr ⇒dmdr
= 4π r2ρ (r) 1( )
ρ( )( )r
m r
a) Mass-continuity equation
stellar radius total mass of star mean density inside r
mean density of star
density at radius r mass enclosed inside r
2
b) Gravitation
Gravitational acceleration inside a spherical body
g = g(r) = Gm(r)r2
=dΦdr
ΦsurfaceGMR
= −
(Po § 2.1.1)
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withΦthegravitationalpotential,and
dPdr
= −Gm(r)ρ(r)
r22( )
1( ) ⇔drdm
=1
4π r2ρ
2( ) ⇔dPdm
= −Gm4π r4
Hydrostatic equilibrium if and only if for all r !!r = 0
This is the equation of hydrostatic equilibrium, which equates the pressure gradient to the gravitational force
Often useful to employ m=m(r) as variable, not r. Then:
c) Hydrostatic equilibrium
Small cylinder of thickness dr, surface dA: mass =ρdrdA
Newton’s equation of motion:
with P the pressure
r = − 1ρdPdr
−Gm(r)r2
KWW § 2.1
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!!rdm= −gdm+P r( )dA−P r +dr( )dA;introduce dP
dr
Solutions of (1) and (2) exist for the special case These are so-called barotropic stars, and include the famous polytropes with (KWW 19; chapter 4.3) But generally with T the temperature; this is called the equation of state; it also depends on composition Example: Ideal gas has
k is Boltzmann’s constant = 1.38062 10-16 erg/K µ is the mean molecular weight (see chapter 3.2)
P K= ργ
Pρ=
kµmH
T
P P= ( )ρ
P P T= ( , )ρ
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d) Time scales
Consider again Newton’s equation:
r = − 1ρdPdr
−Gm(r)r2
Two extreme cases:
Dynamical time scale
Explosion time scale
P = 0 : r = g =: Rτ dyn2 ⇒ τ dyn =
Rg
g = 0 : r ≈ PρR
=: Rτ expl2 ⇒ τ expl = R
ρP≈
Rυsound
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Stars are in hydrostatic equilibrium
τ dyn ≈ 0.02 ρo / ρ days ≈1600 RRO
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3/2MO
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1/2
sec
3 sec ≤ τdyn ≤ 1000 days
Free-fall time scale:
In Solar units:
Use range of mass and radii for stars
This is similar to the periods of pulsating stars
τ ff = 3π / 32Gρ
This is an example of a period-mean-density relation
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e) Potential (gravitational) energy Eg
Eg = the amount of work needed to bring stellar matter from infinity to the present configuration Let m(r) be present inside radius r, and add mass dm between r and r + dr. Then:
dEg = − −Gmr2
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∞
r
∫ dr dm = −Gmrdm
Eg = −Gmr0
M
∫ dm =12
Φ dm0
M
∫
The gravitational potential energy Eg of a star is:
Eg = −qGM 2
Rq =O(1)
Where the precise value of q depends on the density distribution 8
Hydrostatic equilibrium
dPdr
= −Gm(r)ρ(r)
r22( )
Multiply by 4πr3dr and integrate
rhs = − Gm(r)r
4πr2ρ(r)dr∫ = Eg
lhs = 4π r30
M
∫ dP = 4π r3P#$ %&0M−12π Pr2 dr = −3 PdV = −3 P
ρ0
M
∫0
M
∫0
M
∫ dm
Eg = −3Pρ0
M
∫ dm
For a γ-law equation of state, with u the internal energy per unit mass:
P = γ −1( )ρu9
It follows that
star is unstable
Obvious example: ideal gas: Internal energy: Define: Then: This is the Virial Theorem
Pρ=
kµmH
T = (cP − cV )T = (γ −1)cVT = (γ −1)u
Ei = udm0
M
∫
Et = Eg +Ei =(3γ − 4)3(γ −1)
Eg
⇒ Et = Eg +Ei > 0 for γ < 4 / 3
Eg = −3(γ −1)Ei
γ =1Ei
γ u0
M
∫ dm
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γ = cP cV
! Ideal monatomic gas: γ=5/3
! For an ideal monatomic gas, in hydrostatic equilibrium, half of the gravitational energy goes into internal energy (heat) and half is radiated away
! A more tightly bound star is hotter
! Quasi-static contraction leads to internal heating
Ei = −12Eg
!Ei = −12!Eg
Et =12Eg L ≡ − !Et = −
12!Eg
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g) Order of magnitude estimates
P = 1M
Pdm =0
M
∫ −1M
m dP = G4πM0
M
∫ m2
r40
M
∫ dm
Pc = P(0)−P(R) = − dP =G4π0
M
∫ mr40
M
∫ dm
g = 1M
gdm =0
M
∫ =GM
mr20
M
∫ dm
T = 1M
T dm =0
M
∫ 1MµmH
kPρdm =
0
M
∫ −µmH
3MkEg
Eg = = −G mr0
M
∫ dm
Mean pressure
Central pressure
Potential energy
Mean grav. acceleration
Mean temperature (for ideal gas)
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Iσ ,ν =Gmσ
rν0
M
∫ dm =4π3
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&'ν /3
G ρν /3
0
M
∫ mσ−ν /3dm
The integrals on the right-hand side are all of the form
where we have used
Assume the density ρ(r) does not increase outwards, Then: and It follows that:
with rc defined by
r m r r3 3 4= ( ) / ( )π ρ
d r drρ( ) / ≤ 0
M rc c= 43
3π ρ
ρπ
ρ ρ ρ( ) ( )M MR c= ≤ ≤ =3
403
3G3σ +3−ν( )
M σ+1
Rν≤ Iσ ,ν ≤
3G3σ +3−ν( )
M σ+1
rcν
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Physical interpretation
Consider a mass distribution with total mass M, radius R and arbitrary ρ(r), which does not increase outwards (II) Consider two related configurations with ρ(r) constant (I&III):
P P P
E E E
g g g
T T T
cI
cII
cIII
gI
gII
gIII
I II III
I II III
≤ ≤
≤ ≤
≤ ≤
≤ ≤
Then:
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3GM 2
20π R4≤ P ≤ 3GM
2
20π rc4
3GM 2
8π R4≤ Pc ≤
3GM 2
8π rc4
3GM 2
5R≤ Eg ≤
3GM 2
5rc3GM4R2
≤ g ≤3GM4rc
2
µmH
5kGMR
≤ T ≤ µmH
5kGMrc
GM 2
R4=1.118×1016 M
MO
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2ROR
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dyne / cm2
GM 2
R= 3.791×1048 M
MO
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2ROR
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&' erg
GMR2
= 2.739×104 MMO
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&'ROR
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&'2
cm / sec2
µmH
kGMR
= 2.293×107 MMO
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&'ROR
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&'µ K
Numbers
Enormous pressures, densities and temperatures!
Specifically, this gives:
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