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Stellar Structure & Evolution

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Lecture 2: The Equations of Stellar

Evolution

Literature: MWW chapters 1–3

1

Spherical stars

r RM m R

r m rrMR

==

=

=

( )

( ) ( )ρπ

ρπ

3434

3

3

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)

3

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

4

!!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

6

Stars are in hydrostatic equilibrium

τ dyn ≈ 0.02 ρo / ρ days ≈1600 RRO

"

#$

%

&'

3/2MO

M"

#$

%

&'

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

7

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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%

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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

10

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)

12

Iσ ,ν =Gmσ

rν0

M

∫ dm =4π3

"

#$

%

&'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

"

#$

%

&'

2ROR

"

#$

%

&'4

dyne / cm2

GM 2

R= 3.791×1048 M

MO

"

#$

%

&'

2ROR

"

#$

%

&' erg

GMR2

= 2.739×104 MMO

"

#$

%

&'ROR

"

#$

%

&'2

cm / sec2

µmH

kGMR

= 2.293×107 MMO

"

#$

%

&'ROR

"

#$

%

&'µ K

Numbers

Enormous pressures, densities and temperatures!

Specifically, this gives:

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