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Lecture 5-3: Applications polytrope

models

Literature: MWW Chapter 19

1

g) Eddington’s standard model

In this model, the energy equation and equation of diffusive radiative transfer are included (approximately) to arrive at simple solutions. In essence, this model boils down assuming a constant b throughout the star. We looked at this in section 5a. From a slightly different angle, with b is constant we have Prad/P is constant:

2

∇ =d lnTd lnP

=3

64πσPκT 4

LrGMr

with Prad = 4σT 4 3c we can write: ∇ =P

4PraddPraddP

dPraddP

=κL

4πcGMLr LMr M

Introduce, ε ≡ dLr dMr

ε(r) = ε dMr0

r

∫ Mr =Lr Mr

η(r) ≡ ε(r) ε(R) =Lr LMr M

dPraddP

=L

4πcGMκ (r)η(r)

3

Prad r( ) = L4πcGM

κ r( )η r( ) P r( ) with

κ r( )η r( ) = κηdP0

P r( )

∫ P r( ), and we have,

1−β = L4πcGM

κ r( )η r( )

Now, realize that κ decreases rapidly inwards while η decreases rapidly outwards for main sequence stars. Eddington assumedthat β is constant and that results in a simple T -P or T -ρ relation.

T r( ) = 3ck4σµmu

1−ββ

#

$%

&

'(

1/3

ρ1/3 r( )

4

P r( ) =Pgasβ

=k

µmu

ρTβ=

kµmu

!

"#

$

%&

43a

1−β( )β 4

(

)**

+

,--

1/3

ρ 4/3 r( ), see slide 4 lecture 5-1

The constant in brackets is also equal to the constant K in slide 8 lecture 5-1

K n = 3( ) =4π( )1/3

4GM 2/3

ξ 4 − !θ3( )2(

)*+,-ξ1

1/3 , or

1−ββ 4 = 0.002996µ 4 M

Mo

!

"#

$

%&

2

, and

T r( ) = 4.62×106βµMMo

!

"#

$

%&

2/3

ρ1/3 r( )

but note β and M in this T -ρ relation are not independent5

One-parameter family of models: decreases with increasing M

Other properties: ρc = 54.18ρ

Pc =1.242×1017 MMO

"

#$

%

&'

2ROR

"

#$

%

&'

4

dyne/cm2

Tc =19.72 ×106βµMMO

"

#$

%

&'ROR

"

#$

%

&' K

T = 0.5852TcStandard model provides the run of density, temperature, and pressure. For absolute values, we need to provide stellar mass and radius. i.e., as the constant K in the pressure density relation is not specified aforehand in the standard model. E.g., the combination sets and hence K but then there are still an infinite number of solutions corresponding to different R.

M & R ⇒ ρ ⇒ ρc and hence ρ r( ), T r( ), P r( ) are set

6

β

µ2M β

m≈0.6

Inner region well approximated by a polytrope of index 3. Outer region is convective and better approximated by a polytrope of index 1.5 (but this is only 0.6% of the mass)

7

The Eddington Solar model

8

h) Completely convective stars

Examples: late M dwarfs cool, gravitationally contracting stars

Internal Structure Consider case with everywhere in the star

Then: or with If Γ2 constant completely convective star: polytrope of index n

Useful specific case: ideal gas n = 3/2 (§5-2, slides 8-10)

Radial behavior of

follows from Lane-Emden function:

K = 2.476×1014 MMO

"

#$

%

&'

1/3RRO

"

#$

%

&'

(K =0.0796µ 5/2

MO

M"

#$

%

&'1/2 RO

R"

#$

%

&'3/2

ρc = 5.991ρ

Pc = 8.658×1015 MMO

"

#$

%

&'

2ROR

"

#$

%

&'

4

dyne/cm2

Tc =1.234×107µMMO

"

#$

%

&'ROR

"

#$

%

&' K

T = 0.5306Tc

P = Kρ1+1n P = !K T n+1 n = 1

Γ2 −1

∇ =∇ad = (Γ2 −1) /Γ2

θ3/2

(effective polytropic index)

9

⇒⇒⇒

ρ ,P ,T

i) Hayahsi track

Hayashi track: Location of low-mass (<3 Mo), fully convective stars in the Hertzsprung-Russell diagram. Relevant for low-mass protostars, red giants, and asymptotic red giants. Match fully convective stellar structure models with boundary conditions for the photosphere. Assume the star is fully convective (polytrope with n=3/2) with a thin radiative outer layer where the opacity is dominated by H–.

10

dPdr

= −gρ and dτdr

= −κρ or dPdτ

=gκ

Assume κ is constant,

P τ = 2 3( ) = 2g3κ

with g = GMR2

with P = kµmu

ρT, κ=κoρ0.5T 7.7, and κo =10−25Z 0.5 we have,

ρ1.5T 8.7 =2µmu

3kGκo

MR2

11

For a polytrope with n = 3 2 we have,

ρρc

=TTc

!

"#

$

%&

3/2

, ρc = 5.99ρ = 5.99 3M4πR2 , and Tc = 0.539 µmu

kGMR

T10.95 =Tc

2.25

ρc1.5

2µmu

3kGκo

MR2 ≈

0.1κo

µmuGk

!

"#

$

%&

3.25

M 1.75R0.25

We assume that convection starts at the photosphere (T = Teff )

and use L = 4πR2σTeff4 to replace R

T11.45 ≈0.07κoσ

1/8µmuGk

!

"#

$

%&

3.25

M 1.75L1/8

Teff ≈ 2000 MMo

!

"#

$

%&

0.15LLo

!

"#

$

%&

0.010.02Z

!

"#

$

%&

0.04

K; actually, ≈ 3000 K

12

Summary

Fully convective stars (with a radiative atmosphere) of a given mass and composition in hydrostatic equilibrium lie at a constant (low) effective temperature independent of luminosity. Conversely, the effective temperature is nearly independent of how the luminosity is generated. Objects to the right of these Hayashi tracks have too steep a temperature gradient and convection will set in. Convection quickly sets the adiabatic temperature gradient (lecture 3-3, slide 14). That quickly rearranges the stellar structure and the star will settle on the Hayashi track.

13

j) protostars & the Hayashi track

Premain sequence evolution: Molecular cloud core collapses on a slow timescale (t~1/(Gr)1/2). When the core becomes optically thick, the internal temperature rises (virial theorem) and molecules (e.g., H2) will dissociate, and then H and He will ionize. No nuclear reactions. High luminosity and high opacity. Protostar is fully convective except for an H– dominated atmosphere. Star will start high up on the Hayashi track. As it contracts, the radius will decrease but the effective temperature stays constant.

Rps

Ro≈ 50 M

Mo

, and T ≈105 K (virial theorem)

14

Et = −Ei = Eg / 2 and L = !Et = − !Eg 2 ≈ 3GM 2

7R2dRdt

with L = 4πR2σTeff4 , we then have:

1R4

dRdt

=28πσTeff

4

3GM 2 , which integrates to (Teff ≈ cst):

t = GM 2

28πσTeff4 R3 and L = L0

τ KH3t

#

$%

&

'(

2/3

where Ro = R(t = τ KH 3) with τ KH = 3GM 2 7RoL0 and L0 = 4πR02σTeff

4

15

Radiative (Henyey) protostellar tracks

Convection will stop when ∇rad =d lnTd lnP

=3κRP

64GπσT 4LM

< 0.4

As the star contracts, the luminosity will drop and radiative energy transport will take over in the core The star evolves now towards the left in the HR diagram with a slowly increasing luminosity.

Note thatPc Tc4 does not decrease (slide 10 of section 5.2).

T ∝M R and ρ∝M R3 and κ ∝ρT −3.5 and adopt dT dr = Tc RL∝M 5.5R−0.5

16

Summary

A protostar will initially be fully convective and contract on a Hayashi track of (almost) constant effective temperature. Eventually, the core will become radiative, the star will switch to a Henyey track and the luminosity will (moderately) increase. As the star contracts, the central temperature will increase (Virial theorem). Eventually, H-burning will be ignited in the core and the star will settle on the main sequence. The timescale for this is set by the Kelvin Helmholtz timescale and this is set for when the star evolves the slowest (eg., near the main sequence).

17

18

19

Protostellar tracks (solid lines) for different protostellar stellar masses. Isochrones are indicated by dashed lines. The solid lines crossing the tracks indicate the region of D-burning and Li-burning.

20

k) White dwarfs

A white dwarf is the stellar remnant of a low mass star with a mass typically half that of the Sun but a radius comparable to the Earth, corresponding to a density of 106 g/cm3. The star is pressure-supported by a degenerate electron gas with (cf., Lecture 3-2 slide 14): The mechanical properties are controlled by the electrons while the thermal energy is stored in the ions. We can use polytropes of index 3/2 and 3 to describe them.

P = KNR ρ µe( )5/3 non-relativistic

P = KER ρ µe( )4/3 extremely relativisticwhere the constants Ki only depend on atomic physics

21

Mass-radius relation

For the non-relativistic equation of state, we find (Lecture 5-2, slide 8): Thus, a more massive white dwarf has to be smaller and will be denser than a less massive white dwarf. Eventually when the density is high enough, the electron gas will become relativistic (cf., lecture 3-2 slide 25) and we have:

M 1/3R = cst and ρ ∝M R3 ∝M 2

M ≈KER

0.3639µe4/3G

"

#$

%

&'

3/2

≈ 0.197 hcGmu

4/3

"

#$

%

&'

3/2

1µe

2

Chandrasekhar mass

Mch =5.86µe

2 Mo ≈1.46Mo for C, O, ... composition

22

Heuristically: Mass-radius relation

When analyzing stars using the virial theorem, we discovered that stars are unstable when g<4/3 (lecture 2 slide 11). Evaluate hydrostatic equilibrium Non-relativistic: For any given mass, star can adjust its radius to reach equilibrium. Extreme-relativistic: Internal energy and gravity have same dependence on radius but different dependence on mass. So, if mass is too large, gravity will win.

dPdr

= −qGMρR2

∝M2

R5

dPdr

≈Pc

R∝M

5/3

R6(non-relativistic)

dPdr

≈Pc

R∝M

4/3

R5(extreme-relativistic)

23

Rad

ius [

km] (

R�

)

0 0.5 1.0 1.5 2.0

25000(0.036)

20000(0.039)

15000(0.022)

10000(0.014)

5000(0.007)

0

25000(0.036)

20000(0.039)

15000(0.022)

10000(0.014)

5000(0.007)

0.5 1.0 1.5 2.00 CM

Non-relativistic Fermi gasRelativistic Fermi gas

Mass [M�]

Ultr

a-re

lativ

istic

lim

it

White dwarfs that reach the Chandrasekhar mass – either through accretion from a companion or through merging of binary WDs – will explode as SN Ia.

24

Observed mass-size relation for White Dwarfs

25

Radiative cooling

The mechanical and thermal structure are decoupled. Hence, we have to relate the temperature to the luminosity and mass to calculate the cooling rate and timescale. Or equivalently, we will use the luminosity to calculate the internal temperature. Conduction keeps the interior at (almost) uniform temperature. We will assume that the photosphere is radiative (only true for white dwarfs with SpT earlier than A) and is described by the ideal gas law. Use the radiative zero conditions (lecture 5-1, slide 5/6) with Kramers opacity, κ =κoρT

−3.5

P = 28.516σ3

kµmu

4πGMκ0L

!

"#

$

%&

1/2

T 4.25

L = 5.7×105 µµe2

Ttr7/2

Z(1+ X)MMO

Ideal gas: (i) Transition from non-degenerate to degenerate gas at

with:

(ii)

Use (i) and (ii), and for bound-free transitions Since interior is isothermal: Ttr is interior temperature Integrate radiative gradient between T=0 and T=Ttr

ρ =6451

µmu

k4πσ GMκ0L

!

"#

$

%&

1/2

T 3.25

P(degenerate,NR) = P(ideal)⇔ 9.91×1012 ρtrµe

#

$%

&

'(

5/3

=k

µemu

ρtrTtr

⇒ ρtr = 2.40×10−8Ttr

3/2

ρ = ρtr

Ttr =µmuGM4.25k

1rtr−1R

"

#$

%

&' rtr = r(ρ = ρtr )

(iii) κ 0 ⇒

⇒

Cooling time scale can be evaluated from:

(iv) Differentiate eqn (iii), substitute in eqn (iv) and use eqn (iii) to eliminate Ttr:

L= − dE

dt= −c

VMdT

tr

dt&c

V= 32

kµAmu

dLdt

=72LTtr

dTtrdt

& Ttr =L

C2M!

"#

$

%&

2/7

1L12/7

dLdt

= −72

C22/7

CVM5/7

tcool =25cVM

5/7

C2

L−5/7 − L05/7( ) = 2×108 M Mo

µA

yr

( mean molecular weight per nucleus; 4.4 for Y=0.9, Z=0.1, 12 for C white

dwarf)

Cooling time: tcool>1010 yr when L<10-4L#

Evolution along line of constant radius in HRD

µA

29

All energies sources have to be taken into account: neutrino losses & nuclear energy left-overs (early on), gravitational contraction (early on and in the last stages) crystallization energy, followed by Debye cooling.

30

l) Neutron stars

Neutron stars pack the mass of the Sun in a radius of some 10 km (e.g., the size of Amsterdam), corresponding to a density of ~1014 g/cm3. Nuclear energy generation stops after Fe-core formation. The core will collapse and neutronization occurs (Lecture 4-2 slide 20): If the core is not too massive, collapse will be halted by the degeneracy pressure due to neutrons. The equation of state is complex but let’s just scale the Chandrasekhar mass: Actually, have to take into account that neutrons can couple and form a boson. Limiting mass ~3-4 Mo. Cores more massive than this will collapse and form a black hole.

e− + p+ → n0 +νe

Mch,n =1.46 2µn

!

"#

$

%&

2

≈ 5.8 Mo