Lecture 5-3: Applications Polytrope Models

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 lnT 3 Pκ Lr ∇ = = 4 d ln P 64πσ T GMr

4 P dPrad with Prad = 4σT 3c we can write: ∇ = 4Prad dP dP κL L L rad = r dP 4πcGM Mr M

Introduce, ε ≡ dLr dMr r ε(r) = ∫ ε dMr Mr =Lr Mr 0 L L η(r) ≡ ε(r) ε(R) = r Mr M dP L rad = κ(r)η(r) dP 4πcGM 3 L Prad (r) = κ (r)η(r) P(r) with 4πcGM P(r) κ (r)η(r) = ∫ κη dP P(r), and we have, 0 L 1− β = κ (r)η(r) 4πcGM Now, realize that κ decreases rapidly inwards while η decreases rapidly outwards for main sequence stars. Eddington assumed that β is constant and that results in a simple T - P or T - ρ relation. 1/3 # 3ck 1− β & 1/3 T (r) = % ( ρ (r) $ 4σµmu β '

4 4 1/3 P k T (! k $ 3 1− β + gas ρ * ( )- 4/3 P(r) = = = # & 4 ρ (r), see slide 4 lecture 5-1 m m a β µ u β )*"µ u % β ,- The constant in brackets is also equal to the constant K in slide 8 lecture 5-1 1/3 4π GM 2/3 K n = 3 = ( ) , or ( ) 2 1/3 4 ( 4 ! + *ξ (−θ3 ) - ) ,ξ1 2 1− β 4 ! M $ 4 = 0.002996µ # & , and β " Mo % 2/3 ! M $ T (r) = 4.62 ×106 βµ# & ρ1/3 (r) " Mo % but note β and M in this T - ρ relation5 are not independent One-parameter family of models: β decreases with increasing M

Other properties: ρc = 54.18ρ " %2 4 17 M " RO % 2 Pc =1.242 ×10 $ ' $ ' dyne/cm # MO & # R & " % 6 M " RO % Tc =19.72 ×10 βµ $ '$ ' K # MO &# R &

T = 0.5852Tc Standard model provides the run of density, temperature, and pressure. For absolute values, we need to provide stellar mass and radius. i.e., M & R ⇒ ρ ⇒ ρ and hence ρ r , T r , P r are set c ( ) ( ) ( ) as the constant K in the pressure density relation is not specified aforehand in the standard model. E.g., the 2 combination µ M sets β and hence K but then there are still an infinite number of solutions corresponding6 to different R. 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 ∇ = ∇ a d = ( Γ 2 − 1 ) / Γ 2 everywhere in the star 1 1+ 1 Then: P = K ρ n or P = K ! T n + 1 with n = (effective polytropic index) Γ2 −1 If Γ2 constant ⇒ completely convective star: polytrope of index n Useful specific case: ideal gas ⇒ n = 3/2 (§5-2, slides 8-10) ⇒ 1/3 " M % " R % ρc = 5.991ρ K = 2.476 ×1014 $ ' $ ' 2 4 # MO & # RO & " M % " R % P = 8.658×1015 O dyne/cm2 1/2 3/2 c $ ' $ ' 0.0796 " M % " R % # MO & # R & K( = $ O ' $ O ' µ 5/2 # M & # R & " % 7 M " RO % Tc =1.234 ×10 µ$ '$ ' K # MO &# R & Radial behavior of ρ,P,T T = 0.5306Tc

9 follows from Lane-Emden function: θ3/2 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 dP dτ dP g = −gρ and = −κρ or = dr dr dτ κ Assume κ is constant, 2g GM P(τ = 2 3) = with g = 2 3κ R

k 0.5 7.7 −25 0.5 with P = ρT, κ=κoρ T , and κo =10 Z we have, µmu

1.5 8.7 2µmu G M ρ T = 2 3k κo R

11 For a polytrope with n = 3 2 we have, ! $3/2 ρ T 3M µmu GM = # & , ρc = 5.99ρ = 5.99 2 , and Tc = 0.539 ρc "Tc % 4π R k R 2.25 3.25 10.95 Tc 2µmu G M 0.1!µmuG $ 1.75 0.25 T = 1.5 2 ≈ # & M R ρc 3k κo R κo " k %

We assume that convection starts at the photosphere (T = Teff ) 2 4 and use L = 4π R σTeff to replace R 3.25 11.45 0.07 !µmuG $ 1.75 1/8 T ≈ 1/8 # & M L κoσ " k % 0.15 0.01 ! M $ ! L $ ! 0.02 $0.04 Teff ≈ 2000# & # & # & K; actually, ≈ 3000 K " Mo % " Lo % " Z % 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.

Rps M 5 ≈ 50 , and T ≈ 10 K (virial theorem) R M o o 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.

14 3GM 2 dR E = −E = E / 2 and L = E! = −E! 2 ≈ t i g t g 7R2 dt 2 4 with L = 4π R σTeff , we then have: 1 dR 28πσT 4 = eff , which integrates to (T ≈ cst): R4 dt 3GM 2 eff 2 2/3 GM #τ KH & t = 4 3 and L = L0 % ( 28πσTeff R $ 3t ' 2 2 4 where Ro = R(t = τ KH 3) with τ KH = 3GM 7RoL0 and L0 = 4π R0σTeff

15 Radiative (Henyey) protostellar tracks

d lnT 3κ P L Convection will stop when ∇ = = R < 0.4 rad d ln P 64GπσT 4 M

As the star contracts, the luminosity will drop and radiative energy transport will take over in the core 4 Note that P T does not decrease (slide 10 of section 5.2). c c 3 −3.5 T ∝ M R and ρ ∝ M R and κ ∝ ρT and adopt dT dr = Tc R L ∝ M 5.5R−0.5

The star evolves now towards the left in the HR diagram with a slowly increasing luminosity.

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 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. 19 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): 5/3 P = K ρ µ non-relativistic NR ( e ) 4/3 P = KER (ρ µe ) extremely relativistic

where the constants Ki only depend on atomic physics 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.

20 Mass-radius relation

For the non-relativistic equation of state, we find (Lecture 5-2, slide 8): 1/3 3 2 M R = cst and ρ ∝ M R ∝ M 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: " %3/2 " %3/2 KER hc 1 M ≈ $ 4/3 ' ≈ 0.197$ 4/3 ' 2 # 0.3639µe G & #Gmu & µe Chandrasekhar mass 5.86 Mch = 2 Mo ≈ 1.46Mo for C, O, ... composition 21 µe 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 dP GMρ M2 = −q ∝ dr R2 R5

dP P M5/3 c ≈ ∝ 6 (non-relativistic) dr R R 4/3 dP Pc M ≈ ∝ 5 (extreme-relativistic) dr R R 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.22 0 0.5 1.0 1.5 2.0

25000 25000 (0.036) (0.036)

20000 20000 (0.039) (0.039)

15000 15000 (0.022) (0.022) )

Non-relativistic Fermi gas Ultra-relativistic limit 10000 10000 (0.014) (0.014) Rel ativ ist ic F erm Radius [km] (R i gas

5000 5000 (0.007) (0.007)

0 0 0.5 1.0 CM 1.5 2.0 Mass [M] White dwarfs that reach the Chandrasekhar mass – either through accretion from a companion or through merging of binary WDs – will explode as23 SN Ia. Observed mass-size relation for White Dwarfs

24 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) −3.5 with Kramers opacity, κ = κoρT

1/2 ! 2 16σ k 4π GM $ 4.25 P = # & T "8.5 3 µmu κ 0L %

1/2 ! 64 µm 4πσ GM $ 3.25 Ideal gas: ρ = # u & T (i) " 51 k κ 0L %

Transition from non-degenerate to degenerate gas at ρ = ρtr # &5/3 12 ρtr k with: P (degenerate, NR) = P(ideal) ⇔ 9.91×10 % ( = ρtrTtr $ µe ' µemu − 8 3 /2 (ii) ⇒ ρtr = 2.40 ×10 Ttr κ Use (i) and (ii), and 0 for bound-free transitions ⇒ 7/2 5 µ Ttr M L = 5.7×10 2 (iii) µe Z(1+ X) MO

Since interior is isothermal: Ttr is interior temperature

Integrate radiative gradient between T=0 and T=Ttr ⇒ " % µ muGM 1 1 Ttr = $ − ' rtr = r(ρ = ρtr ) 4.25k # rtr R & Cooling time scale can be evaluated from:

dE dT 3 k L = − = −c M tr & c = (iv) dt V dt V 2 µ m A u Differentiate eqn (iii), substitute in eqn (iv) and use eqn (iii) to eliminate Ttr: ! $2/7 dL 7 L dTtr L = & Ttr = # & dt 2 Ttr dt "C2M % 2/7 1 dL 7 C2 12/7 = − 5/7 L dt 2 CV M

5/7 2 cV M −5/7 5/7 8 M Mo tcool = (L − L0 ) = 2 ×10 yr 5 C2 µA

µ ( A mean molecular weight per nucleus; 4.4 for Y=0.9, Z=0.1, 12 for C white dwarf) 10 -4 Cooling time: tcool>10 yr when L<10 L# Evolution along line of constant radius in HRD 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. 29 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 − + 0 20): e + p → n +νe 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: 2 ! 2 $ Mch,n =1.46# & ≈ 5.8 Mo µ " n % 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. 30