What is ?

Convection and • A highly efficient way of transporting energy in the Other Instabilities stellar interior

• Heat carried by , Prialnik 6 not , Depends linearly Glatzmair and Krumholz 11 on fluid speed, not on random Pols 3, 10 Highly idealized schematic. walk Real convection is not so ordered, but mass is conserved. Mass going up = mass going • Occurs when the heat gradient down. exceeds some critical value required for

Solar surface

Kerr (1996) Filamentary flows in Rayleigh-Benard convection (high Rayleigh number 2 x 107) Volume rendering of structure Consider the radial Adiabatic expansion displacement of a blob of Recall from Lecture 5 (equation of state) that the first gas. As it rises its internal law of thermodynamics can be used to define an density will decrease in "adiabatic expansion" and an adiabatic exponent. accordance with the decrease ⎛ 1⎞ in in the surrounding du + Pd = 0 medium. In the absence of ⎝⎜ ρ ⎠⎟ heat exchange with its which given that surroundings it will expand P adiabatically. u =φ ρ

leads to The density in d logP φ +1 = ( ) = γ the surrounding medium will d log ρ φ ad Which is denser after a decrease too in accordance and a displacement? The blob with hydrostatic equilibrium. φ+1 φ+1 or its surroundings? φ φ γ P∝ ρ or P=Kρ =Kρ a

Consider a rising “blob” expanding adiabatically Such a relation between pressure and density is called "adiabatic". If the gas expands due to decreased pressure on its extremities or is compressed by increased pressure Because the hydrodynamic without heat flow across its boundary, this expression gives time is much shorter than the change in internal pressure relative to the change the float time (v << csound) in density the pressure inside the floating blob at a given radius is always equal to From our previous discussion of φ, for ideal gas γ =5 / 3; a r the external pressure at the same radius. P = P , but for radiation, γ a = 4/3. e 2 e may not equal 2

Note that the gamma here is a local gamma γ a(r), unlike the gamma that characterizes a polytrope “e” stands here for “expanded” ρe = ρ1 +δρe ρ = ρ +δρ d log P = γ d log ρ (adiabatic) δρ > δρ for stability 2 1 a e P δ e δρe dρ ⇒ = γ a r or P δρe > Δ e ρe dr

where δPe is determined by the ρe 1 dPe dρ dPe dP existing pressure gradient in the Δr > Δr = Pe γ a dr dr dr dr ρ 1 dP star. So δρ = e Δr. e P dr 1 dρ 1 dP 1 e γ a < dρ and dP Background gradients Background gradients dr P dr This will be stable against ρ γ a dP dρ dP dρ , convection if , are both dr dr dr dr δρ < δρ = ρ − ρ r e 1 2 r negative so

but δρe and δρ are both negative numbers so stability ⇒ δρ > δρ e d log ρ 1 stable if > That is, if adiabatic expansion leads d logP γ P P and a to a greater density (less density e = ρe ≈ ρ if Δr is small ((ρ − ρ) / ρ d log ρ 1 decrease) than the surroundings, there e unstable if < 8 is no convection − d logP γ a might be  10 )

d log ρ 1 log ρ S so things stable if > , but in general we don't know the ρ < ρ S: stable gradient d logP γ e 2 ad a U: unstable gradient dT dP density gradient. Instead we have information about and adiabatic U dr dr expansion Expand dP in terms of partial derivatives. P is a function of ρ P ρe Pe = P2 2 2 T, ρ, and µ (for any EOS!) ρ1 ∂P ⎞ ∂P ⎞ ∂P ⎞ dP = dT + dρ + dµ ∂T ⎟ ∂ρ ⎟ ∂µ ⎟ ρ ⎠ ρ,µ ⎠ T,µ ⎠ ρ,T ρ > ρ 2,u e 2 ρe dP ⎛ ∂logP ⎞ dT ⎛ ∂logP ⎞ dρ ⎛ ∂logP ⎞ dµ = + + ρ2,s P ⎜ logT ⎟ T ⎜ log ⎟ ⎜ log ⎟ Pols 5.3 ⎝ ∂ ⎠ ρ,µ ⎝ ∂ ρ ⎠ µ,T ρ ⎝ ∂ µ ⎠ ρ,T µ dT dρ dµ upward displacement = χ + χ + χ T T ρ ρ µ µ These values "χ " are dimensionless numbers which ρ1 P1 ρ1 P1 basically give the powers of the given quantities to which PP21 log P P is sensitive. ρT d log ρ 1 e.g., for ideal gas P ∝ so χ = 1; χ = 1;χ = −1 Figure 5.3. Schematic illustration of the Schwarzschild criterion for ststabilityable if against convection.> A gas µ T ρ µ d logP γ a element is perturbed and displaced upwards from position 1 toposition2,whereitexpandsadiabaticallyto d logP = d logT + d log ρ + − d logµ maintain pressure equilibrium with its surroundings. If itsdensityislargerthantheambientdensity,itwill sink back to its original position. If its density is smaller,however,buoyancyforceswillaccelerateitupwards: convection occurs. On the right the situation is shown in a density-pressure diagram. A layer is stable against convection if the density varies more steeply with pressure than for an adiabatic change.

The expansion of the gas element as it rises over ∆r occurs on the local dynamical timescale (i.e. with the speed of ), which is typically much shorter than the local timescale for heat exchange, at least in the deep interior of the star. The displacement and expansion of the gas element will therefore be very close to adiabatic. We have seen in Sec. 3.4 that the adiabatic exponent γad defined by eq. (3.56) describes the logarithmic response of the pressure to an adiabatic change in the density. Writing as δρe and δPe the changes in the density and pressure of the element when it is displaced over a small distance ∆r, we therefore have

δPe δρe = γad . (5.40) Pe ρe

Here δPe is determined by the pressure gradient dP/dr inside the star because Pe = P2, i.e. δPe = P P = (dP/dr) ∆r. Therefore the change in density δρ follows from eq. (5.40) 2 − 1 e ρe 1 dP δρe = ∆r. (5.41) Pe γad dr

We can write ρe = ρ1 + δρe and ρ2 = ρ1 + (dρ/dr) ∆r, where dρ/dr is the density gradient inside the star. We can then express the criterion for stability against convection, ρe > ρ2, as dρ δρ > ∆r, (5.42) e dr which combined with eq. (5.41) yields an upper limit to the density gradient for which a layer inside the star is stable against convection, 1 dρ 1 dP 1 < , (5.43) ρ dr P dr γad where we have replaced Pe and ρe by P and ρ, since the perturbations are assumed to be very small. Remember, however, that both dρ/dr and dP/dr are negative. Therefore, in absolute value the sign

65 continuing: ∇ is usually a positive quantity. It is related to the radial d logP d logT d log d log temperature derivative by: = χT + χ ρ ρ + χ µ µ 1 − H d log ρ d logT d logµ d logT P dT P dT ⎛ dP ⎞ P dT χ = 1 − χ − χ ∇ = = = ⎜ ⎟ = − =∇ ρ d logP T d logP µ d logP d logP T dP T dr ⎝ dr ⎠ T dr −1 χ ⎛ 1 dP ⎞ d log ρ 1 χT µ where H is the pressure "scale helight", = − ∇ − ∇ P =⎜ − ⎟ d logP µ ⎝ P dr ⎠ χ ρ χ ρ χ ρ the distance over which pressure declines by 1 e-fold d logT d logµ where ∇≡ ; ∇ ≡ d logP µ d logP For ideal gas (only) in hydrostatic equilibrum ∇ here (not to be confused with the "grad" operator) is dP ρN kT 1 N kT H N k H P / A A ; P A a dimensionless describing how T P = − = = = dr µ ρg µg T µg behaves with pressure.

d log ρ 1 χ χ ⎛ d log ρ ⎞ 1 Previously, = − T ∇ − µ ∇ and = d logP µ ⎜ d logP ⎟ continuing: χ ρ χ ρ χ ρ ⎝ ⎠ ad γ ad 1 χT χ µ 1 1 χT For an adaibatic expansion at consant composition(∇ =0; µ − ∇ − ∇µ > = − ∇ad in the displaced gas the composition does not change) χ ρ χ ρ χ ρ γ ad χ ρ χ ρ

χ ρ 1− χT χ µ χT ⎛ d log ρ ⎞ 1 1 χ γ γ ad − χ − ∇ − ∇ > − ∇ = = − T ∇ ⇒ ∇ = ad = ρ = ∇ µ ad ⎜ d logP ⎟ ad ad ad χ χ χ ⎝ ⎠ ad γ ad χ ρ χ ρ χT γ ad χT ρ ρ ρ This is the "dimensionless adiabatic temperature gradient" d logT 5 / 3 −1 ⎛ d log ρ ⎞ 1 −χT ∇ − χ ∇ >− χT ∇ad ∇ ≡ For ideal gas 0.4. For stability , so µ µ ∇ad = = ⎜ ⎟ > d logP 5 / 3 ⎝ d logP ⎠ star γ ad 1 χ χ 1 χT ∇ + χ µ∇µ < + χT ∇ad T µ but − ∇ − ∇µ > χ ρ χ ρ χ ρ γ ad χ µ γ − χ 1 1 ⎛ χ ⎞ ⎛ χ ⎞ 1 1 ⎛ χ ⎞ for stabiity ∇ < ∇ − ∇ Ledoux ad ρ = ∇ ⇒ − ρ = ∇ and T − = T ∇ ad µ ad ⎜ ⎟ ad ⎜ ⎟ ⎜ ⎟ ad χT γ ad χT χT γ ad ⎝ χT ⎠ ⎝ χ ρ ⎠ χT γ ad ⎝ χ ρ ⎠ and if 0(or is ignored) Schwarzschild 1 1 χT ∇µ = ∇ < ∇ad = − ∇ad γ ad χ ρ χ ρ

The dimensionless radiative temperature gradient Maximum luminosity for convective stability: is given by the chain rule and hydrostatic equilibrium and radiative diffusion ⎛ ⎞ For stability: ∇ < ∇ dT dP dT Gm T d logT 3 L(r) rad ad − κ ⎜ ⎟ = i = 4 = − 3 2 dm dm dP 4πr P d logP 4ac T ⎜ 2 ⎟ 3 P κL(r) ⎝ (4πr ) ⎠ < ∇ ~ 0.4 (ideal gas) 16πacG T 4 m ad d logT ⎞ 3 κL(r)P ∇ = = rad ⎟ 4 3 N kρ κL(r) d logP ⎠ rad 16πacG mT A 3 < 0.4 The Schwarzschild condition for stability becomes, e.g. 16πacG µT m ∇ < ∇ 3 rad ad 16πacGµT m(r) For ideal gas for example 5 / 3, 1, 1 L(r) 0.4 γ a = χ ρ = χT = < 3NAkρκ γ − χ 2 / 3 ad ρ 3 ∇ad = = =0.4 γ χ 5 / 3 −18 µT −1 ad T < 1.22×10 m(r) erg s For radiation 4 / 3, 0, 4 γ a = χ ρ = χT = κρ 6 33 −1 γ ad − χ ρ 4 / 3 e.g. 0.61, 1,T 10 , M M L 1.5 10 erg s µ = κρ = = =  < × ∇ad = = =0.25 γ χ 16 / 3 ad T

Where is convection important? µT 3 L(r) > 1.22×10−18 m(r) erg s−1 κρ Convection will be important in regions where the opacity is high or the energy generation is concentrated in a small mass giving a large ratio of L(r)/m(r). Examples: • The cores of massive stars powered by the CNO cycle which is very temperature sensitive.

• Regions of high opacity especially in The sun – Manfred Schussler ionization zones and near stellar surfaces. Max Plank Insitut fur Aeronomie Kramers opacity increases at low T.

• The interiors of massive stars where L/M is large Convection in the sun and massive stars

Pols, Fig 5.4

Main sequence stars Heat transport by convection and mixing length theory

In a region where the temperature gradient is even slightly superadiabatic, there will be efficient The sun transport of energy by advection. Carrying entrained energy and dispersing it after some distance is much more efficient than radiative diffusion or conduction.

It turns out that the efficiency of the convection can be simply and fairly accurately (based upon 3D simulations) characterized by a representative length scale called the “mixing length”. This is how far a relatively intact plume or bubble of hot gas will rise adiabatically before dispersing and depositing its energy by diffusive and turbulent processes. To conserve mass, there will also be matter falling Heat transport = C ΔT back down from large radii to smaller ones, but the down- P flowing matter now carries less energy than when it went up, ⎡⎛ dT ⎞ ⎛ dT ⎞ ⎤ ⎛ dT ⎞ ΔT=⎢ − ⎥ = Δ  > 0 the difference having now been dissipated. The net heat ⎜ dr ⎟ ⎜ dr ⎟ m ⎜ dr ⎟ m ⎣⎢⎝ ⎠ ad ⎝ ⎠ actual ⎦⎥ ⎝ ⎠ being transported up (per gram) is approximately the dT d logT d logT d logP d logP T T T ⎛ ∂u ⎞ = = = ∇ heat capacity, c = , times the temperature difference, dr dr d logP dr dr P ⎝⎜ ∂T ⎠⎟ P T ΔT, between the gas inside the convective plume and = − ∇ as previously discussed (slide 14) HP outside of it evaluated at r + where r is the radius m Note: where the transport starts and m is the mixing length.

1) The "mixing length",  , is a characteristic scale That is, ΔT is given by the difference between the temperature m over which a given convection element dissipates. The gradient in the external medium times m, and the temperature change that happens to a rising adiabatic bubble after floating total extent of a convective region can itself be greater (or even less) than the mixing length. the same distance.

1 ⎛ dP ⎞ d logP 3) We defined the reciprocal of = as the P ⎜ dr ⎟ dr 2) ∇connects directy to the equation of state, ⎝ ⎠ pressure is a better variable here than radius. ∇ is "pressure scale height" (the radial distance over which dimensionless and the log's are base "e". the pressure declines by a factor of e). Almost universally

the mixing length m is taken to be a (parametric) multiplier, α, times the pressure scale height. ⎡⎛ dT ⎞ ⎛ dT ⎞ ⎤ ΔT=⎢ − ⎥ −1 ⎜ dr ⎟ ⎜ dr ⎟ m ⎛ d logP ⎞ ⎣⎢⎝ ⎠ ad ⎝ ⎠ actual ⎦⎥  =− α = αH m ⎝⎜ dr ⎠⎟ p dT d logT d logT d logP d logP = T =T =T ∇ 0.3 < α <1.5 dr dr d logP dr dr d logP is < 0 dr d logP H = − is > 0 P dr Then −1 ⎛ d logP ⎞  The heat being transported per unit area is then = H m m − α ⎜ ⎟ = α p ⇒ = − α ⎝ dr ⎠ HP

⎡⎛ dT ⎞ ⎛ dT ⎞ ⎤ F=vc ρ Δu = vc ρcP ΔT ΔT=⎢ − ⎥  ⎜ dr ⎟ ⎜ dr ⎟ m ⎣⎢⎝ ⎠ ad ⎝ ⎠ actual ⎦⎥  That is the flux is the mass flux in a given convective = T m T (∇ad − ∇)=α (∇ − ∇ad ) HP element, vc ρ, times the excess heat content (erg/gm) Where in the element compared with its surroundings. d logT ∇≡ actual in the star d logP cP , the heat capacity,can be obtained from the

5N k d logT equation of state but is A for an ideal gas. ∇ ≡ adiabatic from the EOS 2 ad d logP

It remains to determine vc ,the typical convective speed. The acceleration, a, is given by Δρ ΔT Gm(r) Δρ a=− g = g where g = a =− g ρ T r 2 ρ Recall also ΔT= αT ∇ − ∇ where Δρ is the density contrast between the floating ( ad ) plume and its surroundings, Δρ = ρinside − ρoutside < 0 2 The typical velocity while moving a distance m= 1/2 at As the plume floats (v <

dP dρ dT dµ m =αHP =0= + − so, if Δµ=0, -Δρ/ρ ≈ ΔT/T. P p T µ l αg α 2H g How superadiabatic is the temperature v ≈ m ∇ − ∇ = P ∇ − ∇ c 2 ( ad ) 2 ( ad ) in the convection region? H g 3/2 L(r) = 4πr 2ρc Tα 2 P ∇ − ∇ P 2 ( ad ) and F= ρc ΔT v ΔT= αT ∇ − ∇ Evaluate approximately for the solar convective zone, r ≈ R , P c ( ad )  −3 6 = ρc Tα ∇ − ∇ v ρ ≈ ρ = 1.4 g cm , T from Virial theorem about 3 × 10 K P ( ad ) c −1 ⎛ 1 dP ⎞ ρN kT N kT For ideal gas and HE, H = = A = A g cancels in H g 3/2 P ⎝⎜ P dr ⎠⎟ µgρ µg c T 2 P = ρ P α (∇ − ∇ad ) 1/2 2 2 ⎛ N kT ⎞ 3/2 L 4 R c T A  ≈ π ρ P (∇ − ∇ad ) ⎝⎜ 2µ ⎠⎟ H g 3/2 2 2 2 P 1/2 L(r) = 4 r F 4 r c T 33 π = π ρ P α (∇ − ∇ad ) 3/2 ((2)(0.6)) 3.9 ×10 2 ∇ − ∇ = ( ) ( ad ) (4)(π )C (6.9 ×1010 )2(1.4)(3 ×106 )[(6.02 ×1023 )(1.38 ×10−16 )(3 ×106 )]1/2 Given the degree to which the actual temperature P 0.0011 = = 3.1 ×10−12 gradient is superadiabatic, , this gives 5 (∇ − ∇ad ) N k 2 A

the convective heat flux. Typically α =1/3 to 1.5. ΔT 8 ∇ − ∇ ≈ 10− for =1; T= T ( ad )  α Δ α (∇ − ∇ad ) T

How superadiabatic is a convection region? The small degree of super-adiabicity also While one might quibble about factors of 2, the deviation implies that the required convective speeds from the adiabatic gradient is clearly very small. This is are quite modest, very subsonic. generally true in other places where convection happens as well. Exceptions: near stellar surfaces, in the outer α 2H g v = P ∇ − ∇ layers of red supergiants, and in the final stages of c 2 ( ad ) massive star evolution. N kT ≈ A ∇ − ∇ Examining the various terms, ρ, T, R, L, we see that the 2µ ( ad ) excess scales as 5/3 ΔT 2/3 ⎛ R ⎞ (Pols p 70) 1/2 ∝ L ⎛ N kT ⎞ T ⎜ M ⎟ −4 A ⎝ ⎠ = 10 ⎜ ⎟ ~ 15 meters/s 6 2µ At very high L, say 10 L and large R the temperature ⎝ ⎠  1/2 gradient could start to become substantially superadiabatic. ⎛ P ⎞ Note that this is also ~10−4 ≈ 10−4 c This rarely happens (SN progenitors), but the theory can ⎜ ⎟ sound ⎝ ρ ⎠ also break down near stellar photospheres.

Stars may become unstable for a variety of reasons

• Insensitivity of the pressure to temperature Stellar Stability as in a degenerate gas with a temperature sensitive energy generation rate

Glatzmaier and Krumholz 11 (p 127) • Too great a component of radiation or relativistic degeneracy pressure

• Burning in too thin a shell

• Opacity and recombination driven pulsations

First, why are they usually stable in the first place? Non-degenerate gas with some small amount of radiation But so long as ideal gas remains a significant contributor

M ⎛ 2 1 ⎞ ρN kT 1 4 3 P P 3 u u dm 2U U A Ω = − ⎜ gas + rad ⎟ = − gas + rad P = + aT ugas = urad = 3 ∫ 3 3 ( ) µ 3 2 ρ ρ 0 ⎝ ⎠ The Virial theorem says that the pressure and binding 1 Ugas = − (Ω +Urad ) energy are related by 2 M 1 P E = Ω +U +U = (Ω +U ) = −U Ω= − 3 dm rad gas rad gas ∫ ρ 2 0 What does this say about the mass-averaged temperature? We can note immediately the neutral stability of a star 3 N k M ρT 3 N kMT supported entirely by radation or relativistic particles. Then U = A dm = A gas 2 ∫ 2 E = U + Ω µ 0 ρ µ M P M P where the ρ in the denominator is because u is the internal = 3∫ dm − 3∫ dm =0 0 ρ 0 ρ energy per gram and The star has no net energy and its radius is undefined. 1 M T = T dm Such is the case with the Chandrasekhar mass white M ∫ 0 dwarf but also stars that are too massive and β too great. Conservation of energy then requires that Instability of a degenerate gas

dE 3 NAk dT Lnuc − L = = − M Obviously if the pressure is insensitive to the temperature dt 2 µ dt and the energy generation is very insensitive to the In thermal equilibrium the left hand side is zero, temperature, a very unstable situation exists that is but assume a small positive imbalance L=Lnuc - L prone to runaway. Temperature rises but there is no expansion and cooling. Energy generation rises and dT 2 µ δL = − increases the temperature still more, etc. dt 3 N k M A The negative sign is important here and relates again To examine this in a bit more detail, consider the to the negative heat capacity of stars as imposed by relation between central pressure and density we the Virial theorem. derived for polytropes.

2/3 4/3 (L − L) ↑ ⇒ T goes down and vice versa Pc = CnGM ρc nuc

where Cn is a slowly varying function of polytropic The star is stable. index, but we might have n = 3 in mind.

Taking the time derivative and dividing by the original A small perturbation in pressure thus perturbs the equation one has temperature and density according to dP 4 dρ dP P a a−1T bd b aT b−1dT 2/3 1/3 c c = 0 ( ρc c ρc + ρc c ) = ⎡CnGM ⎤ ρc dt ⎣ ⎦ 3 dt a b which when divided by Pc = P0 ρcTc gives 2/3 4/3 P = ⎡C GM ⎤ ρ dP dρ dT ⎣ n ⎦ c c = a c +b c P ρ T dividing gives c c c dP 4 dρ Substituting the results for the polytrope, c c , dP 4 dρc = = Pc 3 ρc P 3 ρc 4 dρc dρc dTc a b = a +b Now assume the pressure is of the form P= P0 ρ T where 3 ρc ρc Tc 4 5 for ideal gas a = b = 1, for fully degenerate gas a = to , 3 3 and dT 1 ⎛ 4 ⎞ dρ b = 0, and so on. c = − a c T b ⎜ 3 ⎟ ρ c ⎝ ⎠ c Thin shell instability dTc 1 ⎛ 4 ⎞ dρc Cases = ⎜ − a⎟ Tc b ⎝ 3 ⎠ ρc Let rshell = r0 + dr dr << r0 δm <

1) Ideal gas - a = b = 1 and δm <4/3 Change the pressure in the thin zone slightly, e.g., by burning.

dTc 1 ⎛ 4⎞ dρc = − a − This leads to a new outer radius rshell = r0 + dr + δr T b ⎝⎜ 3⎠⎟ ρ c c = old r + δr (δr still <

δP δρ dr δP δr = 4 = − 4 The mass does not change P ρ r P r shell shell when the zone expands Using the same representation of the EOS as before, 2 2 but δm = 4πr0 ρ dr = 4πr0 (ρ + δρ)(dr + δr) a b δP δρ δT δρ dr −1 P=P0 ρ T = a + b = 4 ρ dr ⎛ δr ⎞ ⎛ δr ⎞ P ρ T ρ rshell ρ + δρ = = ρ 1+ ≈ ρ 1− dr + δr ⎝⎜ dr ⎠⎟ ⎝⎜ dr ⎠⎟ So that δρ δr δr r 1 δP r δρ ⎛ dr ⎞ δT So ≈ − = − shell = shell ⎜ a − 4 ⎟ = − b ρ ⎝ r ⎠ T ρ dr rshell dr 4 P dr shell δP δρ dr δT 1 ⎛ dr ⎞ = 4 = ⎜ 4 − a⎟ T b ⎝ r ⎠ P ρ rshell shell Since a and b are positive numbers and dr << r δP δρ shell dr <

The thermally pulsing AGB phase As the He-burning shell approaches the H-He discontinuity, its luminosity decreases as it runs out of fuel. The layers above then contract somewhat in response, thus heating the extinguished H-burning shell until it is re-ignited. Both shells now provide energy and a phase of double shell burning begins. However, the shells do not burn at the same pace: the He-burning shell becomes thermally unstable and undergoes periodic thermal pulses, discussed in detail in Sec. 11.1.1. This phase is thus referred to as the thermally pulsing AGB (TP-AGB). The structure of a star during the TP-AGB phase is schematically depicted in Fig. 11.2. The thermally pulsing phase of the AGB has a number of salient properties: The periodic thermal pulses alternate with mixing episodes and give rise to a unique nucleosyn- • thesis of (among others) 12C, 14N, and elements heavier than iron (Sec. 11.1.2). This process gradually makes the stellar envelope and more carbon-rich. Dynamical instability Examples:Similar to the RGB, the stellar properties mainly depend on the size of the degenerate CO core. • In particular there is a tight core mass-luminosity relation, M L = 5.9 104L c 0.52 , (11.1) × " M − • Asymptotic Giant! " "Branch Stars The instabilities discussed so far manifest over which is not as steep as the RGB relation (10.2). a long time – typically the nuclear time, but there CS envelope Figure 11.2. Schematic structure of an AGB star during its thermally puls- are other more violent instabilities that affect ing phase. The CO core is degenerate stellar and very compact, and is surrounded by two burning shells very close to- hydrostatic equilibrium. gether in mass coordinate. The con- vective envelope by contrast is very ex- convective tended and tenuous, having a radius envelope 104–105 times the size of the core. This loosely bound envelopePrialnik is gradu- p 163 11.2 Quite basically, when a star, or a portion of a star H burning ally eroded by the strong stellar wind, which forms a dusty circumstellar enve- He burning lope out to several hundreds of stellar contracts, its pressure goes up but so too does the H, He radii. The convective envelope, stellar deg. core He atmosphere and circumstellar envelope C, O force of gravity. If gravity goes up sufficiently faster have a rich and changing chemical com- position driven by nucleosynthesis pro- cesses in the burning shells in the deep than pressure, then an unbalanced state remains 0.01 R . 100−500 R . ~1 pc? interior. 0.5−1.0 M . 0.1−few M . ~0.05 R . unbalanced and may collapse even faster. If on the 0.001−0.02 M . other hand pressure goes up faster, the star when • Classical novae 163 compressed will expand again and perhaps oscillate, but it is stable against collapse. • X-ray bursts on neutron stars

Very roughly: Now consider an adiabatic compression:

dP Gm(r) 2 −4 = − ⇒ P ∝m r γ dm 4πr 4 P=Kρ

3 2/3 4/3 γ Also ρ ∝ m / r so for hydrostatic equilibrium P ∝ m ρ . ⎛ P '⎞ ⎛ ρ '⎞ So = ρ ' > ρ If the density changes due to expansion or contraction, ⎝⎜ P ⎠⎟ ⎝⎜ ρ ⎠⎟ γ 4/3 leading to new pressure and density, P' and ρ', hydrostatic ⎛ ρ '⎞ ⎛ ρ '⎞ If, for a given mass, > one has stability, equilibrum demands ⎝⎜ ρ ⎠⎟ ⎝⎜ ρ ⎠⎟ 4/3 ⎛ P '⎞ ⎛ ρ '⎞ hydrostatic equilibrium is satisfied. If not, things are unstable. = ⎜ P ⎟ ⎜ ρ ⎟ 4 4 ⎝ ⎠ ⎝ ⎠ Thus if γ > the star is stable and if γ < it is not. If the pressure increases more than this, 3 3 4/3 This is a global analysis and doesn't necessarily apply ⎛ P '⎞ ⎛ ρ '⎞ i.e., > there will be a restoring to small regions of the star but illustrates the importance ⎝⎜ P ⎠⎟ ⎝⎜ ρ ⎠⎟ of γ =4/3. We will return to this later. force that will lead to expansion. If it is less, the contraction will continue and perhaps accelerate. Glatzmaier and Krumholz treat this more correctly using perturbation n Keep only first order terms (1+ ε) ≈ 1 + nε theory. 2 d 2 ⎡ ⎛ δr ⎞ ⎤ Gm ⎡ ⎛ δr ⎞ ⎤ ⎡ ⎛ δP ⎞ ⎤ dm ⎢r 1+ ⎥ = - dm − 4π ⎢r 1+ ⎥ d ⎢P 1+ ⎥ dt 2 0 ⎜ r ⎟ 2 0 ⎜ r ⎟ 0 ⎜ P ⎟ The force equation is ⎣⎢ ⎝ 0 ⎠ ⎦⎥ ⎡ ⎛ δr ⎞ ⎤ ⎣⎢ ⎝ 0 ⎠ ⎦⎥ ⎣⎢ ⎝ 0 ⎠ ⎦⎥ ⎢r 1+ ⎥ 0 ⎜ r ⎟ Gm 1 dP ⎣⎢ ⎝ 0 ⎠ ⎦⎥ r = - 2 − r ρ dr ⎛ δr ⎞ Gm ⎛ δr δP ⎞ 2 dm δr = 1− 2 dm − 1+ 2 + 4πr dP ⎜ r ⎟ 2 ⎜ r P ⎟ 0 0 Now multiply by dm ⎝ 0 ⎠ r0 ⎝ 0 0 ⎠ Gm Gm 2 but 0=- dm 4 r 2 dP so dmr = - dm 4 r dP 2 − π 0 0 2 − π r r 0 Initially forces are balanced and 2Gm ⎛ δr δP ⎞ dm δr = δr dm − 2 + 4πr 2 dP 3 ⎜ r P ⎟ 0 0 Gm r0 ⎝ 0 0 ⎠ 0=- dm − 4πr 2 dP r 2 To go further we must assume a behavior for P(ρ). Assume, But perturb and include r . A change in r by δr causes changes in P and because the time scale is too short for heat transport, that the compression is adiabatic, i.e., P = K ργ . ρ so that r0 → r0 +δr, P0 → P0 +δP, ρ0 → ρ0 +δρ, so 2 Then since log P = log K+γ log ρ d 2 ⎡ ⎛ δr ⎞ ⎤ Gm ⎡ ⎛ δr ⎞ ⎤ ⎡ ⎛ δP ⎞ ⎤ dm ⎢r 1+ ⎥ = - dm − 4π ⎢r 1+ ⎥ d ⎢P 1+ ⎥ dP dρ dt 2 0 ⎜ r ⎟ 2 0 ⎜ r ⎟ 0 ⎜ P ⎟ d logP = = γ d log ρ = γ ⎣⎢ ⎝ 0 ⎠ ⎦⎥ ⎡ ⎛ δr ⎞ ⎤ ⎣⎢ ⎝ 0 ⎠ ⎦⎥ ⎣⎢ ⎝ 0 ⎠ ⎦⎥ P ρ ⎢r 1+ ⎥ 0 ⎜ r ⎟ dP dρ ⎣⎢ ⎝ 0 ⎠ ⎦⎥ = γ P ρ

Also linearizing δP 3γ δr = − P r dm= 4πr 2ρ dr 0 0 2Gm ⎛ δr δP ⎞ 2 dm δr = δr dm − 2 + 4πr 2 dP ⎡ ⎛ δr ⎞ ⎤ ⎛ δρ ⎞ ⎛ δr ⎞ r 3 ⎜ r P ⎟ 0 0 dm 4 r 1 1 dr 1 0 ⎝ 0 0 ⎠ = π ⎢ 0 ⎜ + ⎟ ⎥ ρ0 ⎜ + ⎟ 0 ⎜ + ⎟ ⎝ r ⎠ ⎝ ρ ⎠ ⎝ r ⎠ ⎣⎢ 0 ⎦⎥ 0 0 2Gm 2 ⎛ δr 3γ δr ⎞ = δr dm − 4πr 2 − dP r 3 0 ⎜ r r ⎟ 0 ⎡ ⎛ 2δr ⎞ ⎤ ⎛ δρ ⎞ ⎛ δr ⎞ 0 ⎝ 0 0 ⎠ 4 r 2 1 1 dr 1 ≈ π ⎢ 0 ⎜ + ⎟ ⎥ ρ0 ⎜ + ⎟ 0 ⎜ + ⎟ ⎡ ⎤ r ρ r 2Gm 2 δr ⎣⎢ ⎝ 0 ⎠ ⎦⎥ ⎝ 0 ⎠ ⎝ 0 ⎠ = ⎢ dm − 4πr 2 − 3γ dP ⎥ r 2 0 ( ) 0 r ⎣⎢ 0 ⎦⎥ 0 ⎛ 2δr δr δρ ⎞ ⎛ 3δr δρ ⎞ 4 r 2 dr 1 .... 4 r 2 dr 1 But for the unperturbed configuration we had ≈ π 0 ρ0 0 ⎜ + + + + ⎟ = π 0 ρ0 0 ⎜ + + ⎟ ⎝ r0 r0 ρ0 ⎠ ⎝ r0 ρ0 ⎠ Gm 2 2 2 dm = − 4πr0 dP0 But dm is precisely 4πr0 ρ0dr0 so r0 δρ 3δr 1 δP So = − = 2 δr ρ0 r0 γ P0 dm δr = ⎡ − 4πr 4 − 3γ dP ⎤ ⎣ 0 ( ) 0 ⎦ r δP 3γ δr 0 = − ⎡ Gm ⎤ δr P0 r0 = ⎢ dm 4 − 3γ ⎥ r 2 ( ) r ⎣⎢ 0 ⎦⎥ 0 ⎡ Gm ⎤ δr dm δr = ⎢ dm 4 − 3γ ⎥ r 2 ( ) r ⎣⎢ 0 ⎦⎥ 0 Gm r 3 4 r δ = −( γ − ) 3 δ r0 This is an equation for a harmonic oscillator. It has solution Kippenhahn and Gm Weigert 34.1 r Aeiωt 3 4 δ = ω = ± ( γ − ) 3 r0 4 If ω is real, get oscillatory motion. If ω is imaginary (γ < ) get 3 exponential growth or decay on a time scale 1 1 τ   iω Gρ i.e., a hydrodynamic time scale

Pulsational instability by the κ mechanism requires that Dynamic Instabilities (Pols Chap 10, esp p 158) d logκ • Especially for stars with  near 4/3, the nuclear >0 d logP energy generation may oscillate periodically. Compression leads to heating which leads to or expansion and cooling but the net energy yield d logκ ⎞ ∂logκ ⎞ ∂logκ ⎞ d logT = + in a cycle is positive so the instability persists. ⎟ ⎟ ⎟ d logP ⎠ ad ∂logP ⎠ T ∂logT ⎠ P d logP This is called the epsilon instability

=κ + κ ∇ > 0 • Heat transport depends on the temperature P T ad gradient and opacity. If this combination behaves in such a way as to trap heat when the gas in the −3.5 For Kramer's opacity κ ∝ ρT and if P ∝ ρT κ P = 1 outer layers of a star is compressed and heated and κ = − 4.5, ∇ ≈ 0.4, so the condition is and release it when the gas expands and cools, T ad then the star may be subject to the kappa not generally satisfied. instability Two notable exceptions:

H− opacity where the opacity depends on a Pols 3.5 – suppression positive power of the temperature (Mira variables) of the adiabatic index

d logT Δ = Kramer's opacity but ∇ad <0.23 as in ionization zones ad d logP (Cepheid variables and RR-Lyrae stars) by partial ionization of a pure hydrogen There are two important ionization zones: gas. D returns to its standard ideal gas H I →H II and He I→HeII both near 15,000 K value when the hydrogen is either neutral or fully ionized ++ He II→HeIII (i.e., He ) near 40,000 K

If the ionization zone lies very close to the surface – and generally in a region that is already convective, it has little effect. Not enough mass (or heat capacity) Ionization in participates. This is the case for T > 7500 K the outer layers eff

of the sun. The If the ionization zone is located too deep, at high second shallow density, the effect of partial ionization is diminished. depression is The adiabatic gradient is not greatly supressed and + He turning into the heat trapped is not enough to drive an He++ oscillation of the heavy overlying material.

This happens if Teff < 5500 K

Between 5500 and 7500 K, instability can ccur.

Period-luminosity relation for Cepheids

As we showed previously when γ <4/3, the instability is dynamical −1/2 ⎛ Gm⎞ 1 = 3 4 ~ τ ⎜( γ − ) 3 ⎟ ⎝ r0 ⎠ Gρ For more massive Cepheids the radius is larger and the average density less, hence the period is longer

Also Cepeids obey a mass luminosity relation somewhat like main sequence stars. So brighter Cepheids have longer periods.

Obtaining a good match with the observations remains a challenge. http://www.atnf.csiro.au/outreach/education/senior/astrophysics/variable_pulsating.html

2001 A&A Equivalence to Glatzmaier Chap 11

3/2 F= c T 2 1/ 2gH ρ P α p (∇ − ∇ad ) −1 ⎛ 1 dP ⎞ N kT Substituting, for ideal gas and HE, H = = A P ⎝⎜ P dr ⎠⎟ µg H ⎛ dT ⎞ N k ⎛ dT ⎞ ∇= P ∇ − ∇ = A δ T ⎝⎜ dr ⎠⎟ ( ad ) µg ⎝⎜ dr ⎠⎟

3/2 F c T 2 1/ 2gH = ρ P α p (∇ − ∇ad ) 1/2 3/2 ⎛ N kT ⎞ ⎛ N k ⎛ dT ⎞ ⎞ c T 2 A A = ρ P α ⎜ δ ⎟ ⎝⎜ 2µ ⎠⎟ ⎝ µg ⎝⎜ dr ⎠⎟ ⎠ 2 3/2 3/2 2 ⎛ NAk ⎞ ⎛T ⎞ α ⎡ ⎛ dT ⎞ ⎤ = ρcP ⎢δ ⎥ ⎜ µ ⎟ ⎜ g ⎟ ⎜ dr ⎟ ⎝ ⎠ ⎝ ⎠ 2 ⎣ ⎝ ⎠ ⎦ α 2H g v ≈ P ∇ − ∇ c 2 ( ad ) H ⎛ dT ⎞ ∇ − ∇ = P δ ad T ⎝⎜ dr ⎠⎟ α 2H g H ⎛ dT ⎞ v = P P δ c 2 T ⎝⎜ dr ⎠⎟ g ⎛ dT ⎞ =αH δ P 2T ⎝⎜ dr ⎠⎟ For an ideal gas and hydrostatic equilibrium ⎛ dP ⎞ ρN kT N kT H = P / = A = A P ⎝⎜ dr ⎠⎟ µgρ µg α 2H g H ⎛ dT ⎞ N k T ⎛ dT ⎞ v = P P δ = α A δ c 2 T ⎝⎜ dr ⎠⎟ µ 2g ⎝⎜ dr ⎠⎟ which is Glatzmaier if β=1/2