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Astronomy 201a: Stellar Astrophysics . . . . . . . . . . September 30, 2014 Stellar Interiors Modeling This is where we take everything we’ve covered so far and fold it into models of spherically symmetric, time-steady stellar structure. (We’ll emphasize insight over rigorous accuracy.) Recap the 5 basic interiors equations: dMr dP GMr = 4πr2 ρ = − 2 ρ dr dr r dT (dT /dr)rad , if conv. stable = (dT /dr)ad − ∆∇T , if unstable dr dLr = 4πr2 ρ ǫ dr P = P (ρ, T, µ) Let’s assume that µ is known, either as an arbitrary input state or on the basis of pre-computed evolutionary history. Other knowns: ǫ(ρ, T, µ) , κ(ρ, T, µ) , ionization state The 5 main unknowns are Mr , ρ, P , T , Lr . Each of them has a full radial dependence from core to surface. The whole thing is a two-point boundary value problem. The values at the boundaries are integration constraints: Mr ρ P T Lr center 0 ρc Pc Tc 0 surface (r = R∗ ) M∗ 0 ? ρphoto ? 0 ? Pphoto ? 0 ? Teff ? L∗ For a lot of interiors modeling, ρ, P , T at the surface are not much different from zero. Computing them in detail is needed for atmospheres modeling. They can be determined in terms of other unknowns. But there are 6 boxed unknowns above! 1 With only 5 basic equations, one of the unknowns must just be a free parameter. −→ M∗ is usually chosen. Vogt-Russell “theorem:” Specifying M∗ and µ (i.e., mass & composition) uniquely determines all other stellar parameters. Of course, this is only strictly true in the sense that these (time-steady, spherically symmetric) equations are true. This leaves out magnetic fields, rotation, pulsation.... .............................................................................. Typically, one must integrate the differential equations numerically. Collins § 4.7 talks about the Henyey relaxation method that treats the derivatives as finite differences, & iterates on small changes to the interiors variables to both −→ find a steady-state solution, −→ compute a step in an evolutionary calculation. For insight into the physics, we’ll look at a couple of approximations that let us reach analytic solutions... (1) the constant-density model (2) polytropes 2 The constant density model Assuming ρ = ρ0 throughout the whole star is a huge approximation, but it will end up allowing us to do a huge number of useful things. It will allow us to obtain all of the results in various sections of your textbooks titled “homology relations” or “homology transformations.” (Usually those are introduced as something separate, but we get them all “for free.”) Start with mass conservation: dMr = 4πr2 ρ0 dr Z Easy! ρ0 = M∗ dMr = 4πρ0 0 Z R∗ dr r2 0 M∗ = hρi (4/3)πR∗3 so if we know any two of {ρ0, M∗ , R∗} this gives the third. If we stop the integration part-way, we get Mr (r) = 4 πρ0 r3 3 and this (black curve) often doesn’t look very much like what one gets numerically for more centrally condensed stars (red curve). Putting aside that bit of incongruity, we can solve the hydrostatic equilibrium equation for P (r), dP GMr 4πGρ20 = − 2 ρ0 = − r dr r 3 Keeping track of limits, Z Z P 4πGρ20 r ′ ′ ′ dr r dP = − 3 0 Pc 2 P (r) − Pc = − π Gρ20 r2 3 If we set P = 0 at r = R∗ , we can solve for the central pressure: 3 2π 2 2 3 GM∗2 Gρ0 R∗ = 3 8π R∗4 r2 And also, P (r) = Pc 1 − 2 R∗ Pc = Note that these results are independent of the equation of state! But we do see from the equation of state diagram (T vs. ρ) that stars typically fall under the curve at which Pgas = Prad . Thus, let’s assume (for now) that P ≡ Ptotal ≈ Pgas. So if we know P (r) and ρ0 , we can use the ideal gas equation of state to solve for T (r), GM∗ kB T c r2 ρ0 kB T (r) , so... T (r) = Tc 1 − 2 = P (r) = µmH R∗ µmH 2R∗ .............................................................................. This looks very fundamental. It is essentially an alternate “proof” of the virial theorem. Work out the gravitational potential energy... Z GMr 3 GM∗2 EG = − dMr =;;= − r 5 R∗ and the integrated thermal energy (assuming ideal gas)... Z Z R∗ 3 3 GM∗2 EG dr 4πr2 P (r) = ; ; = EK = dV U = = − ! 2 10 R 2 ∗ 0 .............................................................................. Moving forward, next is energy conservation. Let’s make use of the power-law parameterization ǫ = ǫ0 ρT ν . dLr = 4πr2 ρ0 ǫ(ρ, T ) dr 4 Let’s not worry too much about getting the full Lr (r). Energy generation is peaked near the core, so the star should reach its full L∗ in the ∼deep interior. L∗ = Z 0 L∗ dLr = 4πρ20 ǫ0 Z 0 R∗ dr r2 Tcν r2 1− 2 R∗ Defining x = r/R∗ and making the integral dimensionless, Z 1 2 ν 3 dx x2(1 − x2)ν L∗ = 4πρ0 ǫ0 Tc R∗ {z } |0 ν √ π Γ(ν + 1) a pure # · · · 4 Γ(ν + 52 ) Now, let’s start looking more at proportionalities (i.e., neglecting the constants out front) and just keep in mind that we can compute the complete expressions whenever we want. (Note, though, that the constant-density model gives ∼okay absolute numbers for pressure & temperature, but not for luminosity.) L∗ ∼ ρ20 Tcν R∗3 Thus... M∗2 ∼ R∗6 µM∗ R∗ ν R∗3 L∗ ∼ µν M∗2+ν R∗−3−ν Let’s also look at the case of a star in radiative equilibrium. In this case, dT 3κρ Lr = − dr 4acT 3 4πr2 It’s tricky because our model has T → 0 at the surface. We can choose an intermediate depth (like r ≈ R∗ /2) where we can be confident that Lr has reached L∗. 5 Neglecting constants again, let’s solve for Lr ≈ L∗. First note that, at this intermediate point, T ≈ Thus, T 3 r2 L∗ ∼ κρ 3 Tc 4 dT dr dT 2 r Tc Tc = − 2 ≈ − dr R∗ R∗ Tc4 R∗ ∼ ∼ κ ρ0 µM∗ R∗ 4 R∗ κ (M∗/R∗3) µ4 M∗3 Noting that all R∗ ’s cancel out, L∗ ∼ κ It’s interesting that we got this result without any reference to nuclear energy generation. Over ∼short time-scales (i.e., K-H), a star’s luminosity depends on how efficiently it transports energy out through the interior (not so much on how rapidly it generates that energy). Higher κ → radiative interior acts like a “blanket” keeping the energy in → lower L∗ . Over longer time-scales, the energy generation does matter. Let’s match the above scaling with what we got from energy conservation, and we can eliminate a variable... µ4 M∗3 ∼ µν M∗2+ν R∗−3−ν κ We can solve for the main-sequence mass–radius relation, as well as some other useful things. What about κ? It depends on ρ & T , so it is likely to depend on radial distance r (and also on R∗ ). For now, let’s assume κ is a constant. Absorbing it into the other constants, we get (ν−1)/(ν+3) R∗ ∼ µ(ν−4)/(ν+3) M∗ L∗ ∼ µ4 M∗3 4/(ν+3) Tc ∼ µ7/(ν+3) M∗ R∗ ∼ µ0.36 M∗0.64 L∗ ∼ µ4 M∗3 Tc ∼ µ0.64 M∗0.36 For PP chain (M∗ < ν∼4 ∼ 1.5 M⊙), For CNO cycle (M∗ > ∼ 2 M⊙ ), ν ∼ 15–20 6 տ choose intermediate ν = 8 We can also now determine the position of the main sequence in the H–R diagram. x i.e., we can solve for x in L∗ ∼ Teff Keeping µ fixed and just looking at the M∗ dependence for ν = 8, 4 σTeff = L∗ so 4πR∗2 Teff ∼ L∗1/4 R∗−1/2 ∼ M∗3/4 M∗−0.32 ∼ M∗0.43 Thus, x L∗ ∼ Teff corresponds to M∗3 ∼ (M∗0.43)x so x ≈ 6.97 . How does this compare to observations? Taking the whole main sequence into account (0.1 < ∼ M∗ /M⊙ < ∼ 100), a rough fit gives R∗ ∼ M∗0.67 L∗ ∼ M∗3.37 6.54 L∗ ∼ Teff See also the H-R diagrams given out in the 1st lecture, and Pols’ Figure 1.3: Handouts: ZAMS tabular data for stars & brown dwarfs (online). Also, see my own mass–radius diagram for stars, planets, asteroids, etc. 7 Mass-radius diagram for stars and planets. ZAMS star/brown-dwarf radii from Baraffe et al. (2003), Girardi et al. (2000), & Ekstrom et al. (2012). Exoplanet radii from exoplanet.eu. WD/NS model curve from Burrows & Liebert (1993). QCD strange-matter curve from Franzon et al. (2012). BH curve from Schwarzschild radius. Idealized degenerate model (yellow dash-dotted curve) from Thomas-Fermi theory → n=1.5 polytrope → n=3 polytrope. Solar system objects from various JPL web pages. Parallel black lines in lower-left are for constant-density models of metallic H, H2 O ice, silicates, and iron (top to bottom). 8 Note: Our assumption of constant density may make you think that we 1/3 should have gotten M∗ ∝ volume, i.e., R∗ ∼ M∗ . However, that would only be true if ρ0 was a universal constant for all stars... which we never assumed. (This would also occur for ν = 3, too.) .............................................................................. Above, we assumed that ∂T /∂r was dominated by the radiative gradient. What about convection? Let’s assume ∂T T ∂P ∂T = = ∇ad ∂r ∂r ad P ∂r µmH ∂T GMr ρ = ∇ad T For an ideal gas, − 2 ∂r ρkB T r Note that both T and ρ cancel, and we can use Mr = (4π/3)ρ0r3 and integrate: Z T (r) Z r 4π µmH ′ G ρ0 dr′ r′ dT = 3 kB ∇ad Tc 0 This looks very similar to the pressure integral. Using the surface boundary condition (T = 0 at r = R∗ ) to solve for Tc , like before, we get r2 ∇ad GM∗ kB T c T (r) = Tc 1 − 2 = and R∗ µmH 2 R∗ For ∇ad = 2/5, the coefficient is 1/5. Inconsistent with the earlier coefficient of 1/2, which came purely from hydrostatic equilibrium & the ideal gas law! What this tells us is that a constant-density, fully convective star is internally inconsistent as a concept. However, it’s only off by about a factor of two... & we’ve never let a little inconsistency stop us before. Let’s use this model to estimate the convective velocity vc, as well as ∆∇T , in an unstably stratified region. 9 Assuming efficient convection, Thus, vc ≈ L∗ 3πρ0 R∗2 1/3 L∗ Lr ≈ 4πr2 πR∗2 Fconv ≈ C1 ρ0 vc3 , But also, Fconv = ≈ 4 L∗R∗ 9 M∗ 1/3 ≈ (at r ≈ R∗/2). with C1 ∼ 3 0.04 km/s (for the Sun) Estimating the superadiabatic gradient, 2 kB T vc |∆∇T | where c2s = ≈ . S≈ |∇T |ad cs µmH r 3 R∗ 3GM∗ , T = Tc , and cs ≈ ≈ 270 km/s (for the Sun) At r = 2 4 8R∗ (If we used the convective Tc , then cs ≈ 170 km/s) Thus, S ≈ 10−4 ≪ 1 , and the actual temperature gradient is ≈ the adiabatic gradient in an efficient convective zone. Lastly, we can also see that convection mixes a star very quickly & efficiently. ℓ ≈ ∼1 week, for the Sun vc ≪ τK−H (thermal diffusion timescale ∼ 107 yr) ≪ τnuc (evolutionary timescale ∼ 109 yr) τmix ≈ 10 Let’s move on from the constant-density model to a slightly more realistic description... Polytropes Nowadays, they’re far from being state-of-the-art stellar models, but they’re still extremely useful at conveying insight about stellar physics. We’ve explored how adiabatic variations can be expressed using γ exponents... P ∼ ργ T ∼ ργ−1 Let’s take from that the idea that the actual equation of state might be written as P = K ργ which is no longer a proportionality for adiabatic changes, but now codified into the constituitive relation for P (ρ). (We’ve seen actual examples of this: electron degeneracy, and the Thomas-Fermi model. Not the ideal gas.) The traditional thing to do is combine the mass conservation & hydrostatic equilibrium equations into a single 2nd order ODE: dMr = 4πr2ρ dr dP GMr = − 2 ρ dr r GMr dρ = − 2 ρ dr r dρ GMr ργ−2 r2 = − dr Kγ Taking the derivative of both sides and using mass conservation, d dρ G 4πr2 ρ ργ−2 r2 = − dr dr Kγ 1 d 4πG γ−2 2 dρ which is a 2nd order ODE for ρ(r). ρ r = − r2 ρ dr dr Kγ K γ ργ−1 Solving it needs 2 boundary conditions: ρ(0) ≡ ρc 11 dρ dr = 0 . r=0 Lane & Emden found a useful change of variables: s ρ = ρc θ n r = αξ α= where n = polytropic index n= (1/n)−1 (n + 1)K ρc 4πG 1 γ−1 γ =1+ 1 n A lot will cancel out & simplify. n d ρ dθ 4πG 1 n c ρc(1/n)−1 θ1−n ξ 2α2 = − 2 2 n ξ α ρc θ α dξ α dξ K n+1 Above: α2 ρc cancels. (1/n)−1 ρc d 2 1−n n−1 dθ 4πG n ξ θ nθ = − α2 ξ 2 θn dξ dξ K n+1 Above: n cancels, and θ1−nθn−1 = 1. 4πG α2 1 d 2 dθ ξ = − (1/n)−1 ξ 2 θn dξ dξ K(n + 1)ρc Recalling definition of α, the whole RHS = −1. Thus, the standard form of the Lane-Emden equation is d 2 dθ ξ = −ξ 2 θn dξ dξ θ = 1 with boundary conditions at ξ = 0 . dθ/dξ = 0 The free parameter n determines the shape of the solution θ(ξ). Next time we’ll discuss analytic & numerical solutions for θ(ξ), and how they are useful in specifying stellar parameters. 12