ρ dr = GM r r 2 Let s assume that µ(r) is known, either as an arbitrary input state or on the basis of pre-computed evolutionary history.

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1 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: dm r dr dt dr = = 4πr 2 dp ρ dr = GM r ρ r 2 { (dt/dr)rad, if conv. stable (dt/dr) ad T, if unstable dl r dr = 4πr2 ρǫ P = P(ρ,T,µ) Let s assume that µ(r) 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 M r, ρ, P, T, L r. 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: center (r = 0) surface (r = R ) M r 0 M ρ ρ c 0? ρ photo? P P c 0? P photo? T T c 0? T eff? L r 0 L At the surface, ρ photo, P photo, and T eff are all their central values. Thus, for a lot of interiors modeling, it s not terrible to assume they re 0. Computing them in detail is needed for atmospheres modeling. They can be determined in terms of other unknowns. Note there are 6 boxed unknowns above, not

2 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. For interesting history, see Gautschy (arxiv: ). 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. (most recent update: arxiv: ) 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 4.2

3 The constant density model Assuming ρ = ρ 0 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: dm r dr = 4πr 2 ρ 0 Separable... ( ) R 3 M = 4πρ 0 3 M i.e., ρ 0 = (4/3)πR 3 so if we know any two of {ρ 0,M,R } this gives the third. 0 R dm r = 4πρ 0 dr r 2 M 0 = ρ (Ultimately, we d like to know everything as a function of M only... recall Vogt-Russell... but we ll get there.) If we stop the integration part-way, we get M r (r) = 4 3 πρ 0r 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), Keeping track of limits, P dp = 4πGρ2 0 P c 3 dp dr = GM r ρ r 2 0 = 4πGρ2 0 r 3 r 0 dr r P(r) P c = 2 3 πgρ2 0r 2 (though we don t know P c yet) If we set P = 0 at r = R, we can solve for the central pressure: 4.3

4 P c = 2π 3 Gρ2 0R 2 = 3GM2 8πR 4 ( ) And also, P(r) = P c 1 r2 (inverted parabola) R 2 The constant-density model is sometimes called the homogeneous compressible model, because (like water in the ocean) there can still be a gravitational buildup of pressure, despite ρ = constant. The above result is independent of the equation of state! However, we do see from the equation of state diagram (T vs. ρ) that stars typically fall under the curve at which P gas = P rad. Thus, let s assume (for now) that P P total P gas. So if we know P(r) and ρ 0, we can use the ideal gas equation of state to solve for T(r), P(r) = ρ ) 0k B T(r), so... T(r) = T c (1 r2 k B T c = GM µm H µm H 2R... This looks very fundamental. It means that the Sun s core temperature is NOT set by nuclear burning! Fusion is what allows this high T c to be maintained for much longer than would otherwise be possible. This relation is also an alternate proof of the virial theorem (for a bound star). Work out the gravitational potential energy... GM r E G = dm r = = 3 GM 2 (α = 0.6) r 5 R and the integrated thermal energy (assuming ideal gas)... R E K = dv U = dr 4πr P(r) = = R 2 GM 2 R = E G 2...! 4.4

5 Moving forward, next is energy conservation. Let s make use of the power-law parameterization ǫ = ǫ 0 ρt ν, which applies when there is one dominant nuclear burning process. dl r dr = 4πr2 ρ 0 ǫ(ρ,t) L = L 0 dl r = 4πρ 2 0 ǫ 0 Let s not worry too much about getting the fulll r (r). Energygenerationispeaked near the core, so the star should reach its full L in the deep interior. R 0 dr r 2 T ν c (1 r2 Defining x = r/r and making the integral dimensionless, R 2 ) ν 1 L = 4πρ 2 0 ǫ 0Tc ν R3 dx x 2 (1 x 2 ) ν } 0 {{} a pure # π 4 Γ(ν +1) Γ(ν ) 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 ρ 2 0T ν c R 3 M2 R 6 Thus... L µ ν M 2+ν ( µm R R 3 ν ) ν R 3 Let s also look at the case of a star in radiative equilibrium. (Most stars have a big interior radiative zone, somewhere.) 4.5

6 In this case, dt dr = 3κρ 4acT 3 L r 4πr 2 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 L r (r) has reached L. We won t integrate, just evaluate quantities there. Neglecting constants again, let s solve for L r L. First note that, at this intermediate point, Thus, L T3 r 2 κρ T 3 4 T c ( ) dt dr dt dr = 2rT c R 2 T4 c R κρ 0 ( µm R T c R ) 4 R κ (M /R 3 ) Noting that all R s cancel out, L µ4 M 3 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... µ 4 M 3 µ ν M 2+ν R 3 ν κ We can solve for the main-sequence mass radius relation, as well as some other useful things. 4.6

7 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 R µ (ν 4)/(ν+3) M (ν 1)/(ν+3) R µ 0.36 M 0.64 L µ 4 M 3 L µ 4 M 3 T c µ 7/(ν+3) M 4/(ν+3) T c µ 0.64 M 0.36 For PP chain (M < 1.5M ), ν 4 For CNO cycle (M > 2M ), ν } տ choose intermediate ν = 8 We can also now determine the shape of the main sequence in the H R diagram. i.e., we can solve for x in L T x eff Keeping µ fixed and just looking at the M dependence for ν = 8, (and using L M 3 and R M 0.64 ), Thus, σt 4 eff = L 4πR 2 so T eff L 1/4 R 1/2 M 3/4 M 0.32 M 0.43 L T x eff corresponds to M 3 (M0.43 ) x so x 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 L Teff

8 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 M / R / ρ diagram for stars, planets, asteroids, etc. 4.8

9 Mass radius and Mass density plots for stars and planets. Stellar data from Torres et al. (2010) (red symbols), and model ZAMS (red curve) from Baraffe et al. (2003), Girardi et al. (2000), & Ekstrom et al. (2012). Exoplanet & brown dwarf radii from Hatzes & Rauer (2015). 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. 4.9

10 Note: Our assumption of constant density may make you think that we should have gotten M Volume, i.e., R M 1/3. However, that would only be true if ρ 0 was a universal constant for all stars... which we never assumed. Note: this model gave us both ρ 0 and R for a given M... Vogt-Russell theorem has been upheld!... Above, we assumed that T/ r was dominated by the radiative gradient. What about an entire star dominated by convection? Let s assume ( ) T T r = T P = ad r ad P r T For an ideal gas, r = adt µm ( H GM ) rρ ρk B T r 2 Note that both T and ρ cancel, and we can use M r = (4π/3)ρ 0 r 3 and integrate: T(r) T c dt = 4π 3 µm H ad k B Gρ 0 This looks very similar to the pressure integral. r 0 dr r Using the surface boundary condition (T = 0 at r = R ) to solve for T c, like before, we get ( ) T(r) = T c 1 r2 k B T c and = ad GM R 2 µm H 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! Thus, a constant-ρ, fully convective star is internally inconsistent as a concept. If ρ = constant, then blobs can t be buoyant (in 1/r 2 gravity)!? 4.10

11 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 v c, as well as T, in an unstably stratified region. Assuming efficient convection, F conv = L r 4πr 2 L πr 2 (at r R /2). Thus, But also, F conv C 1 ρ 0 v 3 c, with C 1 3 v c ( L 3πρ 0 R 2 ) 1/3 ( 4 9 Estimating the superadiabatic gradient, ) 1/3 L R 0.04 km/s (for the Sun) M S = T T ad = ad ad ( vc c s ) 2 where c 2 s = k BT µm H. At r = R 2, T = 3 4 T 3GM c, and c s 270 km/s (for the Sun) 8R (If we used the convective T c, then c s 170 km/s) Thus, S , 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. τ mix l v c 1 week, for the Sun t K H t MS (thermal diffusion timescale 10 7 yr) (evolutionary timescale yr) 4.11

12 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 ρ γ T ρ γ 1 Let s take from that the idea that the actual equation of state might be written as P = Kρ γ 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. Strictly speaking, not the ideal gas.) The traditional thing to do is combine the mass conservation & hydrostatic equilibrium equations into a single 2nd order ODE: dm r dr = 4πr 2 ρ dp dr = GM r r 2 ρ γ 1 dρ Kγρ dr = GM r ρ r 2 ρ γ 2 r 2 dρ dr = GM r Kγ Taking the derivative of both sides and using mass conservation, [ d ρ γ 2 r 2 dρ ] = G ( 4πr 2 ρ ) dr dr Kγ [ 1 d ρ γ 2 r 2 dρ ] = 4πG which is a 2nd order ODE for ρ(r). r 2 ρ dr dr Kγ Solving it needs 2 boundary conditions: ρ(0) ρ c ( ) dρ dr r=0 =

13 Note: we don t yet know the value of ρ c... that s okay. Also, you may wonder why we impose (dρ/dr) = 0 at the origin? Recall that dp/dr = 0 at the origin from hydrostatic equilibrium (since M r 0 there, and that dominates L Hopital s rule in the GM r /r 2 term. Polytropes have P = Kρ γ, so dρ/dr = 0. (i.e., the central core isn t a sharp density cusp ) Lane & Emden found a useful change of variables: ρ = ρ c θ n (n+1)kρ c (1/n) 1 r = αξ α = 4πG where n = polytropic index n = 1 γ 1 γ = 1+ 1 n A lot will cancel out & simplify. [ 1 d ρ ξ 2 α 2 ρ c θ n c (1/n) 1 αdξ θ 1 n ξ 2 α 2 ρ cdθ n ] αdξ = 4πG K ( ) n n+1 Above: α 2 ρ c cancels. ρ (1/n) 1 c α 2 ξ 2 θ n d dξ [ ξ 2 θ 1 n nθ n 1 dθ dξ ] = 4πG K ( ) n n+1 Above: n cancels, and θ 1 n θ n 1 = 1. [ 1 d ξ 2 dθ ] ξ 2 θ n dξ dξ = Recalling definition of α, the whole RHS = 1. 4πGα 2 K(n+1)ρ (1/n) 1 c Thus, the standard form of the Lane-Emden equation is [ d ξ 2 dθ ] = ξ 2 θ n dξ dξ with boundary conditions { θ = 1 dθ/dξ = 0 } at ξ =

14 The free parameter n determines the shape of the solution θ(ξ). For a given n, we ll call the first zero ξ n (see black circles), and consider that to be the stellar surface. Note that K, α, & ρ c don t appear in the Lane-Emden eqn itself, so they re only used for translating θ(ξ) back into real units.... One more bit of physics: note that ρ T n. If we consider T(r) known, then the higher the n, the more strongly ρ(r) varies from core to surface. n is a kind of compressibility index.... Handout: Instead of θ(ξ), plot ρ(ξ/ξ n ) in units of ρ(0) = ρ c. Note there are analytic solutions for n = 0, 1, 5. We ve got a table of ξ n and other useful scalars vs. n (online). There s a whole book about these functions: Polytropes: Applications in Astrophysics and Related Fields, by Georg P. Horedt (Springer, 2004). 4.14

15 Scaled radial dependence of density in polytropes with a range of indices n = 1/(γ 1). Also shown are models of internal density structure for the Sun (red) and Earth (blue). The solar model comes from Eggleston s STARS code ( stars/), and the terrestrial model is known as the Preliminary Reference Earth Model (PREM), from Dziewonski & Anderson (1981, Phys. Earth Plan. Int., 25, 297). Interesting Polytropic Indices n γ Description 0 Incompressible gas; constant density /3 Thomas-Fermi EOS 1 2 analytic Lane-Emden solution; constant R ; Jeans polytrope 1.5 5/3 ideal monatomic gas EOS; convective; non-relativistic degenerate ϕ ϕ n = γ when it is the golden ratio (ϕ = ) 2 3/2 Holzer & Axford s maximum γ for an accelerating solar wind 2.5 7/5 ideal diatomic gas EOS 3 4/3 Eddington s standard model; ultra-relativistic degenerate; constant M /13 Chandrasekhar s constant-ǫ Kramers model 5 6/5 Schuster sphere of infinite radius 1 Isothermal gas; Bonnor-Ebert sphere n = 0 n = 1 θ(ξ) = 1 ξ2 6 θ(ξ) = sinξ ξ ξ 0 = 6 ξ 1 = π n = 5 θ(ξ) = ) 1/2 (1+ ξ2 ξ 5 =

16 On the plot, see the list of interesting values of n... n = 0: incompressible, constant-density sphere. This is our earlier model. planets: n = stars: n = n = 5: ξ n, so this is formally an infinite-radius star. n : corresponds to γ = 1, an isothermal sphere (solution not shown).... How do we start converting information about θ(ξ) into useful stellar parameters? Radius: if r = αξ, then R = αξ n. However, we do need to know K and ρ c to specify α. Mass: M = R 0 ξn dr 4πr 2 ρ(r) = 4πα 3 ρ c dξ ξ 2 [θ(ξ)] n But the integrand is the right-hand side of the Lane-Emden equation itself! Thus, ξn [ M = 4πα 3 ρ c dξ d ( ξ 2 dθ )]. dξ dξ 0 The integral of a derivative is just the argument itself, evaluated at the proper limits. ( ξ 2 dθ ) ( = 0 at ξ = 0, and Θ n ξ 2 dθ ) dξ dξ ξ n with Θ n being a numerically computed scalar function of n (see table). Thus, M = 4πα 3 ρ c Θ n. Validate earlier results for n = 0: we can compute Θ n 4.9, so α R /2.4, and thus M = (4π/3)R 3 ρ c as it should

17 Note that we can also use above definitions of M and R to write ρ = M (4π/3)R 3 and the dimensionless ratio ρ c ρ = ξ3 n 3Θ n (α s cancel out). This is a good measure of the central condensation of a star (also tabulated as a function of n). n = 0 ρ c / ρ = One can work out a few other pure functions of n (see Pols, Chapter 4): P c = W n GM 2 R 4 and k B T c µm H = H n GM R (for ideal gas) Recall that for the n = 0 constant-density model, we had W n = 3/(8π) and H n = 1/2. (Interestingly, H n stays close to 1/2 all the way up to n 2.5.)... We can do a bit more to explore polytropic mass radius relationships. Even though α (i.e., the K constant) dropped out of the ratio ρ c / ρ, it s still true that both R and M depend on it. If we knew R and M, we can solve for K. R = αξ n = [ ] 1/2 (n+1)k ρ (1 n)/2n 4πG c ξ n [ ] 3/2 (n+1)k M = 4π ρ (3 n)/2n c 4πG Solve one of these for ρ c & plug into the other, to get [ ] (4π) 1/n K = n+1 Θ(1 n)/n n ξ n (n 3)/n }{{} N n Θ n GM (n 1)/n R (3 n)/n 4.17

18 If K was a known universal constant, this gives a definite mass radius relation. (But, of course, much like our use of ρ 0 in the const-density model, it s not necessarily a constant from star to star.) But if it is, then R M (1 n)/(3 n) Note that for the incompressible case (n = 0), we get R M 1/3, which is what we expected from a constant-density rubble pile. Other interesting limits: n = 1: R M 0 one universal radius, no matter the mass! n = 3: M R (3 n)/(1 n) R 0 one universal mass, no matter the radius! In both cases, these universal values still depend on K.... There s one interesting case in which K really is a known constant: Electron degeneracy: In the non-relativistic limit, recall that ( ) 2/3 P = h2 3 ne 5/3 = Kρ 5/3 20m e π since we can write n e = ρ/(µ e m H ). Recall that µ e is the plasma s mean mass per free electron, in units of a hydrogen atom. Thus, in this case, γ = 5 3 or n = 1.5 and R M 1/3. Counter-intuitive! As mass is added, the star compresses (R and ρ c ). 4.18

19 Ultimately, as ρ c increases (see the T-vs-ρ diagram), degenerate electrons will become relativistic, and γ = 4 3 n = 3 i.e., converges to a unique M. What is that unique mass? In the ultra-relativistic limit, P = hc ( ) 1/3 ( ) 4/3 3 1 ρ 8 π m 4/3 µ e }{{ H} Thus, (cgs) { K ( )µ 4/3 e Θ } M 5.8M µ 2 e Recall that mean molecular weight per free electron is µ e 2 1+X For present-day solar composition (X 0.73), µ e But for white dwarf cores, X 0, so µ e 2. Thus, M 1.4M is the Chandrasekhar mass. This is typically the upper limit to WD masses. Most WDs have masses in the range of M. I ve seen one with 1.28 M called ultramassive. Recent WD models that include magnetic fields (and/or rapid rotation) may allow maximum masses up to M (arxiv: )

20 If we start at the non-relativistic white dwarfs, and go to even lower masses (i.e., ρ c, T c ), we get into the regime of cold, low-density degenerate gases in which the ions become crystallized. The system converts from γ = 5/3 to a Thomas-Fermi equation of state (γ = 10/3), i.e., n = 3 7 and R M 2/9. Going even more to the left in the above diagram, the curve steepens from M to the solid rock agglomeration model (n = 0) of R M 1/3. The polytropic index varies (right to left) as n = The maximum radius occurs around { M 1M Jupiter R 1R Jupiter } 4.20

21 Digression on solid rock When degenerate bodies become even less massive, gravity starts being less important than Coulomb forces in holding the thing together. (e.g., solid rock doesn t need gravity to hold itself together!) In planetary science textbooks, you can find arguments that use the virial theorem, but replace E G by E Coulomb, and you can derive n 0 by balancing it with E K (i.e., nearly incompressible crystalline matter.) Instead of writing P(ρ), planetary scientists switch it around, H 2 O ice: ρ SiC silicate rock: ρ ρ(p) = ρ 0 +δρ(p) Mg/Si silicate rock: ρ iron ores: ρ g/cm 3 where ρ 0 is a universal laboratory density and δρ is a small, pressure-dependent perturbation, usually P 0.5 or so. Self-gravity is still moderately strong for planets, though. For example, gravity is what keeps planets spherical. Moons & asteroids: not so much ( potatos rubble piles). What dominates? (1) interior pressure (due to self-gravity) or (2) the tensile strength or rigidity S of the rock? (i.e., its ability to resist external forcing) { } compressive, S = strength, in units of force/area. or crushing If P > S, then gravity wins and it s spherical. For the constant-ρ model, P(r) = 2π 3 Gρ2 (R 2 r 2 ) and at r R 2, P π 2 Gρ2 R

22 ice/sandstone: S 10 7 silicate rock: S 10 8 Examples: iron ores: S 10 9 dynes/cm 2 stainless steel: S carbon nanotubes: S S Thus, for sphericity, one needs R > πgρ 2 ices: km. iron cores: km. See, e.g., Lineweaver & Norman (2010, arxiv: ): 4.22

23 Before we re done with this section on classical/polytropic stellar interiors, let s go from the least massive bodies to the most massive (i.e., M > 5M ). Radiation pressure & the Eddington Limit: We ve assumed quite frequently that P tot = (P gas +P rad ) is dominated by P gas. How do stars behave when P rad = 1 3 at4 is important? Let s look at hydrostatic equilibrium: P r = ρg (That s P tot on the left side!) P gas r at3 T r = ρg And if radiation dominates the interior energy transport, ( ) P gas 4 3κρ = ρg + r 3 at3 4aT 3 F rad c [ = ρ GM ] r +a r 2 rad where a rad = κf rad c a rad is the radiative acceleration on matter.. Near the surface, M r M and L r L, so the total acceleration on the right-hand side is often written [ GM + κl ] r 2 4πcr 2 Notice that both terms drop off as 1/r 2. The outward radiation force cancels out some fraction of gravity. Define g eff = GM r 2 (1 Γ) so that P gas r = ρ g eff and the Eddington factor is defined as Γ = Γ must be < 1 for a star to remain hydrostatic & bound. κl 4πcGM 4.23

24 Look back to the Rosseland κ diagram. Above T 10 4 K, the lower limit is from the Thomson scattering of free electrons. That s often used as a baseline, and the constants work out to Γ e ( )(1+X) L /L M /M For the Sun & less massive stars, radiation pressure isn t so important. Recall our constant-density model, for which L µ4 M 3 κ so that Γ κl M µ 4 M 2. ( Scaling it with the Sun, we get Γ e M M for the H-burning main sequence. When M > 200M, we get Γ e > 1 and gravity can t hold the star together any more! If κ were higher, this limit would be reached at a lower mass. Observationally, there s the Humphreys-Davidson limit, above which no stars seem to exist. )

25 This is surely related to the Eddington limit, but stars up in this part of the H-R diagram are luminous blue variables (LBVs): strong pulsations, strong winds. Lots of complex dynamics.... Note: We should pause and be a bit skeptical of using our earlier L M 3 scaling, since it was based on a model where P tot = P gas. That s NOT true for the hottest, most massive stars! Thus, let s develop a model with P gas +P rad. The reason that Γ 1 is called the Eddington limit is that A. S. Eddington worked on it in the context of a specific model of luminous stars: Eddington s standard model It s really just an n = 3 polytrope, but with specific extra meaning for the K constant. Why n = 3? Recall the initial lectures about the adiabatic exponents. For a pure photon gas in radiative (Bose-Einstein) equilibrium, γ = Γ 1 = Γ 2 = Γ 3 = 4 3 which is equivalent to n = 3. Eddington started with a key assumption, which we ll prove later: β P { } gas constant everywhere = (free param.) P tot inside the model star But is this consistent with an n = 3 polytrope? Let s see. Under this assumption, we can write P tot in two equivalent ways. First, P tot = P gas β = ρk BT βµm H. (*) Second, since 1 β = P rad P tot, then P tot = Keeping in mind that β is just a constant, solve for at 4 3(1 β). T = [ 3(1 β)ptot a ] 1/4 4.25

26 and plug into the first (*) expression for P tot... P tot = ρk [ B 3(1 β)ptot βµm H a There s a P tot on both sides. Divide by P 1/4 tot to get P 3/4 tot = ρk [ ] 1/4 B 3 βµm H a (1 β) and P tot = ] 1/4 [ ] 3k 4 1/3 B (1 β) ρ 4/3 = Kρ 4/3 aµ 4 m 4 H β4 which confirms that this idealized star is an n = 3 polytrope!... What can we do with this model? Recall that for an n = 3 polytrope, K sets the mass. ( ) 3/2 K M = 4π Θ 3 Θ πg so if we know β and µ, that uniquely determines the stellar mass! We ll solve for it in a minute, but first let s realize that β is expressible in terms of another parameter. Let s write two differential equations that we know: dp tot dr = GM rρ r 2 dp rad dr = κρ c Both contain 1/r 2. Divide one by the other, and we get dp rad dp tot = κl r 4πcGM r Γ L r 4πr 2 (near surface) But we also can write P rad = (1 β)p tot. If β really is constant, we get dp rad dp tot = P rad P tot = 1 β Thus, in the Eddington standard model, Γ = 1 β. 4.26

27 Working out the n = 3 polytrope expression for mass, we get M (18M ) Γ 1/2 µ 2 (1 Γ) 2 At low values of Γ & M, we recover Γ M 2 as before (where P tot P gas ). The full model says that M can be increased indefinitely. Really, though, once Γ gets to be of order 0.5 or higher, stellar interiors start to behave less like n = 3 polytropes. β constant. LBVs are strange & just barely confined by gravity. The huge P rad cancels out a lot of gravity, and the star isn t so centrally condensed. There may even be regions where (vector) g eff points outward, then in again... Lots more in Stan Owocki s review (arxiv: ) and Eliot Quataert s recent work (arxiv: ). 4.27

28 One more interesting thing: fully convective stars are n = 1.5 polytropes. How? Recall the internal temperature gradient for a fully convective star, ( ) 1 T For = ad, T r = Γ2 1 1 P Γ }{{ 2 P r. } ad But also consider how T depends on P for an ideal gas (P = ρk B T/ m ) that is also a polytrope (P = Kρ γ ). First solve ideal gas law for T, T = m P ρk B then solve the polytrope law for ρ and plug that into the ideal gas law, which means that and it s clear that γ = Γ 2 in this case. T = m K1/γ k B P (γ 1)/γ, 1 T T r = ( γ 1 γ ) 1 P P r For a fully ionized interior, Γ 2 = 5/3, so the polytropic n =