Lecture 7a Polytropes Prialnik Chapter 5 Glatzmaier and Krumholz Chapter 9 Pols 4 We have reached a point where we have most of the informaon required to solve realis3c stellar models. A cri3cal piece s3ll mixing is the nuclear physics, but it turns out that many of the observed proper3es of stars, except their life3mes and radii, reflect chiefly the need to be in hydrostac and thermal equilibrium, and not the energy source. Historically, prior to computers, stellar structure was o@en calculated using polytropes. Even today. they provide a valuable tool for understanding many stellar proper3es. The Stellar Structure Equations dm dr 1 = 4πr 2 ρ = dr dm 4πr 2 ρ dP Gm dP Gm = − ρ = − dr r 2 dm 4πr 4 dT 3 κρ L(r) dT 3 κ L(r) = − = − dr 4ac T 3 4πr 2 dm 4ac T 3 (4πr 2 )2 dL(r) dL(m) = 4πr 2 ρε = ε dr dm S3ll to be defined - In addition we have or will have the physics equations ρ ⎞ 1 4 P = P(T, , Y ) N kT P aT ρ { i } = A ⎟ + e + µ ⎠ ions 3 (T, , Y ) aT b κ = κ ρ { i } ≈ κ 0 ρ (T, , Y ) mT n (TBD) ε = ε ρ { i } ≈ ε0 ρ and the boundary conditions at r = 0 L(r)=0, m = 0 1/4 ⎛ L(R) ⎞ at r = R P(R) ≈ 0, m= M, T= ≈0 ⎝⎜ 4πR2σ ⎠⎟ These are 7 equations in 7 unknowns: ρ, T, P, L, ε, κ, and r which, in principle, given M and initial composition, can be solved as a function of time for the boundary conditions. Most of our theoretical knowledge of stellar evolution comes from doing just that. Aside: Boundary Conditions l Radiative Zero BC: Ideally we would use some atmospheric model to tell us what the temperature BC is at the surface The T change over the whole star is so large that the difference between 0 and the real Teff at surface is small – except when computing a luminosity. To get effective T at the end, use: Polytropes We assume a global relaon between pressure and density: P ∝ ργ γ =(n +1) / n n is called the polytropic index. Don’t confuse this with the adiabatic index which is local. 5/3 4/3 Some examples: P ∝ ρ or ρ n =constant l Fully convective (adiabatic): l White dwarfs (completely degenerate): l Pressure is a mix of gas + radiation, but the ratio is constant throughout Polytropes l Consider HSE dP Gm(r) = − ρ r = 0 dr r 2 r 2 dP = − Gm(r) ρ dr 2 l Differentiating again and dividing by r : d ⎛ r 2 dP ⎞ dm(r) = − G = − 4πGr 2ρ(r) dr ⎜ dr ⎟ dr ⎝ ρ ⎠ 1 d ⎛ r 2 dP ⎞ = − 4πGρ(r) r 2 dr ⎜ ρ dr ⎟ ⎝ ⎠ Polytropes l Now make this dimensionless Central density: Define: Then: P(r)=Kργ (r) n +1 1 n γ = = 1+ γ γ n n = K ρc θ (r) 1+1/n n+1 = K ρc θ (r) Given n, K, and c these two equations define the distribution of pressure and density in the star Use these assumptions in the equation for hydrostatic equilibrium 1 d ⎛ r 2 dP ⎞ = − 4πGρ P =Kργ =K ργ θ γ n =K ργ θ n+1 r 2 dr ⎝⎜ ρ dr ⎠⎟ c c K ργ d ⎛ r 2 dθ n+1 ⎞ c 4 G n n 2 ⎜ n ⎟ = − π ρcθ ρ = ρcθ r dr ⎝ ρcθ dr ⎠ 1 d ⎛ r 2 dθ n+1 ⎞ P = − 4πGρ θ n c r 2 dr ⎝⎜ θ n dr ⎠⎟ c P (n +1) d ⎛ r 2θ n dθ ⎞ c 4 G n 2 2 ⎜ n ⎟ = − π θ ρc r dr ⎝ θ dr ⎠ ⎡P (n +1)⎤ 1 d ⎛ 2 dθ ⎞ n ⎢ c ⎥ r = −θ 4πGρ 2 r 2 dr ⎝⎜ dr ⎠⎟ ⎣⎢ c ⎦⎥ ⎡P (n +1)⎤ 1 d ⎛ 2 dθ ⎞ n ⎢ c ⎥ r = −θ 4 G 2 r 2 dr ⎜ dr ⎟ ⎣⎢ π ρc ⎦⎥ ⎝ ⎠ Note that the quantity in brackets has units length2 Define it to be α 2 then 1/2 α 2 d ⎛ dθ ⎞ ⎡P (n +1)⎤ r 2 n with = c 2 ⎜ ⎟ = −θ α ⎢ 2 ⎥ r dr ⎝ dr ⎠ ⎢ 4πGρc ⎥ ⎣ ⎦ 2 6 dyne gm cm 2 2 2 2 2 = cm for the dimension of α cm dynecm gm Now define r =αξ with ξ a dimensionless radius- like variable. Then 1 d ⎛ dθ ⎞ ξ 2 = −θ n ξ 2 dξ ⎝⎜ dξ ⎠⎟ This is the Lane Emden equation. It involves only dimensionless quantities and can be solved for a given n to give θ(ξ). n does not have to be an integer. Basically all that went into it was hydrostatic equilibrium, mass conservation, and a power law equation of state Polytrope boundary condions ρ r 1) θ n = So at = ξ =0 θ =1 ρc α r 1 dθ dθ 2) ξ ≡ so dξ= dr ⇒ = α α α dξ dr dρ dθ but ρ ≡ ρ θ n so =nρ θ n−1 c dr c dr dθ 1 dρ dθ α dρ and ⇒ = n−1 = n−1 dr nρc θ dr dξ nρc θ dr but in spherical symmetry for hydrostatic equilibrium dρ ρ is a local maximum at r = 0, so = 0 dr dθ So at at ξ =0 = 0 dξ That is dP Gm(r)ρ = − → 0 as r → 0 dr r 2 because m(r) ∝ r 3 approaches 0 faster than r 2 dP dρ And since P=K ργ = γ Kργ −1 dr dr dρ so also →0. The density and pressure have local dr maxima at the center of the star - no "cusps" P or r Polytrope boundary condions 3) At the surface r = R the density (first) goes to zero so R =αξ where ξ is where θ(ξ) first reaches 0 * 1 1 For any value of n, the mass R ξ1 r = αξ R = αξ M = 4πr 2ρ dr = 4πα 3 ρ ξ 2θ n dξ 1 ∫ c ∫ n 0 0 ρ = ρ θ c But the Lane Emden equation says 1 d ⎛ dθ ⎞ d ⎛ dθ ⎞ ξ 2 = −θ n or ξ 2θ n = − ξ 2 ξ 2 dξ ⎝⎜ dξ ⎠⎟ dξ ⎝⎜ dξ ⎠⎟ ξ 1 d ⎛ dθ ⎞ So M = − 4πα 3 ρ ξ 2 dξ c ∫ ⎜ ⎟ 0 dξ ⎝ dξ ⎠ ξ1 dθ 3 ⎛ 2 dθ ⎞ 3 2 dθ ⎞ =0 at ξ=0 = − 4πα ρc d ξ = − 4πα ρc ξ1 = M dξ ∫ ⎜ ⎟ ⎟ 0 ⎝ dξ ⎠ dξ ⎠ ξ1 dθ ⎞ The quantity ξ 2 is uniquely determined by n and 1 dξ ⎠⎟ ξ1 is given in tables. So we have an equation connecting M, α. and ρc for a given polytropic index, n. Cox and Guilli (1965) 2 dθ ρ ξ ξ D = c 1 1 n dξ ρ Further α can be expressed in terms of ρc and K 1/2 1/2 n 1 1 n 1/2 + − ⎡ n ⎤ ⎡ n ⎤ ⎡Pc (n +1)⎤ ⎢Kρc (n +1)⎥ ⎢Kρc (n +1)⎥ α = ⎢ ⎥ = = 4 G 2 ⎢ 4 G 2 ⎥ ⎢ 4πG ⎥ ⎣⎢ π ρc ⎦⎥ π ρc ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥ 3/2 1−n ⎡ n ⎤ 3 2 dθ ⎞ ⎢Kρc (n +1)⎥ 2 dθ ⎞ M = − 4πα ρ ξ = − 4π ρ ξ c 1 d ⎟ ⎢ 4 G ⎥ c 1 d ⎟ ξ ⎠ ξ π ξ ⎠ ξ 1 ⎣⎢ ⎦⎥ 1 3/2 3/2 3−3n n +1 +1 ( ) 2 dθ ⎞ ⎛ K ⎞ 2n = − ξ1 ⎟ ⎜ ⎟ ρc 4 dξ ⎠ ⎝ G⎠ π ξ1 3/2 3/2 n +1 dθ ⎞ ⎛ K ⎞ 3−n M ( ) 2 2n = − ξ1 ⎟ ⎜ ⎟ ρc 4 dξ ⎠ ⎝ G⎠ π ξ1 which is another useful form which gives M in terms of ρc (or vice versa) for a given K and n. Note that M is independent of ρ if n = 3. c Another interesting quantity that can be obtained from the tables is the ratio of central density to average density - how compact the core of the star is. See the previous table. The case n = 0, Dn = 1 is the sphere of constant density case Analytic solutions For n = 0, 1, 5, and only for these values, there exist analytic solutions to the Lane Emden equation. Unfortunately none of them correspond to common stars, but the solutions help to demonstrate how polytropes can be solved and they can used for interesting approximate cases. n = 0 Technically the case n = 0 is a singularity P = Kρ1+1/n diverges as n → 0 This reflects the fact that constant density can only be maintailed in the face of gravity if the fluid is incompressible (like the ocean; P can vary but not ρ). Nevertheless, some interesting properties of polytropes can be illustarted with n = 0 so long as we don't use the pressure-density relation explicitly n = 0 - the constant density case 1 d ⎛ dθ ⎞ ξ 2 = −θ n = −1 ξ 2 dξ ⎝⎜ dξ ⎠⎟ dθ ξ 3 ξ 2 = − ξ 2 = − + C dξ ∫ 3 dθ ξ C 1 Cdξ = − + ⇒ dθ = − ξ dξ + dξ 3 ξ 2 ∫ 3 ∫ ∫ ξ 2 ξ 2 C θ = − − + D but θ=1 at ξ=0 6 ξ So C = 0 and D =1 and ξ 2 θ = 1 − 6 hKp://www.vistrails.org/index.php/User:Tohline/SSC/Structure/Polytropes 6 The first zero of θ = 1- 2 is at ξ1 = 6 r ξ ξ = α so R = 6 α n+1 P(r) = Pcθ (r)=Pcθ(r) for n = 0 ⎛ ξ 2 ⎞ ⎛ r 2 ⎞ P(r) = P 1− =P 1− c ⎝⎜ 6 ⎠⎟ c ⎝⎜ α 2 6⎠⎟ R R ⎛ r 2 ⎞ but = so P(r) P 1 α = = c ⎜ − 2 ⎟ ξ1 6 ⎝ R ⎠ For ideal gas pressure and constant composition T would have the same dependence (ρ=constant) ρN kT ⎛ r 2 ⎞ P = A ⇒ T(r) = T 1− c ⎜ R2 ⎟ µ ⎝ ⎠ 1/2 1/2 ⎡P (n +1)⎤ ⎡ P ⎤ R but also α =⎢ c ⎥ = ⎢ c ⎥ = so 4 G 2 4 G 2 ⎣⎢ π ρc ⎦⎥ ⎣⎢ π ρc ⎦⎥ 6 ⎛ 4 ⎞ G R3 2 2 ⎜ π ρc ⎟ ρc 2πR Gρc ⎝ 3 ⎠ GMρ Pc = = = 3 2R 2R This agrees with what can be obtained by integration of the equation of hydrostatic equilibrium.