Physics 576 Stellar Astrophysics Prof. James Buckley. Lecture 11 Polytropes and the Lane Emden Equation

Transcription

1 Physics 576 Stellar Astrophysics Prof. James Buckley Lecture 11 Polytropes and the Lane Emden Equation

2 Reading/Homework Assignment Makeup class tomorrow morning at 9:00AM Read sections 2.5 to 2.9 in Rose over spring break! Midterm Exam, April 5 (special time? take home?) Final Project, May 4 (end of reading period)

3 Stellar Structure

4 Equations of Stellar Structure 1 Hyostatic Equilibrium : dp 2 Mass Continuity : dm(r) (r) =4πr 2 ρ(r) = ρ(r) GM(r) r 2 3 Eqn. of State : P gas = nkt(r) = ρ(r) µm u kt(r) P rad (r) = 1 at (r)4 3 4 Radiative transport equation : T 3 κρ L r = 16πac r 2 T 3 5 Energy dl(r) Production/Conservation= T γ 4πr 2 ρ(r)ɛ(r) Convective Energy Transport: P dt dp = γ 1

5 Convection For a stellar atmosphere where radiation pressure is negligible, the equation of state is given by the ideal gas law: Differentiating both sides gives P = ρ m kt ln P = ln ρ + ln T + constant dp P Starting with the expression for stability = dρ ρ + dt T Po ρ o dρo dp o < 1 γ or dρ o ρ o 1 γ dp o P o < dpo γ P o Stability condition : P T dt o T o < 0 dt dp > γ 1 γ

6 Convective Transport Stability condition : P T dt dp > γ 1 γ In your textbook, the stability condition is derived in terms of the temperature gradient giving Stability condition : dt < g c p The δt and velocity (from the boyant force) of the rising elements are determined by the difference between the temperature gradient (outside) and the adiabatic gradient (inside), any difference will result in convective transport that tends to reduce the outside temperature gradient. So, to a reasonable degree of accuracy we can assume that in a convective zone, the temperature gradient takes on almost exactly the adiabatic value, turning our inequality into an equality: P T dt dp = γ 1 γ Giving us the equation for convective transport

7 Polytropes In general we have a system of equations to solve for 3 variables (the temperature, pressure and density gradients). Solving this equation depends on the details of the equation of state, opacity and other physics that may change throughout the star. If we have a relationship that instead gave a relationship between pressure and density P = P (ρ) (with no explicit temperature dependence) we can solve for the structure of the star using the equations of hyostatic equilibrium and mass conservation alone. We call hypothetical stars with a simple relationship between pressure and density polytropes, where P = Kρ 1+1/n The equations of hyostatic equilibrium and mass continuity are (i) (ii) dp = ρgm(r) r 2 dm(r) = 4πr 2 ρ combining (i) and (ii) we obtain 4πr 2 ρ = d r2 ρg dp

8 Polytropes substituting ρ(r) = ρ c θ n (r) P = Kρ (1+1/n) c θ (n+1) dp d r 2 ρ dp r 2 = 4πGρ = Kρ(1+1/n) c (n + 1)θ n (r) dθ Define a new dimensionless variable corresponding to the radial distance from the center of the star: r = aξ r d 2 dp ρ d a 2 ξ 2 ρ c θ Kρ (1+1/n) r 2 = n c (n + 1)θ n dθ adξ a 3 ξ 2 dξ if we choose a = (n + 1)Kρ (1/n 1) c 4πG = 4πGρ c θ n 1/2 Lane-Emden equation of index n 1 ξ 2 d dθ ξ2 dξ dξ = θn

9 Polytropes Analytic solutions of the Lane-Emden equation exist only for a few values of n ξ 2 n = 0, θ = 1 6 n = 1, n = 5, θ = For other values of n, we can solve the equations numerically, starting with the boundary conditions ρ = ρ c at r = 0 so θ(0) = 1 dp = 0 at r = 0 dθ dξ = 0 ξ=0 Then, rewrite the Lane-Emden equation 1 d dθ ξ 2 ξ2 dξ dξ = θn 2ξ dθ dξ + ξ2 d2 θ dξ + θn ξ 2 = 0 d 2 θ dξ 2 = 2 ξ θ = sin ξ ξ ξ2 0.5 dθ dξ θn

10 Polytropes d 2 θ dξ 2 = 2 ξ dθ dξ θn Now convert this differential equation to a finite difference equation dθ dξ 1 ξ (θ i+1 θ i ) d 2 θ dξ 2 1 ξ 2 (θ i+1 2θ i + θ i 1 ) ξ i ξ θ n θ n i Then, starting at the middle of the star with i = 0, θ = 1, ξ = 0 and dθ/dξ = 0, solve for θ i+1 and repeat until the surface is reached (when θ ops below zero, and hence density ops below zero) The surface of the star R = aξ s can be found by finding the root, θ(ξ s )=0

11 Numerical Solutions θ(ξ) ξ Hint: Start integration at ξ =, i.e., some small positive number not exactly equal to zero!

12 Polytropes We can derive some useful physical quantities from the solution of the Lane- Emden equation: Stellar radius : R = aξ s = (n + 1)K 4πG 1/2 ρ (1 n)/2n c ξ s Total mass : M =4πa 3 ρ c ξs 0 θ n ξ 2 dξ using 1 ξ 2 d dθ ξ2 dξ dξ = θn we see that M =4πa 3 ρ c ξs o d ξ 2 dθ dξ and thus M =4π (n + 1)K 4πG 3/2 ρ (3 n)/2n c ξ 2 dθ dξ ξ s