Physics 576 Stellar Astrophysics Prof. James Buckley. Lecture 10

Transcription

1 Physics 576 Stellar Astrophysics Prof. James Buckley Lecture 10

2 Reading/Homework Assignment Makeup class tomorrow morning at 9:00AM Read sections 2.5 to 2.9 in Rose over spring break!

3 Stellar Structure

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

5 Convection ρ i,p i,t i ρ o (r + r), P o (r + r), T o (r + r) P i = P o,t i >T o ρ i <ρ o ρ i,p i,t i ρo(r + r), Po(r + r), To(r + r) ρ i,p i,t i ρ o (r), P o (r), T o (r)

6 Adiabatic Expansion C v dq dt Enthalpy : H U + PV V = U T dh = du +(PdV + V ) H C p T Using the ideal gas law PV = NkT dh =(α + 1)(P dv + V ) V Nk C p =(α + 1)P =(α + 1)P T P For adiabatic expansion, dq =0=dU + P dv, substituting for du gives: P du = αnk dt C v = αnk d(pv)=nkdt du = αnkdt = αd(pv)=α(p dv + V ) P α(p dv + V )= P dv (α + 1)P dv = αv V =(α + 1)Nk (α + 1) dv V = α P

7 Adiabatic Expansion (α + 1) dv V = α P ln Integrating both sides: = α +1 α ln Pb P a Pb P a = Vb V a α+1 α Vb V a PV γ = constant where γ α +1 α = C p C v γ mn PV γ = const P = const ρ P ρ γ = const Substituting P = ρ m kt gives PTγ P γ = const P 1 γ T γ = const

8 Convection r + δr ρ i (r + δr) P i (r + δr) T i (r + δr) r ρ i,p i,t i ρ o (r + δr) P o (r + δr) T o (r + δr) Assume no heat flows across the boundary adiabatic expansion Assume that pressure balance is always established by free expansion/contraction of the element P o (r) =P i (r) Start by considering the adiabatic expansion of the element: P o (r + δr)ρ i (r + δr) γ = P o (r)ρ i (r) γ (Note P i = P o ) Expanding : P o (r)+ δr ρ i (r) γ 1 γ 1 dρ i ρ i (r) δr = P o (r)ρ i (r) γ dρ i = ρ i(r) γp o (r) o Now, we compare this with the outside density gradient to see if the object continues to be boyant (convective instability) or sinks back down (convective stability). Stability condition : ρ i (r) γp o (r) o > dρ o to first order, we can assume that ρ i (r) ρ o (r), substitute, divide both sides by o / to obtain Po dρo < 1 o γ ρ o

9 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 P Starting with the expression for stability = dρ ρ + dt T Po ρ o dρo o < 1 γ or dρ o ρ o 1 γ o P o < o γ P o Stability condition : P T dt o T o < 0 dt > γ 1 γ

10 Stability Condition Stability condition : P T dt > γ 1 γ dt < γ 1 γ T P Hyostatic Equilibrium : defining g GM r 2 (r) PV = NkT P = ρkt m = ρ(r) GM(r) r 2 = ρ g substituting for / and P dt < dt < γ 1 T ( ρ g) γ ρkt/m γ 1 gm γ k

11 Convective Transport Stability condition : P T dt > 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 = γ 1 γ Giving us the equation for convective transport

12 Convective Energy Flux In your textbook, Rose calculates the difference in potential and kinetic energy over a characteristic length scale, i.e., the scale height λ of the stellar atmosphere, defined to be the height at which the pressure ops by 1/e δr = λ P / I prefer another model, as described in Kippenhahn and Weigert. Here, we assume that convective elements are constantly being born out of small temperature perturbations with initial velocity v c = 0. They start there motion at a range of different distances from the center of the star, and pick uo (or loose) speed as they rise (or sink) until after moving through a distance l m they mix with the surroundings and loose their identity. δr = l m /2 The change in kinetic energy of the rising element (per unit volume) is v c = kinetic energy density = 1 2 ρv2 c Equating this to the change in potential energy, and setting δr = l m /2 1 ρ 1 2 ρv2 c = ρ GM(r) r 2 δr 1/2 1 GM(r) P r 2 dρ 1 γ 1/2 l m 2

13 Convective Energy Flux δv Stability condition : ρ i (r) γp o (r) Adiabatic density gradient : o > dρ o dρ i = ρ γp δv δr External density gradient : dρ The deficiency of the density of the bubble (or a little differential volume element δv thereof) compared with the surroundings is: ρ dρ δr dρ i dρ δr = 1 ρ δr γ P The change in the potential energy density is then GM(r) GM(r + δr) GM(r) u ρ r (r + δr) r 2 δr This expression could be equated to the change in kinetic energy density, and we could solve for some sort of velocity. Then we need to do some sort of very complicated average over elements throughout the star. Doing this correctly is very difficult, so we can use one of a couple of simple models that get the basic physics...

14 Convective Flux δv δv δr P (r)+δr 1 + (1 γ)δr 1 P P 1 γ T γ = const P (r + δr) 1 γ T i (r + δr) γ = P (r) 1 γ T i (r) γ 1 + (1 γ)δr 1 P 1 γ T i (r)+δr dt γ i P (r) 1 γ T i (r) γ 1+γδr 1 dt i T i 1 + γδr 1 T i dt i + O(δr2 ) 1 dt i = γ 1 γ T P The difference in temperature of the element and the surroundings is given by: T dt δr dt i γ 1 T δr = γ P dt δr T δr u = c p T And the convective energy flux is F c = L c 4πr 2 = c p T δr ρv c u mass flux

15 Convective flux F c = L c 4πr 2 = c p T δr ρv c u mass flux v c = 1 ρ dρ 1 γ 1 P 1/2 GM(r) r 2 1/2 l m 2 and we can show this can be rewritten as v c =( T ) 1/2 GM(r) Tr 2 1/2 l m 2 Combining these equations, and dividing by two (to account for both rising and falling convective blobs???) F c = L c 4πr 2 = c pρ( T ) 3/2 GM(r) Tr 2 1/2 l 2 m 4

16 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) = ρgm(r) r 2 dm(r) = 4πr 2 ρ combining (i) and (ii) we obtain 4πr 2 ρ = d r2 ρg

17 Polytropes d r 2 ρ r 2 = 4πGρ substituting ρ(r) = ρ c θ n (r) P = Kρ (1+1/n) c θ (n+1) = 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 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

18 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, θ = sin ξ ξ 1 n = 5, θ = ξ2 0.5 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 = 0 at r = 0 dθ dξ = 0 ξ=0 Then, rewrite the Lane-Emden equation 1 ξ 2 d dθ ξ2 dξ dξ = θn 2ξ dθ dξ + ξ2 d2 θ dξ + θn ξ 2 = 0 d 2 θ dξ 2 = 2 ξ dθ dξ θn

19 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

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

21 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