Physics 556 Stellar Astrophysics Prof. James Buckley

Transcription

1 hysics 556 Stellar Astrophysics rof. James Buckley Lecture 8 Convection and the Lane Emden Equations for Stellar Structure

2 Reading/Homework Assignment Read sections 2.5 to 2.9 in Rose over spring break!

3 Stellar Structure

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

5 Convection ρ i, i,t i ρ o (r + r), o (r + r), T o (r + r) i = o,t i >T o ρ i <ρ o ρ i, i,t i ρo(r + r), o(r + r), To(r + r) ρ i, i,t i ρ o (r), o (r), T o (r) The gas bubble in the star moves fast enough that there is not time to exchange heat at the boundary adiabatic expansion

6 Solar Granulation 30 minute time lapse images of the solar granulation made with the SVST (La alma) (G. Scharmer, et al.). Hot gas rises in center of bright regions, is diverted horizontally (radiates) and sinks back into the Sun in the darker intergranular lanes. The size of the granules range from 250 km (resolution limit) to 2000~km with an average of 1300 km.

7 Adiabatic Expansion Specific heat or heat capacity is defined as the amount of heat that must be added to a substance to raise its temperature by 1 deg K either at constant pressure C or constant temperature C T Enthalpy : H U + V dh = du +(dv + V d ) =(δq dv )+(dv + V d )=δq + Vd δq H C p = dt T Using the ideal gas law V = NkT du = δq dv δq U C v = dt T du = αnk dt C v = αnk d(v)=nkdt du = αnkdt = αd(v)=α( dv + V d ) dh =(α + 1)( dv + V d ) V Nk C p =(α + 1) =(α + 1) T The adiabatic index γ is defined to be: V V and U = α NkT for 2α D.O.F. =(α + 1)Nk γ C /C V =(α + 1)/α

8 Adiabatic Expansion d(v)=nkdt du = αnkdt = αd(v)=α(dv + V d ) For adiabatic expansion, δq =0=dU + dv, substituting for du gives: α( dv + V d )= dv (α + 1) dv = αv d ln b a = α +1 α (α + 1) dv V ln Vb V a = αd b a = Vb V a α+1 α V γ = constant where γ α +1 α = C p C v V γ = const γ mn = const ρ ρ γ = const Substituting = ρ m kt gives Tγ γ = const 1 γ T γ = const

9 Convection r + δr ρ i (r + δr) i (r + δr) T i (r + δr) ρ o (r + δr) 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 o (r) = i (r) Start by considering the adiabatic expansion of the element: o (r + δr)ρ i (r + δr) γ = o (r)ρ i (r) γ (Note i = o ) r ρ i, i,t i d( 0 ρ γ i )=d(const) = 0 d 0 ρ γ i dρ i d 0 = ρ i γ 0 dρ i = dρ i d 0 d 0 = ρ i(r) γ o (r) γ 0 ρ γ 1 i dρ i =0 d 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 : dρ i < dρ 0 dρ i < dρ 0 ρ i(r) d o γ o (r) > dρ o to first order, we can assume that ρ i (r) ρ o (r), substitute, divide both sides by d o / to obtain o dρo Stability condition: < 1 d o γ ρ o

10 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 = ρ m kt ln = ln ρ + ln T + constant d Starting with the expression for stability = dρ ρ + dt T o ρ o dρo d o < 1 γ or dρ o ρ o 1 γ d o o < do γ o Stability condition : T dt o T o < 0 dt d > γ 1 γ

11 Stability Condition T Stability condition : T dt/ d/ > γ 1 γ Since dt/ = dt/ and d/ = d/ dt γ 1 T < d dt γ 1 T γ < γ Hyostatic Equilibrium : defining g GM r 2 dt d > γ 1 γ dt γ 1 T > γ d (r) V = NkT = ρkt m = ρ(r) GM(r) r 2 d = ρ g d substituting for d/ and dt γ 1 < T (ρ g) γ ρkt/m Condition for stability : dt γ 1 gm < γ k d

12 Convective Transport γ 1 γ Stability condition : T = 1+1/α 1 (α + 1)/α = 1 α +1 dt d > γ 1 γ dt or < dt γ 1 gm < γ k gmn (α + 1)N k = gmn C p = gc p (In chemistry, specific heat c p is defined as heat capacity per unit mass C p /m N) Which matches the stability condition given in some textbooks (e.g., Rose): 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: Giving us the equation for convective transport T dt d = γ 1 γ

13 Mixed Convection and Radiation Let s assume that the atmosphere starts out in radiative equilibrium, i.e F rad = σt 4 everywhere and 4πr 2 F rad (r) =const = L where L is the total luminosity of the star. Together with the other equations of stellar structure, this determines T (r), (r) and ρ(r) Now say that some distance from the center of the star r 0 the atmosphere becomes unstable to convection - we need to calculate F (r) =F rad + F conv, solve for energy balance, solve for T (r), (r) and ρ(r), check the conditions for convective stability again, and iterate. T i T l m At each step, we can start with the external values T,,ρ and the internal values T i,,ρ i and calculate quantities like: ρ(r) =ρ i (r) ρ(r), T (r) =T i (r) T (r), ρ i ρ Then the flux = mass flow rate excess heat per gram: F conv = ρ v c p T r r 0 To find the velocity of the element v we assume that the convective cell rises through a distance l m (the mixing length) where it looses its identity. The average velocity is found by equating the kinetic energy at r = r 0 + l m with the work done by the boyant force acting through l m on an element with under-density ρ

14 Convective Energy Flux Flux = mass flow rate excess heat per gram: F conv = ρ v c p T Recall that for adiabatic expansion 1 γ T γ i = const T i ρ i ρ T l m d( 1 γ T γ i ) = (1 γ) γ T γ i multiplying by γ T γ i d + γ 1 γ T γ 1 i dt i =0 (1 γ)d + γ dt i =0 d ln T i T i d ln = γ 1 γ r r 0 dt i = γ 1 γ T d The difference in temperature of the element and the surroundings is given by: T dt i δr dt δr = γ 1 γ T d dt δr T δr The excess internal energy per unit mass carried by the gas is 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 Energy Flux Now let s find v using the conversion of work to kinetic energy First, lets scale l m by some characteristic scale. The scale height λ of the stellar atmosphere is defined to be the height at which the pressure ops by 1/e Assume that convective elements are constantly being born out of small temperature perturbations with initial velocity v c = 0 at some distance r 0 from the center of the star (the position at which the atmosphere becomes convectively unstable. The elements pick up (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. We can write l m = αλ where α 0.5 to 1.5 The change in kinetic energy of the rising element (per unit volume) is kinetic energy density = 1 2 ρv2 c The boyant force does work per unit volume of dw = ρ g δr in moving through a distance δr. Integrating this and equating this to the change in kinetic energy density gives 1 2 ρv2 c = λ r0 +l m r 0 d/ ρ GM(r) r 2

16 Convective Energy Flux To solve for ρ we use the equation of state: = ρ kt µm u d = dρ ρ + dt T dµ µ (where, to be completely general, we consider the case that even the ionization state or composition of the gas may be different inside and outside of the convective element, i.e. µ = 0) If we have pressure balance between the inside and outside of the gas element, then d =0=dρ ρ + dt T dµ µ dt dρ = ρ T dµ = ρ dt µ T η where η 1 d ln µ d ln T 1 2 ρv2 c = r0 +l m r 0 ρ GM(r) r 2 ρ = η ρ T T 1 2 η ρ dti T dt r0 +l m l m r 0 v = ηg T ( T )1/2 l m g η 2 g ρ T T l2 m

17 Convective Energy Flux substituting v = ηg T ( T )1/2 l m into F conv = L c 4πr 2 = c p T δr ρv c and taking δr l m F conv = ρ c p η g T ( T )3/2 l 2 m Finally, we demand that π(f rad + F conv )=σt 4 and iterate.

18 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 d 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

19 olytropes 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 = (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 = Kρ 1+1/n The equations of hyostatic equilibrium and mass continuity are (i) (ii) d = ρgm(r) r 2 dm(r) = 4πr 2 ρ combining (i) and (ii) we obtain 4πr 2 ρ = d r2 ρg d

20 olytropes d r 2 ρ d r 2 = 4πGρ substituting ρ(r) = ρ c θ n (r) = Kρ (1+1/n) c θ (n+1) d = 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 d 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

21 olytropes 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 d = 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

22 olytropes 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

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

24 olytropes 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