We#have#reached#a#point#where#we#have#most#of#the## informa3on#required#to#solve#realis3c#stellar#models.# Lecture 7a A#cri3cal#piece#s3ll#mixing#is#the#nuclear#physics,#but#it# turns#out#that#many#of#the#observed#proper3es#of#stars,# Polytropes except#their#life3mes#and#radii,#reflect#chiefly#the#need#to## be#in#hydrosta3c#and#thermal#equilibrium,#and#not#the#energy## source.## Prialnik Chapter 5 # Glatzmaier and Krumholz Chapter 9 Historically,#prior#to#computers,#stellar#structure#was#o@en# Pols 4 calculated#using#polytropes.#Even#today.#they#provide#a## valuable#tool#for#understanding#many#stellar#proper3es.## In addition we have or will have the physics equations The Stellar Structure Equations ρ ⎞ 1 4 P = P(T, , Y ) N kT P aT ρ { i } = A ⎟ + e + µ ⎠ ions 3 dm 2 dr 1 4 r a b = π ρ = 2 κ = κ (T, ρ, Y )≈ κ ρ T dr dm 4πr ρ { i } 0 ε = ε(T, ρ, Y )≈ ε ρmT n (TBD) dP Gm dP Gm { i } 0 = − ρ = − and the boundary conditions dr r 2 dm 4πr 4 at r = 0 L(r)=0, m = 0 dT 3 κρ L(r) dT 3 κ L(r) 1/4 = − 3 2 = − 3 2 2 ⎛ L(R) ⎞ dr 4ac T 4πr dm 4ac T (4πr ) at r = R P(R) ≈ 0, m= M, T= ≈0 ⎝⎜ 4πR2σ ⎠⎟ dL(r) dL(m) = 4πr 2 ρε = ε These are 7 equations in 7 unknowns: ρ, T, P, L, ε, κ, and r dr dm which, in principle, given M and initial composition, can be solved as a function of time for the boundary conditions. S3ll#to#be#defined#D#" Most of our theoretical knowledge of stellar evolution comes from doing just that. Aside: Boundary Conditions Polytropes* We#assume#a#global#rela3on#between#pressure#and# ! Radiative Zero BC: density:# Ideally we would use some atmospheric model to tell us what the temperature BC is at the surface # γ P ∝ ρ γ =(n +1) / n The T change over the whole star is so large that the difference between 0 and the real Teff at surface is small – n is called the polytropic index. Don’t confuse this except when computing a luminosity. with the adiabatic index which is local. To get effective T at the end, use: 5/3 4/3 Some examples: P ∝ ρ or ρ n =constant ! Fully convective (adiabatic): ! White dwarfs (completely degenerate): ! Pressure is a mix of gas + radiation, but the ratio is constant throughout Polytropes Polytropes* ! Consider HSE dP Gm(r) ! Now#make#this#dimensionless# # = − 2 ρ r = 0 dr r Central#density:# # r 2 dP = − Gm(r) Define:## # ρ dr Then:# 2 ! Differentiating again and dividing by r : # d ⎛ r 2 dP ⎞ dm(r) γ # = − G = − 4πGr 2ρ(r) P(r)=Kρ (r) dr ⎝⎜ ρ dr ⎠⎟ dr n +1 1 # γ nγ γ = = 1+ 1 d ⎛ r 2 dP ⎞ = K ρ θ (r) n n 4 G (r) c 2 ⎜ ⎟ = − π ρ r dr ⎝ ρ dr ⎠ 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 ⎡P (n +1)⎤ 1 d ⎛ 2 dθ ⎞ n ⎢ c ⎥ r = −θ 2 2 2 dr ⎜ dr ⎟ 1 d ⎛ r dP ⎞ γ γ γ n γ n+1 ⎢ 4πGρc ⎥ r ⎝ ⎠ = − 4πGρ P =Kρ =K ρ θ =K ρ θ ⎣ ⎦ r 2 dr ⎝⎜ ρ dr ⎠⎟ c c K ργ d ⎛ r 2 dθ n+1 ⎞ Note that the quantity in brackets has units length2 c = − 4πGρ θ n ρ = ρ θ n 2 dr ⎜ n dr ⎟ c c 2 r ⎝ ρcθ ⎠ Define it to be α then 2 n 1 1/2 + 2 1 d ⎛ r dθ ⎞ n ⎡P (n 1)⎤ P = − 4πGρ θ α d ⎛ 2 dθ ⎞ n c + c 2 ⎜ n ⎟ c r = −θ with α =⎢ ⎥ r dr ⎝ θ dr ⎠ r 2 dr ⎝⎜ dr ⎠⎟ 4πGρ 2 ⎣⎢ c ⎦⎥ P (n +1) d ⎛ r 2θ n dθ ⎞ c 4 G n 2 2 ⎜ n ⎟ = − π θ ρc r dr ⎝ θ dr ⎠ 2 6 ⎡P (n +1)⎤ 1 d d dyne gm cm 2 2 c ⎛ 2 θ ⎞ n = cm for the dimension of α ⎢ ⎥ r = −θ 2 2 2 4πGρ 2 r 2 dr ⎝⎜ dr ⎠⎟ cm dynecm gm ⎣⎢ c ⎦⎥ Polytrope*boundary*condi2ons* Now define r =αξ with ξ a dimensionless radius- ρ r 1) θ n = So at = ξ =0 θ =1 like variable. Then ρc α 1 d ⎛ 2 dθ ⎞ n 2 ⎜ξ ⎟ = −θ ξ dξ ⎝ dξ ⎠ r 1 dθ dθ 2) ξ ≡ so dξ= dr ⇒ = α α α dξ dr This is the Lane Emden equation. It involves only dρ dθ but ρ ≡ ρ θ n so =nρ θ n−1 dimensionless quantities and can be solved for a given c dr c dr n to give θ(ξ). n does not have to be an integer. dθ 1 dρ dθ α dρ and ⇒ = n−1 = n−1 dr nρc θ dr dξ nρc θ dr Basically all that went into it was but in spherical symmetry for hydrostatic equilibrium hydrostatic equilibrium, mass conservation, dρ ρ is a local maximum at r = 0, so = 0 dr and a power law equation of state dθ So at at ξ =0 = 0 dξ That is Polytrope*boundary*condi2ons* dP Gm(r)ρ = − → 0 as r → 0 dr r 2 3) At the surface r = R the density (first) goes to zero so because m(r) ∝ r 3 approaches 0 faster than r 2 dP dρ R =αξ where ξ is where θ(ξ) first reaches 0 And since P=K ργ = γ Kργ −1 * 1 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 For any value of n, the mass Cox and Guilli (1965) R ξ1 r = αξ R = αξ M = 4πr 2ρ dr = 4πα 3 ρ ξ 2θ n dξ 1 ∫ c ∫ n 0 0 ρ = ρcθ 2 dθ ρ But the Lane Emden equation says ξ ξ D = c 1 1 d n ρ 1 d ⎛ dθ ⎞ d ⎛ dθ ⎞ ξ 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. Further α can be expressed in terms of ρc and K 1/2 1/2 n 1 1 n 1/2 + − Another interesting quantity that can be obtained ⎡ n ⎤ ⎡ n ⎤ ⎡Pc (n +1)⎤ ⎢Kρc (n +1)⎥ ⎢Kρc (n +1)⎥ α = ⎢ ⎥ = = from the tables is the ratio of central density to average 4 G 2 ⎢ 4 G 2 ⎥ ⎢ 4πG ⎥ ⎣⎢ π ρc ⎦⎥ π ρc ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥ density - how compact the core of the star is. 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 See the previous table. The case n = 0, Dn = 1 is the independent of ρ if n = 3. sphere of constant density case c n = 0 Analytic solutions Technically the case n = 0 is a singularity For n = 0, 1, 5, and only for these values, there exist analytic solutions to the Lane Emden P = Kρ1+1/n diverges as n → 0 equation. This reflects the fact that constant density can only Unfortunately none of them correspond to be maintailed in the face of gravity if the fluid is common stars, but the solutions help to incompressible (like the ocean; P can vary but not ρ). demonstrate how polytropes can be solved and they can used for interesting approximate Nevertheless, some interesting properties of polytropes cases. can be illustarted with n = 0 so long as we don't use the pressure-density relation explicitly n = 0 - the constant density case 6 The first zero of θ = 1- 2 is at ξ1 = 6 r 1 d ⎛ 2 dθ ⎞ n ξ ξ = 1 2 ⎜ξ ⎟ = −θ = − α ξ dξ ⎝ dξ ⎠ so R = 6 α 3 2 dθ 2 ξ ξ = − ξ = − + C P(r) P n+1(r) P (r) for n = 0 dξ ∫ 3 = cθ = cθ 2 2 dθ ξ C 1 Cdξ ⎛ ξ ⎞ ⎛ r ⎞ = − + 2 ⇒ dθ = − ξ dξ + 2 P(r) = P 1− =P 1− dξ 3 ξ ∫ 3 ∫ ∫ ξ c ⎝⎜ 6 ⎠⎟ c ⎝⎜ α 2 6⎠⎟ 2 ξ C 2 D but =1 at =0 R R ⎛ r ⎞ θ = − − + θ ξ but = so P(r) P 1 6 ξ α = = c ⎜ − 2 ⎟ ξ 6 ⎝ R ⎠ So C = 0 and D =1 and 1 2 For ideal gas pressure and constant composition ξ θ = 1 − T would have the same dependence ( =constant) 6 ρ 2 ρNAkT ⎛ r ⎞ hKp://www.vistrails.org/index.php/User:Tohline/SSC/Structure/Polytropes# P = ⇒ T(r) = Tc 1− 2 µ ⎝⎜ R ⎠⎟ 1/2 1/2 ⎡Pc (n +1)⎤ ⎡ Pc ⎤ R but also α = = = so 3 2 dθ ⎞ ⎢ 2 ⎥ ⎢ 2 ⎥ Recall M= 4 4πGρ 4πGρ − πα ρc ξ1 ⎣⎢ c ⎦⎥ ⎣⎢ c ⎦⎥ 6 dξ ⎠⎟ ξ1 ⎛ 4 3 ⎞ G R 2 dθ ⎞ 2 2 π ρc ρc This last quantity, , is a number that 2 R G ⎜ 3 ⎟ ξ1 π ρc ⎝ ⎠ GMρ dξ ⎠⎟ Pc = = = ξ1 3 2R 2R depends on the poytropic index n.