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Lecture 5-2: Polytropes

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Lecture 5-2: Polytropes Lecture 5-2: Polytropes Literature: MWW chapter 19 !" 1 Preamble The 4 equations of stellar structure divide into two groups: Mass and momentum describing the mechanical structure and thermal equilibrium and energy transport describing the temperature structure. These two sets of equations are linked through the equation of state. Under certain circumstances, pressure is independent of temperature and then solving the equations of stellar structure simplify enormously. Define: P = Kργ = Kρ1+1/n with n the polytropic index 2 a) Decoupling examples: i) Stars supported by electron degeneracy: 5/3 Pe = Knrρ non-relativistic Fermi electron gas (n = 3 / 2) 4/3 Pe = Krρ relativistic Fermi electron gas (n = 3) The constants, K & K , depend only on µ & atomic physics nr r e ii) Convective energy transport & radiation pressure unimportant: ! $ d logT 1/∇ad # & = ∇ad ⇒ P ∝T " d log P %ad ρ 1/(1−∇ad ) P = kT = Kcρ µmu Now, the constant, Kc, depends on the star's boundary conditions. For a completely ionized gas, ∇ad 3= 2 5, and n = 3 / 2 iii) Ratio of (ideal) gas to radiation pressure is constant: 4 Pgas ρ P aT P = = kT & P = rad = m (1 ) 3(1 ) β µ uβ − β − β 1/3 4/3 1/3 " 3% " k % "1− β % 4/3 4/3 P = ρ = K ρ n = 3 $ ' $ ' $ 4 ' β # a & #µmu & # β & The value of the constant, Kb , depends now on b. Consider: P dT 1 3P κL(r) ∇ = = rad T dP 16πcG aT 4 M(r) 3P dPrad 1 P κL(r) dPrad 1 κL(r) 4 = ⇒ = 4aT dP 16πcG Prad M(r) dP 16πcG M(r) −3.5 3.5 with Kramer's law opacity, κ ∝T and pp-chain ε pp ∝ L M ∝T 4 b) Definition P = P ρ (1) & (2) If pressure is a function of density only, ( ), then eqs can be treated separately. Use r as independent variable: dm(r) " = 4π r2ρ(r) $ dr $ 1 d r2 dP # 2 = −4πGρ(r) (1) dP(r) Gm(r)ρ(r) $ r dr ρ(r) dr = − 2 dr r %$ (Poisson eqn) dP Boundary conditions: (i) r = 0 : = 0 dr (ii) either r = R : P = 0 (zero conditions) or r = 0 : P = P c (R: value of r where P=0) 5 c) Lane-Emden equation 1 1+ Write: P = K ρ n = K ρ γ with K = polytropic constant n = polytropic index γ = polytropic exponent n Define dimensionless function θ = θ ( r ) by ρ = ρcθ ⇒ 1 ⇒ n+1 1+ (n 1) P 1 d d P P n + c 2 θ n = c θ with P c = K ρ c and (1) 2 2 r = −θ 4πG ρc r dr dr 1 −1 2 (n +1)K n Define dimensionless radius ξ via r = rnξ rn = ρc 4πG with rn = Emden length 1 d 2 dθ n Then (1) becomes the Lane-Emden equation ξ = −θ ξ 2 dξ dξ dθ Boundary conditions: ξ = 0 : θ =1 and = 0 dξ Configuration has radius: R = r n ξ 1 with ξ 1 first zero of θ Solutions θ n ( ξ ): Lane, Ritter, Emden,6 Chandrasekhar, … d) Solutions 1 n = 0 : θ (ξ) =1− ξ 2 ξ = 6 Homogeneous sphere 0 6 1 sinξ n =1: θ1(ξ) = ξ1 = π ξ 1 n = 5 : θ5 (ξ) = ξ1 = ∞ " 1 %1/2 Plummer sphere $1+ ξ 2 ' # 3 & For other n: numerical solution of Lane-Emden equation Properties: θ n ( ξ ) decreases monotonically with increasing ξ > 0 n < 5: ξ1 < ∞ n ≥ 5: configuration extends to infinity Series expansion: 2 1 2 n 4 n(8n − 5) 6 n(122n −183n + 70) 8 θ (ξ) =1− ξ + ξ − ξ + ξ −... n 6 120 15120 3265920 7 e) Physical properties Possible to obtain much insight in the properties of polytropes without numerically solving the Lane-Emden equation 1/2 ! $1/2 ! 2 $ (n +1)Pc N0k (n +1)Tc Radius: R = rnξ1 = # 2 & ξ1 = # & ξ1 " 4πGρc % µ " 4πGPc % rnξ ξ ξ 2 3 2 n 3 d # 2 dθ & 3 2 dθ Mass: m(ξ) = ∫ 4π r ρ(r)dr = 4πρcrn ∫ ξ θ dξ = −4πρcrn ∫ %ξ (dξ = −4π rn ρcξ 0 0 0 dξ $ dξ ' dξ 3/2 ⇒ ! $ 3−n ( + (n +1)K 2n 2 dθ M = 4 π # & ρ c * − ξ - This is finite for n ≤ 5 " 4πG % ) dξ , ξ=ξ1 ρ Mass-radius relation: eliminate c from expressions for R and M: 1 1 3 n 1− −1 1 (4π ) n n K = 1 GM R n +1 n−1 n ξ n+1 "−θ! $ { # n % } ξ=ξ1 8 Lane-Emden Solutions 9 Ratio of mean and central density: 3m(ξ) 3 dθ 1# ξ & ρ(ξ) = = − ρ ⇒ ρ = − ρ 3 3 c c % ! ( 4π rn ξ ξ dξ 3 θn (ξ) $ 'ξ=ξ1 1 2 1+ n 1 GM Central pressure: Pc = Kρc = ! ! #2 R4 4π (n +1)"θn (ξ)$ ξ=ξ1 Central temperature: use T = T c θ and assume ideal gas µK 1 1n+1 GM µ T = ρ n = cP =c K 'T" ! $ N0k (n +1) −ξθnn (ξ) R N0k # %ξ=ξ1 Mass-radius-chemical composition relation: ideal gas: P = K 'T n+1 n−1 n ( ! ( )* n 1 )−θn ξ + Pc + ! n +1$ n+1 ξ=ξ1 1 K ' = n+1 = (N0k) # & ξ1 n+1 n−1 3−n Tc " G % 4π µ M R 10 Recap: For a given polytrope index, n, we have a set of models that are specified by two parameters, Pc and rc, (or equivalently M and R) which yield the constant K and the run of density and temperature as well as the mass and radius (or equivalently, the central density and pressure). 11 Constants and ratios that appear in above expressions Compare with slide 16 Lecture 2 Results for n= 0, 1 and 5 follow from analytical solutions given earlier Other values derived from numerical solutions 12 f) Potential energy Write Φ = gravitational potential. Then 1 dP dΦ * H.E. = − , ρ dr dr , P P GM 1 + ⇒ Φ = −(n +1) + ΦI = −(n +1) − 1+ $ ' n 1 dP d P , ρ ρ R P = Kρ ⇒ = (n +1) & ) ρ dr dr % ρ ( , - where P / ρ ∝ θ Also: R R R 1 1 P 1 GM Eg = Ω = ∫ Φdm = − (n +1) ∫ dm − ∫ dm 2 0 2 0 ρ 2 R 0 1 R GM 2 1 GM 2 = − (n +1) ∫ P dV − = (n +1)Ω− 2 0 2R 6 2R 2 3 GM so that we obtain: Ω = n − 5 R 13 .
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