Summary of Results. 1. Mass Conserva-on. dr dm = 1. dm dr = 4πr 2 ρ. 4πr 2 ρ
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1 Summary of Results 1. Mass Conserva-on dm dr = 4πr 2 ρ eulerian dr dm = 1 4πr 2 ρ lagrangian
2 Summary of Results 2. Hydrosta-c Equilibrium dp dr = ρ Gm eulerian r 2 dp dm = Gm 4πr 4 lagrangian Need equa-on of state: P = f ( ρ,t, X ) i
3 Summary of Results 3. Energy Genera-on dl r dr = 4πr 2 ρε eulerian dl m dm = ε lagrangian Need nuclear physics: ε = f ( ρ,t, X ) i
4 Summary of Results 4. Energy Flow: Radia-on dt dr = 3κρL r 16πacr 2 T 3 dt dm = 3κ L m 64π 2 acr 4 T 3 eulerian lagrangian Need opacity: κ = f ( ρ,t, X ) i
5 Summary of Results 4. Energy Flow: Convec-on dt dr = 1 1 Γ 2 GmρT r 2 P dt dm = 1 1 Γ 2 GmT 4πr 4 P eulerian lagrangian Need adiaba-c exponent: Γ 2 = f ( ρ,t, X ) i
6 Summary of Results Equa-on of State (non-degenerate maker) P = 1 3 at 4 + N A k µ ρt Mean molecular weight: 1 µ = 1 µ i + 1 µ e = i X i A i + i Z i X i y i A i
7 Summary of Results Ioniza-on frac-ons y i Saha equa-on : n + n e = 2G + n 0 G 0 ( 2πm e kt ) 3 2 h 3 exp χ kt e.g., Hydrogen only : y 2 1 y = 1 ( 2πm e kt ) 3 2 N A ρ h 3 exp χ H kt
8 Summary of Results Thermodynamics Adiaba-c index : Γ 2 Γ 2 1 = ln P lnt ad Γ 2 = 32 24β 3β β 3β 2 β = P g P tot (assuming neutral or fully ionized gas)
9 Summary of Results Opacity: approxima-ons 1 κ 1 κ H + 1 κ e +κ ff +κ bf κ e = 0.2( 1+ X)cm 2 g 1 (assuming fully ionized and no metals) κ ff ( X + Y )( 1+ X)ρT 3.5 cm 2 g 1 κ bf Z ( 1 + X)ρT 3.5 cm 2 g 1 κ H ( Z 0.02)ρ 1 2 T 9 cm 2 g 1
10 Summary of Results Opacity: tables Interpolate in density, temperature, and composi-on. logt logr logt = logr = -5.83
11 Summary of Results Convec-on Convec-on happens when: dt dr rad > dt dr ad L m > 16πacG 1 1 3κ R Γ 2 T 4 m P
12 Summary of Results Energy genera-on ε = ε pp + ε CNO + ε 3α ρx 2 ε pp = T exp ( 3.38 T )ergs 1 g 1 9 ρxz ε CNO T exp ( T )ergs 1 g 1 9 ρ 2 Y 3 ε 3α = exp 3 ( 4.4 T 9 )ergs 1 g 1 T 9
13 Stellar Models Given M and X,Y,Z, solve structure equa-ons dr dp dt dl m dm dm dm dm to get r, P, T, L m as a func-on of m Need to know: ρ P,T, X,Y,Z ( ) κ ( R ρ,t, X,Y,Z ) ε ( ρ,t, X,Y,Z ) Γ ( 2 ρ,t, X,Y,Z )
14 Stellar Models CENTER SURFACE m = 0 m = M r = 0 r = R P = P c T = T c P 0 T 0 L m = 0 L m = L
15 dr dm = 1 4πr 2 ρ dp dm = Gm 4πr 4 dl m dm = ε dt dm = 3κ L m 64π 2 acr 4 T 3 dt dm = 1 1 Γ 2 Stellar Models GmT 4πr 4 P Some of these equa-ons are indeterminant at center. It is beker to start the integra-on at a very small m>0.
16 Stellar Models Central boundary condi-ons At a very small m>0: r 0 m = 4 3 πr 3 ρ c r = 3m 4πρ c 1 3 L m 0 L m = ε c m
17 Stellar Models Central boundary condi-ons We can get boundary condi-ons for P and T by expanding and demanding that their deriva-ves are zero at r=0 P( r) = P( 0) + P ( )( r 0) P ( 0) r 0 P 0 = P c + P ( 0) r 2 2 ( r) = Gmρ r 2 0 = Gρ 4 r 2 3 πr3 ρ c = 4 3 πgρ 2 c r ( )2 P( r) = P c 2 3 πgρ 2 c r 2
18 Stellar Models Central boundary condi-ons T Similarly, T ( r) = T ( 0) + T 0 0 = T c + T ( 0) r 2 2 ( r) = 3κ RρL r 16πacr 2 T 3 T ( r ) = 1 1 Γ 2 T P ( )( r 0) T ( 0) r 0 dp dr T r ( )2 T ( r) = T c κ cρ 2 c ε c 8acT r 2 3 c Radia-ve Convec-ve ( ) = T c 1 1 Γ 2,c 2πGρ c 2 T c 3P c r 2
19 Stellar Models Surface boundary condi-ons We can do beker than T=P=0 at surface. guess R and L: L = 4π R 2 σt s 4 T s = L 4π R 2 σ 1 4 Guess ρvery small, e.g., g cm-3: P s = 1 3 at 4 s + N k A µ ρ T s s
20 Stellar Models Steps in construc-ng a stellar model 1. Compute as a func-on of DENSITY Beware of P dropping below OPACITY Use approxima-ons for crude results Interpolate using tables (use approxima-ons outside table bounds) ENERGY GENERATION Use formulas for pp, CNO, and 3 THERMODYNAMICS 5/3 for ideal gas, 4/3 for radia-on pressure departs from these values when Mixture of ideal gas and radia-on Ioniza-on zones Γ 2 = Γ 2 ρ, κ, ε, Γ at 4 α P, T, X, Y, Z
21 Stellar Models Steps in construc-ng a stellar model 2. Use four structure equa-ons to compute vs. given star-ng values for these (boundary condi-ons). m Given values of at shell compute 4 deriva-ves: Use these deriva-ves to compute at shell i e.g., Shell size near center and surface. i + 1 m, r, P, T, L m, ρ, κ, ε, Γ 2 dm < 10 4 M dr dm, dp dm, dl m dm, dt dm r, P, T, L m r[i + 1] = r[i] + dr dm r, P, T, L m [ i] dm
22 Stellar Models Steps in construc-ng a stellar model 3. Deal with lack of complete boundary condi-ons at center or surface. At center, have r, L m, but not P, T At surface, have P, T, but not r, L m General approach: guess values for missing condi-ons at one end, run model, and compare boundary condi-ons at other end. PROBLEM: small changes in condi-ons at center can cause large differences at surface à difficult to reach convergence. SOLUTION: Shoot from both center and surface and meet halfway through star.
23 Stellar Models m = 0.5
24 Stellar Models Guess values for and and integrate outwards from to m = 0 T c m = M 2 P c R m = M 2 Guess values for and and integrate inwards from to m = M L Compute discrepancies at Δr, ΔP, ΔT, ΔL m m = M 2 Work in log space: logp, logt, logr, logl m
25 guess Stellar Models guess logp Δ log P logt Δ logt guess guess logr Δ log r logl m Δ log L m 0 M/2 M 0 M/2 M
26 Stellar Models
27 Stellar Models Boundary Conditions log P c log R Δ log P Δ logt Δ log r logt c log L Δ log L m
28 Repeat using new trial values: Stellar Models ( logt c d logt c ) and logt c + d logt c ( log P c d log P c ) and log P c + d log P c ( log R d log R) and log R + d log R ( ) with logp c, logr, logl ( ) with logt c, logr, logl ( ) with logp c, logt c, logl ( log L d log L) and ( log L + d log L) with logp c, logt c, logr Compute new discrepancies in each case. Δ logr, Δ log P, Δ logt, Δ log L m Get 16 deriva-ves. e.g., ( Δ log r) log P c ( ) log P c +d log P c ( Δ log r) log P d log Pc = Δ logr 2d log P c
29 Stellar Models Use these deriva-ves to calculate improved boundary condi-ons. General idea: ( Δ log r) Δ logr = δ log P c log P c log P c = log P c + δ log P c log T c = logt c + δ logt c log R = log R + δ log R log L = log L + δ log L More complicated with 4 variables! ( ) δ log P c = Δ logr Δ log r log P c 1
30 Stellar Models ( Δ log P) log P c ( ) log P c Δ logt ( ) log P c Δ logr ( ) log P c Δ log L m ( Δ log P) logt c ( ) logt c Δ logt ( ) logt c Δ log r ( ) logt c Δ log L m ( Δ log P) log R ( ) log R Δ logt ( ) log R Δ log r ( ) log R Δ log L m ( Δ log P) log L ( ) log L Δ logt ( ) log L Δ log r ( ) log L Δ log L m δ log P c δ logt c δ log R δ log L = Δ log P Δ logt Δ logr Δ log L m Invert matrix to solve for (actually take smaller steps ~0.1x) δ log P c, δ logt c, δ log R, δ log L Iterate un-l convergence is reached: discrepancies vanish (i.e., drop below a threshold value)
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