#### Transcript Static Stellar Structure

Static Stellar Structure Static Stellar Structure Most of the Life of A Star is Spent in Equilibrium Evolutionary Changes are generally slow and can usually be handled in a quasistationary manner We generally assume: Hydrostatic Equilibrium Thermodynamic Equilibrium The Equation of Hydrodynamic Equilibrium d r Gm( r ) P 2 2 dt r r 2 Static Stellar Structure 2 Limits on Hydrostatic Equilibrium If the system is not “Moving” - accelerating Gm( r ) P dP in reality - then d2r/dt2 = 2 r r dr 0 and then one recovers the equation of hydrostatic equilibrium: If ∂P/∂r ~ 0 then 2 d r Gm ( r ) which is just the freefall 2 2 condition for which the dt r time scale is tff (GM/R3)-1/2 Static Stellar Structure 3 Dominant Pressure Gradient When the pressure gradient dP/dr dominates one gets (r/t)2 ~ P/ρ This implies that the fluid elements must move at the local sonic velocity: cs = ∂P/∂ρ. When hydrostatic equilibrium applies V << cs te >> tff where te is the evolutionary time scale Static Stellar Structure 4 Hydrostatic Equilibrium Consider a spherical star Shell of radius r, thickness dr and density ρ(r) Gravitional Force: ↓ (Gm(r)/r2) 4πr2ρ(r)dr Pressure Force: ↑ 4r2dP where dP is the pressure difference across dr Equate the two: 4πr2dP = (Gm(r)/r2) 4πr2ρ(r)dr r2dP = Gm(r) ρ(r)dr dP/dr = -ρ(r)(Gm(r)/r2) The - sign takes care of the fact that the pressure decreases outward. Static Stellar Structure 5 Mass Continuity r m(r) = mass within a shell = m( r ) 0 4 r2 (r)dr This is a first order differential equation which needs boundary conditions We choose Pc = the central pressure. Let us derive another form of the hydrostatic equation using the mass continuity equation. Express the mass continuity equation as a differential: dm/dr = 4πr2ρ(r). Now divide the hydrostatic equation by the masscontinuity equation to get: dP/dm = Gm/4πr4(m) Static Stellar Structure 6 The Hydrostatic Equation in Mass Coordinates dP/dm = Gm/4πr4(m) The independent variable is m r is treated as a function of m The limits on m are: 0 at r = 0 M at r = R (this is the boundary condition on the mass equation itself). Why? Radius can be difficult to define Mass is fixed. Static Stellar Structure 7 The Central Pressure Consider the quantity: P + Gm(r)2/8πr4 Take the derivative with respect to r: d Gm( r )2 dP Gm( r ) dm Gm( r )2 P 4 4 dr 8 r dr 4 r dr 2 r 5 But the first two terms are equal and opposite so the derivative is -Gm2/2r5. Since the derivative is negative it must decrease outwards. At the center m2/r4 → 0 and P = Pc. At r = R P = 0 therefore Pc > GM2/8πR4 Static Stellar Structure 8 The Virial Theorem dP mG dm 4 r ( m ) 4 mG 3 dP 4 r dm r d dr mG 3 2 (4 r P ) 4 r 3P dm dm r M 3P M mG 3 M (4 r P ) |0 dm dm 0 0 r Remember: 4 r 2 rdr dm Static Stellar Structure 9 The Virial Theorem The term (4 r3P) |0M is 0: r(0) = 0 and P(M) = 0 Remember that we are considering P, ρ, and r as variables of m For a non-relativistic gas: 3P/ = 2 * Thermal energy per unit mass. M 0 3P dm 2U for the entire star GM dm the gravitional binding energy r Static Stellar Structure 10 The Virial Theorem -2U = Ω 2U + Ω = 0 Virial Theorem Note that E = U + Ω or that E+U = 0 This is only true if “quasistatic.” If hydrodynamic then there is a modification of the Virial Theorem that will work. Static Stellar Structure 11 The Importance of the Virial Theorem Let us collapse a star due to pressure imbalance: If hydrostatic equilibrium is to be maintained the thermal energy must change by: This will release - ∆Ω ∆U = -1/2 ∆Ω This leaves 1/2 ∆Ω to be “lost” from star Normally it is radiated Static Stellar Structure 12 What Happens? Star gets hotter Energy is radiated into space System becomes more tightly bound: E decreases Note that the contraction leads to H burning (as long as the mass is greater than the critical mass). Static Stellar Structure 13 An Atmospheric Use of Pressure We use a different form of the equation of hydrostatic equilibrium in an atmosphere. The atmosphere’s thickness is small compared to the radius of the star (or the mass of the atmosphere is small compared to the mass of the star) For the Sun the photosphere depth is measured in the 100's of km whereas the solar radius is 700,00 km. The photosphere mass is about 1% of the Sun. Static Stellar Structure 14 Atmospheric Pressure Geometry: Plane Parallel dP/dr = -Gm(r)ρ/r2 (Hydrostatic Equation) R ≈ r and m(R) = M. We use h measured with respect to some arbitrary 0 level. dP/dh = - gρ where g = acceleration of gravity. For the Sun log(g) = 4.4 (units are cgs) Assume a constant T in the atmosphere. P = nkT and we use n = ρ/μmH (μ is the mean molecular weight) so P = ρkT/μmH Static Stellar Structure 15 Atmospheric Pressure Continued Or P= dP = dP = dP/P = ρkT/μmH -g ρ dh -g P(μmH/kT) dh -g(μmH/kT) dh Integrate : P P0e 0e gmH h kT gmH h kT Where: P0 = P and ρ0 = ρ at h = 0. Static Stellar Structure 16 Scale Heights H (the scale height) is defined as kT / μmHg It defines the scale length by which P decreases by a factor of e. In non-isothermal atmospheres scale heights are still of importance: H ≡ - (dP/dh)-1 P = -(dh/d(lnP)) Static Stellar Structure 17 Simple Models To do this right we need data on energy generation and energy transfer but: The linear model: ρ = ρc(1-r/R) Polytropic Model: P = Kργ K and γ independent of r γ not necessarily cp/cv Static Stellar Structure 18 The Linear Model r c (1 ) R dP Gm ( r ) r c (1 ) 2 dr r R Defining Equation Put in the Hydrostatic Equation Now we need to deal with m(r) Static Stellar Structure 19 An Expression for m(r) dm r 4 r 2 c (1 ) dr R 4 r 3 c 2 4 r c R 4 r 3 c r 4 c m( r ) 3 R R 3 c Total Mass M m( R ) 3 4r 3 3r 4 So m( r ) M 3 4 R R Static Stellar Structure Equation of Continuity Integrate from 0 to r 20 Linear Model • We want a particular (M,R) • ρc = 3M/πR3 and take Pc = P(ρc) • Substitute m(r) into the hydrostatic equation: dP G 2 4r 3 r 4 r 2 c 1 dr r R R 3 2 3 4 r 7 r r 2 G c 2 3 3R R So 2 3 4 2 r 7 r r 2 P Pc G c 2 3 9 R 4 R where Pc is the central pressure. Static Stellar Structure 21 Central Pressure At the surface r R and P 0 5 G R P( R ) Pc 0 36 2 2 2 2 5 G c R 5 GR 9 M Pc 2 6 36 36 R 2 5GM Pc 4 4 R 2 c Static Stellar Structure 2 22 The Pressure and Temperature Structure The pressure at any radius is then: 2 3 4 5 24 r 28 r 9 r 2 2 P( r ) G c R 1 4 2 3 36 5R 5R 5R Ignoring Radiation Pressure: mH P T k 2 3 5 G mH r 19 r 9 r 2 c R 1 2 3 36 k 5R R 5R Static Stellar Structure 23 Polytropes P = Kργ which can also be written as P = Kρ(n+1)/n N ≡ Polytropic Index Consider Adiabatic Convective Equilibrium Completely convective Comes to equilibrium No radiation pressure Then P = Kργ where γ = 5/3 (for an ideal monoatomic gas) ==> n = 1.5 Static Stellar Structure 24 Gas and Radiation Pressure Pg = (N0k/μ)ρT = βP Pr = 1/3 a T4 = (1- β)P P= P Pg/β = Pr/(1-β) 1/β(N0k/μ)ρT =1/(1-β) 1/3 a T4 Solve for T: T3 = (3(1- β)/aβ) (N0k/μ)ρ T = ((3(1- β)/aβ) (N0k/μ))1/3 ρ1/3 Static Stellar Structure 25 The Pressure Equation P N 0k T 1 3 N k 4 3 1 4 0 3 a True for each point in the star If β ≠ f(R), i.e. is a constant then P = Kρ4/3 n = 3 polytrope or γ = 4/3 Static Stellar Structure 26 The Eddington Standard Model n = 3 For n = 3 ρ T3 The general result is: ρ Tn Let us proceed to the Lane-Emden Equation ρ ≡ λφn λ is a scaling parameter identified with ρc - the central density φn is normalized to 1 at the center. Eqn 0: P = Kρ(n+1)/n = Kλ(n+1)/nφn+1 Eqn 1: Hydrostatic Eqn: dP/dr = -Gρm(r)/r2 Eqn 2: Mass Continuity: dm/dr = 4πr2ρ Static Stellar Structure 27 Working Onward r 2 dP Solve 1 for m( r ) : m( r ) G dr Put m( r ) in 2 : 1 r2 d r 2 dP 4 G dr dr n 1 dP d K n ( n 1) n dr dr n 1 1 d r2 n d n n K ( n 1) 4 G r 2 dr n dr Static Stellar Structure 28 Simplify 1 r2 n 1 d r2 n d n n n K ( n 1) 4 G dr dr 1 n K ( n 1) K 1 n n 1 r2 ( n 1) d 2 d n dr r dr 4 G 1 1 d 2 d n r 2 4 G r dr dr 1 2 K ( n 1) Define a 4 G r so ad dr a 1 d 2 d n So : 2 d d 1 n n Static Stellar Structure 29 The Lane-Emden Equation 1 d 2 d n 2 d d Boundary Conditions: Center ξξ = 0; φ = 1 (the function); dφ/d ξ = 0 Solutions for n = 0, 1, 5 exist and are of the form: φ(ξ) = C0 + C2 ξ2 + C4 ξ4 + ... = 1 - (1/6) ξ2 + (n/120) ξ4 - ... for ξ =1 (n>0) For n < 5 the solution decreases monotonically and φ → 0 at some value ξ1 which represents the value of the boundary level. Static Stellar Structure 30 General Properties For each n with a specified K there exists a family of solutions which is specified by the central density. For the standard model: N k 3 1 K 0 4 a 4 1 3 We want Radius, m(r), or m(ξ) Central Pressure, Density Central Temperature Static Stellar Structure 31 Radius and Mass R a1 1 2 1 n (n 1) K 2 n 1 4 G d M ( ) 4 a 3 d 2 0 d d 4 a 3 2 d Now substitute for a and evaluate at 1 (boundary ) a M ( ) 4 r 2 dr 0 4 a 3 1 d 2 d 2 d n d 0 n 2d M ( ) 4 a 3 0 d d 3 2 3 n (n 1) K 2 n M 4 4 G 2 d d 1 2 d d d Static Stellar Structure 32 The Mean to Central Density c 3 2 M 4 R3 3 3 n ( n 1) K 2 n 4 4 G 3 2 2 d d 1 4 ( n 1) K 3 4 G 3(1 n ) 2n 13 d 1 d 1 3 3 d c 1 d 1 Static Stellar Structure 33 The Central Pressure At the center ξ = 0 and φ = 1 so Pc = Kλn+1/n This is because P = Kρn+1/n = Kλ(n+1)/nφn+1 Now take the radius equation: 1 2 1 n ( n 1) K 2 n R 1 4 G 1 2 ( n 1)12 1nn K 4 G So K 1 n n 1 2 4 GR 2 n 1)12 Static Stellar Structure 34 Central Pressure Pc K 1 n n 2 K 1 n n c2 2 1 1 2 3 d / d 1 4 16 2 6 2 3 2 M R so M R 3 9 4 GR ( n 1) 2 2 But 1 1 3 d / d 1 2 GM 2 1 4 4 R ( n 1) d / d 1 2 4 GR 2 Pc (n 1) 2 Static Stellar Structure 9M 2 16 2 R 6 35 Central Temperature Pg N 0k cTc c Pc which means c Pc Tc N 0k c Static Stellar Structure 36