Static Stellar Structure
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2 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 dr Gmr () ρ P = ρ dt r r
3 Limits on Hydrostatic Equilibrium If the system is not Moving - accelerating in reality - then d r/dt = 0 and then one recovers the equation of hydrostatic equilibrium: If P/ P/ r r ~ 0 then which is just the freefall condition for which the time scale is t ff (GM/R 3 ) -/ Gm() r ρ P dp = = r r dr dr dt Gmr () = r 3
4 Dominant Pressure Gradient When the pressure gradient dp/dr dominates one gets (r/t) ~ P/ρ This implies that the fluid elements must move at the local sonic velocity: c s = P/ P/ ρ. When hydrostatic equilibrium applies V << c s t e >> t ff where t e is the evolutionary time scale 4
5 Hydrostatic Equilibrium Consider a spherical star Shell of radius r, thickness dr and density ρ(r) Gravitional Force: (Gm(r)/r ) 4πr4 ρ(r)dr Pressure Force: 4r dp where dp is the pressure difference across dr Equate the two: 4πr dp = (Gm(r)/r ) 4πr4 ρ(r)dr r dp = Gm(r) ρ(r)dr dp/dr = -ρ(r)(gm(r)/r ) The - sign takes care of the fact that the pressure decreases outward. 5
6 Mass Continuity m(r) ) = mass within a shell = mr () = 4 πr ρ( r ) dr 0 This is a first order differential equation which needs boundary conditions We choose P c = 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 dr = 4πr4 ρ(r). Now divide the hydrostatic equation by the mass- continuity equation to get: dp/dm = Gm/4πr 4 (m) r 6
7 The Hydrostatic Equation in Mass dp/dm = Gm/4πr 4 (m) Coordinates 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. 7
8 The Central Pressure Consider the quantity: P + Gm(r) /8πr 4 Take the derivative with respect to r: d Gm() r dp Gm() r dm Gm() r P dr + 8πr = + dr 4πr dr πr But the first two terms are equal and opposite so the derivative is -Gm /r 5. Since the derivative is negative it must decrease outwards. At the center m /r 4 0 and P = P c. At r = R P = 0 therefore P c > GM /8πR 4 8
9 The Virial Theorem dp mg = 4 dm 4 π r( m) 3 dp mg 4π r = dm r d 3 dr mg (4 πrp) 4πr3P = dm dm r M 3 3 M M P mg (4 π r P) 0 dm = dm 0 ρ 0 r Remember: 4π r rdr = dm 9
10 The Virial Theorem 3 The term (4 π rp) M 0 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/ = * Thermal energy per unit mass. M 3 Pdm = U for the entire star 0 ρ GM dm =Ω r the gravitional binding energy 0
11 The Virial Theorem -U = Ω U + 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.
12 The Importance of the Virial Theorem Let us collapse a star due to pressure imbalance: This will release - Ω If hydrostatic equilibrium is to be maintained the thermal energy must change by: U U = -/ Ω This leaves / Ω to be lost from star Normally it is radiated
13 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). 3
14 An Atmospheric Use of Pressure We use a different form of the equation of hydrostatic equilibrium in an atmosphere. The atmosphere s 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 00's of km whereas the solar radius is 700,00 km. The photosphere mass is about % of the Sun. 4
15 Atmospheric Pressure Geometry: Plane Parallel dp/dr = -Gm(r) Gm(r)ρ/r (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/μm H (μ is the mean molecular weight) so P = ρkt/ kt/μm H 5
16 Atmospheric Pressure Continued P = ρkt/ kt/μm H dp = -g ρ dh dp = -g P(μm H /kt) ) dh Or dp/p = -g( g(μm H /kt) ) dh Integrate : P = P e 0 ρ = ρ e 0 gmh μh kt gmh μh kt Where: P 0 = P and ρ 0 = ρ at h = 0. 6
17 Scale Heights H (the scale height) is defined as kt / μm H g 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) - P = -(dh/d(lnp)) 7
18 Simple Models To do this right we need data on energy generation and energy transfer but: The linear model: ρ = ρ c (-r/r) Polytropic Model: P = Kρ γ K and γ independent of r γ not necessarily c p /c v 8
19 The Linear Model r ρ = ρc( ) R dp Gm() r r = ρ ( ) c dr r R Defining Equation Put in the Hydrostatic Equation Now we need to deal with m(r) 9
20 An Expression for m(r) dm r = 4 πr ρc( ) dr R 3 4π r ρc = 4πr ρc R 3 4 4π r ρc πr ρc mr () = 3 R 3 π R ρc Total Mass M = m( R) = 3 4r 3r So m() r = M R R Equation of Continuity Integrate from 0 to r 0
21 Linear Model We want a particular (M,R) ρ c = 3M/πR 3 and take P c = P(ρ c ) Substitute m(r) into the hydrostatic equation: dr r 3 R R 3 4 dp πg 4r r r = ρ c 4r 7r r = πgρc + 3 3R R r 7r r So P = Pc πgρc + 3 9R 4R where P is the central pressure. c
22 Central Pressure At the surface r = R and P = 0 4 R c 5 G PR ( ) = P = 0 c 36 c 6 5 G R 5 GR 9M P = = c π R P c = 5GM 4 R π ρ π ρ π π
23 The Pressure and Temperature Structure The pressure at any radius is then: Pr () = G R R 5R 5R 3 4 π r r r ρ c 3 4 Ignoring Radiation Pressure: T μmh P = kρ 5π Gμm r 9r 9r 36 k R 5R 5R 3 H = ρcr
24 Polytropes P = Kρ γ which can also be written as P = Kρ K (n+)/n N Polytropic Index Consider Adiabatic Convective Equilibrium Completely convective Comes to equilibrium No radiation pressure Then P = Kρ γ where γ = 5/3 (for an ideal monoatomic gas) ==> n =.5 4
25 Gas and Radiation Pressure P g = (N 0 k/μ)ρt T = βp P r = /3 a T 4 = (- β)p P = P P g /β = P r /(-β) /β(n 0 k/μ)ρt = /(-β) ) /3 a T 4 Solve for T: T 3 = (3(- β)/aβ) ) (N 0 k/μ)ρ T = ((3(- β)/aβ) ) (N 0 k/μ)) )) /3 ρ /3 5
26 The Pressure Equation P = Nk 0 μ ρt β β 3 Nk = μ a β True for each point in the star If β f(r), i.e. is a constant then P = KρK 4/3 n = 3 polytrope or γ = 4/3 ρ 6
27 The Eddington Standard Model n = 3 For n = 3 ρ T 3 The general result is: ρ T n Let us proceed to the Lane-Emden Equation ρ λφ n λ is a scaling parameter identified with ρ c - the central density φ n is normalized to at the center. Eqn 0: P = Kρ K (n+)/n = Kλ K (n+)/n φ n+ Eqn : Hydrostatic Eqn: dp/dr = -Gρm(r)/r Eqn : Mass Continuity: dm/dr dr = 4πr4 ρ 7
28 Working Onward Solve for m( r): m( r) = d r dp Put m() r in : 4πGρ = r dr ρ dr dp dr n+ n n = Kλ ( n+ ) φ r dp ρgdr dφ dr φ n Kλ ( n+ ) φ = 4πGλφ n+ d r n n d n r dr dr λφ 8
29 Simplify n+ d r n dφ n n ( ) 4 Kλ n+ φ n = πgλφ r dr λφ dr d n dφ n Kλ ( n+ ) r 4πGλφ r dr dr = n d n dφ n Kλ ( n+ ) r φ 4 πg r = dr dr n n Kλ ( n+ ) Define a 4π G r ξ so adξ = dr a d dφ n So : ξ ξ dξ dξ = φ 9
30 The Lane-Emden Equation Boundary Conditions: ξ d = ξ d ξ d dφ ξ φ n Center ξξ = 0; φ = (the function); dφ/d ξ = 0 Solutions for n = 0,, 5 exist and are of the form: φ(ξ) = C 0 + C ξ + C 4 ξ = - (/6) ξ + (n/0) ξ for ξ = (n>0) For n < 5 the solution decreases monotonically and φ 0 at some value ξ which represents the value of the boundary level. 30
31 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: K 4 3 Nk 0 3 β = 4 We want Radius, m(r), or m(ξ) Central Pressure, Density Central Temperature μ a β 3
32 Radius and Mass R = aξ ( n+ ) K = 4π G aξ M( ξ) = 4πr ρdr 0 ξ 3 n 0 λ n n = 4πa λφ ξ dξ d dφ ξ ξ dξ = dξ d dφ ξ π λ ξ ξ dξ dξ ξ 3 M ( ) = 4 a d 0 ξ n φ ξ 3 dφ M( ξ) = 4πa λ d 0 ξ dξ 3 dφ = 4πa λξ dξ Now substitute for a and evaluate at ξ = ξ ( boundary) M 3 3 n ( n+ ) K n dφ = 4π λ ξ 4πG dξ ξ= ξ 3
33 The Mean to Central Density ρ c = λ 3 3 n ( n+ ) K dφ n 4π λ ξ 4πG dξ ξ = ξ 3 3 3( n) R 4 ( n+ ) K n 3 π λ ξ M ρ = = 4 π 3 3 4π G 3 dφ ρ = λ ξ dξ ρ 3 dφ = ρc ξ dξ ξ= ξ ξ= ξ 33
34 The Central Pressure At the center ξ = 0 and φ = so P c = KλK n+/n This is because P = KρK n+/n = Kλ K (n+)/n φ n+ Now take the radius equation: ( n+ ) K R = 4π G λ n n n n ( n + ) ξ = Kλ 4π G 4π GR So Kλ n n = n + ) ξ ξ 34
35 Central Pressure c n n n n c P = Kλ λ = Kλ ρ 4πGR ξ = ρ ( n+ ) ξ 3 ( dφ/ dξ ) ξ But M = πr ρ so M = π R ρ 3 9 P c ξ M 6 4πGR 9 = ( n+ ) ξ 3 ( dφ/ dξ) 6π R ξ GM = 4 4 πr ( n+ ) ( dφ/ dξ ) ξ 35
36 Central Temperature Nk P T P = 0 ρ = β g c c c c which means T c = β μ c 0 P Nk c μ ρ c 36
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