Static Stellar Structure Static Stellar Structure Most of
- Slides: 36
Static Stellar Structure
Static Stellar Structure Most of the Life of A Star is Spent in Equilibrium n n Evolutionary Changes are generally slow and can usually be handled in a quasistationary manner We generally assume: n n n Hydrostatic Equilibrium Thermodynamic Equilibrium The Equation of Hydrodynamic Equilibrium Static Stellar Structure 2
Limits on Hydrostatic Equilibrium n n If the system is not “Moving” - accelerating in reality - then d 2 r/dt 2 = 0 and then one recovers the equation of hydrostatic equilibrium: If ∂P/∂r ~ 0 then which is just the freefall condition for which the time scale is tff (GM/R 3)-1/2 Static Stellar Structure 3
Dominant Pressure Gradient n When the pressure gradient d. P/dr dominates one gets (r/t)2 ~ P/ρ n n This implies that the fluid elements must move at the local sonic velocity: cs = ∂P/∂ρ. When hydrostatic equilibrium applies V << cs n te >> tff where te is the evolutionary time scale n Static Stellar Structure 4
Hydrostatic Equilibrium n Consider a spherical star n n n n Shell of radius r, thickness dr and density ρ(r) Gravitional Force: ↓ (Gm(r)/r 2) 4πr 2ρ(r)dr Pressure Force: ↑ 4 r 2 d. P where d. P is the pressure difference across dr Equate the two: 4πr 2 d. P = (Gm(r)/r 2) 4πr 2ρ(r)dr r 2 d. P = Gm(r) ρ(r)dr d. P/dr = -ρ(r)(Gm(r)/r 2) The - sign takes care of the fact that the pressure decreases outward. Static Stellar Structure 5
Mass Continuity n n m(r) = mass within a shell = This is a first order differential equation which needs boundary conditions n n We choose Pc = the central pressure. Let us derive another form of the hydrostatic equation using the mass continuity equation. n n Express the mass continuity equation as a differential: dm/dr = 4πr 2ρ(r). Now divide the hydrostatic equation by the masscontinuity equation to get: d. P/dm = Gm/4πr 4(m) Static Stellar Structure 6
The Hydrostatic Equation in Mass Coordinates n d. P/dm = Gm/4πr 4(m) The independent variable is m n r is treated as a function of m n The limits on m are: n 0 at r = 0 n M at r = R (this is the boundary condition on the mass equation itself). n n Why? Radius can be difficult to define n Mass is fixed. n Static Stellar Structure 7
The Central Pressure n n n Consider the quantity: P + Gm(r)2/8πr 4 Take the derivative with respect to r: But the first two terms are equal and opposite so the derivative is -Gm 2/2 r 5. Since the derivative is negative it must decrease outwards. At the center m 2/r 4 → 0 and P = Pc. At r = R P = 0 therefore Pc > GM 2/8πR 4 Static Stellar Structure 8
The Virial Theorem Static Stellar Structure 9
The Virial Theorem n n n The term 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: 3 P/ = 2 * Thermal energy per unit mass. Static Stellar Structure 10
The Virial Theorem -2 U = Ω n 2 U + Ω = 0 Virial Theorem n Note that E = U + Ω or that E+U = 0 n This is only true if “quasistatic. ” If hydrodynamic then there is a modification of the Virial Theorem that will work. n Static Stellar Structure 11
The Importance of the Virial Theorem n Let us collapse a star due to pressure imbalance: n n If hydrostatic equilibrium is to be maintained thermal energy must change by: n n This will release - ∆Ω ∆U = -1/2 ∆Ω This leaves 1/2 ∆Ω to be “lost” from star n Normally it is radiated Static Stellar Structure 12
What Happens? Star gets hotter n Energy is radiated into space n System becomes more tightly bound: E decreases n Note that the contraction leads to H burning (as long as the mass is greater than the critical mass). n Static Stellar Structure 13
An Atmospheric Use of Pressure We use a different form of the equation of hydrostatic equilibrium in an atmosphere. n 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) n n 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 n n Geometry: Plane Parallel d. P/dr = -Gm(r)ρ/r 2 (Hydrostatic Equation) n n R ≈ r and m(R) = M. We use h measured with respect to some arbitrary 0 level. d. P/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 = nk. T and we use n = ρ/μm. H (μ is the mean molecular weight) so P = ρk. T/μm. H Static Stellar Structure 15
Atmospheric Pressure Continued n n n P d. P = Or d. P/P = ρk. T/μm. H = -g ρ dh -g P(μm. H/k. T) dh = -g(μm. H/k. T) dh Where: P 0 = P and ρ0 = ρ at h = 0. Static Stellar Structure 16
Scale Heights n H (the scale height) is defined as k. T / μm. Hg n n It defines the scale length by which P decreases by a factor of e. In non-isothermal atmospheres scale heights are still of importance: n H ≡ - (d. P/dh)-1 P = -(dh/d(ln. P)) 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) n Polytropic Model: P = Kργ n K and γ independent of r n γ not necessarily cp/cv n Static Stellar Structure 18
The Linear Model n n n Defining Equation Put in the Hydrostatic Equation Now we need to deal with m(r) Static Stellar Structure 19
An Expression for m(r) Equation of Continuity Integrate from 0 to r Static Stellar Structure 20
Linear Model • We want a particular (M, R) • ρc = 3 M/πR 3 and take Pc = P(ρc) • Substitute m(r) into the hydrostatic equation: Static Stellar Structure 21
Central Pressure Static Stellar Structure 22
The Pressure and Temperature Structure n The pressure at any radius is then: n Ignoring Radiation Pressure: Static Stellar Structure 23
Polytropes P = Kργ which can also be written as n P = Kρ(n+1)/n n N ≡ Polytropic Index Consider Adiabatic Convective Equilibrium Completely convective n Comes to equilibrium n No radiation pressure n Then P = Kργ where γ = 5/3 (for an ideal monoatomic gas) ==> n = 1. 5 n Static Stellar Structure 24
Gas and Radiation Pressure Pg = (N 0 k/μ)ρT = βP n Pr = 1/3 a T 4 = (1 - β)P n P = P n Pg/β = Pr/(1 -β) n 1/β(N 0 k/μ)ρT = 1/(1 -β) 1/3 a T 4 n Solve for T: T 3 = (3(1 - β)/aβ) (N 0 k/μ)ρ n T = ((3(1 - β)/aβ) (N 0 k/μ))1/3 ρ1/3 n Static Stellar Structure 25
The Pressure Equation n n 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 n n For n = 3 ρ T 3 The general result is: ρ Tn Let us proceed to the Lane-Emden Equation ρ ≡ λφn n n 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: d. P/dr = -Gρm(r)/r 2 Eqn 2: Mass Continuity: dm/dr = 4πr 2ρ Static Stellar Structure 27
Working Onward Static Stellar Structure 28
Simplify Static Stellar Structure 29
The Lane-Emden Equation n Boundary Conditions: n n n 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 ) = C 0 + C 2 ξ 2 + C 4 ξ 4 +. . . = 1 - (1/6) ξ 2 + (n/120) ξ 4 -. . . for ξ =1 (n>0) For n < 5 the solution decreases monotonically and φ → 0 at some value ξ 1 which represents the value of the boundary level. Static Stellar Structure 30
General Properties n For each n with a specified K there exists a family of solutions which is specified by the central density. n n For the standard model: We want Radius, m(r), or m(ξ) Central Pressure, Density Central Temperature Static Stellar Structure 31
Radius and Mass Static Stellar Structure 32
The Mean to Central Density Static Stellar Structure 33
The Central Pressure n n n 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: Static Stellar Structure 34
Central Pressure Static Stellar Structure 35
Central Temperature Static Stellar Structure 36
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