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 Static Stellar Structure

Limits on Hydrostatic Equilibrium If the system is not “Moving” - accelerating in reality - then d2r/dt2 = 0 and then one recovers the equation of hydrostatic equilibrium: If ∂P/∂r ~ 0 then we have the freefall condition for which the time scale is tff (GM/R3)-1/2 Static Stellar Structure

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

Static Stellar Structure FYI: Sound Speed Static Stellar Structure

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: ↑ 4πr2dP 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

Static Stellar Structure Mass Continuity m(r) = mass within a shell = 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 mass-continuity equation to get: dP/dm = Gm/4πr4(m) Static Stellar Structure

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

Static Stellar Structure The Central Pressure Consider the quantity: P + Gm(r)2/8πr4 (~pressure at center of a constant density sphere) Take the derivative with respect to r: 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 assume m2/r4 → 0 and P = Pc. At r = R P = 0 therefore Pc > GM2/8πR4 For the Sun this gives 4.5(10^13) Pa actual is = 2.5(10^16) Pa = 2.5(10^17) dynes / cm^2 Static Stellar Structure

Central Pressure of a Constant Density Sphere Static Stellar Structure

Static Stellar Structure The Virial Theorem Static Stellar Structure

Static Stellar Structure The Virial Theorem 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: 3P/ρ = 2 * Thermal energy per unit mass. Static Stellar Structure

Static Stellar Structure 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

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 = -1/2 ∆Ω This leaves 1/2 ∆Ω to be “lost” from star Normally it is radiated Static Stellar Structure

Static Stellar Structure 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

Static Stellar Structure 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

Static Stellar Structure The Linear Model Defining Equation Put in the Hydrostatic Equation Now we need to deal with m(r) Static Stellar Structure

Static Stellar Structure An Expression for m(r) Equation of Continuity Integrate from 0 to r Static Stellar Structure

Static Stellar Structure Linear Model We want a particular (M,R) ρc = 3M/πR3 and take Pc = P(ρc) Substitute m(r) into the hydrostatic equation: Static Stellar Structure

Static Stellar Structure Central Pressure Static Stellar Structure

The Pressure and Temperature Structure The pressure at any radius is then: Ignoring Radiation Pressure: Static Stellar Structure

Static Stellar Structure 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

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

Static Stellar Structure The Pressure Equation 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

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

Static Stellar Structure Working Onward Static Stellar Structure

Static Stellar Structure Simplify Static Stellar Structure

The Lane-Emden Equation Boundary Conditions: Center ξ = 0; φ = 1 (the function); dφ/dξ = 0 Explicit Solutions for n = 0, 1, 5 For the general case: Assume that φ(ξ) can be expressed as a power series (NB: φ(ξ) and φ(-ξ) both solve the equation) 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. Why? Static Stellar Structure

Static Stellar Structure 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: We want Radius, m(r), or m(ξ) Central Pressure, Density Central Temperature Static Stellar Structure

Static Stellar Structure Radius and Mass Static Stellar Structure

The Mean to Central Density Static Stellar Structure

Static Stellar Structure 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: Static Stellar Structure

Static Stellar Structure Central Pressure Static Stellar Structure

Static Stellar Structure Central Temperature Static Stellar Structure