1. Static Stellar Structure

2. 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

3. 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

4. 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

5. Static Stellar Structure
FYI: Sound Speed
Static Stellar Structure

6. 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

7. 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

8. 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

9. 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

10. Central Pressure of a Constant Density Sphere
Static Stellar Structure

11. Static Stellar Structure
The Virial Theorem
Static Stellar Structure

12. 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

13. Static Stellar Structure
The Virial Theorem
-2U = Ω
2U + Ω = 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

14. 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

15. 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

16. 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

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

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

19. 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

20. Static Stellar Structure
Central Pressure
Static Stellar Structure

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

22. 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

23. 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

24. 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

25. 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

26. Static Stellar Structure
Working Onward
Static Stellar Structure

27. Static Stellar Structure
Simplify
Static Stellar Structure

28. 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 ξ
= 1 - (1/6) ξ2 + (n/120) ξ 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

29. 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

30. Static Stellar Structure
Radius and Mass
Static Stellar Structure

31. The Mean to Central Density
Static Stellar Structure

32. 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

33. Static Stellar Structure
Central Pressure
Static Stellar Structure

34. Static Stellar Structure
Central Temperature
Static Stellar Structure