Static Stellar Structure

2 Evolutionary Changes are generally slow and can usually be handled in a quasistationary manner Evolutionary Changes are generally slow and can usually be handled in a quasistationary manner We generally assume: We generally assume: Hydrostatic Equilibrium Hydrostatic Equilibrium Thermodynamic Equilibrium Thermodynamic Equilibrium The Equation of Hydrodynamic Equilibrium The Equation of Hydrodynamic Equilibrium Most of the Life of A Star is Spent in Equilibrium

3Static Stellar Structure Limits on Hydrostatic Equilibrium 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 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 t ff  (GM/R 3 ) -1/2 If ∂P/∂r ~ 0 then which is just the freefall condition for which the time scale is t ff  (GM/R 3 ) -1/2

4Static Stellar Structure Dominant Pressure Gradient When the pressure gradient dP/dr dominates one gets (r/t) 2 ~ P/ρ 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: c s = ∂P/∂ρ. This implies that the fluid elements must move at the local sonic velocity: c s = ∂P/∂ρ. When hydrostatic equilibrium applies When hydrostatic equilibrium applies V << c s V << c s t e >> t ff where t e is the evolutionary time scale t e >> t ff where t e is the evolutionary time scale

5Static Stellar Structure Hydrostatic Equilibrium Consider a spherical star Consider a spherical star Shell of radius r, thickness dr and density ρ(r) Shell of radius r, thickness dr and density ρ(r) Gravitional Force: ↓ (Gm(r)/r 2 ) 4πr 2 ρ(r)dr Gravitional Force: ↓ (Gm(r)/r 2 ) 4πr 2 ρ(r)dr Pressure Force: ↑ 4r 2 dP where dP is the pressure difference across dr Pressure Force: ↑ 4r 2 dP where dP is the pressure difference across dr Equate the two: 4πr 2 dP=(Gm(r)/r 2 ) 4πr 2 ρ(r)dr Equate the two: 4πr 2 dP=(Gm(r)/r 2 ) 4πr 2 ρ(r)dr r 2 dP=Gm(r) ρ(r)dr r 2 dP=Gm(r) ρ(r)dr dP/dr=-ρ(r)(Gm(r)/r 2 ) dP/dr=-ρ(r)(Gm(r)/r 2 ) The - sign takes care of the fact that the pressure decreases outward. The - sign takes care of the fact that the pressure decreases outward.

6Static Stellar Structure Mass Continuity m(r) = mass within a shell = m(r) = mass within a shell = This is a first order differential equation which needs boundary conditions This is a first order differential equation which needs boundary conditions We choose P c = the central pressure. We choose P c = the central pressure. Let us derive another form of the hydrostatic equation using the mass continuity equation. Let us derive another form of the hydrostatic equation using the mass continuity equation. Express the mass continuity equation as a differential: dm/dr = 4πr 2 ρ(r). Express the mass continuity equation as a differential: dm/dr = 4πr 2 ρ(r). Now divide the hydrostatic equation by the mass- continuity equation to get: dP/dm = Gm/4πr 4 (m) Now divide the hydrostatic equation by the mass- continuity equation to get: dP/dm = Gm/4πr 4 (m)

7Static Stellar Structure The Hydrostatic Equation in Mass Coordinates dP/dm = Gm/4πr 4 (m) dP/dm = Gm/4πr 4 (m) The independent variable is m The independent variable is m r is treated as a function of m r is treated as a function of m The limits on m are: The limits on m are: 0 at r = 0 0 at r = 0 M at r = R (this is the boundary condition on the mass equation itself). M at r = R (this is the boundary condition on the mass equation itself). Why? Why? Radius can be difficult to define Radius can be difficult to define Mass is fixed. Mass is fixed.

8Static Stellar Structure The Central Pressure Consider the quantity: P + Gm(r) 2 /8πr 4 Consider the quantity: P + Gm(r) 2 /8πr 4 Take the derivative with respect to r: Take the derivative with respect to r: But the first two terms are equal and opposite so the derivative is -Gm 2 /2r 5. But the first two terms are equal and opposite so the derivative is -Gm 2 /2r 5. Since the derivative is negative it must decrease outwards. Since the derivative is negative it must decrease outwards. At the center m 2 /r 4 → 0 and P = P c. At r = R P = 0 therefore P c > GM 2 /8πR 4 At the center m 2 /r 4 → 0 and P = P c. At r = R P = 0 therefore P c > GM 2 /8πR 4

9Static Stellar Structure The Virial Theorem

10Static Stellar Structure The Virial Theorem The term is 0: r(0) = 0 and P(M) = 0 The term is 0: r(0) = 0 and P(M) = 0 Remember that we are considering P, ρ, and r as variables of m 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. For a non-relativistic gas: 3P/ = 2 * Thermal energy per unit mass.

11Static Stellar Structure The Virial Theorem -2U = Ω -2U = Ω 2U + Ω = 0 Virial Theorem 2U + Ω = 0 Virial Theorem Note that E = U + Ω or that E+U = 0 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. This is only true if “quasistatic.” If hydrodynamic then there is a modification of the Virial Theorem that will work.

12Static Stellar Structure The Importance of the Virial Theorem Let us collapse a star due to pressure imbalance: Let us collapse a star due to pressure imbalance: This will release - ∆Ω This will release - ∆Ω If hydrostatic equilibrium is to be maintained the thermal energy must change by: If hydrostatic equilibrium is to be maintained the thermal energy must change by: ∆U = -1/2 ∆Ω ∆U = -1/2 ∆Ω This leaves 1/2 ∆Ω to be “lost” from star This leaves 1/2 ∆Ω to be “lost” from star Normally it is radiated Normally it is radiated

13Static Stellar Structure What Happens? Star gets hotter Star gets hotter Energy is radiated into space Energy is radiated into space System becomes more tightly bound: E decreases 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). Note that the contraction leads to H burning (as long as the mass is greater than the critical mass).

14Static Stellar Structure An Atmospheric Use of Pressure We use a different form of the equation of hydrostatic equilibrium in an atmosphere. 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) 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. 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.

15Static Stellar Structure Atmospheric Pressure Geometry: Plane Parallel Geometry: Plane Parallel dP/dr = -Gm(r)ρ/r 2 (Hydrostatic Equation) dP/dr = -Gm(r)ρ/r 2 (Hydrostatic Equation) R ≈ r and m(R) = M. We use h measured with respect to some arbitrary 0 level. 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) 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 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/μm H P = ρkT/μm H

16Static Stellar Structure Atmospheric Pressure Continued P=ρkT/μm H P=ρkT/μm H dP=-g ρ dh dP=-g ρ dh dP=-g P(μm H /kT) dh dP=-g P(μm H /kT) dh Or dP/P=-g(μm H /kT) dh Or dP/P=-g(μm H /kT) dh Where: P 0 = P and ρ 0 = ρ at h = 0. Where: P 0 = P and ρ 0 = ρ at h = 0.

17Static Stellar Structure Scale Heights H (the scale height) is defined as kT / μm H g 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. It defines the scale length by which P decreases by a factor of e. In non-isothermal atmospheres scale heights are still of importance: In non-isothermal atmospheres scale heights are still of importance: H ≡ - (dP/dh) -1 P = -(dh/d(lnP)) H ≡ - (dP/dh) -1 P = -(dh/d(lnP))

18Static Stellar Structure Simple Models The linear model: ρ = ρ c (1-r/R) The linear model: ρ = ρ c (1-r/R) Polytropic Model: P = Kρ γ Polytropic Model: P = Kρ γ K and γ independent of r K and γ independent of r γ not necessarily c p /c v γ not necessarily c p /c v To do this right we need data on energy generation and energy transfer but:

19Static Stellar Structure The Linear Model Defining Equation Defining Equation Put in the Hydrostatic Equation Put in the Hydrostatic Equation Now we need to deal with m(r) Now we need to deal with m(r)

20Static Stellar Structure An Expression for m(r) Equation of Continuity Integrate from 0 to r

21Static Stellar Structure 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:

22Static Stellar Structure Central Pressure

23Static Stellar Structure The Pressure and Temperature Structure The pressure at any radius is then: The pressure at any radius is then: Ignoring Radiation Pressure: Ignoring Radiation Pressure:

24Static Stellar Structure Polytropes P = Kρ γ which can also be written as P = Kρ γ which can also be written as P = Kρ (n+1)/n P = Kρ (n+1)/n N ≡ Polytropic Index N ≡ Polytropic Index Consider Adiabatic Convective Equilibrium Consider Adiabatic Convective Equilibrium Completely convective Completely convective Comes to equilibrium Comes to equilibrium No radiation pressure No radiation pressure Then P = Kρ γ where γ = 5/3 (for an ideal monoatomic gas) ==> n = 1.5 Then P = Kρ γ where γ = 5/3 (for an ideal monoatomic gas) ==> n = 1.5

25Static Stellar Structure Gas and Radiation Pressure P g = (N 0 k/μ)ρT = βP P g = (N 0 k/μ)ρT = βP P r = 1/3 a T 4 = (1- β)P P r = 1/3 a T 4 = (1- β)P P=P P=P P g /β=P r /(1-β) P g /β=P r /(1-β) 1/β(N 0 k/μ)ρT =1/(1-β) 1/3 a T 4 1/β(N 0 k/μ)ρT =1/(1-β) 1/3 a T 4 Solve for T: T 3 =(3(1- β)/aβ) (N 0 k/μ)ρ Solve for T: T 3 =(3(1- β)/aβ) (N 0 k/μ)ρ T=((3(1- β)/aβ) (N 0 k/μ)) 1/3 ρ 1/3 T=((3(1- β)/aβ) (N 0 k/μ)) 1/3 ρ 1/3

26Static Stellar Structure The Pressure Equation True for each point in the star True for each point in the star If β ≠ f(R), i.e. is a constant then If β ≠ f(R), i.e. is a constant then P = Kρ 4/3 P = Kρ 4/3 n = 3 polytrope or γ = 4/3 n = 3 polytrope or γ = 4/3

27Static Stellar Structure The Eddington Standard Model n = 3 For n = 3 ρ  T 3 For n = 3 ρ  T 3 The general result is: ρ  T n The general result is: ρ  T n Let us proceed to the Lane-Emden Equation Let us proceed to the Lane-Emden Equation ρ ≡ λφ n ρ ≡ λφ n λ is a scaling parameter identified with ρ c - the central density λ is a scaling parameter identified with ρ c - the central density φ n is normalized to 1 at the center. φ n is normalized to 1 at the center. Eqn 0: P = Kρ (n+1)/n = Kλ (n+1)/n φ n+1 Eqn 0: P = Kρ (n+1)/n = Kλ (n+1)/n φ n+1 Eqn 1: Hydrostatic Eqn: dP/dr = -Gρm(r)/r 2 Eqn 1: Hydrostatic Eqn: dP/dr = -Gρm(r)/r 2 Eqn 2: Mass Continuity: dm/dr = 4πr 2 ρ Eqn 2: Mass Continuity: dm/dr = 4πr 2 ρ

28Static Stellar Structure Working Onward

29Static Stellar Structure Simplify

30Static Stellar Structure The Lane-Emden Equation Boundary Conditions: Boundary Conditions: Center ξξ = 0; φ = 1 (the function); dφ/d ξ = 0 Center ξξ = 0; φ = 1 (the function); dφ/d ξ = 0 Solutions for n = 0, 1, 5 exist and are of the form: Solutions for n = 0, 1, 5 exist and are of the form: φ(ξ)=C 0 + C 2 ξ 2 + C 4 ξ φ(ξ)=C 0 + C 2 ξ 2 + C 4 ξ =1 - (1/6) ξ 2 + (n/120) ξ for ξ =1 (n>0) =1 - (1/6) ξ 2 + (n/120) ξ 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. For n < 5 the solution decreases monotonically and φ → 0 at some value ξ 1 which represents the value of the boundary level.

31Static 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 each n with a specified K there exists a family of solutions which is specified by the central density. For the standard model: For the standard model: We want Radius, m(r), or m(ξ) We want Radius, m(r), or m(ξ) Central Pressure, Density Central Pressure, Density Central Temperature Central Temperature

32Static Stellar Structure Radius and Mass

33Static Stellar Structure The Mean to Central Density

34Static Stellar Structure The Central Pressure At the center ξ = 0 and φ = 1 so P c = Kλ n+1/n At the center ξ = 0 and φ = 1 so P c = Kλ n+1/n This is because P = Kρ n+1/n = Kλ (n+1)/n φ n+1 This is because P = Kρ n+1/n = Kλ (n+1)/n φ n+1 Now take the radius equation: Now take the radius equation:

35Static Stellar Structure Central Pressure

36Static Stellar Structure Central Temperature