Lecture 7a Polytropes

Prialnik Chapter 5 Glatzmaier and Krumholz Chapter 9 Pols 4 We have reached a point where we have most of the informaon required to solve realisc stellar models. A crical piece sll mixing is the nuclear physics, but it turns out that many of the observed properes of stars, except their lifemes and radii, reflect chiefly the need to be in hydrostac and thermal equilibrium, and not the energy source.

Historically, prior to computers, stellar structure was oen calculated using polytropes. Even today. they provide a valuable tool for understanding many stellar properes. The Stellar Structure Equations dm dr 1 = 4πr 2 ρ = dr dm 4πr 2 ρ dP Gm dP Gm = − ρ = − dr r 2 dm 4πr 4 dT 3 κρ L(r) dT 3 κ L(r) = − = − dr 4ac T 3 4πr 2 dm 4ac T 3 (4πr 2 )2 dL(r) dL(m) = 4πr 2 ρε = ε dr dm

Sll to be defined -  In addition we have or will have the physics equations ρ ⎞ 1 P = P(T, , Y ) N kT P aT 4 ρ { i } = A ⎟ + e + µ ⎠ ions 3 (T, , Y ) aT b κ = κ ρ { i } ≈ κ 0 ρ (T, , Y ) mT n (TBD) ε = ε ρ { i } ≈ ε0 ρ and the boundary conditions at r = 0 L(r)=0, m = 0 1/4 ⎛ L(R) ⎞ at r = R P(R) ≈ 0, m= M, T= ≈0 ⎝⎜ 4πR2σ ⎠⎟ These are 7 equations in 7 unknowns: ρ, T, P, L, ε, κ, and r which, in principle, given M and initial composition, can be solved as a function of time for the boundary conditions. Most of our theoretical knowledge of stellar evolution comes from doing just that. Aside: Boundary Conditions

l Radiative Zero BC: Ideally we would use some atmospheric model to tell us what the temperature BC is at the surface The T change over the whole star is so large that the

difference between 0 and the real Teff at surface is small – except when computing a luminosity.

To get effective T at the end, use: Polytropes We assume a global relaon between pressure and density:

P ∝ ργ γ =(n +1) / n

n is called the polytropic index. Don’t confuse this with the adiabatic index which is local.

5/3 4/3 Some examples: P ∝ ρ or ρ n =constant

l Fully convective (adiabatic):

l White dwarfs (completely degenerate):

l Pressure is a mix of gas + radiation, but the ratio is constant throughout Polytropes l Consider HSE dP Gm(r) = − ρ r = 0 dr r 2 r 2 dP = − Gm(r) dr ρ 2 l Differentiating again and dividing by r :

d ⎛ r 2 dP ⎞ dm(r) = − G = − 4πGr 2ρ(r) dr ⎜ dr ⎟ dr ⎝ ρ ⎠ 1 d ⎛ r 2 dP ⎞ = − 4πGρ(r) r 2 dr ⎜ ρ dr ⎟ ⎝ ⎠

Polytropes

l Now make this dimensionless Central density: Define: Then:

P(r)=Kργ (r) n +1 1 n γ = = 1+ γ γ n n = K ρc θ (r) = K ρ1+1/n θ n+1(r) c

Given n, K, and c these two equations define the distribution of pressure and density in the star Use these assumptions in the equation for hydrostatic equilibrium

1 d ⎛ r 2 dP ⎞ = − 4πGρ P =Kργ =K ργ θ γ n =K ργ θ n+1 r 2 dr ⎝⎜ ρ dr ⎠⎟ c c K ργ d ⎛ r 2 dθ n+1 ⎞ c 4 G n n 2 ⎜ n ⎟ = − π ρcθ ρ = ρcθ r dr ⎝ ρcθ dr ⎠ 1 d ⎛ r 2 dθ n+1 ⎞ P = − 4πGρ θ n c r 2 dr ⎝⎜ θ n dr ⎠⎟ c P (n +1) d ⎛ r 2θ n dθ ⎞ c 4 G n 2 2 ⎜ n ⎟ = − π θ ρc r dr ⎝ θ dr ⎠

⎡P (n +1)⎤ 1 d ⎛ 2 dθ ⎞ n ⎢ c ⎥ r = −θ 4πGρ 2 r 2 dr ⎝⎜ dr ⎠⎟ ⎣⎢ c ⎦⎥ ⎡P (n +1)⎤ 1 d ⎛ 2 dθ ⎞ n ⎢ c ⎥ r = −θ 4 G 2 r 2 dr ⎜ dr ⎟ ⎣⎢ π ρc ⎦⎥ ⎝ ⎠

Note that the quantity in brackets has units length2 Define it to be α 2 then 1/2 2 α d ⎛ 2 dθ ⎞ n ⎡P (n +1)⎤ r = −θ with α =⎢ c ⎥ r 2 dr ⎝⎜ dr ⎠⎟ 4πGρ 2 ⎣⎢ c ⎦⎥

2 6 dyne gm cm 2 2 2 2 2 = cm for the dimension of α cm dynecm gm Now define r =αξ with ξ a dimensionless radius- like variable. Then 1 d ⎛ dθ ⎞ ξ 2 = −θ n ξ 2 dξ ⎝⎜ dξ ⎠⎟

This is the Lane Emden equation. It involves only dimensionless quantities and can be solved for a given n to give θ(ξ). n does not have to be an integer.

Basically all that went into it was hydrostatic equilibrium, mass conservation, and a power law equation of state Polytrope boundary condions ρ r 1) θ n = So at = ξ =0 θ =1 ρc α

r 1 dθ dθ 2) ξ ≡ so dξ= dr ⇒ = α α α dξ dr dρ dθ but ρ ≡ ρ θ n so =nρ θ n−1 c dr c dr dθ 1 dρ dθ α dρ and ⇒ = n−1 = n−1 dr nρc θ dr dξ nρc θ dr but in spherical symmetry for hydrostatic equilibrium dρ ρ is a local maximum at r = 0, so = 0 dr dθ So at at ξ =0 = 0 dξ That is dP Gm(r)ρ = − → 0 as r → 0 dr r 2 because m(r) ∝ r 3 approaches 0 faster than r 2 dP dρ And since P=K ργ = γ Kργ −1 dr dr dρ so also →0. The density and pressure have local dr maxima at the center of the star - no "cusps"

P or 

r Polytrope boundary condions

3) At the surface r = R the density (first) goes to zero so

R =αξ where ξ is where θ(ξ) first reaches 0 * 1 1 For any value of n, the mass R ξ1 r = αξ R = αξ M = 4πr 2ρ dr = 4πα 3 ρ ξ 2θ n dξ 1 ∫ c ∫ n 0 0 ρ = ρ θ c But the Lane Emden equation says 1 d ⎛ dθ ⎞ d ⎛ dθ ⎞ ξ 2 = −θ n or ξ 2θ n = − ξ 2 ξ 2 dξ ⎝⎜ dξ ⎠⎟ dξ ⎝⎜ dξ ⎠⎟ ξ 1 d ⎛ dθ ⎞ So M = − 4πα 3 ρ ξ 2 dξ c ∫ ⎜ ⎟ 0 dξ ⎝ dξ ⎠

ξ1 dθ 3 ⎛ 2 dθ ⎞ 3 2 dθ ⎞ 0 at =0 = − 4πα ρ d ξ = − 4πα ρ ξ = M = ξ c ∫ ⎜ ⎟ c 1 ⎟ dξ 0 ⎝ dξ ⎠ dξ ⎠ ξ1 dθ ⎞ The quantity ξ 2 is uniquely determined by n and 1 dξ ⎠⎟ ξ1 is given in tables. So we have an equation connecting

M, α. and ρc for a given polytropic index, n.

Cox and Guilli (1965)

2 dθ ρ ξ ξ D = c 1 1 n dξ ρ Further α can be expressed in terms of ρc and K 1/2 1/2 n 1 1 n 1/2 + − ⎡ n ⎤ ⎡ n ⎤ ⎡Pc (n +1)⎤ ⎢Kρc (n +1)⎥ ⎢Kρc (n +1)⎥ α = ⎢ ⎥ = = 4 G 2 ⎢ 4 G 2 ⎥ ⎢ 4πG ⎥ ⎣⎢ π ρc ⎦⎥ π ρc ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥ 3/2 1−n ⎡ n ⎤ 3 2 dθ ⎞ ⎢Kρc (n +1)⎥ 2 dθ ⎞ M = − 4πα ρ ξ = − 4π ρ ξ c 1 d ⎟ ⎢ 4 G ⎥ c 1 d ⎟ ξ ⎠ ξ π ξ ⎠ ξ 1 ⎣⎢ ⎦⎥ 1 3/2 3/2 3−3n n +1 +1 ( ) 2 dθ ⎞ ⎛ K ⎞ 2n = − ξ1 ⎟ ⎜ ⎟ ρc 4 dξ ⎠ ⎝ G⎠ π ξ1 3/2 3/2 n +1 dθ ⎞ ⎛ K ⎞ 3−n M ( ) 2 2n = − ξ1 ⎟ ⎜ ⎟ ρc 4 dξ ⎠ ⎝ G⎠ π ξ1 which is another useful form which gives M in terms of ρc (or vice versa) for a given K and n. Note that M is independent of ρ if n = 3. c Another interesting quantity that can be obtained from the tables is the ratio of central density to average density - how compact the core of the star is.

See the previous table. The case n = 0, Dn = 1 is the sphere of constant density case Analytic solutions

For n = 0, 1, 5, and only for these values, there exist analytic solutions to the Lane Emden equation.

Unfortunately none of them correspond to common stars, but the solutions help to demonstrate how polytropes can be solved and they can used for interesting approximate cases. n = 0

Technically the case n = 0 is a singularity

P = Kρ1+1/n diverges as n → 0

This reflects the fact that constant density can only be maintailed in the face of gravity if the fluid is incompressible (like the ocean; P can vary but not ρ).

Nevertheless, some interesting properties of polytropes can be illustarted with n = 0 so long as we don't use the pressure-density relation explicitly n = 0 - the constant density case 1 d ⎛ dθ ⎞ ξ 2 = −θ n = −1 ξ 2 dξ ⎝⎜ dξ ⎠⎟ dθ ξ 3 ξ 2 = − ξ 2 = − + C dξ ∫ 3 dθ ξ C 1 Cdξ = − + ⇒ dθ = − ξ dξ + dξ 3 ξ 2 ∫ 3 ∫ ∫ ξ 2 ξ 2 C θ = − − + D but θ=1 at ξ=0 6 ξ So C = 0 and D =1 and ξ 2 θ = 1 − 6 hp://www.vistrails.org/index.php/User:Tohline/SSC/Structure/Polytropes 6 The first zero of θ = 1- 2 is at ξ1 = 6 r ξ ξ = α so R = 6 α

n+1 P(r) = Pcθ (r)=Pcθ(r) for n = 0 ⎛ ξ 2 ⎞ ⎛ r 2 ⎞ P(r) = P 1− =P 1− c ⎝⎜ 6 ⎠⎟ c ⎝⎜ α 2 6⎠⎟ R R ⎛ r 2 ⎞ but = so P(r) P 1 α = = c ⎜ − 2 ⎟ ξ1 6 ⎝ R ⎠ For ideal gas pressure and constant composition T would have the same dependence (ρ=constant) ρN kT ⎛ r 2 ⎞ P = A ⇒ T(r) = T 1− c ⎜ R2 ⎟ µ ⎝ ⎠ 1/2 1/2 ⎡P (n +1)⎤ ⎡ P ⎤ R but also α =⎢ c ⎥ = ⎢ c ⎥ = so 4 G 2 4 G 2 ⎣⎢ π ρc ⎦⎥ ⎣⎢ π ρc ⎦⎥ 6 ⎛ 4 ⎞ G R3 2 2 ⎜ π ρc ⎟ ρc 2πR Gρc ⎝ 3 ⎠ GMρ Pc = = = 3 2R 2R

This agrees with what can be obtained by integration of the equation of hydrostatic equilibrium. dP GM = − ρ dr r 2 surf R dr GMρ dP = P = − GMρ = ∫ c ∫ 2 c 0 r 2R but from the polytropic equation we learn how P varies with r dθ ⎞ Recall M=− 4πα 3 ρ ξ 2 c 1 dξ ⎠⎟ ξ1 dθ ⎞ This last quantity, ξ 2 , is a number that 1 dξ ⎠⎟ ξ1 depends on the poytropic index n. For n = 0 it is just -2 6 (see table) and so

R3 4π M = 4π 2 6 ρ = R3 ρ 6 6 c 3 as one would expect for constant density. So we understand sars of constant density quite well. Other analytic solutions exist for n = 1 and 5

n = 1

1 d ⎛ dθ ⎞ ξ 2 = −θ ξ 2 dξ ⎝⎜ dξ ⎠⎟

The solution is order zero Spherical Bessel function Sinξ R θ ξ = ⇒ ξ = π at θ=0 α = ( ) ξ 1 π Sinξ nb. lim =1 (l'Hopital's rule) ξ→0 ξ n=1 connued

1/2 1/2 ⎡P (n +1)⎤ ⎡ P ⎤ R α =⎢ c ⎥ = ⎢ c ⎥ = 4 G 2 2 G 2 π ⎣⎢ π ρc ⎦⎥ ⎣⎢ π ρc ⎦⎥ 2G So P = ρ 2R2 c π c

πr Sin(ξ) R πr Also ξ ≡ αr = ⇒ θ = = Sin( ) so R ξ πr R

2 ⎛ R ⎛ πr ⎞ ⎞ P P 2 P Sin = cθ == c ⎜ ⎟ ⎝ πr ⎝⎜ R ⎠⎟ ⎠ ⎛ R ⎛ πr ⎞ ⎞ ρ = ρ θ = ρ Sin c c ⎜ r ⎜ R ⎟ ⎟ ⎝ π ⎝ ⎠ ⎠ n=1 connued

dθ ⎞ M = − 4πα 3 ρ ξ 2 c 1 dξ ⎠⎟ ξ1 dθ ⎞ ξ 2 = − π for n = 1 1 dξ ⎠⎟ ξ1 3 dθ ⎞ ⎛ R ⎞ 4 so M =− 4πα 3 ρ ξ 2 = 4π 2ρ = ρ R3 c 1 dξ ⎠⎟ c ⎝⎜ π ⎠⎟ π c ξ1 3/2 2G ⎛ πP ⎞ but P 2R2 so R3 c c = ρc = ⎜ 2 ⎟ π ⎝ 2Gρc ⎠ 3/2 n+1 ⎛ πK ⎞ and P = Kρ n = Kρ 2 so R3 = c c c ⎝⎜ 2G ⎠⎟ 1/2 ⎛ πK ⎞ R = independent of M! ⎜ 2G ⎟ ⎝ ⎠ n=1 continued

The density adjusts to keep the same radius no matter what M is. R is independent of M.

This may seem a bit strange but actally neutron stars are approximately polytropes with index 0.5 < n < 1. Central density rises with M but radius does not vary much. n =5

1 d ⎛ dθ ⎞ ξ 2 = −θ 5 ξ 2 dξ ⎝⎜ dξ ⎠⎟ −1/2 ⎡ 1 2 ⎤ θ = ⎢1+ ξ ⎥ which →0 only asξ → ∞ ⎣ 3 ⎦ −5/2 4 3 ρ ⎡ 1 ⎤ ⎡2i3 K ⎤ see 12 = 1+ ξ 2 M = ρ −1/5 ⎢ 3 ⎥ ⎢ 3 ⎥ c pages back ρc ⎣ ⎦ ⎣ πG ⎦ 1/3 2 3 ⎡πM G ⎤ 4/3 Pc = ⎢ 4 ⎥ ρc 2• 3 ⎣ ⎦ This configuration, which is not very physical has an infinite radius and a finite mass, central density and central pressure. It is infinitely centrally concentrated and we will see later has infinite binding energy. There are no solutions, analytic or otherwise for n > 5. Most physically relevant polytropes have 1.5 < n < 3. Analyc Soluons Note tendency of all to agree with n = 0 at the origin

θ(ξ) The General Case (0 < n < 5)

The general solution requires numerical integration of the Lane Emden equation. There are tools available for this purpose. A simple program is also given at the class website http://nucleo.ces.clemson.edu/home/online_tools/polytrope/0.8/ From Pols (2011) ⎛ dθ ⎞ In his notation z = ξ θ = − ξ 2 n 1 n 1 ⎝⎜ dξ ⎠⎟ ξ1 In the general case, we know M, n, and K (e.g. from the EOS):

First get α from the given mass n/(n−1) ⎡(n +1)K ⎤ ⎛ dθ ⎞ M =-4 3 2 as function of M, K, and n πα ⎢ 2 ⎥ ξ1 ⇒ α 4πGα ⎝⎜ dξ ⎠⎟ ⎣ ⎦ ξ1

Where we have replaced ρc using the definition of α 1/2 1/2 1/2 ⎡P (n +1)⎤ ⎡ Kρ(n+1)/n(n +1)⎤ ⎡ Kρ(1−n)/n(n +1)⎤ α =⎢ c ⎥ = ⎢ c ⎥ = ⎢ c ⎥ 4 G 2 4 G 2 4 G 2 ⎣⎢ π ρc ⎦⎥ ⎣⎢ π ρc ⎦⎥ ⎣⎢ π ρc ⎦⎥ n/(n−1) ⎡(n +1)K ⎤ ρ = ⎢ ⎥ c 4πGα 2 ⎣⎢ ⎦⎥ Once one has n, α and K, ρ ,ρ, and R easily follow. c

n/(n−1) ⎡(n +1)K ⎤ ρ = ⎢ ⎥ c 4πGα 2 ⎣⎢ ⎦⎥ ρ c = D ρ n

1/3 ⎛ 3M ⎞ R = ⎜ 4 ⎟ ⎝ πρ ⎠ Note the existence of a mass-radius relation n/(n−1) ⎡(n +1)K ⎤ ⎛ dθ ⎞ M =-4 3 2 πα ⎢ 2 ⎥ ξ1 4πGα ⎝⎜ dξ ⎠⎟ ⎣ ⎦ ξ1 (n−1) ⎡ ⎤ n M (n−1) ⎡(n +1)K ⎤ ⎢ ⎥ = 4πα 3 ⎢ 2 d / d ⎥ ( ) ⎢ 4 G 2 ⎥ −ξ1 ( θ ξ) ⎣ π α ⎦ ⎣ ξ1 ⎦ (n−1) ⎡ ⎤ n M −1 ⎡(n +1)K ⎤ ⎢ ⎥ = 4π α 3n−3−2n ⎢ 2 d / d ⎥ ( ) ⎢ G ⎥ −ξ1 ( θ ξ) ⎣ ⎦ ⎣ ξ1 ⎦ (n−1) move  to other side ⎡ ⎤ n−1 M 1 ⎛ 1 ⎞ n ⎢ ⎥ α 3−n = (n +1)K ⎢ 2 d / d ⎥ 4πG ⎜ G⎟ ( ) −ξ1 ( θ ξ) ⎝ ⎠ ⎣ ξ1 ⎦ (n−1) ⎡ ⎤ n−1 M 1 ⎛ 1 ⎞ n ⎢ ⎥ 3−n (n 1)K 2 α = ⎜ ⎟ ( + ) ⎢ d / d ⎥ 4πG G move Gn-1 to other side −ξ1 ( θ ξ) ⎝ ⎠ ⎣ ξ1 ⎦ (n−1) n ⎡ ⎤ ⎡ n +1 K ⎤ GM 3 n ( ) ⎢ ⎥ α − = ⎣ ⎦ ⎢ 2 d / d ⎥ 4πG −ξ1 ( θ ξ) ⎣ ξ1 ⎦

Substituting the value of R α=R/ξ1 (n−1) 3−n n ⎡ GM ⎤ ⎛ R ⎞ ⎡(n +1)K ⎤ ⎢ ⎥ = ⎣ ⎦ ⎢ 2 d / d ⎥ ⎜ ξ ⎟ 4πG −ξ1 ( θ ξ) ⎝ 1 ⎠ ⎣ ξ1 ⎦

For a given n and K, the right hand side is constant and (n−3)/(n−1) (n−1)/(n−3) MR or R  M (n−1) 3−n n ⎡ GM ⎤ ⎛ R ⎞ ⎡(n +1)K ⎤ ⎢ ⎥ = ⎣ ⎦ ⎢ 2 d / d ⎥ ⎜ ξ ⎟ 4πG −ξ1 ( θ ξ) ⎝ 1 ⎠ ⎣ ξ1 ⎦

Note the existence of "singularities" at n = 1 and n =3 For n = 1 the mass dependence drops out and the radius is independent of the mass (the central density just adjusts). Even more interesting for n = 3, the radiis drops out and one just a critical mass whose radius is undefined. This will have important implications for the maximum mass of white dwarfs and for massive stars.

Mass radius relation for white dwarfs:

For a non-relativistic white dwarf γ =5/3 which implies n = 3/2 1/2 3/2 3/2 ⎡ GM ⎤ ⎛ R ⎞ ⎡(5 / 2)K ⎤ ⎢ ⎥ = ⎣ ⎦ ⎢ 2 d / d ⎥ ⎜ ξ ⎟ 4πG −ξ1 ( θ ξ) ⎝ 1 ⎠ ⎣ ξ1 ⎦

3/2 −1/2 3/2 ⎛ R ⎞ ⎡(5 / 2)K ⎤ ⎡ GM ⎤ = ⎣ ⎦ ⎢ ⎥ ⎜ ξ ⎟ 4πG ⎢ 2 d / d ⎥ ⎝ ⎠ −ξ1 ( θ ξ) ⎣ ξ1 ⎦ −1/3 ⎡ 5 / 2 K ⎤ ⎡ ⎤ ⎛ R ⎞ ( ) M 13 5/3 = ⎣ ⎦ ⎢ ⎥ K =1.00 ×10 Y ⎜ ξ ⎟ 2/3 ⎢ 2 d / d ⎥ e ⎝ ⎠ 4π G −ξ1 ( θ ξ) ( ) ⎣ ξ1 ⎦ 13 ξ1 =3.654 3.654 ⎡ 2.5 1.00 ×10 ⎤ 1/3 ⎣( )( )⎦ 5/3 −1/3 2 R = 2.714 Y M −ξ dθ / dξ = 2.714 2/3 ( ) e 1 ( )ς ⎡ 4π 6.67 ×10−8 ⎤ 1 ⎣⎢( ) ( )⎦⎥ 5/3 1/3 5/3 1/3 ⎛ Y ⎞ ⎛ M ⎞ ⎛ Y ⎞ ⎛ M ⎞ 8800 km e  0.0127 R e  = ⎜ ⎟ ⎜ ⎟ =  ⎜ ⎟ ⎜ ⎟ ⎝ 0.5⎠ ⎝ M ⎠ ⎝ 0.5⎠ ⎝ M ⎠ hp://en.wikipedia.org/wiki/White_dwarf The Chandrasekhar Mass

(n−1) 3−n n ⎡ GM ⎤ ⎛ R ⎞ ⎡(n +1)K ⎤ ⎢ ⎥ = ⎣ ⎦ ⎢ 2 d / d ⎥ ⎜ ξ ⎟ 4πG −ξ1 ( θ ξ) ⎝ 1 ⎠ ⎣ ξ1 ⎦ If fully relativstic throughout, P= Kρ 4/3 and n = 3 2 3 ⎡ GM ⎤ ⎡4K ⎤ ⎢ ⎥ = ⎣ ⎦ ⎢ 2 d / d ⎥ 4πG −ξ1 ( θ ξ) ⎣ ξ1 ⎦ 3/2 M (4K ) = 1/2 2.01824 (4π ) G3/2 3/2 2.01824 4 1.2435 1015Y 4/3 ( ) ( ) × e M ( ( )) Ch = 1/2 3/2 (4π ) (6.67 ×10−8 ) 2 ⎛ Y ⎞ 1.456 e M = ⎜ ⎟  ⎝ 0.5⎠

Another useful expression is for the central pressure (n+1)/n Pc = Kρc Use the previous equation for the mass-radius relation to solve for K

as a function of R and M and put it in the equation for Pc (n−1) 3−n n ⎡ GM ⎤ ⎛ R ⎞ ⎡(n +1)K ⎤ ⎢ ⎥ = ⎣ ⎦ ⎢ 2 d / d ⎥ ⎜ ξ ⎟ 4πG −ξ1 ( θ ξ) ⎝ 1 ⎠ ⎣ ξ1 ⎦ (n−1)/n (3−n)/n ⎡(4 G)1/n ⎤ ⎡ GM ⎤ ⎛ R ⎞ (n+1)/n π ⎢ ⎥ (n+1)/n Pc =Kρc = ⎢ ⎥ 2 ρc (n +1) ⎢ d / d ⎥ ⎜ ξ ⎟ ⎣ ⎦ −ξ1 ( θ ξ) ⎝ 1 ⎠ ⎣ ξ1 ⎦ −1 3M ⎛ 3 ⎛ dθ ⎞ ⎞ But D so we have R in terms of and M ρc = ρ n = 3 ⎜ − ⎜ ⎟ ⎟ ρc 4πR ⎝ ξ1 ⎝ dξ ⎠ 1⎠ (3−n)/3n ⎡ ⎤ 3 (3−n)/n ⎢ ⎥ ⎛ R ⎞ 3M ⎛ R ⎞ ⎢ 3M ⎥ ⎜ ⎟ = ⎜ ⎟ = ⎢ ⎥ ⎝ ξ1 ⎠ 3 ⎡−3 ⎛ dθ ⎞ ⎤ ⎝ ξ1 ⎠ 3 ⎡−3 ⎛ dθ ⎞ ⎤ 4πρ ξ ⎢ ⎥ ⎢4πρ ξ ⎢ ⎥ ⎥ c 1 ξ ⎝⎜ dξ ⎠⎟ ⎢ c 1 ξ ⎝⎜ dξ ⎠⎟ ⎥ ⎣ 1 1⎦ ⎣ ⎣ 1 1⎦ ⎦ So we can obtain Pc as a function of just M and ρc 1 1/n−1/n+1/3 P = 4π G1/n+1−1/n M1−1/n+1/n−1/3ξ −2+2/n−2/n+2/3 c n +1( ) 1 −1+1/n−1/n+1/3 ⎛ dθ ⎞ ρ1+1/n−1/n+1/3 ⎝⎜ dξ ⎠⎟ c ξ1 −2/3 (4π )1/3G ⎛ dθ ⎞ P = M 2/3 ρ 4/3 ξ −4/3 c n +1 c 1 ⎝⎜ dξ ⎠⎟ ξ1 2/3 4/3 Pc =CnGM ρc

1/3 −2/3 4π ⎡ ⎛ dθ ⎞ ⎤ C ( ) ⎢ 2 ⎥ Pols. p 49 (4.18) n = ξ1 ⎜ ⎟ n +1 ⎢ ⎝ dξ ⎠ ⎥ ⎣ ξ1 ⎦ which is a slowly varying function of n

C 0.542 C 0.431 n=1 = n=2 = C 0.364 n=3 =

2/3 4/3 Pc =CnGM ρc 3 Pc 1) For a given polytropic index and mass the ratio 4 ρc is a constant as the star expands or contracts. 2) For a given polytropic index this ratio increases as M2 3) For a given mass this ratio does not vary much across a reasonable range of polytropic indices 1.5≤n≤3 ρN kT Important example: Suppose P = P = A ideal µ T 3 Then c in a contracting polytrope is a constant ρc (that increases with mass). Stellar cores will evolve 3 trying to keep ρc ∝ Tc and higher mass stars will have a higher central temperature at a given central density 2/3 4/3 Pc =CnGM ρc 3 3 3 2 4 Pc =CnG M ρc 3 ⎛ N kT ⎞ A c M 2 ⎜ ⎟ = ρc ⎝ µCnG ⎠ T 3 ∝ M 2ρ c c M1 > M 2 Actual models Not polytropes No burning Include radiaon

Ne-O cores

CO-cores

He-cores Crical Masses

0.08 M Lower limit for hydrogen ignition

0.45 M helium ignition

7.25 M carbon ignition

9.25 M neon, oxygen, silicon ignition (off center) ~11 M ignite all stages at the stellar center 

These are for models that ignore rotaon. With rotaon the numbers may be shied to lower values. Low metallicity may raise the numbers slightly since less inial He means a smaller helium core. ρ ∝ T 3 Temperature as functions of c One can solve explicitly for the central temperature in terms of the density if the pressure is entirely due to ideal gas

ρ N kT P =GM 2/3 ρ 4/3 C = P = c A c c c n ideal µ 2/3 1/3 GM ρc µCn Tc = Cn = 0.36 for n = 3 NAk 2/3 ⎛ M ⎞ ⎛ µ ⎞ ⎛ C ⎞ T =1.51×107 K n ρ1/3 c ⎜ M ⎟ ⎜ 0.61⎟ ⎜ 0.36⎟ c ⎝  ⎠ ⎝ ⎠ ⎝ ⎠ −3 if ρc = 160 g cm for the sun (actually 15.7 M K today from neutrinos) The structure of the sun and earth compared with polytropes of various indices (from M. Zingale). The sun is about an n = 3 polytrope.

Lecture 7b

Polytropes: Binding Energies and the “Standard Model” (n = 3)

Prialnik Chapter 5 Glatzmaier and Krumholz Chapter 10 Pols 4 The gravitational binding energy of polytropes First a useful identity: n+1 1 1+ For a polytrope P = Kργ = Kρ n = Kρ n n +1 dP = K ρ1/n dρ n

P but = K ρ1/n ρ multiply and divide by (n+1) ⎛ P ⎞ K ⎛ n +1 ⎞ 1 1 d = ρ(1−n)/n dρ = K ρ1/n dρ ⎝⎜ ρ ⎠⎟ n ⎝⎜ n ⎠⎟ ρ (n +1) 1 dP = (n +1) ρ

The gravitational binding energy of polytropes

So now let's calculate the gravitational binding energy: R Gm 1 surf G Ω = -∫ dm = − ∫ d(m2 ) 0 r 2 cent r Integrate by parts ∫ d(uv)=uv = ∫ udv + ∫v du ∫ udv =uv − ∫v du surf ⎡Gm2 ⎤ 1 surf Gm2 d ⎛ G⎞ G Ω = − ⎢ ⎥ − 2 dr since = − 2 2r 2 ∫ r dr ⎜ r ⎟ r ⎣ ⎦cent cent ⎝ ⎠ GM 2 1 surf Gm2 = − − dr 2R 2 ∫ r 2 cent At the surface the first term is zero Now use the equation of hydrostatic equilibrium dP=-(−Gm / r )(ρ / r )dr GM 2 1 surf Gm2 Ω= − − dr from previous page ∫ 2 2R 2 cent r GM 2 1 surf dP = − + ∫ m from hydrostatic equilibrium 2R 2 cent ρ GM 2 n +1 surf ⎛ P ⎞ = − + ∫ m d ⎜ ⎟ 2R 2 cent ⎝ ρ ⎠ dP ⎛ P ⎞ since = (n +1)d ⎜ ⎟ from ID 2 pages ago ρ ⎝ ρ ⎠ GM 2 n +1 surf ⎛ P ⎞ Ω = − + ∫ m d ⎜ ⎟ from previous page 2R 2 cent ⎝ ρ ⎠ surf GM 2 ⎡n +1 P ⎤ n +1 surf P = − + m − dm 2R ⎢ 2 ⎥ 2 ∫ ⎣ ρ ⎦cent cent ρ at the center m = 0; at the surface P= 0; drop 2nd term GM 2 n +1 surf P Ω = − − ∫ dm 2R 2 cent ρ but by the Virial Theorem surf P Ω ∫ dm = − cent ρ 3 GM 2 n +1 So Ω = − + Ω 2R 6 ⎛ n +1⎞ ⎛ 6 − n −1⎞ GM 2 Ω 1− = Ω = − ⎝⎜ 6 ⎠⎟ ⎝⎜ 6 ⎠⎟ 2R ⎛ 6 ⎞ GM 2 Ω = − ⎜ 5 n⎟ 2R ⎝ − ⎠ ⎛ 3 ⎞ GM 2 Ω = − ⎜ ⎟ ⎝ 5 − n⎠ R

This is the gravitational binding energy of a polytrope of index n with mass M and radius R. Note the singularity at n = 5 and also the correct n = 0 limit. Note also that the polytrope of mass M and radius R is more tighly bound if it is more centrally condensed, i.e., n is larger For example if the sun could be characterized by a polytropic index n = 3, its binding energy would be

2 −8 33 3 GM 2 6.67x10 1.99x10 Ω =  = 1.5 ( )( )  2 R 6.96x1010  ( ) = 5.7×1048 erg Accurate stellar models give 6.9×1048 erg To the extent that any star is supported by ideal gas pressure, the Virial theorem holds and the total energy of the star Ω ⎛ 3 ⎞ GM 2 E = U + Ω= = − tot 2 ⎝⎜ 10 − 2n⎠⎟ R

(note typo with “–” sign fixed from earlier version) Kelvin Helmholz time scale (again)

The time scale required for an adjustment of stellar structure in the absence of energy sources other than gravitation is the Kelvin-Helmholtz time 3 GM 2 τ = KH 10 − 2n RL where L is the luminosity of the star (or region of the the star in light (or neutrinos).

3 GM 2 3 GM 2 For the sun τ ≈ =  KH 10 2n L 4 R L (n = 3) −   7.4 1014 sec = 23 million years = × which is close to correct Eddington’s standard model (n=3) Consider a star in which radiation pressure is important (though not necessarily dominant) dP d ⎛ 1 ⎞ 4 dT rad = aT 4 = aT 3 dr dr ⎝⎜ 3 ⎠⎟ 3 dr dT 3κρ L(r) But for radiative diffusion, = so dr 16πacT 3 r 2 dP κρ L(r) rad = − dr 4πc r 2 but hydrostatic equilibrium requires dP Gmρ = − dr r 2 Recall the definition of the Eddington lumnosity and divide 2 eqns 4πGMc L = ⇒ Ed κ dP κL(r) L(r) rad = = dP 4πGmc L Edd P P Define β = gas = 1- rad where P= P +P , P P gas rad

dP κL(r) L(r) rad = (1− β)= = dP 4πGmc LEdd

If, and it is a big IF, β (or 1-β) were a constant throughout the star, then one could write everywhere

L(r) = 1− β L ( ) Ed β = constant would imply that the star was an n=3 polytrope!

4 1/4 Prad aT ⎡3P(1− β)⎤ P= = ⇒T = ⎢ ⎥ (1− β) 3(1− β) ⎣ a ⎦ P 1/4 gas NAk NAk ⎡3P(1− β)⎤ P = = ρT ⇒ P = ρ ⎢ ⎥ β µβ µβ ⎣ a ⎦ 1/4 3/4 NAk ⎡3(1− β)⎤ hence P = ρ ⎢ ⎥ and µβ ⎣ a ⎦ 1/3 4 ⎡3(NAk) (1− β)⎤ 4/3 P = ⎢ 4 ⎥ ρ ⎣⎢ a(µβ) ⎦⎥ If β = constant thoughout the star, this would be the equation 4/3 for an n = 3 polytrope and the multiplier of ρ is K. 3/2 3/2 n +1 dθ ⎞ ⎛ K ⎞ 3−n Recall M ( ) 2 2n = − ξ1 ⎟ ⎜ ⎟ ρc 4 dξ ⎠ ⎝ G⎠ π ξ1 1/3 4 ⎡3(NAk) (1− β)⎤ K = ⎢ 4 ⎥ ⎣⎢ a(µβ) ⎦⎥ For n = 3, this becomes 3/2 3/2 4 ⎛ dθ ⎞ ⎛ K ⎞ ⎛ K ⎞ M = − ξ 2 = 4.56 1 ⎝⎜ dξ ⎠⎟ ⎝⎜ G⎠⎟ ⎝⎜ G⎠⎟ π ξ1 1/2 4 ⎡3(NAk) (1− β)⎤ M = 4.56 ⎢ 4 3 ⎥ lim M →0 ⎣⎢ a(µβ) G ⎦⎥ β→0 1/2 lim M →∞ Eddington’s quartic 18.1M ⎛ 1− β ⎞ β→1 M =  equation µ2 ⎝⎜ β 4 ⎠⎟

Alternatively For each β there is a unique ⎛ aG3 ⎞ π 1 4 4 M 2 M. R drops out. Like a Chandrasekhar − β = ⎜ 4 ⎟ 2 µ β 3(N k) mass, but M = f ⎝ A ⎠ ⎛ dθ ⎞ crit (β ) 16ξ 4 1 ⎝⎜ dξ ⎠⎟ For stars of constant β, Eddington's quartic equation says how β varies with mass and composition, µ.

X Prialnik 5.3 This is wrong!!! Correct figures from Clayton p. 163

µ2 ≈ 0.37 for main sequence stars 2 4 ⎛ M ⎞ ⎛ µ ⎞ 1− β = 4.13×10−4 β 4 and since ⎜ M ⎟ ⎜ 0.61⎟ ⎝  ⎠ ⎝ ⎠

L(r) = (1- β ) LEdd ⎛ aG3 ⎞ π 4πGc L = µ4 β 4M 2 M ⎜ 4 ⎟ 4 2 ⎝ 3(NAk) ⎠ 16 d / d κ ξ1 ( θ ξ) ξ1 π 2 ⎛ acG4 ⎞ ⎛ µ4β 4 ⎞ = M 3 4 2 ⎜ 4 ⎟ ⎜ ⎟ 12 d / d ⎝ (NAk) ⎠ ⎝ κ ⎠ ξ1 ( θ ξ) ξ1 4 3 2 −1 Mass luminosty 4 ⎛ µ ⎞ ⎛ 1 cm g ⎞ ⎛ M ⎞ = 5.5β L Relation ⎜ 0.61⎟ ⎜ ⎟ ⎜ M ⎟  ⎝ ⎠ ⎝ κ surf ⎠ ⎝  ⎠ where κ surf is the value of the opacity near the surface. This was obtained with no mention of nuclear reactions. For M not too far from M is close to 1 and L M3.  β ∝

At higher masses however the mass dependence of β 2 4 ⎛ M ⎞ ⎛ µ ⎞ becomes important. 0.0004 β 4 eventually ⎜ M ⎟ ⎜ 0.61⎟ ⎝  ⎠ ⎝ ⎠ dominates and β 4 ∝ M −2 so that L∝M. In fact, as we have discussed, the luminosity of very massive stars approaches the Eddington limit as 0 β → This was all predicated on a very doubtful postulate however that  is a constant throughout the star (or L(r)/Ledd is a constant). Why or when might this be approximately true?

dP κL(r) L(r) Recall rad = = dP 4πGmc LEdd dL dL and the energy equation = 4πr 2ρε(r) or =ε(m) dr dm where ε is the nuclear energy generation in erg g−1 s−1 r ∫ ε(m)dm 0 ε(r) = r and clearly ε(R) =L / M ∫ dm 0 ε(r) L(r) m(r) define η(r)= = / ε(R) L M * * dP κ (r)L(r) L rad = = κ (r)η(r) dP 4πGm(r)c 4πGMc

Integrating from the surface, where both Prad and P are assumed to be zero, inwards L 1 r P = κη(r) P where κη(r) = κη dP ' rad 4 GMc ∫ π P (r ) 0(surf ) which yields L 1− β = κη(r) 4πGMc L 1− β = κη(r) 4πGMc The condition for the right hand side being a constant, which leads to  being a constant, is that, at each r, the pressure weighted average product of opacity times average energy generation interior to r be a constant. Since the energy generation is centrally concentrated .

L(r) m(r) / is slowly decreasing with m(r) except L* M* at the center. κ (r) on the other hand is slowly increasing with r. Either it is a constant (electron scattering) or Kramer's like proportional to ρT−7/2. But roughly ρ ∝ T3. So the product is approximately constant especially where P is large. This was Eddington's reasoning. For n = 3 one can also derive useful equations for the central conditions based upon the original polytropic equation for mass dθ ⎞ M = − 4πα 3 ρ ξ 2 = 2.01824 (4πα 3 ρ ) c 1 dξ ⎠⎟ c ξ1 1/2 1/2 ⎡P (n +1)⎤ ⎡ P ⎤ and the definitions α = ⎢ c ⎥ = ⎢ c ⎥ 4 G 2 G 2 ⎣⎢ π ρc ⎦⎥ ⎣⎢π ρc ⎦⎥ P ρ N kT ρ 4πR3 ρ and P = ideal = c A c and c = c = 54.18 c β µβ ρ 3M

2 ⎛ M / M ⎞ 17 (  ) Pc =1.242×10 ⎜ 4 ⎟ ⎜ R / R ⎟ ⎝ (  ) ⎠ ⎛ M / M ⎞ 6 (  ) T =19.57×10 βµ ⎜ ⎟ K c R / R ⎝ (  ) ⎠ 2/3 T = 4.62×106 βµ M / M ρ1/3 K c (  ) c For example, 2/3 T = 4.62×106 βµ M / M ρ1/3 K c (  ) c

The central density of the zero age sun (when its composition was constant) was 82 g cm−3 and its temperature was 13.6 MK (Bohm-Vitense, Sellar Structure and Evolution, Vol 3, p 156). Its radius was 0.884 times its present radius. The formula gives for β ≈ 1, µ = 0.61

Tc = 19.57 (0.61)(1/ 0.884) =13.5 MK in very good agreement In order to specfify a the radius (or equivalently the mean density)on the on the main sequence, it is necessary to specify and energy source, e.g., nuclear reactions, which we have not done so far. But taking the 2/3 R ⎛ M ⎞ empirical (relation on the MS) that = one has R ⎜ M ⎟  ⎝ sun ⎠ for β ≈ 1 (a better job could be done with the quartic equation) and the ZAMS value for the solar radius

⎛ M / M ⎞ 1/3 6 (  ) T =19.6×10 (0.61) ⎜ ⎟ = 13.5 M / M MK c R / R (  ) ⎝ (  ) ⎠ These two relations for R and T predict correctly that more massive stars should have higher central temperatures and lower central densities on the main sequence.

Comparison of the n=3 polytrope of the Sun versus the Standard Solar Model.

The surface is not well fit because the surface of the sun is convective

74 l Trends: Massive stars have more radiaon pressure dominance Massive stars have higher T l Note: this is best for a ZAMS star—structure changes as the star evolves