sign in sign up
<<
Home , Polytrope

We#have#reached#a#point#where#we#have#most#of#the## informa3on#required#to#solve#realis3c#stellar#models.# Lecture 7a A#cri3cal#piece#s3ll#mixing#is#the#nuclear#physics,#but#it# turns#out#that#many#of#the#observed#proper3es#of#,# Polytropes except#their#life3mes#and#radii,#reflect#chiefly#the#need#to## be#in#hydrosta3c#and#thermal#equilibrium,#and#not#the#energy## source.## Prialnik Chapter 5 # Glatzmaier and Krumholz Chapter 9 Historically,#prior#to#computers,#stellar#structure#was#o@en# Pols 4 calculated#using#polytropes.#Even#today.#they#provide#a## valuable#tool#for#understanding#many#stellar#proper3es.##

In addition we have or will have the physics equations ρ ⎞ 1 The Equations P = P(T, , Y ) N kT P aT 4 ρ { i } = A ⎟ + e + µ ⎠ ions 3 dm 2 dr 1 4 r a b = π ρ = 2 κ = κ (T, ρ, Y )≈ κ ρ T dr dm 4πr ρ { i } 0 ε = ε(T, ρ, Y )≈ ε ρmT n (TBD) dP Gm dP Gm { i } 0 = − ρ = − and the boundary conditions dr r 2 dm 4πr 4 at r = 0 L(r)=0, m = 0 dT 3 κρ L(r) dT 3 κ L(r) 1/4 = − 3 2 = − 3 2 2 ⎛ L(R) ⎞ dr 4ac T 4πr dm 4ac T (4πr ) at r = R P(R) ≈ 0, m= M, T= ≈0 ⎝⎜ 4πR2σ ⎠⎟ dL(r) dL(m) = 4πr 2 ρε = ε These are 7 equations in 7 unknowns: ρ, T, P, L, ε, κ, and r dr dm which, in principle, given M and initial composition, can be solved as a function of time for the boundary conditions. S3ll#to#be#defined#D#" Most of our theoretical knowledge of stellar evolution comes from doing just that. Aside: Boundary Conditions Polytropes* We#assume#a#global#rela3on#between##and# ! Radiative Zero BC: :# Ideally we would use some atmospheric model to tell us what the BC is at the surface # γ P ∝ ρ γ =(n +1) / n The T change over the whole is so large that the

difference between 0 and the real Teff at surface is small – n is called the polytropic index. Don’t confuse this except when computing a luminosity. with the adiabatic index which is local.

To get effective T at the end, use: 5/3 4/3 Some examples: P ∝ ρ or ρ n =constant

! Fully convective (adiabatic):

! White dwarfs (completely degenerate):

! Pressure is a mix of gas + radiation, but the ratio is constant throughout

Polytropes Polytropes*

! Consider HSE dP Gm(r) ! Now#make#this#dimensionless# # = − 2 ρ r = 0 dr r Central#density:# # r 2 dP = − Gm(r) Define:## # ρ dr Then:# 2 ! Differentiating again and dividing by r : # d ⎛ r 2 dP ⎞ dm(r) γ # = − G = − 4πGr 2ρ(r) P(r)=Kρ (r) dr ⎝⎜ ρ dr ⎠⎟ dr n +1 1 γ nγ γ = = 1+ # 2 n n 1 d ⎛ r dP ⎞ = K ρc θ (r) = − 4πGρ(r) r 2 dr ⎜ ρ dr ⎟ 1 1/n n 1 # ⎝ ⎠ + + = K ρc θ (r) # 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 ⎡P (n +1)⎤ 1 d ⎛ 2 dθ ⎞ n ⎢ c ⎥ r = −θ 2 2 2 dr ⎜ dr ⎟ 1 d ⎛ r dP ⎞ γ γ γ n γ n+1 ⎢ 4πGρc ⎥ r ⎝ ⎠ = − 4πGρ P =Kρ =K ρ θ =K ρ θ ⎣ ⎦ r 2 dr ⎝⎜ ρ dr ⎠⎟ c c K ργ d ⎛ r 2 dθ n+1 ⎞ Note that the quantity in brackets has units length2 c = − 4πGρ θ n ρ = ρ θ n 2 dr ⎜ n dr ⎟ c c 2 r ⎝ ρcθ ⎠ Define it to be α then 2 n 1 1/2 + 2 1 d ⎛ r dθ ⎞ n ⎡P (n 1)⎤ P = − 4πGρ θ α d ⎛ 2 dθ ⎞ n c + c 2 ⎜ n ⎟ c r = −θ with α =⎢ ⎥ r dr ⎝ θ dr ⎠ r 2 dr ⎝⎜ dr ⎠⎟ 4πGρ 2 ⎣⎢ c ⎦⎥ P (n +1) d ⎛ r 2θ n dθ ⎞ c 4 G n 2 2 ⎜ n ⎟ = − π θ ρc r dr ⎝ θ dr ⎠ 2 6 ⎡P (n +1)⎤ 1 d d dyne gm cm 2 2 c ⎛ 2 θ ⎞ n = cm for the dimension of α ⎢ ⎥ r = −θ 2 2 2 4πGρ 2 r 2 dr ⎝⎜ dr ⎠⎟ cm dynecm gm ⎣⎢ c ⎦⎥

Polytrope*boundary*condi2ons* Now define r =αξ with ξ a dimensionless radius- ρ r 1) θ n = So at = ξ =0 θ =1 like variable. Then ρc α 1 d ⎛ dθ ⎞ 2 n 2 ⎜ξ ⎟ = −θ ξ dξ ⎝ dξ ⎠ r 1 dθ dθ 2) ξ ≡ so dξ= dr ⇒ = α α α dξ dr This is the Lane Emden equation. It involves only dρ dθ but ρ ≡ ρ θ n so =nρ θ n−1 dimensionless quantities and can be solved for a given c dr c dr dθ 1 dρ dθ α dρ n to give θ(ξ). n does not have to be an integer. and ⇒ = n−1 = n−1 dr nρc θ dr dξ nρc θ dr Basically all that went into it was but in spherical symmetry for hydrostatic equilibrium hydrostatic equilibrium, mass conservation, dρ ρ is a local maximum at r = 0, so = 0 dr and a power law dθ So at at ξ =0 = 0 dξ That is Polytrope*boundary*condi2ons* dP Gm(r)ρ = − → 0 as r → 0 dr r 2 3) At the surface r = R the density (first) goes to zero so because m(r) ∝ r 3 approaches 0 faster than r 2 dP dρ R =αξ where ξ is where θ(ξ) first reaches 0 And since P=K ργ = γ Kργ −1 * 1 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

For any value of n, the mass Cox and Guilli (1965) R ξ1 r = αξ R = αξ M = 4πr 2ρ dr = 4πα 3 ρ ξ 2θ n dξ 1 ∫ c ∫ n 0 0 ρ = ρcθ 2 dθ ρ But the Lane Emden equation says ξ ξ D = c 1 1 d n ρ 1 d ⎛ dθ ⎞ d ⎛ dθ ⎞ ξ 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.

Further α can be expressed in terms of ρc and K 1/2 1/2 n 1 1 n 1/2 + − Another interesting quantity that can be obtained ⎡ n ⎤ ⎡ n ⎤ ⎡Pc (n +1)⎤ ⎢Kρc (n +1)⎥ ⎢Kρc (n +1)⎥ α = ⎢ ⎥ = = from the tables is the ratio of central density to average 4πGρ 2 ⎢ 4πGρ 2 ⎥ ⎢ 4πG ⎥ ⎣⎢ c ⎦⎥ ⎢ c ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ density - how compact the core of the star is. 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 See the previous table. The case n = 0, Dn = 1 is the independent of ρ if n = 3. sphere of constant density case c

n = 0 Analytic solutions Technically the case n = 0 is a singularity For n = 0, 1, 5, and only for these values, there exist analytic solutions to the Lane Emden P = Kρ1+1/n diverges as n → 0 equation. This reflects the fact that constant density can only Unfortunately none of them correspond to be maintailed in the face of gravity if the fluid is common stars, but the solutions help to incompressible (like the ocean; P can vary but not ρ). demonstrate how polytropes can be solved and they can used for interesting approximate Nevertheless, some interesting properties of polytropes cases. can be illustarted with n = 0 so long as we don't use the pressure-density relation explicitly n = 0 - the constant density case 6 The first zero of θ = 1- 2 is at ξ1 = 6 r 1 d ⎛ dθ ⎞ ξ ξ = 2 n 1 2 ⎜ξ ⎟ = −θ = − α ξ dξ ⎝ dξ ⎠ so R = 6 α 3 2 dθ 2 ξ ξ = − ξ = − + C P(r) P n+1(r) P (r) for n = 0 dξ ∫ 3 = cθ = cθ 2 2 dθ ξ C 1 Cdξ ⎛ ξ ⎞ ⎛ r ⎞ = − + 2 ⇒ dθ = − ξ dξ + 2 P(r) = P 1− =P 1− dξ 3 ξ ∫ 3 ∫ ∫ ξ c ⎝⎜ 6 ⎠⎟ c ⎝⎜ α 2 6⎠⎟ 2 ξ C 2 D but =1 at =0 R R ⎛ r ⎞ θ = − − + θ ξ but = so P(r) P 1 6 ξ α = = c ⎜ − 2 ⎟ ξ 6 ⎝ R ⎠ So C = 0 and D =1 and 1 2 For ideal gas pressure and constant composition ξ θ = 1 − 6 T would have the same dependence (ρ=constant) ρN kT ⎛ r 2 ⎞ P = A ⇒ T(r) = T 1− hKp://www.vistrails.org/index.php/User:Tohline/SSC/Structure/Polytropes# c ⎜ R2 ⎟ µ ⎝ ⎠

1/2 1/2 ⎡Pc (n +1)⎤ ⎡ Pc ⎤ R but also α = = = so 3 2 dθ ⎞ ⎢ 2 ⎥ ⎢ 2 ⎥ Recall M= 4 4πGρ 4πGρ − πα ρc ξ1 ⎣⎢ c ⎦⎥ ⎣⎢ c ⎦⎥ 6 dξ ⎠⎟ ξ1 ⎛ 4 3 ⎞ G R 2 dθ ⎞ 2 2 π ρc ρc This last quantity, , is a number that 2 R G ⎜ 3 ⎟ ξ1 π ρc ⎝ ⎠ GMρ dξ ⎠⎟ Pc = = = ξ1 3 2R 2R depends on the poytropic index n. For n = 0 it is just -2 6 This agrees with what can be obtained by (see table) and so integration of the equation of hydrostatic equilibrium. dP GM R3 4π = − ρ M = 4π 2 6 ρ = R3 ρ dr r 2 6 6 c 3 surf R dr GMρ dP = P = − GMρ = ∫ c ∫ 2 c 0 r 2R as one would expect for constant density. So we understand but from the polytropic equation we learn how sars of constant density quite well. P varies with r Other analytic solutions exist for n = 1 and 5 n=1*con2nued*

1/2 1/2 ⎡P (n +1)⎤ ⎡ P ⎤ R n = 1 α =⎢ c ⎥ = ⎢ c ⎥ = 4 G 2 2 G 2 π ⎣⎢ π ρc ⎦⎥ ⎣⎢ π ρc ⎦⎥ 2G So P = ρ 2R2 1 d ⎛ dθ ⎞ c c ξ 2 = −θ π ξ 2 dξ ⎝⎜ dξ ⎠⎟ πr Sin(ξ) R πr Also ξ ≡ αr = ⇒ θ = = Sin( ) so R ξ πr R The solution is order zero Spherical Bessel function

2 Sinξ R ⎛ R ⎛ πr ⎞ ⎞ at =0 P P 2 P Sin θ ξ = ⇒ ξ1 = π θ α = = cθ == c ⎜ ⎟ ( ) ξ π ⎝ πr ⎝⎜ R ⎠⎟ ⎠ ⎛ R ⎛ πr ⎞ ⎞ Sinξ Sin ρ = ρcθ = ρc ⎜ ⎟ nb. lim =1 (l'Hopital's rule) ⎝ πr ⎝⎜ R ⎠⎟ ⎠ ξ→0 ξ

n=1*con2nued* n=1 continued

dθ ⎞ M = − 4πα 3 ρ ξ 2 c 1 d ⎟ ξ ⎠ ξ 1 The density adjusts to keep the same radius no matter dθ ⎞ ξ 2 = − π for n = 1 what M is. R is independent of M. 1 dξ ⎠⎟ ξ1 3 dθ ⎞ ⎛ R ⎞ 4 so M =− 4πα 3 ρ ξ 2 = 4π 2ρ = ρ R3 This may seem a bit strange but actally neutron stars c 1 dξ ⎠⎟ c ⎝⎜ π ⎠⎟ π c ξ1 are approximately polytropes with index 0.5 < n < 1. 3/2 2G ⎛ πP ⎞ Central density rises with M but radius does not vary much. 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 =5 Analy3c#Solu3ons# Note tendency of all to agree with n = 0 at the origin 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 2 M −1/5 = ⎢ + ξ ⎥ = ⎢ 3 ⎥ ρc ρ 3 πG pages#back# c ⎣ ⎦ ⎣ ⎦ θ(ξ) 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.

From Pols (2011) ⎛ dθ ⎞ The General Case In his notation z = ξ θ = − ξ 2 (0 < n < 5) n 1 n 1 ⎝⎜ dξ ⎠⎟ ξ1

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/ In the general case, we know M, n, and K (e.g. from the EOS): Once one has n, α and K, ρ ,ρ, and R easily follow. c First get α from the given mass n/(n−1) ⎡ n +1 K ⎤ n/(n−1) ( ) ⎡(n +1)K ⎤ ⎛ dθ ⎞ ρc = ⎢ 2 ⎥ M =-4 3 2 as function of M, K, and n 4πGα πα ⎢ 2 ⎥ ξ1 ⇒ α ⎢ ⎥ 4πGα ⎝⎜ dξ ⎠⎟ ⎣ ⎦ ⎣ ⎦ ξ1 Where we have replaced ρ using the definition of α c ρc 1/2 1/2 1/2 = Dn ⎡P (n +1)⎤ ⎡ Kρ(n+1)/n(n +1)⎤ ⎡ Kρ(1−n)/n(n +1)⎤ ρ α =⎢ c ⎥ = ⎢ c ⎥ = ⎢ c ⎥ 4 G 2 4 G 2 4 G 2 ⎣⎢ π ρc ⎦⎥ ⎣⎢ π ρc ⎦⎥ ⎣⎢ π ρc ⎦⎥ n/(n−1) 1/3 ⎡(n +1)K ⎤ ⎛ 3M ⎞ ρ = ⎢ ⎥ R = ⎜ ⎟ c 4πGα 2 ⎝ 4πρ ⎠ ⎣⎢ ⎦⎥

Note the existence of a mass-radius relation (n−1) ⎡ ⎤ n−1 n/(n−1) n M 3−n 1 ⎛ 1 ⎞ 3 ⎡(n +1)K ⎤ 2 ⎛ dθ ⎞ ⎢ ⎥ (n 1)K M =-4 2 α = ⎜ ⎟ ( + ) πα ⎢ 2 ⎥ ξ1 ⎢ d / d ⎥ 4πG G move#GnD1#to#other#side# ⎜ d ⎟ −ξ1 ( θ ξ) ⎝ ⎠ 4πGα ⎝ ξ ⎠ ξ1 ⎣ ⎦ ξ1 ⎣ ⎦ (n−1) (n−1) n n ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ (n−1) GM n +1 K M 3 ⎡(n +1)K ⎤ ⎢ ⎥ 3−n ⎣( ) ⎦ ⎢ ⎥ = 4πα 2 α = 2 ( ) ⎢ 2 ⎥ ⎢ d / d ⎥ 4πG ⎢−ξ dθ / dξ ⎥ 4πGα −ξ1 ( θ ξ) 1 ( ) ⎣ ⎦ ξ1 ⎣ ξ1 ⎦ ⎣ ⎦ (n−1) Substituting the value of R α=R/ξ ⎡ ⎤ n 1 1 M − 3n−3−2n ⎡(n +1)K ⎤ (n−1) ⎢ ⎥ 4 3−n n 2 =( π ) α ⎢ ⎥ ⎡ ⎤ ⎡ n 1 K ⎤ ⎢−ξ dθ / dξ ⎥ G GM ⎛ R ⎞ ⎣( + ) ⎦ 1 ( )ξ ⎣ ⎦ ⎢ ⎥ ⎣ 1 ⎦ 2 ⎜ ⎟ = move##to#other#side# ⎢−ξ dθ / dξ ⎥ ⎝ ξ ⎠ 4πG (n−1) 1 ( )ξ 1 ⎡ ⎤ n−1 ⎣ 1 ⎦ M 1 ⎛ 1 ⎞ n ⎢ ⎥ 3−n (n 1)K 2 α = ( + ) ⎢−ξ dθ / dξ ⎥ 4πG ⎝⎜ G⎠⎟ 1 ( )ξ For a given n and K, the right hand side is constant and ⎣ 1 ⎦ (n−3)/(n−1) (n−1)/(n−3) MR or R  M Mass radius relation for white dwarfs: (n−1) 3−n n ⎡ ⎤ ⎛ ⎞ ⎡ n +1 K ⎤ For a non-relativistic =5/3 which implies n = 3/2 ⎢ GM ⎥ R ⎣( ) ⎦ γ = 1/2 2 ⎜ ⎟ 3/2 3/2 ⎢−ξ dθ / dξ ⎥ ⎝ ξ ⎠ 4πG ⎡ ⎤ ⎡ 5 / 2 K ⎤ 1 ( )ξ 1 GM ⎛ R ⎞ ⎣( ) ⎦ ⎣ 1 ⎦ ⎢ ⎥ = ⎢ 2 d / d ⎥ ⎜ ξ ⎟ 4πG −ξ1 ( θ ξ) ⎝ 1 ⎠ ⎣ ξ1 ⎦ Note the existence of "singularities" at n = 1 and n =3 3/2 −1/2 3/2 ⎛ R ⎞ ⎡(5 / 2)K ⎤ ⎡ GM ⎤ For n = 1 the mass dependence drops out and the radius = ⎣ ⎦ ⎢ ⎥ ⎜ ξ ⎟ 4πG ⎢ 2 d / d ⎥ ⎝ ⎠ −ξ1 ( θ ξ) is independent of the mass (the central density just adjusts). ⎣ ξ1 ⎦ −1/3 Even more interesting for n = 3, the radiis drops out and ⎡ 5 / 2 K ⎤ ⎡ ⎤ ⎛ R ⎞ ( ) M 13 5/3 = ⎣ ⎦ ⎢ ⎥ K =1.00 ×10 Y ⎜ ξ ⎟ 2/3 ⎢ 2 d / d ⎥ e one just a critical mass whose radius is undefined. This ⎝ ⎠ 4π G −ξ1 ( θ ξ) ( ) ⎣ ξ1 ⎦ 13 ξ1 =3.654 will have important implications for the maximum mass of 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 white dwarfs and for massive stars. ⎡ 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 ⎠

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⎠ hKp://en.wikipedia.org/wiki/White_dwarf# So we can obtain P as a function of just M and ρ Another useful expression is for the central pressure c c 1/n 1/n 1/3 (n+1)/n 1 − + 1/n+1−1/n 1−1/n+1/n−1/3 −2+2/n−2/n+2/3 Pc = Kρc P 4 G M c = ( π ) ξ1 Use the previous equation for the mass-radius relation to solve for K n +1 −1+1/n−1/n+1/3 ⎛ d ⎞ as a function of R and M and put it in the equation for Pc θ 1+1/n−1/n+1/3 ρc (n−1) ⎜ ⎟ 3−n n ⎝ dξ ⎠ ⎡ ⎤ ⎡ n 1 K ⎤ ξ1 GM ⎛ R ⎞ ⎣( + ) ⎦ ⎢ ⎥ = −2/3 ⎢ 2 d / d ⎥ ⎜ ξ ⎟ 4πG (4 )1/3G ⎛ d ⎞ −ξ1 ( θ ξ) ⎝ 1 ⎠ π 2/3 4/3 −4/3 θ ξ1 P M ⎣ ⎦ c = ρc ξ1 ⎜ ⎟ (n−1)/n n +1 ⎝ dξ ⎠ (3−n)/n ξ1 1/n ⎡ ⎤ ⎡(4πG) ⎤ GM ⎛ R ⎞ P K (n+1)/n ⎢ ⎥ (n+1)/n 2/3 4/3 c = ρc = ⎢ ⎥ 2 ⎜ ⎟ ρc P =C GM ρ (n +1) ⎢−ξ dθ / dξ ⎥ ⎝ ξ ⎠ c n c ⎣ ⎦ 1 ( )ξ 1 ⎣ 1 ⎦ 1/3 −2/3 −1 4π ⎡ ⎛ dθ ⎞ ⎤ 3M ⎛ 3 ⎛ dθ ⎞ ⎞ C ( ) 2 Pols. p 49 (4.18) But D so we have R in terms of and M n = ⎢ξ1 ⎥ ρc = ρ n = 3 ⎜ − ⎜ ⎟ ⎟ ρc n 1 ⎜ d ⎟ 4 R ξ dξ + ⎢ ⎝ ξ ⎠ ξ ⎥ π ⎝ 1 ⎝ ⎠ 1⎠ ⎣ 1 ⎦ (3−n)/3n ⎡ ⎤ which is a slowly varying function of n 3 (3−n)/n ⎢ ⎥ ⎛ R ⎞ 3M ⎛ R ⎞ ⎢ 3M ⎥ = = ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ C 0.542 C 0.431 ⎝ ξ1 ⎠ 3 ⎡−3 ⎛ dθ ⎞ ⎤ ⎝ ξ1 ⎠ 3 ⎡−3 ⎛ dθ ⎞ ⎤ n=1 = n=2 = 4πρ ξ ⎢ ⎥ ⎢4πρ ξ ⎢ ⎥ ⎥ c 1 ⎜ d ⎟ c 1 ⎜ d ⎟ ξ1 ⎝ ξ ⎠ 1 ⎢ ξ1 ⎝ ξ ⎠ 1 ⎥ 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. P C GM 2/3 4/3 2) For a given polytropic index this ratio increases c = n ρc 2 as M P 3 =C3G3M 2ρ 4 3) For a given mass this ratio does not vary much c n c 3 across a reasonable range of polytropic indices ⎛ N kT ⎞ 1.5≤n≤3 A c = M 2ρ ρN kT ⎜ ⎟ c Important example: Suppose P = P = A ⎝ µCnG ⎠ ideal µ 3 3 2 Tc T ∝ M ρ Then in a contracting polytrope is a constant c c 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 Actual#models# Not#polytropes# No#burning# Include#radia3on# M1 > M 2 Ne-O cores

CO-cores

He-cores

Cri3cal#Masses#

0.08 M Lower limit for ignition

0.45 M 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#rota3on.#With##rota3on#the# ρ ∝ T 3 numbers#may#be#shi@ed#to#lower#values.#Low#metallicity#may# raise#the#numbers#slightly#since#less#ini3al#He#means#a#smaller# helium#core.# 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 (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 The gravitational binding energy of polytropes First a useful identity: n+1 1 1+ So now let's calculate the gravitational binding energy: γ n n For a polytrope P = Kρ = Kρ = Kρ R surf Gm 1 G 2 n +1 1/n Ω = - dm = − d(m ) dP = K ρ dρ ∫ r 2 ∫ r n 0 cent Integrate by parts

P 1/n d(uv)=uv = udv + v du but = K ρ ∫ ∫ ∫ ρ multiply and divide udv =uv − v du by (n+1) ∫ ∫ surf ⎛ P ⎞ K ⎛ n +1 ⎞ 1 1 ⎡Gm2 ⎤ 1 surf Gm2 d ⎛ G⎞ G d = ρ(1−n)/n dρ = K ρ1/n dρ Ω = − ⎢ ⎥ − 2 dr since = − 2 ⎝⎜ ρ ⎠⎟ n ⎝⎜ n ⎠⎟ ρ n +1 2r 2 ∫ r dr ⎜ r ⎟ r ( ) ⎣ ⎦cent cent ⎝ ⎠ 1 dP 2 surf 2 = GM 1 Gm n 1 ρ = − − dr ( + ) 2R 2 ∫ r 2 cent At the surface the first term is zero

GM 2 n +1 surf ⎛ P ⎞ Ω = − + m d from previous page 2R 2 ∫ ⎜ ⎟ Now use the equation of hydrostatic equilibrium cent ⎝ ρ ⎠ surf GM 2 ⎡n +1 P ⎤ n +1 surf P dP=-(−Gm / r )(ρ / r )dr = − + m − dm 2R ⎢ 2 ⎥ 2 ∫ ⎣ ρ ⎦cent cent ρ GM 2 1 surf Gm2 Ω= − − dr from previous page at the center m = 0; at the surface P= 0; drop 2nd term 2R 2 ∫ r 2 cent GM 2 n +1 surf P 2 surf Ω = − − dm GM 1 dP 2R 2 ∫ ρ = − + ∫ m from hydrostatic equilibrium cent 2R 2 cent ρ but by the Virial Theorem surf GM 2 n +1 surf ⎛ P ⎞ P Ω = − + m d ∫ dm = − ∫ ⎜ ⎟ cent ρ 3 2R 2 cent ⎝ ρ ⎠ GM 2 n +1 dP ⎛ P ⎞ So Ω = − + Ω since = (n +1)d ⎜ ⎟ from ID 2 pages ago 2R 6 ρ ⎝ ρ ⎠ ⎛ n +1⎞ ⎛ 6 − n −1⎞ GM 2 Ω 1− = Ω = − ⎝⎜ 6 ⎠⎟ ⎝⎜ 6 ⎠⎟ 2R ⎛ 6 ⎞ GM 2 Ω = − ⎜ 5 n⎟ 2R ⎝ − ⎠ For example if the sun could be characterized by a polytropic index n = 3, its binding energy would be 2 ⎛ 3 ⎞ GM 2 Ω = − GM 2 6.67x10−8 1.99x1033 ⎜ ⎟ 3  ( )( ) ⎝ 5 − n⎠ R Ω = = 1.5  2 R 6.96x1010  ( ) 48 This is the gravitational binding energy of a polytrope of index = 5.7×10 erg n with mass M and radius R. Note the singularity at n = 5 and Accurate stellar models give 6.9×1048 erg also the correct n = 0 limit. Note also that the polytrope of To the extent that any star is supported by ideal gas pressure, mass M and radius R is more tighly bound if it is more the Virial theorem holds and the total energy of the star centrally condensed, i.e., n is larger 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) Eddington’s standard model (n=3) Consider a star in which radiation pressure is important The time scale required for an adjustment of (though not necessarily dominant)

stellar structure in the absence of energy sources dPrad d ⎛ 1 4 ⎞ 4 3 dT = ⎜ aT ⎟ = aT other than gravitation is the Kelvin-Helmholtz time dr dr ⎝ 3 ⎠ 3 dr 2 dT 3κρ L(r) 3 GM But for radiative diffusion, = 3 2 so dr 16πacT r τ KH = 10 − 2n RL dP κρ L(r) rad = − where L is the luminosity of the star (or region of the dr 4πc r 2 the star in light (or neutrinos). but hydrostatic equilibrium requires dP Gmρ = − dr r 2 2 3 GM 2 3 GM Recall the definition of the Eddington lumnosity and divide 2 eqns For the sun τ ≈ =  KH 10 2n L 4 R L 4πGMc (n#=#3)# −   L = ⇒ Ed κ 7.4 1014 sec = 23 million years = × dP κL(r) L(r) rad = = dP 4πGmc L which is close to correct Edd P P gas rad β = constant would imply that the star was an n=3 polytrope! Define β = = 1- where P= Pgas +Prad , P P 4 1/4 Prad aT ⎡3P(1− β)⎤ P= = ⇒T = ⎢ ⎥ (1− β) 3(1− β) ⎣ a ⎦ dP 1/4 rad κL(r) L(r) P = (1− β)= = gas NAk NAk ⎡3P(1− β)⎤ dP 4πGmc L P = = ρT ⇒ P = ρ ⎢ ⎥ Edd β µβ µβ ⎣ a ⎦ 1/4 3/4 NAk ⎡3(1− β)⎤ hence P = ρ ⎢ ⎥ and If, and it is a big IF, β (or 1-β) were a constant throughout µβ ⎣ a ⎦ 1/3 the star, then one could write everywhere 4 ⎡3(NAk) (1− β)⎤ 4/3 P = ⎢ 4 ⎥ ρ ⎣⎢ a(µβ) ⎦⎥ L(r) = 1− β L ( ) Ed 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.

Alternatively 3/2 3/2 For each β there is a unique 3−n 3 n +1 dθ ⎞ ⎛ K ⎞ ⎛ aG ⎞ π 4 4 2 ( ) 2 2n 1 M M. R drops out. Like a Chandrasekhar Recall M = − ξ ρ − β = ⎜ 4 ⎟ 2 µ β 1 ⎟ ⎜ ⎟ c 3(N k) mass, but M = f β 4π dξ ⎠ ⎝ G⎠ ⎝ A ⎠ 4 ⎛ dθ ⎞ crit ( ) ξ1 16 ξ1 ⎜ ⎟ 1/3 ⎝ dξ ⎠ ⎡3(N k)4(1 )⎤ A − β For stars of constant , Eddington's quartic equation K = ⎢ ⎥ β a( )4 ⎣⎢ µβ ⎦⎥ says how β varies with mass and composition, µ. For n = 3, this becomes 3/2 3/2 4 ⎛ dθ ⎞ ⎛ K ⎞ ⎛ K ⎞ M 2 4.56 = − ξ1 ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ X# ⎝ dξ ⎠ ⎝ G⎠ ⎝ G⎠ Prialnik 5.3 π ξ1 1/2 This is wrong!!! 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 ⎠⎟

2 4 ⎛ M ⎞ ⎛ µ ⎞ 1− β = 4.13×10−4 β 4 and since ⎜ M ⎟ ⎝⎜ 0.61⎠⎟ Correct figures ⎝  ⎠

from Clayton p. 163 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 2 4 3 µ ≈ 0.37 for main 2 −1 Mass luminosty 4 ⎛ µ ⎞ ⎛ 1 cm g ⎞ ⎛ M ⎞ sequence stars = 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 ∝ This was all predicated on a very doubtful postulate however that  is a constant throughout the star (or L(r)/L is a constant). Why or when At higher masses however the mass dependence of edd β might this be approximately true? 2 4 ⎛ M ⎞ ⎛ ⎞ µ 4 dP becomes important. 0.0004 β eventually rad κL(r) L(r) ⎜ M ⎟ ⎝⎜ 0.61⎠⎟ Recall = = ⎝  ⎠ dP 4πGmc LEdd 4 −2 dL dL dominates and β ∝ M so that L∝M. In fact, as we have and the energy equation = 4πr 2ρε(r) or =ε(m) dr dm discussed, the luminosity of very massive stars approaches 1 1 where ε is the nuclear energy generation in erg g− s− the Eddington limit as β → 0 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 * * L 1− β = κη(r) 4πGMc dP κ (r)L(r) L The condition for the right hand side being rad = = κ (r)η(r) dP 4πGm(r)c 4πGMc a constant, which leads to  being a constant, is that, at each r, the pressure weighted Integrating from the surface, where both Prad and P average product of opacity times average energy are assumed to be zero, inwards generation interior to r be a constant. Since the energy generation is centrally concentrated . L 1 r P = κη(r) P where κη(r) = κη dP ' L(r) m(r) rad 4 GMc ∫ / is slowly decreasing with m(r) except π P (r ) 0(surf ) L* M* which yields at the center. κ (r) on the other hand is slowly increasing L with r. Either it is a constant (electron scattering) or Kramer's 1− β = κη(r) like proportional to ρT−7/2. But roughly ρ ∝ T3. So the 4πGMc 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 For example, mass 2/3 T = 4.62×106 βµ M / M ρ1/3 K dθ ⎞ c (  ) c M = − 4πα 3 ρ ξ 2 = 2.01824 (4πα 3 ρ ) c 1 dξ ⎠⎟ c ξ1 1/2 1/2 ⎡Pc (n +1)⎤ ⎡ Pc ⎤ and the definitions α = ⎢ 2 ⎥ = ⎢ 2 ⎥ The central density of the zero age sun (when its ⎢ 4πGρc ⎥ ⎢πGρc ⎥ ⎣ ⎦ ⎣ ⎦ composition was constant) was 82 g cm−3 and its P ρ N kT ρ 4πR3 ρ and P = ideal = c A c and c = c = 54.18 temperature was 13.6 MK (Bohm-Vitense, Sellar Structure c β µβ ρ 3M and Evolution, Vol 3, p 156). Its radius was 0.884 times its

2 present radius. The formula gives for β ≈ 1, µ = 0.61 ⎛ M / M ⎞ P 1.242 1017 ⎜ (  ) ⎟ c = × 4 Tc = 19.57 (0.61)(1/ 0.884) ⎜ R / R ⎟ ⎝ (  ) ⎠ =13.5 MK in very good agreement ⎛ 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 In order to specfify a the radius (or equivalently the mean density)on the on the , it is necessary to specify and energy source, e.g., nuclear Comparison of the n=3 polytrope reactions, which we have not done so far. But taking the of the Sun versus 2/3 R ⎛ M ⎞ the Standard Solar empirical (relation on the MS) that = one has Model. R ⎜ M ⎟  ⎝ sun ⎠ for 1 (a better job could be done with the quartic equation) The surface is not β ≈ well fit because the and the ZAMS value for the solar radius # surface of the sun is convective ⎛ M / M ⎞ 1/3 # 6 (  ) Tc =19.6×10 (0.61) ⎜ ⎟ = 13.5 M / M MK ## R / R (  ) ⎝ (  ) ⎠ These two relations for R and T predict correctly that more massive stars should have higher central and lower central on the main sequence. 74#

! Trends:# Massive#stars#have#more#radia3on#pressure#dominance# Massive#stars#have#higher#T#

! Note:#this#is#best#for#a#ZAMS#star—structure# changes#as#the#star#evolves#