4. Polytropes 4.1Definitions An equation of state of the form P= where γ and κ are constants, is called polytropic. It is cutomary to define the corresponding polytropic index, denoted by n, as: 1 =1 n Note that n does not have to be integer. In the case that γ=5/3 we get n=3/2. There are several types of stars that can be approximated well by a polytrope: convective cores of stars fully convective low-mass stars white dwarfs neutron stars 4.2The Lane-Emden Equation 1 dP GM dM 2 Recall = and =4 r d r2 dr r2 dP hence: =GM dr 2 d r dP dM 2 and =G =4 G r (*) dr dr dr 1 1 Now we add the equation of state for a polytrope: P= n , giving us a single equation for r : 1 2 1 d r d 2 n =4 r G dr dr We can try to simplify this equation: θ = n Define a new variable , defined as c θ is called the Lane-Emden function. θ=1 in the centre of the star and 0 at the surface. So, in terms of θ we can write P as: 1 1 1 1 = n n 1 = n 1 , where = n P c P c P c c Hence: 2 2 r dP r P c n d 2 d P c = n1 = r n1 dr n dr dr c c So the derivative with this with respect to r : 2 d r dP P c d 2 d 2 n = n1 r = 4 G r c using formula (*) dr dr c dr dr So we have: 4 G 2 d 2 d = c 2 n r r dr dr n 1 P c Now let's introduce a dimensionless radius ξ, so that r= , where α has the dimension of length. With ξ we have: 4 G 2 2 d 2 d = c 2 n (**) d d n 1 P c n1 P 2 = c Now choose 2 4 G c So that equation (**) becomes: 1 d 2 d = n 2 d d This is the Lane-Emden equation. There are exact solutions of this equation for only 3 examples: 2 n=0: =1 6 = sin n=1: = 1 n=5: 2 1 3 The solutions look like: Draw graph. First zero ( 1 ) correponding to the radius of the star corresponds to d = 6 n=0 , = n=1 , = n=5 Also, at =0:=1 ; =0 d 4.3Scaling Relations 1. Central Density R S = 2 Consider M S 4 r r dr 0 = = n Set r , c This gives: 1 = 2 n 3 M S 4 d c 0 1 d 2 d n but: = , so 2 d d 1 = 3 d 2 d M S 4 c d 0 d d hence = 3 2 d M S 4 c 1 d = 1 which implies for the central density: 3 M a = 1 = S a n c 3 n , where d 4 RS 3 d = 1 and this means that we can write the relation for c : M = S = a , where 4 3 c n R 3 S 2. Central Pressure from the definition of α we have: 4 G 2 2 P = c c n1 2 a 3 M R2 = 4 G n S S n1 3 2 4 RS 1 reshuffle: 2 1 GM S 1 P = c 4 n1 R4 2 S d d = 1 hence 1 c = GM 2 n 2 = S P c cn 4 with d RS 4 n1 d = 1 3. Central Temperature P 4 c GM 2 R3 T = c = n S S c R R 3 a 4 M c n RS S GM 4 c = S = n bn where bn R RS 3 an 4.4 A Mass-Radius relation 1 1 Recall P= n 1 1 so that = n P c c This means that given c and P c we can calculate κ: P GM 4 = c = S 1 cn 1 1 1 n 3 a M n c R4 n S S 3 4 RS simplifying this gives: 1 1 1 1 n 1 3 n GM 1 1 c 4 = S = n n = n dn 3 dn M S RS and dn 1 1 1 n n RS 3 an Since κ is a constant, this leads to the Mass-Radius relation: n 3 n 1 M S RS 5. Summary M GM 2 GM S S S 3 , P c 4 and T RS RS R RS 4.5White dwarfs 1. In the non-relativistic case we have already seen (degenerate): 2 5 2 5 h 3 3 1X 3 P = 3 , where = e 20 m m 2 e H mH 3 For a polytrope with index n= we have such a relation between P and ρ with 2 P GM 2 = c = S 5 c3 5 3 2 3 c 3 a 3 M R4 2 S S 3 4 RS These constants an , bn and cn in practise can be lookd up in tables. Doing this we get: 2 5 1 2 h 3 3 1X 3 =0.424 GM 3 R = S S 20 me mH mH 2 putting in the numbers and X=0 (White dwarfs = Helium star): 1 3 R M S =0.8710 2 S RSun M Sun so in case of a solar mass white dwarf, the radius would be about Earth-size. Its density (see exercises week 1) would be about ~10 9 kg m 3 . One matchbox of such material weighs 1 ton. 2. In the relativistic case we have a different polytrope: (as M S rises P e rises as well, until the electrons become relativistic). 1 4 4 h c 3 3 1X 3 = 3 with = P 8 m 2 H mH This means that we have n=3. As before: P GM 2 = c = S 4 c3 4 3 3 a M 3 c R4 3 S S 3 4 RS 4 2 4 3 = 3 GM S c3 3 a3 This does not depend on RS , which means that the mass is uniquely determined: 2 3 1 2 3 2 2 h c 3 1 X 3 a3 M = c 2 S 3 8Gm H mH 2 4 = = If X=0 (pure Helium) M S M Crit 1.44 M Sun If M M Crit there will be no stable, degenerate, relativistic solution, so the star will undergo further collapse, and become a neutron star, or a Black Hole. 1 Since RS me a white dwarf is about a factor 1000 larger than a neutron star. 4.6A practical case: the Sun In general we need to integrate the Lane-Emden equation numerically. This goes as follows: We can write ther Lane-Emden equation as: 2 d = 2 d n d 2 d We would like to integrate by starting in the centre and going outward in steps of . We can write: = = d i i 1 i d = i In the same way we can write for the first derivative: 2 d = d d 2 d i 1 d i d Now use the Lane-Emden equation: d = d 2 d n d i 1 d i d i d =1 =0 In the centre we know that 0 and , so we start from those and d 0 integrate outward until reaches 0. Now, for the Sun and should be converted to density and radius: R = = Sun r , so at the photosphere of the Sun we haveL 1 We also had: 3 M a = 1 = S a n c 3 n with d 4 RS 3 d = 1 We now input M Sun and RSun from observations, and then have a solution for : = n c d We had: M=4 3 2 , through which we have the mass profile. c d P and T can be calculated as before. In the figure profiles are plotted calculated in this way for a n=3 polytrope model. A comparison is made with the Standard Solar Model (Bahcall 1998; Physics Letters B, 433, 1), which is the most up-to-date solution to the equations of stellar structure currently available. It can be seen that the polytrope model does remarkably well, considering how simple the physics is - we have used only the mass and radius of the Sun and an assumption about the relationship between internal pressure and density as a function of radius. The agreement is particularly good at the core of the star, where the polytrope gives a central density of 7.65 x 104 kg m-3, a central pressure of 1.25 x 1016 N m-2 and a central temperature of 1.18 x 107 K, in comparison with the SSM values of 1.52 x 105 kg m-3, 2.34 x 1016 N m-2 and 1.57 x 10 7 K. Only in the outer regions of the Sun, where convection takes place, do the two solutions deviate significantly from one another. .