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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 = dM GM =4 r2 and 2 Recall 1 dP d hence: r 2 dP = G M dr and d r 2 dP dr dr dr r = 4 G r =G dM dr 2 (*) Now we add the equation of state for a polytrope: r : d r2 dr 1 1 d n dr = 4 r 2 P= 1 1 n , giving us a single equation for G We can try to simplify this equation: Define a new variable θ, defined as = c θ is called the Lane-Emden function. θ=1 in the centre of the star and 0 at the surface. n So, in terms of θ we can write P as: 1 P = c 1 n 1 1 n 1 = P c n 1 , where P c = c n Hence: r 2 dP dr = r2 P c c n n 1 n d 2 d = r dr dr n 1 Pc c So the derivative with this with respect to r : = d r 2 dP dr dr Pc d 2 d r dr c dr n 1 = 4 G c r 2 n using formula (*) So we have: d d r2 dr dr = 4 G c r2 n n 1 P c 2 Now let's introduce a dimensionless radius ξ, so that r = , where α has the dimension of length. With ξ we have: 4 G 2c 2 = n1 P 2 n (**) c d d 2 d d 2 Now choose = n1 P c 4 G c 2 So that equation (**) becomes: d 1 d 2 2 d d = n This is the Lane-Emden equation. There are exact solutions of this equation for only 3 examples: n=0: = 1 n=1: = 2 6 sin = n=5: 1 1 2 3 The solutions look like: Draw graph. First zero ( 1 ) correponding to the radius of the star corresponds to = 6 n= 0 , = n= 1 , Also, at = 0 : = n= 5 4.3Scaling Relations 1. Central Density RS Consider M S = 4 r r dr 2 0 Set r = , = c This gives: MS = 4 n 1 2 d c n 3 0 but: d 1 d 2 2 d d M S = 4 c 3 = n , so 1 dd 2 0 d d d hence M S = 4 3 c 12 d d = 1 which implies for the central density: c = 3MS 4 R3S a n , where 1 a n = 3 d d = 1 and this means that we can write the relation for c : = 1 ; dd = 0 c = a = , where n MS 4 R3S 3 2. Central Pressure from the definition of α we have: 4G c Pc = n 1 2 2 2 = 4nG1 an 3 M S R 2S 4 RS 12 3 reshuffle: G M 2S 1 Pc = 4 n 1 R 4S 1 2 d d = 1 hence P c = cn cn = 2 G MS R with 4 S 1 2 4 n 1 d d = 1 3. Central Temperature Tc= Pc R =b c 2 3 4 cn G M S R S =R 3 an G MS n R 4S MS where bn = R RS 4 cn 3 an 4.4 A Mass-Radius relation Recall P = so that P c = 1 1 n 1 1 n c This means that given c and P c we can calculate κ: = Pc 1 c 1 n =c 4 G MS n 3 an M S 4 RS 1 1 n 4 RS 3 simplifying this gives: =d 1 G MS n 1 RS 1 n =d 3 n 1 MS n 1 n R 3 1 n S and d n = cn 4 1 1 3 an 1 n 1 n Since κ is a constant, this leads to the Mass-Radius relation: MS R n 3 n 1 S 5. Summary MS R , 3 S GM and T R R S G M 2S Pc R 4 S S 4.5White dwarfs 1. In the non-relativistic case we have already seen (degenerate): Pe = = 20 mh m 3m e H H 2 5 3 , where 2 3 1 X 2 5 3 For a polytrope with index n= 2 we have such a relation between P and ρ with 3 = P =c c 5 3 c G M 2S 3 2 3 a3 R 4 S 2 5 3 MS 4 R 3S These constants a n , b n and cn in practise can be lookd up in tables. Doing this we get: 1 3 S 2 = 0.424 G M RS = 20 mh m 3 mH e H 2 3 1 X 2 5 3 putting in the numbers and X=0 (White dwarfs = Helium star): RS = 0.8710 R Sun 1 3 MS M Sun 2 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 ~109 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). P = 4 3 with = hc 8 mH 1 X 2 1 3 3 mH 4 3 This means that we have n=3. As before: = Pc = 4 3 c =c 2 G MS 3 4 RS 2 3 S G M c3 3 a3 M S 4 3 4 R 3S 4 3 a3 4 3 This does not depend on RS , which means that the mass is uniquely determined: MS = hc 8 Gm H 3 2 If X=0 (pure Helium) 3 mH 1 2 1 X 2 2 2 c3 3 2 3 a3 4 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. Since R S m e 1 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: d 2 d 2 = 2 d d n We would like to integrate by starting in the centre and going outward in steps of . We can write: i = i 1 = i d d = i In the same way we can write for the first derivative: d d i 1 = d d d2 d 2 i Now use the Lane-Emden equation: d d i 1 = d d i In the centre we know that integrate outward until Now, for the Sun 2 d d 0 =1 n i d d and =0 , so we start from those and 0 reaches 0. and should be converted to density and radius: r = , so at the photosphere of the Sun we haveL = R Sun 1 We also had: c = 3MS 4 R3S a n with 1 a n = 3 d d = 1 We now input M Sun and RSun from observations, and then have a solution for : = c n 3 2 We had: M = 4 c d d , through which we have the mass profile. 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.