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Uses of Polytropes

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Uses of Polytropes The Mass-Radius Relation for Polytropes The mass within any point of a polytropic star is 0 R ξ M 2 3 2 n = 4πr ρ dr = 4πrnρc ξ θ dξ (16.1.1) Z0 Z0 This can be simplified by substituting the left side of the Lane- Emden equation for θn 0 ξ M 0 3 − d 2 dθ 3 02 dθ (ξ ) = 4πrnρc ξ dξ = 4πrnρc ξ dξ dξ dξ 0 Z0 µ ¶ µ ¶ξ 3 2 3−3n (n + 1)K 2n +1 02 dθ = 4π ρc ξ − 4πG dξ 0 · ¸ µ ¶ξ 3 2 (3−n) (n + 1)K 2n 02 dθ = 4π ρc ξ − (16.1.2) 4πG dξ 0 · ¸ µ ¶ξ Now compare this to the radius 1 (n + 1)K 2 (1−n) r(ξ0)= r ξ0 = ρ 2n ξ0 (16.1.3) n 4πG c · ¸ If you solve for ρc in the mass equation (16.1.2) 2n 3 −1 3−n 2 M 4πG 02 dθ ρc = −ξ 4π (n + 1)K " dξ ξ0 # · ¸ · ¸ µ ¶ and eliminate it from the radius equation (16.1.3), you get n n−1 3−n 3−n 1 (n + 1)K 02 dθ 1−n r = (4π) n−3 −ξ M 3−n (16.1.4) G dξ · ¸ · ¸ In other words, the radius of a polytropic star is related to its mass by ∝ M(1−n)/(3−n) R T Note what this means: for a polytropic index of 3/2 (the γ = 5/3 case), R ∝ M−1/3. Thus, for a set of stars with the same K and n (i.e., white dwarfs, or a fully convective star that is undergoing mass transfer), the stellar radius is inversely proportional to the mass. Equation (16.1.4) can also be re-written to give the polytropic constant, K, in terms of the star’s total mass and radius. The expression is 1−n 1/n 4π −dθ G M1−1/n −1+3/n K = n+1 T R (16.1.5) "ξ dξ # n + 1 µ ¶ ξ1 The Chandrasekhar Mass Limit For n = 3, (γ = 4/3), the total mass of a polytropic star is independent of its central density. This is important, for if you substitute in the equation of state for a completely relativistic electron gas (K = 1.2435 × 1015 µ−4/3 cgs) into (16.1.2), and do the math, the total mass of the star becomes M 5.82M Ch = 2 ¯ (16.1.6) µe This is the Chandrasekhar mass limit. For M < MCh, a star can adjust its structure and back away from the completely rela- tivistic limiting case. However, since (16.1.4) already represents the limit to electron pressure, degenerate stars cannot exceed MCh. If we assume such a star has no hydrogen, then, by (5.1.6), µe = 2, and MCh = 1.456M¯. The Central Temperature and Density With a little algebra, equations (16.1.2) and (16.1.3) can be ma- nipulated to yield the conditions in the center of a polytropic star. By combining the two equations, while eliminating K, the expression for central density as a function of MT and R becomes − 1 dθ 1 M ρ = ξ − T (16.2.1) c 4π 1 dξ R3 µ ¶ξ1 The central pressure then follows, by combining this with the equation for K given in (16.1.5) − 1 dθ 2 GM2 P = Kρ1+1/n = − T (16.2.2) c c 4π(n + 1) dξ R4 µ ¶ξ1 Finally, the central temperature can be found from (16.2.1), (16.2.2), and the ideal gas law −1 µ ma Pc µ ma dθ GMT Tc = = −ξ (16.2.3) k ρc (n + 1)k dξ R µ ¶ξ1 Numerically, these expressions evaluate to −1 M −3 dθ T R −3 ρc = 0.47 ξ1 − g cm dξ M¯ R¯ µ ¶ξ1 µ ¶ µ ¶ × 14 −2 M 2 −4 8.97 10 dθ T R −2 Pc = − dyne cm (n + 1) dξ M¯ R¯ µ ¶ξ1 µ ¶ µ ¶ 7 −1 −1 2.29 × 10 dθ MT R Tc = −ξ K (n + 1) dξ M¯ R¯ µ ¶ξ1 µ ¶ µ ¶ The Hayashi Track We can locate polytropic-like stars in the HR diagram by match- ing fully convective models to boundary conditions for cool, con- vective stars. To do this, we first assume that the star is entirely convective, except for a thin radiative region at the surface, and that the star is cool, so that H− opacity dominates. From our boundary condition analysis 2/3 Pt = 2 Pp (9.26) Now recall that the pressure at the photosphere is 2 GM P = T (9.12) p 3 R2κ and that κ itself is a function of pressure 0 α β κ = κ0P T So, M M 2 G T 1 α+1 2 G T Pp = =⇒ P = 3 R2 0 α β p 3 2 0 β κ0Pp Teff R κ0 Teff L 2 4 or, if we substitute for R using T = 4πR σTeff , 2 4πσ GM 1/(α+1) P = T T 4−β (16.3.1) p 3 κ0 L eff ½ 0 T ¾ The pressure at the top of the convective polytrope is therefore 1/(α+1) M − P = c T T (4 β)/(α+1) (16.3.2) t 1 L eff µ T ¶ where 2 4πσG 1/(α+1) c = 22/3 (16.3.3) 1 3 κ0 µ 0 ¶ Now, consider the polytropic equation k n+1 P = Kρ1+1/n = K−n T n+1 (16.3.4) µ m µ a ¶ where the constant K is given by (16.1.5), i.e., 1 dθ n + 1 n K−n = −ξn+1 M1−nRn−3 4π dξ G T µ ¶ξ1 µ ¶ − n (n 3) 1 dθ n + 1 L 2 = −ξn+1 T M1−n T 6−2n 4π dξ G 4πσ T eff µ ¶ξ1 µ ¶ µ ¶ At the top of the convective layer, 8 2/9 T = T (9.24) t 5 eff µ ¶ so if we substitute this in (16.3.4), the pressure becomes M1−nL(n−3)/2 7−n Pt = c2 T T Teff (16.3.5) where 2n+2 3−n n+1 n 9 2 k n+1 dθ n + 1 8 (4πσ) c2 = −ξ µ ma dξ G 5 4π µ ¶ µ ¶ξ1 µ ¶ µ ¶ If we now set (16.3.2) equal to (16.3.5), the relation between MT , LT , and Teff for fully convective stars becomes − − 7−n+ β 4 c1 n−1+ 1 3 n − 1 T α+1 = M α+1 L 2 α+1 (16.3.6) eff c2 T T µ ¶ Or, for the case of fully convective (n = 3/2) stars with H− opacity (α = 1/2, β = 17/2), M 7/51 L 1/102 ∼ 13/51 T T Teff 2600 µ M L K µ ¯ ¶ µ ¯ ¶ This is the Hayashi track (although, due to our boundary condi- tion assumptions, the constant is a factor of 1.5 too small). Note that the temperature is nearly independent of luminosity, and is only weakly dependent on mass. All fully convective stars will therefore fall in the same cool area of the HR diagram..
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