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THE CHANDRASEKHAR LIMIT for WHITE DWARFS 1. Introduction

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THE CHANDRASEKHAR LIMIT for WHITE DWARFS 1. Introduction THE CHANDRASEKHAR LIMIT FOR WHITE DWARFS NICOLAS RUBIDO AND SEBASTIAN BRUZZONE Abstract. The aim of this work is to solve the set of differential equations that rules the behavior of white dwarfs and to find the limit mass for such peculiar stars. To solve this set of two non-linear differential equations we used the fourth order Runge-Kutta Method running under Matlab 6. In order to solve properly this equations we wrote them in terms of the dimensionless J J J J variables µ = ρ/ρc, ξ = r/R and ζ = m/M where ρc, R and M stands as the Sun’s central density, radius and mass respectively. In this report we are going to find how the density and mass behaves as functions of the radial coordinate for a given value of its central density ρc, the radius of such stars, the relationship between their Radius and total Mass and as a final result, the limit mass known as the Chandrasekhar Mass. Founded by Subrahmanyan Chandrasekhar in the early 30’s this limit mass of about 1.4MJ became fun- damental for the understanding of the final stages of massive stars and gave a glimpse over new fields of astrophysics. 1. Introduction Stars with masses that are not to large compared with the mass of the Sun end their lives as white dwarfs when they run out of their thermonuclear fuel. The core of such stars consist mainly of carbon and oxygen due to the outcomes of the triple- alpha process and by radiative capture of 4He and by 12C respectively. The pressure needed to prevent the gravitational collapse can be assumed to come from the pressure of the degenerate electron gas in the interior of the star. The white dwarfs are the last stage of stars with masses close to the mass of the sun, e.g., Sirius B and Eri B. This compact objects are the remanent core of red giant stars when they get rid of their external layers in a stage known as planetary nebulae. Containing central densities of the order 108-1012 kgm−3 this stars present a radius of the order 10−2RJ. We start by considering the basic set of equations needed to make a simple model of a white dwarf. This simplified model assumes a core consisted of completely ionized 12C, a spherical non-rotating star and neglects effects due to the presence of magnetic fields. As before we also assume that the pressure is due to the degenerate electrons. If the a star is at Hydrostatic equilibrium the gravitational force on each mass element should be equal to the pressure gradient again at each mass element. The modulus of the gravitational force acting at distance r per volume of mass is given by: Gmρ (1.1) F = − grav r2 Where G, m and ρ stands as the gravitational constant, mass and density respec- tively. 1 2 NICOLAS RUBIDO AND SEBASTIAN BRUZZONE As said before, at hydrostatic equilibrium conditions this gravitational force should be equal to the pressure gradient. Thus using (1.1) we get the equation for Hydrostatic equilibrium dP Gmρ (1.2) = − dr r2 The mass within a sphere of radius r is given by: Z r (1.3) m(r) = 4π ρ(r0)r02dr0 0 Taking the derivative of this expression we get the mass continuity equation: dm (1.4) = 4πr2ρ(r) dr The pressure due to the degenerate electrons would depend if one consider a electron gas of Non Relativistic orUltra Relativistic particles. Therefore we need a model of a white dwarf that consider both extreme cases. To do so we express the kinematics of the degenerate electron gas as: 2 2 4 2 2 εp = mec + p c then expressing the pressure in terms os the density of states g(p) as V/h34πp2 we get 1 Z PF p2c2g(p) (1.5) P = dp 3V 0 εp one obtains Z PF 8π 4 2 2 4 2 2 −1/2 P = 3 p c (mec + p c ) dp 3h 0 Z PF 8π 4 2 p 2 1/2 2 (1.6) = 3 p c [1 + ( ) ] mec dp 3h 0 mec applying a variable change: x = p , dx = dp yields mec mec 8πm4c5 Z xF x4 (1.7) P = e dx 3 2 1/2 3h 0 (1 + x ) Some integration and a bit of tidying-up, leads to the following expression for the pressure of a degenerate electron gas 4/3 (1.8) P = KURne I(xf ) where 3 I(x) = [x(1 + x2)1/2(2/3x2 − 1) + ln[x + (1 + x2)1/2]] 2x4 Here hc 3 K = ( )1/3 UR 4 8π A SIMPLE MODEL FOR WHITE DWARFS 3 where xF is the dimensionless Fermi momentum: PF 3ne 1/3 h xF = = ( ) mec 8π mec and neis the number of electrons per unit volume. This number is related to the density using Ye as the number of electrons per nucleon Yeρ ne = mH Finally we express the Fermi momentum as a function of the density 3Ye 1/3 h 1/3 (1.9) xF = ( ) ρ 8πmH mec This pressure is now expressed in terms of the dimensionless Fermi momentum xF , which depends on the density as shown. When considering higher densities 4/3 xF 1 then I(xF ) → 1, thus P → KURne , at low densities, xF 1 and then 4/3 I(xF ) → 4/5xF thus P → KNRne Now we are ready to find how ρ should change as a function of the radial coor- dinate and xF . We start by taking the radial derivative of the equation (1.8) dP d (1.10) = K n4/3I(x ) dr dr UR e F 4/3 d Ye 4/3 = KUR ρ(r) I(x) dr mH 4/3 Ye 1/3 dρ 4/3 dI dxF = KUR (4/3ρ I(x) + ρ |xF ) mH dr dx dr Where dI 4x2 + 6 6ln[(1 + x2)1/2 + x] = − dx 2x4(1 + x2)1/2 x5 and dx dρ F = 1/3K ρ−2/3 dr F dr 12 From here we assume that Ye = 0.5, the star consists of completely ionized C. Writing again this pressure gradient we find the following equation: 1/3 dP 0.5 4/3 4ρ dρ 2/3 KF dρ dI (1.11) = KUR( ) ( I(x) + ρ ) dr mH 3 dr 3 dr dx Where 3 ∗ 0.5 1/3 h KF = ( ) 8πmH mec Now, being EXTRA CAREFUL, one can rewrite this equation by just trying to find some hidden xF by using the density and KF . After this considerations and 4 NICOLAS RUBIDO AND SEBASTIAN BRUZZONE having close to us the Derive5 we can find the following friendly expression for the pressure gradient. dP hc 3 0.5 4x dρ (1.12) = ( )1/3( )4/3ρ1/3 2 1/2 dr 4 8π mH 3(1 + x ) dr Tidying up we get the pressure gradient for a degenerate electron gas. dP 0.5m c2 x2 dρ (1.13) = e 2 1/2 dr 3.mH (1 + x ) dr With this pressure gradient and the mass continuity equation we get the set of equations that will rule the behavior of a white dwarf. dρ −Gm(r)ρ(r) mH (1.14) = 2 2 dr r Υ(x) 0.5c me dm (1.15) = 4πr2ρ(r) dr Where x2 Υ(x) = 3(1 + x2)1/2 The reader probably would be wondering why we bother to present Υ in this fashion. The reasons is because we still want the factor 0.5 coming from Ye. So it could be easier to change this factor if we are asked to perform the same procedures for a star of different composition. dζ dµ 2. Obtaining the Dimensionless Equations dξ , dξ In order to solve the set of equations it is advisable to write them in terms of dimensionless quantities. Therefore let us introduce the following dimensionless variables. ρ (2.1) µ = ρc r (2.2) ξ = RJ m (2.3) ζ = MJ With this in mind let us rewrite the former set of equations by using our new set of variables µ, ξ and ζ. Lets start with the mass continuity equation. It follows directly from the definition of the Fermi momentum the following 1/3 (2.4) xF = KF (ρcµ) then we can directly obtain from the former(2.1), our first dimensionless equation: 2 J3 dζ 4πξ µR ρc (2.5) = dξ MJ A SIMPLE MODEL FOR WHITE DWARFS 5 or it can be introduced as dζ (2.6) = C ξ2µ dξ m where J3 4πR ρc C = m MJ For the remaining equation let us consider (1.14) dρ −Gm(r)ρ(r) mH = 2 2 dr r Υ(x) 0.5c me from there we start by considering the pressure gradient in (1.13) 2 2 dP 0.5mec x dρ = 2 2 dr 3.mH (1 + x ) dr in similar way featuring Υ(x) dP 0.5m c2 dρ (2.7) = e Υ(x) dr mH dr By a quick inspection we can pass from Υ(x) to Υ(ρ), K2 ρ2/3 (2.8) Υ(ρ) = F 2 2/3 1/2 3(1 + KF ρ ) From here we get K2 ρ2/3µ2/3 (2.9) Υ(µ) = F c 2 2/3 2/3 1/2 3(1 + KF ρc µ ) Thus the pressure gradient goes from (1.13) to dP 0.5m c2Υ(µ) dρ (2.10) = e dξ mH dξ From the condition of Hydrostatic equilibrium, equation (1.2) we can easily get the same equilibrium condition in terms of µ and ξ.
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