Chapter 16 White Dwarfs The end product of stellar evolution depends on the mass of the initial configuration. Observational data and theoretical calculations indicate that stars with mass M < 4M after ejecting part of their mass in the form of a planetary nebula give birth to a white dwarf, with typical mass, radius and density M 1M,R 5 km, and ρ 1 6 gr/cm. White dwarfs are composed largely of helium, carbon and oxygen, because the progenitors masses are such that the temperature never becomes high enough to burn much beyond carbon, and even if burning may, in principle, proceed all the way to iron the time needed to complete the process would be longer than the Universe age. As we shall later show, white dwarfs of mass exceeding the critical value M CH 1.4M cannot exist. Neutron stars or black holes are thought to be the leftover of the gravitational collapse, following a supernova explosion, of stars whose mass is greater than 4M, but the mechanism that may produce one or the other is still unclear. (For a review, see for instance A. Heger, C. L. Fryer, S. E. Woosley, N. Langer, D.H. Hartmann How massive single stars end their life, The astrophysical Journal 59, 288-, 2. Also available at http://www.journals.uchicago.edu/apj/journal/issues/apj/v591n1/57419/57419.html). Numerical simulations indicate that if the mass of the progenitor star is smaller than [2, ]M, a neutron star should form, whereas bigger masses would produce a black hole. As for white dwarfs, a critical mass exists also for neutron stars. The absolute upper limit is in the range 2 M ; the value of the critical mass depends on the equation of state which is chosen to describe matter at supranuclear densities, as those prevailing in the core of a neutron stars. Neutron stars have been observed in binary systems or as isolated objects. Typical parameters are M 1 M,R 1 km, and ρ 1 12 gr/cm. Black holes of astrophysical origin can have very different masses, ranging from a few solar masses of the stellar black holes, born in the gravitational collapse of big stars or in the coalescence or accretion iven processes in binary systems, to supermassive black holes, with masses M 1 6 1 8 M, which sit at the center of several galaxies. In this chapter we shall focus on the study of white dwarfs, whose structure can be described using the equations of newtonian gravity; in the next chapter we shall derive the equations of stellar structure in general relativity, needed to describe neutron stars. 216
CHAPTER 16. WHITE DWARFS 217 16.1 The discovery of white dwarfs The first white dwarf, Sirius B, was observed in 1915 by Adams. He found that the spectrum of the stellar object orbiting around Sirius, named Sirius B was that of a white star, not very different from the spectrum of Sirius. The mass of the newly discovered star was found by applying third Kepler s law ω 2 r = GM SB, r 2 and it was estimated to be in the range.75.95m. Knowing the distance of the system from Earth, from the observed flux of radiation it was possible to estimate the effective temperature, that in this case was 8 K. Since for a black-body emission L R 2 Teff, 4 from spectral measurements it was then possible to estimate the radius of the star, which was, surprisingly, R SB = 18.8 km, much smaller than that of the Sun! The actual values of the mass and radius are M SB =1.4 ±.26 M and R SB =.84 ±.25 R (i.e. R SB 585 km). At that time this result was really a surprise because a star having a mass comparable to that of the Sun but a radius nearly forthy times smaller had never been observed. In addition, although the gravitational redshift predicted by Einstein s theory of Relativity had already been measured in the famous Eddington expedition in 1919, the redshift of spectral lines of Sirius B measured by Adams in 1925 provided a much better verification of the theory, and in fact in his book The internal constitution of stars Sir Arthur Eddington wrote Professor Adams has killed two birds with one stone: he has carried out a new test of Einstein s general theory of relativity, and he has confirmed our suspicion that matter 2 times denser than platinum is not only possible, but it is actually present in our universe. The discovery of such an extremely dense star raised a main question: how can this white dwarf, as it was named, support its matter against collapse? Indeed, if the matter composing the star were a perfect gas its temperature would be too low to prevent the collapse, i.e. the corresponding pressure gradient would not be sufficient to balance the gravitational attraction. About this problem Eddington wrote It seems likely that the ordinary failure of the gas laws due to finite sizes of molecules will occur at these high densities, and I do not suppose that the white dwarfs behave like perfect gas. What is then that keeps white dwarfs in equilibrium? The answer to this question came a few years later, when Dirac formulated the Fermi-Dirac statistics (August 1926), R.H. Fowler identified the pressure holding up a white dwarf from collapsing with the electron degeneracy pressure (December 1926). This was the crucial step toward the formulation of a consistent theory of these stars that led S. Chanasekhar to predict the existence of a critical mass above which no stable white dwarf could exist. In order to formulate the theory, let us briefly recall some basic equations of degenerete gases.
CHAPTER 16. WHITE DWARFS 218 16.1.1 Degenerate gas in quantum mechanics A perfect gas is said to be degenerate if its behaviour differs from the classical behaviour due to the quantum properties of the system of particles. Since degenerate gases are important in the study of the internal structure of compact stars, we shall outline some basic elements of the theory. Consider a gas composed by particles all belonging to the same species. In general, the system will be completely described if we assign the number of particles per unit phase-space volume, i.e. the number density in the phase space dn d xd p = g f(x, p), (16.1) h where h is the volume of a cell in the phase-space, g =2s + 1 is the number of states of a particle with a given value of the -momentum p, s is the spin, and f(x, p) is the probability density function, i.e. the probability of finding a particle at a position between x and x + dx and with a -momentum between p and p + dp. 1 If the rest mass of a particle is m, its total energy is E =[p 2 c 2 + m 2 c 4 ] 1 2 and the total energy density of the gas is E = E dn d xd p d p = g Ef(x, p) d p. (16.4) h The distribution function for an ideal gas of fermions or boson in equilibrium is f = 1 e Ec µ kt ± 1, (16.5) where the + sign holds for fermions (Fermi-Dirac statistics) and the - for bosons (Bose- Einstein statistics). 1 Some useful relations: The 4-momentum of a relativistic particle is p α =(mcγ, p), where p = mγv is the -momentum. Moreover, remember that the total energy of the particle is E = p c. Since p α p α = m 2 c 2, it follows that E2 c 2 + p2 = m 2 c 2, where p 2 is the norm of the -momentum, and consequently the total energy of the particle can be written as E = [ p 2 c 2 + m 2 c 4] 1/2. (16.2) From this equation it follows that, since E = mc 2 γ γ = [ p 2 c 2 + m 2 c 4] 1/2 mc 2 and since the norm of the particle velocity is v = p/(mγ), pc 2 v =. (16.) [p 2 c 2 + m 2 c 4 1/2 ]
CHAPTER 16. WHITE DWARFS 219 In eq. (16.5) E c is the particle kinetic energy E c =[p 2 c 2 + m 2 c 4 ] 1 2 mc 2 and µ is the chemical potential, which is the partial derivative of any thermodynamical potential of the system (the enthalpy, the internal energy, etc.) with respect to the number of moles, keeping fixed the number of moles of the other species of particles if present, and the state parameters in terms of which the potential is expressed. For example µ i = { } H n i S, P, n k =const = { } U n i S, V, n k =const, (16.6) where H is the enthalpy and U the internal energy. From eq. (16.5) we see that, since f must be positive, the chemical potential of fermions can take any real value, either positive or negative, whereas that of bosons is bounded to be µ<e c. If the temperature is high, or the energy is low (E << kt ) the Bose-Einstein and the Fermi-Dirac distribution tend to the classical Maxwell-Boltzmann distribution f e Ec µ kt. (16.7) Since f given in (16.5) only depends on E c, i.e. it only depends on the norm of the - momentum p, the distribution of momenta is isotropic and we can write d p =4πp 2 dp. Thus, eq. (16.4) becomes E = 4πg Ep 2 dp h kt ± 1. (16.8) The pressure can be written as P = 1 e Ec µ pv dn d xd p d p = 4πg h vp dp e Ec µ kt ± 1, (16.9) 1 where v is the particles velocity and the factor comes from the hypothesis of isotropy. This equation defines the pressure as the momentum flux. Furthermore the total number of particles and the internal energy of the system can be written as N = U = dn d xd p d xd p = 4πgV h and dn E c d xd p d xd p = 4πgV h 16.1.2 A criterion for degeneracy p 2 dp e Ec µ kt ± 1, (16.1) E c e Ec µ p 2 dp kt ± 1. Let us consider the non-relativistic limit when variables E c 1 2 mv2 = p2 2m. ξ = e µ/kt and x 2 = p2 2mkT If we introduce the
CHAPTER 16. WHITE DWARFS 22 it is easy to see that eqs. (16.1) reduce to N = 4πgV (2mkT ) /2 h and U = 4πgV (2m) /2 (kt) 5/2 h x 2 dx ξ 1 e x2 ± 1, (16.11) x 4 dx ξ 1 e x2 ± 1. In principle, these integrals can be solved and ξ can be found as a function of the thermodynamical variables. Here we shall consider explicitely the limit when ξ << 1, i.e., for the Fermi-Dirac statistics in which we are primarily interested, when µ is negative and much bigger than kt. In this case the integrals become x 2 dx ξ 1 e x2 ± 1 ξ x 2 e x2 dx x 4 dx ξ 1 e x2 ± 1 ξ x 4 e x2 dx; thus, combining the expressions of N and U given in eqs. (16.11) we find U N =(kt) x 4 e x2 dx x 2 e x2 dx and since x 4 e x2 dx = 8 π and x 2 e x2 dx = 1 4 π we find U = NkT, which is the 2 classical expression of the internal energy of a perfect gas. Thus ξ << 1 corresponds to the classical limit. In this limit, from the first eq. (16.11) we find ξ = Nh gv (2πmkT ) /2. (16.12) If we now put n = N/V, where n is the number of particles per cm, and define a degeneracy temperature ( ) 2/ T deg = h2 2πmk n, (16.1) g eq. (16.12) can be rewritten as ( ) /2 Tdeg ξ =. (16.14) T Thus, T deg T, then ξ << 1 and the gas behaves as a classical gas; Conversely a perfect gas is said degenerate if T deg T (i.e. ξ >> 1). When h the degeneracy temperature tends to zero, showing that the degeneracy of a gas is of a quantum nature. Degeneracy sets in at high densities or low temperatures. Eq. (16.1) shows that at a given density n,t deg is higher for particles with smaller mass m. Thus, electrons becomes degenerate earlier than heavier particles. EXAMPLES
CHAPTER 16. WHITE DWARFS 221 For a hyogen gas in normal condition, i.e. T = K and n 1 19 cm ξ 1.5 1 5 and the corresponding degeneracy temperature is T.18K, thus it behaves as a classical perfect gas. For gases heavier then hyogen ξ and T are even smaller, and consequently at ordinary pressures and temperatures they are non-degenerate. A gas of photons is always degenerate because m = and T =. Electrons in metals are degenerate, due to their small mass (m =9.1989 1 28 g) and high density (n 1 2 cm ). Indeed in this case T 75.4 1 K, and if, for example, T = K ξ.99 1. Let us now go back to white dwarfs. As we said before, they are mainly composed of helium, carbon and oxygen, with heavier elements in the inner core. When the nuclear material in the core has been burnt, the core contracts up to a point when the distance between two nuclei becomes comparable with the dimensions of the nuclei (this happens when ρ 5z 2 g/cm and d r Bhor z 1 where z is the nuclear charge). In this situation, there is no more space left for the external orbits of the electrons which are squeezed off starting a pressure iven ionization process which proceeds as the density increases, progressively involving the innermost orbits. As a consequence of this process a dense core of nucleons forms, immersed in a degenerate gas of free electrons. At the same time the shells of lighter elements that surround the nucleus continue their nuclear evolution until all nuclear fuel is exhausted, and contraction and ionization processess take place also in the more exterior layers; the star then radiates its residual thermal energy and cools down. A more accurate description of white dwarfs should take into account other effects, like for example electrostatic corrections due to the fact that the positive charges are concentrated in individual nuclei rather than being uniformly distributed. 2 However, in what follows we shall neglect these effects. We shall consider a white dwarf at the endpoint of the evolution, assuming that the ionization process has been completed throughout the configuration and that the star has radiated away its thermal energy, so that it is composed exclusively of a dense core of nucleons, immersed in a gas of electrons that behave as a degenerate gas at zero temperature. To describe the structure of a white dwarf we do not need General Relativity. Indeed, for a typical white dwarf the surface gravity is quite small GM c 2 R M (in km) R 1.5 km 5 = 1 4. Thus, we shall use the newtonian equations of stellar structure, which can easily be found as follows. Let us consider a shell of matter of radius r and thickness. Be dv = da the volume of a fluid element belonging to the shell, where da is its section (orthogonal to r), and be dm = ρ dv its mass. The forces acting on the fluid element are the gravitational 2 Electrostatic corrections have been considered by Hamada and Salpeter in 1961. (T. Hamada, E.E. Salpeter Astrophys. J. bf 14, 68, 1961).
CHAPTER 16. WHITE DWARFS 222 attraction exerted by the sphere of mass M(r) and the gradient of pressure across the shell; if the fluid element is in equilibrium they balance, i.e. GM(r) da = dm r 2 The mass contained within a sphere of radius r is = GM(r)ρ(r) r 2. (16.15) M(r) = r ρ(r) 4πr 2, dm(r) =4πr 2 ρ(r). (16.16) Equations (16.15) and (16.16) can be solved only if we assign a further equation which relates pressure and density, i.e. and equation of state P = P (ρ). Finally, the equilibrium equations to be solved are dm(r) =4πr 2 ρ(r), = GM(r) ρ(r), r 2 P = P (ρ). (16.17) We shall now determine the equation of state (EOS) of a degenerate gas. 16.1. The equation of state of a degenerate gas When T the Fermi-Dirac distribution function becomes { 1 for E EF (or p p f(e) = F, ) for E > E F, (16.18) where E F and p F are the Fermi energy and momentum. Since the temperature is zero, the particles have zero kinetic energy. If they were bosons they would occupy the lowest energy level E =, as it happens in Bose condensation. But fermions cannot do this, since Pauli s exclusion principle states that in each energy level there can be at most two electrons, one with spin up and one with spin down. Thus, electrons will fill all states with energy lower than E F. An expression of p F as a function of the density can be found as follows. The number of levels with momenta between p and p + dp per unit volume is dχ = number of levels unit volume = 4πp2 dp h. (16.19) Since Pauli s principle establishes that two spin states are available, there are two electrons in each level; thus the number of electrons per unit volume is pf pf n =2 dχ = 8πp 2 dp h = 8π h p F. (16.2)
CHAPTER 16. WHITE DWARFS 22 If there are κ nucleons for each electron (κ 2 for stars that have used their hyogen fuel) the mass density is ρ = κnm N, (16.21) where m N =1.67 1 24 g is the mass of the nucleons. The electrons contibution to the mass density is negligible since m e << m N. From eqs. (16.2) and (16.21) we can find p F as a function of the density ( ) 1 p F = h ρ. (16.22) 8πκm N Knowing p F, we can determine the kinetic energy-density ɛ and the pressure P of the gas as follows ɛ = U V = 8π pf {[p 2 c 2 + m 2 h ec 4 ] 1 2 me c 2 }p 2 dp, (16.2) where E c =[p 2 c 2 + m 2 e c4 ] 1 2 m e c 2 is the kinetic energy of each electron, and using eq. (16.9) and (16.) P = 8π pf p 4 c 2 dp. (16.24) h [p 2 c 2 + m 2 e c4 ] 1 2 These equations can be easily integrated in two regimes: 1) the non-relativistic and 2) the ultrarelativistic regime. To this purpose, it is useful to define a critical density, ρ crit, as the density at which the Fermi momentum becomes equal to m e c; using eq. (16.22) ρ crit = κ 8π m N ( ) me c =.98 16 κ g/cm. (16.25) h 1) If ρ << ρ crit, cp F << m e c 2 and the electrons are non relativistic. In this case eq. (16.24) gives P 8π pf p 4 dp = 8π h m e 15h p5 F, (16.26) m e and using eq. (16.22) P = h2 5m e ( ) 2/ ( 1 8π κm N ) 5 ρ 5. (16.27) Thus, the gas of degenerate electrons behaves as a perfect gas with a polytropic equation of state K = h2 5m e ( ) 2/ ( 1 8π κm N P = Kρ γ, where (16.28) ) 5, and γ = 5. Moreover, from eq. (16.2) the kinetic energy-density is ɛ 8π pf {m h e c 2 (1 + 1 2 p 2 c 2 m ) m ec 2 }p 2 dp = 4π 2 e c4 h pf p 4 m e dp, (16.29)
CHAPTER 16. WHITE DWARFS 224 and using eq. (16.26) ɛ = 4π p 5 F = P. (16.) 5h m e 2 2) If ρ >> ρ crit, cp F >> m e c 2 and the electrons are ultra-relativistic. In this case from eq. (16.24) we find and using eq. (16.22) P = 8π pf p cdp = 2πc h h p4 F, (16.1) P = ch 8 ( π ) 1/ ( 1 m N κ ) 4 ρ 4. (16.2) Again, the degenerate gas of electrons behaves as a perfect gas with a polytropic equation of state K = ch 8 P = Kρ γ, where (16.) ( ) 1/ ( ) 4 1, and γ = 4 π m N κ. Moreover i.e. ɛ = 8π pf p cdp, (16.4) h ɛ =P. (16.5) SUMMARY: We have shown that a degenerate gas of electrons can be described by a polytropic equation of state P = Kρ γ in two different regimes: non relativistic regime ρ << ρ crit, K = h2 5m e ( ) 2/ ( 1 8π κm N ) 5 ultra-relativistic regime ρ >> ρ crit, K = ch ( ) 1/ ( 1 8 π m N κ where ) 4 ρ crit = κ 8π m N =9.9156 1 12 κ 5/ erg 2 /g 8/, and γ = 5, (16.6) =1.216 1 15 κ 4/ erg 2 /g 7/, and γ = 4, ( ) me c =.98 16 κ g/cm. h (16.7)
CHAPTER 16. WHITE DWARFS 225 From these expressions we see that, in a completely degenerate gas, pressure depends only on density. As the density increases, degeneracy pressure increases as well, and the pressure gradients which develops inside the star is sufficient to support the equilibrium against gravitational contraction. This is true, as we shall later see, if the mass does not exceed a critical value. It should also be noted that, either in the non releativistic and in the highly relativistic regime, a degenerate gas behaves as a perfect gas with a polytropic equation of state. This clearly contradicts Eddinghton s idea that in the high density regime typical of the interior of a white dwarf, stellar matter should not behave as a perfect fluid. 16.1.4 The structure of a White Dwarf We shall now find the equilibrium configuration of a white dwarf solving the newtonian equations of hyostatic equilibrium (16.17) and using the results obtained in the previous section. As mentioned in section 16.1.2, in order to solve eqs. (16.17) we need to know the equations of state of matter, i.e. an equation which relates pressure to density; since we are interested in the two regimes described in section 16.1., i.e. the non relativistic (ρ << ρ crit ), and the relativistic regimes (ρ >> ρ crit ), we shall assume that the EOS has a polytropic form; thus the complete set of equations to solve by imposing appropriate boundary conditions is dm(r) =4πr 2 ρ = GM(r) ρ r 2 P = Kρ γ. (16.8) It is easy to see that the first two equations can be combined into the following second order equation (hint: differentiate the second equations and replace the expression of dm(r) given by the first) ( ) 1 d r 2 = 4πGρ. (16.9) r 2 ρ Be ρ = ρ(r = ), the central density; by putting γ = 1 + 1 n, where n is called polytropic index ρ = ρ Θ n (r) (16.4) eq. (16.9) becomes P = Kρ 1+ 1 n Θ (n+1) (r), (n + 1)K ρ ( 1 n 1) ( 1 d r 2 dθ ) = 4πG Θ n. (16.41) r 2
CHAPTER 16. WHITE DWARFS 226 If we now introduce the following dimensionless radial coordinate ξ = r 1 α, where α = (n + 1)K ρ( 4πG n 1) 1 2, (16.42) eq. (16.41) becomes ( 1 d ξ 2 dθ ) = Θ n, (16.4) ξ 2 dξ dξ known as the Lane-Emden equation. It should be noted that this is a dimensionless equation, which depends only on the polytropic index n. The physical boundary conditions that have to be imposed to solve the structure equations are that at r = the density has some assigned value ρ and that at the surface of the star, r = R, the pressure vanishes, i.e.: ρ() = ρ, p(r) =. (16.44) Since ρ = ρ Θ n, the first condition implies that Θ() = 1; moreover, since the mass goes to zero as M(r) 4π ρ r, from eq. (16.8) it follows that 4πG rρ 2, i.e. it goes to zero as r. From the EOS P = Kρ γ we find = Kγ ργ 1 dρ from which it follows that if dρ tends to zero must tend to zero as well. Thus, a further condition to impose on Θ is theta Θ (r = ) =. In conclusion the Lane-Emden equation (16.4) must be integrated by imposing that at the center of the star { Θ() = 1, Θ (16.45) () =. It can be shown that if γ> 6 5, Θ(ξ) vanishes for some ξ = ξ 1. When Θ = both the density and the pressure vanish, therefore ξ 1 is the boundary of the star, which can be determined numerically. The procedure to find the stellar structure can be summarized as follows. Choose a value of γ (for instance γ = 5 or γ = 4 ), find the corresponding polytropic index n = 1, and integrate numerically eq. (16.4) with the initial conditions (16.45) γ 1 up to the value ξ = ξ 1 where Θ =. For instance, for γ = 5 and γ = 4 we would find γ = 5 n = 2 ξ 1 =.6575 ξ 2 1 Θ (ξ 1 )= 2.7146 (16.46) γ = 4 n = ξ 1 =6.89685 ξ 2 1 Θ (ξ 1 )= 2.1824 (16.47) It should be noted that Θ is a monotonically decreasing function of ξ, that is why its first derivative at the boundary is negative.
CHAPTER 16. WHITE DWARFS 227 Assign a value to κ, i.e. the number of nucleons per free electrons, then find K from eq. (16.6) or (16.7). Choose a central density ρ. Knowing K and ρ the radius of the star can be found using the definition of ξ given in eqs. (16.42) [ ] 1 (n + 1)K 2 1 n R = αξ 1 R = ξ 1 ρ 2n. (16.48) 4πG The mass of the star can now be determined as follows M = R 4πr 2 ρ(r) =4πα ρ ξ1 = 4πα ρ ξ1 = 4πα ρ ξ 2 1 Θ (ξ 1 ) ( d ξ 2 dθ dξ dξ ) dξ ξ 2 Θ n dξ where use has been made of eq. (16.4). Finally, the value of M as function of K and ρ can be found by using the expression of α given in (16.42) M =4πξ 2 1 Θ (ξ 1 ) [ (n + 1)K 4πG ] 2 ρ n 2n. (16.49) Let us define so that and [ ] (n + 1)K A =, B =4πξ1 2 4πG Θ (ξ 1 ), (16.5) R = ξ 1 A 1/2 ρ 1 n 2n, (16.51) M = B A /2 ρ n 2n. (16.52) Combinig eqs. (16.51) and (16.52), a relation between M and R can easily be derived ( ) M = B A n n n 1 ξ 1 n R n 1 n. (16.5) From the procedure outlined above we understand that, having fixed the number of nucleons per free electrons, κ, and the polytropic index n, once we have found ξ 1 and Θ (ξ 1 ) by numerical integration of the Lane-Emden equation we obtain a family of solutions parametrized with different values of the central density ρ, the radii and masses of which are given by (16.48) and (16.49). Conversely, if we change the number of nucleons per free electrons, the new configuration can easily be obtained by rescaling the various quantities in the following way 1 ρ = κ κ ρ, P = P, (16.54) M(r) = ( κ κ ) 2 M(r), r = κ κ r.
CHAPTER 16. WHITE DWARFS 228 16.1.5 A note on the numerical integration of eq. (16.4) Although the initial conditions (16.45) are correct, it would be impossible to integrate eq. (16.4) numerically starting from ξ = with these conditions. Indeed, since ξ = is a singular point, running the code we would get immediately an overflow. However this problem can be overcome if we start the numerical integration at some small, but finite, value of ξ = ξ start and use as initial values for the function Θ(ξ) a suitable Taylor expansion. Let us do it step by step. Since we know from (16.45) that Θ() = 1 and Θ () =, we can write the approximate solution near ξ = as a power series Θ(ξ) 1+Θ 2 ξ 2 +Θ ξ +Θ 4 ξ 4 + O(ξ 5 ), (16.55) (we can keep as many terms we want, but let us stop here). Θ 1, Θ 2, Θ.. are the constants we need to find using eq. (16.4), therefore we also need to Taylor-expand the function Θ n on the right hand side, i.e. Θ n 1+nΘ 2 ξ 2 + O(ξ ); (16.56) by substituting in eq. (16.4) the expansions (16.55) and (16.56) we find 6Θ 2 + 12Θ ξ + 2Θ 4 ξ 2 +... = [1 + nθ 2 ξ 2 ]+... (16.57) and this equation is satisfied only if the coefficients of the same power of ξ vanish, i.e. 1= 6Θ 2 Θ 2 = 1 6 Θ = 2 Θ 4 = nθ 2 Θ 4 = n 12 ; the expansion has only even powers of ξ (this is true also at higher order). Thus the approximate solution for Θ and Θ near the origin is Θ(ξ) 1 1 6 ξ2 + Θ (ξ) 1 ξ + n ξ + O(ξ 5 ). n 12 ξ4 + O(ξ 6 ) (16.58) We now have all we need to numerically integrate the Lane-Emden equation, because we can start at, say, ξ start = 1 4 using as initial values the functions (16.59) computed at ξ start. 16.2 The Chanasekhar limit In section 16.1. we have shown that if the density is much smaller than the critical density, electrons behave as a polytropic gas with γ = 5. In this regime eqs. (16.5) give A =2.9562 1 19 κ 5/, B = 4.159
CHAPTER 16. WHITE DWARFS 229 and using eq.(16.25) and (16.52) we can write the mass of the star in this form ( ) 1/2 M =2.7 κ 5/2 ρ M, (16.59) ρ c where M =1.989 1 g is the mass of the Sun. This equation shows that the mass of the star increases with the central density. As the central density increases above the critical density, the electrons start to behave as a relativistic gas with a polytropic equation of state with γ = 4. Equation (16.49) shows that in this limit the mass becomes independent of the central density ρ and takes the value M = M CH =5.74 κ 2 M. (16.6) This is a critical mass above which no stable configuration for a white dwarf can exist, and it is called the Chanasekhar limit, as it was derived by Subrahmanyan Chanasekhar in 191. It should be noted that the information on the internal composition is contained entirely in the parameter κ. For instance, if we set κ = 2 we find M CH =1.45 M. (16.61) The fact that a critical mass should exist can also be understood from the following qualitative considerations. A given configuration of matter will be in equilibrium if the gradient of pressure is balanced by the gravitational attraction. In the non relativistic case a) P ρ 5 P M 5 In the ultra-relativistic case b) P ρ 4 P M 4 The gravitational force per unit volume behaves like R 5 R 4 M 5 R. (16.62) 6 M 4 R. (16.6) 5 Gm(r)ρ r 2 M 2 R 5. (16.64) If the star is in equilibrium = Gm(r)ρ ; r 2 in the non-relativistic case the gradient of pressure (16.62) and the gravitational force (16.64) depend on the radius to a different power thus, for a given value of the mass, the star can The concept of a limiting mass for white dwarfs was first introduced by Chanasekhar in a paper published in 191: The Maximum Mass of Ideal White Dwarfs, in The Astrophysical Journal, 74 n.1, 81. The problem was subsequently investigated in a series of papers, and a complete account can be found in the book Chanasekhar wrote on the subject in 199: An introduction to the study of stellar structure, University of Chicago Press, Chicago Illinois
CHAPTER 16. WHITE DWARFS 2 adjust the radius until the two forces are equal. Conversely, in the ultra-relativistic case the gradient of pressure (16.6) and the gravitational force (16.64) have the same dependence on the radius, and therefore the equilibrium is possible only for one value of the mass, i.e. for the critical mass. If M > M CH the gravitational attraction exceeds the gradient of pressure and stable configurations are no longer possible. Although the existence of a critical mass for white dwarfs seems an obvious consequence of the theory today, it was not accepted when Chanasekhar found it. The prejudice at that time was that white dwarfs do represent the final state of a star, and that they could have any mass (neutron stars were discovered much later in 1965). The famous astronomer Sir Arthur Eddington was the strongest opponent to the new theory, and called it a stellar buffonery. Nobody at that time gave to Chanasekhar any public support, although a few, as for example Rosenfeld, told him in private that they thought his result was correct 4. It should be stressed that this limit is a static limit, i.e. it refers only to the equilibrium configuration. It says that stars with a mass exceeding the critical mass cannot exist. However, even if a star is in equilibrium it may become unstable against small perturbations. In this case we would call it a dynamical instability. A second point which should be noted is that in the derivation of the critical mass general relativity plays no role. The basic ingredients are special relativity and the Fermi-Dirac statistics. 4 An interesting account of the controversy between Eddington and Chanasekhar on white dwarfs maximum mass can be found in the book Chana: a biography of S. Chanasekhar, University of Chicago Press 1991