Ay123 Set 3 solutions

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1 Ay123 Set 3 solutions Mia de los Reyes October Equation of state and the Chandrasekhar mass (a) Using the Fermi-Dirac distribution for non-relativistic electrons, derive the relationship between density and pressure, and hence appropriate value of γ and K. The Fermi-Dirac distribution is given by Then pressure is given by n e = 8π h 3 P e = 1 3 = 8π dp p 2 [exp ( ) 1 ɛ µ + 1] (1) vp dn d 3 xd 3 p d3 p (2) ( ) 1 ɛ µ dp vp [exp 3 + 1] (3) We can rewrite this in terms of momentum using the fact that the Fermi-Dirac distribution is essentially a step function: P e = 8π pf dp vp 3 (4) ( ) 1/3 3h where p F = 3 n e 8π is the Fermi momentum. In the non-relativistic limit, substitute v p/me and the definition of the Fermi momentum into equation (4): Note that electron number density n e = electron pressure as P e = P e = 8π m e pf = 8π p 5 F m e 5 dp p 4 (5) (6) ρ µ em p, where µ e = 2 for 4 He. Then we can write the ( ) 2/3 ( ) ( ) 5/3 8π 1 1 ρ 5/3 (7) 5m e µ e m p Then γ = 5/3 and K = ( ) ( ) ( ) 8π 2/3 5/ m e µ em p = cm 4 g 2/3 s 2 in the polytropic equation P = Kρ γ. (b) Using the mass-radius relations we derived for polytropes, derive the mass-radius relation for a white dwarf. Calculate the radius of a 1 M white dwarf. From the discussion of polytropes (HKT Eq. 7.4): [ 4π K = ξ n+1 ( θ n ) n 1 1 ] 1/n ξ 1 G n + 1 M 1 1/n R 1+3/n (8)

2 Here, n = 1 γ 1 = 3/2 for a γ = 5/3 polytrope. From HKT Table 7.1, ξ 1 = and θ n (ξ 1 ) =.233 for n = 3/2. Set this equal to the equation for K from part (a) and solve for R: R = ( cm g 1/3 ) M 1/3. For a 1 M white dwarf, this yields R = cm.1 R. (c) Now assume the helium white dwarf is supported by highly relativistic degeneracy pressure. Use the Fermi-Dirac distribution to derive the appropriate values of γ and K, and then derive its mass. Express the mass in units of M. Now consider relativistic electrons, so v = c. Then P e = 8π Again, plugging in the definition for n e = P e = = 8πc pf pf = 8πc p 4 F 4 = 8πc c 4 dp vp 3 (9) dp p 3 (1) ( n e 8π (11) ) 4/3 (12) ρ µ em p, we find that the pressure goes as ( ) 1/3 8π ( c ) ( ) 4/3 1 ρ 4/3 (13) 4 µ e m p So γ = 4/3 and K = ( ) 8π 1/3 ( c ) ( ) 4/3= µ em p g 1/3 cm 3 s 2 (again assuming µ e = 2 for 4 He). To get the mass, start with the equation given in class for a polytrope: where n = 1 γ 1 = 3 for a γ = 4/3 polytrope. Then ( ) 3/2 ( n + 1 M = 4π 4πG K 3 n ρ c ξ 2 dθ ) (14) 2n dξ ξ 1 ( ) 3/2 K M = 4π (ξ 1 ) 2 ( θ n ) ξ1 (15) πg Use values from HKT Table 7.1: ξ 1 = and θ n (ξ 1 ) =.4243 for n = 3 to find M = 1.43 M. This is very close to the Chandrasekhar mass! (d) Recalculate this mass for a relativistic white dwarf made of pure gold ( 197 Au). Gold is currently worth $39.47/g. What is the value of this golden dwarf? The only thing that changes here is the value of µ e, which is the number of baryons per electron. For 197 Au, which has atomic number Z = 79, this yields µ e = = Then use the formula from part (c) to find K = g 1/3 cm 3 s 2. Plug this into equation (15) for M to find M = g=.94 M. The cost of this golden dwarf is $ ! (e) If the golden dwarf has a radius of 3 km and an internal temperature of 1 7 K, will most of its mass be liquid gold or solid gold? To determine if the gold is solid (crystallized) or not, calculate the ratio of Coulomb to thermal energy Γ c = Z2 e 2 a. 2

3 ( ) 1/3, 3 The separation between gold ions is given by a = 4πn ion where nion is the number density of ions. The number density is given by n ion = This yields 3M 197m p(4πr 3 ). ( ) 1/3 ( ) 1/3 Γ c = Z 2 e 2 3M 3 197m p (4πR 3 ( ) 1 (16) ) 4π Plugging in values gives Γ c = , which is well into the regime of crystallization. Most of the mass of the white dwarf will be solid gold. (f) If the surface layers of the golden dwarf have an opacity of κ = 1 cm 2 /g, what is the surface pressure at the photosphere? If the gold at the photosphere can be treated as an ideal gas and has a temperature of 1 4 K, what is its density? The radius of a gold atom is a Au = cm. Is the gold at the surface likely to be pressure ionized? Recall (from, e.g., the last HW set) that the surface pressure at the photosphere is given by P (τ s ) = 2gs 3κ. Plugging in the surface gravity g s = GM R yields P (τ 2 s ) = 2GM 3R 2 κ. Use κ s = 1 cm 2 /g and the values of M and R from the previous parts to find the surface pressure at the photosphere P (τ s ) = erg cm 3. Now, assuming the photosphere is an ideal gas, P = ρk BT µm p so ρ = P µmp. Recall that µ is the mean molecular weight µ = # baryons # particles = = Plug in values to find ρ = g/cm 3. Pressure ionization occurs when ρ m N 4πa 3 /3. For 197 Au, m N = 197m p and a N = cm. N m This yields N 4πa 3 N /3 = 14.9 g/cm3. So ρ < m N 4πa 3 N /3, and pressure ionization is unlikely. 2. Nuclides and kilonova event rates (a) Consult the table of nuclides. Identify two stable, pure s-process nuclides and two stable, pure r-process nuclides. Note that the nuclides must be stable (or at least long-lived)! Some examples of pure s-process elements: 92 Mo, 12 Pd Some examples of pure r-process elements: 76 Ge, 94 Zr (b) In the Sun, the mass fraction of r-process nuclides is X rp 1 7. The stellar mass of the Milky Way is M MW 1 11 M. Assuming similar abundances in other stars, estimate the total r-process mass within the Milky Way. M rp,mw = X rp M MW = 1 4 M. (c) Assuming r-process elements are produced solely in neutron star mergers, and that M rp.3 M of r-process nuclides were expelled from GW17817, estimate the number of neutron star mergers that have occurred in the Milky Way, and the average neutron star merger rate over the τ MW 1 Gyr lifetime of the Milky Way. The number of neutron star mergers is given by N NSM =(mass of r-process elements)/(mass of r-process elements per NSM). So N NSM = Mrp,MW M rp = The average rate of neutron star mergers is just N NSM /τ MW = yr 1. (d) Type Ia supernovae synthesize M Fe.5 M of iron, whose mass fraction in the Sun is X Fe 1 3. Estimate the average Type Ia supernovae rate of the Milky Way. As in part (b), first compute the total mass of iron in the Milky Way: M Fe,MW = 1 8 M. Then follow the procedure in part (c) to compute the number of Type Ia SNe: N Ia = MFe,MW M Fe = Divide this by the lifetime of the Milky Way to get the average Type Ia SNe rate: N Ia /τ MW =.2 yr 1. 3

4 Figure 1: Plot of the ionization fraction n e /n tot as a function of temperature for a pure hydrogen gas. (e) Estimate the number of neutron star mergers that have occurred in a dwarf galaxy with metallicity 1 1 that of the Milky Way, and stellar mass M gal 1 6 M. Do we expect all dwarf galaxies to be enriched in r-process elements such as Europium? As in part (b), first compute the total mass of r-process elements in the dwarf galaxy: M rp,gal = X rp M gal = 1 2 M. Then as in part (c), compute the number of neutron star mergers: N NSM = M rp,gal M rp do not expect all dwarf galaxies to be enriched in r-process elements. 3. Free electrons at low temperatures =.33. We (a) Use the Saha equation to calculate the fraction of free electrons n e /n tot for a pure hydrogen gas. Plot your result as function of temperature. Assume n tot = 1 17 cm 3 as is appropriate for the solar photosphere, where n tot is the total number of nuclei (ionized + neutral). The Saha equation is n e n H+ n H ( 2πme = h 2 ) 3/2 e χ H (17) We want y = n e /(n H+ + n H ), which is simply the degree of ionization. The Saha equation can then be written as y 2 1 y = 1 ( ) 3/2 2πme n tot h 2 e χ H (18) This can then be solved numerically for y as a function of T using the quadratic formula (note that y must be positive, so use the positive solution). The result is shown in Figure 1. (b) Calculate the fraction of free electrons n e /n tot,h as a function of temperature for the combined hydrogen + metal gas. As in (a), assume n tot,h = 1 17 cm 3. Make a plot showing the ratio of n e for part (b) to n e for part (a) as a function of temperature. Below what temperature does the assumption of a pure hydrogen gas start to become inaccurate for predicting n e? Using ( ) h 2 1/2 λ 2πm e kt we can write Saha for hydrogen (rewriting equation 17): n H n H+ n e = λ 3 e χ1/kt (19) 4

5 Figure 2: Plot of the ionization fraction n e /n tot,h as a function of temperature for a metal and hydrogen gas. For the metal we have We also have the relations and Using 2 and 22 we can solve for n m+ in terms of n e only: n m n m + = n tot,m 1 = n e λ 3 e χm/kt n m+ n tot,m n m+ = n e λ 4 e χm/kt We can likewise use (2) and (4) to solve for n H+ : n m n m+ n e = λ 3 e χm/kt (2) n tot,h = n H+ + n H = 1 17 (21) n tot,m = n m+ + n H = 1 11 (22) n H+ + n m+ = n e (23) n H+ = n tot,h n e λ 4 e χ1/kt Therefore, we can write down an expression for the electron density by (6): n e = n tot,m n e λ 4 e + n tot,h χm/kt n e λ 4 e χ1/kt We use n e = n tot,h y (where y is the ionization fraction we want) to rewrite this as a function of y: y = n tot,m 1 n tot,h yn tot,h λ 3 e χm/kt yn tot,h λ 4 e χ1/kt + 1 I couldn t be bothered to actually solve this, but the person who created Figure 2 used IDL s F Z ROOT S. The ratio of the ionization fractions is shown in Figure 3. This shows that metals are important at T < 45 K. 5

6 Figure 3: Ratio of ionization fractions for metal+hydrogen gas and pure hydrogen gas. 6