Solution: HW 4 AY 13, Fall 007 by Fabio Altenbach, Melissa Enoch, and Matthew Stevenson December, 007 Problem 1: 1a: To approximate the Gamow peak by a gaussian, first rewrite the < σv > equation as a function of energy. To do so, simply substitute v = E and m dv = de me. Then, assuming S is an experimentally derived constant near the Gamow peak, ( ) 8 1/ [ < σv >= (kt) 3/ S 0 Exp E πm 0 kt b ] de (1) E 1/ where b is given in the problem. We wish to compare the integrand to a gaussian, [ (E E0 ) ] f Gauss (E) = C Exp ( /) The peak E 0 is given by minimizing E kt + b E 1/, so 1 kt E 0 = b E 3/ 0 () = 0 (3) ( ) 3/ ktb (4) 1
Clearly, for the Gaussian to have the same peak value as the Gamow integrand, ( ) 8 1/ [ C = (kt) 3/ S 0 Exp E 0 πm kt b ] (5) E 1/ 0 Now compute the second derivatives at E 0 of the Gamow integrand and the gaussian to get the wih (the first derivative at E 0 is zero because it is at a maximum): f Gamow (E 0) = 8 C (6) f Gauss (E 0) = 4E 0 4kTE 3/ 0 3k T be 1/ 0 + k T b C (7) 4k T E0 3 Substituting in our value of b and equating yeilds: So, 8 = 3 kte 0 (8) = 4 3 kte 0 (9) 1b: From our calculations in part a, [ (E E0 ) ] < σv > = C Exp 0 ( /) de (10) ( ) [ T 3/ S m 1/ 0 Exp E kt b ] ( ) e x dx (11) E 1/ ( T 1 ) [ E 0 S m 1/ 0 Exp E 0 kt b ] (1) E 1/ 0 T 1 E 1/ [ 0 S m 1/ 0 Exp 3E ] 0 (13) kt So, let τ = 3E 0 kt, then < σv > T 1 E 1/ 0 m 1/ S 0 e τ τ ( ) kt (14) E 0
TE 3/ 0 m 1/ ( T btm 1/ S 0 e τ τ (15) ) S 0 e τ τ (16) 1 Z A Z B m S 0τ e τ (17) 3
Problem : a: By HKT 6.76, the energy generation rate for the p + p reaction is: So, n = 1. To find η, b: ǫ ρ T /3 9 η = 1/3 at e 9 a = 3.380 (18) = For solar temperature, η 3.9. We are looking at the reaction ( ) lnr lnt a 3 3T 1/3 9 = τ 3 3 (19) (0) (1) 7 Be + p 8 B + γ, () which begins the PP-III chain. The reaction rate for this reaction is r7 Beγ = ρ X7 BeX p < σv m u A7 BeA >7 Beγ cm 3 s 1 (3) p The temperature dependence comes in from the < σv >: with We write the rate as 1/3 r < σv >7 Beγ e at 6 a = 4.49(Z 7 BeZ pµ) 1/3 and µ = T /3 6, (4) A7 BeA p A7 Be + A p. (5) r T ν (6) 4
and thus we can solve for ν: ν = ( ) lnr lnt = a 3T 1/3 3 6 (7) We have the following values: A7 Be = 7, A p = 1, Z7 Be = 4 and Z p = 1. We estimate µ as µ = 7/8 1. Thus at a temperature of T = 1.5 10 7 = 15 10 6 K, we have: ν 4.49(16)1/3 3(15) 1/3 3 = 13.8 14 (8) So we indeed find that the reaction rate is approximately proportional to T 14. How much would the centeral temperature of the sun need to change to solve the solar nuetrino problem? The PP-III chain continues as follows: 8 B 8 Be + e + + ν e (9) creating an electron nuetrino. Assuming the rate of the chain is dominated by the first reaction ( 8 B has a half life of 0.8 s), the rate of production of nuetrinos from this chain is T 14. Now, Davis experiment can only detect nuetrinos from the PP-II and PP-III chains, and the majority are expected to come from this B inverse beta decay in PP-III. So let s assume that all the neutrinos detected are from the above reaction. The observed nuetrino flux (from HK Section 8.3) is.07 ± 0.3 SNU, while the predicted flux is 7.9 ±.4 SNU. So to explain the discrepency we will need to lower the predicted number of nuetrinos produced by approximately 4: r 1 4 r 1 T 14 1 4 T 14 1 (30) T 0.9T 1 (31) So we would have to lower the central temperature of the sun by 10%, or 1.5 10 6 K. Such a change would require a drastic reworking of stellar models. Instead, it is explained through neutrino oscillations, in which part of the electron neutrinos produced change into µ or τ neutrinos on the way to earth. These are not detected by the Ray Davis experiment. 5
Problem 3: We are concerned with the PP-I chain, given as follows: p + p d + e + + ν e (3) d + p 3 He + γ (33) 3 He + 3 He 4 He + p + p (34) The reaction rate for the reaction α+x Y +β is given by (e.g. HKT) R αβ = 1 (1 + δ αx ) n αn X < σv > αβ, (35) the units of this rate being cm 3 s 1 (Note that there is a degeneracy factor of if α and X are the same element. < σv > is given by < αβ > < σv > αβ, which we obtain from (CF1988) at T c, = 1.5 10 7. We let: C 1 = < p, e + >= 1.36 10 43 (36) C = < p, γ >=.19 10 6 (37) C 3 = < 3 He, p >= 3.89 10 34 (38) The rate of change of any given species in the PP chain is given by the rate at which it is created minus the rate at which it is destroyed (R created R destroyed ). The reaction rates for the three steps in the PP chain are given by Eq. 35: = n p < p, e+ > (39) R pγ = n p n d < p, γ > (40) R pe + R3 He,p = n3 He < 3 He, p > (41) (4) The above PP reactions give four equations for the abundances of H, D, 3 He and 4 He: ( ) ( ) dn p n p n = < p, 3 e+ He > n p n d < p, γ > + < 3 He, p > (43) 6
dn d dn3 He = n p < p, e+ > n p n d < p, γ > (44) ) = n p n d < p, γ > ( n 3 He < 3 He, p > (45) dn4 He = n3 He < 3 He, p > (46) The < σv > s are constant in time, dependent only on temperature and the S 0 s. In order to find the time when the first reaction rate Eq. 39 becomes longer than the other two, we need the abundances of of H, D, and 3 He as a function of time (and while we re at it we ll calculate the abundance of 4 He too as we need to plot it later). To find these abundances we will have to solve the system of differential equations Eqs. 43-46. If we replace all the n s with n i = ρx i /A i m u so that things can be written in terms of ratios X n, the equations we need to simultaneously solve are: dx p dx d dx3 He dx4 He = = = = ( ) [ ρ C 1 Xp C mu X px d + C 3 ( ) ρ [C1 ] Xp C X p X d mu ( ) [ ] ρ 3C mu ( ρ mu X px d C 3 ) [ ] C3 9 X3 He 3 X3 He 9 X3 He ] (47) (48) (49) (50) (51) subject to the constraint that X p + X d + X3 He + X4 He = 1 at all times, and the boundary conditions X p (0) = 1 and X d (0) = X3 He(0) = X4 He(0) = 0, which apply to a star initially composed of pure hydrogen. Our constraint is satisfied by the initial conditions because d (X p + X d + X3 He + X4 He) = 0. Simplifying: dx p dx d dx3 He = αx p βx px d + γx 3 He (5) = αx p βx p X d (53) = 3βX p X d 3γX 3 He (54) 7
dx4 He = γx 3 He (55) (56) where α = 1.19 10 17, β = 0.964, γ = 3.80 10 9. We now want to find the equilibrium timescale. This occurs when deuterium production (the fastest) saturates. Also, during these early times, X p remains constant at about 1. So, we have, which has the solution dx d = α βx d (57) X d (t) = α β ( 1 e βt ) (58) α β so it has a timescale (β) 1 0.5 seconds and saturates at a value of 6.17 10 18. Now we use this saturation value in the third equation to see when H and D come into equilibrium with 3 He. which has the solution dx3 He X3 He(t) = = 3α 3γX3 He (59) [ ] α 3 αγ γ tanh t (60) so it has a timescale ( 3 αγ ) 1 7049 years and saturates at a value of α γ 1.57 10 9. Below we plot the X p and X3 He fractions for the first million years, and the all the abundances over 10 10 years. 8
Log X X_H - -4-6 D 3He -10 Log t s 4 6 8 10 1 14-1 -14-16 -18 Figure 1: This plot shows the X p and X3 He fractions for the first 3 years. The solid line, deuterium, comes into equilibrium at abou 0.5 seconds and saturates at a value of 6.17 10 18. The dashed line, helium 3, takes closer to 70 kyr to reach equilibrium and saturates at a value of 3.96 10 5 9
X 1 0.8 0.6 H D 10^17 3HE 10^4 4He 0.4 0. 1 3 4 5 6 7 Log t kyr Figure : Over the next 10 Gyr, things build up and change, yielding 4 He. So, long timescales( 10 Myr) will upset the equilibrium set up on short timescales ( 70 kyr). Note that the D and 3 He fractions have been multiplied by a constant factor so they will show up on the graph. 10
Problem 4: 4a: In problem set #1, we arrived derived the scaling solutions for ρ c and T c. Using solar units for M and R, and the emperical scaling relation R M 3/4, ρ c 5.57M 5/4 (61) T c 7 10 6 M 1/4 (6) Now, using solar values of X =.7 and Z = 0.0, we find that HKT 6.76-6.77 reduce to ǫ pp ǫ CNO.39 106 19.766M 1/1 e M 17/1 1.173 106 e 89.405M 1/1 M 17/1 (63) (64) 4b: The fracion of energy comming from the pp chain is ǫ pp /ǫ. Similarly for the fraction of energy produced by the CNO cycle. The plots comparing them to the total and each other follow. 4c: We would like to know when the CNO cycle saturates. This occures when all the protn capture rates are faster than the slowest decay rate in the chain. Looking at the class notes, 13 N(e +,ν e ) 13 C is the slowest beta decay at 7 minuts. Now we use CF88 to look at what temperatures the other timescales ((ρxf CF88 ) 1 ) are shorter than this. Therefore, we need to find where f CF88 > 3.4 10 9 for each reaction and take the largest temperature: 1 C(p,γ) 13 N T crit,9 0.047 13 C(p,γ) 14 N T crit,9 0.040 14 N(p,γ) 15 O T crit,9 0.060 15 N(p,α) 1 C T crit,9 0.031 So, T sat = 6 10 7 K. 11
Log Ε cgs -0.6-0.8-1 Ε Εpp Εcno -1. -1.4-1.6-1.8-1 -0.5 0.5 1 1.5 Log M M_sun Figure 3: The transition from pp to cno in our model star occurs at a slightly larger mass than in real stars, which are more centrally concentrated than our model. 1
fraction of Ε 1 0.8 0.6 0.4 Εpp Ε Εcno Ε 0. -1-0.5 0.5 1 1.5 Log M M_sun Figure 4: The fraction of energy produced by ǫ pp and ǫ CNO 13
Log Εpp Εcno 1 10 8 6 4-1 -0.5 0.5 1 1.5 Log M M_sun Figure 5: The fraction of energy produced by ǫ pp comparted to ǫ CNO 14
Problem 5: The class notes for nuclear reactions give the following equation for the energy generation rate of the triple-α process: ǫ 3α = 5.1 108 ρ Y 3 e 4.40/T 9 ergs g 1 s 1 (65) T9 3 Evaluating this for ρ = 10 5 g cm 3, T = 10 8 K, and Y = 1 (pure helium): ǫ 3α = 397 ergs g 1 s 1 (66) Now, if the excitation energy of 1 C is 7.664MeV instead of 7.654MeV, then the exponent of 4.40/T 9 will change. The 4.40 comes from the mass difference between 1 C and the three α s. From Fig. 6.11 of HKT, the true m = 0.3818MeV, which our hypothetical value is m = 0.3918MeV, an increase of 1.06 times. The 4.40 then becomes 4.5. We then find: ǫ 3α = 119 ergs g 1 s 1 (67) 15
Problem 6: For this problem, all units are cgs. We are looking at 1 gm of material, so the energy generation rate per unit volume per density, for V ρ = 1 gm, de = ǫ = 5.1 Y 3 108ρ e 4.407/T 9 (68) T 3 9 where de = c Vρ dt (69) and the specific heat is the sum of the (degenerate) electron and ideal (ion) specific heats c Vρ = 1.35 10 5T ρ x ( 1 + x ) 1/ + and x is the normalized fermi momentum For our ionized helium core, x 3 = 3k µ I m u (70) ρ Bµ e (71) ρ = 10 5 (7) T = 1.5 10 8 (73) Y = 1 (74) µ I = 4 (75) µ e = (76) Putting this all together and converting time into days yields dt 9 d = (77) ǫ 4 3600 c Vρ 10 9 (78) = 5.05 106 e 4.407 T 9 T9 3 (T 9 + 0.0893) (79) 16
T 10^9 K 0.8 0.6 0.4 0. 1 3 4 5 t days 0.18 0.16 Figure 6: As is seen in the graph, the Helium Flash occurs in about 5 to 6 days. This time, however, is very sensitive to the intial conditions. 17