VIEWS: 0 PAGES: 5 CATEGORY: Education POSTED ON: 3/26/2010
Homework 9 Solutions for Physics 133 with Prof. Crystal Martin These solutions were prepared by Amanda Fournier on March 5, 2010 using L TEX A Prob. 1 Ryden 10.4 Helium Production by Stars The energy emitted in time ∆t is ∆E = ∆t · L: ∆E = 2.8 × 1054 J = 1.7 × 1067 MeV The number of helium nuclei that would have to be formed to release this much energy is ∆NHe = ∆E/EHe , EHe = 28.4 MeV: ∆NHe = 6.1 × 1065 helium nuclei If the baryonic mass of our galaxy is Mbary = 1011 M , and the primordial mass fraction of helium is Y = 0.24, the mass of helium in our galaxy was originally MHe,primordial = Y × Mbary = 2.4 × 1010 M If we convert available hydrogen into ∆NHe helium nuclei, we add ∆MHe = ∆NHe · mHe ∆MHe = 4.1 × 1042 g = 2.1 × 109 M So the current mass fraction of helium would be MHe,primordial + ∆MHe = Ycurrent = 0.26 Mbary i.e., Y has increased by about 0.02. 1 Prob. 2 Ryden 10.5 Primordial Helium Fraction The “helium mass fraction” is Y = ρHe /ρbaryons = ρHe /(ρp + ρn ). ρp = mp np , the mass of a single proton times the number density of protons, so Y = mHe nHe /(mp np + mn nn ). But mHe ≈ 2mp + 2mn and mp ≈ mn , so 4mp nHe 4nHe Y ≈ = m p np + m p nn np + nn Every helium nucleus is composed of two neutrons and two protons. If all possible neutrons combine into helium, and we assume that there is no shortage 1 of protons to form helium, then nHe = 2 nn . We now have the maximum value of Y : 2nn Ymax = np + nn Using f = nn /np : 2f Ymax = 1+f QED. Prob. 3 Ryden 11.1 Early Universe World Models If the pre-inﬂation universe was dominated by radiation + curvature, and Ω was greater than one, then the mathematics similar to those of the “Big Crunch” model apply. We’ll start by building an equation analogous to 6.12 for this universe. We ﬁnd that, for Ω0 = Ω0,rad , H2 Ω0 1 − Ω0 a2 ˙ 2 = a4 + H0 a 2 = 2 2 a H0 Rearranging, H0 Ω0 + (1 − Ω0 )a2 = aa ⇒ ˙ t2 a2 1 ada dt = t1 H0 a1 Ω0 + (1 − Ω0 )a2 2 Setting (1 − Ω0 )a = x, (1−Ω0 )a2 1 2 dx t2 − t1 = √ ⇒ 2H0 (1 − Ω0 ) (1−Ω0 )a2 1 Ω0 + x 1 t2 − t1 = Ω0 + (1 − Ω0 )a2 − 2 Ω0 + (1 − Ω0 )a2 1 H0 (1 − Ω0 ) 2 We now have a useful relation between Ω0 , H0 , a, and time. We want to know about the universe’s behavior between the times tp and ti , so we choose those times as our limits of integration. 1 ti − tp = Ω0 + (1 − Ω0 )a2 − i Ω0 + (1 − Ω0 )a2 p H0 (1 − Ω0 ) The maximum possible value of Ω0 is the value that causes a Big Crunch exactly at ti , so the boundary condition for the collapse of the universe before inﬂation begins is ai = 0: 1 ti − tp = Ω0,max − Ω0,max + (1 − Ω0,max )a2 p H0 (1 − Ω0,max ) To further simplify these equations, we must recognize that we are free to choose any time as our reference time t0 , so long as we deﬁne H0 as H(t0 ), Ω0 as Ω(t0 ), and a(t0 ) as one. Since we want to know about Ω(tp ), tp is the logical choice for the reference time t0 . We set H0 = Hp , Ω0 = Ωp , and ap = 1: Ωp,max − 1 ti − tp = Hp (1 − Ωp,max ) Now the question arises - what is the value of Hp ? To answer this question, recognize that for a universe with a life expectancy of ti = 10−36 s, the time tp = 10−44 s is very early, a mere 10−8 of the time it takes the universe to expand, turn around, and crunch again. At such an early stage, and hence at such a small scale factor compared to that it will achieve at turnaround, it is likely to be strongly dominated by radiation, since the radiation term dominates the curvature term in the Friedmann equation for small a. Given the assumption that it is indeed radiation dominated at this stage, we can use the single-component universe relations we found in chapter 5 to describe the universe at this stage, namely 1 H0 = 2t0 and so Ωp − 1 ti − tp = 2tp 1 − Ωp At last, we can solve for Ωp . This expression can be manipulated and solved as a quadratic equation, yielding √ X 2 + 2X + 2 ± 2X 3 + 4X 2 + 8X + 4 Ωp,max = X2 where ti − tp ti X= ≈ 1 tp tp 3 Since X 1, the equation above is well approximated by √ X 2 ± 2X 3 Ωp ≈ X2 And since we know Ωp > 1, √ X 2 + 2X 3 Ωp ≈ =1+ 2/X X2 which yields the numeric values Ωp ≈ 1 + 1.4 × 10−4 if ti = 10−36 s Ωp ≈ 1 + 1.4 × 10−9 if ti = 10−26 s Prob. 4 Ryden 11.3 Decay of the Vacuum Energy? Then What? This problem is an exercise in approximations. For each epoch we will deal with, there will be two or more components of interest, but we will calculate the evolution of the universe as if there were only one, the “dominant” component at that time. At the present epoch, the universe’s dynamics are well-described by H2 Ωm,0 2 = ΩΛ,0 + a3 H0 where ΩΛ,0 = 0.7 and Ωm,0 = 0.3. When the vacuum energy component becomes strongly dominant (i.e., much greater than the matter component), the Hubble parameter will be given by 2 HΛ 2 = ΩΛ,0 = 0.7 ⇒ H0 −1 HΛ = 58.6 km s−1 Mpc−1 or HΛ = 16.7 Gyr After ﬁfty Hubble times of expansion, the universe will be very chilly indeed, and the energy contributed by the decay of the false vacuum will be far and away the largest component of energy around. We can ﬁnd the appropriate temperature for the new bath of radiation using eqn.s 2.26 and 2.27: = αT 4 α = 7.56 × 10−16 J m−3 K−4 4 We also know that the energy density of the vacuum energy component before decay is = 0.7 c,0 = 3600 MeV m−3 so the temperature after decay will be Tf = 29.6 K To ﬁnd the density of matter at tf , we recall that for a cosmological-constant dominated universe, a ∝ exp(HΛ t). Therefore, using a(t0 ) = 1, we ﬁnd the expression for a during the lambda-dominated phase: a(t) = eHΛ (t−t0 ) Using t0 = 13.5 Gyr (Ryden gives this value during her discussion of the Benchmark model) and tf = 50t0 , we ﬁnd a(tf ) = e49HΛ t0 = 1.59 × 1017 So the density of matter at tf is −3 m,0 a = m,f = 3.94 × 10−49 MeV m−3 We can easiest see how the universe will evolve in its new, radiation-dominated stage by setting t0 = tf , a (t0 ) = 1, etc. The scale factor will evolve as t a = t0 and radiation-matter equality will once again be reached when m,0 r,0 = ⇒ a3 a4 2 r,0 t0 = trm ⇒ m,0 trm = 4.17 × 10105 t0 = 5.63 × 10106 Gyr After this time, the universe will be dominated by matter. 5