Homework 9 Solutions Homework 9 Solutions for by qfs14600

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									                       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.




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