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How do I solve this cosmological model FRW?

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data

    (a)Find how ##\rho## varies with ##a##.
    (b) Show that ##p = \frac{2}{\lambda^2}##. Find ##B## and ##t_0##.
    (c) Find ##w## and ##q_0##. What values of ##\lambda## makes the particle horizon infinite? Find the event horizon and age of universe.
    (d) Find luminosity distance ##D_L## in terms of redshift ##z##. Find ##q_0## by expanding.


    2013_B5_Q4.png
    2. Relevant equations


    3. The attempt at a solution

    Part(a)
    For ##V_0 = 0##, we can see that ##\rho = P = \frac{1}{2} \dot \phi^2##. Thus ##w=1##. For dependency on ##a##:
    [tex]\ddot \phi + 3(\frac{\dot a}{a})\dot \phi = 0[/tex]
    [tex]\dot \rho + 12(\frac{\dot a}{a}) \rho = 0[/tex]
    [tex]a^{-12}\frac{d}{dt}(\rho a^{12}) = 0 [/tex]
    [tex] \rho \propto a^{-12}[/tex]

    Part (b)
    I'll replace the ##p## by ##x## to avoid confusion with pressure ##P##. Given ##a(t) = t^x## and ##\phi = BM ln(\frac{t}{t_0})##, we have ##\frac{\dot a}{a} = \frac{x}{t}## and ##\dot \phi = \frac{BM}{t}## and ##\ddot \phi = -\frac{BM}{t^2}##.

    Substituting into equation of motion:
    [tex]\frac{-BM}{t^2} + 3\left(\frac{x}{t}\right)\left(\frac{BM}{t}\right) - \frac{\lambda V_0}{M} \left( \frac{t}{t_0} \right)^{-\lambda B}[/tex]
    [tex]BM(3x - 1) - \frac{\lambda}{M} V_0 t^2 \left( \frac{t}{t_0} \right)^{-\lambda B} = 0 [/tex]

    Substituting into FRW equation:
    [tex]\frac{x^2}{t^2} = \frac{8\pi G}{3} \left[ \frac{2}{2}\left(\frac{BM}{t}\right)^2 + V_0 \left( \frac{t}{t_0} \right)^{-\lambda B} \right] [/tex]
    [tex]x^2 = \frac{B^2}{6} + \frac{8 \pi G}{3}V_0 t^2 \left( \frac{t}{t_0} \right)^{-\lambda B} [/tex]
    Using our result from the equation of motion:
    [tex]x^2 = \frac{B^2}{6}+ \frac{8 \pi G}{3} \left[ \frac{BM^2}{\lambda} (3x-1) \right] [/tex]
    [tex]x^2 - \left(\frac{B}{\lambda}\right)x + \left( \frac{B}{3\lambda} - \frac{B^2}{6} \right) = 0 [/tex]

    Can't seem to get ##x## solely in terms of ##\lambda##, am I doing something wrong?

    Part(d)
    The metric for a flat, isotopic and homogeneous universe is given by
    [tex]ds^2 = -c^2 dt^2 + a(t)^2 \left[ d\chi^2 + S^2(\chi) \left( d\theta^2 + sin^2\theta d\phi^2 \right) \right] [/tex]

    Flux is given by ##F = \frac{L}{4\pi D_L^2}##. From the metric, proper area is given by ##A = 4\pi(a_0 \chi)^2 = 4\pi \chi^2##. But due to redshift, photons are delayed by ##\nu_0 = \frac{\nu_e}{1+z}##. Thus we have
    [tex]D_L = \chi(1+z)[/tex]
    where ##\chi## is the comoving distance.
    This is only in first order, how do I expand it in 2nd order?!
     
    Last edited: Apr 19, 2015
  2. jcsd
  3. Apr 20, 2015 #2
    Would appreciate help on parts (b) and (c), but I made a slight bit of progress on part (d).

    Part (d)
    We know that ##D_L = \chi (1+z)##. I now need to re-express ##\chi## in terms of ##z##. For a light-like geodesic,
    [tex]\chi = c \int \frac{1}{a(t)} dt [/tex]
    I read a useful trick is ##dz = d(1+z) = - \frac{\dot a}{a^2} dt = -(1+z) H(z) dz##. Substituting in,
    [tex] \chi = c \int_0^z \frac{1}{H(z)} dz [/tex]
    [tex]D_L = c(1+z) \int_0^z \frac{1}{H(z)} dz [/tex]
    Using ##H(z) = H_0 \left[ 1 + (1+q_0)z + \cdots \right] ##:
    [tex]D_L \approx \frac{c(1+z)}{H_0} \int_0^z \frac{1}{1 + (1+q_0)z} dz [/tex]
    [tex]D_L \approx \frac{c(1+z)}{H_0} \int_0^z \frac{1}{1 + (1+q_0)z} dz [/tex]
    [tex]D_L = \frac{c(1+z)}{H_0(1+q_0)} ln \left[ 1 + (1+q_0)z \right] [/tex]
    [tex] D_L \approx \frac{c(1+z)}{H_0(1+q_0)} \left[ (1+q_0)z - \frac{\left[ (1+q_0)z \right]^2}{2} \right] [/tex]
    [tex] D_L \approx \frac{c}{H_0}z \left[1 + \frac{z}{2}(1-q_0) \right] [/tex]
     
    Last edited: Apr 20, 2015
  4. Apr 22, 2015 #3
    bump on part (b)..
     
  5. Apr 23, 2015 #4
    bump part (b)
     
  6. Apr 24, 2015 #5
    Bump on part (b) - How do I get ##x## in terms of ##\lambda##?
     
  7. Apr 25, 2015 #6
    Would appreciate help with part (b) please
     
  8. Apr 26, 2015 #7
    any luck with solving ##x(\lambda)##?
     
  9. Apr 30, 2015 #8
    anyone had a go with part (b)?
     
  10. May 4, 2015 #9
    Still can't see how you can find ##p(\lambda)##..
     
  11. May 7, 2015 #10
    anyone else tried part (b) yet?
     
  12. May 10, 2015 #11
    bumpp part (b)
     
  13. May 14, 2015 #12
    bump on (b) and (c)
     
  14. May 16, 2015 #13
    bump on part (b) first
     
  15. May 18, 2015 #14
    bump on (b)
     
  16. May 22, 2015 #15
    part (b) bumping
     
  17. May 23, 2015 #16
    bump on part (b)
     
  18. May 25, 2015 #17
    bumpp on part (b)
     
  19. May 31, 2015 #18
    bumpp - I think there is a trick somewhere (dimensional analysis or something)
     
  20. Jun 1, 2015 #19
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