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Horizon size of a photon in a matter dominated universe

  1. Jun 11, 2013 #1
    1. The problem statement, all variables and given/known data

    The maximum proper distance a photon can travel in the interval (0,t) is given by the horizon size

    h(t) = R(t) ∫0t dt' / R(t')

    Show that, for a matter dominated universe

    h(z) = H0-1(1+z)-1(Ω-1)-1/2cos-1(1-2(Ω-1)/(Ω(1+z))) for Ω>1
    = 2H0-1(1+z)-3/2 for Ω=1
    = H0-1(1+z)-1(Ω-1)-1/2cosh-1(1+2(Ω-1)/(Ω(1+z))) for Ω<1

    Also show that dH ≈ 3H0-1Ω0-1/2(1+z)-3/2 for (1+z)>>Ω-1

    2. Relevant equations

    As is a matter dominated universe, can assume:
    Ωm = 1, Ωr = 0, Ωλ = 0

    dH(z) = c H0-1((1-Ω)(1+z)2m(1+z)3)-1/2

    dt = -dH(z)dz/(c(1+z))

    cosh(ix) = cos(x)

    cosh(x) = (ex+e-x)/2

    3. The attempt at a solution

    For the Ω=1 case, which seems to me as if it should be the simplest, my main problem seems to be with the R(t'). I don't really know how I'm supposed to integrate that! Anyway, this is what I've tried thus far:

    dH(z) = c H0-1(1+z)-3/2

    0t dt = -H0-1z (1+z)-5/2
    = 2(1+z)-3/2/(3H0)

    problem is I haven't taken into consideration R(t') as I really don't know how. I know R(t0)/R(te) = z+1 but I don't think that can be used here. And even if it was I end up with:

    h(z) = 2H0-1 (1+z)-1/2

    which is wrong anyway.

    At this point in time I'm only trying to get this part of the question out, as I believe it will make the other parts easier.

    If you have any ideas on what I'm doing wrong or what I seem to be missing, that would be greatly appreciated.

  2. jcsd
  3. Jun 11, 2013 #2
    Did you remember the extra factor of a in the front of the integral?
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