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Homework Help: Change in Redshift vs. Obs. Time

  1. May 31, 2006 #1
    Greetings--this is from Ryden's Introduction to Cosmology, question # 5.2. Let me restate the question:

    Here w is the equation of state parameter for whatever the universe is made of (e.g. w = 0 for matter, 1/3 for radiation).

    Here is my best attempt so far:
    Simplify notation by introducing [tex]y=\frac{2}{3(1+w)}[/tex]

    Then (eq. 5.51 in Ryder):
    [tex]1+z = \left(\frac{t_0}{t_e}\right)^y[/tex]
    Where [tex]t_e[/tex] is the time at which the light was emitted.

    Also (eq. 5.48 in Ryder):
    [tex]t_0 = \frac{y}{H_0}[/tex]

    So differentiating the first equation w/rt [tex]t_0[/tex], we get:

    [tex]\frac{dz}{dt_0} = yt_0^{y-1}t_e^{-y} - y t_e^{-y-1}t_0^y \frac{dt_e}{t0}[/tex]

    [tex]\frac{dz}{dt_0} = H_0(1+z) - y t_e^{-1} (1+z) \frac{dt_e}{t0}[/tex]

    But [tex]\frac{dt_e}{t0} = (1+z)^{-1/y}[/tex] from our equation (5.51) above. Equation (5.52) in Ryder also tell us:

    [tex]t_e = \frac{y}{H_0}(1+z)^{-1/y}[/tex]

    So plugging these two in:
    [tex]\frac{dz}{dt_0} = H_0(1+z) - y\left(\frac{H_0}{y}(1+z)^{1/y}\right)(1+z)(1+z)^{-1/y}[/tex]

    Which becomes identically zero! Where am I making my mistake?

  2. jcsd
  3. Jun 10, 2006 #2


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    Homework Helper

    Is this one of those derivatives like you get in thermodynamics where the answer depends on whether you are holding [tex]t_e[/tex] constant? And you end up using something like

    [tex]\frac{dz}{dt_0}\frac{dt_0}{dt_e}\frac{dt_e}{dz} = -1 ?[/tex]

    These always confuse me.

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