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General Relativity

  1. Nov 6, 2009 #1
    I'm reading Chapter 5.3b in Wald's "General Relativity".

    (i) On p105, it says that if the integral [itex]t=\int \frac{d \tau}{a(\tau)}[/itex] converges, the Robertson Walker will be conformally related to a section of Minkowski spacetime above a t=constant hypersurface. This makes sense to me...it's kind of like saying that you can't see beyond a certain point in the history of the universe because the light from there hasn't reached you yet.

    However, in Figure 5.6, which supposedly demonstrates this, I can't figure out where the t=constant surface would be drawn on that diagram. As far as I can tell the observer would be able to see everything in their past light cone right down to the big bang singularity. The section of the universe invisible to them is described not by a t=constant surface but rather by the particle horizon defined by the light cone boundary.

    I'm clearly missing something here, because both of these can't be right.

    (ii) At the bottom of the last paragraph on p105 it says that [itex]a(\tau} \alpha \tau^{\frac{2}{3}}[/itex] for dust and [itex]a(\tau)[/itex] is larger for P>0 i.e. radiation. However in Table 5.1 on p98, [itex]a(\tau) \alpha \tau^{\frac{1}{2}}[/itex] for radiation and so since [itex]\tau^{\frac{1}{2}} \leq \tau{\frac{2}{3}}[/itex] isn't this statement false?
    Then at the top of p106 it says taht because of this the integral will be convergent. However on p105 we were told that the integral would be divergent if [itex]a(\tau) \leq k \tau[/itex] for some constant k. Clearly, [itex]\tau^{\frac{2}{3}},\tau^{\frac{1}{2}} \leq k \tau[/itex] so surely the integral should diverge?

    thanks.
     
  2. jcsd
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