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Light cones in an expansive Universe

  1. Oct 8, 2014 #1
    Hi everyone!
    I'm solving some GR problems and I have a question.
    The problem is that we have a metric like the FRW metric but in 1+1 dimensions, i.e.:
    [tex] ds^2 = -dt \otimes dt + a(t)^2 dx \otimes dt[/tex]
    Where [itex]a(t)=t^{1/\epsilon}[/itex] (for [itex] t>0[/itex]) and [itex]a(t)=(-t)^{1/\epsilon}[/itex] (for [itex]t<0 [/itex]).
    We take a vector [itex] V^\mu = dx^\mu / d\lambda [/itex] and apply it to the metric as [itex] ds^2 (V,V)=0[/itex].
    For [itex]t>0 [/itex] we get:
    [tex]dt = \pm a(t) dx \rightarrow x(t) = \pm \frac{t^{1-1/\epsilon}}{1-1/\epsilon}[/tex]
    Ok, having all this said (until this point everything is correct, is part of the exercise) here's my problem.
    For [itex]\epsilon >1 [/itex] we have a decelerating expansion and we get, for instance for [itex] \epsilon = 2 [/itex]:
    [tex] t = x^2 [/tex]
    (Up to some constant I don't care)
    I can more or less understand this result, it's similar to the light cone in Minkowski's, and it gets more stretched with time as the expansion gets decelerated.

    Let's go to the case [itex] 0 < \epsilon < 1[/itex]. In this case, for instance for [itex] \epsilon = 1/2 [/itex], we get
    [tex] t=x^{-1} [/tex]
    I don't understand this case, the light-cone from the future and the past are not even conected.
    http://upload.wikimedia.org/wikipedia/commons/4/43/Hyperbola_one_over_x.svg
    Imagine this image but with the red lines in both side of the axis.

    Does anyone have an explanation to this fact? I don't see why the accelerating and the decelerating expansions give us so different light-cones!
     
  2. jcsd
  3. Oct 8, 2014 #2
    At the beginning I thought that the second case could be because a(0)=0 (even though t=0 is not defined...), so every observer is like conected at t=0... but then I wouldn't understand the first case! Because both are expansions, they should look like similar.
     
  4. Oct 8, 2014 #3

    Orodruin

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    Did you notice that you also happened to change sign on the ##\pm##? If you draw the light cone from any given point (including the constant you neglected), all of the past at t = 0 is going to be included. Similarly, for t smaller than 0, all space at t=0 will be in the future light cone. In particular, for t > 0, this means all of space for t < 0 is within the past light cone.
     
  5. Oct 8, 2014 #4
    Yeah, but why that doesn't happen in the first case? :(

    In addition, if we take [itex] \epsilon =1 \rightarrow a(t) = t[/itex], then:
    [tex] dt = \pm t dx \rightarrow \ln (t)=\pm x \rightarrow t = \exp (\pm x)[/tex]
    That case is also reaaally weird, as you have line cossing in the future and in the past...

    Thanks for the answer ;)
     
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