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FLRW metric

  1. Dec 18, 2013 #1
    The generic FLRW metric is dS^2 = a^2.(dx^2 + dy^2) - (c.dt)^2. Is it equivalent to the metric dS^2 = (dx^2 + dy^2) - (c.dt/a)^2 with the scale factor in the denominator of the time dimension? (I suppressed one dimension just for simplicity).

    Thanks for the help.
     
  2. jcsd
  3. Dec 18, 2013 #2

    PAllen

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    That's not the generic FLRW metric. That's only for the metric for the spatially flat case. Please specify whether you want to restrict attention to this special case.
     
  4. Dec 19, 2013 #3
    Yes, it can be only for the flat case. My point is: in order to get the metric growing either I can keep the space component growing or the time component decreasing and that is the reason of my question if the are equivalent or not.
     
  5. Dec 19, 2013 #4

    fzero

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    A metric of the form

    $$ds^2 = - \left( \frac{c dt}{A(t)}\right)^2 + ds_3^2 ~~~~(*)$$

    is a flat metric, which you can see by defining a new variable

    $$ t' = \int^{t'} \frac{c dt}{A(t)},$$

    so that

    $$ds^2 = - (dt')^2 + ds_3^2. $$

    Therefore there is no choice of ##A(t)## such that this metric is equivalent to the FLRW metric.

    $$ ds^2 = - c^2 dt^2 + a(t)^2 ds_3^2 .$$

    If we were dealing with the open or closed versions of ##ds_3^2## in (*), we'd find that the curvature of the 4-metric was independent of ##A(t)## and was just given by the spatial curvature.

    An equivalent form to the FLRW metric is

    $$ ds^2 = a(\tau)^2 \left( - d\tau^2 + ds_3^2 \right),$$

    which can be obtained by defining the so-called conformal time

    $$ \tau = \int^{\tau} \frac{c dt}{a(t)}.$$

    The problem with the form that you postulate is that the coordinate transformation you have to make to remove the ##a(t)## from the spatial part involves mixing the time variable with the radial variable:

    $$ r' = a(t) r.$$

    However, now ##dr'## involves a term with ##dt## and you'll inevitably find cross terms in the transformed metric of the form ##dt' dr'## that don't vanish (these terms preserve the curvature). There doesn't appear to be an appropriate choice of ##t'## such that you end up with the simple expression that you're hoping for, even if we were willing to let the function ##A=A(r)##.
     
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