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What is my proper time?

  1. Oct 27, 2012 #1
    As an observer who is simply travelling with the Hubble flow is my proper time the same as the cosmological time or is it equal to the conformal time?
     
  2. jcsd
  3. Oct 27, 2012 #2

    mfb

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    If you are at rest relative to the CMB, otherwise you will get a different value. For typical velocities relative to the CMB (just the motion of galaxies, not relativistic spacecrafts), the difference is very small compared with the current timing uncertainties.
     
  4. Oct 28, 2012 #3
    I think I am a free-falling co-moving observer rather than simply a co-moving observer.

    My local spacetime should therefore be flat - in other words my local frame is inertial.

    Starting with the FRW metric with cosmological time [itex]t[/itex] and co-moving spatial co-ordinates:

    [itex] \large ds^2 = -dt^2 + a(t)^2 [ \frac{dr^2}{1-kr^2} + r^2(d\theta^2+\sin^2\theta d\phi^2)] [/itex]

    I rewrite the FRW metric with conformal time [itex]\tau[/itex] and co-moving spatial co-ordinates:

    [itex] \large ds^2 = a(t)^2(-d\tau^2 + \frac{dr^2}{1-kr^2} + r^2(d\theta^2+\sin^2\theta d\phi^2)) [/itex]

    where an element of conformal time [itex]d\tau[/itex] is given by

    [itex] \large d\tau = \frac{dt}{a(t)} [/itex]

    The worldline of a radial lightbeam according to the re-written metric is given by

    [itex] \large d\tau = \frac{dr}{\sqrt{1-kr^2}}[/itex]

    For small [itex]r[/itex] this metric describes a locally flat spacetime in which light travels on diagonals on a spacetime diagram. This is consistent with the co-ordinate system of a free-falling observer with a local inertial frame.

    Therefore I think my proper time is conformal time.
     
    Last edited: Oct 28, 2012
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