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FRW Metric: time component

  1. Apr 7, 2015 #1
    Taken from Hobson's book:

    frw2.png

    Metric is given by
    [tex]ds^2 = c^2 dt^2 - R^2(t) \left[ d\chi^2 + S^2(\chi) (d\theta^2 + sin^2\theta d\phi^2) \right] [/tex]

    Thus, ##g_{00} = c^2, g_{11} = -R^2(t), g_{22} = -R^2(t) S^2(\chi), g_{33} = -R^2(t) S^2(\chi) sin^2 \theta##.

    Geodesic equation is given by:
    [tex] \dot u_\mu = \frac{1}{2} \left( \partial_\mu g_{v\sigma} \right) u^v u^\sigma [/tex]

    The coordinates are given by ##u^0 = \dot t, u^1 = \dot \chi, u^2 = \dot \theta, u^3 = \dot \phi##.

    For the temporal component,
    [tex]\dot u_0 = \frac{1}{2} (\partial_0 g_{v\sigma})u^v u^\sigma[/tex]

    Photons

    [tex]u^0u_0 = 0[/tex]
    [tex]u^0 g_{00} g^0 = 0 [/tex]
    [tex]g_{00}\dot t^2 = 0 [/tex]
    [tex]\dot t = 0[/tex]

    This doesn't make any sense. For massive particles, ##\dot t = 1##.
     
  2. jcsd
  3. Apr 7, 2015 #2

    Orodruin

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    Why do you think ##u_0 u^0 = 0##? It is not true. Remember the Einstein summation convention.
     
  4. Apr 7, 2015 #3
    For a photon, it is 0, as shown in the text. (null vector)
     
  5. Apr 7, 2015 #4

    Orodruin

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    No, it is not ##u^\mu u_\mu = 0## does not imply ##u^0 u_0 = 0##.
     
  6. Apr 7, 2015 #5
    So, ##u^0u_0 + u^1u_1 + u^2u_2 + u^3u_3 = 0##?
     
  7. Apr 7, 2015 #6

    Orodruin

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    Yes.
     
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