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Origin of spatial metric

  1. Aug 21, 2008 #1
    Can anyone explain to me the origin of the spatial metric for measuring distances in non-inertial frames?

    [itex]d\ell^2 = [(g_{0i}g_{0j})/g_{00} - g_{ij}]dx^i dx^j[/itex].

    I've heard it quoted but never seen it derived. I believe it works on the assumption that distance is half the proper time for return of an em signal (whatever that means).

  2. jcsd
  3. Aug 21, 2008 #2
    My best guess:

    [itex]ds^2 = g_{00} c^2 dt^2 + g_{ij}dx^{i} dx^{j} + 2g_{0i} cdt dx^i [/itex]

    [itex]c^2 d\tau^2 = g_{00} dt^2[/itex]

    On null geodesics

    [itex]0 = g_{00} c^2 dt^2 + g_{ij}dx^{i} dx^{j} + 2g_{0i} cdt dx^i \implies[/itex]
    [itex] c dt = -\frac{g_{0i} dx^i}{g_{00}} \pm \frac{\sqrt{g_{0i}g_{0j}dx^i dx^j - g_{00} g_{ij} dx^i dx^j }}{g_{00}}[/itex] (quadratic equation)

    Now [itex]d\ell =c\frac{1}{2}d\tau = c\frac{1}{2}\sqrt{g_{00}}(dt_++dt_-) = \frac{g_{0i}dx^i}{g_{00}} \implies[/itex]

    [itex]d\ell^2 = \frac{g_{0i}g_{0j}dx^i dx^j}{g_{00}}[/itex]

    which is wrong.
  4. Aug 21, 2008 #3


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    I've never seen this before, but a little thought suggests the following.

    I think we have to assume that the curves [itex]dx^1=dx^2=dx^3=0[/itex] represent worldlines of observers and I think the metric you quoted measures distance within the surfaces that are orthogonal to those worldlines.

    If [itex]U^{\alpha}[/itex] is parallel to the 4-velocity of such an observer, with components (1, 0, 0, 0), consider decomposing [itex]dx^{\alpha}[/itex] as

    [tex]dx^{\alpha} = dP^{\alpha} + dQ^{\alpha}[/tex]​

    where [itex]dP^{\alpha}[/itex] is parallel to [itex]U^{\alpha}[/itex] and [itex]dQ^{\alpha}[/itex] is orthogonal to it. The orthogonality ensures that

    [tex]dx_{\alpha}dx^{\alpha} = dP_{\alpha}dP^{\alpha} + dQ_{\alpha}dQ^{\alpha}[/tex]​

    The required projection is given by

    [tex]dP^{\alpha} = \frac{U_{\beta}U^{\alpha}}{U_{\gamma}U^{\gamma}}dX^{\beta}[/tex]​

    and then I think it all follows from that, noting that

    [tex]U_{\beta} = g_{\beta \alpha} U^{\alpha} = g_{\beta 0}[/tex]​

    and the answer you want is [itex]d\ell^2 = dQ_{\alpha}dQ^{\alpha}[/itex].

    That would be "radar distance" which amounts to the same thing infinitesimally, but not over larger distances (except in flat spacetime).
  5. Aug 21, 2008 #4


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    [tex]d\ell^2 = [(g_{0i}g_{0j})/g_{00} - g_{ij}]dx^i dx^j[/tex]

    It is the radar infinitismal spatial distance. There's a good derivation in appendix A of the attached.

    DrGreg's derivation is along the right lines but I haven't checked it.


    Attached Files:

  6. Aug 21, 2008 #5
    Hi DrGreg,

    You're correct, the metric is actually the same thing as the projection operator [itex]U^\alpha U^\beta - g^{\alpha\beta}[/itex], written in the frame where the observer has vanishing 3-velocity.

    The error in my derivation of radar distance was to add [itex]dt_+,dt_-[/itex] when I should have subtracted them. This is because one of them gives a time into the past which is negative.
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