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Second order expansion of metric in free-fall

  1. Dec 22, 2012 #1

    I have read that, in a freely-falling frame, the metric/ interval will be of the form:

    ds2 = -c2dt2(1 + R0i0jxixj) - 2cdtdxi([itex]\frac{2}{3}[/itex] R0jikxjxk) + (dxidxjij - [itex]\frac{1}{3}[/itex] Rikjlxkxl)

    to second order.

    Does anyone know where I could find a derivation of this result?
  2. jcsd
  3. Dec 22, 2012 #2


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    InsertName, That's an interesting formula! It's derivation is almost immediate, except I have doubts about the absence of the time coordinate, and the factors of 1/3.

    Expand the metric in a Taylor's series:
    gμν = Aμν + Bμνσxσ + Cμσντxσxτ + ...
    It's always possible to choose coordinates such that Aμν = ημν and Bμνσ = 0, and in these coordinates the Christoffel symbols vanish. Then the formula for the Riemann tensor reduces to
    Rμσντ = ½(gμτ,σν + gσν,μτ - gμν,στ - gστ,μν) = Cμστν + Cτνμσ - Cμσντ - Cσμτν.
    If you assume C to have the same symmetry as the Riemann tensor, then this is 4Cμστν, showing that Cμστν = (1/4)Rμσντ
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