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Rabcd = 0 proof

  1. Apr 18, 2007 #1
    "Show that if a space time metric admits three linearly independent 4 vector fields with vanishing covariant derivatives then Rabcd = 0"

    We can set the three vectors as (1,0,0,0), (0,1,0,0) and (0,0,1,0). Use covariant derivative of vector field X^b is:

    d(X^b)/d(x^a) + (Christoffel symbol with superscript b and subscripts a, c)* (X^c)

    where the derivative above is partial.

    Therefore the following Christoffel symbols are zero:

    (superscript b, subscripts a,0)
    (superscript b, subscripts a,1)
    (superscript b, subscripts a,2)

    Assume that the Christoffel symbols are symmetric (for a symmetric gab), therefore we know that only the Christoffel symbol with both subscripts equal to 3 can be non zero, i.e

    (superscipt b, subscripts 3,3)

    At this point I get stuck.
  2. jcsd
  3. Apr 18, 2007 #2


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    Science Advisor
    Homework Helper

    You are being really casual about just slinging around vector components for someone who hasn't even said what the coordinate system is. You should also know by now that it's not true that any set of vector fields can be treated as corresponding to the partial derivatives of some coordinate system. (The partial derivatives commute - general vector fields don't). So stop being sloppy and use a coordinate independent definition of the Riemann tensor - one where it is defined in terms of the covariant derivative.
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