Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Riemann curvature tensor as second derivative of the metric

  1. Aug 23, 2008 #1
    It is a standard fact that at any point [itex]p[/itex] in a Riemannian space one can find coordinates such that [itex]\left.g_{\mu\nu}\right|_p = \eta_{\mu\nu}[/itex] and [itex]\left.\partial_\lambda g_{\mu\nu}\right|_p[/itex].

    Consider the Taylor expansion of [itex]g_{\mu\nu}[/itex] about p in these coordinates:

    [itex]g_{\mu\nu} = \eta_{\mu\nu} + \frac{1}{2!} (\partial_\lambda\partial_\sigma g_{\mu\nu})(x^\lambda - x_p^\lambda)(x^\sigma-x^\sigma_p) + \cdots[/itex].

    The claim is that in fact [itex]R_{\mu\lambda \nu\sigma} = \partial_\lambda\partial_\sigma g_{\mu\nu}[/itex]. The problem is that I'm not sure the Riemann curvature tensor has these symmetries.
     
  2. jcsd
  3. Aug 23, 2008 #2
    The expression I gave before is not quite complete, I should have written

    [itex](\partial_\lambda\partial_\sigma g_{\mu\nu})_p = \frac{2}{3} R_{\mu\lambda \nu\sigma}|_p[/itex]
     
  4. Aug 23, 2008 #3
    After thinking about this a little more I realize that the correct expression is in fact

    [itex](\partial_\lambda\partial_\sigma g_{\mu\nu})_p = \frac{1}{3} (R_{\mu\lambda \nu\sigma}+ R_{\mu\sigma\nu\lambda})|_p[/itex]

    which implies

    [itex]g_{\mu\nu} = \eta_{\mu\nu} + \frac{1}{3} (R_{\mu\lambda\nu\sigma})_p(x^\lambda - x_p^\lambda)(x^\sigma-x^\sigma_p) + \cdots[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Riemann curvature tensor as second derivative of the metric
Loading...