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Lorentz invariance of an equation (metric)

  1. May 23, 2013 #1
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  2. jcsd
  3. May 24, 2013 #2

    fzero

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    I'm not sure what you mean by "metric is invariant." The metric transforms like a tensor, covariant or contravariant depending on whether the indices are up or down. The line element ##ds^2 = g_{\mu\nu} dx^\mu dx^\nu## is Lorentz invariant, but that's not what appears in your formula.

    You should write down the explict transformation rules for ##V^\mu##, ##\partial/\partial x^\mu##, ##g^{\nu\sigma}##, and ##g_{\nu\sigma}##. You will need to combine them in your expression and carefully apply the derivatives with the appropriate product rules for differentiation.
     
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