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Covariant derivative of covector

  1. Sep 28, 2014 #1
    I was trying to see what is the covariant derivative of a covector. I started with

    $$ \nabla_\mu (U_\nu V^\nu) = \partial_\mu (U_\nu V^\nu) = (\partial_\mu U\nu) V^\nu + U_\nu (\partial_\mu V^\nu) $$ since the covariant derivative of a scalar is the partial derivative of the latter.

    Then I know $$ \nabla_\mu (U_\nu V^\nu) = \nabla_\mu (U_\nu) V^\nu +U_\nu \nabla_\mu V^\nu$$ according to Leibniz rule.

    Setting them equal to each other I get $$ \nabla_\mu (U_\nu) V^\nu = (\partial_\mu U\nu) V^\nu - U_\nu (\Gamma ^\nu _{\mu\lambda} V^\lambda) $$

    Can this last term be equally written as $$U_\lambda(\Gamma ^\lambda _{\mu\nu} V^\nu)$$?? Because the proof was to get finally this form: $$ \nabla_\mu U_\nu = \partial_\mu U_\nu - \Gamma^\lambda_{\mu\nu}U_\lambda $$ I just got stuck at the last part. Thank you.
     
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  3. Sep 28, 2014 #2

    Fredrik

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    Yes, because both notations mean exactly the same thing. Compare this to the slightly simpler result ##\sum_{n=1}^2 x_n=x_1+x_2=\sum_{k=1}^2 x_k##.
     
  4. Sep 28, 2014 #3
    Thank you for replying to the thread. I didn't quite get the comparison you were pointing at. Is there some concrete property that allows us to do so?
     
  5. Sep 28, 2014 #4

    vanhees71

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    In your formula [itex]\lambda[/itex] and [itex]\nu[/itex] are both indices that are summed over, often called "dummy indices". So you can just rename them as you please. Then you use the fact that your formula holds for all vector-field components [itex]V^{\nu}[/itex], and thus you get the covariant derivative for the co-vector field as you wrote in the final step.
     
    Last edited: Sep 28, 2014
  6. Sep 28, 2014 #5
    Thank you, that was exactly what I wanted to make sure of.
     
  7. Sep 28, 2014 #6

    Fredrik

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    It follows immediately from the definition of the notation. ##U_\nu \Gamma ^\nu _{\mu\lambda} V^\lambda## is a sum of 16 terms that don't contain either of the indices ##\nu## or ##\lambda##.
    $$ U_\nu \Gamma ^\nu _{\mu\lambda} V^\lambda =U_0\Gamma^0_{\mu 0} V^0+U_1 \Gamma^1_{\mu 0} V^0+\cdots+U_3 \Gamma^3_{\mu 2} V^2+U_3 \Gamma^3_{\mu 3} V^3.$$ I illustrated this idea with an example with 2 terms instead of 16, because it's a pain to type 16 terms.
     
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