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

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.
  2. jcsd
  3. Sep 28, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    2016 Award

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Covariant derivative of covector
  1. Covariant derivative (Replies: 13)

  2. Covariant Derivative (Replies: 1)