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General relativity, geodesic, KVF, chain rule covariant derivatives

  1. Jun 25, 2017 #1
    1. The problem statement, all variables and given/known data

    To show that ##K=V^uK_u## is conserved along an affinely parameterised geodesic with ##V^u## the tangent vector to some affinely parameterised geodesic and ##K_u## a killing vector field satisfying ##\nabla_a K_b+\nabla_b K_a=0##

    2. Relevant equations
    see above
    3. The attempt at a solution

    Proof:

    So the proof is to use the chain rule that ##\frac{d}{ds}= ## , where ##s## is some affine parameter:

    ## \implies \frac{dK}{ds}=V^{\alpha}\nabla_{\alpha}(K_uV^u)= V^u V^{\alpha}\nabla_{\alpha}K_u + K_u V^{\alpha}\nabla_{\alpha}V^u## ; first term is zero from KVF equation - antisymetric tensor multiplied by a symmetric tensor, and the second term is zero from the geodesic equation

    MY QUESTION - this may be a stupid question, but concerning the order of the chain rule application since the covariant derivative operates on everything to the right...

    How do you know to write ##\frac{d}{ds}=V^{u}\nabla_u## as a pose to ##\frac{d}{ds}=\nabla_uV^u##

    In normal calculus when you use the chain rule, the order doesn't matter does it? For e.g ## \frac{d}{ds}=\frac{dx}{ds}\frac{d}{dx} = \frac{d}{dx}\frac{dx}{ds} ## ?

    But if i try to apply the above proof writing ##\frac{d}{ds}=\nabla_uV^u## I get an extra non-zero term : ## K_uV^u\nabla_{\alpha}V^{\alpha}##

    so the proof fails.

    Many thanks
     
  2. jcsd
  3. Jun 26, 2017 #2

    Orodruin

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    Yes it does.
     
  4. Jun 26, 2017 #3
    I never knew this, what is the order?
     
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