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What valence tensor produced by the absolute derivative?

  1. Apr 17, 2015 #1
    Acting upon a vector say,
    so it is defined as:
    ##\frac{d}{d\lambda}V^{u}+\Gamma^{u}_{op}\frac{dx^{o}}{d\lambda}V^{p}=\frac{DV^{u}}{D\lambda}##

    And this can also be written in terms of the covariant derivative, ##\bigtriangledown_{k}## by ##\frac{DV^{u}}{D\lambda}=\frac{d x^{k}}{d \lambda} \bigtriangledown_{k}V^{u}## [1]

    So I know that the covariant derivaite takes a tensor of valance ##(p,q)## to ##(p,q+1)##

    So using this and the fact that the RHS of [1] has a vector multiplied by a covariant derivative acting upon a vector I would conclude that it takes a ##(1,0)## to a ##(1,1)##. Is this correct?

    However my textbook states '##\frac{DV^{a}}{Du}## transforms as a covariant vector, which I dont see,

    Thanks.
     
  2. jcsd
  3. Apr 17, 2015 #2

    Orodruin

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    ##\nabla_k V^\mu## is a type (1,1) tensor, but in ##DV^\mu/D\lambda## it is contracted with the vector ##\dot x^k## and therefore becomes a (1,0) tensor.
     
  4. Apr 17, 2015 #3
    Ahh I see thanks. And this is a contravariant not a covariant though? And typo in the book?
     
    Last edited: Apr 17, 2015
  5. Apr 17, 2015 #4

    Orodruin

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    Yes, it is contravariant. You can check this by checking the transformation properties.
     
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