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The covariant derivative of a contravariant vector

  1. Jan 21, 2005 #1
    Since there are some equations in my question. I write my question in the following attachment. It is about the covariant derivative of a contravariant vector.

    Thank you so much!
     

    Attached Files:

    Last edited: Jan 22, 2005
  2. jcsd
  3. Jan 22, 2005 #2
    Can anyone give me some guidance?
    Thanks
     
  4. Jan 22, 2005 #3

    dextercioby

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    Okay.
    1.Those Gamma's components are not zero...Not in the general case,anyway...

    2.I'll use the column-semicolumn notation (though we physicst are not really fond of it...)
    In the following,"g" is the determinant of the metric tensor:
    [tex] g_{,i}=g \ g^{kl} g_{kl,i} [/tex] (1)
    (1):This is the rule as how to differentiate the determinant of a matrix...

    [tex] A^{i}_{;i}=A^{i}_{,i}+\Gamma^{i} \ _{ij}A^{j} [/tex](2)

    (2):The covariant divergence (the one u're interested in).

    [tex] \Gamma^{i} \ _{ij} =\frac{1}{2}g^{ki}(g_{kj,i}+g_{ik,j}-g_{ji,k})
    =\frac{1}{2}g^{ki}g_{ki,j}=\frac{1}{2g}g_{,j} [/tex] (3)

    In getting (3) I made use of the definition of 2-nd rank Christoffel symbols (mannifold with both connection & metric) and of relation (1).

    Use (3) and (2) and the fact that:
    [tex] g=(\sqrt{g})^{2} [/tex] (4)

    to get your result.

    Report any problems...

    Daniel.
     
    Last edited: Jan 22, 2005
  5. Jan 24, 2005 #4
    Thank you for your reply.
    But from your result (3), there should be 3 terms in the following equation's second part of right hand side.
    [tex] A^{i}_{;i}=A^{i}_{,i}+\Gamma^{i} \ _{ij}A^{j} [/tex]
    Then the result is not the same with my results.
     
  6. Jan 24, 2005 #5

    dextercioby

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    What 3 terms are u talking about??The ones in the definition of Christoffel's symbols...??

    Daniel.
     
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