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Covariant derivative transformation

  1. Jun 21, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem concerns how to transform a covariant differentiation. Using this formula for covariant differentiation and demanding that it is a (1,1) tensor:
    [tex]
    \nabla_cX^a=\partial_cX^a+\Gamma^a_{bc}X^b
    [/tex]
    it should be proven that
    [tex]
    \Gamma'^a_{bc}=
    \frac{{\partial}x'^a}{{\partial}x^d}
    \frac{{\partial}x^e}{{\partial}x'^b}
    \frac{{\partial}x^f}{{\partial}x'^c}
    \Gamma^d_{ef}
    -
    \frac{{\partial}x^d}{{\partial}x'^b}
    \frac{{\partial}x^e}{{\partial}x'^c}
    \frac{{\partial}^2x'^a}{{{\partial}x^d}{{\partial}x^e}}
    [/tex]

    2. Relevant equations
    [tex]
    \partial'_cX'^a=
    \frac{{\partial}x^f}{{\partial}x'^c}\frac{\partial}{{\partial}x^f}
    (
    \frac{{\partial}x'^a}{{\partial}x^d}X^d
    )=
    \frac{{\partial}x'^a}{{\partial}x^d}
    \frac{{\partial}x^f}{{\partial}x'^c}
    \frac{\partial}{{\partial}x^f}
    X^d
    +
    \frac{{\partial}^2x'^a}{{\partial}x^f{\partial}x^d}
    \frac{{\partial}x^f}{{\partial}x'^c}
    X^d
    [/tex]

    [tex]
    X^d
    =
    \frac{{\partial}x^d}{{\partial}x'^b}
    \frac{{\partial}x'^b}{{\partial}x^d}
    X^d
    =
    \frac{{\partial}x^d}{{\partial}x'^b}
    X'^b
    [/tex]
    [tex]
    X^e
    =
    \frac{{\partial}x^e}{{\partial}x'^b}
    X'^b
    [/tex]

    3. The attempt at a solution
    I thought I'd simply look at the primed coordinates from two directions and that that should do it... didn't quite happen as planned though.

    In primed coordinates
    [tex]
    \nabla_cX^a'
    =
    {\partial}'_cX'^a
    +
    \Gamma^a_{bc}'X'^b
    [/tex]

    Transformed using the equations under 2.
    [tex]
    \nabla_cX^a'
    =
    \frac{{\partial}x'^a}{{\partial}x^d}
    \frac{{\partial}x^f}{{\partial}x'^c}
    \frac{\partial}{{\partial}x^f}
    X^d
    +
    \frac{{\partial}x'^a}{{\partial}x^d}
    \frac{{\partial}x^f}{{\partial}x'^c}
    \Gamma^d_{ef}X^e
    =
    {\partial}'_cX'^a
    -
    \frac{{\partial}^2x'^a}{{\partial}x^f{\partial}x^d}
    \frac{{\partial}x^f}{{\partial}x'^c}
    X^d
    +
    \frac{{\partial}x'^a}{{\partial}x^d}
    \frac{{\partial}x^f}{{\partial}x'^c}
    \Gamma^d_{ef}X^e
    =
    {\partial}'_cX'^a
    -
    \frac{{\partial}^2x'^a}{{\partial}x^f{\partial}x^d}
    \frac{{\partial}x^f}{{\partial}x'^c}
    \frac{{\partial}x^d}{{\partial}x'^b}
    X'^b
    +
    \frac{{\partial}x'^a}{{\partial}x^d}
    \frac{{\partial}x^f}{{\partial}x'^c}
    \Gamma^d_{ef}
    \frac{{\partial}x^e}{{\partial}x'^b}
    X'^b
    [/tex]

    So my thought was that anything related to X'^b would be the what I seek which gives

    [tex]
    \Gamma'^a_{bc}=
    \frac{{\partial}x'^a}{{\partial}x^d}
    \frac{{\partial}x^e}{{\partial}x'^b}
    \frac{{\partial}x^f}{{\partial}x'^c}
    \Gamma^d_{ef}
    -
    \frac{{\partial}x^d}{{\partial}x'^b}
    \frac{{\partial}x^f}{{\partial}x'^c}
    \frac{{\partial}^2x'^a}{{\partial}x^f{\partial}x^d}
    [/tex]

    This i kind of close but it's different in the last part where I have f:s instead of e:s. I would really appreciate some help, I've been stuck for a long time.
     
  2. jcsd
  3. Jun 21, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi faklif! :smile:

    Do you mean you're worried about having ∂xf and /∂xf instead of ∂xe and /∂xe ?

    They're "dummy variables" … it doesn't matter what they are! :wink:
     
  4. Jun 21, 2009 #3
    Hi and thank you!

    Now that I think some more about it (should have done that before spending the better part of the weekend :wink:, having to much spare time makes me stupid...) it makes sense. For a given a,b,c you'd still get the same implied summations. Thanks again for rescuing my monday!
     
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