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Some Basic Stuff

  1. Feb 27, 2008 #1
    Can someone explain to me how to go from

    [tex]\Gamma ^k_{ij}=-\bold{e}_j \cdot D_i \bold{e}^k[/tex]


    [tex]D_i \bold{e}^k = \Gamma ^k_{ij} \bold{e}^j \ \cdot \ \ [/tex]
  2. jcsd
  3. Feb 28, 2008 #2
    Try this:
    [tex]\Gamma^k_{ij} = -\textbf{e}_j \cdot D_i\textbf{e}^k[/tex]
    [tex]\Gamma^k_{ij} = -D_i\textbf{e}^k \cdot \textbf{e}_j [/tex] because [tex] \textbf{v} \cdot \textbf{u} = \textbf{u} \cdot \textbf{v} [/tex]
    [tex]\Gamma^k_{ij} \textbf{e}^j = -D_i\textbf{e}^k \cdot \textbf{e}_j \textbf{e}^j[/tex]
    [tex]\Gamma^k_{ij} \textbf{e}^j = -D_i\textbf{e}^k \cdot \textbf{1}[/tex] because [tex] \textbf{e}_j \textbf{e}^j = \textbf{1} [/tex]
    [tex]\Gamma^k_{ij} \textbf{e}^j = -D_i\textbf{e}^k [/tex] because [tex] \textbf{e} \cdot \textbf{1} = \textbf{e} [/tex]
    Well, I got close, but I don't know how to drop the minus sign. For the identity tensor, I believe [tex] \textbf{1} \cdot \textbf{v} = \textbf{v} \cdot \textbf{1} [/tex] is true (though not true for other second rank tensors)
    Last edited: Feb 28, 2008
  4. Mar 16, 2008 #3
    I think you are right, Davidcantwell!
    \Gamma^k_{ij} \textbf{e}^j = -D_i\textbf{e}^k
    \Gamma^k_{ij} = -\textbf{e}_j \cdot D_i\textbf{e}^k
    The original statement seems not correct.
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