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Help deriving Lagrange's Formula with the levi-civita symbol

  1. Feb 3, 2008 #1
    Ok, so I'm really at a loss as to how to do this. I can prove the formula by just using determinants, but I don't really know how to do manipulations with the levi-civita symbol.
    Here's what I have so far:
    [tex]
    (\vec{B} \times \vec{C})_{i} = \epsilon_{ijk}(B_{j}C_{k})\vec{e_{i}}
    [/tex]

    And I'm trying to get to:
    [tex]
    \vec{A} \times (\vec{B} \times \vec{C}) = B(A \bullet C) - C(A \bullet B)
    [/tex]

    Does anyone have any suggestions?
    Thanks
     
  2. jcsd
  3. Feb 4, 2008 #2
    Ok So I figured it out, I'll just post the answer for the sake of completeness:

    [tex]
    (\vec{B} \times \vec{C})_{k} = \epsilon_{kmn} (B_{m} C_{n})
    [/tex]

    [tex]
    let (\vec{B} \times \vec{C}) = \vec{N}
    [/tex]

    [tex]
    \vec{A} \times (\vec{B} \times \vec{C}) = \vec{A} \times \vec{N}
    [/tex]
    [tex]
    (\vec{A} \times \vec{N})_{i} = \epsilon_{ijk} A_{j} N_{k}
    [/tex]
    [tex]
    = \epsilon_{ijk} A_{j} (\epsilon_{kmn} B_{m} C_{n}) [/tex]
    [tex]
    = \epsilon_{ijk} \epsilon_{kmn} (A_{j} B_{m} C_{n}) [/tex]

    [tex]
    \epsilon_{ijk} \epsilon_{mnk} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}
    [/tex]

    [tex]
    \vec{A} \times (\vec{B} \times \vec{C}) = (\delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}) A_{j} B_{m} C_{n}
    [/tex]
    [tex]
    = B_{i} A_{j} C_{j} - A_{j} B_{j} C_{i}
    [/tex]
    [tex]
    = \vec{B}(\vec{A} \bullet \vec{C}) - \vec{C}(\vec{A} \bullet \vec{B}) [/tex]
     
  4. Feb 4, 2008 #3
    It's threads like these that seem to be causing PF to accumulate helpful Google searches. :)
     
  5. Feb 6, 2009 #4

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    [tex]

    (\vec{B} \times \vec{C})_{i} = \epsilon_{ijk}(B_{j}C_{k})\vec{e_{i}}

    [/tex]

    Should be:

    [tex]

    (\vec{B} \times \vec{C})_{i} = \epsilon_{ijk}(B_{j}C_{k})

    [/tex]

    And start using \cdot instead of that big black ball :-D
     
  6. Feb 7, 2009 #5
    Haha wow this seems like so long ago. I couldn't find the dot for dot product, so thanks for that :D
     
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