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Is this correct? (Tensor)

  1. Oct 11, 2007 #1
    delta x (a x b) = (b . delta) a - b (delta . a) + a (delta . b) - (a . delta) b
    all terms are in vectors, so delta x means curl.
    Can anybody prove that?
    Because I have tried to prove it, but it keeps failing.
    Please help me to figure this out...
    Thanks a lot...
  2. jcsd
  3. Oct 11, 2007 #2
    Do it in component form :

    [tex]\big( \nabla \times A \big)_{i} = \epsilon_{ijk} \partial_{j} A_{k}[/tex] where [tex]A_{k} = \epsilon_{klm} a_{l}b_{m}[/tex], so

    [tex]\big( \nabla \times ( a \times b) \big)_{i} = \epsilon_{ijk} \partial_{j} (\epsilon_{klm} a_{l}b_{m})[/tex]

    Remember the identity [tex]\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}[/tex] and just do some clearing up to get your answer.
  4. Oct 11, 2007 #3
    But, with that, there will be only 2 terms left, instead of 4...
    Is that true?
    And what will happen with the other 2 terms?
    I'm confused.
    Thanks a lot...
  5. Oct 11, 2007 #4
    Use the Leibnitz rule on [tex]\partial_{j}(a_{l}b_{m})[/tex], that'll double the number of terms.
  6. Oct 11, 2007 #5
    I've never heard about that. Can you please explain a bit?
  7. Oct 11, 2007 #6


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    With ordinary functions it's also called the product rule:
    [tex]\frac{d}{dx}(f g) = g \frac{df}{dx} + f \frac{dg}{dx}.[/tex]
    In this case,
    [tex]\partial_j(a_l b_m) = a_l \partial_j b_m + b_m \partial_j a_l[/tex]
  8. Oct 11, 2007 #7
    If you're being asked to do vector calculus identities, you must have been taught the Liebnitz rule?! It's one of the most important properties of a derivative.

    Are you learning just from books? Because if you're just picking up a random book on calculus and geometry, you might be missing the essential requirements by skipping over the prerequesite books.

    There's no point doing calculus in n dimensions if you can't do it in 1.
  9. Oct 11, 2007 #8
    Oh yes, I forgot the name, but I remember that equation...
    Thanks... :)
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