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I Prove ∇(A x B) = (∇ x A)⋅B - (∇ x B)⋅A where A,B are vectors

  1. Oct 22, 2016 #1
    I can prove this relationship by defining A = (A1,A2,A3) and B=(B1,B2,B3) and expanding but I tried another approach and failed.

    I read that for any 3 vectors,
    a⋅(b x c) = (a x b)⋅c
    and thus applying this to the equation, I only get
    (A x B) = ( x A)⋅B
    Can anyone explain why this is so?
     
  2. jcsd
  3. Oct 22, 2016 #2

    Charles Link

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    That should be the divergence operator ## \nabla \cdot (A \times B ) ## ,and the divergence operator behaves quite differently from a vector dot product.
     
  4. Oct 22, 2016 #3

    fresh_42

    Staff: Mentor

    Your mixture of products is wrong. Either use
    ##\vec{a} \times (\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\cdot\vec{b}-(\vec{a}\cdot\vec{b})\cdot\vec{c}\;## (Graßmann-identity) or
    ##\vec{a}\times(\vec{b}\times\vec{c})+\vec{b}\times(\vec{c}\times\vec{a})+\vec{c}\times(\vec{a}\times\vec{b})=0 \;## (Jacobi-identity).
     
  5. Oct 22, 2016 #4
    exact, it's not the same operation and then we cannot speak of associativity. ∇ is itself an operator noted as 3 derivators components. See its definition, its components are not numbers
     
  6. Oct 22, 2016 #5

    robphy

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    Derivative operators satisfy the Leibniz rule.
     
  7. Oct 22, 2016 #6

    fresh_42

    Staff: Mentor

    ... which is closely related to the Jacobi-identity or likewise the definition of a derivation.
     
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