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Nabla operator and working with it

  1. Jun 28, 2013 #1
    While using the ∇ operator, most of the times we can treat it as a vector. I came across a few formulae(basically product rules)..
    Code (Text):
    ∇×([B]A[/B]×[B]B[/B])=([B]B[/B].∇)[B]A[/B]-([B]A[/B].∇)[B]B[/B]+[B]A[/B](∇.[B]B[/B])-[B]B[/B](∇.[B]A[/B])
    where A and B are vectors

    I wanted to know if there is any direct way of deriving it. By direct I mean assuming the basic vector identity
    Code (Text):
    [B]C[/B]×([B]B[/B]×[B]A[/B])=[B]B[/B]([B]C[/B].[B]A[/B])-[B]A[/B]([B]C[/B].[B]B[/B])
    . Any other better way is also fine. I derived it by splitting each of the vectors A, B, and ∇(although this is not truly a vector) in their orthogonal components and then doing the appropriate cross products.
     
  2. jcsd
  3. Jun 28, 2013 #2
    The best way to derive these results is to use summation convention. Everything drops out quite nicely.

    The vector identity you've stated is also derived in the same way but you'll notice the subtleties of the fact that you're dealing with an operator when doing the derivation.
     
  4. Jun 28, 2013 #3
    Thanks for the idea. I have never really used summation notation, always expanded terms and got my way through. Will try summation from now on.

    I realised one thing that ∇.A=A.∇ doesn't hold in this case while it holds with vectors.

    yeah thats fine, I just wanted to think if there is any result which we can exploit using the fact that ∇ behaves as a vector(well, almost). I was just worried that incase I forget the long identity, and I want to use it, I won't be able to derive it quickly.
     
  5. Jun 28, 2013 #4

    SteamKing

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    I realised one thing that ∇.A=A.∇ doesn't hold in this case while it holds with vectors.


    That's because Del isn't really a vector; it's a set of operators which work on vectors.

    http://en.wikipedia.org/wiki/Del
     
  6. Jun 28, 2013 #5
    Yeah right.
     
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