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Product Rule

  1. May 15, 2008 #1
    Hello everyone. I'm trying to get my head around this product rule:

    [tex] \nabla \times (A\times B) = (B\cdot \nabla )A - (A\cdot \nabla )B + A(\nabla \cdot B) - B(\nabla \cdot A) [/tex]

    Ok, we have this

    [tex] \nabla = (\partial /\partial x,\partial/\partial y,\partial /\partial z) [/tex]

    and for dot products

    [tex] a\cdot b = b\cdot a [/tex]

    Therefore in the product rule given above, is it not the case

    [tex] (B\cdot \nabla )A = A(\nabla \cdot B) [/tex]

    and similarly, the other two terms on the RHS are equal?
    Thank-you for your help.
  2. jcsd
  3. May 15, 2008 #2


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    Be careful. [tex]\nabla[/tex] is a vector operator not a vector. It will not commute the way you expect it to.
  4. May 15, 2008 #3


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    In fact, I would much prefer the notation [itex]B\cdot(\nabla A)[/itex] to [itex](B\cdot\nabla)A[/itex].
  5. May 15, 2008 #4
    Aha! I think I get it now! The brackets were confusing, because usually we have to evaluate the stuff in the brackets first right?

    Does this mean [tex]\nabla [/tex] always acts on the vector directly to its right?
  6. May 15, 2008 #5

    Hold on, this can't be right can it? Then we would have
    [tex]\nabla \times (A\times B) = 0 [/tex]

    wouldn't we? Can someone please tell me what [tex](B\cdot\nabla)A[/itex] is?
  7. May 15, 2008 #6


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    First you evaluate [tex] \nabla A [/tex]. You get a vector field, i.e. a vector at every point of space. Then [tex] B \cdot(\nabla A) [/tex] would be vector field whose value at any point is the dot of B with vector [tex] \nabla A [/tex] at that point.
    Last edited: May 15, 2008
  8. May 15, 2008 #7
    I'm sorry for being extremely thick, but then doesn't that mean that

    [tex] \nabla \times (A\times B) = 0 [/tex]?

    This makes no sense, because it means the cross product of any two vectors has zero curl? Surely [tex](B\cdot \nabla)\cdot A[/tex] is not the same thing as [tex]B\cdot (\nabla \cdot A)[/tex]?
  9. May 15, 2008 #8


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    why does it mean that?
  10. May 15, 2008 #9


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    The second operation not a dot product. del.A is a scalar. So (del.A)B is the scalar (d/dxAx + d/dyAy + d/dzAz) times the vector B = d/dxAx*B + d/dyAy*B + d/dzAz*B

    Also, remember the del is an operator so del.A is not the same as A.del. A.del is still a scalar though being applied to B to it gets pretty messy looking.

    Last edited: May 15, 2008
  11. May 15, 2008 #10


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    No, what I said was [tex] B \cdot (\nabla A) [/tex] is the same thing as [tex](B \cdot \nabla)A[/tex]
  12. May 15, 2008 #11
    Er, ok, what's the difference? A and B are vectors.
  13. May 15, 2008 #12

    Oh, ok, this site explains it to me. Thanks for all your help everyone, taking time to answer my stupid questions :biggrin:
  14. May 15, 2008 #13


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    Some people do not like to take the gradient of a vector since it is a dyad, and it makes them feel ickky inside.
  15. May 15, 2008 #14


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    This is a confusing aspect of vector calculus.
    [tex]\nabla [/tex] acts to the right only.
    One may define a bidirectional del/nabla
    consider this bidirectional derivative
    it is a good and correct habit when working with vectors to switch to bidirectional form
    -hold all function left of operators constant
    -change operators to birectional
    -perfom manipulations ending with a form easy to conver to unidirectional form

    recall this identity when working with products

    [tex]\mathbf{(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b}[/tex]
  16. May 15, 2008 #15


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    It's only ickky if you don't know that there are things which aren't vectors or scalars. There are also tensors (or dyads). Then it could make you queasy.
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