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Homework Help: Gradient of a dot product identity proof?

  1. Jan 28, 2013 #1
    Gradient of a dot product identity proof???

    1. The problem statement, all variables and given/known data
    I have been given a E&M homework assignment to prove all the vector identities in the front cover of Griffith's E&M textbook. I have trouble proving:

    (1) ∇(A[itex]\bullet[/itex]B) = A×(∇×B)+B×(∇×A)+(A[itex]\bullet[/itex]∇)B+(B[itex]\bullet[/itex]∇)A

    2. Relevant equations
    (2) A×(B×C) = B(A[itex]\bullet[/itex]C)-C(A[itex]\bullet[/itex]B)

    3. The attempt at a solution
    I applied the identity in equation (2) to the first two terms on the right hand side of equation (1), and that allowed me to cancel out 4 terms. Yet, I end up with:

    ∇(A[itex]\bullet[/itex]B) = ∇(A[itex]\bullet[/itex]B)+∇(B[itex]\bullet[/itex]A)

    ....which I know cannot be correct. How can I prove this identity in a relatively straightforward way? I have seen other pages asking this yet they all involved the use of Levi-Cevita symbols and the Kronecker Delta, which I am trying not to use because we haven't learned them. I would gladly appreciate anyone's effort to help me out.
  2. jcsd
  3. Jan 28, 2013 #2


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    Re: Gradient of a dot product identity proof???

    The only sensible way I can see is to do it by hand for let's say the <x> component in both sides and show they are the same.
  4. Jan 28, 2013 #3
    Re: Gradient of a dot product identity proof???

    I was hoping I can get around the long calculations lol
  5. Jan 28, 2013 #4


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    Re: Gradient of a dot product identity proof???

    You have to be careful with the ∇ operator in vector identities, as it is not commutative. I think this caused the problem here.
    The long calculation will work, and it is sufficient to consider one component.
  6. Jan 28, 2013 #5


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    Re: Gradient of a dot product identity proof???

    I know you wanted avoid them, but it's definitely worth learning about the Kronecker delta and Levi-Civita symbol. Using them makes verifying the identity much easier.
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