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Homework Help: Vector identity involving grad and a function

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    The question is to use index notation to show that the following is true, where a is a three-vector and f is some function.

    2. Relevant equations


    3. The attempt at a solution

    Hmmmm . . . I haven't really got anything to put here!

    I am starting to get to grips with the basics of index notation, and using the Levi-Civitas identities for other identity proofs. I haven't ever worked with grad before though, and I can't find any help online because I can't find any other identities that look like this one. Does it have a name or something to help me search?

    The closest I have found is "The product of a vector and a scalar" on wikipedia


    but my question states that f is a function, not a scalar, which must make some difference I guess! I just don't know where to start here, do any of the operations within my question have an index-notation version using the LC tensor or similar?

    Any help much appreciated :)

    PS: You will need to click the thumbnails in the post to see the full pictures of the equations. I am not much of a computer guy. Or a maths guy, it seems :(
  2. jcsd
  3. Oct 20, 2011 #2


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    f is a scalar function, as opposed to, say, a vector function.

    The i-th component of [itex]\nabla f[/itex] is [itex]\partial_i[/itex]f. Use that and the Levi-Civita symbol to prove the identity.
  4. Oct 20, 2011 #3
    I think the root of my problem here is that I have no idea how to write out the starting equation in ordinary longhand notation (as components of vectors), let alone in index notation.

    Is [itex]\nabla f[/itex] equal to [itex] f \nabla[/itex]?

    And what do I get from multiplying [itex]\nabla[/itex] with fa?
  5. Oct 20, 2011 #4


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    If you don't understand the notation or what the gradient is, a good place to start is looking up what it means.


    Your textbooks will probably have a more accessible discussion.
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