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Indicial Notation-Proving grad(u.v) and EijkEkpq

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove grad(u.v)=(u.grad)v+(v.grad)u+ux(curlv)+vx(curlu) [x's are cross product, .'s are dot product]
    or better seen as: [tex]\nabla[/tex](u[tex]\bullet[/tex]v)=(u[tex]\bullet[/tex][tex]\nabla[/tex])v+(V[tex]\bullet[/tex][tex]\nabla[/tex])u+u[tex]\wedge[/tex]([tex]\nabla[/tex][tex]\wedge[/tex]v)+v[tex]\wedge[/tex]([tex]\nabla[/tex][tex]\wedge[/tex]u)


    2. Relevant equations



    3. The attempt at a solution
    I am new to this notation, and I know that the gradient multiplied by a scalar (from the dot product) will be a vector, but I don't know how to show it is this identity. I have racked my brain!!!

    1. The problem statement, all variables and given/known data
    Show that [tex]\epsilon[/tex][tex]_{}kij[/tex][tex]\epsilon[/tex][tex]_{}kpq[/tex]=[tex]\delta[/tex][tex]_{}ip[/tex][tex]\delta[/tex][tex]_{}jq[/tex]-[tex]\delta[/tex][tex]_{}iq[/tex][tex]\delta_{}jp[/tex]

    2. Relevant equations


    3. The attempt at a solution
    Again, I haven't a clue!
     
  2. jcsd
  3. Sep 1, 2009 #2
    For the second problem:

    [tex]\epsilon_{ijk}[/tex] is called the Levi-Civita symbol. Basically, you should form a matrix of Kronecker deltas and then compute the determinant of the matrix. It should look something like this:

    http://i29.tinypic.com/13ydzl0.png

    You should be able to prove the identity once you've written the determinant out once or twice.
     
  4. Sep 1, 2009 #3
    For 1:

    Simplest possible method I know of is to write u as (u1 u2 u3) and v as (v1 v2 v3) and then solve each part.

    e.g. For left part,

    it is

    (d/dx,d/dy,d/dz) (k)

    k = scalar quantity from the dot product = u1.v1 + u2.v2 + u3.v3

    so

    (dk/dx,dk/dy,dk/dz)

    (bit lengthy and messy)
     
  5. Sep 1, 2009 #4

    cristo

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    Isn't the idea to use index notation?
     
  6. Sep 1, 2009 #5
    Yes, this must be proved using index notation.
     
  7. Sep 2, 2009 #6
    I'm actually stuck on this too if anybody knows how to do it that'd be fantastic.
     
  8. Sep 3, 2009 #7

    cristo

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    Why have neither of you made any attempt at solving the problem? For the first one, what is the dot product of two vectors and the cross product of two vectors in index notation?
     
  9. Sep 3, 2009 #8
    Becuase I don't know how to work with gradients in index notation..
     
  10. Sep 4, 2009 #9
    I have been working on this problem all week and there's still no light at the end of the tunnel...
     
  11. Sep 4, 2009 #10

    cristo

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    If you show your work, then you will receive help. How about, say, answering my questions above?
     
  12. Sep 4, 2009 #11
    I gave you a starting point. Look at the link that I posted. It is an image showing how to relate the epsilon notation to the Kronecker delta notation from first principles. It is the cross-product of the matrix in the picture.

    1) Write out the equation in that link.

    2) Take the determinant of that matrix and set the epsilon notation equal to it.

    3) Simplify the resulting algebra.

    When you start simplifying the algebra, it should be apparent how to prove it.

    I don't think there is an easier way to do this problem.
     
  13. Sep 4, 2009 #12
    My main problem was I didn't really know the rules with the del operator and indicial notation. I think I got it now. Using V as the del operator:

    I solved out uX(VXv) and vX(VXu) which ended up cancelling out the first two terms on the right side and left me with the expanded version of the LHS. The only reservation I had about the way I solved it is that it left me with a kronecker delta at the beginning of both of the chain ruled out expressions on the LHS (not actually on the LHS, but when I expanded the two RHS expressions until they were the same as the LHS except fot the two kroneckers if that makes sense) I just ended up saying they were equal to 1 because it seems like that's something my teachers do sometimes without really giving justification.

    Sorry if that's a bit confusing but I don't know a good way of writing up all the expressions with subscripts and whatnot on here.

    edit: Actually after thinking about it you probably just solve for the case of the two being equal (K.D. being 1) and unequal (K.D. being 0) and the 0 case just never matters because it zeroes everything out, so you can usually just say that the indices are equal as long as they're free indices?

    Sorry for not answering your question but I didn't really have time to try to write it out in this as I had to figure it out for today, especially because my only problem was really how to work with the del operator.
     
    Last edited: Sep 4, 2009
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