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Vector Calculus Identities

  1. Dec 15, 2009 #1
    1. The problem statement, all variables and given/known data

    For arbitrary vector fields A and B show that:


    ∇.(A ∧ B) = B.(∇∧A) - A.(∇∧B)





    3. The attempt at a solution

    I considered only the 'i'-axis, by saying that it is perpendicular with A and B and then I expanded both the left and right side out. The working is too much to post here..I didn't manage to prove it. I was hoping someone would know if there's a useful webpage where I can find out more about vector calculus identities.

    Thanks!
     
  2. jcsd
  3. Dec 15, 2009 #2
    You can use the fact that

    [tex][\mathbf A \times \mathbf B]_i = \epsilon_{ijk} A_j B_k[/tex]

    Where there is a summation over repeated indices and [tex]\epsilon_{ijk}[/tex] is completely antisymmetric under an exchange of indices (Levi-Civita symol). That way it's not difficult to prove. Don't know about web sites though.
     
  4. Dec 15, 2009 #3
    Yea, although I'm being asked to prove it without using suffix notation..which is much more tedious
     
  5. Dec 15, 2009 #4
    It is, but it's still a straightforward calculation, just remember the Leibniz rule and keep track of the indices.
     
  6. Dec 23, 2009 #5
    Ok I completed the identity without using suffix notation, but now I will to use it:

    My working out is the following:

    I take the ith component of both sides, I first start with the left side:


    [tex]\epsilon[/tex]ijk δ/δx[tex]_{}i[/tex](A_j B_k)


    And then for the right side:

    (B_i ε_ijk δA_k/δx[tex]_{}j[/tex]) - (A_i ε_ijk δB_k/δx[tex]_{}j[/tex])

    As you can see the differentials are different, for the left side it is w.r.t the 'i' component while for the right side it is w.r.t 'j' component
     
    Last edited: Dec 23, 2009
  7. Dec 24, 2009 #6
    There are no components on either side, the expression is a scalar since it's a dot product. There is a summation over i,j and k so the left hand side is

    [tex]\nabla \cdot (\mathbf A \times \mathbf B) = \sum_{i=1}^3 \partial_i (\mathbf A \times \mathbf B)_i = \sum_{i=1}^3 \partial_i (\sum_{j,k=1}^3\epsilon_{ijk} A_j B_k) = \sum_{i,j,k=1}^3 \epsilon_{ijk} \partial_i (A_j B_k)[/tex]

    Calculate the derivative and rearrange the terms to get the expression on the right hand side (Hint: use the fact that epsilon is antisymmetric under an exchange of indices).
     
  8. Dec 24, 2009 #7
    Oh ok! why is k=1 though? Im a bit confused. So you have summed the left side over i, will this mean :

    (A_j )(B_z) will be differentiated w.r.t to dx_i and then I can use the product rule.
     
  9. Dec 24, 2009 #8
    No, the summation is over all three indices, it has to be because the end result is a scalar so there can't be any indices remaining.What the above means is that i goes from 1 to 3, j goes from 1 to 3 and k goes from 1 to 3.

    In any case, yes, differentiate w.r.t dx_i inside the sum and you will get two terms. Rearrange them using the antisymmetricity of epsilon and you will get the two terms of the right hand side.
     
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