Vector Calculus Identities

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!
 
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.
 
Yea, although I'm being asked to prove it without using suffix notation..which is much more tedious
 
It is, but it's still a straightforward calculation, just remember the Leibniz rule and keep track of the indices.
 
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
 
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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).
 
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).
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.
 
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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