I have been asked to prove:

If $\mathbf{A}$ and $\mathbf{B}$ are constant vectors, then $\nabla(\mathbf{A} \cdot \mathbf{B} \times \mathbf{r}) = \mathbf{A} \times \mathbf{B},$

using the Levi-Civita formalism. I tried doing it but I get zero as amy asnwer. I suspected the question is wrong so I tried replacing $\mathbf{A} \cdot \mathbf{B}$ with $\mathbf{A} \times \mathbf{B}$, but I again get zero.

Any help on the initial question?


So the $i$th entry of the right hand side is

$$ (\nabla(\mathbf A\cdot \mathbf B\times\mathbf r))_i = \partial_i (A_j(\epsilon_{jk\ell} B_k r_\ell)) $$

since $\mathbf A$ and $\mathbf B$ are constant, I'm assuming that the gradient will act only on $\mathbf r$ therefore

$$ \partial_i (A_j(\epsilon_{jk\ell} B_k r_\ell)) = \epsilon_{jk\ell} A_j B_k \delta_{i\ell} $$

replacing $\ell$ by $i$ as a result of the $\delta_{i\ell}$ and swapping the indices of the Levi-Civita symbol twice, the left hand side then reads

$$ \epsilon_{i jk} A_jB_k = (\mathbf A\times\mathbf B)_i$$

Adding clarifications

Just in case, I've used above:

  • components of $\nabla$: $ \partial_i = {\partial\over\partial r_i}$
  • definition of the cross product using the Levi-Civita symbol $ (\mathbf X \times \mathbf Y)_i = \epsilon_{ijk} X_jY_k $
  • the derivative is non zero only if it corresponds to the appropriate component: $ \partial_i r_\ell = \delta_{i\ell} $
  • permutation of the indices of the Levi-Civita symbol: $ \epsilon_{ijk} = -\epsilon_{jki} = \epsilon_{kij}$
  • 1
    $\begingroup$ I did the same working as you did but I guess I make the wrong inferences /played around with the Kronecker deltas wrongly. Will look into it. Thanks. $\endgroup$ Sep 15 '16 at 19:58
  • $\begingroup$ if this solves your question, consider marking it as such? $\endgroup$
    – tibL
    Sep 15 '16 at 20:06

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