$$\vec{B}\cdot \operatorname{curl} \vec{A}-\vec{A}\cdot \operatorname{curl} \vec{B} = \operatorname{div}(\vec{A}\times \vec{B})$$

So I started from the LHS as follows:

$$=B_i\epsilon_{ijk}\frac{\partial}{\partial x_j} A_k-A_i\epsilon_{ijk}\frac{\partial}{\partial x_j}B_k$$

$$= \epsilon_{ijk}\left(B_j\frac{\partial A_k}{\partial x_j} - A_i\frac{\partial B_k}{\partial x_j}\right)$$

I'm not sure if I've done these steps correctly, but assuming I have; I can't see where to go from here.

I know that $$\operatorname{div}\vec{A}\times\vec{B} = \frac{\partial}{\partial x_i}(\epsilon_{ijk}A_jB_k)$$

By the way I'm new to tensor notation, so please explain things you might think are obvious (because they probably aren't obvious to me). Thanks

  • $\begingroup$ is it $div \vec {A} \times \vec{B}$ or $div.(\vec {A}\times\vec {B})$ ? $\endgroup$ Commented Oct 15, 2017 at 17:05
  • $\begingroup$ @TheDeadLegend the second one $\endgroup$
    – mrnovice
    Commented Oct 15, 2017 at 17:22
  • $\begingroup$ Differentiate using the product rule, flip a couple of indicies and you are there. $\endgroup$ Commented Oct 15, 2017 at 18:04
  • $\begingroup$ @DonaldSplutterwit I tried what you suggested, but I'm not sure how to flip the indices on terms. $\endgroup$
    – mrnovice
    Commented Oct 15, 2017 at 18:27

1 Answer 1


The Levi-Cevati symbol is antisymmetric in the interchange of any pair of indicies $ \epsilon_{ijk}= - \epsilon_{kji}$ and symmetric for a cycling of the indicies $ \epsilon_{ijk}= \epsilon_{kij}$ Using the product rule \begin{eqnarray*} \operatorname{div}\vec{A}\times\vec{B} = \frac{\partial}{\partial x_i}(\epsilon_{ijk}A_jB_k) = \epsilon_{ijk}\frac{\partial A_j}{\partial x_i} B_k+ \epsilon_{ijk}A_j\frac{\partial B_k}{\partial x_i}. \end{eqnarray*} In the first term use $ \epsilon_{ijk}= \epsilon_{kij}$ and rename the indicies $k \rightarrow i $ ,$i \rightarrow j $ and $j \rightarrow k $.

In the second term interchange the indicies $ \epsilon_{ijk}= - \epsilon_{jik}$ nd rename the indicies $i \rightarrow j $ and $j \rightarrow i $ . \begin{eqnarray*} = \epsilon_{ijk} B_i \frac{\partial A_k}{\partial x_j} -\epsilon_{ijk} A_i\frac{\partial B_k}{\partial x_j}. \end{eqnarray*}


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