Vector proof using Levi-Civita notation

Question

Can anyone expand/simplify this proof: I am unsure why it is the partial derivative with respect to $x_i$, is there no $x_j$ or $x_k$?

$$ \begin{align} \underline{\nabla} \cdot (\underline{A} \times \underline{B}) &= \frac{\partial}{\partial x_i} \epsilon_{ijk} A_j B_k \\ &= \epsilon_{ijk} \left(\frac{\partial A_j}{\partial x_i}\right) B_k + \epsilon_{ijk} A_j \left(\frac{\partial B_k}{\partial x_i}\right) \\ &= B_k \epsilon_{kij} \frac{\partial A_j}{\partial x_i} - A_j \epsilon_{jik} \frac{\partial B_k}{\partial x_i} \end{align} $$

Any help is much appreciated!

Answer 1

The reason the derivatives are with respect to $x_i$ only, is that we write $(\nabla)_i =\hat x_i\cdot \nabla=\partial_i$.

We can proceed using this notation by writing

$$\nabla \cdot (\vec A\times\vec B)=(\hat x_i\partial_i)\cdot (\hat x_jA_j\times\hat x_kA_k)$$

where the summations over $i,j,k$ are implied.

Then, continuing we have

$$(\hat x_i\partial_i)\cdot (\hat x_j A_j\times\hat x_k A_k)=\hat x_i \cdot (\hat x_j\times\hat x_k)(A_j\partial_i(B_k)+B_k\partial_i(A_j))$$

The scalar triple product is invariant under a "circular shift" and hence

$$\begin{align} \hat x_i \cdot (\hat x_j\times\hat x_k)(A_j\partial_i(B_k)+B_k\partial_i(A_j))&=-\hat x_j \cdot (\hat x_i\times\hat x_k)(A_j\partial_i(B_k)) + \hat x_k \cdot (\hat x_i\times\hat x_j)B_k\partial_i(A_j))\\\\ &=\vec B\cdot \nabla \times \vec A-\vec A\cdot \nabla \times \vec B \end{align}$$

as expected.