triple vector product: vector vs gradient

Question

I think there's a simple explanation for this, but could not find one from a few online searches. The triple vector product and the curl of $\mathbf{A}\times \mathbf{B}$ have very similar forms, however there are additional terms in the differentiation case:

$ \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\mathbf{A} \bullet \mathbf{C}) - \mathbf{C} (\mathbf{A} \bullet \mathbf{B}) \\ \nabla \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\nabla \bullet \mathbf{C}) - \mathbf{C} (\nabla \bullet \mathbf{B}) + (\mathbf{C} \bullet \nabla)\mathbf{B} - (\mathbf{B} \bullet \nabla)\mathbf{C} $

Can someone explain why? I'm familiar with Einstein notation if that helps. Thank you for your help!

Answer 1

I thought it might be useful to present way forward that I prefer. Here, we begin by introducing notation.

Let $(x_1,x_2,x_3)$ be Cartesian coordinates. We designate by $\hat x_i$, a unit vector along the $x_i$ axis and by $\partial_i$ the partial derivative with respect to $x_i$.

Then, using the convention of summing over repeated indices, the $i$'th component of the vector $\nabla \times (\vec B\times \vec C)$ can be written as

$$\begin{align} \hat x_i\cdot \left(\nabla \times (\vec B\times \vec C)\right)&=\hat x_i\cdot \left(\hat x_j\times (\hat x_k\times \hat x_\ell)\right)\partial_j(B_kC_\ell)\tag 1\\\\ &=\left(\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk}\right)\left(B_k \partial_j(C_\ell)+C_\ell \partial_j(B_k)\right)\tag 2\\\\ &=B_i\partial_j(C_j)-B_j\partial_j(C_i)+C_j\partial_j(B_i)-C_i\partial_j(Bj)\tag 3\\\\ &=B_i(\nabla \cdot \vec C)-(\vec B\cdot \nabla)C_i+(\vec C\cdot \nabla)B_i-C_i(\nabla \cdot \vec B) \tag 4 \end{align}$$

In going from $(1)$ to $(2)$, we used the vector triple product rule (presumed established) while in going from $(2)$ to $(3)$ we used the sifting property of the Kronecker Delta.

Since $(4)$ is true for all $i$, then upon multiplying by $\hat x_i$ and summing over $i$, we find the coveted identity

$$\nabla \times (\vec B\times \vec C)=\vec B(\nabla \cdot \vec C)-(\vec B\cdot \nabla)\vec C+(\vec C\cdot \nabla)\vec B-\vec C(\nabla \cdot \vec B)$$

Answer 2

As stated above, the first expression given is simply product of vectors, which can be expressed in terms of the dot product. The second involves differentiation, acting on a product. The product rule for vector differentiation will inevitably lead to the extra terms.

Answer 3

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$\ds{\epsilon_{abc}}$ is the Levi-Civita Symbol or/and the Levi-Civita Tensor.

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\begin{align} \color{#f00}{\vec{A}\times\pars{\vec{B}\times\vec{C}}} & = \sum_{i}\hat{e}_{i}\bracks{\vec{A}\times\pars{\vec{B}\times\vec{C}}}_{i} = \sum_{i}\hat{e}_{i}\sum_{jk}\epsilon_{ijk}\,A_{j}\pars{\vec{B}\times\vec{C}}_{k} \\[5mm] & = \sum_{ijk}\hat{e}_{i}\,\epsilon_{ijk}\,A_{j} \sum_{\ell m}\epsilon_{k\ell m}\,\,B_{\ell}\,C_{m} = \sum_{ij\ell m}\hat{e}_{i}\,A_{j}\,B_{\ell}\,C_{m}\ \underbrace{\sum_{k}\epsilon_{ijk}\,\,\epsilon_{\ell mk}} _{\ds{\delta_{i\ell}\,\delta_{jm} - \delta_{im}\,\delta_{j\ell}}} \\[5mm] & = \sum_{ij}\hat{e}_{i}A_{j}B_{i}C_{j} - \sum_{ij}\hat{e}_{i}A_{j}B_{j}C_{i} \\[5mm] & = \pars{\sum_{i}\hat{e}_{i}B_{i}}\pars{\sum_{j}A_{j}C_{j}} - \pars{\sum_{i}\hat{e}_{i}C_{i}}\pars{\sum_{j}A_{j}B_{j}} \\[5mm] & = \color{#f00}{\vec{B}\pars{\vec{A}\cdot\vec{C}} - \vec{C}\pars{\vec{A}\cdot\vec{B}}} \end{align}

The other one is somehow similar. You can take the opportunity to 'play' with the Levi-Civita $\ds{\epsilon_{abc}}$ Symbol.