A friend asked how to show that
$$(r\times\nabla)\cdot(r\times\nabla)=r\cdot[\nabla\times(r\times\nabla)]$$
$r$ is a position vector, $\nabla$ is the grad operator, and $\cdot$ and $\times$ are the dot and vector product signs.
Can this be proven from the BAC-CAB identity?
Best Answer
Here's how I would prove it: $$ \begin{eqnarray} \left({\bf r} \times \nabla\right) \bullet \left({\bf r} \times \nabla\right) &=& \left({\bf r} \times \nabla\right)_k \left({\bf r} \times \nabla\right)_k\\ &=& \left(r_i \partial_j \epsilon_{ijk}\right) \left(r_m \partial_n \epsilon_{mnk}\right)\\ &=& r_i \left[\partial_j \left( r_m \partial_n \epsilon_{mnk}\right)\epsilon_{jki}\right]\\ &=& r_i \left[\partial_j \left({\bf r} \times \nabla\right)_k\epsilon_{jki}\right]\\ &=& r_i \left[\nabla \times \left({\bf r} \times \nabla\right)\right]_i\\ &=& {\bf r} \bullet \left[\nabla \times \left({\bf r} \times \nabla\right)\right]\\ \end{eqnarray} $$ Repeated indices are summed, and I used 16 here and a property of the Levi-Civita symbol, namely that it is invariant with respect to cyclic permutations of its indices.
Note that, in general, you cannot always substitute $\nabla$ in a vector identity because $\nabla$ is an operator. For example, take the "identity" $$ {\bf A} \times \left(b {\bf C}\right) = b {\bf A} \times {\bf C}, $$ where $\bf A\left({\bf r}\right)$, $\bf B\left({\bf r}\right)$, and $c\left({\bf r}\right)$ are fields (vector, vector, and scalar, respectively) that depend on position. Now substitue $\nabla$ for $\bf A$: $$ \begin{eqnarray} \nabla \times \left(b {\bf C}\right) &=& \partial_i \left(b C_j\right) \epsilon_{ijk} {\bf e}_k \\ &=& \left[\left(\partial_i b\right)C_j + b \partial_i C_j \right] \epsilon_{ijk} {\bf e}_k \\ &=& \nabla b \times {\bf C} + b \nabla \times {\bf C}\\ &\ne& b \nabla \times {\bf C} \end{eqnarray} $$