# How to show that $(r\times\nabla)\cdot(r\times\nabla)=r\cdot[\nabla\times(r\times\nabla)]$?

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?

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Yes you are right.I think if you set a = (r × ∇), you can use the identity of crossproducts/bac-cab:
a(b x c) =b (c x a) with b=r and c=∇... Then you "only" have to prove this by writting out the grad and the vector r and calculating this in "vector-form".

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So it's basically just a variable substitution from the old bac-cab that we already proven. I can follow the substitutions and I think it is correct. Thanks a lot for a very prompt proof. –  909 Niklas Dec 13 '12 at 2:46
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 $\bf A$: $$\begin{eqnarray} \nabla \times \left(b {\bf C}\right) &=& \partial_i \left(b C_j\right) \epsilon_{ijk} {\bf e}_k \\ &=& \left[\left(\partial_i b\right)C_j + b \partial_i C_j \right] \epsilon_{ijk} {\bf e}_k \\ &=& \nabla b \times {\bf C} + b \nabla \times {\bf C}\\ &\ne& b \nabla \times {\bf C} \end{eqnarray}$$