Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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".

share|improve this answer
    
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. –  Niklas Rtz 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 $\bf A$: $$ \begin{eqnarray} \nabla \times \left(b {\bf C}\right) &=& \partial_i \left(b C_j\right) \epsilon_{ijk} {\bf e}_k \\ &=& \left[\left(\partial_i b\right)C_j + b \partial_i C_j \right] \epsilon_{ijk} {\bf e}_k \\ &=& \nabla b \times {\bf C} + b \nabla \times {\bf C}\\ &\ne& b \nabla \times {\bf C} \end{eqnarray} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.