How to Prove Vector Calculus Identity Involving Cross Product and Gradient?

lylos
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Homework Statement


Prove the following:
(\vec{r}\times\nabla)\cdot(\vec{r}\times\nabla)=r^2\nabla^2-r^2 \frac{\partial^2}{\partial r^2}-2r\frac{\partial}{\partial r}

Homework Equations


(\hat{e_i}\times\hat{e_j})=\epsilon_{ijk}
(\hat{e_i}\cdot\hat{e_j})=\delta_{ij}


The Attempt at a Solution


(r_i\nabla_j\epsilon_{ijk}r_l\nabla_m\epsilon_{lmn})(\hat{e_k}\cdot\hat{e_n})
(r_i\nabla_j\epsilon_{ijk}r_l\nabla_m\epsilon_{lmn}\delta_{kn})
(r_i\nabla_j\epsilon_{ijk}r_l\nabla_m\epsilon_{lmk})
(r_i\nabla_jr_l\nabla_m)(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})
(r_i\nabla_jr_i\nabla_j)-(r_i\nabla_ir_j\nabla_j)

At this point, I'm lost. Does the gradient operator work on all terms, should I rearrange?
 
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One of your equations is wrong,
<br /> (\hat{e_i}\times\hat{e_j})=\epsilon_{ijk}\hat{e}_{k}<br />
I personally would write:
<br /> \mathbf{r}=x_{i}\hat{e}_{i},\quad\nabla =\hat{e}_{i}\partial_{i}<br />
Hopefully this should help.

Mat
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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