How to prove this vector identity $({\bf r}\times\nabla)\cdot ({\bf r}\times\nabla)$?

Question

Let ${\bf r}$ a position vector. How can I prove this?

$$({\bf r}\times\nabla)\cdot ({\bf r}\times\nabla)=r^2\nabla\cdot\nabla-r^2\frac{\partial^2}{\partial^2r}-2r\frac{\partial}{\partial r}$$

My efforts:

First note that $({\bf r}\times\nabla)\cdot ({\bf r}\times\nabla)={\bf r}\cdot(\nabla\times ({\bf r}\times\nabla))$. Now, for BAC-CAB identity we have $${\bf r}\cdot(\nabla\times ({\bf r}\times\nabla))={\bf r}\cdot[\nabla\cdot\nabla {\bf r}-\nabla(\nabla\cdot{\bf r})] $$.

What are ${\bf r}\cdot\nabla\cdot\nabla {\bf r}$ and ${\bf r}\cdot\nabla(\nabla\cdot{\bf r})? $

Answer 1

It always help to use tensor notation and the completely antisymmetric tensor to deal with cross products. If you don't know what I am talking about, I think you will benefit very much from a quick study about this convention.

That done, you can write

$ (r \times \nabla) \cdot (r \times \nabla) $

as

$ e^{ijk} r_j \partial_k e^{ilm} r_l \partial_m $

And use the always helpful identity:

$ e^{ijk} e^{ilm} = \delta_j^l \delta_k^m - \delta_j^m \delta_k^l $

Relating the product of the completely antissimetric tensors with the kronecker delta. Now:

$ e^{ijk} r_j \partial_k e^{ilm} r_l \partial_m = e^{ijk} e^{ilm} r_j \partial_k (r_l \partial_m) = e^{ijk} e^{ilm} [ r_j (\partial_k r_l) \partial_m + r_j r_l \partial_k \partial_m ] $

$ (\delta_j^l \delta_k^m - \delta_j^m \delta_k^l) [ r_j (\partial_k r_l) \partial_m + r_j r_l \partial_k \partial_m ] = r_j (\partial_k r_j) \partial_k + r_j r_j \partial_k \partial_k - r_j (\partial_k r_k) \partial_j - r_j r_k \partial_k \partial_j $

Now, we use that $ \partial_i r_j = \delta_i^j $:

$ r_j (\partial_k r_j) \partial_k + r_j r_j \partial_k \partial_k - r_j (\partial_k r_k) \partial_j + r_j r_k \partial_k \partial_j = r_j \partial_j + r_j r_j \partial_k \partial_k - 3 r_j \partial_j - r_j r_k \partial_j \partial_k $

$ = r_j r_j \partial_k \partial_k - 2 r_j \partial_j - r_j r_k \partial_j \partial_k $

Rewriting that in vector notation would depend on some rather awkward notational definitions, but using the ones you wrote, this is exactly

$ r^2 \nabla^2 - 2 r \partial_r - r^2 \partial_r ^2 $

Answer 2

Note that we have using Einstein's summation convention

$$\begin{align} (\vec r\times \nabla)\cdot (\vec r\times \nabla)&=(x_i\partial_j) (\hat x_i\times \hat x_j)\cdot (\hat x_k\times \hat x_\ell)(x_k\partial_\ell)\\\\ &=(x_i\partial_j )\hat x_i\cdot (\hat x_j\times(\hat x_k\times \hat x_\ell))(x_k\partial_\ell)\\\\ &=(x_i\partial_j)(\delta_{j\ell}\delta_{ik}-\delta_{jk}\delta_{i\ell})(x_k\partial_\ell)\\\\ &=(x_i\partial_j)(x_i\partial_j)-(x_i\partial_j)(x_j\partial_i)\\\\ &=x_i^2\partial^2_j-x_ix_j\partial_i\partial_j-x_i\partial_i\\\\ &=r^2\nabla\cdot \nabla-(\vec r\cdot \nabla)^2-\vec r\cdot \nabla\\\\ &=r^2\nabla\cdot \nabla-r^2\partial^2_r-\vec r\cdot \nabla-\vec r\cdot\nabla\\\\ &=r^2\nabla\cdot \nabla-r^2\partial^2_r-2r\partial_r \end{align}$$

as expected!