Curl of $({\bf b}\cdot{\bf r}){\bf b}$?

Question

I’m currently attempting to calculate $({\bf b}\cdot{\bf r}){\bf b}$. My attempt with index notation did not go far:

$$\nabla\times({\bf b}\cdot{\bf r}){\bf b} = \epsilon_{ijk} \partial_j b_k b_l r_l.$$

I don’t really know how to approach the dot product with my curl operator since it’s a scalar. Help is much appreciated, thank you!

Answer 1

\begin{align}\epsilon_{ijk} \partial_j b_k b_l r_l &= \epsilon_{ijk} b_kb_l\partial_j r_l \\ &= \epsilon_{ijk} b_kb_l \delta_{jl} \\ &= \epsilon_{ijk} b_kb_j\\ &= 0. \end{align} So, $\nabla\times({\bf b}\cdot{\bf r}){\bf b} = \bf 0$.

Answer 2

$\require{cancel}$If you want to use an index-free notation, do $$\nabla\times ({\bf b}\cdot {\bf r}){\bf b} = ({\bf b}\cdot {\bf r}) \cancelto{{\bf 0}}{\nabla\times {\bf b}} + \underbrace{\nabla({\bf b}\cdot {\bf r})}_{={\bf b}} \times {\bf b} = {\bf 0}+ {\bf 0} = {\bf 0}.$$