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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!

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  • $\begingroup$ If you think about what the curl of a vector field measures, you should expect this to be zero: the vector field is everywhere parallel to the fixed vector $\mathbf b$. $\endgroup$ – amd May 2 '18 at 0:37
  • $\begingroup$ @amd: So is the vector field $(x,y,z)\mapsto(y,0,0)$, but its curl is nonzero. $\endgroup$ – user856 May 2 '18 at 3:49
  • $\begingroup$ @Rahul Good point. It’s not simply that everything’s parallel to $\mathbf b$ but that the vector field picks out a constant multiple of the orthogonal projection onto $\mathbf b$. After a coordinate rotation, it’s basically $(x,y,z)\to(cx,0,0)$. $\endgroup$ – amd May 2 '18 at 5:24
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\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, $\nabla\times({\bf b}\cdot{\bf r}){\bf b} = \bf 0$.

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  • $\begingroup$ Hello, why are we allowed to reposition the differential operator as shown in the first line? $\endgroup$ – user107224 May 2 '18 at 0:12
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    $\begingroup$ Great question. Remember that $b$ is just a constant vector, and constants pass through differential operators safely. $\endgroup$ – B. Mehta May 2 '18 at 0:13
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$\require{cancel}$If you want to use an index-free notation, do $$\nabla\times ({\bf b}\cdot {\bf r}){\bf b} = ({\bf b}\cdot {\bf r}) \cancelto{{\bf 0}}{\nabla\times {\bf b}} + \underbrace{\nabla({\bf b}\cdot {\bf r})}_{={\bf b}} \times {\bf b} = {\bf 0}+ {\bf 0} = {\bf 0}.$$

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