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$\space f:\mathbb{R}^3\rightarrow \mathbb{R}\space$ is a differentiable scalar field and $\space\mathbf{g}:\mathbb{R}^3\rightarrow \mathbb{R}^3\space$ is a differentiable vector field.

I have been asked to simply the following using any rules of computation:

$\space$ $\space\mathbf{g}\times(\nabla \times f \space\mathbf{g})$

I know that $ ∇ × (f{\bf g }) = (∇f) × {\bf g }+ f(∇ × \bf g)$

But I don't know how to then compute $ \mathbf {g} \times $ that.

Is the vector field ${\bf g }$ always perperdicular to $ ∇ × (f{\bf g })$?

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$\def\e{\varepsilon}$ Define $\,h=fg,\;$ then use the triple product rule in such a way that $\nabla$ stays to the left of $h$ and never operates on $g$ $$\eqalign{ a &= g\times (\nabla\times h) \\ &= \nabla(h\cdot g) - (g\cdot \nabla)h \\ &= (\nabla h)\cdot g - g\cdot(\nabla h) \\ }$$ Substitute $\,h=fg\;$ and expand $$\eqalign{ a &= (\nabla fg)\cdot g - g\cdot(\nabla fg) \\ &= (\nabla f)(g\cdot g) + f(\nabla g)\cdot g - (g\cdot\nabla f)g - (g\cdot\nabla g)f \\\\ }$$ An alternative method is to use the Levi-Civita symbol and index notation. $$\eqalign{ a_i &= \e_{ijk}\;g_j(\e_{k\ell m}\;\partial_\ell h_m) \\ &= (\e_{ijk}\e_{k\ell m})\;(g_j\;\partial_\ell h_m) \\ &= (\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell}) \;(g_j\;\partial_\ell h_m) \\ &= (g_m\;\partial_i h_m) - (g_\ell\;\partial_\ell h_i) \\ &= g_m\;\partial_i(fg_m) - g_\ell\;\partial_\ell(fg_i) \\ &= g_mg_m\;\partial_if + f\;(\partial_ig_m)g_m - g_ig_\ell\;\partial_\ell f - fg_\ell\;\partial_\ell g_i \\\\ }$$ This is the same result as before, but does not require ad hoc rules about $\nabla$ staying to the left of $h$ and not (initially) operating on $g$.

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  • $\begingroup$ would the expansion of $a$ then simplify to give: $a=f\textbf{g}\times (\nabla \times \textbf{g})$ $\endgroup$
    – user827887
    Oct 9, 2020 at 13:08
  • $\begingroup$ No, $\;a=\textbf{g}\times (\nabla \times (f\textbf{g}))\;$ $\endgroup$
    – greg
    Oct 9, 2020 at 17:01

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