# finding the vector product of a vector field and the curl of fg

$$\space f:\mathbb{R}^3\rightarrow \mathbb{R}\space$$ 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 $$∇ × (f{\bf g })$$?

• Oct 9, 2020 at 13:32

$$\def\e{\varepsilon}$$ 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 $$g$$.
• would the expansion of $a$ then simplify to give: $a=f\textbf{g}\times (\nabla \times \textbf{g})$
• No, $\;a=\textbf{g}\times (\nabla \times (f\textbf{g}))\;$