I have looked through several lists of curl identities but cannot find anything for this.

$\nabla \times$ denotes the curl, and $\mathbf{\vec F}$ is just some arbitrary vector field in $\Bbb R^3$


1 Answer 1


We can use Einstein summation and the Levi-Civita tensor $\varepsilon_{ijk}$ to write the $i$-th component of a more general expression as $$[(\vec{\nabla}\times\vec{a})\times\vec{b}]_i=\varepsilon_{ijk}(\varepsilon_{j\ell m}\partial_\ell a_m)b_k,$$ now use the relation $\varepsilon_{ijk}\varepsilon_{j\ell m}=-\varepsilon_{jik}\varepsilon_{j\ell m}=\delta_{im}\delta_{k\ell}-\delta_{i\ell}\delta_{km}$ and write (note that $\partial_i$ only acts on $a_k$ and not $b_k$) $$\varepsilon_{ijk}\varepsilon_{j\ell m}(\partial_\ell a_m)b_k=(b_k\partial_k)a_i-(\partial_i a_k)b_k.$$ Letting the sum run over all $i$ gives the result: $$(\vec{\nabla}\times\vec{a})\times\vec{b}=(\vec{b}\cdot\vec{\nabla})\vec{a}-\sum_i\sum_k(\partial_i a_k)b_k\hat{e}_i.$$ Looking at the last part, this is equivalent to $$\sum_i\sum_k(\partial_i a_k)b_k\hat{e}_i=\sum_n\delta_{nk}b_n\sum_i\sum_k(\partial_i a_k)\hat{e}_i\hat{e}_k$$ but this is just a "dot product" between a vector and the Jacobian of $\vec{a}$, so we have $$(\vec{\nabla}\times\vec{a})\times\vec{b}=(\vec{b}\cdot\vec{\nabla})\vec{a}-\vec{b}\cdot\boldsymbol{\mathrm{J}}_{\vec{a}},$$ where the last part is in the sense of performing the dot product column wise and use the result as new row entry.

For the special case of $\vec{a}=\vec{b}=\vec{F}$ note that $\vec{F}\cdot\boldsymbol{\mathrm{J}}_{\vec{F}}$ equals half of the gradient of the squared norm (i.e. of $F_1^2+F_2^2+\dotsm$), so this becomes: $$\boxed{(\vec{\nabla}\times\vec{F})\times\vec{F}=(\vec{F}\cdot\vec{\nabla})\vec{F}-\frac{1}{2}\vec{\nabla}\lVert\vec{F}\rVert^2}$$


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