I'm trying to compute $u \times(\nabla \times v)$. My solution so far is below:

\begin{align*} [u \times (\nabla \times v)]_i& = \epsilon_{ijk}u_j\epsilon_{klm}\partial_lv_m \\ & =\epsilon_{kij}\epsilon_{klm}u_j\partial_lv_m \\ &=u_j\partial_lv_m(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}) \\ & =u_j\partial_iv_j-u_j\partial_jv_i \\ \end{align*}

I can see that the second term will give me $(u\cdot\nabla)v$, but I'm not sure what to do with the first term.


1 Answer 1


The formula you derived reads $$u \times (\nabla \times v) = \nabla_v(u \cdot v) - (u \cdot \nabla)v$$ where the notation $\nabla_v$ is called Feynman notation and should indicate that the derivative is applied only to $v$ and not to $u$.


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