# Proving a vector calculus identity

Right under equation 40.7 in the feynman lecture (here), this identity is written:

$$(v \cdot \nabla) v = (\nabla \times v) \times v + \frac{1}{2} \nabla (v \cdot v)$$

I seek a proof for this identity/ an intuitive proof for why it is true. I'm not sure how I'd even start the derivation but I think this identity is the same as the one under the 'special sections' part of this wiki page

An attempt:

By the vector triple product identity

$$a \times b \times c = (b ) c \cdot a - ( c ) b \cdot a$$

Now this gave me zero when applied to $$\nabla \times v \times v$$ and that doesn't look right..

Hints would also be appreciated :)

In the equation you have, in the $$(\nabla \times \ \textbf{v}) \times \textbf{v}$$ term, there are parantheses around the first two terms, so it's not necessarily zero since first you do $$\nabla \times \ \textbf{v}$$, which is perpendicular to $$\textbf{v}$$, and then you cross it with $$\textbf{v}$$.
Starting with the vector triple product seems like a good approach. You can always write it out in tensor notation with the Levi-Civita symbol to get the $$n^{th}$$ component: $$\left((\partial_i v_j) \epsilon_{ijk}\right) v_m \epsilon_{kmn},$$ where $$\left((\partial_i v_j) \epsilon_{ijk}\right)$$ is the $$k^{th}$$ component of $$(\nabla \times \ \textbf{v})$$.
Since $$\epsilon_{kmn} = \epsilon_{mnk}$$ and $$\epsilon_{ijk}\epsilon_{mnk}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$$, the $$n^{th}$$ component is equal to $$\begin{eqnarray}(\partial_i v_j) v_m \epsilon_{ijk}\epsilon_{mnk}\\ &=&(\partial_i v_j) v_m (\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm})\\ &=&(\partial_i v_j) v_m \delta_{im}\delta_{jn}-(\partial_i v_j) v_m\delta_{in}\delta_{jm}\\ &=& v_m \partial_m v_n -\partial_n v_m v_m. \end{eqnarray}$$ Now, you can use what @PAM1499 noted (the chain rule) to write $$\partial_n v_m v_m = \frac{1}{2} \partial_n (v_m v_m)$$.
Putting everything back into vector form, you get: $$(\nabla \times \ \textbf{v}) \times \textbf{v} = (\textbf{v} \cdot \nabla ) \textbf{v}- \frac{1}{2} \nabla(\textbf{v} \cdot \textbf{v}).$$ and all you need to do is rearrange the terms to get the identity.