1
$\begingroup$

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 :)

$\endgroup$
1
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Now why did I get a zero when I tried it? $\endgroup$
    – Buraian
    Nov 16 '20 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.