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I've set out to prove $\vec{u}\times \vec{\omega}=\vec{\nabla}(\frac{1}{2}\vec{u}\cdot \vec{u})-\vec{u}\cdot \vec{\nabla}\vec{u}.$ This is my try: $$\vec{u}\times \vec{\omega}=\frac{1}{2}\vec\nabla\times\vec{u}\times\vec{u}$$ \begin{align} \vec\nabla\times\vec{u}\times\vec{u} &= e^{lmn}\vec{e}_l(e_{mjk}\frac{\partial}{\partial x^j}u^k)u_n\nonumber\\ &= -\vec{e}_le_{lnm}e^{jkm}\frac{\partial}{\partial x^j}u^ku_n\nonumber\\ &= -\vec{e}_l\delta^{ln}_{jk}\frac{\partial}{\partial x_j}u_ku_n\nonumber\\ &= -\vec{e}_l\left(\delta^l_j\delta^n_k\frac{\partial}{\partial x^j}u^ku_n-\delta^l_k\delta^n_j\frac{\partial}{\partial x^j}u^ku_n\right)\nonumber\\ &= -\vec{e}_l\left(\frac{\partial}{\partial x^l}u^ku_k-\frac{\partial}{\partial x^j}u^lu_j\right)&&\text{what's next?}\nonumber\\ &= -\vec{u}\cdot\vec\nabla\vec{u}+ \end{align} And here i stuck. Could you help me?

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  • $\begingroup$ what is $\omega$? $\endgroup$ – Daniel Cunha Jun 27 '17 at 11:57
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    $\begingroup$ @DanielCunha It's vorticity, $\omega = \nabla \times u$. $\endgroup$ – mattos Jun 27 '17 at 11:58
  • $\begingroup$ Also, up to your second last line is correct. Now just write it in operator form instead of summation form. I have no idea what the last line is supposed to mean. $\endgroup$ – mattos Jun 27 '17 at 12:00
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$\vec{\omega} = \vec{\nabla}\times\vec{u} = -u_{i,j}\,\varepsilon_{ijk}$

$\vec{u}\times \vec{\omega}=u_i\,w_j\,\varepsilon_{ijk} = -u_i\,(u_{a,b}\,\varepsilon_{abj})\,\varepsilon_{ijk} = -u_i\,u_{a,b}\,(\varepsilon_{jab}\,\varepsilon_{jki}) = -u_i\,u_{a,b}\,(\delta_{ak}\delta_{bi}-\delta_{ai}\delta_{bk})=\\\boxed{u_i\,u_{i,k}-u_i\,u_{k,i}}$


$\vec{u}\cdot\vec{u} = u_i\,u_i$

$\vec{\nabla}\vec{u} = u_{i,j}$

$\frac{1}{2}\vec{\nabla}(\vec{u}\cdot\vec{u}) = \frac{1}{2}u_{i,k}\,u_i+\frac{1}{2}u_i\,u_{i,k}=\boxed{u_i\,u_{i,k}}$


$\vec{u}\cdot\vec{\nabla}{\vec{u}} = (\vec{u}\cdot\vec{\nabla})\,\vec{u}= \boxed{u_i\,u_{k,i}}$


$\varepsilon$ is the Levi-Civita Symbol

$\delta$ is the Kronecker Delta

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You have a couple of problems here: firstly, the cross product is not associative, so you have to put brackets around it. \begin{align} \big[ u\times (\nabla \times u) - \nabla(\tfrac{1}{2}u \cdot u) + u \cdot \nabla u \big]_i &= \epsilon_{ijk} u_j \epsilon_{klm} \partial_l u_m - \tfrac{1}{2}\partial_i(u_ju_j) + u_j \partial_j u_i \\ &= (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})u_j \partial_l u_m - u_j \partial_i u_j + u_j \partial_j u_i \\ &= u_j \partial_i u_j - u_j \partial_j u_i - u_j \partial_i u_j + u_j \partial_j u_i \\ &=0 \end{align}


I think you're mostly correct, but I don't think it's a good idea to use contra- and covariant indices for cross products: $u=u^ie_i$ is a vector, so should have upper indices on its components. The cross product sends vectors to (pseudo)vectors, so it is actually $$ e_i \cdot (u \times v) =\epsilon_{ijk} u^j v^k \implies u \times v = e_i(\epsilon^{i}{}_{jk} u^j v^k). $$ Also, in this formalism, $\nabla$ does not behave like a vector: it carries a covariant index, and so "$\nabla \times u$" can't be interpreted as a literal cross product: you can see here that the cross product is defined in terms of components of the covariant derivative of a vector, which is less sensitive to index positions than the ordinary partial derivative. At the end of the day, in flat space one can manage with only lower indices, while in non-flat space $\partial_i$ is not tensorial when acting on non-scalars anyway, so would still be incorrect in this calculation.

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