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I'm having difficulty proving the formula: $$u\times\omega = \nabla\ (\frac{ u\cdot\ u}{2}) - u\cdot\nabla\ u$$ I should be using tensor notation. Given is that: $$\omega\ = \nabla\times\ u$$ and $$\nabla\cdot\ u\ = 0$$

I've done this so far: $$ (u\times\omega)_i = (u \times\ (\nabla\times\ u))_i = \epsilon_{ijk} u_j(\epsilon_{klm}\frac{\partial}{\partial\ x_l}u_m)=\epsilon_{ijk}\epsilon_{klm}\ u_j\frac{\partial}{\partial\ x_l}u_m=\epsilon_{kij}\epsilon_{klm}\ u_j\frac{\partial}{\partial\ x_l}u_m=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})u_j\frac{\partial}{\partial\ x_l}u_m=u_j\frac{\partial}{\partial\ x_i}u_j-u_j\frac{\partial}{\partial\ x_j}u_i $$

But that is as far as I come. I could really need some help+input, thanks on beforehand.

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If $\vec w=\nabla \times \vec u$, then using implied summation notation reveals

$$\begin{align} \vec u\times \vec w&=\vec u\times \nabla \times \vec u\\\\ &=u_i\hat x_i\times \partial_j(\hat x_j\times \hat x_ku_k)\\\\ &=(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k)u_i\partial_j(u_k)\\\\ &=\hat x_ju_i\partial_j(u_i)-\hat x_ku_i\partial_i(u_k)\\\\ &=\frac12\nabla (|\vec u|^2)-(\vec u\cdot \nabla)\vec u \end{align}$$

as was to be shown!



Alternativley, using the Levi-Civita notation, we can write

$$\begin{align} (\vec u\times \vec w)_i&=(\vec u\times \nabla \times \vec u)_i\\\\ &=\epsilon_{ijk}u_j(\nabla \times \vec u)_k\\\\ &=\epsilon_{ijk}u_j\epsilon_{k\ell m}\partial_\ell (u_m)\\\\ &=(\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell})u_j\partial_\ell (u_m)\\\\ &=u_j\partial_i(u_j)-u_j\partial_j(u_i)\\\\ &=\frac12\partial_i(u_j u_j)-(u_j\partial_j)(u_i)\\\\ &=\left(\frac12 \nabla(\vec u\cdot \vec u)-(\vec u\cdot \nabla)\vec u \right)_i \end{align}$$

Hence, we conclude that

$$\vec u\times \vec w=\frac12 \nabla(\vec u\cdot \vec u)-(\vec u\cdot \nabla)\vec u$$

as expected!

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  • $\begingroup$ From line 2 to line 3, is it a version of the double cross product formula ? $\endgroup$ – Jean Marie Aug 28 '20 at 21:36
  • $\begingroup$ It is the vector triple product. $$\vec A\times (\vec B\times \vec C)=(\vec A\cdot \vec C)\vec B-(\vec A\cdot \vec B)\vec C$$Now, let $\vec A=\hat x_i$, $\vec B=\hat x_j$, and $\vec C=\hat x_k$. The inner product of two unit vectors $\hat x_i$ and $\hat x_j$ is equal to $\delta_{ij}$, where $\delta$ is the Kronecker Delta. $\endgroup$ – Mark Viola Aug 28 '20 at 21:38
  • $\begingroup$ @MarkViola thank you very much for that fast answer! However, I'm not familiar with the use of $\hat{\mathbf{x}}$, unit vectors, in the solving of problems like this. I've received a list of problems like this one from my teacher that we should do in order to learn tensor/index notation and the Einstein convention. I use my course literature as a reference, and in them the use of $\epsilon$ (the permutation symbol) seems to be the convention. Is there anyway I could expand the work I've done to get the right answer? (without the explicit use of unit vectors?) Hope my answer makes any sense. $\endgroup$ – flme79 Aug 29 '20 at 7:01
  • $\begingroup$ @flme79 The Levi-Civita symbol, $\epsilon_{ijk}$ is equivalent to the scalar triple product $$\epsilon_{ijk}=\hat x_i \cdot (\hat x_j\times \hat x_j)$$I'll edit the answer to use this notation. $\endgroup$ – Mark Viola Aug 29 '20 at 16:34

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