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.


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!

  • $\begingroup$ 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

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.