$\nabla \times (v \nabla)v = – \nabla \times[v \times (\nabla \times v)]$

Question

I need to show that $\nabla \times (v \nabla)v = – \nabla \times[v \times (\nabla \times v)]$.

So what I did was :

$[v \times (\nabla \times v)]_i = \epsilon_{ijk} v_j(\epsilon_{klm} \partial_l v_m) $. Using the relation between the Levi Civita Symbol product and Kronecker delta I got :

$[v \times (\nabla \times v)]_i = v_j \partial_i v_j -v_j \partial_j v_i$ = $v_j \partial_i v_j -(v \cdot \nabla)v_i$. Using the product rule we have that $v_j \partial_i v_j = \frac{1}{2} \partial_i (v^2)$.

So then I tried applying the curl :

$\nabla \times[v \times (\nabla \times v)]_i$ = $\epsilon_{ijk}\partial_j (\frac{1}{2} \partial_i (v^2)-(v \cdot \nabla)v_i)$

My doubt is how can I proceed from here ? I see that I am close to the result ,but I'm stuck.

Any help will be appreciated

Answer 1

You want to prove $\nabla\times X=0$ with $X:=(v\cdot\nabla)v+v\times(\nabla\times v)$, i.e.$$X_i=v_j\partial_jv_i+\underbrace{\epsilon_{ijk}\epsilon_{klm}}_{\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}}v_j\partial_lv_m=v_j\partial_jv_i+v_j\partial_iv_j-v_j\partial_jv_i=v_j\partial_iv_j=\partial_i(\tfrac12v_j^2).$$In other words, $X=\nabla(\tfrac12v^2)$, making the result trivial by @MarkViola's hint.