Is it true that $\vec{\nabla}\times[(\vec{a}\cdot\vec{\nabla})\vec{a}]=\vec{\nabla}\times[\vec{a}\times(\vec{\nabla}\times\vec{a})]$?

Question

I'm reading some notes on particle physics by a university professor, and after doing calculations I've reached the conclusion that in order for one of his claims to be true, this equality would have to hold. I assume that it does, but I would like to know how to prove it mathematically.

I suppose that I could attempt to expand both sides of the equality, calculating the products involved, and see if the results are equal, but that seems too rudimentary, so I'm looking for a more elegant way.

Thank you very much in advance!

Answer 1

You are off by a sign.

Recall $$ V\times(\nabla\times V)=\underbrace{V\cdot (\nabla V)^T}_{=\nabla(\tfrac12 V\cdot V)} - (V\cdot\nabla)V $$ (This is the usual "BAC-CAB" identity for triple vector product except because we are dealing with operators we need to preserve the order. The transpose here on $\nabla V$ make sure the $\cdot$ is contracting the two $V$s) and apply curl, remembering curl of a gradient vanishes.

Answer 2

In component Einstein notation $$\varepsilon_{i\ j\ k} \partial_i ( a_l \partial_l a_l) = \varepsilon_{i\ j\ k} \partial_i \varepsilon_{j\ l \ m} a_l \varepsilon_{m \ s\ t} \partial_s a_t\ \text{?}$$

but Mathematica confirms

   a = {a1[x, y, z], a2[x, y, z], a3[x, y, z]}

    Curl[ Cross[ a, Curl[a, {x, y, z}]], {x, y, z}]==
   -  Curl[ ( a[[1]]D[#,x]+a[[2]]D[#,y] + a[[3]]D[#,z]&)/@ a,{x,y,z}] 

   True

So there seems to be an order error

Answer 3

start from the right hand side's inner cross product:

$a\times (\nabla \times a)$ It is best to use tensor algebra here:

$\epsilon_{ijk}\epsilon_{klm}a_j\frac{\partial a_m}{\partial x_l}=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})a_j\frac{\partial a_m}{\partial x_l}$

$=(a_j\frac{\partial a_j}{\partial x_i}-a_j\frac{\partial a_i}{\partial x_j})$

implying: $a\times(\nabla \times a)=\frac{1}{2}\nabla(a\cdot a)-(a\cdot\nabla)a$

So now the first term is a gradient, taking the cross product of it is 0. $\nabla\times(a\times(\nabla \times a))=-\nabla\times(a\cdot \nabla)a)$