Grassmann identity with Nabla operator in fluid dynamics (Euler equation)

Question

I stumbled across these lines / identity I don’t understand. To derive an alternative form of the Euler equation including the vorticity vector $\vec \omega = \vec \nabla \times \vec{\mathsf v}$, the author of the following lines states:

To my mind the Grassmann identity $$\vec a \times \left(\vec b \times \vec c\right) = \vec b \left( \vec a \cdot \vec c \right) - \vec c \left( \vec a \cdot \vec b \right) $$ was performed “in reverse order”. However, to me it is unclear why the author’s first term on the RHS is $$\vec \nabla \left( \frac{\vec{\mathsf v} ^2}{ 2}\right)$$ In my opinion $$\vec \nabla \left( \vec{\mathsf v} ^2\right)$$ would be the correct term.

I am specifying my question: If I simply set $\vec a = \vec c = \vec{\mathsf v}$ and $\vec b = \vec \nabla $ and naively apply the identity it will yield: \begin{align} \vec{\mathsf v} \times \left(\vec \nabla \times \vec{\mathsf v} \right) &= \vec \nabla \left( \vec{\mathsf{v}} \cdot \vec{\mathsf v}\right)-\vec{\mathsf{v}}\left(\vec{\mathsf v}\cdot \vec{\nabla}\right) \\ &= \vec{\nabla}\left(\vec{\mathsf{v}}^2\right) - \left(\vec{\mathsf v}\cdot\vec{\nabla}\right) \vec{\mathsf v} \end{align}

What's the error in the prior proceeding?

Answer 1

$$\let\eps\epsilon (v\times( \nabla\times v))_i= \eps_{ijk}v_j(\eps_{klm} \partial_l v_m) = \eps_{kij}v_j(\eps_{klm} \partial_l v_m)=v_j (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\partial_lv_m=v_j \partial_i v_j -v_j\partial_jv_i$$ So the correct formula is $$ v\times (\nabla \times v) = \sum_{j=1}^3 v_j \nabla v_j - (v\cdot\nabla )v=\nabla |v|^2/2 - (v\cdot \nabla )v$$

Note that this first term in the RHS is different from $\nabla (v_j v_j)$ by exactly a factor of 2. I believe the error was rearranging a differential operator like a vector.

Answer 2

While vector identities can be a useful mnemonic for vector calculus identities, one must still remember that $\nabla$ is not a vector and does not have to obey normal vector identities. The relevant identity is the product rule for the gradient of a dot product: $$ \nabla(\mathbf{A} \cdot\mathbf{B}) = \mathbf{A}\cdot\nabla\mathbf{B} + \mathbf{B}\cdot\nabla\mathbf{A} + \mathbf{A}\times(\nabla\times\mathbf{B}) + \mathbf{B}\times(\nabla\times\mathbf{A}). $$ Just let $\mathbf{A} = \mathbf{B} = \mathbf{v}$ and solve for $\mathbf{v}\times(\nabla\times\mathbf{v})$.