$\nabla \, \times ( {\bf u} \times {\bf v} ) = (\nabla . {\bf v}) \, {\bf u} - (\nabla {\bf v}) \, {\bf u}$???

Question

One of the problems (#7) in the following notes appears to suggest that

$ \nabla \, \times ( {\bf u} \times {\bf v} ) = (\nabla . {\bf v}) \, {\bf u} - (\nabla {\bf v}) \, {\bf u} $

(it is the second term that I am struggling with), instead of

$ \nabla \, \times ( {\bf u} \times {\bf v} ) = (\nabla . {\bf v}) \, {\bf u} - (\nabla . {\bf u}) \, {\bf v} $

Note the second term, as suggested by the triple vector product formula. Why is the form given by author correct? If indeed it is.

Answer 1

Notice that $\nabla v$ is not the same as $\nabla\cdot v$. Do you know how $\nabla v$ is defined?

Answer 2

Note that $\nabla v$ is actually a matrix, since $v$ has three components, and there are three variables $x, y, z$ to differentiate with, you get 9 numbers, which can be neatly arranged into a 3x3 matrix. $\nabla \cdot v$ is exactly $\mathrm{Tr}(\nabla v)$.

Answer 3

Note that you can not apply the vector triple product identity since differentiation is involved. Switching to ESN you get, \begin{align} \nabla \times ( \mathbf{u} \times \mathbf{v} ) &\stackrel{esn}\equiv \epsilon_{ijk}\partial_{x_{j}}( \epsilon_{klm} u _{l} v_{m})\\ &=-\epsilon_{ijk}\epsilon_{mlk}\partial_{x_{j}}( u _{l} v_{m})\\ &=-(\delta_{im}\delta_{jl} - \delta_{il}\delta_{jm})\partial_{x_{j}}(u _{l} v_{m})\\ &=-\partial_{x_{l}}(u_{l} v_{i}) + \partial_{x_{m}}(u _{i} v_{m})\\ &=-(\partial_{x_{l}}u _{l})v_{i} - u_{l}(\partial_{x_{l}}v _{i}) + (\partial_{x_{m}}u_{i})v_{m} + u_{i}(\partial_{x_{m}}v_{m})\\ &\stackrel{vect}\equiv -(\nabla \cdot \mathbf{u})\mathbf{v} - (\nabla \mathbf{v})\mathbf{u} + (\nabla\mathbf{u})\mathbf{v}+(\nabla \cdot \mathbf{v} )\mathbf{u} \qquad. \end{align} So we see that there is actually more terms that come out of that expression than the paper seems to show. However since the author states that the vector $\mathbf{u}$ is constant, the terms with $\nabla \cdot \mathbf{u}$ and $\nabla \mathbf{u}$ are zero. So indeed it follows that, for constant $\mathbf{u}$, $$ \nabla \times ( \mathbf{u} \times \mathbf{v} ) = (\nabla \cdot \mathbf{v} )\mathbf{u} - (\nabla \mathbf{v})\mathbf{u}\qquad . $$