Dot product of vector and del.

[pyroknife]

I'm not sure which section is best to post this question in.

I was wondering if the expression (u $ ∇) is the same as (∇ $ u).
Here $ represents the dot product (I couldn't find this symbol.
∇=del, the vector differentiation operator
and u is the velocity vector or any other vector

 

[lurflurf]

The usual convention is that ∇ acts to the right so (u $ ∇) and (∇ $ u) are not equal.

This is analogous to asking if uD is equal to D u where D is the differentiation operator.

 

[HallsofIvy]

QUOTE=pyroknife;4654871]I'm not sure which section is best to post this question in.

I was wondering if the expression (u $ ∇) is the same as (∇ $ u).
Here $ represents the dot product (I couldn't find this symbol.
∇=del, the vector differentiation operator
and u is the velocity vector or any other vector[/QUOTE]
Before anyone can answer that question, you will have to tell us what you mean by "(u $ ∇). The reason I say that is that things like $\nabla\cdot u$ and $\nabla\times u$ are mnemonics for $\partial u_x/\partial x+ \partial u_y/\partial y+ \partial u_z/\partial$ and $(\partial u_z/\partial y- \partial u_y/\partial z)\vec{i}+ (\partial u_x/\partial z- \partial u_z/\partial x)\vec{j}+ (\partial u_y/\partial x- \partial u_x/\partial y)\vec{k}$. In particular "$\nabla$" is NOT a real vector and you cannot combine it with vector functions without saying HOW that is to be done.