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Dot product of vector and del. 
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#1
Feb814, 02:14 PM

P: 398

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 


#2
Feb814, 03:05 PM

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P: 2,263

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. 


#3
Feb814, 05:40 PM

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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 [itex]\nabla\cdot u[/itex] and [itex]\nabla\times u[/itex] are mnemonics for [itex]\partial u_x/\partial x+ \partial u_y/\partial y+ \partial u_z/\partial [/itex] and [itex](\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}[/itex]. In particular "[itex]\nabla[/itex]" is NOT a real vector and you cannot combine it with vector functions without saying HOW that is to be done. 


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