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Dot product of vector and del.

by pyroknife
Tags: product, vector
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pyroknife
#1
Feb8-14, 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
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lurflurf
#2
Feb8-14, 03:05 PM
HW Helper
P: 2,264
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
#3
Feb8-14, 05:40 PM
Math
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PF Gold
P: 39,491
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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