Dot product of vector and del.

  1. 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. jcsd
  3. lurflurf

    lurflurf 2,327
    Homework Helper

    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.
     
  4. HallsofIvy

    HallsofIvy 40,544
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted