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How do these operations with Del operator work?

  1. Aug 20, 2013 #1
    How do these operations with Del operator work??

    1. The problem statement, all variables and given/known data
    Let's say A and B are expressed by their cartesian components as:
    A = <P, Q, R> and B = <M, N, O>

    what would be the differente between (A.∇)B and B(∇.A) ?


    2. Relevant equations



    3. The attempt at a solution
    I tried to expand the expression (A.∇)B as:
    [itex](P\partial_{x} + Q\partial_{y} + R\partial_{z})B[/itex]

    Where [itex]\partial_{x}[/itex] is an abbreviation for [itex]\partial/\partial_{x}[/itex] but I don't know how to proceed next. Should I multiply the inside of the parenthesis with every component o B? That looks strange.
     
  2. jcsd
  3. Aug 20, 2013 #2

    jedishrfu

    Staff: Mentor

    Think of it like a dot product with the B vector. You have B as a component vector so it should be obvious.
     
  4. Aug 20, 2013 #3
    But that's what I don't understand. I already made the dot product, and now I have to multiply the scalar
    [itex](P\partial_{x} + Q\partial_{y} + R\partial_{z})[/itex] by B? My confusion is created because now I see the product like this:

    [itex](P\partial_{x} + Q\partial_{y} + R\partial_{z})B =
    (P\partial_{x} + Q\partial_{y} + R\partial_{z})<M, N, O>[/itex]

    Since the outside is scalar, we don't have a dot product, so it would be expanded like this?:
    [itex]=<P\partial_{x}M + Q\partial_{y}M + R\partial_{z}M, P\partial_{x}N + Q\partial_{y}N + R\partial_{z}N, P\partial_{x}O + Q\partial_{y}O + R\partial_{z}O>[/itex]

    or should I expand it something like this?
    [itex]=P\partial_{x}M + Q\partial_{y}N + R\partial_{z}O[/itex]
     
  5. Aug 20, 2013 #4
    Are you sure that B is indeed a vector function? "[itex](A\cdot\nabla)[/itex]" is usually "applied" to a scalar function f so that

    [tex](\textbf{A}\cdot\nabla)f=P\partial_xf+Q\partial_yf+R\partial_zf[/tex]

    If B is actually vector field, then I'd say (in cartesian coordinates, where the scale factors are constant) that what you initially wrote is correct,

    [tex]\nabla \textbf{B} = \sum_{ij} \partial_j B_i \textbf{e}_i \textbf{e}_j \implies ((\textbf{A}\cdot\nabla)\textbf{B})_i=\sum_j A_j\partial_j B_i [/tex]

    However, I don't recall seeing that exact expression anywhere, so I might be mistaken. It does look like the second term of the material derivative, but I'm not sure why the parentheses would be written like that.

    The second expression evaluates to

    [tex]\textbf{B}(\nabla\cdot \textbf{A})=\textbf{B}(\partial_x P+\partial_y Q+\partial_z R)[/tex]

    Where you simply multiply every component of B with the divergence of A.

    EDIT: I'd say it's reasonable for you to be confused, especially if you don't know the definition of a gradient of a vector field. Even if you do, the notation used is not the clearest, in my opinion.

    EDIT2: That is to say, that this
    expression is the correct one from what I can tell.
     
    Last edited: Aug 20, 2013
  6. Aug 20, 2013 #5
    Oh, sorry for my notation.

    I was trying to do the following exercise:

    http://2.bp.blogspot.com/-ymI_Ifgrtl4/UhPUvmQnboI/AAAAAAAAAT0/hPkqWoo3OMA/s1600/dipole.png
    where r = x i + y j + z k

    when I tried to evaluate ∇x(m x r) i used the identity:

    ∇x(m x r) = (r . ∇)m - (m . ∇)r - r(∇ . m) + m(∇ . r)

    and that's where I got confused about how (r . ∇)m and m(∇ . r) worked

    EDIT: Ok, I'm gonna try to finish it now. Thank you :)
     
  7. Aug 20, 2013 #6
    I wasn't really blaming you, and I'm not even sure if it's a valid complaint : P

    Ahh, that's what it was? Sure, what I (and you) said earlier is definitely the correct way to expand them. My reply would've been much shorter if I knew that ∇×(a×b) was what you were calculating.

    Do remember that these only work "well" in cartesian coordinates; for example with strong spherical symmetries simply writing out the cross products is often the quicker way of doing things.
     
  8. Aug 20, 2013 #7
    Thank you :) I'll see how far i'll get with this problem.
     
  9. Aug 20, 2013 #8

    lurflurf

    User Avatar
    Homework Helper

    By the usual convention ∇ acts to the right so
    (r . ∇)m and m(∇ . r)
    have m and r being acted on.
     
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