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Dot and Cross Products (of Gradients)

  1. Jul 23, 2009 #1
    Statement:
    I was wondering if the following are identical,
    [tex]\nabla \bullet \nabla \times \vec{A} = \nabla \bullet (\nabla \times \vec{A}) ?[/tex] (#1)

    Also, more importantly I was wondering if someone could explain to me why the following is zero for any vector,

    [tex]\nabla \bullet \nabla \times \vec{A} = 0?[/tex]

    Reasoning:
    If we look at [tex]\nabla \bullet \nabla \times \vec{A}[/tex], Isn't [tex]\nabla \bullet \nabla[/tex] the laplacian, or [tex]\nabla^{2}?[/tex] Is that the reason we cannot perform the operation in equation (#1) above- the laplacian is a function that needs a vector field to be operated on, and there is none. So [tex]\nabla \bullet \nabla = 0.[/tex]


    Thanks, alot,


    JL
     
  2. jcsd
  3. Jul 23, 2009 #2
    Yeah they're identical. Check out http://en.wikipedia.org/wiki/Triple_product" [Broken] on what's called the "triple product."
     
    Last edited by a moderator: May 4, 2017
  4. Jul 23, 2009 #3
    [tex]\nabla \bullet \nabla \times \vec{A} = 0 = div(curl(A))[/tex]

    I mean if you think of it in terms of geometrical vectors, then if we're looking at the divergence of the amount of A which is contained it sort of intuitively seems like it should be zero.

    If it's too hard to visualize then just work it out. Assume A is a vector function and take the curl of A then the divergence of the curl of A and by Clairaut's theorem you should see that it all cancels out.

    Secondly [tex](\nabla \bullet \nabla) \times \vec{A}[/tex] makes no sense you cannot take the cross product of a scalar operator (which the laplacian is).

    Finally the laplacian doesn't equal zero. It's a linear operator, it doesn't equal anything be itself. In fact it really doesn't have any meaning by itself if it's not being applied to a function in euclidean space.
     
  5. Jul 23, 2009 #4

    Pengwuino

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    It's more complicated then that. Work it out and you'll see the derivatives canceling.
     
  6. Jul 24, 2009 #5
    I tried, for example I let [tex]\vec{H} = 2xy\hat{x} + x^2y^2z^2\hat{y} + x^3y^3z^3\hat{y}.[/tex]

    When I take the curl,
    [tex]\nabla \times \vec{H} = [3x^3y^2z^3 - 2x^{2}y^{2}z]\hat{x} + [3x^{2}y^{3}z^{3} - 0]\hat{y} + [2xy^{2}z^{2} - 2x]\hat{z}[/tex] and let that whole thing equal [tex]\alpha[/tex]

    So now when I take the divergence,
    [tex]\nabla \bullet \alpha = (9x^{2}y^{2}z^{3} - 4xy^{2}z) + 9(x^{2}y^{2}z^{3}) + (4xy^{2}z) = 18(x^{2}y^{2}z^{3}) \neq 0[/tex]

    It seems like it almost worked out, the second term above needed a sign change, but I've reviewed the calculation, and it doesn't seem that I've made a mistake.

    That was along the lines I was thinking, the laplacian requires a vector to be operated on, and it doesn't have one.
     
    Last edited: Jul 24, 2009
  7. Jul 24, 2009 #6

    alphysicist

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    I think your second term (the y-hat term) is missing a minus sign here.
     
  8. Jul 24, 2009 #7

    tiny-tim

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    Hi jeff1evesque! :smile:

    (have a del: ∇ and use \cdot, not \bullet:wink:)
    You're very confused. :redface:

    Learn the following:

    div curl = curl grad = 0

    div(curlA) = ∇.(∇xA) = 0

    curl(gradφ) = ∇x(∇φ) = 0

    div(gradφ) = ∇2φ = Laplacian.

    ∇.(∇xA) is zero for the same reason that B.(BxA) is zero. :wink:
     
  9. Jul 24, 2009 #8
    The y-hat consists of the following,
    [tex][\frac{\partial}{\partial x}(x^{3}y^{3}z^{3}) - \frac{\partial}{\partial z}(2xy)]\hat{y} = [3x^{2}y^{3}z^{3} - 0]\hat{y}[/tex]

    Am I incorrect?
     
  10. Jul 24, 2009 #9

    alphysicist

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    That is incorrect; it is the negative of that.
     
  11. Jul 24, 2009 #10
    Thanks, I totally forgot (the notation is similar to taking determinants).
     
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