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Quick question about gradients

  1. Jul 25, 2010 #1
    No, this isn't for homework. I'm trying to brush up on gradients for the Physics GRE. Let's say I have vector V.

    What is V(gradient squared)V ?

    In english, if I multiply the vector to the gradient squared of the same vector, what is that? I'm thinking that it'd just be the gradient of V, but I can't seem to prove it.

  2. jcsd
  3. Jul 26, 2010 #2


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    Do you mean
    [tex]\vec{V}\nabla^2\vec{V}[/tex]? That is commonly called "del squared". "gradient" is specifically [itex]\nabla f[/itex] where f is a scalar valued function. "curl f" is [itex]\nabla\times \vec{f}[/itex] where f is a vector valued function, and "div f" is [itex]\nabla\cdot \vec{f}[/itex].

    Now, [itex]\nabla^2= \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}+ \frac{\partial }{\partial z^2}[/itex] (in Cartesian coordinates) is also, strictly speaking, applied to a scalar function but it is not uncommon to see it applied to a vector meaning [itex]\nabla^2[/itex] applied to each component.

    That is, if
    [tex]\vec{v(x,y,z)}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}[/tex] then

    [tex]\nabla^2 \vec{v}(x,y,z)= \nabla^2f \vec{i}[/tex][tex]+ \nabla^2g\vec{j}[/tex][tex]+ \nabla^2h \vec{k}[/tex]

    a vector. But to say what "[itex]\vec{v}\nabla^2\vec{v}[/itex]" means we would still have to know what kind of vector product that is- dot product or cross product?

    If that is dot product and [itex]\vec{v}(x,y,z)= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}[/itex] then
    [tex]\vec{v}\cdot \nabla^2\vec{v}= f(x,y,z)\nabla^2 f(x,y,z)+ g(x,y,z)\nabla^2 g(x,y,z)+ h(x,y,z)\nabla^2h(x, y, z)[/tex]

    If, instead, it is the cross product then
    [tex]\vec{v}\times\nabla^2\vec{v}= \left(g(x,y,z)\nabla^2h(x,y,z)- h(x,y,z)\nabla^2g(x,y,z)\right)\vec{i}- \left(f(x,y,z)\nabla^2h(x,y,z)- h(x,y,z)\nabla^2f(x,y,z)\right)\vec{i}+ \left(f(x,y,z)\nabla^2g(x,y,z)- g(x,y,z)\nabla^2f(x,y,z)\right)\vec{k}[/tex]
    Last edited by a moderator: Jul 26, 2010
  4. Jul 26, 2010 #3


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    Dearly Missed

    It might also, Halls, be an outer product, i.e a matrix
  5. Jul 26, 2010 #4


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    Good point. In that case it would be
    [tex]\begin{bmatrix}f(x,y,z)\nabla^2f(x,y,z) & f(x,y,z)\nabla^2g(x,y,z) & f(x,y,z)\nabla^2h(x,y,z) \\ g(x,y,z)\nabla^2f(x,y,z) & g(x,y,z)\nabla^2g(x,y,z) & g(x,y,z)\nabla^2h(x,y,z) \\ h(x,y,z)\nabla^2f(x,y,z) & h(x,y,z)\nabla^2g(x,y,z) & h(x,y,z)\nabla^2h(x,y,z)\end{bmatrix}[/tex]
    in the x, y, z, coordinate system.
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