Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

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


    User Avatar
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook