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Gradient of a vector

  1. Jun 28, 2011 #1
    Just a simple, quick question:
    Does the gradient of a vector give a scalor or a vector?
  2. jcsd
  3. Jun 28, 2011 #2


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    Last edited by a moderator: Apr 26, 2017
  4. Jun 28, 2011 #3
    Thank you very much for your quick reply. :)
  5. Jun 28, 2011 #4


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    You can, of course, have
    [tex]\nabla\cdot \vec{\phi}(x, y, z)= \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}[/tex]
    the "divergence" of the vector valued function, [itex]\vec{\phi}(x, y, z)[/itex], which is a scalar, or
    [tex]\nabla\cdot \vec{\phi}= \left(\frac{\partial g}{\partial z}- \frac{\partial h}{\partial y}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{i}[/tex]
    the "curl" of the vector valued function, [itex]\vec{\phi}(x, y, z)[/itex], which is a vector.

    Perhaps that is what you are thinking of. There are three kinds of vector "multiplication" and so three ways we can attach the "del" operator to a function.
  6. Jun 28, 2011 #5
    I'd say there's a perfectly good definition for the gradient of a vector.
    it's a rank-2 cartesian tensor.

    For the vector [tex]\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}[/tex]

    we have the beast [tex]\nabla \vec{A}
    =\nabla A_x \hat{i} + \nabla A_y \hat{j} + \nabla A_z \hat{k}[/tex]

    such that for any vector [itex] \vec{v}[/itex]
    [tex]\vec{v} \cdot \nabla \vec{A}
    = (\vec{v} \cdot \nabla A_x) \hat{i} + (\vec{v} \cdot \nabla A_y) \hat{j}
    + (\vec{v} \cdot \nabla A_z) \hat{k}

    This is more clear in component notation
    [tex]\vec{A} \rightarrow A^i[/tex]
    [tex] \nabla \vec{A} \rightarrow (\nabla A)^i_j
    = \frac{\partial A^i}{\partial x^j}. [/tex]

    where the product with the vector [itex]\vec{v}[/itex] above is really
    contraction on the second index.
    [tex] \vec{v} \cdot \nabla \vec{A} \rightarrow
    \sum_j \quad \left( v^j \frac{\partial A^i}{\partial x^j} \right) [/tex]
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