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Nabla operator

  1. Dec 6, 2008 #1

    [tex]\vec{\nabla} \cdot \vec{E} = 0[/tex]


    [tex]\vec{\nabla}^2 \cdot \vec{E} = 0[/tex]


    Is this true:

    [tex]\vec{\nabla}^2 \cdot \vec{E} = \vec{\nabla}(\vec{\nabla} \cdot \vec{E})[/tex]
    Last edited: Dec 6, 2008
  2. jcsd
  3. Dec 6, 2008 #2


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    No. If you work out grad(div(E)) you are going to get other terms as well. The correct relation is laplacian(E)=grad(div(E))-curl(curl(E)).
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