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About tensorial calculus

  1. Jan 4, 2008 #1

    I have a doubt because I don't find the general definition of the nabla operator in order to solve my matter.

    I am working with dyadic analysis and I have to prove that

    [tex]\nabla^{2}F = \nabla \nabla \bullet F[/tex]

    where F is a symmetric dyadic function.

    My problem is when I have to get [tex]\nabla^{2}F[/tex]

    because I don't know how to calculate the gradient of a dyadic. I know the definition of the gradient of a vector what results in a dyadic, but no idea about the general definition of the gradient in cartesian coordinates for whatever order of cartesian tensors.

    I 'd be very thankful for a little of help.
    Last edited: Jan 4, 2008
  2. jcsd
  3. Jan 5, 2008 #2


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    As far as i know, the gradient of a (n,0) tensor is a (n+1,0) tensor. Let's say you have a (3,0) tensor T

    [tex] T=T_{ijk}e_{i}\otimes e_{j}\otimes e_{k} [/tex] (cartesian tensor and coordinates)


    [tex] \nabla T= \partial_{l}T_{ijk} e_{l}\otimes e_{i}\otimes_{j}\otimes e_{k} [/tex]

    is the gradient of the tensor T.
  4. Jan 5, 2008 #3

    I only need to relate the gradient of a (n+1,0) tensor with its divergence now. Because I have to get the divergence of the gradient of a (2,0) tensor, and comparate with the gradient of the divergence of the same one.

    I had no too problems with the gradient of the divergence, but when I put in reverse order the operators I had some troubles because I didn't get the same result and I have to prove it's the same.

    I'll put my results in a few minutes, I have to leave now.

    Thank you
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