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Curl of the transpose of a gradient of a vector: demonstration of an identity

  1. Dec 19, 2009 #1
    I would like to demonstrate an identity with the INDICIAL NOTATION. I have attached my attempt. Please let me know where I made mistakes. Any suggestion? I am trying to understand tensors all by myself because they are the keys in continuum mechanics

    Attached Files:

    Last edited: Dec 19, 2009
  2. jcsd
  3. Dec 19, 2009 #2
    What sort of vector product are you using here [itex]\hat{e}_i \hat{e}_j[/itex]?
  4. Dec 19, 2009 #3
    tensor product
  5. Dec 21, 2009 #4
    Is this question so difficult? Please help me: I am trying to learn tensors and I would like to know what my mistake is. Thanks!
  6. Jan 6, 2010 #5
    Any suggestion?
  7. Jan 10, 2010 #6
    Can anyone suggest a forum to post my question? Thanks
  8. Jan 13, 2010 #7
    anything? please help!
  9. Jan 24, 2010 #8
  10. Jan 26, 2010 #9
    Is my question too difficult? Please advise.
  11. Jan 26, 2010 #10


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    I don't really understand what is meant by
    [tex]\nabla \mathbf{u}[/tex].

    For example, if [tex]\mathbf{u}=u_j\hat{e}_j[/tex], then [tex]\nabla \mathbf{u}=(\partial_i\hat{e}_i)(u_j\hat{e}_j)=\partial_iu_j\hat{e}_i\hat{e}_j[/tex].

    But what is [tex]\hat{e}_i\hat{e}_j[/tex]; the inner product between the unit basis vectors? Then the result would be a scalar instead of a vector.
    Last edited: Jan 26, 2010
  12. Jan 29, 2010 #11
  13. Feb 6, 2010 #12
    Any other input?
  14. Feb 11, 2010 #13
  15. Feb 11, 2010 #14


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    Staff Emeritus
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    Gold Member

    You should at least explain how you define [itex]\nabla u[/itex] when u is a vector.
  16. Feb 22, 2010 #15
    The problem is at the very bottom line in the definition of a curl of a tensor. I found 2 definitions which contradict to each other. Mine is one of them. I will email the authors.
  17. Apr 3, 2010 #16
    I asked an expert. The question was not trivial. After a while I found out that there are different definitions of curl of a tensor.
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