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Product of a third- and second-order tensor.

  1. Mar 4, 2012 #1
    Hi,
    I have read that vector of polarisation [itex]\vec{P}[/itex] and stress tensor T are linked with equation
    [itex]\vec{P}=\underline{\underline{d}} \: \underline{T}[/itex]
    where d is a third-order piezoelectric tensor. Can anyone explain how you multiply a third and a second-order tensor to get a vector? A good link will also help.
    Thank you,
    N.
     
  2. jcsd
  3. Mar 5, 2012 #2

    Fredrik

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    I'm not familiar with this equation, but I'm guessing that it's a contraction: ##A_{i}=\sum_{j,k}B_{ijk}C_{jk}##.

    Also, it's usually a good idea to include a better reference than "I have read...". It might be easier to answer if we knew where you read it. If possible, you should link directly to the relevant page at Google Books.
     
  4. Mar 5, 2012 #3
    So if I understand this correctly, in order to get "i-th" component of a vector you have to take i-th "slice" of third-order tensor and multiply it with second order tensor: not in the standard way we multiply matrices, but rather just multiply j,k-th component of both tensors and sum over all nine components (assuming that both tensors are three dimensional)?
    I would have included a reference, but I only have it in printed version in Slovenian. :blushing:
     
  5. Mar 5, 2012 #4

    Fredrik

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    You certainly don't have to do this to obtain the ith component of an arbitrary vector. What I'm saying is if you take a tensor with 3 indices and one with 2, and "contract" two indices as I did, the result is a component of a vector.
     
  6. Mar 5, 2012 #5
    Does contracting over two indices mean that you make a sum of nine products?
    [itex]B_{i11}C_{11}+B_{i12}C_{12}+B_{i13}C_{13}+...+B_{i32}C_{32}+B_{i33}C_{33}[/itex]
     
  7. Mar 5, 2012 #6

    Fredrik

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    Yes, that's what it means.
     
  8. Mar 5, 2012 #7
    Thank you very much!
     
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