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I Is Second rank tensor always tensor product of two vectors?

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  1. Apr 7, 2017 #1
    Suppose a second rank tensor ##T_{ij}## is given. Can we always express it as the tensor product of two vectors, i.e., ##T_{ij}=A_{i}B_{j}## ? If so, then I have a few more questions:
    1. Are those two vectors ##A_i## and ##B_j## unique?
    2. How to find out ##A_i## and ##B_j##
    3. As ##A_i## and ##B_j## has ##3+3 = 6## components in total (say, in 3-dimension), it turns out that we need only ##6## quantities to represent the ##9## components of the tensor ##T_{ij}##. Is that correct?
     
  2. jcsd
  3. Apr 7, 2017 #2

    fresh_42

    Staff: Mentor

    No, it is a sum of such products.
    No. Even for dyadics ##A_i \otimes B_j## you always have ##A_i \otimes B_j = c \cdot A_i \otimes \frac{1}{c}B_j## for any scalar ##c \neq 0##.
    ##T_{ij}## is basically any matrix and ##A_i \otimes B_j## a matrix of rank ##1##. So write your matrix as a sum of rank-##1## matrices and you have a presentation.
    No. See the rank explanation above.
     
  4. Apr 8, 2017 #3

    haushofer

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    Science Advisor

    To add to Fresh's answer: this last remark already should make you suspicious. An arbitrary (!) second-rank tensor in three dimensions has 3*3=9 components, but two vectors have 2*3=6 components. What you can do, is to decompose a second rank tensor like [itex]T_{ij}[/itex] as

    $$T_{ij} = T_{[ij]} + T_{(ij)} $$

    where [ij] stands for antisymmetrization, whereas (ij) stands for symmetrization. Both parts transform independently under coordinate transfo's. The antisymmetric part has 3 independent components, whereas the symmetric part has 6 components. You can even go further in this decomposition, because the trace of the tensor components also does not change under a coordinate transformation.
     
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