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Basic tensor properties

  1. Jul 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Task 1. Show that if components of any tensor of any rank vanish in one particular coordinate system they vanish in all coordinate systems.

    Task 2. The components of tensor T are equal to the corresponding components of tensor W in one particular coordinate system; that is, [tex]T^0_{ij} = W^0_{ij}[/tex].
    Show that tensor T is equal to tensor W, [tex]T_{ij} = W_{ij}[/tex] in all coordinate systems.

    2. Relevant equations

    3. The attempt at a solution
    task 1. I have no idea how to start
    task 2. transforming to the any other coordinate system I obtain:

    [tex]T_{i'j'} = A_{i'}^i A_{j'}^j T^0_{ij} = A_{i'}^i A_{j'}^j W^0_{ij} = W_{i'j'}[/tex] is it ok?
  2. jcsd
  3. Jul 25, 2009 #2


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    If you can do 2 you can do 1 by taking [itex]W_{ij}= 0[/itex]]!

    But it is easiest to prove 2 by proving 1 first:
    You apparently know that if [itex]T^0_{ij}[/itex] are the components of T in one coordinate system, then the coordinates [itex]T_{i'j'}[/itex] in any coordinates system are given by [itex]T_{i'j'}= A^i_{i'}A^j_{j'}T^0_{ij}[/itex]. Okay what if all components of [itex]T^0_{ij}[/itex] are 0?

    And if [itex]T^0_{ij}= W^0_{ij}[/tex], then [itex]T^0_{ij}- W^0_{ij}[/itex] is also a tensor, with all components 0.
    Last edited by a moderator: Jul 27, 2009
  4. Jul 25, 2009 #3
    Well the fact from the 1st task is for me intuitive but I don't understand it's formal proof. Because when some components of [tex]T_{ij}[/tex] are 0 (but not all) than the [tex]T_{i'j'}[/tex] can have no 0 components. So why when all components of [tex]T_{ij}[/tex] are 0 then [tex]T_{i'j'}[/tex] are 0 as well? Is it because changing coordinate system is a linear transformation and so 'A(0)=0'?
  5. Jul 27, 2009 #4
    Am I right?
  6. Oct 25, 2011 #5
    stress is an example of a tensor....If we have a bar (cuboid with a length much greater than other dimensions), and we apply axial forces to it, then there are only normal stresses...however, if we now cut off a piece from it diagonally, then on the inclined surface so obtained, we have both normal and shear stresses.....so the same thing appears different in two different perspectives....is this also a basic property of tensors??
  7. Oct 27, 2011 #6
    Please help!!
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