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Tensor problem

  1. Nov 5, 2007 #1
    I hope no one minds but I'm gonna keep all my questions and solutions here and then write it down for the report =)

    1. The problem statement, all variables and given/known data
    1 Prove the following identities using index notation:
    a) For vectors, u, v, w,
    (u cross v) dot w = u dot (v cross w) XsolvedX

    b)
    For two second order tensors, show that trace(AB) = trace(BA) using index notation

    c)
    For two second order tensors, show that trace(AB) = trace(BA) using index notation
    (AB)^t = B^t A^t

    2 Consider a tensor V and a vector w related as follows
    V_kj = e_ijk*w_i

    a) Write out V as a 3 x 3 matrix

    b) Show, using index notation, that V is a skew symmetric tensor (V = -V^t)

    c) Show, using index notation or otherwise, that for any vector, v,
    Vv = w x v

    d) Show, using index notation, that the vector w is written in terms of V by,
    w_i = 1/2 * e_ijk * V_kj

    The vector w is known as the vector dual of the skew symmetric tensor V

    Note: For part (d) you may make use of one or the both identities:
    e_ijk * e_ipq = delta_jp * delta_kq - delta_kp * delta_jq
    e_ijk * e_jkr = e_ijk * e_rjk = 2delta_ir

    2. Relevant equations
    a)
    (u dot v) = uivj
    (u cross v) = eijkujvk
    (eijk is the permutation tensor)

    b)
    trace(A) = Aii


    3. The attempt at a solution
    (u cross v) dot w= eijkujvk * wi =
    eijkwiujvk =
    ejkiuivjwk =
    (as eijk = ejki) number of permutations are even:
    eijkvjwk * ui=
    u dot (v cross w)

    solved

    b)
    trace(AB) = trace(ABij) = trace(AikBkj) = AikBki = BkiAik = trace(BkiAjk) =
    trace(BAij) = trace(BA)

    c)
    A=Aik, A^t = Aki
    B=Bkj, B^t = Bjk
    (AB)^t = (AikBkj)^t .. not quite sure how the transpose operation would work on a multiplication

    2
    a)
    V_kj =
    w1, w1, w1
    w2, w2, w2
    w3, w3, w3

    No idea what to do with e_ijk...

    b)
    For A to be skew symmetric, A^t = -A

    V_kj^t = ... Do not know how to transpose e_ijk * w_i
    possibly: e_ikj * w_k
    Then permutation k <-> j:
    -e_ijk * w_k = -V_kj

    Is this allowed?
     
    Last edited: Nov 5, 2007
  2. jcsd
  3. Nov 5, 2007 #2

    Meir Achuz

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    Gold Member

    a) your cyclic permuttation of ijk is correct because e_ijk is invarinat ulnder cyclic permutation.

    b) recall (AB)_ij=AikBkj. Now the trace sets i=j and sums.
     
  4. Nov 5, 2007 #3
    edit: check first post
     
    Last edited: Nov 5, 2007
  5. Nov 7, 2007 #4
    Comon people I'm really struggling with this, no one done tensors?
     
  6. Nov 7, 2007 #5

    cristo

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    Staff Emeritus
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    Sorry, I forgot that we were at your beck and call :rolleyes:
     
  7. Nov 7, 2007 #6
    Helpful, I hope you can do tensors and aren't just wasting my time
     
  8. Nov 7, 2007 #7

    cristo

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    Staff Emeritus
    Science Advisor

    For (c) you know how to write a product, you know how to write a transpose, and so you should be able to write the transpose of that product.

    The parts you've answered from 2 look correct.
     
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