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Isotropic tensor functions

  1. Nov 4, 2006 #1
    Hello,

    I consider only Cartesian tensors in the following. The definition of
    isotropic tensor function I know is

    1) T = F ( G )

    such that, for any rotation ( ' = transpose),

    2) O F( G ) O' = F( O G O' )

    But, if I change to component notation, it seem to me that any tensor
    function is isotropic, which cannot obviously be. Denoting the
    components in the new basis with ^*, I have

    3a) T_ij^* = O_ir T_rs (O_sj)'

    3b) G_ij^* = O_ir G_rs (O_sj)'

    since T and G are tensors. Then, by 1),

    4a) T_rs = F_rs ( G_mn )

    4b) T_rs^* = F_rs^* ( G_mn^* )

    Then, substituting 3a) and 3b) into 4b), I get

    5) O_ir T_rs (O_sj)' = F_rs^* ( O_mk G_kl (O_ln)' )

    Finally, substituting 4a) into 5), I have

    6) O_ir F_rs ( G_mn ) (O_sj)' = F_rs^* ( O_mk G_kl (O_ln)' )

    that is,

    O F( G ) O' = F (O G O' )

    So any tensor function would be isotropic. Clearly that's false, but I
    don't see where the error is. Can you help me find it? Thanks,

    Andrea
     
  2. jcsd
  3. Feb 11, 2009 #2
    Here's an example of a tensor valued, anisotropic function

    [tex]H = g(T) = (T\cdot{n\otimes{n}})n\otimes{n}[/tex]
    where n is a fixed unit vector.

    Then
    [tex]
    g(QTQ^T) = (QTQ^T\cdot{n\otimes{n}})n\otimes{n} = (T \cdot (Q^Tn \otimes Q^Tn)) n\otimes{n}
    [/tex]
    which is not the same as
    [tex]
    Qg(T)Q^T = (T\cdot{n\otimes{n}})Qn\otimes{Qn}
    [/tex]
    for general orthogonal Q, only if n is a proper vector of Q.
    Im basically extrapolating from Jaunzemis' Continuum Mechanics book, pg 287.
     
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