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Transformation of Rank 2 mixed tensor

  1. Dec 12, 2007 #1
    Thanks for the help on the other questions.

    I am having trouble with another derivation. Unlike the others, it's not abstract whatsoever.

    Okay I wish to find the transformation Law for the components of a rank 2 tensor.

    Easy, I know: [tex]T: V^* \times V \mapsto \mathbb{R}[/tex]

    So

    [itex]T = T^i_{\phantom{i} j} e_i \otimes e^j[/itex]

    I wish to find

    [itex]T^i'_{\phantom{i'} j'}[/itex]

    Where the following hold:

    [tex]e_{i'}=a^{J}_{\phantom{J} i'} e_J[/tex]

    and

    [tex]e^{i'}=b^{i'}_{\phantom{i'} J} e^J[/tex]

    where, the coefficients are just real numbers. Now [itex]T^i_{\phantom{i} j}=T(e^i, e_j)[/itex] so that:

    [itex]T^{i'}_{\phantom{i'} j'}=T(e^{i'}, e_{j'})[/itex]

    [itex]=T(b^{i'}_{\phantom{i'} J} e^J, a^{L}_{\phantom{L} j'} e_L)[/itex]

    By linearity of [itex]\otimes[/itex] we have:

    [itex]=b^{i'}_{\phantom{i'} J} a^{L}_{\phantom{L} j'} T( e^J, e_L)[/itex]

    [itex]=b^{i'}_{\phantom{i'} J} a^{L}_{\phantom{L} j'} T^{J}_{\phantom{J} L}[/itex]

    Now my lecturer done a funny thing and said,

    "it may be shown that [itex]b=a[/itex]"

    which confuses the hell outta me because,

    [tex]e_{i'}=a^{J}_{\phantom{J} i'} e_J[/tex]

    [tex]e^{i'}=b^{i'}_{\phantom{i'} J} e^J[/tex]

    abide by the duality relation so that:

    [tex]e^{i'} e_{j'}=b^{i'}_{\phantom{i'} J} e^J a^{K}_{\phantom{K} j'} e_K=\delta^{i'}_{j'}[/tex]

    So do the original bases obey their own set of duality relations so that:

    [tex]e^{i'} e_{j'}=b^{i'}_{\phantom{i'} J} a^{K}_{\phantom{K} j'} \delta^J_K=\delta^{i'}_{j'}[/tex]

    So

    [tex]e^{i'} e_{j'}=b^{i'}_{\phantom{i'} K} a^{K}_{\phantom{K} j'} =\delta^{i'}_{j'}[/tex]

    Is this not the definition of [itex]a[/itex] being the inverse of [itex]b[/itex]. Of course, I know a priori I am wrong as this would give [itex]T^{i'}_{\phantom{i'} j'}=T^{i}_{\phantom{i} j}[/itex] regardless of transformation.

    In my definition of [itex]a[/itex] and [itex]b[/itex], the superindex refers to row and the lower index refers to column.

    My gut feeling is that I am using the matrix notation all wrong.

    Any takers?
     
  2. jcsd
  3. Jan 5, 2008 #2

    dextercioby

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    Science Advisor
    Homework Helper

    My take is this one:

    [tex] T=T_{i}{}^{j}e^{i}\otimes e_{j}=T_{i'}{}^{j'} e^{i'}\otimes e_{j'} [/tex]

    Assume

    [tex] e^{i'}= A^{i'}{}_{i} e^{i} [/tex] and

    [tex] e_{j'}= B_{j'}{}^{j} e_{j} [/tex] .

    You know that [tex] e_{j}(e^{i})=\delta_{j}^{i} [/tex].

    Now, if you want to have [tex] e_{j'}(e^{i'})=\delta_{j'}^{i'} [/tex], that is keep the orthonormality relation, they you must necessarily have

    [tex] A^{-1}=B^{T} [/tex]

    which means that the components of T are invariant under a transformation, bacause the basis in the tensor product space is invariant.
     
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