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  1. Aug 2, 2010 #1
    From Box 3.3, p. 85:

    Since [tex]
    S^{\alpha}_{\phantom{\alpha}\beta\gamma} = S(\omega^\alpha, e_\beta, e_\gamma)


    since S[tex]=S^{\alpha}_{\phantom{\alpha}\beta\gamma}e_\alpha\otimes\omega^\beta\otimes\omega^\gamma[/tex]

    is it then true that

    S[tex]=S(\omega^\alpha, e_\beta, e_\gamma)e_\alpha\otimes\omega^\beta\otimes\omega^\gamma\ ?[/tex]

    Also, to get a new tensor from an old tensor, one of the techniques is to contract two of the indexes with each other. Is this another form of contraction, namely:

    [tex]T_\gamma = S^{\alpha}_{\phantom{\alpha}\alpha\gamma} = S^{\alpha}_{\phantom{\alpha}\beta\gamma}\eta^{\beta}_{\phantom{\beta}\alpha} = S^{\alpha}_{\phantom{\alpha}\beta\gamma}\eta^{\beta\lambda}\eta_{\lambda\alpha}\ ?[/tex]

    Finally, why is the 1st term on the rhs of this equation transposed??

    [tex]\nabla([/tex]R[tex]\otimes[/tex]M[tex]) = (\nabla[/tex]R[tex]\otimes[/tex]M[tex])^T\ +\ [/tex]R[tex]\otimes\nabla[/tex]M
    Last edited: Aug 2, 2010
  2. jcsd
  3. Aug 2, 2010 #2

    yes. but the reason is easy if you remember [itex]\eta^\alpha_\beta = \delta^\alpha_\beta[/itex]. where [itex]\eta^\alpha_\beta[/itex] is defined
    by your second equality.

    Just a reminder of the way the indices line up. look at the lhs
    (Ra Mb),c = Ra,c Mb + RaMb,c.

    The order of indices is abc. So how do you make the 1st term on the right have
    the same order? Transpose the last two entries.
  4. Aug 3, 2010 #3
    Thanks for the explanation. So ok, my transpositioning skillz are weak... is it:

    (Ra,c Mb)T = MbT(Ra,c) T = Mb Rc,a?

    But then the order is bca != abc.
    Last edited by a moderator: Aug 3, 2010
  5. Aug 3, 2010 #4
    no. they're using transpose differently here.

    (and not in an especially great way - in my oppinion
    /see earlier in the chapter where they only transpose
    the last two indices of a rank 3 tensor/ )

    let's go slowly and see how this all works.

    start with two rank 1 tensors, R and M. They act on vectors to give numbers.
    We can form a rank two tensor R [itex] \otimes [/itex] M which "eats" two vectors and spits out a number. from this we can form a rank 3 tensor by using the "gradient".

    ok. say we had a rank two tensor S. the definition for the gradient
    says given 3 vectors "u,v,w" we have
    [tex] \nabla S (u, v, w) = \frac{\partial S_{ab}}{\partial x^c} u^a v^b w^c [/tex]
    or in the case of the S = R [itex] \otimes [/itex] M
    we have
    [tex] \nabla (R\otimes M) (u, v, w) = \frac{\partial (R_a M_b)}{\partial x^c} u^a v^b w^c = \frac{\partial R_a}{\partial x^c}M_b u^a v^b w^c + R_a \frac{\partial M_b}{\partial x^c} u^a v^b w^c [/tex]

    Now we want to make sense of these coordinate independently
    The second term is: [itex](R \otimes \nabla M )(u, v, w)[/itex].
    But the first term is: [itex](\nabla R \otimes M )(u, w, v)[/itex].

    Notice we've switched the order only of the last two slots. so we define a new
    tensor Transpose [itex] (\nabla R \otimes M) [/itex] such that
    for any three vectors u, v, w
    Transpose [itex] (\nabla R \otimes M) [/itex] (u,v,w) = [itex] (\nabla R \otimes M) [/itex] (u, w, v).

    That's it.
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