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A proof about tensor invariants

  1. Nov 13, 2017 #1
    1. The problem statement, all variables and given/known data
    How to proof the following property of tensor invariants?
    Where:
    ##[\mathbf{a\; b\; c}]=\mathbf{a\cdot (b\times c)} ##,
    ##\mathbf{T} ##is a second order tensor,
    ##\mathfrak{J}_{1}^{T}##is its first invariant,
    ##\mathbf{u, v, w}## are vectors.

    2. Relevant equations
    $$[\mathbf{T\cdot u\; v\; w}]+[\mathbf{u\; T\cdot v\; w}]+[\mathbf{u\; v\; T\cdot w}]=\mathfrak{J}_{1}^{T}[\mathbf{u\; v\; w}]$$

    3. The attempt at a solution
    $$T^{l}{ }_{i}u^{i}v^{j}w^{k}\epsilon_{ljk}+T^{l}{ }_{j}u^{i}v^{j}w^{k}\epsilon_{ilk}+T^{l}{ }_{k}u^{i}v^{j}w^{k}\epsilon_{ijl}
    $$$$=1/6(T^{l}{ }_{i}u^{i}v^{j}w^{k}\epsilon_{ljk}\epsilon_{\alpha \beta \gamma }\epsilon^{\alpha \beta \gamma }+T^{l}{ }_{i}u^{i}v^{j}w^{k}\epsilon_{ilk}\epsilon_{\alpha \beta \gamma }\epsilon^{\alpha \beta \gamma }+T^{l}{ }_{i}u^{i}v^{j}w^{k}\epsilon_{ijl}\epsilon_{\alpha \beta \gamma }\epsilon^{\alpha \beta \gamma })=???$$
     
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  3. Nov 14, 2017 #2

    Orodruin

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    Trying to blindly chop away with tensor calculus is not the most straightforward way here. I would rather suggest that you look at the properties of the left-hand side to draw conclusions on how it has to be expressed.

    Alternatively you could write out some of the sums explicitly and see several terms cancel to finally collect everything.
     
  4. Nov 14, 2017 #3
    Many thanks for the hint and I've worked out this problem, but is there a more concise and elegant way to give out a proof using tensor calculus?
     
  5. Nov 14, 2017 #4

    Orodruin

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    By forum rules, full solutions cannot be provided unless the OP (in this case you) have shown that they have solved the problem (i.e., have presented their solution), so I cannot answer that question until you post your solution.
     
  6. Nov 16, 2017 #5
    $$T_i^1 u^i v^2 w^3 + T_i^1 u^1 v^i w^2 + T_i^1 u^1 v^3 w^i=$$
    \begin{equation*}
    \left.\begin{aligned}
    T_1^1 u^1 v^2 w^3 + T_1^1 u^3 v^1 w^2 + T_1^1 u^2 v^3 w^1 \\
    T_2^1 u^2 v^2 w^3 + T_2^1 u^3 v^2 w^2 + T_2^1 u^2 v^3 w^2 \\
    T_3^1 u^3 v^2 w^3 + T_3^1 u^3 v^3 w^2 + T_3^1 u^2 v^3 w^3 \\
    \end{aligned}\right.
    \end{equation*}
    There're terms such as ##-T_1^1 u^1 v^3 w^2## which can be obtained by those terms by exchanging two indices once.
    So terms like ##-T_2^1 u^2 v^3 w^2##, which is obtained by changing the indices of v and w in ##T_2^1 u^2 v^3 w^2## will cancel out with ##T_2^1 u^2 v^3 w^2##
    and terms such as ##T_2^1 u^3 v^2 w^2## with repeated indices of u, v and w will cancel out by their own "negative" terms.
    Thus the terms
    \begin{equation*}
    \left.\begin{aligned}
    T_2^1 u^2 v^2 w^3 + T_2^1 u^3 v^2 w^2 + T_2^1 u^2 v^3 w^2 \\
    T_3^1 u^3 v^2 w^3 + T_3^1 u^3 v^3 w^2 + T_3^1 u^2 v^3 w^3 \\
    \end{aligned}\right.
    \end{equation*}
    will cancel out and the terms left are
    T_1^1 u^1 v^2 w^3 + T_1^1 u^3 v^1 w^2 + T_1^1 u^2 v^3 w^1 - T_1^1 u^1 v^3 w^2 - T_1^1 u^3 v^2 w^1 - T_1^1 u^2 v^1 w^3 \\
    That's exactly $$[T_1^1\left(u^i v^j w^k \varepsilon_{ijk}\right)$$
    Thus $$\mathbf{T\cdot u\; v\; w}]+[\mathbf{u\; T\cdot v\; w}]+[\mathbf{u\; v\; T\cdot w}]=$$
    \begin{equation*}
    \left.\begin{aligned}
    T_i^1 u^i v^2 w^3 + T_i^1 u^1 v^i w^2 + T_i^1 u^1 v^3 w^i= \\
    T_i^2 u^i v^2 w^3 + T_i^2 u^1 v^i w^2 + T_i^2 u^1 v^3 w^i= \\
    T_i^3 u^i v^2 w^3 + T_i^3 u^1 v^i w^2 + T_i^3 u^1 v^3 w^i=
    \end{aligned}\right.
    \end{equation*}
    $$=\left(T_1^1+T_2^2+T_3^3\right)\left(u^i v^j w^k \varepsilon_{ijk}\right)=
    \mathfrak{J}_{1}^{T}[\mathbf{u\; v\; w}]
    $$
     
  7. Nov 16, 2017 #6

    Orodruin

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    Ok, so the easier argument is to note that your entire expression is invariant and completely anti-symmetric under exchange of the vectors ##\vec u##, ##\vec v##, and ##\vec w##. Due to this, it must be on the form ##S_{ijk} u^i v^j w^k## where ##S_{ijk}## is a completely anti-symmetric tensor. Due to this, ##S_{ijk} = \alpha \epsilon_{ijk}## for some scalar ##\alpha##. Writing down the expression for ##S## leads to
    $$
    \alpha \epsilon_{ijk} = T_i^\ell \epsilon_{\ell jk} + T_j^\ell \epsilon_{i\ell k} + T_k^\ell \epsilon_{ij\ell}
    $$
    and contraction with ##\epsilon^{ijk}## now directly leads to
    $$
    6\alpha = 6T^\ell_\ell,
    $$
    i.e., ##S_{ijk} = T^\ell_\ell \epsilon_{ijk}##, and therefore
    $$
    S_{ijk}u^i v^j w^k = T^\ell_\ell \epsilon_{ijk} u^i v^j w^k = \mathfrak I^T_1 [\vec u \vec v \vec w].
    $$
     
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