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Tensor differentiation (element-by-element)

  1. Jun 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Proof the following:
    [tex]\frac{\text{d}\boldsymbol\{\mathbf{I}\boldsymbol\}}{\text{d}t} \, \boldsymbol\omega = \boldsymbol\omega \times (\boldsymbol\{\mathbf{I}\boldsymbol\}\,\boldsymbol\omega)[/tex]

    where [itex]\boldsymbol\{\mathbf{I}\boldsymbol\}[/itex] is a tensor: [tex]\boldsymbol\{\mathbf{I}\boldsymbol\} = \begin{bmatrix}
    I_{1,1} & I_{1,2} & I_{1,3} \\
    I_{2,1} & I_{2,2} & I_{2,3} \\
    I_{3,1} & I_{3,2} & I_{3,3} \\
    \end{bmatrix}[/tex]

    with [itex]\text{I}_{i,j}[/itex] as the following:
    [tex]\text{I}_{i,j} = \delta_{i,j} \sum_k r_k^2 - r_i\,r_j[/tex]

    and [itex]\boldsymbol\omega[/itex] is a vector:

    [tex]\begin{bmatrix}\omega_1 \\ \omega_2 \\ \omega_3\end{bmatrix}[/tex]

    2. Relevant equations

    The relevant equation should be on how to perform [itex]\frac{\text{d}}{\text{d}t}\left(\begin{bmatrix}
    I_{1,1} & I_{1,2} & I_{1,3} \\
    I_{2,1} & I_{2,2} & I_{2,3} \\
    I_{3,1} & I_{3,2} & I_{3,3} \\
    \end{bmatrix}\right)[/itex].
    But, I haven't found an online resource that shows the way. Wikipedia on tensor derivative does not touch on how to perform tensor derivative element-by-element.

    3. The attempt at a solution

    [tex]\begin{align}
    \frac{\text{d}}{\text{d}t}\left(\begin{bmatrix}
    I_{1,1} & I_{1,2} & I_{1,3} \\
    I_{2,1} & I_{2,2} & I_{2,3} \\
    I_{3,1} & I_{3,2} & I_{3,3} \\
    \end{bmatrix}\right) &= \begin{bmatrix}
    \frac{\text{d}\,I_{1,1}}{\text{d}t} & \frac{\text{d}\,I_{1,2}}{\text{d}t} & \frac{\text{d}\,I_{1,3}}{\text{d}t} \\
    \frac{\text{d}\,I_{2,1}}{\text{d}t} & \frac{\text{d}\,I_{2,2}}{\text{d}t} & \frac{\text{d}\,I_{2,3}}{\text{d}t} \\
    \frac{\text{d}\,I_{3,1}}{\text{d}t} & \frac{\text{d}\,I_{3,2}}{\text{d}t} & \frac{\text{d}\,I_{3,3}}{\text{d}t} \\
    \end{bmatrix}
    \end{align}[/tex]

    Then, I calculate the above one as follows:
    [tex]\begin{align}
    \frac{\text{d}\text{I}_{i,j}}{\text{d}t} &= \frac{\text{d}}{\text{d}t}\,(\delta_{i,j} \sum_k r_k^2 - r_i\,r_j) \\
    &= \frac{\text{d}}{\text{d}t}\,\delta_{i,j} \sum_k r_k^2 - \frac{\text{d}}{\text{d}t}\,r_i\,r_j \\
    &= \delta_{i,j} \frac{\text{d}}{\text{d}t}\, \sum_k r_k^2 - \frac{\text{d}}{\text{d}t}\,r_i\,r_j \\
    &= \delta_{i,j} \sum_k \frac{\text{d}}{\text{d}t}\,r_k^2 - \frac{\text{d}}{\text{d}t}\,r_i\,r_j \\
    &= \delta_{i,j} \sum_k \frac{\text{d}\,r_k^2}{r_k}\frac{\text{d}\,r_k}{\text{d}t} - \frac{\text{d}}{\text{d}t}\,r_i\,r_j \\
    &= \delta_{i,j} \sum_k 2\,r_k\,\omega_k - (\omega_i\,r_j + r_i\,\omega_j) \\
    \frac{\text{d}\text{I}_{i,j}}{\text{d}t} &= \begin{bmatrix}
    2\,r_2\,\omega_2 + 2\,r_3\,\omega_3 & -(r_1\,\omega_2 + r_2\,\omega_1) & -(r_1\,\omega_3 + r_3\,\omega_1) \\
    -(r_2\,\omega_1 + r_1\,\omega_2) & 2\,r_1\,\omega_1 + 2\,r_3\,\omega_3 & -(r_2\,\omega_3 + r_3\,\omega_2) \\
    -(r_3\,\omega_1 + r_1\,\omega_3) & -(r_3\,\omega_2 + r_2\,\omega_3) & 2\,r_1\omega_1 + 2\,r_2\,\omega_2
    \end{bmatrix}
    \end{align}
    [/tex]

    But, multiplying the result of the differentiation with [itex]\boldsymbol\omega[/itex] does not yield [itex]\boldsymbol\omega \times (\boldsymbol\{\mathbf{I}\boldsymbol\}\,\boldsymbol\omega) = \begin{bmatrix}
    0 & -\omega_3 & \omega_2 \\
    \omega_3 & 0 & -\omega_1 \\
    -\omega_2 & \omega_1 & 0
    \end{bmatrix}\left(\begin{bmatrix}
    r_2^2 + r_3^2 & -r_1 r_2 & -r_1 r_3 \\
    -r_2 r_1 & r_1^2 + r_3^2 & -r_2 r_3 \\
    -r_3 r_1 & -r_3 r_2 & r_1^2 + r_2^2
    \end{bmatrix}\begin{bmatrix}
    \omega_1 \\ \omega_2 \\ \omega_3
    \end{bmatrix}\right)[/itex]:
    [tex]
    \begin{align}
    \frac{\text{d}\text{I}_{i,j}}{\text{d}t} \, \boldsymbol\omega &= \begin{bmatrix}
    2\,r_2\,\omega_2 + 2\,r_3\,\omega_3 & -(r_1\,\omega_2 + r_2\,\omega_1) & -(r_1\,\omega_3 + r_3\,\omega_1) \\
    -(r_2\,\omega_1 + r_1\,\omega_2) & 2\,r_1\,\omega_1 + 2\,r_3\,\omega_3 & -(r_2\,\omega_3 + r_3\,\omega_2) \\
    -(r_3\,\omega_1 + r_1\,\omega_3) & -(r_3\,\omega_2 + r_2\,\omega_3) & 2\,r_1\omega_1 + 2\,r_2\,\omega_2
    \end{bmatrix}\begin{bmatrix}
    \omega_1 \\ \omega_2 \\ \omega_3
    \end{bmatrix} \\
    &= \begin{bmatrix}
    (2\,r_2\,\omega_2 + 2\,r_3\,\omega_3)\,\omega_1 - (r_1\,\omega_2 + r_2\,\omega_1)\,\omega_2 - (r_1\,\omega_3 + r_3\,\omega_1)\,\omega_3 \\
    -(r_2\,\omega_1 + r_1\,\omega_2)\,\omega_1 + (2\,r_1\,\omega_1 + 2\,r_3\,\omega_3)\,\omega_2 - (r_2\,\omega_3 + r_3\,\omega_2)\,\omega_3 \\
    -(r_3\,\omega_1 + r_1\,\omega_3)\,\omega_1 - (r_3\,\omega_2 + r_2\,\omega_3)\,\omega_2 + (2\,r_1\omega_1 + 2\,r_2\,\omega_2)\,\omega_3
    \end{bmatrix}
    \end{align}
    [/tex]

    I also have tried several factorizations on paper but to no avail.
    But, I may miss some wonderful factorization tricks.
    Or, should a tensor be differentiated element-by-element in another way?

    Thank you.
     
  2. jcsd
  3. Jun 3, 2014 #2

    CAF123

    User Avatar
    Gold Member

    Do you have to write it out explicitly using components?
    Perhaps if you write it like $$\left(\frac{\text{d}}{\text{d}t} I_{ij} \right)w_j = \epsilon_{ijk}w_j I_{kl}w_l,$$ (summation convention understood) where ##I_{ij} = \delta_{ij}\sum_k r_k^2 - r_i r_j## and show the two sides are equivalent. Have not tried it myself, but it might save you the mess of writing out all the components.
     
  4. Jun 3, 2014 #3
    Is $$\omega$$ the time derivative of r?
     
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