# Tensor differentiation (element-by-element)

1. Jun 2, 2014

1. The problem statement, all variables and given/known data

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

where $\boldsymbol\{\mathbf{I}\boldsymbol\}$ is a tensor: $$\boldsymbol\{\mathbf{I}\boldsymbol\} = \begin{bmatrix} I_{1,1} & I_{1,2} & I_{1,3} \\ I_{2,1} & I_{2,2} & I_{2,3} \\ I_{3,1} & I_{3,2} & I_{3,3} \\ \end{bmatrix}$$

with $\text{I}_{i,j}$ as the following:
$$\text{I}_{i,j} = \delta_{i,j} \sum_k r_k^2 - r_i\,r_j$$

and $\boldsymbol\omega$ is a vector:

$$\begin{bmatrix}\omega_1 \\ \omega_2 \\ \omega_3\end{bmatrix}$$

2. Relevant equations

The relevant equation should be on how to perform $\frac{\text{d}}{\text{d}t}\left(\begin{bmatrix} I_{1,1} & I_{1,2} & I_{1,3} \\ I_{2,1} & I_{2,2} & I_{2,3} \\ I_{3,1} & I_{3,2} & I_{3,3} \\ \end{bmatrix}\right)$.
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

\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}

Then, I calculate the above one as follows:
\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}

But, multiplying the result of the differentiation with $\boldsymbol\omega$ does not yield $\boldsymbol\omega \times (\boldsymbol\{\mathbf{I}\boldsymbol\}\,\boldsymbol\omega) = \begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix}\left(\begin{bmatrix} r_2^2 + r_3^2 & -r_1 r_2 & -r_1 r_3 \\ -r_2 r_1 & r_1^2 + r_3^2 & -r_2 r_3 \\ -r_3 r_1 & -r_3 r_2 & r_1^2 + r_2^2 \end{bmatrix}\begin{bmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \end{bmatrix}\right)$:
\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}

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. Jun 3, 2014

### CAF123

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.

3. Jun 3, 2014

### bloby

Is $$\omega$$ the time derivative of r?