- #1
TadeusPrastowo
- 21
- 0
Homework Statement
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]
Homework 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.
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.