# Deriving tensor in Euler's equations for rigid body rotation

The answer to physics.stackexchange.com/questions/104513 gives the following derivation of tensor $I$:

\begin{align} \frac{\text{d}}{\text{d}t} I &= \frac{\text{d}}{\text{d}t} (E\,I_{\text{body}}\,E^\top) \\ &= \frac{\text{d}E}{\text{d}t}\,I_{\text{body}}\,E^\top + E\,I_{\text{body}}\,\frac{\text{d}E}{\text{d}t}^\top \\ &= ([\omega]_\times\,E)\,I_{\text{body}}\,E^\top + E\,I_{\text{body}}\,([\omega]_\times\,E)^\top \\ \frac{\text{d}}{\text{d}t} I&= [\omega]_\times\,I_{\text{body}} + I_{\text{body}}\,[\omega]_\times \end{align}

But, I want to see the derivation happens element-by-element according to the definition of a tensor as I attempted below.

\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 calculated the above one as follows: \begin{align} \frac{\text{d}I_{i,j}}{\text{d}t} &= \frac{\text{d}}{\text{d}t}\left(\sum_p m_p\,\left(\delta_{i,j} \sum_k r_k^2 - r_i\,r_j\right)\right) \\ &= \sum_p m_p \frac{\text{d}}{\text{d}t}\,\left(\delta_{i,j} \sum_k r_{p,k}^2 \right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \frac{\text{d}}{\text{d}t}\, \left(\sum_k r_{p,k}^2\right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}}{\text{d}t}\,(r_{p,k}^2) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}\,r_{p,k}^2}{\text{d}r_{p,k}}\frac{\text{d}\,r_{p,k}}{\text{d}t} - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k 2\,r_{p,k}\,\omega_k - (\omega_i\,r_{p,j} + r_{p,i}\,\omega_j) \\ \frac{\text{d}\text{I}_{i,j}}{\text{d}t} &= \sum_p m_p \begin{bmatrix} 2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & -(r_{p,1}\,\omega_2 + r_{p,2}\,\omega_1) & -(r_{p,1}\,\omega_3 + r_{p,3}\,\omega_1) \\ -(r_{p,2}\,\omega_1 + r_{p,1}\,\omega_2) & 2\,r_{p,1}\,\omega_1 + 2\,r_{p,3}\,\omega_3 & -(r_{p,2}\,\omega_3 + r_{p,3}\,\omega_2) \\ -(r_{p,3}\,\omega_1 + r_{p,1}\,\omega_3) & -(r_{p,3}\,\omega_2 + r_{p,2}\,\omega_3) & 2\,r_{p,1}\omega_1 + 2\,r_{p,2}\,\omega_2 \end{bmatrix} \end{align}

But, after trying some factorizations, I can't seem to get to the following:

$[\omega]_\times\,I_{\text{body}} + I_{\text{body}}\,[\omega]_\times$ $= \sum_p m_p \left( \begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix}\begin{bmatrix} r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\ -r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\ -r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2 \end{bmatrix} + \begin{bmatrix} r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\ -r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\ -r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2 \end{bmatrix}\begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix} \right)$ $= \sum_p m_p \left( \begin{bmatrix} r_{p,1}\,r_{p,2}\,\omega_3 - r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} + \begin{bmatrix} -r_{p,1}\,r_{p,2}\,\omega_3 + r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} \right)$ $= \sum_p m_p \begin{bmatrix} 0 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} \ne \sum_p m_p \begin{bmatrix} 2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & \ldots \\ \vdots & \ddots \\ \end{bmatrix}$

So, I should have made a mistake in the differentiation. But, where? And, how to proceed?

Or, is it because the definition of tensor cannot be used in performing the derivation? Why?

• Hint: Use the vector form of the parallel axis theorem $$I = I_{cm} - [\vec{r}]\times [\vec{r}]\times$$ and note that your big matrix of $[ r_{p,2}^2 + \ldots]$ is really $[r_p]\times$. Jun 2, 2014 at 18:56
• Title should be deriving the rate of change of the inertia tensor. Jun 2, 2014 at 18:59
• Funny how MIT just glosses over this part of the derivation in ocw.mit.edu/courses/aeronautics-and-astronautics/… Jun 2, 2014 at 19:10
• By your hint, should I have started my derivation from the following? $\frac{\text{d}I}{\text{d}t} = \frac{\text{d}}{\text{d}t} (I_{\text{CM}} - [\vec{r}_p]_\times[\vec{r}_p]_\times) = \frac{\text{d}I_{\text{CM}}}{\text{d}t} - \frac{\text{d}}{\text{d}t} ([\vec{r}_p]_\times[\vec{r}_p]_\times)$ Jun 2, 2014 at 19:41
• [Renewed] By your hint, should I have started my derivation from the following? $\frac{\text{d}I}{\text{d}t} = \frac{\text{d}}{\text{d}t} (I_{\text{CM}} - [\vec{r}_p]_\times[\vec{r}_p]_\times) = \frac{\text{d}I_{\text{CM}}}{\text{d}t} - \frac{\text{d}}{\text{d}t} ([\vec{r}_p]_\times[\vec{r}_p]_\times) = \frac{\text{d}I_{\text{CM}}}{\text{d}t} - (\frac{\text{d}[\vec{r}_p]_\times}{\text{d}t}[\vec{r}_p]_\times + [\vec{r}_p]_\times \frac{\text{d}[\vec{r}_p]_\times}{\text{d}t})$ Jun 2, 2014 at 19:48