Inertia tensor of a rigid body rotating about an axis in body-frame vs space-fixed frame Consider a rigid body rotating about a fixed axis $\vec{OQ}=|\vec{OQ}|\hat{n}$ passing through two points $O$ and $Q$ in the body with a uniform angular velocity $\vec{\omega}=\omega\hat{n}$. With O as the origin of coordinates, the angular momentum about $O$ is $$\vec{L}=\sum_im_i\vec{r}_i\times(\vec{\omega}\times\vec{r}_i)=\sum_im_i[r_i^2\vec{\omega} -{\vec r}_i({\vec r}_i\cdot\vec{\omega})]   \qquad\qquad (1)$$ where $\vec{r}_i$ is the position vector of any point in the body w.r.t O.
I think that this expression $(1)$ is valid both in the body-fixed coordinate system (say, $OX'Y'Z'$) as well as in the space-fixed coordinate system (say, $OXYZ$) with both of their origins at the common point O. But if we consider the body-frame, the components of the moment of inertia tensor $I_{ij}$'s will be independent of time since the coordinates of any point inside the body, does not change w.r.t time and if we consider the space-fixed frame, the components of the moment of inertia tensor $I_{ij}$'s will be time-dependent.
Since we write the moment of inertia tensor often with time-independent components, it means that we implicitly use the body-frame when we write $$\vec{L}=\stackrel{\leftrightarrow}{I}\vec{\omega}, ~ {\rm or}\quad L_i=I_{ij}\omega_j.$$ Am I right?
 A: Given two reference frames, one of which is rotating at some 3D angular rate $\vec \omega$ with respect to the other, the time derivatives of some 3D vector quantity $\vec q$ in those two frames of reference are related by
$$\dot{\vec q}_A = \dot{\vec q}_B + \vec \omega \times \vec q$$
where

*

*$\dot{\vec q}_A$ is the time derivative of the vector quantity $\vec q$ as observed in frame $A$,

*$\dot{\vec q}_B$ is the time derivative of the vector quantity $\vec q$ as observed in frame $B$,

*$\vec \omega$ is the rotation rate of frame $B$ with respect to frame $A$, and

*$\vec q$ is the vector quantity in question.

This means that the time derivative of angular momentum as calculated from the perspective of an inertial frame and the time derivative of angular momentum as calculated from the perspective of a rotating frame via $\vec L_{\text{rotating}} = \boldsymbol I \,\vec \omega$ is
$$\dot L_{\text{inertial}} = \dot L_{\text{rotating}} + \vec \omega \times (\boldsymbol I\, \vec \omega) = \boldsymbol I \,\dot{\vec \omega} + \vec \omega \times (\boldsymbol I\, \vec \omega)$$
The inertial frame time derivative of the angular momentum vector is the sum of the external torques. Note that the time derivative of angular velocity, $\dot{\vec\omega}$, is the same vector in both the inertial frame and rotating frame as $\omega \times \omega$ is identically zero.
A: Yes, you're right:

*

*Expression (1) is a vector formula, and thus it doesn't depend on the coordinate system used.


*It's often useful to write inertia tensor in a material base, often a Cartesian basis $\{ \mathbf{\hat{E}}_i \}_i$, $\mathbb{I}_H = I_{H,ij} \mathbf{E}_i \otimes \mathbf{E}_j$, so that:

*

*components only depends on the geometry of the body, that thus don't depend on time;


*when we need to evaluate the time derivative, we only need to evaluate the time derivative of the vectors of the material basis. It we use a ortonormal basis, these time derivatives read
$\dot{\mathbf{\hat{E}}_i} = \boldsymbol{\omega} \times \mathbf{\hat{E}}_i$,
so that
$\dfrac{d \mathbb{I}_H}{dt} = I_{H,ij} \dfrac{d}{dt}\mathbf{E}_i \otimes \mathbf{E}_j + I_{H,ij} \mathbf{E}_i \otimes \dfrac{d}{dt} \mathbf{E}_j = \\
\quad \ \ \ = \boldsymbol{\omega} \times \left( I_{H,ij}\mathbf{E}_i \otimes \mathbf{E}_j\right) - I_{H,ij} \mathbf{E}_i \otimes \left( \mathbf{E}_j \times \boldsymbol{\omega} \right) = \\
\quad \ \ \ = \boldsymbol{\omega} \times \mathbb{I}_H - I_{H,ij} \mathbf{E}_i \otimes \left( \mathbf{E}_j \times \boldsymbol{\omega} \right)$
and, since $\left( \mathbf{E}_j \times \boldsymbol{\omega} \right) \cdot \boldsymbol{\omega} = 0$,
$\dfrac{d}{dt} \left( \mathbb{I}_H \cdot \boldsymbol{\omega} \right) = \mathbb{I}_H \cdot \boldsymbol { \dot{\omega}} + \boldsymbol{\omega} \times \left( \mathbb{I}_H \cdot \boldsymbol{\omega} \right)$.
