# Angular velocity in body-frame vs inertial frame

This question is based on the nomenclature and definitions in Structure and Interpretation of Classical Mechanics.

In Section 2.2, we start with a rotation $$\mathscr{M}(q(t))$$ where $$q$$ is the path describing the motion of the body. The rotation is later represented by the rotation matrix $$\mathbf{M}$$.

To paraphrase:

Let $$\vec{\xi}_\alpha(t)$$ be the vector to some constituent particle with the body in the orientation specified by q(t) given by $$q(t)$$, and $$\vec{\xi'}_\alpha$$ be the vector to the same constituent with the body in the reference orientation, then:

$$\vec{\xi}_\alpha(t) = \mathscr{M}(q(t)) \vec{\xi'}_\alpha\tag{2.11}$$

Through a series of steps, it shows how the angular velocity vector consists of the components that can be extracted from the skew-symmetric matrix $$D\mathbf{M} \mathbf{M}^\intercal$$. Or

$$\mathbf{\omega} = \mathscr{A}^{-1}( D\mathbf{M} \mathbf{M}^\intercal )$$

$$\mathscr{A}^{-1}$$ is defined as the function that can extract the components of the angular velocity vector from an anti-symmetric matrix. In Eq. 2.20, it defines the derivative of the position of any constituent particle in the rigid body as:

$$\dot{\vec{\xi}}_\alpha = \vec{\omega} \times \vec{\xi}_\alpha\tag{2.20}$$

While it is not explicitly mentioned anywhere, I assume that the angular velocity components mentioned above are in terms of the inertial reference frame. So after this, it goes on to state the following:

The components $$\omega'$$ of the angular velocity vector on the body axes are $$\omega'= \mathbf{M}^\intercal \omega$$, so

$$\mathbf{\omega}' =\mathbf{M}^\intercal \mathscr{A}^{-1}( D\mathbf{M} \mathbf{M}^\intercal )\tag{2.21}$$

So here is my question:

In Eq. 2.11, $$\mathbf{M}_\alpha$$ seems to transform $$\vec{\xi}$$ from its reference frame components $$\vec{\xi}'$$ to its body-fixed frame components at time $$t$$. But in the last equation, angular velocity is transformed from reference to body-frame components as $$\omega'= \mathbf{M}^\intercal \omega$$. Why is it $$\mathbf{M}^\intercal$$ for angular velocity and not $$\mathbf{M}$$?

Is this somehow inferred from the prior steps?

• skew-symmetric matrix of the angular velocity $\mathbf{\dot{M}} \mathbf{M}^\intercal$ . Matrix components of the angular velocity are given in inertial system. The transformation matrix transformed the vector components of from body to internal system.
– Eli
Nov 1, 2022 at 8:04

The angular velocity skew matrix is:

$$\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]_I= \mathbf{\dot{M}}\,\mathbf{M}^T\tag 1$$

where $$~\mathbf M~$$ is the transformation matrix between body system (B-system) and inertial system (I-system)

from equation (1) you obtain the components of the angular velocity vector in inertial system$$~(\vec\omega)_I~$$ , the components of the angular velocity vector in B-system are $$(\vec\omega)_B=\mathbf M^T\,(\vec\omega)_I$$

the components of the angular velocity vector in B-system can also obtained with this equation

\begin{align*} \mathbf M^T\,&\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]_I\,\mathbf M= \mathbf M^T\,\mathbf{\dot{M}}\,\mathbf{M}^T\,\mathbf M\quad \Rightarrow\\\\ &\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]_B= \mathbf M^T\,\mathbf{\dot{M}}\end{align*}

• How did you get $\mathbf{M}^T$ <matrix> $\mathbf{M}$ ? Why is it not the other way around ($M$ <matrix> $M^T$) ? Nov 1, 2022 at 16:04
• Which equation?
– Eli
Nov 1, 2022 at 16:48
• In your comment: " the components of the angular velocity vector in B-system are ...", how was that derived? Why is it $\mathbf{M}^T$ and not $\mathbf{M}$ that is used to transform $\omega_I$? And similarly, when transforming the skew-symmetric matrix, why do you left multiply by $\mathbf{M}^T$ and right-multiply by $\mathbf{M}$ instead of the opposite? Nov 1, 2022 at 16:53
• \begin{align*} &{_B^I}\mathbf{M}\quad,\text{ M is the transformation matrix from B-Frame to I-Frame and}\\ &{_I^B}\mathbf{M}=\mathbf M^T\,\quad,\text{ $~M^T~$ is the transformation matrix from I-Frame to B-Frame where}\\ &{_B^I}\mathbf{M}\,{_I^B}\mathbf{M}=\mathbf{I} \end{align*} thus vector transformation \begin{align*} &{_B^I}\mathbf{M} \vec{v}_B=\vec{v}_I\\ &{_I^B}\mathbf{M}\,\vec{v}_I=\vec{v}_B \end{align*}
– Eli
Nov 1, 2022 at 17:09
• Matrix transformation \begin{align*} &{_I^B}\mathbf{M}\, \mathbf A_I\,{_B^I}\mathbf{M}=\mathbf{A}_B\quad, \mathbf M^T\,\mathbf A_I\,\mathbf M=\mathbf A_B\\ & {_B^I}\mathbf{M}\, \mathbf A_B\,{_I^B}\mathbf{M}=\mathbf{A}_I\quad, \mathbf M\,\mathbf A_B\,\mathbf M^T=\mathbf A_I \end{align*}
– Eli
Nov 1, 2022 at 17:25