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?

  • $\begingroup$ 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. $\endgroup$
    – Eli
    Nov 1, 2022 at 8:04

1 Answer 1


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*}

  • $\begingroup$ How did you get $\mathbf{M}^T$ <matrix> $\mathbf{M}$ ? Why is it not the other way around ($M$ <matrix> $M^T$) ? $\endgroup$ Nov 1, 2022 at 16:04
  • $\begingroup$ Which equation? $\endgroup$
    – Eli
    Nov 1, 2022 at 16:48
  • $\begingroup$ 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? $\endgroup$ Nov 1, 2022 at 16:53
  • $\begingroup$ \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*} $\endgroup$
    – Eli
    Nov 1, 2022 at 17:09
  • $\begingroup$ 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*} $\endgroup$
    – Eli
    Nov 1, 2022 at 17:25

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