When we have a rigid body, the rigidity constraint allows us to write the trajectory $\mathbf{r}_i$ of the $i$-th particle as

$$\mathbf{r}_i(t) = R(t)\mathbf{b}_i + \mathbf{w}(t),$$

where we are using a $\mathbf{b}_i$ as the initial position of the $i$-th particle with respect to one stationary frame and where $R(t)\in SO(3)$ is a path of rotations.

In that case we have the velocity


thus recognizing that $R'(t)R^{-1}(t)\in \mathfrak{so}(3)$ and remembering this is the Lie Algebra of antisymmetric $3\times 3$ matrices, there are numbers $\omega_1,\omega_2,\omega_3$ so that

$$R'(t)R^{-1}(t)=\begin{pmatrix}0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 &-\omega_1 \\ -\omega_2 & \omega_1 &0\end{pmatrix},$$

we are able to write the action $R'(t)R^{-1}(t)$ as $\omega(t)\times$ where $\omega(t)=(\omega_1(t),\omega_2(t),\omega_3(t))$. Thus we find

$$\mathbf{v}_i(t)=\omega(t)\times \mathbf{r}_i(t)-\omega(t)\times \mathbf{w}(t)+\mathbf{w}'(t),$$

and we recognize $\omega(t)$ as the angular velocity. This constructions is quite nice as to explain where the angular velocity comes from and its properties.

On the other hand, I don't know how to properly identify what reference frame the components $\omega_i$ are with respect to. The obvious thing would be to say that it is with respect to the stationary system, but I don't know how to justify this.

Are these components really with respect to the stationary frame? If they are, how do we justify it? If not, these components are with respect to what frame?

  • $\begingroup$ I have no time to properly answer, but the answer is that every reference frame is OK. You fix a reference frame where you describe everything and write down your equations in components. It may not coincide with the one at rest with the body or at rest with the laboratory. $R(t)$ denotes the rotation matrices referred to the axes of that reference frame (exactly as the components of all the involved vectors). The components of $\omega$ appearing in your equation are referred to that basis. $R'$ is the matrix appearing in the equation of the velocity, equation written in the basis you fixed. $\endgroup$ Oct 1, 2016 at 7:13
  • $\begingroup$ The velocity is that referred to the laboratory (not the basis your arbitrarily fixed). I know it is a bit complicated, I hope to have time to write an extended answer later. $\endgroup$ Oct 1, 2016 at 7:18
  • $\begingroup$ My answer here may be helpful : physics.stackexchange.com/questions/67053/…. Be careful with my Symbol Conventions therein. $\endgroup$
    – Frobenius
    Oct 1, 2016 at 8:13

1 Answer 1


The answer is within your question. "... the ith particle with respect to a stationary frame..." When you select the stationary frame, then the equations are instantiated (valid) for that particular frame.


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