Velocity addition as a special case of change of reference frame

Question

In this question, I want to restrict the discussion to classical mechanics as understood before 1900; that is, to exclude any discussion of relativity (however, if there is a neat generalization I would be eager to hear about it).

As I go back and reread a classical mechanics textbook, I am again struck by how opaque solutions involving relative velocities and velocities in different frames are. As is well-known, if $\textbf{V}$ is the velocity of a particle in frame $S$ and $\textbf{V}_0$ is the velocity of frame $S'$ which moves rigidly relative to $S$, then the relative velocity of the particle in $S'$ is $$\textbf{v}' = \textbf{V}-\textbf{V}_0$$

My question is, is this just a special case of how we "translate" between different frames? As per this question/answer, the most general relationship that we can have between the position of a particle as expressed in two different frames (neither of which need be inertial -- this is just a statement about how different vectors are related) is \begin{equation} \mathbf{R}(t)=\mathbf{R}_{0}(t)+\mathbf{r}(t)=\mathbf{R}_{0}(t)+\Bbb{S}(t)\:\mathbf{r}^{\prime}(t) \end{equation} where I here use the notation of the linked answer.

Then do I obtain the rule for addition of velocities by simply differentiating the above and solving for $\textbf{v}' \equiv d\textbf{r}'/dt$? That is, using the standard dot notation for time derivatives, \begin{equation} \dot{\mathbf{R}}(t)=\dot{\mathbf{R}}_{0}(t)+\dot{\Bbb{S}}(t)\:\mathbf{r}^{\prime}(t)+\Bbb{S}(t)\:\dot{\mathbf{r}}^{\prime}(t) \end{equation} and solving for $\dot{\mathbf{r}}^{\prime}(t)$? The presence of $\mathbf{r}^{\prime}(t)$ seems to obscure this.

In the special case I mentioned at the start of the question wherein $S'$ moves rigidly so that $\Bbb{S} \equiv id$ I seem to recover the right answer, but I ask this question because I'm not sure if I'm missing the bigger picture somehow. These details don't seem to be mentioned as explicitly/mathematically as I would like in my textbook (Taylor's Classical Mechanics).

Answer 1

To link it with the common formula, since $S$ is orthogonal, it satisfies: $$ SS^T=1 $$ Taking the derivative: $$ \dot SS^T+S\dot S^T=0 $$ so $\dot SS^T$ is skew symmetric. In 3D, a skew symmetric operator can be uniquely represented as the cross product by a vector, so there exists $\omega$ such that: $$ \dot Sr’=\omega\times r $$ Your formula is therefore: $$ \dot R=V_0+\omega\times r+v’ $$ where I’ve set $v’=S\dot r’$ the velocity of the particle in the second frame (converted to the first frame), and $V_0=\dot R_0$.

You then define $\omega$ as the angular velocity of the second frame with respect to the first frame. This is a standard definition of angular velocity (perhaps not for engineers but for mathematicians/physicists it is and is related to the more general study of Lie algebra and Lie groups). Intuitively, $\dot S$ captures the velocity part, while the $S^T$ allows you to stay in the first frame. Note that $S^T\dot S$ is also skew symmetric. It corresponds to the angular velocity in the second frame $\Omega$: $$ \dot S r’=S(\Omega \times r’) $$ Note that both antisymmetric operators are related by a conjugation via $S$, or equivalently $\omega=S\Omega$.

Hope this helps.