# Velocity addition as a special case of change of reference frame

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 $$$$\mathbf{R}(t)=\mathbf{R}_{0}(t)+\mathbf{r}(t)=\mathbf{R}_{0}(t)+\Bbb{S}(t)\:\mathbf{r}^{\prime}(t)$$$$ 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, $$$$\dot{\mathbf{R}}(t)=\dot{\mathbf{R}}_{0}(t)+\dot{\Bbb{S}}(t)\:\mathbf{r}^{\prime}(t)+\Bbb{S}(t)\:\dot{\mathbf{r}}^{\prime}(t)$$$$ and solving for $$\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).

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 $$\omega=S\Omega$$.
• Just following up here. Thanks so much for the lovely answer. I am curious about "You then define $\omega$ as the angular velocity of the second frame with respect to the first frame." Is this standard? Is there another definition which we can show is equivalent to this one? I think this question is related to my question (4) on Frobenius's answer here (physics.stackexchange.com/questions/67053/…).