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I'm trying to understand an argument in Taylor's Classical Mechanics section 9.3.

Consider two frames, 1 and 2, and a body, 3. Let $\mathbf v_{21}$ be the relative velocity of frame 2 w.r.t. frame 1, and let $\mathbf v_{3i}$ be the relative velocity of body 3 w.r.t. frame $i$. Taylor then simply states that

$$ \mathbf v_{31} = \mathbf v_{32} + \mathbf v_{21}. $$

Then he considers the case where the two frames have the same origin, and frame 2 is rotating relative to frame 1 around this origin. He next uses the well-known result (which he gives a geometric argument for) that $\mathbf v = \boldsymbol\omega \times \mathbf r$, where $\boldsymbol \omega$ is the angular velocity of the rotating frame. He uses this to show (by linearity of the cross product) that

$$ \boldsymbol \omega_{31} = \boldsymbol \omega_{32} + \boldsymbol \omega_{21}, $$

where the indices refer to the same things as before, and body 3 also rotates around the common origin.

First of all, what does it mean for a body to have a velocity "relative to a frame"? It makes sense to me to talk about velocity relative to a point, but surely the way we choose to orient our axes doesn't change the velocity vector, only the coordinates.

Secondly, what does it mean for two frames to have a relative velocity? If one frame is rotating, it seems like the relative velocity of the two frames would depend on which points you measure it relative to. But the only non-arbitrary way to measure it would be the relative velocity of their origins, but that is zero if their origins coincide (as they do here).

I guess I'm also concerned about what relative angular velocity means, but it seems less problematic in this case, since the origins (around which everything rotates) coincide.

Optimally, I'm looking for precise definitions of terms and rigorous arguments for the above identities, relying as little as possible on my physical intuition.

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First of all, what does it mean for a body to have a velocity "relative to a frame"? It makes sense to me to talk about velocity relative to a point, but surely the way we choose to orient our axes doesn't change the velocity vector, only the coordinates.

This just means that if you were to be at rest relative to the frame you would observe the body to have that velocity.

Secondly, what does it mean for two frames to have a relative velocity? If one frame is rotating, it seems like the relative velocity of the two frames would depend on which points you measure it relative to. But the only non-arbitrary way to measure it would be the relative velocity of their origins, but that is zero if their origins coincide (as they do here).

I think here you just need to think of it point by point. Certainly since one frame is rotating relative to the other, each point in one frame has a different relative velocity. $\mathbf v_{21}$ depends on what coordinate you are looking at. This is shown in your expression $\mathbf v = \boldsymbol\omega \times \mathbf r $. In other words, $\mathbf v_{21}$ is a function of $\mathbf r$ in the case of the rotating frames.

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  • $\begingroup$ But I can be at rest relative to many different frames at once, so presumably the velocities relative to all of those would be the same. I guess that makes sense. Only relative to frames that are accelerating relative to each other would the velocity be different. And what is a "point in a frame"? I am imagining a point in space that is fixed relative to a set of axes. So the velocity of frame 2 w.r.t. frame 1 is in fact a (generally) time-dependent function that maps points in frame 2 to their velocity as measured by an observed at rest relative to frame 1. Would that be correct to say? $\endgroup$ – Danny Hansen Aug 4 at 14:55
  • $\begingroup$ @DannyHansen There is only one frame you are at rest in. It is called your "rest frame". This frame may or may not be inertial. And your second point is right. For example, in the rotating case you ask about $\mathbf v_{21}$ of a point would be time dependent. $\endgroup$ – Aaron Stevens Aug 4 at 15:55
  • $\begingroup$ I think that uses a different definition of reference frame than the one I was taught (which seems to consider an origin and a set of axes as intrinsic to the frame), but then I don't think I've ever seen a definition of reference frames that I was comfortable with, so never mind. The difference in definition certainly doesn't matter when only considering velocities and not positions. I may have to come back to you after I've thought this through properly. But thank you so far for your help! $\endgroup$ – Danny Hansen Aug 4 at 16:42
  • $\begingroup$ Just to belabour the point, Taylor himself defines/characterises a reference frame as "a choice of spatial origin and axes to label positions [...] and a choice of temporal origin to measure times". I'm much more comfortable with a definition like this than one that e.g. refers to an abstract rigid body that moves along with an observer. The latter can then be constructed from the former by partitioning the set of reference frames by the obvious equivalence relation. $\endgroup$ – Danny Hansen Aug 4 at 16:48
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This notation doesn't not make sense for me:

$$\vec{v}_{31}=\vec{v}_{32}+\vec{v}_{21}$$

how can you add the vector components if they are in different frames ?

so I prefer this notation :

a vector from point 1 to point 2 where the arrow is at point $2$ is:

$$\vec{v}_{12}$$

if you change the indices you change the sign :

$$\vec{v}_{21}=-\vec{v}_{12}$$

I need additional index for the frame in which the vector components are given.

$$\left(\vec{v}_{12}\right)_B$$

mean the vector components are given in coordinate system $B$

vector addition:

$$\left(\vec{v}_{12}\right)_B=R_{CB}\left(\vec{v}_{13}\right)_C+R_{DB}\left(\vec{v}_{32}\right)_D$$

Where $R$ is the transformation matrix between the frames

you can omit the frame index if all the components are in the same frame

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  • $\begingroup$ It's pretty standard notation for relative velocities actually. $\endgroup$ – Aaron Stevens Aug 4 at 14:56
  • $\begingroup$ but how can you add those vectors in different frame ? any why this is just notation and we speak also different languages $\endgroup$ – Eli Aug 4 at 14:58
  • $\begingroup$ That's the point of relative velocities. The velocity of $3$ as observed by $1$ is the velocity of $2$ as observed $1$ plus the velocity of $3$ as observed by $2$. $\endgroup$ – Aaron Stevens Aug 4 at 15:01
  • $\begingroup$ In the first example, the two frames are parallel but not coincident. $\endgroup$ – ja72 Aug 4 at 22:25
  • $\begingroup$ This is not the general case $\endgroup$ – Eli Aug 5 at 7:45

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