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Velocity reciprocity means that the velocity $v$ of a reference frame $R$ with respect to another reference reference frame $R'$ is the opposite of the velocity $v'$ of $R'$ with respect to $R$, i.e. $v'=-v$.

I would like to learn about the experimental evidence - in particular, is it really possible to compare the directions of $v$ and $v'$ or can we only conclude that \begin{equation} \|v\|=\|v'\| \end{equation} (i.e. that the speeds associated to $v$ and $v'$ are equal)?

Filippo
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  • I don't really get what you think the problem here is - if you give both frames a reference direction (like via a compass or south-pointing chariot), why would they not be able to "compare directions"? – ACuriousMind Sep 25 '22 at 15:13
  • @ACuriousMind I am not sure I understand the concept of a "reference direction". Could you please elaborate on how a compass or a south-pointing chariot allows us to establish that $v'=-v$? (Of course you are welcome to post an answer...) – Filippo Sep 25 '22 at 16:00

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If we want to be pedantic about it (and if we don't want to say that the question is trivial), the statement of "velocity reciprocity" does not make sense on its face in a fully general setting: Mathematically, positions are points in the space(time) manifold, and velocities are tangent vectors attached to these points; saying that $v_1 = -v_2$ for two velocity vectors only makes sense when they're attached to the same point, otherwise these are simply incomparable objects. When you attach reference frames to two moving different objects, unless the objects are currently colliding, we therefore first need to establish what the statement that their velocities are "equal" actually means:

What allows us to compare tangent vectors at different points is parallel transport. In general, this transport depends on a path, but in flat space, it does not, and this explains why all this stuff about tangent vectors usually doesn't play a role in introductory physics course: In flat space, we can just identify all the tangent spaces with each other, since there's a unique flat parallel transport operator between them.

In curved space, the question becomes ill-defined as there is no unique definition of relative velocities in that case since parallel transport varies with the path we might choose (see e.g. this excellent answer by A.V.S).

So, assuming we're in flat space, there isn't really any problem: The simplest way to implement parallel transport there is just to draw a grid of straight lines (i.e. geodesics) on the floor. Then both your frames have a coordinate system given by the tangent vectors of these geodesics (and additionally the grid makes measuring velocity easy with a stopwatch and some geometry), and you can tell the people in the frames to measure velocity and you'll get $v_1 = -v_2$.

hft
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ACuriousMind
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  • Maybe I have misunderstood your answer, but I tried to work out the details and I ended up with the result that$$v'=-v$$is equivalent to$$\begin{pmatrix}0\v\end{pmatrix}=\gamma\begin{pmatrix}\frac vc\cdot\frac vc\v\end{pmatrix}$$Did you expect something like that? This seems like a weird result... – Filippo Sep 26 '22 at 11:42
  • @Filippo I don't really know what details you think there are to work out. It seems you have a much more specific question in mind than "how can two different observers compare directions", but you didn't actually ask that question! – ACuriousMind Sep 26 '22 at 11:47
  • "But you didn't actually ask that question!" - Indeed, I asked for experimental evidence, but I thought that you were saying that we can give some sort of proof for a Minkowski spacetime. Here is how I interpreted your answer: – Filippo Sep 26 '22 at 12:32
  • If we define a reference frame as an equivalence class of charts, then each frame defines a decomposition of the tangent space $V^4$ into a $1$-dim. space $V^1$ and a $3$-dim. space $V^3$ (parallel transport allows me to act as if there is only one tangent space) and it is not difficult to define the velocity of $R'$ w.r.t. $R$ as a vector in $V^3_R$. I thought that maybe we can show that $v,v′\in V^3_R\cap V^3_{R'}$ and $v'=−v$. But this doesn't seem to work... – Filippo Sep 26 '22 at 12:37
  • @Filippo you're overcomplicating things: The abstract mathematical reason relative velocities are reciprocal is just that the inverse of the Lorentz boost with velocity $v$ is the Lorentz boost with velocity $-v$, which you can prove with elementary linear algebra. – ACuriousMind Sep 26 '22 at 13:02
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What do you call "experimental evidence" in an experiment you dicide what is the positiv direction. So if you move say in a train with v in positive direction , the trees along the road move from your standpoint in negative direction (or backwards) with the same speed so train R, trees R' than v'=-v

trula
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Consider two railway carriages, $A$ and $B$ which were reference frames $A$ and $B$ respectively, moving towards one another on a straight railway track aligned in $\hat x$ direction.

By using a radar type device situated on carriage $A$, the speed of approach of carriage $B$ was measured to be $v_{\rm BA}$ in the $-\hat x$ direction so the velocity of carriage $B$ relative to carriage $A$ is $v_{\rm BA}(-\hat x) = -v_{\rm BA}\hat x$.
While the measurements were going on in carriage $A$ the same set of measurements were also going on with in carriage $B$ with the result that the velocity of carriage $A$ relative to carriage $B$ was found to be $v_{\rm AB}\hat x$.

Those doing the experiment then compared results and found that $v_{\rm BA} = v_{\rm AB}$ which is the experimental evidence that you required.

Farcher
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  • The equation $v_{\rm BA} = v_{\rm AB}$ is the equality of the speeds (i.e. the magnitudes of the velocities), right? – Filippo Sep 21 '22 at 11:49
  • @Filippo I suddenly realised that I had set up an experiment in which the speed of the carriages relative to the ground were being measured! I think that is what you were pointing out to me? – Farcher Sep 21 '22 at 13:17
  • Indeed you implicitly consider a third reference frame $C$ (the ground), don't you? By assumption, $A$ and $B$ move in opposite directions in $C$, but this is different from "velocity reciprocity", isn't it? In fact, $A$ and $B$ do not necessarely move in opposite directions in a fourth frame $D$, do they? – Filippo Sep 21 '22 at 13:42
  • @Filippo The arguement is the same if the carriages are moving away from one another. – Farcher Sep 21 '22 at 15:15
  • This doesn't change the fact that they have opposite velocities in $C$. Or what do you mean? – Filippo Sep 21 '22 at 16:08
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I think I know now, what your concern is. In the claim v'=-v it is taken for granted that both systems have the same coordinate system, so for both for example on earth the direction going north is positiv, going south is negativ. If R and R' have different definition of positiv direction , one can not compare the sign of velocity. So first the two moving systems have to agree what is the positiv direction, only than the sign of velocity makes sense. And than it is easy to see that the have opposite signs.

trula
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