B V=dx/dt (relative velocity between two inertial observers)

[robphy]

Geometrically speaking, reduce the complexities of a (3+1)d-Spacetime by working in the plane made by the two worldlines and define their x-axes to lie on that plane. (This is the Spacetime interpretation of “let’s assume that the observers have a relative velocity along their x-axes”.)

So, we can draw the usual (1+1)d-spacetime diagrams so that each observer can find the spatial and time components of a displacement along the other worldline to define that worldline’s spatial velocity in the observer’s frame.

Yes, this can be formalized by writing projections operations (which is what we mean when we say “find the components” in some frame).

 

[Ibix]

So the point you are making is that the two inertial frames have different notions of "space", which are non-parallel 3d planes in 4d spacetime. Since my measurement of your three-velocity lies in my spatial plane, and your measurement of my three-velocity lies in your spatial plane, those vectors cannot be parallel.

For simplicity's sake, perhaps we could just observe that if I define my +x direction to be the direction you are going, and you define your -x direction to be the direction that I am going then: our y and z directions coincide (or can be made to coincide); your x and t directions lie in the plane defined by my x' and t' directions and vice versa; we measure equal three velocity magnitudes for each other; we regard those velocities to have opposite signs.

Let's expand on this a bit.

If I aim a laser pointer in the direction that I say you are travelling, and you aim one in the direction you say that I am travelling, every frame will agree that the three-velocities of the two beams are anti-parallel (actually this is restricted to frames moving in the $\pm x$ direction, although similar statements can be constructed in any other frame). This is the sense that people mean when they say that the two velocities are opposite. We are thinking of something like a pair of trains traveling away from each other along a straight track.

However, as @etotheipi and @robphy will agree, the three-velocities of those laser beams are not the three-velocities of the two observers. If we state the (correct) usual claim formally, it says that if I define your three-velocity to be $(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt})=(v,0,0)$ then you will define my three velocity to be $(\frac{dx'}{dt'},\frac{dy'}{dt'},\frac{dz'}{dt'})=(-v,0,0)$. This is correct. However, despite the components being the same with opposite signs, it does not mean that the velocities are opposite because we are not using the same coordinate system. We need to find a way to express your measure of my three-velocity in my coordinate system if we want to compare them.

The way to do that is to note that a spacelike three-vector is three components of a four-vector with the fourth component being zero (this is the same as a two-vector $(x,y)$ in a plane being expressible as a three-vector $(x,y,0)$). My measure of your three-velocity can be written in my un-primed coordinates as a four-vector $(0,v,0,0)$ (this is a purely spacelike vector lying in my "space" plane, not your four-velocity). Your measure of my three-velocity can be written in your primed coordinates as the four-vector $(0,-v,0,0)$. Now I should apply the inverse Lorentz transforms to your measure, which maps $(0,-v,0,0)$ to $(-\gamma v^2/c,-\gamma v,0,0)$. This latter form is in my coordinate system, and is clearly not parallel to $(0,-v,0,0)$.

So, to summarise, there are good senses in which you and I can be said to be traveling in opposite directions. It is also true that if I say that you have a speed of $v$ in the $+x$ direction then you will say I have a speed of $v$ in the $-x$ direction. However, since we don't share a common definition of "space", we cannot correctly say that our three-velocities (even extending them to four vectors) are anti-parallel. We can only say that they lie in the plane defined by our two four-velocities.

 

[etotheipi]

That was nicely explained; it's not such an easy thing to get used to. We can imagine again a train traveling along some straight tracks with constant speed, and consider the usual platform-fixed $O_p$ and train-fixed $O_t$ observers, who have established co-ordinate charts relative to which the other observer is moving parallel to their respective $x$-axes.

The restricted Lorentz transformation that maps $O_p$'s basis $(\mathbf{e}_{\mu})$ to $O_t$'s basis $(\tilde{\mathbf{e}}_{\mu})$ leaves the plane $\pi = \mathrm{span} (\mathbf{e}_2, \mathbf{e}_3)$ invariant, whilst its orthogonal complement $\pi^{\bot} = \mathrm{span}(\mathbf{e}_0, \mathbf{e}_1)$ is the time-like plane of the Lorentz boost within which $\mathbf{e}_0$ and $\mathbf{e}_1$ have been rotated into $\tilde{\mathbf{e}}_0 = \Lambda(\mathbf{e}_0)$ and $\tilde{\mathbf{e}}_1 = \Lambda(\mathbf{e}_1)$ respectively.

Although our Galilean thinking might initially lead us to think that $\mathbf{e}_1 = \tilde{\mathbf{e}}_1$, that's actually not the case precisely because in space-time these vectors are rotated with respect to each-other. Thus $v \mathbf{e}_1$ is not anti-parallel to $-v\tilde{\mathbf{e}}_1$ in space-time.

Luckily I don't think there's too much potential for confusion, because it's pretty obvious what someone means when they say the relative velocities are opposite, i.e. that $\mathrm{d}x/\mathrm{d}t = - \mathrm{d}x' / \mathrm{d}t'$. But I think it's rather nice to think about Lorentz boosts in terms of space-time rotations and planes.

 

[robphy]

It's important to realize that this issue about working in the plane of these worldlines
and these non-parallel spatial axes
is NOT just about Minkowski spacetime.
This is what we do in Euclidean space!
(Look at the x and x' axes in Euclidean space.)

From this unified [Cayley-Klein geometry] point of view (admittedly viewed in historical hindsight),
one should realize that
it is the Galilean case that is the special-case, the exception, the peculiarity!
But our everyday low-relative-speed "common sense" has
misled us to believe in the Galilean structure
...and makes it hard for us
to accept the Einstein-relativity/Minkowski-spacetime viewpoint [which agrees with experiment].

 

[vanhees71]

In SRT, the "relative velocity" usually denotes a "three-velocity" like $\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t$, which is a pretty complicated object, because it's not covariant. Particularly it is NOT the spatial part of a four-vector wrt. some Minkowski-orthonormal basis.

Suppose you have two particles. Then the relative velocity is the three-velocity of particle 2 wrt. the rest frame of particle 1. The problem is to get this relative velocity when the three-velocities $\vec{v}_j=\vec{\beta}_j c$ ($j \in \{1,2\}$) of the two particles are given wrt. some arbitrary inertial reference frame.

It's of course much more convenient to answer this question, using covariant objects, i.e., scalars, vectors, and tensors. Here we just use the (normalized) four-velocities, which are defined as
$$u_1=\gamma_1 \begin{pmatrix} 1 \\ \vec{\beta}_j \end{pmatrix}, \quad j \in \{1,2 \}.$$
To find the relative velocity we just need to boost into the reference frame, where particle 1 is at rest. The corresponding Lorentz transformation is of course with the boost-velocity $\vec{v}_1$, i.e.,
$$\hat{\Lambda}=\begin{pmatrix} \gamma_1 & -\gamma_1\vec{\beta}_1^{T} \\ -\gamma_1 \vec{\beta}_1 & I+(\gamma_1-1) \frac{\vec{\beta}_1 \otimes \vec{\beta}_1}{\beta_1^2} \end{pmatrix}.$$
It's easy to check that then indeed $u_1'=\hat{\Lambda} u_1=(1,0,0,0)^T$. So we really boost into the rest frame of particle 1. Then the relative four-velocity is simply given by $u_2'=\hat{\Lambda} u_2$. After some algebra (nothing really complicated but a bit of somewhat tedious algebraic work ;-)) you find for the relative three-velocity
$$\vec{\beta}_{\text{rel}2}=\frac{\vec{u}_2'}{u_2^{\prime 0}}=\frac{1}{1-\vec{\beta}_1 \cdot \vec{\beta}_2} \left [ \vec{\beta}_2-\vec{\beta}_1 + \frac{\gamma_1}{\gamma_1+1} \vec{\beta_1} \times (\vec{\beta_1} \times \vec{\beta}_2) \right].$$
Note that the corresponding relative velocity of particle 1 is NOT the negative of this, except when $\vec{\beta}_1 \parallel \vec{\beta}_2$.

 

[etotheipi]

There's also a nice way to find the magnitude $V$ of the relative velocity between two particles of masses $m_1$ and $m_2$ and 4-momenta $p_1^a$ and $p_2^a$. If we work in the rest frame of $m_1$, then $p_1^a = (m_1, \vec{0})$ and $p_2^a = (\frac{m_2}{\sqrt{1-V^2}}, \vec{p}_2')$ the product of the two 4-momenta can be written\begin{align*}
(p_1)_a (p_2)^a &= \frac{m_1 m_2}{\sqrt{1-V^2}} \implies V = \sqrt{1- \frac{(m_1m_2)^2}{[(p_1)_a (p_2)^a]^2}}

\end{align*}Let's now switch to any arbitrary frame of reference, in which $p_1^a = (E_1, \vec{p}_1)$ and $p_2^a = (E_2, \vec{p}_2)$. We can write $(p_1)_a (p_2)^a$ as\begin{align*}

(p_1)_a (p_2)^a &= E_1 E_2 - \vec{p}_1 \cdot \vec{p}_2 \\ \\

&= \frac{m_1 m_2}{\sqrt{1-v_1^2}\sqrt{1-v_2^2}} - \frac{m_1 m_2 \vec{v}_1 \cdot \vec{v}_2}{\sqrt{1-v_1^2}\sqrt{1-v_2^2}} \\ \\

&= \frac{m_1m_2 (1-\vec{v}_1 \cdot \vec{v}_2)}{\sqrt{(1-v_1^2)(1-v_2^2)}}

\end{align*}Hence, we can determine $V$ in terms of $\vec{v}_1$ and $\vec{v}_2$,\begin{align*}

V = \sqrt{1- \frac{(1-v_1^2)(1-v_2^2)}{(1-\vec{v}_1 \cdot \vec{v}_2)^2}} = \frac{\sqrt{v_1^2 -2 \vec{v}_1 \cdot \vec{v}_2 + v_2^2 - \left( v_1^2 v_2^2 - [\vec{v}_1 \cdot \vec{v}_2]^2 \right) }}{1-\vec{v}_1 \cdot \vec{v}_2}

\end{align*}However, $v_1^2 -2 \vec{v}_1 \cdot \vec{v}_2 + v_2^2 = (\vec{v}_1 - \vec{v}_2)^2$, and from vector analysis we also know that\begin{align*}
(\vec{v}_1 \times \vec{v}_2)^2 = (\vec{v}_1 \times \vec{v}_2) \cdot (\vec{v}_1 \times \vec{v}_2) &= (\vec{v}_1 \cdot \vec{v}_1)(\vec{v}_2 \cdot \vec{v}_2) - (\vec{v}_1 \cdot \vec{v}_2)(\vec{v}_2 \cdot \vec{v}_1) \\

&= v_1^2 v_2^2 - [\vec{v}_1 \cdot \vec{v}_2]^2

\end{align*}which results in\begin{align*}
V = \frac{\sqrt{(\vec{v}_1 - \vec{v}_2)^2 - (\vec{v}_1 \times \vec{v}_2)^2 }}{1-\vec{v}_1 \cdot \vec{v}_2}

\end{align*}which is, of course, symmetric in the two particles!

 

[vanhees71]

The magnitude ("relative speed") is symmetric, the vectors are not, except for $\vec{v}_1 \parallel \vec{v}_2$.

 

[Kairos]

The effect of velocity on time dilation is orientation-independent but the Doppler effect depends on orientation, so when the vectors are not parallel, can the two inertial observers perceive a different Doppler effect?

 

[Sagittarius A-Star]

The effect of velocity on time dilation is orientation-independent but the Doppler effect depends on orientation, so when the vectors are not parallel, can the two inertial observers perceive a different Doppler effect?

As you can see at the example of the moving light clock, the angle between the vectors is in general differently in the 2 frames. This effect is called "aberration".

If the angle is for example 90° in the receivers frame, then there is a Doppler redshift by $1/\gamma$.
But if the angle is 90° in the senders frame, then there is a Doppler blueshift by $\gamma$.

So, that in each frame the moving clock ticks slower than the clock at rest does not lead to a contradiction.

 

[PeroK]

As you can see at the example of the moving light clock, the angle between the vectors is in general differently in the 2 frames. This effect is called "aberration".

If the angle is for example 90° in the receivers frame, then there is a Doppler redshift by $1/\gamma$.
But if the angle is 90° in the senders frame, then there is a Doppler blueshift by $\gamma$.

So, that in each frame the moving clock ticks slower than the clock at rest does not lead to a contradiction.

That's not the question. The question is: if relative motion is not symmetric, then can we see differences in the Doppler shift?

 

[Sagittarius A-Star]

That's not the question. The question is: if relative motion is not symmetric, then can we see differences in the Doppler shift?

Sorry, I did not read the complete thread :-(

 

[PeroK]

The effect of velocity on time dilation is orientation-independent but the Doppler effect depends on orientation, so when the vectors are not parallel, can the two inertial observers perceive a different Doppler effect?

If we imagine that object $B$ has coordinates $(t, x_0 + v_xt, y_0 + v_yt, z_0 + v_zt)$ in object $A$'s rest frame. The components $(v_x, v_y, v_z)$ define a three-vector, which we can take to be the new x-axis. Likewise we can take the new y-axis so that object $B$ has coordinates $(t, x_0 + vt, y_0, 0)$, where we take $v > 0$ and $y_0 > 0$..

Now object $B$ can do the same and choose axes so that $A$ has coordinates $(t', v't', y'_0)$, where again we take $v' > 0, y'_0 > 0$.

The question is: do we necessarily have $v = v', x_0 = x'_0, y_0 = y'_0$?

My argument is that, by the isotropy and homogeneity of spacetime, we must. We must have complete symmetry of these coordinates. Relative motion must be physically completely symmetric.

Moreover, we could take the point of closest approach to define $t = 0$ and $t' = 0$ in each case.

Note that I've changed strategy to make the displacements and velocities the same, rather than negatives of each other, which further emphasises the symmetry.

By this argument, there must also be completely symmetric Doppler shift.

 

[vanhees71]

The effect of velocity on time dilation is orientation-independent but the Doppler effect depends on orientation, so when the vectors are not parallel, can the two inertial observers perceive a different Doppler effect?

Concerning the Doppler effect, I've just written a short note on the general case for waves with arbitrary phase velocity ($<c$, i.e., in the region of "normal dispersion"):

https://itp.uni-frankfurt.de/~hees/pf-faq/rela-waves.pdf

 

[vanhees71]

That's not the question. The question is: if relative motion is not symmetric, then can we see differences in the Doppler shift?

Yes, as was nicely explained in #39, the angles between the wave vector and the velocity of the source and/or observer are frame dependent. If you have in addition a medium, where the waves propagate in (e.g., a fluid for sound waves or a dielectric for electromagnetic waves), then everything also depends on the velocity of the medium too. You find the kinematics of the Doppler effect in all generality here:

https://itp.uni-frankfurt.de/~hees/pf-faq/rela-waves.pdf

 

[Kairos]

Thank you, I already knows that and #39, but that's not the point:

Is the "non-reciprocity" of relative velocity (introduced above in this thread) compatible with the reciprocity of the Doppler effect. If not, is the admitted reciprocity of the Doppler effect in SR also matter of debate.

 

[vanhees71]

What do you mean with "reciprocity"? There is no matter of debate concerning the Doppler effect in SR (nor in GR). For em. waves in vacuo you find the complete explanation of the Doppler effect already in Einstein's paper of 1905!

 

[PeroK]

Thank you, I already knows that and #39, but that's not the point:

Is the "non-reciprocity" of relative velocity (introduced above in this thread) compatible with the reciprocity of the Doppler effect. If not, is the admitted reciprocity of the Doppler effect in SR also matter of debate.

The apparent non-reciprocity of relative velocity is a coordinate effect, resulting by considering the velocities in a wider context, with a necessary convention for spatial coordinate axes.

Post #42 shows that if we consider only the two objects in relative motion, then we can choose coordinates so that the relative motion is seen to be physically completely symmetric. Using those coordinates, we can see that any effects, such as the Doppler effect,must be symmetric.

Note that if either A or B uses a different coordinate system than that proposed in post #42, then the complete physical symmetry of the situation may not be immediately apparent. In that case, it may take yet more mathematical bravura to show the complete physical symmetry of relative inertial motion.

 

[Kairos]

Many thanks! It is a coordinate effect but not a physical effect! I was destabilized by post #9:

There are some subtleties to the subject of the reciprocity of the relative velocity. It does not hold true in special relativity! But we ought to put discussion of that on hold until you have a firmer grasp of the underlying machinery.

 

[etotheipi]

Sigh! This quote refers to the fact that the relative velocities each measures for the other are elements different local rest spaces (see the image in #30), and so when considered as vectors in $\mathbf{R}^4$ they obviously cannot be parallel not least because although the intersection $E_{u_1} \cap E_{u_2}$ of those rest spaces is still non-trivial, the relative velocities do not lie within that intersection.

Even the three dimensional relative velocities are not simply the negations of each other except in the special case where the three velocities of the two particles in some frame are parallel as showed in #35.

 

[PeroK]

Sigh! This quote refers to the fact that the relative velocities each measures for the other are elements different local rest spaces (see the image in #30), and so when considered as vectors in $\mathbf{R}^4$ they obviously cannot be parallel not least because although the intersection $E_{u_1} \cap E_{u_2}$ of those rest spaces is still non-trivial, the relative velocities do not lie within that intersection.

Even the three dimensional relative velocities are not simply the negations of each other except in the special case where the three velocities of the two particles in some frame are parallel as showed in #35.

The relative velocity (and indeed everything else about the relative motion) is completely symmetric, as it must be. However $A$ describes the motion of $B$, $B$ must be able to describe the motion of $A$ identically.

The relative motion of $B$ with respect to $A$ must be physically indistinguishable from the relative motion of $A$ with respect to $B$.

This was the original point at issue.

 

[etotheipi]

The relative velocity (and indeed everything else about the relative motion) is completely symmetric, as it must be. However $A$ describes the motion of $B$, $B$ must be able to describe the motion of $A$ identically.

This is not possible in the relativistic framework, again because the relative velocities are elements of different vector spaces. The actual relationship I derived in #28! You must use projection operators.

 

[vanhees71]

The apparent non-reciprocity of relative velocity is a coordinate effect, resulting by considering the velocities in a wider context, with a necessary convention for spatial coordinate axes.

Post #42 shows that if we consider only the two objects in relative motion, then we can choose coordinates so that the relative motion is seen to be physically completely symmetric. Using those coordinates, we can see that any effects, such as the Doppler effect,must be symmetric.

Note that if either A or B uses a different coordinate system than that proposed in post #42, then the complete physical symmetry of the situation may not be immediately apparent. In that case, it may take yet more mathematical bravura to show the complete physical symmetry of relative inertial motion.

It's important to keep in mind that this is a special case, i.e., the velocities of the two points are collinear (then they are collinear in any inertial frame!). The additional rotation between the two relative velocities in the general case must be kept in mind. It's due to the fact that the composition of two rotation-free Lorentz boosts in non-collinear directions is not again a rotation-free Lorentz boost but a boost followed by a rotation (or a rotation followed by a boost). That's related to the Thomas precession.

It's also related to the fact that the "addition of velocities" is not a commutative operation. That's also discussed on the Wikipedia page:

https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

 

[PeroK]

It's important to keep in mind that this is a special case, i.e., the velocities of the two points are collinear (then they are collinear in any inertial frame!).

You must be talking about something different from me. If we take an inertial reference frame - e.g. the rest frame of an inertial observer $A$, then there is only the three-velocity of object $B$ under discussion here. There is no concept of $B$'s velocity being "collinear" with $A$ or anything else. $A$ has zero velocity in this reference frame.

Likewise, if we change to the rest frame of $B$, there is only $A$'s velocity under discussion. $B$ has zero velocity in that frame, so again there is no concept of collinearity.

The motion of $B$ relative to $A$ must be physically identical to the motion of $A$ relative to $B$.

This should be an elementary observation.

 

[etotheipi]

What is means is that, take an arbitrary reference system in which the two particles possesses three-velocities $\vec{v}_{\mathcal{A}}$ and $\vec{v}_{\mathcal{B}}$, then the relative three-velocity of each particle with respect to the other satisfies $\vec{V}_{\mathcal{AB}} = - \vec{V}_{\mathcal{BA}}$ iff $\vec{v}_{\mathcal{A}} \parallel \vec{v}_{\mathcal{B}}$. Otherwise you also need to take into account a rotation (cf. 35), although of course $||\vec{V}_{\mathcal{AB}}|| = ||\vec{V}_{\mathcal{BA}}||$ is still always satisfied (cf. 36).

 

[vanhees71]

You must be talking about something different from me. If we take an inertial reference frame - e.g. the rest frame of an inertial observer $A$, then there is only the three-velocity of object $B$ under discussion here. There is no concept of $B$'s velocity being "collinear" with $A$ or anything else. $A$ has zero velocity in this reference frame.

Likewise, if we change to the rest frame of $B$, there is only $A$'s velocity under discussion. $B$ has zero velocity in that frame, so again there is no concept of collinearity.

The motion of $B$ relative to $A$ must be physically identical to the motion of $A$ relative to $B$.

This should be an elementary observation.

I'm talking about the general case that $A$ moves with some velocity $\vec{v}_1$ and $B$ with some velocity $\vec{v}_2$ wrt. an inertial frame of reference. By definition the relative velocity of particle 2 wrt. particle 1, $\vec{v}_{\text{rel}2}$, is the velocity of particle 2 in the inertial frame, where particle 1 is at rest. To calculate this you need a Lorentz boost with $\vec{v}_1$ (as detailed in one of my postings above).

Correspondingly the relative velocity of particle 1 wrt. particle 2, $\vec{v}_{\text{rel}1}$ is the velocity of particle 1 in the rest frame of particle 2, to which you get by the Lorentz boost with $\vec{v}_2$. As the above explicit calculation unambigously shows in general $\vec{v}_{\text{rel}2} \neq -\vec{v}_{\text{rel}1}$, and the explanation is that rotation-free Lorentz boosts in different directions don't form a group but their composition is not a rotation-free boost.

That's obviously if $u_1$ and $u_2$ are the four-velocity vectors of particle 1 and 2, respectively in the original frame you have
$$u_{\text{rel}2}=\hat{\Lambda}(\vec{v}_1) u_{2}, \quad u_{\text{rel} 1}=\hat{\Lambda}(\vec{v}_2) u_1.$$
Obviously this implies that
$$u_{\text{rel}2}=\hat{\Lambda}(\vec{v}_1) \hat{\Lambda}(-\vec{v}_2) u_{\text{rel} 1},$$
where I have used that $\hat{\Lambda}^{-1}(\vec{v}_2)=\hat{\Lambda}(-\vec{v}_2)$. The additional term with the double-cross product derived above is thus due to the fact that the decomposition of the above two rotation free Lorentz boosts only lead to a rotation free Lorentz boost if and only if $\vec{v}_1 \parallel \vec{v}_2$ and then and only then the additional cross-product terms cancel. These additional cross-product terms arise because the composition of two rotation free Lorentz boost in non-collinear directions leads to a Lorentz transformation which is composed of some rotation-free Lorentz boost followed by a rotation (or a rotation followed by a rotation-free Lorentz boost). These rotation(s) are called "Wigner rotations".

That the relative speeds are the same comes from the much simpler calculation that
$$\gamma_{\text{rel}1}=\gamma_{\text{rel}2}=u_1 \cdot u_2,$$
which is symmetric under exchange of particles 1 and 2. For details, see

https://en.wikipedia.org/wiki/Wigner_rotation

 

[PeroK]

I'll leave it to the OP to ask any further questions if he has them.

 

[Kairos]

Many thanks. I thought that in special relativity in absence of any reference to a fixed medium, the reciprocity of a relative velocity between two inertial objects A and B interchangeable is just natural. As a theoretical observer, you can change your point of view as you wish, invert it, rotate it, change your origin, assign the velocity entirely to one of them, or split it into v1 and v2 using velocity composition rules, and so on. This gives you some excellent mathematical exercises, but in no way question the intrinsic reciprocity of the relative velocity between A vs B and B vs A. I have the impression that the claim:

There are some subtleties to the subject of the reciprocity of the relative velocity. It does not hold true in special relativity! But we ought to put discussion of that on hold until you have a firmer grasp of the underlying machinery.

is not rigorous and can mislead interested non-specialist readers, like me, asking for help on this forum. Please correct me if I am wrong!

 

[vanhees71]

I think the calculation in #35 is not so complicated. You can do it knowing matrix multiplication and some vector algebra. It is thus a bit surprising, how much confusion exists about it even among practitioners of relativity theory. Where is something unclear with this calculation?

 

[PeroK]

Many thanks. I thought that in special relativity in absence of any reference to a fixed medium, the reciprocity of a relative velocity between two inertial objects A and B interchangeable is just natural.

You can, of course, consider the Centre of Momentum frame for two objects of nominally the same mass. In that frame, the three-momenta and three-velocities are reciprocal.

 

[vanhees71]

Of course, because in the CM system you have $\vec{p}_1=-\vec{p}_2$ by definition (no matter whether the particles have the same or different masses; mass is always invariant mass!), and there $\vec{p}_1 \parallel \vec{p}_2$ in this specific frame. The same holds true for the "lab frame", where by definition $\vec{p}_1=0$.