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

[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 $\vec{p}_1=0$.

 

[PeroK]

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 $\vec{p}_1=0$.

If I am not wrong, this is precisely what the OP believes that you have been denying all through this thread!

 

[Ibix]

I thought I'd have a shot at some Minkowski diagrams, which may be helpful. I'm restricting motion to one dimension, but this is actually completely general for intersecting inertial worldlines, since in either one's rest frame we can simply define the x direction as "$\pm$ the direction in which the other guy is moving". So these diagrams are all drawn in the hyperplane @etotheipi shaded in #30.

So we have two observers, red and blue. Here's a Minkowski diagram in red's frame - red's worldline is straight up the page and blue's is slanted because they're doing 0.6c.

Red can drop a perpendicular to the x-axis and draw blue's three-velocity, the heavy dotted blue line. This is what red calls "blue's velocity through space", so it must be a vector in the spatial direction. Blue's four-velocity is, of course, parallel to blue's worldline.

(Apologies that there's no arrowhead on the three-velocity - I didn't add the capability to draw arrowheads when I wrote my Minkowski diagram software.)

Now we can do the same thing in blue's frame:

Aside from the colours, this is an exact mirror image of the first diagram. This is the "reciprocity" between the red and blue's three-velocities that we've been talking about.

But the complexity comes in if we draw what red called "blue's velocity through space" on the diagram above:

I've included red's axes and the dropped perpendicular and three-velocity from the first diagram. And here we can see the problem with comparing three-velocities in different frames - three velocities in the red frame lie along the red spatial axis and three-velocities in the blue frame lie along the blue spatial axis. You can't compare them because they lie in different spatial planes. You can "promote" them to four-vectors (which I have kind of implicitly done by drawing them both on one diagram here), and then you can compare them, but they aren't parallel and boosting the blue dotted line to be parallel to the blue axis would not yield blue's three velocity in this frame (a problem @vanhees71 alluded to in #35).

So, the "reciprocity" of the three-velocities each observer assigns to the other in their respective rest frames just means that we can draw the first two Minkowski diagrams and they are mirror-symmetric. It follows from the symmetry of the situation - it must be true unless there's some difference between the red and blue observers, and there isn't. However, the three velocities don't lie in the same space and can't be meaningfully compared as vectors - you are, of course, free to compare their magnitudes.

I hope that's helpful, anyway.

 

[robphy]

It might be helpful to repeat this for Euclidean case, again to emphasize that (in this geometric view) the Galilean geometry is the special (atypical) case.

 

[Ibix]

It might be helpful to repeat this for Euclidean case, again to emphasize that (in this geometric view) the Galilean geometry is the special (atypical) case.

You mean, two intersecting lines shown on two rotated maps with dropped perpendiculars? I wonder if I can just replace the Lorentz transforms in my code...

 

[Ibix]

So that was simple: add a flag saying to use Minkowski or Euclidean geometry and change the transform function to look at the flag and use the appropriate coordinate transform. I'm almost tempted to change the Lorentz transform code to use cosh and sinh instead of $v$ and $\gamma$ just for the elegance, but it ain't broke so I'm going to resist the urge to fix it.

Here are the three graphs using Euclidean geometry. In this case, "horizontal lines" are the frame-dependent phenomenon and you can't compare "the x component of the direction vector of the line" from the two frames without doing something like promoting the line to a two-vector (but it still can't be interpreted as "the horizontal component of the line" in any frame except the one we started in). Note that, apart from the axis labels, the first two diagrams appear identical to the first two diagrams in my previous post #62.

I'm wondering if I can add a third option that does Galilean transforms now. Would that be called Cartan geometry?

 

[Ibix]

Well, whatever it's called, I went ahead and did it. So here are the equivalent Newtonian displacement-time graphs, if Newtonian physics were remotely valid at 0.6c. Note how, once again, the first two diagrams are identical to the first two diagrams in my recent posts. It's only the third diagram where things differ - and in this case the three-velocities do share a plane and can be directly compared in a meaningful way.

Note the messy horizontal axis labelling in the third diagram - this is because the $x$ and $x'$ axes coincide in this version.

 

[robphy]

I'm wondering if I can add a third option that does Galilean transforms now. Would that be called Cartan geometry?

I would call it Galilean geometry.
(Newton-Cartan would also be okay but maybe more appropriate... if considering constant curvature space[times].)

Check out my Spacetime diagrammer
for a variable parameter “E”
https://www.desmos.com/calculator/kv8szi3ic8

 

[Ibix]

Check out my Spacetime diagrammer
for a variable parameter “E”
https://www.desmos.com/calculator/kv8szi3ic8

That $E$ slider is neat. I couldn't see documentation (I found lots for the Desmos tool, but not for your application), but I think that the $E$ parameter works like this: $$
\left(\begin{array}{c}x'\\t'\end{array}\right)=\frac 1{\sqrt{1-Ev^2}}\left(\begin{array}{cc}1&-Ev\\-v&1\end{array}\right)\left(\begin{array}{c}x\\t\end{array}\right)$$So $E=1$ gives you the Lorentz transforms and $E=-1$ gives you Euclidean rotations. Is there a name for the "intermediate" geometries when $-1<E<1$?

 

[robphy]

That $E$ slider is neat. I couldn't see documentation (I found lots for the Desmos tool, but not for your application), but I think that the $E$ parameter works like this: $$
\left(\begin{array}{c}x'\\t'\end{array}\right)=\frac 1{\sqrt{1-Ev^2}}\left(\begin{array}{cc}1&-Ev\\-v&1\end{array}\right)\left(\begin{array}{c}x\\t\end{array}\right)$$So $E=1$ gives you the Lorentz transforms and $E=-1$ gives you Euclidean rotations. Is there a name for the "intermediate" geometries when $-1<E<1$?

Everything in desmos is a definition of a slider, point, or function or an equation... not necessarily presented in any order... they are simultaneous equations. One just has to open everything up and track everything down, unfortunately.

There are really three cases, +1, 0, -1.
The fractional values are really just scaling factors (like varying the maximum signal speed) that interpolate the three cases.

 

[Ibix]

There are really three cases, +1, 0, -1.

Ah! In that case I have my matrix transposed, and it should be$$
\left(\begin{array}{c}x'\\t'\end{array}\right)=\frac 1{\sqrt{1-Ev^2}}\left(\begin{array}{cc}1&-v\\-Ev&1\end{array}\right)\left(\begin{array}{c}x\\t\end{array}\right)$$Then $E=1$ corresponds to Lorentz, $E=0$ corresponds to Galileo, and $E=-1$ corresponds to Euclid. Neat.

 

[robphy]

Here’s a poster of the ideas.
I really should write it up.
https://www.aapt.org/doorway/Posters/SalgadoPoster/Salgado-GRposter.pdf
The contact information was when I was a graduate student. It’s not current.

 

[vanhees71]

If I am not wrong, this is precisely what the OP believes that you have been denying all through this thread!

No, I've given the general formula and always stressed in which special cases the naive relative velocity comes out and when the two relative velocities are opposite to each other, namely precisely when $\vec{v}_1 \parallel \vec{v}_2$ :-).

 

[Kairos]

I thought I'd have a shot at some Minkowski diagrams, which may be helpful. I'm restricting motion to one dimension, but this is actually completely general for intersecting inertial worldlines, since in either one's rest frame we can simply define the x direction as "$\pm$ the direction in which the other guy is moving". So these diagrams are all drawn in the hyperplane @etotheipi shaded in #30.

So we have two observers, red and blue. Here's a Minkowski diagram in red's frame - red's worldline is straight up the page and blue's is slanted because they're doing 0.6c.
View attachment 282015
Red can drop a perpendicular to the x-axis and draw blue's three-velocity, the heavy dotted blue line. This is what red calls "blue's velocity through space", so it must be a vector in the spatial direction. Blue's four-velocity is, of course, parallel to blue's worldline.

(Apologies that there's no arrowhead on the three-velocity - I didn't add the capability to draw arrowheads when I wrote my Minkowski diagram software.)

Now we can do the same thing in blue's frame:
View attachment 282016
Aside from the colours, this is an exact mirror image of the first diagram. This is the "reciprocity" between the red and blue's three-velocities that we've been talking about.

But the complexity comes in if we draw what red called "blue's velocity through space" on the diagram above:
View attachment 282017
I've included red's axes and the dropped perpendicular and three-velocity from the first diagram. And here we can see the problem with comparing three-velocities in different frames - three velocities in the red frame lie along the red spatial axis and three-velocities in the blue frame lie along the blue spatial axis. You can't compare them because they lie in different spatial planes. You can "promote" them to four-vectors (which I have kind of implicitly done by drawing them both on one diagram here), and then you can compare them, but they aren't parallel and boosting the blue dotted line to be parallel to the blue axis would not yield blue's three velocity in this frame (a problem @vanhees71 alluded to in #35).

So, the "reciprocity" of the three-velocities each observer assigns to the other in their respective rest frames just means that we can draw the first two Minkowski diagrams and they are mirror-symmetric. It follows from the symmetry of the situation - it must be true unless there's some difference between the red and blue observers, and there isn't. However, the three velocities don't lie in the same space and can't be meaningfully compared as vectors - you are, of course, free to compare their magnitudes.

I hope that's helpful, anyway.

Thank you for your diagrams. Your first two diagrams represent exactly what I was attempting to say in my post #8.. before the debate on the non reciprocity of relative velocity.

When you write "But the complexity comes in if we draw ..", do you draw an alternative representation of the same thing or of an another thing? supposedly a different configuration of inertial observers?

 

[Ibix]

When you write "But the complexity comes in if we draw ..", do you draw an alternative representation of the same thing or of an another thing? supposedly a different configuration of inertial observers?

The third diagram is simply the second diagram with some extra lines copied and transformed from the first diagram. But it's clear from this that the three-velocities of each object in the rest frame of the other are not opposites, despite what the first two diagrams (or your intuition) might suggest.

The Euclidean diagrams are analogous - the x extent of one line and the x' extent of the other are reciprocal but not opposite.

 

[Kairos]

Thank you, indeed it is difficult to fight against a strong intuition with mathematics. I'm not sure I grasp the meaning of opposite because 'reciprocal' means for me that the perception of the other is identical for each. One last attempt if you will: imagine a theoretical elastic thread (without restoring force) linking the two inertial objects. Do you agree with me that the rate of shortening or extension of this thread is necessarily reciprocal for both objects since it is unique? It seems to me that whatever the chosen reference frame, the relative velocity vector can be univocally defined as a function of this unique thread rate change.

 

[Ibix]

It seems to me that whatever the chosen reference frame, the relative velocity vector can be univocally defined as a function of this unique thread rate change.

You can certainly define a three-velocity in each frame totally unambiguously, yes. The problem is that you have two reference frames with different definitions of space, so those three vectors don't lie in the same 3d subspace of 4d spacetime, so they cannot be compared (except comparing magnitudes, which are equal). You can see that in the third diagram, reproduced here:

The red dotted line is "red's three-velocity according to blue" and it lies along the x' axis. The blue dotted line is "blue's three-velocity according to red" and it lies along the x axis. These axes are not parallel, so these vectors (treated as three vectors) do not exist in the same space, or ("promoted" to four vectors) are not parallel so aren't opposite.

The principle of relativity requires that if I measure your velocity relative to me to be $v$ then you must measure my velocity relative to you to be $-v$. So yes, to answer the question in your post, we will agree that a thread connecting us would need to be payed out at the same rate. But since we don't agree what we mean by "space" that does not mean that we are measuring opposite velocities, oddly enough.

The Euclidean analogy is useful. I reproduced the analogous diagram below:

Here the red dotted line is the blue frame's notion of the x'-component of a vector along the solid red line, and the blue dotted line is the red frame's notion of the x-component of a vector along the solid blue line. These components are the same length, but they aren't opposites because the directions they lie in are rotated with respect to one another.

Hope that makes sense.

 

[Kairos]

Very clear, thank you. I wonder if the origin of the misunderstanding is that we didn't agree on a mathematical definition of reciprocity from the beginning. My view of reciprocity is illustrated below using one of your diagrams and simply swapping blue and red:

The points of view of A and B are the same, not opposite but just the same. I believe that this definition of reciprocity in this context allows that, if A and B are both sources and observers, they see exactly the same Doppler effects.
I don't know if there is an official definition of reciprocity but from your considerations on opposite diagrams, I have the impression that you are rather talking about symmetry. By construction of the SR spacetime, everything depends on the point of view (except the spacetime intervals between events), so one cannot expect a single symmetric diagram for A and B if the viewpoint is asymmetric with respect to the system A,B. But let's introduce a third inertial observer standing forever in the middle of the elastic thread postulated above (#75). Seen from his reference frame, A and B would have symmetric spacetime trajectories on a single diagram.