Are the Lorentz transformation formulas on wikipedia correct?

[Jeronimus]

They seem to defy the most fundamental principle of SR. The first postulate/equivalence principle.

According to wikipedia, we get

Lorentz boost (x direction)

and slightly different formulas for the inverse Lorentz boost

"This "trick" of simply reversing the direction of relative velocity while preserving its magnitude..." - Wikipedia *snap*

You cannot reverse the direction of relative velocity. The whole point is that it is *relative*

Taking two observers A and A' which move at v relative to each other, with A measuring an even e1(x1,t1) and A' measuring e1'(x2,t2) then it SHOULD follow by the equivalence principle,
that if A' was to measure an event e2 with the same values of e1, hence e2'(x1,t1) then A has to necessarily measure this even happening with the SAME values as e1', hence e2(x2,t2).

Only then can we talk about this being equivalent.

But this is NOT the case, as according to wikipedia we have to use different formulas when transforming events from A to A' compared to transforming events from A' to A

x=γ(x'+vt') should be x=γ(x'-vt') instead OR the former for both cases.

What changes is not the velocity sign, since the velocity is relative.
What changes are the signs of the events' coordinates depending on how the observers are facing in relation to each other.

Of course, i might be wrong, but my brain is starting to wobble, so i will have to stop here before serious damage. I might pick up on it later.

 

[Orodruin]

No, you are wrong. If A' moves with velocity $v$ relative to A, then A moves with velocity $-v$ with respect to A', hence the sign in the inverse Lorentz transformation. You can also easily check that this is the case by inverting the transformation.

Note that this is the Lorentz transformation in standard configuration, i.e., when the x and x' axes are pointing in the same direction, which is usually the case by convention. In general you can write down Lorentz transformations that boost along arbitrary axes and include rotations.

 

[Jeronimus]

No, you are wrong. If A' moves with velocity $v$ relative to A, then A moves with velocity $-v$ with respect to A',

But then they wouldn't be equivalent. You are probably right, but it certainly does not click with me.

 

[PeroK]

The Lorentz Transformation has an assumption that the origins of the coordinates coincide and the axes point in the same physical direction.

If the velocity were the same in each frame, then the x-axes would be pointing in opposite directions. The Lorentz Transformation would then be more complicated to take account of the x-axis being in the opposite direction from the x'-axis.

Although, then it would be exactly the same transformation for each.

It might be an interesting exercise to work this out for yourself!

 

[Orodruin]

I think you have a misconception about what the equivalence principle states. It states that physical laws will take the same form in all inertial systems. Both the Lorentz transform in standard configuration and its inverse preserve the speed of light.

 

[Orodruin]

Also, the change in sign is not something peculiar to Lorentz transformations. It is already present when inverting the classical Galilean transformations.

 

[Ibix]

It's a fairly straightforward requirement of the equivalence principle that the forward and reverse transforms should be identical except for the sign on the velocity. After all, if we both agree which way forwards is, and I think you are moving forwards you must think I'm moving backwards. Thus my velocity term and yours must have opposite signs.

Note that this discussion is about the transform from some frame S to S' and its inverse. The transforms aren't different except in the sense that one is translating to a frame going in the +x direction at speed v and the other to a frame going in the -x direction at speed v.

 

[Jeronimus]

It's a fairly straightforward requirement of the equivalence principle that the forward and reverse transforms should be identical except for the sign on the velocity. After all, if we both agree which way forwards is, and I think you are moving forwards you must think I'm moving backwards. Thus my velocity term and yours must have opposite signs.

Note that this discussion is about the transform from some frame S to S' and its inverse. The transforms aren't different except in the sense that one is translating to a frame going in the +x direction at speed v and the other to a frame going in the -x direction at speed v.

I don't think so.

If we were both to meet in empty space, flying by each other and facing each other, we would not be able to tell who is the moving one and who is standing still. At some point we would have to agree that none of us is special, hence we are equivalent.

From my perspective, you are coming from the front, pass me, and then you move backwards away of me.
From your perspective, i am doing pretty much the same.

So why would i assume that if i see an event happening at x1,t1 from my perspective and this event was observed by you at x2,t2 i could compute using the Lorentz transformation formulas, that the other way around would be different?

To put it differently. If this was some virtual reality with virtual astronauts and i was able to place you into any of the two astronauts as they are passing by, and given the astronauts were also inside exactly the same spaceships with the same events happening, you would not be able to tell which of the two astronauts you are watching at any given point.

You would have a very hard time to decide on if to use the "standard" or the inverse Lorentz transformation formulas if i was to ask you how the other astronaut would see a given event, you see happening at x1,t1. But you shouldn't have to be puzzled and i will show (hopefully) that this can be resolved without the astronauts having to agree prior to who is the one moving forward and who is the one moving backwards.

 

[PeterDonis]

If we were both to meet in empty space, flying by each other and facing each other, we would not be able to tell who is the moving one and who is standing still.

No, but we can tell that your direction, relative to me, must be opposite to my direction, relative to you. And that is what the sign change from the Lorentz transformation to its inverse is representing.

From my perspective, you are coming from the front, pass me, and then you move backwards away of me.
From your perspective, i am doing pretty much the same.

Only if "pretty much the same" takes into account that, since we are facing each other (at least, you appear to assume that we are), "front" and "back" relative to me are opposite directions from "front" and "back" relative to you.

why would i assume that if i see an event happening at x1,t1 from my perspective and this event was observed by you at x2,t2 i could compute using the Lorentz transformation formulas, that the other way around would be different?

Because of the opposite directions. From the perspective of a faraway observer watching both of us, if our coordinate axes both point in the same directions--which is what the standard Lorentz boost formulas we are using assume, they assume no spatial rotation, just a change in velocity--then the signs of our velocities, relative to each other, must be opposite.

What you are imagining is that you are using coordinates in which the direction "towards me" is the positive $x$ direction, but I am using coordinates in which the direction "towards you" is the positive $x$ direction. But these are opposite directions, so a simple Lorentz transformation like the ones we are discussing cannot transform between our coordinates in this case. You would also have to include a spatial rotation by 180 degrees, so that the direction of the $x$ axis flips. But that would require different (and more complicated) formulas for the transformations; the ones given in this thread won't work. It's much simpler to have our coordinate axes point in the same direction and let the sign change in the transformation vs. its inverse take care of things.

 

[Ibix]

From my perspective, you are coming from the front, pass me, and then you move backwards away of me.
From your perspective, i am doing pretty much the same.

Then we are disagreeing on which way is forwards, and we can't use this form of the Lorentz transforms. You can derive a version of them for this setup where my +x direction is your -x direction (just flip the signs on x' and introduce a v') but it seems unnecessarily complicated.

I think you are not thinking clearly about the physics here. You are correct that there is no global sense in which either of us is moving. However one of us is always moving and setting up an experiment to show that in my rest frame you are moving in the opposite direction from the way I am moving in yours is fairly trivial.

Edit: beaten to it by Peter, I see.

 

[Orodruin]

I think the OP is assigning too much importance to what is the "forward" direction. This is a matter of definition for both observers, but is well defined for the Lorentz transformation in standard configuration. You can also define Lorentz transformations with the coordinate axes pointing in arbitrary directions for either observer, but it will no longer be in standard configuration.

 

[ogg]

Generally, coordinates should be distinguished by the f.o.r. they are relative to (if they are relative coordinates). "Taking two observers A and A' which move at v relative to each other, with A measuring an even[t] e1(x1,t1) and A' measuring e1'(x2,t2) then..."
The event E1 which is measured by both A and A' is better signified by e1(x1,t1) and e1'(x1', t1') {or, equivalently, e1'(x'1,t'1) or even e1(x1',t1')}. Unless the frames of reference are identical, the coordinates will not be identical, but they're both describing the SAME event, E1, which must, obviously, not change with a change in coordinates. Writing the way you did suggests that x1 & x2 are different but IN the same frame of reference (coordinates) which is just plain wrong. The event is invariant, but the values of the coordinates (given different f.o.r.s) can not be.) As far as the written equations, if for A the closing velocity is +v then for A' the closing velocity (IF you assume they share the same x direction!) is -v and using those values in the equations will give identical results. But note the problem (ambiguity) that is injected by not clearly specifying the difference between the velocity in a f.o.r. (which should be v and v' respectively) and its VALUE in those frames (+v, and -v, resp.). if v' = -v, then v = -v'. Also of note is that the word "coordinates" may designate the spatial coordinates, the space-time coordinates, or the more abstract coordinates of phase space (eg. 6N elements for N objects; 3 for xyz and 3 for vx,vy,vz for each object) - you need to get on the same page as the author using the terminology. (Also note that time isn't included as a "coordinate" in the (elementary) classical treatment of dynamical systems)...

 

[Dale]

But then they wouldn't be equivalent.

This isn't the standard for equivalence. Two frames are equivalent if the laws of physics are the same in them.

 

[PeroK]

You would have a very hard time to decide on if to use the "standard" or the inverse Lorentz transformation formulas if i was to ask you how the other astronaut would see a given event, you see happening at x1,t1. But you shouldn't have to be puzzled and i will show (hopefully) that this can be resolved without the astronauts having to agree prior to who is the one moving forward and who is the one moving backwards.

No, because the two astronauts have to agree on a common set of axes. If they agree that the positive x-axis is in the forward direction for astronaut A, then he will use the "normal" transformation and astronaut B will use the inverse transfomation. And, if they agree that the positive x-axis is in the forward direction for astronaut B, then he will use the normal transformation and the astronaut B will use the inverse.

And, if they agree to use different coordinate systems, then they will both have to apply an additional transformation on top of the Lorentz. Just as they would if they were at rest with respect to each other.

The Lorentz transformation (it often neglected to say) assumes this common agreement of the origin and orientation of the axes. In particular, $x = 0, t = 0$ must represent the same spacetime event as $x' = 0, t' = 0$ AND the $x, y, z$ and $x', y', z'$ axes must coincide.

 

[Jeronimus]

No, because the two astronauts have to agree on a common set of axes. If they agree that the positive x-axis is in the forward direction for astronaut A, then he will use the "normal" transformation and astronaut B will use the inverse transfomation. And, if they agree that the positive x-axis is in the forward direction for astronaut B, then he will use the normal transformation and the astronaut B will use the inverse.

And, if they agree to use different coordinate systems, then they will both have to apply an additional transformation on top of the Lorentz. Just as they would if they were at rest with respect to each other.

The Lorentz transformation (it often neglected to say) assumes this common agreement of the origin and orientation of the axes. In particular, $x = 0, t = 0$ must represent the same spacetime event as $x' = 0, t' = 0$ AND the $x, y, z$ and $x', y', z'$ axes must coincide.

What i am questioning is if them having to "meet"/communicate and agree on anything is at all necessary, which is why i used a scenario which looks exactly identical, no matter whose astronaut's perspective you observe it from.

 

[PeroK]

What i am questioning is if them having to "meet"/communicate and agree on anything is at all necessary, which is why i used a scenario which looks exactly identical, no matter whose astronaut's perspective you observe it from.

It's clear that if someone (anyone) is trying to transform between two coordinate systems, then they need to know them both.

 

[Dale]

What i am questioning is if them having to "meet"/communicate and agree on anything is at all necessary, which is why i used a scenario which looks exactly identical, no matter whose astronaut's perspective you observe it from.

Coordinate systems are not objects in nature waiting to be found. They are mathematical conventions used by physicists to make it easier to analyze a situation. As PeroK succinctly put it, if you want to transform between two conventions then obviously you must know both conventions.

The Wikipedia article carefully explains the convention used. Under that convention the transformation equations are correct. There are certainly other conventions you could adopt and then other transformation equations would be required.

I hope that is clear to you now.

 

[Mister T]

If we were both to meet in empty space, flying by each other and facing each other, we would not be able to tell who is the moving one and who is standing still. At some point we would have to agree that none of us is special, hence we are equivalent.

The only reason the velocity of an object is deemed negative is because it's moving in a direction of decreasing position. So if your x-axis points away from you, with the values of x increasing in the direction you are facing, then in the scenario you describe you will conclude that my velocity is negative. Likewise I will conclude that your velocity is negative. On the other hand, if I were instead moving in the opposite direction you would conclude that my velocity is positive and I will conclude that your velocity is positive.

Nothing about this allows either of us to conclude that we the one who's at rest and it's the other who's moving, so nothing about this violates the Principle of Relativity. Each of us is at rest in an inertial frame of reference, and those two frames of reference are equivalent to each other!

 

[Jeronimus]

The only reason the velocity of an object is deemed negative is because it's moving in a direction of decreasing position. So if your x-axis points away from you, with the values of x increasing in the direction you are facing, then in the scenario you describe you will conclude that my velocity is negative. Likewise I will conclude that your velocity is negative. On the other hand, if I were instead moving in the opposite direction you would conclude that my velocity is positive and I will conclude that your velocity is positive.

Nothing about this allows either of us to conclude that we the one who's at rest and it's the other who's moving, so nothing about this violates the Principle of Relativity. Each of us is at rest in an inertial frame of reference, and those two frames of reference are equivalent to each other!

Correct, since this is a symmetrical scenario, both astronauts facing each other (or both having their backs turned towards each other) what one measures, the other will measure as well. Now if one of them was to turn, hence one watching the other astronaut's back, while the other astronaut was "looking away", the scenario would not be symmetric any more. The astronaut who was looking away, having his back turned, would consider himself at rest, with someone approaching from his back(negative x axis) and then moving away from him. It would be a matter of switching the signs of the x axes on the diagram.
But that is not the point. I used the symmetric scenario to make the point clearer.
More important, using this symmetric scenario is something that can be used as a standard, which requires no communication at all once declared as such.

Since none of the astronauts can tell who is the moving one, and the scenario is symmetric, several statements can be made by logic alone IMO.

For example, what you said about velocities. If one astronaut sees the other approaching from the front, hence negative velocity, the other will see the same from his perspective. Hence, also negative velocity.

If astronaut A measures a ruler at length L but astronaut A' measures that ruler which is moving at Vrel to be shorter by a factor 1/γ, hence L' = L*(1/γ) then a ruler which A' measures to be of the length L2' = L will be measured by astronaut A to be shorter by the same factor 1/γ. Hence L2 = L2' * (1/γ) = L * (1/γ) = L'
In other words. If A measures a non-moving ruler that at 10m from his perspective and A' measures this ruler to be 8m from his perspective, then the other way around HAS to be also true. If A' measures a non-moving ruler at 10m from his perspective then A would measure this ruler at 8m from his perspective.

If A measures moving clocks which are at rest relative to A' to be "ticking slower" by some factor 1/γ, then A' will measure moving clocks at rest relative to A to be "ticking slower" by the same factor 1/γ.

Continuing this line of reasoning, it follows (unless my mind plays tricks on me) that

If A measures an event e1 at x=5 and t=10 and A' measures this event to be "happening" at let's say e1' at x'=3 t'=11 (no calculations made, just fictive numbers to make the point), then it would follow that if A' was to measure an event e2' at x'=5 and t'=10, A would measure this event to be happening at e2 with coordinates x=3 t=11.
For this to be the case, we could not possibly use the two _different_ formulas we get on wikipedia. As we would use the Lorentz transformation formula on e1 but we would have to use the inverse Lorentz transformation formula on e2'.

What would also have to be correct, if i am not mistaken, would be for the formula i am looking for to have the property of mapping e1' x'=3 t'=11 right back to e1 x=5 t=10. The formula would be the same going from e1 to e1' and back from e1' to e1.

The astronatus would not have to communicate anything at all in order to get their values right and decide who is A and who is A'. And neither would have any of the other observers potential moving relative to A and A'.

I think the correct term for the formula i am looking for would be an involution.

as a simple example:
x' = 1/x x=3 we would get x'= 1/3 and with x= 1/x' we would get back to x=3. The Lorentz transformation formulas for the symmetric case of the two astronauts approaching each other and facing each other or having both their backs turned against each other should be involutions.

 

[Ibix]

You can use that transformation if you wish. No one will stop you. What will you do when there are three astronauts moving at different velocities? Who will face who? Or will you rotate the coordinate system of the middle astronaut when transforming to the frame of the astronaut behimd him but facing him?

 

[Jeronimus]

As to how i would prove that the Lorentz transformation formulas for the symmetric case of the astronauts would have to be involutions,

i would draw the two diagrams as i did here

...however, even though i initially intended to draw the symmetric case, i failed when drawing the diagram for System B.
For diagram B to be correct, the positive x-axis would have to become negative and vice versa. It would have to be flipped around the t-axis.
The scenario i drew, an observer A' at Eo' (0,0) at rest relative to the blue ruler/rocket would see an observer A at rest relative to the green ruler/rocket fly away of him towards the positive axis while A' is watching his back. Not a symmetric scenario.

Furthermore, to arrive at the most general case, i would have to start with diagram A having two differently sized rockets/rulers which would complicate the whole thing quite a lot.
If done right however, i believe that i would arrive at Lorentz transformation formulas which are involutions for both x, x' and t, t'

 

[Ibix]

The easiest way to derive your revised LTs is to write the regular ones then note that you want to use $x''=-x'$. You might also need to use $-v$ in the inverse transforms depending on how you define $v$ - I haven't thought that through completely.

 

[Jeronimus]

Should you succeed in deriving the Lorentz transformation formulas which are involutions, you should see that if for example:

e1 x=5 t=10 , e1' x'=3 t'=11 and e1'' x''=1 t''=13 (again, fictive values), it won't matter if you apply the involution on e1 or e1' or e2'' you will arrive at the correct values as long as you use the proper vrel between e1 and e1' or e1 and e1'' etc.

following would be true:
x * <involution(<Vrel between e1&e1''>)> = x''
x'' * <involution(<Vrel between e1&e1''>)> = x

x'* <involution(<Vrel between e1&e1'>)> = x
x* <involution(<Vrel between e1&e1'>)> = x'

x'* <involution(<Vrel between e1'&e1''>)> = x''
x''* <involution(<Vrel between e1'&e1''>)> = x'

<involution(<Vrel between e1'&e1''>) would be a single function and would only depend on vrel

Your derivation will be based on two imaginary observers facing each other(or having their backs turned against each other) and moving relative to each other at vrel. It's a derivation between TWO systems, not three. Observers will never again have to agree on who is A or A' or A'' etc.

 

[PeterDonis]

As to how i would prove that the Lorentz transformation formulas for the symmetric case of the astronauts would have to be involutions

The transformations you are talking about are not Lorentz transformations in the usual meaning of that term. A given transformation of the type you are talking about is a Lorentz transformation plus a parity reversal. The parity reversal is required because each astronaut defines the direction "towards the other astronaut" as having the same sign, but these are opposite directions. Think of one astronaut trying to move into the exact orientation of the other at the instant they pass each other. He has to flip himself around to face in the opposite direction. The flip is the parity reversal.

So what you are proving (assuming your proof is valid; I haven't tried to check it) is that a Lorentz transformation + a parity reversal is an involution. However, there is a very good reason why these transformations are not included in the usual category of "Lorentz transformations": they aren't continuously connected to the identity. For example, suppose the two astronauts were at rest relative to each other. They would still be facing each other, so to transform from the frame of one to the frame of the other, you would still have to flip by 180 degrees, i.e., you would still have to do the parity reversal. This is not an identity transformation--it does not correspond to no change at all. But for two observers at rest relative to each other, the standard Lorentz transformation between their frames is the identity transformation--no change at all. This is a very important property for the group of Lorentz transformations to have, and your scheme destroys it.

So the problem with your scheme is not that it is incorrect mathematically; it is that it does not describe Lorentz transformations, it describes Lorentz transformations + parity reversals, which is a group of transformations which lacks a key property that the group of standard Lorentz transformations has.

 

[PeterDonis]

Btw, this discussion has clearly gone beyond the "B" level, so I have changed the level of the thread to "I".

 

[Jeronimus]

Think of one astronaut trying to move into the exact orientation of the other at the instant they pass each other. He has to flip himself around to face in the opposite direction. The flip is the parity reversal.

The involution formulas i am imagining will be universal and will only require to be fed with the proper vrel between the TWO systems you want to transform.

Unfortunately i was sloppy in explaining the "facing" part.

The derivation for one of the symmetric cases will be astronauts facing each other as they approach each other and once they pass each other, they will of course have their backs turned against each other.

The other symmetric case would be astronauts having their backs turned against each other as they measure each other to be approaching from the negative x-asis and then when they just pass each other they will be facing each other, measuring each other to be moving away towards the positive axis.

No matter which of the two scenarios you solve for, as they are symmetric you should get involution formulas which whenever you apply them on an arbitrary amount of systems, as long as you use the correct vrel between the two systems you want to tranform the coordinates of, you should always arrive at the correct values.

No communication will be needed at all between any of the observers to decide who is A, A', A'' etc.

 

[Ibix]

But your transforms can't be trivially combined. With the standard setup you can transform from frame A to frame B, from frame B to frame C, from frame C to frame D and from frame D to frame E, and you inevitably get the same thing as if you'd gone straight from A to E. Whereas with yours it depends which way round each pair of your coordinate systems were defined whether you'll get the same answer or the negative thereof. Unless you rotate your coordinate systems as necessary each time you consider a transform.

As I say, you can do this. But it's just making your life difficult in all but the most trivial of cases.

Edit: In addition to the points PeterDonis made, your transforms are not well defined in the case where $v=0$. The astronauts aren't moving so cannot unambiguously define which way is forwards for them. Have they passed each other or not?

 

[Jeronimus]

But your transforms can't be trivially combined. With the standard setup you can transform from frame A to frame B, from frame B to frame C, from frame C to frame D and from frame D to frame E, and you inevitably get the same thing as if you'd gone straight from A to E. Whereas with yours it depends which way round each pair of your coordinate systems were defined whether you'll get the same answer or the negative thereof. Unless you rotate your coordinate systems as necessary each time you consider a transform.

As I say, you can do this. But it's just making your life difficult in allbut the most trivial of cases.

There are only two symmetric cases as i described them above. One of those cases would become the "standard" all observers would assume. Then life gets way easier, because the observers do not have to communicate with each other or having to decide on if to use the inverse or non-inverse version of the transformations.

Either way, i would like to see those involution formulas (for x,x' and t,t') i am imagining exist, even if you believe that it would make our lives harder. So if anyone has them handy, feel free to post :)

 

[PeroK]

The involution formulas i am imagining will be universal and will only require to be fed with the proper vrel between the TWO systems you want to transform.

Unfortunately i was sloppy in explaining the "facing" part.

The derivation for one of the symmetric cases will be astronauts facing each other as they approach each other and once they pass each other, they will of course have their backs turned against each other.

The other symmetric case would be astronauts having their backs turned against each other as they measure each other to be approaching from the negative x-asis and then when they just pass each other they will be facing each other, measuring each other to be moving away towards the positive axis.

No matter which of the two scenarios you solve for, as they are symmetric you should get involution formulas which whenever you apply them on an arbitrary amount of systems, as long as you use the correct vrel between the two systems you want to tranform the coordinates of, you should always arrive at the correct values.

No communication will be needed at all between any of the observers to decide who is A, A', A'' etc.

No communication is needed in any case. You are largely misunderstanding the point of the transformations. In a typical situation each observer would use the LT independently to calculate what another observer would measure.

The point you're missing is that the coordinates themselves are not important but things like lengths, elapsed times, velocity of particles etc. are, and these are coordinate independent - for a given inertial reference frame.

So you can happily calculate, say, the energy of a particle in someone else's frame. Even if there is no one in that frame to measure it. If there is, then they could use any coordinate system. You need never know the coordinates they used but you can use the LT to determine what energy they would measure in their frame.

You need to take a step back as you are descending into a rabbit hole now!

 

[Jeronimus]

No communication is needed in any case. You are largely misunderstanding the point of the transformations. In a typical situation each observer would use the LT independently to calculate what another observer would measure.

The point you're missing is that the coordinates themselves are not important but things like lengths, elapsed times, velocity of particles etc. are, and these are coordinate independent - for a given inertial reference frame.

So you can happily calculate, say, the energy of a particle in someone else's frame. Even if there is no one in that frame to measure it. If there is, then they could use any coordinate system. You need never know the coordinates they used but you can use the LT to determine what energy they would measure in their frame.

You need to take a step back as you are descending into a rabbit hole now!

Of course something as energy and lengths does not require any communication because a length is always positive and so is energy, but you are wrong in that there is no communication required between two observers traveling at vrel when handling it with the LTs given by wikipedia. How would an Observer who has not communicated with the other to decide who is A and who is A' know to use the inverse or non inverse LT formula when transforming an event? He couldn't, so you must be wrong.

 

[robphy]

Sorry for jumping in late and being too lazy to read everything up to this point...
But isn't the essence of the issue already seen in the Galilean Transformation?
$\left(\begin{array}{rr} 1 & 0 \\ -v & 1 \end{array}\right)^{-1} = \left(\begin{array}{rr} 1 & 0 \\ v & 1 \end{array}\right)$
(Trying to resolve it with Lorentz Transformations adds unnecessary complication to the issue.)
Implicitly, the observers face the "same x"-direction and agree that that direction is positive.

The issue also arises with rotations.

Maybe $v$ needs some extra notation to clarify its meaning and who is doing the measuring.
$\left(\begin{array}{r} t^{Bob} \\ x^{Bob} \end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ -v_{Bob}{}^{Alice} & 1 \end{array}\right) \left(\begin{array}{r} t^{Alice} \\ x^{Alice} \end{array}\right)$

Maybe I'm missing some aspect of the issue at hand...

 

[Nugatory]

How would an Observer who has not communicated with the other to decide who is A and who is A' know to use the inverse or non inverse LT formula when transforming an event? He couldn't, so you must be wrong.

You are still misunderstanding what a transformation does, perhaps because so many presentations introduce an observer into every frame. Those observers are just there to make the implications of the math a bit easier to visualize - neither observers nor communication between observers is necessary.

Here's the observer-free language:
We have a coordinate system that labels every point in flat spacetime with coordinates x, y, z, and t. We have a second coordinate system that labels every point in the same spacetime with coordinates x', y', z', and t'. How do we write x', y', z', and t' as functions of x, y, z, and t and vice versa? The Lorentz transformations in the wikipedia article are the answer to that question for the case in which the origin of the second coordinate system is moving with speed v along the x-axis (and some other conditions are met).

 

[stevendaryl]

Of course something as energy and lengths does not require any communication because a length is always positive and so is energy, but you are wrong in that there is no communication required between two observers traveling at vrel when handling it with the LTs given by wikipedia. How would an Observer who has not communicated with the other to decide who is A and who is A' know to use the inverse or non inverse LT formula when transforming an event? He couldn't, so you must be wrong.

I'm not exactly sure what point you are making. But you are right--If I see an alien creature hurtling through space at some velocity $\vec{v}$, there is no way for me to write down a transformation relating my coordinate system to the coordinate system of the alien. To figure out the coordinate system of the alien, we need to know:

1. Is he using isotropic inertial coordinates? (Isotropic for our purposes just means for these purposes that the coordinates are such that the speed of light is the same in every direction)
2. What is the origin of his coordinate system?
3. What directions does he consider his x-axis, y-axis and z-axis?
4. Does he consider himself to be at rest? (That's the usual assumption)
5. What units is he using for distance and time?

You probably can't answer those questions without talking to the alien. But given the answers to those questions, you can uniquely determine a transformation between his coordinates and yours. And given the same information about you, he can uniquely determine a transformation between your coordinates and his. What's not necessary is for the two of you to decide whose coordinates get a "prime" on them.

 

[stevendaryl]

Sorry for jumping in late and being too lazy to read everything up to this point...
But isn't the essence of the issue already seen in the Galilean Transformation?

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The issue also arises with rotations.

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Maybe I'm missing some aspect of the issue at hand...

No, you're not. The issues brought up by Jeronimus are just as relevant (and as trivial) for Galilean transformations and rotations.

 

[PeterDonis]

The involution formulas i am imagining will be universal and will only require to be fed with the proper vrel between the TWO systems you want to transform.

Suppose this is true (again, I haven't tried to verify any of your proofs). That still doesn't change anything I said, or address any of the issues I raised. Your transformations are not Lorentz transformations; they are Lorentz transformations + parity reversals. Those are perfectly good transformations mathematically speaking, but they do not have certain properties that the group of Lorentz transformations alone has. Most physicists want to use a group of transformations that has those properties; that's why they use the standard Lorentz transformations instead of yours.