Light clock moving to demonstrate time dilation

[JesseM]

Yes, I am saying if there was a hypothetical system where the the speed of light is (c+v) where v is the velocity of the source, then a MM type experiment would give the required null result, without requiring length contraction. To an observer co-moving with the MM apparatus the speed of light from the source mounted on the apparatus would be c. I know that real experiments rule out this possibility but I was wondering if it could be ruled out on purely theoretical considerations such as violation of causality or the relativity principle?

Are you assuming the "c" also represents the value of the invariant speed in the transformation, or can it be different? You could have a universe with Galilei-symmetric laws where light was emitted at a constant finite speed relative to the emitter, for example.

 

[yuiop]

... You could have a universe with Galilei-symmetric laws where light was emitted at a constant finite speed relative to the emitter, for example.

Yep, I think that is it. It might be clearer with a hypothetical static light medium that represents an absolute reference. There is no time dilation or length contraction ( t=t' and x=x' so yes, Galiliei symmetric). Light moves at c+v relative to this medium where v is the speed of the emitter. As far as I can tell, an observer moving relative to the absolute medium would not be able to determine their absolute motion, because when the source and the observer are co-moving the observer measures the speed of light as c and an MM type experiment would yield a null result whatever your absolute motion. I am just wondering why that type of universe does not come up as a possible solution in the possible transformations which all seem to assume the speed of light has to be infinite in a Galilei system.

 

[JesseM]

Yep, I think that is it. It might be clearer with a hypothetical static light medium that represents an absolute reference. There is no time dilation or length contraction ( t=t' and x=x' so yes, Galiliei symmetric). Light moves at c+v relative to this medium where v is the speed of the emitter. As far as I can tell, an observer moving relative to the absolute medium would not be able to determine their absolute motion, because when the source and the observer are co-moving the observer measures the speed of light as c and an MM type experiment would yield a null result whatever your absolute motion. I am just wondering why that type of universe does not come up as a possible solution in the possible transformations which all seem to assume the speed of light has to be infinite in a Galilei system.

Well, I think the point is that if you assume just the first postulate you can prove the laws of physics must be invariant under a general coordinate transformation that includes an invariant speed that can be either infinite or finite (or perhaps imaginary if you allow the causality-violating case?) But there's no logical reason that the invariant speed that appears in the transformation has to have any relation to the speed of light, even if it turns out that the speed of a given light ray is always the same in the frame of the emitter of that ray. It might be helpful to use different symbols for the two, like C for the invariant speed in the coordinate transformation and c for the speed of light relative to the emitter. In the Galilean case, C could be infinite while c could have some finite value. I don't think any of the proofs of the coordinate transformation that follows from the first postulate claim that the invariant speed that appears in the transformation equations necessarily has anything to do with the physical speed of light, they don't even address the issue of light AFAIK.

 

[Saw]

These derivations don't have anything to do with light. They simply show that there is some invariant velocity c.

I really appreciate the idea of introducing an invariant velocity c in the derivation, as a didactic approach. Thus you end up with a range of possibilities (GT, LT and CVT) and, what is more, the merits of each option shine up and hence the choice among them becomes more meaningful. I would even daresay that, in the absence of experimental evidence, if I had to make an intelligent guess about which option is preferable, I would opt for the LT.

But that would still be that, just a bold guess. After acknowledging the enormous interest of this sort of derivation, bcrowell, I still think that the substance of the debate has not changed and it deserves the same answer: the derivations do NOT show that “there is some invariant velocity c”, they just show that you can introduce the hypothesis about an invariant c in an equation and then end up with a set of theoretical solutions about the nature of such c. And then you choose between one or the other either randomly (and trust the issue will be settled by experiment) or dare to make a choice on the basis of some hypothesis about how the underlying physical processes operate (and trust that your guess will be confirmed by experiment). But you do not “prove” the LTs just with mathematics. Do you agree to that?

Well, I think the point is that if you assume just the first postulate you can prove the laws of physics must be invariant under a general coordinate transformation that includes an invariant speed that can be either infinite or finite (or perhaps imaginary if you allow the causality-violating case?) But there's no logical reason that the invariant speed that appears in the transformation has to have any relation to the speed of light, even if it turns out that the speed of a given light ray is always the same in the frame of the emitter of that ray. It might be helpful to use different symbols for the two, like C for the invariant speed in the coordinate transformation and c for the speed of light relative to the emitter. In the Galilean case, C could be infinite while c could have some finite value. I don't think any of the proofs of the coordinate transformation that follows from the first postulate claim that the invariant speed that appears in the transformation equations necessarily has anything to do with the physical speed of light, they don't even address the issue of light AFAIK.

Yes, but I am thinking… You will probably agree that, if light conformed to a classical ballistic model (where speeds are simply additive), you would not need any invariant speed in your derivations. I am not sure if you could still introduce it, but anyhow it would be a nuisance, because in that case the invariant thing that makes the transformation successful would simply be time. Of course, nature does not conform to that model. But again it is experiment what leads you to that conclusion or, for someone who is smart enough, some intuition about the underlying physical processes, but not purely mathematics.

 

[JesseM]

I really appreciate the idea of introducing an invariant velocity c in the derivation, as a didactic approach.

As I understand it, it's not just that they "introduce" it as a new postulate, they actually prove that if you start with the postulate that the laws of physics are invariant under a coordinate transformation that gives a set of coordinate systems moving at constant coordinate velocity relative to one another, a consequence of this postulate is that the coordinate transformation must take a certain general form that includes this type of invariant velocity. I haven't looked over the proof in great detail so I could be wrong, but that's what they seem to be saying.

 

[bcrowell]

As I understand it, it's not just that they "introduce" it as a new postulate, they actually prove that if you start with the postulate that the laws of physics are invariant under a coordinate transformation that gives a set of coordinate systems moving at constant coordinate velocity relative to one another, a consequence of this postulate is that the coordinate transformation must take a certain general form that includes this type of invariant velocity. I haven't looked over the proof in great detail so I could be wrong, but that's what they seem to be saying.

Yes, this is correct.

 

[Saw]

As I understand it, it's not just that they "introduce" it as a new postulate, they actually prove that if you start with the postulate that the laws of physics are invariant under a coordinate transformation that gives a set of coordinate systems moving at constant coordinate velocity relative to one another, a consequence of this postulate is that the coordinate transformation must take a certain general form that includes this type of invariant velocity. I haven't looked over the proof in great detail so I could be wrong, but that's what they seem to be saying.

Ah... Maybe in Rindler's text, I don’t know (google does not include page 57 in my preview form). But at least the article you link to does not end up with an invariant speed shared by all relativity models. It ends up with a constant K. And then it warns that (page 5):

Specific theories of relativity, of course, have to make extra assumptions in order to determine the value of K. In the case of Galilean relativity, this extra assumption shows up in the form of the universality of time, which means t' = t for any v. Obviously, this requires K = 0. The extra assumption for Einstein’s theory of relativity is the constancy of the speed of light in vacuum.

(Emphasis is mine)

You can note that, strictly speaking, in Galilean relativity you don't need that K becomes zero by equating it with 1/c², where c is infinite. Before that, you can simply affirm that K = 0 because there is no such thing as a frame-invariant speed, which by the way is a logical impossibility in a universe where distances traversed by a moving body are relative but time is absolute.

A different thing is that introducing such invariant speed may be an attractive idea if you reason on the basis of some guess about the underlying physical processes. For example, think of two mechanical clocks in their respective frames (eg: bouncing balls). Imagine there is neither external nor internal damping (vacuum + elastic collision). If the mechanism applied on each ball to make it oscillate and act as a clock is the same in both cases, it follows that they should tick simultaneously... except for one “little” problem: when the balls collide with the walls of their respective clocks, i.e. when they accelerate, is that phenomenon identical in both cases? Galilean relativity (to hold that time is universal) has to argue that the acceleration of the balls is the same in both cases. But if the acceleration is due to an electromagnetic interaction and light (unlike the balls) does not take the motion of the source, the only possibility to make the accelerations identical (and hence preserve the idea of a transformation) is to consider the speed of light as infinite. Since this is not the case, Galilean relativity has to be discarded. Well, sorry, I couldn't help commenting this idea that was nagging me, I know the forum is not the place for speculations...

 

[JesseM]

Ah... Maybe in Rindler's text, I don’t know (google does not include page 57 in my preview form). But at least the article you link to does not end up with an invariant speed shared by all relativity models. It ends up with a constant K.

Yes, but the constant K has units of 1 over speed squared, nothing about the proof is changed if you substitute K=1/S^2, where S is a speed. And note that equation 29 of the paper is a generalized version of velocity addition, w = (u + v)/(1 + Kuv), so if you substitute K=1/S^2, it becomes w = (u + v)/(1 + uv/S^2). This implies that if v=S, then w=S as well, even if S is infinite (or in the limit as S approaches infinity, if you prefer). So, it seems reasonable to call S an invariant speed.

And then it warns that (page 5):

Specific theories of relativity, of course, have to make extra assumptions in order to determine the value of K. In the case of Galilean relativity, this extra assumption shows up in the form of the universality of time, which means t' = t for any v. Obviously, this requires K = 0. The extra assumption for Einstein’s theory of relativity is the constancy of the speed of light in vacuum.

(Emphasis is mine)

But the extra assumptions here are about what is needed to decide whether the invariant speed S is finite or infinite (and if it's finite, what its value is), not about deriving that there is an invariant speed S in the first place.

You can note that, strictly speaking, in Galilean relativity you don't need that K becomes zero by equating it with 1/c², where c is infinite. Before that, you can simply affirm that K = 0 because there is no such thing as a frame-invariant speed

I suppose it's just a matter of semantics, but it's certainly true that if anything is moving at infinite speed in one frame (i.e. if any two events on its worldline are simultaneous) it is moving at infinite speed in every other frame according to the Galilei transformation (which wouldn't be true in the Lorentz transformation, for example), and that is directly implied by the fact that the Galilei transformation is what you get when you plug K=0 into the general transformation derived in that paper.

which by the way is a logical impossibility in a universe where distances traversed by a moving body are relative but time is absolute.

Only if you define "frame-invariant speed" in a way that presupposes it must be finite! As I said, it is a nontrivial feature of the Galilei transformation that anything moving at infinite speed in one frame is moving at infinite speed in every other frame.

A different thing is that introducing such invariant speed may be an attractive idea if you reason on the basis of some guess about the underlying physical processes. For example, think of two mechanical clocks in their respective frames (eg: bouncing balls). Imagine there is neither external nor internal damping (vacuum + elastic collision). If the mechanism applied on each ball to make it oscillate and act as a clock is the same in both cases, it follows that they should tick simultaneously... except for one “little” problem: when the balls collide with the walls of their respective clocks, i.e. when they accelerate, is that phenomenon identical in both cases? Galilean relativity (to hold that time is universal) has to argue that the acceleration of the balls is the same in both cases. But if the acceleration is due to an electromagnetic interaction and light (unlike the balls) does not take the motion of the source, the only possibility to make the accelerations identical (and hence preserve the idea of a transformation) is to consider the speed of light as infinite. Since this is not the case, Galilean relativity has to be discarded. Well, sorry, I couldn't help commenting this idea that was nagging me, I know the forum is not the place for speculations...

I don't follow. If we are using an aether theory, then the two mechanical clocks would be moving at different speeds relative to the physical medium that light is a vibration is, so there is no requirement that they would see the same results in their own frames since the physical components of the experiment aren't the same in each frame. It would be equivalent to two experimenters doing an experiment with sound waves, one experimenter at rest relative to the air while the other is on an open-air platform of a train moving relative to the air (unless you could detach a piece of the aether and cause it to move along with one clock, like a sealed train compartment filled with air at rest relative to the train car even though the car is moving relative to the external air). On the other hand, with a ballistic theory of light, if the light was emitted inside the clock it would have the same speed relative to the clock in both cases, so it would cause the same acceleration for the balls. So either way, it's possible for light to have a finite speed in a universe with Galilei-invariant fundamental laws.

 

[sneh]

The thought experiment assumes there is a single photon emitted, say from a train and observed in the train and ground frame, or if you prefer two photons, i.e. train photon emitted from the train and observed in both train and ground frame and ground photon emitted from the ground and observed in both ground and train frame.

Let us call the frame from which each photon is emitted the "local" frame and the other the "foreign frame".

What is clear and unavoidable is that the local frame observes that its local photon traverses from bottom to top mirror a vertical trajectory, while the foreign frame observes a diagonal pat. Obviously, the first is shorter.

...

I don't see how it is actually shorter (though I understand it is observed to be shorter)

So from here I would say that the photon is obviously moving diagonally through space no matter which perspective -- The observer on the train has matching velocity in one direction to the photon, therefore relative to them they don't see the movement of the photon in that direction... big deal?

So again, I still cannot see how the photon would actually be moving any further or less, nor taking any more time or less, from either frame of reference.

 

[Saw]

I don't see how it is actually shorter (though I understand it is observed to be shorter)

So from here I would say that the photon is obviously moving diagonally through space no matter which perspective -- The observer on the train has matching velocity in one direction to the photon, therefore relative to them they don't see the movement of the photon in that direction... big deal?

In terms of space traversed, Newton himself would not doubt that the path of the train photon is observed or measured to be shorter in the train frame. A different thing is that he postulated that there is an absolute space where the photon might be really moving a longer or a shorter path. But he never drew any consequences out of that and he accepted that in practical terms you have to adhere to the relative measurement.

So again, I still cannot see how the photon would actually be moving any further or less, nor taking any more time or less, from either frame of reference.

Time is a different thing. With a ballistic theory about light, you'd assume that the two photons, train and ground photon, take the same time to do their jobs. With an aether theory, it's a little long to explain, but I suppose that one could say that there is a "real" time and a "local" time, the first depending on the speed of the train relatve to the aether but not measurable, the second quite equivalent to SR's relative time. In SR, yes, the train photon is observed or measured to be slower than the ground photon in the ground frame, just like the ground photon is measured to be slower than the train photon in the train frame.

How do you get that result? I was arguing that only after assuming that the speed of light is the same in both frames. It appears, however, that the existence of an invariant speed is somethig that flows from the principle of relativity itself, although after that realization I suppose you have to assume that such invariant speed is the speed of light, if you want to arrive at the above conclusions. By the way, thanks for the clarifications, JesseM. I'll come back to them later.

 

[Saw]

Coming back to the derivation of this paper and trying to clarify a little further:

the constant K has units of 1 over speed squared, nothing about the proof is changed if you substitute K=1/S^2, where S is a speed.

Ok. Accepted. This derivation ends up with this constraint: under any model of relativity, you must have an invariant speed.

extra assumptions here are about what is needed to decide whether the invariant speed S is finite or infinite (and if it's finite, what its value is), not about deriving that there is an invariant speed S in the first place.

Ok. I see the point: with this derivation, your choice (to be exercised on the basis of physical assumptions, to be contrasted with experiment) is narrowed down: there is an invariant speed and you can then assume it is infinite or finite (and choose a value for it), but not leave it out. In other words, the principle of relativity (which is the only pre-derivation constraint) equals an invariant speed, whatever it is.

it's certainly true that if anything is moving at infinite speed in one frame (i.e. if any two events on its worldline are simultaneous) it is moving at infinite speed in every other frame according to the Galilei transformation (which wouldn't be true in the Lorentz transformation, for example), and that is directly implied by the fact that the Galilei transformation is what you get when you plug K=0 into the general transformation derived in that paper.

Only if you define "frame-invariant speed" in a way that presupposes it must be finite! As I said, it is a nontrivial feature of the Galilei transformation that anything moving at infinite speed in one frame is moving at infinite speed in every other frame.

Ok, I gather you mean that, in the Galilean option (the one that ends up with a transformation for the combination of velocities w = u + v), the invariant speed that must forcefully exist is an infinite one, which is the only one that is shared by all frames. Is this correct?

And if you now choose a finite speed for the role of invariant one and you want to fix its value, what is the criterion that orientates this choice?

 

[Saw]

Well, I didn't give you time to answer the first comments and here are a few more. I’ve been thinking about the derivation of this paper and my impression is that:

- It does not cover the (faulty, as proved by experiment, but theoretically possible) most obvious model of relativity, which is what we could call the “ballistic” or “nothing changes” model (I explain later this expression). As kev noted, this model does not come out of the derivation. And I think that is because the latter contains an implicit assumption that excludes it.
- Due to this, it only addresses a subset of models and you cannot say that, strictly speaking, the principle of relativity itself, without further qualification, forcefully requires an invariant speed.

For orientation, let us take the simpler example of two observers (S and S’) in the same frame but with different positions, who measure the distance (x and x’) to a third point. In order to make the transformation between x and x’ feasible, you need two things: that they measure with the same length units and that they also agree on the relative distance between the two of them. In turn, whenever one talks about units, it seems unavoidable to refer to the measurement instrument. If those observers say that they share the same units, they mean that they both use identical rods labelled as 1-metre units, for example, and that their diverse perspectives (i.e. positions) do not have any impact on the relevant physical characteristics of their instruments.

Now, in a new scenario where the two observers are moving wrt each other, everything is the same but dynamic: they have to agree on the variation of distance across time between them (relative speed) and… on which units?

The first evident possibility is that both observers agree on both length and time units. In turn, this means that a certain one-metre rod at rest in frame S is still a 1-metre rod as measured in frame S’ and that 1 s as measured by a clock in S is also 1 s as measured by (forcefully two) clocks in S’. Thus if a ball flies across the distance x’ in S’ (L’ length units) during time t’, you can admit in S that it has flown L’ plus (vt) or (vt’) length units, with the tranquillity that you are not combining apples with pears.

The ballistic model assumes this. You have two balls at rest in their respective inertial ships, sharing the “basic” states of motion of the latter. They are physically identical. “Nothing changes” because of relative motion. Now each ball is accelerated in its rest frame by some procedure, which is identical in both places. Is the acceleration of ball B wrt S strictly identical to the acceleration of B’ wrt S’? You have to make a choice, an assumption. If you say “yes”, if “nothing changes”, either, because of the acceleration (= if the “additional” states of motion acquired by the balls wrt to their respective ships are the same), then you are inferring that time (lapses and simultaneity) and length units are also absolute for S and S’. (This can be discussed more calmly, if you have any objection, though for brevity I suppose you agree.)

But you may doubt that the effects of the acceleration are identical. For example, because you check that light, unlike a ball, does not share a “basic” state of motion with its source and you also discard the air-drag model. Furthermore, you realize that the acceleration procedure of mechanical objects is either electromagnetic or due to other force that may have a behaviour akin to light. Hence you do not have identical length and time units any more. (Again this can be developed, if needed.) Solution: you may not get that but you still may do well enough for transformation purposes if at least you get, due to a compensation of effects, the same… speed units, in the form of a frame-invariant speed?

Coming back to the derivation: It’s important to note that it allows time (t) to be different from time (t’). Once you make this mathematical choice, you are banning the “nothing changes” model (where t = t’) from the picture, you are narrowing down the choice to relativity models where “something changes” and hence t≠t. Given this, the only way to restore the equality t = t’ and come back to the Galilean Transformation is to annul the constant (to make k = 0) by making the invariant speed infinite.

 

[JesseM]

Ok. I see the point: with this derivation, your choice (to be exercised on the basis of physical assumptions, to be contrasted with experiment) is narrowed down: there is an invariant speed and you can then assume it is infinite or finite (and choose a value for it), but not leave it out. In other words, the principle of relativity (which is the only pre-derivation constraint) equals an invariant speed, whatever it is.

Yup, exactly.

Ok, I gather you mean that, in the Galilean option (the one that ends up with a transformation for the combination of velocities w = u + v), the invariant speed that must forcefully exist is an infinite one, which is the only one that is shared by all frames. Is this correct?

Yes, in order to get the generalized coordinate transformation to reduce to the Galilei transformation you have to pick K=0 which is equivalent to S=infinity, and if you plug that into the generalized velocity addition equation w = (u + v)/(1 + Kuv) = (u + v)/(1 + uv/S^2), you do get back the Galilean formula w = (u + v).

And if you now choose a finite speed for the role of invariant one and you want to fix its value, what is the criterion that orientates this choice?

If you can determine the equations for some of the basic laws of physics in one frame (like Maxwell's equations or the equations of quantum electrodynamics), you can see whether they would be invariant under one of the two transformations, and if it's the transformation with the finite invariant speed, you can see what value that speed must take in order to get back exactly the same equations when you apply the coordinate transformation.

 

[JesseM]

- It does not cover the (faulty, as proved by experiment, but theoretically possible) most obvious model of relativity, which is what we could call the “ballistic” or “nothing changes” model (I explain later this expression). As kev noted, this model does not come out of the derivation. And I think that is because the latter contains an implicit assumption that excludes it.

But what do you mean by "does not cover"? It doesn't deal with light one way or another, the coordinate transformation is certainly not incompatible with a ballistic model of light, as long as light moves ballistically at a speed c that's different from the invariant speed S that must appear in the coordinate transformation.

For orientation, let us take the simpler example of two observers (S and S’) in the same frame but with different positions, who measure the distance (x and x’) to a third point. In order to make the transformation between x and x’ feasible, you need two things: that they measure with the same length units and that they also agree on the relative distance between the two of them. In turn, whenever one talks about units, it seems unavoidable to refer to the measurement instrument. If those observers say that they share the same units, they mean that they both use identical rods labelled as 1-metre units, for example, and that their diverse perspectives (i.e. positions) do not have any impact on the relevant physical characteristics of their instruments.

Now, in a new scenario where the two observers are moving wrt each other, everything is the same but dynamic: they have to agree on the variation of distance across time between them (relative speed) and… on which units?

The first evident possibility is that both observers agree on both length and time units. In turn, this means that a certain one-metre rod at rest in frame S is still a 1-metre rod as measured in frame S’ and that 1 s as measured by a clock in S is also 1 s as measured by (forcefully two) clocks in S’. Thus if a ball flies across the distance x’ in S’ (L’ length units) during time t’, you can admit in S that it has flown L’ plus (vt) or (vt’) length units, with the tranquillity that you are not combining apples with pears.

That would be true under the Galilei transformation with the invariant speed set to infinity.

The ballistic model assumes this. You have two balls at rest in their respective inertial ships, sharing the “basic” states of motion of the latter. They are physically identical. “Nothing changes” because of relative motion. Now each ball is accelerated in its rest frame by some procedure, which is identical in both places. Is the acceleration of ball B wrt S strictly identical to the acceleration of B’ wrt S’? You have to make a choice, an assumption. If you say “yes”, if “nothing changes”, either, because of the acceleration (= if the “additional” states of motion acquired by the balls wrt to their respective ships are the same), then you are inferring that time (lapses and simultaneity) and length units are also absolute for S and S’. (This can be discussed more calmly, if you have any objection, though for brevity I suppose you agree.)

I don't think the ballistic model requires the acceleration should be the same in each frame, it just requires that the acceleration and velocity of the thing that's shot out (whether a ball or a photon) is always the same in the rest frame of the apparatus that emitted it, regardless of how the thing that emitted it is moving. As long as the velocity of the thing that's shot out is not equal to the invariant speed built into the laws of physics (and it's been shown that any laws of physics that respect the principle of relativity must have such an invariant speed built in), then different frames will see it shot out with different velocities, this can still be true if the invariant speed is finite (and if it's finite, then different frames will also disagree about the acceleration).

But you may doubt that the effects of the acceleration are identical. For example, because you check that light, unlike a ball, does not share a “basic” state of motion with its source and you also discard the air-drag model. Furthermore, you realize that the acceleration procedure of mechanical objects is either electromagnetic or due to other force that may have a behaviour akin to light.

That may be true of acceleration in our universe, but you're free to imagine any force laws you like that are invariant under the coordinate transformation...for example, you could imagine a universe where springs were continuous indivisible objects (a bit like strings in string theory) and thus the Newtonian spring force law was a fundamental property of springs, not dependent on electromagnetic interactions between atoms mediated by photons as in our universe.

Hence you do not have identical length and time units any more. (Again this can be developed, if needed.) Solution: you may not get that but you still may do well enough for transformation purposes if at least you get, due to a compensation of effects, the same… speed units, in the form of a frame-invariant speed?

I don't really understand what you're saying/asking here.

Coming back to the derivation: It’s important to note that it allows time (t) to be different from time (t’). Once you make this mathematical choice, you are banning the “nothing changes” model (where t = t’) from the picture, you are narrowing down the choice to relativity models where “something changes” and hence t≠t. Given this, the only way to restore the equality t = t’ and come back to the Galilean Transformation is to annul the constant (to make k = 0) by making the invariant speed infinite.

Yes, according to the generalized transformation you can only have t = t' if K=0.

 

[Saw]

Well, we should be sure we talk about the same thing when referring to a ballistic relativity model. I don’t like the term because it suggests it’s a model about light, a model that makes it analogous to a mechanical bullet. And then the question would be, “what is your model for a bullet?” And the answers could be different: there would be a Galilean answer (simple law of combination of velocities) and another conforming to the relativistic law of combination of velocities. If the answer is the second one, then we still have two possibilities: light speed is the invariant speed of the derivation (= the pure SR model) or a peculiar system where both mechanical objects and light behave SR-wise but light speed is not the invariant speed. I think this is what you are referring to when you say:

I don't think the ballistic model requires the acceleration should be the same in each frame, it just requires that the acceleration and velocity of the thing that's shot out (whether a ball or a photon) is always the same in the rest frame of the apparatus that emitted it, regardless of how the thing that emitted it is moving. As long as the velocity of the thing that's shot out is not equal to the invariant speed built into the laws of physics (and it's been shown that any laws of physics that respect the principle of relativity must have such an invariant speed built in), then different frames will see it shot out with different velocities, this can still be true if the invariant speed is finite (and if it's finite, then different frames will also disagree about the acceleration).

If I catch your point, under this theoretical possibility: a train photon emitted from the train would be measured to have speed c in the train frame; a ground photon emitted by a source on the ground would also have speed c in the ground frame; but since the invariant speed S would not be c, in the train frame the ground photon would not travel at c, just like in the ground frame the train photon would not move at c…

Yes, that would be a ballistic model “covered” by the derivation, but I don’t refer to that. I refer to a model where:

the acceleration (not in the technical sense of dv/dt, because we have not yet talked about time, but in an intuitive sense of the physical effect generated by some agent) of ball or photon B emitted from frame S wrt S =

the acceleration of B’ emitted from S’ wrt S’,

so that, if those balls or photons are used to measure time lapses or establish simultaneity, you get frame-invariant results. To put it simply, a model where time is universal. So probably “ballistic“ is not the best adjective to define this model. A model where “nothing changes” due to relative motion is the expression I had proposed, but it’s a little long. Shall we call it conventionally the “classical relativistic” model?

If we agree on the definition of this model, then the question was: is it “covered” by the derivation? By this I mean: if a proponent/defender of such model came to you and you told him “I am sorry, Sir, if you are classical but you still want to be relativist, you have to admit that, for mathematical reasons, the universe is built in with an invariant speed”, could he argue that he doesn’t need such speed and that he is still entitled to be relativist without it?

As you are aware of, I am not a classical relativist myself, thanks to belief in experiments, but the question here is a conceptual and pedagogical one. The issue is whether just out of respect for algebra, without assuming any axiom except his own relativism, the classical relativist must bow to the idea that there is such invariant speed. And I think he shouldn’t. I would readily accept the defence of the classical relativist . Why? Because otherwise the game would not be fair…

You say: “Well, if you are a classical relativist, you can still accommodate in the generalised coordinate transformation, in your preferred shape,

under the Galilei transformation with the invariant speed set to infinity.

Yes, my client could accept that, but what if tomorrow he wanted to defend the superiority of the GT over the LT? For that purpose, he would have to prove that the universe contains an infinite speed, which could be the speed of light or whatever, but infinite after all. Unfortunately, my client does not believe in infiniteness. He’s got a dislike for that. And he claims he is entitled to be relativist + classical + infiniteness-denier. Why shouldn’t he? After all, from the LT perspective, you yourself do without an infinite speed, why shouldn’t he have the same privilege, if you are both to play on level fields?

Certainly, you can prove that his model is wrong, but not so easily: please prove empirically that light does not behave as the classical relativist argues and, furthermore, that mechanical objects don’t, either. With those weapons (physical assumptions + experimental test), you win. But only with algebra…, it’s a draw.

Well, this is a sort of colourful argument, just for fun. (Hope you enjoyed it). But taking again a more rigorous stance, I had said:

It’s important to note that it allows time (t) to be different from time (t’). Once you make this mathematical choice, you are banning the “nothing changes” model (where t = t’) from the picture, you are narrowing down the choice to relativity models where “something changes” and hence t≠t. Given this, the only way to restore the equality t = t’ and come back to the Galilean Transformation is to annul the constant (to make k = 0) by making the invariant speed infinite.

I think you are not appreciating the idea that, even in terms of mathematical rules, by taking the choice in the derivation that t can be different from t’, you are playing a nasty trick on the classical relativist model. Before the authors of the paper thought hard, the classical relativist had a theory that was no doubt relativist, was perfectly consistent (leaving experiments aside) and didn’t need the existence of any invariant speed, least of all an infinite one. The world was peacefully divided in two fields: those for whom t = t’ and those for whom t ≠ t’. Now you decide that t = t’ must change camps and be forced into the house of the t ≠ t’ subset. Obviously, since there is an intrinsic contradiction in this move, you can only do it at the expense of introducing an absurd element. In the new house (since t≠t’) you need something else that is invariant, otherwise there is no possible transformation and this invariant element is, logically, a speed. Since the newcomer has to adhere to this foreign creed but at the same time keep his local faith in t = t’, the only way to reach this contradictory result is to introduce another queerest element: for the newcomer the invariant speed must be infinite…! Good trick! But beware: I am not good at derivations, but I wonder whether someone could play at you the opposite trick: build a derivation where t is kept invariant and the generalised transformation reduces to the LT… only if time is set to infinity!

No need to insist that the jokes are for common fun. I appreciate much the exchange. There are more things to comment, but this is already too long for today...

 

[JesseM]

Well, we should be sure we talk about the same thing when referring to a ballistic relativity model. I don’t like the term because it suggests it’s a model about light, a model that makes it analogous to a mechanical bullet. And then the question would be, “what is your model for a bullet?” And the answers could be different: there would be a Galilean answer (simple law of combination of velocities) and another conforming to the relativistic law of combination of velocities. If the answer is the second one, then we still have two possibilities: light speed is the invariant speed of the derivation (= the pure SR model) or a peculiar system where both mechanical objects and light behave SR-wise but light speed is not the invariant speed.

What do you mean by "behave SR-wise"? They certainly can't obey the usual SR equations in every frame if light speed is not the invariant speed, because by definition only things traveling at the invariant speed will have the same speed in all frames. If you want light to behave SR-wise in just one preferred frame, that would be more like an aether model than a ballistic model, where the speed of light waves is independent of the speed of the emitter, but a frame moving relative to the preferred frame will say light waves move faster in one direction than the other.

If I catch your point, under this theoretical possibility: a train photon emitted from the train would be measured to have speed c in the train frame; a ground photon emitted by a source on the ground would also have speed c in the ground frame; but since the invariant speed S would not be c, in the train frame the ground photon would not travel at c, just like in the ground frame the train photon would not move at c…

Right, that's what I was thinking would be meant by "ballistic model".

Yes, that would be a ballistic model “covered” by the derivation, but I don’t refer to that. I refer to a model where:

the acceleration (not in the technical sense of dv/dt, because we have not yet talked about time, but in an intuitive sense of the physical effect generated by some agent) of ball or photon B emitted from frame S wrt S =

the acceleration of B’ emitted from S’ wrt S’,

I don't understand how you can use "acceleration" in an "intuitive sense" rather than a quantitative one, but then write an equation suggesting equality between the accelerations in each frame. If you don't specify what "physical effect" you mean when you talk about acceleration, how is this meaningful? Or do you mean that all physical properties of photon B as seen in frame S would be identical with the physical properties of B' as seen in frame S', including other things like their coordinate velocities? If so, since we have already specified that we are interested in a universe where the laws of physics are invariant under the coordinate transformation in question, it's automatically going to be true that if you do any experiment using an apparatus at rest in S and repeat the experiment with the same type of apparatus at rest in S', each frame will measure exactly the same thing about the results of their own experiment, including the coordinate velocity/acceleration/whatever of any particles or waves emitted by the apparatus (assuming the apparatus isn't interacting with anything external to itself like an aether filling space...assume the experiment is being done in a sealed-off room moving inertially, and everything inside the room is considered part of the apparatus).

A model where “nothing changes” due to relative motion is the expression I had proposed, but it’s a little long.

Does "nothing changes due to relative motion" express the same idea as above, that repeating the same experiment in different frames will yield the same results as seen in that frame? If so, again, that is automatically implied by the starting assumption that the laws of physics are invariant under the coordinate transformation. If not, can you clarify what is it is exactly that shouldn't change?

If we agree on the definition of this model, then the question was: is it “covered” by the derivation? By this I mean: if a proponent/defender of such model came to you and you told him “I am sorry, Sir, if you are classical but you still want to be relativist, you have to admit that, for mathematical reasons, the universe is built in with an invariant speed”, could he argue that he doesn’t need such speed and that he is still entitled to be relativist without it?

Since I'm not clear about what you mean by "this model" or "classical" I can't answer the question yet.

As you are aware of, I am not a classical relativist myself, thanks to belief in experiments, but the question here is a conceptual and pedagogical one. The issue is whether just out of respect for algebra, without assuming any axiom except his own relativism, the classical relativist must bow to the idea that there is such invariant speed. And I think he shouldn’t. I would readily accept the defence of the classical relativist . Why? Because otherwise the game would not be fair…

Are you disputing the following if-then premise that I used to summarize the results of the paper earlier?

IF you start with the postulate that the laws of physics are invariant under a coordinate transformation that gives a set of coordinate systems moving at constant coordinate velocity relative to one another,

THEN the coordinate transformation must take a certain general form that includes an invariant velocity S.

(with the understanding that the postulate that the laws of physics are invariant under a coordinate transformation, that means that if you do two experiments which are 'equivalent' when viewed in terms of the coordinates of two different frames, i.e. the initial coordinate positions, velocities etc. of all the components of the two apparatuses are the same in each frame, then the results of the experiments will be 'equivalent' in the same sense)

Yes, my client could accept that, but what if tomorrow he wanted to defend the superiority of the GT over the LT? For that purpose, he would have to prove that the universe contains an infinite speed, which could be the speed of light or whatever, but infinite after all. Unfortunately, my client does not believe in infiniteness. He’s got a dislike for that.

He can determine the fundamental equations of the laws of physics in one frame, then see if the equations are invariant under the Galilei transformation. If he then wants to deny that the Galilei transformation contains an infinite invariant speed because he "does not believe in infiniteness", it seems like he's just playing semantic games, presumably he has no disagreement that the Galilei transformation is a special case of the generalized coordinate transformation in which K is set to 0, or that K has units of one over a speed squared.

And he claims he is entitled to be relativist + classical + infiniteness-denier. Why shouldn’t he? After all, from the LT perspective, you yourself do without an infinite speed, why shouldn’t he have the same privilege, if you are both to play on level fields?

Certainly, you can prove that his model is wrong, but not so easily: please prove empirically that light does not behave as the classical relativist argues and, furthermore, that mechanical objects don’t, either.

What is his "model"? Does "classical" mean he is proposing that the fundamental laws of physics are Galilei-invariant? If so, does he accept that the dynamics of charged objects and electromagnetic waves obey Maxwell's equations in at least one frame? Maxwell's equations are not invariant under the Galilei transformation, so the only way to reconcile this with the postulate of Galilei-invariant fundamental laws would be say Maxwell's laws are not really fundamental. Perhaps one could say that they only apply in an aether rest frame, but that this frame is not picked out as special by the fundamental laws because the aether is an actual physical substance which can be accelerated so any other inertial frame can become its new rest frame. But if the fundamental laws were Galilei-invariant it should be possible to use them to design some sort of physical rulers and clocks that would measure all electromagnetic waves to move at c when they were at rest relative to the aether, but when moving at v relative to the aether they'd measure electromagnetic waves to move at c+v in one direction and c-v in the other.

With those weapons (physical assumptions + experimental test), you win. But only with algebra…, it’s a draw.

But algebra alone proves that if you assume the first postulate, then the coordinate transformation that the laws of physics are invariant under must take the generalized form discussed in the paper, in which case it's just a matter of semantics whether you say that the K=0 case still includes an "invariant speed". That is all I have been claiming, no experimental tests are needed to prove it, experiment is only needed to decide which of the special cases of this generalized coordinate transformation actually apply (as well as whether the fundamental laws respect the first postulate in the first place)--do you disagree?

I think you are not appreciating the idea that, even in terms of mathematical rules, by taking the choice in the derivation that t can be different from t’, you are playing a nasty trick on the classical relativist model.

That's not a "choice" at all if you are starting from the initial postulate I described, namely "the laws of physics are invariant under a coordinate transformation that gives a set of coordinate systems moving at constant coordinate velocity relative to one another". It's just a mathematical matter to show that this postulate is consistent with coordinate transformations where t is different than t', so it would be mathematically incorrect to rule such a transformation out without adding additional postulates. Of course there is no a priori reason to say that this postulate is physically plausible, or that it's physically plausible that the laws of physics could be invariant under a transformation where t is not equal to t', but I'm simply not talking about physical plausibility at all here, just about what conclusions follow mathematically from assuming that postulate (which is just a way of restating the first postulate of SR).

Before the authors of the paper thought hard, the classical relativist had a theory that was no doubt relativist, was perfectly consistent (leaving experiments aside) and didn’t need the existence of any invariant speed, least of all an infinite one.

Again, just semantic games. Mathematically the Galilei transformation's velocity addition formula is a special case (or a limit case) of the generalized formula w = (u + v)/(1 + uv/S^2), which implies that if v=S then w=S too, regardless of whether anyone had recognized that fact or chosen to speak of it in terms of the Galilei transformation having an "invariant speed". Likewise, mathematically it is true that if "speed" is defined as distance/time between two points on a line through spacetime, then if in one frame you have a line where two points have a time-interval of zero but a nonzero distance-interval, then under the Galilei transform this will be true of the time and distance intervals between those points in other frames.

Now you decide that t = t’ must change camps and be forced into the house of the t ≠ t’ subset.

By saying "house of the t ≠ t’ subset" do you mean the "house" (camp) of people who would agree with the fact that the laws of physics are symmetric under a transformation which includes an invariant speed? But do you suppose that any of the physicists who would have advocated t = t' would have disagreed with the mathematical statements in my previous paragraph? If not, then once again you are just talking about semantics, nothing more.

On the other hand, if by "house of the t ≠ t’ subset" you literally mean the house of people who accept that simultaneity is relative and a time-interval in one frame can be different than a time-interval in another, then clearly this is wrong; the generalized coordinate transformation found in the paper is compatible either with relative or absolute simultaneity, you can't decide between them without fixing the value of K/S.

Obviously, since there is an intrinsic contradiction in this move

How is there an "intrinsic contradiction"? Are you conflating the two meanings of "house of the t ≠ t’ subset" which I suggested above? Obviously no one is suggesting that those who believe t=t' are literally forced to believe t=t' as in the second paragraph above. And if you are conflating the two meanings, then again it looks like you are playing semantic games.

you can only do it at the expense of introducing an absurd element. In the new house (since t≠t’) you need something else that is invariant, otherwise there is no possible transformation and this invariant element is, logically, a speed. Since the newcomer has to adhere to this foreign creed but at the same time keep his local faith in t = t’

Don't know what you mean by "local faith"--you often seem to express yourself in a very opaque manner, hinting at things without really spelling them out. Are you somehow suggesting that if all we assume/know is the first postulate and the derivation showing the generalized transformation that follows from it, it is somehow more natural to then believe t≠t', and more contrived or requiring of "faith" to believe t=t'? If so then of course this makes no sense at all, the generalized transformation alone does not favor either possibility in the slightest.

the only way to reach this contradictory result

"Contradictory" only if you play odd semantic games and use phrases like "house of the t ≠ t’ subset" with ambiguous meanings! I see nothing contradictory about the fact that the first postulate leads you to the generalized transform, and that one possible case of the generalized transform is the one where K=0 in which case t=t'.

is to introduce another queerest element: for the newcomer the invariant speed must be infinite…! Good trick! But beware: I am not good at derivations, but I wonder whether someone could play at you the opposite trick: build a derivation where t is kept invariant and the generalised transformation reduces to the LT… only if time is set to infinity!

There is no "trick" in the derivation, it's just exploring what can be concluded by assuming only the first postulate. If you want to imagine some alternate derivation where it's assumed from the outset that time is invariant, then you'd have to specify what other starting assumptions you want (first postulate + invariant time would just give the Galilei transformation of course, so you couldn't use the first postulate as a starting assumption if you want to get a generalized transformation as the result). I don't know what it would even mean to say "time is set to infinity"--"time" in what equation? Would the coordinate transformation itself say something like t' = t*infinity? Honestly it seems like you lead yourself up a lot of blind alleys by using evocative-sounding phrases that don't necessarily correspond to any well-defined mathematical/physical meanings in your head (which may be true of earlier phrases I asked about like "behave SR-wise" or "acceleration in an intuitive sense" or "does not believe in infiniteness" or "house of the t ≠ t' " or "local faith" or "intrinsic contradiction"), I'd suggest you try to construct your arguments around terms that you could define very clearly if asked, and avoid more vague verbal formulations that don't correspond to some very clear math or image of a physical scenario...

 

[Saw]

Since I'm not clear about what you mean by "this model" or "classical"

It’s one where you assume that simultaneity is absolute and a time-interval for a physical process (no matter whether mechanical or electromagnetic) measured in one frame is equal to the time-interval measured for the same process in another frame. As to the physical reason for that, in as much as light is concerned, it might be that: there is an aether but it is dragged with the source or that there is no aether at all and light can be accelerated as if it were a massive particle… Both things are ruled out by experiments. But precisely our discussion requires that we don’t make initial physical assumptions and don’t look at experiments. The reasons, therefore, can be left out of the discussion. Hence, to be neutral, we can call it simply the “t=t’ model”.

Next step: is this model “relativist”?

Your definition of a model complying with the principle of relativity is:

the laws of physics are invariant under a coordinate transformation that gives a set of coordinate systems moving at constant coordinate velocity relative to one another

(with the understanding that the postulate that the laws of physics are invariant under a coordinate transformation, that means that if you do two experiments which are 'equivalent' when viewed in terms of the coordinates of two different frames, i.e. the initial coordinate positions, velocities etc. of all the components of the two apparatuses are the same in each frame, then the results of the experiments will be 'equivalent' in the same sense)

I will develop to be sure that we agree on this definition:

First requirement: identical experiments give identical results = if an experiment with a bullet being shot in frame S shows that the bullet has traveled 2 m in 1 s in S coordinates (for light, just change the numbers), then the same experiment carried out in S’ will also give 2m in 1s in S’ coordinates.

Second requirement: for a given unique experiment carried out in one frame, the two frames may obtain different coordinates; however, you can relate the coordinates in frame S with the coordinates in S’ for the same object… through a transformation equation.

Well, in order to make the t=t’ model compliant with these conditions, we just need to stipulate that it does comply with them, since we have decided to leave physical reasons aside. In particular, in principle, the second requirement is met through the GT… or do you dispute that?

Maxwell's equations are not invariant under the Galilei transformation, so the only way to reconcile this with the postulate of Galilei-invariant fundamental laws would be say Maxwell's laws are not really fundamental. Perhaps one could say that they only apply in an aether rest frame, but that this frame is not picked out as special by the fundamental laws because the aether is an actual physical substance which can be accelerated so any other inertial frame can become its new rest frame. But if the fundamental laws were Galilei-invariant it should be possible to use them to design some sort of physical rulers and clocks that would measure all electromagnetic waves to move at c when they were at rest relative to the aether, but when moving at v relative to the aether they'd measure electromagnetic waves to move at c+v in one direction and c-v in the other.

Here I don’t follow you. Maxwell's equations are of course fundamental. What do they require? That light travels at c. In a model that is not relativist, that might mean that light only travels at c in a preferred aether frame. But if that were the model we talk about, it’d not be relativist and the discussion would be over. However, the t = t’ model can claim to be relativist since it complies the two requirements:

First, in the t=t’ model, light travels at c in all frames, if we decide it does. I understand that you find it difficult to visualize the reasons. With the aether-drag model, it’s possible, although you need that the conditions for aether-drag (a sealed compartment trapping and dragging aether) are met. But, in the absence of that, think of the theory accepting that photons can be accelerated. I know it’s absurd, but again I insist: we are supposed *not* to be allowed to make *any* physical assumptions at this stage.

Second, if a light beam is projected in frame S, in frame S it will travel at c, but in frame S’ the same light beam will travel at c-v or c+v, depending on the direction of projection. Does that make the model non-relativist? It doesn’t. We do not need that the two frames get the same coordinates. They agree on time but they disagree on distance. No problem. They can still relate their coordinates through the GT.

So we have a well-defined model that is relativist, don’t we?

Now we face it with another that claims that t≠t’. And we want to know whether we can find a generalized transformation that comprises the two models, without making any further assumption.

We take the route of the paper and we get what we get.

I asked if the meaning is what bcrowell was suggesting:

with this derivation, your choice (to be exercised on the basis of physical assumptions, to be contrasted with experiment) is narrowed down: there is an invariant speed and you can then assume it is infinite or finite (and choose a value for it), but not leave it out. In other words, the principle of relativity (which is the only pre-derivation constraint) equals an invariant speed, whatever it is.

And you answered:

Yup, exactly.

And have also said:

(and it's been shown that any laws of physics that respect the principle of relativity must have such an invariant speed built in)

But you have also said:

But algebra alone proves that if you assume the first postulate, then the coordinate transformation that the laws of physics are invariant under must take the generalized form discussed in the paper, in which case it's just a matter of semantics whether you say that the K=0 case still includes an "invariant speed". That is all I have been claiming, )

So you have often declared (also in other places) that the derivation DOES include an invariant speed but in the latter passage you hint that the important thing is that the generalised transformation includes a constant K, which can be K = 0 and it’s “just a matter of semantics whether” that case includes (or NOT?) an invariant speed.

What do you mean by “semantics”? You seem to have a peculiar understading of the word: if something falls into the concept of semantics, it’s unimportant, it doesn’t have a clear meaning. It’s just the opposite: “semantics” is the discipline that studies the meaning of language. So discussing about semantics equates to clarifying what somebody means when he says something. For example, it’s not the same to simply declare that the derivation has a constant K and stating that it has an invariant speed. You clarified to me that, because the units of K are 1/speed squared, the derivation *means* (yes, semantics) that any relativity model that deserves that name must have an invariant speed. Do you confirm that?

If so, then, obviously, it’s not unimportant. It *means* that the t=t’ model can only be relativist at the expense of accepting that the invariant speed is infinite. Or not?

If we found an understanding on these matters, we could keep discussing the core of the issue: whether the derivation actually makes or not any implicit extra-assumption from scratch, which is extraneous to the relativity principle itself.

 

[Saw]

I will come back now to the main point.

It should be highlighted that who is claiming something new and against established thought is you and the authors you cite, not me. Einstein did not follow your line of reasoning but the following:

- The principle of relativity was well established for experiments with mechanical objects, on the basis of the idea that they share the state of motion of their source.
- Then it was found out (experimentally!) that light is different, in the sense that its speed is independent of the motion of the source, like it happens with sound.
- In principle, the latter represented a threat for the principle of relativity: experiments with light would not give the same results in different frames.
- Einstein’s solution to reconcile the principle of relativity with light experimentally’s observed behaviour was to postulate that in a given experiment not only the frame of the source but any other observing frame would measure a frame-invariant speed for light, which sounds odd but is possible if you abandon the postulate that time is absolute and adhere instead to the postulates of relativity of simultaneity (RS), time dilation (TD) and length contraction (LC), which are contained in the Lorentz Transformation (LT).
- Experiments have proved that Einstein’s postulates were a good intuition.

To sum up, focusing on the subject under discussion here:

- Einstein’s frame-invariant speed was the speed of a physical phenomenon, the speed of light.
- Einstein did not claim that such invariant speed stemmed from the principle of relativity alone. He introduced that as an additional postulate and at the same time admitting that the very introduction of such new postulate required a change of mental framework (from absolute to relative) with regard to time and length: both things go together = you cannot mathematically derive RS, TD or LC without the frame-invariant c and, vice versa, you do not get RS, TD or LC without c.

Now we have apparently some authors like Rindler or Pal here taking one bold step further. Basically, their “discovery” is that Einstein could have been much more aggressive and state that:

- The principle of relativity alone implies (by rigorous mathematical demonstration) a frame-invariant speed, without any further assumption (i.e. without assuming the relativity of time).
- There is no need to identify this invariant speed with the speed of light or anything physical whatsoever. It’s a logical need that must exist, as long as : so even if tomorrow experiments get playful and start proving that the speed of light is not c in every frame, there will .

Do you really think this is right…? I find this hard to believe… I have already given all the arguments against that idea. But it seems you are not fond of the metaphoric language I used. I’ll try now to put it very simply in the hope of being more didactic:

what a coordinate transformation requires is not an invariant speed, but an invariant “something”:

- In the variant time model (t≠t’), that ”something” is speed, not time.
- In the invariant time model (t=t’), that “something” is time, not speed.

What you call the “generalised transformation” (as if it were a super-expression of both GT and LT) is simply the Lorentz Transformation with some cosmetics: instead of 1/c^2 you have written a constant K, as if it that were the way to accommodate the GT.

Now I will retake metaphoric language, because I like it. All this reminds me of the Spanish saying, “aunque la mona se vista de seda, mona se queda” (= even if you dress the monkey in silk clothes, it is still a monkey and not something else). That we are in front of the LT, purely and simply, is evidenced by the fact that you take off the silk clothes, you lift the veil of the constant K and what do you find? Figuratively, big teeth, whih is Ok for a monkey, not for a lady. Literally, an invariant speed, which is OK for the LT, not for the GT.

Convince yourself. It’s the choice of t≠t’, not the principle of relativity, what leads to the invariant speed.

That's not a "choice" at all if you are starting from the initial postulate I described, namely "the laws of physics are invariant under a coordinate transformation that gives a set of coordinate systems moving at constant coordinate velocity relative to one another". It's just a mathematical matter to show that this postulate is consistent with coordinate transformations where t is different than t', so it would be mathematically incorrect to rule such a transformation out without adding additional postulates. Of course there is no a priori reason to say that this postulate is physically plausible, or that it's physically plausible that the laws of physics could be invariant under a transformation where t is not equal to t', but I'm simply not talking about physical plausibility at all here, just about what conclusions follow mathematically from assuming that postulate (which is just a way of restating the first postulate of SR).

Here you are playing with words, isolating them from their context and meaning. Certainly, if we are discussing what relativity is, t≠t’ is not a option that you can leave out. Of course, who doubts that? What I am saying is that your so called generalised transform makes the choice of t≠t’ and leaves the option t = t’ out. Well, it can still accommodate t = t’, but only after introducing two requirements that are strange and unnecessary in this model: invariant speed + its infinity. Come on… Are you serious about that? Isn’t it simpler to frankly admit that t = t’ is *also* a valid relativity model and that the only invariant thing it requires to exist is an invariant time, as its definition proclaims?

The best interpretation that we can make of this generalised transformation is that it means nothing, nothing new. If you start attributing to it any meaning, any additional meaning, it starts lying conspicuously.

 

[bcrowell]

I will come back now to the main point.

It should be highlighted that who is claiming something new and against established thought is you and the authors you cite, not me.

I don't think "new" is the right word here. The earliest derivation along these lines is W.v.Ignatowsky, Phys. Zeits. 11 (1910) 972. So Einstein's approach is 105 years old, and this one is 100 years old. Another old reference along these lines is L.A. Pars, Philos. Mag., 42 (1921) 249.

The other question is whether the approach taken by Rindler et al. is controversial. The answer is no. If you want to convince us that the answer is yes, please show us some published reference to that effect. Yes, it is possible that Ignatowsky, Pars, Rindler, Morin, and Pal are all wrong, and that you are right. Rindler is a very well known relativist. (He introduced the term "event horizon.") But of course it's still possible that he's made an elementary mistake, and you just need to point it out to him. Here is his faculty web page, with his contact information: http://www.utdallas.edu/physics/faculty/wolfgang.html

Well, it can still accommodate t = t’, but only after introducing two requirements that are strange and unnecessary in this model: invariant speed + its infinity.

These derivations do not introduce new "requirements" such as the existence of an invariant speed. They derive the existence of an invariant speed from other axioms involving the symmetry properties of spacetime.

 

[JesseM]

Since I'm not clear about what you mean by "this model" or "classical"

It’s one where you assume that simultaneity is absolute and a time-interval for a physical process (no matter whether mechanical or electromagnetic) measured in one frame is equal to the time-interval measured for the same process in another frame.

And does "this model" also assume the first postulate which says that the laws of physics are invariant under some set of coordinate systems moving at constant velocity relative to one another? If so, then your model can just be summed up as the idea that the laws of physics are Galilei-invariant (since the proof shows there is no other coordinate transformation that has both the property of obeying the first postulate and the property of absolute simultaneity). But I guess you address this question below:

Next step: is this model “relativist”?

Your definition of a model complying with the principle of relativity is:

the laws of physics are invariant under a coordinate transformation that gives a set of coordinate systems moving at constant coordinate velocity relative to one another

(with the understanding that the postulate that the laws of physics are invariant under a coordinate transformation, that means that if you do two experiments which are 'equivalent' when viewed in terms of the coordinates of two different frames, i.e. the initial coordinate positions, velocities etc. of all the components of the two apparatuses are the same in each frame, then the results of the experiments will be 'equivalent' in the same sense)

I will develop to be sure that we agree on this definition:

First requirement: identical experiments give identical results = if an experiment with a bullet being shot in frame S shows that the bullet has traveled 2 m in 1 s in S coordinates (for light, just change the numbers), then the same experiment carried out in S’ will also give 2m in 1s in S’ coordinates.

Yes, assuming they were both shot by identical guns and that the description of the gun used in each experiment is the same in each coordinate system...for example, if the bullet measured in frame S was shot by a gun at rest in frame S then it must also be true that the bullet measured in frame S' was at rest in frame S' (likewise if the bullet measured in frame S was shot by a gun moving at 1 meter/second in the +x direction in frame S, then the bullet measured in frame S' was shot by a gun moving at 1 meter/second in the +x' direction of frame S'...there is no requirement that all parts of the apparatus be at rest in the frame you're doing the experiment, just that all the initial velocities, accelerations etc. be 'equivalent' in both frames).

Second requirement: for a given unique experiment carried out in one frame, the two frames may obtain different coordinates

No, not if the coordinate description of the initial setup was identical (including details like the initial distance of various components of the apparatus from the origin). In this case the coordinate description of the results must be identical too.

however, you can relate the coordinates in frame S with the coordinates in S’ for the same object… through a transformation equation.

No! If that was all that was required, then even in a Galilean aether model where light moved at c in all directions in a preferred frame but moved at c+v in one direction and c-v in the other direction in some other inertial frame, then your version of the "first postulate" would still be satisfied since we could related the coordinates back to the preferred frame via a coordinate transformation. In fact this would make the first postulate into a tautology, since all coordinate transformations are just different ways of labeling the same physical events, so if you have the description of some results in coordinate system A, and you transform these events into coordinate system B and look at the description in B, it's automatically going to be true that if you perform the inverse transformation on these results in B you'll get back the original description you had in A.

In reality, the scenario I described above where light always moves at c in the aether rest frame but can move at other speeds in different frames would not satisfy the first postulate as it is understood by all physicists (at least not unless the aether was was an actual physical substance which could be moved around, so the aether rest frame was not really preferred by the fundamental laws of physics...see the discussion about air below).

Well, in order to make the t=t’ model compliant with these conditions, we just need to stipulate that it does comply with them, since we have decided to leave physical reasons aside.

You are certainly free to stipulate that the laws of physics obey the first postulate and that the coordinate transformation satisfies t=t', but then if you want to talk about some kind of "ballistic model" of light you have to make sure your ballistic model is logically compatible with these stipulations. Just as an example, you would not be allowed to assume in your ballistic model that a light ray traveling at some finite speed c in one frame will also be traveling at c in other frames, since the proof shows that first postulate plus t=t' implies the laws of physics are invariant under the Galilei transformation, and the Galilean velocity addition formula would not be compatible with this notion of light traveling at the same finite speed in all frames. If you give me some details of what you meant by "ballistic model" we can see if it's compatible with those two stipulations about the laws of physics.

Maxwell's equations are not invariant under the Galilei transformation, so the only way to reconcile this with the postulate of Galilei-invariant fundamental laws would be say Maxwell's laws are not really fundamental. Perhaps one could say that they only apply in an aether rest frame, but that this frame is not picked out as special by the fundamental laws because the aether is an actual physical substance which can be accelerated so any other inertial frame can become its new rest frame. But if the fundamental laws were Galilei-invariant it should be possible to use them to design some sort of physical rulers and clocks that would measure all electromagnetic waves to move at c when they were at rest relative to the aether, but when moving at v relative to the aether they'd measure electromagnetic waves to move at c+v in one direction and c-v in the other.

Here I don’t follow you. Maxwell's equations are of course fundamental. What do they require? That light travels at c. In a model that is not relativist, that might mean that light only travels at c in a preferred aether frame.

If you have a frame that is "preferred" by the fundamental laws of physics then that violates the first postulate, and the fundamental laws of physics will not be Galilei-invariant (since Galilei-invariance just means the equations of the laws of physics are exactly the same in all the coordinate systems given by the Galilei transformation). What I was getting at in the paragraph above is that if you treat the aether as a dynamical entity whose rest frame can be changed, then you could have an aether theory that was Galilei-invariant. Consider an analogy between air and aether. Air is a physical medium such that, in the rest frame of a volume of air, sound waves all travel at the same speed in all directions regardless of the velocity of the emitter. One could write down a set of equations describing how physical substances other than air are able to interact with one another at a distance via movements of air, like one object making a sound and a distance object starting to vibrate in response. If these equations were written down from the perspective of the air's rest frame, they would include the notion that sound waves always travel at the speed of sound s. So if you were to ignore the fact that air is itself a physical substance and treat these equations as fundamental, they would be incompatible with Galilei-invariance. But of course we know that the rest frame of a volume of air can be changed and so the fact that sound waves all travel at s in this frame need not be incompatible with Galilei-invariance (or Lorentz-invariance), as long as your equations for the fundamental laws of physics treat air (or the particles it's made of) as a dynamical entity in its own right. Going back to the notion of equivalent experiments in different inertial frames yielding the same results in the coordinate of each frame, we could imagine two sealed containers moving inertially relative to one another, each filled with air that is at rest relative to the container. If we do equivalent experiments involving sound waves within each container, and use the Galilei transformation to define the coordinates of the rest frame of each container, then if the laws of physics are Galilei-invariant the coordinate description of the results of the experiment will look the same in each frame. In principle the same could be true for an aether theory, if you could have two different containers in motion relative to one another but with the aether within each container being at rest relative to that container. In this case the idea that light always moves at c in the aether rest frame could be compatible with the idea of the fundamental laws of physics being Galilei-invariant (or Lorentz invariant).

However, the t = t’ model can claim to be relativist since it complies the two requirements:

First, in the t=t’ model, light travels at c in all frames, if we decide it does. I understand that you find it difficult to visualize the reasons.

It's mathematically impossible that the same light beam could be moving at c in all frames under the t=t' model as you've described it. The proof shows it's impossible to write down a set of equations for hypothetical laws of physics that have all the following properties:

1. The equations are invariant under a coordinate transformation involving coordinate systems moving at constant velocity relative to one another
2. Under this coordinate transformation, t=t'
3. Under this coordinate transformation, anything moving at c in one frame is also moving at c in other frames

The proof shows that 1 and 2 logically imply the coordinate transformation in question must be the Galilei transformation, it's mathematically impossible for it to be anything else. And the Galilei transformation implies the Galilean velocity addition rule w = v + u, which demonstrates that something moving at v=c in one frame must be moving at c + u in a second frame which sees the first frame moving at u in the same direction.

With the aether-drag model, it’s possible, although you need that the conditions for aether-drag (a sealed compartment trapping and dragging aether) are met.

That sounds like the Galilei-invariant dynamical aether model I imagined above, which is also what I was talking about in the previous post when I said:

Maxwell's equations are not invariant under the Galilei transformation, so the only way to reconcile this with the postulate of Galilei-invariant fundamental laws would be say Maxwell's laws are not really fundamental. Perhaps one could say that they only apply in an aether rest frame, but that this frame is not picked out as special by the fundamental laws because the aether is an actual physical substance which can be accelerated so any other inertial frame can become its new rest frame.

But in that case it's not that any single light wave moves at c in all frames, it's just that any frame can see light moving at c in its frame if it is using a chunk of aether at rest in that frame. If we have a sealed container of aether S and another sealed container S' moving at v relative to the first (with the aether at rest relative to its container in both cases), then a light wave inside S will be moving c in the rest frame of S, and at c+v (or c-v depending on direction) in the rest frame of S'; likewise, a light wave inside S' will be moving at c in the rest frame of S', and c+v (or c-v) in the rest frame of S. So both will agree that not all light rays move at c, only light rays which are vibrations in chunks of aether that are at rest in that frame.

But, in the absence of that, think of the theory accepting that photons can be accelerated. I know it’s absurd, but again I insist: we are supposed *not* to be allowed to make *any* physical assumptions at this stage.

Of course, there is nothing "absurd" about the idea that photons can be accelerated if we are starting only from the assumptions of the first postulate and t=t', but that statement is too vague to give me an idea of what your model is for photon behavior in this case. Are you talking about a model where the photons in some sense start out at rest inside the emitter and are then all shot out with the same acceleration, which under the Galilei transformation means they will always have the same speed relative to their emitter. This is what I originally understood by "ballistic model", but then we had the following exchange:

If I catch your point, under this theoretical possibility: a train photon emitted from the train would be measured to have speed c in the train frame; a ground photon emitted by a source on the ground would also have speed c in the ground frame; but since the invariant speed S would not be c, in the train frame the ground photon would not travel at c, just like in the ground frame the train photon would not move at c…

Right, that's what I was thinking would be meant by "ballistic model".

Yes, that would be a ballistic model “covered” by the derivation, but I don’t refer to that.

So apparently by ballistic model you did not mean a model where the acceleration of the photons was constant and thus (under Galilei-invariant laws) there speed relative to the emitter would be constant. So were you thinking of the kind of "moveable aether" model above instead? If not, I don't really see a third option, so if you do you'll have to explain in more detail.

Second, if a light beam is projected in frame S, in frame S it will travel at c, but in frame S’ the same light beam will travel at c-v or c+v, depending on the direction of projection. Does that make the model non-relativist? It doesn’t.

I agree it doesn't in either the "moveable aether" or the "constant speed relative to the emitter" model, but I don't see any other option.

We do not need that the two frames get the same coordinates. They agree on time but they disagree on distance. No problem. They can still relate their coordinates through the GT.

As I pointed out above, the first postulate requires that they actually get identical results for any experiments when expressed in their own coordinates, not just that they "can still relate their coordinates through the GT" which would make the first postulate into a tautology since it's impossible to come up with any possible laws of physics where they couldn't relate their results in this way. However, the fact that they must get identical results for equivalent experiments does not preclude the fact that for a given light ray, one will say it travels at c while the other says it travels at c-v, since the circumstances surrounding how this ray was emitted might not be "equivalent" in the two systems. For example, one might say it was emitted in a chunk of aether that was at rest in his frame, while the other might say it was emitted in a chunk of aether that was moving at speed v in his frame. So, in that case they could both still agree that the fundamental laws of physics as described in one's own frame imply that light rays emitted in a chunk of aether at rest will move at c, while light rays emitted in a chunk of aether moving at speed v will move at c+v or c-v (depending on whether they're emitted in the same direction as the chunk or the opposite direction).

Now we face it with another that claims that t≠t’. And we want to know whether we can find a generalized transformation that comprises the two models, without making any further assumption.

The logic of the paper was not to start with any assumptions about t and t' at all, but just to start with the first postulate and see what conclusions could be drawn from that. I suppose if by "the two models" you mean "a model which assumes the first postulate along with t=t', and another model which assumes the first postulate along with t≠t' ", then looking at what general conclusions you can draw from the assumption that one of these models must be correct is equivalent to looking at what general conclusions you can draw from the first postulate alone (which is what the paper actually did), since it's just a tautology that any possible coordinate transformation satisfies either t=t' or t≠t'.

 

[JesseM]

(continued from previous post)

So you have often declared (also in other places) that the derivation DOES include an invariant speed but in the latter passage you hint that the important thing is that the generalised transformation includes a constant K, which can be K = 0 and it’s “just a matter of semantics whether” that case includes (or NOT?) an invariant speed.

What do you mean by “semantics”? You seem to have a peculiar understading of the word: if something falls into the concept of semantics, it’s unimportant, it doesn’t have a clear meaning. It’s just the opposite: “semantics” is the discipline that studies the meaning of language. So discussing about semantics equates to clarifying what somebody means when he says something.

My point was not that semantics is unimportant, but that in physics it's only important insofar as people are trying to communicate about things that are relevant to physics, namely 1) mathematical truths about the mathematical models used in physics, and 2) predictions/observations about quantitative empirical measurements. If we both agree on that stuff, any further disagreement about what English words should be used to describe some set of mathematical conclusions is then unimportant, just "semantics", it is not any sort of dispute about the science or the math.

For example, it’s not the same to simply declare that the derivation has a constant K and stating that it has an invariant speed. You clarified to me that, because the units of K are 1/speed squared, the derivation *means* (yes, semantics) that any relativity model that deserves that name must have an invariant speed. Do you confirm that?

As long as you agree mathematically that the derivation still works fine if you substitute K=1/S^2 with S having units of speed, and that the derivation then logically implies that any laws of physics which satisfy the first postulate must be invariant under a coordinate transformation which has the velocity addition formula (u+v)/(1 + Kuv) = (u+v)/(1 + uv/S^2), then I don't mind if you say (semantics) that the K=0, S=infinity case does not include an "invariant speed" because you define an "invariant speed" to be a finite one, and no object with finite speed will have the same speed in all frames according to the Galilean velocity addition formula w=u+v. Something like this sort of finite definition of "invariant speed" seemed to be implied in your earlier suggestion that your hypothetical classical relativist "claims he is entitled to be relativist + classical + infiniteness-denier". If this classical relativist accepts the purely mathematical validity of the proof but wants to define his English-language words in this way, I have no problem with that!

If we found an understanding on these matters, we could keep discussing the core of the issue: whether the derivation actually makes or not any implicit extra-assumption from scratch, which is extraneous to the relativity principle itself.

Are you talking about implicit mathematical assumptions, or just implicit assumptions about word-definitions implied by the verbal conclusion that "the first postulate implies an invariant speed"? If the latter I agree, and as I said one can redefine the words "invariant speed" in such a way that the proof does not imply an invariant speed. But if you're implying there's a problem in the math itself, independent of the English words we use to talk about the initial assumptions and the conclusions, then you'll have to be more specific about what this problem is.

 

[Saw]

Bcrowell, JesseM… it’s not that I do not like language precisions and counter-precisions… I find that exercise most helpful to clarify ideas. But with so many branches, we do not see the forest. Please look at what I said in post #88 and take position:

* On the one hand, we have Einstein with this simple approach: if you combine a first postulate (relativity) with a second postulate (the invariance of the speed of light), you get the LT, which implies RS, TD and LC.

Or alternatively we can phrase it this other way, with the same meaning: if you combine a first postulate (relativity) with a second postulate (t≠t’ and if necessary RS, TD and LC) you get the LT, which implies an invariant c.

* On the other hand, we have the contribution of those other authors. Their math (algebra) is of course correct. I have never disputed that. The question is only whether that math brings some “added meaning”. With JesseM’s definition, “meaning” is new predictions. I would also accept that “added meaning” can be a more didactic approach: we would be in face of a valuable contribution if it better expressed (more elegantly, more clearly…) the logic of the reasoning.

There are two possible judgments:

FIRST: No added meaning, no contribution at all. What these authors are doing is just repeating Einstein’s words, but pretending they are saying something different, which only creates confusion!

SECOND: Added meaning! Which one?

I think bcrowell is clearly taking the SECOND option and saying that this meaning is the following: first postulate alone (relativity), without any choice for a second postulate, leads to the generalised transform, where (i) there is forcefully an invariant speed and (ii) the GT only shows up if that invariant speed is infinite. That is consistent. I disagree with him, but one knows where he is.

But with you, JesseM, I don’t know where you are. Lately you are saying that I can choose to read the derivation with these “English words”: I can perfectly adhere to a relativity model where t = t’ and where the universe does not include any invariant speed, least of all an infinite one. Is that true? Then in what sense does this “choice of English words” differ from the FIRST possibility: there is no added meaning at all, we were doing better with Einstein’s explanation and this derivation only introduces confusion.

(Edit: well, for clarity, what introduces confusion is not the derivation itself, which is algebraically correct, but claiming that it carries an added meaning wrt to Einstein's explanation)

I insist, “semantics” is not the art of hiding what you really mean or being ambiguous about what you mean. That is the misuse of semantics. Semantics is taking position about what symbols (whether mathematical or linguistic) mean in practical terms. What is your position, FIRST or SECOND? I am afraid you are doing acrobacies to find some room in between, but there is no such room. It’s black or white:

For Einstein, with the 1st postulate alone, you do not have any transformation. For a transformation to exist, you need to make a choice about an “invariant something”. If your choice of second postulate is t=t’ but c≠c’, you get the GT. If your choice of second postulate is t≠t but c = c’, you get the LT.

For the other authors, with the 1st postulate alone, you go somewhere. If you define that “somewhere” as I did before (in a manner that clearly “favors, yes, the LT over the GT”), that’s wrong, but at least clear. What is “somewhere” for you? You should define it in a precise manner, you should spell out what it means. Otherwise it means nothing.

 

[Saw]

To complete my point, I have been reading once more, all the "corrections" you make to my words in #90 and #91. My meditated best answer is that I FULLY agree with ALL your corrections! I just find it surprising that you deem those corrections necessary. In my opinion, they are only if you read some words as if 100 other surrounding words didn't exist. You don't seem to like the principle that one must interpret words in their context, the so called systematic interpretation... But I am not going to repeat or try to summarize the agreement, because in that case you would have 10 reasons for disagreement and we could never make any progress.

So we have two perfectly relativist models, one where t=t' and one where t≠t'... Can you answer what I ask in #92? JesseM, what is your clear and unambiguous position with regard to the claim that bcrowell is making? Do you believe that the principle of relativity implies a generalised transform where there is an invariant speed and the LT results if such invariant speed is finite while the GT only results if such invariant speed is infinite?

 

[JesseM]

Bcrowell, JesseM… it’s not that I do not like language precisions and counter-precisions… I find that exercise most helpful to clarify ideas. But with so many branches, we do not see the forest. Please look at what I said in post #88 and take position:

* On the one hand, we have Einstein with this simple approach: if you combine a first postulate (relativity) with a second postulate (the invariance of the speed of light), you get the LT, which implies RS, TD and LC.

Or alternatively we can phrase it this other way, with the same meaning: if you combine a first postulate (relativity) with a second postulate (t≠t’ and if necessary RS, TD and LC) you get the LT, which implies an invariant c.

* On the other hand, we have the contribution of those other authors. Their math (algebra) is of course correct. I have never disputed that. The question is only whether that math brings some “added meaning”. With JesseM’s definition, “meaning” is new predictions.

That's not quite what I said:

My point was not that semantics is unimportant, but that in physics it's only important insofar as people are trying to communicate about things that are relevant to physics, namely 1) mathematical truths about the mathematical models used in physics, and 2) predictions/observations about quantitative empirical measurements.

Purely mathematical truths that bring no new added predictions are certainly relevant to physics; for example, if someone were to come up with an elegant new way of deducing the entire structure of quantum mechanics from a few simple information-theoretic axioms, this would be a result worth publishing! I'd say it is also interesting to know what can be deduced from the first postulate; the result shows, for example, that it would be futile to look for a new theory of physics that is neither Lorentz-symmetric nor Galilei-symmetric but which still obeys the same equations in all inertial frames.

FIRST: No added meaning, no contribution at all. What these authors are doing is just repeating Einstein’s words, but pretending they are saying something different, which only creates confusion!

Obviously from a mathematical point of view they are not just repeating Einstein, since Einstein did not derive what could be deduced from the first postulate without the second.

But with you, JesseM, I don’t know where you are. Lately you are saying that I can choose to read the derivation with these “English words”: I can perfectly adhere to a relativity model where t = t’ and where the universe does not include any invariant speed, least of all an infinite one. Is that true?

Sure, if you choose to define "invariant speed" in a way that presupposes it must be finite.

Then in what sense does this “choice of English words” differ from the FIRST possibility: there is no added meaning at all

What do you mean by "meaning"? According to my definitions above, any valid statement about the mathematical models used in physics or their predictions is in some sense "meaningful" in physics, though obviously some such statements may be less interesting than others.

we were doing better with Einstein’s explanation and this derivation only introduces confusion.

By "Einstein's explanation" do you just mean Einstein's mathematical derivation, or do you imply something more?

I insist, “semantics” is not the art of hiding what you really mean or being ambiguous about what you mean.

I didn't say it was. I just said that if you agree on all the mathematical/physical details, but still insist that you "disagree" somehow on how to describe these details in words, then you are having a purely semantic debate which is irrelevant to physics (aside from pedagogy). On the other hand, if you want to debate whether a given result is "interesting" in some sense that might be worth discussing in a physics context.

What is your position, FIRST or SECOND?

Your "SECOND" is too vague because you have not explained what you mean by "meaning". If you are talking about "meaning" in any sense separate from facts about mathematical models or physical predictions, or possibly whether or not such facts are "interesting" or "trivial" in some sense from a physicist's point of view, then I don't really see how your discussion is relevant to physics.

For Einstein, with the 1st postulate alone, you do not have any transformation.

Einstein simply didn't explore what would follow from the 1st postulate alone, he did not make any positive claim that if you do start with just that postulate, then "you do not have any transformation" (obviously from a mathematical perspective such a positive claim would be nonsense).

For a transformation to exist, you need to make a choice about an “invariant something”.

What does "need to make a choice" mean mathematically? If you don't deny that mathematically it is valid to derive a general transformation without making such a choice, then again it seems like you have left the realm of mathematical truths and are using some other mysterious criteria to make your judgments.

For the other authors, with the 1st postulate alone, you go somewhere. If you define that “somewhere” as I did before (in a manner that clearly “favors, yes, the LT over the GT”), that’s wrong, but at least clear. What is “somewhere” for you? You should define it in a precise manner, you should spell out what it means. Otherwise it means nothing.

I suppose "somewhere" is a mathematically valid description of the complete set of possible coordinate transformations the laws of physics could be invariant under, assuming the laws of physics respect the first postulate. Again I am not sure what kind of "meaning" you are looking for if it goes beyond just valid statements about the mathematical models used in physics or the empirical predictions of these models.

 

[JesseM]

To complete my point, I have been reading once more, all the "corrections" you make to my words in #90 and #91. My meditated best answer is that I FULLY agree with ALL your corrections! I just find it surprising that you deem those corrections necessary. In my opinion, they are only if you read some words as if 100 other surrounding words didn't exist.

Can you give some examples? I wonder if in some cases you were not interpreting my words as "corrections" when I was just explaining something without necessarily implying you had said anything different.

You don't seem to like the principle that one must interpret words in their context

Hmm, what possible words of mine, interpreted in their context, could have given you the idea that I don't like the principle that one must interpret words in their context?

So we have two perfectly relativist models, one where t=t' and one where t≠t'... Can you answer what I ask in #92? JesseM, what is your clear and unambiguous position with regard to the claim that bcrowell is making? Do you believe that the principle of relativity implies a generalised transform where there is an invariant speed and the LT results if such invariant speed is finite while the GT only results if such invariant speed is infinite?

This just depends on the boring semantic question of whether you define "invariant speed" in a way that presupposes it must be finite (in which case 'if such invariant speed is infinite' would be a contradiction in terms) or whether you say that the GT does have an infinite invariant speed since any pair of events with distance/time = infinity (because difference in position is finite while difference in time is 0) in one frame will also have distance/time = infinity in every frame. If you define "invariant speed" in the second way, then of course it's true that "the principle of relativity implies a generalised transform where there is an invariant speed and the LT results if such invariant speed is finite while the GT only results if such invariant speed is infinite", that's just a straightforward description of the mathematical result which you've already said you don't dispute.

 

[Saw]

Obviously from a mathematical point of view they are not just repeating Einstein, since Einstein did not derive what could be deduced from the first postulate without the second.

That’s obvious in bcrowell’s interpretation. It was not so obvious in yours, because when I ask you:

But with you, JesseM, I don’t know where you are. Lately you are saying that I can choose to read the derivation with these “English words”: I can perfectly adhere to a relativity model where t = t’ and where the universe does not include any invariant speed, least of all an infinite one. Is that true?

You still reply:

Sure, if you choose to define "invariant speed" in a way that presupposes it must be finite.

… which is not easy to interpret… After quite a few minutes scratching my head, I’ve thought that you might mean this, but I am not sure at all:

If someone adheres to a relativity model where t = t’, she has two options:

- Take the route of bcrowell and arrive at a generalised transform that only reduces to the GT if a certain invariant speed that the equation contains is infinite.
- Take another route and arrive at the pure and simple GT without any invariant speed?

 

[Saw]

If you wish, disregard the previous post and look at this one, which may be more clarifying:

This just depends on the boring semantic question of whether you define "invariant speed" in a way that presupposes it must be finite (in which case 'if such invariant speed is infinite' would be a contradiction in terms)

Well, personally, it’s true I tend to think that an infinite speed (whether variant or invariant) does not make physical sense. But we are discussing about a level where that sort of assumptions are excluded. So please forget that.

Thus I don’t define "speed" in a way that presupposes it must be finite = a speed can be finite or infinite = an infinite speed is not a contradiction in terms. And if the speed in question is frame-invariant, that doesn’t change the conclusion = an invariant speed can be infinite.

(By the way, as to what “frame-invariant” means… I suppose we have agreed on this: We say that the speed of an object is invariant if the *same* object in the *same* experiment is measured to have the same speed in all frames. For example, light in SR model, where the speed of a single light beam is said to be identical for all observers)

or whether you say that the GT does have an infinite invariant speed since any pair of events with distance/time = infinity (because difference in position is finite while difference in time is 0) in one frame will also have distance/time = infinity in every frame.

Yes, I didn’t take the first meaning, I do take this second one. Having accepted that an infinite speed is a possible speed and, furthermore, that there can be an invariant infinite speed, I do concur that the GT is compatible with an invariant speed, as long as the same is infinite, which (as noted above) would be a theoretically valid possibility.

So far, what I have accepted can be summarized as follows: the t = t’ model can be relativist in a universe where there is an invariant speed that is infinite.

But… but… then the question is… would you accept that the t=t’ model can also be relativist in a universe where there is no finite invariant speed.

You are aware for sure that these are different things. The derivation admits a strong interpretation and a weak one.

With the strong version, which is apparently the prevailing one, we are saying that:

- A relativistic universe MUST contain an invariant speed.
- The t=t’ model can only enter this club (= relativity) if such invariant speed happened to be infinite.

With the weak one, we would be saying:

- A relativistic universe MAY BUT DOES NOT HAVE TO contain an invariant speed.
- The t = t’ model can enter this club (=relativity) (i) if there is an invariant speed and the latter is infinite BUT ALSO (ii) if there is no invariant speed at all and the only invariant thing is time.

Which one is yours, the "strong" or the "weak" version?

 

[JesseM]

That’s obvious in bcrowell’s interpretation. It was not so obvious in yours, because when I ask you:



You still reply:



… which is not easy to interpret… After quite a few minutes scratching my head, I’ve thought that you might mean this, but I am not sure at all:

If someone adheres to a relativity model where t = t’, she has two options:

- Take the route of bcrowell and arrive at a generalised transform that only reduces to the GT if a certain invariant speed that the equation contains is infinite.
- Take another route and arrive at the pure and simple GT without any invariant speed?

I'm just talking about the words used to describe the proof, not the math of the proof itself. You could take the exact same math in that paper, with the exact same equations, but define the words "invariant speed" to mean a finite invariant speed, and thus the K=0, S=infinity case would not include an "invariant speed" in your English-language description of the proof. Look at my last two paragraphs from post #91, hopefully it will be clear that there I was saying you could change the verbal description without changing the math, and I commented that perhaps this was what you might be talking about when you described a "classical relativist" who "claims he is entitled to be relativist + classical + infiniteness-denier". Surely if the classical relativist does allow some notion of an infinite "invariant speed", then mathematically he cannot deny that something with infinite speed in one Galilean frame would have infinite speed in any other--this would be true even if he had never seen the proof which shows that the Galilei transform is a special case of the more general transform derived from the first postulate.

 

[JesseM]

(By the way, as to what “frame-invariant” means… I suppose we have agreed on this: We say that the speed of an object is invariant if the *same* object in the *same* experiment is measured to have the same speed in all frames. For example, light in SR model, where the speed of a single light beam is said to be identical for all observers)

Right, that's what I mean when I say "invariant speed".

Yes, I didn’t take the first meaning, I do take this second one. Having accepted that an infinite speed is a possible speed and, furthermore, that there can be an invariant infinite speed, I do concur that the GT is compatible with an invariant speed, as long as the same is infinite, which (as noted above) would be a theoretically valid possibility.

OK, but just to make sure we're on the same page, when you say "compatible with", do you just mean it's an option, or do you mean that GT does have an invariant speed, period? If we define "invariant speed" in this way that allows it to be infinite, it seems to me that there's no way to deny that the GT has an invariant speed, hopefully you'd agree.

So far, what I have accepted can be summarized as follows: the t = t’ model can be relativist in a universe where there is an invariant speed that is infinite.

But… but… then the question is… would you accept that the t=t’ model can also be relativist in a universe where there is no finite invariant speed.

no finite speed? Sure, why wouldn't I accept that? A Galilei-symmetric model would have t=t' and no finite invariant speed.

You are aware for sure that these are different things. The derivation admits a strong interpretation and a weak one.

With the strong version, which is apparently the prevailing one, we are saying that:

- A relativistic universe MUST contain an invariant speed.
- The t=t’ model can only enter this club (= relativity) if such invariant speed happened to be infinite.

With the weak one, we would be saying:

- A relativistic universe MAY BUT DOES NOT HAVE TO contain an invariant speed.
- The t = t’ model can enter this club (=relativity) (i) if there is an invariant speed and the latter is infinite BUT ALSO (ii) if there is no invariant speed at all and the only invariant thing is time.

Are we still defining "invariant speed" in a way that allows it to be infinite? If so I would disagree with the weak version, since the proof shows that there is no possible coordinate transformation with t=t' that satisfies the first postulate except for the Galilei transformation, and it's undeniable that if you define "invariant speed" in this way, the Galilei transformation definitely has an invariant speed of infinity.

 

[Saw]

OK, but just to make sure we're on the same page, when you say "compatible with", do you just mean it's an option, or do you mean that GT does have an invariant speed, period? If we define "invariant speed" in this way that allows it to be infinite, it seems to me that there's no way to deny that the GT has an invariant speed, hopefully you'd agree.

Yes, I agree that the GT does have an invariant infinite speed (a).

But, as you comment, it also has (b), the denial of a

finite invariant speed.

So we have a man that has two legs, (a) and (b). He wants to enter a club (relativity). The security service (the generalised transform) checks the man and says: “ok, you can enter the club but only with (a); you must leave (b) out, because here we only admit transforms containing an invariant speed; your first leg is that sort of transform, your second leg isn't”. So we can say that the GT embedded in the generalised transform is a “mutilated” one: it has (a) but not (b). I think you adhere to that position.

I myself would not agree with that. I would say that the relativity principle allows that man to go into the club with both legs, (a) and (b).

 

[JesseM]

So we have a man that has two legs, (a) and (b). He wants to enter a club (relativity). The security service (the generalised transform) checks the man and says: “ok, you can enter the club but only with (a); you must leave (b) out, because here we only admit transforms containing an invariant speed; your first leg is that sort of transform, your second leg isn't”. So we can say that the GT embedded in the generalised transform is a “mutilated” one: it has (a) but not (b). I think you adhere to that position.

I myself would not agree with that. I would say that the relativity principle allows that man to go into the club with both legs, (a) and (b).

I really don't understand at all. Who said you must "leave (b) out", where (b) stands for no finite invariant speed, to qualify as a coordinate transformation that satisfies the principle of relativity (i.e. the first postulate)? Clearly the GT does have no finite invariant speed and it does satisfy the first postulate.

 

[Saw]

Clearly the GT does have no finite invariant speed and it does satisfy the first postulate.

Yes! The stand-alone GT! But what about the GT embedded in the generalised transform? Once the GT is dressed in this costume, it only shows up if K = 0 and K is 0, if and only if there is a REAL-LIFE and MEASURABLE infinite invariant speed? As I said, I have no problem to admit the THEORETICAL possibility of an infinite speed. Something theoretical is something that may exist or not. If it does exist, it will be invariant for the both the LT and the GT. However, the way that the generalised transform is mathematically worded, seems to require that such possibility becomes actual, for the GT to peep out of its hole, doesn't it?

The LT embedded in the generalised form is also subject to a similar, although maybe less demanding requrement. If we want to use it in real life, we need to know the value of s or at least to associate it with some real physical object. For example, light. Once s becomes c, the speed of light, the LT becomes usable, no matter if c is finite or infinite.

What about the GT embedded in the generalised form? If we wanted to use it in real life, it seems we would also need to identify the invariant speed s with some real phenomenon. If unfortunately we do not find one that is infinite, what happens?

I am afraid that for our friends holding the "strong interpretation", that is the end of the story for the GT. It's wrong and period.

If instead you agree with me that the GT is still valid, that the GTist can still claim to have a valid point, unless some other axiom (beyond relativity principle) is taken, that is what I am calling the weak interpretation.

 

[JesseM]

Yes! The stand-alone GT! But what about the GT embedded in the generalised transform? Once the GT is dressed in this costume, it only shows up if K = 0 and K is 0, if and only if there is a REAL-LIFE and MEASURABLE infinite invariant speed?

Don't understand what "dressed in this costume" means mathematically. The GT has the mathematical properties that it has, they don't somehow change just because you consider that the GT can be treated as a special case of a more general transformation. I also don't understand what it would mean to say the invariant speed should be "real-life and measurable", isn't the invariant infinite speed just a mathematical property of the Galilei transformation?

Consider an analogy: suppose we lived in a universe with Lorentz-symmetric fundamental universe, but unlike in the real universe, in this all particles have nonzero rest mass and it is impossible for any real physical effect to travel at c. Would you say in this case that the invariant speed c is not "real-life and measurable"? But it is still perfectly measurable in the sense that if we write down the equations of the laws of physics in one inertial frame, we can see that they will remain unchanged under a particular unique coordinate transformation which involves the constant c (whose value can be determined from the equations), one which has the velocity addition law w = (u + v)/(1 + uv/c^2). We will also still observe that a clock moving at velocity v in a given frame is slowed down by a factor of sqrt(1 - v^2/c^2), which would be another empirical way to determine the value of c.

I suppose it is again just a semantic matter whether you choose to define "real-life and measurable invariant speed" in such a way that you need to be able to find actual physical entities that travel at that speed, or whether you define it in such a way that if one can empirically determine the coordinate transformation that the laws of physics are invariant under, and mathematically speaking this coordinate transformation includes an invariant speed, then that counts as a "real-life and measurable invariant speed". I would find the second definition more natural (and more reflective of how physicists would likely talk about these issues), but if you prefer the first definition, then under that definition we can say that the Galilei transformation does not include a "real-life and measurable invariant speed" if no actual physical object or effect (like instantaneous action-at-a-distance) travels at infinite speed. But again playing around with different definitions like this seems like a waste of time to me, since whatever words you choose to use, at the level of mathematics or of physical predictions you will not be saying anything different than bcrowell is saying when he argues that the first postulate implies an invariant speed, you're just quibbling about how he chooses to define the words "invariant speed".

As I said, I have no problem to admit the THEORETICAL possibility of an infinite speed. Something theoretical is something that may exist or not. If it does exist, it will be invariant for the both the LT and the GT. However, the way that the generalised transform is mathematically worded, seems to require that such possibility becomes actual, for the GT to peep out of its hole, doesn't it?

Are you saying that the mathematics of the proof (independently of what words you use to describe it) somehow implies that if the laws of physics are Galilei-symmetric there must be actual physical objects or effects traveling at infinite speed? If so that makes no sense, the proof only implies that the laws of physics are Galilei-symmetric in the K=0 case, but it's quite possible to have Galilei-symmetric laws in a universe where nothing physical actually travels infinitely fast.

What about the GT embedded in the generalised form? If we wanted to use it in real life, it seems we would also need to identify the invariant speed with some real phenomenon. If unfortunately we do not find one whose speed is infinite, what happens?

Again I think you're confusing yourself with vague language like "the GT embedded in the generalised form" or "dressed in this costume". The same physical equations don't somehow make different physical predictions just because they were derived in a different way! The physical meaning of "the laws of physics are invariant under the Galilei transformation" would always be the same regardless of what set of postulates you used to derive this, it makes no sense to imagine that it will lead to one set of predictions if you derive it by starting with the principle of relativity and then postulating that K=0 or if you derive it in some totally different manner.

I am afraid that for our friends holding the "strong interpretation", that is the end of the story for the GT. It's wrong and period.

Well I would say this imaginary character is deeply confused. Is he really claiming that that we can empirically determine the equations of the fundamental laws of physics in one inertial frame, and see mathematically that these equations will be unchanged under the GT, and yet say that Galilei-invariance (or 'the principle of relativity' which is what you said before the 'strong interpretation' was about) is "wrong" because he can't measure any physical object or effect traveling at infinite speed?

If instead you agree with me that the GT is still valid, that the GTist can still claim to have a valid point, unless some other axiom (beyond relativity principle) is taken, that is what I am calling the weak interpretation.

"unless some other axiom (beyond relativity principle) is taken", what? What is the GTist's "valid point" about what happens if we don't take any other axiom? Does he disagree with bcrowell about any substantive (mathematical or physical) matter, or just about bcrowell's choice of words to describe the results of the paper's proof?

 

[Saw]

You keep accusing me of “playing semantic games”. Again, as if semantics were a sort of idle good-for-nothing and “boring” activity, instead of what it is, the art of understanding what the heck it means what you are saying…! I suppose that what you actually mean is what you have just said, i.e. that I am “quibbling” about words. I had to look that up in my dictionary. It happens to mean (you can see here the utility of semantics) “to evade the truth or importance of an issue by raising trivial distinctions and objections”. Ah, that is exactly how I feel treated!

I am not going to invent more expressions, more metaphors to try to illustrate my point. To me it is clear that the derivation we talk about:

- Does not lead to a super-transformation that contains both the LT and the GT. It gives out the LT with some make-up, in the form of a constant (“aunque la mona se vista de seda, mona se queda”, see translation above) plus a mutilated and unrecognizable GT.

- And that is so because the derivation has made an extra implicit assumption from the outset: it has chosen that t≠t’ and that is why it simply gives out the LT (where the existence of an invariant speed is a must) and it only accommodates the GT in a very exceptional case (where such invariant speed is infinite).

So we are not in face of any valuable contribution. In fact, the derivation is based on two postulates. By pretending, however, that it uses only one, it lies and only creates confusion.

But I am not going to convince you and you are not going to convince me. Shall we leave it like that and discuss on some technical associated issues?

For example, I had said:

For Einstein, with the 1st postulate alone, you do not have any transformation. For a transformation to exist, you need to make a choice about an “invariant something”. If your choice of second postulate is t=t’ but c≠c’, you get the GT. If your choice of second postulate is t≠t but c = c’, you get the LT.

To which you answered:

Einstein simply didn't explore what would follow from the 1st postulate alone, he did not make any positive claim that if you do start with just that postulate, then "you do not have any transformation" (obviously from a mathematical perspective such a positive claim would be nonsense).

Well, of course Einstein didn’t explore what would follow from the first postulate alone. That is precisely what I am arguing: He knew that with the aid of the 1st postulate alone, you derive nothing, unless you combine it with a 2nd postulate (eg, there is an invariant time or there is an invariant speed). That doesn’t mean that he positively and expressly claimed so. But he implicitly counted on it, that is why he used two postulates.

But leaving historical Einstein aside, what was important in my comment is that for any transformation to be constructed you need an invariant something. I don’t think that is mathematical “nonsense”, as you said. I think that is how a transformation operates.

Take the example I gave many posts ago about two observers, two coordinate systems at rest with each other, but whose origins are in different positions. S measures that the origin of S’ is 30 m away from the origin of S. Now S’ declares that a third point S’’ is situated at x’ = 20 mjxfs. What is x, the distance from S to S’’ in S frame? S argues that S’ should clarify what “mjxfs” means, whether it is units of length measured with a 1-metre rod or what, so as to be able to determine whether the 30 m that he has measured as distance from S to S’ can be combined or not with the 20 mjxfs. Unfortunately, S’ refuses to answer: “we don’t need to share any invariant concept, your m does not have to be equal to my mjxfs, that is mathematical nonsense”. How do you build then the transformation?

Edit: Well, I reviewed what I said here and admit that it's to a good extent imprecise. I'll try to give some more detail. I had said before that I understand that the relativity principle has two components:

1) given two experiments E and E', result of experiment E with ball B wrt frame S = result of identical experiment E' with object B' wrt frame S' and
2) in a single experiment E, the coordinates of ball B in S can be related through a transformation with the coordinates of that same B in S'.

It's my impression that 1) is the condition for 2) to exist, since it permits to establish common measurement units which are used in the transformation. Frame S decomposes a problem in two component vectors and it turns out that for a certain component the measurements of S' are valid for S's purposes, due precisely to 1). What may vary, however, in different models of relativity is the vector that is considered as usable inter-frames. In the t=t' model, it is time. In the c=c' model, it is a speed.

Of course, you will attack these comments mercilessly, but I would appreciate that.

 

[JesseM]

An earlier post I had gotten distracted from responding to:

I will come back now to the main point.

It should be highlighted that who is claiming something new and against established thought is you and the authors you cite, not me. Einstein did not follow your line of reasoning but the following:

- The principle of relativity was well established for experiments with mechanical objects, on the basis of the idea that they share the state of motion of their source.
- Then it was found out (experimentally!) that light is different, in the sense that its speed is independent of the motion of the source, like it happens with sound.
- In principle, the latter represented a threat for the principle of relativity: experiments with light would not give the same results in different frames.
- Einstein’s solution to reconcile the principle of relativity with light experimentally’s observed behaviour was to postulate that in a given experiment not only the frame of the source but any other observing frame would measure a frame-invariant speed for light, which sounds odd but is possible if you abandon the postulate that time is absolute and adhere instead to the postulates of relativity of simultaneity (RS), time dilation (TD) and length contraction (LC), which are contained in the Lorentz Transformation (LT).
- Experiments have proved that Einstein’s postulates were a good intuition.

To sum up, focusing on the subject under discussion here:

- Einstein’s frame-invariant speed was the speed of a physical phenomenon, the speed of light.
- Einstein did not claim that such invariant speed stemmed from the principle of relativity alone. He introduced that as an additional postulate and at the same time admitting that the very introduction of such new postulate required a change of mental framework (from absolute to relative) with regard to time and length: both things go together = you cannot mathematically derive RS, TD or LC without the frame-invariant c and, vice versa, you do not get RS, TD or LC without c.

Now we have apparently some authors like Rindler or Pal here taking one bold step further. Basically, their “discovery” is that Einstein could have been much more aggressive and state that:

- The principle of relativity alone implies (by rigorous mathematical demonstration) a frame-invariant speed, without any further assumption (i.e. without assuming the relativity of time).
- There is no need to identify this invariant speed with the speed of light or anything physical whatsoever. It’s a logical need that must exist, as long as : so even if tomorrow experiments get playful and start proving that the speed of light is not c in every frame, there will .

Do you really think this is right…? I find this hard to believe… I have already given all the arguments against that idea.

To me it seems like you are guilty of a physics analogue of the intentional fallacy in literary criticism--instead of just focusing on the text of the proof itself, looking at whether it's valid mathematically and so forth, you are coming up with some extra-mathematical ideas about the "intentions" behind the proof, like that the proof shows Einstein took the "wrong approach" somehow in deriving the LT (even though his approach was perfectly valid mathematically), that he "could have been much more aggressive" and that it was a mistake for him to make the assumptions he did. But the proof is just a proof, in assessing it's validity I don't care about the author's secret motivations in coming up with it. Suppose two authors independently came up with the proof in that paper, but using our handy cerebroscope we were able to peek inside their brains and see their hidden motivations--

Author #1: "Jeez, everyone from Galileo through Einstein sure was STUPID! Galileo came up with the idea that the laws of physics didn't look any different from the perspective of different observers in uniform motion relative to one another way back in 1632, but no one ever thought of looking at what could be derived mathematically from this assumption! If they had, they would have seen how OBVIOUS it is that the laws of physics must be invariant under either the Galilei transformation or one like the Lorentz transformation...they could have come up with the possibility of the LT much earlier, and without Einstein's CLUMSY proof which started out assuming an invariant speed of c, instead of my more elegant approach of showing that the first postulate alone gives a general transformation with an invariant speed, and that if this invariant speed is set to infinity you get the GT, while if it's set to c you get the LT. Physics could have progressed a lot faster if someone had noticed this before...too bad they were all such IDIOTS!"

Author #2: "Oh, I do love these whimsical explorations of what conclusions about physics can be derived from various possible starting axioms! Today I think I'll look at what can be derived from Einstein's first postulate without the second one, then tomorrow I'll look at what can be derived from the second postulate without the first one, then the next day I'll take a look at what happens if we drop one of the http://www.stanford.edu/~mukul/tutorials/Quantum_Mechanics.pdf ...what fun! But today, the first postulate...I wonder what I'll find? (several hours later) Isn't that interesting, a generalized transformation that reduces to the LT if you substitute K=1/c^2, and reduces to the GT if you substitute K=0! Still, it's not surprising that no one came up with the idea that the laws of physics might be invariant under the LT before Einstein--the notion of an 'invariant speed' is a very strange and counterintuitive one, it took a genius like Einstein to see that it could make actual physical sense! I suppose technically one could say that the GT features an 'invariant speed' of infinity, but this is a very unnatural way of thinking about it, one that only makes sense to us in retrospect since Einstein has already done the hard work of making us understand what it means to have a finite invariant speed. Oh well, the proof was an interesting exercise anyway even if it has no great significance!"

Would you somehow then say that author #1's proof is invalid while author #2's proof is valid, despite the fact that they are mathematically identical?

But it seems you are not fond of the metaphoric language I used. I’ll try now to put it very simply in the hope of being more didactic: what a coordinate transformation requires is not an invariant speed, but an invariant “something”:

- In the variant time model (t≠t’), that ”something” is speed, not time.
- In the invariant time model (t=t’), that “something” is time, not speed.

It's not obvious a priori that a coordinate transformation requires an invariant "something" at all, much less that it requires one of these two possible invariants. You only know that this is true because you have seen the proof! What's more, your statement is only true if we define the English phrase "invariant speed" in a way that requires it be finite, since if we allow the phrase to include the possibility of an infinite invariant speed it's clear from a mathematical perspective that the GT does feature an invariant speed.

What you call the “generalised transformation” (as if it were a super-expression of both GT and LT)

What does "super-expression of both GT and LT" mean? It must mean something other than "an equation which reduces to the GT if you plug in one value for an arbitrary constant, and reduces to the LT if you plug in a different value for that constant", since mathematically it is undeniable that the "generalized transformation" does have that property!

is simply the Lorentz Transformation with some cosmetics: instead of 1/c^2 you have written a constant K, as if it that were the way to accommodate the GT.

Again I don't know how to interpret this, you certainly aren't saying anything that I can translate into mathematically meaningful statements. What does "the Lorentz Transformation with some cosmetics" mean? Mathematically it is an equation that can reduce to either the LT or the GT depending on the value of some constant, so you don't seem to be making a mathematical judgment here, maybe it's more like a kind of aesthetic judgment that it "looks more like" the LT. Or maybe it's an argument about secret motivations again, that the authors of the proof were secretly "favoring" the LT even though they were pretending to be even-handed by giving an equation that could "technically" reduce to either the LT or the GT (the last phrase, 'as if that were the way to accommodate the GT', may also suggest this interpretation). But really I have no clear idea what you meant, since you aren't doing a good job of explaining it in clear terms that don't involve metaphors. You aren't even giving any suggestion as to the nature of the http://findarticles.com/p/articles/mi_m2346/is_n421_v106/ai_19793598/ in which you are attacking the proof--since you're clearly not making mathematical arguments against the proof, are you attacking the proof on the basis of the perceived motivations of the authors, or from an "aesthetic" perspective, or are you attacking the choice of words people like bcrowell have used to describe the proof, or what? Whatever your basis for attacking it, it seems like it's coming from a mode of thinking that's alien to the mode of thinking physicists normally use to discuss results in physics, which is perhaps why no one here can understand what you are trying to say.

Now I will retake metaphoric language, because I like it.

"Retake" it? Your comments above seem 100% metaphoric to me--is "the Lorentz Transformation with some cosmetics" not a metaphor? It certainly has no mathematical meaning.

All this reminds me of the Spanish saying, “aunque la mona se vista de seda, mona se queda” (= even if you dress the monkey in silk clothes, it is still a monkey and not something else). That we are in front of the LT, purely and simply, is evidenced by the fact that you take off the silk clothes, you lift the veil of the constant K and what do you find?

What does "lift the veil of constant K" mean? Do you deny that mathematically you are free to plug in either K=1/c^2 or K=0? If so, are you making some suggestion about motivations, that the authors were trying to imply something about how obvious it should have been that it was more "natural" to have a transformation like the LT if you started from the first postulate? Again I can't even understand what domain of discourse your metaphors are supposed to be pointing towards...

That's not a "choice" at all if you are starting from the initial postulate I described, namely "the laws of physics are invariant under a coordinate transformation that gives a set of coordinate systems moving at constant coordinate velocity relative to one another". It's just a mathematical matter to show that this postulate is consistent with coordinate transformations where t is different than t', so it would be mathematically incorrect to rule such a transformation out without adding additional postulates. Of course there is no a priori reason to say that this postulate is physically plausible, or that it's physically plausible that the laws of physics could be invariant under a transformation where t is not equal to t', but I'm simply not talking about physical plausibility at all here, just about what conclusions follow mathematically from assuming that postulate (which is just a way of restating the first postulate of SR).

Here you are playing with words, isolating them from their context and meaning.

No, I'm just talking about mathematical results. You seem to have a problem differentiating math from words, and judging whether a particular mathematical result is true or false mathematically without getting hung up on the English words we might use as shorthand to describe the result. If the "domain of discourse" is math, then semantics is only important insofar as the wrong choice of words might lead to confusion about what the other person meant in a mathematical sense--as long as we do understand what the other person means mathematically by any particular phrase, it's totally irrelevant whether the words in the phrase are used in the same way they would be in everyday English. Do you disagree?

Certainly, if we are discussing what relativity is, t≠t’ is not a option that you can leave out. Of course, who doubts that? What I am saying is that your so called generalised transform makes the choice of t≠t’ and leaves the option t = t’ out. Well, it can still accommodate t = t’, but only after introducing two requirements that are strange and unnecessary in this model: invariant speed + its infinity.

No, the proof only involves math, not word-definitions. The english phrase "invariant speed" appears nowhere in the proof. You are free to define "invariant speed" in a way that rules out all but finite speeds, and then say that the proof can accommodate t=t' without an "invariant speed". Mathematically the proof accommodates t=t' in the case of K=0, the language you use as English shorthand for the K=0 case is totally irrelevant as long as we both understand the mathematical meaning of any such shorthand. For example, if we both agree in advance to denote the K=0 case with the English shorthand "cows develop a sudden hunger for human flesh", then it would be perfectly correct to say that "the generalized transform can only accommodate the t=t' case if we also assume that cows develop a sudden hunger for human flesh". We would both understand what this means mathematically, and if the domain of discourse is math, then semantics is only important insofar as it's important that we all have the same understanding of what any English phrases used are supposed to denote mathematically.

Are you serious about that? Isn’t it simpler to frankly admit that t = t’ is *also* a valid relativity model and that the only invariant thing it requires to exist is an invariant time, as its definition proclaims?

I'm happy to admit anything you like about English phrases like "invariant speed" and "invariant time" as long as you explain in advance what they are supposed to denote on a purely mathematical level. But if you think that it matters whether we choose to define English phrases in one way as opposed to another way, then apparently you have left the "domain of discourse" of mathematics, and are using some non-mathematical criteria to judge which definitions are "better", in which case you need to actually explain the details of what these criteria are, and why they are relevant to a discussion about physics.