Acceleration and the twin paradox

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CKH

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I think that it was mentioned earlier in this thread that the traveler has a free choice of inertial frames.
I suppose, but not if the traveler wants his position to remain fixed in his frame and his clock time to be continuous in his frame. We are not picking some random choice of the new inertial frame at turn around, we are choosing the unique inertial frame that is consistent with the traveler's local frame (which is continuous).

Real on-board clocks will not "jump", and consequently the home clock will also not jump to a later time if the traveler passively uses an extension of his on-board clocks to determine time at home.
No, the traveler's clock never jumps in the traveler's frame, but the home clock does jump in the traveler's frame. Nothing is magical here, the traveler's frame abruptly changes its plane of simultaneity from the outbound IRF to the inbound IRF. So we are forced to admit that the simultaneous reading on the home clock "jumps" in the travelers frame.

Instead, the traveler will first have to synchronize the on-board clocks to the new rest frame. Thus the simultaneous reading on the home clock is adjusted by the traveler, according to his/her choice of instruments and "maps".
No, I intend to do it the other way around, the on-board clock defines the current time in the new rest frame. There is no free choice for time in the new rest frame because we have constrained it to be consistent with the traveler's non-inertial frame. The requirements for the traveler's frame are:

1) Orientation in space is fixed (the traveler is not rotating).
2) The traveler's position in the frame is fixed. (He is at rest in his own frame).
3) The traveler's clock is continuous (it can't jump in it's own frame, his clock is a good clock).

Finally just to be absolutely clear the "jump" in the home clock is not physical; it is rather a "jump" in the perspective in the traveler's frame. I can make the distance stars move faster than the speed of light in my frame just by rotating my body. It is not physical however.
 
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The thoughts of this whole picture becomes quite impossible the more you think about it. For instance take one single leg of this trip and look around. Bob and Alice agree what the universe looks like around them. Alice leaves and is time dilated on her trip and stops. Doesn't the universe around her look younger than what Bob experiences at this point in time? When she returns Bob looks older and so should his entire universe. How can her view of the universe, which you say doesn't deviate from her "rest frame" observations that for her Bob's universe evolves slower when the omnipotent chart of the universe around them can't possibly deviate at any point in time!
 
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Alice leaves and is time dilated on her trip and stops. Doesn't the universe around her look younger than what Bob experiences at this point in time?
No. They are both seeing the same universe, with the same age, because they are both at the same event in spacetime. They just took different paths to get to that event, and those paths have different lengths.
 
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Then they could never reunite in reality because those realities can never coexist. If she traveled that speed in orbit around Earth and Bob sat still where she could fly by him periodically very close to Bob and see him age and he see her they would both appear younger to each other? When she stops nothing could have changed timewise.
 
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[..] the traveler's clock never jumps in the traveler's frame, but the home clock does jump in the traveler's frame. Nothing is magical here, the traveler's frame abruptly changes its plane of simultaneity from the outbound IRF to the inbound IRF. So we are forced to admit that the simultaneous reading on the home clock "jumps" in the travelers frame.
[.. + rearranging:]
Finally just to be absolutely clear the "jump" in the home clock is not physical; it is rather a "jump" in the perspective in the traveler's frame. I can make the distance stars move faster than the speed of light in my frame just by rotating my body. It is not physical however.
Such a frame of varying velocity and acceleration can be constructed in SR but does not correspond to a valid reference system for physics in SR; as a result it leads to unphysical statements. But happily you finally clarified that you did not mean it the way you made it sound - although, I'm still not totally assured:
[..+ more rearranging, and emphasis mine:]

[..] we are choosing the unique inertial frame that is consistent with the traveler's local frame (which is continuous).
[..] the on-board clock defines the current time in the new rest frame. There is no free choice for time in the new rest frame because we have constrained it to be consistent with the traveler's non-inertial frame. [..]
3) The traveler's clock is continuous (it can't jump in it's own frame, his clock is a good clock).
SR uses non-local (universal) inertial frames in which nothing can "jump". Thus I had the impression that perhaps you try to establish something that is neither SR nor GR - and possibly with an inherent mistake. But you missed the pertinent part of how I tried to check if this is just a matter of phrasing (somewhat a matter of taste), or if there is more to it. I'm still not sure!
Once more, your phrasing sounded - and it still does - as if you think that you can set up a "local" physical frame such as a rocket (note: in SR a "frame" is extended and equipped with clocks) that can be automatically consistent with both inertial frames in which the traveller is in rest most of the time. I stressed that for such a purpose you can not use a 3D "frame" that is equipped with multiple clocks at different positions. Of course you can use a single clock (a point!), but that does not define a "frame". As a reminder, I stated:
"the home clock will also not jump to a later time if the traveler passively uses an extension of his on-board clocks to determine time at home. Instead, the traveler will first have to synchronize the on-board clocks to the new rest frame."
 
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Then they could never reunite in reality because those realities can never coexist.
Sure they can. Since it is apparent that relativity is a new subject to you, you should avoid making such categorical statements based on your current (highly incomplete) understanding. I assure you, everything I am saying comes from a perfectly self-consistent model of a single spacetime (i.e., a single "reality"), and none of it is in any way difficult or contentious from the standpoint of relativity. This is information you can find in any good relativity textbook.

If she traveled that speed in orbit around Earth and Bob sat still where she could fly by him periodically very close to Bob and see him age and he see her they would both appear younger to each other?
No. She would be younger and Bob would be older. Note that this scenario is not set in flat spacetime, so it can't really be analyzed using SR. However, there is a similar scenario which is set in flat spacetime and can be analyzed using SR: for example, suppose Bob sits at the edge of a big merry-go-round, and Alice goes around with it. Then Alice will be younger than Bob when they meet again. (The key difference between the two scenarios is that, in the latter one--the merry-go-ground--Alice is accelerated and Bob is inertial; whereas, in the Earth-orbit scenario, it's the other way around--Bob is accelerated and Alice is inertial.)
 

CKH

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And the answer is "mu": the question is not well-posed, because it assumes that there is a unique, physically given answer to the question. There isn't. Simultaneity is not a physically meaningful thing; it's just a convention. The way you are defining simultaneity is one way to do it, but it's not the only way, it has issues (like not assigning a unique time to all events), and it's no more meaningful, physically, than other ways of doing it.
It is well-posed if you consider all the details of how simultaneity is defined for the traveler. It is definitely not the same for all observers (but neither is the length of an object which is "physical"). I believe simultaneity is well-defined for an inertial frame. MCIFs along the path of a non-inertial object can be well-defined also by imposing time continuity and fixed spatial coordinates (to not do so yields meaningless results). I'll call this construction the "rest frame of that object". Given all that, you can consistently define simultaneity for non-inertial travelers in SR.

Maybe I forgot to note this; it's obvious but important. If you want to ask "in my rest frame what is the simultaneous reading on a clock at rest in a different frame", you must first synchronize clocks in those frames at some common event (such as departure in the example).

Concerning clock synchronizing conventions in SR. Bear in mind that if you pick one you must use it consistently. By convention I'm assuming the definition in Einstein's 1906 paper. Will the simultaneous readings on the home clock in the traveler's frame be different with a different choice of clock synchronization convention? If you don't already know the answer we can pick a different convention and try it out. I suspect it doesn't matter, because I don't see where that convention matters in the scenario of the twins.

Any clock synchronization scheme has to obey some rules to be consistent, so they are not arbitrary.

Einstein said "simultaneity is relative [to the observer's frame]". Measurements in general are relative but have meaning so long as you declare how the measurement is made.

Can you elaborate on "Simultaneity is not a physically meaningful thing; it's just a convention"? In particular, what criteria apply to "physically meaningful things".

Simultaneity is simply fixed time in an inertial coordinate system, a hyperplane. You have to pick such a system to talk about it. Would you go further and say that coordinates in general are not physically meaningful? They are a choice, but we have decided how we make that choice in an given situation, and thus we can make well-defined measurements of "physical things" in a coordinate system.
 
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No it's not unique, but the relative time results are the same whatever free choices you make for frames, given that you synchronize the traveler's clock with the home clock at the departure event.
I am not sure what you mean by "relative time results". If you mean the final numbers at the end, then yes, I agree. Those are invariant quantities so it doesn't matter what chart you use to calculate them (inertial or not), you will always get the same answer.

The MCIFs of the traveler along the path then have no free choices.
Yes, but the MCIF's are not the same as the traveler's rest frame.

Then let's not call it a chart if it has to satisfy such criteria.
The criteria are not optional if you want to use it to do physics. Chapter 2 explains why.

Just using the definitions provided in SR for inertial frames and simultaneity, I see no way of avoiding the fact the traveler's worldline has a tangent time axis at each point and the time on this axis at each point is the traveler's clock time.
If you use the SR definition for simultaneity on a non-inertial worldline then you get the radar coordinates of Dolby and Gull, not the kind of coordinates you have been describing so far.


The author complains that there are events to which multiple values of time can be assigned by the traveler and that the home clock can jump backward and what not. Isn't this exactly what the definition of simultaneity and proper time in SR require?
No. Each inertial chart assigns one and only one time to every event.

This "method" has little or nothing to do with simultaneous readings by the traveler of the time on the home clock as defined by SR.
It has everything to do with it. That is the way that Einstein defines simultaneity in his 1905 paper, but he used only inertial observers. The only thing that they did was to simply say "let's apply the standard SR method to a non-inertial observer".

On the other hand, a global coordinate system for the travelers frame can be defined using the MCIFs on the world line. But in that case you have to accept that events do not a have unique coordinates in this frame. That's not the fault of the method, it simply a consequence of SR.
It is not even a coordinate system (aka chart) let alone a global one, the problem is not a consequence of SR, it is the fault of the method, and there exist other methods without that problem.

I may not have much time for replying the rest of this week, so I would recommend that you really focus on reading and learning the material suggested.
 
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They can both be outside the vicinity of gravity, in deep interstellar space between two stars, I'll diagram it:
revolving dilation.png

Well Bob sits still (blue circle) and Alice plays the hadron which definitely has an experimentally proven time dilation. They both see the same stars, and SR says Alice's stars should evolve slower, as her existence is slowed. If Bob really is older he could say " a solar flare happened at my "t=?" time and Alice could say, at some point in her slower time line, "Oh, yeah! Now I see it!".
 

CKH

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Proper time for a particular observer is only meaningful for events on that observer's worldline. If the observer wants to assign a "time" to events not on his worldline, he has to pick a simultaneity convention, and there is no one unique way of doing that. Even the obvious way for an inertial observer, namely just using the coordinate time in a coordinate chart in which he is always at rest, is not the only possible way for that observer. Simultaneity is simply not a physically meaningful thing; it's a convention.
OK. Can you give me an example of another way (in SR) to define what is simultaneous with an observer's proper time? Hopefully I'll finally get what you mean by "convention" here.
 

pervect

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This may or may not be an issue, depending on what you're trying to do. See below.

This depends on the spacetime. In flat spacetime, it is always possible to construct a valid chart (i.e., one meeting all of DaleSpam's requirements) that covers the entire spacetime, with any chosen timelike curve as its "time axis". The Dolby & Gull paper shows how to do this.
Are you sure? I was just quoting this paper in another thread (or earlier in this one, I forget where exactly), and my recollection is that Dolby & Gull only claimed that their coordinate system would cover the region of space-time that the observer on the timelike curve could send and receive signals from. Since the method is radar based, I don't see how the method could assign coordinates to regions of space-time that the observer could not both send and receive signals from. It's good - but not THAT good.

Unfortunately, when we consider the specific example of a uniformly accelerated observer, said observer can't send and receive signals from all of space-time :(.
 
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Can you give me an example of another way (in SR) to define what is simultaneous with an observer's proper time?
The Dolby & Gull paper gives one, as DaleSpam has repeatedly pointed out to you.

Another one would be to just pick a particular inertial frame (say Bob's in the twin paradox scenario under discussion), and use that frame's simultaneity convention for everybody, regardless of their state of motion. For the scenario we've been discussing, this won't seem very natural, because only linear motion is involved. But, for example, if we have a bunch of observers at various points on a rotating disk, it can be very useful for everybody to adopt the simultaneity convention of the observer at the center of the disk (who is the only one who is at rest in an inertial frame), even though that convention does not match up with your definition for any other observer on the disk.
 
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SR says Alice's stars should evolve slower, as her existence is slowed.
SR says no such thing. Alice's state of motion does not affect the evolution of the stars, except as that evolution is perceived by Alice. And the effect of Alice's motion will be that she perceives the stars to be evolving faster, compared to the rate of her clock, than Bob sees them evolving compared to the rate of his clock.

For example, if Bob sees each star emit a flare once per hour by his clock, Alice will see the same flares happening less than an hour apart by her clock, i.e., she will see them happening at a faster rate.

If Bob really is older he could say " a solar flare happened at my "t=?" time and Alice could say, at some point in her slower time line, "Oh, yeah! Now I see it!".
No, that's not what would happen. Suppose Alice and Bob use some particular flare to synchronize their clocks; they pick a flare whose image (the light signal emitted from the flare) just happens to reach both of them at the same instant they are passing each other. They will both see this flare at the same instant, and at that instant, they both set their clocks to time ##t = 0##.

Now suppose that the interval between the flares is just right so that the next flare's image reaches Alice and Bob the next time they happen to be passing each other (i.e., Alice has completed half of an orbit while Bob sat still). Then they will see the next flare, once again, at the same instant; but at that instant, Bob's clock will read a later value than Alice's clock does. (For example, if Bob's clock reads exactly 1 hour, Alice's will read less than 1 hour.) That's how they know that Alice's clock is the one that is running slower: they have the same common pair of events (seeing the two flares), at which they are both co-located, to compare to, and Alice's clock shows less elapsed time between those two events.
 
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my recollection is that Dolby & Gull only claimed that their coordinate system would cover the region of space-time that the observer on the timelike curve could send and receive signals from. Since the method is radar based, I don't see how the method could assign coordinates to regions of space-time that the observer could not both send and receive signals from.
Hm, you're right, their basic method is radar-based. I thought I remembered that there was a way of extending it into the rest of spacetime, but re-reading the paper there doesn't seem to be.

One can still construct a coordinate chart with the properties I described, without using the Dolby & Gull method, if one is willing to drop orthogonality. Such a chart would still be one-to-one and diffeomorphic to other charts; it just would not have orthogonal axes at all events. (Also, of course, the physical interpretation of such a chart in a region which could not exchange light signals with one's chosen worldline would be problematic.)
 
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They can both be outside the vicinity of gravity, in deep interstellar space between two stars, I'll diagram it:
View attachment 75385
Well Bob sits still (blue circle) and Alice plays the hadron which definitely has an experimentally proven time dilation. They both see the same stars, and SR says Alice's stars should evolve slower, as her existence is slowed. If Bob really is older he could say " a solar flare happened at my "t=?" time and Alice could say, at some point in her slower time line, "Oh, yeah! Now I see it!".
Not necessarily. Alice might be closer the the flare and see it first.

Think about another scenario. Alice is moving very fast relative to Bob and as Alice passes Bob, the light from a solar flare hits their common area and both see it at the same time. Why do you think Alice's speed alone dictates when she sees things?

"Alice's existance is slowed" is meaningless. It only has meaning relative to Bob, not to Alice.
 
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PAllen

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OK. Can you give me an example of another way (in SR) to define what is simultaneous with an observer's proper time? Hopefully I'll finally get what you mean by "convention" here.
Dalespam gave you a link to a very well known alternative back in post #141 (http://arxiv.org/abs/gr-qc/0104077). Did you look at it? It directly extends the Einstein simultaneity convention to non-inertial frames, and the result is quite different from MCIF simultaneity. Both are just definitions, with no physical content. The one in the referenced paper, though, is actually measurable, while the MCIF simultaneity is solely a computed quantity that cannot be measured.
 
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I imagine we could have 2 way video links between A and B which would allow Bob to see Alice moving slower while Alice would see Bob move faster... just like the pulses of a 5 second pulsar for Bob would be every 4 seconds for Alice spinning circles around Bob at .6c. Would the 60Hz video sent by Bob be recieved as 72hz by Alice and her 60hz be received by Bob at 48hz (frames per second)?

I've been thinking about this so much I'm confusing myself, lol. Good night.
 

stevendaryl

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Then they could never reunite in reality because those realities can never coexist. If she traveled that speed in orbit around Earth and Bob sat still where she could fly by him periodically very close to Bob and see him age and he see her they would both appear younger to each other? When she stops nothing could have changed timewise.
I really do think it's worth while to explore how the questions about mutual time dilation and the twin paradox have very close analogies in Euclidean geometry.

As for mutual time dilation: You have two roads that run between points A and B. One road, R, goes straight from A to B, and the other road, R', takes a round-about path. The length of R between [itex]A[/itex] and [itex]B[/itex] is [itex]L[/itex]. How do you compute the length of R'?

Well, let [itex]x[/itex] measure the distance along R, with A being at [itex]x=0[/itex]. At every point along R, you can look perpendicular to the road to see a corresponding point on R'. Let [itex]x'[/itex] be the distance along R' from A to this point. Then Euclidean geometry tells us that

[itex]\delta x' = \sqrt{1+m^2} \delta x[/itex]

where [itex]m[/itex] is the relative slope between the two roads (slope = the tangent of the angle between them). So to compute the length of a path along R', just integrate: [itex]L' = \int \sqrt{1+m^2} dx[/itex]. As long as [itex]m^2 > 0[/itex], [itex]L' > L[/itex]. So the nonstraight road is longer.

At this point, you could point out a "road paradox": Someone traveling along road R' could just as well look perpendicular to the road to see a corresponding point on R. The distance [itex]x[/itex] along R is related to the distance [itex]x'[/itex] along R' via the formula:

[itex]\delta x = \sqrt{1+m^2} \delta x'[/itex]

So the traveler along road R' would see that [itex]\delta x > \delta x'[/itex]. So he should conclude that the distance along R is longer than the distance along R'. He should conclude [itex]L > L'[/itex]. Paradox!

The resolution of the paradox is to realize that the recipe of finding a point [itex]x'[/itex] on road R' that corresponds to a point [itex]x[/itex] on road R gives different answers for R and for R'. If the line between [itex]x'[/itex] and [itex]x[/itex] is perpendicular to road R, then it WON'T be perpendicular to road R'. The two roads disagree about which points correspond. This is the Euclidean equivalent of the "relativity of simultaneity" in SR. R uses one convention for setting up corresponding points, and R' uses a different convention. The two conventions have different notions of which points correspond. They both correctly compute the relative lengths of segments of the other road, but they compare DIFFERENT segments.

So why does the calculation of relative length by R turn out to be right, and the calculation of relative length by R' turn out to be wrong? It's because R' is not straight. The convention that R' uses to figure out a corresponding point on R has sudden jumps whenever R' makes a turn. Because of these jumps, either a segment of R is skipped (when R' turns away from R), or a segment of R is counted twice (when R' turns toward the road). Because of these jumps, the length calculation by R' does not give the right answer. You can't use a nonstraight road to compute the length of a straight road (at least not in a straight-forward way). This is exactly analogous to the twin paradox: you can't use a noninertial path to compute the proper time for an inertial path without a lot of extra work.

The above is all about roads on a flat Earth, which is analogous to paths through spacetime in flat spacetime. Your question about orbits brings up curved spacetime, so the analogy would be roads on a curved Earth. Consider two roads that both run straight from the North Pole to the South Pole. Near the North Pole, the length of one road can be approximately computed in the same way as for a flat Earth, because the curvature is only noticeable when you've traveled a long distance. So right near the North Pole, you can calculate the length of one road relative to the other, and use the formula [itex]\delta x' = \sqrt{1+m^2} \delta x[/itex], and each road will conclude that the other road must be longer. But the resolution here is that the formula [itex]\delta x' = \sqrt{1+m^2} \delta x[/itex] is only good for small distances, where you can ignore the curvature of the Earth. It can't be used to compute the length of a road running all the way from the North Pole to the South Pole.
 
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I imagine we could have 2 way video links between A and B which would allow Bob to see Alice moving slower while Alice would see Bob move faster... just like the pulses of a 5 second pulsar for Bob would be every 4 seconds for Alice spinning circles around Bob at .6c. Would the 60Hz video sent by Bob be recieved as 72hz by Alice and her 60hz be received by Bob at 48hz (frames per second)?

I've been thinking about this so much I'm confusing myself, lol. Good night.
No, that doesn't make sense. WHATEVER the speed of Bob relative to Alice it is exactly the same as the speed of Alice relative to Bob.
 
I did a search on this forum for "Brian Greene", together with the key words "accelerating viewpoint", "chart" and "co-moving", and I found an old thread that has a lot of discussion about some of the issues that have been raised on this thread.

Here's a link to one of the posts on that thread that talks about charts (it is post #14):

https://www.physicsforums.com/threads/no-definite-viewpoint-for-the-accelerating-traveler.671804/#post-4278370


Here's a link to one of the posts that talks about how to do the co-moving method (it is post #34):

https://www.physicsforums.com/threads/no-definite-viewpoint-for-the-accelerating-traveler.671804/page-2#post-4293784


Here's a link to a post that describes how to get the result that Brian Greene talked about on that NOVA show (it is post #289 of a different thread):

https://www.physicsforums.com/threads/speed-of-the-light-and-dilation-of-time.605080/page-15#post-4050221
 

CKH

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Dalespam gave you a link to a very well known alternative back in post #141 (http://arxiv.org/abs/gr-qc/0104077). Did you look at it? It directly extends the Einstein simultaneity convention to non-inertial frames, and the result is quite different from MCIF simultaneity. Both are just definitions, with no physical content. The one in the referenced paper, though, is actually measurable, while the MCIF simultaneity is solely a computed quantity that cannot be measured.
I did go back and read that article about the "radar conventions" some more, but the full significance of how this (radar time) convention is "different" has not sunk in yet. In all honesty the light hasn't dawned yet on what the "conventionality" you refer to is about, how it affects our analysis of motion (I obviously need to study it more). You say radar time directly extends Einstein's convention (meaning it matches in any inertial frame and so it is the same convention in that context) but then you say the result is different from MCIF simultaneity. That sounds paradoxical because any MCIF already defines planes of simultaneity. Can two different definitions coexist in a consistent system?

There is also a "radar distance" defined in that paper. It's interesting that Einstein didn't explicitly establish a convention (AFAIK) for spatial measurement (he refers only to normal Euclidean measurement of space using rulers) but he did define a convention for time measurements in different locations. It hardly a convention at all in Einstein's description of SR, since he takes the one-way constancy of c as a given; no different convention that makes sense with those givens.

Simultaneity of two events means that they have the same time coordinate value in some "valid" 4-D coordinate system, agreed? So time coordinates must also be conventional. Is the Lorentz transformation just "conventional" whatever that means? The transformation (as I've seen it written) transforms time and space coordinates in different inertial frames. If the transformations are actually independent of these "conventions" that implies that all simultaneity conventions must be consistent in some way for all inertial frames.

What about Minkowski space, does that define simultaneity as Einstein's convention? Do you break Minkowski space by adopting some other convention?

By the way, I don't believe that simultaneity cannot be measured in a MCIF. It is nothing more that a certain inertial frame. In principle, a physical frame can be pre-equipped with synchronized clocks linked in a rigid framework at any required points in space. At the spatial position of an event, the clock can be read, when the event arrives. Hence we can measure when events in an MCIF occur and directly determine whether two of them were simultaneous by experiment. So doing this mathematically could be confirmed by experiment.

I believe that in the simplified twin paradox scenario, that simultaneous readings on the home clock and traver's clock can be made in an experiment. I'll elaborate if necessary, but to summarize, comoving (3D) frames for the outbound and inbound journeys can be pre-equipped with instruments that record both the home clock and traveler clocks readings simultaneously in each frame near the turn around. One recording is made just before turnaround in the outbound frame and one just after in the inbound frame. In the travel's 3D frame these two readings are made as close as you like to simultaneously.

The resulting records will show a jump in the home clock reading before and after the turn around while the traveler's time has barely changed. These frames move inertially with the traveler on each leg of the journey, thus simultaneity for the travel is defined by each in each leg. Their recording for the traveler's clock and the home clock all occur simultaneously with the turn-round event. Simultaneity in these frames is as described by Einstein's 1905 SR. Thus it seems that in SR (at least as described by Einstein), we are stuck with the result that home clock is discontinuous with the traveler's clock time at the turn around, and in principle it can be tested.

Does this jump mean anything non-physical occurred? No, the observation can be explained without any magic. Blame relativity.
 
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pervect

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Simultaneity of two events means that they have the same time coordinate value in some "valid" 4-D coordinate system, agreed? So time coordinates must also be conventional. Is the Lorentz transformation just "conventional" whatever that means?
The Lorentz transforms assumes that one is using Einstein clock synchronization in an inertial frame. It requires conventions to define, conventions which Einstein gave. Whether or not you want therefore to call it "conventional" is a semantic issue.

The transformation (as I've seen it written) transforms time and space coordinates in different inertial frames. If the transformations are actually independent of these "conventions" that implies that all simultaneity conventions must be consistent in some way for all inertial frames.
See above.

What about Minkowski space, does that define simultaneity as Einstein's convention? Do you break Minkowski space by adopting some other convention?
You would change the metric (usually diagonal -1, 1, 1, 1) if you changed the coordinates or the conventions on which they were based, but I would say that you wouldn't change the space itself. It'd still be Minkowskii space, but with unusual coordinates. the different coordinates would imply a different metric. Exactly what the metric would be would depend on the coordinate choice.

By the way, I don't believe that simultaneity cannot be measured in a MCIF. It is nothing more that a certain inertial frame. In principle, a physical frame can be pre-equipped with synchronized clocks linked in a rigid framework at any required points in space. At the spatial position of an event, the clock can be read, when the event arrives. Hence we can measure when events in an MCIF occur and directly determine whether two of them were simultaneous by experiment. So doing this mathematically could be confirmed by experiment.
Certainly simultaneity can be measured in a MCIF. Every different inertial frame has a different notion of simultaneity, though. I am suspecting from some of your later comments you might not be aware of this. Are you familiar with Einstein's train experiment?

See for instance http://www.bartleby.com/173/9.html

I'll omit quoting the whole page, and just stress the important part of the conclusion. I would recommend reading the original , my summary is a bit choppy. Though if your time is limited, I'd just skim over the reference by Einstein, and study the Scherr et al paper below instead.

Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result: 3
Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity).
The following paper may be useful as well, it is oriented towards the teacher but it has some expereince behind it on the best way to present the issue.

If you only have time to study one paper in depth, I'd suggest studying this one, assuming you have the time to study any at all.

"The challenge of changing deeply held student beliefs about the relativity of simultaneity", Scherr et al, http://arxiv.org/abs/physics/0207081

The original version of the train paradox used in tutorial differs slightly from the one
described above. Students are told that two sparks occur at either end of a train that
moves with relativistic speed relative to the ground. The sparks leave char marks on
the ground and on the train.

The ground-based observer, Alan, who is at rest midway between the marks on the
ground, receives the wavefronts from the sparks at the same time. (See Fig. 2(a).)
The question is what order Beth, who is on the train, and in the middle of the train, receives the events. Without quoting the paper in full, I'll state that Beth, in the center of the train, does not receive the signals at the same time. Due to student resistance in drawing the correct conclusion from the above facts, the following reformulation was used by the authors to try to drive the point home.

We decided to modify the tutorial to help students recognize the ‘paradox’ in the
train paradox. The approach we took was to shift the focus from the time order of two
events (the reception of each wavefront) to whether or not a single event occurs.

In the modified tutorial, students are told that Beth has a tape player that operates as
follows. When wavefront F reaches the tape player, it starts to play music at top volume.
When wavefront R reaches it, the tape player is silenced. If both wavefronts reach
the tape player at the same instant, it remains silent. Students are asked whether
the tape player plays (i) in Alan’s frame and (ii) in Beth’s frame.
 
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At this point there is no reason to continue this thread. Plenty of references have been provided. There is no value in continuing to address the same basic misconceptions that have already been addressed multiple times.

If any of the participants feel they have questions remaining then they should read the provided references carefully prior to posting a specific question in a new thread.
 
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