# Symmetrical motion and resolving differences in FoR

1. Dec 30, 2013

### coktail

Hi All,

Long time no post! I've asked a highly-related question to the one I'm about to ask here, but there's a specific aspect of it that I'd love your input on.

I understand that if two observers in the same location were to move away from each other symmetrically at a significant fraction of c, they would both report each other's clocks as "slowing down" relative to their own, as well as each other aging more slowly than their self. I also understand that if they each were to symmetrically reverse direction back towards each other, the differences in clocks and age would be resolved by the time they were back in the same location/FoR. I hope I got all that right. It's been a while.

What I'm unclear on is how the differences are resolved. Would they each see each other's clocks ticking more quickly than their own on the return trip? This seems like the only way to me, but I thought that things always slowed down with time dilation, regardless of the direction of movement (see this thread). Does this have to do with the change in inertia/FoR of the two observers?

As a side note, I'm bringing this up due to a conversation/debate I just got into with a couple co-workers who think that relativity is "broken" and "impossible," in which I vehemently defended it, and I managed to actually kind of know what I was talking about thanks to all I learned with help from you guys! So, thanks!

2. Dec 30, 2013

### Staff: Mentor

Yes, you did. But note that both of them would have aged *less* (by the same amount) than a third observer who stayed at rest at the starting location the whole time.

If by "see" you mean what each one actually observes via light signals from the other, then yes; during the outward trip each one sees the other's light signals Doppler redshifted (meaning each one will see the image of the other's clock running slow), and during the return trip each one sees the other's light signals Doppler blueshifted (meaning each one will see the image of the other's clock running fast).

(Actually, to be precise, there will be a period after each observer turns around during which he sees *no* Doppler shift at all in the other's light signals--this is because he has turned around, but the light he is seeing from the other observer was emitted before the other turned around. Once he starts receiving light from *after* the other turns around, that light will be Doppler blueshifted. This is not an unimportant correction, btw, because it's necessary to explain why the other observer's clock ends up reading the same, instead of *more*, elapsed time when they meet up again, just using the Doppler shifted images.)

This is true *after* correcting for the light travel-time delay; i.e., if each observer takes the actual Doppler-shifted images he receives, and corrects the observed clock rate in those images based on the light travel time from the other observer, he will end up calculating that the other's clock is running slow the whole time. However, to predict what the other's clock will read when they meet up again based on this, he will also have to include the effect of the turnaround: see below.

Yes. When each observer turns around, the event on the other observer's worldline that he considers to be "at the same time" jumps forward. So in order to predict what the other observer's clock will read when they meet up again, he has to add in this forward jump to the elapsed time he calculates based on time dilation, as described just above. (Note that if he doesn't bother doing any of that calculation, but just uses the actual Doppler-shifted images of the other's clock, he doesn't have to go through any of these gyrations: he just watches the images. So using the Doppler-shifted data directly is actually *easier* than trying to figure out all this stuff about time dilation.)

3. Dec 30, 2013

### coktail

Thanks very much for your response, Peter!

Regarding the quote above, does the actual reconciliation occur during the turn-around itself, making what they visually observe as each other's clocks being sped up on the return trip merely a visual effect? Apologies if I'm confusing things.

4. Dec 30, 2013

### Staff: Mentor

I'm not sure what you mean by "reconciliation". The calculations based on time dilation are just calculations; there's no point at which they "become real". See further comments below.

Certainly not; in fact, as I noted in my previous post, the actual Doppler-shifted images allow each observer to directly *see* the other's clock readings first run slow, then run fast (with a period of neither slow nor fast in between), so that the other's clock ends up reading the same as his when they meet up again--without having to calculate anything at all. So if anything is "less real" in this scenario, it's the calculated, time-dilated clock rates and times, not the actually observer Doppler-shifted clock rates.

5. Dec 31, 2013

### coktail

By "reconciliation" I'm referring to the event or process in which their clocks go from disagreeing to agreeing with each other once more. This being a thought experiment with symmetrical motion, the clocks of the two observers should match by the time they return to a common FoR. That's what I mean when I refer to them being "reconciled." Apologies if my use of that non-technical term has confused things.

It seems like we're talking about two different ways for our observers to report what they witnessed during our experiment:

1. What they see with their eyes.
2. What they calculate independent of their visual experience.

From what I've learned on these forums and other exploration, what we see with our eyes (red/blue shifts and moving towards/away from propagating light) is interesting, but doesn't have much to do with what's *actually* happening regarding time dilation (or length contraction, for that matter) of objects in spacetime.

It seems like what you're saying is that what they *see* is each other's clocks slowing down and red-shifted on the away trip and speeding up and blue-shifted on the return trip so they match once their back together, but what they calculate once they take the propagation of light into account is each other's clocks being slowed down the entire time. I'm stumped on this. Shouldn't their calculations show what's happening in objective reality independent of their senses?

For example, we see light from distant galaxies that left them many years ago, but that doesn't mean that they're actually in the past. If we headed towards those galaxies, we'd start "catching up" with their light, and the image stream we saw would be blue-shifted and appear to be moving quickly, but that's just a result of heading towards their light, and independent of the time dilation taking place.

I've rewritten this a few times to try to be clear, but this is a tough on for me. I truly appreciate your time and patience!

6. Dec 31, 2013

### PAllen

'Now' at a distance is not a feature of reality but a feature of convention subject to constraints on reasonability. In SR it is easy to attach excess objectivity to 'now' at a distance because spacetime is considered flat and non-dynamic, and there really is only one sensible convention for inertial frames.

For your galaxies, we certainly presume there is something there following the era of the light we receive. However, exactly what is there is unknown, and which point in the distant galaxy's evolution is 'now' for us, is very much a matter of convention.

I think you missed that Peter said that there both of your symmetric observers must model a forward jump in time for the distant clock during the turnaround, in addition to the running slow before and after this. This is a required part of the reconciliation.

Unlike for inertial observers, there is no unique best way to model simultaneity during the turnaround. However, if the turnaround is slightly smoothed, as it must be in reality, any plausible approach will have a period where the other clock is considered fast, and also some portion of the received history will be considered to come from this fast period on the distant clock.

Using simultaneity defined by an instantly co-moving inertial observer, you would model that rapid distant clock period all corresponded to your period of acceleration; however you don't start seeing this period of the distant clock's history for some time after your acceleration. In particular, using this convention, part of the time you see the distant clock running the same rate as your, and part of the blue shifted history, would correspond to delayed reception of this fast history on the distant clock.

Using a different simultaneity convention (radar simultaneity - send a signal out, get reflection back, assume the reflection event is simultaneous with half way between sending and receiving the signal), you get a somewhat different description. Again, there is a fast period on the distant clock, but it is considered to begin before your acceleration, and end some time after your acceleration, and it is much less extreme a speedup than with the prior convention. Using this convention, all of the time when the distant clock is seen running the same as yours plus a chunk of the blueshift period are considered to correspond to the fast period on the distant clock. Thus, as soon as you are done accelerating, and see matching time rate, you start considering this delayed reception of the speed up that started before your acceleration. This just falls out of the way radar simultaneity works.

There is no objective (verifiable) reason to prefer one of these conventions over the other, or over various alternatives.

Last edited: Dec 31, 2013
7. Dec 31, 2013

### Staff: Mentor

A better way to say this would be, when they return to the same spatial location. If they are spatially separated, their clocks may not agree even if they are (perhaps temporarily) at rest relative to each other.

Only the first is what they "witness" in the sense of direct observations (note that you could supplement #1 with what they measure with whatever instruments are at their disposal, such as spectrometers to measure Doppler shift). They calculate the second.

I would put it almost exactly the opposite: what's "actually" happening is what can be directly observed or measured, or more generally what is invariant--what doesn't depend on who calculates it or what coordinates they use to do so. Time dilation and length contraction are not invariants; they depend on who is calculating them and what coordinates they are using. So they are not good examples of what is "actually happening", IMO.

Yes, but also with a period of no Doppler shift at all in between (when an observer has turned around, but is still seeing light emitted by the other *before* they turned around).

Yes, except for the period of no Doppler shift (see above)--*but* in addition, each one calculates that the other's clock "jumps forward" when he turns around (so his surfaces of simultaneity shift). They have to include this "jump" in order to correctly predict what the other's clock will read when they meet up again.

It depends on what you mean by "objective reality". For example, when each observer calculates that the other's clock "jumps forward", is that happening in "objective reality"? No: it's an artifact of the way each observer is assigning coordinates to distant events. Spacetime itself, and the events within it, are certainly "objective reality", yes; but when you talk about calculations of things like length contraction and time dilation, that objective reality gets mixed in with coordinates--arbitrary labels that people put on events. The same objective reality--the same spacetime--can be described by many *different* coordinate charts, and if you take what a given coordinate chart says too literally, you can get mixed up.

Sure they are; how could they emit the light we see if they weren't actually there?

What I think you mean to say is something like: "the galaxies aren't actually in the past now". But that assumes that "now" has some invariant meaning. It doesn't; it depends on how you set up your coordinates, because surfaces of simultaneity--surfaces containing events that all happen "now"--depend on how you set up your coordinates. Spacetime itself doesn't pick out any particular "now" as being the "right" one.

True.

What time dilation? Time dilation is relative; it depends on how you set up your coordinates. Someone living in the distant galaxy wouldn't experience any "time dilation"; to him, his clocks are behaving perfectly normally, and *your* clocks are the ones that are slowed (according to his calculations, if he chooses to set up his coordinates that way--what he actually observes is blueshifted light from you, so that your clocks appear to be running fast).

8. Jan 1, 2014

### ghwellsjr

You have described a symmetrical scenario which requires symmetry in your initial Inertial Reference Frame (IRF). Here is a spacetime diagram depicting an example where the two observers move away from their initial location at 0.6c in opposite directions. The dots mark off one-year increments of their own time. After 16 years of their own time, they reverse direction and in 16 more years of their own time, they rejoin at their initial location:

Note that their own times are dilated by the same amount during the entire scenario and as such they remain in sync in this IRF and there is no problem that needs to be resolved or mystery why they have the same time when they reunite. The Time Dilation factor is equal to gamma which at 0.6c is 1.25. This means that it takes 1.25 years of Coordinate Time for each year of their own Proper Time.

But the observers have no awareness of their own or the other observer's Time Dilation because Time Dilation is a coordinate effect. I would suggest that you print this diagram and draw in light signals at 45 degree angles going between them to see the Doppler effect of what they actually see of the other observer's clock. You will see that it is symmetrical and that at the beginning, until 16 years when they turn around, they both see the other clock ticking at one quarter the rate of their own, then for 12 years at the same rate, and finally, for the last four years, at four times the rate of their own. This is what PeterDonls described in post #2.

If we transform to another IRF moving at some speed with respect to the original IRF, then both observers will have different speeds and thus different Time Dilations but their own Doppler observations will remain the same. Here is a spacetime diagram for the IRF in which the blue observer is at rest at the beginning and the red observer is at rest at the end by transforming to -0.6c:

You can verify that both observers see the other ones clock ticking the same way as in the original IRF.

Now we transform to the IRF in which the blue observer is at rest at the end and the red observer is at rest at the beginning:

Still, you can verify that both observers' observations are the same as in the previous two IRF's.

Now if we want to construct a frame in which one of the observers is at rest during the entire scenario, it will have to be a non-inertial frame. There is no standard way to do this but I'm going to show you the way that PAllen mentioned in post #6 which he called radar simultaneity. I have added into the original diagram some significant radar signals that the blue observer sends in blue and the returns that he gets in red. He is also continually sending and receiving additional radar signals during the entire scenario but I don't show them as it would be a very cluttered diagram:

But from all his measurements, he keeps a log and after the scenario is over, he can construct this non-inertial reference frame:

Note that the red observer could have been doing the same thing and constructing his own non-inertial reference frame that would be a mirror image of blue's. The advantage of this type of reference frame is that you can continue to verify the Doppler signals that you verified in the previous IRF's. In addition, there is no issue of reconciliation that you are concerned about.

A fundamental precept of Special Relativity is that all frames are just as valid and none is preferred. They all contain exactly the same information so you could have analyzed the scenario in the initial IRF and be done with it.

Does this all make sense to you? Any questions?

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9. Jan 2, 2014

### coktail

Thanks you so much, all! There's a lot here for me to digest and learn from, but I don't have further questions at this time.

10. Jan 2, 2014

### Sugdub

As usual your diagrams are extremely powerful and your explanations crystal clear. Still I'm quite puzzled with several statements of yours like the one quoted above, since they seem to mean that “somebody” holds a kind of “awareness” about relativistic effects which the “observers” involved in the experiment have no access to. However SR applies to any physical context including those where nobody is there to “observe” or measure what happens in the world. Whose “awareness” do you refer to?

I would like to suggest a different wording which: i) replaces “somebody” with the role/function exerted by the theoretician (as opposed to the “observer”), and ii) replaces the “awareness” with the formal representation of the scenario in whichever IRF selected by the theoretician. It reads as follows...

The time dilation and length contraction properties of SR manifest themselves when the theoretician arbitrarily selects a particular inertial reference frame for his/her formal representation of the events occurring in the world, irrespective of anybody taking care of observing those. Obviously nothing changes in the world as a consequence of the decision by the theoretician to represent an event in a given inertial reference frame (i.e. according to a convention assigning the rest state to a given object) better than in another inertial reference frame (assigning the rest state to another object). Simply the coordinates (position, date) describing the event must change, in compliance with the Lorentz transformation, according to the reference frame in which this event is formally represented, in order to account for the invariance of the speed of light and therefore of the time synchronization convention it enables.

SR deals with different formal representations for the same scenario and therefore the Lorentz transformation (which merely deals with a coordinate effect) must be invoked for assigning the correct value to the relevant physical quantities (space and time inter alia) depending on the inertial reference frame selected by the theoretician for analyzing this scenario. The predicted outcome of the scenario is independent on the selected IRF, provided the laws of physics are expressed in a Lorentz-covariant form. The “relativistic effects” originate in the fact that the usual value of physical quantities (length, duration, …) attached to different objects are expressed in their rest frame. These values must be Lorentz-transformed into their “coordinate” value in an unique IRF before one can analyze the outcome of a scenario involving these objects in high-speed relative motion.

Example: the theoretician comes to the same conclusion (a fast muon can reach the ground before decaying) via two different but fully equivalent theoretical explanations, depending on the inertial reference frame he/she decides to use for the representation of this single scenario. The Lorentz transformation controls the change of values in the pair (distance to be covered, time it takes) according to the selected inertial reference frame, irrespective of any observation or measurements: in the reference frame assuming the Earth is at rest, the thickness of the atmosphere is assigned its proper length and the muon's lifetime is assigned a dilated value; conversely in the reference frame where the muon is assumed to be at rest, the muon lifetime is assigned its proper value and the thickness of the atmosphere is assigned a contracted value. And of course nothing has changed in the world: the same muon behaves the same way through the same atmosphere.

IMHO the Lorentz transformation deals with a change in the formal representation of the world by the theoretician, not with a change in the world. Hence, time dilation deals with a change of our formal representation of the time flow and length contraction deals with our formal representation of space extension. These are dual properties.

I hope you will comment on the above and explain if and why you disagree with this suggestion formulated by a non-physicist. Thanks.

11. Jan 2, 2014

### ghwellsjr

Thanks.

I never said they have no access to the relativistic effects, I said they have to continually send and receive radar signals and after the scenario is over, they can construct a frame. My point is that they cannot directly observe any coordinate effects in real time but they can certainly collect the necessary data and apply a timing convention for a particular frame and then have all the awareness that anyone else has. Please note that important phrase "apply a timing convention for a particular frame". If they don't know to do this or how to do this, the "awareness" will be unavailable to them.

Anyone who is there to observe or measure or who can learn about such observations or measurements from somebody else. We, of course, believe that things happen in the world that we are not aware of, either because we aren't paying attention or because they happen far away and the light signals or other information hasn't reached us yet.

Hopefully, my previous comments will show you that the observers in the scenario can also be the theoretician that you describe as long as they do the same things the theoretician must do.

Don't limit yourself to just two IRF's. You can have IRF's where both the muon and the Earth are Time Dilated and Length Contracted.

Don't limit yourself to just inertial bodies. For accelerating bodies, there is a change in the world.

12. Jan 4, 2014

### Sugdub

So far in all presentations of SR I could access, the statements made by physicists indicated their belief that SR (the Lorentz transformation, time dilation, length contraction,...) deals with a change in the world. You are the first one I read of who stated several times that SR deals with a “coordinate effect”, and for me this means a change of the formal representation of physical quantities (space and time, and subsequently of all physical quantities which are dependent on space and time variables), according to the inertial reference frame chosen by the theoretician to describe the world and what happens there. This is why I suggested the wording: “The Lorentz transformation deals with a change in the formal representation of the world by the theoretician, not with a change in the world.”

It is not yet clear for me whether you agree with this wording. If not, I haven't understood what you actually mean with a “coordinate effect”.

13. Jan 4, 2014

### ghwellsjr

I just said in the previous post "For accelerating bodies, there is a change in the world." The most obvious case of this is where we take a rod of a given length as measured according to the international standard (such as a commercial laser rangefinder would measure) and we accelerate it along its length to a very high speed and measure it again according to the same standard, it may be a different length, either shorter or longer, but Special Relativity has no way of predicting exactly how long it will be. But this change is not Length Contraction.

If we did this and it turned out the rod measured the same length after as before, we might be tempted say that according to SR the rod underwent Length Contraction because in its original rest frame it had one length and after acceleration it is Length Contracted according to Special Relativity but that is merely a Coordinate Effect. We could just as legitimately say that according to its final rest state it underwent length expansion. Both viewpoints justify "a change in the world" but we can't uncategorically say that it is Length Contraction when it is accelerated.

So if you can specify the lengths of objects, even accelerating ones, in one Inertial Reference Frame, then the Lorentz Transformation process will tell you exactly what their lengths are in any other IRF you choose.

But there is another application of the Lorentz Transformation which is that the laws of physics must remain unchanged after undergoing the Lorentz Transformation. This guarantees that accelerated objects will approximately undergo the length changes as predicted by SR but it can't even predict if the acceleration forces will cause the object to shatter or warp or otherwise permanently deform so that they end up with the changes that I mentioned at the beginning of this post.

14. Jan 4, 2014

### WannabeNewton

If a rod is initially at rest in an inertial frame and the rod is given a proper acceleration longitudinally that is constant across the rod then the proper length of the rod will increase until the resulting internal stresses break the rod. This is not a "coordinate effect".

However it is without a doubt obvious that length contraction and time dilation of spatial and temporal intervals between events in space-time obtained from Lorentz boosting from one inertial frame to another are "coordinate effects".

15. Jan 4, 2014

### pervect

Staff Emeritus
I would say that the Lorentz transformation is expressed in terms of coordinates, but it represents a difference in the group structure of space-time that can be expressed without coordinates.

The Lorentz transformation implies that the group structure of space-time is the Poincare group.
The continuous isometries of space-time in special relativity consist of:

space translation invariance (3 degrees of freedom)
time translation invariance (1 degree of freedom)
space rotational invariance (3 degrees of freedom)
boost invariance (3 degrees of freedom)

The above set of symmetries are the generators of the group. Generators can be thought of as those infinitesimal transformations that repeated, "generate" the group. I find it physically and intuitively meaningful to discuss the difference in the groups by discussing the difference in their generators, as the generators are sufficient to define the group -

I believe the correct description for the classical group structure of Newtonian space + time would be the E(3) x E(1) the tensor product of an Euclidean space group of dimension 3 for space, with an Euclidean group of dimension 1 to represent time. (Alas, I don't have a reference on this point to cross-check and make sure I haven't made a blunder here :-().

The point is that it is a different group than the Poincare group, and that this difference is independent of the choice of coordinates.

Let's look at and compare the groups by looking at their generators of the group for classical Newtonian space + time. We should have (and I dont' have a reference for this point either):

space translation invariance (3 degrees of freedom)
time translation invariance (1 degree of freedom)
space rotation invariance (3 degrees of freedom).

In classical Newtonian physics we don't expect to have the same boost symmetry that we have in SR, i.e. in classical Newtonian physics we expected (but failed to find via experiment!) an ether.

http://en.wikipedia.org/wiki/Euclidean_group
http://en.wikipedia.org/wiki/Poincaré_group
http://en.wikipedia.org/wiki/Generating_set_of_a_group

16. Jan 7, 2014

### phyti

View attachment 65392

In the U frame light is emitted for an interval of length t. A is any observer moving toward the source as A1, with the source as A2, and away from the source as A3. Since the wave number for the interval is constant, one wave will be sufficient. Without considering any relativistic effects, it is obvious that t1 < t2 < t3, and t2 = t. In doppler terms, A1 sees blue shift, A2 sees no shift, and A3 sees red shift.

View attachment 65393

This drawing illustrates the text of posts 2 thru 6.
Biff and Fritz move in opposite directions following the same speed profile out and the reverse profile back. A continuous velocity change occurs for the interval t4 to t5. B and R represent blue and red, + and - represent increasing and decreasing. The signed values along Biff's path are what Fritz observes for the indicated path segments between light signals (blue). Biff would observe the same effects for Fritz, since the scenario is symmetrical by definition. Since clocks are frequencies, red and blue shifts will correspond to slow and fast clock rates respectively.
If an observer can alter the frequency of a signal by changing his speed, he is in reality altering his perception of the signal, not the signal.
The “jump” in time is a misleading term since it’s just a poor substitute for omitting the non-inertial portion of the motion.

Keep in mind this is a trivial case, and the 'twins' a special/simple case with one vs two segments, with the returning twin aging less. In the general and more realistic case with two clocks separating, then rejoining along independent paths, the total time for each path can be determined if the speed profile is known. You could approximate the path with small inertial segments, or integrate a highly curved path, which is just what a moving clock does. Then the simplest solution (for your reconciliation) is to compare the clocks when reunited.

Notice that the paths for the two clocks form a closed boundary. All signals from one will be detected by the other. Assuming a standard clock if one clock accumulates less ticks than the other it will be evident at the reunion.

Last edited: Mar 9, 2014
17. Jan 7, 2014

### ghwellsjr

If I understand what you mean by your last commit--that making the turn-around be a gradual acceleration period rather than an instantaneous one would eliminate the "jump", then all you are doing is changing it from a "jump" to a "slide" but the issue remains--that the time on the other observer's clock has to progress more rapidly during some portion of the trip as stated by the OP:

The posts that you referred to that discuss the "jump" were talking about the "explanation" that would depict the other observer's clock always running slower as shown in this "hybrid" diagram made by taking the bottom part of my second diagram in post #8 with the top part of the third diagram:

This is the way the OP expressed the problem, namely that during both halves of the trip, the other observer's clock is running slower, so how do they end up with the same time? He thought maybe it occurred during the return part of the trip.

Your suggestion to make the acceleration be gradual doesn't answer the question--it just changes the problem from one where the other clock is not running slower during the entire trip and instead expands the cluster of dots indicating times 8 through 24 into a finite interval and maybe even encroaching on the dots below and above, but as you pointed out, the problem is symmetrical which means that the Time Dilation in the rest frame that you drew is always the same for both observers and so the answer is that both observer's clocks are always in sync--the problem goes away once you see express it that way.

But the radar solution that PAllen introduced in post #6 and I depicted in post #8 will work just at well with any gradual acceleration you want to propose and will end up with a non-inertial diagram for each observer that looks approximately like the one I presented already:

Instantaneous accelerations are rarely a problem in presenting or analyzing scenarios in Special Relativity, and stretching the accelerations out just makes the problem more complicated without adding any understanding.

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18. Jan 9, 2014

### phyti

The 2nd drawing shows there are no missing events that require a "jump" in time, and that there is no discontinuous (non physical) motion. It shows the op that each observer sees both a slow and a fast clock for the other, and graphically demonstrates post#2 by PeterDonis, and that the observations are symmetrical.

19. Jan 9, 2014

### coktail

Thanks, ghwellsjr!

20. Jan 10, 2014

### Sugdub

I'm still unsure having properly understood the explanations provided in this thread, in particular distinguishing the respective contributions of the laws involved in the background of your diagrams (time dilation and Doppler?), so based on your inputs I tried to imagine what could be my answer to the question initially raised by the OP. Please correct me if I'm wrong.

Due to the full symmetry of the experimental pattern, both clocks will indicate the same time when they are collocated back at their original position. No need to invoke SR and/or Doppler to reach that conclusion. On top of this, both clocks will actually remain synchronous all along the trip and both travelers will actually be equally aged at all times during the experiment. Any alternative would breech the symmetry of the problem.

In line with the question raised by the OP, the analyzes presented in this thread actually deal with the “observation” of the other traveler's clock by each of the travelers (this assumes that a signal propagates the “ticks” of the remote clock). If during the trip a traveler “observes” the remote clock owned by the other traveler, then the remote clock is effectively “observed” running at a different path as compared to the local clock of the observer. This is physically caused by the Doppler effect which affects the period of the received ticking signal. It should be noted that in order to determine the difference in the “observed” aging of the travelers, the periods of both the remote clock and the local clock must be described in the same inertial reference frame, I mean the Doppler effect must be applied to the Lorentz-transformed period of the remote clock before the output is compared to the Lorentz-transformed period of the local clock.

So the SR time dilation property and the Doppler effect both contribute to the analysis of the “observed” path of the remote clock. The SR time dilation depends on the speed of each clock in the selected IRF, irrespective of the direction of motion (toward or away), whereas the Doppler effect depends on the relative speed between both travelers and changes sign according to the direction of motion (toward or away). Globally each traveler will “observe” the other traveler's clock slowing down when flying away and then catching up when flying back, but such observations have no impact whatsoever on the actual path of the travelers' clocks, which remain synchronous all along the trip.

Please correct me if I'm wrong.