Undergrad Simultaneity and the Twin Paradox

[PeterDonis]

He calculates how long it has been traveling away from A.

How?

Now B says: "At the same time that we were X days from our conjunction with A, A sent this frame." This seems to me where I am assuming something about simultaneity.

No, you did it in the step before (that I quoted above and asked "how?"). You need to assume a simultaneity convention in order to calculate how long the light has been traveling. There is no absolute answer to that question; it is frame-dependent because time is frame-dependent (which is equivalent to being simultaneity convention dependent).

B calculates the travel time of the video frame (2.4 years)

This is an example of what I just said above. It's 2.4 years in B's frame, but not in A's frame. So the "how long" calculation itself is frame-dependent.

 

[Freixas]

This is an example of what I just said above. It's 2.4 years in B's frame, but not in A's frame. So the "how long" calculation itself is frame-dependent.

I would say, of course, that's the whole point. I am trying to calculate the light delay in B's frame. For example, astronomers see a star blow up and tell you that, although we see it happen now, it actually happened some calculable time in the past. All the calculations are in our frame--our distance to the star and our time line.

The mysteries of my approach are not just that I got what I think is a wrong answer, but that the answer is exactly the inverse of what I would expect (B thinks his clock runs 0.6 slower than A's), that the answer when looked at from A's point of view wasn't also reversed and that A's answer is what I would have expected.

I just remembered that one of my neighbors is a retired physics professor, so maybe I can get him to walk me through this. Thanks for your help, though.

 

[PeterDonis]

I am trying to calculate the light delay in B's frame.

Yes, but you are then combining it incorrectly with a clock time in A's frame, because you are leaving out relativity of simultaneity--or, to use the term that seemed to work better for you in another thread, clock offset.

Let me rephrase things explicitly from the point of view of B's frame:

A departs B at a speed of 0.8; both A's and B's clocks read zero. M is 2.4 light-years away and is moving towards B at the same speed, 0.8. But at this time in B's frame, M's clock does not read zero. It reads 3.2 years. (I'll leave it to you to figure out how to get that number; the easiest way to visualize it is to draw a spacetime diagram.)

3 years later by B's clock, M arrives at B. M's clock reads 5 at this event. That means that, in B's frame, during the time between meeting A and meeting M, M's clock ticked away 1.8 years. Since B's clock ticked 3 years, M's clock is ticking at a rate 1.8/3 = 0.6 relative to B's clock--precisely the expected time dilation factor. And since A's clock and M's clock are synchronized, A's clock must be ticking at the same rate.

At the same event where B meets M, B receives a light pulse from A showing A's clock reading 1 year. B correctly calculates that this light pulse took 2.4 years to travel according to his frame. [Edit--this is incorrect--see follow-up post #34.] However, in order to calculate how fast A's clock is ticking relative to his own, he has to know what event on A's worldline is simultaneous, in his frame, with the event where he meets M, and what A's clock reads at that event. The light pulse's travel time in B's frame does not give any information about that. In fact, the single light pulse itself is irrelevant to the calculation of A's tick rate during the time between the two meetings.

The correct answer is that A's clock reads 1.8 years at the event on A's worldline that is simultaneous with B meeting M (note that this calculation is very similar to the calculation above that told us that M's clock reads 3.2 years at the event on M's worldline which is simultaneous, in B's frame, with B and A's meeting); and, again, that means A's clock ticked 1.8 years while B's clock ticked 3 years, just like M's clock, so again we get the expected time dilation factor of 0.6.

In other words, the short answer is that, if all you have is a single light pulse showing a single reading of A's clock, you cannot tell anything about the tick rate of A's clock relative to B's clock. You need at least two light pulses, if light pulses are what you want to use.

 

[PeterDonis]

B correctly calculates that this light pulse took 2.4 years to travel according to his frame.

Actually, this is wrong! See below.

the short answer is that, if all you have is a single light pulse showing a single reading of A's clock, you cannot tell anything about the tick rate of A's clock relative to B's clock.

Actually, this is a bit pessimistic. We can use the light pulse's travel time in B's frame if we figure out at what time, in B's frame, the pulse was emitted. In other words, we need B's clock reading at the event on B's worldline which is simultaneous with A's emitting the light pulse (when A's clock reads 1). That is not the event at which B's clock reads 0.6 (which is what your calculation implicitly assumed)--that event is simultaneous to A's clock reading 1 in A's frame, not B's frame. (This is why I said in my previous post that your calculation failed to take clock offset into account.)

If we run the numbers, we find that the time, in B's frame, when A emits the light pulse (i.e., when A's clock reads 1) is 5/3 (fractions turn out to be easier than decimals for these numbers). In other words, this time is 4/3 earlier than the time B meets M (which is time 3 by B's clock). Which in turn means that the light pulse took 4/3 years to travel, not 2.4 years, so it covered 4/3 light years, not 2.4 light years.

Where did we go wrong before? By assuming that, since M is 2.4 light years from B when B meets A, A must be 2.4 light years from B when A emits the light pulse that arrives at B when B meets M. But that is not valid logic. The only valid reasoning we can do to obtain the distance the light pulse covered is the reasoning I did above, which starts from A's known clock reading when the light pulse was emitted and calculates what B's clock read at the same time, in B's frame, as that emission event. And, of course, knowing that the light pulse covered 4/3 light years, we can see that A, traveling at speed 0.8, took 4/3 divided by 0.8 = 5/3 years in B's frame to travel from A's meeting with B to the point where the light pulse was emitted--and we already saw above that the time of emission, in B's frame, was 5/3 years. So A's clock ticked 1 year while 5/3 years elapsed in B's frame, for a time dilation factor of 0.6, again as expected.

 

[PeterDonis]

you are then combining it incorrectly with a clock time in A's frame, because you are leaving out relativity of simultaneity--or, to use the term that seemed to work better for you in another thread, clock offset.

Actually, this was a bit pessimistic too. With the correct light travel time in B's frame--4/3 years--you can indeed subtract that time from 3 years (B's clock reading when he receives the light pulse), obtain 5/3 years for the time of emission according to B's frame, and compare that with A's clock reading of 1 year shown in the video frame to obtain the correct time dilation factor of 0.6, as I showed in the previous post.

The miscalculation of the light travel time as 2.4 years is really more a matter of not realizing that A is 2.4 light years from B when B receives the light pulse, not when A emits it, because A is moving away from B while the light travels.

 

[Dale]

B receives a frame from A and completely ignores the contents. He calculates how long it has been traveling away from A. Then he places the frame in a new video, where the time at which the frame is positioned is B's current time minus how long it has been traveling. This is the correction for light delay.

Wait, I thought A was doing all of the corrections at the end:

A can calculate the light delay for every frame and adjust as needed.

 

[Dale]

I just took a look at Carroll's first two chapters in Lecture Notes on General Relativity. If understanding that is what it takes, then it's definitely not going to happen. I'm not 100% sure that my mistake couldn't be explained to me in a way that I could understand, but I'm not 100% it could, either.

I am sorry to hear that you are feeling that way. My advice is to drastically simplify things. When you are learning a new concept you should start with the simplest example that captures the essence of the concept.

You are feeling frustrated, in my opinion, because you don’t understand the relativity of simultaneity* between inertial frames but instead of analyzing a scenario with only that, you are analyzing a scenario with non inertial frames, Doppler shifts, multiple observers, length contraction, and time dilation.

A simpler scenario that isolated the problematic concept (as much as possible) would help; one where there were fewer messy distractions.

*I am not sure if it is the relativity of simultaneity or the Lorentz transform that is the obstacle

 

[Freixas]

Actually, this was a bit pessimistic too. With the correct light travel time in B's frame--4/3 years--you can indeed subtract that time from 3 years (B's clock reading when he receives the light pulse), obtain 5/3 years for the time of emission according to B's frame, and compare that with A's clock reading of 1 year shown in the video frame to obtain the correct time dilation factor of 0.6, as I showed in the previous post.

The miscalculation of the light travel time as 2.4 years is really more a matter of not realizing that A is 2.4 light years from B when B receives the light pulse, not when A emits it, because A is moving away from B while the light travels.

@PeterDonis: You rock! Thanks! This is pretty much the same problem I had in my other thread Reconciling two observers in two frames, but I was having trouble seeing how I could apply it in this situation.

I'm hesitant to bring this up, but it looks like we could produce a video. For B's portion of the trip, the video would last 5/3 years and show A's clock going from 0 to 1. Maybe we'll leave C's video for another day...

 

[Freixas]

Wait, I thought A was doing all of the corrections at the end:

I said: A can calculate the light delay for every frame and adjust as needed.

What I meant is that anyone can do the calculations after the fact. PeterDonis can do the calculations and we don't know what frame he's in relative to A, B or C. :-) I did not mean that the calculations were relative to A's frame.

Sorry for the confusion.

 

[Freixas]

You are feeling frustrated...

Well, maybe a bit frustrated. Mostly obsessed, elated and fascinated...

but instead of analyzing a scenario with only that, you are analyzing a scenario with non inertial frames

A small correction: the scenario with A, B and C includes only inertial frames.

A simpler scenario that isolated the problematic concept (as much as possible) would help; one where there were fewer messy distractions.

You may be right, but the problem is that there are already a lot of simpler scenarios out on the web. For instance, John D. Norton has some of the best explanations of relativity I've seen that avoid math. When he explains the simple scenarios, it all makes sense, but when I try to apply them, it's not so easy. One learns more from trying, getting it wrong and having the error pointed out. However, since the same error has bitten me twice, I really should work on some simpler cases. :-) I'm sure there must be some problem cases with answers somewhere on the web.

By the way, this all started when I was thinking of slowing down light speed to make things clearer. The scenario was a husband and wife having breakfast. The husband goes to get some coffee and returns. Presto—the twin paradox! The husband aged less than his wife. (This is true even if you don't slow down light!) I tried to describe what the husband saw and what the wife saw, but then I realized that I was assuming that they could see as though light speed were normal and not slowed down. So then I started wondering if there would be any way to simulate what one would see if the light delay weren't there.

Some physicist wrote a book I read a long time ago in which light was slowed. As one bicycled across town, the harder on pedaled, the smaller the distances became. I'm not sure if he wrote his story assuming that vision was unaffected. It seems unlikely, but I don't remember the book's name or the author. I think he might have changed various "constants" and then described the effects.

 

[PeroK]

Some physicist wrote a book I read a long time ago in which light was slowed. As one bicycled across town, the harder on pedaled, the smaller the distances became. I'm not sure if he wrote his story assuming that vision was unaffected. It seems unlikely, but I don't remember the book's name or the author. I think he might have changed various "constants" and then described the effects.

That might have been Gamov, writing about Mr Tompkins:

 

[Dale]

I did not mean that the calculations were relative to A's frame

Oh, my mistake. That is exactly what I had assumed. It is not important who does the calculations, but the calculations require the clear specification of the reference frame in which the calculations are to be made.

A small correction: the scenario with A, B and C includes only inertial frames.

Sure, but it also includes 3 inertial frames. Only two are needed to understand either the Lorentz transform or the relativity of simultaneity. Plus all of the additional complications.

For instance, John D. Norton has some of the best explanations of relativity I've seen that avoid math. When he explains the simple scenarios, it all makes sense, but when I try to apply them, it's not so easy. One learns more from trying, getting it wrong and having the error pointed out. However, since the same error has bitten me twice, I really should work on some simpler cases.

Fair enough, but let's try for some middle ground then. Something more complicated than John D. Norton's stuff, but less complicated that this three person corrected movie hand-off scenario.

Do you have a feel for whether your confusion is the relativity of simultaneity or the Lorentz transform?

 

[PeterDonis]

For B's portion of the trip, the video would last 5/3 years and show A's clock going from 0 to 1.

You could do this, yes, but note that it would require post-processing of the raw video. If B has a telescope receiving light signals from A for the entire time between meeting A and meeting M, that telescope will show A's clock going from 0 to 1 while B's clock goes from 0 to 3--in other words, B's clock ticks off 3 years, not 5/3 years, while B actually sees (through the telescope) A's clock going from 0 to 1. That is because the Doppler shift factor of 1/3 is not the same as the time dilation factor of 0.6. The Doppler shift factor is what tells you what B will actually see through the telescope. So making a video which lasts 5/3 years and shows A's clock going from 0 to 1 requires changing the rate of the raw video that was obtained through the telescope.

 

[Freixas]

You could do this, yes, but note that it would require post-processing of the raw video.

Post-processing was always in the works, which is why I said A did the calculations after C provided him with the video. For B's leg, once post-processed, you only get 5/3 years of video. To get the full six years of video, you will need C's portion, which is where it gets interesting.

Using some similar calculation, C could produce a video running from his 3 year to 6 year mark showing A's clock going from 8.2 to 10. Using the same mapping, he could continue to produce the video for the first three years, showing A's clock counting down from 8.2 to 6.4. This overlaps and conflicts with B's mapping (no surprise, they are in different frames). With some clever work, I'm sure one of you could figure out a different mapping that would allow C's raw recording to continue the mapped recording from the 5/3 mark to the 3 year mark (with A's clock going from 1 to 1.8). We would then be left with a lot of frames on the cutting room floor--this would be exactly as expected. There's nothing wrong with assembling the frames some other way. They are just choices of whose frame of view you want to pick.

My numbers might be off, sorry. But you get the idea...

So with A, B, and C, you have two frameworks to choose from. B gets short-shrift, because A will never receive video recorded after B passes M. C has a choice of using two different mappings for filling the missing 1 1/3 years of B's outgoing trip. This implies, that for an accelerated framework, one might have all sorts of equally valid mappings, specially on the return leg.

 

[Freixas]

That might have been Gamov, writing about Mr Tompkins:

View attachment 228738

This looks familiar and the name "Gamov" is familiar. I'm pretty sure this was it. I want to re-read it now.

 

[Freixas]

This looks familiar and the name "Gamov" is familiar. I'm pretty sure this was it. I want to re-read it now.

It's Gamow.

 

[Ibix]

It's Gamow.

You see both Gamov and Gamow. It depends on how you choose to transliterate the Cyrillic alphabet into the Latin one, I believe.

 

[PeterDonis]

Using some similar calculation, C could produce a video running from his 3 year to 6 year mark showing A's clock going from 8.2 to 10.

No, he can't. B meets C at the same time he meets M. If C synchronizes his clock to read 3 years at the meeting, and starts his video at that instant, he will see A's clock, through his telescope, reading 1 year, and that's what his video will show. So C's raw video will show A's clock going from 1 year to 10 years, while C's clock goes from 3 years to 6 years. In other words, the Doppler shift factor for C watching A is 3--C sees, through his telescope, A's clock running 3 times as fast as his own.

There isn't any post-processing that C can do on this video to show A's clock running slow relative to his, by the time dilation factor of 0.6. Think about it: no matter what subset of the frames in this raw video C takes, that subset will show C's clock changing by a smaller amount than A's clock. But to produce a video, by post-processing this raw video, that showed A's clock running slow relative to C, you would have to find some subset of frames that showed A's clock changing by a smaller amount than C's clock. There is no such subset of frames in the raw video.

A will never receive video recorded after B passes M. C has a choice of using two different mappings for filling the missing 1 1/3 years of B's outgoing trip.

I don't understand what you are referring to here. B and C are receiving telescopic images of A's clock and recording them. A isn't recording anything. And the raw video covers B's entire outgoing trip from A to M; there is no "missing 1 1/3 years". Post-processing B's raw video to decrease the "play time" from 3 years to 5/3 years does not make that portion not cover B's entire outgoing trip; it just changes the rate at which B's clock goes from 0 to 3 years in the video, relative to the "play rate" of the video itself.

 

[Freixas]

You see both Gamov and Gamow. It depends on how you choose to transliterate the Cyrillic alphabet into the Latin one, I believe.

Ahh... When I looked up Gamov on my library's site, I got nothing. So I looked up Mr. Tompkins and it showed Gamow. The Wikipedia page confuses this further by saying "George Gamow, born Georgiy Antonovich Gamov..." :-)

 

[Freixas]

There isn't any post-processing that C can do on this video to show A's clock running slow relative to his, by the time dilation factor of 0.6.

Of course there's a mapping. On the mapped video, A's 10-year frame gets mapped to the 6 year position. A's 1-year frame gets mapped to the -9 position. Cut this off at the 3 year mark (where A reads 8.2).

Peter, I'm confident that if you could adjust for light delay on B's outgoing trip, you could adjust for light delay on C's incoming trip. Other than the change in direction, all the factors look pretty much the same. You still have distance, time (and one can be calculated from the other), the speed of c and your beautiful adjustment for the relativity of simultaneity.

The trickier part is that, since B didn't get enough video to fill out its 3-year segment, we'd like to use C's data to calculate B's missing portion. This means treating C as intercepting data meant for B and determining when B would receive it. Calculating this might cause heads to explode, but once calculated, the problem is reduced to the original light-delay adjustment done for B's leg.

I don't understand what you are referring to here. B and C are receiving telescopic images of A's clock and recording them. A isn't recording anything.

I never said A was recording anything. However, A receives B's video as delivered by C in a nicely wrapped package. What I meant is that A cannot receive video recorded by B after B's meeting with C (unless we postulate a UPS carrier D further along B's path).

 

[PeterDonis]

On the mapped video, A's 10-year frame gets mapped to the 6 year position. A's 1-year frame gets mapped to the -9 position. Cut this off at the 3 year mark (where A reads 8.2).

And doing this maps the frame taken when C's clock reads 6 years to the 6-year position, and the frame taken when C's clock reads 3 years to the -9 position. So the frame at the 3-year position will be the frame taken when C's clock reads 5.4 years. In other words, this portion of the video will have C's clock ticking off 0.6 years while A's clock ticks off 1.8 years, for a Doppler factor of 3, as I said. The fact that the video takes 3 years of "play time" to play doesn't change any of that.

I'm confident that if you could adjust for light delay on B's outgoing trip, you could adjust for light delay on C's incoming trip.

You can always calculate coordinate times and positions by adjusting for light delay from the raw data, yes. But you don't do that by taking raw video frames and just picking out some subset of them and adjusting the playback rate. The relationship between the Doppler shift in the raw data and the time dilation factor in the calculated coordinates is not that simple.

since B didn't get enough video to fill out its 3-year segment

I don't understand what you mean by this. B's raw video, which shows A's clock going from 0 to one year, is captured during B's entire 3-year journey--i.e., the frames showing A's clock going from 0 to one year are captured over a range of B clock readings from 0 to 3 years. There is nothing missing.

A receives B's video as delivered by C in a nicely wrapped package. What I meant is that A cannot receive video recorded by B after B's meeting with C

Ah, ok.

 

[Freixas]

And doing this maps the frame taken when C's clock reads 6 years to the 6-year position, and the frame taken when C's clock reads 3 years to the -9 position. So the frame at the 3-year position will be the frame taken when C's clock reads 5.4 years. In other words, this portion of the video will have C's clock ticking off 0.6 years while A's clock ticks off 1.8 years, for a Doppler factor of 3, as I said. The fact that the video takes 3 years of "play time" to play doesn't change any of that.

Did I get the backwards? Sorry, I do that a lot. Let me try again. A's 10 year frame gets mapped to the 6 year position. A's 7.6 year frame gets mapped to the 3 year position. Since the original, long-forgotten intent of the video was to show what happens during C's leg of the trip if C could instantaneously see A's clock, that's as far back as we need to go.

For B's leg, we only get the first 5/3's of the mapped video. If we wanted to do it the easy way, B would continue mapping frames until a mapped frame landed at the 3-year mark and then would then ship the entire mapped 3 years back to A (via UPS Interstellar). If we wanted to do it the hard way, we would use the video intercepted by C (but sent during B's outbound leg) to do the same mapping. You'd know if you did it right (for some meaning of "right") if B's 3-year section ended with A's clock reading of 2.4.

Since both B and C map a frame to the 3-year mark (the way I've defined it), we would have a "double-exposure" with A's clock reading both 2.4 and 7.6 years.

Again, while the recorded video is 6 years and the mapped video is 6 years, the two are different and I get the feeling they are being confused. The mapped video is trying to capture (from B's or C's perspective) what is happening at A at any point during their trip. While there will be disagreement about whether two events are simultaneous or not when viewed by different observers, there are never any disagreements when viewed by just one observer.

By (properly) correcting for light delay, B (or C) can view what happened at A simultaneous with any point on B's (or C's) journey. We already know the answer, but I find the approach of correcting for light delay appealing. Also, if the answer doesn't come out as expected, it can reveal a lack of understanding about how to properly calculate light delay adjustments.

 

[Herman Trivilino]

For instance, John D. Norton has some of the best explanations of relativity I've seen that avoid math. When he explains the simple scenarios, it all makes sense, but when I try to apply them, it's not so easy.

That's because you have to learn the math to be able to do what he's doing.

 

[PeterDonis]

A's 10 year frame gets mapped to the 6 year position. A's 7.6 year frame gets mapped to the 3 year position.

First, by the implicit method you are using here, it's A's 8.2 year frame (1.8 years before the 10 year frame, which happens when C meets A).

Second, if by "the 3 year position" you mean 3 years in "video play time", that's fine. But the video frame in which A's clock reads 8.2 years is not received by C when C's clock reads 3 years. It is received by C when C's clock reads 5.4 years.

Since the original, long-forgotten intent of the video was to show what happens during C's leg of the trip if C could instantaneously see A's clock

Once more: "instantaneously" is frame dependent. There is no absolute meaning to "instantaneously". Even the implicit definition you are using here, which is "instantaneously" in the inertial frame in which C is at rest, does not generalize to any situation where either C has nonzero proper acceleration, or spacetime is curved (i.e., any gravitating masses are present), because in either of those situations, there is no global inertial frame in which C is at rest.

For B's leg, we only get the first 5/3's of the mapped video.

No, you don't. The frame in which A's clock reads 1 year is the frame received by B when B meets M, i.e., when B's clock reads 3 years, not 5/3 years. The fact that you have jiggered the "video play time" of this segment of video so it only takes 5/3 years to play doesn't change that.

If we wanted to do it the easy way, B would continue mapping frames until a mapped frame landed at the 3-year mark

B already has a frame at his 3-year mark: the frame received when he meets M. That frame is included in the video that C brings back with him and gives to A when they meet. What you are saying here makes no sense.

Since both B and C map a frame to the 3-year mark (the way I've defined it), we would have a "double-exposure" with A's clock reading both 2.4 and 7.6 years.

Even with the implicit frames you are using, this is wrong. At the event on A's worldline which is simultaneous, in B's frame, with B meeting M (and C), A's clock reads 1.8 years, not 2.4. And as above, at the event on A's worldline which is simultaneous, in C's frame, with C meeting M (and B), A's clock reads 8.2 years, not 7.6.

The mapped video is trying to capture (from B's or C's perspective) what is happening at A at any point during their trip.

And even by that criterion, it is failing, because B and C have different simultaneity conventions and so they do not agree on what is happening at A "at the same time" as a point on their respective trips.

While there will be disagreement about whether two events are simultaneous or not when viewed by different observers, there are never any disagreements when viewed by just one observer.

So who is the one observer? You have two, B and C, and their views, even when corrected as you describe, are not compatible.

By (properly) correcting for light delay, B (or C) can view what happened at A simultaneous with any point on B's (or C's) journey.

If you really think this is what you are doing, I can only say that I think you are greatly mistaken. What you are trying to construct makes no sense to me, and does not convey any useful information as far as I can see. It is just a way of deepening your confusion. I'm sorry if that sounds harsh, but that's the way I see it.

We already know the answer

No, we do not, because there is no unique answer. I think you should read that sentence again and again until it sinks in.

 

[PeterDonis]

B (or C) can view what happened at A simultaneous with any point on B's (or C's) journey.

Here is another reason why what you are trying to do here makes no sense. When C meets up with A and delivers the final video, C's clock reads 6 years and A's clock reads 10 years. There's no way around that: that is an invariant fact which is independent of any choice of frames or any simultaneity convention or any jiggering about with videos. So any video that is going to tell a true and useful story about "what is happening at A while B and C are traveling" has to end up showing A's clock ticking off 10 years while the B/C clock (it's effectively one clock since C sets his clock to B's clock reading when they pass) ticks off 6 years. But you are trying to construct a video in which A's clock ticks slower than the B/C clock, as if that somehow "truly" represents "what is happening at A". That can't possibly be correct.

 

[PeroK]

While there will be disagreement about whether two events are simultaneous or not when viewed by different observers, there are never any disagreements when viewed by just one observer.

That's only true if you are careful to define "observer", "simultaneous" and "reference frame". In practical terms, it may not be at all clear to one observer whether two events are simultaneous in their frame. For example:

https://www.physicsforums.com/threa...ological-order-of-events.946250/#post-5988680

 

[MikeGomez]

My favorite version of the twin paradox involves no acceleration.

I've read articles and watched videos that have reinforced that the fact that one twin experiences acceleration is not relevant.

No acceleration would mean no age differential.

The twin paradox is a thought experiment involving two participants. One twin accelerates and the other does not. The traveling twin accelerates away from the stay at home twin, then re accelerates to switch from outbound to inbound direction, then accelerates again (deceleration) to come to a stop upon returning. Trying to change the scenario by removing acceleration of the traveling twin is not only nonphysical, but adds unnecessary confusion instead of helping.

Introducing a third participant on the inbound leg of the journey in order to record the traveling twins clocks is not a valid step in this thought experiment. The paradox is based on one twin returning to Earth to find his/her twin has aged much more, and that can’t happen if the traveling twin perpetually tours the cosmos instead of returning home. Even more problematic is that introducing the non-accelerating inbound traveler is exactly equivalent to the inertial-less drive, and I agree with Dale that this is not physical.

I think it is much simpler to stop adding more than two participants , and stop trying to invent non-physical solutions like removing acceleration.

This link does a pretty good job ( read up until the Asteroids section)

 

[PeroK]

No acceleration would mean no age differential.

The twin paradox is a thought experiment involving two participants. One twin accelerates and the other does not. The traveling twin accelerates away from the stay at home twin, then re accelerates to switch from outbound to inbound direction, then accelerates again (deceleration) to come to a stop upon returning. Trying to change the scenario by removing acceleration of the traveling twin is not only nonphysical, but adds unnecessary confusion instead of helping.

Introducing a third participant on the inbound leg of the journey in order to record the traveling twins clocks is not a valid step in this thought experiment. The paradox is based on one twin returning to Earth to find his/her twin has aged much more, and that can’t happen if the traveling twin perpetually tours the cosmos instead of returning home. Even more problematic is that introducing the non-accelerating inbound traveler is exactly equivalent to the inertial-less drive, and I agree with Dale that this is not physical.

I think it is much simpler to stop adding more than two participants , and stop trying to invent non-physical solutions like removing acceleration.

I don't agree with any of that. Understanding relativity is about understanding the nature of spacetime. And, in particular, spacetime distances (proper time) along any timelike worldline. That understanding allows one to handle problems where one, both or neither worldline involves acceleration. The acceleration/no acceleration view is a black and white view that only covers a few simple cases and cannot be generalised.

Putting acceleration at the heart of spacetime physics is a false step.

 

[Ibix]

Introducing a third participant on the inbound leg of the journey in order to record the traveling twins clocks is not a valid step in this thought experiment.

It's perfectly fine. The outbound leg is done by one person, who reports their clock reading to the (different) inbound traveller when they meet. The inbound traveller starts their clock with that setting, and then compares the final time to the stay-at home when they meet. The point is to compare elapsed times along different paths, not get bogged down in the details of how the paths are defined.

You personally may find this more confusing, but there's nothing wrong with it. A triangle drawn with three straight strokes of a pen is just as much a triangle as one drawn with one straight stroke and one > shaped stroke. The geometry is exactly the same.

stop trying to invent non-physical solutions

Aside from the bit with the inertialess drive, which is indeed non-physical, acceleration-free versions of the twin paradox are fine. They're just slightly mis-named because you need three players.

 

[Dale]

The twin paradox is a thought experiment involving two participants. One twin accelerates and the other does not.

It's perfectly fine.

I have to admit that my feeling is somewhat in between. The triplet version is physically and mathematically fine, but in my opinion is loses all of the paradoxical character.

The “paradox” part is not a logical contradiction, but a common misunderstanding produced by the usual pedagogy. Students learn that all frames are equivalent (instead of learning that all inertial frames are equivalent), so they see the twins as symmetric. I don’t know of anyone who sees one frame as symmetric with two frames.

So the triplet version is mathematically and physically fine, but IMO it misses the point since it loses even the false symmetry of the traditional twin paradox.