# Acceleration is not important in the twin paradox

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I'll try this again!

What's important in the twin paradox, which results in the asymmetry between the twins, is that one of the twins remains in an inertial frame while the other uses primarily two different inertial frames. It is tempting then to think that a mechanical switch between the frames somehow "causes" the relativistic effects---and further that the only way to switch frames is to accelerate---but this is not true.

This can be shown by running the experiment with 3 moving clocks, none of which needs to accelerate during the experiment.

Start with 2 passing clocks, A and B, which are each set to zero at passing. Let B travel some distance at velocity v, where it passes clock C traveling in the opposite direction at the same speed, and have C set its clock to match B's as they pass. When C and A meet they'll record the same difference in proper time as if the experiment was run with clock B instantly turning around as it passes clock C.

What this shows is that the effect of time dilation does not depend on physically turning around, it only depends on what happens while in the various inertial frames.

If B and C pass at a distance of 0, their meeting can be considered a single event that is simultaneous according to all observers. However at that moment, their different frames are important (not any mechanical effect of switching anything between their frames) as they have very different measures of simultaneity relative to A. C will have measured A's aging as greater than B has measured A's aging, even though they are at the same place at the moment of their passing.

Because there is no difference in relativistic effects whether we abstractly switch from B to C, or physically cause B to turn around, it cannot be said that the mechanics of turning around has any physical bearing on the paradox. If we use a single traveling observer, then the resulting frame switch is what's important, but any acceleration used to cause the frame switch should not be confused as being the cause of the relativistic effects.

Edited by md65536

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I think you are missing the point here.

The twin paradox has three elements :

1. Both twins start off at rest in the same frame of reference

2. One twin remains at rest throughout the experiment, the other twin undergoes phases of acceleration and uniform relative motion along a closed path which eventually returns him to the first ( stationary ) twin

3. The twins reunite once again at rest in the same frame of reference, and find that their clocks do not agree

Let us take a look at their proper times; in general terms proper time is defined as the arc length of an observer's worldline, which is

$\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{g_{\mu \nu }dx^{\mu }dx^{\nu }}}$

where C is the path taken.

Without even needing any specific figures we can immediately evaluate what happens for our twins; the stationary twin travels only through time and does not experience acceleration, so his proper time is simply

$\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{\eta _{\mu \nu }dx^{\mu }dx^{\nu }}=\int_{C}dt}$

On the other hand, the travelling twin moves through time and space, and does experience acceleration; his proper time is therefore

$\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{g_{\mu \nu }dx^{\mu }dx^{\nu }}=\int_{C}\sqrt{g_{00}dt^2-\sum_{i=1}^{3}g_{ii}(dx^{i})^2}}$

which is always less than the stationary twin. Due to the equivalence principle the metric tensor in the above expression encapsulates the acceleration information. This shows us that stationary observers not subject to acceleration always experience the longest proper time.

It remains to be noted that trying to consider this scenario without acceleration is not physically meaningful, because the travelling observer would not be able to return to his stationary twin, so their clocks could not be compared in any meaningful way. Bear in mind that any deviation from uniform relative motion always involves acceleration.

Edited by Markus Hanke
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I do get the point of the twin paradox and am making a further point. The "simplest" version of the experiment is set up to show the effects of time dilation in a humanly understandable situation. It does not mean however that ALL OF the details of the simplest version are in any way essential. Some are there just to make it easier to comprehend.

1) The twins don't have to start at rest.

3) They don't have to end at rest.

2) Forcing one twin to turn around is no different than using 2 observers who each implement one leg of the traveling twin's trip.

Here's another way of putting this:

Suppose that A, B, and C are all inertial observers.

Let there be an event E1 where A and B are coincident, a later event E2 where B and C are coincident, and a later event E3 where A and C are coincident.

(We could run this experiment in parallel with a normal twin experiment, where E1 is "twins part", E2 is "traveler instantly turns around", and E3 is "twins reunite".)

Even if no clocks are synchronized and no information is passed between observers, anyone who calculates A's proper time between E1 and E3 will find it is greater than the sum of B's proper time between E1 and E2 plus C's proper time between E2 and E3.

This is no longer exactly the twin paradox, but it uses the exact same time dilation amounts, and has the "switch between inertial frames" without needing to switch anything physical.

The paradox is clearly there in the modified experiment. It is false to claim that any missing aspects (eg. having one observer physically turn around) are the "reason" for the paradox, when the exact same effects can be found without those aspects.

In other words, if we modify the experiment to remove any non-essential aspects, and get the same exact result, then those aspects are not the reason for the effect. I don't think it is possible to fully understand an experiment if one is unable to separate the material and immaterial aspects of it.

Edited by md65536
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I think you are missing the point here.

The twin paradox has three elements :

1. Both twins start off at rest in the same frame of reference

2. One twin remains at rest throughout the experiment, the other twin undergoes phases of acceleration and uniform relative motion along a closed path which eventually returns him to the first ( stationary ) twin

3. The twins reunite once again at rest in the same frame of reference, and find that their clocks do not agree

Let us take a look at their proper times; in general terms proper time is defined as the arc length of an observer's worldline, which is

$\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{g_{\mu \nu }dx^{\mu }dx^{\nu }}}$

where C is the path taken.

Without even needing any specific figures we can immediately evaluate what happens for our twins; the stationary twin travels only through time and does not experience acceleration, so his proper time is simply

$\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{\eta _{\mu \nu }dx^{\mu }dx^{\nu }}=\int_{C}dt}$

On the other hand, the travelling twin moves through time and space, and does experience acceleration; his proper time is therefore

$\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{g_{\mu \nu }dx^{\mu }dx^{\nu }}=\int_{C}\sqrt{g_{00}dt^2-\sum_{i=1}^{3}g_{ii}(dx^{i})^2}}$

which is always less than the stationary twin. Due to the equivalence principle the metric tensor in the above expression encapsulates the acceleration information. This shows us that stationary observers not subject to acceleration always experience the longest proper time.

It remains to be noted that trying to consider this scenario without acceleration is not physically meaningful, because the travelling observer would not be able to return to his stationary twin, so their clocks could not be compared in any meaningful way. Bear in mind that any deviation from uniform relative motion always involves acceleration.

You do realize that the OP outlines a perfectly possible (and commonly used to illustrate this exact point) version of the twin experiment that involves no acceleration. Saying that it's not physically meaningful to describe the experiment without acceleration doesn't make a whole lot of sense when you're responding to a physically meaningful version of the experiment that doesn't involve acceleration.
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You do realize that the OP outlines a perfectly possible (and commonly used to illustrate this exact point) version of the twin experiment that involves no acceleration. Saying that it's not physically meaningful to describe the experiment without acceleration doesn't make a whole lot of sense when you're responding to a physically meaningful version of the experiment that doesn't involve acceleration.

If there is only uniform relative motion between the three reference frames, then the frames are symmetric, and the clocks can't disagree, because you can exchange any two of them via Lorentz transformations without effecting the physical outcome. Then what is the point of the exercise ? The "twin paradox" is specifically designed to illustrate what happens when frames are not symmetric; if you eliminate acceleration you eliminate the asymmetry. It is a trivial matter to show mathematically that there will not be any "disagreement" between clocks if there is no acceleration.

Can the OP elaborate exactly what it is you are trying to show ?

it cannot be said that the mechanics of turning around has any physical bearing on the paradox

That is plain wrong. "Turning around" is equivalent to the presence of acceleration, which of course renders the frames asymmetric, leading to the clocks "disagreeing" at the end of the experiment. That is definitely physically relevant ! Without the turning around symmetry is maintained ( provided there is no other source of acceleration ), and the clocks remain in agreement.

Edited by Markus Hanke
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If there is only uniform relative motion between the three reference frames, then the frames are symmetric, and the clocks can't disagree, because you can exchange any two of them via Lorentz transformations without effecting the physical outcome. Then what is the point of the exercise ? The "twin paradox" is specifically designed to illustrate what happens when frames are not symmetric; if you eliminate acceleration you eliminate the asymmetry. It is a trivial matter to show mathematically that there will not be any "disagreement" between clocks if there is no acceleration.

Can the OP elaborate exactly what it is you are trying to show ?

What I'm trying to show is that some of the aspects that some people think are important in the twin paradox---even to the point that they "cause" the paradox---are not essential to the phenomenon.

An example from your quote here is "if you eliminate acceleration you eliminate the asymmetry." Maybe that's true in the typical twin experiment, but it is not true in general because there are other ways to achieve asymmetry.

I elaborated in post #3 at "Here's another way of putting this", using a 3-observer setup where there is neither symmetry nor acceleration.

That is plain wrong. "Turning around" is equivalent to the presence of acceleration, which of course renders the frames asymmetric, leading to the clocks "disagreeing" at the end of the experiment. That is definitely physically relevant ! Without the turning around symmetry is maintained ( provided there is no other source of acceleration ), and the clocks remain in agreement.

Yes, but the turn around doesn't cause the time dilation, it causes the frame switch, which doesn't actually require that a physical object switch frames.

In the experiment, using a single traveling observer, having the observer switch frames is essential, and so acceleration is essential. However that's essential to implement the experiment, not the effect. If you change the experiment, you can reproduce the same effect without the non-essential aspects.

Here's another example of my point:

You might say the twins must come together at rest to meaningfully compare clocks. And that might be true in some sense but comparing clocks is not essential to the effect, it just makes the experiment reliable. For example, if you run the experiment in two parts, where twin A is stationary for a day, then a year later you have twin B travel at near c and return, B will age less than a day per one Earth day, just as if the two twins ran the experiment together and actually compared their ages. You don't have to have the twins actually together, IF you can reliably measure proper times in a properly run experiment.

So poor reasoning might go like this: The clocks have to start synchronized, and that's what having twins accomplishes. If the clocks aren't sync'd, there's no meaningful measurement that can be made between the clocks. Therefore the experiment doesn't work except for twins. Then you can keep going: There must be some physical difference between twins and non-twins. So I will try to explain/figure out the twin paradox by speculating that twins are somehow "entangled", and that causes time dilation. This is an extreme exaggeration but it's the type of reasoning that leads to pseudoscience, astrology etc. The mix between causation and correlation. In the twin paradox, turning around correlates with the effect, but it doesn't physically cause it.

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I elaborated in post #3 at "Here's another way of putting this", using a 3-observer setup where there is neither symmetry nor acceleration.

Can you show why these frames are not symmetric, as you say ?

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Ok, let me try to restate the initial example in more concrete terms, and see if that helps.

You hav three clocks. Clock one is at rest in our solar system. Clock two is traveling at an appreciable percentage of the speed of light from our solar system toward Alpha Centauri. Clock three is traveling from Alpha Centauri toward the solar system at an appreciable percentage of the speed of light. All three clocks are identical.

Clock two syncs with clock one as it leaves the solar system. At some arbitrary point A, clock two passes clock three, at which point they sync with each other. When clock three reaches clock one, it will show the combined elapsed time that it took for clock two to travel to point A from the solar system and clock three to travel from point A to the solar system. When clock one and three are brought together, this total time will be less than that experienced by clock one, just as if clock two had gone to point A and then returned itself. At no point do any of the clocks accelerate.

md65536, did I get that right?

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Well, this is a common fallacy believed by generations of naive students. There is acceleration in the scenario that you depicted (it is a well known scenario). Except: IT IS HIDDEN! When B and C pass each other you need to transfer the information (clock reading) from B to C, i.e. you need to JUMP from the frame comoving with B to the frame comoving with C. As Markus pointed out, acceleration is key in creating the necessary conditions for the paradox to exist.

I'll try this again!

Let B travel some distance at velocity v, where it passes clock C traveling in the opposite direction at the same speed, and have C set its clock to match B's as they pass.

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To be clearer, a better title for this topic might be "Acceleration is not essential in the twin paradox effect".

md65536, did I get that right?

Yes, that all sounds right.

If clocks are "synchronized" does that mean they remain synchronized for some non-zero duration? I avoided using the word because I thought that, though from your description it should be clear that the synchronization is only at a single instant.

Well, this is a common fallacy believed by generations of naive students. There is acceleration in the scenario that you depicted (it is a well known scenario). Except: IT IS HIDDEN! When B and C pass each other you need to transfer the information (clock reading) from B to C, i.e. you need to JUMP from the frame comoving with B to the frame comoving with C. As Markus pointed out, acceleration is key in creating the necessary conditions for the paradox to exist.

No, the clock reading doesn't have to be transferred from B to C.

All 3 clocks will measure proper time correctly, whether they're ever synchronized or not. The 3 observers don't even have to know about each other, they just have to properly time the events (at the right locations).

For example, the 3 proper times could be measured independently, and then later the results are sent to another observer to compare. These could be sent by light signal, and I don't think you can say that the photons switch frames or accelerate in order to do this??

Can you show why these frames are not symmetric, as you say ?

Do you mean that because no observer changes frame, each pair of observers has a reciprocal relationship, or whatever? That's true, but...

To elaborate further on where the asymmetry might be "seen", suppose that in the example observer A (clock one) ages 4 years before meeting C, while B (clock two) ages 1 year before meeting C, and C (clock 3) ages 1 year from there until meeting A (gamma=2).

Then, when B and C meet, B can say "I have aged 1 year, while A has aged 0.5 years," but C will say "I agree B has aged 1 year, but A has aged 3.5 years since B left it."

The difference in their measurements is the same as the difference in measurements before and after a twin's instantaneous turnaround.

Edited by md65536
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You hav three clocks. Clock one is at rest in our solar system. Clock two is traveling at an appreciable percentage of the speed of light from our solar system toward Alpha Centauri. Clock three is traveling from Alpha Centauri toward the solar system at an appreciable percentage of the speed of light. All three clocks are identical.

These three clocks are all in uniform unaccelerated motion relative to each other, so all three of these clocks represents inertial frames, and they are thus perfectly symmetric. It is not possible to get any disagreements in proper times between the clocks. If you look over on the other thread "Paradox in Relativity", you will find the mathematical proof of this there - I have provided it there in a different context, but it holds nonetheless.

Also, in order to compare proper times, you have to either bring the clocks to rest in the same frame of reference ( which involves acceleration ), or you calculate the outcomes using relativity of simultaneity. You cannot, however, directly compare clock readings so long as they are in relativistic motion relative to each other.

So once again - acceleration is a crucial element in the twin paradox scenario. If it is absent then you will get no disagreements between the clocks.

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(...)You cannot, however, directly compare clock readings so long as they are in relativistic motion relative to each other. (...)

Yes.

And that happens 3 times in the paradox.

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So once again - acceleration is a crucial element in the twin paradox scenario. If it is absent then you will get no disagreements between the clocks.

There are 3 events: E1: "A and B pass" (twins part), E2: "B and C pass" (equivalent to turnaround), and E3: "C and A pass" (equivalent to twins reunite).

No pair of clocks passes through the same two events. There is nothing to disagree on. There are 3 clocks measuring 3 individual proper times; everyone agrees on all 3 proper times measured.

In the example I've used, with gamma=2, A measures the proper time between E1 and E3 as 4 units, B measures E1 to E2 as 1 unit, and C measures E2 to E3 as 1 unit. The sum of B's and C's proper times is 2 units, corresponding to a traveling twin aging half as much as an inertial twin.

Where is the error in that?

Relativity of simultaneity doesn't come into consideration, because we're not comparing the simultaneity of any separated events. We're treating "B passes C" and "C passes B" as one event, ie. simultaneous for all observers, which is acceptable if they pass at a negligible distance.

Yes.

And that happens 3 times in the paradox.

In the example, the proper time measured by A is 4 units,

the proper time measured by B is 1,

and the proper time measured by C is 1.

How is it not possible to compare these values? Eg. 4 > 1, all agree.

-------

Perhaps to HELP explain this version of the paradox, or perhaps it will only make it more complicated, note that A ages half as fast according to B and C, and vice versa.

So according to A, while A ages 4 units, C ages only 2.

According to C, A ages 4 over the experiment while C ages 8.

This is perfectly consistent with SR. A and C begin the experiment separated and with a relative velocity, so they don't agree on the timing of the start of the experiment (however we don't care, since it doesn't matter how much C has aged before it passes B).

Using composition of velocities, the Lorentz factor between B and C is 7.

This is consistent: C measures A aging 3.5 units between events E1 and E2, while it ages 7 units itself, and thus it measures B aging 1 unit -- in agreement with the proper time that B measures between E1 and E2.

Edited by md65536
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No, the clock reading doesn't have to be transferred from B to C.

False, you forgot what you wrote earlier:

Let B travel some distance at velocity v, where it passes clock C

traveling in the opposite direction at the same speed, and have C set

its clock to match B's as they pass.

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Let us take a look at their proper times; in general terms proper time is defined as the arc length of an observer's worldline, which is

$\displaystyle{\tau =\int_{C}d\tau =\int_{C}\sqrt{g_{\mu \nu }dx^{\mu }dx^{\nu }}}$

where C is the path taken.

Integration is essentially an infinite sum of parts. The parts are independent. In this case it is the sum of infinitesimal proper times along the path of the clock. Those parts are independent, meaning a clock has no memory. You could swap in another clock for part of the path, and the two clocks would behave identically over that part.

If you split a path into N parts, the proper time of one clock following all N parts is the same as the sum of the proper times of N clocks each following one of the N parts.

False, you forgot what you wrote earlier:

That also doesn't matter in the effect, as per:

Suppose that A, B, and C are all inertial observers.

Let there be an event E1 where A and B are coincident, a later event E2 where B and C are coincident, and a later event E3 where A and C are coincident.

(We could run this experiment in parallel with a normal twin experiment, where E1 is "twins part", E2 is "traveler instantly turns around", and E3 is "twins reunite".)

Even if no clocks are synchronized and no information is passed between observers, anyone who calculates A's proper time between E1 and E3 will find it is greater than the sum of B's proper time between E1 and E2 plus C's proper time between E2 and E3.

No information is passed here.

The proper times between events measured by any of the observers will not change based on whether or not B sends any information to C.

A clock will age 1 year per year, regardless of when it was last set, or what value it was set to.

Edited by md65536
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md65536,

I see you are hell bent on promovating your crackpot ideas. As always, it is useless trying to convince a crackpot of his wrong ways. Bye.

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md65536,

I see you are hell bent on promovating your crackpot ideas. As always, it is useless trying to convince a crackpot of his wrong ways. Bye.

I... am really not understanding why what he said is crack pottery, but would actually like to know if it is. If you can't convince him, could you explain it to me?

For me it boils down to a few points that seem fine to me, but would muck things up if I'm missing something.

It seems like two clocks passing each other would agree on what time the other one reads at the moment of passing. It would also seem that any other observer would agree on the reading of each clock at the moment that they past each other. Is there anything wrong there?

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I... am really not understanding why what he said is crack pottery, but would actually like to know if it is. If you can't convince him, could you explain it to me?

For me it boils down to a few points that seem fine to me, but would muck things up if I'm missing something.

It seems like two clocks passing each other would agree on what time the other one reads at the moment of passing. It would also seem that any other observer would agree on the reading of each clock at the moment that they past each other. Is there anything wrong there?

No, he needs to transfer the reading on clock B to clock C. He even admits to that in the OP (it is a well known scenario, he copied it from somewhere without fully understanding it), nevertheless he later denies that such a transfer of information from the frame of B to the frame of C is necessary.

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No, he needs to transfer the reading on clock B to clock C. He even admits to that in the OP (it is a well known scenario, he copied it from somewhere without fully understanding it), nevertheless he later denies that such a transfer of information from the frame of B to the frame of C is necessary.

Well, in the way that he first formulated it, with setting the clocks to each others' time when they passed, that happened, but it's not actually a necessary part of the experiment and could be removed as well, which I think he explained.

Everyone will agree on what time the two passing clocks read at the moment they pass each other. The passing clocks do not then need to read each others' time or alter their own in response. A third party can observe the entire experiment by noting the time on the "rest" clock and the clock departing it at the moment of departure, the time on each of the passing clocks at the moment of passing, and the time on the "rest" clock and the clock that returns to it at the moment they pass.

At no point do any of the clocks accelerate nor di they exchange information, and the third party's observations are based only on information that is agreed upon in all frames of reference. That is, the time on each pair of clocks at the three events where two of the three clocks are in proximity.

As far as I'm aware, this experiment is not physically impossible to perform, and would yield the same difference in elapsed time as if a twin had gone out and returned.

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Then we are done. He denied it, that was my point.

Well, in the way that he first formulated it, with setting the clocks to each others' time when they passed, that happened,

Everyone will agree on what time the two passing clocks read at the moment they pass each other.

Incorrect, there is an infinity of observers that will actually disagree that B and C show the same time. Only one observer agrees, that observer is the one that measures B and C to move with equal and opposite velocities wrt him.

Edited by xyzt
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He denied that it was necessary and then explained why it wasn't. He didn't deny that it was included in what he first said, but I'm still not actually clear on why it's even relevant.

I know that different frames will not necessarily agree on the order of events, but they do agree on what happens at the event. They'll be able to say that at some point "Clock two reads 1 minute and has not passed clock three" and they'll agree that at some point "clock two reads 2 minutes and has passed clock three." They may not agree on the duration of that "clock two minute" but no one is going to see clock two reading 10 minutes before it passes clock three.

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He denied that it was necessary and then explained why it wasn't. He didn't deny that it was included in what he first said, but I'm still not actually clear on why it's even relevant.

I know that different frames will not necessarily agree on the order of events, but they do agree on what happens at the event. They'll be able to say that at some point "Clock two reads 1 minute and has not passed clock three" and they'll agree that at some point "clock two reads 2 minutes and has passed clock three." They may not agree on the duration of that "clock two minute" but no one is going to see clock two reading 10 minutes before it passes clock three.

This is not what you said initially, you claimed that "Everyone will agree on what time the two passing clocks read at the moment they pass each other." When I showed you that the claim is false, you changed your tune. Besides, now you are talking about "order of events" and this exercise is about elapsed proper time. I cannot teach you, I am sorry.

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This is not what you said initially, you claimed that "Everyone will agree on what time the two passing clocks read at the moment they pass each other." When I showed you that the claim is false, you changed your tune.

Another way to put it is that the relative simultaneity of an event at B and one at C depends on gamma (ie velocity) and on separation distance. If the separation distance * gamma is negligible, the disagreement of simultaneity is negligible. If B and C are at the same location as an event at the moment of the event occurs, everyone will agree on that.

Besides, now you are talking about "order of events" and this exercise is about elapsed proper time. I cannot teach you, I am sorry.

We could defer to an expert as referee perhaps?

Yes, you're correct, elapsed proper time is all that matters in this experiment. If we remove the "transfer of information" described in post #1, then I am claiming that with gamma=2, observer A measures 4 units of proper time between its two events, B measures 1 unit between its two events, and C measures 1 unit between its two events, and this is the same value calculated in a similar twin paradox experiment, using the same calculations of the Lorentz transformation. If this is wrong, where is it wrong? I'm calculating 3 proper times between 3 events; are the 3 events valid, and if so, what are the conflicting proper times that you calculate between those events?

In my opinion, claiming that transferring information from B to C changes the proper time that one of the observers measures, is a crackpot theory. Do you suggest that C's clock is invalid until it gets information from B, or that the information physically changes the clock, or other?

Edited by md65536
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Another way to put it is that the relative simultaneity of an event at B and one at C depends on gamma (ie velocity) and on separation distance.

this exercise has nothing to do with "relative simultaneity"

If the separation distance * gamma is negligible, the disagreement of simultaneity is negligible. If B and C are at the same location as an event at the moment of the event occurs, everyone will agree on that.We could defer to an expert as referee perhaps?

This exercise has nothing to do with "events", it has everything to do with "total elapsed proper time".

Yes, you're correct, elapsed proper time is all that matters in this experiment. If we remove the "transfer of information" described in post #1, then I am claiming that with gamma=2, observer A measures 4 units of proper time between its two events, B measures 1 unit between its two events, and C measures 1 unit between its two events, and this is the same value calculated in a similar twin paradox experiment, using the same calculations of the Lorentz transformation. If this is wrong, where is it wrong? I'm calculating 3 proper times between 3 events; are the 3 events valid, and if so, what are the conflicting proper times that you calculate between those events?

You are correct on this. You are incorrect on the issue of the "absence" of acceleration, the acceleration is present, underneath the skin of the"gedank'. You need to keep switching frames in order to compare the "total elapsed proper time" of the moving clocks.

In m

y opinion, claiming that transferring information from B to C changes the proper time that one of the observers measures,

I am not claiming that, you seem unable to follow the argument.

is a crackpot theory. Do you suggest that C's clock is invalid until it gets information from B, or that the information physically changes the clock, or other?

No, I am not claiming any of this, you seem unable to follow the argument.

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-----

To show that transfer of information is not important, we can further modify the experiment by putting markers (flags, or planets) at the location where A and B meet, and where B and C meet.

Then B and C are measuring the proper time between passing the two markers, which they could do reliably. (Perhaps observing the markers counts as transfer of information, but it doesn't matter; whether an event is observed or not does not change the prediction of the event.)

Observer A is measuring the proper time between B passing, and then C passing. Without those events, there is no other meaning to the proper time that A is measuring here. However, again, SR does not care whether an event is observed, in its prediction of timing.

So all these details that I'm arguing are unnecessary for the EFFECT of twin paradox time dilation, such as synchronizing clocks, transferring information, observing events, these are important for RELIABLY MEASURING an experiment to confirm the twin paradox... but the effect still occurs exactly the same whether it is measured properly or not.

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