Homework Helper
When you see the title of this thread, I'm sure you will either be bored or excited because this is one of those famous 'paradoxes' in SR.
So I read the book by Schutz on the part of explanation behind the twin paradox. There he gave an example of two sisters Diana and Artemis involved in the experiment. In particular Diana leaves her sister on earth with a rocket travelling with close-to-light velocity. Then at certain moment Diana turns around and go back to the earth. The question is, as is very obvious already, which one is older upon the reunion.
The author explains it using the spacetime diagram of Artemis, the sister which stays on earth. I can understand the explanation that the reason Diana is still much younger than Artemis is that because they move with very high speed with each other and Diana undergoes abrupt change of direction. But then one can argue why we hadn't used the spacetime diagram of Diana in the rocket? This could have resulted in the opposite conclusion. Up to now I'm still unsure why we should use Artemis's spacetime as the reference, but I think it's because Artemis is inertial while Diana is not, is that the true reason? I am not sure with my own conclusion because it seems that the author doesn't mention that. He only puts a stress on the explanation why there can be such a large difference in the ages, not why we hadn't use the other sister's spacetime to draw the diagram. I think the reason why this gedanken has become a 'paradox' is that because there can be an ambiguity in the final conclusion of who's older when different references are considered, not because there is a large age difference which the author emphasized. Even when the rocket moves with everyday velocity, Artemis is still the older one at the reunion.

Nugatory
Mentor
Up to now I'm still unsure why we should use Artemis's spacetime as the reference, but I think it's because Artemis is inertial while Diana is not, is that the true reason?

No. You can use any spacetime diagram and you will get the same answer: Artemis's path through spacetime between the point of departure and reunion is "longer" (covers more proper time) than Diana's so more time passes for Artemis and she ages more. It's mathematically convenient to use Artemis's spacetime diagram, but we could just as easily use any other: for example, the one in which Artemis is moving and Diana is at rest on the outbound leg of her trip, or the one in Artemis is moving and Diana is at rest on the inbound leg of her trip, or one in which both twins are moving. We'll get the same answer in all cases.

The calculations are always simpler and more straightforward if you use (close to) an inertial frame. That is why the calculations are normally presented from the stay-at-home frame of reference. The traveling twin makes the trip using two different frames (at the very least). This is analogous to the case where the shortest line between two points is the straight line. Any other path - such as a line from A to a distant C and then from C to B will be longer. These are the cases where there is less aging. The greater amount of space traversed amounts (seemingly paradoxically) to less time traversed. Right?

Homework Helper
You can use any spacetime diagram and you will get the same answer.
I'm just guessing the reason there is asymmetry between Artemis and Diana is because one stays inertial but the other ever undergoes acceleration?

The greater amount of space traversed amounts (seemingly paradoxically) to less time traversed.
From where?

I'm just guessing the reason there is asymmetry between Artemis and Diana is because one stays inertial but the other ever undergoes acceleration?

From where?

As measured from the inertial twin, after the non-inertial twin (who has traveled in 2 or more inertial frames) returns home.

Nugatory
Mentor
I'm just guessing the reason there is asymmetry between Artemis and Diana is because one stays inertial but the other ever undergoes acceleration?[

No, that's not right. There is asymmetry between them, but the asymmetry is that they take different paths through spacetime. Acceleration only comes into the picture because we have to do something to at least one of the twins to set them on divergent paths through spacetime and accelerating them is one of the easier ways of doing that.

Nugatory
Mentor
If you are not already familiar with this FAQ, you will want to read it. Much of what you're asking about here is already covered by the FAQ.

To add to the above posts :

One of the things that repeatedly crops up, is that, SR treats all inertial frames as equal. This causes an apparent problem ... where for example, two objects are spaced apart on earth (say at the poles) and both are put in motion by the same acceleration boost so after a brief period of acceleration they travel toward each other at same constant velocity wrt the earth. We can put them on parallel tracks that follow the curvature of the earth so they pass one another in the vicinity of the equator. If they are trains we can put clocks in the engine and caboose of both trains and since each train is entitled to consider itself as an inertial frame, each will measure the clocks in the other train as running slow. Obviously, when the trains are stopped by equally applied forces, and the clocks are brought together for comparison, they will both read the same because the situation is symmetrical.

The reality of real time dilation can only be experienced if the spacetime intervals are different. In the case of the symmetrical trains, each has experienced the same space increment and the same time increment, As between two objects in relative uniform motion, the spacetime interval is invariant - thus if one clock travels a greater distance in space than the other, the other must travel a different distance in time (a simply Pythagorean relationship). When a 1st clock is taken as inertial, it travels zero distance in space during the experiment - so the time logged by the first clock must be large enough to account for the combined space and time distances traveled by the relatively moving second clock. In the symmetrical train thought experiment, each train travels the same space increment relative to the earth and the clocks log the same time. The frequent dismissal of an earnest question with: "each clock runs slower than the other" obviously leads to be embellished upon to avoid confusion.

PeterDonis
Mentor
2020 Award
SR treats all inertial frames as equal

SR does not apply in the presence of gravity. Your trains on Earth scenario involves gravity, so SR can't be applied. That's why SR scenarios usually involve spaceships, so they can be placed far out in empty space where gravity is negligible.

The reality of real time dilation can only be experienced if the spacetime intervals are different for two observers between the same pair of events.

The bolded part that I added is crucial; to make an invariant comparison between two observers' clocks, the two observers need to start out together, then separate, and then come back together again (as in the standard twin paradox). If they just keep moving away from each other, there is no invariant way to compare their clocks.

"The bolded part that I added is crucial; to make an invariant comparison between two observers' clocks, the two observers need to start out together, then separate, and then come back together again (as in the standard twin paradox). If they just keep moving away from each other, there is no invariant way to compare their clocks"

They don't need to start at the same place nor do they need to return to the same place - all clocks on earth can by synchronized - North pole, South pole, Pasadena, whatever. Likewise any other clock that is not moving wrt the earth reference frame can by synchronized. Obviously there is a bit of curvature in using the earth as bases - but that does not invalidate the thought experiment - .... any 2 points far removed from one another can be used as starting points of a flat space and as long as they are not moving with respect to each other - in theory, they can be synced - likewise there is no need to bring clocks together to verify the amount of time dilation at the end of the experiment - the clock on Alpha that is running in sync with the earth clock is a perfectly good extension of the reference frame. We can imagine another planet Beta twice the distance from alpha in the same direction - and when the one way spaceship arrives at Beta it reads the Beta clock and the traveling twin immediately knows his brother back on earth has long since died of old age. The clock on Beta in sync with earth and alpha will read the same as though the traveler turned around at Alpha and returned home.

Ibix
2020 Award
Likewise any other clock that is not moving wrt the earth reference frame can by synchronized.
Not without choosing a synchronisation convention. Which means that any time dilation effect you measure is just a matter of the choice of synchronisation.

Response to 11:

Simply use the Einstein convention for syncing all clocks at rest in the same inertial frame. If you want to sync with a moving clock on the fly, e.g., to eliminate the error introduced during the brief acceleration phase, the traveling clock is first boosted to its cruising speed and its time is then set equal to a local clock as the traveling clock passes close-by. .

[..] Up to now I'm still unsure why we should use Artemis's spacetime as the reference, but I think it's because Artemis is inertial while Diana is not, is that the true reason? [..].
Apparently I understand your question different than Nugatory, for I answer Yes.
The Laws of SR are meant for inertial frames, applying it to an accelerating frame leads to error - it can be put as simple as that. That's also how Einstein replied in his 1918 paper:
"according to the special theory of relativity the coordinate systems K and K' are by no means equivalent systems. Indeed this theory asserts only the equivalence of all Galilean (unaccelerated) coordinate systems, that is, coordinate systems relative to which sufficiently isolated, material points move in straight lines and uniformly. K is such a coordinate system, but not the system K', that is accelerated from time to time."

And theNewsnet physics faq puts it like this:
"Perhaps we can make short work of the "travelling Earth" argument. SR does not declare that all frames of reference are equivalent, only so-called inertial frames. Stella's frame is not inertial while she is accelerating".
Also this analysis in the same FAQ may be useful: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gap.html

MacroMeasurer
ShayanJ
Gold Member
Eric Gourgoulhon's treatment of the paradox in his book "Special Relativity in general frames" is really nice and I think it will clear your mind on this. In fact the whole book is really nice and, in my idea, is a must-read for every one who is serious in learning SR but only after reading some more elementary books on SR.

PeterDonis
Mentor
2020 Award
They don't need to start at the same place nor do they need to return to the same place - all clocks on earth can by synchronized - North pole, South pole, Pasadena, whatever. Likewise any other clock that is not moving wrt the earth reference frame can by synchronized.

Not in the sense of SR; as I said, SR is not valid in the presence of gravity, and there is gravity on the Earth. Also, the Earth is rotating, which would prevent clocks on Earth from being synchronized in the sense of SR even in the absence of gravity.

any 2 points far removed from one another can be used as starting points of a flat space

No, they can't. The surface of the Earth is not a flat space.

there is no need to bring clocks together to verify the amount of time dilation at the end of the experiment - the clock on Alpha that is running in sync with the earth clock is a perfectly good extension of the reference frame.

Note the phrase I bolded. If the clocks are not brought together at the start and end of the experiment, then the time dilation is frame-dependent; it changes if you change frames. Frame-dependent quantities do not have physical meaning; only invariants--things that are independent of the choice of frame--do.

(Also, this is a different scenario from the Earth one, correct? Here Alpha and Beta are assumed to be points in a flat spacetime. If not--if gravity is present--then, as I said, SR is not valid and there are no global inertial frames.)

PeterDonis
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2020 Award
Simply use the Einstein convention for syncing all clocks at rest in the same inertial frame.

Clocks at different points on the Earth's surface are not at rest in the same inertial frame.

Nugatory
Mentor
Apparently I understand your question different than Nugatory, for I answer Yes.
The Laws of SR are meant for inertial frames, applying it to an accelerating frame leads to error - it can be put as simple as that.

That is simply not true (unless, perhaps, you're using an unreasonably restrictive definition of "the laws of SR").
It is true that misapplying the laws of SR will lead to error, and it is true that it's easier to apply them correctly when you're working with an inertial frame - which is why I said that it's convenient to use Artemis's frame - it's the inertial frame that's sitting under our nose just begging to be used.

But it's certainly not necessary to use such a frame. I can calculate the differential ageing of the two twins using the frame of someone passing by and accelerating at a constant rate, as long as both twins' worldlines remain outside the Rindler horizon, and I'll get the right answer. Or I can find a coordinate transformation such that Diana is at rest and use that frame to calculate the differential ageing, and I'll get the right answer.

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Peter, you are obscuring the thought experiment with nitpicking objections .....of course, in a real experiment there will always be errors and practical problems with altitudes and curvature and precise synchronization - .....these are totally insignificant to the objective of simplifying the round trip experiment to two one-way journeys. The goal should be conceptual clarification. One of the reasons the Twin paradox comes up again and again is that the experts don't agree - Those who think GR is required to resolve the TP, are in good company. Three Nobel prize winners Einstein, Born and Feynman + a host of others) all had the same sentiments, but they are wrong. The round trip is most simply analyzed by two one way trips, each corrected for the time accumulated during the brief acceleration periods. Einstein's original paper "On the Electrodynamics of Moving Bodies - Part IV" was and is still a correct description of Einstein's revelation that "time is not the same for all observers." There is no need to muck up the tutorial with confusing statements which portray it as a General Relativity problem by dwelling upon the infinitesimal turn around error involved in making a round trip flight to a distant target. Sadly, Einstein himself is the cause of the confusion - specifically his 1918 paper. In a long trip to a distant star, a time difference of 1000 years is not going to be affected by initial sync errors or the fact that the path geometry is not perfectly straight.

Ibix
2020 Award
Response to 11:

Simply use the Einstein convention for syncing all clocks at rest in the same inertial frame.
Exactly - a convention. So your results will depend on that convention, and will not be universally agreed upon.

One of the reasons the Twin paradox comes up again and again is that the experts don't agree - Those who think GR is required to resolve the TP, are in good company. Three Nobel prize winners Einstein, Born and Feynman + a host of others) all had the same sentiments, but they are wrong. .
I do not believe this. The elapsed time on any clock is the proper distance it has travelled and this can be calculated if the worldlines and metric are known. This is true in GR and SR. It resolves any possible paradoxes.

The round trip is most simply analyzed by two one way trips, each corrected for the time accumulated during the brief acceleration periods.
No it is not. If the proper length of the worldlines is calculated the elapsed times are known. Why mess about with calculations that depend on which twin does what ?

Frankly, I think you are utterly confused. There is no paradox to be resolved...

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[...]
But then one can argue why we hadn't used the spacetime diagram of Diana in the rocket?
[...]

It IS possible to work the problem from Diana's "perspective". You can indeed ask the question: "At each instant of Diana's life during the trip, what does she conclude about Artemis' current age at that instant?". You can get that answer (for each instant "t" in Diana's life) by using the "co-moving inertial frame" at that instant. The Lorentz equations can be used to relate that "co-moving inertial frame" and Artemis' inertial frame ... i.e., the Lorentz equations can tell you what Artemis' current age is when Diana's age is "t", ACCORDING TO DIANA. That capitalized phrase is needed, because Artemis will NOT agree ... his conclusions about the relationship between their ages during (most of) the trip are quite different from Diana's conclusions. The whole process for carrying out the above procedure is well described in the webpage

The short answer is that, according to Diana, Artemis ages slowly during Diana's constant-speed outbound and inbound portions of her trip, but during her accelerated turnaround, Artemis ages VERY quickly. And the TOTAL of those three components of Artemis' aging is such that Artemis' and Diana's "perspectives" both agree at the end of the trip (but they generally disagree during the trip).

One of the reasons the Twin paradox comes up again and again is that the experts don't agree - Those who think GR is required to resolve the TP, are in good company.

I'm not finding a link to this right now - but there is a version of this which uses three clocks and two spaceships.

Clock A stays at home.
Clock B was in a ship in motion before the experiment began. At the moment clock B passes very close to clock A, information from A is sent to B and B becomes a synched "twin" of A.
Later, at some distance from A, B passes clock C in a ship which has been heading toward A since before the experiment began. At the moment they pass, information is sent from B to C and C becomes a synched twin of B, but this time, heading back toward A without any forces or accelerations. When clock C arrives back at A, less time has passed on the combination of B/C outgoing+return than the time which passed at A.

Mentz114
I'm not finding a link to this right now - but there is a version of this which uses three clocks and two spaceships.

Clock A stays at home
...
I like that a lot so I made s-t diagram. The travelling phases add up to 15 and the at home is 16.25. No accelerations.

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One of the reasons the Twin paradox comes up again and again is that the experts don't agree
You are wrong and misleading. Please take the time to study this as penance. The twin "paradox" is the second simplest worked example in SR, because it involves three frames of reference; the simplest just uses two.

PeterDonis
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2020 Award
of course, in a real experiment there will always be errors and practical problems with altitudes and curvature and precise synchronization - .....these are totally insignificant to the objective of simplifying the round trip experiment to two one-way journeys.

I'm not talking about those sorts of issues. The fact that SR is not valid in the presence of gravity is not a "practical problem"; it's a fundamental limitation on the theory. You are trying to use SR in a domain in which it is not valid.

The goal should be conceptual clarification.

And trying to set what is fundamentally an SR thought experiment in a scenario where gravity is not negligible does not help with that goal; it hinders it. Why can't you just put the two twins, or two trains, or whatever, far out in empty space away from all gravitating bodies? Then you could use an actual inertial frame to do the analysis, instead of using an obviously non-inertial frame and then handwaving about how "it doesn't really matter".