Okay, this may sound quite stupid, but I am currently learning special relativity in my physics class (2301 OSU). So since both frames assume that their frame is stationary and that time in all other frames is dilated. So if Twin A stays on earth and Twin B travels at constant velocity to another planet, both can say that the other person is aging slower. But since the twin that’s traveling goes through some sort of acceleration to return back to earth, they do age slower (due to effects of general relativity according to my friend). But what if that twin could return to compare clocks without having to accelerate (I know its not possible but its just a what if situation), how would their clocks compare?

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    $\begingroup$ It is very good question! In my assumption, such world have non-zero space-time curvature, and this will affect clocks of flying twin.. $\endgroup$ – Nikita Feb 27 at 21:03
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    $\begingroup$ @Nikita you can have a cylinder with zero intrinsic curvature and as such no gravitational time dilation. However in this case distinct paths will be distinguishable by winding number (not sure how one physically measures this - but the spacetime is weird and unphysical so I'm not sure we need a physical interpretation). $\endgroup$ – jacob1729 Feb 27 at 21:59
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    $\begingroup$ You are asking about the twin paradox in a closed universe, this question, which was closed and has another marked for it as an answer should start you off right. physics.stackexchange.com/q/353216/23615 $\endgroup$ – Triatticus Feb 27 at 22:00
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    $\begingroup$ @jacob1729 can you define an "absolute" winding number for a single worldline? It looks like you need absolute velocity (at least in the compact direction) to do that $\endgroup$ – fqq Feb 27 at 22:49
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    $\begingroup$ It turns out that indeed, there is an absolute frame in the compact direction physics.stackexchange.com/a/362/18688 $\endgroup$ – fqq Feb 27 at 22:59

The time that passes on your clock between event E and event F is given by the length (in the Minkowski metric) of the path you followed to get from E to F. You can compute this length by integrating the spacetime interval along this path.

If I remain on earth while you head outward along a path that eventually brings you back to earth again without accelerating, we can take E to be the event of your departure and F to be the event of your return. You and I both traveled from E to F but along different paths. Those paths have different lengths, and therefore our clocks can show different amounts of time passed. Exactly what that difference is depends on the details of the spacetime geometry and the path you followed.

  • $\begingroup$ i dont really understand what you said, but you say that one of the clock would read a different value compared to the other?! Its so crazy to think about $\endgroup$ – Narutachi Feb 27 at 23:10
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    $\begingroup$ @Narutachi : Is it crazy to think about the fact that if you and I both drive from New York to Los Angeles along different routes, our odometers will differ? A clock is an odometer for spacetime. Two clocks that travel along different routes in spacetime will show different elapsed time for the same reason that two odometers that travel along different routes in space will show different elapsed mileage. $\endgroup$ – WillO Feb 27 at 23:43
  • $\begingroup$ that makes a lot more sense. Thanks a lot. So time is the same as space? $\endgroup$ – Narutachi Feb 28 at 17:36

The technical term for this sort of universe is a torus. It is possible for a three torus to be flat.

In such a universe you can have a set of observers that are all stationary with respect to each other, this is the same as in a standard flat universe, however, in a torus observer A is to the left of observer B who is to the left of observer ... Z who is to the left of observer A.

Now, just as in our universe, you can have multiple sets of such observers, say the ones described and another set A’, B’, ... Z’. All of the primed observers are at rest to each other and all of them measure the unprimed observers as having the same speed. Locally everything follows the standard rules of SR.

However, there is one big difference: if the unprimed observers measure the distance between A and B and add it to the distance between B and ... add it to the distance between Z and A they will get a number for the size of the universe. If the primed observers do the same procedure they will get a different number for the size of the universe. If many different sets of observers do this they will find that there is one unique set of observers for whom the size of the universe is maximum.

So although such a universe behaves locally like we expect, globally it violates the first postulate. There is a uniquely identifiable reference frame that can be singled out by physical measurements.

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    $\begingroup$ I keep reading this again and again and I still do not understand. Thanks for responding tho $\endgroup$ – Narutachi Feb 27 at 23:10
  • $\begingroup$ You are not alone. $\endgroup$ – Albert Feb 28 at 6:00
  • $\begingroup$ @Narutachi@Albert Imagine the universe is a cone. For observers whose motion make the circumference of the cone spatial, they will measure a universe of minimal size in periodicity. If the observer moves such that the spatial part will be a "diagonal", they will measure a larger size for the universe. The more inclined the larger, no maximum. $\endgroup$ – Wolphram jonny Feb 28 at 6:21
  • $\begingroup$ @Narutachi I have added an answer that elaborates on this answer. In that answer I linked to an article by a physicist called Olaf Wucknitz. In that article Wucknitz argues that understanding of the Sagnac effect carries over to understanding of closed Minkowski spacetime. In my answer I discuss the operating principle of the Sagnac effect. $\endgroup$ – Cleonis Feb 28 at 7:34

This answer elaborates on the answer given by stackexchange contributor Dale

The case of a closed Minkowski space-time is discussed by Olaf Wucknitz, in a 2004 article titled: Sagnac effect, twin paradox and space-time topology - Time and length in rotating systems and closed Minkowski space-times

Wucknitz argues that understanding of the Sagnac effect carries over to understanding of a closed Minkowski space-time.

The Sagnac effect is mostly known for its essential role in ring interferometry.

To explain the Sagnac effect I will use the case of a series of relay stations, positioned around the equator. Let's say the equator of the Earth. The number of relay stations is arbitrary, let's put it at twelve. The relay stations are in radio contact. These relay stations proceed as follows: they start two counterpropagating relay transmissions, consisting of pulses. Let's say twelve pulses. Each time a pulse is received it is retransmitted to the next relay station.

Each pulse train, the clockwise and the counter-clockwise, has the same amount of pulses. The timing of the pulses is adjusted as follows: the measured time interval between the twelve clockwise propagating pulses is the same, and the measured time interval between the twelve counterclockwise propagating pulses is the same.

Under those circumstances the clockwise and counter-clockwise pulse trains will not have the same time interval, as measured by the relay stations. Here is why: as a matter of principle the two counterpropagating pulses are moving at the same speed: the speed of radio transmission. The relay stations are co-rotating with the Earth, clockwise as seen from the south pole. Each pulse takes some time to travel from one relay station to the next, so in the time between a relay pulse being emitted and a relay pulse being received the relay stations have moved in the clockwise direction. This lengthens the measured time interval between the pulsed of the clockwise pulse train, and it shortens the measured time interval between the pulses of the counter-clockwise pulse train.

Note especially that this lengthening/shortening is enforced by the constraint that the clockwise and counterclockwise pulse train must consist of the same number of pulses. You have to count the number of pulses.

Returning to the relay stations positioned around the equator: the difference in measured time interval between the clockwise and counterclockwise propagating pulses indicates the rotation rate of the Earth. More specifically, it allows you to identify the non-rotating cordinate system.

Again: the principle that gives rise to the Sagnac effect is that in all directions of propagation the speed of electromagnetic radiation is the same.

The Sagnac effect comes into play when you close a loop. In Minkowski space-time, when you close a loop very interesting things happen.

If you don't close the loop then you don't get the Sagnac effect. Imagine a setup where the loop isn't quite closed, but that instead there are two stations that do not relay to the next station, but they bounce the pulse back. Then you get the standard Einstein synchronisation. When you do close the loop, and you compare clockwise and counter-clockwise propagation, you get access to information that you otherwise would not have access to.

A closed Minkowski spacetime

Olaf Wucknitz argues that in a closed Minkowski space-time one can set up the same procedure as in the above explanation of the Sagnac effect. That will establish a universal reference frame.

In general: in Minkowski spacetime closing a loop is significant. By contrast: in Newtonian space & time closing a loop isn't particularly interesting. But in Minkowski spacetime closing a loop makes all the difference.

  • $\begingroup$ This question was closed [duplicate] I think this answer is relevant, so my intention is to move this answer to the 'already has answers here' location. So if you find this answer gone, that is where it went to. $\endgroup$ – Cleonis Feb 28 at 12:24

First of all, it is a myth that General Relativity is needed to handle accelerations. Special Relativity can handle acceleration just fine.

Second, in order for Twin B to return back to Twin A, Twin B must accelerate in some form. They would at the very least need to turn around, even if you approximated their velocity as constant while they’re travelling away from Twin A, they would have to instantly flip around the other way to make the trip back.

The difference in time between the observers arises from the fact that there is no SINGLE inertial frame that can represent the motion of Twin B’s entire trip. Even if the velocity was constant on the way out and the way back from Twin A, you would still have to “piece” two inertial frames together for the whole trip. This leads to a discrepancy between the observers.

So essentially, asking “what if the twin could return to compare clocks without having to accelerate” is kind of an ill-posed question. But to be clear, the “acceleration” is what causes the discrepancy between the observers so if Twin B could somehow not accelerate then you would be in the same situation where each observer measures the other to have time running slower.

I recommend watching this video on the twin paradox: https://youtu.be/0iJZ_QGMLD0

EDIT: As other comments here have noted, this is not the case if the universe is closed and you can loop back around to where you started. In that case, you can return to the twin without acceleration. Check this post out if you want to learn more about that: Twin paradox in closed universe

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    $\begingroup$ Question about special relativity in cyclic world. Did you understand this? $\endgroup$ – Nikita Feb 27 at 21:45
  • $\begingroup$ @Nikita What does a “cyclic” or “periodic” universe even mean? I was addressing the content of the post. $\endgroup$ – Thatpotatoisaspy Feb 27 at 22:35
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    $\begingroup$ The choice of the word "cyclic" or "periodic" might not be optimal, but the meaning was crystal clear from the content of the post. This answer completely ignores the question. $\endgroup$ – WillO Feb 27 at 22:44
  • $\begingroup$ @WillO I don’t see how that’s the case at all. OP never even mentions the concept of a closed universe, and even states that they “know it’s impossible” (returning without accelerating). I merely gave an answer that was in the usual framework of a spatially flat, relativistic universe. So hopefully you can see why it was a bit opaque to me. $\endgroup$ – Thatpotatoisaspy Feb 27 at 22:49
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    $\begingroup$ The acceleration is a bit of a red herring, IMHO. The thing causing the age difference (in the standard twin paradox) is that B must occupy 2 (or more) inertial frames. The acceleration is "merely" the mechanism by which B changes frames. $\endgroup$ – PM 2Ring Feb 28 at 13:16

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