Special relativity implies the space cannot be "closed"?

[Mister T]

I don't see how that's relevant to what you quoted me saying. I never contradicted the above statement. In fact I have mentioned that each observer will measure the other's clock to tick slower when passing the same position again.

It's not possible to compare clock rates with a single clock reading.

But how do you determine which observer is the one that made the round trip when no acceleration is involved (since both observers are in inertial reference frames)?

Acceleration is not relevant. Differences in paths through spacetime is. And the one who makes the round trip is the one who has the shorter path.

 

[Flatland]

It's not possible to compare clock rates with a single clock reading.

And can't you make multiple clock readings when passing each other?

Acceleration is not relevant. Differences in paths through spacetime is. And the one who makes the round trip is the one who has the shorter path.

Yes this has been explained.

 

[Ibix]

So are you saying that even if observer S' accelerated and decelerated it could still be determined that observer S was the one that made the round trip based on topology?

Yes. Here's a Minkowski diagram in the rest frame of the "straight up the cylinder" observer (black line). It's just a sketch, but if I did it right and you print it and roll it into a cylinder the red lines should meet up - this is the worldline of the circumnavigating observer.

Note that I have not shown any history before the experiment or future after the experiment. Feel free to extend the lines in any way you choose - let them continue as they are, extend them both parallel to the black line, extend them both parallel to the red line, something else. This does not change the experiment in the slightest.

You can draw an analogous diagram in the frame of the red observer. It looks much like the above diagram, but must be wrapped around into a cylinder at an angle so that the black observer goes straight up. But if you add lines of simultaneity (for red's frame) to this diagram, you'll find that you can't make the black line meet up and be straight and have the lines of simultaneity meet up at the edges.

 

[Mister T]

And can't you make multiple clock readings when passing each other?

Passing each other is a single event. It occurs at a single location at a single clock reading. So, no. When a second clock reading is taken they will no longer have the same position.

 

[Flatland]

Passing each other is a single event. It occurs at a single location at a single clock reading. So, no. When a second clock reading is taken they will no longer have the same position.

I guess my point was that you can continue to make clock readings even after you've passed each other.

 

[Hiero]

My point is that the observer that circumnavigates the cylinder is determined by which observer is the one that decelerates. The OP suggests that there is no acceleration/deceleration and when the two observers passes each other again they will measure each other's clock to be slower causing some kind of paradox.

It doesn't. This is no different then what two observers would see passing each other at relativistic speeds even when no circumnavigation is involved.

I have to say, you seem to not have appreciated my paradox, for it most certainly not the same as two observers passing by a single time in 'regular' space.

Two observers passing once is a single event. In order to compare clocks at a future time, they would have to do so at a separate location. It is by virtue of this separation that the Lorentz transforms work out.

It's not about the clocks seeming to go slower... it's about the actual elapsed time on the clocks between two events. The proposed situation (two passings) is incompatible with the Lorentz transformations alone.

My problem was not "oh how paradoxical they both see time go slower," my problem was that the Lorentz transforms would give {Δt=γΔt', Δt'=γΔt} which is obviously inconsistent (since ϒ≠1). The Lorentz transformation breaks down when we no longer have Δx=vΔt between the two events.

[I'll show why we don't have this problem in the 'regular space' ... if the two events happen in the same place in S' (Δx'=0) then S' will say Δt=ϒ(Δt'-vΔx'/c²) --->Δt = ϒΔt' ... but if they happen in the same place on S' then the separation in S will be Δx=-vΔt, so that S will say Δt'=ϒ(Δt+vΔx/c²)=ϒΔt(1-(v/c)²)=ϒΔt/ϒ^2 ---> Δt' = Δt/ϒ ... so you see everyone agrees on the time elapsed by each clock even though they both see the other clock running slow.]

 

[Flatland]

I have to say, you seem to not have appreciated my paradox, for it most certainly not the same as two observers passing by a single time in 'regular' space.

Two observers passing once is a single event. In order to compare clocks at a future time, they would have to do so at a separate location. It is by virtue of this separation that the Lorentz transforms work out.

It's not about the clocks seeming to go slower... it's about the actual elapsed time on the clocks between two events. The proposed situation (two passings) is incompatible with the Lorentz transformations alone.

My problem was not "oh how paradoxical they both see time go slower," my problem was that the Lorentz transforms would give {Δt=γΔt', Δt'=γΔt} which is obviously inconsistent (since ϒ≠1). The Lorentz transformation breaks down when we no longer have Δx=vΔt between the two events.

[I'll show why we don't have this problem in the 'regular space' ... if the two events happen in the same place in S' (Δx'=0) then S' will say Δt=ϒ(Δt'-vΔx'/c²) --->Δt = ϒΔt' ... but if they happen in the same place on S' then the separation in S will be Δx=-vΔt, so that S will say Δt'=ϒ(Δt+vΔx/c²)=ϒΔt(1-(v/c)²)=ϒΔt/ϒ^2 ---> Δt' = Δt/ϒ ... so you see everyone agrees on the time elapsed by each clock even though they both see the other clock running slow.]

From what I gathered in this thread, correct me if I'm wrong (I probably am), is that the Lorentz Transformation would not apply since the observer that circumnavigates is determined by topology. As in, there are no Lorentz Transformations you can make where observer S' is stationary and that observer S is the one that circumnavigates the cylinder since circumnavigation around the cylinder is absolute. This seems more apparent to me now when you see their worldlines traced out on a cylinder.

 

[Hiero]

From what I gathered in this thread, correct me if I'm wrong (I probably am), is that the Lorentz Transformation would not apply since the observer that circumnavigates is determined by topology. As in, there are no Lorentz Transformations you can make where observer S' is stationary and that observer S is the one that circumnavigates the cylinder since circumnavigation around the cylinder is absolute. This seems more apparent to me now when you see their worldlines traced out on a cylinder.

That is about what I gathered as well, although I don't entirely understand the resolution. [I will relegate that understanding to my future studies of GR though, as there are more fundamental topics I should focus on first.]

 

[Mister T]

I guess my point was that you can continue to make clock readings even after you've passed each other.

Yes, but those additional clock readings are needed for them to compare clock rates. That was and is my point.

 

[Battlemage!]

Yes it can. The OP is suggesting a cylinder-like universe. Such a universe is not necessarily curved. It has a non-trivial global topology so it is not globally Minkowski space, but it is flat.It is not a circular path. A cylinder has no intrinsic curvature and yet you can draw a straight line on it that closes.

Is this because you can cut the cylinder and have a rectangle? (I'm taking it we're ignoring the ends of the cylinder here?)

 

[Orodruin]

Is this because you can cut the cylinder and have a rectangle?

It is because you can cut out any (simply connected) part of the cylinder and put it flat on the ground. You can do this with a cone as well as long as you exclude the apex. You cannot do this with a sphere for example, which is the reason map projections have to make concessions with either shapes or areas.

 

[Battlemage!]

It is because you can cut out any (simply connected) part of the cylinder and put it flat on the ground. You can do this with a cone as well as long as you exclude the apex. You cannot do this with a sphere for example, which is the reason map projections have to make concessions with either shapes or areas.

Does this have to do with functions between the two maps? That is, a cylinder and some flat square or whatever would have a one-to-one relationship for each point, while a sphere would not?

 

[Orodruin]

Does this have to do with functions between the two maps? That is, a cylinder and some flat square or whatever would have a one-to-one relationship for each point, while a sphere would not?

No. What you describe is about the global topology of the space, not its flatness.

 

[Victor Ray Rutledge]

This occurred to me, while reading, (with almost no understanding) the thread. http://www.businessinsider.com/do-astronauts-age-slower-than-people-on-earth-2015-8 Which explains(?) the gravity time-dilation and the velocity time-Dilation, in uncertain terms. (humor referencing the uncertainty principle). In this, and other, explanations, the 'traveler' must accelerate, to return to the original location. In a non-infinite universe, this would not be true, since the curvature of space would eventually return the traveler to the original location, where the acceleration to speed occurred. In the example used, there is no acceleration needed, as the curvature of the proposed space causes the 'straight line' to coincide with a line created by acceleration around the surface. In this situation, the paradox would not occur, since there would be no time-dilation from velocity. The 'ship' carrying the traveler would have a change caused by initial acceleration, which would be apparent, to the 'at rest' observer. This difference would be the time-dilation of the system. Only if the 'traveler' needed no acceleration, to move through space, would there be no time-dilation. This, to my limited intelligence, is the explanation. Please feel free to tell me how wrong I am, {laughing}

 

[Ibix]

Time dilation has nothing to do with acceleration, only velocity. So yes, there would be time dilation. The Business Insider article does not, on a quick read, claim otherwise and would be wrong if it did.

 

[Mister T]

This occurred to me, while reading, (with almost no understanding) the thread. http://www.businessinsider.com/do-astronauts-age-slower-than-people-on-earth-2015-8 Which explains(?) the gravity time-dilation and the velocity time-Dilation, in uncertain terms.

Well, time dilation due to velocity has nothing to do with the curvature of spacetime, so that is a misconception. One that's probably not relevant to this discussion. But the fact that time dilation is symmetrical is something they ignore in that article, and that is relevant. They state that astronaut Scott Kelly aged 0.01 seconds less than his brother Mark Kelly who stayed on Earth because Scott was moving at a speed of 5 mi/h. Mark is moving at 5 mi/h relative to Scott, and the author of the article doesn't explain why that doesn't make Mark the one who is younger. This is the part that's relevant to this discussion: Scott is the one who takes the shorter path through spacetime, and that is the reason he ages less.

In this, and other, explanations, the 'traveler' must accelerate, to return to the original location.

And that is the misconception that underlies the OP's original query: The traveler takes the shorter path through spacetime to get to the reunion. Whether that happens because of acceleration or not.