B Special relativity implies the space cannot be "closed"?

[Hiero]

I think special relativity would disallow our universe from having the structure similar to an ant on a sphere. What I mean is that it can't be possible to travel in a constant direction and to come back to your original location.

Suppose there is an observer S on a planet and an observer S' in a spaceship having a relative speed v. (These are the only things in this hypothetical universe and the gravity is ignored.) Suppose when S' moves past S they synchronize their clocks, so they both read zero at the instant of the first passing. Now suppose S' moves "around the surface of the sphere" and comes back to S (without ever needing to accelerate). Then we will have a paradox, because these two events will occur at the same place in both frames, and so each frame will say they measured the proper time and the other frame's clock was ticking more slowly between the events. (There is no acceleration as in the resolution of the twin "paradox"). We can't have both clocks show a lower reading than each other, so this set up seems paradoxical.

Is this a justified reason to disallow universes with such structure?

 

[phinds]

If you had a railroad track that went around the Earth on a Great Circle, you would get exactly the same results. Do you think THAT situation is impossible?

 

[Drakkith]

If you had a railroad track that went around the Earth on a Great Circle, you would get exactly the same results. Do you think THAT situation is impossible?

Perhaps the example should do away with a planet and instead simply say that space is set up in such a way as to curve back on itself, so no acceleration or gravity is needed.

 

[Nugatory]

Is this a justified reason to disallow universes with such structure?

No, because special relativity starts with the assumption that the universe does not have such a structure. Thus, any contradictions that you find from applying SR to a universe that does have such a structure are a result of having started with inconsistent premises and tells you nothing except that your premises are inconsistent.

General relativity resolves the problem by saying that SR only applies locally, across volumes of spacetime that are small enough that curvature effects can be ignored. Clearly this isn't the case for a round trip around a sphere.

 

[Battlemage!]

Isn't such a universe automatically curved? If they follow a circular path but don't accelerate, how can it be flat spacetime?

 

[Nugatory]

Isn't such a universe automatically curved? If they follow a circular path but don't accelerate, how can it be flat spacetime?

It cannot be. And therefore it cannot be described SR, except as an approximation valid only across regions small enough that the curvature can be neglected.

 

[Orodruin]

It cannot be.

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.

Isn't such a universe automatically curved? If they follow a circular path but don't accelerate, how can it be flat spacetime?

It is not a circular path. A cylinder has no intrinsic curvature and yet you can draw a straight line on it that closes.

 

[Hiero]

No, because special relativity starts with the assumption that the universe does not have such a structure. Thus, any contradictions that you find from applying SR to a universe that does have such a structure are a result of having started with inconsistent premises

My understanding is that the principle of relativity (that all inertial frames are equivalent) implies the Lorentz transforms with a general limiting speed, and all special relativity does is postulate this speed is that of light. Those are the only 2 assumptions I know which underly SR but I lack GR knowledge so perhaps you can give your perspective on the assumptions underlying SR.

General relativity resolves the problem by saying that SR only applies locally, across volumes of spacetime that are small enough that curvature effects can be ignored.

Orodruin says a cylindrical like structure would give no intrinsic curvature. How would general relativity treat such a universe without seeing this paradox?

 

[Orodruin]

Orodruin says a cylindrical like structure would give no intrinsic curvature. How would general relativity treat such a universe without seeing this paradox?

There is no paradox. The only paradox arises when you try to apply concepts derived in Minkowski spacetime to a nontrivial global setting. If you have the full spacetime structure it is a simple matter of computing the proper times of the world lines. The result is only dependent on the spacetime geometry, not on any local choice of coordinates.

 

[Ibix]

In a cylindrical universe there's a time-like direction parallel to the axis of the cylinder and many that aren't. That direction is detectable in universe, so frames aren't interchangeable as they are in a non-cylindrical spacetime. So there's no particular issue with the two inertial clocks having different readings when they meet up again.

The reason it seems paradoxical is that only the observer moving along the cylinder can draw a Minkowski-like chart covering the whole of spacetime. Other observers can't get it to mesh on the other side, so naive time dilation calculations don't quite work. The relativity of simultaneity bites you if you don't work it out carefully.

 

[Nugatory]

The OP is suggesting a cylinder-like universe.

Is he? The original post says "sphere", twice. But of course you're right about the surface of a cylinder.

 

[Orodruin]

Is he? The original post says "sphere", twice. But of course you're right about the surface of a cylinder.

Indeed the word used is "sphere", but I think the tone of the post suggests something more general and there are flat spacetimes with the properties relevant to the OP, cylinders, tori, etc. I do not think it is desirable to hide this from the OP just because he happened to use the word "sphere".

 

[martinbn]

I think he means spatially a sphere, which can be a cylinder space-time with spherical space-like slices.

 

[Hiero]

The word sphere was just an analogy, the point was just that an inertial frame comes back to an old location.
Now that it has been emphasized so much though, let me say, I actually do not understand the issue with the sphere; isn't the 'inertial path' on the surface of a sphere along a great circle? So in a universe restricted to the surface of a (higher dimensional) sphere, an object traveling with no acceleration-(at least none "inside" the universe?)* will come back to an old location, which is the point of my setup.
*[I'm thinking in analogy to particles restricted to surfaces; there will be some acceleration "outside" (normal) the universe, and I'm not sure where that fits. But I would also say that exists for the cylindrical case, so I'm still not sure of the sphere's problem.]

Anyway thank you for the insightful replies. General Relativity will likely be a fun one to study in the future.

 

[Gigaz]

Generally, it is invalid to argue that because the curvature is extremely, the universe must be extremely large. Proof: Pacman ;)

 

[Flatland]

Wouldn't one of the observers have to "stop" relative to the other since you can only compare clocks and extrapolate any relativistic time dilation when both observers are in the same inertial frame? Just a thought.

 

[Ibix]

Wouldn't one of the observers have to "stop" relative to the other since you can only compare clocks and extrapolate any relativistic time dilation when both observers are in the same inertial frame? Just a thought.

No. You only need two instantaneous clock comparisons at first and second meeting. Why would you need to be at rest with respect to a clock to read it?

 

[Flatland]

No. You only need two instantaneous clock comparisons at first and second meeting. Why would you need to be at rest with respect to a clock to read it?

What I mean is that according to what the OP proposed, when observer S' circumnavigates the Universe and passes by observer S again, observer S' will observe the clock tick of S to be slower than S' and vice versa. One of the observers would have to decelerate in order to compare clocks and see who's clock actually ran slower. In other words there is no paradox in what the OP proposes.

 

[Herman Trivilino]

Now suppose S' moves "around the surface of the sphere" and comes back to S (without ever needing to accelerate). Then we will have a paradox,

I think the flaw in your logic is thinking that one of them must accelerate in order for them to experience different amounts of proper time. It's not the acceleration of one twin that explains the difference in ages. It's the fact that they take different paths through spacetime. There are versions of the twin paradox where neither twin experiences nonzero proper acceleration.

 

[Ibix]

observer S' will observe the clock tick of S to be slower than S' and vice versa

This is time dilation in action, and you seem to be accepting this. So I don't understand why you then say:

One of the observers would have to decelerate in order to notice any time dilation.

...since that seems to directly contradict your previous sentence.

In other words there is no paradox in what the OP proposes.

There is no paradox because the topology of the cylindrical universe picks out a frame that is special in a global sense, although local measurements are completely in line with the principle of relativity.

 

[Flatland]

This is time dilation in action, and you seem to be accepting this. So I don't understand why you then say:
...since that seems to directly contradict your previous sentence.

I'm saying that because one of the observers would have to stop and compare clocks with the other to see which clock ran "absolutely" slower a la symmetry breaking.

 

[Ibix]

I'm saying that because one of the observers would have to stop and compare clocks with the other to see which clock ran "absolutely" slower a la symmetry breaking.

Each observer takes a photo of their own clock as they pass the first time, and another when they pass the second time. Then they transmit the photos to the other. The difference in the readings of their own photos tells them their elapsed time; the difference in the readings of the other observer's photos tells them the other's elapsed time. Why would slowing down be necessary?

 

[Flatland]

Each observer takes a photo of their own clock as they pass the first time, and another when they pass the second time. Then they transmit the photos to the other. The difference in the readings of their own photos tells them their elapsed time; the difference in the readings of the other observer's photos tells them the other's elapsed time. Why would slowing down be necessary?

Let me use a different example. Let's suppose observer S' decelerates and stops next to observer S (after circumnavigating the Universe) and compared clocks. What would their clocks read?

 

[Ibix]

Let me use a different example. Let's suppose observer S' decelerates and stops next to observer S (after circumnavigating the Universe) and compared clocks. What would their clocks read?

At the moment of meeting, exactly the same as if they didn't stop. After that the results would start to diverge from the other scenario, of course.

Edit: The only thing that matters is the intervals along the two different paths. Saying they need to stop is like saying you can't compare the lengths of two lines unless they're parallel for at least art of their length.

 

[Flatland]

At the moment of meeting, exactly the same as if they didn't stop. After that the results would start to diverge from the other scenario, of course.

That's exactly what I was trying to get at. After S' stops, his clock will indeed appear to be slower when compared to S. The OP seems to suggest that this would happen even if one of them didn't stop.

 

[Ibix]

That's exactly what I was trying to get at. After S' stops, his clock will indeed appear to be slower when compared to S.

What? When they come to relative rest their clocks will tick at the same rate. While they are not at rest, they will both say that the other's clock ticks slowly. None of this changes the fact that the one who went round the cylinder will show a lower elapsed time than the one who did not. This is because the interval along the path around the cylinder is shorter than the interval along the cylinder. This is closely analogous to drawing a line up a Euclidean cylinder and another spiralling round it. The spiral line is longer than the straight one, whether or not I continue it past the straight line.

The OP seems to suggest that this would happen even if one of them didn't stop.

Correctly so, given the later clarification of what was meant by a spherical universe.

 

[Flatland]

What? When they come to relative rest their clocks will tick at the same rate. While they are not at rest, they will both say that the other's clock ticks slowly.

What I'm saying is that less time would have appeared to have passed not that one clock will actually tick slower compared to other when at rest.

None of this changes the fact that the one who went round the cylinder will show a lower elapsed time than the one who did not

How would you determine who went around the cylinder and who didn't? Observer S' can just as well claim that it was observer S who circumnavigated the cylinder. The OP specifically mentioned there were no accelerations involved.

 

[Ibix]

What I'm saying is that less time would have appeared to have passed not that one clock will actually tick slower compared to other when at rest.

Right. But that's true whether they stop or not, as my photos experiment shows. How could the photos be altered by whether or not one of the ships accelerates immediately after the photo was taken?

 

[Flatland]

Right. But that's true whether they stop or not, as my photos experiment shows. How could the photos be altered by whether or not one of the ships accelerates immediately after the photo was taken?

Yes but if one of them stopped relative to the other AND THEN compared clocks, BOTH observers would agree that less time has passed on the clock that decelerated and stopped.

 

[Ibix]

Yes but if one of them stopped relative to the other AND THEN compared clocks, BOTH observers would agree that less time has passed on the clock who decelerated and stopped.

I fail to see your point. Both observers would agree that less time has elapsed on the one who circumnavigates the cylinder, whether or not either of them stopped.

 

[Flatland]

I fail to see your point. Both observers would agree that less time has elapsed on the one who circumnavigates the cylinder, whether or not either of them stopped.

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.

 

[Herman Trivilino]

My point is that the observer that circumnavigates the cylinder is determined by which observer is the one that decelerates.

Not true. As has been stated they just need to share the same position to make the comparison of their clock readings. They do not need to share the same velocity.

The one that made the round trip has the smaller clock reading.

This is no different then what two observers would see passing each other at relativistic speeds even when no circumnavigation is involved.

Insofar as all they can do is compare clock readings, that is correct.

You are somehow mixing up a comparison of clock readings with a comparison of clock rates. To compare clock rates they need more than just the one event where they pass each other. They need a second event so that they can compare their elapsed times (difference in clock readings on S clock compared to difference in clock readings on S' clock).

 

[Flatland]

Not true. As has been stated they just need to share the same position to make the comparison of their clock readings. They do not need to share the same velocity.

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.

The one that made the round trip has the smaller 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)?

 

[Ibix]

My point is that the observer that circumnavigates the cylinder is determined by which observer is the one that decelerates.

No. It's determined by the topology of the cylinder. I've drawn the analogy of a Euclidean cylinder before: draw a straight line up it and a spiral line around it. Can you change the spiral into the straight line by what you draw further up the cylinder? Of course not.

Alternatively you can see it from my experiment with photos. Zero the clocks when they meet; photo the clocks just before they meet again, then decide who decelerates. According to you, deciding who decelerates must influence what the photos we've just taken show...

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.

As long as they remain in motion they will see the other's clock ticking slow, because special relativity is valid locally. But the one who went round the cylinder will have the lower clock reading. The apparent paradox is resolved by the fact that only one family of observers can wrap their Minkowski chart around the cylinder without there being a mismatch of time coordinates at the seam.

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)?

Each observer sends out two light pulses in opposite directions. An observer who is not moving around the cylinder will receive the pulses back simultaneously. An observer who has any motion around the cylinder will receive the backwards-emitted pulse first.

As has been mentioned several times upthread, the topology of this case picks out a special frame, while the geometry remains Minkowski.

 

[Flatland]

No. It's determined by the topology of the cylinder. I've drawn the analogy of a Euclidean cylinder before: draw a straight line up it and a spiral line around it. Can you change the spiral into the straight line by what you draw further up the cylinder? Of course not.

Alternatively you can see it from my experiment with photos. Zero the clocks when they meet; photo the clocks just before they meet again, then decide who decelerates. According to you, deciding who decelerates must influence what the photos we've just taken show...
As long as they remain in motion they will see the other's clock ticking slow, because special relativity is valid locally. But the one who went round the cylinder will have the lower clock reading. The apparent paradox is resolved by the fact that only one family of observers can wrap their Minkowski chart around the cylinder without there being a mismatch of time coordinates at the seam.
Each observer sends out two light pulses in opposite directions. An observer who is not moving around the cylinder will receive the pulses back simultaneously. An observer who has any motion around the cylinder will receive the backwards-emitted pulse first.

As has been mentioned several times upthread, the topology of this case picks out a special frame, while the geometry remains Minkowski.

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? I will have to look more into this.

 

[Herman Trivilino]

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.

 

[Herman Trivilino]

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.]

 

[Herman Trivilino]

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.