1. Nov 3, 2004

### Garth

There have been plenty of posts about the Twin Paradox. I don't think this version has been aired before except in my post #47 on the "Is Age Relative" thread.

The Twin Paradox in a closed universe.

If cosmic expansion slows down and reverses it would become hypothetically possible to circumnavigate the universe.

Take two twins, one twin stays put and the other is accelerated to
9.999…%c she passes her sister and they synchronise clocks. She continues at constant velocity along a geodesic path and circumnavigates the universe. She eventually meets up with her sister again.

She has aged only 50 years in this near light-speed voyage. However her sister has aged 10 billion years!

Or is it the other way round? How do you tell?

Just a thought - Garth

Last edited: Nov 3, 2004
2. Nov 3, 2004

3. Nov 3, 2004

### Garth

Wow - straight in at the deep end! Thank you robphy.
It is important to see the consequences of this version of the paradox, and I quote from the Barrow and Levin paper "The twin paradox in compact spaces" http://arxiv.org/abs/gr-qc/0101014
and from the Uzan et al. paper "Twin paradox and space topology"
http://arxiv.org/abs/physics/0006039
So a closed universe has a preferred frame of reference! It has so by virtue of its topology, which is finite yet unbounded.

In our discussions in the thread "Is age relative?" I argued
If these authors are right, and actually it is necessary for Quantum Gravity to have a preferred frame of reference, then perhaps by reversing the above conclusion, we might further conclude that the universe has to be closed.

As I have posted several times
This would seem to agree with the conclusions of the above two papers.

Just a thought.

Garth

Last edited: Nov 3, 2004
4. Nov 3, 2004

### Garth

The more I think about it the more the topological argument seems unsatisfactory.

Two observers with a mutual velocity of 0.999...c with clock-calendars closely pass each other when they start their clock-calendars. Say a flash bulb goes off midway between them and they start their clocks on receiving the pulse of light.

After a long time they circumnavigate the universe and pass each other again and record the time as they pass.

Each observer will believe they are at rest and the other has circumnavigated the universe.
Each will believe that they have a winding number of zero and the other has a winding number of one.

[Imagine the universe is a cylinder with time parallel to the axis. In each observer's frame of reference their world-line is straight up the cylinder and the other's winds around it. The winding number is the number of times the world-line is wrapped around the cylinder when the two encounter points are brought together and the "world-lines pulled taut". This is the topological argument.

Do the experiment from the point of view of one observer, now do it for the other, the cylinder itself rotates, exchanging the winding number values. How do we 'tie down' the cylinder so it doesn't rotate?]

There must be something else that distinguishes between the two observers.
I think it can only be the matter in the universe and the Machian frame of reference that matter defines that gives you something to 'hold onto' and enables you to 'tie the cylinder down'!

What do you think?
Garth

Last edited: Nov 3, 2004
5. Nov 3, 2004

### robphy

Here's an argument based on the article by Weeks (http://www.maa.org/pubs/monthly_aug_sep01_toc.html).

The twin with winding number zero is distinguished because his "surfaces of simultaneity" are closed curves. Twins with nonzero winding numbers have helical "surfaces of simultaneity".

6. Nov 4, 2004

### Garth

Yes, that was the argument in Barrow and Levin's paper. However I still do not think it answers the paradox.

Take my idealised thought experiment. Not two twins, one of whom has to be accelerated, and thereby infers a state of rest, but two independent inertial observers with clocks, with a high mutual velocity, who happen to pass by each other at which event they set their clocks. They are in a homogeneous and isotropic compact space universe. After a long time their geodesic paths cross again.

According to the Principle of Relativity, i.e. the doctrine of "no preferred frames of reference", each observer will consider that they are the one at rest and the other is moving, each will think that they have closed surfaces of simultaneity and the other has helical surfaces of simultaneity. Each will think their clock has recorded the longest duration between the two encounters.

This is of course nonsense.

What breaks down, and this is what the argument using surfaces of simultaneity depends on, is the Principle of Relativity. In fact all frames are not equivalent. The curvature of space-time that is required to close the universe and create a compact space requires the presence of a homegeneous ' representative gas' throughout the universe. The preferred frame, the one that does have a surface of closed simultaneity rather than any other frame, is that at rest with that representative gas, co-moving with the Hubble flow.

Does this not question the theoretical basis of GR?
Garth

Last edited: Nov 4, 2004
7. Nov 4, 2004

### robphy

I believe that the winding number is a topological invariant... so this shouldn't be possible. I don't have a proof right now.

The "Principle of Relativity" is really only a local statement... that physical laws can be written down in a tensorial form, independent of the choice of coordinate axes.

Recall that the spacetime of "Special Relativity" is not merely flat... but is also topologically $R^4$.

Consider a strip... $R\times I$. Now, make the identification so that the manifold is now $R\times S^1$. This manifold is flat, i.e., has zero Riemann curvature. On this manifold, we draw geodesics as straight lines for our inertial observers. If I'm not mistaken, no Lorentz Transformation can transform the geodesic with winding number 1 into a vertical line (parallel to the axis of the cylinder) with winding number 0. Remember, the lightlike direction is unchanged. [Analogously, no rotation on that cylinder can make that geodesic vertical.]

Here's a physical experiment to consider in this closed cylindrical universe.
The "twin at rest" can exchange light signals with the "twin in inertial motion".
For a short enough time (short compared to the time for a light signal to circumnavigate the closed universe), the two twins are indistinguishable. However, consider light signals emitted to the right and to the left at the first meeting event. When those signals are received by the "twin at rest", those receptions coincide. However, those analogous reception signals for the "twin in inertial motion" are not coincident. Thus, the frames are not equivalent... but one had to wait until the topological features at the "large-scale" are encountered.

8. Nov 5, 2004

### Garth

robphy - Thank you for your post - it "has me thinking"!

And thank you too for that very illuminating thought experiment of the circumnavigating light signals. However are they not an example of the same paradox?

By the Relativity Principle each observer would think they are stationary and believe it is the other one who is moving, and each would expect to receive the two opposite circumnavigating light signals simultaneously. That they cannot is part of the same paradox as before; i.e. they cannot both have the longest proper times between their two meeting events.

Take simply a cylindrical space-time, at first empty of all matter apart from the two observers, i.e. test particles with their inertial frames of reference.
The topological system is the cylinder with the two world-lines, one straight (A) with a winding number of zero, and one wound round the cylinder (B) with a winding number of one. Represent the system as a flexible cylinder with two elastic strings A and B, fixed at the two meeting events.

Is not the topological invariant the winding number of the total system?
i.e. by continuous deformations one cannot 'unwind' both strings. However we can continuously deform the system to transform from A's frame of reference to B's by twisting the cylinder so that B becomes straight and A becomes wound round. The total winding number of the system is still one and that number has remained invariant under the transformation.

I believe I have shown the topological argument, on its own, does not work and leads to a nonsense! This cannot be right. The two frames of reference cannot be equivalent. One is a preferred frame.

The mistake in the above illustration is to assume the topological space is empty, in fact it is not and indeed it cannot be, matter, i.e. the cosmological representative gas, is essential to obtain a closed compact space, that is finite yet unbounded.

I believe the presence of the matter reasserts the importance of Mach's Principle. We can identify the 'stationary' observer, the one in the preferred frame of reference, as that observer who is stationary w.r.t. the representative gas, co-moving with the Hubble flow, in whose frame of reference the CMB is globally isotropic.

Now, is not the Principle of Relativity in SR is a necessary and sufficient condition for the Principle of Equivalence in GR? If so, in the cosmological twin paradox does not GR contain the seeds of its own destruction, i.e. it is internally inconsistent?

Garth

Last edited: Nov 5, 2004
9. Nov 5, 2004

### robphy

I think that this operation cannot be performed while keeping the light-cones (light-like directions) invariant. So, the winding number for a given curve cannot be changed from, say, 1 to 0.

10. Nov 6, 2004

### Garth

Then the question remains;
"According to the Principle of Relativity (PR), how do you distinguish between the two observers?"
They both think the other is circumnavigating the universe and they themselves are stationary.

As Barrow and Levin said in the paper you linked to:
So
PR -> Equivalence Principle -> GR -> Friedmann Cosmology -> A possible compact space -> Twin Paradox -> preferred frame which contradicts the PR!

Is this therefore logically inconsistent?

Garth

11. Nov 7, 2004

### robphy

As I alluded to before, the Principle of Relativity is a local statement.
That is to say, "experiments done in an inertial `lab' can't distinguish themselves".

Since the "universe" in this scenario is not topologically R4, we are not dealing with Special Relativity any more. We should not expect everything from SR to carry over to here.

So, the existence of a preferred [inertial] frame (due to a global [i.e. non-local] topological condition) in this non-SR spacetime does not contradict the "PR".

Last edited by a moderator: May 1, 2017
12. Nov 7, 2004

### Garth

Thank you I do understand, but I'm still labouring over the point. Given the idealised scenario above, a empty compact space with two idealised test particles and their inertial frames of reference that happen to closely pass each other at speed. As they pass we are dealing with a local situation and are saying that one will end up with a winding number of zero and the other with a winding number of one.

Which is which?

What is it that gives a particular observer her winding number?

Put it another way, as they pass each other the first time can we predict which one will have the longer proper time interval between the two meeting events?

Do you see my problem?

Garth

Last edited: Nov 7, 2004
13. Nov 9, 2004

### robphy

Take again the experiment of sending out a light signal in both directions when they first separate. I have drawn the universe below... essentially unrolling the cylinder. (I think I might have missed an observer's worldline in the upper left corner.)

$$\begin{picture}(140.00,150.00)(0,0) \multiput(0.00,100.00)(0.12,-0.12){833}{\line(1,0){0.12}} \multiput(0.00,100.00)(0.12,0.12){417}{\line(1,0){0.12}} \multiput(0.00,140.00)(0.12,-0.12){1167}{\line(1,0){0.12}} \multiput(0.00,20.00)(0.12,-0.12){167}{\line(1,0){0.12}} \multiput(0.00,20.00)(0.12,0.12){1083}{\line(1,0){0.12}} \multiput(0.00,30.00)(0.12,0.18){667}{\line(0,1){0.18}} \multiput(0.00,60.00)(0.12,-0.12){500}{\line(1,0){0.12}} \multiput(0.00,60.00)(0.12,0.12){750}{\line(1,0){0.12}} \multiput(100.00,0.00)(0.12,0.12){333}{\line(1,0){0.12}} \multiput(100.00,0.00)(0.12,0.18){333}{\line(0,1){0.18}} \multiput(110.00,150.00)(0.12,-0.12){250}{\line(1,0){0.12}} \multiput(20.00,0.00)(0.12,0.12){1000}{\line(1,0){0.12}} \multiput(20.00,0.00)(0.12,0.18){833}{\line(0,1){0.18}} \multiput(30.00,150.00)(0.12,-0.12){917}{\line(1,0){0.12}} \multiput(60.00,0.00)(0.12,0.12){667}{\line(1,0){0.12}} \multiput(60.00,0.00)(0.12,0.18){667}{\line(0,1){0.18}} \multiput(70.00,150.00)(0.12,-0.12){583}{\line(1,0){0.12}} \put(20.00,0.00){\line(0,1){150.00}} \put(60.00,0.00){\line(0,1){150.00}} \put(100.00,0.00){\line(0,1){150.00}} \put(140.00,0.00){\line(0,1){150.00}} \put(35.99,24.05){\circle{2.00}} \put(51.97,48.03){\circle{2.00}} \put(67.89,72.11){\circle{2.00}} \put(83.95,96.05){\circle{2.00}} \put(100.26,119.47){\circle{2.00}} \end{picture}$$

As the signals travel around the closed universe,
one observer receives the incoming signals at the same event(s).
The other observer does not.

When they meet at the first reunion, they can compare the number of times they received that signal. In this example, the first observer received "a single signal from both directions", while the other received "two from the forward direction, but none from the backward direction".

When they meet at the second reunion, they can compare again. In this example, the first observer received "three signals from both directions", while the other received "four from the forward direction, then a fifth from both directions".

Clearly they disagree.

One can probably work out the number of signal receptions for arbitrary observer velocities and radii of the closed universe and relate that to the winding number.

14. Nov 9, 2004

### Alkatran

Of course, this argument assumes it is the speed in the space you exist on that matters, maybe your 'acceleration' to go around the sphere matters?

15. Nov 9, 2004

### Garth

Yes, I have no problem with that, and as I have said there is clearly a preferred frame of reference, but can you tell which observer is in that frame? (Without waiting for the second encounter)

Consider my case presented above.

Take a topological compact space, a cylinder, empty except for two test particles in inertial frames of reference, with an arbitrary mutual velocity and which encounter each other at least twice.

Take the local space-time volume around the first encounter. Within that volume each test particle is equivalent, and the question is, "Is it possible to predict which one will experience the longest possible proper time between the two encounters, i.e. the one in a ‘preferred frame of reference’?" I would argue that in this case the answer is, “No, you have to wait to perform an experiment to find out, such experiments as the light ray experiment or the comparison of clocks at the second encounter.” This is because, as a simple topological space, otherwise empty of physics, the global structure can be continuously deformed (twisted) to give each observer the zero winding number.

Now convert to the real universe, add in the physics and consider two observers. Suppose one observer is more or less at rest w.r.t. the surface of last scattering of the CMB, and the other has a very high relative velocity. Wait and then perform the above experiments. I would find it incomprehensible to imagine that the first observer is not in the preferred frame of reference.

If this must be the case then the cosmological distribution of matter and energy has not only determined the global geometry but also locally affected the frames of reference, selecting one as preferred against all others.

But does this not undermine the Principle of Relativity?

Garth

Last edited: Nov 9, 2004
16. Nov 9, 2004

### abbott

Observation is the core necessity of an existence that operates as a balance of opposites. Our powers of observation will always end in unpercieveability (time or space "before" the big bang, the meaning of time or space in the face of the infinity we observe them approach).

17. Nov 9, 2004

### abbott

Observation is the core necessity of an existence that operates as a balance of opposites. Our powers of observation will always end in unpercieveability (time or space "before" the big bang, the meaning of time or space in the face of the infinity we observe them approach).

18. Nov 9, 2004

### abbott

We (classical philosophy "who am I?" and "what is everything?") must be the knowable universes understanding of itself, after all what is anything we have observed other than a reflection of the energies we've used to detect it. And what is a living thing? A converter of energies, we sustain ourselves by intaking things and changing them into other things (sounds like a process of observation to me).

I've just recently suggested to www.superstringtheory.com that perhaps light/energy (infinity in every "direction," and a string without concievable begining or end) is oscilating through its opposite, nothingness. I am very curious to discover if this idea makes sense to superstring theory and was very dissopointed to find that shortly after my post the forum stoped being accesible. So if anyone here knows anything about string theory and can shed some light on the way I'm trying to understand it I would be very pleased to recieve that feedback.

19. Nov 9, 2004

### abbott

The paradox is maybe believing that the twins are in fact not the same thing. If they are just different versions of the same perspective of knowable existence, then differences they experience in time or anything else only serve to illuminate that the only things that change are the opinions of what is percieved.

20. Nov 9, 2004

### Garth

Are all observers then "the same thing"? That you and I are the same? I think not.

Garth