# A variation on the twin paradox

1. Feb 10, 2006

### dicerandom

OK, so I'm pretty familliar with the regular twin paradox and the explinations of how you decide which twin is younger by noticing which one accelerated, and the geometrical view of this using k-calculus and minkowsky diagrams, etc. I just thought of something though, and I guess my relativity isn't strong enough to think of a solution, at least not in the period of time it took me to go make some tea

Suppose that we live in a closed universe, for simplicity's sake let's also assume that universe is balanced with a cosmological constant so that it is neither expanding nor contracting (ala Einstein's static universe model). If I'm in a spaceship and I go flying by my friend at a very high velocity, and then wait for some time, I should wrap around and go flying past him a second time. If we set both our clocks to zero at the first flyby, what will our clocks read when we pass eachother the second time?

I'm at a loss to try and explain any differences in the clocks without attaching a prefered reference frame to "the surface of the universe" which my friend is allegedly still with respect to, which of course sets off all of my in-built relativity alarms. I'm tempted to say that we'll both have the same number on our watches when we pass the second time and ascribe it to a general relatavistic "you can't compare time or space intervals at a distance" effect.

2. Feb 10, 2006

### dicerandom

And I'm bothered by this paragraph:

Couldn't we make the same argument if we chose to draw Betty's spacetime diagram instead?

3. Feb 10, 2006

### JesseM

If you try to draw each observer's line of simultaneity in the usual way, the line of simultaneity in the +x direction won't necessarily match up with the line of simultaneity in the -x direction. So if one twin tried to use a badly-behaved coordinate system like this, she might conclude that her twin was "currently" at position x=8 light years and aged 35 years, but also "currently" at position x=-5 light years and aged 32 years, something like that. So maybe she's heading off in the +x direction and thinking that both "copies" of her twin are aging slower than she is, but the "copy" that already lies in her +x direction (as opposed to the one she just left behind in the -x direction) already has a "head start" in age, so even though that copy is aging slower too he can still be older when they reunite. On the other hand, there will be a "preferred frame" where the lines of simultaneity in both directions will meet up smoothly, so you don't have to worry about such problems in this coordinate system.

4. Feb 10, 2006

### Hurkyl

Staff Emeritus
You are getting close.

It is correct to note that the laws of special relativity are only guaranteed to hold on short scales, and that reference frames are only really meaningful at infinitessimal distances.

Special relativistic time dilation is, of course, a measurement that depends on a choice of reference frame.

But you can't just do that. The problem with this scenario is that there isn't a shortcut you can use. You (more or less) actually have to do the calculations to compare.

To restate what JesseM said, what is happening is this:

The argument assumes that space-time is shaped like a cylinder -- i.e. if we took a flat (1+1)-dimensional space-time (i.e. the space-time of special relativity), and then we rolled it up into a cylinder.

In some sense, cylinders do have a distinguished direction: along the axis. Of course, this direction can only be distinguished with a global experiment: one that actually involves one observer having gone one more loop around the universe than another observer.

In other words, this distinguished direction has nothing to do with physics: it is simply related the shape of the universe.

For them to draw the space-time diagrams they did, they assumed that Albert was travelling in the distinguished direction. And if Albert is travelling in the distinguished direction, then Betty clearly is not, so the problem is not symmetrical. Of course, it could have been Betty (or neither) who is travelling in the distinguished direction.

5. Feb 11, 2006

### dicerandom

Thanks for your comments everyone, I think I have a better grasp on this now. I do have one more related question though.

In the example discussed in the article I linked above, and more generally in the Robertson-Walker metric, space is treated as being curved whereas the temporal direction has no curvature. For reference the line element in the Robertson-Walker metric is:

$$ds^2 = -dt^2 + a(t)^2 \left[ dr^2 + S_{\kappa}(r)^2 d\Omega^2 \right]$$

And $a(t)=1$ for a static configuration, of course. Extending the arguments posed for the 1D spatial case I believe that this entire class of metrics will exhibit identical behavior, at least in the case when $\kappa=1$ (positive uniform curvature, closed universe).

Is it possible to have a metric for a closed universe where the temporal dimension is included on the surface of the hypersphere, and if so would this eliminate the prefered coordinate system?

6. Feb 11, 2006

### Garth

This is the cosmological twin paradox, which has already been discussed several times.

It is not so easily dismissed. For example Barrow and Levin discuss it in The twin paradox in compact spaces in which they say:
and Uzan et al. in Twin paradox and space topology in which they say:
So a closed universe has a preferred frame of reference! It has so by virtue of its topology, which is finite yet unbounded. That frame of reference is defined by the 'cosmic fluid'.

The paradox has arisen because we have treated the compact space as an empty topological space, in fact to get the universe to be closed it will have to have matter within it. That matter itself can be used to define the preferred frame of reference, the one co-moving with the cosmic fluid; however, you may feel that to do so relies on a use of Mach's Principle that GR cannot bear.

The link in the second post #2 above, contains the statement:
Yet GR takes this local symmetry of the equivalence of all inertial frames and through the conservation of energy-momentum and the stress-energy tensor treats it globally. Perhaps this is where the point is stetched too far?

Garth

Last edited: Feb 11, 2006
7. Feb 13, 2006

### Zanket

I assume that you and your friend were once at rest with respect to each other, and then you accelerated to make the flybys. Had you synchronized your watches before you initially accelerated, then just before re-synchronizing your watches on the first flyby, your watch will show less time elapsed. It will show less time elapsed on the second flyby too, because the cause of the difference in elapsed time is the same for both flybys—it is your acceleration.

Last edited: Feb 13, 2006
8. Feb 13, 2006

### dicerandom

I don't think what you've written here is in agreement with the solutions which have been discussed both here and in the papers linked, at least not in the cosmological version.

For instance, supposing my twin and I were initially in a frame such that our time axis was not aligned with the coordinate time, i.e. the vertical axis of the cylinder in the first paper I linked. If I accelerate until my time axis is aligned with the coordinate time, and then synchronize watches with my twin when we pass, he will then read a smaller proper time when we pass again.

9. Feb 14, 2006

### Garth

No, they do not have to have been at rest.

Two inertial observers, moving mutually at high speed, happen to pass each other when they exchange signals and synchronise clocks.

A very long time later they pass close by each other again and compare clocks, one of them having circumnavigated the universe, but which one?

Garth

10. Feb 14, 2006

### Zanket

Sorry, talk of spacetime diagrams makes my eyes glaze over. I assume you’re talking about a situation where both you and your friend accelerated to attain high velocity relative to the more or less fixed galaxies, but stay at rest with respect to each other, and then you decelerated relative to the galaxies to make the flybys relative to your friend. Yes, in that case your friend’s clock elapses less time than yours does between flybys, because between flybys the friend travels less distance relative to you than you travel relative to your friend, and your velocities relative to each other are identical. When you decelerated relative to the galaxies then distances (like those between galaxies) along your axis of motion uncontracted from a contracted state, whereas distances remain contracted for your friend. This situation is just a fancy way of making the “stay at home” twin look like the traveling twin and vice versa.

If you’re talking about some other situation, can you put it in plain English?

Last edited: Feb 14, 2006
11. Feb 14, 2006

### pervect

Staff Emeritus
The cosmological twin paradox has been discussed here before, and in certain hypothetical space-times (which are not expected to be our own) it is possible to circumnavigate the universe without ever accelerating. These sorts of hypothetical universe are where the paradox arises, and acceleration is not the solution to the paradox in these instances.

Thus I have to agree with Garth and Dicerandom.

12. Feb 14, 2006

### Zanket

That depends on what or who you choose the circumnavigation to be relative to. As Einstein said, there’s no universal hitching post. Regardless who is declared to be circumnavigating, the driver of the spaceship or the driver’s friend, the one who observes the greatest length contraction of distances between galaxies elapses less time between flybys.

Last edited: Feb 14, 2006
13. Feb 14, 2006

### Hurkyl

Staff Emeritus
There's no such thing as "experiencing" length contraction.

14. Feb 14, 2006

### Zanket

Like it better now?

15. Feb 14, 2006

### JesseM

There's no reason in principle that all the galaxies in this universe couldn't have some substantial velocity relative to the coordinate system whose time axis is the vertical axis of the cylinder. In this case, one twin could be at rest relative to the galaxies and the other could be in motion relative to them, yet it would be the one who was at rest relative to the galaxies who would elapse less time between flybys.

16. Feb 14, 2006

### Hurkyl

Staff Emeritus
Nobody in that example would "observe" length contraction -- for each observer, the distance between galaxies would remain constant.

Also, it is certainly possible for the twin who perceives the greater distance between galaxies to be the one who experiences less time between flybys.

17. Feb 14, 2006

### Zanket

I agree with that. Good luck accelerating all those galaxies though!

18. Feb 14, 2006

### Zanket

How did you gather that from Garth's example?

Yes, if the galaxies are accelerated along with the traveling twin, as JesseM points out.

19. Feb 14, 2006

### JesseM

Why would they need to be accelerated? Why do you assume that their "initial" velocity would be at rest relative to the vertical cylinder axis coordinate system?

20. Feb 14, 2006

### Zanket

Because if that assumption were false then, in tests of the twin paradox in special relativity, objects accelerated in some directions would show more time elapsed on their clocks relative to the unaccelerated clock, rather than the less time they do elapse when accelerated in any direction.

21. Feb 14, 2006

### JesseM

Only if they circumnavigated a closed universe. Any non-circumnavigating trip would still work the same way as in SR, regardless of the earth-twin's velocity relative to the vertical cylinder axis coordinate system.

22. Feb 14, 2006

### Zanket

I disregard the "vertical cylinder axis coordinate system" stuff--makes my eyes glaze over. Whether circumnavigating the universe or a short trip, it is always possible in principle that all benchmark reference points, like the observable universe, can be accelerated beforehand so that a test of the twin paradox has opposite results (i.e. traveling twin elapses more time) when the traveling twin accelerates in the the direction opposite to that of the direction in which the benchmarks were accelerated.

23. Feb 14, 2006

### JesseM

No. In flat spacetime, whether closed like the cylindrical universe or open and infinite like in SR, any twin experiment which doesn't involve circumnavigating the universe will have the same result (the inertial twin will always have aged more), regardless of each twin's velocity relative to "benchmark reference points".

24. Feb 14, 2006

### Zanket

If that were true then real paradoxes would arise, so it cannot be true.

25. Feb 14, 2006

### JesseM

Like what? Consider the possibility that you are just misunderstanding something here.
It is.