A variation on the twin paradox

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 :wink:

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.
 
I found this article: http://www.findarticles.com/p/articles/mi_qa3742/is_200108/ai_n8954244

And I'm bothered by this paragraph:

The statement of the closed universe twin paradox is completely symmetrical-- Albert and Betty travel in opposite directions from one another until they reunite-- yet the resolution is asymmetrical-Albert's erstwhile twin is now a little sister 10 years younger than he is. How can this be? The answer is that we broke the problem's symmetry when we constructed the spacetime diagram. Albert occupies a very special inertial frame. His is the only frame in which lines of constant time are closed circles. In all other frames the lines of constant time are helices. The faster an observer is moving relative to Albert, the steeper the pitch of the helix.
Couldn't we make the same argument if we chose to draw Betty's spacetime diagram instead?
 

JesseM

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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.
 

Hurkyl

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a general relatavistic "you can't compare time or space intervals at a distance" effect.
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.


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
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.


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

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

And [itex]a(t)=1[/itex] 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 [itex]\kappa=1[/itex] (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?
 

Garth

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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:
In a compact space, the paradox is more complicated. If the travelling twin is on a periodic orbit, she can remain in an inertial frame for all time as she travels around the compact space, never stopping or turning. Since both twins are inertial, both should see the other suffer a time dilation. The paradox again arises that both will believe the other to be younger when the twin in the rocket flies by. The twin paradox can be resolved in compact space and we will show that the twin in the rocket is in fact younger than her sibling after a complete transit around the compact space. The resolution hinges on the existence of a preferred frame introduced by the topology.
and Uzan et al. in Twin paradox and space topology in which they say:
Thus in Friedmann–Lemaıtre universes, (i) the expansion of the universe and (ii) the existence of a non–trivial topology for the constant time hypersurfaces both break the Poincare invariance and single out the same “privileged” inertial observer who will age more quickly than any other twin: the one comoving with the cosmic fluid – although aging more quickly than all her travelling sisters may be not a real privilege!
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:
We have seen that the traditional lesson of special relativity--that all inertial frames are equivalent--applies only locally.
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
 
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dicerandom said:
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 each other the second time?
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.

The asymmetry in the twin paradox, including this cosmological twin paradox, is acceleration. Whoever accelerates experiences length contraction. Distances contract along your axis of motion when you accelerate in your spaceship, and expand from their contracted state when you decelerate to come to rest with respect to your friend. During your trip, both you and your friend are moving, relative to each other. But your friend does not experience length contraction of distances. Both of you travel at the same velocity relative to each other, but you travel less distance. Traveling less distance between flybys than your friend does, at the same velocity your friend has, takes less time for you than it does for your friend. It’s no more complicated than that.
 
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Zanket said:
The asymmetry in the twin paradox, including this cosmological twin paradox, is acceleration. Whoever accelerates experiences length contraction. Distances contract along your axis of motion when you accelerate in your spaceship, and expand from their contracted state when you decelerate to come to rest with respect to your friend. During your trip, both you and your friend are moving, relative to each other. But your friend does not experience length contraction of distances. Both of you travel at the same velocity relative to each other, but you travel less distance. Traveling less distance between flybys than your friend does, at the same velocity your friend has, takes less time for you than it does for your friend. It’s no more complicated than that.
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.
 

Garth

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Zanket said:
I assume that you and your friend were once at rest with respect to each other,
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
 
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dicerandom said:
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.
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?
 
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pervect

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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.
 
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Garth said:
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?
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.
 
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Hurkyl

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the one who experiences the greatest length contraction
There's no such thing as "experiencing" length contraction.
 
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Like it better now?
 

JesseM

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Zanket said:
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.
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.
 

Hurkyl

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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.
 
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JesseM said:
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.
I agree with that. Good luck accelerating all those galaxies though! :smile:
 
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Hurkyl said:
Nobody in that example would "observe" length contraction -- for each observer, the distance between galaxies would remain constant.
How did you gather that from Garth's example?

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

JesseM

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Zanket said:
I agree with that. Good luck accelerating all those galaxies though! :smile:
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?
 
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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.
 

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