The Twin Paradox in a Closed Universe

[Souvik]

Twins A and B.
A stays put.
B flies off in a straight line with uniform velocity -- keeps flying
straight on along a geodesic -- till he meets A again because the
universe curves and is closed!
Now who's the younger of the two? How do we tell?

Confused again I am.
-Souvik

[Uncle Al]

The twin that traverses the most space accumulates the least time.
Acceleration is irrelevent. One can trivially formulate the Triplet
Paradox and eliminate acceleration entirely during the course of the
experiment and for arbitrary lengths of time before adn after,

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--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf

[Perspicacious]

> Twins A and B.
> A stays put.

> B flies off ... till he meets A again because the

> universe curves and is closed!
> Now who's the younger of the two? How do we tell?

Souvik,

The paper, "On the Twin Paradox in a Universe with a Compact
Dimension" presents a very clear answer to your question:

"We consider the twin paradox of special relativity in a
universe with a compact spatial dimension. Such topology
allows two twin observers to remain inertial yet meet
periodically. The paradox is resolved by considering
the relationship of each twin to a preferred inertial
reference frame which exists in such a universe because
global Lorentz invariance is broken. The twins can perform
'global' experiments to determine their velocities with
respect to the preferred reference frame (by sending
light signals around the cylinder, for instance)."
http://arxiv.org/PS_cache/gr-qc/pdf/0503/0503070.pdf

See these references also:

http://physics.ucr.edu/Active/Abs/abstract-13-NOV-97.html
http://www.everythingimportant.org/viewtopic.php?t=79
http://cornell.mirror.aps.org/abstract/PRD/v8/i6/p1662_1
http://arxiv.org/PS_cache/gr-qc/pdf/0101/0101014.pdf
http://arxiv.org/PS_cache/physics/pdf/0006/0006039.pdf
http://www.everythingimportant.org/viewtopic.php?t=605
http://www.everythingimportant.org/relativity/simultaneity.htm

All this analysis has a pointed answer. When the two twins
meet again, the youngest, least-aged twin will be the one
who is moving the fastest with respect to the absolute
frame of reference.

[Souvik]

Uncle Al wrote:
> The twin that traverses the most space accumulates the least time.

So how do we know who traverses the most space?
Given that there are no accelerations in the system, each can consider
herself to be in a state of rest and claim that the other is moving
relative to her. I don't see any physical way that one can claim to
have traversed more space than the other.

Thanks Perspicacious. That is exactly one of the possibilities I was
juggling in my head! That there therefore must be an absolute
four-velocity in a closed universe.
I kept thinking it would be a rather far-fetched conclusion from such a
simple thought experiment.
Cool! I'll go through the papers you referred to.

Thanks,
Souvik

[a student]

Perspicacious wrote:
> > Twins A and B.
> > A stays put.
> > B flies off ... till he meets A again because the
> > universe curves and is closed!
> > Now who's the younger of the two? How do we tell?

The way we resolve all twin paradoxes - by going to the principle of
equivalence, which specifies what happens to clocks (and what is meant
by an inertial frame) no matter what the motion is. In fact special
relativity in its purest form suffers from two mild forms of
incompleteness: it doesn't tell us (i) which of two relatively
accelerating frames is inertial (if either); and (ii) how a clock
behaves if accelerated relative to an inertial frame. This is probably
what generates so much debate on twin paradoxes.

For a given spacetime point, one physically specifies a family of local
inertial frames, as those relative to which a freely moving particle
passing through the point has zero acceleration (to first order).
These frames thus differ by constant relative speeds, and the principle
of equivalence says that the local physics is invariant under the
corresponding set of Lorentz transformations between these frames.

This gives the recipe for finding out what happens to any clock -go to
an inertial frame in which it is freely moving, and calculate the
proper time relative to that frame. But in fact, having the recipe,
one can then transform back to any other coordinate system, and rewrite
the proper time as
ds^2 = g_uv (dx^u/dt) (dx^v/dt)
where g_uv is the induced metric tensor in the general coordinate
system (and reduces to the Minkowski tensor in the local inertial
frames by construction).

In your form of the twin paradox, you are clearly assuming that the
spacetime is locally flat, like a cylinder, so that both twins are
freely moving, and g_uv can be taken to be the Minkowski tensor for
both twins. If you take one to be "motionless", and one to wind around
the cylinder at speed v, (these are topologically distinct and
invariant notions), the above formula gives proper times T and
T/sqrt(1-v^2/c^2) respectively, where T is the time measured by the
motionless twin until the rendezvous. No paradox. The circumnavigator
is (deservedly) younger!

One moves on from (the weak form of) the principle of equivalence
above, to general relativity, by (i) asserting that freely falling
particles are freely moving particles (i.e., gravity doesn't count as a
force), leading to the geodesic equation of motion, and (ii) being
incredibly clever like Einstein and finding the simplest way of
specifying the metric tensor from the energy-momentum tensor, by
equating the simplest possible divergence-free function of the former
with a conservation for the latter.

[Bossavit]

This was very well treated in

J.R. Weeks: "The Twin Paradox in a Closed Universe", Am. Math. Monthly,
108, 7 (2001), pp. 585-90.

As Weeks writes,
"The traditional lesson of special relativity-that all inertial frames
are equivalent-applies only locally. Globally the symmetry is broken in
any universe that is finite, or began with a big bang."