Twin Paradox in S^3.

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  • #1
KingOrdo
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X and Y reside in the 3-sphere.

Y is accelerated to near the speed of light; say, 0.9c. He does not ever change direction. In a little while, he meets X, who happens to be residing on the great circle upon which Y is traveling. When they meet, they give each other high-fives. At that moment, the two are identical twins (Y was younger prior to his acceleration).

Y continues alone his great circle, unaccelerated, and therefore in an inertial frame. X is similiarly in an inertial frame. In a little while, they meet again. When they high-five for a second time, which is younger?

From the moment they high-fived for the first time, neither has undergone any accelerations; yet my understanding is that the reason the Twin Paradox can be resolved in the canonical case is that one of the twins underwent acceleration (when his spaceship turned around), and that is why there is an asymmetry between the two twins. But in this case there seems to be no difference, and it really is as accurate to say that X's time dilates with respect to Y as it is to say that Y's time dilates with respect to X.
 

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  • #2
StatusX
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I'm pretty sure this has to do with the question of absolute angular inertial frames, ie, determining whether a particle is spinning or not. Think of it this way: if two buckets of water are at the center of this great circle, one points towards X and one points towards Y, which will have a curved surface of water from the centrifugal force? If the one pointing towards Y has a flat surface, then Y is the one truly at rest. To figure out which it will be, you need to specify a metric, and since this is an artificial model with no masses, it's completely arbitrary. You can make anybody you want be the one at rest. But it's been a little while since I've done general relativity, so maybe someone else should verify (or refute) this.
 
  • #3
Meir Achuz
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"Y continues alone his great circle, unaccelerated"
That can't happen. In circular motion there is centripetal acceleration.
 
  • #4
yenchin
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I remember we have a thread on this sometime ago. This is known as twin paradox in compact spaces. Do a search. It has something to do with topology of the spaces concerned, more specifically, on "winding number".

"Y continues alone his great circle, unaccelerated"
That can't happen. In circular motion there is centripetal acceleration.

No. Travelling in a great circle, which is a geodesic, has no acceleration tangential to the spaces. That is what counts. For a two-dimensional surface, for example S^2, a geodesic is a curve in the surface whose acceleration is perpendicular to the surface. But in dealing with abstract manifold, there is no "outside" of the surface for it to be embedded in. So any component outside of tangent spaces do not count. So travelling along geodesic entails unaccelerated frame.
 
  • #5
George Jones
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Elapsed proper time is found by integrating the metric (which, in general relativity, is a function of spacetime "position") along a worldline.Consequently, if X and Y are coincident at events p and q, there is no reason to expect the same elapsed proper times if they followed different worldlines between p and q, even if both observers have zero 4-acceleration.
 
  • #7
yogi
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All twin paradox(s) disappear (whether cosmological or otherwise) by a straight forward application of the "invariance of the interval." Acceleration has nothing to do with any differential aging problem in SR. In any situation where observers in one frame set up an experiment involving two synchronized clocks in their frame to measure the time logged by a single clock in another frame as it journeys from one clock synchronized clock to the other, the single clock will always log less time than the two synchronized clocks. How the frames got into relative motion and how the single clock returns to the first clock (if it does) is immaterial - neither start-up acceleration nor turn around acceleration plays any part in determining the time difference. Einstein had in right in 1905 - he screwed it up later in 1918 by trying to make his principle theory into a constructive theory
 
  • #8
KingOrdo
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All twin paradox(s) disappear (whether cosmological or otherwise) by a straight forward application of the "invariance of the interval." Acceleration has nothing to do with any differential aging problem in SR. In any situation where observers in one frame set up an experiment involving two synchronized clocks in their frame to measure the time logged by a single clock in another frame as it journeys from one clock synchronized clock to the other, the single clock will always log less time than the two synchronized clocks. How the frames got into relative motion and how the single clock returns to the first clock (if it does) is immaterial - neither start-up acceleration nor turn around acceleration plays any part in determining the time difference. Einstein had in right in 1905 - he screwed it up later in 1918 by trying to make his principle theory into a constructive theory

Yes, roger that; but I don't think that answers the question.

When X and Y high-five the first time, they are identical entities (by the definition of the thought experiment)--equivalently, you could consider them one body with a certain symmetry. When they high-five the second time, time dilation tells us that that symmetry is broken; i.e. one of the bodies will be 'older' than the other. But how can this be without postulating a preferred frame of reference, which is disallowed by the equivalence principle?
 
  • #9
yogi
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The paradox arises because it is mistakenly presumed that SR requires the moving clock to be running slow - it logs less time because it is measured to travel a spatial distance between the two synchronized clocks - it still runs at one second per second in its own frame - there are fewer seconds accumulated because part of the trip involves spatial motion and part of it involves temporal distance ct - SR does not say that the moving clock runs slower than either of the two clocks doing the measuring in the frame that was arbitrarily selected to be at rest - turn the situation around - put the two clocks in the other frame and measure how much time it takes for the remaining clock to pass between the two clocks - it is always the single clock that travels the spatial distance between two fixed clocks that logs less time - its a misnomer to say the moving clock runs slow - it runs the same - but it accumlates less time during the particular experiment.
 
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  • #10
KingOrdo
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The paradox arises because it is mistakenly presumed that SR requires the moving clock to be running slow - it logs less time because it is measured to travel a spatial distance between the two synchronized clocks - it still runs at one second per second in its own frame - there are fewer seconds accumulated because part of the trip involves spatial motion and part of it involves temporal distance ct - SR does not say that the moving clock runs slower than either of the two clocks doing the measuring in the frame that was arbitrarily selected to be at rest - turn the situation around - put the two clocks in the other frame and measure how much time it takes for the remaining clock to pass between the two clocks - it is always the single clock that travels the spatial distance between two fixed clocks that logs less time - its a misnomer to say the moving clock runs slow - it runs the same - but it accumlates less time.

Yes, I understand that. So then would you claim that both X and Y are identical at the time of the second high-five?

And if not, if one has 'aged' with respect to the other, which one is aged, and why?
 
  • #11
yogi
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A local demonstration of the great circle scenero is played out with each GPS satellite - ignoring the height correction - a GPS satellite in orbit experiences no acceleration - each time it passes over a particular location on the earth, it will appear to have lost some time (we can erect a tower to be at the same height of the orbiting satellite to avoid any height difference.) and we will find that the single clock in the tower serves the same function as two clocks in a one way journey - part of the orbiting clock's motion is temporal and part is spatial - the circumference of the orbit and the time lapse of the on board GPS clock have to combine in a pythagorean way to equal a single temporal value ct on the clock at the top of the tower.
 
  • #12
yogi
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Yes, I understand that. So then would you claim that both X and Y are identical at the time of the second high-five?

And if not, if one has 'aged' with respect to the other, which one is aged, and why?

If you know that two clocks are initially at rest, and one is then put in motion on a great circle, the one that has been put in motion will have aged the least as per Einstein 1905 part IV (analogous to the GPS satellite)
 
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  • #13
KingOrdo
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If you know that two clocks are initially at rest, and one is then put in motion on a great circle, the one that has been put in motion will have aged the least as per Einstein 1905 part IV (analogous to the GPS satellite)

The difference is that the GPS satellite is not in an inertial frame; it's orbiting, after all.

And motion is relative; in the thought experiment I provided neither of the two twins were put in motion (Y was not X's twin prior to Y's acceleration); they both were moving at constant velocity.

So I must ask again: "So then would you claim that both X and Y are identical at the time of the second high-five?

And if not, if one has 'aged' with respect to the other, which one is aged, and why?"
 
  • #14
yogi
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As you probably know, the traveler that circumnavigates the universe doesn't live any longer - if his round trip takes one week then he will experience 7 nights of sleep, 21 meals etc - he just covers a lot of territory in a short amount of time on his clock.
 
  • #15
KingOrdo
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As you probably know, the traveler that circumnavigates the universe doesn't live any longer - if his round trip takes one week then he will experience 7 nights of sleep, 21 meals etc - he just covers a lot of territory in a short amount of time on his clock.

Indeed. "So then would you claim that both X and Y are identical at the time of the second high-five?"
 
  • #16
yogi
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Your post 13 - an orbiting satellite is a perfect example of an inertial frame - it feels no acceleration.

In order to properly answer the other questions, one would need to know what you mean by who is younger - In the usual twin paradox, it is clear which twin moved and which did not - and how you are going to measure age - usually it is done with two separated clocks or one clock that remains at rest to which the traveler returns - so in your story it could be that x is also traveling on a geodesic since there is no absolute rest frame in SR.
 
  • #17
yogi
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In other words, in your first post there is an ambiguity as to the initial conditions and as to how one measures which is older. Einstein created confusion when he said moving clocks run slow - you have to look at the experiment and see which observer is measuring age and how it is done. If you rely on Einstein 1905 part IV you could say only that if one of two initially synchronized clocks were accelerated to a high velocity and returned because of space curvature, the non accelerated clock would age most.

Have to Go

Yogi
 
  • #18
KingOrdo
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Your post 13 - an orbiting satellite is a perfect example of an inertial frame - it feels no acceleration.

In order to properly answer the other questions, one would need to know what you mean by who is younger - In the usual twin paradox, it is clear which twin moved and which did not - and how you are going to measure age - usually it is done with two separated clocks or one clock that remains at rest to which the traveler returns - so in your story it could be that x is also traveling on a geodesic since there is no absolute rest frame in SR.

My understanding is that an inertial frame is a non-accelerating coordinate system. The satellite certainly is accelerating; that people living aboard the satellite aren't aware of this is of no concern.

And, Yogi: you're bringing up exactly my point. First, it's not that in the "usual twin paradox, it is clear which twin moved and which did not"--it's that it's clear which twin *accelerated* and which did not. In my example, neither accelerated; they're both candidates for time dilation modulo the other. And yes, they're both on the same geodesic.
 
  • #19
K.J.Healey
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But there is not even a "first time" they're twins.

If they're both holding a digital clock to their chest with big bold letters:
When X see's Y right next to him its possible that their clocks both read 0.
But when Y see's X right next to him, Ys clock isnt at 0 yet, and X's is perhaps already past 0.

There is no simultaneity. (my numbers are not right, maybe even in the wrong order) But the point remains. Who is viewing them as twins (that their clocks are the same) X or Y? Cause by the second time they see eachother, they will both see eachother as OLDER than the other see's themselves, right? It has completely to do with relativity and WHEN you consider them to be "passing" eachother.
 
  • #20
KingOrdo
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But there is not even a "first time" they're twins.

If they're both holding a digital clock to their chest with big bold letters:
When X see's Y right next to him its possible that their clocks both read 0.
But when Y see's X right next to him, Ys clock isnt at 0 yet, and X's is perhaps already past 0.

There is no simultaneity. (my numbers are not right, maybe even in the wrong order) But the point remains. Who is viewing them as twins (that their clocks are the same) X or Y? Cause by the second time they see eachother, they will both see eachother as OLDER than the other see's themselves, right? It has completely to do with relativity and WHEN you consider them to be "passing" eachother.

No, they are twins: that's the whole point of the thought experiment. When X and Y meet the first time, they are physically identical. That's the stipulation; indeed, as I mentioned, one could consider their first high-five as a single body with a symmetry.

And I'm quite aware that there's no simultaneity--indeed, that's what I'm arguing. But in classic relativity, we distinguish the 'younger' twin because of the accelerations he's undergone (the speeding up of the spaceship, its turning around, changes of direction, etc.) In this example, since the space is S^3 and BOTH X and Y are ALWAYS in inertial frames, what is the basis for making that judgment?
 
  • #21
George Jones
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And yes, they're both on the same geodesic.

They most certainly do not move along the same geodesic. The question is: Do they move along equivalent geodesics? Without doing a calculation, I can't say for sure, but I suspect the answer is no. In any case, there is no paradox.

If they move along different geodesics, then there can be an accumulated time difference because the two clocks experience the gravitational field in different ways.

The equivalence principal notwithstanding, the situation you described has to be analysed using general relativity; the concepts and intuition fron special relativity do not suffice.

It is an interesting calculation (not paradox), though.
 
  • #22
KingOrdo
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They most certainly do not move along the same geodesic. The question is: Do they move along equivalent geodesics? Without doing a calculation, I can't say for sure, but I suspect the answer is no. In any case, there is no paradox.
Why can't they be on the same geodesic? Just imagine them in S^2: one is Cairo, one is Mexico City (assume both Cairo and Mexico City are on the Equator).

If they move along different geodesics, then there can be an accumulated time difference because the two clocks experience the gravitational field in different ways.
Again, that's the whole point: there is no gravitational field. From the moment of the first high-five (when the twins are identical) to the moment of the second high-five, both are in inertial frames of reference.

The equivalence principal notwithstanding, the situation you described has to be analysed using general relativity; the concepts and intuition fron special relativity do not suffice.

It is an interesting calculation (not paradox), though.
Well, no one seems to be able to say *why* it's not a paradox. Indeed, the calculation has to be done in GR because the spacetime's not flat. And you're begging the question: yes, elapsed proper time is different for X and Y. But *why* is that the case? It can be resolved in the actual Universe (apparently) because of the acceleration asymmetries undergone by the twin in the spaceship. But that would not be the case in either a matter-free compact space or, say, S^3 with a totally homogeneous distribution of matter. There would be NO grounds for distinguishing X from Y unless you posit--in direct opposition to the equivalence principle--that there exists a preferred frame of reference.
 
  • #23
StatusX
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The paradox is in the apparent symmetry between the two twins. I pointed out in the second post how this symmetry is broken.
 
  • #24
KingOrdo
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The paradox is in the apparent symmetry between the two twins. I pointed out in the second post how this symmetry is broken.

There is no "center of [a] great circle" in S^3. We're not talking about a higher-order embedding. Think of it this way: is there a center to the surface of the Earth (NOT the ball of the Earth)? There is not.
 
  • #25
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No unique center, but there are centers. The centers of the equator are the north and south poles. It is just a point that is the same distance to each point on the great circle. There is a little ambiguity about measuring distance, but you can always take the shortest path between two points.
 
  • #26
KingOrdo
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No unique center, but there are centers. The centers of the equator are the north and south poles. It is just a point that is the same distance to each point on the great circle. There is a little ambiguity about measuring distance, but you can always take the shortest path between two points.

Indeed: and that's exactly the problem. It's just a pure matter of stipulation which person, X or Y, is at rest. So the Twin paradox is not resolved, unlike the R^3 case in which the symmetry is broken when one person's reference frame becomes non-inertial.
 
  • #27
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No, that's not what I meant. Think of it on a 2D sphere. X is on the equator, and Y is going around the equator at a constant velocity. But to Y, it is X that is moving. How do we determine which is really moving? (there must be an asymmetry here, unlike the case in flat space, because they will meet again and we'll be able to compare clocks, the whole point of this thread) Place an object at the north pole. Then this object is at rest with respect to X and Y, but they will disagree as to how it is spinning. We can set it to spin so that Y observes it to be at rest, or that X does. However, only in (at most) one of these cases will it have no centrifugal effects, and so be truly not spinning. This determines which of X and Y is really at rest, and so which ages more.

There may be some subtelty with observing this object across curved space, which is why I'm still not completely convinced this is the correct explanation.
 
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  • #28
KingOrdo
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No, that's not what I meant. Think of it on a 2D sphere. X is on the equator, and Y is going around the equator at a constant velocity. But to Y, it is X that is moving. How do we determine which is really moving? (there must be an asymmetry here, unlike the case in flat space, because they will meet again and we'll be able to compare clocks, the whole point of this thread) Place an object at the north pole. Then this object is at rest with respect to X and Y, but they will disagree as to how it is spinning. We can set it to spin so that Y observes it to be at rest, or that X does. However, only in (at most) one of these cases will it have no centrifugal effects, and so be truly not spinning. This determines which of X and Y is really at rest, and so which ages more.
I just don't see how this could possibly make a difference . . . use primitive test particles instead of extended entities (viz. people). Or, use a geometry in which the distance to circumnavigate the great circle is sufficiently less than the distance from the circle to the 'center' such that no signal can go from the center to the circle, even at c, in the time our sub-c test particle can circumnavigate the circle (imagine a squished globe, for the S^2 case).

And anyway, it doesn't resolve the paradox: the whole point is that it can't be just a stipulation of coordinates that makes one entity younger than the other. In the canonical example, one twin returns to Earth and occupies the same frame of reference as his 'twin'; yet he *still* is younger. It's not that he *appears* younger to some observers: he *is* younger in that frame of reference; it's coordinate-invariant. And your approach is tantamount to just stipulating a coordinate system, which, by the principles of relativity, can't work.
 
  • #29
pervect
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This has been discussed here before, as several people have mentioned. Unfortunately, I don't recall all the details of the previous discussions.

One of several threads on this was https://www.physicsforums.com/archive/index.php/t-51197.html

The literature seems to have some conflicting claims.

http://arxiv.org/abs/physics/0006039 Eur.J.Phys. 23 (2002) 277

for instance, claims that the winding number (homotopy class) can distinguish between the twins.

http://arxiv.org/abs/astro-ph/0606559 published in
http://adsabs.harvard.edu/abs/2006astro.ph..6559R claims otherwise

But there is broad agreement on the general results (one twin will be older) though there is some debate on whether this can be explained by homotopy alone or whether one needs the metric.

Note that one does not need to have a compact universe or a multiply connected topology to have two twins, both following geodesics, have different ages.

A simple case would be if one twin left earth at a high starting velocity, took a loop around a black hole, then came back and passed the Earth again at a high velocity.

That twin would have aged less than the twin who stayed on Earth.

Geodesics are only a local maximum of proper time. To be a global maximum of proper time, one must not only follow a geodesic, but pick the "right one".

To use an analogy I've used before, it's similar to the way that the shortest distance between two points on a curved surface is always a straight line, but a given straight line on a curved surface is not always the shortest distance between two points.

Consider

x ^^^^^ y

where x and y are separated by a tall mountain.

The path directly over the top of the mountain is a straight line. However, it is not the shortest distance between points x and y. The shortest distance is a different straight line - one that goes around the mountain, rather than over it.
 
  • #30
StatusX
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And anyway, it doesn't resolve the paradox: the whole point is that it can't be just a stipulation of coordinates that makes one entity younger than the other. In the canonical example, one twin returns to Earth and occupies the same frame of reference as his 'twin'; yet he *still* is younger. It's not that he *appears* younger to some observers: he *is* younger in that frame of reference; it's coordinate-invariant. And your approach is tantamount to just stipulating a coordinate system, which, by the principles of relativity, can't work.

My point is that while there are no preferred velocities, there are preferred states of rotational motion. That is, no person can claim to be absolutely at rest, but they can claim to not be spinning. Determining this in an abstract situatuation like you've described does entirely come down to a (arbitrary) specification of coordinates. It's difficult to make precise, but I believe this is how you distinguish two people who undergo trips around closed loops.
 
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  • #31
KingOrdo
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Ahh . . . very interesting: so this is by no means an open-and-shut case; there's no consensus even in the literature.

There seems to be no dispute that one twin *will* be older in this experiment. However, no one is quite sure *why* that's the case. We know why--or, at least, we thought we knew why--in the canonical experiment (viz. non-inertial reference frames), but that can't be the explanation here. 'pervect''s black hole example could even be explained that way, since an orbit is, by definition, non-inertial (though it is freely falling).

But that can't be the case in a matter-free S^3 universe, since there's no gravitational force at all (make the test particles vanishingly small). Or, in a universe with perfectly homogeneous distribution of matter (which affects both test particles identically).

So it appears we're left with one of two options: either
(1) we posit that physical laws are not invariant modulo the geometry in which they are instantiated, or
(2) our currented accounting of the twins paradox is wrong.

Neither is attractive, but I see no other options. (2) also seems to imply that foundational relativistic principles (e.g. the equivalence principle) are wrong.

Thoughts?
 
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  • #32
JustinLevy
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'pervect''s black hole example could even be explained that way, since an orbit is, by definition, non-inertial (though it is freely falling).
I think you are still missing something here. Free fall, or a gravitational orbit, etc IS inertial motion. I think you may be confusing "coordinate" acceleration with physical acceleration.
 
  • #33
KingOrdo
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I think you are still missing something here. Free fall, or a gravitational orbit, etc IS inertial motion. I think you may be confusing "coordinate" acceleration with physical acceleration.
No. Imagine you're on the Shuttle, orbiting the Earth. It may *appear* inertial to you, and indeed if you don't need hyper-precise measurements you can *assume* it's inertial, but it's not: it's an accelerated frame, which you can tell from the tidal forces due to the presence of the Earth. It's especially obvious if you substitute "neutron star" for "Earth".

Now, we're talking about curved spacetime, in which GR--not SR--holds, and global inertial frames are nonexistent. However, locally both X and Y are in inertial frames of reference. And hence, the paradox.
 
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  • #34
pervect
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Ahh . . . very interesting: so this is by no means an open-and-shut case; there's no consensus even in the literature.

While there appears to be some disagreement in the literature, it's about rather fine details.

There is a general agreement about the broad details, which is that the two twins won't be the same age.

The disagreement as I read it is how to explain the age difference, i.e. to point to a particular mechanism. The first papers say that it can be addressed in terms of winding number (which is a topological property that doesn't even need a metric). The second papers say that this is not correct, that one needs more than the winding number to properly explain the age difference, that one needs the metric information.

Everyone agrees that there should be an age difference AFAIK.

As I said, I haven't really stuidied this particular matter very closely.
 
  • #35
KingOrdo
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While there appears to be some disagreement in the literature, it's about rather fine details.

There is a general agreement about the broad details, which is that the two twins won't be the same age.

Right; and this was never in question in the first place--like I said, "There seems to be no dispute that one twin *will* be older in this experiment. However, no one is quite sure *why* that's the case."

I am curious to hear what people think about a potential resolution; the options I cited above seem to me to be the only way to patch a pretty glaring lacuna in our physics. I've always thought intuitively that the obstacles to quantum gravity were on the QM side-of-the-house . . . but perhaps relativity theory is the problem?
 

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