1. On 2002-08-23 12:00, Zathras wrote:
The difference between this example and the standard Twin Paradox is that no acceleration, and hence no change in inertial frames, occurs in this example.
Your example seems to be one where space is so curved that the path comes back on itself, even though the object is moving at a speed of c/2--I assume you are thinking of an orbit, perhaps? The gravity would probably be quite strong for such high velocity. If the first one is in orbit, how does the second one maintain its position?

I realize that this is just a thought experiment, but I am trying to point out that your assumption that they are in inertial reference frames may not be valid--so you'll have to be more specific.

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Good question. I was not thinking of an orbit, but perhaps more of a space with a uniform distribution of mass, so that the space has a uniform curvature. This way, space (I think) would have the topology of a hypersphere. Since the distribution of mass is constant, there would not be any gravitational "force."

I am of course thinking of massless twins here (or is it "test twins?")

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I don't think you'll be able to find a mass distribution in which the worldline of the moving twin crosses the same point in space without some acceleration being involved.

Don

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I would think that many such universe topologies should be able to admissible. The simplest one is the one I gave, the hypersphere. The 1-dimensional analog is a bead which is constrained to move in a circle. Without friction, it experiences no acceleration in its embedded space (the operative distinction here). If a simultaneous slice of space had the topology of a 3-sphere, it would seem that no acceleration would require an object in motion, relative to one frame, to return to its previous spatial position. Saying that it has the topology of a 3-sphere is not the same as saying it in an actual 3-sphere embedded in a 4-d space, as the topology is instrinsic. In this situation no acceleration is required to return.

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Hmmm... Don sees your point. Much apologizings. Don was trapped in linear thinking. Poor Don.

No one listen to poor Don...

Won't talk. Not the One.

Don

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spaced out

you wrote:

The slowing of time is strictly a function of speed and direction of travel does not factor into it, therefore the twin in motion will age more slowly from the perspective of the twin who is not in motion.
-------------------------------------------
I say

due to relativity which twin is moving and which twin is stationary?

anyone would say in this universe - Where is the stationary fixed refence point? - there is none

therefore which twin ages more than the other one

I think I am still correct in thinking when two objects travel towards each other time goes quicker and when two objects travel away from each other time slows
for both in equal amounts

<font size=-1>[ This Message was edited by: 99homer99 on 2002-08-23 17:56 ]</font>

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On 2002-08-23 17:54, 99homer99 wrote:
due to relativity which twin is moving and which twin is stationary?
anyone would say in this universe - Where is the stationary fixed refence point? - there is none
This is correct. That's why it's called the twin paradox.

I think I am still correct in thinking when two objects travel towards each other time goes quicker and when two objects travel away from each other time slows
for both in equal amounts
No, you are wrong.

As I and others have been trying to tell you: direction is irrelevant, as the velocity is squared. The symmetry is broken by the acceleration that one twin undergoes to get up to speed and then turn around. That resolves the paradox.

But you don't have to take my word for it; just read a book on relativity.

Don

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homer, you are right to feel the way you do. We think that it should be symmetric. This is where the paradox comes into play. Indeed, on the way out the twin on Earth will receive fewer "ticks of the light clock" than the rocket-twin when leaving and more "ticks of the light clock" than the rocket-twin when coming home. However, if you add up all the ticks seen by the twin on Earth and all the ticks seen by the twin on the spaceship, they are not the same! This paradox has, at its root, the fact that we're not dealing with inertial reference frames. You are right, homer, when you feel that there is symmetry in relativity. However, once accleration occurs, the symmetry is broken. This is a profound effect, and is dealt with more carefully in another Equivalence Principle of Einstein's. For more info, homer, please go to the following website:

http://www.physics.hku.hk/~tboyce/mo...ics/twins.html

9. On 2002-08-23 23:59, JS Princeton wrote:
This paradox has, at its root, the fact that we're not dealing with inertial reference frames.
But, as JW reminded us, the paradox appeared in Einstein's first paper on relativity--using nothing more than his special relativity. The whole paradox can be expressed with nothing more than inertial reference frames.

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Actually, having the twins go in an orbit gives another possibility for my question about a GR twin paradox. Suppose the two twins are in the same circular orbit about a black hole, going in oppposite directions. You can have them going around at r=3m(*G/c^2), so that their speed relative to each other is certainly relativistic. Their perspectives are certainly indistinguishable, so how would time dilation play out in this circumstance?

This problem started badly, and it will likely end badly. At least there is symmetry.
-Zathras

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On 2002-08-24 09:50, Zathras wrote:
Actually, having the twins go in an orbit gives another possibility for my question about a GR twin paradox. Suppose the two twins are in the same circular orbit about a black hole, going in oppposite directions. You can have them going around at r=3m(*G/c^2), so that their speed relative to each other is certainly relativistic. Their perspectives are certainly indistinguishable, so how would time dilation play out in this circumstance?

This problem started badly, and it will likely end badly. At least there is symmetry.
-Zathras
My guess is that since neither of them is accelerating relative to the other, they are both subjected to the same time dialation. There is no paradox in this situation, as all effects are equal.

12. On 2002-08-24 11:29, David Hall wrote:
My guess is that since neither of them is accelerating relative to the other, they are both subjected to the same time dialation.
I think that's the whole point, the symmetry isn't broken. But I think zathras's question is, aren't they experiencing a relative difference in speed (or velocity), so wouldn't their perception of time be different? And how can those differences be reconciled when they hook up each time around.

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On 2002-08-24 12:17, GrapesOfWrath wrote:
On 2002-08-24 11:29, David Hall wrote:
My guess is that since neither of them is accelerating relative to the other, they are both subjected to the same time dialation.
I think that's the whole point, the symmetry isn't broken. But I think zathras's question is, aren't they experiencing a relative difference in speed (or velocity), so wouldn't their perception of time be different? And how can those differences be reconciled when they hook up each time around.
Right, that is the whole point of my question. How does time dilation work in the examples I gave previously?

14. Right, that is the whole point of my question. How does time dilation work in the examples I gave previously?

The time dilation calculation uses the square of the velocity, so the direction component essentially drops out. As long as the two are moving at the same velocity they would experience the same time dilation relative to the same reference frame.

t=t'L=t'(1-(v/c)<sup>2</sup>)<sup>-1/2</sup>
where t=reference frame time

If that's true, then they experience no time dilation relative to each other.

(Hit submit too soon.)

<font size=-1>[ This Message was edited by: Jim on 2002-08-26 12:32 ]</font>

15. On 2002-08-26 12:25, Jim wrote:
The time dilation calculation uses the square of the velocity, so the direction component essentially drops out. As long as the two are moving at the same velocity they would experience the same time dilation relative to the same rest frame.
That's not what zathras is asking. Do they view each other as different? Obviously not, since they experience the same dilation, according to you. Why not?

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Zathras,

First, thanks for the interesting question.

Second, a disclaimer, I am not a astrophysicist but a poor schlub teaching himself general relativity.

This problem has been analyzed for orbits and other cyclic problems. The major problem is that the "moving" twin will not be able to synchronize his or her clock. Observers at the same point in spacetime can have clocks that read different times. The paradox is resolved because the moving twin agrees that he is younger than the stationary twin. All this is tantamount to a "preferred" frame (Lord, I hope the geocentrist are not reading this.) which is set by the geometry of spacetime.

Judging from your previous posts, you are not unaquainted with mathematics. Barrow and Levin did a paper on this: The twin paradox in compact spaces. I'm still going through the details of this paper so don't ask me any hard questions. [img]/phpBB/images/smiles/icon_smile.gif[/img]

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On 2002-08-26 13:51, Wiley wrote:
All this is tantamount to a "preferred" frame (Lord, I hope the geocentrist are not reading this.) which is set by the geometry of spacetime.
An intuitive way of thinking about this is to consider what happens if an observer fires off photons in all directions. Those photons will travel around the compact space and meet up again at their starting point. The observer, whether moving or stationary, will encounter the photons at some point. If he encounters them all together, he knows he is at the same location as when he fired them. Thus establishing an absolute rest frame. I don't think this will help the Geocentrists any, though.

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On 2002-08-26 14:28, MartinM wrote:
Those photons will travel around the compact space and meet up again at their starting point. The observer, whether moving or stationary, will encounter the photons at some point. If he encounters them all together, he knows he is at the same location as when he fired them. Thus establishing an absolute rest frame.
Doesn't this imply that that the space around a "preferred" observer is symmetric? The geodesic path length in each direction is the same? I don't see that.

If an observer emits photons at a constant interval dT. When she recieves the photons, if they are at the same interval spacing, dT, then she is in a "preferred" frame. I can see that.

I don't think this will help the Geocentrists any, though.
You may be surprised at the geocentrist's ability to twist legitmate science to fit his agenda. Read a few of Dunash's and Prince's posts. (shudders)

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On 2002-08-26 15:03, Wiley wrote:
On 2002-08-26 14:28, MartinM wrote:
Those photons will travel around the compact space and meet up again at their starting point. The observer, whether moving or stationary, will encounter the photons at some point. If he encounters them all together, he knows he is at the same location as when he fired them. Thus establishing an absolute rest frame.
Doesn't this imply that that the space around a "preferred" observer is symmetric? The geodesic path length in each direction is the same? I don't see that.

If an observer emits photons at a constant interval dT. When she recieves the photons, if they are at the same interval spacing, dT, then she is in a "preferred" frame. I can see that.

I don't think this will help the Geocentrists any, though.
You may be surprised at the geocentrist's ability to twist legitmate science to fit his agenda. Read a few of Dunash's and Prince's posts. (shudders)
To take the example of the 3-sphere, anyone who stays an inertial frame for the duration of the photons' trip around would see the photons simultaneously. I guess if your space had less symmetry, then there might be only isolated pockets where the above situation might occur.

BTW, thanks for the Barrow article. It looks interesting and on-point, but I haven't taken the time to digest it yet.

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A spacetime diagram affords an easy way to visualize the lack of symmetry in the Twin Paradox. It also simplifies things if we treat the spaceship as traveling so close to the speed of light that the difference is undetectable as far as space and time increments are concerned. The fact that the spaceship mass and kinetic energy relative to earth are both eseentially infinite is not relevant to the thought experiment.

Visualize a horizontal x axis extending to the right as the spatial direction of spaceship travel to a point one light year from earth and a vertical upward t axis as the direction toward future time. Let one light year in distance and one year in time be marked off as equal distances along the x and t axes, respectively. Light traveling to the right has a world line that bisects the angles between the x and t axes, that is, slants upward to the right at a 45-degree angle from the x axis. Light traveling in the opposite direction slants upward to the left at a 45-degree angle from the -x axis.

Let the spaceship travel at so close to the speed of light that for practical purposes we can treat it as traveling at exactly the speed of light. Relativity says it would take an infinite amount of energy to accelerate it to that speed, but this only a thought experiment, so this approximation does not affect the validity of the results of the experiment as far as kinematics are concerned.

The world line of the spaceship is therefore indistinguishable from that of a light ray leaving earth in the direction of travel of the space ship simultaneously with the spaceship.

An observer on earth watching the spaceship will see it arrive at the point one light year from earth for practical purposes precisely TWO years after it left earth, ignoring the slight difference between the spaceship's speed and the speed of light. This is because it takes one year for the space ship to reach its turnaround point and another year for light that left the space ship as it's turning around to reach earth. Let the turnaround be instantaneous. From that point, it takes the spaceship another year to return to earth, and it arrives at essentially the same instant as does the light ray that left it as it at the instant when it turned around. Since the observer on earth saw the spaceship traveling at essentially the speed of light both going and returning, time on it appeared to be essentially standing still both going and returning. Thus, two years after the spaceship departed earth, it was back on earth with its occupants essentially the same ages that they were when they started out while the observers on earth are essentially two years older.

Incidentally, it will appear to observers on earth that the space ship took TWO years to reach the point ONE light year away, then to make the trip back to earth INSTANTLY -- in zero time!

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On 2002-08-26 15:03, Wiley wrote:
Doesn't this imply that that the space around a "preferred" observer is symmetric? The geodesic path length in each direction is the same? I don't see that.
You're right, that was sloppy.

Consider a closed null geodesic passing through the observers location. Then two photons fired along this geodesic in opposite directions will meet up again in only two locations - the other side of the compact space, and the original starting point. Since the observer cannot reach the other side of the compact space before the photons, the only way he can re-encounter both simultaneously is if he is still at the same location. Similarly for pairs of photons on any other closed null geodesic.

There. I think that's better.

You may be surprised at the geocentrist's ability to twist legitmate science to fit his agenda. Read a few of Dunash's and Prince's posts. (shudders)
Heh. I've spent the last few months debating Creation vs Evolution on a fundie board. I know exactly what you mean [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

22. On 2002-08-26 12:33, GrapesOfWrath wrote:
On 2002-08-26 12:25, Jim wrote:
The time dilation calculation uses the square of the velocity, so the direction component essentially drops out. As long as the two are moving at the same velocity they would experience the same time dilation relative to the same rest frame.
That's not what zathras is asking. Do they view each other as different? Obviously not, since they experience the same dilation, according to you. Why not?
I think they would see each other as aging at the same rate.

They are both in motion wrt the same reference frame; for all relativistic purposes, their motion is the same (the only diference is the velocity vector, but the absolute value is the same).

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On 2002-08-26 17:30, Jim wrote:
On 2002-08-26 12:33, GrapesOfWrath wrote:
On 2002-08-26 12:25, Jim wrote:
The time dilation calculation uses the square of the velocity, so the direction component essentially drops out. As long as the two are moving at the same velocity they would experience the same time dilation relative to the same rest frame.
That's not what zathras is asking. Do they view each other as different? Obviously not, since they experience the same dilation, according to you. Why not?
I think they would see each other as aging at the same rate.

They are both in motion wrt the same reference frame; for all relativistic purposes, their motion is the same (the only diference is the velocity vector, but the absolute value is the same).
This is the result that I would have expected, but the Barrow article referred to earlier states that the result is not the same in the instance of a compact space. It reaches this result even though there would seem to exist a CM frame in which they are both moving in equal but opposite directions.

Now I'm more confused than when I first asked the question. I hope that's a sign of progress.

24. On 2002-08-26 17:30, Jim wrote:
I think they would see each other as aging at the same rate.

They are both in motion wrt the same reference frame; for all relativistic purposes, their motion is the same (the only diference is the velocity vector, but the absolute value is the same).
That would be the natural conclusion, but what about the usual point of view of one which assumes itself is stationary while the other is moving--and then from the POV of the other? How do you treat those POVs?

25. Guest
I did watch TV When the Pz came to PDX to poise 4 City Mayor
Standard Arms race stuff {Yes & nO}
anyway later in the news cast they showed the Governer of the State
{Yes, Yes} ive run for Governor as a LIBERTERIAN
never mind politicks
anyway it looked to me as though the Governer
age like 100 years well at least 10
you know Standard symptoms Gray Hair
wrinkles & just plain aged look
of course i dnt rule out makeup artist in this?

26. On 2002-08-27 10:24, HUb' wrote:
you know Standard symptoms Gray Hair
wrinkles & just plain aged look
of course i dnt rule out makeup artist in this?
Another conspiracy, so they can play golf more often

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On 2002-08-26 17:22, MartinM wrote:
Consider a closed null geodesic passing through the observers location. Then two photons fired along this geodesic in opposite directions will meet up again in only two locations - the other side of the compact space, and the original starting point. Since the observer cannot reach the other side of the compact space before the photons, the only way he can re-encounter both simultaneously is if he is still at the same location. Similarly for pairs of photons on any other closed null geodesic.

There. I think that's better.
Yeah, that makes sense. The simplest case of an orbit or a 3-sphere that Zathras suggest would have symmetry in all directions. Now does this stationary frame always measure the smallest volume? Thus satisfying Barrow & Levin's definition of a "preferred" frame.

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On 2002-08-23 23:59, JS Princeton wrote:
homer, you are right to feel the way you do. We think that it should be symmetric. This is where the paradox comes into play. Indeed, on the way out the twin on Earth will receive fewer "ticks of the light clock" than the rocket-twin when leaving and more "ticks of the light clock" than the rocket-twin when coming home. However, if you add up all the ticks seen by the twin on Earth and all the ticks seen by the twin on the spaceship, they are not the same! This paradox has, at its root, the fact that we're not dealing with inertial reference frames. You are right, homer, when you feel that there is symmetry in relativity. However, once accleration occurs, the symmetry is broken. This is a profound effect, and is dealt with more carefully in another Equivalence Principle of Einstein's. For more info, homer, please go to the following website:

http://www.physics.hku.hk/~tboyce/mo...ics/twins.html
I'm sorry this web page doesn't prove anything to me
except on the way out Dick receives a signal every three years from Jane

and on the way back Jane receives 3 signals from Dick every year

and it took Dick 15 years to get to a planet that was 20 light years away

at 0.8c it should have taken him 20 devided by 0.8 = 25 years

is it me?

<font size=-1>[ This Message was edited by: 99homer99 on 2002-08-29 17:52 ]</font>

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On 2002-08-29 17:49, 99homer99 wrote:

and it took Dick 15 years to get to a planet that was 20 light years away

at 0.8c it should have taken him 20 devided by 0.8 = 25 years

is it me?
It does take Dick 25 years to get there, according to the way Jane measures time. From Dick's point of view, it only took 15 years because the space in Jane's frame is contracted. So the star looks 20 light years away to Jane, but for Dick, moving at 0.8c, it's only 15 light years away. Equivalently, Jane expects Dick to arrive looking 50 years older, but he is only 30 years older, because his time ran slower than hers.

The space-time diagram in that web page shows how this is true, even when you take light travel time into account. In other words, the time dilation is not due to a simple delay of signal propagation due to increasing distance, and it does not "undo itself" when Dick switches direction.

Don

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sorry you can't convince me
that Dick is travelling faster than the light is, when he is travelling at 0.8c

you said:
So the star looks 20 light years away to Jane, but for Dick, moving at 0.8c, it's only 15 light years away.

You are actualy saying that dick gets to the planet before the light from earth has.

(he gets there in fifteen years and the light gets there in 20 years)

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