Twin paradox in a closed universe

[Daniel42]

I`ve thought about a special sort of twin paradox.
I know the usual explanation of the twin paradox but give me please the answer to this special case:

Imagine:
A static universe (non-expanding) with a closed geometry and a circumference of one lightyear. The twins start their journey in different direction from their planet (EARTH2) with nearly light speed.


<-------- [TWIN1] [EARTH2] [TWIN2] -------->


Here is my question:
When they will met again after one year on EARTH2 --
which twin is the younger one?

The answer to my question isn`t so easy to give as it seems.
Please think correctly.

You may try this answer:

Twin 1 travels in the system of twin 2 and therefore he ages less because of the time-slowing, conclusion:
he has a different age!
This answer is in contradiction with the principle of relativity!


You may try perhaps this answer:

They both travel the same journey, the conclusion is:
- they both have the same age.

Now you have a serious problem.
Twin 1 had placed 1000 clocks along the circumference of the universe.
The clocks have all the same distance between them - and they also move exactly in the same direction and with the same velocity like twin 1 does.
Imagine now twin 1 synchronises all these moving clocks with his own clock. They all show the same time in the system of twin 1.

Twin 1 says:
I see my twin moves forward from clock1 to clock2 and so on and so I can see that his own slow-down-time conforms to Einsteins theory of relativity.
When my twin finally reaches clock999 to clock1000 and EARTH2 he is finally jounger than me.

Twin 2 makes the same experiment with another set of 1000 clocks.

Twin 2 says:
I see my twin moves forward from clock1 to clock2 and so on and so I can see that his own slow-down-time conforms to Einsteins theory of relativity.
When my twin finally reaches clock999 to clock1000 and EARTH2 he is finally younger than me too!

What are you thinking now?
Who`s one is right?


Please don`t give the usual answer to me I have to search the solution in the general theory of relativity.

That`s no answer at all.
My question is:
Which twin is actually the younger one?
- or have both the same age?
What answer gives us the general theory to this question?
Is there no time-slowing at all?!
Please answer the question.


In my opinion there is no solution to this question.
This paradox shows:
In our universe the special theory of relativity is right. I agree.
But in a closed universe it can`t be right,
conclusion:

we don`t live in a closed universe.

Think you got to know this.
That`s all!

 

[Tao-Fu]

Why are you assuming that special relativity is valid in a situation with such high curvature? You also haven't exactly described the large scale structure. A toroid has an intrinsic curvature in only one direction, for example.

 

[Dale]

You would have to give the metric of that universe. Once you have done so each twin simply integrates the metric along the worldlines and they come up with a number that they both agree on.

 

[Daniel42]

Why are you assuming that special relativity is valid in a situation with such high curvature? You also haven't exactly described the large scale structure. A toroid has an intrinsic curvature in only one direction, for example.

The metric is of no importance. Choose the metric you want with the circumference of one lightyear. and -
you didn`t answer the question cause you can`t.
That`s it is as I said.

I`m not assuming that special relativity is valid in a situation with such high curvature.
But what is the answer?

The answer is:
We did not live in a closed universe.
Give me some other answer...

 

[Daniel42]

You would have to give the metric of that universe. Once you have done so each twin simply integrates the metric along the worldlines and they come up with a number that they both agree on.

Choose the metric you want with the circumference of one lightyear.
Answer the question.

You can`t do this.
I know it.

 

[atyy]

http://arxiv.org/abs/gr-qc/0101014

http://arxiv.org/abs/gr-qc/0503070

 

[Dale]

Choose the metric you want with the circumference of one lightyear.
Answer the question.

OK, if I am free to choose then I will chose a cylindrical universe with circumference in x of 1 light year such that $(t,x,y,z) = (t,x+n,y,z)$ in units where c=1. The metric on a cylinder is flat so:
$$ds^2=dt^2-dx^2$$

If the stay at home twin is at rest such that light pulses sent in opposite directions at the same time will be received at the same time and the moving twin has a speed of 0.6 c relative to him then they will meet again every 1.66... years according to the stay at home twin's clock. Their worldlines are:
$$w_s=(t,0,0,0)$$
and
$$w_m=(t,mod(.6t),0,0)$$

Integrating the metric along each worldline from t=0 to t=1.66... gives
$$s_s= \int_0^{1.66...} \sqrt{1^2-0^2} \, dt =1.66...$$
$$s_m= \int_0^{1.66...} \sqrt{1^2-.6^2} \, dt =1.33...$$

You can`t do this.
I know it.

I can do it for any case in which you give me a topology a metric and two worldlines.

 

[JustinLevy]

This paradox shows:
In our universe the special theory of relativity is right. I agree.
But in a closed universe it can`t be right,
conclusion:

we don`t live in a closed universe.

Just to make it ultra-clear in case you don't see it from the previous posts, you cannot make that claim.

In Dalespam's example, the spacetime was still everywhere flat, so we can still use SR to make calculations. There is no local preferred frame, but there is now a global preferred frame. This is because the global structure of the universe itself isn't invariant to poincare symmetry in that example. It is not even invariant to rotational symmetry (there is a clear global difference between the two spatial directions that aren't closed and the one that is closed). So yes, if you could send things all the way around the universe, we could measure our speed with respect to this "global preferred frame" even though we cannot do so locally.

In summary, because there is no local preferred frame, and spacetime is everywhere flat, we will be blissfully unaware of any problems unless an experiment involves a patch of spacetime large enough that it connects around the closed universe. As all our experiments are "local" in this sense, you cannot claim to have ruled out a closed topology for the universe.

 

[Dmitry67]

Even in open universe, there is a globelly preferred... well, it is not a frame, but in every point there is a preferred frame, which is at rest relative to CMB. In different points these frames are different, but in some sense Lorentz invariance is broken in Cosmology.

 

[Daniel42]

Thank you very much for this answer.

If the stay at home twin is at rest ...

Unfortunally my twin paradox is a little bit different as the usual one.

Actually no twin stay at home.
Both twins takes the journey.
Only EARTH2 stay at home.

Please tell me how much older twin1 is when he is back at EARTH2 and how much older twin2
is, when he is back at EARTH2.

Which twin is older?
Or have twin1 and twin2 the same age?

That was the question.


(Both answers are in contradiction to the SRT.
Before I disprove your answer as wrong you have to give one).

Every answer is welcome.

 

[atyy]

Thank you very much for this answer.



Unfortunally my twin paradox is a little bit different as the usual one.

Actually no twin stay at home.
Both twins takes the journey.
Only EARTH2 stay at home.

Please tell me how much older twin1 is when he is back at EARTH2 and how much older twin2
is, when he is back at EARTH2.

Which twin is older?
Or have twin1 and twin2 the same age?

That was the question.


(Both answers are in contradiction to the SRT.
Before I disprove your answer as wrong you have to give one).

Every answer is welcome.

So what theory do you use to get your answer? Neither the special nor general theory of relativity, I presume, since both are wrong?

 

[Ich]

It should be obvious from DaleSpam's answer that both twins have the same age.

And before you start to "disprove": think about two satellites going round the Earth in opposite directions. That's proof that SR is invalid, even without closed universes.

 

[Daniel42]

So what theory do you use to get your answer? Neither the special nor general theory of relativity, I presume, since both are wrong?

I don`t think, that the theory of relativity is wrong at all.

My argument is the following:
1. The SR is right.
2. It is wrong in a closed universe.
3. We don`t live in a closed universe.


Please use the special theory to calculate the twins aging.
I agree with the statement of JustinLevy:

> In Dalespam's example, the spacetime was still everywhere flat, so we can still use SR to make calculations. <

 

[Ich]

1. The SR is right.
2. It is wrong in the presence of gravitation.
3. There is no gravitation.

Great.

 

[JesseM]

I don`t think, that the theory of relativity is wrong at all.

My argument is the following:
1. The SR is right.
2. It is wrong in a closed universe.
3. We don`t live in a closed universe.

GR allows for arbitrary topologies, so it is possible to have a flat spacetime where space is nevertheless closed, a bit like the video game "Asteroids" where if you disappear off the top part of the screen you'll reappear on the bottom, and if you disappear off the right side you'll reappear on the left (technically this corresponds to the topology of a torus--see this page). In any small region of this spacetime, the laws of physics are exactly as they are in SR (with no locally preferred frames), but in a global sense there will be a preferred pseudo-inertial frame (by 'pseudo-inertial frame' I mean a global coordinate system that in any local region looks just like an inertial coordinate system in SR). This will be the frame where if you draw lines of simultaneity from a given point in spacetime, the lines will wrap around the spacetime in such a way that they return to that same point, as opposed to wrapping around it in a "slanted" way like the stripes on a candy cane. In a closed universe there is also a "hall of mirrors" effect where you see copies of every object in regular intervals in different directions, and the globally preferred frame will also have the property that observers at rest in this frame will see the nearest copies of themselves to the left and right as both being the same age, and both appear younger than the observer by an amount corresponding to their distance in the observer's frame (so if I see a copy of myself 3 light years away, his visual image will appear 3 years younger than me), while this is not true in other frames. Anyway, the answer to all twin paradox questions involving inertial twins circumnavigating the universe is that whichever of the two inertial twins is closer to being at rest in this globally preferred frame, that will be the twin who's aged more on the second of two times they cross paths.

A previous thread on this topic:

https://www.physicsforums.com/showthread.php?t=110172

And here's a paper:

http://arxiv.org/abs/gr-qc/0101014

 

[A.T.]

It should be obvious from DaleSpam's answer that both twins have the same age.

Anyway, the answer to all twin paradox questions is that whichever of the two inertial twins is closer to being at rest in this globally preferred frame, that will be the twin who's aged more on the second of two times they cross paths.

Now I'm confused

 

[Ich]

Daniel42 called one of his three twins "earth". The other two are on the same footing.

 

[A.T.]

Daniel42 called one of his three twins "earth". The other two are on the same footing.

If we assume that "earth" is at rest in the "globally preferred frame"?

 

[LukeD]

In a closed loop with circumference R, there is a preferred rest frame. Special relativity predicts that someone moving with respect to the loop will measure its circumference as being smaller than R, so the preferred rest frame is one that views the closed loop with its maximum circumference.

This means that special relativity predicts that one twin will indeed age more than the other. (The only way to say otherwise is to insist that everyone sees the loop with circumference R, but then they no longer agree on ages and positions and the such. In other words, you have a true contradiction then)

Special relativity does not say that it is impossible to tell if you are moving. It says that you cannot tell if you are moving by some experiment involving only your local frame. There are other experiments that might tell you that you are moving, and special relativity can still be used in these cases.

 

[Ich]

If we assume that "earth" is at rest in the "globally preferred frame"?

Yes.

If the stay at home twin is at rest such that light pulses sent in opposite directions at the same time will be received at the same time

That's earth. The moving ones (+/- 0.6 c) are the twins.

 

[atyy]

My argument is the following:
1. The SR is right.
2. It is wrong in a closed universe.
3. We don`t live in a closed universe.

Now reading your OP, I see you meant the Principle of Relativity. In a flat cylindrical universe, we can still use special relativity, but the Principle of Relativity will not hold. That's true.

However, the Principle of Relativity also does not hold globally in our universe, which is well described by general relativity on large scales. The Principle of Relativity only holds locally in our universe.

So the failure of the Principle of Relativity in our universe is not because it is flat and closed (where special relativity still holds), but because of gravity (where special relativity holds only locally, and general relativity must be used for large distances).

 

[bcrowell]

In a closed loop with circumference R, there is a preferred rest frame. Special relativity predicts that someone moving with respect to the loop will measure its circumference as being smaller than R, so the preferred rest frame is one that views the closed loop with its maximum circumference.

Actually I think it's the other way around. The way you're saying it is the way Ehrenfest originally thought it would be when he posed Ehrenfest's paradox (P. Ehrenfest, Gleichförmige Rotation starrer Körper und Relativitätstheorie, Z. Phys. 10 (1909) 918). By the time Einstein published GR ( "The Foundation of the General Theory of Relativity," Annalen der Physik 49 (1916) 769, http://hem.bredband.net/b153434/Works/Einstein.htm ), he'd turned it around. That is, Ehrenfest figured that a rotating disk would have a circumference $C < 2\pi r$, due to length contraction, whereas Einstein decided it would be $C>2\pi r$, because the comoving rulers used to measure the circumference would be contracted. The general consensus today is that Einstein was correct and Ehrenfest was wrong. The reason is that Ehrenfest imagined a rigid disk being put into rotational motion, whereas today we know that's impossible.

Anyway, I don't think that invalidates the application of your argument, which I think is very interesting.

Now reading your OP, I see you meant the Principle of Relativity. In a flat cylindrical universe, we can still use special relativity, but the Principle of Relativity will not hold. That's true.

Hmm...this confuses me.

In the case of a $\Lambda=0$ closed universe, I think it is possible to circumnavigate the universe n times, and I suspect that there is a limit on n (1?2?). Suppose different observers set out from event A, and are reunited at event B after moving inertially. An n=0 observer will have recorded the most proper time, so it seems to me that this closed universe has to have a preferred frame. And of course it does ... there is a preferred time coordinate, which is defined simply by looking out the window and seeing the average mass density. Given a preferred time coordinate, you should be able to define a preferred rest frame. This should be the same as the frame that's at rest with respect to the CMB.

In the case of a vacuum-dominated universe, with no matter or radiation, I think it is impossible to circumnavigate the universe, so the argument for a preferred frame fails. This matches up with the standard interpretation of de Sitter universe as more symmetric than other cosmological solutions, and it matches up with the fact that there is no CMB in this universe that you can use to measure your motion with respect to. LukeD's extremal-circumference argument fails here, because you can't measure the circumference -- it's expanding too fast.

But the flat, cylindrical universe is a horse of a different color. I'm a little suspicious of it because it doesn't seem to correspond to any solution of the field equations for any reasonable equation of state.

[EDIT] Hmm...I think my objection to the flat, cylindrical universe was wrong. It's a perfectly good vacuum solution, just one with unusual topological properties. It's empty, but that shouldn't prevent us from measuring its circumference. ... still confused

 

[Dale]

Unfortunally my twin paradox is a little bit different as the usual one.

Actually no twin stay at home.
Both twins takes the journey.
Only EARTH2 stay at home.

Please tell me how much older twin1 is when he is back at EARTH2 and how much older twin2
is, when he is back at EARTH2.

Which twin is older?
Or have twin1 and twin2 the same age?

That was the question.

Adding more twins doesn't make it any more difficult at all.

$$w_{m2}=(t,-mod(.6t),0,0)$$

Integrating the metric along the worldline from t=0 to t=1.66... gives
$$s_{m2}= \int_0^{1.66...} \sqrt{1^2-(-.6)^2} \, dt =1.33...$$

I think you should be able to add any additional twins on your own.

 

[atyy]

[EDIT] Hmm...I think my objection to the flat, cylindrical universe was wrong. It's a perfectly good vacuum solution, just one with unusual topological properties.

That's what I assumed - naively though - do you see any problem such an assumption?

 

[bcrowell]

That's what I assumed - naively though - do you see any problem such an assumption?

No, as I said in #22, I'm just confused about it.

 

[atyy]

No, as I said in #22, I'm just confused about it.

Hmm, apparently even more bizarre things are allowed. This guy (http://books.google.com/books?id=d6...urve+special+relativity&source=gbs_navlinks_s, p7) draws the cylinder in time, so that there are closed timelike curves in flat spacetime! I wonder why he says 2D spacetime, whether there is any obstacle to having a 4D flat spacetime with closed timelike curves.

 

[Dale]

There is no obstacle to it. You can have very strange topologies and still be generally flat. I think that the use of 2D spacetime is just for illustration and visualization purposes.

 

[yuiop]

It should be obvious from DaleSpam's answer that both twins have the same age.

And before you start to "disprove": think about two satellites going round the Earth in opposite directions. That's proof that SR is invalid, even without closed universes.

I agree with Ich and Dalespam that the twins will be the same age on the second meeting (providing Earth 2 is not moving).

We can do a similar analogy with two cars driving at relativistic speeds in opposite directions around a large circle in flat spacetime. No differential ageing.

 

[atyy]

There is no obstacle to it. You can have very strange topologies and still be generally flat. I think that the use of 2D spacetime is just for illustration and visualization purposes.

Do you know whether the manifold must be orientable?

 

[Dale]

Do you know whether the manifold must be orientable?

There are some flat spacetimes which are closed in more than one dimension. For those situations I think in terms of "tiles" rather than in terms of cylinders. You can even have spacetimes which are "tiled" in the time direction. However, I don't know if a tiled spacetime can be made non-orientable since the tiles always form a sort of grid, even if you can't exactly see the "grout".

Obviously you can have positively curved spacetimes which are not orientable (spherical), but I don't know about flat spacetimes.

 

[Daniel42]

Thanks to everybody who show me that this is not a new problem.
There is allready an answer to the question.

There is one more thing I didn`t understand.

It should be obvious from DaleSpam's answer that both twins have the same age.

Ok.
Let`s say that both twins have the same age.
Let`s look what this means in detail in the system of twin1:


Twin 2 moves from EARTH2 to clock1, clock2 and so on to clock999 and clock1000 and back to EARTH2.

This looks like this:

[TWIN 2] ---> [clock1] ---> [clock2] ---> [clock3] --->...

... ---> [clock999] ---> [clock1000] and EARTH2 and twin1

The movement of twin2 along the clocks is symmetric.
There is no difference between the journey from clock2 to clock 3 and the journey from clockn to clockn+1 .

The jump from clock to clock is equal along the whole circumference of the universe.
If twin 2 ages 10 seconds and the clocks ages 100 seconds during the jump from clock to clock - then this must be the case on the whole journey.
The situation is totally symmetric and there is no difference between any clock---->clock jump.

If the twins have the same age back again at EARTH2, then there is no time slowing at all of twin 2 in the system of twin 1.
There is no time difference between the twin-ages only if there is no time difference between any clock--->clock jump.


But this is NEWTON and NOT Einstein!

Now I`m puzzled.
Can anybody explain me this?

(Can anybody please answer the question?...)

 

[bcrowell]

After thinking some more about the cylindrical universe argument atyy gave, I'm feeling a little more comfortable with it. It really surprised me that you could find a preferred rest frame based on a global topological property like this. There's also the issue of whether you want to think of postulates like "no preferred rest frame" as local statements or global ones. And I wasn't sure whether the preferred frame was a property only of the cylindrical model in some particular number of dimensions.

Now that I think I've understood it, I've written it up in the form of an end-of-chapter exercise here: http://lightandmatter.com/html_books/genrel/ch03/ch03.html The exercise tries to guide the student through the issues that I had trouble with, and I've given a solution in the back of the book. The solution ends up by giving an argument based on atyy's, and atyy is given credit for suggesting the argument. Thanks, atyy!

 

[atyy]

The solution ends up by giving an argument based on atyy's, and atyy is given credit for suggesting the argument. Thanks, atyy!

I can't remember where I learned this from, the links I gave in post #6 are the best Kosher references I know of presently - anyway, I'm happy to take credit where none is due

 

[atyy]

There are some flat spacetimes which are closed in more than one dimension. For those situations I think in terms of "tiles" rather than in terms of cylinders. You can even have spacetimes which are "tiled" in the time direction. However, I don't know if a tiled spacetime can be made non-orientable since the tiles always form a sort of grid, even if you can't exactly see the "grout".

Obviously you can have positively curved spacetimes which are not orientable (spherical), but I don't know about flat spacetimes.

I was thinking orientable as eg. a Moebius strip being non-orientable. I think one can put a flat Euclidean metric on such a space, since I can make it out of paper - can a Minkowski metric be put on it, and what would the observable consequences be? Googling yields http://arxiv.org/abs/gr-qc/0202031, which I haven't read.

 

[JesseM]

Ok.
Let`s say that both twins have the same age.
Let`s look what this means in detail in the system of twin1:


Twin 2 moves from EARTH2 to clock1, clock2 and so on to clock999 and clock1000 and back to EARTH2.

This looks like this:

[TWIN 2] ---> [clock1] ---> [clock2] ---> [clock3] --->...

... ---> [clock999] ---> [clock1000] and EARTH2 and twin1

The movement of twin2 along the clocks is symmetric.
There is no difference between the journey from clock2 to clock 3 and the journey from clockn to clockn+1 .

The jump from clock to clock is equal along the whole circumference of the universe.
If twin 2 ages 10 seconds and the clocks ages 100 seconds during the jump from clock to clock - then this must be the case on the whole journey.
The situation is totally symmetric and there is no difference between any clock---->clock jump.

If the twins have the same age back again at EARTH2, then there is no time slowing at all of twin 2 in the system of twin 1.
There is no time difference between the twin-ages only if there is no time difference between any clock--->clock jump.


But this is NEWTON and NOT Einstein!

Now I`m puzzled.
Can anybody explain me this?

(Can anybody please answer the question?...)

One useful way to think of a finite but flat spacetime is in terms of an infinite flat spacetime where all the matter just happens to repeat itself at regular intervals, as with a hall of mirrors--see fig. 8 on this page. Any predictions you make about the finite one should be replicated by the infinite-but-repeating one and vice versa. So in this case, imagine such a repeating universe, and focus on 3 copies of the Earth, each with its own copies of two twins A and B departing from them. Let's say the one on the left is Earth 1 (with twins A1 and B1 leaving it), the one on the middle is Earth 2 (with twins A2 and B2), and the one on the right is Earth 3 (with twins A3 and B3). So if Earth is at rest in the preferred frame (so in this frame, identical events on each Earth such as the twins departing happen simultaneously), we can imagine that at t=0 in this frame all the twins depart from their respective Earths at 0.5c, with each Earth 10 light years from its nearest neighbor. On each Earth, twin A heads left, and twin B heads right. So traveling at 0.5c, each twin takes 10/0.5 = 20 years to return to Earth. But from the perspective of this repeating universe, the copy of Earth that each twin departs from is different (though physically identical) from the copy they left--for example, twin B1 heads right from Earth 1 and 20 years later ends up at Earth 2, where he meets with twin A3 who was heading left from Earth 3. In the Earth rest frame, each twin is only aging at 0.866 the normal rate due to time dilation, so if they were both 40 years old when they left Earth they will be 40 + 0.866*20 = 57.32 when they meet a copy of their twin at Earth.

Now you can look at the same series of events in the frame of one of the twins, say twin B1 who departs from Earth 1. In his frame, all the A-twins are moving at (0.5c + 0.5c)/(1 + 0.5*0.5) = 1c/1.25 = 0.8c according to the relativistic velocity addition formula, which means in his frame the aging of the A-twins is slowed down by a factor of 0.6. So twin B1 leaves Earth 1 at aged 40, then after 17.32 years in his frame, he arrives Earth 2 at age 57.32 years old. In his frame twin A1 left Earth 1 at the same moment, also at aged 40, so at the moment twin B1 reaches Earth 2, twin A1 has only aged 0.6*17.32 = 10.39 years, and is only 51.39 years old. But, the copy of twin A that twin B1 meets at Earth 2 is not twin A1, but rather twin A3. And because of the relativity of simultaneity, in frames other than the preferred Earth frame, the event of different copies of twin A leaving their respective Earths do not happen at the same moment in twin B1's frame. Instead, in twin B1's frame each copy leaves his own Earth 5.7735 years before the copy to the left, so twins A2 and B2 leave Earth 2 5.7735 years before A1 and B1 leave Earth 1, and twins A3 and B3 leave Earth 3 5.7735 + 5.7735 = 11.547 years before A1 and B1 leave Earth 1 (and all twins are age 40 at the moment they leave their own Earth). So although A3 is aging slower than B1 in B1's frame, when A3 and B1 meet at Earth 2, in B1's frame B1 has been traveling 17.32 years while A3 has been traveling 17.32 + 11.547 = 28.867 years. So, if A3 has been aging at 0.6 the normal rate than A3 will have aged 28.867*0.6 = 17.32 years since departing Earth 3 at age 40, so A3 will be 57.32 years upon reaching Earth 2 just like B1, in spite of the fact that in B1's frame A3 has been aging more slowly than B1 throughout the journey.