Twin paradox in a closed universe

1. Feb 4, 2010

Daniel42

Ive 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 isnt so easy to give as it seems.

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!

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?
Whos one is right?

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

Thats 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?!

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

we dont live in a closed universe.

Think you got to know this.
Thats all!

Last edited: Feb 4, 2010
2. Feb 4, 2010

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.

3. Feb 4, 2010

Staff: Mentor

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.

4. Feb 4, 2010

Daniel42

The metric is of no importance. Choose the metric you want with the circumference of one lightyear. and -
you didnt answer the question cause you cant.
Thats it is as I said.

Im not assuming that special relativity is valid in a situation with such high curvature.

We did not live in a closed universe.

Last edited: Feb 4, 2010
5. Feb 4, 2010

Daniel42

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

You cant do this.
I know it.

Last edited: Feb 4, 2010
6. Feb 4, 2010

7. Feb 4, 2010

Staff: Mentor

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
$$s_s= \int_0^{1.66...} \sqrt{1^2-0^2} \, dt =1.66...$$
$$s_m= \int_0^{1.66...} \sqrt{1^2-.6^2} \, dt =1.33...$$

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

Last edited: Feb 4, 2010
8. Feb 5, 2010

JustinLevy

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.

9. Feb 5, 2010

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.

10. Feb 5, 2010

Daniel42

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.

Before I disprove your answer as wrong you have to give one).

11. Feb 5, 2010

atyy

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

12. Feb 5, 2010

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.

13. Feb 5, 2010

Daniel42

I dont 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. <

14. Feb 5, 2010

Ich

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

Great.

15. Feb 5, 2010

JesseM

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:

And here's a paper:

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

Last edited: Feb 5, 2010
16. Feb 5, 2010

A.T.

Now I'm confused

17. Feb 5, 2010

Ich

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

18. Feb 5, 2010

A.T.

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

19. Feb 5, 2010

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.

20. Feb 5, 2010

Ich

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

21. Feb 5, 2010

atyy

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).

22. Feb 5, 2010

bcrowell

Staff Emeritus
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 [Broken] ), 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.

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

Last edited by a moderator: May 4, 2017
23. Feb 5, 2010

Staff: Mentor

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
$$s_{m2}= \int_0^{1.66...} \sqrt{1^2-(-.6)^2} \, dt =1.33...$$

24. Feb 5, 2010

atyy

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

25. Feb 5, 2010

bcrowell

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