# Closed flat space twin paradox

1. Feb 1, 2013

### Ibix

I'm not sure if this belongs here or in Cosmology - possibly either would do, I think. I was reading the thread about open/closed/flat universes that's currently ongoing in Cosmology, and a related question occurred to me.

According to post #7 you can have geometrically flat universes that are finite and unbounded if they have a non-trivial topology, such as a torus. Fair enough. Could someone explain the resolution to the twin paradox in this environment? It seems to me that I could sync watches with my twin as I pass him at constant velocity, circumnavigate the closed universe and return without having accelerated in either a proper or coordinate sense, since the geometry is flat.

Appreciating that there's ten thousand twin paradox threads here I had a quick search. This one seems to me to be saying that a flat FLRW spacetime is not a flat spacetime in the Minkowski sense of the word. I'm not sure if I've understood that right, though.

What don't I understand? For some context, I took a course in GR about fifteen years ago and remember not a lot and I'm currently on my second time through the first chapter of Carroll's lecture notes. "A lot" is a correct answer, but some precision would be appreciated.

2. Feb 1, 2013

### Mentz114

The resolution of the twin paradox in any environment is to calculate the proper length τ of the worldlines of the twins between the first and second meeting. The one with the longest proper interval has aged more. In fact the proper interval is exactly the time elapsed on the clocks.

c22 = c2dt2 - dx2 -dy2 - dz2

Simple innit ?

3. Feb 1, 2013

### Fredrik

Staff Emeritus
Interesting problem. I have never thought of this. I don't immediately see the solution. I don't have a problem with the existence of two timelike geodesics between the same two events, but I can make a naive argument for why each of them has a greater proper time than the other. I don't have time to think it through now. I will return later to see if someone else has posted the resolution.

I think it's sufficient to consider 1+1-dimensional flat spacetime $\mathbb R\times [-1,1]$ with -1 and 1 identified. The spacetime diagram from one twin's point of view would look something like this:

The vertical line is his own world line. The other two lines are actually just one line, because of the identification of 1 and -1. That line is the other twin's world line.

It's flat, but it's expanding.

File size:
223 bytes
Views:
291
4. Feb 1, 2013

### Fredrik

Staff Emeritus
But isn't the other twin's spacetime diagram just a reflection of the one I drew? Then this argument gives the same proper time for both.

5. Feb 1, 2013

### Mentz114

I'm not sure what you mean, because you said both worldlines are shown on your diagram. But I don't see a problem with the twins having equal proper time.

Last edited: Feb 1, 2013
6. Feb 1, 2013

### Ibix

Mentz - my problem is that, by my understanding, that if you transform Fredrik's spacetime diagram into the frame of the moving twin, you get an identical diagram, mirror-reversed. This is not the case in the regular twin paradox.

Put another way, since neither twin changes frame, both can claim proper time T and expect the other twin to have a lower proper time T', $T'=\sqrt{T^2-D^2}$, where D is the circumference of the universe.

Obviously something's wrong with that reasoning, but I don't know what.

7. Feb 1, 2013

### Fredrik

Staff Emeritus
Both lines are drawn in the diagram that shows the first guy's point of view, but I didn't draw the diagram showing the other guy's point of view.

I think I do, but the naive argument doesn't suggest that they're the same. It suggests that both are strictly greater than the other, i.e. $\tau_A>\tau_B$ and $\tau_B>\tau_A$.

8. Feb 1, 2013

### someGorilla

I think this shows that a flat spatially closed universe can't be locally Minkowskian if the principle of relativity is to be preserved. Or it can be locally Minkowskian but with a special frame of reference in which the closed dimension's length is maximum.
Off the top of my head, maybe both principles could be maintained if the closed dimension's length is not constant?

9. Feb 1, 2013

### A.T.

10. Feb 1, 2013

### Mentz114

So, we calculated the proper intervals. It was easy, wasn't it.

I still can't make sense of this, and this

11. Feb 1, 2013

### HallsofIvy

Staff Emeritus
This problem, having a twin move away and back along a curved geodesic, so that he does not accelerate, is called the "cosmic twin paradox". Of course, to have a curved geodesic- and especially a closed geodesic so that the twin can return- requires so very strong gravitational fields. It is the "equivalence" of gravitational fields and acceleration that resolves the paradox.

12. Feb 1, 2013

### stevendaryl

Staff Emeritus
The resolution in the case of a torus is that space is only a torus in one particular rest frame.

If the universe is cylindrical in the x-direction, that means that looking out at the universe with a telescope, you see the Earth at x=0, and then it looks like a second, identical Earth at x=L (where L = the distance "around" the universe), and another one at x=2L, etc. So there is no observable difference between having a "cylindrical" universe and having a universe with repeating pattern of objects.

Now, let's consider what things look like in another reference frame. An observer in this frame will also see a repeating pattern of copies of the Earth. However, because the Lorentz transformations mix up space and time, the second observer doesn't see exactly identical Earths. (I actually shouldn't say "see" here, because what you "see" is light that left a long time ago. So I should really say "deduces", because what's relevant is what the observer deduces taking into account the transit time of light.) In the new reference frame, there appears to be:

• One copy of Earth that is at $x'=0$
• A second copy of Earth at $x'=\dfrac{L}{\gamma}$ that is older than the first copy, by an amount $\dfrac{vL}{c^2}$
• A third copy of Earth at $x'=\dfrac{2L}{\gamma}$ that is older than the first copy, by an amount $\dfrac{2vL}{c^2}$
• etc.

How do I know this? Well, let $(x_1',t_1')$ be the coordinates of some event at the first copy of the Earth. Let $(x_2',t_2')$ be the coordinates of some event at the second copy of the Earth. By the Lorentz transformations:

$\delta x' = \gamma (\delta x - v \delta t)$
$\delta t' = \gamma (\delta t - \dfrac{v}{c^2} \delta x)$

where $\delta x' = x_2' - x_1'$, $\delta t' = t_2' - t_1'$, $\delta x = x_2 - x_1$, $\delta t = t_2 - t_1$

Since the distance between copies of the Earth in the unprimed frame is $L$, we have

$\delta x = L$

If the events are simultaneous in the primed frame, then that means

$\delta t' = 0$

So that implies
$\delta t = \dfrac{v}{c^2} \delta x = \dfrac{vL}{c^2}$

So from the point of view of the traveler, there appears to be a line of Earths, where each one is older than the last, and each one is a distance of $\dfrac{L}{\gamma}$ from the next. The people on these copies of the Earth appear to age slower by a factor of $\gamma$.

So to the traveler, it takes time $T' = \dfrac{L}{v \gamma}$ to travel from the first Earth to the next. The people on this second Earth start off older by an amount $\dfrac{vL}{c^2}$. During the trip, the people on this second Earth age by amount $\dfrac{T'}{\gamma} = \dfrac{L}{v \gamma^2}$, which means that by the time the traveler reaches the second copy of the Earth, the people will be older than the people on the first copy when he left by an amount:

$\dfrac{vL}{c^2} + \dfrac{L}{v \gamma^2} = \dfrac{vL}{c^2} + (1-\dfrac{v^2}{c^2})\dfrac{L}{v} = \dfrac{L}{v}$

(where I used $\dfrac{1}{\gamma^2} = 1 - \dfrac{v^2}{c^2}$)

So when the traveler goes from one Earth to the next, he ages by $\dfrac{L}{v \gamma}$ while the people on the second Earth end up older than those on the first Earth by an amount $\dfrac{L}{v}$. So the traveler ends up younger than the people on the second Earth.

13. Feb 1, 2013

### Fredrik

Staff Emeritus
I don't understand why. A diagram shows the coordinate assignments made by one coordinate system. There are two twins, and therefore two comoving inertial coordinate systems. So there are two diagrams to be drawn, and I only drew one.

14. Feb 1, 2013

### Mentz114

Nice, but I think I prefer the paradox.

15. Feb 1, 2013

### Mentz114

I thought your diagram is a space-time plot. In which case one can put as many worldlines as as needed.

You've really baffled me now, we must be talking about different things.

OK, 'Stationary' frame
OK...
You just described a diagram with 2 worldlines on it . Unless 'that' line is somewhere else.

16. Feb 1, 2013

### Fredrik

Staff Emeritus
Yes, of course.

I'm just as baffled.

So? There's still another diagram to be drawn.

17. Feb 1, 2013

### Staff: Mentor

18. Feb 1, 2013

### Staff: Mentor

This is the point of the Barrow & Levin paper I just linked to. It also means that that particular rest frame is the only one where Einstein clock synchronization can be done between objects at rest in the frame. The difference in apparent age of the copies of Earth that you describe is due to a failure of clock synchronization in any frame moving relative to the "preferred" frame that is picked out by the topology of the spacetime as a whole.

19. Feb 1, 2013

### Ibix

Thanks to all for your answers. I'm a little annoyed I didn't spot the thread A.T. linked to, since the title is practically the same as the title of this one.

I had actually thought of a preferred reference frame as a solution, from a length-contracted circumference angle. I discarded it because it seemed such an anathema to Relativity. The "Hall of Mirrors" approach is a compelling argument that I hadn't quite clocked. A regular grid like that can't be invariant under Lorentz transformation, is the short way of putting it, I think.

Thank you all.

20. Feb 1, 2013

### Mentz114

I don't see why. I'm impressed with the speed you put up your diagram, and I understand it represents an observer at rest and one looping around the closed dimension. Surely that's enough to read off the proper times of both observers ?