Special Relativity in a closed universe

[kochanskij]

TL;DR

If a spacecrafts travels close enough to C in a closed finite unbounded universe, length contraction will make the circumference of the universe smaller than the length of the craft, in its own reference frame. Its nose will bump into its tail. But, in another frame, its length is far less than the universe. Its nose and tail don't bump. Is this a contradiction? What did I miss?

In this thought experiment, I am assuming a finite but unbounded flat space with a 3-torus topology. There is no concern about curvature. This problem is completely special relativity. No GR needed.
When the spacecraft travels very near C, the universe contracts, relative to the craft's frame. If its circumference becomes less than the length of the craft, its nose will crash into its tail. But in earth's frame, this never happens, since it is the craft length that contracts.
Is this a physical contradiction? Does the nose of the craft get crushed or not?

 

[robphy]

In this thought experiment, I am assuming a finite but unbounded flat space with a 3-torus topology. There is no concern about curvature. This problem is completely special relativity.

The underlying manifold of special relativity is $R^4$.
Flatness is not enough.

 

[Dale]

In such a universe the first postulate does not hold. So you have to be careful with any "these frames contradict" type of reasoning. They can not possibly contradict, but the rules may be different in the different frames.

When the spacecraft travels very near C, the universe contracts, relative to the craft's frame.

In particular, this is incorrect. In a closed universe the preferred (stationary) observers measure the smallest universe. The moving observers all measure a universe that is larger. So if the proper length is smaller than the length of the universe according to the preferred frame then it is impossible for the nose to crash into the tail.

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

 

[PeterDonis]

In this thought experiment, I am assuming a finite but unbounded flat space with a 3-torus topology. There is no concern about curvature.

But the change in topology does create something that is not present in SR on $R^4$: a preferred frame.

To see how this works, consider a simpler example: SR in two spacetime dimensions on a cylinder, $S \times R$, instead of a flat plane. (The paper @Dale referenced does this with two more spatial dimensions added, but those extra two dimensions don't really play any role in the analysis.) There will be one particular inertial frame in which the time axis is parallel to the cylinder; that's the preferred frame. In any other frame, the time axis winds around the cylinder in a helix, and things get more complicated. In particular, you can't assume that things like length contraction work the same as you're used to in SR on $R^4$.

 

[kochanskij]

In such a universe the first postulate does not hold. So you have to be careful with any "these frames contradict" type of reasoning. They can not possibly contradict, but the rules may be different in the different frames.

In particular, this is incorrect. In a closed universe the preferred (stationary) observers measure the smallest universe. The moving observers all measure a universe that is larger. So if the proper length is smaller than the length of the universe according to the preferred frame then it is impossible for the nose to crash into the tail.

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

Yes, there is a preferred at-rest reference frame in a closed universe. But special relativity still applies locally, doesn't it? In such a universe, if a ship goes to a star 10 ly away at .99C, relative to earth, it will take less than 10 years of proper time for the astronaut, (1.42 yrs) won't it? So, the distance between earth and the star is contracted, relative to the ship, to 1.41 ly, isn't it? What if the star is half way around the universe and speed is much closer to C? Are you saying the SR breaks down? Why? How? At what speed or at what distance?

 

[kochanskij]

But the change in topology does create something that is not present in SR on $R^4$: a preferred frame.

To see how this works, consider a simpler example: SR in two spacetime dimensions on a cylinder, $S \times R$, instead of a flat plane. (The paper @Dale referenced does this with two more spatial dimensions added, but those extra two dimensions don't really play any role in the analysis.) There will be one particular inertial frame in which the time axis is parallel to the cylinder; that's the preferred frame. In any other frame, the time axis winds around the cylinder in a helix, and things get more complicated. In particular, you can't assume that things like length contraction work the same as you're used to in SR on $R^4$.

Yes, there is a preferred at-rest frame. A person in that frame will experience the most proper time compared to a person who circumnavigates the universe at any relativistic speed and returns to their start point. So the moving person must measure a smaller universe circumference, right? The closer to C he travels, relative to the at-rest frame, the less proper time he measures, so the less distance he measures. Correct? Taking this to the extreme, the circumference must shrink to less than the length of the ship, doesn't it? When does something go wrong with length contraction?

 

[Ibix]

But special relativity still applies locally, doesn't it?

It applies everywhere at all scales. There are just some consequences of the topology. The easiest way to understand it, IMO, is in a (1+1)d universe where you can imagine a spacetime diagram drawn on a cylinder. Then you can simply slit the diagram parallel to the cylinder axis, unroll it, and lay it out flat. This is a normal Minkowski diagram except that lines that exit either timelike edge enter the other one at the same time coordinate.

You can then see what's going on almost immediately. You can draw two vertical lines representing two stars, a shaded diagonal band representing the rocket, and the x' coordinate axis of the rocket. For a sufficiently high speed, the x' and t' axes are close enough to parallel (in this Euclidean representation!) so that the x' axis doesn't exit the shaded band before it crosses the next star ("both stars are simultaneously between nose and tail"), but the nose never catches the tail because the tail worldline is slanted, unlike the stars, so the spatial plane wraps many times round the cylinder before reaching the tail again.

 

[robphy]

Here's an ancient thread with more references

Graduate Thread 'The Twin Paradox Revisited: A Local Case Study'

Nov 3, 2004

There have been plenty of posts about the Twin Paradox. I don't think this version has been aired before except in my post #47 on the "Is Age Relative" thread.

The Twin Paradox in a closed universe.

If cosmic expansion slows down and reverses it would become hypothetically possible to circumnavigate the universe.

Take two twins, one twin stays put and the other is accelerated to
9.999…%c she passes her sister and they synchronise clocks. She continues at constant velocity along a geodesic path and circumnavigates the universe. She eventually meets up with her...

• Garth
• Cosmological Paradox Twin paradox
• Replies: 44
• Forum: Special and General Relativity

 

[Dale]

Are you saying the SR breaks down? Why?

Yes. I already said why: The first postulate fails in such a universe.

How?

This is a much better question. We know why this scenario breaks with standard SR, but what exactly is the consequence of this?

If you look at the paper that I linked earlier (https://arxiv.org/abs/gr-qc/0503070 ) it shows how this works out. First, for notation I will use unprimed coordinates for the coordinates in the preferred frame, and for simplicity I will use units where $c=1$ and a single spatial dimension.

We can express relativity in the metric used: $$ds^2=-dt^2+dx^2$$

Then the fact that the universe is closed is expressed by the equivalence relation $$(t,x)=(t,x+nL)$$ where $n$ is an integer and $L$ is the size of the closed universe in the preferred frame. These two equations allow us to analyze the situation as follows. We can use completely standard special relativity due to the metric, including the Lorentz transform with the resulting length contraction, time dilation, etc. And we can envision the closed universe condition as the same as a standard open universe with an infinitely repeating set of any events or worldlines where the integer $n$ identifies which repetition we are indicating.

But your question in the OP involves both $n=0$ and $n=1$. So you have to use the equivalence relation above, not just the Lorentz transform. And thus there is no way for the nose to hit the tail (assuming the proper length of the craft is less than $L$).

Now, to see what happens we can use the Lorentz transform on the equivalence relation to get the corresponding equivalence relation for a primed frame moving with velocity $v$ relative to the unprimed (preferred) frame. $$(t',x')=(t'-\gamma v n L, x' +\gamma n L)$$ where $\gamma=(1-v^2)^{-1/2}$.

So the key is that everything works as normal in this universe as long as we can deal with a single $n$, say $n=0$. But as soon as we introduce multiple $n$, then SR no longer applies.

But special relativity still applies locally, doesn't it?

As @Ibix said, it isn't exactly about locality. It is about the topology. You can have as large an experiment as you like. As long as your experiment doesn't involve multiple $n$ then all of the usual open-universe SR principles apply. You can travel multiple times around the universe, and as long as you are still only interacting with the $n=0$ copies, then all is standard.

But your OP is explicitly about a scenario involving the $n=0$ nose reaching the $n=1$ tail. So you cannot merely apply special relativity. You must apply both special relativity and the equivalence relation above. With that we see that there is simply no way for the $n=0$ nose to reach the $n=1$ tail assuming that the proper length of the ship is less than $L$.

 

[PeterDonis]

special relativity still applies locally, doesn't it?

If by "special relativity" you mean "my own personal intuition about length contraction", yes, but looking at the whole circumference isn't "locally".

A person in that frame will experience the most proper time compared to a person who circumnavigates the universe at any relativistic speed and returns to their start point.

Yes.

So the moving person must measure a smaller universe circumference, right?

No. See my first comment above.