# I Motion in relativity

1. Mar 2, 2016

### Alberto87

Hi I´ve got some question about this paradox (and the whole relativity theory) too but first of all I want to apologize for my poor english.

If there's one thing I've understood (reading here and there on the web) it is that it is impossible to determine who is at rest and who is in motion. An observer can perceive only the accelerations (or decelerations) but also in this case he cannot know if he passed from a rest condition to motion or vice versa, or if he simply changed speed within a motion condition. The whole universe is at the same time moving at constant speed and at rest (unless it is accelerating). We see a cosmic ray that travels almost at the speed of light but for the cosmic ray we are traveling almost at the speed of light.

Could you please confirm this first part?

Now comes the question about the twin paradox, since I still have not figured out if you get old slower when traveling fast (compared to what has to be clarified) or when accelerating or both, I want to propose a modified version:

Imagine an "empty" universe, no important masses and only two observers each with its own clock. These two observers move towards one another with relative speed next to the speed of light. Obviously it makes no sense to say that one is in motion and one is stationary but for simplicity we imagine that the observer A is "stationary" and sees B approach at nearly the speed of light. The moment they cross they synchronize the clocks. At this point A will see B receding.
Now imagine for simplicity of depiction that the universe in which the observers are located do not have 3 spatial dimensions but only 2. In particular we imagine that this two-dimensional universe (three-dimensional with time) is the surface of a sphere. Consequently B continuing on his way sooner or later will meet with A again after traveling a "full circle" of the spherical universe.
My question is what will the observers see when they will compare the clocks? there was no acceleration. According to each observer they were stationary and saw the other going away (or getting closer) at the speed of light. Who will be older than the other?

I hope that people will be able to clarify these doubts.

2. Mar 2, 2016

### Staff: Mentor

A better way of putting it would be that "being at rest" and "being in motion" are relative; there is no absolute rest or absolute motion, only rest or motion relative to something else.

With the clarification above, yes.

No. "The whole universe" is not a single object; it consists of many, many objects, many of which are moving relative to each other. So there is no single state of motion for the whole universe.

Yes. Note, however, that there is a key caveat to all of the above, which is that, unless spacetime is globally flat, all of the above statements are only true locally. Globally, in a general curved spacetime, the concept of "relative velocity" doesn't even have a well-defined meaning for objects that are spatially separated. See below.

Making the spatial geometry a 2-sphere changes things, because it means spacetime has to be curved, not flat, and in a curved spacetime, as above, all of the statements about relativity of motion are only true locally. So when A and B pass each other, neither one is "at rest" or "moving" in any absolute sense. But when they meet up again, after one of them has circumnavigated the 2-sphere, they will, in general, have aged differently. This can happen because, while they are spatially separated, they do not have any well-defined "relative velocity", so the relativity of motion when they pass each other does not allow either one of them to deduce that their motion is symmetric everywhere.

In fact, the issue is even a bit more complicated than I have said, because even a spacetime that has zero intrinsic curvature can still lack a well-defined relative velocity for spatially separated objects, if it has a non-standard topology. For example, imagine a 2-dimensional spacetime (1 space dimension, 1 time dimension), where the space dimension is rolled up into a cylinder. Then, if observer A and B are in relative motion when they pass each other, they will meet up again after one of them has gone around the cylinder, and again, they will have aged differently. This spacetime has zero intrinsic curvature; the only difference between it and standard flat Minkowski spacetime is its unusual topology.

3. Mar 2, 2016

### Alberto87

Yes that´s what I meant, I meant that everything within the universe is moving and at rest at the same time. I didn´t express myself properly

Unfortunately I don´t have enough knowledge on the matter to understand this. Do you mean that a fixed curvature of the spacetime (cylinder or sphere) is just like a curvature generated by masses and therefore the observers are "accelerating" when they move within this kind of space-time and therefore their perception of motion is affected in the same way? (This is the only way I can give a sense to what you said)
If this is true then in such a universe (imagining that we know the "shape" of the universe and it´s curvature) is it actually possible to know if I´m moving or I´m at rest (by checking the clocks as in the example above)?

4. Mar 2, 2016

### Staff: Mentor

First, only the sphere has curvature; the cylinder is not curved at all, it just has a non-standard topology.

If we are going to obey the physical laws of GR, then any curvature of spacetime (such as for the sphere) has to be generated by masses; no masses, no curvature. So the "sphere universe", according to the laws of GR, can only exist if there are masses in it.

In the case of the cylinder, it's not quite so clear, because the laws of GR don't talk about the topology of spacetime, only about its curvature. So strictly speaking, the "cylinder universe" would be consistent with the laws of GR with no masses present. But it certainly is physically different from an ordinary flat spacetime, so there ought to be some physical law that accounts for the difference--or it could be that unusual topologies like the "cylinder universe" simply can't exist physically, even though we can model them mathematically.

You have to be very careful with the term "accelerating". In the sense of proper acceleration, acceleration that is actually felt and measured by accelerometers, none of the observers we've been talking about are accelerating; they're all in free fall, feeling no acceleration, and accelerometers attached to them would read zero, just like observers in free fall around the Earth.

In the sense of coordinate acceleration, i.e., the second derivative of position with respect to time in some coordinate chart, as the term makes clear, this kind of acceleration depends on how you choose your coordinates. You can always make it zero locally by choosing appropriate coordinates. So it's not really a good thing to focus on if you're trying to understand the physics.

I'm not sure what you're referring to here.

"Moving" and "at rest" are still relative, locally, in such a universe; but globally, the geometry of the spacetime does pick out a particular class of observers who can be viewed as being "at rest" relative to the geometry, so to speak. These would be observers who remain at the same spatial point on the 2-sphere at all times. These observers will be the ones whose clocks read the most elapsed time between a given pair of events--for example, if A is one of these observers, and B passes A, circumnavigates the universe, and comes back to A again, B's clock will show less elapsed time than A's clock.

5. Mar 2, 2016

### PAllen

I would say relative velocity is well defined for such a universe because parallel transport is path independent. What distinguishes it from trivial topology is that two points can have multiple geodesics connecting them, so geodesic motion no longer guarantees a globally extremal interval.

6. Mar 2, 2016

### olgerm

Whether something is moving or is in rest depends on frame of reference. That is also true in classical physics.
You can describe any physical system (universe) in frame of reference where its total momentum is zero (its masscenter is in rest) or in frame of reference where its total momentum is not zero (its masscenter is moving), but system can not have zero total momentum and not-zero total momentum in same frame of reference.

Last edited: Mar 2, 2016
7. Mar 2, 2016

### Alberto87

thanks a lot for all your answers, I definitely have a more clear idea on this matter.
Ok, I understand the difference

I had the wrong idea that masses only curved the space-time locally and not globally and that the global shape of the universe was independent from the energy inside.

That's what I meant, is it possible somehow (maybe knowing the topology and the curvature of the universe and by comparing different clocks whose dynamic history is known) to figure out if one of the clocks was "at rest" in respect to the spacetime itself? Or would it still be out of what we are allowed to know?

8. Mar 3, 2016

### Staff: Mentor

Yes, I described a way of doing it. But it involves at least one observer (B in the case I described) having to circumnavigate the universe--when B and A meet up again, B can tell that he is the one that circumnavigated, and A didn't, because B has aged less.

9. Mar 4, 2016

### Alberto87

But will the observer know if he was at rest (to the spacetime itsef) or he can just know that he was moving slower than the other observer?

10. Mar 4, 2016

### Staff: Mentor

A will know that he didn't circumnavigate the universe and B did. If there are only two observers in the entire universe, that's the best A can do. However, if there is a third observer, C, who flies off from A in the opposite direction from B, circumnavigates the universe in that direction, and comes back to meet A, then A can tell if he is actually at rest relative to the spacetime--if he is, B and C will return to him at the same instant (assuming that they both fly off at the same constant speed relative to him).

Note that a better term way of saying "at rest relative to the spacetime" would be "at rest relative to the average matter distribution in the spacetime". If A has a way of measuring the matter distribution, he can tell that he is at rest relative to it by seeing whether it appears homogeneous and isotropic to him--i.e., the same everywhere and in all directions.