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I Does global topology lead to a preferred frame in SR?

  1. Nov 16, 2016 #1
    According to this author, http://www.math.uic.edu/undergraduate/mathclub/talks/Weeks_AMM2001.pdf, a locally Minkowski spacetime with a nontrivial global topology may have a preferred inertial frame, in the sense that hypersurfaces of constant time can only be defined using particular time coordinates. Is he right? Is there any reason such spaces should be considered unrealistic? This seems to cast doubt on the basic (at least the philosophical) premise of relativity...
    Also, does this point have relevance within GR, in particular for our universe? Does it say something important about the CBMR frame, as the writer claims?
     
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  3. Nov 16, 2016 #2

    PeterDonis

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    Yes.

    Mathematically, they're perfectly well defined. Physically, we expect any spacetime to be a solution of the Einstein Field Equation with some physically reasonable stress-energy tensor. There are such solutions with nontrivial global topology, so the possibility at least has to be considered. Other issues come into play as well, though, such as singularities.

    Why?

    Only in the general sense that special relativity no longer holds in a curved spacetime, such as the spacetime of our actual universe. That means that the SR property of "no global preferred frame" no longer applies; only locally can we apply the "no preferred frame" principle. It's worth noting, though, that our current best-fit spacetime model describing our actual universe does not have nontrivial global topology. The only reason there is a "preferred frame" is the curved geometry of the spacetime.
     
  4. Nov 24, 2016 #3

    stevendaryl

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    It's enlightening (maybe) to consider the simplest nontrivial topology: A cylinder.

    So our cylindrical spacetime is 2D, and can be parametrized by two coordinates: [itex](x,t)[/itex] but with the constraint that the point [itex](x+L,t)[/itex] is the same point as [itex](x,L)[/itex]. So flying a distance [itex]L[/itex] in the x-direction brings you back to where you started.

    For all experiments taking place in a small region (a region of size [itex]\ll L[/itex]), the topology doesn't matter---this looks just like ordinary Minkowsky spacetime. So locally, you can perform Lorentz transformations and all frames look equivalent. But for large-scale phenomena, things look very weird in a boosted frame.

    Let me illustrate. Suppose you're an astronomer in this cylindrical universe. You're on Earth (assumed to be at rest in the coordinates x,t), and looking in the +x direction through a powerful telescope. What you would see, at a distance [itex]L[/itex] away, is another copy of Earth (it's actually your Earth, but it looks like a copy). Not only that, you will see another copy of Earth at a distance [itex]2L[/itex] away, and another copy at a distance of [itex]3L[/itex] away, etc. After correcting for the travel time of light, you discover that these Earths are all the same age. You can interpret this as good evidence that the universe is cylindrical, and these are not "copies" at all, but the real thing. But observationally, the situation is indistinguishable from an infinite universe that happens to be periodic: all conditions repeat in space with period [itex]L[/itex].

    Now, hop into a rocket that is moving at speed [itex]v[/itex] relative to the Earth in the +x direction. Now, when you look through a telescope, you will again see a line of copies of Earth. But in this frame:
    • The distance between copies is [itex]L/\gamma[/itex], rather than [itex]L[/itex] (since Earth is moving relative to this frame, distances are Lorentz-contracted).
    • The copies of Earth are not the same age! After taking into account the travel time for light, the astronomer on the rocket would conclude that there is a second Earth that is a distance [itex]L/\gamma[/itex] away, and that this Earth is older than ours by an amount [itex]\frac{vL}{c^2}[/itex] (assuming I've applied the Lorentz transformations correctly). At a distance of [itex]2L/\gamma[/itex], there is yet another Earth, and this one is [itex]\frac{2vL}{c^2}[/itex] older than ours. So all the infinitely many copies of Earth are different ages.
    So that's the sense in which there is no global notion of time for the rocket frame. He can set up a coordinate system locally that works out fine. But if he tries to have a global notion of time, then he'll find that if some event happens at time [itex]t'[/itex], it also happens at infinitely many other times: [itex]t' + \frac{n vL}{c^2}[/itex] for every integer value of [itex]n[/itex]. So there is no unique time for events.
     
    Last edited: Nov 24, 2016
  5. Nov 24, 2016 #4

    Dale

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    Except in one frame, the frame where all the copies are the same age.
     
  6. Nov 24, 2016 #5

    Ibix

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    Does this also happen for a closed FRW universe? Or does the expansion make it impossible to get a light ray all the way round in finite time?
     
  7. Nov 24, 2016 #6

    martinbn

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    Like they do in the paper from the first post.
     
  8. Nov 24, 2016 #7

    PeterDonis

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    In the case of a closed universe that recollapses, yes, there is not sufficient time for a light ray to get all the way around before the Big Crunch (a light ray is actually just completing one circumnavigation when the crunch is reached).

    When you add dark energy to the mix, I believe it is possible in a closed model (which doesn't have to recollapse if dark energy is present) to allow a light ray to circumnavigate the universe. In such a case, yes, there would be a global "preferred frame" in the same sense as for the cylindrical spacetime that stevendaryl describes.
     
  9. Nov 25, 2016 #8

    timmdeeg

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    I think whether or not astronomers living in a cylindrical universe would see copies of Earth depends on its size and the value of ##\ddot{a}##.
     
  10. Nov 25, 2016 #9

    stevendaryl

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    What is [itex]\ddot{a}[/itex]?
     
  11. Nov 25, 2016 #10

    timmdeeg

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    The second derivative of the scale factor ##a##.
     
  12. Nov 25, 2016 #11

    stevendaryl

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    Okay, what do you mean by the scale factor?
     
  13. Nov 25, 2016 #12

    timmdeeg

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    The scale factor stands for the relative expansion of the universe, if its size has doubled the scale factor has doubled.

    https://en.wikipedia.org/wiki/Friedmann_equations

    The second Friedmann equation shows under which conditions the universe expands (or contracts) accelerated (##\ddot{a}>0##). So, I think this together with the size of the universe puts certain constraints on the possibility to see copies of Earth.
     
  14. Nov 26, 2016 #13
    What would characterize the preferred frame in this case, given that GR has no global inertial frames?
     
  15. Nov 26, 2016 #14

    PeterDonis

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    The frame in which the universe appears homogeneous and isotropic; i.e., the frame of the standard FRW coordinates that are used in cosmology.
     
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