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Rotation in special relativity

  1. Nov 1, 2015 #1
    [this thread was split off from https://www.physicsforums.com/threads/confusion-in-general-relativity.840710/] [Broken]
    I read that Einstein gave the following argument:
    If we look at a rotating (and hence accelerated) frame from an Inertial frame, which is supposed to be away from all gravitational forces, then looking from the inertial frame we can see that any circle drawn on the rotating plane will have its circumference contracted (due to lorentz contraction) but not is diameter so we are not going to see the ratio of circumference to diameter to be pi. Doesn't this mean that space is curved?
     
    Last edited by a moderator: May 7, 2017
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  3. Nov 1, 2015 #2
    And another related question:
    As above, suppose we are observing a rotating frame from an inertial frame, free from gravity, and try to measure the circumference of a circle drawn in the rotating frame. Since our measuring rod would be contracted if we measure the circumference but it would not if we are measuring the diameter, according to Einstein we would conclude that the circumference to diameter ratio will be greater than pi, since we would measure circumference greater.

    But shouldn't the circumference itself contract so that we would get the same ratio pi in both the rotating as well as stationary case, where the circle is drawn in the inertial frame itself?
     
    Last edited by a moderator: Nov 1, 2015
  4. Nov 1, 2015 #3

    PeterDonis

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    Yes. (Note that there are a lot of complications lurking beneath seemingly simple question and answer; for a taste of them, see the links below. Note that Einstein himself was not aware of many of these complications, so you can't really take what he says as accurate according to our modern view.) But it does not mean that spacetime is curved. Whether or not spacetime is curved is an invariant; it's the same for all observers. But whether or not "space" is curved can depend on the observer, because different observers can see spacetime split up into "space" and "time" in different ways, and some of those ways can make "space" appear curved to those observers even if spacetime as a whole is flat. (Or "space" as seen by some observers might not even be a "slice" cut out of spacetime at all, but something else--that is what actually happens for observers at rest on a rotating disk.)

    No, it wouldn't. If we're at rest in an inertial frame, so is our measuring rod. The circle itself is moving (since it's rotating), so it would appear shorter compared to our measuring rod, at least locally. (Again, there are a lot of complications lurking here.)

    The circumference contracts according to the inertial observer, but the diameter does not, so the ratio of circumference to diameter would be less than pi for the inertial observer (at least, it would looking at it in this simple way and ignoring all the complications I've mentioned that are lurking).

    The following links go into some of the complications:

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

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html
     
  5. Nov 1, 2015 #4

    stevendaryl

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    Yes, but what Einstein taught us is that space + time are joined together into a single entity, spacetime. The splitting of spacetime into space + time is frame-dependent. A particular choice can make the spatial part curved even when the full spacetime part is not.

    An analogy: The full 3-D Euclidean space is flat. But we can split it up into a one-dimensional space, described by the coordinate [itex]r[/itex], plus a two-dimensional space, described by the coordinates [itex]\theta, \phi[/itex]. For a fixed value of [itex]r[/itex], the 2-D space described by [itex]\theta, \phi[/itex] is curved (it's a sphere of radius [itex]r[/itex]), even though the full 3-D space is not curved.

    A similar thing happens when you use a rotating reference frame. The full 4-D spacetime is flat, but the rotating frame splits it up into a 1-D space described by a time coordinate [itex]t[/itex] plus a 3-D space described by [itex]r, \theta, \phi[/itex]. That 3-D space is curved.
     
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  6. Nov 1, 2015 #5
    Hi PeterDonis,
    The situation which you are talking about is called Ehrenfest Paradox but what is Einstein said was something different where he said not the circumference but the measuring rod contracts. This has something to do with the Born's rigidity and I myself is not much clear about it.
    http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf
     
  7. Nov 1, 2015 #6

    PeterDonis

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    As I said, Einstein himself wasn't aware of all the issues involved. The two links I gave will give you a start at seeing the issues. The Ehrenfest paradox, what Einstein said, and Born rigidity are all, in the modern view, aspects of the same issue, which is how to model the physical experience of non-inertial observers in relativity. It is definitely not as simple as it might seem at first glance, or even at second, third, fourth, or fifth glance.
     
  8. Nov 2, 2015 #7

    A.T.

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    What do you mean by "circumference itself"? The circumference in a certain frame is simply what rulers at rest in that frame measure.
     
  9. Nov 2, 2015 #8
    If we assume drawing a circle in the rotating frame, then when observing from the inertial frame the circumference should get contracted like the measuring rods themselves. I thought since both the measuring rods and the circumference are getting contracted by same amount so the measured value should not be affected and hence the ratio should remain pi.
     
  10. Nov 2, 2015 #9

    PeterDonis

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    The measuring rods at rest in the rotating frame are contracted in the circumferential direction, but not in the radial direction, as viewed from an inertial frame. Since you're trying to measure the ratio of circumference to radius, not just the circumference, you have to look at how measuring rods behave in the rotating frame in both directions.

    Also, as I said before, there are a lot of complications lurking here, and I don't think you are considering them.
     
  11. Nov 2, 2015 #10

    A.T.

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    There is no "both" here. Circumference is the number of unit rods at rest that you can place tangentially end-to-end at distance R from the center. For the rotating frame this number is higher than for the inertial frame, while R is the same for both frames.
     
  12. Nov 2, 2015 #11
    So say Alice is standing on a merry-go-round rotating at a constant velocity and tries to measure the approximate circumference by placing small rods of equal lenght end-to-end along the edge. Bob is on the ground beside the merry-go-round. If I understand you correctly then the rods would contract along their length, approximately, as that's their instantaneous direction of motion. So Bob would see gaps between the rods while for Alice they're end-to-end?
     
  13. Nov 2, 2015 #12

    PeterDonis

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    No, that's not possible. If the rods are touching in one frame, they must be touching in every frame.

    However, there is a key distinction here between what Alice and Bob see locally, as Alice and the rods right next to her pass Bob, and what they conclude globally, by looking at the whole merry-go-round. Locally, Bob sees Alice and her rods pointed along her direction of motion (but not her rods placed radially, perpendicular to her direction of motion) length contracted; but Alice also sees Bob and his rods pointed along his direction of motion (but not his rods placed perpendicular to his direction of motion) length contracted. Length contraction is frame-dependent. So you can't conclude anything about the whole circumference of the circle, as measured by either Alice or Bob, just by looking at this local comparison. You have to look globally.

    Globally, however, the whole scenario is under-specified. To fully specify it, you are going to have to dig into the complications I have mentioned in previous posts in this thread.
     
  14. Nov 3, 2015 #13

    Demystifier

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  15. Nov 3, 2015 #14
    Right, my bad. For some reason I was thinking of relativity of simultaneity while forgetting the whole "spatially separated" part. Please don't ask why...

    Maybe I wasn't being specific enough in my scenario, I was writing it on my phone which is very tedious so I kept it short. Sorry about that.

    The merry-go-round is firmly anchored to the ground, so the axis of rotation is at rest relative to Bob, who is standing next to the merry-go-round. Bob has no rods, he's just observing Alice. As such I do not understand the parts I highlighted above.

    However, I will read the paper posted by Demystifier first, it sounds very relevant indeed.
     
  16. Nov 3, 2015 #15

    PeterDonis

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    And Alice is also observing Bob; there's nothing special about Bob that makes him capable of observations where Alice is not.

    I gave Bob some rods just in order to emphasize the fact that locally, Alice sees Bob length contracted, just as Bob sees Alice length contracted. There is no absolute sense in which Alice is length contracted but Bob is not. Even if Bob doesn't have any rods, he still has some length himself, and Alice will see that length contracted.
     
  17. Nov 3, 2015 #16
    Certainly.

    Right, I just wanted to make sure there was no confusion regarding the scenario.
     
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