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Ehrenfest's paradox

  1. May 22, 2013 #1
    I'm reading about Ehrenfest's paradox where one considers a rotating disk. If one let r' be the radius of the disc in an inertial frame and r be the radius of the disc when it is at rest. Then the periphery must be Lorentz' contracted such that ##2\pi r' < 2\pi r##, but since the radial line is perpendicular to the direction of motion of the disk it is not Lorent'z contracted so that ##r'=r##. Thus the paradox.

    The supposed kinematical solution is to consider the disk being accelerated up to a given angular velocity. By an analysis involving the relativity of simultaneity it is found that it is inconsistent to require that the disk be accelerated up to the angular velocity and additionally require it being done while keeping the periphery 'Born rigid'. I.e. accelerated in such a way that the rest length of each line element along the periphery remains constant.

    This is claimed to resolve the paradox and my question is why it does that.
    More specifically; why does one need to assume a 'Born rigid' acceleration of the disk in order to accept the paradox?
     
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  3. May 22, 2013 #2

    PeterDonis

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    But what, exactly, is the paradox? Is it that the actual, physical circumference of the disk is not equal to 2 pi times its actual, physical radius? That's an interesting fact, certainly, but why is it a paradox? There's nothing self-contradictory about it that I can see; it just shows that the spatial geometry of the rotating disk is not Euclidean. (Note that there are issues with even defining what "the spatial geometry of the rotating disk" means; we'll see if we end up opening that can of worms in this thread, as it has been in previous ones on this topic. :wink:)

    Because it shows how the spatial geometry of the rotating disk can change as it is spun up.

    Because that amounts to assuming that the spatial geometry of the disk (again, with caveats as to how that actually is defined) remains unchanged as it spins up. In other words, it amounts to assuming that either (a) there are no stresses induced in the disk as it spins up, or (b) whatever stresses are induced in the disk as it spins up don't deform the disk. When put this way, of course, the resolution is obvious: the act of spinning up the disk does induce stresses in the disk, and the stresses do deform the disk, changing its spatial geometry.
     
  4. May 22, 2013 #3

    Ibix

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    In a Newtonian world you can simply spin a disc faster and faster. There's a limit to a real disc, though. It'll fly apart under its centrifugal stresses at some speed.

    The Special Relativistic case adds stresses due to the different Lorentz contractions experienced by the different parts of the disc. The disc will deform and eventually fly apart under these forces.

    There is only a paradox if you have an ideal material that can neither deform nor disintegrate. If you can find such a material, I'd like a suit of armour... :wink:
     
  5. May 22, 2013 #4
    Ah, so in order to even write that ##2\pi r' < 2 \pi r## one are implicitly assuming that the spatial geometry of the disk have not changed after it was accelerated up to a given angular velocity. Otherwise it would not be generally true that the circumference were ##2\pi## times the radius.

    Thus, if I understand it correctly, the kinematical resolution is actually stating that the spatial geometry of the disk must be deformed under the acceleration process? And relativity is also telling us what that geometry must be at a given angular velocity?

    Namely a physical circumference of

    $$ \frac{2\pi r}{\sqrt{1 - \frac{\omega^2 r^2}{c^2}}}$$

    I'm not sure if I've totally missed the point when I say that I'm a bit surprised that relativity gives a requirement on what shape the disk deform into without an input theory of how the parts of the disk are 'bound together' (elasticity etc.)
     
  6. May 22, 2013 #5

    Bill_K

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    The input is Born Rigidity. That is we assume the disk does not stretch, consequently its elastic properties do not enter.
     
  7. May 22, 2013 #6
    Where does that enter? What parts of the disk are required to be Born rigid? As I understand it, it was inconsistent to require elements on the periphery be Born rigid under the acceleration.

    Btw, just to make sure: Even though the spatial geometry of the disk is non-Euclidean, it does not mean that spacetime is curved, right? One can't change the geometry of the space by a coordinate transformation.
     
  8. May 22, 2013 #7

    PeterDonis

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    But isn't that the "input" that creates the paradox? In other words, the point of the paradox, as I understand it, is that the disk cannot be spun up without deforming, so the assumption of Born rigidity cannot be true for an actual disk. Put another way, for the case of angular acceleration (unlike the case of linear acceleration), it is impossible to induce Born rigid motion even with a "conspiracy of forces" applied to each individual point of the disk.
     
  9. May 22, 2013 #8

    PeterDonis

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    Correct. It is assumed in this scenario that the disk has negligible mass and therefore negligible gravity, so spacetime as a whole is flat. As you say, that's true regardless of which coordinates we adopt.
     
  10. May 22, 2013 #9
    That was my current understanding as well.. :)
     
  11. May 22, 2013 #10

    Bill_K

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    Yes, exactly.
     
  12. May 22, 2013 #11

    pervect

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    There are a ton of papers out there on the Erhenfest paradox. I've found http://arxiv.org/abs/gr-qc/9805089 to be useful.

    A key insight here is the following:

    So you can't Einstein synchronize all the clocks on a rotating disk - or a rotating globe. There are systems that by convention synchronize clocks on a rotating disk (or globe), but they aren't Einstein synchronized.

    If you can't agree on a concept of "physical space", it becomes obvious that it's confusing to talk about it's circumference.

    I hope this somewhat different perspective helps. But at eny event, with a little googling, you should be able to find a TON of papers on the rotating disk paradox, enough to give you much more specific questions than the ones you've been asking.
     
  13. May 22, 2013 #12

    Ben Niehoff

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    I disagree. The disk is spinning at relativistic speeds. Its angular momentum is not negligible; it contributes to the energy-momentum tensor, and thus to the local curvature of spacetime. This is precisely why the geometry of the disk becomes non-Euclidean.

    Rotation is an absolute motion and not merely a coordinate transformation. Every point on the disk experiences proper acceleration.
     
  14. May 22, 2013 #13

    PeterDonis

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    The angular *velocity* of the disk is certainly not negligible; but does that automatically imply that its angular momentum is not negligible? What if we make the disk out of test particles, whose mass is so small that the disk's effective moment of inertia is small enough to make the angular momentum negligible even if the angular velocity is not?

    Also, if you are right and it is impossible for the disk's angular momentum to be non-negligible, wouldn't that mean that it's impossible to realize the Ehrenfest paradox in flat spacetime? It seems to me that a lot of the literature on the paradox uses flat spacetime to model it.

    I'm not sure I agree with this, in the sense that there is certainly a congruence of timelike worldlines in flat spacetime, the Langevin congruence, for which Einstein clock synchronization can't be used to define surfaces of simultaneity, as pervect pointed out. The "spatial geometry" associated with such a congruence can only be defined as a quotient space, not as a foliation of identical spacelike slices; and this spatial geometry will be non-Euclidean. So it's certainly possible to have an object in flat spacetime with non-Euclidean spatial geometry.

    Yes, agreed; this is true regardless of how the angular momentum issue comes out. The proper acceleration vector associated with the Langevin congruence is nonzero.
     
  15. May 23, 2013 #14

    Ben Niehoff

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    Hmm...you're probably right.
     
  16. May 23, 2013 #15
    According to an argument given by Øyvind Grøn it seems like it's precisely the fact that standard measuring rods remain Born rigid, while the disk can not, that makes the geometry non-Euclidean. As the disk is accelerated there are space for more standard measuring rods and hence the circumference around the periphery will appear to be longer.

    From that point of view it does not seems like the resulting geometry of the disk is a consequence of Born rigidity of the disk.


    The argument is given here at page 35 under 'A rotating disk with angular acceleration':
    http://www.google.no/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CDUQFjAB&url=http%3A%2F%2Fareeweb.polito.it%2Fricerca%2Frelgrav%2Fsolciclos%2Fgron_d.pdf&ei=ddWdUauVIYXl4QS834FY&usg=AFQjCNFhAQw918KWnBS7B2Xl-3NOECKsrg&sig2=SsZ0Z0-A01kbTT22mM0iqA&bvm=bv.46865395,d.bGE&cad=rja
     
  17. May 23, 2013 #16
    According to Nathaniel I. Reden there has not yet been found a definition of time for which the time curvature of the Riemann tensor for the spacetime geometry of the rotating frame disappear.

    https://www.google.no/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=0CEYQFjAC&url=https%3A%2F%2Fwww.amherst.edu%2Fmedia%2Fview%2F10267%2Foriginal%2Freden05.pdf&ei=Nu6dUbDtA8S24AS3hoCoCQ&usg=AFQjCNG6x91ToivTx2HJK1UhQmhj41Dpsg&sig2=9DAQIwiovmKpA5u_rO-OwA&bvm=bv.46865395,d.bGE&cad=rja
     
  18. May 23, 2013 #17

    pervect

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    Where does he say that exactly?

    At the moment, I'm thinking that if the components of any tensor vanish in one coordinate system, they must vanish in all coordinate systems.

    And in a non-rotating coordinate system with no matter, the Riemann is zero.

    So if you have an empty rotating frame with no matter, it must also have a zero Riemann.
     
  19. May 23, 2013 #18

    Mentz114

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    In the cited paper Reden seems to be working in 3 dimensions. It is true that the 3 dimensional sub-manifold of the rotating frame is curved. But it is not spacetime curvature.
     
    Last edited: May 23, 2013
  20. May 23, 2013 #19
    At page 25 under summary he states:

    'Moreover, none of the three results in a Riemann curvature tensor with all
    components zero, so we have not solved the problems mentioned in section 1.7.
    '

    I would think that some of the components being non-vanishing implies a non-vanishing Riemann tensor and thus a non vanishing curvature.
     
  21. May 23, 2013 #20
    It is indeed working in three dimensions in section one, but in section two he includes the time coordinate and still gets non vanishing Riemann tensor components.
     
  22. May 23, 2013 #21

    Mentz114

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    I've looked a section 2 in Reden's paper. None of the metrics he writes has zero Einstein tensor and they are *not* physical solutions of the EFE ( in my estimation ). So they don't tell us anything useful.

    The Langevin frame is the correct treatment and shows that only a 3D submanifold is curved, not the global spacetime.

    For rotating disks in GR we need something like the Neugebauer-Meinel disks*, see for instance arXiv:gr-qc/9703077v1 26 Mar and references therein.

    *Neugebauer & Meinel, Phys. Rev. Lett. 75 (1995) 3046
     
    Last edited: May 23, 2013
  23. May 23, 2013 #22

    PeterDonis

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    Can you clarify what you mean by "Langevin frame"? My understanding is that there is *no* 3D submanifold of the spacetime as a whole that corresponds to "the spatial geometry of the rotating disk". This is because of the failure of Einstein clock synchronization for the worldlines of the Langevin congruence. The "spatial geometry" of the rotating disk is the quotient space of the Langevin congruence over proper time for Langevin observers (I think I've stated that correctly--the idea is that each point of the quotient space corresponds to a particular worldline in the Langevin congruence). This quotient space is not a submanifold of the spacetime, is it?
     
  24. May 23, 2013 #23

    Mentz114

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    In answer to the question I've highlighted I can only quote the Wiki article http://en.wikipedia.org/wiki/Born_coordinates

    which shows I've mis-read it. The quotient manifold is not a submanifold of the global spacetime.
     
  25. May 23, 2013 #24

    WannabeNewton

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    The wiki paragraph quoted is very incomplete as far as I can tell. Why is the quotient space a topological manifold? The quotient is a very ill-behaved operation in the category of topological spaces and in general does not take a Hausdorff space to a Hausdorff space or a locally euclidean space to a locally euclidean space etc. Moreover, why is it a smooth manifold? The article assumes, on top of the topological structure that it doesn't prove it has (and, as far as I can tell, fails to give reference to something that does), that the quotient also has a smooth structure because it freely uses the theorem that any smooth manifold can be given a Riemannian metric.

    I checked the Landau & Lifgarbagez reference as well in my copy of classical theory of fields, and they don't prove it either.

    EDIT: Never-mind, I found an exercise in one of my manifolds texts that deals with the issue of quotient manifolds obtained from vector fields :)
     
    Last edited: May 23, 2013
  26. May 23, 2013 #25

    Mentz114

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    In view of the posts above, can we say that the circumference problem is explained by the non-Euclidean nature of the 3D space transported along a worldline of the Langevin congruence ?
     
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