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Circular barn/pole?

  1. Dec 7, 2015 #1
    This is a follow on to a prior thread, Another circular twin paradox

    Consider a train of proper length 100 at rest in an inertial ground frame. It is accelerated Born rigidly to the right on a straight track to 0.6c in the ground frame, after which is still has proper length 100 but has now ground length 80. It is then shunted onto a circular track of ground circumference 80. The front grazes the rear as the front exits the circle and returns to moving inertially to the right in the lab frame.

    Is this just a circular version of the barn/pole paradox? The entire train is in the 80 circumference circle simultaneously in the ground frame. It seems consistent with Einstein's view of the circumference in the Ehrenfest paradox. On the other hand, if I analyze it from the train frame, an issue arises.

    Say I treat the front as if it were momentarily at rest in a series of co-moving inertial frames (as if the circle were an polygon, with inertial frames parallel to the sides of the polygon). When the front changes direction to be at rest in the co-moving frame parallel to the first side of the polygon, the front undergoes infinite acceleration at a slight angle relative to the rest of the train. This should cause the train to deform (the front should stretch out away from the train) as in Bell's spaceship paradox. That is, the train does not just deform in the simple sense of bending. It also stretches out, as in Bell's paradox.

    Would this occur? Or, by making the sides of the polygon smaller and more numerous could I reduce the stretching continually until in an actual circle there would be no stretching, so that the train would not be deformed? Then the scenario should be a circular version of the barn/pole paradox.
     
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  3. Dec 7, 2015 #2

    A.T.

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    Yes. And unlike in the linear case, you cannot resolve it via relativity of simultaneity, because all frames agree that the whole train is in the loop at some time point (when front and end meet). The resolution is that space is non-Euclidean in rotating frames, so there is more circumference than 2*pi*r to fit the proper length of the train:
    http://www.projects.science.uu.nl/igg/dieks/rotation.pdf [Broken]
     
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  4. Dec 7, 2015 #3

    PeterDonis

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    Any time you find yourself thinking of "the train frame" when the train is moving non-inertially, beware. You are extremely likely to be making invalid assumptions.

    The Bell spaceship paradox involves continuous, finite acceleration, not discrete, infinite bursts of acceleration. I don't see how there is any analogy here.

    The question you are trying to answer is whether the train would, at some instant in the ground frame, fit exactly in the circular shunt with circumference 80 in the ground frame. In your previous thread, we assumed that the answer to that question was yes, and I don't see any reason why it wouldn't be. The proper acceleration of the train is perpendicular to its velocity at every point, so there will not be any "stretching" of the train; all the acceleration will do is change the train's direction.

    No, it isn't, because the barn/pole paradox relies on inertial frames, and the train, while any portion of it is in the shunt, is not at rest in any inertial frame. (A particular point on the train can always be viewed as being momentarily at rest in some inertial frame, but there is no inertial frame in which the entire train is at rest while any portion of it is in the shunt.) So you can't just say "well, the train should be of length 100 in its rest frame, and the shunt should be length contracted in that frame, so how can the train fit into the shunt?" The train has no "rest frame".
     
  5. Dec 7, 2015 #4

    A.T.

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    It has no inertial rest frame. But there is a rotating frame, where the train is at rest and has a length of 100, while forming a circle of radius 80 / 2pi.
     
  6. Dec 7, 2015 #5

    pervect

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    You've already run into trouble with your assumptions. While it's possible to accelerate the train, it's only possible to do so in a Born rigid manner on a straight track, not on a circular one. The wiki article on Born rigidity mentions this, https://en.wikipedia.org/w/index.php?title=Born_rigidity&oldid=615087209, oddly enough the current Wiki article on the Ehrnfest paradox does not, though as the Wiki article on Born Rigidity notes, Ehrnfest was the one who first noted the impossibility of making a disc rotate rigidly - which is why the "paradox" is known as the Ehrnfest paradox.

    Trying to argue that the disk is a "train" doesn't solve the paradox, the motion is still rotational, and rotational motion cannot be Born rigid.

    See also the sci.physics.faq on the issue, http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html. (I don't particularly care for the editors notes that were tacked onto the end of the original, though. The main article (sans notes) is a fair summary of the usual resolution, though it may be worth noting that there is a surplus of explanations of various degrees of popularity in the literature. Gron has several articles on the topic, the one cited in the sci.physics.faq, O. Gron, AJP Vol. 43 No. 10 pg 869 (1975) is probably the best-written and most-cited one, if you can track it down.



     
  7. Dec 7, 2015 #6

    A.T.

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    If I understand the OP correctly the acceleration along the track is on the straight part. On the circle there is only centripetal acceleration, perpendicular to the track. Why would that radial acceleration change the proper length of the train cars?
     
  8. Dec 7, 2015 #7

    PeterDonis

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    I assume you mean Born coordinates?

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

    If you mean Born coordinates, then this is not correct. The train is "at rest" in Born coordinates, but its circumference is 80, not 100--i.e., the same as its circumference in cylindrical coordinates in the ground frame. This is easily verified from the line elements given in the Wikipedia page linked to above.

    Rotational motion with a time-varying angular velocity cannot be Born rigid. But this scenario technically doesn't have time-varying angular velocity, in the sense of angular acceleration. It just has two points--entry into the shunt and exit from the shunt--where there is a discontinuity in angular velocity between zero and a fixed nonzero value. But this discontinuity can be made arbitrarily small, and I don't think it has a significant effect on the overall scenario. The constant angular velocity portion of the motion--i.e., the portion within the shunt, between the two discontinuities--can be Born rigid.
     
  9. Dec 7, 2015 #8

    A.T.

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    I mean a rotating frame, with the circular track center as origin, and the same angular velocity as the train on the circular track. The entire train is at rest in that frame, so its length in that frame is its proper length.
     
  10. Dec 7, 2015 #9

    PeterDonis

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    Please give the explicit math for the frame you're talking about. Did you look at the Born coordinates page that I linked to? It describes a frame that appears to meet your description; but the length of the train in that frame (meaning, the circumference of the circle it occupies) is what I said. Non-inertial frames don't work the same as inertial frames.
     
  11. Dec 7, 2015 #10

    A.T.

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    http://www.projects.science.uu.nl/igg/dieks/rotation.pdf [Broken] (Chapter 2)
     
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  12. Dec 7, 2015 #11

    PeterDonis

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    These are Born coordinates; the length of the train in these coordinates will be 80, not 100. More precisely:

    The train at a single instant of time (the instant at which the front of the train is just at the shunt exit and the rear of the train is just at the shunt entrance) in Born coordinates (the line element in the Wikipedia page, which is also given on page 4 of the paper you linked to) is described by a line element with constant ##r## and ##dt = dr = dz = 0##; the line element is then integrated over the range ##0 \le \varphi \lt 2 \pi## to obtain the length of the closed spacelike curve ##s## that describes the train.

    The only nonzero term in the line element is then ##ds^2 = r^2 d\varphi^2##, and ##r## here is the radius of the circular shunt in the ground frame (because the ##r## coordinate is unchanged by the transformation from the ground frame to Born coordinates--we have placed the spatial origin of the ground frame at the center of the shunt and used cylindrical coordinates in the ground frame). So the length of the train is just ##2 \pi r##, which is 80.

    What may be confusing you is the spatial metric given in equation (5) of the paper you linked to, which has a "length contraction" factor in the ##r^2 d\varphi^2## term. However, this metric, as the paper carefully notes, describes "the spatial distance between two infinitesimally near points". It can be interpreted as "the spatial geometry seen by a rotating observer", but you have to be very careful about what that means. It does not mean that you can integrate this metric around a complete circle ##0 \le \varphi \lt 2 \pi## to get "the length of the train in the train frame". (The discussion of distances in the Wikipedia page I linked to earlier also goes into this.)
     
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  13. Dec 7, 2015 #12

    pervect

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    Ah, I should have read the post more carefully, sorry. The train can accelerate rigidly on the straight track. The train (as a whole) obviously has to flex and bend when it makes the transition to the curved track, so that part of the motion isn't rigid (at least not in the Born sense, and probably not in any other sense). But as you point out the proper length of the train doesn't change in spite of the necessity for it's motion to be non-rigid.

    It's rather interesting how a train manages to go around a curve and keep it's wheels on the track when you think about it. But it's quite off-topic :-), I did get distracted by the question though.
     
  14. Dec 7, 2015 #13

    The consensus seems to be that the train fits into the 80 ground circumference circle without being deformed -- other than the non-relativistic bending caused by following the curved track.

    So the scenario is like the barn/pole paradox in the inertial ground frame. But it is not like the barn/pole paradox in two other ways: (a) the resolution does not depend on the relativity of simultaneity, because there is no shared simultaneity for the parts of the train that are within the circle, and (b) there is no equivalent for the train of the fact that in the pole's reference frame the barn is length contracted.

    Here is another hopefully clearer explanation of my concern about whether there might be some relativistic deformation as the train enters the circle (some deformation other than the non-relativistic bending). When the front of the train enters the circle, the rest of the train behind it is still moving to the right inertially. So by the "train frame" I meant the inertial frame in which the remainder of the train is at rest. Alternatively, one can say the inertial reference frame moving to the right at 0.6c that is co-moving with the rest of the train.

    So in that inertial frame in which all parts of the train other than the front are at rest, the front of the train undergoes acceleration and moves away at the angle of the first side of the polygon.

    I just mean that if the front accelerates away from the rest of the still-inertial train, then perhaps it stretches out or deforms the train. If this occurs when both the front and rear accelerate at the same rate in the ground frame (as occurs in Bell's paradox), then presumably it should also happen when only the front accelerates (in fact, it should stretch out even more, because the part of the train immediately behind the front is not accelerating at all). Put another way, if instead of having a train we have a row of point masses, what happens when the first point mass accelerates to move along the first side of the polygon? In the inertial frame in which all of the other points are at rest, does the second point in the row measure an increase in distance between it and the first point mass?

    Or does that acceleration of the first point maintain the same distance between it and the second point but change their relative positions, thus causing the non-relativistic bending? That seems unlikely to be the case because the first point cannot change position instantaneously (or it would be moving faster than light).
     
  15. Dec 7, 2015 #14

    PeterDonis

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    Thinking of it as a polygon, as I've already said, is not necessary. What's more, it may be misleading, because if the front undergoes a discontinuous acceleration for an instant to change direction by a finite angle, its acceleration is not perpendicular to its velocity. But if it undergoes a discontinuous change in acceleration, from zero to the nonzero value that is just right to move it around the circle, its acceleration is always perpendicular to its velocity. That is what allows the front to accelerate while the rest of the train is still moving inertially, without causing any stretching of the train (at least to a first approximation--as I said, there will be some effect due to the discontinuity in acceleration, but I think it can be kept small enough to not change the overall behavior significantly).

    No. In the Bell spaceship paradox, the acceleration is parallel to the velocity for both ships. In this case, the acceleration is perpendicular to the velocity. That makes a big difference.
     
  16. Dec 8, 2015 #15

    A.T.

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    It doesn't have to be single instant of time. The train can stay in the circle forever, so all the timing issues become irrelevant. You can even connect the front to the back of the train, so you have a rotating ring, which is forever at rest in the rotating frame. The length of each part of the train in the rotating frame is its proper length, so how can the total length of the train not be its proper length?

    So in the rotating frame I have N train parts (all at rest forever), each is 100/N long (its proper length). But total length of the train is not N * 100/N = 100 ?
     
  17. Dec 8, 2015 #16

    PeterDonis

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    It does if you're trying to define the "length" of the train. That is the spacelike interval occupied by the train at a single instant of time.

    Because you can't add all those infinitesimal proper lengths; they are not spacelike intervals all lying in a single surface of constant time. Each of those infinitesimal proper lengths is in a different spacelike hypersurface. The only global "length" you can define for the train as a whole is the one I computed in Born coordinates, which is 80.

    Yes. Your implicit assumption that you can always add the N lengths to get a total length is not valid in the rotating frame. It is only valid in a frame in which all of the N lengths lie in the same spacelike hypersurface. An inertial frame covering the entire train meets this criterion; the rotating frame does not.
     
  18. Dec 8, 2015 #17

    pervect

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    The length will still be 100, regardless of coordinates. But the method of calculating the length may not be obvious. I've seen two correct techniques. (There may be more). One of these techniques, used by AT's paper, and also by Ruggerio in his "Relative Space: Space Measurements on a Rotating Platform" http://arxiv.org/abs/gr-qc/0309020 involves the exchange of light signals. The approaches are very similar but the details vary. Ruggerio's method involves sending a light signal from one nearby observer "at rest" in the Born coordinate system to another nearby observer, and computing the proper time difference for a round trip. This is also called the "radar method" for obvious reasons. It should be noted that in generalized coordinates (such as in the rotating frame) the radar method only works for infinitesimal distances.

    The end result is that the distance then is ##(c/2)\, d\tau##, where ##d\tau## is the proper time for the round trip exchange of light signals between two nearby observers "at rest" (i.e. with constant Born coordinates).. This is probably the simplest technique, it does require one to know what proper time is and how to calculate it from coordinate time (something that I'm sure Peter knows, but I'm not so sure if the OP knows). I believe I've mentioned the Ruggerio paper to the OP in another thread. I'm willing to attempt to explain more to the OP about this approach if there are some specific questions, though I suspect that the references will do a better job than I will.

    The less-simple method is to use projection operators. This is the approach that Wald uses, for instance, but Wald's exposition is so terse it's not very clear. It's also used by Eric Poisson in "A Relativists Toolkit", his exposition is clearer, but very abstract. There is also a PF thread on this, https://www.physicsforums.com/threads/origin-of-spatial-metric.251228/#post-1843733

    The end result is that with a metric signature of (-+++), you get the induced spatial metric ##h_{ab}## by the equation ##h_{ab} = g_{ab} + u_a u_b##, where ##u_a## and ##u_b## are the 4-velcoties of some observer with constant Born coordinates. ##u^a## will have components ##(1/\sqrt{g_{00}},0,0,0)##, so ##u_a## will have components ##g_{ab} \, u^b##. You let ##a,b## take on the range ##0,3## in peforming the sum, but throw away the ##h_{0i}## and ##h_{i0}## terms. The PF thread, I believe, uses a (+---) signature, just to make things a bit more confusing.

    In any event, all the authors mentioned to date winds up with the same answer for the induced spatial line element, given by (5) in AT's reference and (14) in Ruggerio's reference, namely

    $$ds^2 = dr^2 + \frac{r^2 d\varphi^2}{1-\omega^2 r^2/c^2}$$
     
  19. Dec 8, 2015 #18

    PeterDonis

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    Please define exactly what "length" you are computing. Ideally, I would like to see what spacelike curve it corresponds to in Born coordinates. The spacelike curve I gave (##t, r, z## constant, ##0 \le \varphi \lt 2 \pi##) has a length of 80, as I've already shown.

    Please note the key word "nearby". I am not disputing that nearby measurements of length, using the methods you describe, give a "proper length" between nearby observers on the train which, if added up around the entire circumference of the train, would sum to 100. I am only disputing the adding up process, not the individual pieces. The adding up process does not make physical sense, because you are adding up a large number of spacelike line elements that are not collinear; they each lie in a different spacelike hypersurface. So calling the sum a "length" is, IMO, an abuse of terminology; it is not a physical "length" in any meaningful sense.

    But this line element does not describe global "lengths" on the train. It only describes an "apparent" geometry formed by piecing together locally measured lengths in a way that does not correspond to any actual global length. (In more technical language, this line element describes the geometry of the quotient space of the congruence of worldlines describing the "rotating ring" formed by the train--i.e., by assuming that it is a ring rotating at the same angular velocity indefinitely. But that quotient space does not correspond to any spacelike hypersurface in the actual spacetime, so, as I said above, calling the circumference of the circle in the quotient space a "length" is, IMO, an abuse of terminology.)
     
  20. Dec 8, 2015 #19

    DrGreg

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    I disagree with your interpretation. The length of 100 in the quotient space is exactly the length you would get if a person on the train used a tape measure (assuming the train continues to circle indefinitely). I wouldn't say that's an "apparent" length but the true "rest length". No, it doesn't correspond to the length of a spacetime curve in a surface of Born-coordinate simultaneity, but why should it? Those surfaces aren't orthogonal to the congruence of worldlines of points in the train. Simultaneity of infinitesimal intervals only becomes an issue if you're trying to measure a distance that is changing over time. The train is at rest, and the length of each segment is constant over time, so whether you can choose a set of synchronised segments or not is irrelevant.

    On the other hand, the length of 80 obtained from a surface of constant Born-time, is a coordinate-dependent quantity. The spacelike curve being measured isn't even locally compatible with (infinitesimal) Einstein synchronisation, so even a tiny section of this curve has a spacetime-length that doesn't at all match the local space-length than a local observer (at rest in the train) would measure.
     
  21. Dec 8, 2015 #20

    PeterDonis

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    I'm not sure I agree. Again, this is the "adding up" issue. Also, there is the issue that, as soon as we move beyond infinitesimal distances in the rotating frame, there is no unique notion of "distance"; different methods of measurement give different answers.

    There is no surface that is orthogonal to the congruence of worldlines of points in the train. Each individual worldline is orthogonal to a different surface.

    I'll have to think about this.

    No, it isn't. It is picked out by the symmetry of the scenario. In any other spacelike hypersurface, the curve occupied by the train will not be a circle; it will be an ellipse.
     
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