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Ehrenfest Paradox + Twin Paradox

  1. Oct 4, 2012 #1
    1. The Ehrenfest Paradox
    After watching some of the posts here, I am not sure if I can conclude -

    The circumference-radius ratio is INDEED not 2π, but with a factor γ (which means Euclidean geometry is not necessarily correct in relativity, and the radius 'cuts through the circumference' ?) , Or

    The 'inelastic' radius 'shortens' to fulfill the Euclidean geometry (which is not correct, I supposed. The radius is perpendicular to the velocity, so it should not experience Lorentz contraction.)

    2. The Twin Paradox
    Time dilation tells us, in both of the frames of the twins, they 'believed' that the his brother is experiencing 'slower time' relative to himself, and both of their measurements of time must be correct, despite the results might seem to contradict to another.

    Problem is -
    When the twin travelling in spaceship with relativistic speed, gets back to earth to meet his brother, the contradiction arises. Who is older?

    If one argues -
    The 'relativistic' twin experiences a change of velocity and the direction of returning journey is opposite to the starting one. The calculation will not be that naive - substituting the speed to the equation of time dilation.

    One could suggest a modified version of twin paradox -
    The twin A travels at v~c, in a 'straight' path. But, the space is closed and curved. A gets back to earth. Same result arises in the original paradox.
    It seems to be hard to combine the 'contradictory' results into one at their reunion. Can anyone get an explanation of non-inertial frame about the twin paradox? Is the modified one still involving in acceleration?

    Thanks for answering.
     
    Last edited: Oct 4, 2012
  2. jcsd
  3. Oct 5, 2012 #2

    PeterDonis

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    Correct; to the extent that a "geometry of the rotating disk" can be defined, that geometry is not Euclidean. However, to really obtain a conceptual framework that generalizes, you need to think of the geometry of *spacetime*, not just space. See further comments below.

    You suppose correctly.

    Yes. For much more detail, check out the Usenet Physics FAQ pages on the twin paradox:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

    Yes, this is possible, because the space (more precisely, the spacetime, as I noted above) is curved. See below.

    If we are allowed to use curved spacetimes, then yes, it is easy to construct "twin paradox" scenarios where the "traveling twin" can be in inertial (free-fall) motion during his entire journey but still end up younger than the "stay at home" twin when they reunite. For example, suppose the universe were spatially closed (our current best-fit model says it isn't, but the spatially closed model is perfectly consistent mathematically and physically). Then one twin could stay on Earth while the other flew all the way around the universe (a higher-dimensional analogue of one person staying at home on Earth while the other flew around the world). The traveling twin in this scenario will have aged less when the two reunite.

    The reason is that, in any curved manifold, different geodesics with different lengths can connect the same pair of points. Here the geodesics are the worldlines of the two twins; they both connect the same pair of points (the events where they separate and reunite), but they are of different lengths (different elapsed proper times) because they connect the two points by different routes (one route wraps around the universe, the other doesn't).

    In fact, we don't even have to go around the whole universe to construct such a scenario. Consider this one: someone on the International Space Station shoots a projectile in such a way that it travels exactly vertically upward, as seen by an observer floating somewhere at a distance (i.e., not rotating with the Earth or with the station), at just the right velocity to come to a halt and then fall back down just in time to meet the station again as it passes the launch point again on its next orbit. Here again the two objects (the station and the projectile) will have "aged" by different amounts between the two events; we could in principle measure this by putting accurate enough atomic clocks on the station and the projectile (I don't think the effect would actually be measurable with today's technology, but I expect it would with near-future technology). In this case the station would actually age slightly less than the projectile.

    The latter scenario more clearly illustrates that it's the curvature of *spacetime*, not just space, that matters; space is very slightly curved around the Earth, but not nearly enough (by orders of magnitude) to account for the difference in aging between the station and the projectile. It's the change in "rate of time flow" with altitude (plus a slight effect from relative velocity) that makes the projectile age more.
     
  4. Oct 5, 2012 #3
    Dear PeterDonis -

    Thanks for answering my problem, and reminding me the important of spacetime, not just space.

    As you said, the travelling twin wil be younger than the one on the earth. However, from the frame of the one on relativistic journey, he will observe that -
    The one on earth experiences slower time.

    From his conclusion before reunion, the one on earth SHOULD be younger instead.

    One can only choose to believe one of the following conclusions -
    1. The travelling one is younger. The observation of the travelling twin is WRONG.
    2. The one on earth is younger. His observation is WRONG.
    while both of the results violates Einstein's postulate -
    Any inertial frame's physics is correct.

    My example of travelling in closed and curved universe, is to avoid the acceleration of travelling twin, and maintain the frame inertial. (I am not sure whether this will still involve in acceleration.)

    Thank you.
     
    Last edited: Oct 5, 2012
  5. Oct 5, 2012 #4

    Demystifier

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    For different observers rotating with the disk, this ratio is not even a constant:
    http://arxiv.org/abs/gr-qc/9904078

    See
    http://arxiv.org/abs/physics/0004024
     
  6. Oct 5, 2012 #5
    Hi simoncks, which thread did you look through? For example there were:

    https://www.physicsforums.com/showthread.php?t=128713
    https://www.physicsforums.com/showthread.php?t=224955
    https://www.physicsforums.com/showthread.php?t=503462

    In particular the last posts of the last thread clarify things a lot.
    See also Lorentz's calculation in Nature (1921) for a massive disk (he used GR but in principle he could have used SR as well). He there finds that the radius will be shortened by 1/8 v2/c2
    - http://en.wikisource.org/wiki/The_Michelson-Morley_Experiment_and_the_Dimensions_of_Moving_Bodies

    I can't parse that... do you understand the solution? Did you read the threads on that topic on this forum in which this is explained?
    E.g. https://www.physicsforums.com/showthread.php?t=607210
    I see a contradiction in your question: you have it "'straight' but curved"... However, Peterdonis seems to have properly answered that one. As a simplified answer, "straight" in SR means that they cannot get back to each other.
     
    Last edited: Oct 5, 2012
  7. Oct 5, 2012 #6

    ghwellsjr

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    I will address only your contention that a paradox arises in the original flat spacetime situation.
    Time dilation is not something that can be measured or observed. It is purely a result of a calculation based on an assignment of remote time within a given reference frame and never will result in any contradiction (unless you do something wrong).
    If we limit our analysis to inertial reference frames, then you can analyze the time dilation that is assigned to each twin within any single inertial reference frame and they all will agree on the final outcome. Remember, the purpose of time dilation is to enable us to calculate the Proper Time on each clock based on its speed according to the selected reference frame.
    In the conventional scenario, the twin on the relativistic journey is not at rest in any single inertial frame for the entire trip. You can pick the inertial frame in which he is at rest during the outbound portion of the trip in which only the earth twin's clock will be time dilated but on the inbound portion of the trip, the traveling twin will have a greater speed and more time dilation than the earth twin and the calculation will come out the same as in any other inertial frame. If you insist in thinking in terms of the traveling twin switching inertial frames when he turns around, then you have the added problem of maintaining a consistent calculation of the times on the remote clocks and the only way to know if you are doing it correctly is to make sure the final outcome is the same. That seems rather artificial to me. Just stick with one inertial frame and there will be no problem.
    Your last statement is the correct one. You need to apply it throughout your analysis. Don't try to misapply Einstein's postulate by employing two different inertial frames for each half of the travelers journey.

    Again, let me state that time dilation is not an observation, it's a calculation. If you want to talk about what each twin actually observes without regard to an assignment based on an arbitrary inertial frame, you should do a Doppler analysis which will also yield the same final result.

    And please note as I stated at the beginning of this post, I am only addressing your contention that the conventional Twin Paradox is a true paradox.
     
    Last edited: Oct 5, 2012
  8. Oct 5, 2012 #7

    PeterDonis

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    Did you read the Usenet Physics FAQ pages on the twin paradox that I linked to? I strongly suggest you do so. It addresses the issue you have just raised in detail (as well as other issues, such as that word "observes", which I don't think you have really thought through).

    The short answer is, as ghwellsjr said, that there is no such thing as "the" frame of the traveling twin. He is in one inertial frame on the way out, and a *different* one on the way back. The implicit argument you are making in the quote above would only be valid if the traveling twin stayed in a single inertial frame the whole time.

    You could try to construct a single non-inertial "frame" for the traveling twin, but then you have to include non-inertial effects, which change what the traveling twin "observes". The Usenet Physics FAQ discusses this too.

    #1 is the correct conclusion, but it does *not* violate Einstein's postulate. See above.

    It won't, as I said in my last post, but it will involve curved spacetime, and in curved spacetime there is no such thing as a global "inertial frame". So any analysis in a curved spacetime, even an analysis from the viewpoint of the stay-at-home twin, is more complicated.
     
  9. Oct 5, 2012 #8

    ghwellsjr

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    Can we be clear here? I did not say "He is in one inertial frame on the way out, and a *different* one on the way back." I don't use that kind of terminology and I disagree with using that kind of terminology. What I said was "the twin on the relativistic journey is not at rest in any single inertial frame for the entire trip". Both twins are always in all inertial frames and will have different speeds and different time dilations in different frames but no matter which one you use, they all calculate the same advancement in the times on the two clocks between the start and the end. EDIT: [They all will calculate the same advancement of the time on the traveling twin's clock at the moment of turn-around too.] This is the real simple theory that Einstein promoted in his 1905 paper where he introduced the so-called Twin Paradox at the end of section 4, except that he did not call it a paradox, because it isn't, and he didn't involve twins, just two clocks.
    Let's be clear here, too. Only the first sentence of #1 is the correct conclusion. The second sentence of #1 is not correct. Both sentences of #2 are wrong. The observations of both twins are correct but this thread is talking about time dilation which is not an observation of either twin. It's a calculation based on a selected inertial frame and it provides us with the means to determine that the traveling twin is younger.
     
  10. Oct 5, 2012 #9

    PeterDonis

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    Sorry if I misrepresented what you were saying. I was trying to say the same thing, but I agree that your version is more precise.

    I certainly didn't mean to imply that this was not true; it is, and I think it's the most important point to grasp in this whole scenario.

    Yes, good point.

    One minor point here: I would say it provides us with the means to *predict* that the traveling twin is younger. It's true that "time dilation" is not a direct observable, but when the twins meet back up, the fact that the traveling twin is younger *is* a direct observable.
     
  11. Oct 5, 2012 #10
    As harrylin corrected me rightly i believe, by saying that all objects are within all inertial frames of reference at the same time, saying a twin switches inertial frames is a nonsensical notion.


    What acceleration does, or a twin being subject to acceleration if you want, is switching which inertial frame of reference the twin is at rest in.

    I would have to actually correct all my posts, to make clear that it is the rest frame acceleration switches and not frames or inertial frames of reference, but i cannot edit my older posts anymore.

    After harrylin's post, it also seems nonsensical to me to say "in the twin's frame". There is no twin's frame, but only a "twin's rest frame". The twin is in all frames at the same time.
     
  12. Oct 5, 2012 #11

    ghwellsjr

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    If by "predict", you mean the same thing as "calculate", then I would agree. My problem with "predict" is that it implies a process to determine a future event based on probability, suggesting that the outcome might be in doubt or might come out differently if we used a different process such as a different frame of reference which is the exact opposite of what we are trying to say here.

    Also, time dilation, as I said before, enables us to calculate what each twin observes at other times of the scenario, such as at the traveling twin's turn-around point, not just at the end when they reunite. It fact, it also allows us to determine what each twin observes during the entire scenario, that is, what each twin sees of the other ones clock compared to their own. And it won't matter which inertial frame we use to make these calculations, they all yield the same answers, even though the time dilations can be different for each twin according to the different frames.
     
  13. Oct 5, 2012 #12

    PAllen

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    The OP hinted at a particularly devious version of differential aging, which was partly dealt with by Peter. I would like to add a bit more on this.

    One can have twin differential aging where the spacetime is flat everywhere (Minkowski geometry), and both paths between events are strictly inertial (geodesic). You simply need non-Minkowski topology. You close the space, say, in the x direction, so x=100 is identified with x=0. Assume c=1.

    It is then true that you have two timelike geodesics between (t,x)=(0,0) and (105,0). One will have less proper time than the other. There is obviously no acceleration, so that is not relevant for this version. There is no change of rest frame either, for the 'traveler'. The two common explanations that continue to apply are the pure metric approach (ultimate cause simply being the non-standard topology), and the doppler explanation.

    The latter is particularly interesting because of the phenomenon of multiple images in such a topology (long and short way, in general). The redshift, then blueshift in the sudden turnaround version in normal Minkowski space become separated into the red image and blue image that each persist the whole trip. The red image remains visually slower the whole way around, while blue image remains visually faster the whole way around (for the circumnavigating twin, viewing the 'stationary' twin). The blue image merges with the co-location at the final meet-up and matches the direct time comparison there. Assuming the 'moving' twin stops, at meet up, the red image remains visible (no longer red) as a long way around image of the other twin, that forever remains less behind your clock than such a long way round image of the other twin was before your journey.

    The 'stationary' twin watching the traveler, sees immediately, a slow red image from one direction, and an old normal image from the other. Much later - only when the traveler is nearly home (almost keeping up with light) do they (home twin) see brief period of blue, fast image. On meet up, the blue image matches what is directly seen on the traveler clock. The red image is still way behind, and remains red and slow for some time after the traveler stops.

    Here, the difference in Doppler is not due to turnaround; it is due to the topology which globally distinguishes a maximum time geodesic from other geodesics. Locally, there is no preference.

    [Edit: Some of the asymmetry in Doppler described above is due so implicit assumption of sudden start and stop, rather than sudden turnaround. If one imagines perpetually inertial paths that intersect e.g. every 105 seconds, the asymmetry is different. The 'stationary' observer sees only red shift images of the traveler for a significant part of each circuit; in the other direction, nothing is seen most of the time, with brief bursts of blue shift image. (In the red shift direction, the 'stationary' observer will see periods of multiple images, after a passing). Meanwhile, the 'traveler' sees red shift images and blue shift images at all times. Further, the blue shift image will be seen to replay part of history on passing; the redshift image will jump ahead on passing. ]
     
    Last edited: Oct 5, 2012
  14. Oct 5, 2012 #13

    PeterDonis

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    Yes, that's what I meant. I agree the calculation is entirely deterministic and frame-independent.
     
  15. Oct 6, 2012 #14
    It has a sensible meaning, but subtly different, I suppose, from what you had in mind. "Switching inertial frames" can be shorthand for selecting a different "frame" to use - just as he can switch maps despite the fact that his location is on all universal maps. It's certainly OK if a space traveller chooses for practical reasons to establish in-flight at appropriate times a "stationary" reference system based on the instruments in the space ship, and one appropriate time for doing so would be just after the turn-around.
    That's (regretfully) the same for all of us; our past errors and imprecisions are for the stay. However, readers will understand this, and it is instructive and nice to see how this forum is actually helpful.
    I was probably one of the initiators of criticism of that kind of shorthand phrasing on this forum, but it's not meant to be a witch hunt. :tongue2:
    For me, the primary reason to try to avoid imprecise jargon is that, IMHO, it is the cause of a large part of the confusions and misunderstandings; so that the economy of leaving out a few words is tenfold revenged by the following damage control (and that may be a conservative estimation :frown:).
     
  16. Oct 6, 2012 #15
    I quickly looked through it; I may have overlooked it but Nikolic's paper doesn't seem to provide a clear answer to the OP's question. Lorentz's 1921 paper (which Nicolic overlooked) does give a clear answer: no nonsensical radius that 'cuts through the circumference".
     
    Last edited: Oct 6, 2012
  17. Oct 7, 2012 #16
    Thank you for providing some much useful information. Finally understand that Ehrenfest Paradox simply demonstrated the invalid direct application of Lorentz transformation to rotating objects, as stated in Ehrenfest's paper - http://en.wikisource.org/wiki/Uniform_Rotation_of_Rigid_Bodies_and_the_Theory_of_Relativity
    I thought it was REALLY a paradox....

    For the twin paradox, the website really helps a lot -
    http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
    The spacetime diagram analysis suits me most =], by using integral of proper time.
    The travelling twin's non-inertial frame seems to be reason why such naive calculation fails.

    Sorry, to ask such redundant question without reading the FAQ section.
     
  18. Oct 7, 2012 #17

    PAllen

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    Can't help thinking about the case of cylindrical, flat, spactime: a key feature being that you may have inertial observers in relative motion that repeatedly meet, and one is aging faster than the other based on comparisons at each meeting. One unique feature of this global topology is that between observers there may be a short and long light signal path. I have concluded a fundamental observable that helps explain this paradoxical situation (inertial twin differential aging) is that the closure (of the universe) picks out a special class of observers for which long and short simultaneity using a chosen convention (e.g. Einstein's) match. That is, for these observers, simultaneity is isotropic (these observers are the ones whose time axis is an axis of symmetry for the closure). For other inertial observers, simultaneity is not isotropic - long and short simulteneity differ. I see this as the root cause of the differential aging for inertial observers in this topology.
     
    Last edited: Oct 7, 2012
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