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B A tale of three clocks

  1. Nov 20, 2017 #1
    The experiment done by Hafele & Keating took clocks around the world in two directions and compared the readout to a static clock to measure time dilation effects. Everything fitted nicely with Einstein’s predictions.

    To avoid confusion let’s move the clocks way out in space to a real inertial frame of reference (no gravitational effects). This will eliminate the gravitational time dilation. One clock stays in a fixed place and two fly around imitating the clocks that flew around the world going east and west. We should get the same amount of time dilation as seen in the H&K experiment without the gravitational correction when the clocks are brought together at the end of the experiment.

    This is not a question of one clock appearing to be running at a different speed compared to a clock flying by at speed. The clocks all start at the same place and come back to the same place so that the readings can be compared at the end of the experiment.

    How do we decide which clocks are moving and which one is staying still?
     
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  3. Nov 20, 2017 #2

    Ibix

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    You pick any coordinate system and integrate ##ds=\sqrt{c^2dt^2-dx^2-dy^2-dz^2}## along the worldline of each clock (Edit: that expression assumes you picked Einstein coordinates - you get something a bit more complex that gives the same result if you pick any other system). That gives you the clock reading on each clock.

    In the scenario you describe, you have one inertial clock and two that travel non-inertially in mirror-symmetric circles. The two non-inertial clocks will have the same lower time.

    Edit 2: I didn't really answer the question - so: Which clock is moving is up to you. Pick your coordinate system so the one you want to be at rest is at rest, then the other two are moving. The maths above, or the more complex variant I alluded to, will tell you the important physics, which is which one(s) is/are younger at the end. That result does not depend on which clock(s) you decided was/were the moving one(s).
     
    Last edited: Nov 20, 2017
  4. Nov 20, 2017 #3

    Orodruin

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    To add to what Ibix said, this

    to me sounds as if you think a position can be an inertial frame. This makes no sense. An inertial frame is a theoretical construct that assigns coordinates to events in a particular fashion. In addition, an object cannot "be" in a particular frame, an object will not disappear because you choose a different description.

    If you want to remain in special relativity (and it seems you do based on your choice of location), then there is no way that the rest frames of all three clocks are inertial frames. You cannot determine if something is "standing still" (there is no such thing as absolute rest - you can only stand still relative to something), but you can determine if a clock has undergone any proper acceleration. If it has, you cannot make the assumption that its rest frame is an inertial frame and hence not analyse the situation as if it was.
     
  5. Nov 20, 2017 #4

    Nugatory

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    You are basically describing the twin paradox: two or more clocks start at rest relative to one another, we set them on different paths through spacetime (which in the absence of gravity requires accelerating at least some of the clocks) that eventually come back together, and we compare their readings when they rejoin.

    In general they will read differently because their different paths through spacetime have different "lengths"; the scare-quotes are because these are paths through spacetime, not just space, so both the time and the distance can be different along the different paths. (The post above by @Ibix explains how to calculate these "lengths"). This, the difference in clock readings is no more surprising (and is analogous to) the way two car odometers can read differently after the cars travel between the same two cities by two different routes.

    Looking at the clocks doesn't tell you much of anything about which clock is "moving" and which is "staying still" - those notions don't really mean much.
     
  6. Nov 20, 2017 #5

    pervect

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    If we ignore the gravitational time dilation effects as you propose, the reason that the pair of clocks moving in different directions on the Earth's surface differ in time reading when they come back is that the Earth is rotating.

    In our imaginary thought experiment, we can imagine a fictitious rotating platform, and one clock that is inertial, the other clock that is on the rotating platform. Then the inertial clock will read the longest proper time, and the clock on the rotating platform will read less time.

    If we imagine two rotating platforms, and one clock on each platform, the clock on the platform rotating more quickly (in the inertial frame) has more time dilation and hence a shorter proper time than the other. Both will still read shorter times than an inertial clock, of course.

    So in the absence of gravity the rule is simple - the inertial clock reads the longest proper time. There is no ambiguity about when a clock is inertial, and when it is not. But perhaps it isn't obvious that there is no ambiguity. If there's a question about the issue, we can discuss it more, I suppose, if there is no question about it then there's no sense in going off on a long tangent about it.
     
  7. Nov 21, 2017 #6
    So we are back at Mach's principle.

    Both of your platforms are rotating in relation to the stars. So not only is the kinetic energy of the clock tied to the stars but also its tick rate.

    Or that is how I see it! IMHO as they say
     
  8. Nov 21, 2017 #7

    Ibix

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    I don't think so. The clocks don't need to rotate to follow a circular path in space, and you will get the same result whether they rotate or not. And you will get similar results from a standard out-and-back twin paradox scenario where there's no rotation or circular motion. And variants of the twin paradox with no acceleration at all are possible.

    This is basically an illustration of the triangle inequality in the x-t plane of Minkowski space. In Euclidean space, the sum of the lengths of the two short sides of a triangle exceeds the length of the long side. In the x-t Minkowski plane the sum of the intervals along the shorter sides is always less than that along the longest side. I don't think the fixed stars need to come into this anywhere.
     
  9. Nov 21, 2017 #8

    pervect

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    I'm not particularly fond of Mach's principle, so I'd disagree with the "we" part, but it gives me some idea of where you are coming from. It only gives me some idea, not an exact idea, as there are many variants of Mach's principle.


    There are situations in General Relativity where force-free gyroscopes rotate (precess) relative to the fixed stars. This is outside the scope of special relativity (and gets into the issues you tried to exclude bye excluding gravity), but then so is Mach's principle outside the scope of SR. I will say that if a force-free gyroscope precesses in some (local) frame of reference, I would not categorize said frame of reference as being "non-rotating", regardless of whether the frame of reference is tied to the fixed stars or not.

    I hope that we've answered the original question, of how things work in SR, though - the whole Mach's principle thing starts to get into an area that goes beyond SR.
     
  10. Nov 22, 2017 #9
    I realize that you and the other leaders & moderators of this forum are not fans of Mach’s principle. This has been made clear to me by the moderators as some of my previous comments were deleted from another conversation.

    I thank you very much for you explanations. I will try to get my head around what you have said and have a good think about the twins paradox.

    But….. Einstein spent most of his life work trying to fit Mach’s principle into his relativity and only gave up at the end of his time. He was definitely a fan of Mach’s principle – he coined it!

    I personally feel that a great and very solid clue to the nature of the Universe is being overlooked because it doesn’t fit the Standard Model.

    In the paper “The Lense— Thirring Effect and Mach's Principle”

    by Hermann Bondi and Joseph Samuel 1996
    They put it very clearly “ Mach's principle (MachO) is the experimental observation that the inertial frame defined by local physics (zero Sagnac shift) coincides with the frame in which the distant objects are at rest.
    MachO is an experimental observation and not a principle.”
    The bold emphasis is my own.
     
  11. Nov 22, 2017 #10

    PeterDonis

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    And this "observation" is not precisely correct. Look up Gravity Probe B. It showed, experimentally, the "frame dragging" effect due to the Earth's rotation, which, in the terms you are using here, makes the "inertial frame defined by local physics" not coincide wtih "the frame in which distant objects are at rest". (This was predicted by GR decades ago.)

    Nor is this "observation" the same as Mach's Principle. (In fact, there is an entire textbook, Cuifolini and Wheeler's Gravitation and Inertia, which argues that "frame dragging", which as just noted contradicts your "observation", is one way in which GR is consistent with Mach's Principle.) In fact, "Mach's Principle" does not name one particular principle or law or anything else. Various authors claim it does, but they each claim it names something different, so taking the literature as a whole, the term "Mach's Principle" does not have a definite meaning. That is a key reason why we "are not fans" of discussions of Mach's Principle.

    What's more, the quote you are taking from the paper you reference (arxiv link below) is taken out of context, and putting the context back in changes the meaning substantially. The paper discussion of the Lense-Thirring effect (another name for frame dragging, the thing Gravity Probe B experimentally confirmed), in Sections 3 and 4, shows that the "experimental observation" you quote is in fact falsified by that effect. The paper also makes quite clear that there are multiple versions of Mach's Principle and they don't all say the same thing. (Section 4 of the paper also argues, similar to the Cuifolini & Wheeler book I referenced above, that the Lense-Thirring effect is "Machian" when an appropriate version of Mach's Principle is adopted.)

    The paper is on arxiv here:

    https://arxiv.org/abs/gr-qc/9607009

    Your personal feelings are up to you, of course, but that doesn't necessarily make them fit subjects for PF discussion. As far as I can see, you are simply misunderstanding the GR literature.
     
    Last edited: Nov 22, 2017
  12. Nov 23, 2017 #11
    The various versions have muddied the waters. The one that needs explanation and inclusion is the real simple one – the one where I spin around, my hands are pulled out and the stars spin around me – or Mach 0.

    As you point out there are dents in Mach’s principle (talking about Mach0 here) – very small dents – where frame dragging occurs or certain types of precession where the is no force on the gyroscope.

    I still can’t understand whether the community thinks that Mach’s principle is included in the SM. Wheeler et al and Bondi et al seem to think so and Sciama spent a lot of energy trying to figure it out. Others think it is of no consequence or passè or maybe too complicated to deal with.

    And after 100 years from Einstein and 150 from Mach it is a shame there is still no clarity. And I don’t really understand why.

    Or maybe you are right and I just don’t know beans about GR.
     
  13. Nov 24, 2017 #12

    PeterDonis

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    And that one is explained perfectly well by GR.

    These aren't "dents" in Mach0; they are ways in which the spacetime geometry is affected by the matter near you, as well as the "distant stars". That is what you should expect from Mach0: Mach0 says that "inertia" (in GR terms, spacetime geometry) is produced by matter. So if there is matter present other than the "distant stars", then it should affect "inertia". And GR says it does, and makes quantitative predictions about it, which have been confirmed.

    Yep.
     
  14. Nov 24, 2017 #13
    Thanks Peter
     
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