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Hafele-Keating with the plane as reference frame?

  1. Aug 28, 2014 #1
    Hello,

    I noticed quite a few questions on the Hafele-Keating experiment here, but I believe none that have my exact question.

    I understand the explanation given for "a frame of reference at rest with respect to the center of the earth", as in the Wikipedia article.

    But what happens if we consider, say, the eastbound plane as the reference frame? Then the clock in there becomes the reference clock. The clock on earth moves away from the plane, in the plane's reference frame. So does the clock on the westbound plane -- even faster. Should we conclude that the clock on earth will tick more slowly than the reference clock, and the clock on the westbound plane more slowly still? But that is not what the experiment showed.

    Furthermore, if you do the same with the westbound plane as the reference frame, then it goes the other way around: now the earth clock and eastbound clock would be expected to be slower.

    (I don't see that either the west or east - or any other - direction is 'preferred' in some sense.)

    Maybe the answer is that, counter-intuitively, everybody is right - those who conclude from the westbound plane, earth, and eastbound plane reference frames are all correct, even though they give conflicting results. But then: surely (??!) the clocks are devices that can only show a single value -- and in the experiment they showed those that correspond to the earth reference frame.

    So... what am I missing? Thanks!
     
  2. jcsd
  3. Aug 28, 2014 #2

    A.T.

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    Then you have a non-inertial reference frame which makes calculations more complicated. But when you do them right, you get the same results.

    The situation between the plane reference frames is not symmetrical. Assuming the same groundspeed, the reference frames planes going east vs. west have different proper acceleration.
     
  4. Aug 28, 2014 #3

    ShayanJ

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    This kind of answer to such questions, always bothers me a little. Because we simply can assume that the plane can change speed instantaneously and so the returning points have almost no effect in the analysis!
     
  5. Aug 28, 2014 #4
    Thanks for answering, but I'm afraid that flies (:wink:) over my head. What makes the plane's reference frame 'non-intertial' (not quite sure what that means) and the earth's not? Naively, I'd say both the airplane clock and the earth clock are moving wrt something else, and are thus subject to inertia... Except if you pick either as the reference frame of course, then they are stationary. But you can do that for either.

    I agree that the situation from the two planes' reference frames is not symmetrical: you'd get different numbers. What I meant with my statement between parentheses, is that I see no reason to prefer either reference frame in terms of correctness.
     
  6. Aug 28, 2014 #5

    russ_watters

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    Answer to both of the last two posts: The plane has engines that it uses to accelerate it up to speed. You cannot just ignore that.
     
  7. Aug 28, 2014 #6

    Nugatory

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    Because HK involves two clocks that begin and end their journeys collocated, i find it easiest to think of it as a twin paradox problem. The key element is that the two paths through spacetime have different lengths.

    This wouldn't be the first time that a realizable experiment is more difficult to analyze and explain than a textbook idealization.
     
  8. Aug 28, 2014 #7

    Nugatory

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    Neither is more "correct" than the other. We have two different paths through spacetime between the same two points - one of these paths is longer (meaning that more time passes along it) than the other.

    You may want to spend some time reading about the "twin paradox", where similar (but more easily analyzed) asymmetries lead to two clocks showing different elapsed times between two events (start of flight and end of flight in both cases).
    http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
     
    Last edited: Aug 28, 2014
  9. Aug 28, 2014 #8

    A.T.

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    I'm not talking about the returning points. A plane in level flight at constant ground speed in not an inertial frame in General Relativity, because an accelerometer on the plane measures upwards proper acceleration. And given equal ground-speed those will be different between planes going west / east.
     
    Last edited: Aug 28, 2014
  10. Aug 28, 2014 #9

    A.T.

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    An accelerometer on the plane measures non-zero proper acceleration.

    You get the same numbers for frame-invariant quantities, like which plane's clock measured more time.
     
  11. Aug 28, 2014 #10

    A.T.

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    You can ignore that if you compare the clocks mid-air, when the planes have already reached their travel speed and pass each-other at close distance.

    What you cannot ignore is the proper acceleration upwards caused by the lift of the wings, even in level flight at constant speed. This happens throughout the whole experiment. And since the required lift for flying west / east is slightly different, so is the proper acceleration.
     
  12. Sep 4, 2014 #11
    Thanks for all the answers. Just found a moment to read up on this a little, particularly the explanation on the Twin Paradox which Nugatory linked to (also: fictitious forces). I wasn't aware of that (pseudo)paradox. I understand a bit more now, but far from all.

    In the Twin Paradox explanation, I understand the path integral and Doppler story, but it doesn't seem to me that that page quite explains why Terence gets a vertical worldline and Stella a curved one in the first place. The 'paradox' is precisely that it feels as though it could just as well be chosen the other way around.

    It seems it comes down to the presence or absence of acceleration. My main struggle is to understand when it is valid to see something as an inertial reference frame. A.T. explains (if I get it right) that the airplane is not an inertial reference frame - not so much because of its acceleration in takeoff and landing (do these cancel out?) but because of its sustained acceleration away from the Earth, resisting gravity, while it is flying at constant horizontal speed. Whereas the Earth, or objects like a clock on its surface, do not have such an acceleration because they 'are where gravity wants them'. Is that correct?

    Though then I wonder about the relay variant of the TP where Stella makes no turnaround, but a third twin flies past her and back to Terence, all at constant velocity. Nothing seems to accelerate there. Thought it looks like the 'time gap' situation is actually just that. This still has to sink in :/

    One detail I don't get:
    Why would it be slightly different (assuming identical atmosphere etc)?

    Thanks again :)
     
  13. Sep 4, 2014 #12

    A.T.

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    An accelerometer will tell you if a frame is inertial:
    http://en.wikipedia.org/wiki/Proper_acceleration
    But in curved space time they are only locally inertial.

    Correct, the planes have proper acceleration upwards at constant horizontal speed. The takeoff and landing accelerations can be worked around as I described. But when they fly east/west at the same groundspeed, they still have slightly different proper accelerations.

    No. A clock resting on the surface does have proper acceleration upwards. A clock resting at the Earth's center has zero proper acceleration.


    No mass accelerates, but the path along which the time is accumulated is not-inertial, because if you would measure the proper acceleration along it, it would be non-zero at some point.

    In the inertial frame of the Earth's center, because of different centripetal coordinate accelerations.
    In the rotating frame of the Earth, because of Coriolis force.
     
    Last edited: Sep 4, 2014
  14. Sep 4, 2014 #13

    Nugatory

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    It seems that wy at first, but it's not. You've already picked up on the way that the relay scenario is acceleration free, and there's another acceleration-free possibility as well: Stella can execute her turnaround by doing a tight hairpin orbit around a suitably placed distant star. That way she's in freefall except at the beginning and end of the journey, and we can eliminate that by synchronizing the clocks after Stella accelerates to start the journey and recording the times before Stella decekerates at the end of the journey.

    Another reason to seriously distrust the acceleration explanation is that the difference in elapsed time depends on the length of Stella's journey. She can travel one light-year or 1000 light-years and it's still the same acceleration to change directions at the turnaround - but with very different results.

    Basically you're asking why we're justified in saying that Stella's path is "really" longer than Terence's path.

    Here's a question back at you: Why are we justified in saying that the path between New York and Miami via Seattle is longer than the direct route down the eastern coast of North America? There are several possible answers, such as:
    - Drive them both, compare the distance recorded on the car's odometers. That's basically what Stella and Terence are doing with their clocks to measure the amount of time on their respective paths through spacetime.
    - It's a property of the locations of New York, Seattle, and Miami on teh surface of the earth. The place that we call "Miami" is positioned in such a way that the shortest path between it and New York doesn' pass through Seattle.
    - We can calculate the distances using what we know about the locations of the three cities and some spherical trignometry. We'll learn that the shortest possible path is the great circle between New York and Miami, and the eastern coast route is much closer to that path than the Seattle detour.

    However, it doesn't take a lot of scrutiny to see that all of these answers come down to "Because it is" - they're just different ways of restating the premise. This doesn't bother us when we're talking about distances in space because we have a lifetime of experience with such distances. There's no logical reason why it should bother us more when we're talking about distances in spacetime.
     
  15. Sep 4, 2014 #14

    A.T.

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    To move on a great circle, you just go straight ahead. In space-time this corresponds to moving inertially, without proper acceleration.
     
  16. Sep 4, 2014 #15

    Dale

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    I don't know why that assumption would be involved for the HK experiment.
     
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