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Asymmetry of SR

  1. May 1, 2013 #1
    Haefele and Keating flew an atomic clock from the west coast to the east coast in 1971 and then they flew it back. The east going clock lost 59 +- 10 ns and the west going clock gained 273 +- 7 ns. What I do not understand is why they used the center of the earth as a reference frame, and not the surface-after all that's where the surface clock is located. They said:

    "The problem encountered with measuring the difference between a surface clock and one on an aircraft is that neither location is really an inertial frame. If we take the center of the earth as an approximation to an inertial frame, then we can compute the difference between a surface clock and the aircraft clock. "

    The need to establish a third frame of reference-the center of the earth-is confusing. Why is neither the surface clock's nor the airborne clock's frame of reference valid? Because the earth's surface is moving? But that's a major point of the einstein equivalence principle-that moving frames of reference have the same physics.

    I understand the equation for the time difference calculation:
    Ta-Ts=To[(2Rwv+v^2)/2c^2]
    where Ta is airborne time, Ts is surface time, To is center of the earth time
    R is earth radius, w is earth angular velocity, v is plane velocity, c is light speed
    And then they go on to use the following approximation:
    Ta-Ts=-Ts[(2Rwv+v^2)/2c^2]
    The difference being they use surface time instead of center of the earth time. They state:
    Note that the "earth center" time has been replaced by the surface time in this expression. This is a valid approximation in this case since the time difference is many orders of magnitude smaller than the time itself, and this allows us to model the difference between two measurable times.
    How can they conclude by using Ts when they started the whole analysis saying that surface time was not a proper inertial frame?

    And another confusing aspect of using the center of the earth as the reference frame is if you change the flight plan a little bit. Suppose the plane flies along a line of longitutude instead of latitude? Since v is orthogonal to earth angular velocity, is it measured as 0? A flight to south pole is the same as a flight to north pole in terms of time dilation? I wonder if that expt has been done?
     
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  3. May 1, 2013 #2
    No, they flew four cesium clocks completely around the world, once in the easterly direction and once in the westerly direction, and compared the results in both cases with surface clocks.

    The quote you mentioned didn't say those frames were not "valid", it said they were not inertial. The reason they are not inertial frames is two-fold. First, they are obviously not moving inertially. (They are going around in a circle.) Second, they are located at different heights within a gravitational field.

    You're mis-informed. The theory of relativity does not assert that the descriptions of phenomena are the same regardless of the system of coordinates in terms of which they are described, let alone that they are the same in terms of non-inertial coordinate systems as they are in terms of inertial coordinate systems. When people say the "physics is the same" for all (diffeomorphically equivalent) systems of coordinates in general relativity they mean that the same equations of physics, i.e., the field equations, are equally applicable to any such coordinates. This is true, but to use simplified equations that are adapted to specialized coordinate systems (e.g., inertial coordinates) we must obviously use the specialized coordinate systems. Otherwise the simplified equations are not applicable.

    This is just simple algebra. The quantity "To" can be written as Ts(1+q), where q is an extremely tiny dimensionless number (with suitable sign). Also the factor in the square brackets is an extremely tiny dimensionless number, let's call it p. So the time difference is

    Ta-Ts = -To[p] = -Ts(1+q)p = -Ts[p] - Ts(qp)

    Obviously the second term in that last expression (containing the factor of both q and p) is tiny compared to the first term, so it's fine to represent Ta-Ts as -Ts[p], provided only that we have an accurate evaluation of the factor p, which of course requires us to correctly account for the difference between To and Ts (and Ta).

    No, v would not be measured as zero. Being orthogonal to the tangential velocity of the Earth's surface does not imply that it is zero.

    Yes, a flight from the north pole to the south pole (along a constant "longitude" line mapped on the Earth's surface) would be symmetrical to a flight from south pole to north pole - at least if you neglect the effect of the Earth's orbital motion around the Sun.
     
    Last edited: May 1, 2013
  4. May 1, 2013 #3
    Tnx samshorn for ur prompt reply.
    1)yes i was inaccurate in describing their procedure. They did fly around the world. I also assumed that the surface clock was at the same latitude as the aircraft which flew circumferentially around the planet at a constant latitude, but I might be wrong there as well.
    2) I am only addressing the kinematic contribution to SR. The flights in fact were at different altitudes and the gravitational contribution to time dilation was different on the different parts of their journey. I am not questioning that. But I do not understand your statement that the surface clock and the airborne clock are in reference frames which " are obviously not moving inertially." What is an example of two reference frames that ARE moving inertially? When Einstein described the light clock on a train and a passenger watching it while it passed him at the station, were the passenger and the train considered to be in inertial frames?
    3) How can we assume that To=Ts(1+q) where q is small? To is the time at the center of the earth. There is no gravity there so time dilation should be greatest there. I would guess that To is radically different than Ts.Consider instead putting the surface clock at the north pole. Ignoring for the moment that the earth is an oblate spheroid and the gravitational effects are slightly different at the poles, consider the kinematic component. Angular velocity of the surface clock at the pole is 0. In this case, I understand how there can be an asymmetry of SR because the aircraft's velocity is relative to ground speed. The total aircraft velocity is therefore radically different depending on its direction.

    As a corollary, I am still pondering the possibility that there is time dilation at the equator due to rotational velocity. So people age faster at the poles?
     
  5. May 2, 2013 #4

    Nugatory

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    I added a few words to your quote, because we should never say "faster" without being clear about what we're comparing with. But with that said, and if we simplify the problem by treating the earth as a perfect sphere (this is a plenty good enough approximation here), the answer is "yes".

    You might try calculating just how much the difference is... you can find the time dilation formula all over the Internet and you can calculate the relative velocity between poles and equator because the circumference of the earth is about 40000 kilometers and it rotates on its axis once every 24 hours.
     
  6. May 2, 2013 #5
    Inertial motion is defined as (locally) unaccelerated motion. Hence anything that is unaccelerated is an example of something that is moving inertially. This is the same as in Newtonian mechanics, so your problem has nothing in particular to do with special or general relativity. I think it's hard for someone who doesn't have a working knowledge of Newtonian mechanics (or quantitative physical reasoning in general) to understand relativity.

    If you mean "were they considered to be at rest in inertial frames", the answer is a qualified Yes, to the level of precision necessary for those scenarios, i.e., over a short and virtually flat span on the Earth's surface, and over an extremely short duration of time (the nanoseconds it takes for light to traverse the train car), the deviations of the objects from inertial motion are negligible. In contrast, if the train continued all the way around the globe in a circle, and if we consider the station over a period of 24 hours circling the center of the earth, the objects are obviously deviating very significantly from being at rest in any single inertial coordinate system for the entire time. Regardless, we can always compute the effects of the non-inertial motion. In some circumstances the effects are negligible, and in others they are not. Exactly the same is true in Newtonian mechanics, so your problem is not with relativity, per se, it is with quantitative reasoning in physics generally.

    Your guess would be radically wrong. Fortunately there's no need to guess. General relativity tells you the difference in time dilation between the surface and center of the Earth, and it is extremely small.

    Well, neglecting differences in gravitational potential, etc., if you just imagine an object at rest (to the necessary precision) in an inertial coordinate system and another object moving around it in a circle, then yes, the object at rest in the inertial frame will accumulate more elapsed time.
     
  7. May 2, 2013 #6

    PeterDonis

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    Not really, because the Earth's oblateness *does* have a significant effect on clock rates for observers on its surface. In fact, it has just enough effect, due to increased altitude from oblateness, to compensate for the effect of rotational velocity, so clocks everywhere on the Earth's surface go at the same rate. So clocks at the poles do *not* run faster (or slower) than clocks at the equator, if both sets of clocks are at rest with respect to the (rotating) Earth.

    Strictly speaking, this is true for an idealized Earth that is a perfect oblate spheroid--the real Earth does deviate from this by small amounts from place to place. But it's still a much better approximation than the spherical one for this question. One reason is that it illustrates an important fact about rotating bodies: the "natural" configuration for such a body is for its surface to be an equipotential surface, i.e., for it to be such that clocks everywhere on the surface go at the same rate. If the surface is not an equipotential surface, there will be stresses within the object that will tend to shift the surface towards being an equipotential surface; i.e., the equipotential surface is the stationary equilibrium state for a rotating object.

    I understand that the questioner was trying to ignore the effects of altitude; but I suspect that is because he thought the effects of altitude would be negligible for the poles vs. the equator. They're not.
     
  8. May 2, 2013 #7

    Nugatory

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    So it does... Next time I'll say "idealization" or "simplification".
     
  9. May 2, 2013 #8
    So here is the nut of the problem

    I appreciate your continued clarifications. So, if I understand this correctly, (and continuing to ignore gravity as a simplification), if I a ringworld spinning at a very high velocity (~ 0.01 c at its equator) and did a similar experiment with a train riding on the inner surface that could reach a speed of 0.000001 c, then I could use a surface clock as an approximation of the inertial rest frame because the difference between the ringworld and the train would be so small?
     
  10. May 2, 2013 #9
    Whether a point on the surface approximates inertial motion depends (by definition) on the acceleration of that point and the relevant duration under consideration, neither of which you've specified. All you have specified is the speed of the surface, which tells us nothing (by itself) about the deviation from inertialness. You need to specify the acceleration (e.g., by specifying the radius of the surface, from which, given the speed, we could infer the acceleration), along with the relevant time duration for the events under consideration.

    Your concluding phrase ("because the difference between the ringworld and the train would be so small") is also puzzling. Again, the deviation from inertialness of any object just depends on its acceleration and the duration. It is not determined by the smallness of the "difference" (whatever kind of difference you have in mind) between two objects - unless you are stipulating that one of those objects is inertial.
     
    Last edited: May 2, 2013
  11. May 2, 2013 #10

    ghwellsjr

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    I've always wondered if this was just a coincidence but you're saying it's not. So the same would be true for Mars? Does the fact that the Moon, Mercury and Venus have synchronous rotations disqualify them from the same effect?
    At least he's in good company. Einstein predicted the same thing at the end of section 4 of his 1905 paper introducing SR:

     
  12. May 2, 2013 #11
    Well, it's a bit unfair to fault Einstein for failing to take into account the effects of general relativity before he discovered it.:biggrin:
     
  13. May 2, 2013 #12

    PeterDonis

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    Yes, at least to a first approximation. As I said, the Earth is not exactly an oblate spheroid, because its substance, being mostly solid (part of the core is liquid), can resist the stresses set up by rotation to some extent. Mars would be the same way.

    First, a note: Mercury actually doesn't keep one face to the Sun; it is in a slightly different tidally locked state, with 3 rotations (relative to the stars) for every 2 revolutions about the Sun. See the Wiki page:

    http://en.wikipedia.org/wiki/Mercury_(planet [Broken])

    That said, the effect I described is due to the rotation of an object relative to an inertial frame, not relative to any object it might be revolving around, so it would apply to the Moon and Venus (as well as Mercury, of course). Of course, the effect is smaller the slower the rotation period is.
     
    Last edited by a moderator: May 6, 2017
  14. May 2, 2013 #13

    ghwellsjr

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    I agree with you but the first time I brought this up I got an argument:

    https://www.physicsforums.com/showthread.php?t=558360
     
  15. May 2, 2013 #14

    ghwellsjr

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    Thanks. I would still call Mercury's rotation synchronous but I was mistaken about Venus'. It's very slow but not synchronized to the sun.

    Your point about the equipotential surface of a planet causing clocks to run at the same rate was also made by PAllen in the thread I linked to in my previous post but I didn't catch the significance of it then. Anyway, thanks, guys.
     
    Last edited by a moderator: May 6, 2017
  16. May 2, 2013 #15

    PeterDonis

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    I don't think that's the standard usage of that term, but I'm not really familiar with the terminology in this area, so I don't know how the various possible tidally locked states are described. Somebody on the Astronomy forum probably does.

    Yes, I missed that too. Venus is even weirder in that its rotation is retrograde (and the period is slower than its orbital period about the Sun).
     
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