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Is SR really internally consistent?

  1. Oct 1, 2013 #1
    Hi all. Hoping someone can give me a brief explanation re: a problem I'm having in trying to understand SR. Not looking for an argument, and won't engage in one. Just seeking a few views on the following issue I'm having with understanding.

    In his 1905 paper on Special Relativity, Einstein says:

    "If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be 1/2tv2/c2 seconds slow. Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."
    (Emphasis mine)

    I note that this was from Einstein's 1905 paper on Special Relativity, not his later work on General Relativity, which incorporates Special Relativity in whole, but in addition covers situations involving acceleration and gravity.

    Can anyone explain this to me: *The first postulate forces us to accept that there are no preferred reference frames (coordinate systems). The second postulate deals with he invariant nature of light velocity, irrespective of the motion of the source. *Given this,

    1. *Why did Einstein predict that the polar clock, and not the equatorial clock, would be faster, since the polar clock is in exactly the same relative motion to the equatorial clock as the equatorial clock is to the polar clock.

    2. *What justifies the assumption, obviously inherent in Einstein's prediction, that it is the equatorial clock that is moving slower than the polar clock? *

    3. *How could we ever test such a proposition, since returning the clocks to a common point where a single observer could make the comparison would invoke acceleration of one or the other clock, thereby causing someone to claim that the system is now one not covered by SR? *

    This seems to me to be sort of an inherent inconsistency between the theory and the above prediction. *The theory (postulate 1) tells us that al motion in such a system is relative, and that the equatorial observer would measure the polar clock as slow, and vice versa. *Why, then, the conclusion that it is the polar clock that is fast?

    Thanks for any clarity you can add to my confusion.
     
  2. jcsd
  3. Oct 1, 2013 #2

    WannabeNewton

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    Who said SR doesn't cover accelerations? SR can perfectly handle accelerations. Also when there is acceleration involved, frames aren't symmetric in the sense prescribed to mutually inertial frames by the principle of relativity.
     
  4. Oct 1, 2013 #3
    Are you saying that Einstein's statement in the 1905 paper leads to or sets out an asymmetric result because in fact acceleration would be involved in the example he gave in his paper, or...?
    Just not following you. The paper posited uniform velocity. Are you saying that his example would of necessity involve acceleration, and this is an unstated assumption, and that is why it is , or appears to be, non-symmetric?
    Sorry, not getting it.
     
  5. Oct 1, 2013 #4
    That's a misunderstanding. Situations involving acceleration are already handled by special relativity. General relativity is needed only to account for gravity.

    Not true at all. To the contrary, the principle of relativity singles out a preferred class of coordinate systems, called the inertial coordinate systems. These - and only these - are the coordinate systems in terms of which the equations of physics take the same homogeneous and isotropic form.

    Please note well that the principle of relativity, which applies only to inertial coordinate systems, does NOT imply that the theory of special relativity applies only to objects that are at rest in an inertial coordinate system, nor does it imply that special relativity requires us to describe things in terms of inertial coordinate systems.

    But only in terms of an inertial coordinate system. The velocity of light (in vacuum) can have any value you like, depending on what coordinate system you choose. But it has the value c in terms of any standard inertial coordinate system (for reasons that are fairly obvious to anyone who understands special relativity).

    Both of your "givens" were wrong, but let's press on with your specific questions.

    First, bear in mind that the variations in gravitational potential on the Earth's non-spherical surface were beside the point of that illustration, which Einstein tried to explain by saying "under otherwise identical conditions", i.e., setting aside any differential effects of gravity or anything else. So essentially he was just saying a clock moving in a circle around some inertial center will show less elapsed time than a clock at rest in the center. As noted above, special relativity clearly distinguishes between inertial coordinate systems and non-inertial coordinate systems. The clock moving in a circle around an inertial center is not moving inertially. It follows unambiguously from the theory that the circling clock will accumulate less elapsed time.

    This is just a re-statement of your previous question. Already asked and answered.

    There are infinitely many ways of testing such a proposition. Note that the reference to "clocks" is really just illustrative, since it applies to ALL physical phenomena involving characteristic time intervals. For example, put a particle with a known characteristic decay rate into circular motion at high speed, and verify that the decay time, as measured by an inertial clock at the center, depends on the particle's speed by precisely the amount predicted by special relativity. Or put clocks on an airplane and fly them around, or a GPS satellite. Confirm the predictions of special relativity. Apply this same kind of test to every other physical phenomenon you can think of, and confirm that they are all Lorentz invariant (so far).

    As already explained, special relativity has no difficulty making unambiguous predictions about accelerated motion. Again, as with all your other "givens", this one is simply a misunderstanding.

    You're welcome!
     
  6. Oct 1, 2013 #5

    PeterDonis

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    As Samshorn noted, this prediction is actually wrong because there is another effect involved: a clock at the poles is closer to the Earth's center than a clock at the equator, and the difference in clock rates due to that altitude difference is exactly equal and opposite to the difference in clock rates because the clock at the equator is moving relative to the center of the Earth and the clock at the poles is not.

    So when you make the prediction using GR, you predict that the two clocks go at the same rate (and in the actual world, the GR prediction is correct). In what follows, I'm ignoring the GR effect, due to gravity, and only talking about the SR effect, due to relative motion.

    The fact that the polar clock is moving inertially, while the equatorial clock is not, it is accelerated.

    As WannabeNewton pointed out, because the equatorial clock is accelerated, the situation is not symmetric. The first postulate only says that all *inertial* frames are equivalent physically; it does not say that inertial motion and accelerated motion are equivalent physically (obviously they're not, since acceleration can be directly measured).

    The equatorial clock's motion is periodic, so we can set up a common standard of time using its periodic motion, even though the two clocks are spatially separated. For example, we could position a radar beacon, at rest relative to the center of the Earth but at some high altitude (obviously it would have to be held in place by rocket thrust), in such a way that it sent out a signal each time the equatorial clock passed directly underneath it. The polar clock would then register more elapsed time (more ticks) between two successive signals from the beacon than the equatorial clock; this is one way of illustrating what "the equatorial clock runs slower" actually means, physically.
     
  7. Oct 1, 2013 #6

    WannabeNewton

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    Yes. A clock moving in uniform circular motion is not described by a single inertial frame. The speed of the clock is constant relative to an inertial frame at the center of the circular motion but the velocity is not. Couple this with the fact that an accelerating frame and an inertial frame are not symmetric in the sense that two inertial frames are symmetric and there are no inconsistencies in the results.

    EDIT: Samshorn and Peter beat me to it! Wee :)
     
    Last edited: Oct 1, 2013
  8. Oct 1, 2013 #7
    That example concerns uniform speed. The translators put "velocity" at a place that may be confusing: the clock moves at constant speed in a circle. Thus it is only at a single instant at rest in any inertial frame. It's certainly clearer if you consider the preceding sentence (which obviously was included for that very purpose!):

    "It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide."

    Each straight section of that line corresponds with a different inertial rest frame.
     
  9. Oct 1, 2013 #8
    I'm not being clear. The Einstein quote was from his first paper, and occurs after his discussion of simultaneity, the moving rod, and so forth. He then goes on to treat uniform movement (non-accelerated) along a curved line as equivalent to a straight line for purposes of the theory. (His words, not mine, but check the paper if you'd like). When he speaks of the clocks, he is clearly speaking of their relative velocity to one another. The one near the pole does not rotate, so relative to the one on the equator (or any other latitude line) it is still, and the other has a velocity. BUT could we not also say, or imagine, that the pole is really doing the moving, and the equatorial clock is not? Just take away the solid earth, and leave both clocks hanging in their respective postitions, dangling in space. We could see a stationary clock, with another clock rotating below it in a circle.
    Einstein says this "circle" could just as easily be a line. So we are back to the point that one can "say" that it is either clock that is stationary, and the other is moving, and that each would measure the other clock as slow.
    So, why is the polar one the one that turns out fast, and the other slow?

    I'm saying, rather badly, I'm sure, that this is just another twin paradox as far as I'm concerned. Is that the case?
     
  10. Oct 1, 2013 #9
    Just an added point: We could look at the polar clock as a point, and other moving past on a nearby line, as E makes clear, OR we could look at the polar clock as moving past a point on the line just described.
     
  11. Oct 1, 2013 #10

    WannabeNewton

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    No, again you cannot make that claim. There is a clear dichotomy between the two clocks that doesn't exist between two inertial clocks. Read the following: http://s10.postimg.org/wl3sjj4xl/nnts.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  12. Oct 1, 2013 #11
    Not seeing it. If we image the clocks in space, and I'm the center clock (polar one) I might see the one circling me, OR I might be moving relative to IT. What if we just look at two particles in space moving as described: One orbits the other, but isn't that situation equally well described by the reverse: A orbits B same as B orbits A.
    Thanks for the link, I'll read it. Computer running out of battery, so it might be a few minutes. Thanks.
     
  13. Oct 1, 2013 #12
    OK, I read the link, thanks. Guess I just don't understand it, probably never will. The theory says it's impossible to tell which is moving, all we can say is that they are in relative motion, i.e, there is a velocity difference between them such that they are moving closer or further away from one another. If we passed by in space, and saw two clocks rotating about one another, which one of the clocks is running slow relative to the other?
     
  14. Oct 1, 2013 #13

    WannabeNewton

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    Choran, I think you're confused about one simple but important thing: proper acceleration is not relative but rather absolute. It can be measured unambiguously. Same goes for rotation; rotation can be measured unambiguously. These two things cannot be treated in the same way as velocity, which is relative. See also this thread: https://www.physicsforums.com/showthread.php?t=711062
     
  15. Oct 1, 2013 #14

    Nugatory

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    The key here is that the two observers are not indistinguishable and equivalent in this case. If each observer reaches into his pocket, removes a coin, holds it out at arm's length and releases it...
    One observer's coin will float motionless; this is the one at the center. The other observer's coin will fall away from him as he follows his circular path while the coin follows a straight line tangent to the circle; he is not in an inertial frame and he can detect his circular acceleration just by observing the behavior of objects such as the coin that are near him.
     
  16. Oct 1, 2013 #15

    Dale

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    They are not symmetric. If you attach accelerometers then one will read zero (inertial) and the other will not (non inertial). The symmetry you are trying to assert does not exist.
     
  17. Oct 1, 2013 #16
    Perhaps my confusion is because the 1905 paper doesn't mention acceleration at all, and simply deals with relative motion. The polar clock example follows a line, as a poster indicated above, that seems to suggest Einstein was giving an example that would be equally true in a non-accelerated situation. You are undoubtedly right, though, so let's take rotation and acceleration out of the vocabulary for a second. I am next to the track, the train passes. OR I am on the train, and pass a guy standing by the track.
    Postulate one, I thought, maintains that these are indistinguishable. So, who's clock ends up slow, if we synchronize at the beginning?
     
  18. Oct 1, 2013 #17

    Dale

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    The theory says no such thing. The theory says that all inertial frames are equivalent, which is a considerably different statement.

    The one with the non zero proper acceleration as measured by an accelerometer.
     
  19. Oct 1, 2013 #18
    Maybe I could ask in this way: I understand in the usual examples given how A measures B's clock as slow, and how B measures A's clock as slow. I understand that, I believe, and the math that goes with it. Now, step two:
    What determines the tie breaker, at the moment of truth: "Gentlemen, show your watches." Trying to get this, really, not trying to be difficult.
     
  20. Oct 1, 2013 #19

    Nugatory

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    In the symmetrical case, where I can equally well say that A is moving relative to B or B is moving relative to A, there is no tie breaker - both observers correctly see the other clock running slow, and there is no paradox.

    In the asymmetrical cases, where one or both of the clocks does not remain in the same inertial frame for the duration (and this includes all the cases in which the two clocks are brought together at the same location so that we can directly compare them without the relativity of simultaneity getting in the way) there is an unambiguous and experimentally verified method of calculating the amount of time experienced (this is called "proper time") by each clock on its path through spacetime to the common meeting point.

    The pole/equator case falls in the latter category - there is an asymmetry between the two clocks because one of them experiences acceleration and moves on a non-inertial path.
     
  21. Oct 1, 2013 #20
    Thanks, Nugatory. Hmm. Think I getcha, but doesn't the verification itself then require application of the theory?
    Just point me to the experiment, I'll go check it out. Is it the one about flying the clocks around the earth in different directions, or...? Thanks again to all of you for taking the time to provide answers.
     
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