# Is SR really internally consistent?

choran
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.

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.

choran
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.

Samshorn
... General Relativity incorporates Special Relativity... but in addition covers situations involving acceleration and gravity.

That's a misunderstanding. Situations involving acceleration are already handled by special relativity. General relativity is needed only to account for gravity.

Can anyone explain this to me: *The first postulate forces us to accept that there are no preferred reference frames (coordinate systems).

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.

The second postulate deals with he invariant nature of light velocity, irrespective of the motion of the source.

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).

*Given this,...

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

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.

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.

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...?

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).

...since returning the clocks to a common point ... would invoke acceleration of one or the other clock, thereby causing someone to claim that the system is now one not covered by SR? *

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.

Thanks for any clarity you can add to my confusion.

You're welcome!

Mentor
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.

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.

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.

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

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? *

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).

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? *

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.

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...?

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 :)

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harrylin
[..] 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.
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.

choran
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?

choran
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.

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]

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choran
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.

choran
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?

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

Mentor
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.

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.

Mentor
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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.
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.

choran
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?

Mentor
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The theory says it's impossible to tell which is moving, all we can say is that they are in relative motion,
The theory says no such thing. The theory says that all inertial frames are equivalent, which is a considerably different statement.

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?
The one with the non zero proper acceleration as measured by an accelerometer.

choran
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.

Mentor
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.

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.

choran
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.

Mentor
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There is a sticky at the top of the forum entitled experimental basis of SR. The experiments are described there.

1 person
The_Duck
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.

To have an unambiguous tiebreaker, you need both clocks to start and end in the same place so that you can compare them side-by-side. In the simplest examples of relative motion, both clocks are moving in a straight line at constant velocity. Thus they can't start and end at the same place. If two observers start out in the same place at t=0, then move away from each other with uniform motion in a straight line at constant velocity, each observer will say the other's clock is running slow, and there can be no tiebreaker because the two clocks will never be in the same place again, so they can't be compared side-by-side.

Einstein in the quote considered circular motion because then you can have both clocks return to their starting point and get an unambiguous comparison of the elapsed time on each. Suppose both clocks start off at the same point; one remains at rest while the other executes a circular trajectory and returns to the initial point. At the end, the two clocks are compared and less time has elapsed on the stationary clock. This is essentially the experiment Einstein considers.

Einstein has no need to discuss acceleration here, because everything is being analyzed from the perspective of an inertial reference frame. The time dilation of a moving clock, as seen from an inertial reference frame, depends only on the speed of the clock, and not on its acceleration.

If we wanted we could analyze the experiment from the perspective of a non-inertial reference frame. For instance we could use the non-inertial reference frame of the moving clock. The problem is that the analysis becomes more complicated. In a non-inertial reference frame, the time dilation of a clock depends not only on its speed but also on its position. So an observer moving with the moving clock would have to include these position-dependent time dilation effects when he calculates how his clock will compare to the stationary clock when they are brought back together. He cannot simply say: "that other clock is moving in my reference frame, so it is running slow, so it will record less elapsed time than my clock when I return to it." His reference frame is non-inertial, so in addition to the time dilation from motion he has to account for the position-dependent time dilation. If he does the calculation correctly, he will get the result that the stationary clock will record more elapsed time than his clock when he returns.

choran
Thanks a million, Duck, I understand your answer. If you don't mind another one, can you shoot me a quick explanation for the east-west difference in the clock experiment I just read? These aren't he actual numbers, but it was, as an example, one clock one minute fast and one clock 40 seconds slow relative to the clock that stayed home. Thanks again.

The_Duck
Thanks a million, Duck, I understand your answer. If you don't mind another one, can you shoot me a quick explanation for the east-west difference in the clock experiment I just read? These aren't he actual numbers, but it was, as an example, one clock one minute fast and one clock 40 seconds slow relative to the clock that stayed home. Thanks again.

Perhaps you are referring to this experiment? The idea here is that the Earth is rotating to the east, so planes that fly east are going faster than planes that fly west. The clock that flies east should therefore experience more time dilation than the clock that flies west, so when the clocks are reunited the east-going clock should show less elapsed time. Indeed, that is what the experiment observed. As the Wikipedia page notes, there is an additional time dilation effect from gravity, but this is approximately the same for both clocks.

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choran
would the result be the same if we sent two light beams around the world at the equator, for example, one traveling east, one west? In other words, set up a a single clock, send off two beams (assuming we could do such a thing, such as through a fiber optic cable or whatever) and see if they arrive together?

Mentor
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The east and west pulses would not arrive together. That is the basis of a ring interferometer. It acts essentially as an optical gyroscope to detect rotation. They are commonly used in inertial navigation systems.

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harrylin
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). [..]
You were clear and I see that several of us of us answered the same in different words. In post #7 I even pointed to the phrase in that text that may have confused you, and explained the misunderstanding with referral to an earlier sentence in that paper that clarifies it. So, apparently I was not clear enough (or you overlooked it?). What was not clear in my answer or in Einstein's clarification??

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choran
My response #8 is not a response to your #7, but to an earlier one. Perhaps it's a problem with simultaneity. Thanks for your response as well.

harrylin
My response #8 is not a response to your #7, but to an earlier one. Perhaps it's a problem with simultaneity. Thanks for your response as well.
You're welcome. :tongue2:
I think that it's first of all a problem with understanding what a reference frame is of SR, that is: a reference system in which according to classical mechanics the laws of Newton hold (see §1 of that 1905 paper; I slightly disagree with the footnote which apparently was inserted by the editor). Usually we say nowadays "inertial frames", but that is somewhat ambiguous in the presence of gravitation.

choran
It is confusing to me that (in the portion of the paper I originally quoted) E speaks about straight lines, then says, of course, this would also apply to constant speed on a curved line, and from that (next line) makes his unqualified and unequivocal statement about the clock on the equator, and the clock at the pole. THEN to add a bit to the confusion, see footnote 7, which I assume was Einstein's:

"7. Not a pendulum-clock, which is physically a system to which the Earth belongs. This case had to be excluded."

For all other clocks than a pendulum clock, then, Einstein rather plainly states that special relativity would apply in the polar clock/equatorial clock example that he used. He excludes a pendulum clock because of effects that the earth's rotation and/or gravity would have on the operation of that clock, and not on his imaginary clock being used in the example. However, you guys point out, correctly, that the equatorial clock that he posits would ALSO be smartly affected due to the fact it is the earth we are talking about, and not some hypothetical reference frame. You guys point out, correctly, I believe, that Einstein was just wrong in his statement, especially if footnote #7, quoted above, is his. I can't say what he meant, but it is clear what he said.

harrylin
It is confusing to me that (in the portion of the paper I originally quoted) E speaks about straight lines, then says, of course, this would also apply to constant speed on a curved line, and from that (next line) makes his unqualified and unequivocal statement about the clock on the equator, and the clock at the pole.
Yes indeed. He simply elaborates in a concrete way on the meaning of doing an integration (Δx -> dx). A circle can be (and always has been) intepreted as an infinite number of infinite straight lines. Thus, if we assume -as he explicitly did- that acceleration itself has no effect on the clock, then the effect of a circular trajectory must be about the same as the effect of many short straight trajectories, and exactly the same as an infinite number of infinitely short straight trajectories.
See in addition post # 4 by samshorn. But probably Einstein did not even consider any eventual effect of gravitation in that paper.
THEN to add a bit to the confusion, see footnote 7, which I assume was Einstein's:
"7. Not a pendulum-clock, which is physically a system to which the Earth belongs. This case had to be excluded." For all other clocks than a pendulum clock, then, Einstein rather plainly states that special relativity would apply in the polar clock/equatorial clock example that he used. He excludes a pendulum clock because of effects that the earth's rotation and/or gravity would have on the operation of that clock, and not on his imaginary clock being used in the example.
Yes that footnote was by Einstein, [edit: OOPS, mistake: I now found back the original and see that that footnote was added later] but you misunderstand his explanation - he says nothing about the Earth's rotation or gravity, that's besides the point. To elaborate: a pendulum clock doesn't work in outer space; the Earth is part of its "spring" mechanism! Thus a moving pendulum clock is only half a moving clock, with the other half of the clock mechanism in rest (approximately).
[..] I believe, that Einstein was just wrong in his statement, especially if footnote #7, quoted above, is his. [..]
See again post # 4. His prediction was effectively wrong but technically correct: at equal temperature, gravitational potential etc. his prediction is surely correct.

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choran
"... says nothing about the earth's gravitation or rotation." True, but that is the reason a pendulum clock would not fit into his example, is it not, because such a clock depends upon gravity to function (and will also be influenced, ala Foucault) by rotation. He's saying, I think, that no clock which depends upon or is influenced by those things would work in his polar/equator example. All others would, according to his implied exclusion of that class of clock. Why did he even use the earth as an example? All clocks on or near the earth's surface are affected by the motion of the earth. Otherwise, no Sagnac effect, right? Or wrong. LOL Not sure which.

choran
Whoops forgot. I have to disagree with one of your statements, "...the pendulum clock doesn't work in outer space." True, but I doubt that Einstein was thinking about outer space at all when he wrote the 1905 paper. The paper's examples are all earth-based. I think there was another reason he mentioned only pendulum clocks in his exclusion. I think he assumed that only the accuracy and/or synchronization of pendulum clocks would be affected by their differing positions on the earth's surface, irrespective of the relativistic effects he was predicting.

choran
Another whoops: I guess what I'm saying is that (unlike in the case of general relativity) there are no "suitable clocks" with which to test special relativity in any earth based system of experiments, for there will always be differential gravitational forces with which to content, and E has, I believed, at least impliedly excluded that type of clock.

choran
Hmm, interesting: I just read that Sommerfield added the "pendulum clock" language in a 1913 edit/reissuance of the paper, and that Einstein's original words had been "balance clock". The Sommerfield language was, some think, added to answer questions of the sort I have asked, and to reinforce (or add the caveat that) we must exclude differential gravitational effects from Einstein's clocks. I don't know one way or the other, just read it in another paper, but didn't want you to think I was sandbagging you or quoting E incorrectly. May have been Sommerfield, don't know.