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Dale

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- #26

Dale

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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??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). [..]

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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.

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"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.

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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.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.

See in addition post # 4 by samshorn. But probably Einstein did not even consider any eventual effect of gravitation in that paper.

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).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.

See again post # 4. His prediction was effectively wrong but technically correct: at equal temperature, gravitational potential etc. his prediction is surely correct.[..] I believe, that Einstein was just wrong in his statement, especially if footnote #7, quoted above, is his. [..]

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- #36

Dale

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See my correction: the footnote was not by Einstein. Einstein made his prediction only for a balance clock, for the reason that Sommerfeld(?) explained and which I elaborated here - the clock prediction is not valid for a partial clock. You can check the original footnotes through the link in the German Wikipedia. http://users.physik.fu-berlin.de/~kleinert/files/1905_17_891-921.pdfHmm, 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". [..].

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The prediction is valid as I stated in post 31. Thus in principle a pure SR test with transporting clocks is possible; but I don't know if anyone ever bothered. All such tests that I know about involved both velocity and height changes, and the GR prediction includes the SR prediction.[..] I wonder then how the theory can be tested in an earth (or near earth) setting, since (I think) that even atomic clocks vary not just according to velocity, but by altitude above the earth. I'm just saying that this makes the theory (SR, not GR) one that may be impossible for us to test. [..].

See also for other tests: http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

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Only atomic clock tests that I know of. Once more, check out the other experiments - there are several mentioned that you overlooked. http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Tests_of_time_dilation

PS: I had forgotten that the Hafele and Keating test involved clocks in airplanes in two directions - thus the difference between these two was largely a test of SR, similar to Einstein's thought experiment (see point 5, with the inappropriate header "Twin paradox").

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- #43

Dale

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Yes, you are wrong. Please read the sticky on experimental tests of SR. First, it is wrong that all tests of SR have involved terrestrial sources. Second, it is wrong that SR is impossible to test, it is emminently falsifiable. There are many possible tests of SR regardless of the fact that the earth has gravity.I'm just saying that this makes the theory (SR, not GR) one that may be impossible for us to test. All of the tests at this point (again, SR, not GR) have of necessity involved tests on or near the earth, and I know of no clock we could use to satisfy the definition of a good clock. Maybe i'm wrong.

SR makes definite predictions about the outcomes of experiments and those predictions can be verified and falsified. It is testable.

- #44

Dale

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To emphasize what The_Duck says here, in a typical partical accelerator the particles travel in a horizontal path, so there is no difference in gravitational potential and therefore no gravitational time dilation, even according to GR. This is just one example of a test that is carried out in a gravitational field that is nonetheless a test of SR. The sticky is full of others.

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I have found an article where the author says the following about einstein's SR: If we have two reference frames, S and S', with S stationary and the S' moving along the 'x' axis with speed 'v' of S, then as per SR there is length contraction in the 'x' direction but no change in the 'y' or the 'z' directions of objects in S'. That is y=y' and z=z'. If we take just the y=y', then the length y2-y1 = y'2-y'1. However, SR says that the time is slower in S' than in S. This means t2-t1 is < t'2-t'1. So, if we have a beam of light traveling along the y axis in S then the speed of the light is C= y2-y1/t2-t1. For the observer in S' who is also looking at the same beam of light it will be C'= y'2-y'1/t'2-t'1. Now, the numerators are equal but the denominator in S' is larger than in S due to time dilation. This means C'<C, which contradicts the constancy of the speed of light postulate of SR!!!! Also, for light traveling in x-axis direction we have from length contraction x2-x1 > x'2-x'1 and from time dilation we have t2-t1 < t'2-t'1, then the speed of light in S is C=x2-x1/t2-t1 and in S' it is C'= x'2-x'1/t'2-t'1. So, for C' we have the numerator that is smaller than in S and the denominator that is greater than that in S. This again leads to C' < C !!!!. So, how do we resolve this? thanks.

In my previous post i should have said that S' is moving along the x-axis of S with constant speed 'v'. thanks.

In my previous post i should have said that S' is moving along the x-axis of S with constant speed 'v'. thanks.

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Could you post a link the the article, assuming it available online? Thank you.

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I think you are incorrectly assuming that the light will be travelling along the y axis in both both S and S'. If it is travelling along the y axis in S, then it will be travelling along a diagonal path in S'. You need use the resultant path length in S' taking the both the horizontal and vertical motion. Do a search for light clock to see what I mean.

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https://www.physicsforums.com/showthread.php?t=229034Could you post a link the the article, assuming it available online? Thank you.

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The author has some misunderstanding of the difference between coordinate effects and invariants.

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