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Atomic clocks were used in Hafele and Keatings experiment which supposedly helped to prove time dilation.

Time Dilation Proof - Hafele and Keating

How can we be sure other forces didn't act upon the clocks as a result of their being at different speeds and elevations? Perhaps the elapsed time may have been the same for all and only appeared different as a result of the clock's atoms behaving differently in different environments.

How might the presence of dark matter (presumably all over the place) impacted the clocks if say the dark matter is stationary relative to the earth's rotation?

What experiments have been done to test the constancy of clocks at different speeds and elevations?

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    $\begingroup$ "What experiments have been done to test the constancy of clocks at different speeds and elevations?" GPS anyone? $\endgroup$ Commented Aug 10, 2012 at 2:19
  • $\begingroup$ @dmckee: according to the caption under the first sidebar imagage on wikipedia's Time Dilation (en.wikipedia.org/wiki/Time_dilation), "GPS satellites work because they adjust for similar bending of spacetime in order to coordinate with systems on earth." Wouldn't this put your GPS proposition under the same assumptions? $\endgroup$ Commented Aug 10, 2012 at 2:40
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    $\begingroup$ The theory makes predictions about how the time kept by clocks on the satellites relate to each other and to that kept by clocks on Earth. These prediction are borne out. Experiment agrees with theory. In any case both gravitation and motion based shifts in the signal frequencies are also measured and also agree with theory. $\endgroup$ Commented Aug 10, 2012 at 2:57
  • $\begingroup$ Your original comment seemed to attempt to point out a verification of the clocks, when in fact it is at best a repeat of the Hafele and Keatings test. Perhaps it is not possible to verify? $\endgroup$ Commented Aug 10, 2012 at 3:02
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    $\begingroup$ It is verified. That's the point. This isn't one experiment at one set of conditions...it is many experiments under lots of conditions. The dependence of the timing difference on both altitude and speed is consistent with the theory. You are left arguing that the theory might be wrong because there might be something else generating exactly the same dependence; at which point some dude named Occam comes along and thwacks you with the giant foam cluebat. $\endgroup$ Commented Aug 10, 2012 at 3:06

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There is always a possibility, in any experiment, that some unknown effect was the main cause for the obtained measurements. This can never be ruled out.

The reasonable way to approach this is to first ask what is our null hypothesis. In this case we would like to test general relativity, so our null hypothesis is that there is no relativity and that Newtonian mechanics -- the previous successful theory -- will be able to explain the experiment. The alternative hypothesis, which is the one we would like to test, is that general relativity is correct.

So now we have two predictions for the shift in the clocks. For this to be interesting the predictions should be very different, compared with the experimental sensitivity. We carry out the experiment, and if the result 'agrees' (in a statistical sense that I will not go into) with general relativity, then this is evidence in favor of general relativity. If the results 'agree' with Newton, this is evidence against general relativity. If the results 'disagree' with both, it is evidence against both theories.

Either way, the experiment does not prove that GR is correct or incorrect, for example because of a possible unknown effect. It can only increase or decrease our confidence in GR, namely it can change the probability that we humans assign to the statement 'GR is correct', and even that is only within the regime of parameters where we can carry out experiments. The best we can hope for in science is to have theories that we are very confident about.

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Newton's predictions are consistent with Einstein's under the appropriate situations. No thwacking. You don't get to adjust a theory for every particular experiment, if your theory is well-constructed. General Relativity starts with a very small number of assumptions, and then creates predictions. These predictions are bourne out by a very large number of experiments. From the time dilation of decaying particles, to the laser ranging of the moon, to gravity probe b, to hartle hawking, to the Eötvos experiment. The effects of time dilation, and general relativity in general, have been measured under too wide a variety of circumstances for concerns about atomic clocks to be a credibe reason for doubt:

http://relativity.livingreviews.org/Articles/lrr-2006-3/

Come up with a simpler, more straightforward theory that does better. If you can, a certain nobel prize awaits you.

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  • $\begingroup$ I appreciate your answer, but it does seem to tacitly admit to there being no verification of clock constancy. I had initially thought to challenge all the proofs one by one beginning with H and K's use of the clocks. Perhaps that is not the best approach. $\endgroup$ Commented Aug 10, 2012 at 3:33
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    $\begingroup$ @CaptainClaptrap: the clock constancy is verified elsewhere. The fact that the clocks then also match the theoretical predictions of GR, which are ALSO verified elsewhere should be sufficient. Unless you're a straight on Humean skeptic, but then why are you doing science at all? There are no absolutes anywhere. $\endgroup$ Commented Aug 10, 2012 at 4:46
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If the effect wasn't due to time dilation, then why would the measured results agree with predictions of relativity so well? On the 25th anniversary of the H-K experiment, the National Physical Laboratory reproduced the experiment. For the clock flying from London to Washington, the relativistic prediction was that the clock would gain 39.8 nanoseconds. The measured result was a gain of 39 $ \pm$ 2 nanoseconds. In 2010, the experiment was done by the NPL once again - this time, around the world. It went from London, to Los Angeles, to Aucklund, to Hong Kong, and then back to London. They measured a gain of 230 $\pm$ 20 nanoseconds, compared to relativity's prediction of 246 $\pm$ 3 nanoseconds.

So, the odds of the clocks being off by just the right amount to account perfectly for relativity are incredibly slim.

In terms of the accuracy of clocks, I don't know if they have been tested in different environments. However, NPL's clocks have been tested to be the most accurate in the world.

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  • $\begingroup$ I'm not suggesting error in the clocks. I'm suggesting variations in the behavior of the atoms due to differing conditions. Do atoms behave differently when traveling around the world? The difference wouldn't have to be much. I think a behavior difference is a possiblity worth exploring. $\endgroup$ Commented Aug 10, 2012 at 3:47
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    $\begingroup$ @CaptainClaptrap: Yes, atoms behave differently when traveling around the world, in exactly the manner in which Mark M describes. The time gains that Mark M reports are not errors in the operation of the clocks, but rather changes in the clocks' behavior due to the atoms behaving differently. $\endgroup$ Commented Aug 10, 2012 at 3:54
  • $\begingroup$ But if the atoms are behaving differently, time is not changing between the clocks. Differences observed do not indicated time dilation. You are on my side if, as you said "the atoms behave differently." $\endgroup$ Commented Aug 10, 2012 at 4:03
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    $\begingroup$ @CaptainClaptrap: I can assure you, I am not your side. "The atoms behave differently" is not a particularly precise phrase, so I suspect you and I are using it differently. What I mean by it is that time dilation causes the atoms to experience time differently. They behave the same relative to the time that they experience. They experience time differently, so they behave differently. What do you mean by "the atoms behave differently"? $\endgroup$ Commented Aug 10, 2012 at 4:15
  • $\begingroup$ I mean that their frequency changes under the varying conditions. The clocks use the frequency of the atoms to count time: An atomic clock is a clock that uses an electronic transition frequency in the microwave, optical, or ultraviolet region of the electromagnetic spectrum of atoms as a freqhuency standard for its timekeeping element. (en.wikipedia.org/wiki/Atomic_clock) If you still disagree, your comment still doesn't make sense. The atoms would behave the same if time dilation were true. They just would have been behaving for a longer period of time than the others. $\endgroup$ Commented Aug 10, 2012 at 4:24
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How can we be sure about the constancy of atomic clocks as in the Hafele and Keating time dilation test?

Of any one clock $\mathfrak A$ it can be measured whether its average rates were equal to each other, or by how much they differed from each other:

In regard to

  • any two (distinct) indications $A_{\circ F}, A_{\circ G} \in \mathcal A$ of an identifiable material participant $A$, with assigned readings $t_{\mathfrak A}[ \, A_{\circ F} \, ], t_{\mathfrak A}[ \, A_{\circ G} \, ] \in \mathbb R$ and

  • any two (distinct) indications $A_{\circ J}, A_{\circ K} \in \mathcal A$ of an identifiable material participant $A$, with assigned readings $t_{\mathfrak A}[ \, A_{\circ J} \, ], t_{\mathfrak A}[ \, A_{\circ K} \, ] \in \mathbb R$

clock $\mathfrak A \equiv (\mathcal A, t_{\mathfrak A})$ had run from $A_{\circ F}$ until $A_{\circ G}$ and from $A_{\circ J}$ until $A_{\circ K}$ at equal average rates if

$$(t_{\mathfrak A}[ \, A_{\circ K} \, ] - t_{\mathfrak A}[ \, A_{\circ J} \, ]) = (t_{\mathfrak A}[ \, A_{\circ G} \, ] - t_{\mathfrak A}[ \, A_{\circ F} \, ]) \, \left( \frac{\tau^A_{[ \, \circ J, \, \circ K \, ]}}{\tau^A_{[ \, \circ F, \, \circ G \, ]}} \right), $$

where $\large \left( \frac{\tau^A_{[ \, \circ J, \, \circ K \, ]}}{\tau^A_{[ \, \circ F, \, \circ G \, ]}} \right)$, i.e. the value of the ratio between $A$'s duration from its indication $A_{\circ J}$ until its indication $A_{\circ K}$ and $A$'s duration from its indication $A_{\circ F}$ until its indication $A_{\circ G}$ can and must be measured beforehand (by the familiar methods of the theory of relativity), of course.

If all average rates of clock $\mathfrak A$ between all (relevant) pairs of its indications were equal, then $\mathfrak A$ is said to have run at constant rate.

Also of interest is the comparison between the average rates of different clocks:

In regard to

  • any two (distinct) indications $A_{\circ J}, A_{\circ K} \in \mathcal A$ of an identifiable material participant $A$, with assigned readings $t_{\mathfrak A}[ \, A_{\circ J} \, ], t_{\mathfrak A}[ \, A_{\circ K} \, ] \in \mathbb R$ and

  • any two (distinct) indications $B_{\circ P}, B_{\circ Q} \in \mathcal B$ of an identifiable material participant $B$, with assigned readings $t_{\mathfrak B}[ \, B_{\circ P} \, ], t_{\mathfrak B}[ \, B_{\circ Q} \, ] \in \mathbb R$

the two clocks $\mathfrak A \equiv (\mathcal A, t_{\mathfrak A})$ and $\mathfrak B \equiv (\mathcal B, t_{\mathfrak B})$ had run at equal average rates if

$$(t_{\mathfrak B}[ \, B_{\circ Q} \, ] - t_{\mathfrak B}[ \, B_{\circ P} \, ]) = (t_{\mathfrak A}[ \, A_{\circ K} \, ] - t_{\mathfrak A}[ \, A_{\circ J} \, ]) \, \left( \frac{\tau^B_{[ \, \circ P, \, \circ Q \, ]}}{\tau^A_{[ \, \circ J, \, \circ K \, ]}} \right), $$

where $\large \left( \frac{\tau^B_{[ \, \circ P, \, \circ Q \, ]}}{\tau^A_{[ \, \circ J, \, \circ K \, ]}} \right)$, i.e. the value of the ratio between $B$'s duration from its indication $B_{\circ P}$ until its indication $B_{\circ Q}$ and $A$'s duration from its indication $A_{\circ J}$ until its indication $A_{\circ K}$ can and must be measured beforehand (by the familiar methods of the theory of relativity).

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  • $\begingroup$ O my! That's way above my head. $\endgroup$ Commented Feb 17, 2022 at 3:40
  • $\begingroup$ Captain Claptrap: "That's way above my head." -- Don't be cute, please. If you like to know anything specific about my answer, please put it in writing; as a comment, or as a separate question. Thank you. $\endgroup$
    – user12262
    Commented Feb 19, 2022 at 15:00

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