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I have several times seen explanations of gravitational free fall (eg, of a small object toward earth with no air resistance) that begin with the following claim about the particles of the free-falling object: its particles that are closer to earth are experiencing greater gravitational time dilation than are the particles that are further from earth.  Example of this claim used in explaining gravity: https://youtu.be/UKxQTvqcpSg

I understand that if a clock is at rest at a fixed distance from earth's center, then the clock exphibits greater time dilation the closer it is to earth's center.  But the latter scenario is not free fall.  And to me, the aforementioned claim seems contrary to the equivalence principle. 

I'll put my question this way: Suppose you have a free-falling lab within a small region of spacetime, at an altitude of several km, with no air resistance, and there are two free-falling clocks in the lab, one of them a few nm closer to earth than the other.  Will the freefalling clocks undergo gravitational time dilation relative to one another such that an observer in the lab will observe the clocks to be ticking at different rates?

I have no formal education in physics, and I would prefer an answer that is more conceptual and less mathematical, because I may not be able to follow the math very far.

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    $\begingroup$ "Will either clock experience time dilation". You need to specify with respect to what. Time dilation isn't an absolute thing. Each other? An observer on the Earth? An observer in the lab? The clocks can't both be in the same inertial frame if they are separated in radius, so of course one expects some time dilation with respect to each other and with respect to observers in other frames of reference. $\endgroup$
    – ProfRob
    Commented Apr 30, 2021 at 19:13
  • $\begingroup$ Although visuals are often considered easier to understand than anything (like print) which has to be deciphered, that's not the case when acceleration (usually indistinguishable from gravity, except for subtle effects curving some trajectories) is concerned, and the video does not directly depict the acceleration of the teapot, leaving the visuals out of sync with the sound-track. Try one of George Gamow's pop-sci books (thru inter-library loan, if necessary), which include pictures & diagrams that do hold still: The Equivalence Principle hasn't changed since he wrote them. $\endgroup$
    – Edouard
    Commented Apr 30, 2021 at 20:08
  • $\begingroup$ Actually, PSE has a Q&A, at physics.stackexchange.com/questions/225761/rocket-or-elevator , which covered, with John Rennie's verbiage alone, an aspect of the EP that I'd found very hard to understand. I don't think even Gamow did it justice. $\endgroup$
    – Edouard
    Commented Apr 30, 2021 at 20:34
  • $\begingroup$ The gray backdrop behind Rennie's answer (already linked) doesn't mean it was a mediocre answer: It means the question's "Originating Person" hasn't (yet) accepted it. (I sure would've: I'd been looking everywhere, for years, to find that answer!) $\endgroup$
    – Edouard
    Commented Apr 30, 2021 at 21:02
  • $\begingroup$ @Bart Wisialowski I have edited my answer. It was not quite correct $\endgroup$
    – Roger Wood
    Commented May 1, 2021 at 21:17

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The equivalence principle dictates that objects in freefall in a gravitational field are not considered to be accelerating. To the extent that the gravitational field is uniform, two clocks, one above the other, will measure time at the same rate. This rate will be slower than a reference clock far above and outside the gravitational field.

In practice, the field may not be exactly uniform. For example, free-falling towards the Earth, the field might be very slightly higher on the lower of the two free-falling clocks. This is a 'micro-gravity' or a tidal effect. In this case, the lower clock would run very very slightly slower. This is not due to the gravitational field and resulting difference in potential since this is cancelled by the freefall acceleration common to both clocks. It is due to the gravitational field gradient (i.e. the fields and thus the effective potentials are not exactly the same on the two clocks). For practical separations, this difference would not be measurable.

[Edit] My comment "the lower clock would run very very slightly slower" is incorrect. It could run slower or faster. Using a clock placed at the center of mass of the free-falling (or orbiting) laboratory as a reference, clocks placed above or below this position will run more slowly. The gravitational potential near a planet is concave downwards. For exactly the same reason, there are two ocean tides each day, not one.

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    $\begingroup$ For further discussion, you can see my question physics.stackexchange.com/questions/569132/… $\endgroup$
    – Mohammad Javanshiry
    Commented May 1, 2021 at 7:48
  • $\begingroup$ Your edit dramatically shifted my perspective on this. “clocks placed above or below will run slower.” 1) So this means, if the lower clock is even with the center of mass, then the lower clock will run faster? 2) Does it follow that the claim/ explanation my original question was about--the particles of a freefalling object that are closer to earth “tick” more slowly--is inaccurate insofar as the particles at the center of mass will “tick” faster than particles just above that? 3) If you have a series of clocks, then at some point the ones further above will run faster though, correct?? $\endgroup$
    – Bart Wisialowski
    Commented May 2, 2021 at 1:22
  • $\begingroup$ @Bart Wisialowski A series of disconnected clocks free-falling towards earth one after the other along the same projectory would tick progressively more slowly. Three adjacent clocks would differ with the lowest clock ticking more slowly and the highest ticking most rapidly. But if you rigidly connect three adjacent clocks, the middle clock won't change much. However, the upper clock will now be slightly slow because it is now accelerating a little faster than it would if it were in independent free-fall. The lower clock will still be slow, but less so. $\endgroup$
    – Roger Wood
    Commented May 2, 2021 at 3:57
  • $\begingroup$ @Bart Wisialowski In freefall you can think of the lab as having a very weak micro-gravitational field diverging from its center. Loose objects will very slowly drift from the middle of the room to either the floor or the ceiling. Both the floor and the ceiling will have very slightly lower gravitational potential than the center. $\endgroup$
    – Roger Wood
    Commented May 2, 2021 at 4:02
  • $\begingroup$ @BartWisialowski this has a picture of the microgravity/acceleration field for the ISS space.stackexchange.com/questions/20356/… $\endgroup$
    – Roger Wood
    Commented May 2, 2021 at 5:38
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If you intend to place the clocks very close to each other, it does not make sense to speak about the gravitational time dilation. However, if you place the freely falling clocks a meter or a kilometer away from each other, the time dilation becomes tangible. Since the gravitational time dilation depends on the gravitational potential of the point at which the clock is located, the time rates of the clocks would be different from the viewpoint of the observer located on the earth whether or not the lab is fixed at a large altitude from the earth or is freely falling at that point.

However, when the lab is fixed, there are solely different gravitational time dilations for the clocks due to different potentials, whereas for a freely falling lab, besides the said time dilation, there is an additional time dilation due to the instantaneous speed of the lab (clocks) relative to the terrestrial observer, which is an SR effect.

If you want to compare the clocks from the standpoint of the lab observer, both the SR and gravitational time dilations seem to happen for the clocks, and the SR effect mainly arises from the tidal forces that tend to accelerate the clocks WRT each other as well as the lab observer.

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  • $\begingroup$ Just for clarification, how are you defining "gravitational potential" here? $\endgroup$
    – m4r35n357
    Commented Apr 30, 2021 at 19:32
  • $\begingroup$ @m4r35n357 It has a clear definition in the context of GR. $\endgroup$
    – Mohammad Javanshiry
    Commented Apr 30, 2021 at 20:23
  • $\begingroup$ @MohammadJavanshiry So would you say that if the region of spacetime is small enough--ie, local enough, an observer in the freefalling lab wouldn't observe the clocks ticking at different rates--just as you wouldn't if the same lab/clocks were moving at constant speed in flat spacetime? But that if the region of spacetime is made large enough, the observer in the freefalling lab would observe the clocks ticking at different rates? $\endgroup$
    – Bart Wisialowski
    Commented May 1, 2021 at 2:34
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    $\begingroup$ @BartWisialowski "So would you say that if ..." Yes I would. "But that if the region of spacetime is made large enough ..." That's correct. For further discussion, you can see my question physics.stackexchange.com/questions/569132/… $\endgroup$
    – Mohammad Javanshiry
    Commented May 1, 2021 at 7:51
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Time dilation does not depend upon acceleration; this is has been experimentally verified in particle accelerators. So whether the clocks are undergoing acceleration or not does not affect their rates relative to distant clocks.

The answer to your question therefore is that clocks closer to the center of the Earth undergo time dilation relative to clocks further away from the center of the Earth, and this is true whether the clocks are in free fall or not.

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  • $\begingroup$ Which experiments in particular are you referring to? Correct, time dilation does occur for two observers at constant relative velocity, but there actually does seem to be some debate or disagreement over whether certain time dilation phenomena, such as one twin being younger than the other, do depend upon acceleration being involved. $\endgroup$
    – Bart Wisialowski
    Commented May 2, 2021 at 2:25
  • $\begingroup$ For example, this answer (linked in other comments above/below) about freefalling clocks also addresses the twin paradox and explains the role of acceleration in that -- more exactly, proper acceleration, as opposed to coordinate acceleration. physics.stackexchange.com/a/569189/297912 $\endgroup$
    – Bart Wisialowski
    Commented May 2, 2021 at 3:13
  • $\begingroup$ Farley et. al, (1966) as quoted in Misner, Thorne, and Wheeler Gravitation, sec. 38.4 p. 1055. MTW says "...elementary particle experiments do suggest that the times measured by atomic clocks depend only on velocity, not on acceleration.". See also N. Ashby "Relativity in the Global Positioning System", where the GPS accounts only for gravitational potential and velocity, not for acceleration in orbit. $\endgroup$
    – Eric Smith
    Commented May 2, 2021 at 13:16
  • $\begingroup$ Acceleration can (indirectly) affect time dilation of course, insofar as a change in velocity means a change in time dilation. But the amount of time dilation itself depends only on the velocity. For example, in the twin paradox the differential aging of the twins does not depend on whether the traveling twin turns around at 1g, 0.5g, 100gs, or whatever; only the actual differences in velocity matter. $\endgroup$
    – Eric Smith
    Commented May 2, 2021 at 13:20
  • $\begingroup$ Without taking a side, there does not appear to be concensus as to whether younger/older twin asymmetry depends upon acceleration. $\endgroup$
    – Bart Wisialowski
    Commented May 3, 2021 at 20:24

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