Time dilation for the Earth's orbit around the Sun

[sweet springs]

we have 2 atomic clocks on Earth's orbit around the Sun, one on Earth's surface, at one pole, and the other on a spaceship, far from Earth, but traveling with the same speed around the Sun, the clocks would suffer the same kinematic time dilation or not?

A simpler case of GPS satellite clock adjustment could help you. Against clock set on Earth’s surface, satellite clocks goes faster by GR gravity potential difference and goes slower by SR going round speed. Actual situation shows the former effect almost doubles the latter so satellite clock goes faster. We have to make periodical adjustment of satellite clock back.

 

[jbriggs444]

A simpler case of GPS satellite clock adjustment could help you. Against clock set on Earth’s surface, satellite clocks goes faster by GR gravity potential difference and goes slower by SR going round speed. Actual situation shows the former effect almost doubles the latter so satellite clock goes faster. We have to make periodical adjustment of satellite clock back.

There are periodic adjustments for other effects. The particular gross effect you mention was compensated for by design. The satellite clocks intentionally tick slowly so that the clocks in orbit and the clocks on the ground will all stay roughly synchronized in an Earth-centered inertial frame.

http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

 

[DanMP]

That is about rotational frame dragging, not linear frame dragging. Are you aware of that? I am not sure if you are deliberately changing topics now or if you think this is the same topic.

Yes, it's about rotational frame dragging, but I think it is helpful in understanding how frame dragging in general may affect timekeeping, so in a way it is the same topic, as you will see soon.

I would have to work the math to be sure, but that sounds right. However, with the Earth surface clock at the pole it doesn’t matter in this scenario.

In this (second) scenario, both clock's are flying around the Earth, over the equator, one of them with the same speed (seen from Sun's frame) as the Earth's dragged frame (but this speed is now significant, not negligible as it is in reality) and the other with the speed needed to appear stationary from Sun's perspective (remember that the Earth is now stationary in Sun's frame).

The rotating worldline is longer than the static worldline.

This would mean that from Sun's frame/perspective the clock flying with the same speed as the Earth's dragged frame should tick slower than the other (static in Sun's frame), contradicting the red text in my quote from Wikipedia. What do you think about this contradiction?

 

[Dale]

In this (second) scenario, both clock's are flying around the Earth, over the equator, one of them with the same speed (seen from Sun's frame) as the Earth's dragged frame (but this speed is now significant, not negligible as it is in reality) and the other with the speed needed to appear stationary from Sun's perspective (remember that the Earth is now stationary in Sun's frame).

I don’t think that this scenario is self consistent. At least, I don’t see a way to translate it from English into math. There are not any “stationary” geodesics that the Earth can follow. Perhaps you intend for the Earth to be kept in place with some gigantic rocket thrusters?

This would mean that from Sun's frame/perspective the clock flying with the same speed as the Earth's dragged frame should tick slower than the other (static in Sun's frame), contradicting the red text in my quote from Wikipedia. What do you think about this contradiction?

It is not a contradiction, it is a mistake on your part. The longer worldline is the “faster” clock. The longer the worldline the more ticks between two given events, and more ticks is a faster clock.

 

[PeterDonis]

The text I colored in red seems to imply that a clock that revolves around a rotating massive object with the same angular speed (seen by a distant observer) as the dragged frame, would have no kinematic time dilation due to the rotation the distant observer sees.

It's more complicated than that.

First, consider the simpler case of a non-rotating, spherically symmetric massive body. Spacetime in the vacuum region outside this body is described by the Schwarzschild metric. This metric has a timelike Killing vector field--the term for this is "stationary"--and this KVF is hypersurface orthogonal--the term for this is "static". Note that in an earlier post, "has a timelike KVF" was called "static", but that's really not correct; the KVF also has to be hypersurface orthogonal, which basically means the gravitating mass can't be rotating. Since you're interested in the case where it is rotating, we have to be careful to distinguish "static" and "stationary".

Now suppose we have a test body in a free-fall circular orbit around the non-rotating massive body. If this body exchanges light signals with an observer at rest at infinity, the observer will calculate that the body's clock runs slow compared to the observer's clock. If we call this "time dilation", we can in the static case, for convenience, split it up into two pieces: gravitational and kinematic. The split is easily visualized: gravitational time dilation depends only on altitude above the object, and kinematic time dilation depends only on speed relative to the object. However, if we want to be rigorous, we have to define exactly what "height" and "speed" mean in terms of the known properties of the spacetime.

We do this as follows: we use the integral curves of the timelike KVF to define "points in space"--each curve corresponds to such a point (more precisely, to the worldline of such a point--you can think of each such curve as the worldline of a hypothetical observer if you like). Each point then has an altitude, defined as its areal radius $r = \sqrt{A / 4 \pi}$, where $A$ is the area of a 2-sphere centered on the massive body and containing the point. Gravitational time dilation then depends only on $r$.

We then define "speed" as the speed of an object relative to the points in space--i.e., relative to the hypothetical observers whose worldlines are the integral curves of the timelike KVF. So, for example, we can imagine a set of such observers that each occupy one of the points along the circular orbit of the test body described above, each of whom is measuring the speed of the test body as it passes by. Since the orbit is circular, all of these observers will measure the same speed. The kinematic time dilation of the body is then just the usual SR time dilation using this speed.

Now suppose the massive body is rotating. We assume that this means spacetime in the vacuum region outside the body is described by the Kerr metric (note that nobody has actually proven that this is the case, but it's a convenient assumption and should be at least reasonably accurate). The Kerr metric is stationary, but not static--it has a timelike KVF, but the KVF is not hypersurface orthogonal. Does this make a difference to the above, and if so, what?

It turns out that, as far as defining "points in space" are concerned, it makes no difference. The integral curves of the timelike KVF are the same as for the non-rotating case: they are worldlines of observers at rest relative to the center of mass of the rotating massive body. And each point in space still has an "altitude" which determines the gravitational time dilation at that point; the only difference is that now the "altitude" depends, not just on the areal radius $r$, but also on the colatitude $\theta$ (basically, $\theta = 0$ denotes the "north pole" of the body, $\theta = \pi / 2$ denotes the "equator", and $\theta = \pi$ denotes the "south pole"). But the key point is that these worldlines are not rotating: they are still at rest relative to the body's center of mass, and relative to the observer at infinity as well.

We can also still define "speed" as we did before--speed relative to observers at rest. So we still have well-defined notions of "altitude" and "speed". But we do have a problem: we can no longer split up the total time dilation into a "gravitational" piece that depends only on altitude and a "kinematic" piece that depends only on speed. And one way of looking at the reason why is the thing you pointed out from the Wikipedia article: the state of motion that has "minimum time dilation", for a given altitude relative to an observer at infinity, is rotating around the massive body (because of frame dragging), so it's not at rest. (Note that in the non-rotating case, the "at rest" state was also the state of "minimum time dilation" for its altitude--only gravitational, no kinematic.)

 

[PeroK]

This would mean that from Sun's frame/perspective the clock flying with the same speed as the Earth's dragged frame should tick slower than the other (static in Sun's frame), contradicting the red text in my quote from Wikipedia. What do you think about this contradiction?

The Wikipedia article relates to the relative time dilation of clocks orbiting the Earth only (as the rotating mass). If one clock is orbiting the Earth at just the right speed to match the "frame dragging", associated with the Earth's rotation, then it will be less time dilated than a clock at rest in the Earth's atmosphere.

This is completely unrelated to the time dilation of a clock at rest relative to the Sun, compared to one in orbit round the Sun - whether that clock is on the Earth and whether the Earth is rotating or not. The Wikipedia analysis has no second massive body about which the rotating mass is orbiting!

There are no contradictions here, only an increasingly elaborate scenario, with several competing factors. To an external observer, we have many factors:

Sun's mass
Angular momentum of the Sun
Earth's orbit round the Sun
Earth's mass
Angular momentum of the Earth
Motion of a clock relative to the Sun
Motion of a clock relative to the Earth

The question is which of these are numerically significant; and, how do you combine these factors (especially if not by linear combination). You don't get a contradiction if you look at only two of these factors and conclude that clock A is more time dilated than B; or, if you consider a different two factors and conlcude that clock A is less time dilated than B.

To compare the time dilation for two clocks you need to take all (seven) factors into account.

 

[DanMP]

Perhaps you intend for the Earth to be kept in place with some gigantic rocket thrusters?

Yes, something like that or not so gigantic if the Earth was orbiting the Sun at much greater distance.

It is not a contradiction, it is a mistake on your part. The longer worldline is the “faster” clock. The longer the worldline the more ticks between two given events, and more ticks is a faster clock.

Sorry for the mistake , you are right, but, on the other hand, if you remove the Earth (but leave the static clock there) and introduce a third clock, orbiting the Sun at the same distance as the static clock, this third clock would have (if I'm not mistaking again) a longer worldline than the static one, but would not be faster. So, if I'm not mistaking again, there is a contradiction.

 

[DanMP]

... the state of motion that has "minimum time dilation", for a given altitude relative to an observer at infinity, is rotating around the massive body (because of frame dragging), so it's not at rest.

So, if the state of motion that has "minimum time dilation" is rotating around the massive body (because of frame dragging), what is happening when the Earth is orbiting the Sun (back to the OP scenario)? If the frame with "minimum time dilation" can be dragged around the Earth, why it would be impossible to be also dragged along the orbit?

 

[PeterDonis]

if you remove the Earth (but leave the static clock there) and introduce a third clock, orbiting the Sun at the same distance as the static clock, this third clock would have (if I'm not mistaking again) a longer worldline than the static one, but would not be faster.

No. The orbiting clock would have a shorter worldline than the static clock between successive points where the two meet. And, correspondingly, the orbiting clock would run slower. Remember that this is spacetime, not space, and your ordinary intuitions about Euclidean geometry do not work. The orbiting worldline seems "longer" to your intuitions (because it spirals around rather than going straight up, so to speak), but your intuitions are wrong.

So, if the state of motion that has "minimum time dilation" is rotating around the massive body (because of frame dragging), what is happening when the Earth is orbiting the Sun (back to the OP scenario)?

If we consider the Sun as rotating (it is, but very slowly, much more slowly than the Earth), the "minimum time dilation" state of motion at the distance of the Earth's orbit will be rotating around it, but much, much slower than the Earth.

 

[Dale]

if you remove the Earth (but leave the static clock there) and introduce a third clock, orbiting the Sun at the same distance as the static clock, this third clock would have

I am sorry but all of these different scenarios are becoming frustrating to me. I think I am done with this discussion at this point. It feels like a moving target.

Best of luck in getting your question answered or at least in figuring out what your question is, but I am not enjoying participating any more.

 

[DanMP]

No. The orbiting clock would have a shorter worldline than the static clock between successive points where the two meet.

Ok, but

The rotating worldline is longer than the static worldline.

If we consider the Sun as rotating (it is, but very slowly, much more slowly than the Earth), the "minimum time dilation" state of motion at the distance of the Earth's orbit will be rotating around it, but much, much slower than the Earth.

This is not what I meant. When I said "If the frame with "minimum time dilation" can be dragged around the Earth, why it would be impossible to be also dragged along the orbit?" I meant some sort of linear frame dragging "done" by the Earth, not a rotational frame dragging due to Sun's rotation. If this "linear" frame dragging does not occur, it would mean that rotational frame dragging around the Earth would seem/be from the Sun's perspective as some kind of fast moving rotational deformation of the spacetime along Earth's orbit.

I am sorry but all of these different scenarios are becoming frustrating to me ...
... at least in figuring out what your question is ...

I'm sorry for these side-scenarios, but they were needed for a better understanding of the OP scenario/problem/question.

The Wikipedia analysis has no second massive body about which the rotating mass is orbiting!

There is such an analysis elsewhere? If not, don't you think that it would be something interesting, important for a better understanding of relativity and a new way to test it?By the way, I also asked about

experimental test/observation similar/related with this scenario

Regarding a possible test, it may be easier in the Earth/Moon system than in Sun/Earth one. Do we have an atomic clock on the moon?

 

[PeterDonis]

The rotating worldline is longer than the static worldline.

This is not correct. The rotating worldline is shorter; hence, the clock following the rotating worldline runs slower. More precisely, the static clock has more elapsed time between two successive meetings of the two clocks than the rotating clock does; hence, the rotating worldline is shorter between those two events than the static worldline is.

 

[PeterDonis]

Ok, but

See my post #47 just now.

When I said "If the frame with "minimum time dilation" can be dragged around the Earth, why it would be impossible to be also dragged along the orbit?" I meant some sort of linear frame dragging "done" by the Earth. If this "linear" frame dragging does not occur, it would mean that rotational frame dragging around the Earth would seem/be from the Sun's perspective as some kind of fast moving rotational deformation of the spacetime along Earth's orbit.

First, if we want to consider both the Sun and the Earth as having non-negligible mass and therefore causing non-negligible spacetime curvature, then we no longer have any exact solutions to use; there is no known exact solution for the case of two massive bodies. The main reason for this is that GR is nonlinear: you can't just add together two solutions and have another solution. Such cases, in general, have to be solved numerically. See further comments below.

For sufficiently weak fields, the nonlinearities should be small, so we could approximate a two-body solution by superposing solutions for the Sun alone and the Earth alone. This appears to be what you are intuitively doing. However, even then there are complications. First, linear frame dragging, as I understand it, requires the massive body doing the dragging to have nonzero proper acceleration. The Earth orbiting the Sun does not; it's in free fall. Second, even if I'm wrong in the previous sentence, linear frame dragging is a tiny, tiny effect, much smaller even than rotational frame dragging, so any such effect on the "minimum time dilation" state of motion for the Sun-Earth system would in all probability be unobservable with the accuracy of measurement we can achieve now or for the foreseeable future.

All that said, the "minimum time dilation" state of motion around the Sun, at the distance of the Earth, is not going to be anywhere near the Earth; it's going to be as far away from the Earth as possible (if we're just considering the Sun and the Earth), because the extra time dilation from the Earth's gravity well is much, much larger than any possible "frame dragging" effect.

don't you think that it would be something interesting, important for a better understanding of relativity and a new way to test it?

GR has already been tested for multi-body systems, using numerical solutions. The solar system and binary pulsar systems are two well known examples.

 

[Dale]

This is not correct. The rotating worldline is shorter;

This is one of the dangers of all of these random scenarios he has been throwing out.

The rotating worldline referred to above is one that is going around a rotating object at the right speed and direction to minimize time dilation. Due to rotational frame dragging this is not a stationary worldline, but one that slowly revolves around the central object. This scenario was identified in the Wikipedia article on frame dragging.

It is not a clock orbiting the sun at 1 AU. So my comment was correct, that worldline is longer, it was just referring to a scenario from a different random tangent. This is why I am done with this conversation.

 

[PeterDonis]

The rotating worldline referred to above is one that is going around a rotating object at the right speed and direction to minimize time dilation.

Ah, ok. You're right, all the different scenarios have gotten very garbled in this thread.

 

[DanMP]

The rotating worldline referred to above is one that is going around a rotating object at the right speed and direction to minimize time dilation. Due to rotational frame dragging this is not a stationary worldline, but one that slowly revolves around the central object. This scenario was identified in the Wikipedia article on frame dragging.

It is not a clock orbiting the sun at 1 AU. So my comment was correct, that worldline is longer,

So, if a clock is seen from the Sun's perspective as revolving around the Earth (at the right speed and direction), its worldline is longer than the one of a stationary clock (in the same frame), but if the clock is revolving around the Sun, its worldline is shorter than the one of a stationary clock. For me it's quite confusing ... (I'm not saying that it's wrong, just that it is confusing), so I'll try again without worldlines:

From Sun's perspective, I see a clock revolving (at the right speed and direction to match the frame dragging) around the stationary/hovering Earth and another clock stationary (in the same - Sun's - frame), and contrary to everything I know, the stationary one is more time dilated than the moving one, while the gravitational time dilation is the same, both clocks being at the same distance from de Earth/Sun. That makes me think that a clock being in that rotational dragged frame is "more stationary" than the one that appears stationary from Sun's perspective. Now, if the Earth is orbiting the Sun, that frame is co-moving with it (while dragged around the Earth) and the question is if a clock stationary in the dragged frame would be influenced by the orbital speed of the Earth around the Sun, or not, being again "more stationary" that it appears to be.

GR has already been tested for multi-body systems, using numerical solutions. The solar system and binary pulsar systems are two well known examples.

Ok, but how are these tests/examples related to my scenario? How can we measure/detect, without atomic clocks, the way is time passing on the surface of a distant massive body?

 

[PeterDonis]

if a clock is seen from the Sun's perspective as revolving around the Earth (at the right speed and direction), its worldline is longer than the one of a stationary clock (in the same frame), but if the clock is revolving around the Sun, its worldline is shorter than the one of a stationary clock. For me it's quite confusing ...

That's because you keep switching scenarios, and the rest of us are having trouble keeping up. From what I can see, you have brought up, at one time or another, at least seven different clocks:

(1) A clock at rest relative to the Sun's center of mass (i.e., not rotating about the Sun at all), at the same distance from the Sun as the Earth's orbit, but not anywhere near the Earth (so it's unaffected by the Earth's gravity well).

(2) A clock moving around the rotating Sun at the "minimum time dilation" angular velocity due to the frame dragging of the rotating Sun, at the same distance from the Sun as the Earth's orbit, but not anywhere near the Earth.

(3) A clock in a free-fall orbit around the Sun at the same distance as the Earth, but not anywhere near the Earth.

(4) A clock orbiting the Sun at the same distance as the Earth, near enough to the Earth to be affected by the Earth's gravity well (for definiteness, say about 100 miles above the Earth's surface, roughly the altitude of low Earth orbit), but not orbiting the Earth at all (i.e., its orbit is the same as that of the Earth's center of mass around the Sun).

(5) A clock moving around the rotating Earth at the "minimum time dilation" angular velocity due to the frame dragging of the rotating Earth, at the same distance from the Earth as clock #4.

(6) A clock at rest relative to the rotating Earth (i.e., it stays above the same point on the Earth as the Earth rotates--think of it as being at the top of a 100 mile tall tower that rotates with the Earth) as it orbits the Sun, at the same distance from Earth as clock #4.

(7) A clock in a free-fall orbit around the Earth at the same distance from Earth as clock #4.

Some of the above you might not have brought up explicitly, but you've used ambiguous language that could imply several of the above possibilities.

To the best of my understanding, the ordering of the time dilation of these clocks, from "minimum" (longest elapsed time) to "maximum" (shortest elapsed time) is as follows:

#2 - #1 - #3 - #5 - #4 - #6 - #7

Note that the ordering of #7, relative to #6, depends on the altitude; as we increase the altitude above the Earth, clock #7's elapsed time will become greater than that of #6. (This happens at an altitude of half the Earth's radius, IIRC.)

 

[PeterDonis]

How can we measure/detect, without atomic clocks, the way is time passing on the surface of a distant massive body?

By looking for spectral lines in the light coming from the distant body, and comparing them with similar spectral lines measured in the lab to obtain a frequency shift. If we know the motion of the distant body relative to Earth, we can subtract out the frequency shift due to the Doppler effect; what is left will tell us the gravitational redshift of the body, which in turn tells us how time is passing on its surface.

This method is not viable for all distant bodies, because we can't always detect spectral lines in the light coming from them. But it's the only method we have. It's been done for the Sun and for many other stars; in fact this method was how the mass of the white dwarf Sirius B was determined (because its radius was known from telescopic observations, and knowing the radius and the redshift let's you calculate the mass).

 

[DanMP]

That's because you keep switching scenarios, and the rest of us are having trouble keeping up.

Sorry about that. I'll try again:

Here (what is between { } was added today):

From Sun's perspective, I see a clock {A1} revolving (at the right speed and direction to match the frame dragging) around the stationary/hovering Earth and another clock {B} stationary (in the same - Sun's - frame), and contrary to everything I know {e.g. Hafele-Keating exp.}, the stationary one {B} is more time dilated than the moving one {A1}, while the gravitational time dilation is the same, both clocks being at the same distance from de Earth/Sun. That makes me think that a clock being in that rotational dragged frame is "more stationary" than the one that appears stationary from Sun's perspective. Now, if the Earth is orbiting the Sun {the second scenario}, that frame is co-moving with it (while dragged around the Earth) and the question is if a clock {A2} stationary in the dragged frame would be influenced by the orbital speed of the Earth around the Sun, or not, being again "more stationary" that it appears to be.

there are 2 scenarios. In both of them the Sun is not rotating, in order to leave out its rotational frame dragging. On the other hand, Earth's rotation (and/or mass) is "enhanced", in order to have a significant rotational frame dragging.

In the first scenario, the Earth is not orbiting the Sun, "hovering" at the same distance from the Sun as it was when orbiting (this may be possible using huge thrusters, at least if the Earth was orbiting the Sun very far from it). Here we have two clocks, both flying around the Earth, over the equator (at the same altitude, say 10 km):
- clock A1, flying at exactly the right speed and direction to match the Earth's rotational frame dragging (seen from the Sun)
- clock B, flying at exactly the right speed and direction to appear stationary from the Sun's perspective, at the same distance from the Sun as clock's A1 average distance from the Sun

In the second scenario, the Earth is orbiting the Sun and we can consider three clocks:
- clock A2, flying around the Earth, over the equator, at exactly the right speed and direction to match the Earth's rotational frame dragging (seen from the Sun)
- clock C, in a free-fall orbit around the Sun at the same distance as the Earth, but not anywhere near the Earth
- clock D, at one of the Earth's poles, at the same height/altitude as clock A2

I added clocks C and D because I mentioned them in the OP, but, if I'm not mistaking, clock D should tick with the same rate as A2, so the OP problem/question is to compare clock A2 with clock C, keeping in mind, from the above first scenario, that a clock matching the Earth's rotational frame dragging seems to have less kinematic time dilation than it would appear from Sun's perspective ...

 

[PeterDonis]

clock A1, flying at exactly the right speed and direction to match the Earth's rotational frame dragging (seen from the Sun)

Ok, so this clock, relative to Earth's center of mass, would be moving in the same direction as the Earth's rotation, but much slower. Relative to someone standing on the surface of the rotating Earth, this clock would be moving in the opposite direction as the Earth's rotation.

clock B, flying at exactly the right speed and direction to appear stationary from the Sun's perspective, at the same distance from the Sun as clock's A1 average distance from the Sun

And this clock, relative to Earth's center of mass, would not be moving at all. Relative to someone standing on the surface of the rotating Earth, this clock would be moving in the opposite direction to the Earth's rotation, somewhat faster than clock A1.

clock A2, flying around the Earth, over the equator, at exactly the right speed and direction to match the Earth's rotational frame dragging (seen from the Sun)

So relative to Earth, this clock moves the same as clock A1, correct? (Its motion relative to the Sun is different since now the Earth is orbiting the Sun.)

clock C, in a free-fall orbit around the Sun at the same distance as the Earth, but not anywhere near the Earth

No problem here.

clock D, at one of the Earth's poles, at the same height/altitude as clock A2

So this clock is not orbiting the Earth, correct? In other words, relative to the Sun, its motion is the same as that of the Earth's center of mass.

For comparison, I will add two more clocks:

First, the one I defined as clock #1 in post #52 (at rest relative to the Sun, at the same distance as the Earth's orbit, but nowhere near the Earth).

Second, a clock moving the same relative to Earth as clock B, but for the case where the Earth is orbiting the Sun. (This clock is not the same as clock D because it is hovering over the equator, not one of the poles.) Call this clock E.

Assuming my statements above are all correct, then the ordering of time dilation of these clocks, from "minimum" (longest elapsed time) to "maximum" (shortest elapsed time) would be, I think:

Clock 1 - Clock A1 - Clock B - Clock C - Clock D - Clock A2 - Clock E

Note that this is just off the top of my head using heuristic estimates of the time dilations involved; I have not done the detailed math. Some key points that I used:

I am estimating that the additional time dilation due to the Earth orbiting the Sun, as compared with being at rest relative to the Sun at the distance of the Earth's orbit, is larger than the time dilation due to the Earth's gravity well at the Earth's surface or in low Earth orbit. The numbers are pretty close here but I think this is right. (This is why clocks A1 and B come between clock 1 and the rest of the clocks.)

For an object above one of the poles of a rotating object, the time dilation is smaller (longer elapsed time) than for any state of motion in the equatorial plane. (This is why clock D comes before clocks A2 and E.)

 

[PeterDonis]

I'm not mistaking, clock D should tick with the same rate as A2

You're mistaken. See the last point in my previous post. The Kerr metric is not spherically symmetric, so being above one of the poles is not the same as being above the equator.

 

[PeterDonis]

a clock matching the Earth's rotational frame dragging seems to have less kinematic time dilation than it would appear from Sun's perspective

This is not correct, because, as I have pointed out before, in the Kerr metric, i.e., in the presence of rotational frame dragging, it is no longer possible to have purely "kinematic" time dilation (i.e., time dilation that depends only on speed relative to observers "at rest").

 

[pervect]

This is not correct, because, as I have pointed out before, in the Kerr metric, i.e., in the presence of rotational frame dragging, it is no longer possible to have purely "kinematic" time dilation (i.e., time dilation that depends only on speed relative to observers "at rest").

The Hafele-Keating experiement [for this post, the HK expriement] experiment is, I think, I good illustration of this point. If one slow-transport a clock around a spinning massive object in different directions (in the HK experiment, this spinning object is the Earth], then reunite the clocks so that they meet again and are at rest relative to each other and to the spinning surface of the object, said clocks will not be synchronized. The clock that moves in the direction of the spin will have a different reading than the clock that moves in the direction opposite to the spin.

This is an experimentally confirmed prediction of special relativity. It'd be overstating the case to say this singe experiment is conclusive, but at this point I don't want to get into a full review of the experimental tests of Special Relativity.

The point I want to make is that if one's understanding of special relativity does not match the results of the HK experiment, if one does not understand why the theory predicts that the clocks will have different readings when re-united, one's understanding of special relativity does not match the current understanding of the professional community.

Presenting the current understanding of the professional community is our mission goal here at PF.

An equivalent experiment to the HK experiment could be created with the Earth, and two spaceships that move in powered trajectories clockwise and anti-clockwise along the Earth's orbit. This modified version detects the frame-dragging effects of the spin of the sun, just as the HK experiment detects the effects of the spin of the Earth.

The approach favored by the OP, where one focuses on "Time dilation", is particularly vulnerable to such confusions, this is a fundamental limitation of the approach which typically neglects the relativity of simultaneity. I suspect that may be an issue for the OP - it's a very common issue. However I have not been following this thread in detail.

 

[PeterDonis]

The Hafele-Keating experiement [for this post, the HK expriement] experiment is, I think, I good illustration of this point.

Not of the point under discussion in this thread, no. The HK experiment was nowhere near sensitive enough to spot the tiny effects on time dilation of the frame dragging due to the Earth's rotation. It was only testing the time dilation effects of the Earth's static gravitational field and of motion relative to an Earth-centered non-rotating frame. The latter type of motion includes the motion of a clock at rest on the surface of the rotating Earth; but the time dilation effect due to this motion is not caused by the frame dragging due to the Earth's rotation. It's caused by motion relative to a non-rotating Earth-centered frame.

To put it another way, to analyze the HK experiment to the current experimental accuracy, it is sufficient to use the Schwarzschild metric (which is static, so a clean split between "gravitational" and "kinematic" time dilation can be made) with some static correction factors due to the Earth's quadrupole moment (which don't change the fact that the metric being used is static, so they don't affect the clean split just mentioned). We would need a number of orders of magnitude better accuracy to spot the time dilation effects of frame dragging, and therefore to require a non-static (but still stationary) metric for the analysis, in which the clean split is no longer possible. I am assuming such a metric in my posts in this thread, but that is only theoretical; experimentally we have no reason to have to do this yet, nor will we for quite some time.

If one slow-transport a clock around a spinning massive object in different directions (in the HK experiment, this spinning object is the Earth], then reunite the clocks so that they meet again and are at rest relative to each other and to the spinning surface of the object, said clocks will not be synchronized.

This is the Sagnac effect, which is a real effect, yes, but is not due to rotational frame dragging. (This is obvious from the fact that the Sagnac effect is present in flat spacetime, where there is obviously no rotational frame dragging.)

The approach favored by the OP, where one focuses on "Time dilation", is particularly vulnerable to such confusions, this is a fundamental limitation of the approach which typically neglects the relativity of simultaneity.

I agree that neglecting relativity of simultaneity can cause problems, but I don't think the OP is making any errors attributable to that in this thread. The "time dilation" the OP is describing is invariant, not frame-dependent (so "time dilation" might be a bad term to use, "differential aging" might be better).