# Hafele and Keating time dilation question

1. Dec 22, 2012

I'm sure the answer to this is obvious, but I'm at a loss. In the Hafele and Keating experiment where atomic clocks are flown east and west and compared to a stationary clock the eastbound clock lost 59 ns and the westbound clock gained 273 ns.

My question is that if both eastbound and westbound flights flew the same speed relative to the clock on the earth (which is rotating with the earth), why were the delta times not the same for both east and west flights. It seems if the reference clock were just sitting in space and not rotating with the earth then of course the two times would be differerent, but with the reference clock on the ground rotating with the earth, shouldn't the delta times be the same for both east and west flights?

2. Dec 22, 2012

### Staff: Mentor

The simple way to look at it is to refer everything to a clock that is not rotating with the Earth, but is at the same altitude as the Earth's surface (so the time dilation factor due to altitude is the same as that of the "reference" clock on the Earth's surface). Relative to this clock, we have:

(1) The "reference" clock, at rest on the Earth's surface (and therefore rotating with it) is moving at some velocity v. So it runs slow relative to the non-rotating clock by gamma(v).

(2) The eastbound clock is moving at velocity v + w, where w is the airplane velocity, and is at altitude h. So it has two effects: it runs slow relative to a non-rotating clock by gamma(v + w), but it runs fast relative to a clock on the Earths' surface by f(h), where f is the gravitational time dilation factor as a function of altitude. (Strictly speaking, it's the *reduction* of the time dilation with increase in altitude.)

(3) The westbound clock is moving at velocity v - w, where w is the airplane velocity, and is at altitude h. So it also has two effects: it runs slow relative to a non-rotating clock by gamma(v - w), but it runs fast relative to a clock on the Earths' surface by f(h), where f is the gravitational time dilation factor as a function of altitude. (Strictly speaking, it's the *reduction* of the time dilation with increase in altitude.)

For the eastbound clock, gamma(v+w) is larger than gamma(v), and this outweighs the reduction in time dilation due to f(h), so the clock, on net, runs slow compared to the reference clock, which is what is observed.

For the westbound clock, gamma(v-w) is smaller than gamma(v), so both effects act to make it run faster than the reference clock, which is what is observed.

Of course this doesn't quite answer the question of how things are calculated in the rotating frame in which the reference clock is at rest. But that frame is non-inertial and non-static, so I don't know of a simple way to model "time dilation" in such a frame. The above analysis, from the point of view of a clock not rotating with the Earth, can use simple results for time dilation due to relative velocity and altitude, which can't be directly applied in a rotating frame.

3. Dec 23, 2012

### arindamsinha

PeterDonis's explanation is very good and detailed.

I will try to give an illustration using the numbers you have used. Note that the numbers I am using are also gross averages, as the experiment itself was within an accuracy level of ~10%, and the numbers quoted in the Hafele-Keating paper had quite wide ranges.

For the eastbound plane:
Gravitational time dilation (TD): +144 ns
Velocity TD: -184 ns
Predicted total TD: -40 ns
Measured total TD: -59 ns

For the westbound plane:
Gravitational TD: +179 ns
Velocity TD: +96 ns
Predicted total TD: +275 ns
Measured total TD: +273 ns

First, take out the gravitational time dilation. It should have been the same for both directions if the planes had flown the same altitude for the same durations. Probably because of their specific flight paths, the westbound plane has a +179ns time dilation, compared to the eastbound plane's +144ns. Both of them should anyway have had a +ve clock rate (faster time) as shown by the experiment.

I think your main questions is around the Velocity TD - why if both of them were traveling at the same speed w.r.t. the Earth clock, should we expect asymmetric figures like -184ns and +96ns? Shouldn't they be of the same magnitude (with opposite signs) since their velocity w.r.t. the Earth clocks is the same?

The reason is that these velocities (incl. satellite velocities) are not computed w.r.t. the stationary Earth surface. The velocities are computed w.r.t. the ECIF, which can be considered to be an axis passing through the center of the Earth, but whose orientation is fixed w.r.t. to distant fixed stars (some say microwave cosmic background). This reference axis does not change orientation with Earth's rotation, or revolution around the Sun. Without this, Earth's rotation itself would have no meaning, in fact, and geostationary satellites overhead would come crashing down on our heads.

Last edited: Dec 23, 2012
4. Dec 23, 2012

Thank you both for your clear and detailed answers. It was the velocity effect only that I was curious about.

The orientation of the clock relative to the axis is an interesting addition to the problem, but I don't see how it answers the question. After all the clocks on the planes are also just moving relative to the axis of the Earth, so the plane clocks are still just moving relative to the clock on the surface of the earth. In addition the orientation of the clock should have no effect on the time it keeps since relativity affects the time of a clock regardless of it's orientation.

To see what I mean, imagine if the Earth were rotating so the surface had a velocity of .999c. the plane going east would have a speed of just over .999c and the one going west would have a speed of just under .999c. If we say that the clock on earth (moving at .999c relative to the fixed stars) can be considered stationary as you suggest because it is oriented to the axis of the Earth, then we would expect that both clocks in the planes would suffer huge losses in time relative to the clock on the Earth even though the clocks in the planes are hardly moving at all relative to the clock on the Earth! In fact we would expect all 3 clocks to all suffer very nearly the great time losses relative to a clock on the sun for example as all 3 clocks are moving at .999 relative to the sun clock (at least 2x a day: at midnight and at noon).

I realize that the fact that the planes and the clock are not really in inertial frames of reference due not only to gravity but also to the fact that they are not traveling in a straight line, but it is all 3 clocks that are rotating around the center of the Earth, but seeing how this rotation is seen as a velocity only (and not an acceleeration) when calculating the results of this experment (gravity part not withstanding) the fact that they are all going around the Earth is irrelevent as far as this experiment is concerned.

5. Dec 23, 2012

### Staff: Mentor

I'm confused; where did either of us say this? The point we are making is that the clock on Earth is *not* stationary in any inertial frame, and you can only calculate time dilation due to velocity relative to an inertial frame; if you try to do it relative to the rotating frame in which the Earth observer is at rest, it won't work. The inertial frame in question does happen to have one spatial axis pointed in the same direction as the Earth's axis, but that doesn't mean the clock on Earth is stationary in that inertial frame; the Earth is rotating and the inertial frame is not.

The acceleration doesn't appear directly in the time dilation formula, but that's not the point. The point is that the fact that acceleration is present at all means that the rotating frame is not an inertial frame, so you can't plug velocities relative to that non-inertial frame into the standard time dilation formula for velocity.

6. Dec 23, 2012

arindamsinha said this in his comment: "The reason is that these velocities (incl. satellite velocities) are not computed w.r.t. the stationary Earth surface."

Are you implying that the clocks in the planes *are* in an inertial frame? Certainly if the earth clock is not stationary in any inertial frame then the clocks on the planes are also not in any inertial frame. If this is what you are saying then how can you calculate any velocity related time dilation effects so that you can come up with an answer that matches the seen results of the experiment? Are they using some other formula beside the standard one used for inertial frames?

I apologize if I'm missing something but I was under the impression that the expected time dilation effects were calculated with 1) the effect of gravity using the GR formula and 2) the effect of velocity using the SR formula. If I am correct in this, then 2) must assume there are inertial frames involved (earth clock and plane clock) if not entirely (because they are accelerating due to the curve in the velocity) then at least partially (because the 2 clocks are moving relative to each other).

7. Dec 23, 2012

### Staff: Mentor

Which equates to "the clock on Earth is not stationary" with respect to the frame in which the velocities are being computed. So he didn't say the clock on the Earth is stationary, which is what you were claiming he said.

No, of course not. *None* of the clocks are stationary in any inertial frame. That doesn't mean you can't use an inertial frame to evaluate their time dilation.

Because, at least in the idealized version of the scenario that I was using for illustration, the *magnitude* of the velocity of each clock is constant; only its direction changes. Since only v^2 appears in the time dilation formula, that formula is only sensitive to the magnitude of the velocity, not its direction.

It's true that in the actual experiment, the magnitudes of the velocities are not constant; the actual numbers computed by Hafele and Keating to compare with their actual measurements were not computed using the simple time dilation formula. They were computed by a more complicated process; but the more complicated process does the same thing, conceptually, that I was describing for the simple idealized case, and it still has to be done with reference to an inertial frame, not a rotating frame. The difference is that the complicated process can deal with the fact that the actual speed and altitude of the clocks on the airplanes varies with time, by applying the standard GR and SR formulas for time dilation due to altitude and velocity along very small segments of the worldlines of each clock, and then integrating over all the segments to get the final answer. But the formula is still sensitive only to the magnitude of the velocity along each small segment, not its direction, because only v^2 appears.

See above.

For the simple idealized case, yes, this is correct. For the more complicated case, this is correct conceptually, but the mathematical details are more complicated because of the variations in altitude and speed. See above.

No, this is wrong. The calculation is done with respect to an inertial frame, but it does not require that any of the clocks are at rest in any inertial frame. See above.

Last edited: Dec 23, 2012
8. Dec 23, 2012

OK so taking into account everything you have said, I'd like to strip this down to a single question. If you have a clock in space just above the Earth and it is not orbiting the Earth, i.e. it is on the dark side of the Earth at all times, and you have a second clock sitting somewhere on the Equator, moving with the rotation of the Earth, is either clock ticking more slowly than the other?

Actually this question is more complicated than I thought now that I've written it so let me change it.

If after several thousand rotations of the Earth is the average elapsed time different on the two clocks? And if so which one is slower? If not why not?

I'm of course subtracting any effects due to the gravitational difference between the two clocks since I'm only focussing on the SR aspect of the experiment.

9. Dec 23, 2012

### Staff: Mentor

Yes. You don't even have to average anything: just compare the clock readings when the rotating clock passes just underneath the hovering clock. The fact that the rotating clock's motion is periodic gives you a common time reference for comparison.

The clock rotating with the Earth is slower.

10. Dec 23, 2012

### pervect

Staff Emeritus
I wouldn't describe a clock permanently on the dark side of the Earth as not orbiting. For instance, if you consider a tide-locked planet (or imagine the Earth was tidelocked), it would still be rotating - just at a rate of 1 rotation per year.

But clearly it will be rotating a lot less, even if it's still rotating...

11. Dec 23, 2012

This was my understanding as well. Just to be sure, (and again ignoring all effects of gravity) is it slower because it is considered to be accelerating relative to the stationary clock in space, or to put it another way is it similar in principle to the twin paradox experiement where the stay at home twin would be the clock in space and the spaceship would be the clock on Earth? (I'm saying this because the clock on Earth is first moving in one direction then changing direction and moving in the other direction, forever oscillating like a sine way, in the same way the space ship accelerates by changing direction)

And to get back to the main discussion, if the Hafele and Keating experiment were done with the stationary clock in space instead of the one on Earth, what velocities would be used to calculate the relative velocities of the two planes relative to the space clock? Would it not be the same forumla, namely v+e and v-e where e is the surface velocity of the Earth relative to the space clock? And if the answer to this is yes, then the result of the experiment using the space clock would be the same as the result of the experment using the ground clock, but since the time between the ground clock and the space clock differs, there is a contridiction here, no?

12. Dec 23, 2012

Yes, that had occurred to me, but I was ignoring its effect because of the low frequency. Thanks.

13. Dec 23, 2012

I did a little hunting around and found the answer I was looking for in this post: (See bcrowells response)

In a nutshell the issue is that all 3 clocks are accelerating and the Earth clock is accelerating more than the clock going west and accelerating less than the clock going east. The clock that is accelerating the most will have the slowest time (the plane going east), followed by the Earth based clock and finally by the west bound clock.

So this definitely cannot be taken as a simple relative-velocity-from-the-surface but instead, as is always the case in SR, must be taken from the point of view of whatever clock in question is experiencing the most acceleration.

Interestingly this is strongly related to the twins experiment where we have an asymetrical situation, namely the ship is accelerating first one way then the other and the Earth is not. Most importantly is not the appearance of acceleration that counts, only the actual acceleration as felt by the twins (or clocks) involved. This is why the spaceship twin cannot say, "Well it looks to me like the Earth is accelerating". Looks don't count, only actual acceleration.

Another way to look at this for the sake of clarity of anyone reading this is that you cannot be standing by the ground based clock and say that the east plane is traveling away at x velocity and the west plane is also traveling away at x volocity but in the opposite direction, but you can be on the slowest accelerating plane, the westbound plane and say that the Earth clock is traveling at x speed and the eastbound plane is traveling at 2x when calculating the result of the experiment and you will have your correct answer.

Thank you very much for your help here. This problem has been haunting me for days and I wouldn't have been able to solve it without this discussion to get my brain in gear and in "fighting" mode to figure this out! You made several statements that were "hinting" at the solution (or maybe stating it, but I couldn't yet see it) and this really helped.

I love physics!!!

14. Dec 23, 2012

### Staff: Mentor

There are several points involved here:

(1) By "acceleration" bcrowell meant proper acceleration; i.e., the acceleration that would be measured by an accelerometer moving with the clock. The eastbound clock's accelerometer will measure a larger acceleration than the reference clock's, which in turn will measure more acceleration than the westbound clock's. (I think you recognize this based on what you say later on in your post, but I wanted to make it explicit.)

(2) The measured acceleration only equates to time dilation if we adopt the view that they are caused by gravitational fields. This is a GR effect, not an SR effect; basically, we are attributing the entire measured acceleration to a gravitational field, instead of just the component that's due to the Earth's static field.

(3) Strictly speaking, saying that measured acceleration is caused by a gravitational field is only valid if the accelerometer, and hence the clock moving with it, is at rest in the field. (It's also only strictly valid if the field is static, but we'll ignore that technicality here.) It's not possible for all three clocks to be at rest in any gravitational field, because they are moving relative to each other. So there is no way to attribute *all* of the measured difference in time dilation to a gravitational field; there has to be some time dilation due to relative velocity as well.

(4) Attributing all of the time dilation to measured acceleration in a gravitational field is an *interpretation*; there is nothing that *requires* you to view things this way. The viewpoint I gave, where everything is referred to an inertial frame, and the only gravitational effect is due to the Earth's static field, is just as valid. There's nothing wrong with the viewpoint you've given, and if it works for you, great! But it's not the only valid one. As usual in relativity, there are multiple ways of getting to the same correct answer.

Not quite; you *can* say this. But you can't plug these velocities into the standard SR formula for time dilation due to velocity and get the right answer. I think this is what you meant; I just wanted to clarify the language.

You're welcome!

15. Dec 23, 2012

### Staff: Mentor

The ones I gave earlier; the analysis I gave earlier *was* relative to a stationary clock in space. The clocks on both planes would run slow compared to the stationary clock in space, at least as far as the SR velocity effect was concerned.

Yes.

No, it wouldn't, because you'd be comparing both planes' clocks with the stationary clock in space, instead of with the ground clock. The elapsed time for both planes' clocks would be the same, but the elapsed time for the "reference" clock would be different because we would be using the stationary clock in space as the reference clock, not the ground clock.

16. Dec 23, 2012

### arindamsinha

I went through exactly the same process myself some years back. Along with the explanation, I also understood things about the IRF of Earth that I didn't know, so let me share it here:
• Earth's IRF (called ECI) is not fixed to the Earth, and does not rotate with it (this is obvious from this whole discussion, and the fact that Earth rotates around it)
• The ECI's orientation does not even revolve around the Sun once a year along with the Earth's revolution. The orientation stays fixed w.r.t. the distant fixed stars at all times (this took me a lot of effort to figure out, as this fine distinction is not very clearly stated in most places)
• There is no clear reason I have come across why the ECI has the orientation it shows, i.e. what fixes it. There have been hints that it may have something to do with Mach's principle, but the principle itself does not appear to have been very well substantiated

17. Dec 23, 2012

### Staff: Mentor

I'm not sure this is correct, at least not for any frame usually referred to as an "Earth-Centered Inertial" frame:

http://en.wikipedia.org/wiki/Earth-centered_inertial

The ECI frames described here are "not truly inertial since the Earth itself is accelerating as it travels in its orbit about the Sun". But for things that happen in a short time compared to a year, such as the Hafele Keating experiment, they can be treated as inertial to a very good approximation.

Note also that there is not one "ECI" frame; there are multiple ones, which differ in the details of how the axes are defined, as explained in the Wikipedia article.

The main thing is convenience; it's nice to have an ECI whose axes match up to an orientation that's intuitive for a particular application. See the Wikipedia page linked to above. Also, see further comments below.

Can you give any references? I haven't come across this in anything I've read about ECI frames.

Bear in mind that, as noted above, ECI frames are not truly inertial, because they stay "attached" to the Earth as it revolves around the Sun. Bear in mind also that, since there is gravity present in the universe, there is no such thing as a global "inertial frame"; any inertial frame can only cover a limited patch of spacetime.

18. Dec 24, 2012

### arindamsinha

Bear with me as I answer different parts. As I mentioned, I could not find any good clear answer on some of this in any one place, but went through a whole lot of material (and that was quite some time back) to gather the information I have put down.

The Wiki article you mentioned was one of my first ports of call at that time, but it did not answer many of my questions - e.g. what axis is used for computing a satellite velocity based on GM/R2 = v2/R = ω2R. So what axis is that v or ω from? Thinking about it, I realized there can be one and only one ECI orientation that will satisfy this.

(Edit: Of course, this is not one particular set of axes projected in 3D space as at this moment. Any other set of axes whose three spatial dimensions do not in any way rotate w.r.t. this is equivalent to this. I presume you understood that, but trying to be clear)

What I scavenged from various sources is that satellite velocities are computed from an ECI whose orientation is fixed w.r.t. distant fixed stars - the one that gives sidereal time I think.

The "not truly inertial" is correct, because even the distant fixed stars do ultimately move over a long period of time. In at least one place it said the CMBR is a better reference frame than fixed stars, though it too may be just a better degree of approximation.

There are other frames also called ECI - some of them a matter of the orientation of the axes rather than rotation w.r.t. the one I am talking about, and some probably other close approximations better suited for other purposes (since the ECI I am refering to is ultimately an approximation as well).

As I have mentioned before, the references are a bit wishy-washy. I haven't kept track of them, and most of them were not research papers. Still, here are some that might be worth a look at. Don't expect the wording I have used based on my summary across many sources, but they do refer to the same thing.

http://en.wikipedia.org/wiki/Mach's_principle[/PLAIN] [Broken]

http://en.wikipedia.org/wiki/Inertial_frame_of_reference

http://www.astro.sunysb.edu/simswg/siswg/node8.html

There were a few others I remember seeing earlier, but cannot find them at short notice.

Absolutely agreed. I discussed this in the last para of the first part of my answer. We are talking about inertial frames that are good enough approximations for the purpose of our use.

Last edited by a moderator: May 6, 2017
19. Dec 24, 2012

### Staff: Mentor

Yes, it's the one with one spatial axis aligned with the Earth's axis of rotation.

Actually, as I understand it, what is done is to compute satellite velocities using the Earth's sidereal period of rotation, yes, but then to use those as approximate velocities relative to the ECI frame that moves with the Earth as it goes around the Sun. Such a frame is not truly inertial, not because the distant stars move, but because the Earth orbits the Sun. However, since the period of rotation of a satellite is much smaller than the Earth's year, for the purpose of studying satellite orbits, or for other similar purposes such as studying the Hafele Keating experiment, such an ECI frame is a good approximation to a frame that is truly inertial with respect to the fixed stars.

Once again, there is *no* truly inertial frame that covers a large region, because spacetime is curved. (I see that you agree with this at the end of your post.) There are only local inertial frames. A reference frame in which the CMBR is isotropic is a better approximation of an inertial frame locally, yes; but because the universe is expanding, such a frame is not a good inertial frame if you are trying to cover a time period that is significant cosmologically.

I don't think any of these issues significantly affect the analysis of the H-K experiment that I gave previously; the ECI frame that I was using is a good enough approximation for that purpose.

20. Dec 24, 2012

### arindamsinha

I have not been able to explain this adequately. We need to look a bit deeper for what I am stating.

Let me take two examples to explain what I mean:

Example 1:
For a satellite path circling the Earth's equator (e.g. geostationary satellite), there are innumerable spatial axes (frames) that satisfy one axis aligned with the Earth's axis of rotation (let us call this the z-axis).
• One example is the ECEF which is fixed to the Earth, with the x and y axes having the same ω around z-axis as Miami beach does. W.r.t. this frame, the ω of a geostationary satellite is zero, so this is not the one we want for satellite velocity computations using the orbital equation
• Next, let us take another frame where the x and y axes do not rotate with the Earth surface, but in the course of a year does have one single rotation w.r.t. the distant fixed stars, because of the revolution of the Earth round the Sun. This axis defines our Solar Day and the calendar year of 365 days that we follow. The ω of a geostationary satellite has a non-zero value in this frame, so this would be a candidate to define ω from the orbital equation. Let us call the value of ω in this frame as ω'
• Lastly, let us take another frame, z-axis still aligned to Earth's axis of rotation, where the x and y axes do not rotate w.r.t. the distant fixed stars at all. This axis defines our sidereal day, which is about 23 hour 56 min long. Let the value of ω of a geostationary satellite be ω'' in this frame. This is also a possible candidate for satellite velocity calculation using the orbital equation
Now ω' and ω'' are different, even if slightly. Both of them cannot satisfy the orbital equation, so only one of them is true. As I understand it, experience has told us that it is the last frame (sidereal) that has proved to be the correct one.

Example 2:
Consider a satellite that does not fly along the equator but has a non-equatorial or polar route. Here, the Earth's axis of rotation has no relevance at all. Still, there are satellites that can and do orbit Earth in non-equatorial trajectory, and will have a calculatable ω, with no relationship to the Earth's axis of rotation.

I believe this again uses the sidereal frame for computations (which can be imagined to be fixed in a 3-D way with distant fixed stars in all possible directions, irrespective of the Earth's rotational axis)

I hope these example paint a better picture of what I am trying to say.

First of all, thanks for confirming the sidereal part of satellite velocity calculation from your experience/understanding. I have worked with skimpy sources, because hardly any of them are explicit about it, and have had little corroboration to my understanding.

Yes, and probably because of the reasons you mention, I have had to struggle hard to figure out the distinction between the solar and sidereal frames from a satellite velocity perspective.

Actually the biggest reason I have found why this is not considered so important is because there are much larger other influences that have to be fended off than this slight velocity difference. Station-keeping is necessary for much more important issues like the Earth's not being a perfect sphere gravitationally, such that if a geostationary satellite was left alone, it would actually leave equator orbit and settle on a trajectory somewhere on top of the Indian Ocean. Given this, that slight velocity difference between the two frames doesn't matter that much in satellite technology station-keeping. With the coordination going on between Earth stations and the satellite to keep it in place, a few m/s this way or that would get worked out automatically to make satellite technology practical.

Still, from a theory perspective, the difference does exist. Small as it may be, a geostationary satellite would have a difference of 1 full revolution around Earth in a year, depending on whether the solar or sidereal frame is used for computing the ω. So it does matter, as only one of them results in a geostationary orbit. I believe experience has taught us (or satellite engineers) to use the sidereal frame as the basis.

We already agreed on this. We are using reasonable approximations over a short period of time, perhaps even just a human lifetime. I hope the first half of my answer provided the clarity I was seeking to put.

Totally agreed. The Mach principle was actually just a footnote in my main message, that we don't know what really fixes the orientation of the ECI we use for satellites etc.

Agreed. Makes little difference to the analysis you provided for H-K. I just thought I will share a few interesting tid-bits that I came across when I learnt the H-K problem resolution the hard way.

Last edited: Dec 24, 2012
21. Dec 24, 2012

### Staff: Mentor

Yes, because it's the closest to a "true" inertial frame, and the orbital equation is only strictly valid in a "true" inertial frame.

Actually, what is going on is that ω is independent of the orientation of the orbit with respect to the sidereal frame's axes. All ω depends on is the orbital radius (for the idealized case of a perfectly circular orbit). The radius is invariant under rotation of the axes, so it doesn't matter how the axes are oriented. (Of course ω also depends on G and M, but those are constant.) For the more realistic case of an elliptical orbit, ω still depends only on orbital parameters that are invariant under rotations of the axes: the semi-major axis and the eccentricity. Those parameters are easier to express in terms of coordinates if the axes are oriented so the orbit is entirely in one coordinate plane, but that's just a matter of mathematical convenience; it doesn't affect the physics.

Yes, definitely.

Once again, as above, this is because the sidereal frame is the closest to a "true" inertial frame, which is the only kind of frame in which the orbital equation that defines ω is valid.

22. Dec 24, 2012

### arindamsinha

Good. I hope I have been able to explain better, and we are now in agreement (let me know if this is not the case).

This discussion was actually very helpful to me, in clarifying some thoughts I have had for a while. I had not had an opportunity to be able to discuss and see if they were correct. Did not seem important enough to start a separate thread on.

23. Dec 24, 2012

### Staff: Mentor

Yes, I think so.

24. Dec 25, 2012

I don't understand why this would be true. Einstein's Equivelence Principle equates acceleration to gravity, so can't we just call the accelerations due to the 3 clocks moving in a circle arount the center of the Earth gravity and add it to the static gravitational field, then plug the whole thing into GR? And of course here I mean the 3 clocks have 3 different accelerations so the 3 additions would all be different.

Well, there does have to be some time dilation due to relative velocity (SR), but why do you say just because all 3 clocks have differing overall gravity (static+acceleration) that this reason alone is why you can't atribute all the measured difference to a gravitational field? (And again by gravitational field I assume you mean static+acceleration)?

Really? I don't know how else you can see this. Wouldn't it be correct to say that what we have here (no interpretation) are 3 factors that contribute to the time dilation, static gravity (GR), equivalent gravity due to acceleration caused by the 3 clocks circling around the Earth (also GR) and finally dilation due strictly to the velocities of the 3 clocks relative to each other (SR)?

If this is the case, then I still have some confusion. If I had my way (in my own personal universe), I would just like to drop the SR addition to the equation and say that the 3 clocks differed in their times due only to the first 2 reasons, gravity and acceleration. It would make my life so much simpler. It's the SR part of this thing that really screws things up for me because by breaking this problem up into 3 components this leaves the SR component as being treated strictly as 3 inertial frames. This means that the equations (for only this SR component) must be created using the Earth based clock as the reference, in which case the planes velocities would have to be v for the eastbound plane and v for the westbound plane and not v+e and v-e.

25. Dec 25, 2012

### pervect

Staff Emeritus
Because the goal is tho have the geodesics of light rays be straight lines, (which is well documented) the ECI pretty much has to be non-rotating with respect to the fixed stars. Actually, the inertial frame will be rotating very, very, slightly with respect to the fixed stars due to the Lense-Thirring effect, I suppose, but it'll certainly be much closer to the "fixed stars" than a "sun-fixed" frame.