Space traveler and time dilation

[JesseM]

A satellite in orbit is a perfectly good inertial frame

The only frames in curved spacetime that qualify as "inertial" are local ones defined on a very small (technically it must be infinitesimally small) patch of spacetime, if you're talking about a coordinate system covering a large spatial or region or a long time interval (like a significant proportion of an orbit), then tidal effects would be detectable in this region so the frame can't be inertial. Do you disagree? This is a very standard idea, any textbook discussing the equivalence principle should make clear it only holds in a very small region of both space and time. For example, read the last section of the txt http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html , the section titled "Tidal forces, and a more precise definition", where they write:

Realizing that what matters are the size of the region, and the duration of our observations, we are led to a formulation in which the equivalence principle is not just a useful approximation, but exactly true: Within an infinitely small ("infinitesimal") spacetime region, one can always find a reference frame - an infinitely small elevator cabin, observed over an infinitely brief period of time - in which the laws of physics are the same as in special relativity. By choosing a suitably small elevator and a suitably brief period of observation, one can keep the difference between the laws of physics in that cabin and those of special relativity arbitrarily small.

 

[yogi]

Hi Jesse, long time no chat. In the context of the twin trip, the analogy is simply a way to illustrate the idea that accelerations (in the fame of the traveler) can be virtually eliminated if the trajectory is bent by a G source that will precisely return the traveler to the starting point). As you will no doubt recognize, this is simply a one orbit version of a GPS satellite - both clocks remain in separate inertial frames during the entire round trip (one orbit). Its true that on a real satellite there are minute tidal affects that would slightly alter the traveling clock frequency - but if the South Pole tower in the thought experiment is extended to 20,000, km, the orbital speed is about 14,000 km/hr, and the frequency shift is about 7 nanosec/day - verified to great accuracy every day.
In one orbit, there is going to be an accurate measurement of the time difference between the two clocks when the traveler returns.

To complete the analogy, one might place a second tower at the North Pole and check the time lapse for the halfway point - in the normal twin scenario, this corresponds to the turn around point which normally involves deceleration and acceleration in the frame of the traveler - but in the orbital version, there are no accelerations except those incidental to tidal affects and the divergence of the Earth's G field.

 

[JesseM]

Hi Jesse, long time no chat. In the context of the twin trip, the analogy is simply a way to illustrate the idea that accelerations (in the fame of the traveler) can be virtually eliminated if the trajectory is bent by a G source that will precisely return the traveler to the starting point).

Yes, but you are wrong in thinking that "no accelerations/G-forces" (i.e., accelerometers floating free at any given point in a room measuring 0 at all times) is a sufficient condition to have an approximately inertial frame in GR. An inertial frame is also one where there are no measurable tidal forces, and over the course of an orbit tidal forces should be measurable, since in general relativity tidal forces only become negligible in small regions of space over short time-intervals (small regions of spacetime, not just small regions of space). I'm not actually sure of the details of how tidal forces would manifest in the scenario of a small room in orbit for a long period of time, but I bet if you set off two balls at slightly different speeds in slightly different directions from one end of the room, such that the time for them to reach the opposite walls at constant velocity would be comparable to the time for an entire orbit, you would in fact observe that the balls' paths would depart appreciably from straight paths at constant velocity over this long time period.

 

[DrGreg]

In general relativity, the south pole pole is accelerating and the satellite is inertial.

A satellite in orbit is a perfectly good inertial frame - the traveling twin stays in orbit - all clocks are at the same gravitational potential (100 miles above the earth) - so there is no general relativity issues and there are no accelerations once the orbit is established and the first measurement is taken on the flyby.

You're missing my point. I did say

"In general relativity, ... the satellite is inertial."​

which agrees with everything you said above (subject to JesseM's correct note about local frames). But I also said

"In general relativity, the south pole pole is accelerating..."​

It is undergoing proper acceleration upwards and therefore is not inertial.

 

[JDługosz]

You don't really need to go around the universe to resolve or simulate the problem - there is one round trip voyage where both twins remain in their own inertial frame for the entire round trip - yet their clocks show different lapsed times when the traveling twin returns - no general relativity involved and no curvature - a polar orbiting satellite will do - simple construct a 100 mile high tower on the South Pole and put a satellite in polar orbit at an elevation of 100 miles - a clock on top of the tower remains fixed in the non rotating Earth centered inertial reference frame and the clock on board the satellite remains in the inertial frame of the orbiting satellite - start the clocks as the satellite passes the tower and stop them when it passes by after completing one orbit - the two clocks will not have logged the same amount of time.

Make it more symmetrical by using two satellites instead, orbiting in opposite directions. Each is traveling through the same curvature and each is accelerating the same. That creates a paradox if you apply only SR. How does it manage to work out using GR? Once I understand that, I can ponder the Universe form of the question again.

 

[yogi]

You're missing my point. I did say

"In general relativity, ... the satellite is inertial."​

which agrees with everything you said above (subject to JesseM's correct note about local frames). But I also said

"In general relativity, the south pole pole is accelerating..."​

It is undergoing proper acceleration upwards and therefore is not inertial.

Yes - things on the surface of the Earth are subject to g - but for purposes of synchronizing GPS satellites, the non-rotating Earth centered reference frame is taken as a basis - in the proposed thought experiment, the affect of satellite height is nullified by the height of the imaginary towers - so perhaps I should have said both clocks are at the same gravitational potential at all times.

The point of the analogy is, it is not necessary to explain the paradox as being the result of the traveling twin feeling "turn-around acceleration" (as is frequently asserted - perhaps even by myself in past posts). This is not what distinguishes the two clocks - they log different amounts of time because of the invariance of the spacetime interval. The traveler travels in both space and time - the tower twin travels in time only.

 

[JesseM]

The point of the analogy is, it is not necessary to explain the paradox as being the result of the traveling twin feeling "turn-around acceleration"

But you can explain it in terms of one moving inertially while the other doesn't, and the question of which one turns can be decided by the turn-around acceleration.

This is not what distinguishes the two clocks - they log different amounts of time because of the invariance of the spacetime interval. The traveler travels in both space and time - the tower twin travels in time only.

But in a GR case, you can come up with a coordinate system where the first remains at a single point in space (traveling 'in time only', not moving in space) and a different coordinate system where the second is the one that remains at a single point in space, both coordinate systems would be equally valid in GR, there'd be no physical basis for saying one "really" moved in space while the other didn't.

 

[Jocko Homo]

Forget about "clocks running slower". That's Lorentz Ether language. Of course, A cannot run slower than B while B runs slower than A. This sort of language is incompatible with SR.
I know that's how they teach it. Forget it. Look at your diagrams. There's a triangle, and there's a triangle inequality (in our case the non-straight path being shorter). That's true no matter in which frame you view it.

Your math is correct. But there's no such things as slowing clocks, contracting metersticks, or planets jumping forward in time. SR is about relations of objects, not changes happening to them.
Take the geometric viewpoint, there are projections, slices, different paths (time dilation, length contraction, twin paradox), not broken clocks.

While Minkowski diagrams provide a powerful abstraction to the physical reality of Special Relativity, I think "clocks running slower" is a fair characterization considering the accelerating twin really does come back younger...

 

[yogi]

Make it more symmetrical by using two satellites instead, orbiting in opposite directions. Each is traveling through the same curvature and each is accelerating the same. That creates a paradox if you apply only SR. How does it manage to work out using GR? Once I understand that, I can ponder the Universe form of the question again.

If you have two circular orbiting satellites at the same height traveling in opposite directions and set both clocks to zero as they pass, when they meet again, both clocks will read the same

 

[yogi]

But you can explain it in terms of one moving inertially while the other doesn't, and the question of which one turns can be decided by the turn-around acceleration.

It will give you the right answer - but you can also get the correct age difference by doing a one way trip - setting the clocks to zero on the initial flyby and stopping the traveling clock at the instant it flies by the destination - you now have a one way trip with no accelerations and no turn around acceleration - to get the result you simple double the time difference for the one way trip - my point is not that other methods give wrong results - but rather, SR age differences are fundamentally a relative velocity problem that has been misappropriated to GR

But in a GR case, you can come up with a coordinate system where the first remains at a single point in space (traveling 'in time only', not moving in space) and a different coordinate system where the second is the one that remains at a single point in space, both coordinate systems would be equally valid in GR, there'd be no physical basis for saying one "really" moved in space while the other didn't.

Agreed, travel with respect to space is meaningless - I guess it should be clarified as travel with respect to space as measured in the fame which is chosen to be at rest

 

[Al68]

IMO - any analysis that results from dependence upon accelerating frames and the like is going to cloud the reality of what is properly explained by SR - this can be done by using the one way trip and doubling the result...

I agree that if the problem was presented and explained as two one way trips, most of the confusion would be completely eliminated. But the ship is still non-inertial for an (entire) one way trip.

Einstein confused a lot of his followers when he published his 1918 article that explained the clock paradox using a pseudo G field that gives the same answer to the aging difference - but for the wrong reason.

Why would the reason be wrong? The reason for the aging difference is exactly the same in Einstein's 1918 paper as in standard resolutions. The only difference is he uses realistic acceleration (with Earth clock running fast in ship frame) instead of an instantaneous turnaround (earth clock "jumps ahead").

 

[matheinste]

In SR if the twins part and reunite then at least one of them has been moving non-inertially at some time.


Matheinste.

If you have two circular orbiting satellites at the same height traveling in opposite directions and set both clocks to zero as they pass, when they meet again, both clocks will read the same

Assuming that they move symmetrically, that is, they meet again for the first time at the opposite "pole" to where they started as referred to the orbited body,then isn't this a case of both objects traveling non-inertially but both traveling equal spacetime paths and so clocks traveling with them will record the same elapsed proper time.

Matheinste.

 

[Ich]

While Minkowski diagrams provide a powerful abstraction to the physical reality of Special Relativity, I think "clocks running slower" is a fair characterization considering the accelerating twin really does come back younger...

IMHO, time dilation is a convenient memory and calculation aid, but has no (or even less) value for understanding SR. It should be used by those only who understand SR and know how, when, and why time dilation can be applied.
If you confront beginners with that phrase ("clocks run slower"), you inevitably push them down the Lorentzian road, where SR is the oncoming traffic. However, it is common praxis in education, and you see the result here every week with a new thread about the twin paradox.

 

[JesseM]

Agreed, travel with respect to space is meaningless - I guess it should be clarified as travel with respect to space as measured in the fame which is chosen to be at rest

OK, but in the GR case where you're dealing with non-inertial frames either way, depending on the frame you choose it may not be the twin that "travels with respect to space" who ages less, it could be the other twin.

 

[Mike_Fontenot]

Here are a couple of my responses to essentially the same question that was asked on an earlier thread:
______________________________________________________

During the constant-speed legs of the trip, BOTH twins conclude that the other twin is ageing slower. But when the trip is over, they both agree that the stay-at-home twin is older. How is that possible?

It's possible because, during the turnaround, the traveler will conclude that the home twin quickly ages, with very little ageing of the traveler. The home twin concludes that neither of them ages much during the turnaround. When you add up all these segments of ageing, you get the result that the home twin is older (and both twins exactly agree on that).

Years ago, I derived a simple equation (called the "CADO" equation) that explicitly gives the ageing of the home twin during accelerations by the traveler (according to the traveler). The equation is especially easy to use for idealized traveling twin problems with instantaneous speed changes. But it also works for finite accelerations. I've got a detailed example with +-1g accelerations on my webpage:

http://home.comcast.net/~mlfasf

And I've published a paper giving the derivation of the CADO equation:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.
____________________________________________

Here's a brief description of my "CADO" equation:
__________________________________________________ __

Years ago, I derived a simple equation that relates the current ages of the twins, ACCORDING TO EACH TWIN. Over the years, I have found it to be very useful. To save writing, I write "the current age of a distant object", where the "distant object" is the stay-at-home twin, as the "CADO". The CADO has a value for each age t of the traveling twin, written CADO(t). The traveler and the stay-at-home twin come to DIFFERENT conclusions about CADO(t), at any given age t of the traveler. Denote the traveler's conclusion as CADO_T(t), and the stay-at-home twin's conclusion as CADO_H(t). (And in both cases, remember that CADO(t) is the age of the home twin, and t is the age of the traveler).

My simple equation says that

CADO_T(t) = CADO_H(t) - L*v/(c*c),

where

L is their current distance apart, in lightyears,
according to the home twin,

and

v is their current relative speed, in lightyears/year,
according to the home twin. v is positive
when the twins are moving apart.

(Although the dependence is not shown explicitely in the above equation, the quantities L and v are themselves functions of t, the age of the traveler).

The factor (c*c) has value 1 for these units, and is needed only to make the dimensionality correct.

The equation explicitly shows how the home twin's age will change abruptly (according to the traveler, not the home twin), whenever the relative speed changes abruptly.

For example, suppose the home twin believes that she is 40 when the traveler is 20, immediately before he turns around. So CADO_H(20-) = 40. (Denote his age immediately before the turnaround as t = 20-, and immediately after the turnaround as t = 20+.)

Suppose they are 30 ly apart (according to the home twin), and that their relative speed is +0.9 ly/y (i.e., 0.9c), when the traveler's age is 20-. Then the traveler will conclude that the home twin is

CADO_T(20-) = 40 - 0.9*30 = 13

years old immediately before his turnaround. Immediately after his turnaround (assumed here to occur in zero time), their relative speed is -0.9 ly/y. The home twin concludes that their distance apart doesn't change during the turnaround: it's still 30 ly. And the home twin concludes that neither of them ages during the turnaround, so that CADO_H(20+) is still 40.

But according to the traveler,

CADO_T(20+) = 40 - (-0.9)*30 = 67,

so he concludes that his twin ages 54 years during his instantaneous turnaround.

The equation works for arbitrary accelerations, not just the idealized instantaneous speed change assumed above. I've got an example with +-1g accelerations on my web page:

http://home.comcast.net/~mlfasf

The derivation of the equation is given in my paper

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

Mike Fontenot

 

[Al68]

Years ago, I derived a simple equation (called the "CADO" equation) that explicitly gives the ageing of the home twin during accelerations by the traveler (according to the traveler). The equation is especially easy to use for idealized traveling twin problems with instantaneous speed changes. But it also works for finite accelerations. I've got a detailed example with +-1g accelerations on my webpage:

http://home.comcast.net/~mlfasf

And I've published a paper giving the derivation of the CADO equation:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

I didn't carefully analyze your web page, but will assume the results match standard lorentz transformations for a co-moving inertial frame for any given time on the accelerated clock. And I did note the phrase "the bizarre behavior of the CADO, for an accelerating observer, must be regarded as being fully meaningful and real."

I would also note that if the traveler is accelerating in a direction away from earth, the results could be even more bizarre. For example, from the accelerated traveler's point of view, the funeral of a person on Earth could be followed by that person's wedding.

I would have to say that people rising from the grave would qualify as not "fully meaningful and real" by any reasonable definition.

Note that I'm not disputing that result. We could obtain the same result by performing lorentz transformations directly during inertial motion before and after acceleration, if the ship were distant from Earth and the acceleration was away from earth. But most would disagree that the result represented reality in any sense other than assigning coordinates.

 

[stevmg]

George Jones and Ich are correct. Attached is a picture I downloaded from among the plethora of explanations on the web. You can see that on the way out, the twin on the rocket sees two years pass on Earth, while four years pass for him. On the way back, he sees 8 years pass on Earth while four years pass for him. So when he returns, 10 years have passed on Earth while 8 years have passed for him, so the twin on the rocket is younger.

JesseM gave an example using SR why the "moving" twin ages less quickly than the earthbound twin. He does it both ways - using the Earth as the inertial frame and then the spaceship as the inertial frame. In both ways the Earth twin aged more than the traveling twin. Here is the post:

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

 

[stevmg]

George Jones and Ich are correct. Attached is a picture I downloaded from among the plethora of explanations on the web. You can see that on the way out, the twin on the rocket sees two years pass on Earth, while four years pass for him. On the way back, he sees 8 years pass on Earth while four years pass for him. So when he returns, 10 years have passed on Earth while 8 years have passed for him, so the twin on the rocket is younger.

How do I post a picture like you did on this post
https://www.physicsforums.com/showpost.php?p=2753642&postcount=13

H-E-L-P

Steve G
Melbourne FL

 

[DrGreg]

How do I post a picture like you did on this post
https://www.physicsforums.com/showpost.php?p=2753642&postcount=13

While you are editing your post (in the "advanced" editor -- which you get automatically if you use the QUOTE button) press the "Manage Attachments" button not far underneath.

 

[Mike_Fontenot]

[...]
I would also note that if the traveler is accelerating in a direction away from earth, the results could be even more bizarre. For example, from the accelerated traveler's point of view, the funeral of a person on Earth could be followed by that person's wedding.

Yes, that behavior is easy to see, directly from the form of the CADO equation itself, when there are instantanious changes in the velocity v, such that v gets more positive, or less negative (v is positive when their separation is increasing). And this effect isn't limited to instantaneous velocity changes...the voyage detailed in the web page includes segments where the traveler must conclude that the home twin gets younger, with only 1g accelerations involved.

But the web page, and my paper, also make it clear that this phenomenon in no way influences the perception, by the home twin, of the normal progression of time. It is somewhat analogous to your ability to reverse the direction of a movie projector...your action doesn't bother the actors, or make any changes to the frames of the film.

I would have to say that people rising from the grave would qualify as not "fully meaningful and real" by any reasonable definition.

The critical point to understand (which is elaborated in detail in the paper), is that the traveler CAN ADOPT NO OTHER CONCLUSION, IF HE WISHES TO AVOID CONTRADICTING HIS OWN ELEMENTARY (AND CORRECT) MEASUREMENTS. It is in that sense that I use the phrase "real and meaningful". If elementary, correct measurements are not "real and meaningful", how can anyone ever hope to do any physics?

Mike Fontenot

 

[ExecNight]

Hello,

Theoretically speaking, if we get a clock that ticks a red light every 1 second on the spaceship or whatever. And it has a quantum entangled pair on earth.

So how would the Twin on the spaceship actually see the Earth clock ticking throughout the journey?

That would be interesting to know maybe explains how this works better in a way.

 

[Al68]

The critical point to understand (which is elaborated in detail in the paper), is that the traveler CAN ADOPT NO OTHER CONCLUSION, IF HE WISHES TO AVOID CONTRADICTING HIS OWN ELEMENTARY (AND CORRECT) MEASUREMENTS. It is in that sense that I use the phrase "real and meaningful". If elementary, correct measurements are not "real and meaningful", how can anyone ever hope to do any physics?

The traveler can certainly recognize that the coordinates he correctly assigns (using Einstein's simultaneity convention) in his frame to events on Earth do not represent a "real and meaningful" sequence of events requiring causality to be explained.

I might even use the word "fictional" in the same way that fictional forces are used to explain the motions of objects when using a non-inertial reference frame. (The traveler must also assign fictional forces to account for Earth's motion). These forces are not "real and meaningful", either, yet physics thrived for hundreds of years using fictional forces.

And those fictional forces arise for the same reason that bizarre time coordinates get assigned to events on earth, so why not call them fictional?

 

[Ich]

CAN ADOPT NO OTHER CONCLUSION, IF HE WISHES TO AVOID CONTRADICTING HIS OWN ELEMENTARY (AND CORRECT) MEASUREMENTS.

Sorry, that's nonsense. Go back one step to the operational meaning of the respective coordinates.

If elementary, correct measurements are not "real and meaningful", how can anyone ever hope to do any physics?

Which measurements? (This is a rhetorical question).
Describe how you take the "measurement" of something going back in time.

There is none.

What you see at this locus will preserve causality.

 

[Mike_Fontenot]

Describe how you take the "measurement" of something going back in time.

If an unaccelerated object (the "home" twin) is periodically transmitting her current age, then an inertial observer (the traveling twin), who is moving at a constant velocity with respect to her, can receive those messages. He knows that, when he receives a message, that the age being reported in the message is not her current age (because she has aged while the message was in transit). From first principles, and elementary calculations, and without knowing anything about special relativity, the traveler can compute what his twin's current age is (by properly allowing for her ageing during the transit of the message).

That calculation, although elementary, is very easy to do incorrectly. The proper way to do it is detailed in my paper. If that process is done correctly, the result is precisely what is given (much more quickly and easily) by my CADO equation.

If you want to know more, you'll have to dig up my paper...most university libraries either have it, or can obtain it from another library.

(And the reason why the above process, which assumes the traveler is also unaccelerated, is of any value in determining the conclusions of an accelerating traveler, about the current age of the home twin, is also detailed in the paper).

Mike Fontenot

 

[yogi]

I agree that if the problem was presented and explained as two one way trips, most of the confusion would be completely eliminated. But the ship is still non-inertial for an (entire) one way trip.

Why would the reason be wrong? The reason for the aging difference is exactly the same in Einstein's 1918 paper as in standard resolutions. The only difference is he uses realistic acceleration (with Earth clock running fast in ship frame) instead of an instantaneous turnaround (earth clock "jumps ahead").

Why is the ship non-inertial?


Yes as to your second comment - it is like many of the standard solutions - those based upon resolving the problem via GR - Max Born and others who thought a reason for the age diffeence other than SR was needed to resolve the clock paradox - that is why after 1918, some authors of repute began saying flatout the problem can only be solved with GR - that is properly refuted by most others - but it all started with the the subtrifuge introduced by the 1918 paper.

 

[yogi]

OK, but in the GR case where you're dealing with non-inertial frames either way, depending on the frame you choose it may not be the twin that "travels with respect to space" who ages less, it could be the other twin.

So, if we hang the satellite clock on a sky hook, and call it a coordinate frame - then rotate the Earth beneath at the speed necessary to eliminate the affects of gravity acting upon the clock at the top of the tower, then it is the tower clock that does the traveling wrt the fixed frame of the satellite clock - so the two clocks are no longer in sync when they meet after one revolution - I guess that is what you are saying - or did I miss your point.

 

[Al68]

I agree that if the problem was presented and explained as two one way trips, most of the confusion would be completely eliminated. But the ship is still non-inertial for an (entire) one way trip.

Why would the reason be wrong? The reason for the aging difference is exactly the same in Einstein's 1918 paper as in standard resolutions. The only difference is he uses realistic acceleration (with Earth clock running fast in ship frame) instead of an instantaneous turnaround (earth clock "jumps ahead").

Why is the ship non-inertial?

Because the ship can't get half way through the twins paradox scenario without accelerating. At the half way point, the ship is at rest with earth.

And if we make that the end of a one-way trip, the ship twin is younger than the Earth twin unambiguously, since both are at rest in the same inertial frame.

As far as Einstein's 1918 paper, he never claimed GR was necessary to resolve the twins paradox. He just showed that it could also be analyzed from the accelerated frame of the ship with the same result for the same reason as the standard SR resolutions.

 

[yogi]

Because the ship can't get half way through the twins paradox scenario without accelerating. At the half way point, the ship is at rest with earth.

And if we make that the end of a one-way trip, the ship twin is younger than the Earth twin unambiguously, since both are at rest in the same inertial frame.

As far as Einstein's 1918 paper, he never claimed GR was necessary to resolve the twins paradox. He just showed that it could also be analyzed from the accelerated frame of the ship with the same result for the same reason as the standard SR resolutions.

Although Einstein started with two clocks at rest in the same frame in his 1905 description -it is not necessary. The one way trip I always thought we had in mind was that of a twin initially accelerated to crusing speed thereafter passing the stay put twin at a close distance at which instant both clocks are set to zero, and the voyage begins. Same thing upon reaching the target -the flyby twin notes the time on a local clock (which we can stipulate to be in the fame of the stay put twin) and the observer at the target notes the time on the flyby clock. This eliminates any acceleration - the voyage from stay-put twin to destination is inertial all the way.

I will agree that Einstein didn't promulgate the 1918 paper as exclusive - but others have taken the position that only GR is the correct reasoning - but this may be a subjective bias of mine - perhaps the issue is at some level, simply a distinction without a difference. - GR deals with potentiall energy, SR with KE differences between moving frames - and we know one can be derived from the other - so maybe I will stop campaigning this issue

 

[stevmg]

JesseM gave an example using SR why the "moving" twin ages less quickly than the earthbound twin. He does it both ways - using the Earth as the inertial frame and then the spaceship as the inertial frame. In both ways the Earth twin aged more than the traveling twin. Here is the post:

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

I repeat, here is JesseM's post. It is short, sweet and simple.

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

The GPS satellites that were sent up thirty years ago confirm this.

 

[JDługosz]

If you have two circular orbiting satellites at the same height traveling in opposite directions and set both clocks to zero as they pass, when they meet again, both clocks will read the same

So I would think. The question is why? Each sees the other running slow as they pass (a SR effect). How does GR resolve this?