# Two travelers both in Inertial Frames

1. Sep 9, 2010

### yogi

The twin thing has been beat to a pulp on these forums - so maybe this got resolved somewhere along the way - if so - someone can point me to the solution. Otherwise here is the set-up. Assume two satellites are placed in orbit around a black hole, one is in circular orbit, the other highly elliptical - we specify observers A and B with clocks in each saterllite in gimbled suspension structures to eliminate any awareness that the satellite is in motion - so since the observers can't see the black hole G source and they do not turn or rotate or tilt during their orbit, they think they are at rest. To complete the illusion, we can blank our the light from the universe so illusion is complete: each observer is firmly convinced he is not moving. The two satellite orbits are adjusted so their paths cross every 100 years as seen from earth - on the first crossing, each sets his clock to zero, when they pass nearby after a 100 years each checks the others time - which clock has logged the larger amount of time and why? Since each observer believes he is in an inertial frame and can observe the other satellite going off doing wild girations including accelerations and decelerations and returning, he would tend to conclude the other dude is younger. From observer A's perspective B has taken a long trip and returned - But B see's A flashing past and watches him return - so B should be able to claim A is the round trip traveler and is younger This presupposes no other visual reference frames to give away the motion and also that any orbit (whether circular or ellipitical) is a good inertial frame.

2. Sep 9, 2010

### yogi

To embellish more on the above, it seems that any difference must be a consequence of the height - in other words a general relativity solution is appropriate and will yield a time difference, but can there be an age difference a la SR that is based only upon the integration of the relative velocities?

3. Sep 9, 2010

Staff Emeritus
Why in heaven's name do you think these frames are inertial?

4. Sep 9, 2010

### yogi

For a quick edification Take at look at Road to Reality - penrose p394

To reduce the complexity introduce by height variations - assume the orbits are circular at the same height - but orthogonal - one polar, one equitorial

5. Sep 9, 2010

### bcrowell

Staff Emeritus
I think the simple answer to the OP's question is that either observer will say that there are both kinematic and gravitational time dilation effects. Each will say that his own kinematic time dilation vanishes, while the other guy's is nonzero. But each will also see a gravitational effect, which breaks the symmetry between observers that led to the original paradox.

I would describe them as inertial, since both objects are free-falling. (They aren't inertial in the sense of inertial frames in Newtonian mechanics, but the OP's question makes it clear that he has the context of relativity in mind.)

6. Sep 9, 2010

### Fredrik

Staff Emeritus
The ages will be the proper times of the world lines. This is true both in SR and GR. You seem to want to separate the proper time calculation into a part due to altitude and a part due to velocity. I don't think that's a meaningful thing to do, and I wouldn't assume that it makes sense for general world lines. (It might, but it would take some work to show it).

Similar questions to the one you asked in #1 have been asked before. I have to go to bed, so I'm not going to search for them now, but perhaps you can find them on your own.

7. Sep 9, 2010

### yogi

Yes there have been - some years ago I recall an example by Garth which involved dropping a clock through a hole in the earth which gains speed until it reaches the center - then slows to a stop at the opposite end of the hole and returns by the same path to the start point assuming no friction etc. Both are in Einstein's equivalence free falling inertial frames. To eliminate the height problem which is going to be a factor in my original post as well as the hole experiment, let both satellites be at the same height with circular orbits in orthogonal planes. Because of the symmetry, no age difference will result. ... but if A observer believes himself to be always inertial, he would calculate that B's spacetime interval is equal to his (A's) temporal interval and therefore B is younger - But from B's perspective, A's spacetime interval will be equal to B's temporal interval, and therefore A is younger. So where is the flaw in the application of interval invariance to this problem?

8. Sep 10, 2010

### Passionflower

Spacetime intervals are identical for all observers. Just because one slices space and time in two different ways does not make the interval any different.

9. Sep 10, 2010

### Staff: Mentor

You are correct, by symmetry both will be the same age. There is no flaw with the application of the interval to this problem, just you have to use a correct metric for the spacetime that they are in. Obviously, if you are in a Schwarzschild spacetime and you use the metric for a Minkowski spacetime you will get incorrect results. This is not a paradox, this is simply a computational error.

10. Sep 10, 2010

### yogi

That sounds like it makes sense - so even though both A and B have a firm conviction they are at rest and no way of proving otherwise, neither is allowed to apply Minkowski's interval for flat space.

Can you embellish upon the mechanism? If the Schwarzschild metric is applied - does the spacetime interval for the other observer turn out to have the correct value to cancel his spatial component so as to leave the temporal intervals equal, or does the proper frame of the observer get modified so that his interval includes a spatial component as well as a temporal component.

11. Sep 11, 2010

### yogi

Looking at the Schwarzschild metric -seems that for two observers in orbit at radius r about a gravitation lump of Schwarzchild radius r_s then relative to a clock at infinity, the relative time dilation for both observers is
(1-r_s/r)^1/2 so there cannot be any relative time dilation between them - i.e., as you said, the Minkowski metric is inapplicable except at very great distances

12. Sep 11, 2010

### JesseM

But a free-falling frame is only like an inertial frame in the limit as you consider an arbitrarily small region of space and time--a free-falling observer who falls through curved spacetime for an extended period of time should be able to detect tidal effects, no matter how small their craft is spatially.

13. Sep 11, 2010

### yogi

But will it be significant compared to the time dilation at the height of the satellite due to the gravitational potential of the central mass. If so, it would seem that GPS satellite clocks would have to be biased with an additional corrective factor prior to launch.

14. Sep 11, 2010

### Staff: Mentor

Yes, since they both are moving inertially and since they both see the other as moving in something besides a straight line they both are aware that they are not in flat spacetime and therefore that the Minkowski metric does not apply.

Essentially yes. It is sometimes a bit sketchy to try to split time dilation up into parts and say that this part is kinematic and that part is gravitational, but assuming that they each chose a coordinate system where you could do so then each observer would find that (on average) the kinematic and gravitational time dilation for the other would cancel out so that the proper time would be equal.

15. Sep 11, 2010

### JesseM

I don't understand the question. What does it mean to ask whether a tidal force is "significant" compared to a time dilation? They are completely different quantities, that's like asking whether an electromagnetic charge is significant compared to a velocity. And anyway it doesn't matter if the tidal forces are "significant" as long as they don't approach zero in the limit as the size of the room approaches zero, either the observer in the room can tell whether his situation is different than that of an inertial observer in flat spacetime or he can't, it's not a question of degree.
Corrective factor for what, tidal forces? Why would you need to correct for tidal forces in order to keep the clock ticking at a constant rate in the Earth-centered coordinate system used by GPS?

16. Sep 12, 2010

### yogi

The subject is time dilation in a proposed thought experiment involving no start-up or turn around accelerations - specifically two satellites in orbit, each viewing the other satellite clock as the traveling twin from their perspective in a proper (free fall frame). Dale Spam answered the question for the symmetrical situation. You raised the issue of tidal forces - and my comment is that the tidal forces do not significantly affect the clock rates - i.e., so for purposes of the thought experiment it is valid to treat an orbit as a good inertial frame, irrespective of tidal affects

17. Sep 13, 2010

### JesseM

You can't treat them as inertial frames if they can do a simple experiment which will give a different result than the one they'd get if they were moving inertially in flat spacetime. That proves they can't apply the laws which work in inertial SR frames, like the SR time dilation equation.