Relative motion and accumulated time dilation

1. May 11, 2012

coktail

I have a few questions about time dilation and how it accumulates.

1a. Say you have a situation where you have two clocks that move away from each other with symmetrical acceleration and relative motion. From what I understand, each clock would appear to have slowed down relative to the other, and this is not a paradox because each is in their own reference frame. Then the two clocks move back towards each other with symmetrical acceleration and relative motion. How would they compare to each other once they were back in the same reference frame?

1b. Same question as above, but assume there is instant acceleration.

1c. What if instead of moving back towards each other, they were teleported back together?

2. Though time dilation at earthly speeds is negligible in our every day lives, it still occurs. How is it that it does not accumulate over time and lead to small differences in our watches and such? Or, does it, but it is so small as to be unnoticeable? This applies to both gravitational and velocity-based time dilation, but I'm mostly interested in velocity for the sake of this question.

Thanks!

2. May 11, 2012

Whovian

1a.
Not an expert on General Relativity, but since this situation is perfectly symmetric, they'll have the same time. General Relativity is required if you want to see how one observes the other over the course of the experiment.

1b.
Same.

1c.
Invalid, one cannot teleport and time travel would necessarily follow from being able to teleport, which raises some paradoxes. This also raises some issues with Relativity of Simultaneity.

2.
Yes, it does lead to tiny differences, but they are so phenomenally tiny that they're unnoticeable.

Also, the differences are small enough that they're within the improbability range of Quantum Mechanics, so you've got some problems there. And before anyone gets mad at me for trying to combine Quantum Mechanics and General Relativity, note that I'm combining Quantum Mechanics with Special Relativity, which has been done and is very well done.

3. May 11, 2012

A.T.

If the paths are symmetrical in an inertial frame, the accumulated times will be same.

Same

Depends on the teleporter specifications.

Yes.

4. May 11, 2012

coktail

Does this mean that they will both dilate, but by the same amount (relative to a third stationary clock, say), or that neither will dilate? Will they dilate on the way away from each other, and then contract on the way back towards each other?

5. May 11, 2012

Whovian

I think it means that both will dilate, but by the same amount, and they'll both dilate both ways. Then again, a perfect explanation requires GR, which I'm no expert at.

6. May 11, 2012

Mentz114

coktail, the reading on a clock between two events can be calculated from the proper time of the clocks worldline between the events. And, most importantly, this time is the same in any frame. It is absolutely invariant. Time dilation is a coordinate effect and is not directly observable, whereas the proper time is ( you can look at the reading on any clock) and is invariant.

No it does not. It is not helpful to bring GR into this. Differential ageing is completely explained by the proper time as I've stated above.

7. May 11, 2012

Whovian

True, but what if we want to figure out what one of the watches observes? This is a change of reference frames ... which is only valid with *gasps* GR, I thought? Or probably my brain isn't working at its best at the moment.

8. May 11, 2012

Staff: Mentor

9. May 11, 2012

coktail

I was under the impression that time dilation was observable, but only from an external reference frame. For example, for clocks observe (if they have eyes or an observer sitting on top of them) each other dilating, but to each clock its own time remains unchanged.

10. May 11, 2012

Mentz114

A change of coordinates does not require GR. A Lorentz transformation in SR is just a change of coordinates from one frame to another.

The accumulated time on any clock between events is easy to calculate using the line element of Minowski spacetime

c22 = c2dt2 - dx2 - dy2 - dz2

11. May 11, 2012

Whovian

Under no acceleration, yes.

Thanks, DaleSpam, I heard accelerating reference frames are treatable under SR, but not necessarily in the same way. Probably got my information from a not so great source.

12. May 11, 2012

coktail

Ah, so with acceleration clock A would see clock B slow down, while clock B would see clock A speed up? What if the acceleration is symmetrical?

13. May 11, 2012

Mentz114

I think if you watch a moving clock the Doppler effect takes over. You would see a receding clock ticking more slowly, and an approaching clock ticking faster.

Yes, an observer will see no change in his clock.

14. May 11, 2012

ghwellsjr

Although we don't notice any difference in the watches we carry around, time dilation, both gravitational and velocity-based, is noticed by the precise atomic clocks that are the international standards and for the GPS clocks in orbit around the earth. They all run at different rates depending mainly on their elevation. So, for purposes of knowing what time it is, we can't rely on our own precise atomic clocks, if we had them, but instead on a co-ordinated time source like GPS.

15. May 11, 2012

coktail

My impression was that since the speed of light is constant, the only thing the Doppler effect would cause is a red/blue shift...

16. May 11, 2012

coktail

Also, in the Twin Paradox, I don't understand why a change in acceleration as the spaceship twin turns around sets off the asymmetry, but the initial acceleration of the spaceship doest not. Sorry for the side tangent.

17. May 11, 2012

ghwellsjr

Even without gravity effects, if clock B is moving at a constant speed but remains a constant distance from clock A so that it is accelerating in a circle around clock A (think orbit), then clock A will observe clock B to be running at a slower rate and clock B will observe clock A to be running at a faster rate. But this is clearly a non-symmetrical acceleration because only clock A is accelerating.

If you are going to talk about a symmetrical acceleration such that the clocks start out together and follow the same acceleration/velocity profile except in opposite directions and they eventually reunite, then each one will view the other ones clock in the same way and they will end up with the same time on them.

18. May 11, 2012

coktail

Thank you, George.
Given this, would clock and A and B detect discrepancies between their times as they travel, but they would resolved themselves by the time they reunited, or would they stay "in sync" the entire time?

19. May 11, 2012

Mentz114

Any oscillatory process is subject to the Doppler effect, including clocks ticking.

20. May 11, 2012

coktail

But the relativistic Doppler effect just affects the perceived frequency of the light (e.g. red/blue shift), not the time dilation itself, correct?

21. May 11, 2012

yuiop

As they accelerate away from each other, they will each see light from the other as red shifted. As soon as they turn around to return, they will each see the red shift of light from the other transition to an increasing blue shift, with the maximum blue shift occurring after they see the the other turn around. (There is a delay between their own turn around and seeing the other turn around, because of light travel times.) If they allow for the classical Doppler effect, they will conclude that the total elapsed time on the others clock is the same as on their own clock. When they return to their starting point they will see equal time has elapsed on their clocks, but more time has elapsed on a clock that remained at the starting point the whole time. In short, they do not see each other as remaining in sync the whole time but the differences resolve after the round trips. An inertial observer that remains at the starting point will see A and B as being in sync the whole time.

22. May 11, 2012

yuiop

Yes, that is sort of correct. The relativistic Doppler effect includes the classical Doppler effect and time dilation. If you can measure the velocity of an object you can calculate the classical Doppler shift and after allowing for this, deduce the time dilation from the observed relativistic Doppler shift.

[EDIT]It is worth adding that if the clocks send out one second signals, they will be subject to relativistic Doppler shift, but the number of one second signals received will agree with the number of seconds elapsed on the other clock when they are back alongside each other again. In the classical and asymmetrical twins paradox, the stay at home twin receives less time signals than the travelling twin, agreeing with the fact the travelling twin experiences less elapsed time.

Last edited: May 11, 2012
23. May 11, 2012

Mentz114

The perceived rate of a clock will also be affected. The formula for the Doppler shift includes the factor γ (the time dilation).

24. May 11, 2012

ghwellsjr

I think your questions have already been answered by others but let me summarize it this way. If each clock (observer) took a video of the other ones clock, assuming, of course, some great telescopic equipment, or if each clock were sending out a radio signal announcing its current time that could be detected by a receiver traveling with the other clock and the information from that signal could be recorded, then each traveler would record exactly the same "program" through their entire trip and when they reunite, they could play them side by side and they would look identical.

I want to also make clear that Relativistic Doppler affects not just the red/blue shift of the frequency of the light, it also affects the actual images such that if you could see the hands move on an analog clock or the numbers displayed on a digital clock, you would see them progressing slower while traveling away and faster while returning. But it involves the motion of both the source and the receiver with the time delay determined by their distance apart.

25. May 11, 2012

coktail

So the relativistic doppler effect *includes* time dilation, and is not a separate cause for seeing the image of the clock slow down/speed up? I just want to make sure I understand that point correctly.