1. Aug 5, 2012

### cragar

Ok I have attached a picture of what I am talking about.
A, B, C, represent people with clocks. And A is at rest with respect to the middle of the planet.
A, B , C all synchronize their clocks . Then B and C start to orbit around the planet with the same speed but in opposite directions. then they will meet on the other side and then eventually meet back where A is. in A's frame wont B and C's clock be dilated the same amount. So A would see B and C's clock reading the same time. In B's frame wouldn't he see
C's clock to be running different than his. Ok now my question is when B and C go around the planet once and meet up back with A. If A see's B and C's clock to be the same, would B see C's clock to be different?

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2. Aug 5, 2012

### bcrowell

Staff Emeritus
It sounds like you're basically describing a variation of the Hafele-Keating experiment. Unless I'm missing something, the answer to your question is yes. At the end, B and C are at the same event, so they agree on simultaneity of that event. They can unambiguously compare their clocks, and must agree on whether the clocks match. By symmetry, the two clocks must match.

3. Aug 5, 2012

### cragar

why would the clocks match if they are moving with respect to one another?

4. Aug 5, 2012

### bcrowell

Staff Emeritus
There is no general rule stating that clocks in motion relative to one another must disagree, regardless of what the motion is.

5. Aug 6, 2012

### Mike Holland

The rotation around the Earh complicates the picture. Rather think about triplets, two of whom travel away from Earth in opposite directions, and then return to compare clocks. While they are in relative motion, each sees the other two time dilated and their clocks running slowly. As they return, the same applies. When they return, the travellers clocks will agree, but they will both be behind the stay-at-home triplet's clock. How come? Because the accelerations to reverse direction and return have a reverse effect on their clocks so that each sees the other clocks running fast during this period.

Your picture is complicated because they are accelerating all the time to counter Earth's gravity and to travel in an orbit. They are travelling away from each other and then towards each other every revolution. But when they stop to compare notes their deceleration balances the books and their clocks agree.

Mike

6. Aug 7, 2012

### jbriggs444

That is a reasonable summary. However the deceleration when they stop to compare notes has absolutely no effect on the "books".

The effect of acceleration manifests as a changing standard of simultaneity. Multiply acceleration by separation and you get a rate of change of clock readings over there, "right now". If the separation is nil, the effect of acceleration is nil.

7. Aug 7, 2012

### Mike Holland

OK, I stand corrected on that detail. I know that it is the acceleration that resolves the space-travelling twin paradox, and blithely assumed that the same appied in this case.
Thanks for the correction.
Mike

8. Aug 7, 2012

### ghwellsjr

Acceleration only resolves which twin ages less if only one twin accelerates. But what do you do if both twins experience exactly the same acceleration? How do you resolve it then?

For example, suppose that some time after the first twin accelerates away, the second twin follows after the first one with exactly the same acceleration but then after a short time turns around and eventually decelerates to rest at the original starting point. Then the first twin turns around in the same way as the second one did and also decelerates identically to rest with the second twin. Now how do you resolve the age issue? Are they the same age since they accelerated identically?

Or suppose their accelerations are not identical but that the second twin takes off with a higher acceleration after the first one, then turns around more quickly and comes to rest with a higher deceleration than the first one will. Does that mean the second one will be younger? How does acceleration resolve the twin paradox in this situation?

9. Aug 7, 2012

### GrammawSally

A++!

10. Aug 7, 2012

### phyti

Imagine the earth moving in the direction of its polar axis. A moves at a constant rate along with the earth center. B & C move in a spiral path centered on the polar axis, but at a constant speed greater than that of A. Same path, same time, more time dilation for B & C.
Less time dilation for A.

11. Aug 7, 2012

### ghwellsjr

Because of what you said here:
Two clocks traveling at the same speed in an inertial frame, no matter what their directions are or how their directions change, even if they are not symmetrical, will be time dilated by exactly the same amount so if they start out with synchronized clocks, they will remain synchronized and therefore will have the same time on them every time they meet up again. All other frames will agree that they will display the same time each time they meet but they may not agree that they display the same time when they are separated.

So, for example, in a non-inertial frame in which clock B is at rest, the time dilation of clock C will be fluctuating all over the place, even becoming time contraction during parts of the orbit but with the net result that every time they rejoin, their two clocks display the same time.

Now if two clocks that start out synchronized and travel at the same speed in one frame but never meet up again, then there are other frames in which their time dilations result in them remaining forever desynchronized, including each others rest frames.

12. Aug 7, 2012

### ghwellsjr

I'm not sure you have this correct and I'm not talking about the deceleration when they stop. You state that while they are traveling away from Earth, "each sees the other two time dilated and their clocks running slowly" and when they are traveling toward the Earth, "each sees the other clocks running fast". Why didn't you also say during the return that "each sees the other two time dilated"? Why did you only say that about the outbound portion of the trip?

13. Aug 8, 2012

### Mike Holland

Sorry, that's not correct. I did say that during the return trip, the same applies. In other words, as long as they are travelling at uniform speed relative to each other they will see each others clocks running slow/time dilated, irrespective of direction.

The only time they see each other's clocks running fast is during acceleration/deceleration. During departure and arrival, when they are close together, the effect is negligible. It is the turnaround when they are far apart that counts. As the stay-at-home guy does not experience accceleration, he does not see the other guy's clocks speed up, but they do see his and each other's clocks speed up.

Mike

14. Aug 8, 2012

### ghwellsjr

Thanks for clarifying. But I would like to ask you what you mean by the word "see". Do you mean literally with their eyes or do you really mean something along the lines of "interpret" or "understand"?

15. Aug 8, 2012

### rickylritter

This is really a simple issue. As long as both clocks are traveling at the same speed for the same amount of time (acceleration and deceleration are irrelevant) they will agree with one another. The clock that has not moved in relationship to the traveling twins will have not changed and thus in relationship to the twins the times will not agree. The reason being time progresses more slowly with change in relationship to position (motion). Thus the twins will have experienced less time in comparison to the stationary clock. The earth may have an effect on the motion; however, as long as the two objects travel the same distance and speed the effect is irrelevant in relationship to the two travelers. If the stationary object truly does not move in relationship to the two moving bodies then their will be no effect by the presence of the Earth. The Earth is irrelevant to the problem all together.

Last edited: Aug 8, 2012
16. Aug 8, 2012

### Mike Holland

It is much more complicated than that. They may be travelling at the same speed (relative to the stay-at-home guy, but if they have some velocity relative to each other then each will see the other's clock running slow, whether they are receding from each other or approaching. This is the relative part of Special Relativity.

George, I am thinking in terms of a space-time diagram that an observer would draw up, knowing SR and GR, to determine times and distances in his time-frame. Actual observations get more complicated because of the time light takes to travel.

Mike

17. Aug 8, 2012

### rickylritter

If their motion over time or change relative position is the same (mirrored) then their clocks will be the same.

18. Aug 8, 2012

### ghwellsjr

The relative part of SR states that we can analyze any scenario in any frame we choose, none is preferred over any other, including frames in which different observers are at rest. So if I choose a frame in which the planet is at rest (because I like simple solutions) you cannot claim that I'm not done and need to do something more complicated.
There is nothing wrong with correctly drawing up multiple space-time diagrams, as long as you realize that they are adding nothing to the analysis, especially that they are not associated with any "actual observations" of any observer. It's nothing more than a fun exercise for you (not for me) and is in no way a requirement of relativity.

19. Aug 8, 2012

### Mike Holland

OK, I made an unstated assumption that each participant in this example would draw up a space-time diagram in which he is at rest, and relate the other's clocks to his diagram.

I don't understand the "not associated with any actual observations" bit. But we normally choose space-time doiagrams to suit the problem and simplify the mathematics. We use a heliocentric frame to work out Mercury's precession, and the results do match with "actual observations". We use a geocentric system when discussing time dilation on Earth relative to satellites, and again get figures that are associated with actual observations. So I don't understand what you are getting at with that comment.

Going back to the original problem of two triplets A and B going on symmetrical journeys in opposite directions while C stays at home, C will see their clocks agreeing all the time, but retarded relative to his. A and B in turn will both see C's clock retarded, and each other's clocks even more retarded as their relative velocity is greater (Relativistic addition of velocities applies, so their relative velocity is less than double their velocity relative to C).

For simplicity, we assume they accelerate, travel at constant velocity for a long time, turn around and return at the same velocity. Assume a velocity that gives a 50% time dilation - I think the figure is around 0.84c. then after 4 years travelling on C's clock, A and B turn around. But their clocks read 2 years (they have travelled half the distand C sees, because of Lorentz contraction of the distance), and they see C's clock reading 1 year - because the velocity and time dilation is relative. So on the return journey they should see C's clock advance another year, and read 2, but when they get there it reads 8! And this is the SR twin paradox.

The paradox is resolved because A and B change direction, and as thery do so they see C's clock accelerate and advance from 1 to 7 years. As ibriggs444 says, the time change is based on acceleration times distance. But the calculation for acceleratiion is a bit too complicated for me, and I like to use Einstein's equivalence principle and look at it from a gravity point of view.

Picture A in his spaceship, now turned to point back to C. He switches his rocket engines on to decelerate and then accelerate back to C. From his point of view, he is motionless and experiences a gravitation acceleration pulling him down to his rockets floorboards, and he sees this same gravity slow down C's recession from him and C starts falling towards him.
Because he and C are now in a gravitational field, and C is much higher, see sees C's clock run faster than his own, just as we on Earth see clocks in orbit above us run faster.

Please note that in all this discussion I have ignored the time for light signals to pass between them, so what they "see" is what they would calculate allowing for this time, or what their space-time diagram would indicate if they drew a straight line across it to denote their "present".

And please accept that my responses on this forum might be time-retarded becauase I live in a different time-frame - Australia!

Mike

20. Aug 9, 2012

### ghwellsjr

We agree that we like to analyze a scenario in a way that simplifies the mathematics. But my point is that after doing that, if someone says we have to do it over again with more complex mathematics, we can tell that person he is all wet. (Do you talk that way in Australia? If not, we can tell the person he doesn't know what he is talking about.) Simply repeating the analysis in different frames does not provide any new knowledge of what is actually happening in the scenario. The parts of the analysis that are different simply because we are using a different reference frame are not "actual observations" because they are not "actual" and they are not "observations". I'll discuss this more at the end.
That may be one way to resolve the paradox but it's not the only way, not even for a frame in which an observer is at rest.
And that's another way.
I hope you are aware that the time it takes for light to travel is what defines a Frame of Reference. Different frames assign different light transit times resulting in different conclusions about what is the "present". This is why I said that different reference frames (or different space-time diagrams) are not associated with any actual observations, rather we get out exactly what our assumptions put in.

21. Aug 9, 2012

### Mike Holland

OK, I think we agree about what goes on, just expressing it differently. My knowledge ibased on reading popular texts, I have not done any courses on GR or SR.
I just hope Cragar gained something from our discussion, and isn't more confused than ever!

Mike

22. Aug 9, 2012

### ghwellsjr

If you hang around here, it'll be just like taking courses on relativity, in fact, better, in my opinion.