1. Hi,

This may have been asked before. I'm pretty new here. It's just bugging me.

If 2 people move away from each other at high speed and each perceives the other's time as moving slower, in layman's terms, what happens when both of them stop moving away from each other? (This is slightly different from the "one twin goes to space and comes back" question)

Should this be in the physics section?

2.

3. If two people stop moving away from each other, then they are at rest in relation to each other, so there is no time-dilation. Each will calculate the others clock to be running at the same speed as their own clock.

4. Originally Posted by SpeedFreek
If two people stop moving away from each other, then they are at rest in relation to each other, so there is no time-dilation. Each will calculate the others clock to be running at the same speed as their own clock.
I agree, provided this is the first stop. It gets more interesting after the two start back. They still each see the other aging more slowly than themselves. So when they get to the starting point and stop again, which is younger? They are in the same inertial frame now, so the question is a valid one. To make the situation really interesting, how old are they relative to an observer who never left the spot?

I am (rather) sure that I know the answer, but I am going to be nasty and not post my argument for a while. To be fair, I promise to post what I thought even if someone proves me wrong. However, I don't think that will happen. If no one posts, I will consider that a general judgement that I should not have withheld my argument and post it for general discussion. I do hope that someone will post more detail than I have available at the moment.

5. Originally Posted by mvb
Originally Posted by SpeedFreek
If two people stop moving away from each other, then they are at rest in relation to each other, so there is no time-dilation. Each will calculate the others clock to be running at the same speed as their own clock.
I agree, provided this is the first stop. It gets more interesting after the two start back. They still each see the other aging more slowly than themselves. So when they get to the starting point and stop again, which is younger? They are in the same inertial frame now, so the question is a valid one. To make the situation really interesting, how old are they relative to an observer who never left the spot?

I am (rather) sure that I know the answer, but I am going to be nasty and not post my argument for a while. To be fair, I promise to post what I though even if someone proves me wrong. However, I don't think that will happen. If no one posts, I will consider that a general judgement that I should not have withheld my argument and post it for general discussion. I do hope that someone will post more detail than I have available at the moment.
It depends, which one was born first?

6. Originally Posted by shlunka
It depends, which one was born first?
Let's assume they were born at the same time!

7. Originally Posted by mvb
Originally Posted by SpeedFreek
If two people stop moving away from each other, then they are at rest in relation to each other, so there is no time-dilation. Each will calculate the others clock to be running at the same speed as their own clock.
I agree, provided this is the first stop. It gets more interesting after the two start back. They still each see the other aging more slowly than themselves. So when they get to the starting point and stop again, which is younger? They are in the same inertial frame now, so the question is a valid one.
Well firstly I would like to make it clear that what each sees is different from what they calculate, due to Doppler effect.

Next I would point out that I never addressed whether they would calculate the others clock to be slower than their own whilst they were moving, or not. I only addressed the question in the OP - what each would calculate for the other when they stop moving away from each other (and thus are at rest in relation to each other and are therefore in the same frame of reference)! :P Sneaky, I know.

So now I will address this. If they both know that they accelerated away from each other by the same amount, then they will calculate that there is no time-dilation between them at all! They both know they are moving and have changed their frame of reference (they each will have felt the acceleration), so neither can consider himself at rest.

If both people are moving away from each other, then stop, then move back towards each other, and they were the same age to begin with, they will be the same age when they meet up again.

Originally Posted by mvb
To make the situation really interesting, how old are they relative to an observer who never left the spot?
Assuming all three started out the same age, the two people who moved away and then came back will be younger than the observer who never left the spot, by the same amount.

The observer who never left the spot will calculate that each of the moving people was time-dilated by the same amount in relation to himself.

The two moving people, knowing they each accelerated by the same amount away from each other, have to base their calculations in an inertial frame, and the only inertial frame here is that of the observer who never left the spot.

In order to get the correct answer, you have to base your calculations in an inertial frame, and know the history of the accelerations involved. This is how we resolve the "paradox of the twins".

8. SpeedFreek has given the argument I had in mind, but he worded it better. Thank you, SpeedFreek.

The only thing that I might add is the symmetry of the problem. The two moving people have moved in a manner that has symmetric accelerations, both required accelerations directly opposite each other. Therefore any answer other than their seeing each other as being the same age would be suspicious.

9. Actually, the question in the OP might be more interesting if we leave out the accelerations, and indeed we might not have answered the question properly, depending on what the OP meant. The way Meowaffles put it, "If 2 people move away from each other at high speed and each perceives the other's time as moving slower..."

So, lets assume we start off with 2 people in inertial frames in fast relative motion. They are coasting along through space, and feel no accelerations. Each can consider themselves to be at rest, and that the other is in motion in relation to themselves.

For the sake of argument, let's say that they each wake up from a long sleep and find themselves in this situation, so they have no record of any previous accelerations. They are just about to pass each other at close range, so they communicate by radio and arrange to synchronise their clocks just as they pass, at closest approach.

Now we have two people performing an experiment which they start when they are next to each other, both having the same reading on their clocks just as they pass each other. They are then moving away from each other at high speed...

In this scenario, each person can measure the speed of the other relative to themselves and will calculate that the others clock is "running slower" than their own clock, just as Meowaffles put it.

The interesting thing here is that both of them will indeed be correct about the time-dilation of the other, since the start of the experiment, and no experiment either can perform can prove otherwise.

The question we have to ask now is how do they come to a stop, relative to each other? Do both accelerate (in a direction opposite to the motion between them) by the same amount, or not, in order to cancel out the motion between them? Remember, in this scenario, whilst there is motion between them, each of them can consider themselves to be at rest and that it is the other is that moving...

10. Originally Posted by SpeedFreek
Actually, the question in the OP might be more interesting if we leave out the accelerations, and indeed we might not have answered the question properly, depending on what the OP meant. The way Meowaffles put it, "If 2 people move away from each other at high speed and each perceives the other's time as moving slower..."

So, lets assume we start off with 2 people in inertial frames in fast relative motion. They are coasting along through space, and feel no accelerations. Each can consider themselves to be at rest, and that the other is in motion in relation to themselves.

For the sake of argument, let's say that they each wake up from a long sleep and find themselves in this situation, so they have no record of any previous accelerations. They are just about to pass each other at close range, so they communicate by radio and arrange to synchronise their clocks just as they pass, at closest approach.

Now we have two people performing an experiment which they start when they are next to each other, both having the same reading on their clocks just as they pass each other. They are then moving away from each other at high speed...
You can analyze the usual twin problem this way as well, omitting the initial acceleration by starting with the second twin already moving. It is generally agreed, although essentially never carefully analyzed, that the change makes no difference. In this problem, that presumption means that neither of the two necessary comparisons with the initial inertial frame change anything, and so this problem still has a symmetry between the twins. With the symmetry still there, the original answer follows.

The usual qualitative arguments suggest that only accelerations at different locations produce differences in the resulting ages of the twins. This is certainly a reasonable outcome. The wikipedia article on the twin paradox has a qualitative analysis of simultaneity in the (three) inertial frames involved in the conventional twin paradox. That analysis looks like it might lead to an analysis that might be more instructive here than the symmetry relation is. I haven't tried to construct the argument, however. I may or may not know enough to do so, since General Relativity is involved when you explicitly analyze the accelerations.

Would all this discussion, and presumably there is more to come, read more easily if we called this the triplet paradox, so that we always had in mind two moving people and one that stayed home? At least we could keep track of when we meant the twin situation and when we had two different people going out and back.

11. Well, we could call our first 2 inertial observers Alice and Bob. We could introduce inertial Carl too if we like, but I'm not sure if Carl will help or not!

So, Carl has large screens on the outside his spaceship, showing the time. Alice and Bob approach him at the same speed from opposite directions and pass him at the same time, each taking the reading from Carl's clock as they pass and setting their own clocks to match. Then they both recede from Carl in opposite directions.

After they pass and synchronise their clocks:

To Carl, who considers himself at rest, both Alice and Bob's clocks will be running slow by the same amount as they recede. Each will be time dilated by the same factor in relation to Carl.

But to Alice, who considers herself at rest, Bob will be receding at twice the speed of Carl. Carl's clock will be running slow, and Bob's clock will be running even slower.

And to Bob, who considers himself at rest, Alice will be receding at twice the speed of Carl. Carl's clock will be running slow, and Alices's clock will be running even slower.

There are many ways to resolve this, and it all depends on what happens next.

If either Alice or Bob turn round and come back to Carl, their clock will show less elapsed time than Carl's clock. If both turn round and come back again, then both their clocks will show less elapsed time than Carl's clock. Their accelerations change their frames of reference, whilst Carl's frame of reference never changes.

But what if Carl decides to accelerate to Alice, while Bob turns round and accelerates even more, so both Carl and Bob reach Alice at the same time? This puts Alice in the position of remaining in her frame of reference and thus she can consider herself to be stationary.

Or what if we want to do away with any accelerations at all? What if we introduce inertial Doug and Eric, both heading towards Carl! They can take the readings from Alice and Bobs clocks as they pass them, and bring them back to Carl! What would happen then?

12. In answer to SpeedFreek's multiple scenarios, which are leaving my head spinning trying to watch Alice and Bob from Carl's position:
It sounds like we really need to know what happens at the acceleration points, rather than arguing from symmetry, if we are going to understand all the scenarios that we can dream up. I'll work on that -- I know that it is available somewhere and I think I worked it out once -- but it may take a while. Maybe somebody else knows?

13. Yes, it can certainly get complicated.

14. Originally Posted by SpeedFreek
Well, we could call our first 2 inertial observers Alice and Bob. We could introduce inertial Carl too if we like, but I'm not sure if Carl will help or not!

So, Carl has large screens on the outside his spaceship, showing the time. Alice and Bob approach him at the same speed from opposite directions and pass him at the same time, each taking the reading from Carl's clock as they pass and setting their own clocks to match. Then they both recede from Carl in opposite directions.

After they pass and synchronise their clocks:

To Carl, who considers himself at rest, both Alice and Bob's clocks will be running slow by the same amount as they recede. Each will be time dilated by the same factor in relation to Carl.

But to Alice, who considers herself at rest, Bob will be receding at twice the speed of Carl. Carl's clock will be running slow, and Bob's clock will be running even slower.

And to Bob, who considers himself at rest, Alice will be receding at twice the speed of Carl. Carl's clock will be running slow, and Alices's clock will be running even slower.
Not quite. Bob will not be receding at twice the speed as Carl for Alice, nor will Alice be receding at twice the speed as Carl for Bob.
The relativistic addition of velocities has to be taken into account. So for instance if Alic and Bob have velocities of 0.6 c Relative to Carl. Bob will being moving at 0.9756c relative to Alice according to Alice.

There are many ways to resolve this, and it all depends on what happens next.

If either Alice or Bob turn round and come back to Carl, their clock will show less elapsed time than Carl's clock. If both turn round and come back again, then both their clocks will show less elapsed time than Carl's clock. Their accelerations change their frames of reference, whilst Carl's frame of reference never changes.

But what if Carl decides to accelerate to Alice, while Bob turns round and accelerates even more, so both Carl and Bob reach Alice at the same time? This puts Alice in the position of remaining in her frame of reference and thus she can consider herself to be stationary.

Or what if we want to do away with any accelerations at all? What if we introduce inertial Doug and Eric, both heading towards Carl! They can take the readings from Alice and Bobs clocks as they pass them, and bring them back to Carl! What would happen then?
Then you have to take into account the Relativity of Simultaneity. According to Doug and Eric, as they pass Alice or Bob, Carl's clock already reads ahead of the clock of whoever they are passing.

15. Thank you, Janus.

I completely forgot about the relativistic addition of velocities in the last two of those scenarios, so Bob will not be receding from Alice at twice the speed of Carl, but as you said he will still be receding from Alice faster than Carl is, and so would still be time-dilated by a greater amount. It sure can make your head spin, trying to work it all out.

16. After looking around the internet trying to remind myself of how I had understood the twin paradox, I found a decent reference,
http://home.sprynet.com/~owl1/twinparadox.pdf, which is well worth taking a look at. In the effort to be thorough, it gets a little complex in places, so I will give my explanation as well.

First, the periods of acceleration do nothing significant to the amount of time on the clocks, as long as the periods are short. General Relativity doesn't do anything to the clocks during uniform acceleration. So we have twins A and S, with A accelerating quickly to an arbitrary speed and S staying put. Then there is a target point [ ] a distance X away, in the S frame, where twin A decelerates, turns around, and accelerates in the opposite direction back towards S. We need to know if A and S agree about the amount of time built up on their own and the other's clock.

The calculation is easy for S. His clock, corrected for the delay of the light showing A's arrival at , gives a travel time of
TS = X/vA
both coming and going, so his clock ticks off
2 TS = 2 X/vA
total. He calculates that A's clock is slowed down by a factor of gamma each way, so that clock shows a time reduced by 1/gamma, or
2TA = (2 X/ vA) / gamma.
Now A measures his clock running at its regular rate, so he gets no factor of gamma from the rate of passage of time. However, the target is coming at him (first leg) or away from him (second leg) at speed vA . So S measures the target to be originally at a distance x/gamma (and on the following leg sees S starting at that distance), so his travel time is
2TA = 2 (X/gamma) / vA .

So the end result is that one twin gets a factor of gamma from the time dilation applied to the other's clock, and that twin gets a factor of gamma from the length contraction of a moving object. Physics is saved.

It seems likely to me that more complicated situations are easily handled a step at a time by this prescription.

17. Yes indeed, that paper is really rather good. Acceleration is not the cause of the difference between the ages of the twins, as acceleration has no effect on time-dilation.

What acceleration does have an effect on is the relativity of simultaneity, which is one way to look at the situation, because the larger the distance where a change in reference frame takes place, the larger the shift in simultaneity across that distance. In my examples above, the transfer of a clock reading from one frame of reference to another is a change in frame of reference.

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