Will two clocks be synchronized after a period of time dialation?

[James Brady]

As 2 time pieces are moving relative to each other in a non accelerated reference frame, they will both see the other as slow. As time goes on, the amount of observed error in the other clock will continue to grow greater and greater. Let's say that after a while both clocks are 10 hours ahead of the other according to itself.

Now what happens when the two time pieces enter the same reference frame? What happens to this 10 hours of observed error? I understand the ticking rate will be synchronized, but will the times displayed by the arms on the clock face be the same as well.

This scenario was set up to be perfectly symetrical. So when entering each other's reference frame, they both accelerate an equal amount to do so.

I appreciate any help in clarifying this matter.

 

[Passionflower]

As 2 time pieces are moving relative to each other in a non accelerated reference frame, they will both see the other as slow. As time goes on, the amount of observed error in the other clock will continue to grow greater and greater. Let's say that after a while both clocks are 10 hours ahead of the other according to itself.

Now what happens when the two time pieces enter the same reference frame? What happens to this 10 hours of observed error? I understand the ticking rate will be synchronized, but will the times displayed by the arms on the clock face be the same as well.

This scenario was set up to be perfectly symetrical. So when entering each other's reference frame, they both accelerate an equal amount to do so.

I appreciate any help in clarifying this matter.

If their worldlines are symmetrical then their clocks should show exactly the same time when they compare them.

 

[James Brady]

But when does this catching up occur? If you were to observe it, would it look like the other clock speeding up to match yours?

 

[bcrowell]

But when does this catching up occur? If you were to observe it, would it look like the other clock speeding up to match yours?

If you're an observer accelerating along with one of the clocks, then you have no way of knowing directly what the other clock is doing "now" (according to your own notion of simultaneity); you can only observe it by receiving signals. If they send you a radio signal every second, then you will notice that the signals are more than one second apart as you recede from one another, and less than one second apart as you approach again. You explain this as a combination of time dilation of their clock plus the time taken for the signals to propagate. If the other observer receives signals from you, the situation will be completely symmetric.

 

[James Brady]

Maybe we can solve this by making millions of synchonized clocks attatched on a long beam. As my clock travels past them, parallel to the beam, I can see that reference rate by looking at the closest time piece.

So after traveling for a while along this line and seeing an accumulated 10 hours of error on the other clocks, what happens to them once I stop abruptly and we enter the same reference frame?

 

[pervect]

If you carefully syncrhonize the millinos of clocks in the frame in which they are at rest (you didn't specify the frame, probably because you didn't realize the need - but this seems like the natural choice to me), when you move relative to them, you'll find that they are no longer synchronized.

 

[James Brady]

Huh, so two synchronous events in the same reference frame can be nonsynchronous as well.

I think I need to just go ahead and read up on the topic.

Appreciate all the help.

 

[JesseM]

Maybe we can solve this by making millions of synchonized clocks attatched on a long beam. As my clock travels past them, parallel to the beam, I can see that reference rate by looking at the closest time piece.

So after traveling for a while along this line and seeing an accumulated 10 hours of error on the other clocks, what happens to them once I stop abruptly and we enter the same reference frame?

When you are traveling relative to them, although each one is individually running slow in your rest frame, they are also out-of-sync, with the ones ahead of you being further ahead than the ones behind you. So for example if you are traveling at 0.8c relative to them and the clocks are 100 light-seconds apart in their rest frame, then each successive clock in the row is ahead of the previous one by (0.8c)*(100 light-seconds)/c^2 = 80 seconds (in general if two clocks are synchronized and at rest a distance L from each other in their own rest frame, then in the frame of an observer who sees them moving at speed v they will be out-of-sync by vL/c^2). So, suppose two successive clocks are labeled A and B, and at t=0 in your frame, you are passing clock A and it reads t_A=0 at that moment, then that means at t=0 clock B must read t_B = 80. The clocks will be 0.6*100 = 60 light-seconds apart in your frame due to length contraction (the length contraction/clock slowdown factor is [tex]\sqrt{1 - v^2/c^2}[/tex], so with v=0.8c this works out to 0.60), so it will take 60/0.8c = 75 seconds for you to reach clock B, meaning you will reach it at t=75. During this time each clock has only ticked forward by 0.6*75 = 45 seconds due to time dilation, so at t=75 clock A reads t_A=0+45=45 seconds, while clock B reads t_B=80+45=125 seconds. So in spite of the fact that each clock was individually ticking slower than your own clock in your frame, it's not true that each successive clock you pass is farther and farther behind your own clock; quite the opposite! When your clock read 0 seconds you were passing clock A which read t_A=0, and when your clock read 75 seconds you were passing clock B which read t_B=125 seconds. Similarly if you then passed another clock C when your clock read 2*75=150 seconds, that clock C would read 2*125=250 seconds when you passed it, and so on.

 

[JesseM]

Huh, so two synchronous events in the same reference frame can be nonsynchronous as well.

Events aren't really "in" any reference frame, reference frames are just ways of labeling events with position and time coordinates. Two events labeled with the same time-coordinate in one frame can be labeled with different time-coordinates in another frame, this is the relativity of simultaneity.

 

[LBrandt]

Events aren't really "in" any reference frame, reference frames are just ways of labeling events with position and time coordinates. Two events labeled with the same time-coordinate in one frame can be labeled with different time-coordinates in another frame, this is the relativity of simultaneity.

Why did you say "two events"? Couldn't a single event also be labeled with coordinates in one frame and be labeled with different coordinates in another frame, such that it could be considered to be in two (or more) reference frames?

 

[JesseM]

Why did you say "two events"? Couldn't a single event also be labeled with coordinates in one frame and be labeled with different coordinates in another frame, such that it could be considered to be in two (or more) reference frames?

I meant that two events can be labeled with time-coordinates the same as one another in one frame and different from one another in another frame, so in one frame they are simultaneous and in another they aren't. Of course a single event can be labeled with different coordinates in two different frames, but this isn't very interesting since it has nothing to do with simultaneity (and if you don't require that the origins of your two coordinate systems coincide, the same could be true of inertial frames in Newtonian physics).

 

[LBrandt]

I meant that two events can be labeled with time-coordinates the same as one another in one frame and different from one another in another frame, so in one frame they are simultaneous and in another they aren't. Of course a single event can be labeled with different coordinates in two different frames, but this isn't very interesting since it has nothing to do with simultaneity (and if you don't require that the origins of your two coordinate systems coincide, the same could be true of inertial frames in Newtonian physics).

Ok, I just wanted to clarify that.

Thanks,
Louis

 

[John232]

The main trick to solving the twin paradox is asking which observer did the most acceleration, since it was solved by considering doppler shift when under acceleration. But in this case they both accelerated by the same amount. They should see each others clock speed up by the same amount when they both meet so that the clocks would read the same when they met, if they both in fact accelerate by the same amount upon reaching each other. So each observer would see that the others clock ran slower until they both came to a complete stop where they met.

 

[JesseM]

The main trick to solving the twin paradox is asking which observer did the most acceleration, since it was solved by considering doppler shift when under acceleration.

Actually, if both accelerate between meetings it doesn't matter who did "the most acceleration", what matter is their overall paths through spacetime. For example, here is a nice diagram DrGreg did a while ago showing three observers A, B, and C who start at the same position in 2000, move apart, then later reunite in 2020; you can see that A and B have an identical set of three accelerations (shown in red), but A will be significantly younger when they reunite because his path diverges more from the inertial path C (and this would still be true even if you altered the diagram so that A spent somewhat less time accelerating than B).

 

[John232]

(A) Would have been traveling a longer distance and then would have to have gone under a higher amount of acceleration during the periods in red if the diagram is to scale. It would take more to make that turn traveling at a greater speed and a greater speed to travel a longer distance.

The reason the clock speeds up in the doppler example is because one observer will be able to receive more pulses of light so then it receives it faster than the information is sent. I wouldn't say that an object went under more or less time dilation being in a situation where the amount of doppler shift was altered by the direction of motion, since it has nothing to do with the amount of dilation put on the object itself. It only allows the observer to contuinuely see the clock run a bit faster so that it is at the correct setting when the two observers meet at a location.

 

[John232]

I don't think relativity states that a relative direction of motion itself causes spacetime dialation, there would have to be other factors causeing it in any case.

 

[JesseM]

(A) Would have been traveling a longer distance and then would have to have gone under a higher amount of acceleration during the periods in red if the diagram is to scale. It would take more to make that turn traveling at a greater speed and a greater speed to travel a longer distance.

Nope, the amount of acceleration has nothing to do with distance. The length and magnitude of the acceleration just determine the change in velocity, and A and B both have the same velocity before acceleration, and they also both have the same velocity after acceleration, so their accelerations can be identical. The reason A travels a greater distance is just that A spends a longer time coasting at constant velocity before accelerating.