Symmetrical acceleration of Synchronised clocks

[james fairclear]

TL;DR

If 2 synchronised clocks at the same location in space (away from any gravitational field) are equally accelerated away from each other and then equally accelerated back together to the original location will they then indicate the same time?

If 2 synchronised clocks at the same location in space (away from any gravitational field) are equally accelerated away from each other and then equally accelerated back together to the original location will they then indicate the same time?

 

[Ibix]

What do you mean by "equally accelerated"? Equal proper acceleration or equal coordinate acceleration? And are both clocks accelerated for the same time as they measure it?

What is the initial velocity of one clock with respect to the other?

Assuming that the answers to these questions are "proper acceleration for equal times" and "they are initially at rest with respect to one another" then you have a scenario where the exact same things happen to two clocks in the exact same order. What do you think the answer will be?

 

[Histspec]

If 2 synchronised clocks at the same location in space (away from any gravitational field) are equally accelerated away from each other and then equally accelerated back together to the original location will they then indicate the same time?

Yes. Let's say that both start at the same time with same acceleration profile (i.e. same proper acceleration in opposite directions and same proper times). So they accelerate away from each other from t=0 to t=5 (with t being the times measured on the rockets), then reverse acceleration from t=5 to t=15, and then again reverse acceleration from t=15 to t=20 until they come to rest again, then both will have experienced 20 seconds in their rockets.
From the thin lines of simultaneity we also notice how their views of simultaneity changed during acceleration, leading to apparent "jumps" in time which they mutually attribute to one another....

 

[james fairclear]

Thank you for your response.

In my view the complete acceleration history of the clocks is required in order to predict their respective indicated times after any subsequent acceleration.

Consider a clock A in deep space with nothing around for reference. The clock is then accelerated in a given direction d. The tick rate of the clock is found to have increased by a factor n.

Clock A is then accelerated equally again in the same direction d. The tick rate of the clock is found to have increased further by a factor m.

Clock A is then accelerated equally in the opposite direction d'. The tick rate of the clock is found to have decreased by a predictable factor m.

A second clock B synchronised with A, stationary with A and at the same location as A is accelerated equally in the direction d' while clock A is equally accelerated in the direction d.

We can predict that clock B is now ticking at a higher rate than Clock A and also indicates a later time than clock A.

Clock A and clock B are each then accelerated equally in the opposite direction until they are again stationary with each other and in the same location.

We can predict that clock B is now ticking at the same rate as Clock A but indicating a later time than clock A.

In summary

2 synchronised clocks equally accelerated in opposite directions and then accelerated equally in the opposite direction until they are stationary with each other in the same location will at the end of the experiment tick at the same rate but not necessarily indicate the same time.

 

[PeterDonis]

The tick rate of the clock is found to have increased

Relative to what?

 

[PeterDonis]

In my view

Your view is not well-defined, because you are failing to realize that many of the quantities you are dealing with are frame-dependent, and you are not specifying a frame. Further, you are assuming that there is some invariant way to compare the "tick rates" of spatially separated clocks, and there isn't. So the questions you are posing are not answerable because you have not specified a well-defined scenario to start with.

Notice, by contrast, that @Histspec specified a scenario purely in terms of invariants--proper accelerations and proper times--and obtained an answer that contradicts your claim here:

2 synchronised clocks equally accelerated in opposite directions and then accelerated equally in the opposite direction until they are stationary with each other in the same location will at the end of the experiment tick at the same rate but not necessarily indicate the same time.

 

[Ibix]

2 synchronised clocks equally accelerated in opposite directions and then accelerated equally in the opposite direction until they are stationary with each other in the same location will at the end of the experiment tick at the same rate but not necessarily indicate the same time.

Assuming that the clocks are initially co-located (which you didn't specify) and have equal and opposite speeds in the frame where they will eventually come to rest (which you didn't specify) and assuming that "equally accelerated" means the same acceleration sequence but not necessarily the same timing between accelerations (which you didn't specify) then this is true.

If they aren't initially co-located then you don't have a unique way to define whether or not they are synchronised.

If they don't have equal and opposite velocities then they won't both come to rest at the end of the experiment.

If "equally accelerated" means that they have equal periods when their acceleration is zero (which is the strict interpretation of the phrase) then they do show the same time.

As Peter points out, your scenario is sloppily specified and the truth or not of your claim depends on what assumptions are made to fill in the blanks.

 

[james fairclear]

you are assuming that there is some invariant way to compare the "tick rates" of spatially separated clocks

Relative to what?

The tick rate of clock A has increased from what it was prior to acceleration.

 

[Ibix]

The tick rate of clock A has increased from what it was prior to acceleration.

Measured by whom? Some frames will say it has increased, others will say it has decreased. It depends whether the clock's speed decreased or increased in the frame of interest.

 

[james fairclear]

Assuming that the clocks are initially co-located (which you didn't specify) and have equal and opposite speeds in the frame where they will eventually come to rest (which you didn't specify) and assuming that "equally accelerated" means the same acceleration sequence but not necessarily the same timing between accelerations (which you didn't specify) then this is true.

If they aren't initially co-located then you don't have a unique way to define whether or not they are synchronised.

If they don't have equal and opposite velocities then they won't both come to rest at the end of the experiment.

If "equally accelerated" means that they have equal periods when their acceleration is zero (which is the strict interpretation of the phrase) then they do show the same time.

As Peter points out, your scenario is sloppily specified and the truth or not of your claim depends on what assumptions are made to fill in the blanks.

The clocks are initially co-located at rest with each other. The acceleration is identical and symmetrical such that the clocks relative motion is symmetrical and that they come to rest with each other at the end of the experiment.

 

[Ibix]

The clocks are initially co-located at rest with each other. The acceleration is identical and symmetrical such that the clocks relative motion is symmetrical and that they come to rest with each other at the end of the experiment.

Then the answer is that the clocks show the same time. This is an obvious result in the frame where they are initially and finally at rest, since their speeds are always equal in this frame and hence so are their instantaneous $\gamma$ factors. And since comparing co-located clocks is a direct observable, the same result will be true regardless of frame (although you may have to work a bit harder to show that if you insist on doing so in another frame).

 

[james fairclear]

Measured by whom? Some frames will say it has increased, others will say it has decreased. It depends whether the clock's speed decreased or increased in the frame of interest.

All non accelerating observers will agree that the accelerated clock is recording increasingly fewer events. Its indicated time will be increasingly less than the indicated time on any non accelerating observer's clock.

 

[Ibix]

All non accelerating observers will agree that the accelerated clock is recording increasingly fewer events.

This is not correct and I have told you why at least twice, including in the post you quoted.

 

[PeterDonis]

The tick rate of clock A has increased from what it was prior to acceleration.

Tick rate relative to what? Clock A itself sees its own tick rate as always the same. Relative to what has it increased?

 

[PeterDonis]

The acceleration is identical and symmetrical such that the clocks relative motion is symmetrical and that they come to rest with each other at the end of the experiment.

You appear to be describing the scenario that @Histspec drew a diagram of. However, as I pointed out in post #6, your claim (which I quoted in that post) is wrong for that scenario.

 

[james fairclear]

Then the answer is that the clocks show the same time. This is an obvious result in the frame where they are initially and finally at rest, since their speeds are always equal in this frame and hence so are their instantaneous $\gamma$ factors. And since comparing co-located clocks is a direct observable, the same result will be true regardless of frame (although you may have to work a bit harder to show that if you insist on doing so in another frame).

My point is that although their clocks will tick at the same rate at the end of the experiment their indicated times may differ. We need to know the prior acceleration history of the clocks. If prior to the experiment they have both been accelerated in a given direction d (which would have affected their tick rates symmetrically) and then during the experiment clock A is further accelerated in direction d while clock B is further accelerated in the opposite direction d' then Clock B will record more events than clock A such that when they are brought together at the end of the experiment their tick rates are once again identical but clock B displays a later time.

 

[Ibix]

My point is that although their clocks will tick at the same rate at the end of the experiment their indicated times may differ.

Not if their accelerations are symmetrical and they were initially synchronised, which you said was the case. As I noted, this is trivial to see in the frame where they are initially and finally at rest.

If prior to the experiment they have both been accelerated in a given direction d (which would have affected their tick rates symmetrically) and then during the experiment clock A is further accelerated in direction d while clock B is further accelerated in the opposite direction d' then Clock B will record more events than clock A such that when they are brought together at the end of the experiment their tick rates are once again identical but clock B displays a later time.

This is just wrong. If they are initially at rest with respect to one another you are free to work in the frame where they are at rest and it is obvious that their experiences are symmetrical so their final times are symmetrical.

You appear to be supposing that there's some absolute sense in which the clocks were at rest in the past and at the start of the experiment they're actually "really" moving. This is not correct - the whole point of relativity is that all inertial frames are equivalent and all are equally good as a definition of "at rest". So the past history of the clocks is utterly irrelevant if they are synchronised at the start of the experiment.

 

[james fairclear]

Tick rate relative to what? Clock A itself sees its own tick rate as always the same. Relative to what has it increased?

What do you mean by "Clock A itself sees its own tick rate as always the same"?

 

[PeterDonis]

What do you mean by "Clock A itself sees its own tick rate as always the same"?

I mean that if all you are looking at is Clock A, there is no basis whatever for saying that its tick rate increases or decreases. The only way to make any such statement well-defined at all is to compare Clock A with something else.

 

[Ibix]

What do you mean by "Clock A itself sees its own tick rate as always the same"?

Clock A's rate divided by clock A's rate is always one, tautologically so. But you have not provided any other standard to which to compare it, so Peter is pointing out a way in which your experiment is ill-defined. I think you think it's OK because you think that the rate of the clock decreases in all frames, but this is not correct. So you need to specify what time standard you are comparing clock A's tick rate to - some inertial frame's time coordinate, for example.

 

[james fairclear]

Not if their accelerations are symmetrical and they were initially synchronised, which you said was the case. As I noted, this is trivial to see in the frame where they are initially and finally at rest.

This is just wrong. If they are initially at rest with respect to one another you are free to work in the frame where they are at rest and it is obvious that their experiences are symmetrical so their final times are symmetrical.

You appear to be supposing that there's some absolute sense in which the clocks were at rest in the past and at the start of the experiment they're actually "really" moving. This is not correct - the whole point of relativity is that all inertial frames are equivalent and all are equally good as a definition of "at rest". So the past history of the clocks is utterly irrelevant if they are synchronised at the start of the experiment.

If we accelerate a clock in direction d it will progressively record less events than it recorded prior to the acceleration. If we then accelerate the same clock in the reverse direction d' it will progressively record more events than it did prior to the acceleration. This is evidenced by comparing clocks that have been variously accelerated all indicating different times at the end of an experiment. Do you not agree with this?

 

[PeterDonis]

Do you not agree with this?

No. You are simply repeating the same wrong statements you have already made. They are not going to get any less wrong by repetition.

This thread is going in circles and is now closed.