Is it possible to have synchronized clocks in a rotating system?

[Foppe Hoekstra]

Only before the train enters the circle. Once parts of the train start accelerating, it not even clear what the "moving frame" is. But whatever frame you construct for the train on the circle, it will have to conclude that the clocks have an offset when they meet, for this simple reason:

They are initially not in sync in the rest frame, and their offset doesn't change until they meet.

This looks to me like circular reasoning.
And the reason why I do not start at the rest frame is because the outcome of that is already clear to me. I just think it should agree with reasoning that starts at the moving frame. But strangely it doesn't.

 

[A.T.]

I just think it should agree with reasoning that starts at the moving frame. But strangely it doesn't.

You haven't presented any analysis of the whole process from the "moving frame". In fact, you haven't even properly defined the "moving frame" for the whole process.

 

[jbriggs444]

I just think it should agree with reasoning that starts at the moving frame. But strangely it doesn't.

It's an accelerating frame. That is a clue that it needs to be analyzed more carefully than "two clocks are moving at the same speed in the same direction at the same event, therefore they are synchronized in spite of their different histories".

 

[Foppe Hoekstra]

The clocks are initially incorrectly zeroed

By whom?

but then have different speeds

?

The rear clock remains stationary while the front clock moves and is time dilated

From the moment the front clock reaches the junction between the straight track and the circle, the front clock is moving with speed v in a circle for the time that is necessary to round the circle and during that time the rear clock is also moving with the same speed and precisely ends up at the junction.

 

[jbriggs444]

moving with speed v in a circle

Relative to the track frame, yes. Relative to the train frame (if it exists), it is stationary. Pick one. Do not pick both.

during that time

A strong hint that the relativity of simultaneity has not been considered.

 

[Foppe Hoekstra]

Relative to the track frame, yes. Relative to the train frame (if it exists), it is stationary. Pick one. Do not pick both.

I thought it is clear that for both the rear and the front clock I chose the track frame (the junction).

 

[PeterDonis]

in the moving frame they are sync

In the inertial frame in which the train is at rest before it reaches the circle, they are in sync. But once the train starts around the circle, that is no longer true.

How is the rest frame to see the difference between the two sync rear clocks and the sync rear and front clock?

By the fact that the two rear sync clocks are always at the same position, so their synchronization is frame independent (they both have the same worldlines). The rear and front clocks are not always at the same position, so their synchronization is frame dependent.

 

[jbriggs444]

I thought it is clear that for both the rear and the front clock I chose the track frame (the junction).

In the track frame, the clocks are not synchronized because they were never properly synchronized. Instead, they were systematically de-synchronized.

 

[PeterDonis]

I thought it is clear that for both the rear and the front clock I chose the track frame (the junction).

No, that's not at all clear. You specified that the train clocks start out synchronized in the frame in which they are all at rest while the train is on the straight track. You cannot specify that and also have the clocks synchronized in the track frame. It's not possible.

 

[hutchphd]

But in the moving frame they are sync. Let's throw in an extra clock at the rear that is sync wtih the other clock at the rear.
How is the rest frame to see the difference between the two sync rear clocks and the sync rear and front clock?

Do you not see that the last car (Homer) and the first car (Romer) have just performed the classic "twin paradox"?? In the (inertial) frame of Homer, Romer just made a high speed voyage with return.

 

[PeterDonis]

In the (inertial) frame of Homer

There isn't one while Homer is moving around the circle. That motion is non-inertial.

 

[Foppe Hoekstra]

Do you not see that the last car (Homer) and the first car (Romer) have just performed the classic "twin paradox"?? In the (inertial) frame of Homer, Romer just made a high speed voyage with return.

Shouldn't Romer return to the same place he started his voyage? In the frame of Homer he made his space voyage plus a voyage from the front to the back of the train.

 

[PeterDonis]

Do you not see that the last car (Homer) and the first car (Romer) have just performed the classic "twin paradox"??

No, they haven't. They didn't start out co-located.

 

[Ibix]

By whom?

By you. You synchronised the clocks in the frame where the trains were initially at rest. Thus they do not show the same time in the track frame.

?

Come on! In the frame where the train is initially at rest, the clocks do not start out in the same position but end up in the same position. One of them must have moved and they were initially at rest.

From the moment the front clock reaches the junction between the straight track and the circle, the front clock is moving with speed v in a circle for the time that is necessary to round the circle and during that time the rear clock is also moving with the same speed and precisely ends up at the junction.

The track is not a circle in this frame. Nor is the speed of the train constant. And for the synchronisation to remain the clock must remain at its initial constant speed - zero in this case.

 

[hutchphd]

No, they haven't. They didn't start out co-located.

OK Romer walked to the front of the train...

 

[hutchphd]

There isn't one while Homer is moving around the circle. That motion is non-inertial.

I thought Homer had just gotten to the circle.

 

[Ibix]

OK Romer walked to the front of the train...

But that would lead to his clock being out of sync with the front-of-train clock.

I agree the situation has similarities to the twin paradox, but there are key differences too.

 

[Foppe Hoekstra]

Come on! In the frame where the train is initially at rest, the clocks do not start out in the same position but end up in the same position. One of them must have moved and they were initially at rest.

Don't blame me for that! That is what the Relativity Theory requires.

 

[hutchphd]

But that would lead to his clock being out of sync with the front-of-train clock.

I agree the situation has similarities to the twin paradox, but there are key differences too.

I was positing that all re-locations on the train were after the speed-up but prior to circular motion. The twin analogy assumes synchronization after the speed-up...perhaps I misread the OP intent.

 

[PeterDonis]

perhaps I misread the OP intent

It seems like you have. None of what you say looks anything like the scenario we are discussing.

 

[PeterDonis]

Don't blame me for that! That is what the Relativity Theory requires

And relativity theory says that, given your statement that the train clocks are initially synchronized in the rest frame of the train while it is on the straight track, when the front and rear clocks meet after the front clock has gone around the circle once, the two clocks do not show the same reading. So if you are accepting what relativity theory requires, then you have to accept that.

I am closing the thread because the correct answer and analysis have been given and there is no point in going around in circles.