# I Is there a solution for the behavioural contradiction in RT?

#### A.T.

Simultaneously in the moving frame (MF).
RS: From the full story: "Every printer has a clock that is synchronized when the train is at rest and thereafter the train is accelerated to a constant velocity v. Thus all clocks are accelerated in the same way and so all clocks will develop the same deviation in relation to SF. So all clocks will still be synchronized in RS when the train is at speed."
So the prints by the circular train are simultaneous in SF, but the prints by the single wagon are not simultaneous in SF.

When the angular and linear motion smoothly transform to each other in the limit, should not the distances between the dots do as well?
The issue here is not angular vs. linear motion.

The issue is that the wagons of the circular train are forced to keep a constant length in SF, so their clocks stay synchronized in SF. Your single wagon is not forced to keep a constant length in SF, so its clocks would desynchronize during acceleration in SF. If you setup your single wagon just like your train wagons (including forced constant length during acceleration) it will print the same dot distances as the train.

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#### Foppe Hoekstra

So you have adopted a coordinate system in which the time coordinate is referenced against a stationary clock in the center of the circular train tracks. That's fine. (The GPS system runs that way). But since you have discarded Einstein clock synchronization, the coordinate speed of light is no longer isotropic in your coordinate system. It is no longer equal to c in every direction.
I deliberatly do not use the Einstein clock synchronization, so I do not have to worry about non-isotropic speeds of light or the Sagnac effect.

#### Foppe Hoekstra

Clearly you break circular symmetry if you Einstein synchronise your wagon clocks. From the track's rest frame, the front clock of any wagon lags behind the back clock if you Einstein synchronise them. So the front clock of the wagon infront is even further behind. What happens when you get all the way round your circle to the first wagon that you synchronised again? There's your break in symmetry.
Nothing is happening to time. It's just that you cannot set clocks consistently in the way you want. The break in clock synchronisation behaves like crossing from one time zone to the next - nothing interesting happens, except that your watch suddenly doesn't agree with the wall clocks.
That is just why I do not use Einstein synchronization. The synchronization used is apllicable to both systems (MF and RS).
Imagine the loose wagon side by side with one of the wagons form the RS. They both have clocks that are synchronised in the same way and both simultaneously print dots on the rail. The only difference between them is that the RS-wagon is stretched by γ. Nevertheless it is the shorter wagon that puts its dots at the larger distance!

#### Foppe Hoekstra

The issue is that the wagons of the circular train are forced to keep a constant length in SF, so their clocks stay synchronized in SF. Your single wagon is not forced to keep a constant length in SF, so its clocks would desynchronize during acceleration in SF. If you setup your single wagon just like your train wagons (including forced constant length during acceleration) it will print the same dot distances as the train.
So if I stretch the single wagon by γ, just because of that the dots will suddenly be put closer to oneanother. I can't believe that.

#### A.T.

That is just why I do not use Einstein synchronization.
Then what do you use for the single wagon?

The only difference between them is that the RS-wagon is stretched by γ.
That difference changes your synchronization, because the stretching happens during the acceleration, which is part of your synchronization procedure: "Every printer has a clock that is synchronized when the train is at rest and thereafter the train is accelerated to a constant velocity v."

So if I stretch the single wagon by γ, just because of that the dots will suddenly be put closer to oneanother. I can't believe that.
If you apply the same synchronization procedure to the single wagon, it will print the same dot distances as the train wagons:
1) Synchronize the single wagon's clocks at rest in SF
2) Accelerate the single wagon, while forcing it to keep a constant length in SF

#### Ibix

Nevertheless it is the shorter wagon that puts its dots at the larger distance!
I don't believe this is correct, assuming you mean "further apart as measured by the track rest frame". It would be correct if you were Einstein synchronising your clocks, but you aren't.

With Einstein-synchronised clocks, the "simultaneous print" of the wagons is non-simultaneous in the track rest frame. Thus the length contracted wagon produces a longer spacing because its back printer prints before its front printer (as described in the track frame).

But you aren't using Einstein synchronisation - all your clocks are synchronised in the track rest frame, even when they are in motion (I assume the clocks in the inertial wagon are not Einstein synchronised either). Thus the length-contracted wagon will print a short spacing.

#### hutchphd

RS: From the full story: "Every printer has a clock that is synchronized when the train is at rest and thereafter the train is accelerated to a constant velocity v. Thus all clocks are accelerated in the same way and so all clocks will develop the same deviation in relation to SF. So all clocks will still be synchronized in RS when the train is at speed."
As soon as the clocks are accelerated they will no longer be synchronized along that direction (your choice of direction spoils any symmetry). I believe they will, however, return to synchronization if the train is brought to rest.

#### jbriggs444

Homework Helper
I believe they will, however, return to synchronization if the train is brought to rest.
Yup. A simple argument from symmetry guarantees this. No clock is treated any differently than any other, so they must all share the same reading when re-united.

#### PeroK

Homework Helper
Gold Member
2018 Award
Which brings us to another point: namely that a consistent theory does not necessarily need to be true.
You're referring to Newtonian mechanics, of course?

Mentor

#### PeterDonis

Mentor
The OP question has been answered in a number of different ways. The thread will remain closed.

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