Is there a solution for the behavioural contradiction in RT?

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  • Thread starter Thread starter Foppe Hoekstra
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Foppe Hoekstra said:
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.
 
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Foppe Hoekstra said:
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.
 
hutchphd said:
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.