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Suppose we have 2 clocks in the same reference system which are synchronized. These two clocks exchange signals every second. Now suppose that one of the two starts to move away (instantly with $||v||=const$), the first signal, after the beginning of the move, will arrive after the last with this time interval: $$\Delta t_1 = 1 s + \Delta x_1/c$$ where $\Delta x_1$ is the distance traveled, after the same space: $$\Delta t_2 = 2 s + 2 \Delta x_1/c.$$ When the moving clock arrive at $s = 3 \Delta x$ sends the last signal and instantly reverses the course by coming back ($||v||= const$): $$\Delta t_3 = 3 s + 3 \Delta x_1/c.$$ The apparent accumulated delay will be: $$D = 3 \Delta x/c \: ?$$ By going back, to the stationary clock it will seem that the moving one is accelerating: $\Delta t_4 = 4 s + 2 \Delta x_1/c$, with an increase of $\Delta x/c$; then $\Delta t_5 = 5 s + \Delta x_1/c$ and $\Delta t_6 = 4 s$.

Will the watches be resynchronized?

Why is this not true for the theory of special relativity?

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Thanks for the suggested answers, which I'm reading.

I accept that time is not absolute, but what does de-synchronize watches? Does the acceleration imprinted on one of them change the mechanics of the watch?

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Mattia Ghio
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    Nothing changes the mechanics of the watch. The traveling watch experiences less time than the stationary watch because the worldline of the stationary watch is straight, but the worldline of the traveling watch is bent. – PM 2Ring Mar 03 '19 at 14:46
  • So my objection is: If the mechanics of the watch does not change then the time mesured by the two clocks would be similar, because the clock does not know if it is moving or not – Mattia Ghio Mar 03 '19 at 16:17
  • A clock is a device that measures proper time. By the geometry of Minkowski spacetime the total proper time measured on the worldline of the stationary clock is greater the proper time on the bent worldline of the moving clock. This is exactly analogous to the fact in Euclidean space that in a triangle ABC the distance AC is shorter than AB + BC. – PM 2Ring Mar 03 '19 at 16:34

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