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?
New updates
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?
