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 Quote by John232 I fail to see how that is true without an explanation.
 Quote by mananvpanchal Can you explain me how acceleration can create desynchronization with respect to O?
Certainly, it follows directly from the Lorentz transform. Let's analyze the scenario from the OP where at t=0 in the original frame A, B, and M all accelerate instantaneously up to v = .6c and furthermore lets use units where c=1 and where the distance from M to A and from M to B is 1 in the original frame.

So, in R's frame the worldline of A, O, and B are:
$$r_d=\left(t,x=\begin{cases} d & \mbox{if } t \lt 0 \\ 0.6 t+d & \mbox{if } t \ge 0 \end{cases} ,0,0\right)$$
where d=-1 for A, d=0 for O, and d=1 for B.

As per the OP, the clocks are initially synchronized in R's frame such that at t=0 they all read 0. So, we can calculate the time displayed on each clock, τ, using the spacetime interval. Solving for t we get:
$$t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}$$

Substituting in to the above we get:
$$r_d=\left( t=\begin{cases} \tau & \mbox{if } \tau \lt 0 \\ 1.25 \tau & \mbox{if } \tau \ge 0 \end{cases}, x=\begin{cases} d & \mbox{if } \tau \lt 0 \\ 0.75 \tau+d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)$$

Noting that τ does not depend on d in this frame we see immediately that the clocks remain synchronized in R's frame.

Now, boosting to the primed frame where O is at rest for τ=t>0, we obtain.
$$r'_d=\left( t'=\begin{cases} 1.25 \tau - 0.75 d & \mbox{if } \tau \lt 0 \\ \tau - 0.75 d & \mbox{if } \tau \ge 0 \end{cases}, x'=\begin{cases} 1.25 d - 0.75 \tau & \mbox{if } \tau \lt 0 \\ 1.25 d & \mbox{if } \tau \ge 0 \end{cases} ,0,0\right)$$

Noting that τ does depend on d in this frame we see immediately that the clocks are desynchronized in O's frame. If there is some step that you do not follow then please ask for clarification. But it is quite clear the the clocks remain synchronized in the unprimed frame and are not synchronized in the primed frame.