Suppose two bodies, A and B, are together in free space. One of them, let’s say- A, accelerates away, then retards, stops, accelerates back and stops again when it reaches B.
Although we can see the objects separating and reuniting, can we be certain about the fact that it was A which really accelerated and not B? Can we determine which object was actually under influence of force? Can any experiment conducted by A or B settle the dispute?
I think that the answer is 'no'. I will explain why I think so presently. But interestingly, if I am correct, and there is no certain way of knowing which object it was that accelerated, then how can one of them end up being younger than the other on their reunion?
So, can we determine which object accelerated? I think no. Suppose A decides to conduct an experiment to find out whether it is under influence of a force or not. To conduct an experiment, A needs an experimental setup. Now, it would be wise for A to keep his experimental setup at rest with him. This means that if A is accelerating, so would his laboratory.
What I mean is this- if a person is confined in a room where all the objects are at rest with respect to him, also meaning that all objects are under influence of same force per unite mass if need be, then no experiment conducted in the room can determine whether the room is accelerating or not.
So, let’s say that A and B are holding a glass of water each. Now, when our experiment starts, one of them accelerates. Let’s say A's glass spills due to inertia while B's does not. B would argue that this is because A has accelerated. A can argue that this is because B and the water in his own glass accelerated together (they were charged, and accelerated when an electric field was introduced). B's glass does not spills because the whole of B, with the glass and the water, can be considered uniformly charged.
Can this dispute be resolved? If not, I think there is symmetry in such situation, and both objects would observe the other object return younger, in parallel universes.