Puh.
The aging process occur during the whole flight
It doesn't, as my example proved. Anyway, if you don't believe me, believe Wikipedia or any text book on that issue.
a) a mission to mars : departure - 1 year flight to go - turning back - 1 year flight back - landing
b) a mission to juppiter : departure - 4 year flight to go - turning back - 4 year flight back - landing
Following your reasoning the time shift ahould be the same but in case b) it is higher than in a)
no, I didn't say that. The inertial systems that B changes are not the same in scenario a) and b).
Once the twin is back, his clock remains desyncronized
Following your reasoning, which is - again to make sure we're talking about the same thing - turning around has nothing to do with it but ONLY the free (no acceleration/turning/whatever) motion of B in space. How do you explain that B is the older one in the end if NOTHING distinguishes their relative motion?
Therefore there is no reason to emphasize the concept of "inertial " system
It is, because without it the twin paradox is a paradox.
Ok but such kind of abstractions are normal in pyisics !
You can figure out a realistic scenario whereas the effect of turning acceleration / deceleration are negligible
Tell us that scenario, I am quite sure that in this case you can't.
The velocity is , in the equation, in term of : V^2 thus the sign does not play any role
What does the sign of v have to do with that?
The " essence" of the twins paradox lies in the special relativity
Maybe you can do such a calculation for us. The steps B moves away from earth, B turns around and B moves towards earth should be done separately. You may neglect the middle step if you think it is not important.
