stevmg
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I think this is getting really complicated when it doesn't have to be so.
This post by JesseM explains the conceptual and simple mathematical approach to this problem
https://www.physicsforums.com/showpost.php?p=2610219&postcount=63
In essence we have twin A and B. If one looks at it from the point of view of twin A's frame of reference (FOR) B moves to the right, turns around and moves to the left. Time in A's FOR is proper time as he ain't movin'. Folks who are moving experience less time because of the motion (you know, Lorentz, et al.) Thus B is moving both away and back and experiences less time. JesseM gives a nice quick calculation.
The supposed symmetrical situation is to look from B's FOR. In this case A moves left - BUT never stops. Twin B starts moving to catch up with A and eventually does. When all the times are added up the elapsed time for B is the same this way as it was looking at it the first way in the above paragraph.
Guess what! This is NOT a symmetrical situation is it? In the first case one of the twins sits still (and gets older) while the other twin moves and gets older slower. The first twin never moves.
In the second case, BOTH twins move although one sits still for a while before mving.
These are NOT symmetrical approaches. There's no way to make them symmetrical. JesseM's calculations and working through the problem is self explanatory.
A symmetrical scenario would be to have both twin A and B depart the reference frame in opposite directions at the same speed for the same time and both turn around and come back to meet. In this case, they both would age at the same rate (though not as fast as their triplet brother who remained on Earth) and be the same age when they rejoined. Their triplet brother who remained on Earth would be older than both.
I am not going through the calculation looking at it from A's FOR or B's FOR but I guarantee you, it would work out. Who cares about Doppler, acceleration, deceleration and all that?
Don't make this more complicated than it is, which it isn't (really.)
This post by JesseM explains the conceptual and simple mathematical approach to this problem
https://www.physicsforums.com/showpost.php?p=2610219&postcount=63
In essence we have twin A and B. If one looks at it from the point of view of twin A's frame of reference (FOR) B moves to the right, turns around and moves to the left. Time in A's FOR is proper time as he ain't movin'. Folks who are moving experience less time because of the motion (you know, Lorentz, et al.) Thus B is moving both away and back and experiences less time. JesseM gives a nice quick calculation.
The supposed symmetrical situation is to look from B's FOR. In this case A moves left - BUT never stops. Twin B starts moving to catch up with A and eventually does. When all the times are added up the elapsed time for B is the same this way as it was looking at it the first way in the above paragraph.
Guess what! This is NOT a symmetrical situation is it? In the first case one of the twins sits still (and gets older) while the other twin moves and gets older slower. The first twin never moves.
In the second case, BOTH twins move although one sits still for a while before mving.
These are NOT symmetrical approaches. There's no way to make them symmetrical. JesseM's calculations and working through the problem is self explanatory.
A symmetrical scenario would be to have both twin A and B depart the reference frame in opposite directions at the same speed for the same time and both turn around and come back to meet. In this case, they both would age at the same rate (though not as fast as their triplet brother who remained on Earth) and be the same age when they rejoined. Their triplet brother who remained on Earth would be older than both.
I am not going through the calculation looking at it from A's FOR or B's FOR but I guarantee you, it would work out. Who cares about Doppler, acceleration, deceleration and all that?
Don't make this more complicated than it is, which it isn't (really.)