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Let there be two children Allen and George .

now Alan boards a train and the train moves at a speed comparable to light. now because George is on the station for George being at rest Allen is moving and its lock is moving slowly so time passes slower for Allen and faster for George. for George after sometime when he will be very old then Allen will be of lesser age.

now but Allen in his own frame in train sees George moving in the opposite direction.so for him George is moving and so the clock of George must be moving slowly hence for Allen George time is running slower than his.

after sometime when he will be old enough then George will be of lesser age as he was standing on the platform but moving for Allen so my question is if George if Georges on the platform and sees that Allen old and, when Allen is in the train thinking he is at rest and observe George he will see George younger

then wont it be a contradiction that both will see each other relatively younger when both meet each other?

Also if there is a third person who is also a twin to George and Allen who is moving in another train but it is lower than alan's train and so he must see George at a different age than Allen right then please tell how will the third person and Allen willbdisagree or agree about George age?

Tanmay Siddharth
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It is true they will both observe the other twin's clock as running slower than their own. But to meet again, the twin on the train moving close to light speed, will have to decelerate, stop, and accelerate back to return. acceleration and deceleration slow time like gravity does. So while Alan is decelerating and accelerating, George will see Alan's clock slow much more, and Alan will see George's clock speed up enough so that when he returns to the station they will both see that George is older than Alan.

Adrian Howard
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    This answer is incorrect. For example, Allen can travel for 1 day or for 1 year. In either case the amount and time period of his acceleration while turning around would be the same. The same acceleration cannot account for the time difference accumulated over one day and over one year. This is not a correct resolution of the twin paradox. – safesphere Jun 12 '19 at 08:42
  • this is correct, I did not say whether George is one day older or one year older, just that they would both agree – Adrian Howard Jun 12 '19 at 08:54
  • They would not agree based on the logic in your answer. Also, please use the proper @ address while replying. – safesphere Jun 12 '19 at 08:57
  • if you disagree with Einstein and Lorentz there is nothing more I can say – Adrian Howard Jun 14 '19 at 00:33
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    I did not see Lorentz or Einstein commenting here, but in all seriousness, you should not be offended by comments posted to help you improve your understanding. I did not downvote your answer, but someone did, because it is not ncorrect. You should search this site for the twin paradox to see that it is resolved by the change of the reference frame, but not by acceleration. And also, once again, please learn to use the proper @ address. Otherwise people don't know that you have replied. – safesphere Jun 14 '19 at 08:42
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    @safesphere Actually, both Adrian's answer is half correct, and your comments are half correct. It is true that, because Alan is in an accelerated reference frame, he will "see" George's clock speed up. The rate of speed-up depends on both the size of the acceleration and the distance between them. But George will not "see" Alan's clock slow down more during the acceleration, the slow-down will be still be according SR, based on the relative velocity. As such, it is possible to resolve the paradox using the acceleration. – fishinear Apr 18 '21 at 13:53
  • @safesphere It is also possible to resolve it using a "change of reference frame" approach. But even then you still need to take into account which twin is accelerating, in order to determine which twin to apply the change of frame to. Without taking the acceleration into account, the symmetry of the paradox cannot be broken. – fishinear Apr 18 '21 at 13:53
  • @fishinear The symmetry is broken by the fact of acceleration, which is equivalent to the fact of changing the frame. Instead of acceleration, consider the third twin in another train moving in the opposite direction. When the trains meet without stopping, the twins in the trains synchronize their clocks. Thus the third twin effectively represents the second twin who instantly reversed his direction of motion without acceleration. And sorry, I will not engage in a discussion in comments on a two-year-old question. Instead, you are welcome to post your own answer to the linked question. – safesphere Apr 18 '21 at 15:59
  • @safesphere My intention was not to engage in a discussion, and there are already plenty of correct answers to the twin paradox on the forum, like the "duplicate answers" ones linked in the question. I was just addressing the misleading points in your comments and Adrian's answer. You cannot break the symmetry if you don't know which twin is accelerating. I thought that should be pretty self-evident. – fishinear Apr 18 '21 at 16:08
  • @fishinear Thank you for sharing your opinion. – safesphere Apr 18 '21 at 16:54