Here is the explanation to the Twin Paradox in Feynman's Lecture on Physics 16–2.

we consider a famous so-called “paradox” of Peter and Paul, who are supposed to be twins, born at the same time. When they are old enough to drive a space ship, Paul flies away at very high speed. Because Peter, who is left on the ground, sees Paul going so fast, all of Paul’s clocks appear to go slower, his heart beats go slower, his thoughts go slower, everything goes slower, from Peter’s point of view. (...) “Heh, heh, heh, from the point of view of Paul, can’t we say that Peter was moving and should therefore appear to age more slowly?" By symmetry, the only possible result is that both should be the same age when they meet. But in order for them to come back together and make the comparison, Paul must either stop at the end of the trip and make a comparison of clocks or, more simply, he has to come back, and the one who comes back must be the man who was moving, and he knows this, because he had to turn around. When he turned around, all kinds of unusual things happened in his space ship—the rockets went off, things jammed up against one wall, and so on—while Peter felt nothing. So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls, and so on, is the one who would be the younger; that is the difference between them in an “absolute” sense, and it is certainly correct

My question is:

1) To make a comparison, why do we have to bring them together? why not just make them have the same velocity thereby sharing the same inertial frame? (not bring them together into the same location!)

2) What if Peter starts to move towards Paul? In that case, Peter would be the one who experiences "the accelerations". Then Paul can be the observer. Is there any error in my argument?

  • $\begingroup$ Does that mean that acceleration makes the time outside as seen in the accelerated frame go faster, so that instead of Paul seeing his brother's time run slower, he would instead see it run faster? $\endgroup$ Jun 20, 2016 at 6:41
  • $\begingroup$ Both of those things defeat the point of the "puzzle". You are supposed to be learning something important here, not fighting it. $\endgroup$
    – m4r35n357
    Jun 20, 2016 at 8:40
  • $\begingroup$ You should probably change the title to a more informative one. $\endgroup$
    – anderstood
    Jun 20, 2016 at 12:42
  • $\begingroup$ @anderstood Thank you for your comment. I've changed it! $\endgroup$
    – Young
    Jun 20, 2016 at 13:43

1 Answer 1


You are quite correct that we don't have to bring the twins together at the same point in space. It's enough to make their relative speed zero i.e. bring them into the same inertial frame. The twin paradox generally involved bringing the twins together just for simplicity. When two observers are at the same point in space they can directly compare their clocks. If they are at different points they would have to use some scheme such as Einstein synchronisation. However this doesn't change the physical principles involved. The accelerating twin is younger whether the two twins meet at the same point or not.

Re your second question, in special relativity although velocity is relative acceleration is absolute i.e. an observer can always tell if they are accelerating using a local experiment. Therefore it is unambiguous which of the twins is accelerating and which is remaining at constant velocity. And it is the accelerating twin who ages less.

If you're interested in learning more about the physics have a look at What is the proper way to explain the twin paradox?. This was written specifically to try and clear up the confusion over the twin paradox.


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