From: hees@fias.uni-frankfurt.de   
      
   One should note that the proper time is an invariant and as such   
   completely independent of the coordinates and the parametrization you   
   choose to calculate it.  You can introduce arbitrary generalized   
   coordinates (most of which then correspond to the description of a   
   non-inertial observer) and an arbitrary parameter to parametrize the   
   worldline.  As in Euclidean space the length of a curve also proper   
   time, which is in a somewhat generalized sense nothing else than the   
   length of a time-like worldline, is independent of these choices.   
      
      
   On 22/05/2020 21:15, Tom Roberts wrote:   
   > On 5/21/20 2:36 AM, Luigi Fortunati wrote:   
   >> [...]   
   >> Where is the conceptual difference that should make the time of a twin   
   >> different from that of the other, if we are talking about Special   
   >> Relativity ONLY?   
   > The difference between the twins comes from the simplicity that applies   
   > only to inertial frames.   
   >   
   > If twin A remains at rest in an inertial frame, it is easy to calculate   
   > the age (elapsed proper time) of each twin: simply integrate   
   >   
   > 	T = \integral sqrt(1-v^2/c^2) dt   
   >   
   > where T is the elapsed proper time, the integral is taken over the path   
   > of the twin relative to A's rest frame, v is the speed of the twin   
   > relative to that frame (as a function of t), and t is the time   
   > coordinate of the frame. Note that neither position nor acceleration   
   > appear in this equation; all that matters is the speed of the twin   
   > relative to the inertial frame being used to calculate.   
   >   
   > For twin A, v = 0, giving an age that is simply the frame's total   
   > coordinate time of the scenario.   
   >   
   > For twin B, 0 clear that for any value of t, B's age will be less than the value of t.   
   > So if twin B follows a path that leaves and then returns to A, when they   
   > rejoin B will have aged less than A.   
   >   
   > If you want to calculate what happens using twin B as a reference, there   
   > is a problem: B does not remain at rest in any inertial frame, and the   
   > above equation does not apply. B must necessarily accelerate in order to   
   > return to A. It is possible to use SR to calculate using the accelerated   
   > coordinates of B; this is done in some textbooks, but is beyond the   
   > scope of a newsgroup post. When done correctly (which is non-trivial),   
   > this calculation yields the same answer as the much easier calculation   
   > using A's inertial frame.   
   >   
   > Tom Roberts   
      
      
   --   
   Hendrik van Hees   
   Goethe University (Institute for Theoretical Physics)   
   D-60438 Frankfurt am Main   
   http://fias.uni-frankfurt.de/~hees/   
      
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