From: nicolaas.vroom@pandora.be   
      
   On Friday, 22 May 2020 21:15:39 UTC+2, 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.   
      
   What is the reason that you cannot use one reference frame for the whole   
   experiment from start to finish i.e. when both clocks again can be   
   compared at point O?   
      
   > 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.   
      
   My understanding is that t represents the time (clock count) of a clock   
   at rest and T the (time) clock count of a moving clock.   
      
   The question is how is v (the speed of the clock) measured?   
   IMO in order to do that, you need a reference rod in the frame at rest with   
   two clocks at backend x1 and the frontend x2 of this rod, also at rest.   
   When you do that you can calculate the speed v by applying   
   the following formula: v = (x2-x1) / (t2-t1)   
   where t1 is the time that the moving clock coincides with the backend x1   
   and t2 the time that the clock coincides with the front end x2 of the   
   clock.   
      
   In fact, if you want to do this whole calculation correct you   
   need clocks all along the path from point O to point +4.   
      
   > 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.   
      
   That is correct.   
   But if the speed varies along this path you need more clocks   
   to calculate the time T correctly.   
   In fact, you need extra clocks near point O and point +4   
      
   But there is another issue:   
   How do you know that the equation to calculate T is correct?   
   The only way is when the calculated T at the end can be verified by   
   means of an actual experiment.   
   IMO experiments are the only way to derive this formula.   
   Using different experiments you can also test if all clocks   
   with a different internal physical construction behave the same   
   i.e. show the same number of clock counts.   
      
   > 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.   
      
   This conclusion follows the same rule I always use when I want to   
   study any system which includes different objects.   
   My golden rule is to use as a reference point a point that moves the least,   
   in order to keep things simple.   
   In practice, this means I study only one reference frame.   
      
   Nicolaas Vroom   
      
   --- SoupGate-Win32 v1.05   
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