From: hitlong@yahoo.com   
      
   On Monday, August 7, 2017 at 1:52:36 AM UTC-6, LuigiFortunati wrote:   
   >   
   > Gary Harnagel alle ore 14:19:01 del 06/08/2017 ha scritto:   
   > >   
   > > > In this discussion there is only the going and not the return.   
   > > >   There is an initial acceleration and then for 5 years the traveling   
   twin   
   > > > travels with engines off and arrives in S where reading S time of the   
   > > > clock without reversing the gear.   
   > > >   What it does next does not concern the present discussion.   
   > >   
   > > But remember that the times at T and S are no longer synchronized according   
   > > to the traveling twin after acceleration.   
   >   
   > How is T and S times different if both are at rest in K?   
      
   The "traveling twin" is not at rest in T and S.   
      
   > Let's take a practical example, acceleration lasts one day of the ship's   
   > time so that at the time of the acceleration the clock of the ship marks   
   > the 24 hour time.   
   >   
   > What could be the times (different) of T and S?   
   >   
   > --   
   > Luigi Fortunati   
   >   
   > Credere e' piu' facile che pensare   
   > Believing is easier than thinking   
      
   You didn't specify a value for the acceleration, but if you expect   
   to have gamma = 8 at the end of one day in the rest frame (i.e.,   
   v/c = .99217), it would have to be about 2805 g's, so there is little   
   point in doing that.   
   >  Why not just make it instantaneous?   
      
   So when the spaceship begins its journey at v/c = 0.99217 at x = 0   
   (i.e/, at T) and time t = 0, the time at S is:   
      
   t' = gamma*(t - v*xs/c^2)   
      
   where xs is the distance to S in the "stationary frame and t' is   
   the time at S in the moving ship's frame.  Since t = 0,   
      
   t' = -gamma*v*xs/c^2)   
      
   which is NOT 0, so the clocks at S and T are no longer synchronized   
   according to the moving ship.   
      
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