From: fortunati.luigi@gmail.com   
      
   Mike Fontenot alle ore 08:47:53 del 04/08/2017 ha scritto:   
   > Assuming that the traveler instantaneously accelerates to gamma = 8, the   
   clock at S will read 40 years when T reaches there at age 5 years.  But T says   
   that the home twin is only 5/8 years old then.  (As soon as T accelerates to   
   gamma = 8, he will no    
   longer regard the clock at S to be synchronized with the home twin's age.  He   
   says that the clock at S reads (40 - 5/8) once he has reached gamma = 8.    
   I.e., he says that the clock at S suddenly changes from reading zero to   
   reading (40 - 5/8) years when    
   he accelerates (instantaneously) from gamma = 0 to gamma = 8.)   
   >   
   > Likewise, if T instantaneously reverses course, and goes back to the   
   stay-at-home twin at gamma = 8, he will conclude that she only ages 5/8 year   
   during the constant-speed portion of his trip back.   
   >   
   > The important question is this: if the home twin only ages 5/8 year during   
   each of the two constant speed portions of T's trip (a total of 10/8 years),   
   how can she be much older (80 years older than T's 10 years) when he returns   
   (as we know she must be)   
   ?   
   >   
   > The answer is that, during the instantaneous turnaround of T, he concludes   
   that her age instantaneously increases by (80 - 10/8) years. THAT is the   
   important resolution of the twin "paradox".   
      
   Yes, this brilliant solution to the paradox tells us that in 10 years of   
   travel (of the traveling twin) the earthly brother ages only one year   
   and little more (1.25) while the two accelerations are enough to make it   
   instantaneously almost 79 years old!   
      
     Accelerations are just lethal!   
      
   --   
   Luigi Fortunati   
      
   Credere e' piu' facile che pensare   
   Believing is easier than thinking   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)