From: mlfasf@comcast.net   
      
   On 6/26/19 1:21 AM, Phillip Helbig (undress to reply) wrote:   
   > In article , Mike   
   > Fontenot  writes:   
   >   
   >> Maybe this will help:   
   >>   
   >> In the classic twin "paradox" with an instantaneous turnaround, the   
   >> apparent paradox arises because the traveling twin incorrectly believes   
   >> that he can use the time dilation result for his ENTIRE trip.  So he   
   >> expects her to be the younger when they are reunited.  When the reunion   
   >> occurs, he is surprised and perplexed that she is the older.  The   
   >> resolution of this apparent paradox is that during his instantaneous   
   >> turnaround, she instantaneously gets much older (according to him, not   
   >> her).  So his acceleration is INDISPENSABLE in the resolution of the   
   >> apparent paradox.   
   >   
   > There is a simple argument to show that the acceleration is not the   
   > cause of the slower aging of the turnaround twin: the magnitude of the   
   > effect, i.e. the difference in ages, depends only on the speed travelled   
   > and the length of the journey: the longer the journey, and the faster   
   > the journey, the greater the difference, even if the acceleration is the   
   > same.   
      
   The material I give below will show that acceleration IS the reason that   
   the traveling twin is the younger at the reunion, not the older.  It   
   will show that, according to the traveler, the home twin instantaneously   
   ages by a large amount during the traveler's instantaneous velocity   
   reversal at the turnaround.  This will be shown from the Minkowski   
   diagram, that I will describe in detail, for a specific example of the   
   classic twin paradox.  The specific example is for a relative speed of   
   0.866 ly/y for the twins, giving a gamma value of 2.0.   
      
   Draw the following diagram, as I go along.  Try to be as accurate as   
   possible.  My words won't be understandable if you don't do that.   
      
   Draw the horizontal axis, and label it T, the age of the home twin, in   
   years.  (I'll always refer to her as "she").  Draw the vertical axis,   
   and label it X, the position coordinate in her inertial reference frame,   
   with units of lightyears (ly).  She is ALWAYS at the position X = 0.  So   
   the horizontal axis is her worldline.  The intersection of the X and T   
   axes (the origin of the diagram) is where the scenario begins: the twins   
   are momentarily co-located there at the instant they are born.   
      
   [Moderator's note: Usually (almost always), such space-time diagrams   
   have time running vertically from bottom to top, and have the one   
   spatial dimension running horizontally.  This is purely convention, of   
   course.  (In other contexts, time is usually plotted horizontally)   
   -P.H.]   
      
   Put a "tic" mark a little less than halfway along her worldline, and   
   underneath the "tic" mark, write the number 40.  That represents the   
   point on her worldline where she is 40 years old. That will be her age   
   when he turns around, ACCORDING TO HER. Then, measure off an equal   
   distance along her worldline, from the 40 year point, and put a "tic"   
   mark there, and write the number 80 below that "tic".  That will be her   
   age at the end of the scenario, when the two twins are reunited.  Both   
   twins will agree about that.   
      
   Immediately after the two twins are born, the traveling twin ("he")   
   instantaneously changes his velocity relative to her, from zero ly/y to   
   0.866 ly/y.  She can therefore compute that when she is 40, he will have   
   traveled a distance of   
      
     L = 40 * 0.866 = 34.64 ly.   
      
   L is the distance between the two twins (ACCORDING TO HER) at the   
   turnaround.   
      
   Locate the point vertically above the T = 40 point on her worldline   
   which is at X = 34.64 ly.  You can do that by measuring with an accurate   
   ruler the length between the T = 0 point and the T = 40 point on her   
   worldline, and then multiply that length by the ratio 34.64 / 40 =   
   0.866, which gives the ruler distance to the turnaround point vertically   
   above the T = 40 point on her worldline.  Mark that point, and draw a   
   straight line between it and the origin of the diagram (the T = X = 0   
   point).  Also, draw a vertical line that passes through that point,   
   starting well above the point, and extending down to the "tic" mark that   
   you previously labeled "40" on the T axis.   
      
   The straight line between the origin of the diagram and point you've   
   just determined is the OUTBOUND segment of HIS worldline.  Just as we   
   used the variable "T" to denote HER age along HER worldline, we will use   
   the variable "t" to denote HIS age along HIS worldline.   
      
    From the famous time-dilation result, with gamma = 2, we know that when   
   she is 40 years old, he is 20 years old, ACCORDING to HER. So put a   
   "tic" mark at the top end of the outbound segment of his worldline,   
   perpendicular to the line, and write "20" above the "tic", oriented   
   parallel to his worldline.   
      
   That initial segment of his worldline has slope 0.866, so its angle with   
   respect to the T axis is   
      
     alpha = arctan (0.866) = 41 degrees.   
      
   Label the angle alpha on your diagram, because that SAME angle will show   
   up later, elsewhere in the diagram, and that is important.   
      
   The above diagram is just a graphical picture of the Lorentz equations,   
   which give HIS outbound inertial reference frame's time ("t") and   
   position ("x") coordinates as functions of HER inertial reference   
   frame's time ("T") and position ("X") coordinates.  He is always at   
   position x = 0 in his reference frame, which corresponds to HIS   
   worldline on the diagram.   
      
   Next, we need to draw HIS line of simultaneity that passes through the   
   end of the outbound segment of his worldline ... i.e., his line of   
   simultaneity that passes through the "tic" mark corresponding to the t =   
   20 point on his worldline.  To find that line, we can use the famous   
   time-dilation result, but this time from HIS perspective, not HER'S.   
   According to HIM, whenever he is not accelerating, she ages half as fast   
   as he does.  So when he is 20 at the end of his outbound segment, he   
   knows she is 10 then. Therefore his line of simultaneity must cross the   
   T axis at T = 10.  So draw a line that starts slightly below the T axis,   
   passes through the T = 10 point on the T axis, and passes through the t   
   = 20 point on HIS worldline.  THAT is his line of simultaneity at the   
   end of the outbound segment of his trip.   
      
   That line of simultaneity slopes upward toward the right, with slope   
   1/0.866 = 1.155.  Its angle with respect to the vertical line that was   
   drawn through the t = 20 point on his worldline is alpha (the SAME angle   
   his worldline makes with respect to her worldline).   (That result comes   
   from the Lorentz equations relating her perpetually inertial reference   
   frame to his outbound inertial reference frame.)  Every point on that   
      
   [continued in next message]   
      
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