From: ram@zedat.fu-berlin.de   
      
   Richard Livingston  writes:   
   >In the modern interpretation gravity is not a force, but what we   
   >attribute to the force of gravity is really a non-Euclidean space-   
   >time in the neighborhood of massive objects.  The only real force   
   >is the one at your feet that is accelerating you upwards at 1 g.   
      
     "Accelerating" means "changing the speed",   
     and "speed" means "change of position", right?   
      
     a = dv/dt, v = dx/dt, a = 1 g ==> dv/dt = 1 g.   
     v = g t + v0, x = (1/2) g t^2 + v0 t + x0.   
     Assuming: v0 = 0 and x0 = 0: x = (1/2) g t^2.   
      
     I assume that the Australians are my antipodes.   
      
     So, I am accelerated since a while with 1 g upwards, as you   
     say above, (=change of my velocity upwards (=change of my   
     location upwards)) and the Australians with 1 g downwards.   
      
     Shouldn't I then be moving further and further away from   
     Australians?   
      
   [[Mod. note -- In order to add/subtract an acceleration vector "here"   
   to/from an acceleration vector "there", we need a common inertial reference   
   frame that contains both "here" and "there".  If were in a flat spacetime,   
   that would be easy, since in a flat spacetime inertial reference frames   
   are of infinite extent.  But we don't live in a flat spacetie, we live   
   in a curved spacetime, and in a curved spacetime an inertial reference   
   frame is only an approximation valid in a small region.  (The precise   
   definition of "small" depends on how good of an approximation you want,   
   i.e., how small you want an acceleration to be in order to call it   
   "negligable".)   
      
   As you've just observed, the opposite sides of the Earth are too far   
   apart to be contained in a common inertial reference frame (assuming   
   that we're not willing to treat +/- 1 g accelerations as "negligable").   
      
   That is your 1 g "up" acceleration is measured with respect to a   
   *different* inertial reference frame from the the Australian's "1 g down",   
   and hence you can't just add/subtract them without taking into account   
   the non-trivial transformation (induced by spacetime curvature) between   
   those two different inertial reference frames.   
   -- jt]]   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)