From: richalivingston@gmail.com   
      
   On Wednesday, September 7, 2022 at 2:43:54 PM UTC-5, Luigi Fortunati wrote:   
   > Richard Livingston mercoledÄ=9B 07/09/2022 alle ore 05:12:12 ha scritto:   
   > >> A geodesic passes from A and also from B.   
   > >>   
   > >> Is the direction from A to B fully equivalent to the direction from B   
   > >> to A?   
   > >>   
   > >> Or can it happen that one of the two directions prevails over the   
   > >> other?   
   > >   
   > > In general relativity, the physical (spatial) distance from A to B is   
   always   
   > > the same as from B to A. This is not true of the time for light travel   
   > > however. If A is higher in a gravitational field than B, the round trip   
   > > light travel time is longer for A to B to A (from the point of view of A)   
   > > than it is for B to A to B (from the point of view of B). For any single   
   > > observer the round trip times are the same either way. That may seem   
   > > paradoxical, but it is actually true and consistent.   
   > I absolutely agree with what you wrote.   
   >   
   > But how does all this affect the direction of motion that a body   
   > decides to take when it is left free to follow its geodesic?   
   >   
   ...   
   > What is it that forces the elevator to always get started downwards and   
   > never upwards?   
   >   
   > What has the down direction more than the up direction?   
   >   
   > In other words, are geodesics open lines or are they vectors with one   
   > direction privileged over the other?   
      
   There are many geodesics passing through any given point (event) in   
   space-time.  What you are asking is how does a velocity 4-vector evolve.   
    The velocity 4-vector for a "stationary" object is (c, 0, 0, 0) in flat   
   space-time.  In the vicinity of a large mass M it is (c(1-2M/r)^-1/2, 0,   
   0, 0) (ref eqn. 20.58 in "Gravity: an introduction to Einsteins general   
   relativity" by J. B. Hartle, 2003).  The most rigorous way to calculate   
   this is to calculate the covariant derivative in the metric you are   
   concerned with.  For the Schwarzschild metric this gives an acceleration   
   vector in the radial direction:.   
      
   (Revtex code:   
   a^\alpha = \GAMMA^{\alpha}_{tt} \left( u^t \right)   
   )   
      
   where \GAMMA is a Christoffel symbol, which gives   
   a = (0, M/r^2, 0, 0)   
      
   This is actually the acceleration necessary to hold the object   
   stationary at this radius.  If the object was released it would be in   
   free fall (i.e. no acceleration in the objects frame).  In terms of the   
   stationary observer the free falling object would accelerate at a = (0,   
   -M/r^2, 0, 0) i.e. a radial acceleration towards the mass M.   
      
   This probably doesn't answer your question (it didn't for me) because   
   the covariant derivative, while precise and correct, is almost opaque to   
   intuitive understanding of the physics, which I believe is what you are   
   asking.   
      
   The covariant derivative is related to the concept of parallel   
   transport. If we introduce a simple idea from quantum mechanics we can   
   more easily visualize the parallel transport of the particle wave   
   function in time, and recognize the resulting acceleration towards the   
   mass.   
      
   In relativity a particle with mass m has energy mc^2.  In QM this is a   
   frequency mc^2/h, which I will call f = 1/T.  For an object at rest, by   
   definition the wave function is in phase over some local region of   
   space. In terms of flat space-time the phase of this wave function   
   advances by 2pi every time interval T.  This happens over the local   
   region at the same time as the g_{00} metric component is a constant   
   everywhere in flat space-time.   
      
   However near a large mass M, due to the gravitational red shift, not   
   every location has the same frequency.  A region closer to the mass M   
   will appear to advance in phase more slowly, and a region further out   
   will advance more quickly.  As a result the wave function for a   
   stationary object of mass m, while initially in phase over some local   
   area, will gradually become more and more out of phase along a radial   
   axis.  This is entirely because the g_{00} metric component varies along   
   the r coordinate direction.  The change in phase along r is, in QM, the   
   momentum of the particle, and this momentum gradually increases with   
   time.  It is easy to show that this gives the same answer as the   
   covariant derivative.   
      
   Does that help?   
      
   Rich L.   
      
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