From: goldenfieldquaternions@gmail.com   
      
   On Thursday, June 27, 2019 at 6:14:52 AM UTC-5, Luigi Fortunati wrote:   
   > A gelatin sphere in remote space, far from any gravitational field,   
   > maintains its spherical shape because it is in an inertial reference   
   > frame where there are no forces that can deform it.   
   >   
   > But if the gelatin sphere approaches a planet or a star and falls free,   
   > it gets longer, it ovalizes.   
   >   
   > Is this deformation of the gelatin sphere in free fall in a   
   > gravitational field due to the action of a force or not?   
      
   There are two ways of looking at this. The Newtonian view is there is a   
   difference if the gravitational force across the sphere of   
   material. There is then a difference of force between the antipodal   
   points given be F - F' = -GMm(1/r^2 - 1/r'^2). Now let r' = r + d, for d   
   the diameter of the sphere and so using an approximation we get   
      
   F - F' = -2GMmd/r^3.   
      
   This is treated as a force.   
      
   In general relativity we have a somewhat different perspective. The geodesic   
   equation   
      
   d^2r/ds^2 + G^r_{ab}U^aU^b = 0 ---> G^r_{ab} Christoffel symbol.   
      
   In a weak gravity field has the proper time s ~= t and U^t ~= 1 while   
   all other spatial U^i ~= 0. So this leads to   
      
   d^2r/ds^2 + GM/r^2 = 0,   
      
   for the Christoffel symbol of the Schwarzschild metric. This however is   
   considered a bit of an illusion, for the geodesic equation is a   
   connection coefficient based equation that is not covariant. It turns   
   out to give the Newtonian result in a weak spherically symmetric   
   case. To get the difference in force across the sphere we need to   
   consider the geodesic deviation equation   
      
   dU^a/ds + R^a_{bcd}U^bV^cU^c = 0.   
      
   I have to cut across stuff, for this editor is not convenient. The V^c   
   is the vector separating two test masses, which we take to be the   
   diameter, and the velocity vectors U^a have only U^t ~= 1 and we then   
   recover   
      
   dU^a/ds + 2GMmd/r^3 = 0.   
      
   This is the same math result, but from the interpretation of there being   
   a separation of neighboring geodesic flows.   
      
   The real force involved is the material resistance to this tidal   
   acceleration. In general relativity gravitation is not a real force. A   
   spring folding two masses will be distended by the different geodesic   
   flows. We call this the tidal acceleration or force, but really any   
   force involved is a resistance to this geodesic separation.   
      
   --- SoupGate-Win32 v1.05   
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