From: goldenfieldquaternions@gmail.com   
      
   [[Mod. note -- I have rewrapped overly-long lines.  -- jt]]   
      
   On Thursday, September 27, 2018 at 1:04:31 AM UTC-5, jacque...@neuf.fr wrote:   
   > Hello   
   >   
   > The timelike geodesic in Schwarzschild spacetime can be written in GR:   
   >   
   > E^2 -1 = (dr/d_tau)^2 -2GM/r +(L/r)^2 - (2GM L^2)/r^3    (1)   
   >   
   > where E is the relativistic conserved energy of a unitary mass, on the   
   > geodesic and L its conserved angular momentum.   
   >   
   > In Newtonian mechanics the Hamiltonian is   
   >   
   > e = 1/2 (dr.dt)^2 - GM/r + (1/2)(l/r)^2      (2)   
   >   
   > where e is the total conserved energy on the geodesic of a unitary mass   
   > and l its conserved angular momentum.   
   >   
   > In weak field, for r >> rs, we may neglect the last term in   
   > (1/r^3). Usually one divide equation 1 by 2 for getting a similar form.   
   >   
   > (E^2 -1)/2 = (1/2)(dr/d_tau)^2 -GM/r + (1/2)(L/r)^2     (3)   
   >   
   >   
   > But for convergence of equations (2) and (3), we have to assume that   
   > e = (E^2 -1)/2.   
   >   
   > That is the point that I do not understand: How an energy can be equal   
   > to a square energy?   
   >   
   > Thanks for help.   
   > jacques   
      
   This is a terribly incovenient format for reading or writing math.   
   However, I will try to do this. Let me start with the Schwarzschild   
   line element   
      
   ds^2 = (1 - 2m/r)dt^2 - dr^2/(1 - 2m/r) - r^2dOmega^2. m = GM/c^2   
      
   I divide through by dt^2 and have, v^r = dr/ds and so forth. The   
   angular part I reduce to a single angle that parameterizes a circular   
   orbit and I call that angle @. Further for a circular orbit dr and   
   v^r = 0. So this becomes   
      
   (ds/dt)^2 = (1 - 2m/r)(U^t)^2 - r^2(v^@)^2,   
      
   In the non-relativistic limit we have U^t = 1 and we can write this as   
      
   (ds/dt)^2 - 1 = 2m/r + r^2(v^@)^2.   
      
   This looks suspiciously similar to a Lagrangian, and by dividing through   
   by 2 and setting rv^@ = v, the orbital velocity, then   
      
   (ds/dt)^2 - 1 = L = 1/2 mv^2 + mc^2/r.   
      
   There is an implicit c^2 with the (1 =E2=80=93 2m/r)dt^2 and this   
   reduces things to   
      
   L = 1/2 mv^2 + GM/r.   
      
   I am presuming all readers know how to work with Lagrangians.   
      
   The line element is formally the same as an action. So this is a   
   quick way to see a connection between general relativity and Newtonian   
   gravity. One does not need to work through all the Christoffel   
   symbols and the rest.   
      
   --- SoupGate-Win32 v1.05   
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