From: PointedEars@web.de   
      
   You ought to trim your quotations to the relevant minimum.   
      
   Ross Finlayson wrote:   
   > You mention least action and it's a pretty reasonable principle,   
      
   Yes, it is.   
      
   > where the theory is sum-of-histories sum-of-potentials   
   > least-action least-gradient a continuum mechanics, that   
   > obviously enough it's a field theory.   
      
   No, no and no.  That's such a nonsense, it's not even wrong.   
      
   > You know, momentum isn't very much conserved in kinematics.   
      
   It is conserved if no force is acting.  Different to Newtonian mechanics,   
   Lagrangian mechanics proves this in a way that does not already presume   
   Newton's Laws of Motion:   
      
   The Euler--Lagrange equation for the coordinate x is   
      
     d/dt ∂L/∂(dx/dt) - ∂L/∂x = 0.   
      
   ("t" could be any parameter, but in physics it is usually taken as time.)   
      
   ∂L/∂(dx/dt) is the *canonical momentum conjugate to x*, a terminology that   
   stems from that for the Newtonian Lagrangian one finds   
      
     ∂L/∂(dx/dt) = m v_x = p_x   
      
   (see below).   
      
   If ∂L/∂x = 0, then trivially   
      
     d/dt ∂L/∂(dx/dt) = 0,   
      
   i.e. ∂L/∂(dx/dt) is conserved.   
      
   The Newtonian Lagrangian is in one dimension   
      
     L = T(dx/dt) - U(x) = 1/2 m (dx/dt)^2 - U(x)   
      
   where T is the kinetic energy and U is the potential energy.  Therefore,   
      
     ∂L/∂x = -∂U/∂x = F_x.   
      
   So   
      
     F_x = d/dt p_x,   
      
   and if F_x = 0, then   
      
     d/dt p_x = 0,   
      
   i.e. the x-component of the linear momentum is conserved.   
      
   This is obtained analogously for the 3-dimensional Lagrangian (here in   
   Cartesian coordinates)   
      
     L = 1/2 m (dX/dt)^2 - U(X)   
       = 1/2 m [(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2] - U(x, y, z),   
      
   and y and z, so   
      
     F = -(∂U/∂x, ∂U/∂y, ∂U/∂z)^T = -∇U = d/dt P   
      
   (Newton's Second Law of Motion).  So if F = 0, then   
      
     d/dt P = 0   
      
   (Newton's First Law of Motion), and the linear momentum is conserved.   
      
   > [pseudo-scientific word salad]   
   --   
   PointedEars   
      
   Twitter: @PointedEars2   
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