From: nospam@de-ster.demon.nl   
      
   Thomas 'PointedEars' Lahn  wrote:   
      
   > 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   
      
   See? This works. (almost, only one ?)   
   BTW, the customaty symbol for momentum is p not P,   
      
   Jan   
      
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