Notes on Classical Dynamics 
                                             
                                             
                                             
                                             
             References: 
              
             Marion, J. B. & Thornton, S. T. (1995)  Classical dynamics of particles and 
                      th
             systems. 4  Ed. Saunders. 
              
                                                         nd
             Shankar, R. (1994) Principles of quantum mechanics. 2  Ed. Plenum. 
                                   
                                     
                                    Newtonian Mechanics (1687) 
                                     
                                    Notation: (d/dt)x(t) = x'(t) 
                                     
                                    F = m x''  , i = 1 to n 
                                       i           i    i
                                     
                                    p = m ×  x'  = Momentum in Cartesian coordinates 
                                       i          i        i
                                     
                                    Then p' = F  
                                                    i           i
                                     
                                     
                                    Example: Simple Harmonic Oscillator 
                                     
                                    x = displacement of pendulum along horizontal axis 
                                     
                                    Hooks law for restoring force (linear approximation for small displacements) 
                                    F = - k x  
                                     
                                                                                              2                                2
                                    m x'' = -k x,    or     x'' + w x = 0 , with w  = (k/m) and with x'(0) = 0 
                                     
                                    The solution is  x(t) = x(0)cos(wt),  
                                     
                                     
                                    We can define a potential U when work done is path independent: 
                                     
                                    dW = increment in work  and U = òdW = òFdx  
                                     
                                    Under this condition, F = - (¶ U /¶ x) 
                                                                                        i                             i
                                     
                                    For the simple harmonic oscillator: 
                                     
                                                                                            2
                                     U= òFdx = òkx dx  =  k x /2 
                  Euler Equation (1744)  
                   
                  A[y(t)] = òL( y , y') dt   for i  < t <  f  . Find Extremum of A. 
                   
                   
                  Suppose x(t) is an extremum. Define y(a,x(t)) = x(t) + a× n(t) for some 
                  constant a and n(t) is an arbitrary function with n(i) = n(f) = 0.  
                   
                  A necessary condition for an extremum is that (¶/¶a)A |a=0 = 0. 
                   
                   
                  (¶/¶a)A = ò(¶/¶a)L( y , y') dt  
                   
                  = ò [ (¶L/¶y)  (¶y/¶a) +  (¶L/¶y') (¶y'/¶a) ] dt 
                   
                  (¶y/¶a) = n(t) and (¶y'/¶a) = n'(t)     so 
                   
                  (¶/¶a)A = ò  [(¶L/¶y)n(t) + (¶L/¶y') n'(t)] dt 
                   
                  Now we use integration by parts on the second term. 
                   
                   
                  Recall  (fg)' = f ' g + f g'  
                   
                  => ò (fg)'  = ò f 'g + òf g'   
                   
                  => òf g'  = fg |    - ò f 'g 
                                  (i,,f)
                   
                  substituting 
                   
                  f = (¶L/¶y') , f ' =  (¶L/¶y')' 
                   
                  g = n(t), g' = n'(t), then 
                   
                  ò (¶L/¶y') n'(t) = (¶L/¶y')n(t)|    - ò (¶L/¶y)'n(t)dt 
                                                  (i,f)
                   
                  (¶L/¶y')n(t)|    = 0 because n(i) = n(f) = 0. 
                                (i,f)
                   
                  Therefore  
                   
                  (¶/¶a)A|     = ò  [(¶L/¶y) - (¶L/¶y')']× n(t) dt |   = 0 
                           a=0                                     a=0
                   
                  for any n(t) implies 
                   
                  [(¶L/¶y) - (d/dt)(¶L/¶y')] = 0   (Euler's Equation).