If a particle of mass m moves in the x- y plane, then its equations of motion are m d 2 x d t 2 = f ( t , x , y ) and m d 2 y d t 2 = g ( t , x , y ) .Here f and g represent the x and y components, respectively, of the force acting on the particle. Replace this system of two second-order equations by an equivalent system of four first order equations of the form: y 1 ′ = f 1 ( x , y 1 , . . . , y n ) y 2 ′ = f 2 ( x , y 1 , . . . , y n ) y n ′ = f n ( x , y 1 , . . . , y n )I understand how replace a differential equation by an equivalent system of first order equations when the differential looks something like x y ″ − x 2 y ′ − x 3 y = 0An equivalent system is: y 0 ′ = y 1 y 1 ′ = x 2 y 0 + x y 1 Therefore, I need someone to point me in the right direction for my question stated at the top.

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If a particle of mass   m moves in the   x-  y plane, then its equations of motion are  m                    d        2            x              d              t        2              =  f  (  t  ,  x  ,  y  )        and        m                    d        2            y              d              t        2              =  g  (  t  ,  x  ,  y  )  .Here   f and   g represent the   x and   y components, respectively, of the force acting on the particle. Replace this system of two second-order equations by an equivalent system of four first order equations of the form:      y    1    ′    =      f    1    (  x  ,      y    1    ,  .  .  .  ,      y    n    )      y    2    ′    =      f    2    (  x  ,      y    1    ,  .  .  .  ,      y    n    )      y    n    ′    =      f    n    (  x  ,      y    1    ,  .  .  .  ,      y    n    )I understand how replace a differential equation by an equivalent system of first order equations when the differential looks something like  x      y    ″    −      x    2        y    ′    −      x    3    y  =  0An equivalent system is:      y    0    ′    =      y    1        y    1    ′    =      x    2        y    0    +  x      y    1  Therefore, I need someone to point me in the right direction for my question stated at the top.

Danika Rojas · Accepted Answer

It&#039;s the same thing, you just state       x    ′    (  t  )  =  r  (  t  ), and       y    ′    (  t  )  =  s  (  t  ) and you&#039;ll have the system                              x          ′                (        t        )                            =        r        (        t        )                                      y          ′                (        t        )                            =        s        (        t        )                                      r          ′                (        t        )                            =                  1          m                f        (        t        ,        x        ,        y        )                                      s          ′                (        t        )                            =                  1          m                g        (        t        ,        x        ,        y        )            Note that if   f  =  f  (  t  ,  x  ,  y  ,      x    ′    ,      y    ′    ) then the substitution would lead to       r    ′    =      1    m    f  (  t  ,  x  ,  y  ,  r  ,  s  ); the same thing for   g.On a side note, if you take                     x        i            ˙        (  t  )  =      1          m      i                          p        i            ˙        (  t  ),   1  ≤  i  ≤  n, you&#039;ll have the problem formulated clearly in the language of classical mechanics. In that case, for a n particle system                                                        x              1                        ˙                          (        t        )                            =                  1                      m            1                                    p          1                (        t        )                                                                              x              2                        ˙                          (        t        )                            =                  1                      m            2                                    p          2                (        t        )                                          ⋮                                                                x              n                        ˙                          (        t        )                            =                  1                      m            n                                    p          n                (        t        )                                                                                            p              1                        ˙                          (        t        )                            =                  f          1                (        t        ,                  x          1                ,        …        ,                  x          n                ,                  p          1                ,        …        ,                  p          n                )                                                                              p              2                        ˙                          (        t        )                            =                  f          2                (        t        ,                  x          1                ,        …        ,                  x          n                ,                  p          1                ,        …        ,                  p          n                )                                          ⋮                                                                p              n                        ˙                          (        t        )                            =                  f          n                (        t        ,                  x          1                ,        …        ,                  x          n                ,                  p          1                ,        …        ,                  p          n                )            This are Newton&#039;s equations rewritten in terms of momenta. The space   {      x    1    ,  …  ,      x    n    ,      p    1    ,  …  ,      p    n    } is called the Configuration Space, and it&#039;s very important for the study of the behavior of the system.

antennense · Accepted Answer

Given the physical flavour of the problem, I&#039;ll explain as follws: If you know the Lagrangian of the system, say   L  ≡  L  (  q  ,            q      ˙        )  =  T  −  V, the equations of motion are (2nd order)      d          d      t                  ∂      L              ∂                                    q            ˙                                =            ∂      L              ∂      q      Now, state the problem in Hamiltonian Mechanics. Define the momenta as  p  =            ∂      L              ∂                        q          ˙                    and solve for             q      ˙        =            q      ˙        (  p  ). Write the Hamiltonian in that new variables:  H  ≡  H  (  p  ,  q  )  =            ∑              i                    p      i                                q          ˙                    i        −    L  Then the equations of motion are (1st order)            p      ˙        =  −            ∂      H              ∂      q        ,            q      ˙        =            ∂      H              ∂      p

If a particle of mass m moves in the x - y plane, then its equations of motion are

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