Why do we use the Lagrangian and Hamiltonian instead of other related functions?












9












$begingroup$


There are 4 main functions in mechanics. $L(q,dot{q},t)$, $H(p,q,t)$, $K(dot{p},dot{q},t)$, $G(p,dot{p},t)$. First two are Lagrangian and Hamiltonian. Second two are some kind of analogical to first two, but we don't use them. Every where they write that using them can cause problems. But why?






classical-mechanics lagrangian-formalism coordinate-systems hamiltonian-formalism







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  • 1




    $begingroup$
    You may find useful to have a look at this related question and answers: physics.stackexchange.com/q/415775
    $endgroup$
    – GiorgioP
    Jan 18 at 18:32












  • $begingroup$
    Hm, my question is different, I want to know, why is L and H and more preferable for us, while solving tasks
    $endgroup$
    – Semen Yurchenko
    Jan 18 at 18:56










  • $begingroup$
    See point 3 in Qmechanic's answer.
    $endgroup$
    – GiorgioP
    Jan 18 at 19:16






  • 6




    $begingroup$
    What are $G$ and $K$?
    $endgroup$
    – DanielSank
    Jan 18 at 21:34
















9












$begingroup$


There are 4 main functions in mechanics. $L(q,dot{q},t)$, $H(p,q,t)$, $K(dot{p},dot{q},t)$, $G(p,dot{p},t)$. First two are Lagrangian and Hamiltonian. Second two are some kind of analogical to first two, but we don't use them. Every where they write that using them can cause problems. But why?






classical-mechanics lagrangian-formalism coordinate-systems hamiltonian-formalism







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You may find useful to have a look at this related question and answers: physics.stackexchange.com/q/415775
    $endgroup$
    – GiorgioP
    Jan 18 at 18:32












  • $begingroup$
    Hm, my question is different, I want to know, why is L and H and more preferable for us, while solving tasks
    $endgroup$
    – Semen Yurchenko
    Jan 18 at 18:56










  • $begingroup$
    See point 3 in Qmechanic's answer.
    $endgroup$
    – GiorgioP
    Jan 18 at 19:16






  • 6




    $begingroup$
    What are $G$ and $K$?
    $endgroup$
    – DanielSank
    Jan 18 at 21:34














9












9








9


3



$begingroup$


There are 4 main functions in mechanics. $L(q,dot{q},t)$, $H(p,q,t)$, $K(dot{p},dot{q},t)$, $G(p,dot{p},t)$. First two are Lagrangian and Hamiltonian. Second two are some kind of analogical to first two, but we don't use them. Every where they write that using them can cause problems. But why?






classical-mechanics lagrangian-formalism coordinate-systems hamiltonian-formalism







share|cite|improve this question











$endgroup$




There are 4 main functions in mechanics. $L(q,dot{q},t)$, $H(p,q,t)$, $K(dot{p},dot{q},t)$, $G(p,dot{p},t)$. First two are Lagrangian and Hamiltonian. Second two are some kind of analogical to first two, but we don't use them. Every where they write that using them can cause problems. But why?






classical-mechanics lagrangian-formalism coordinate-systems hamiltonian-formalism




classical-mechanics lagrangian-formalism coordinate-systems hamiltonian-formalism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 21:32









David Z♦

63.3k23136252




63.3k23136252










asked Jan 18 at 17:39









Semen YurchenkoSemen Yurchenko

584




584








  • 1




    $begingroup$
    You may find useful to have a look at this related question and answers: physics.stackexchange.com/q/415775
    $endgroup$
    – GiorgioP
    Jan 18 at 18:32












  • $begingroup$
    Hm, my question is different, I want to know, why is L and H and more preferable for us, while solving tasks
    $endgroup$
    – Semen Yurchenko
    Jan 18 at 18:56










  • $begingroup$
    See point 3 in Qmechanic's answer.
    $endgroup$
    – GiorgioP
    Jan 18 at 19:16






  • 6




    $begingroup$
    What are $G$ and $K$?
    $endgroup$
    – DanielSank
    Jan 18 at 21:34














  • 1




    $begingroup$
    You may find useful to have a look at this related question and answers: physics.stackexchange.com/q/415775
    $endgroup$
    – GiorgioP
    Jan 18 at 18:32












  • $begingroup$
    Hm, my question is different, I want to know, why is L and H and more preferable for us, while solving tasks
    $endgroup$
    – Semen Yurchenko
    Jan 18 at 18:56










  • $begingroup$
    See point 3 in Qmechanic's answer.
    $endgroup$
    – GiorgioP
    Jan 18 at 19:16






  • 6




    $begingroup$
    What are $G$ and $K$?
    $endgroup$
    – DanielSank
    Jan 18 at 21:34








1




1




$begingroup$
You may find useful to have a look at this related question and answers: physics.stackexchange.com/q/415775
$endgroup$
– GiorgioP
Jan 18 at 18:32






$begingroup$
You may find useful to have a look at this related question and answers: physics.stackexchange.com/q/415775
$endgroup$
– GiorgioP
Jan 18 at 18:32














$begingroup$
Hm, my question is different, I want to know, why is L and H and more preferable for us, while solving tasks
$endgroup$
– Semen Yurchenko
Jan 18 at 18:56




$begingroup$
Hm, my question is different, I want to know, why is L and H and more preferable for us, while solving tasks
$endgroup$
– Semen Yurchenko
Jan 18 at 18:56












$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
Jan 18 at 19:16




$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
Jan 18 at 19:16




6




6




$begingroup$
What are $G$ and $K$?
$endgroup$
– DanielSank
Jan 18 at 21:34




$begingroup$
What are $G$ and $K$?
$endgroup$
– DanielSank
Jan 18 at 21:34










2 Answers
2






active

oldest

votes


















14












$begingroup$

Here is one argument:




  1. Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.


  2. A Legendre transformation $vleftrightarrow p$ to the Hamiltonian $H(q,p,t)$ is well-defined for a wide class of systems because there is typically a bijective relation between velocity $v$ and momentum $p$.


  3. On the other hand, there is seldomly a bijective relation between position $q$ and force $f$ (although Hooke's law is a notable exception). Therefore the Legendre transforms $K(f,v,t)$ and $G(f,p,t)$ are often ill-defined.







share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    To describe the motion of bodies, it is enough to know two variables — the coordinates and velocitys, or the coordinates and momentums, then the initial conditions are determined. In the case of setting other variables, it is necessary to use boundary conditions, which is more complicated. It is necessary to set the initial coordinates and speeds to satisfy the initial values of other variables. Only the initial coordinates and speed determine the movement of bodies. The value of other variables is a function of coordinates and velocity, and the inverse function does not always exist. When the inverse function does not exist or is not unique, the motion is undefined. To get the initial value of variables $dot p,dot q$, it is necessary to calculate the solution with initial variables - coordinate, speed, go back and get the initial value of variables $dot p,dot q$. In addition, by describing the motion in variables $dot p,dot q$, you will not achieve a complete description; instead of derivatives $dot p,dot q$, knowledge of variables $p,q$ is necessary. To find coordinates and impulses by their derivatives, it is necessary to know their initial values. The circle is closed.






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      14












      $begingroup$

      Here is one argument:




      1. Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.


      2. A Legendre transformation $vleftrightarrow p$ to the Hamiltonian $H(q,p,t)$ is well-defined for a wide class of systems because there is typically a bijective relation between velocity $v$ and momentum $p$.


      3. On the other hand, there is seldomly a bijective relation between position $q$ and force $f$ (although Hooke's law is a notable exception). Therefore the Legendre transforms $K(f,v,t)$ and $G(f,p,t)$ are often ill-defined.







      share|cite|improve this answer









      $endgroup$


















        14












        $begingroup$

        Here is one argument:




        1. Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.


        2. A Legendre transformation $vleftrightarrow p$ to the Hamiltonian $H(q,p,t)$ is well-defined for a wide class of systems because there is typically a bijective relation between velocity $v$ and momentum $p$.


        3. On the other hand, there is seldomly a bijective relation between position $q$ and force $f$ (although Hooke's law is a notable exception). Therefore the Legendre transforms $K(f,v,t)$ and $G(f,p,t)$ are often ill-defined.







        share|cite|improve this answer









        $endgroup$
















          14












          14








          14





          $begingroup$

          Here is one argument:




          1. Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.


          2. A Legendre transformation $vleftrightarrow p$ to the Hamiltonian $H(q,p,t)$ is well-defined for a wide class of systems because there is typically a bijective relation between velocity $v$ and momentum $p$.


          3. On the other hand, there is seldomly a bijective relation between position $q$ and force $f$ (although Hooke's law is a notable exception). Therefore the Legendre transforms $K(f,v,t)$ and $G(f,p,t)$ are often ill-defined.







          share|cite|improve this answer









          $endgroup$



          Here is one argument:




          1. Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.


          2. A Legendre transformation $vleftrightarrow p$ to the Hamiltonian $H(q,p,t)$ is well-defined for a wide class of systems because there is typically a bijective relation between velocity $v$ and momentum $p$.


          3. On the other hand, there is seldomly a bijective relation between position $q$ and force $f$ (although Hooke's law is a notable exception). Therefore the Legendre transforms $K(f,v,t)$ and $G(f,p,t)$ are often ill-defined.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 18:57









          Qmechanic♦Qmechanic

          103k121851178




          103k121851178























              0












              $begingroup$

              To describe the motion of bodies, it is enough to know two variables — the coordinates and velocitys, or the coordinates and momentums, then the initial conditions are determined. In the case of setting other variables, it is necessary to use boundary conditions, which is more complicated. It is necessary to set the initial coordinates and speeds to satisfy the initial values of other variables. Only the initial coordinates and speed determine the movement of bodies. The value of other variables is a function of coordinates and velocity, and the inverse function does not always exist. When the inverse function does not exist or is not unique, the motion is undefined. To get the initial value of variables $dot p,dot q$, it is necessary to calculate the solution with initial variables - coordinate, speed, go back and get the initial value of variables $dot p,dot q$. In addition, by describing the motion in variables $dot p,dot q$, you will not achieve a complete description; instead of derivatives $dot p,dot q$, knowledge of variables $p,q$ is necessary. To find coordinates and impulses by their derivatives, it is necessary to know their initial values. The circle is closed.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                To describe the motion of bodies, it is enough to know two variables — the coordinates and velocitys, or the coordinates and momentums, then the initial conditions are determined. In the case of setting other variables, it is necessary to use boundary conditions, which is more complicated. It is necessary to set the initial coordinates and speeds to satisfy the initial values of other variables. Only the initial coordinates and speed determine the movement of bodies. The value of other variables is a function of coordinates and velocity, and the inverse function does not always exist. When the inverse function does not exist or is not unique, the motion is undefined. To get the initial value of variables $dot p,dot q$, it is necessary to calculate the solution with initial variables - coordinate, speed, go back and get the initial value of variables $dot p,dot q$. In addition, by describing the motion in variables $dot p,dot q$, you will not achieve a complete description; instead of derivatives $dot p,dot q$, knowledge of variables $p,q$ is necessary. To find coordinates and impulses by their derivatives, it is necessary to know their initial values. The circle is closed.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  To describe the motion of bodies, it is enough to know two variables — the coordinates and velocitys, or the coordinates and momentums, then the initial conditions are determined. In the case of setting other variables, it is necessary to use boundary conditions, which is more complicated. It is necessary to set the initial coordinates and speeds to satisfy the initial values of other variables. Only the initial coordinates and speed determine the movement of bodies. The value of other variables is a function of coordinates and velocity, and the inverse function does not always exist. When the inverse function does not exist or is not unique, the motion is undefined. To get the initial value of variables $dot p,dot q$, it is necessary to calculate the solution with initial variables - coordinate, speed, go back and get the initial value of variables $dot p,dot q$. In addition, by describing the motion in variables $dot p,dot q$, you will not achieve a complete description; instead of derivatives $dot p,dot q$, knowledge of variables $p,q$ is necessary. To find coordinates and impulses by their derivatives, it is necessary to know their initial values. The circle is closed.






                  share|cite|improve this answer











                  $endgroup$



                  To describe the motion of bodies, it is enough to know two variables — the coordinates and velocitys, or the coordinates and momentums, then the initial conditions are determined. In the case of setting other variables, it is necessary to use boundary conditions, which is more complicated. It is necessary to set the initial coordinates and speeds to satisfy the initial values of other variables. Only the initial coordinates and speed determine the movement of bodies. The value of other variables is a function of coordinates and velocity, and the inverse function does not always exist. When the inverse function does not exist or is not unique, the motion is undefined. To get the initial value of variables $dot p,dot q$, it is necessary to calculate the solution with initial variables - coordinate, speed, go back and get the initial value of variables $dot p,dot q$. In addition, by describing the motion in variables $dot p,dot q$, you will not achieve a complete description; instead of derivatives $dot p,dot q$, knowledge of variables $p,q$ is necessary. To find coordinates and impulses by their derivatives, it is necessary to know their initial values. The circle is closed.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 23 at 15:22

























                  answered Jan 23 at 14:25









                  Evgeniy YakubovskiyEvgeniy Yakubovskiy

                  6416




                  6416






























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