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












6












$begingroup$


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
    9 hours ago












  • $begingroup$
    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
    8 hours ago










  • $begingroup$
    See point 3 in Qmechanic's answer.
    $endgroup$
    – GiorgioP
    8 hours ago






  • 1




    $begingroup$
    What are $G$ and $K$?
    $endgroup$
    – DanielSank
    6 hours ago
















6












$begingroup$


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









New contributor




Semen Yurchenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




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












  • $begingroup$
    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
    8 hours ago










  • $begingroup$
    See point 3 in Qmechanic's answer.
    $endgroup$
    – GiorgioP
    8 hours ago






  • 1




    $begingroup$
    What are $G$ and $K$?
    $endgroup$
    – DanielSank
    6 hours ago














6












6








6


1



$begingroup$


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









New contributor




Semen Yurchenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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









New contributor




Semen Yurchenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Semen Yurchenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









David Z♦

63.3k23136252




63.3k23136252






New contributor




Semen Yurchenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 10 hours ago









Semen YurchenkoSemen Yurchenko

384




384




New contributor




Semen Yurchenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Semen Yurchenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Semen Yurchenko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




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












  • $begingroup$
    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
    8 hours ago










  • $begingroup$
    See point 3 in Qmechanic's answer.
    $endgroup$
    – GiorgioP
    8 hours ago






  • 1




    $begingroup$
    What are $G$ and $K$?
    $endgroup$
    – DanielSank
    6 hours ago














  • 1




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












  • $begingroup$
    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
    8 hours ago










  • $begingroup$
    See point 3 in Qmechanic's answer.
    $endgroup$
    – GiorgioP
    8 hours ago






  • 1




    $begingroup$
    What are $G$ and $K$?
    $endgroup$
    – DanielSank
    6 hours ago








1




1




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






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














$begingroup$
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
8 hours ago




$begingroup$
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
8 hours ago












$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
8 hours ago




$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
8 hours ago




1




1




$begingroup$
What are $G$ and $K$?
$endgroup$
– DanielSank
6 hours ago




$begingroup$
What are $G$ and $K$?
$endgroup$
– DanielSank
6 hours ago










1 Answer
1






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8












$begingroup$

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









$endgroup$













  • $begingroup$
    Sounds reasonable, thank you very much!
    $endgroup$
    – Semen Yurchenko
    8 hours ago











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

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









$endgroup$













  • $begingroup$
    Sounds reasonable, thank you very much!
    $endgroup$
    – Semen Yurchenko
    8 hours ago
















8












$begingroup$

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









$endgroup$













  • $begingroup$
    Sounds reasonable, thank you very much!
    $endgroup$
    – Semen Yurchenko
    8 hours ago














8












8








8





$begingroup$

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 8 hours ago









Qmechanic♦Qmechanic

103k121841172




103k121841172












  • $begingroup$
    Sounds reasonable, thank you very much!
    $endgroup$
    – Semen Yurchenko
    8 hours ago


















  • $begingroup$
    Sounds reasonable, thank you very much!
    $endgroup$
    – Semen Yurchenko
    8 hours ago
















$begingroup$
Sounds reasonable, thank you very much!
$endgroup$
– Semen Yurchenko
8 hours ago




$begingroup$
Sounds reasonable, thank you very much!
$endgroup$
– Semen Yurchenko
8 hours ago










Semen Yurchenko is a new contributor. Be nice, and check out our Code of Conduct.










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