4 forms of mechanics
$begingroup$
There are 4 main functions in mechanics. L(q,q',t), H(p,q,t), K(p',q',t), G(p,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 hamiltonian-formalism
asked 2 hours ago
Semen YurchenkoSemen Yurchenko
183
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add a comment |
$begingroup$
There are 4 main functions in mechanics. L(q,q',t), H(p,q,t), K(p',q',t), G(p,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 hamiltonian-formalism
asked 2 hours ago
Semen YurchenkoSemen Yurchenko
183
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
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– GiorgioP
1 hour 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
1 hour ago
$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
56 mins ago
add a comment |
$begingroup$
There are 4 main functions in mechanics. L(q,q',t), H(p,q,t), K(p',q',t), G(p,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 hamiltonian-formalism
asked 2 hours ago
Semen YurchenkoSemen Yurchenko
183
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,q',t), H(p,q,t), K(p',q',t), G(p,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 hamiltonian-formalism
classical-mechanics lagrangian-formalism hamiltonian-formalism
asked 2 hours ago
Semen YurchenkoSemen Yurchenko
183
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 2 hours ago
Semen YurchenkoSemen Yurchenko
183
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 2 hours ago
Semen YurchenkoSemen Yurchenko
183
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 2 hours ago
Semen YurchenkoSemen Yurchenko
183
asked 2 hours ago
Semen YurchenkoSemen Yurchenko
183
183
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
1 hour 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
1 hour ago
$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
56 mins ago
add a comment |
1
$begingroup$
You may find useful to have a look at this related question and answers: physics.stackexchange.com/q/415775
$endgroup$
– GiorgioP
1 hour 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
1 hour ago
$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
56 mins 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
1 hour ago
$begingroup$
You may find useful to have a look at this related question and answers: physics.stackexchange.com/q/415775
$endgroup$
– GiorgioP
1 hour 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
1 hour 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
1 hour ago
$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
56 mins ago
$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
56 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is one argument:
Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.
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 $p$.
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.
answered 1 hour ago
Qmechanic♦Qmechanic
103k121841172
$endgroup$
$begingroup$
Sounds reasonable, thank you very much!
$endgroup$
– Semen Yurchenko
58 mins ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is one argument:
Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.
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 $p$.
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.
answered 1 hour ago
Qmechanic♦Qmechanic
103k121841172
$endgroup$
$begingroup$
Sounds reasonable, thank you very much!
$endgroup$
– Semen Yurchenko
58 mins ago
add a comment |
$begingroup$
Here is one argument:
Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.
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 $p$.
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.
answered 1 hour ago
Qmechanic♦Qmechanic
103k121841172
$endgroup$
$begingroup$
Sounds reasonable, thank you very much!
$endgroup$
– Semen Yurchenko
58 mins ago
add a comment |
$begingroup$
Here is one argument:
Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.
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 $p$.
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.
answered 1 hour ago
Qmechanic♦Qmechanic
103k121841172
$endgroup$
Here is one argument:
Starting from Newton's 2nd law, the Lagrangian $L(q,v,t)$ is just one step away.
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 $p$.
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.
answered 1 hour ago
Qmechanic♦Qmechanic
103k121841172
answered 1 hour ago
Qmechanic♦Qmechanic
103k121841172
answered 1 hour ago
Qmechanic♦Qmechanic
103k121841172
answered 1 hour ago
Qmechanic♦Qmechanic
103k121841172
103k121841172
$begingroup$
Sounds reasonable, thank you very much!
$endgroup$
– Semen Yurchenko
58 mins ago
add a comment |
$begingroup$
Sounds reasonable, thank you very much!
$endgroup$
– Semen Yurchenko
58 mins ago
$begingroup$
Sounds reasonable, thank you very much!
$endgroup$
– Semen Yurchenko
58 mins ago
$begingroup$
Sounds reasonable, thank you very much!
$endgroup$
– Semen Yurchenko
58 mins ago
add a comment |
Semen Yurchenko is a new contributor. Be nice, and check out our Code of Conduct.
Semen Yurchenko is a new contributor. Be nice, and check out our Code of Conduct.
Semen Yurchenko is a new contributor. Be nice, and check out our Code of Conduct.
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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
1 hour 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
1 hour ago
$begingroup$
See point 3 in Qmechanic's answer.
$endgroup$
– GiorgioP
56 mins ago