Why do we use the Lagrangian and Hamiltonian instead of other related functions?
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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
edited Jan 18 at 21:32
David Z♦
63.3k23136252
asked Jan 18 at 17:39
Semen YurchenkoSemen Yurchenko
584
$endgroup$
add a comment |
$begingroup$
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
edited Jan 18 at 21:32
David Z♦
63.3k23136252
asked Jan 18 at 17:39
Semen YurchenkoSemen Yurchenko
584
$endgroup$
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
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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
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See point 3 in Qmechanic's answer.
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– GiorgioP
Jan 18 at 19:16
6
$begingroup$
What are $G$ and $K$?
$endgroup$
– DanielSank
Jan 18 at 21:34
add a comment |
$begingroup$
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
edited Jan 18 at 21:32
David Z♦
63.3k23136252
asked Jan 18 at 17:39
Semen YurchenkoSemen Yurchenko
584
$endgroup$
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
edited Jan 18 at 21:32
David Z♦
63.3k23136252
asked Jan 18 at 17:39
Semen YurchenkoSemen Yurchenko
584
edited Jan 18 at 21:32
David Z♦
63.3k23136252
asked Jan 18 at 17:39
Semen YurchenkoSemen Yurchenko
584
edited Jan 18 at 21:32
David Z♦
63.3k23136252
edited Jan 18 at 21:32
David Z♦
63.3k23136252
edited Jan 18 at 21:32
David Z♦
63.3k23136252
63.3k23136252
asked Jan 18 at 17:39
Semen YurchenkoSemen Yurchenko
584
asked Jan 18 at 17:39
Semen YurchenkoSemen Yurchenko
584
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
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
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
add a comment |
2 Answers
2
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 Jan 18 at 18:57
Qmechanic♦Qmechanic
103k121851178
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 23 at 14:25
Evgeniy YakubovskiyEvgeniy Yakubovskiy
6416
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 Jan 18 at 18:57
Qmechanic♦Qmechanic
103k121851178
$endgroup$
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 Jan 18 at 18:57
Qmechanic♦Qmechanic
103k121851178
$endgroup$
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 Jan 18 at 18:57
Qmechanic♦Qmechanic
103k121851178
$endgroup$
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 Jan 18 at 18:57
Qmechanic♦Qmechanic
103k121851178
answered Jan 18 at 18:57
Qmechanic♦Qmechanic
103k121851178
answered Jan 18 at 18:57
Qmechanic♦Qmechanic
103k121851178
answered Jan 18 at 18:57
Qmechanic♦Qmechanic
103k121851178
103k121851178
add a comment |
add a comment |
$begingroup$
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.
answered Jan 23 at 14:25
Evgeniy YakubovskiyEvgeniy Yakubovskiy
6416
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 23 at 14:25
Evgeniy YakubovskiyEvgeniy Yakubovskiy
6416
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 23 at 14:25
Evgeniy YakubovskiyEvgeniy Yakubovskiy
6416
$endgroup$
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.
answered Jan 23 at 14:25
Evgeniy YakubovskiyEvgeniy Yakubovskiy
6416
edited Jan 23 at 15:22
answered Jan 23 at 14:25
Evgeniy YakubovskiyEvgeniy Yakubovskiy
6416
answered Jan 23 at 14:25
Evgeniy YakubovskiyEvgeniy Yakubovskiy
6416
answered Jan 23 at 14:25
Evgeniy YakubovskiyEvgeniy Yakubovskiy
6416
6416
add a comment |
add a comment |
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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