Hamiltonian and Lagrangian correspondence
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I'm trying to clarify how we get a Hamiltonian directly from a Lagrangian using the Legendre transform. Let me give some preliminaries for my question to make sense.
A Hamiltonian system is a triple $(M,omega, H)$, where $(M,omega)$ is a symplectic manifold and $Hin C^infty(M)$ is some prescribed smooth function (the symplectic structure isn't important to this question, but since this is a symplectic topic, I'm going to include it).
Now, let $M$ be a smooth manifold and let $omega$ be the standard symplectic form on $T^*M$, and let $Fin C^infty(TM)$ be our Lagrangian, that is, $F_p:=F|_{T_pM}$ is strictly convex for each $pin M$. For the moment, let's fix $pin M$ and focus on $F_p:T_pMtomathbb{R}$. For $alphain T_p^*M$, define
$$F_{p,alpha}(v)=F_p(v)-alpha(v)$$
for $vin T_pM$. The mapping $F_{p,alpha}$ is said to be stable if there exists a unique critical point in $T_pM$ for $F_{p,alpha}$. Then define the stability set
$$S_{F_p}={alphain T_p^*M:F_{p,alpha}text{ is stable}}.$$
Then $S_{F_p}$ is open and convex in $T_p^*M$, and leads to my first question of when do we have that $S_{F_p}=T_p^*M$?
Furthermore, since $F_p$ is strictly convex, we have that the Legendre transform associated to $F_p$, denoted $L_{F_p}:T_pMto S_{F_p}$ is a diffeomorphism. Moreover, with our dual $F_p^*:S_{F_p}tomathbb{R}$, we see that
$$L_{F_p}^{-1}=L_{F_p^*}.$$
Again, with my understanding, don't we need $S_{F_p}=T_p^*M$ for this to make sense?
Combining the fibewise defined maps, we obtain
$$mathcal{L}:TMto T^*M,qquad mathcal{L}|_{T_pM}=L_{F_p},$$
and
$$H:S_Ftomathbb{R},qquad H|_{S_{F_p}}=F_p^*,$$
and $mathcal{L}:TMto S_F$ is a diffeomorphism.
Note that I'm using the notation of
$$S_F=bigcup_{pin M}S_{F_p}.$$
Now, this defined $H$ should yield the Hamiltonian system $(T^*M,omega,H)$, but we don't necessarily have $H$ defined on the whole cotangent bundle, which seems odds since our Lagrangian was defined on the whole tangent bundle. Am I missing something in this construction? I was under the impression that Lagrangians on the tangent bundle were equivalent to Hamiltonians on the cotangent bundle.
I guess all of this reduced to the question: How do we remedy the fact that $S_{F_p}$ and $T_p^*M$ aren't necessarily equal?
As always, any help or references would be greatly appreciated.
differential-geometry differential-topology symplectic-geometry euler-lagrange-equation hamilton-equations
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I'm trying to clarify how we get a Hamiltonian directly from a Lagrangian using the Legendre transform. Let me give some preliminaries for my question to make sense.
A Hamiltonian system is a triple $(M,omega, H)$, where $(M,omega)$ is a symplectic manifold and $Hin C^infty(M)$ is some prescribed smooth function (the symplectic structure isn't important to this question, but since this is a symplectic topic, I'm going to include it).
Now, let $M$ be a smooth manifold and let $omega$ be the standard symplectic form on $T^*M$, and let $Fin C^infty(TM)$ be our Lagrangian, that is, $F_p:=F|_{T_pM}$ is strictly convex for each $pin M$. For the moment, let's fix $pin M$ and focus on $F_p:T_pMtomathbb{R}$. For $alphain T_p^*M$, define
$$F_{p,alpha}(v)=F_p(v)-alpha(v)$$
for $vin T_pM$. The mapping $F_{p,alpha}$ is said to be stable if there exists a unique critical point in $T_pM$ for $F_{p,alpha}$. Then define the stability set
$$S_{F_p}={alphain T_p^*M:F_{p,alpha}text{ is stable}}.$$
Then $S_{F_p}$ is open and convex in $T_p^*M$, and leads to my first question of when do we have that $S_{F_p}=T_p^*M$?
Furthermore, since $F_p$ is strictly convex, we have that the Legendre transform associated to $F_p$, denoted $L_{F_p}:T_pMto S_{F_p}$ is a diffeomorphism. Moreover, with our dual $F_p^*:S_{F_p}tomathbb{R}$, we see that
$$L_{F_p}^{-1}=L_{F_p^*}.$$
Again, with my understanding, don't we need $S_{F_p}=T_p^*M$ for this to make sense?
Combining the fibewise defined maps, we obtain
$$mathcal{L}:TMto T^*M,qquad mathcal{L}|_{T_pM}=L_{F_p},$$
and
$$H:S_Ftomathbb{R},qquad H|_{S_{F_p}}=F_p^*,$$
and $mathcal{L}:TMto S_F$ is a diffeomorphism.
Note that I'm using the notation of
$$S_F=bigcup_{pin M}S_{F_p}.$$
Now, this defined $H$ should yield the Hamiltonian system $(T^*M,omega,H)$, but we don't necessarily have $H$ defined on the whole cotangent bundle, which seems odds since our Lagrangian was defined on the whole tangent bundle. Am I missing something in this construction? I was under the impression that Lagrangians on the tangent bundle were equivalent to Hamiltonians on the cotangent bundle.
I guess all of this reduced to the question: How do we remedy the fact that $S_{F_p}$ and $T_p^*M$ aren't necessarily equal?
As always, any help or references would be greatly appreciated.
differential-geometry differential-topology symplectic-geometry euler-lagrange-equation hamilton-equations
This question has an open bounty worth +50
reputation from Matt ending in 1 hour.
Looking for an answer drawing from credible and/or official sources.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to clarify how we get a Hamiltonian directly from a Lagrangian using the Legendre transform. Let me give some preliminaries for my question to make sense.
A Hamiltonian system is a triple $(M,omega, H)$, where $(M,omega)$ is a symplectic manifold and $Hin C^infty(M)$ is some prescribed smooth function (the symplectic structure isn't important to this question, but since this is a symplectic topic, I'm going to include it).
Now, let $M$ be a smooth manifold and let $omega$ be the standard symplectic form on $T^*M$, and let $Fin C^infty(TM)$ be our Lagrangian, that is, $F_p:=F|_{T_pM}$ is strictly convex for each $pin M$. For the moment, let's fix $pin M$ and focus on $F_p:T_pMtomathbb{R}$. For $alphain T_p^*M$, define
$$F_{p,alpha}(v)=F_p(v)-alpha(v)$$
for $vin T_pM$. The mapping $F_{p,alpha}$ is said to be stable if there exists a unique critical point in $T_pM$ for $F_{p,alpha}$. Then define the stability set
$$S_{F_p}={alphain T_p^*M:F_{p,alpha}text{ is stable}}.$$
Then $S_{F_p}$ is open and convex in $T_p^*M$, and leads to my first question of when do we have that $S_{F_p}=T_p^*M$?
Furthermore, since $F_p$ is strictly convex, we have that the Legendre transform associated to $F_p$, denoted $L_{F_p}:T_pMto S_{F_p}$ is a diffeomorphism. Moreover, with our dual $F_p^*:S_{F_p}tomathbb{R}$, we see that
$$L_{F_p}^{-1}=L_{F_p^*}.$$
Again, with my understanding, don't we need $S_{F_p}=T_p^*M$ for this to make sense?
Combining the fibewise defined maps, we obtain
$$mathcal{L}:TMto T^*M,qquad mathcal{L}|_{T_pM}=L_{F_p},$$
and
$$H:S_Ftomathbb{R},qquad H|_{S_{F_p}}=F_p^*,$$
and $mathcal{L}:TMto S_F$ is a diffeomorphism.
Note that I'm using the notation of
$$S_F=bigcup_{pin M}S_{F_p}.$$
Now, this defined $H$ should yield the Hamiltonian system $(T^*M,omega,H)$, but we don't necessarily have $H$ defined on the whole cotangent bundle, which seems odds since our Lagrangian was defined on the whole tangent bundle. Am I missing something in this construction? I was under the impression that Lagrangians on the tangent bundle were equivalent to Hamiltonians on the cotangent bundle.
I guess all of this reduced to the question: How do we remedy the fact that $S_{F_p}$ and $T_p^*M$ aren't necessarily equal?
As always, any help or references would be greatly appreciated.
differential-geometry differential-topology symplectic-geometry euler-lagrange-equation hamilton-equations
I'm trying to clarify how we get a Hamiltonian directly from a Lagrangian using the Legendre transform. Let me give some preliminaries for my question to make sense.
A Hamiltonian system is a triple $(M,omega, H)$, where $(M,omega)$ is a symplectic manifold and $Hin C^infty(M)$ is some prescribed smooth function (the symplectic structure isn't important to this question, but since this is a symplectic topic, I'm going to include it).
Now, let $M$ be a smooth manifold and let $omega$ be the standard symplectic form on $T^*M$, and let $Fin C^infty(TM)$ be our Lagrangian, that is, $F_p:=F|_{T_pM}$ is strictly convex for each $pin M$. For the moment, let's fix $pin M$ and focus on $F_p:T_pMtomathbb{R}$. For $alphain T_p^*M$, define
$$F_{p,alpha}(v)=F_p(v)-alpha(v)$$
for $vin T_pM$. The mapping $F_{p,alpha}$ is said to be stable if there exists a unique critical point in $T_pM$ for $F_{p,alpha}$. Then define the stability set
$$S_{F_p}={alphain T_p^*M:F_{p,alpha}text{ is stable}}.$$
Then $S_{F_p}$ is open and convex in $T_p^*M$, and leads to my first question of when do we have that $S_{F_p}=T_p^*M$?
Furthermore, since $F_p$ is strictly convex, we have that the Legendre transform associated to $F_p$, denoted $L_{F_p}:T_pMto S_{F_p}$ is a diffeomorphism. Moreover, with our dual $F_p^*:S_{F_p}tomathbb{R}$, we see that
$$L_{F_p}^{-1}=L_{F_p^*}.$$
Again, with my understanding, don't we need $S_{F_p}=T_p^*M$ for this to make sense?
Combining the fibewise defined maps, we obtain
$$mathcal{L}:TMto T^*M,qquad mathcal{L}|_{T_pM}=L_{F_p},$$
and
$$H:S_Ftomathbb{R},qquad H|_{S_{F_p}}=F_p^*,$$
and $mathcal{L}:TMto S_F$ is a diffeomorphism.
Note that I'm using the notation of
$$S_F=bigcup_{pin M}S_{F_p}.$$
Now, this defined $H$ should yield the Hamiltonian system $(T^*M,omega,H)$, but we don't necessarily have $H$ defined on the whole cotangent bundle, which seems odds since our Lagrangian was defined on the whole tangent bundle. Am I missing something in this construction? I was under the impression that Lagrangians on the tangent bundle were equivalent to Hamiltonians on the cotangent bundle.
I guess all of this reduced to the question: How do we remedy the fact that $S_{F_p}$ and $T_p^*M$ aren't necessarily equal?
As always, any help or references would be greatly appreciated.
differential-geometry differential-topology symplectic-geometry euler-lagrange-equation hamilton-equations
differential-geometry differential-topology symplectic-geometry euler-lagrange-equation hamilton-equations
asked Nov 17 at 22:05
Matt
39218
39218
This question has an open bounty worth +50
reputation from Matt ending in 1 hour.
Looking for an answer drawing from credible and/or official sources.
This question has an open bounty worth +50
reputation from Matt ending in 1 hour.
Looking for an answer drawing from credible and/or official sources.
add a comment |
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