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Undefined Hamiltonian for this particular Lagrangian
Adding a total time derivative term to the LagrangianLagrangian to HamiltonianHamiltonian mechanics really useful for numerical integration? Lagrangian can become 1st-orderFinding geodesics: Lagrangian vs HamiltonianInconsistency in Lagrangian vs Hamiltonian formalism?Gauge invariance of the HamiltonianWhat is my state in the context of Hamiltonian mechanics?Can we write a Lagrangian for the classical system with $H=kqp$?Canonical transformations preserve Hamilton's equations. Which transformation preserve the Euler-Lagrange equations?Can all canonical transformations be generated using a generating function?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
So, this a question from a Phd qualifying examination. Given the following Lagrangan $$L=frac12dotqtextsin^2q,$$ what is the Hamiltonian for this system? So, finding the canonical momentum $p$ and then using $H=pdotq -L$, I find that it is identically zero. Now, the answer 'Zero' was a choice in the choices of answers given but the correct answer is that the Hamiltonian is undefined. I'm not sure why this should be the answer, though this is a pretty weird dynamical system. The Euler Lagrange equations just give an identity and the Hamiltonian equations also seem rather off. The Hamiltonian equations would are $$dotp=dotq=0. $$
I'm hoping someone shed some light on this, it seems that any random Lagrangian doesn't constitute a dynamical system, so what I should I be looking for?
homework-and-exercises classical-mechanics lagrangian-formalism hamiltonian-formalism
$endgroup$
add a comment |
$begingroup$
So, this a question from a Phd qualifying examination. Given the following Lagrangan $$L=frac12dotqtextsin^2q,$$ what is the Hamiltonian for this system? So, finding the canonical momentum $p$ and then using $H=pdotq -L$, I find that it is identically zero. Now, the answer 'Zero' was a choice in the choices of answers given but the correct answer is that the Hamiltonian is undefined. I'm not sure why this should be the answer, though this is a pretty weird dynamical system. The Euler Lagrange equations just give an identity and the Hamiltonian equations also seem rather off. The Hamiltonian equations would are $$dotp=dotq=0. $$
I'm hoping someone shed some light on this, it seems that any random Lagrangian doesn't constitute a dynamical system, so what I should I be looking for?
homework-and-exercises classical-mechanics lagrangian-formalism hamiltonian-formalism
$endgroup$
$begingroup$
I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
$endgroup$
– S V
8 hours ago
add a comment |
$begingroup$
So, this a question from a Phd qualifying examination. Given the following Lagrangan $$L=frac12dotqtextsin^2q,$$ what is the Hamiltonian for this system? So, finding the canonical momentum $p$ and then using $H=pdotq -L$, I find that it is identically zero. Now, the answer 'Zero' was a choice in the choices of answers given but the correct answer is that the Hamiltonian is undefined. I'm not sure why this should be the answer, though this is a pretty weird dynamical system. The Euler Lagrange equations just give an identity and the Hamiltonian equations also seem rather off. The Hamiltonian equations would are $$dotp=dotq=0. $$
I'm hoping someone shed some light on this, it seems that any random Lagrangian doesn't constitute a dynamical system, so what I should I be looking for?
homework-and-exercises classical-mechanics lagrangian-formalism hamiltonian-formalism
$endgroup$
So, this a question from a Phd qualifying examination. Given the following Lagrangan $$L=frac12dotqtextsin^2q,$$ what is the Hamiltonian for this system? So, finding the canonical momentum $p$ and then using $H=pdotq -L$, I find that it is identically zero. Now, the answer 'Zero' was a choice in the choices of answers given but the correct answer is that the Hamiltonian is undefined. I'm not sure why this should be the answer, though this is a pretty weird dynamical system. The Euler Lagrange equations just give an identity and the Hamiltonian equations also seem rather off. The Hamiltonian equations would are $$dotp=dotq=0. $$
I'm hoping someone shed some light on this, it seems that any random Lagrangian doesn't constitute a dynamical system, so what I should I be looking for?
homework-and-exercises classical-mechanics lagrangian-formalism hamiltonian-formalism
homework-and-exercises classical-mechanics lagrangian-formalism hamiltonian-formalism
edited 8 hours ago
Qmechanic♦
113k13 gold badges223 silver badges1341 bronze badges
113k13 gold badges223 silver badges1341 bronze badges
asked 8 hours ago
Soumil MaulickSoumil Maulick
938 bronze badges
938 bronze badges
$begingroup$
I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
$endgroup$
– S V
8 hours ago
add a comment |
$begingroup$
I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
$endgroup$
– S V
8 hours ago
$begingroup$
I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
$endgroup$
– S V
8 hours ago
$begingroup$
I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
$endgroup$
– S V
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:
- Finding $p$
- Writing $H=pdot q-L$
- Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.
For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.
However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is
$$
J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
$$
That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.
$endgroup$
add a comment |
$begingroup$
OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.
The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
$$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:
- Finding $p$
- Writing $H=pdot q-L$
- Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.
For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.
However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is
$$
J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
$$
That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.
$endgroup$
add a comment |
$begingroup$
The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:
- Finding $p$
- Writing $H=pdot q-L$
- Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.
For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.
However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is
$$
J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
$$
That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.
$endgroup$
add a comment |
$begingroup$
The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:
- Finding $p$
- Writing $H=pdot q-L$
- Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.
For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.
However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is
$$
J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
$$
That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.
$endgroup$
The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:
- Finding $p$
- Writing $H=pdot q-L$
- Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.
For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.
However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is
$$
J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
$$
That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.
answered 8 hours ago
Jahan ClaesJahan Claes
4,88012 silver badges34 bronze badges
4,88012 silver badges34 bronze badges
add a comment |
add a comment |
$begingroup$
OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.
The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
$$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$
$endgroup$
add a comment |
$begingroup$
OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.
The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
$$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$
$endgroup$
add a comment |
$begingroup$
OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.
The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
$$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$
$endgroup$
OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.
The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
$$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$
edited 7 hours ago
answered 8 hours ago
Qmechanic♦Qmechanic
113k13 gold badges223 silver badges1341 bronze badges
113k13 gold badges223 silver badges1341 bronze badges
add a comment |
add a comment |
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$begingroup$
I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
$endgroup$
– S V
8 hours ago