The physical meaning of a symplectic form.
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So I've studied a bit about symplectic geometry, and I know that phase space is a symplectic manifold, and the symplectic form induces a poisson bracket. However, what is the physical meaning of the symplectic form? Perhaps it's simply the same as asking what the meaning is of the poisson bracket. What does the poisson bracket say about a system? And what is the physical meaning of observables?
differential-geometry classical-mechanics symplectic-geometry
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add a comment |
$begingroup$
So I've studied a bit about symplectic geometry, and I know that phase space is a symplectic manifold, and the symplectic form induces a poisson bracket. However, what is the physical meaning of the symplectic form? Perhaps it's simply the same as asking what the meaning is of the poisson bracket. What does the poisson bracket say about a system? And what is the physical meaning of observables?
differential-geometry classical-mechanics symplectic-geometry
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1
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Have you looked at the first chapter of McDuff and Salamon's book on symplectic topology? I don't know if that will answer your question but it should be useful.
$endgroup$
– Mike Miller
4 hours ago
$begingroup$
a little bit, just scanned through it
$endgroup$
– AkatsukiMaliki
3 hours ago
add a comment |
$begingroup$
So I've studied a bit about symplectic geometry, and I know that phase space is a symplectic manifold, and the symplectic form induces a poisson bracket. However, what is the physical meaning of the symplectic form? Perhaps it's simply the same as asking what the meaning is of the poisson bracket. What does the poisson bracket say about a system? And what is the physical meaning of observables?
differential-geometry classical-mechanics symplectic-geometry
$endgroup$
So I've studied a bit about symplectic geometry, and I know that phase space is a symplectic manifold, and the symplectic form induces a poisson bracket. However, what is the physical meaning of the symplectic form? Perhaps it's simply the same as asking what the meaning is of the poisson bracket. What does the poisson bracket say about a system? And what is the physical meaning of observables?
differential-geometry classical-mechanics symplectic-geometry
differential-geometry classical-mechanics symplectic-geometry
asked 4 hours ago
AkatsukiMalikiAkatsukiMaliki
338110
338110
1
$begingroup$
Have you looked at the first chapter of McDuff and Salamon's book on symplectic topology? I don't know if that will answer your question but it should be useful.
$endgroup$
– Mike Miller
4 hours ago
$begingroup$
a little bit, just scanned through it
$endgroup$
– AkatsukiMaliki
3 hours ago
add a comment |
1
$begingroup$
Have you looked at the first chapter of McDuff and Salamon's book on symplectic topology? I don't know if that will answer your question but it should be useful.
$endgroup$
– Mike Miller
4 hours ago
$begingroup$
a little bit, just scanned through it
$endgroup$
– AkatsukiMaliki
3 hours ago
1
1
$begingroup$
Have you looked at the first chapter of McDuff and Salamon's book on symplectic topology? I don't know if that will answer your question but it should be useful.
$endgroup$
– Mike Miller
4 hours ago
$begingroup$
Have you looked at the first chapter of McDuff and Salamon's book on symplectic topology? I don't know if that will answer your question but it should be useful.
$endgroup$
– Mike Miller
4 hours ago
$begingroup$
a little bit, just scanned through it
$endgroup$
– AkatsukiMaliki
3 hours ago
$begingroup$
a little bit, just scanned through it
$endgroup$
– AkatsukiMaliki
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This topic is incredibly deep so my answer barely skims the surface, however I believe you're looking for an intuitive meaning of the symplectic form. Consider the standard symplectic form
$$
omega = sum_i d x_i wedge d lambda_i
$$
This can also be written as
$$
omega = begin{bmatrix}
mathbf{0} & mathbf{I} \
-mathbf{I} & mathbf{0}
end{bmatrix}
$$
Now consider a rotation matrix
$$
R_theta = begin{bmatrix}
cos(theta) & sin(theta) \
-sin(theta) & cos(theta)
end{bmatrix}
$$
By setting $theta = 90^o$, you find the data of these tensors are equivalent. Ergo the standard symplectic form is very similar to a rotation matrix that rotates vectors by 90 degrees. Consider the symplectic form of Hamiltonian mechanics
$$
dH = iota_{X_H} omega
$$
where $dH$ is the exterior derivative of the Hamiltonian and $iota$ is the interior derivative. This exterior derivative generates a covector field whose covectors point in the direction of steepest descent. Rotate these covectors by 90 degrees and they point tangential to the level sets of $H$. In Hamiltonian mechanics, motion almost always lies on level sets of $H$.
Ultimately, I like to think about the symplectic form as a rotation matrix that properly encodes the structure of mechanics, but be aware that due to its differential geometric nature, it's much deeper than I lead on.
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This topic is incredibly deep so my answer barely skims the surface, however I believe you're looking for an intuitive meaning of the symplectic form. Consider the standard symplectic form
$$
omega = sum_i d x_i wedge d lambda_i
$$
This can also be written as
$$
omega = begin{bmatrix}
mathbf{0} & mathbf{I} \
-mathbf{I} & mathbf{0}
end{bmatrix}
$$
Now consider a rotation matrix
$$
R_theta = begin{bmatrix}
cos(theta) & sin(theta) \
-sin(theta) & cos(theta)
end{bmatrix}
$$
By setting $theta = 90^o$, you find the data of these tensors are equivalent. Ergo the standard symplectic form is very similar to a rotation matrix that rotates vectors by 90 degrees. Consider the symplectic form of Hamiltonian mechanics
$$
dH = iota_{X_H} omega
$$
where $dH$ is the exterior derivative of the Hamiltonian and $iota$ is the interior derivative. This exterior derivative generates a covector field whose covectors point in the direction of steepest descent. Rotate these covectors by 90 degrees and they point tangential to the level sets of $H$. In Hamiltonian mechanics, motion almost always lies on level sets of $H$.
Ultimately, I like to think about the symplectic form as a rotation matrix that properly encodes the structure of mechanics, but be aware that due to its differential geometric nature, it's much deeper than I lead on.
$endgroup$
add a comment |
$begingroup$
This topic is incredibly deep so my answer barely skims the surface, however I believe you're looking for an intuitive meaning of the symplectic form. Consider the standard symplectic form
$$
omega = sum_i d x_i wedge d lambda_i
$$
This can also be written as
$$
omega = begin{bmatrix}
mathbf{0} & mathbf{I} \
-mathbf{I} & mathbf{0}
end{bmatrix}
$$
Now consider a rotation matrix
$$
R_theta = begin{bmatrix}
cos(theta) & sin(theta) \
-sin(theta) & cos(theta)
end{bmatrix}
$$
By setting $theta = 90^o$, you find the data of these tensors are equivalent. Ergo the standard symplectic form is very similar to a rotation matrix that rotates vectors by 90 degrees. Consider the symplectic form of Hamiltonian mechanics
$$
dH = iota_{X_H} omega
$$
where $dH$ is the exterior derivative of the Hamiltonian and $iota$ is the interior derivative. This exterior derivative generates a covector field whose covectors point in the direction of steepest descent. Rotate these covectors by 90 degrees and they point tangential to the level sets of $H$. In Hamiltonian mechanics, motion almost always lies on level sets of $H$.
Ultimately, I like to think about the symplectic form as a rotation matrix that properly encodes the structure of mechanics, but be aware that due to its differential geometric nature, it's much deeper than I lead on.
$endgroup$
add a comment |
$begingroup$
This topic is incredibly deep so my answer barely skims the surface, however I believe you're looking for an intuitive meaning of the symplectic form. Consider the standard symplectic form
$$
omega = sum_i d x_i wedge d lambda_i
$$
This can also be written as
$$
omega = begin{bmatrix}
mathbf{0} & mathbf{I} \
-mathbf{I} & mathbf{0}
end{bmatrix}
$$
Now consider a rotation matrix
$$
R_theta = begin{bmatrix}
cos(theta) & sin(theta) \
-sin(theta) & cos(theta)
end{bmatrix}
$$
By setting $theta = 90^o$, you find the data of these tensors are equivalent. Ergo the standard symplectic form is very similar to a rotation matrix that rotates vectors by 90 degrees. Consider the symplectic form of Hamiltonian mechanics
$$
dH = iota_{X_H} omega
$$
where $dH$ is the exterior derivative of the Hamiltonian and $iota$ is the interior derivative. This exterior derivative generates a covector field whose covectors point in the direction of steepest descent. Rotate these covectors by 90 degrees and they point tangential to the level sets of $H$. In Hamiltonian mechanics, motion almost always lies on level sets of $H$.
Ultimately, I like to think about the symplectic form as a rotation matrix that properly encodes the structure of mechanics, but be aware that due to its differential geometric nature, it's much deeper than I lead on.
$endgroup$
This topic is incredibly deep so my answer barely skims the surface, however I believe you're looking for an intuitive meaning of the symplectic form. Consider the standard symplectic form
$$
omega = sum_i d x_i wedge d lambda_i
$$
This can also be written as
$$
omega = begin{bmatrix}
mathbf{0} & mathbf{I} \
-mathbf{I} & mathbf{0}
end{bmatrix}
$$
Now consider a rotation matrix
$$
R_theta = begin{bmatrix}
cos(theta) & sin(theta) \
-sin(theta) & cos(theta)
end{bmatrix}
$$
By setting $theta = 90^o$, you find the data of these tensors are equivalent. Ergo the standard symplectic form is very similar to a rotation matrix that rotates vectors by 90 degrees. Consider the symplectic form of Hamiltonian mechanics
$$
dH = iota_{X_H} omega
$$
where $dH$ is the exterior derivative of the Hamiltonian and $iota$ is the interior derivative. This exterior derivative generates a covector field whose covectors point in the direction of steepest descent. Rotate these covectors by 90 degrees and they point tangential to the level sets of $H$. In Hamiltonian mechanics, motion almost always lies on level sets of $H$.
Ultimately, I like to think about the symplectic form as a rotation matrix that properly encodes the structure of mechanics, but be aware that due to its differential geometric nature, it's much deeper than I lead on.
answered 3 hours ago
Michael SparapanyMichael Sparapany
715
715
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$begingroup$
Have you looked at the first chapter of McDuff and Salamon's book on symplectic topology? I don't know if that will answer your question but it should be useful.
$endgroup$
– Mike Miller
4 hours ago
$begingroup$
a little bit, just scanned through it
$endgroup$
– AkatsukiMaliki
3 hours ago