The physical meaning of a symplectic form.












5












$begingroup$


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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    $begingroup$
    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
    3 hours ago












  • $begingroup$
    a little bit, just scanned through it
    $endgroup$
    – AkatsukiMaliki
    2 hours ago
















5












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
    3 hours ago












  • $begingroup$
    a little bit, just scanned through it
    $endgroup$
    – AkatsukiMaliki
    2 hours ago














5












5








5





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 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
    3 hours ago












  • $begingroup$
    a little bit, just scanned through it
    $endgroup$
    – AkatsukiMaliki
    2 hours ago














  • 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
    3 hours ago












  • $begingroup$
    a little bit, just scanned through it
    $endgroup$
    – AkatsukiMaliki
    2 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
3 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
3 hours ago














$begingroup$
a little bit, just scanned through it
$endgroup$
– AkatsukiMaliki
2 hours ago




$begingroup$
a little bit, just scanned through it
$endgroup$
– AkatsukiMaliki
2 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

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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    active

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    5












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Michael SparapanyMichael Sparapany

        715




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