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I am reading Chapter 9 of Goldstein. He proves that any time-independent canonical transformations satisfy symplectic condition. And after that, he shows that if we ignore second order small quantity, then, the infinitesimal time-dependent canonical transformation will satisfy symplectic condition. But how can we make sure that finite canonical transformations also satisfy this condition? What I am confused about is that this problem is very similar to infinitesimal rotations, but there, the commutative property is lost when the rotation becomes finite. I think that is because the second order terms can not be ignored, so, in canonical transformation, for same reason, will it also lose the property that it satisfied symplectic condition when canonical transform is finite?

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  1. In differential-geometric terms, there is a bijective correspondence between 1-parameter time-dependent flows $\sigma$ and time-dependent vector fields $X$. In particular, there is a bijective correspondence between 1-parameter time-dependent symplectic flows $\sigma$ and time-dependent symplectic vector fields $X$. See also this related Phys.SE post.

  2. Goldstein's definition $$ (\sum_{i=1}^np_i\mathrm{d}q^i-H\mathrm{d}t) -(\sum_{i=1}^nP_i\mathrm{d}Q^i -K\mathrm{d}t) ~=~\mathrm{d}F\tag{9.11}$$ of a finite (possibly time-dependent) canonical transformation (CT) form a groupoid.

  3. Goldstein shows below eq. (9.66) that a (possibly time-dependent) infinitesimal canonical transformation (ICT) of type 2 is an infinitesimal symplectomorphism (IS).

  4. An application of the Poincare lemma shows that the opposite infinitesimal statement holds locally. (There could be topological obstructions, since not all symplectic vector fields are Hamiltonian vector fields.)

  5. We deduce by integrating the infinitesimal result, that a CT path-connected to the identity transformation is a symplectomorphism. (It follows from the fundamental theorem of calculus that second-order terms in $\epsilon$ can be ignored.)

  6. One could presumably with some work show directly (i.e. without integrating the infinitesimal result) that a (possibly time-dependent) finite CT is a symplectomorphism, although I have not attempted it so far.

  7. Lastly, OP mentions the orthogonal Lie group $$SO(3)~:=~ \{ M\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid M^TM = \mathbb{1}_{3\times 3}, \det(M)>0 \}$$ of 3D rotations, which is non-abelian/non-commutative, i.e. the order matters. Infinitesimal rotations correspond to the Lie algebra $$so(3)~:=~ \{ m\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid m^T = -m \},$$ which consists of antisymmetric $3\times 3$ real matrices. A 1-parameter family of infinitesimal rotations $$[0,1]~\ni~ s~~\mapsto~~ m(s)~\in~ so(3)$$ can be integrated to a finite rotation $$M~=~{\cal P}\exp\left\{\int_0^1\! \mathrm{d}s~ m(s)\right\}~\in ~SO(3),$$ where ${\cal P}$ denotes path-ordering. We stress that the orthogonality condition for the finite rotation $M$ is not spoiled by higher-order contributions, cf. OP's remarks.

References:

  1. H. Goldstein, Classical Mechanics; Chapter 9.
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  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Dec 21, 2019 at 13:37
  • $\begingroup$ Thanks sir, very clear and helpful! The place where I get wrong here is that I miss the truth which orthogonal condition is also be satisfied when rotation is infinitesimal. And I was thinking the commutative property is the same problem with those condition, but obviously it's not. Thanks a lot for your patience! @Qmechanic $\endgroup$
    – GK1202
    Dec 21, 2019 at 17:54

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