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So recently I've been doing some self-study on canonical transformations and relating together different Hamiltonian systems. I've found this paper (PDF) with a remarkable result showing that any two Hamiltonian systems $H(q,p)$ and $K(Q,P)$ with the same degrees freedom, are locally equivalent and connected via specific canonical transformation derived from the solution to the Hamiltonian-Jacobi equations. The construction essentially shows (to my understanding) that the action functional can be looked at as a generating function mapping, which transforms the Hamiltonian to systems where nothing changes. Applying this to both systems $H$ and $K$ and requiring that the constants match links the two systems. However, this construction goes through the HJ canonical transformation, which is time-dependent as it essentially contains information on the dynamics of the original system. Nevertheless, the canonical transformations that connect both systems together in their examples always end up being time-independent.

So my question is for something I did not currently see a clear idea how to prove or disprove - given two Hamiltonian systems $H(q,p)$ and $K(Q,P)$ with the same degrees of freedom is it true that there always exists a time-independent canonical transformation $f: (q,p) \to (Q,P)$ such that the dynamics of $K(f(q,p))$ in terms of $(q,p)$ are the same as the dynamics of $H(q,p)$ (at least locally since the transformation might not be one-to-one)?

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TL;DR: The answer is No, since a time-independent canonical transformation (CT) cannot turn a non-zero Hamiltonian into a vanishing Hamiltonian. However OP's conjecture is true for 2 non-zero sufficiently well-behaved Hamiltonians.

More details:

  1. In the same way as time-dependent Hamilton-Jacobi (HJ) theory, is governed by Hamilton's principal function $S(q,P,t)$, the corresponding time-independent problem is governed by Hamilton's characteristic function $W(q,P)$, cf. e.g. Ref. 1.

  2. The underlying mathematical problem comes down to guaranteeing the local existence of a solution to the time-independent HJ equation, which is a first-order non-linear PDE. To ensure this, OP's 2 Hamiltonians $H$ & $\widetilde{H}$ would have to be sufficiently well-behaved.

  3. One new feature (in the time-independent case) is that the 2 Kamiltonians $K$ & $\widetilde{K}$ (produced by time-independent HJ theory) are not necessarily zero (as in the time-dependent case), so as an extra step one should make CTs to ensure that $K=P_1$ and $\widetilde{K}=\widetilde{P}_1$. (This last step could fail if precisely one of the Kamiltonians vanishes.) $\Box$

References:

  1. H. Goldstein, Classical Mechanics; Section 10.3.
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    $\begingroup$ can you expand a bit on the following obvious example: turn (or at least relate by CT) $H=p^2/2+ x^2/2$ to $\tilde H=p^2/2-x^2/2$? $\endgroup$ Commented Aug 15, 2020 at 14:03

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