1
$\begingroup$

Consider an infinitesimal transformation:

$$(q_{i},p_{j}) \quad\longrightarrow \quad(Q_{i},P_{j}) ~=~ \left(q_{i} + \alpha F_{i}(q,p),~p_{j} + \alpha E_{j}(q,p)\right) $$

where $α$ is considered to be infinitesimally small.

Now, if we construct Jacobian matrix, we will have:

$$ \jmath =\begin{pmatrix} \delta_{ij}+ \alpha{\frac{\partial F_{i} }{\partial q_{j}}} & \alpha{\frac{\partial F_{i} }{\partial p_{j}}} \\ \alpha{\frac{\partial E_{i} }{\partial q_{j}}} & \delta_{ij}+ \alpha{\frac{\partial E_{i} }{\partial p_{j}}} \end{pmatrix}.$$

What functions $F_{i} (q, p)$ and $E_{i} (q, p)$ are allowed for this to be a canonical transformation?

To be canonical transformation, it's required to hold: $$\jmath j \jmath^{T} = j$$ in which $ j = \begin{pmatrix} 0 & 1\\ -1&0 \end{pmatrix}$. To hold the canonical transformation, there should be: $$\frac {\partial F_{i}}{\partial q_{j}} = - \frac {\partial E_{i}}{\partial p_{j}} $$

which is true if

$$F_{i} = \frac {\partial G}{\partial p_{i}} \; \; , \; \; E_{i} = - \frac {\partial G}{\partial q_{i}} $$

for some function $G(q, p)$.

Now my problem is that by calculating everything I can't figure out how to reach to last two formulas. The formulas which shows the possibilities for $F_{i}$ and $E_{i}$?

$\endgroup$
3
$\begingroup$
  1. First of all, be aware that there exist various different definitions of canonical transformations (CT) in the literature, cf. e.g. this Phys.SE post. What OP (v3) above refers to as a CT, we will in this answer call a symplectomorphism for clarity. What we in this answer will refer to as a CT, will just be a CT of type 2.

  2. It is possible to show (see e.g. Ref. 1) that an arbitrary time-dependent infinitesimal canonical transformation (ICT) of type 2 with generator $G=G(z,t)$ can be identified with a Hamiltonian vector field (HVF) $$ \delta z^I~=~\varepsilon\{ z^I,G\}_{PB}~\equiv~ \sum_{K=1}^{2n} J^{IK} \frac{\partial G}{\partial z^K} , $$ $$ X_{-G}~\equiv~-\{G,\cdot\}_{PB}~\equiv~\{\cdot,G\}_{PB},\tag{1} $$ with (minus) the same generator $G$. Here $z^1,\ldots, z^{2n}$, are phase space variables, $t$ is time, $\varepsilon$ is an infinitesimal parameter, and $J$ is the symplectic unit matrix, $$\tag{2} J^2~=~-{\bf 1}_{2n\times 2n}.$$

  3. A general time-dependent infinitesimal transformation (IT) of phase space can without loss of generality be assumed to be of the form $$ \tag{3} \delta z^I~=~\varepsilon \sum_{K=1}^{2n} J^{IK} G_K(z,t) ,\qquad I~\in~\{1,\ldots, 2n\}, $$ because the matrix $J$ is invertible.

  4. It is possible to show that a time-dependent infinitesimal symplectomorphism (IS) [written in the form (3)] satisfies the Maxwell relations$^1$ $$\tag{4} \frac{\partial G_I(z,t)}{\partial z^J}~=~(I \leftrightarrow J),\qquad I,J~\in~\{1,\ldots, 2n\}. $$

  5. Eq. (4) states that the one-form $$\tag{5} \mathbb{G}~:=~ \sum_{I=1}^{2n}G_I(z,t) \mathrm{d}z^I$$ is closed $$\tag{6} \mathrm{d}\mathbb{G}~=~0. $$

  6. It follows from Poincare Lemma, that locally there exists a function $G$ such that $ \mathbb{G}$ is locally exact $$\tag{7} \mathbb{G}~=~\mathrm{d}G. $$ Or in components, $$\tag{8} G_I(z,t)~=~\frac{\partial G(z,t)}{\partial z^I},\qquad I~\in~\{1,\ldots, 2n\} .$$

  7. In summary we have the following very useful theorem for a general time-dependent infinitesimal transformation (IT).

    Theorem. An infinitesimal canonical transformation (ICT) of type 2 is an infinitesimal symplectomorphism (IS). Conversely, an IS is locally a ICT of type 2.

References:

  1. H. Goldstein, Classical Mechanics; 2nd eds., 1980, Section 9.3; or 3rd eds., 2001, Section 9.4.

--

$^1$ OP already listed some (but not all) of the Maxwell relations (4) in his second-last equation. All of the Maxwell relations (4) are necessary in order to deduce the local existence of the generating function $G$.

$\endgroup$
  • $\begingroup$ I'm hopeful to not caught for another mistake, like questioning something that it shouldn't. However if this is ok, please introduce me a book with simple attitude about Symplectomorphism, in level of physics master student who has taken the "group theory" course in undergrad. $\endgroup$ – Sina Aug 4 '15 at 10:26
  • $\begingroup$ One would probably have to study differential geometry to properly understand symplectomorphisms. However, for the case at hand, it enough to know some of its properties, such as OP's third equation. $\endgroup$ – Qmechanic Aug 4 '15 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.