In Goldstein's Classical Mechanics (3rd edition) section 9.1, we introduce the generating function method of describing a canonical transformation. We then introduce four types of generating functions: $$ \begin{align} &F_1(q,Q,t),\\ &F_2(q,P,t),\\ &F_3(p,Q,t), \text{ and}\\ &F_4(p,P,t), \end{align} $$ noting that not all generating functions must conform to such a form but that we can have 'mixed' generating functions, too, with an example of this mixed form in equation 9.20: $$ F'(q_1,p_2,P_1,Q_2,t) \tag{9.20} $$ and 9.21: $$ F = F'(q_1,p_2,P_1,Q_2,t) - Q_1 P_1 + q_2 p_2.\tag{9.21} $$

A key observation is that, for the $i$th degree of freedom, all of the listed generating functions depend on ($q_i$ or $p_i$) and ($Q_i$ or $P_i$); none of the listed generating functions depend on both $q_i$ and $p_i$; none of the listed generating functions depend on both $Q_i$ and $P_i$.

  1. Question: Is this observed functional dependence a requirement? That is, for the $i$th degree of freedom, can we be confident that an arbitrary generating function depends only on one $x\in\{q_i,p_i\}$ and one $X\in\{Q_i,P_i\}$. Or would it be possible to see generating functions with dependence like $F'(q_1, p_1, \dots Q_1)$?

  2. Question: As a corollary, if question #1 is true, is this why we can later say $K = H + \partial F'/\partial t$ (see the line below 9.44 or the paragraph following 9.104) despite only showing that such an equation holds in the four 'standard' cases of generating functions $F_1$, $F_2$, $F_3$, and $F_4$?

As noted in section 9.1, these generating functions can simply be thought of as Legendre transformations of one another. As stated in these lecture notes, "Now we can perform a Legendre transformation on either $q$ or $Q$ to replace them with either $p$ or $P$, so that we can express the same canonical transformation by any of four generating functions, $F=F_1(q,Q,t)$, $F_2(q,P,t)$, $F_3(p,Q,t)$, or $F_4(p,P,t)$."

  1. Question: Is the above quote the reason why we can limit attention to a specific class of generating function ($F_2$) when discussing infinitesimal canonical transformations (ICTs; see equation 9.62), for example? That all representations are equivalent and choice between them can be made by convenience?

  2. Question: How can we say that these generating functions describe the same transformation when even Goldstein specifically notes that Legendre transforms don't give us everything.

As a simple example of the loss of information via Legendre transforms, consider the 'trivial special case' generating function $F_1 = q_i Q_i$. We know $$ \begin{align} F_2 &= F_1 + Q_i P_i = Q_i(q_i + P_i)\\ F_3 &= F_1 - q_i p_i = q_i(Q_i - p_i)\\ F_4 &= F_1 - q_i p_i + Q_i P_i = q_i Q_i - q_i p_i + Q_i P_i. \end{align} $$ Attempting to find the transformation equations using the generating function derivatives (see table 9.1), we see that each Legendre transformed equation gives an incomplete description of the transformation. For example, take $F_2$. One derivative gives $p_i = \partial F_2/\partial q_i = Q_i$ (as expected) but the other gives $Q_i = \partial F_2/\partial P_i = Q_i$, which is identically true. No information on $P_i$ seems to be given from partial derivatives of $F_2$.

A (failed) potential explanation as to why $F_2$ doesn't say anything about $P_i$ is because, since both $q$ and $P$ are arguments of $F_2$, it'd be nice to claim that these arguments must be independent of each other. This explanation fails because, as demonstrated for the generating function of type $F_2$ in equation 9.28: $$ F_2 = f_i(q_1,\dots,q_n;t) P_i + g_i(q_1,\dots,q_n;t),\tag{9.28} $$ we find $P_i$ is dependent on both the old position and momentum.

  1. Question: If dependence between 'old' and 'new' variables cannot be used to decide between the form of the generating function and if a generic transformation can be written in any form, do we simply pick between the different representations by convenience (as in question #3)?

1 Answer 1

  • Geometrically a canonical transformation (CT) carves out a codimension-$2n$ (or a $2n+1$-dimensional) submanifold in the $4n+1$ space ${\cal M}$ with local coordinates $(q,p,Q,P,t)$.

  • A generating function $F(q,p,Q,P,t)$ can in principle depend on all $4n+1$ variables.

  • The 4 types of generating functions $F_1(q,Q,t)$, $F_2(q,P,t)$, $F_3(p,Q,t)$, $F_4(p,P,t)$ reproduce the above imbedding as graphs.

  • Geometrically, some of the 4 types might not exist, as OP's example shows. See also this related example.

  • An infinitesimal canonical transformation (ICT) can always be realized as a type-2 or a type-3 CT, because we can expand $$F_2(q,P,t)~=~qP+f_2(q,P,t),$$ $$F_3(p,Q,t)~=~-pQ+f_3(p,Q,t),$$ where the first terms generate the identity CT, while $f_2(q,P,t)$ & $f_3(q,P,t)$ generate the infinitesimal modification, respectively. It's important that the $2n+1$ arguments of the graph are independent.

  • $\begingroup$ In the last sentence, you state the $2n+1$ arguments must be independent. Isn't eq 9.28, $F_2 = f_i(q_1,\dots,q_n;t)P_i + g(q_1,\dots,q_n;t)$, a counterexample of such claim? This equation is a $F_2$ generating function, so, as you state, we'd expect $P_i$ to be independent of $q_i$ but, upon solving for $P_i$, we find it to be a function of both $q_i$ and $p_i$. This seems to be problematic from the proposed graph viewpoint, but maybe I'm wrong. --- Also, I don't see how the canonical transformation generated by $F_1 = q_i Q_i$ can be made up from ICTs using $F_2$ and $F_3$. Can you explain? $\endgroup$ Commented Mar 20, 2022 at 22:25
  • $\begingroup$ More scarily to me, it seems like $P_i=p_i$ and $Q_i=-q_i$ would be a valid canonical transformation (it preserves phase space area, albeit inverting orientation in the $(q_i,p_i)$ plane), but this transformation appears to be impossible to create using any of the four types of generating functions (all of them preserve orientation). Do we also require the preservation of orientation? If so, what justification is there for such a requirement? $\endgroup$ Commented Mar 20, 2022 at 22:30
  • $\begingroup$ Canonical transformations preserve orientation by definition. $\endgroup$
    – Qmechanic
    Commented Mar 21, 2022 at 3:29
  • $\begingroup$ I see - that makes sense. Sorry for that naïve question! Can you provide clarity on the previous comment? 1. How eq. 9.28 implies dependence of $P_i$ on $q_i$ and $p_i$, seemingly contrasting your stated independence of the arguments to $F_2$? 2. Also clarity on how one would generate the transformation $Q_i=p_i$, $P_i=-q_i$ using an ICT? It is represented by $F_1=q_i Q_i$ or $F_4=p_i P_i$, but I don't see how to integrate an $F_2$ or $F_3$ ICT into those forms. $\endgroup$ Commented Mar 21, 2022 at 13:19
  • $\begingroup$ 1. The independence statement was made in the context of an ICT. For a finite CT there may be more conditions. 2. The $90^\circ$ rotation of type 1 can e.g. be composed of two $45^\circ$ rotations. The latter is of type 2. $\endgroup$
    – Qmechanic
    Commented Mar 22, 2022 at 2:07

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