In Goldstein's Classical Mechanics (2nd ed.), Section 9-1, pgs. 382-385, the generating functions (hereafter denoted $F$) for canonical transformations are introduced. From here on out, I'll refer to the original canonical coordinate variable and canonical momenta variable as $q$ and $p$, while the "new" canonical coordinate and momenta variables which I'm transforming to will be $Q$ and $P$. I believe this is standard notation.

There are four separate cases enumerated in the text - when the canonical transformation is best described by a generating function dependent on only on the mixed coordinates:

  1. $q$ and $Q$ ~ the transformation relations can be found by setting $F=F_1(q,Q,t)$
  2. $q$ and $P$ ~ the transformation relations can be found by setting $F=F_2(q,P,t)-QP$
  3. $p$ and $Q$ ~ the transformation relations can be found by setting $F=F_3(p,Q,t)+qp$
  4. $p$ and $P$ ~ the transformation relations can be found by setting $F=F_4(p,P,t)-QP+qp$

From what I've read online, the four types of variable-dependencies for the generating function enumerated above are known as the four-kinds of the generating function (i.e. it seems well-defined as standard nomenclature in the literature).

My question is, how come the cases where the canonical transformations are best described by a generating function of non-mixed coordinates covered? (i.e. $F_5(q,p,t)$ or $F_6(Q,P,t)$) Are there no canonical transformations best described by such generating functions? Am I overseeing something trivial?


Suppose there was a canonical transformation that was best described by a generating function dependent only on $q$, $p$, and $t$. Then for the generating function I could use


I could plug that into the fundamental relation between the original and newly transformed-to quantities to then get the transformation relations.

$$\begin{align*}\require{\cancel} p\dot{q}-H&=P\dot{Q}-K+\frac{dF}{dt}\\ &=P\dot{Q}-K+\frac{\partial F_5}{\partial t}+\frac{\partial F_5}{\partial q}\dot{q} +\frac{\partial F_5}{\partial p}\dot{p} - \dot{p}Q - p\dot{Q}\\ &\\ &\rightarrow \left(H-K+\frac{\partial F_5}{\partial t}\right)+\left(\frac{\partial F_5}{\partial q}-p\right)\dot{q}+\left(P-p\right)\dot{Q}+\left(\frac{\partial F_5}{\partial p}-Q\right)\dot{p}=0 \end{align*}$$

I can now read off the transformation relations as:

$$Q=\frac{\partial F_5}{\partial p}\tag{1}$$

$$p=\frac{\partial F_5}{\partial q}\tag{2}$$


$$K=H+\frac{\partial F_5}{\partial t}\tag{4}$$

I can see that (1) and (3) define the canonical transformation, whereas (2) puts a restriction on what functions $F_5(q,p,t)$ will work. Nevertheless, this still defines a family of canonical transformations, doesn't it?

  • $\begingroup$ How does this have a net downvote? What's wrong with this question? $\endgroup$ Commented May 2, 2016 at 22:55

2 Answers 2


Two reasons

  1. A generating function by definition generates canonical transformations. If you take terms like $F_{5}(p,q,t)$ then it doesn't relate coordinates involving two separate canonical transformations.

  2. The area preserving property of canonical transformations imply the existence of generating function. Then there is no way you can construct generating functions like $F_{5}(p,q,t)$. This is because if you take the generating function to be $F_{5}(p,q,t)$ then there is no way in which

$$\oint_{C}\dfrac{\partial F_{5}(p,q,t)}{\partial q} dq + \oint_{C}\dfrac{\partial F_{5}(p,q,t)}{\partial p} dp$$ where C is an arbitrary closed curve in the phase space can be tinkered to give the differences in the phase space volume

$$ \oint_{C}p dq - \oint_{C}P dQ$$

by adding and subtracting $$ \oint_{C} d(pq) = 0$$ or $$ \oint_{C} d(PQ) = 0.$$

  • $\begingroup$ So is it possible to prove that all generating functions have to be of type 1-4? $\endgroup$
    – lalala
    Commented Mar 16, 2021 at 15:49
  • $\begingroup$ @lalala That is what I have shown in the answer. $\endgroup$
    – Bruce Lee
    Commented Mar 29, 2021 at 10:13
  • $\begingroup$ Yes. I meant more like can all canonical transformations be written in such a way. I do not see why canonical implies existence of generating functions. $\endgroup$
    – lalala
    Commented Mar 29, 2021 at 13:43
  • $\begingroup$ Why does it not relate the old and new coordinates? $\endgroup$
    – Cheng
    Commented Nov 25, 2022 at 23:53
  1. Because of the extra condition (3) OP's type 5 canonical transformation (CT) is the same as a type 2 CT with an extra condition (3), and hence nothing new.

  2. The 4 types of CTs are locally sufficient (via the implicit function theorem) in a $2n$-dimensional phase space if $n=1$. For $n>1$ there are various hybrid types.


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