In Goldstein's Classical Mechanics (2nd ed.) on section 9-1 page 382, there is a discussion about finding a canonical transformation $$(q_i,p_i)\rightarrow (Q_j(q_i,p_i,t),P_j(q_i,p_i,t))$$ from a given generating function $F=F_1(q_i,Q_i,t)$. The following is written on the page:

Equation (9-11) then takes the form, $$\begin{align*} p_i\dot{q}_i-H&=P_i\dot{Q}_i-K+\frac{dF_1}{dt},\\ &=P_i\dot{Q}_i-K+\frac{\partial F_1}{\partial t}+ \frac{\partial F_1}{\partial q_i}\dot{q}_i+\frac{\partial F_1}{\partial Q_i}\dot{Q}_i \tag{9-13} \end{align*}$$ Since the old and the new coordinates, $q_i$ and $Q_i$, are separately independent, Eq. (9-13) can hold identically only if the coefficients of $\dot{q}_i$ and $\dot{Q}_i$ each vanish:

I don't understand the bold text above. How are $q_i$ and $Q_i$ separately independent? I thought $Q_i$ was, in general, a function of the original canonical coordinate variables $q_i$, thereby making it explicitly dependent on $q_i$ (not independent). Could somebody explain how these two canonical coordinate variables are separately independent?

  • 2
    $\begingroup$ For the purpose of $F_1$, they are independent - you are right they aren't independent as actual coordinates, but as "slots" for the function $F_1$, they are. $\endgroup$
    – ACuriousMind
    Commented Jan 12, 2016 at 15:05
  • $\begingroup$ @ACuriousMind Oh I see. Also, I made a mistake in saying that $Q_i$ is a function of $q_i$, because in general it's a function of $q_i$, $p_i$, and $t$ (and it's certainly possible for it to not depend on one or more of those variables). $\endgroup$ Commented Jan 12, 2016 at 17:48
  • 2
    $\begingroup$ The question concerning the text in bold has been real issue for me. Searching through books and online resources only adds to the confusion, as does the above answer, I'm sorry to say. I hope there's a real expert out there who can shed light on this. $\endgroup$
    – user116174
    Commented May 2, 2016 at 15:28
  • $\begingroup$ @Alan I suggest rereading the accepted answer. If one starts off with $q$ and $p$ as their initial coordinates, then of course the text in bold wouldn't be valid, as $\dot{Q}_i$ could be put in terms of, in general, $\dot{q_i}$ and $\dot{p}_i$ (which wouldn't lead us anywhere because we still have $\dot{Q}_i$ in (9-13)). However, what if we assumed that $q$ and $Q$ are the original coordinates from which we find the separate "functions"/"coordinates" $p$ and $P$ which, by hypothesis, are separate from each other? Then we get the $F_1$ type generating function. $\endgroup$ Commented May 2, 2016 at 23:02
  • $\begingroup$ @Alan Suppose $f(x_0,y_0)$ is an extremum of $f(x,y)$ (a suitably differentiable function) over some domain. The obviously $f_x(x_0,y_0)=f_y(x_0,y_0)=0$. If we slap a new coordinate system on our domain, $(u(x,y),v(x,y))$, and that same extremum in the new coordinates is $f(u_0,y_0)$, this wouldn't change the fact that $f_x(x_0,y_0)=f_y(x_0,y_0)=0$ - $x$ and $y$ are still independent. To take it further, suppose we can "invert" the coordinates in such a way that we get the coordinates $(x(y,v),u(y,v))$. Then $f_x(x_0,u_0)=f_u(x_0,u_0)=0$, as now $(y,v)$ is a different coordinate system! $\endgroup$ Commented May 2, 2016 at 23:09

1 Answer 1


Let us suppress $t$-dependence in this answer for simplicity. A canonical transformation (CT)$^1$ $$(q,p)\quad\longrightarrow\quad (Q(q,p),P(q,p))$$ can be viewed as a graph, thereby yielding a $2n$-dimensional submanifold $M$ embedded inside a $4n$-dimensional total manifold $N$. The total manifold $N$ has $4n$ local coordinates $(q,p,Q,P)$. The submanifold $M$ can in certain cases be viewed as a graph of a function $$(q,Q)\quad\longrightarrow \quad(p(q,Q),P(q,Q)),$$ where $q$ and $Q$ are independent variables, and where the functions $$ p(q,Q)~=~\frac{\partial F_1(q,Q)}{\partial q} , \tag{9-14a}$$ $$ P(q,Q)~=~-\frac{\partial F_1(q,Q)}{\partial P} , \tag{9-14b}$$ comes from an $F_1$-type generating function $F_1(q,Q)$.


  1. H. Goldstein, Classical Mechanics.

$^1$ Whatever a CT means.

$^2$In more detail: Given a point on the differential submanifold $M$, a [sufficiently small] open neighborhood $U\subseteq M$ of the point can generically be viewed as a graph of $(q,Q)$ [i.e. almost always, except for cases with special tangent directions].


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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