Independent canonical coordinate variables?

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?

• 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. – ACuriousMind Jan 12 '16 at 15:05
• @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). – Arturo don Juan Jan 12 '16 at 17:48
• 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. – user116174 May 2 '16 at 15:28
• @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. – Arturo don Juan May 2 '16 at 23:02
• @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! – Arturo don Juan May 2 '16 at 23:09

Let us suppress $t$-dependence in this answer for simplicity. A canonical transformation$^1$ $$(q,p)\quad\longrightarrow\quad (Q(q,p),P(q,p))$$ 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 $F_1(q,Q)$.
$^1$ Whatever that means.