Derive the generating function for canonical transformation of type $F_3$

Question

I'm working on some practice questions and I am a bit confused with this one:

Generating functions of the type $F_1(q,Q)$ satisfy the condition: $$pdq-PdQ = dF_1$$ Starting from this condition derive the analogous result for the generating functions of type $F_3(q,P)$.

I must admit, I am a little confused with the Legendre transform and how the other generating functions are derived.

Here is what I have tried by following my textbook,

Applying the Legendre transform with respect to the new coordinates and momenta to $F_1$:

$$ d(\sum QP + F_1) = PdQ + QdP + pdq-PdQ = QdP + pdq = dF_2 $$

Which seems correct for $F_2$. (I'm not even sure if I have done this correctly, I'm pretty much just following my textbook). To get $dF_3$, my textbook just says perform the Legendre transform with respect to the 'other' variables?

However, I'm not sure what this means?. I know I am expecting to get $-qdp-PdQ$, but just not too sure about what I am doing.

ADDED:

I think I may have gotten a little further with my understanding. If I take the legendre transform of $F_1$ with respect to the old coordinate $q$ and it's conjugate momenta, then:

$$F_1 - pq = F_3$$ so, $$d(F_1 - pq) = pdq-PdQ-qdp-pdq = -qdp-PdQ = dF_3$$

Which seems to be correct. But then isn't

$$d(F_1 - QP) = pdq-PdQ-QdP-PdQ $$ which is not $dF_2$. Where am I going wrong? How would I say, derive $dF_4$? By definition of the legendre transform, is it true that for instance if $G$ is the legendre transform of $F$, i.e. $G=F-wy$, the pair $w,y$ must always be a conjugate pair (i.e $w=\frac{\partial f}{\partial y})$?

Answer 1

Intuitively, Legendre transform is just "integrate by parts". So, from $pdq - PdQ =F_1$, we have $pdq -d(PQ)+QdP =dF_1$, i.e., $pdq +QdP =d(F_1+PQ) \equiv dF_2$. The "Transform" means after "integration by parts", we need to change the independent variables. For example, in $F_1$, $p=p(q,Q)$ but in $F_2$, $p=p(q,P)$. So we need to solve $Q = Q(q,P)$ and insert in to $p=p(q,Q(q,P))$ to get $p(q,P)$.

Answer 2

This here is not exactly an answer, but a clarification of why Legendre transforms are not needed to get this result. Using exterior derivatives is much easier.

If you have $pdq - PdQ = dF$, take the exterior derivative on both sides to get $dp \wedge dq - dP \wedge dQ = 0$ (the exterior derivative of a differential is zero). The wedge product is antisymmetric, so $dp \wedge dq - dP \wedge dQ = dp \wedge dq + dQ \wedge dP =0$. Take now the 1-form whose exterior derivative gives us $dp \wedge dq + dQ \wedge dP =0$, that is, $pdq + QdP = dG$. This means that

$$\partial_q G(q,P) = p \\ \partial_P G(q,P) = Q ,$$

and that's the answer. The Legendre transform is just a cumbersome way of transforming variables in 1-forms instead of using the much simpler structure of the 2-forms.