# Canonical Transformations

1. Apr 24, 2007

### Moose352

I'm not sure I understand the use of generating functions in canonical transformations. In particular, why are there four basic canonical transformations? It isn't true that any canonical transformation is one of the four basic types, so what makes them special over any other transformation. Also, why is the generating fuction written in terms of both the new and the old coordinates? Since the old and new are related by the transformation, shouldn't it be possible to write the generating function solely in terms of the old or the new?
Also, is there any way to find a generating function given the transformation explicitly?

Last edited: Apr 24, 2007
2. Apr 24, 2007

### vanesch

Staff Emeritus
Canonical transformations are smooth mappings from the (p,q) into the (P,Q) space which satisfy certain properties (the deeper reason behind is the "conservation of symplectic structure", but I only write that here to show off somewhat ). So not just all mappings from (p,q) to (P,Q) will do.
Now, of course, by the implicit function theorem, if you define a mapping from, say, (p,Q) into (P,q), you can partly inverse this relation, and you ALSO define a mapping from (p,q) into (P,Q).
So the "4 different types" are simply different ways of looking upon mappings from (p,q) into (P,Q) through the implicit-function theorem. The mapping itself is not different, we've just written it implicitly.
The nice thing about these 4 "types" is that they give us a simple way to generate canonical transformations, and we have 4 different ways of doing so. But the canonical transformations themselves couldn't care less of how they are written down, in a way.

Doesn't work, unfortunately. You will have to inverse some equations in order to find the mapping (p,q) into (P,Q) in all 4 cases.

I'm not sure about this. I would think that if you write the new Hamiltonian as a function of, say, p and Q, that you obtain the generating function, but typing from the top of my head here, this might as well not be correct.

3. Apr 25, 2007

### Moose352

Regarding writing the generating in terms of the old or new, I still don't understand why it doesn't work. Sure, I will have to inverse the equations, but what difference does it make? Given a generating function, I know the explicit transformation, so why can't I just substitute. Is this wrong for some reason?

To find the generating function given the transformation (ie. P = P(p,q), Q = Q(p,q)): assuming the Hamiltonians (old and new) are the same, I can just write pq' = PQ' + dF/dt solely in terms of p and q, and get a partial differential equation for F. Problem is, then F is solely in terms of p and q. Is that wrong? The math works out all the same, right?