General Form of Canonical Transformations

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kolawoletech
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Homework Statement


How do I go about finding the most general form of the canonical transformation of the form
Q = f(q) + g(p)
P = c[f(q) + h(p)]
where f,g and h are differential functions and c is a constant not equal to zero. Where (Q,P) and (q,p) represent the generalised cordinates and conjugate momentum in the new and old system

Homework Equations


{Q,Q}={P,P}=0 {Q,P}=1

The Attempt at a Solution


I arrived at a function

c.f'(q)[h'(p)-g'(p)]=1

I don't know how to get further to prove canonicity
 
kolawoletech said:
c.f'(q)[h'(p)-g'(p)]=1
There are two independent variables in this equation: p and q, and we require it to hold for all values of p and q. That should enable us to radically narrow down the possibilities for functions f, g and h.

Try partial differentiating both sides of the equation with respect to q to get one equation and then wrt p to get another.
What do those two equations tell you about the functions f and (h-g)?
 
I did that and got to "see the attachment"
But I am not so sure if I did it right
The rest of the question see solve the inverse of the canonical transformation: express q, p in terms of Q and P. The actual question is attached see attachment(it is the second question)
 
Attachments
  • Handwritten whiteboard derivation for a canonical transformation: Q = f(q,p) + g(q,p), P = c[f(q)+h(p)], Poisson bracket conditions, then f(q) = K/c and derivation of P = cq^2 + ... expressions in q, p, Q and P.
    IMG-20160412-WA0001.jpg
    24.1 KB · Views: 598
  • Photo of a printed homework sheet for PHY 702 Classical Mechanics Project 3, showing a canonical transformation problem with matrix equations and the transformation Q = f(q) + g(p), P = c[f(q) + h(p)].
    HjIkBnu9.jpg
    25.9 KB · Views: 532
I'm sorry but that photo is way too hard to read. Try typing it in using latex. This post is a primer to get you started. If you're studying physics then any time spent learning latex is a very good investment.