# Canonical transformation

by Logarythmic
Tags: canonical, transformation
P: 282
1. The problem statement, all variables and given/known data

Consider a canonical transformation with generating function

$$F_2 (q,P) = qP + \epsilon G_2 (q,P)$$,

where $$\epsilon$$ is a small parameter.
Write down the explicit form of the transformation. Neglecting terms of order $$\epsilon^2$$ and higher,find a relation between this transformation and Hamilton's equations of motion, by setting $$G_2=H$$ (why is this allowed?) and $$\epsilon = dt$$.

2. The attempt at a solution

I think the transformation equations are

$$\delta p = P - p = -\epsilon \frac{\partial G_2}{\partial q}$$

and

$$\delta q =Q-q=\epsilon \frac{\partial G_2}{\partial q}$$

 Quote by vanesch I guess there's a typo here: $$\delta q =Q-q=\epsilon \frac{\partial G_2}{\partial P}$$
but how can I solve the last part? Can I just say that with the use of H and dt the equations can be written as

$$\dot{p}=-\frac{\partial H}{\partial q}$$

and

$$\dot{q}=\frac{\partial H}{\partial P}$$

which are the Hamiltonian equations of motion? And why is this allowed?

 Quote by vanesch The idea is that we work in first order in $$\epsilon$$, and that you can hence replace everywhere $$P$$ by $$p$$ as the difference will introduce only second-order errrors.
 Emeritus Sci Advisor PF Gold P: 6,236 Ooops ! I'm terribly sorry, instead of replying, I erroneously edited your post !
 P: 282 Yes it should be a P, not a q. I get the relations $$Q = q + \delta q$$ and $$P = p + \delta p$$ but why is it allowed to use G = H?
Emeritus
 Quote by Logarythmic Yes it should be a P, not a q. I get the relations $$Q = q + \delta q$$ and $$P = p + \delta p$$ but why is it allowed to use G = H?