How Does a Canonical Transformation Relate to Hamilton's Equations of Motion?

[Logarythmic]

Homework Statement

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

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

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?

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.

 

[vanesch]

Ooops !

I'm terribly sorry, instead of replying, I erroneously edited your post !

 

[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?

 

[vanesch]

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?

You have the choice ! Every function G(q,P) (which is smooth enough) will generate a canonical transformation. So you may just as well use H(q,P), and then - that's the whole point - the transformation equations from (q,p) into (Q,P) give you simply the genuine time evolution where epsilon is the small time step. For an arbitrary G that isn't the case of course, you've just transformed your coordinates (q,p) in some other (Q,P). But for G = H, you've transformed the coordinates (q,p) in what they will be, a small moment later !
The reason for that is that your transformation equations you've found for an arbitrary G are what they are, and become the Hamilton equations of motion when you pick G to be equal to H.

Now, strictly speaking we should write H(q,P) instead of H(q,p), but we can replace the P by p here, because they are only a small amount different.

 

[Marco_84]

and it also preserve the fundamental poisson brackets...

its just a quick calcolus {Q,P}