1. Sep 23, 2010

javiergra24

Hi everybody

I've got a problem related to canonical transformations that I can`t solve:

Given the expression of the canonical transformation

$$Q=3q\cdot\big[ \exp\big((p+q)^5\big)+1\big] +3p\cdot \big[\exp((p+q)^5)+1\big]+p$$
$$P=p+q$$
I have to calculate an associated canonical transformation. Anybody can help me?

Thanks

2. Sep 23, 2010

Are you sure you have stated exactly your problem?

3. Sep 23, 2010

javiergra24

Exact problems is (from exam):
Given the transformation

$$Q=3q\cdot\big[ \exp\big((p+q)^5\big)+1\big] +p\cdot \big[2\exp((p+q)^5)-1\big]$$

$$P=q+p$$

Modify it slightly in order to be canonical

Answer. After imposing the condition for the Poisson bracket (equal to one) we get the result:
$$\boxed{Q=3q\cdot\big[ \exp\big((p+q)^5\big)+1\big] +3p\cdot \big[\exp((p+q)^5)+1\big]+p}$$

In part two we're asked to obtain an associated canonical transformation. But after reading my books and papers about mechanics I still don't know what's an "associated trasformation mean". Is it the inverse transformation?

Last edited: Sep 23, 2010
4. Sep 24, 2010

OK. Now part of the problem is clear - supposing it is indeed a canonical transformation (I didn't check). But what the author of this exercise means by an "associated canonical transformation" - that I don't know.

5. Sep 24, 2010

gabbagabbahey

Same here. I suppose the questioner might just be looking for an equivalent transformation, but written in a different functional form... something like

$$Q=Q(q,P)=3P\left(e^{P^5}+1\right) +P-q$$ and $$P=q+p$$
$$Q=Q(q,p)=3q\left(e^{(p+q)^5}+1\right)+3p\left(e^{(p+q)^5}+1\right)+p$$ and $$P=q+p$$