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 $$P=q+p$$