# Proving that two variables are canonical

1. Oct 8, 2009

### quasar_4

1. The problem statement, all variables and given/known data

Show that the polar variables $$\rho = \sqrt{x^2+y^2}$$ and $$\phi = tan^{-1}(y/x)$$, with $$p_{rho} = \frac{x p_x +yp_y}{\rho}$$ and $$p_{phi} = xp_y-yp_x$$, are canonical.

2. Relevant equations

To be canonical, a set of variables must satisfy {qi,qj}={pi,pj} = 0 and {qi, pj} = dij (the Kronecker delta)

3. The attempt at a solution

I'm just a wee bit confused about the brackets here. The thing is, the way I've learned Poisson brackets is that given functions w=w(p,q) and z=z(p,q), the bracket is $$\sum_{i} (\frac{dw}{dq_i}\frac{dz}{dp_i} - \frac{dw}{dp_i}\frac{dz}{dq_i})$$. But this definition gives w=w(p,q) and z=z(p,q). My problem gives me rho = rho (x,y) and phi =phi(x,y). But these are both positions. So, doesn't the bracket have to be

$$\{\rho,\phi\} = (\frac{d\rho}{dx}\frac{d\phi}{dy} - \frac{d\rho}{dy}\frac{d\phi}{dx}) + (\frac{d\rho}{dy}\frac{d\phi}{dx} - \frac{d\rho}{dx}\frac{d\phi}{dy})$$? Or am I supposed to have derivatives of rho and phi with respect to px and py, even though the positions aren't explicitly functions of momentum?? help!:yuck:

2. Oct 9, 2009

### fantispug

"Or am I supposed to have derivatives of rho and phi with respect to px and py, even though the positions aren't explicitly functions of momentum??"

Dead right - in a way they are functions of momentum, just trivial functions. In the same way f(x)=1 is a function of x. You just want to follow your prescription exactly with your phase variables $$q_1=x, q_2=y, p_1=p_x,p_2=p_y$$. See if you get the right numbers out.