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Proving that two variables are canonical

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the polar variables [tex] \rho = \sqrt{x^2+y^2} [/tex] and [tex] \phi = tan^{-1}(y/x) [/tex], with [tex] p_{rho} = \frac{x p_x +yp_y}{\rho} [/tex] and [tex] p_{phi} = xp_y-yp_x [/tex], 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 [tex] \sum_{i} (\frac{dw}{dq_i}\frac{dz}{dp_i} - \frac{dw}{dp_i}\frac{dz}{dq_i}) [/tex]. 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

    [tex] \{\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}) [/tex]? 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. jcsd
  3. Oct 9, 2009 #2
    "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 [tex]q_1=x, q_2=y, p_1=p_x,p_2=p_y[/tex]. See if you get the right numbers out.
     
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