# Show that a transformation is canonical

by fluidistic
Tags: canonical, transformation
 PF Gold P: 3,207 1. The problem statement, all variables and given/known data Show that the following transformation is canonical: $Q=\ln \left ( \frac{\sin p }{q} \right )$, $P=q \cot p$. 2. Relevant equations A transformation is canonical if $\dot Q=\frac{\partial H'}{\partial P}$ and $\dot P =-\frac{\partial H'}{\partial Q}$. H' is the Hamiltonian in function of Q and P. 3. The attempt at a solution I've calculated $\dot Q = \dot p \cot p - \frac{\dot q \sin p }{q}$ and $\dot P=\dot q \cot p - \frac{q \dot p}{\sin ^2 p}$. Now I guess I must check out if the conditions about the partial derivatives of the Hamiltonian are satisfied but I do not know either H(q,p) nor do I know H'(Q,P). I'm stuck here.
 PF Gold P: 3,207 Ok guys, discard my attempt in my last post. I've found in some class notes and in Landau&Lifgarbagez's book the necessary condition for a transformation to be canonical. If I'm not wrong, I must show that I)$[Q,P]_{q,p}=1$ II)$[Q,Q]_{q,p}=0$ III)$[P,P]_{q,p}=0$. Where $[f,g]_{p,q}$ denotes the Poisson brackets of f and g with respect to p and q. In other words, this is worth $\frac{\partial f}{\partial p} \frac{\partial g}{\partial q}-\frac{\partial f}{\partial q} \frac{\partial g}{\partial p}$. Now I have that $Q=\ln \left ( \frac{\sin p }{q} \right )$ and $P=q \cot p$. I tried to be careful in doing the partial derivatives. I found out that: (1)$\frac{\partial Q}{\partial p}=q \cot p$ (2)$\frac{\partial P}{\partial q}= \cot p$ (3)$\frac{\partial Q}{\partial q}=-\frac{1}{q}$ (4)$\frac{\partial P}{\partial p}=q(-1- \cot ^2 p)$ Using this, I found out I) to be worth $(1-q) \cot ^2 p-1$. Unfortunately this isn't worth 1. What am I doing wrong? Edit: I just found a mistake in my definition of Poisson's brackets. But now I find I) to be worth $1+ \cot ^2 p-q \cot ^2 p$ which is still wrong. I've rechecked the partial derivatives, I do not see any mistake, yet my result keeps being wrong. Edit 2: This can be rewritten as $\frac{1}{\sin ^2 p}-q \cot ^2 p$. This is not worth 1 so that the transformation isn't canonical, which is absurd. I do not see any mistake. Edit 3 : I am nut guys! I found out the mistake in a derivative. I now reach all the results I should, problem solved!