
#1
Feb2912, 04:13 PM

PF Gold
P: 3,173

1. The problem statement, all variables and given/known data
Show that the following transformation is canonical: [itex]Q=\ln \left ( \frac{\sin p }{q} \right )[/itex], [itex]P=q \cot p[/itex]. 2. Relevant equations A transformation is canonical if [itex]\dot Q=\frac{\partial H'}{\partial P}[/itex] and [itex]\dot P =\frac{\partial H'}{\partial Q}[/itex]. H' is the Hamiltonian in function of Q and P. 3. The attempt at a solution I've calculated [itex]\dot Q = \dot p \cot p  \frac{\dot q \sin p }{q}[/itex] and [itex]\dot P=\dot q \cot p  \frac{q \dot p}{\sin ^2 p}[/itex]. 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. 



#2
Mar212, 03:57 PM

PF Gold
P: 3,173

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)[itex][Q,P]_{q,p}=1[/itex] II)[itex][Q,Q]_{q,p}=0[/itex] III)[itex][P,P]_{q,p}=0[/itex]. Where [itex][f,g]_{p,q}[/itex] denotes the Poisson brackets of f and g with respect to p and q. In other words, this is worth [itex]\frac{\partial f}{\partial p} \frac{\partial g}{\partial q}\frac{\partial f}{\partial q} \frac{\partial g}{\partial p}[/itex]. Now I have that [itex]Q=\ln \left ( \frac{\sin p }{q} \right )[/itex] and [itex]P=q \cot p[/itex]. I tried to be careful in doing the partial derivatives. I found out that: (1)[itex]\frac{\partial Q}{\partial p}=q \cot p[/itex] (2)[itex]\frac{\partial P}{\partial q}= \cot p[/itex] (3)[itex]\frac{\partial Q}{\partial q}=\frac{1}{q}[/itex] (4)[itex]\frac{\partial P}{\partial p}=q(1 \cot ^2 p)[/itex] Using this, I found out I) to be worth [itex](1q) \cot ^2 p1[/itex]. 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 [itex]1+ \cot ^2 pq \cot ^2 p[/itex] 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 [itex]\frac{1}{\sin ^2 p}q \cot ^2 p[/itex]. 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! 


Register to reply 
Related Discussions  
Canonical Transformation  Advanced Physics Homework  7  
Canonical Transformation  Classical Physics  0  
Canonical Transformation  Advanced Physics Homework  1  
Canonical transformation  Advanced Physics Homework  2  
canonical transformation  Introductory Physics Homework  1 