# Canonical transformations, poisson brackets

1. Mar 11, 2015

### Coffee_

Three questions

1) Let's say that N $q_i$ and $p_i$ are transformed into $Q_k$ and $P_k$, so that:

$q_i = q_i(Q_1,Q_2,. ... , P_1,P_2, .... )$ and $p_i=p_i((Q_1,Q_2,. ... , P_1,P_2, .... )$

We have proved that these transformations are canonical only and only if $\forall i$

$\{Q_k(q_i,p_i),H(q_i,p_i,t)\}_{p,q} = \{Q_k, H(Q_k,P_k,t)\}_{Q,P}$ (1)

Until now I understand why this is the case. Next step in the reasoning was saying :

And thus, for any functions $f(q_i,p_i)$ and $g(q_i,p_i)$ must hold $\{f,g\}_{q,p}= \{f,g\}_{Q,P}$ (2)

How to make this final step from (1) to (2) ?

2) We also use the term ''transformed Hamiltonian'' $K(Q_i,P_i,t)$ for which the Hamilton equations hold for $Q_i$ and $P_i$

I'm pretty sure that it's basically the old hamiltonian $H(q_i,p_i,t)$ with all the $q_i$ and $p_i$ expressed in function of the new variables but the new letter $K$ makes me doubt it a little. If it's the same thing why use a different letter? So just asking to be sure here.

3) Why all of a sudden restrictions on variables we can transform into? In Lagrangian mechanics one can transform in any generalized coordinates ( as long as they are time independent if you want to have nice properties ) but basically no real restrictions.

Suddenly in Hamiltionian mechanics you need to check what I mentioned above. I find it weird since Lagrangian and Hamiltonian formalisms should be equivalent in power. Is there an intuitive reason why one has no restrictions on ( basic ) transformations and the other does?

Last edited: Mar 11, 2015
2. Mar 11, 2015

### Orodruin

Staff Emeritus
Have you tried simply using the chain rule for derivatives and the fact that canonical transformations preserve the Poisson brackets?

You have less restrictions on canonical transformations than you do on the general coordinate transformations in Lagrangian mechanics. The reason is that you can always find a canonical transformation corresponding to the Lagrangian change of variables $q_i \to Q_i(q)$. In the Lagrangian setting, the derivatives of $q_i$ are not independent from the $q_i$ in the same way as the $p_i$ are in Hamiltonian mechanics.