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I was trying to prove, that for a transformation to be Canonical, one must have a relationship:

$$ \left\{ Q_a,P_i \right\} = \delta_{ai} $$

Where $Q_a = Q_a(p_i,q_i)$ and $P_a = P_a(p_i,q_i)$.

Now to do the proof I started with $\dot{Q_a}$:

  1. Chain rule and Hamilton's equation for initial coordinates $q_i,p_i$ $$ \dot{Q_a} = \frac{\partial Q_a}{\partial q_j} \dot{q_j} + \frac{\partial Q_a}{\partial p_j} \dot{p_j} = \frac{\partial Q_a}{\partial q_j} \frac{\partial H_a}{\partial p_j} - \frac{\partial Q_a}{\partial p_j} \frac{\partial H_a}{\partial q_j} $$

  2. Then I apply chain rule for the Hamiltonian derivatives: $$ \dot{Q_a} = \frac{\partial Q_a}{\partial q_j} \left( \frac{\partial H}{\partial Q_i} \frac{\partial Q_i}{\partial p_j} + \frac{\partial H}{\partial P_i} \frac{\partial P_i}{\partial p_j} \right) - \frac{\partial Q_a}{\partial p_j} \left( \frac{\partial H}{\partial Q_i} \frac{\partial Q_i}{\partial q_j} + \frac{\partial H}{\partial P_i} \frac{\partial P_i}{\partial q_j} \right) $$

  3. Now reordering the terms yields us: $$ \dot{Q_a} = \frac{\partial H}{\partial Q_i} \left\{ Q_a,Q_i \right\} + \frac{\partial H}{\partial P_i} \left\{ Q_a,P_i \right\} $$

Now here the problem, for the transformation from a coordinate system $(q_i,p_i)$ to a coordinate system $(Q_a(q_i,p_i), P_a(q_i,p_i))$ to be canonical we require:

$$\left\{ Q_a,P_i \right\} = \delta_{ai}$$

But why we have as well the following requirement, or is it just too obvious or true because of some property of any coordinate transformation?

$$\left\{ Q_a,Q_i \right\} = 0$$


The problem I am having is as follows. I agree, that the following two are true (if I used the covariant notation correctly):

$$ \left\{ q^i,q_j \right\}_{q,p} = \frac{\partial q^i}{\partial q^k} \frac{\partial q_j}{\partial p_k} - \frac{\partial q^i}{\partial p^k} \frac{\partial q_j}{\partial q_k} = 0 $$

$$ \left\{ Q^i,Q_j \right\}_{Q,P} = 0 $$

But why its the case that the following is also true?

$$ \left\{ Q^i,Q_j \right\}_{q,p} = 0 $$

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1 Answer 1

up vote 3 down vote accepted

A canonical transformation $(q^i,p_j) \to (Q^i,P_j)$ preserves the form of Hamilton's equations.

Similarly, a symplectic transformation$^1$ $(q^i,p_j) \to (Q^i,P_j)$ preserves the Poisson structure, aka. as a symplectomorphism. In other words, all the fundamental Poisson brackets (PB)

$$ \{ q^i,p_j \} ~=~ \delta^i_j, \qquad \{q^i,q^j \}~=~0, \qquad \{ p_i,p_j \} ~=~ 0,\qquad i,j \in\{1, \ldots, n\},$$

have the same form in the new coordinates

$$ \{ Q^i,P_j \} ~=~ \delta^i_j, \qquad \{Q^i,Q^j \}~=~0, \qquad \{ P_i,P_j \} ~=~ 0,\qquad i,j \in\{1, \ldots, n\}. $$

In particular, to answer OP's question(v2), the relations $\{Q^i,Q^j \}=0$ and $\{P_i,P_j\} = 0$ are only trivial if $n=1$, because of skewsymmetry of PB.

As is well-known, canonical and symplectic transformations are the same. For a proof [at least in the case of restricted transformations, i.e. transformations without explicit time dependence], see e.g. Ref.1, which uses so-called symplectic notation. An important point is that the Jacobian matrix of a symplectic transformation must be a symplectic matrix.

References:

  1. H. Goldstein, Classical Mechanics, Section 9.4 in eds. 3 or Section 9.3 in eds. 2.

--

$^1$ In this answer we will for simplicity only discuss non-degenerate Poisson brackets in finite dimensions using globally defined coordinates.

share|improve this answer
    
I see. But I'm still not 100% convinced why the structure is preserved... I will ammend my question. –  gns-ank Jan 8 '13 at 14:50
    
Thanks a lot for the answer and the reference. –  gns-ank Jan 8 '13 at 15:39

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