# Invariant Phase space volume under canonical transfromation

I have a volume element in phase space:

$$d\omega = \prod _{i=1}^{N}(dq_{i},dp_{i})$$

Now I should show the invariance of this product under canonical transformations. I think first I would have to write down a general equation for the canonical transformation then compute the total differential and plug this into the equation given. My problem is now I can't figure out how the equation for a general canonical transformation would look like. Does anybody have a hint how to start? The proof should be done without the concept of Symplectomorphisms like it was already shown in this question Which transformations are canonical?

• This question (v2) is a duplicate of the second subquestion in this Phys.SE post. – Qmechanic May 11 '18 at 6:50
• @Qmechanic I read this question but it does not really answer the question to me since I am trying to find this property without the concept of Symplectomorphisms. – zodiac May 11 '18 at 7:08
• So which definition of CT do you use then? – Qmechanic May 11 '18 at 7:49
• I was thinking to define the CT as a transformation that preserves the poisson bracket. – zodiac May 11 '18 at 8:45
• That is the defining property of a symplectomorphism. – Qmechanic May 11 '18 at 22:15

Consider a phase space volume $d\omega = \prod dq^idp_i$. Time evolution can be viewed as the unfolding of canonical transformations, hence, we shall evolve this system $t\rightarrow t+dt$. We wish to show invariance of the measure under this canonical transformation $(q,p) \rightarrow (Q,P)$ where: $$Q^i = q^i + \dot q^id t, \qquad P^i = p_i + \dot p_idt$$ Compute the new volume $d\omega' = \prod dQ^idP_i$:
\begin{align} d\omega ' &= \prod d( q^i + \dot q^id t, p_i + \dot p_id t)\\ & = \prod ( dq^i + \frac{\partial \dot q^i}{\partial q^i}dt, dp_i + \frac{\partial \dot p_i}{\partial p_i}d t) \end{align} Let us now expand this and retain terms linear in $dt$ we obtain $$d\omega' \approx \sum_i \bigg( \frac{\partial \dot q^i}{\partial q^i} + \frac{\partial \dot p_i}{\partial p_i}\bigg) = \bigg\{\frac{\partial }{\partial q^i}\frac{\partial H}{\partial p_i} - \frac{\partial }{\partial p_i}\frac{\partial H}{\partial q^i}\bigg\} = 0$$ Therefore, we have shown invariance of the measure under a canonical transformation.