Volume in phase space- Louviles theorem

[QuasarBoy543298]

I was looking at the following proof of Louviles theorem :
we define a velocity field as V=(dp_i/dt, dq_i/dt). using Hamilton equations we find that div(V)=0. using continuity equation we find that the volume doesn't change.
I couldn't figure out the following :

1- the whole point was to show that volume in phase space acts as an incompressible fluid,
but the proof assumed that the current field J is equal to v*constant -assumed ρ=const

2- after that using the continuity equation we said that dρ/dt = 0 so the volume must not change.
I couldn't understand the logical jump over here

I hope someone here can help me understand, thanks in advance!

 

[vanhees71]

The problem is often the statement of the theorem. Here's as I understood it's content best (I don't remember, where I've found this derivation).

First of all a set of phase-space variables $(q_i,p_i)$ define a $2f$-dimensional manifold, called phase space. The Hamilton canonical equations of motion
$$\dot{q}_i = \partial_{p_i} H, \quad \dot{p}_i=-\partial_{q_i}H$$
define a flow in phase space.

Now consider each point in some volume of phase space as a particle of a fluid (moving not in usual configuration space but in $2f$-dimensional phase space!). Now consider an infinitesimal volume element in phase space at time $t=0$ and think about how the volume changes when we let each of its points "flow" according to the Hamiltonian equations. The volume change involves the Jacobian
$$J(t)=\mathrm{det} \frac{\partial(q,p)}{\partial(q_0,p_0)}.$$
The infinitesimal change when going from $t$ to $t+\mathrm{d} t$ is given by
$$J(t+\mathrm{d} t)=\det \frac{\partial(q+\mathrm{d} q,p+\mathrm{d} p)}{\partial(q_0,p_0)} = J(t) \mathrm{det} \left (\hat{1} + \frac{\partial(\mathrm{d} q,\mathrm{d} p)}{\partial(q_0,p_0)} \right)= J(t) \left [1+\frac{\partial \mathrm{d} q_j}{\partial q_j} + \frac{\partial \mathrm{d} p_j}{\partial p_j} \right ].$$
In the last step we have expanded the determinant up to order $\mathrm{d} q,\mathrm{d} p$. Now this gives together with
$$\frac{\partial \mathrm{d} q_j}{\partial q_j} = \mathrm{d} t \frac{\partial^2 H}{\partial p_j \partial q_j}, \quad \frac{\partial \mathrm{d} p_j}{\partial p_j}=-\mathrm{d} t \frac{\partial^2 H}{\partial q_j \partial p_j}.$$
This gives
$$J(t+\mathrm{d} t)-J(t)=0 \; \Rightarrow \; \dot{J}=0.$$
The Jacobian thus doesn't change due to the flow, and this means that the phase-space volume element $\mathrm{d}^f p_0 \mathrm{d}^f q_0=\mathrm{d}^f p \mathrm{d}^f q$. The flow of the particles in phase space is thus an incompressible flow, and that's the intuitive content of Liouville's theorem.