Why Liouville's theorem is obvious?

Question

In Florian Scheck's Mechanics, he stated the local form of Liouville's theorem as follows:

Let $\Phi_{t,s}(x)$ be the flow of the differential equation $-J\frac{d}{dt}x=H_{x}$. Then for all $x,t,s$ for which the flow is defined, we have $$D\Phi_{t,s}(x)\in Sp_{2f}.$$

In his proof he claimed that $$\frac{d}{dt}[D\Phi_{t,s}(x)^{T}JD\Phi_{t,s}(x)]=0$$ Thus since $D\Phi_{t,s}(x)^{T}JD\Phi_{t,s}(x)=J$ when $t=s$, we proved the theorem. My question is:

Why at $t=s$, $D\Phi_{s,s}(x)^{T}JD\Phi_{s,s}(x)=J$ holds? The author claimed that this is `obvious', but it is not obvious to me. Mathematically $t=s$ just means the flow starts at the time $t=s$ for which it is defined. So we can use $s=0$ without losing generality. But why would Hamilton's equation $$-J\frac{dD\Phi_{t,s,t=s}(x)}{dt}=DH_{x}\circ D\Phi_{t,s,t=s}(x)$$

imply $$D\Phi_{s,s}(x)^{T}JD\Phi_{s,s}(x)=J?$$ Writing out this in block matrix form we should have this equal to $det[D\Phi_{s,s}]J$ instead. And we do not know as prior $det[D\Phi_{s,s}]=1$.

Answer 1

It seems like nobody answered your specific question, so here goes. $\Phi_{s,s}$ is just the identity map on the phase space (see section 1.20 in the third edition of Scheck's book if this isn't clear to you). Therefore $D\Phi_{s,s}(x)=I_{2n}$, the $2n\times2n$ identity matrix.

Answer 2

The reason its obvious is because the Hamiltonian flow is a canonical transformation on phase space, and this means that the Jacobian of the Hamiltonian flow, which performs a linear transformation on the tangent space of $D\Phi$, preserves the symplectic form.

The way $D\Phi$ transforms the symplectic form J is the thing he wrote down, and the fact that it preserves J, implies Liouville's theorem. But it is difficult to argue which implies which, since they are so simply equivalent to each other. To see that Hamiltonian flow is a canonical transformation, choose canonical coordinates, and evolve x and p by an infinitesimal amount dt to new coordinates:

$$ x_i + {\partial_{p_i} H} dt$$ $$ p_j - {\partial_{x_j} H} dt$$

then check that the Poisson bracket of these new coordinates ${x'_i,p'_j}$ (using the old coordiantes to compute the Poisson bracket) is still $\delta_{ij}$, so they are still canonical and J hasn't changed. This follows from the cancelling second partial derivative of H in the Poisson bracket calculation, and this shows you that J is preserved at each time step, so it must be preserved be integrating the differential equation to finite time. There are a million ways to say the same thing, some superficially more rigorous, but this is good enough.

Answer 3

There are better ways to do this but I am trying to compute using Ron Maimon's suggestion. So to make matters simpler I will only assume two variables(hopefully other variables will not matter). Then we have $$q_{i}'=q_{i}+\frac{d}{dp_{i}}Hdt$$ and $$p_{i}'=p_{i}-\frac{d}{dq_{i}}Hdt$$

Thus $$\{q_{1}',p_{2}'\}=[\frac{dq_{1}'}{dq_{1}}\frac{dp_{2}'}{dp_{1}}-\frac{dq_{1}'}{dp_{1}}\frac{dp_{2}'}{dq_{1}}]+[\frac{dq_{1}'}{dq_{2}}\frac{dp_{2}'}{dp_{2}}-\frac{dq_{1}'}{dp_{2}}\frac{dp_{2}'}{dq_{2}}]$$

Notice that $$\frac{dq_{i}'}{dq_{j}}=\delta_{ij}+\frac{d^{2}}{dp_{i}dq_{j}}Hdt;\frac{dq_{i}'}{dp_{j}}=\frac{d^{2}}{dp_{i}dp_{j}}Hdt$$

We should have the first term to be $$(1+\frac{d}{dp_{1}dq_{1}}Hdt)(-\frac{d}{dq_{2}p_{1}}Hdt)+(\frac{d^{2}}{dp_{1}^{2}}Hdt)(\frac{d}{dq_{2}dq_{1}}Hdt)$$ after expansion and cancellation this leaves us with $$-\frac{d}{dq_{2}dp_{1}}Hdt$$

The second term after expansion and cancellation this leaves us with $$\frac{d}{dq_{2}dp_{1}}Hdt$$

So they indeed cancel out each other and we verified $\{q_{1}',p_{2}'\}=0$. The other calculation $\{q_{i}',p_{i}'\}=1$ should be largely similar.