# Why Liouville's theorem is obvious?

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 $\det[D\Phi_{s,s}]=1$.

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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.

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Great as usual; I would like to append the comment that to some extent we define a canonical coordinate system by the fact that it is preserved under Hamiltonian flow. Alternatively, one can take the covariant view and say that we are trying to put coordinates on the space of physical motions, which are constants such as the position and momentum of a particle at some particular time; if the system has some preferred reference coordinate such as time, one would require a volume preserving transform between coordinatisations related by transforms in time. – genneth Jun 2 '12 at 14:09
@Ron Maimon: This is helpful, but not so transparent to a layman like me (who learned Possion bracket yesterday). So I have to spend some time to understand this. Thanks for the help. – Bombyx mori Jun 2 '12 at 16:55
@user32240: It isn't transparent if you're not familiar with this stuff--- you should explicitly do the PB calculation I gave--- it's not difficult at all. The point here is that it's like doing a rotation in small increments that preserve dot products. The PB is a mathematical structure, like a dot-product, except not a dot-product, encoded by J. When time-evolution preserves the dot-product you get rotation, when time-evolution preserves J, it is symplectic. You need to internalize this, and then all this stuff is obvious, but it requires going through it carefully once. – Ron Maimon Jun 2 '12 at 19:04
@genneth: I agree, but it isn't obvious a-priori that such a thing is possible, so you need an explicit demonstration that it works with the usual canonical coordinates. If you have a volume destroying differential equation (like with fricion) any attempt to find canonical coordinates would be counterproductive. – Ron Maimon Jun 2 '12 at 20:24
@RonMaimon: Agreed. I've never really thought that hard about it, but in principle it would be sufficient to require that the dynamics is reversible? But even that isn't necessary since we postulate that there exists trajectories, and we just need to choose coordinates for them. The existence of the canonical pairs is non-obvious, and fundamentally requires the Lagrangian to be first order; but more generally it is not trivial in the presence of gauge which prevents a global choice (ncatlab.org/nlab/show/phase+space). – genneth Jun 2 '12 at 23:43

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.

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Thanks! (I realized this myself). – Bombyx mori Jun 3 '12 at 14:46
@user32240: The determinant is obviously 1 at t=s (at what other people call t=0), it is not still obviously 1 at other times. This requires showing the derivative is zero, and this is the commutation business. This is the irreducible content of the Liousville theorem (or the symplectic character of time evolution) and there is no simplification possible that avoids this. – Ron Maimon Jun 4 '12 at 4:12
@RonMaimon: A double-check of the author's proof verified your claim, thanks. – Bombyx mori Jun 5 '12 at 0:44

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.

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The only part of this which is wrong is that there are better ways to do this. There aren't better ways to do this, they are all equivalent to this deep down. This is not an answer, and should be removed, anyone can reproduce it, this is the point. – Ron Maimon Jun 3 '12 at 18:17
@RonMaimon: I mean by realizing the determinant is 1, the proof just follows. It is easier than working with the mixed partial derivatives. – Bombyx mori Jun 3 '12 at 22:54
Absolutely not! The determinant is not obviously one at all times, that's not what is going on, it is only obviously 1 at t=s, when there is no transformation. The derivative of the determinant with respect to time is zero and showing this statement however way you do it contains the calculation with mixed partial derivatives. This calculation is the irreducible minimum for understanding the result, all proofs contain it, and this is everything nontrivial that they contain. There is no other way, only this way. – Ron Maimon Jun 4 '12 at 4:10