According to Liouville's theorem, why is the measure on an energy-surface different from the measure on the phase space in general

Question

I recently read Khinchin's derivation of Liouville's theorem. I was able to follow the math for the most part, however I was hoping for an intuitive understanding about why the form of the measure on an energy surface in phase space is different than the form of the measure on the whole phase space

If we have an $N$-dimensional phase space then the measure on that phase space is simply $dV$. I was able to follow the proof about why that measure is conserved under the natural motion. However, if we restrict our analysis to an energy surface to that phase space then the measure on that surface becomes $\frac{d \Sigma}{| \nabla H |}$, and not just $d \Sigma,$ the area element on that energy surface. From what I understand, the stated reason for this is that $d \Sigma \cdot dn = \frac{d \Sigma}{| \nabla H |}$ (where $dn$ is the normal to the energy surface) is just a volume element in the phase space, and we can then appeal to the fact that we already know that a differential volume element in phase space is conserved. However, if we view the energy surface itself as an $N-1$ dimensional phase space then your old $d \Sigma$ essentially becomes a volume element in this new $N-1$ dimensional space. Why, then, does Liouville's theorem as derived for the higher dimensional phase space not also apply to the $N-1$ dimensional space, thus making $d \Sigma$ the correct measure?

Edit to clarify what my quantities mean: $dV$ is a differential volume element in the phase space. $d \Sigma$ is a differential area element on an energy surface in the phase space. $H$ is my Hamiltonian, and for my $N$ dimensional phase space

$$| \nabla H | = \sqrt{ \sum_{i=1}^{N/2} \left( \frac{\partial H}{\partial q_i} \right)^2 + \left( \frac{\partial H}{\partial p_i} \right)^2}.$$

$dn$ is a differential normal vector to the energy surface.

Answer 1

The intuitive picture is that phase points move more quickly through regions of phase space where $|\nabla H|$ is higher. As a result if you have a constant-energy ensemble of phase points (a flat "packet"), their phase space area enlarges as they move through high $|\nabla H|$ regions.

For the full Liouville's theorem this is not a problem. A constant-energy ensemble is flat and thus always has zero phase space volume; this zero volume is conserved.

Now, if you now have a "thick" packet which does have a volume --- an ensemble encompassing different energies. The points move more rapidly in high $|\nabla H|$ regions, spreading out in area, but the packet gets thinner. In this way the volume is conserved.

I made an animation for wikipedia that might help:

The upper plot shows a 2D phase space (the phase space "volume" is the displayed area; a constant-energy surface would be a 1D curve). The mechanics are simply that of a particle moving in the potential well plotted in red on the bottom graph.

The ensemble of blue phase points is a volume ensemble. Oberve the green point in particular. Notice how the phase points slow down on the left side of the well where $|\nabla H|$ is lower, and speed up on the steep potential slope on the right side.

Now back to the constant-energy Liouville's theorem. The whole point is we want a measure where the phase area is conserved in evolution. We thus need to compensate for the tendency of phase points to vary their phase space speed.

Answer 2

One way to understand it is to write it using the Dirac measure to express the phase space in the microcanonical ensemble (because that's what it is about).

In this ensemble the idea is to say that the energy $H(\textbf{r})$ (where $r$ is a point in phase space) is fixed at some value $E$ (actually it belongs to a very small interval $[E,E+\delta E]$).

One to express that in the sense of distributions is to write that the measure will look like:

$$\mu = \int \:\mathrm{d}^{6N}r \:\delta(H(\textbf{r})-E) $$

where $\mathrm{d}^{6N}r$ is the Lebesgue measure $\mathrm{d}V$ you were referring to.

Now there is an identity about the Dirac delta function that says that

$$\delta(f(x)) = \frac{\delta(x-x_0)}{|f'(x_0)|}$$

where $x_0$ is the point (assumed unique here in the domain of integration) that satisfies $f(x_0) = 0$.

In our multidimensional case the remaining $\delta(\textbf{r}-\textbf{r}_0)\mathrm{d}V \equiv d\Sigma$ and $|f'(x_0)| \rightarrow |\nabla H|$ and it comes:

$$\mu = \int \:\frac{\mathrm{d}\Sigma}{|\nabla H|} \;.$$