Time dependence in Gibbs ensemble density

Question

I am trying to understand why there is time dependence in a Gibbs ensemble density. I am loosely following Kardar's Statistical Mechanics of Particles, but here is the argument as I understand it.

Let $\Gamma$ be the phase space of our system, let M be a macrostate of our system with $\mathscr{N}$ microstates corresponding to it. We define $d\mathscr{N}(g)$ to be the number of such microstates in some small neighborhood of $g \in \Gamma$ so that $\int d\mathscr{N} = \mathscr{N}$. We then define $\rho (g) d \Gamma = d\mathscr{N}/\mathscr{N} $ so that $\rho (g) \geq 0 $ everywhere and $\int \rho d \Gamma = 1$. This tells us that $\rho$ is a probability density function.

Now I am going to work out what I think is a generic and representative case, and it leads me to a conflict with what I am reading everywhere.

Let's assume I have a dilute, non-interacting gas consisting of N particles. I will consider a macrostate corresponding to a fixed energy E. The microstates corresponding to this are all {$ p_i $} such that

$$E = \sum_i \frac{p_i ^2}{2m_i}$$

is fixed at some fixed value E. So if I am trying to imagine what my $\rho$ looks like in this case, I would imagine that it has some non-zero value for any point in phase space where the momenta satisfy the above equation and zero everywhere else. Crucially, this has no time dependence whatsoever, this is just a function of the position and momenta.

Now what every book does at this point is the following. They say that yes, $\rho$ only depends on p and q, but they themselves depend on time. To which I say that's only because you are sloppy with your notations.

$\Gamma$ is some space with a density function on it which we call $\rho$, it is only a function of the points in $\Gamma$. Period. Now, we also look at curves in $\Gamma$ i.e $\gamma : \mathbb{R} \to \Gamma$, these trajectories have to satisfy Hamilton's equations, etc.

Now, sometimes we are sloppy and instead of writing $\gamma(t)$ we write its image in $\Gamma$ as $p_i(t),q_i(t)$. Thats all fine, but how does putting a curve in your space satisfying Hamilton's equations suddenly make $\rho$ depend on time?

Where am I going wrong?

Answer 1

Any particular microstate belonging to the ensemble is itself a dynamical system. Over time, each microstate traces out its own trajectory in phase space. With this in mind, let's imagine a differential element of phase space that spans the volume $[\vec\Gamma, \vec\Gamma+\vec{d\Gamma}]$. In a short time $dt$, some number of microstates will enter this little volume element and some number will exit. A net flow of $\rho$ through this element can come from two possible sources: explicit dependence on time in $\rho$ and implicit dependence through the coordinates and momenta (i.e., deformation of the volume element $\rho$ passes through). That is, the total rate of change of $\rho(\vec \Gamma, t)$ with respect to time is $$ \frac{d\rho}{dt} = \underbrace{\frac{\partial\rho}{\partial t}}_{\text{explicit}}+\underbrace{\frac{d\vec\Gamma}{d t}\cdot\frac{\partial\rho}{\partial\vec\Gamma}}_{\text{implicit}}. \tag{$\star$} $$ On the other hand, there is a flux of $\rho$ out the volume element given by $$ \vec J =\frac{\partial}{\partial\vec\Gamma}\cdot\left(\rho\frac{d\vec\Gamma}{dt}\right) = \frac{\partial \rho}{\partial\vec \Gamma}\cdot\frac{d\vec\Gamma}{dt} +\rho\frac{\partial}{\partial\vec\Gamma}\cdot\frac{d\vec\Gamma}{dt} $$ and since what goes in must come out, $$ \frac{\partial\rho}{\partial t} + \vec J = 0 \implies \boxed{\frac{d\rho}{dt} = -\rho\frac{\partial}{\partial\vec\Gamma}\cdot\frac{d\vec\Gamma}{dt}} $$ This is the Liouville equation, and it states that the probability density along dynamical trajectories depends on the compressibility of phase space.

For Hamiltonian systems, the Liouville equation takes a very special form. This is because $d\vec\Gamma/dt$ is given by Hamilton's equations of motion: $$ \frac{d\vec p}{dt} = -\frac{\partial H}{\partial\vec q} \quad \frac{d\vec q}{dt}=\frac{\partial H}{\partial\vec p}. $$ If you substitute these into the right-hand side of the Liouville equation, you find that it becomes zero. That is, Hamiltonian systems obey $$ \frac{d\rho}{dt}=0 $$ which says that along system trajectories, $\rho$ for an ensemble of Hamiltonian systems is indeed constant.