# What does Liouville's Theorem actually mean?

Basically, the mathematical statement of Liouville's theorem is:

$$\frac{\partial \rho }{\partial t}= -\sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$

While I could comprehend the derivation which is nicely done in Reif's Fundamentals of Statistical and Thermal Physics, I could not get what this theorem actually wants to imply.

The Wikipedia article mentions:

It asserts that the phase-space distribution function is constant along the trajectories of the system [...]

What does this mean?

What does the word trajectory mean in the present context?

Is $\rho$ not a function of time?

Can anyone please clarify what that quoted line actually means?

• I wrote something about this here: http://physics.stackexchange.com/a/177972/59023. At the very least, I listed several useful references that have rather lengthy discussions about Liouville's theorem. Mar 19, 2016 at 13:15
• Probability conservation. Jun 10 at 12:39

$$\frac{\partial \rho }{\partial t}= -\sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$

This means that if we have a function of $t, p, q$ namely $\rho(t,\vec p,\vec q)$ and we have a trajectory that is a curve in $(p,q)$ space, namely $q_i(t), p_i(t), i=1\ldots N,$ then:

$$\frac{\mathrm d}{\mathrm dt} \rho(t, \vec p(t), \vec q(t)) =\frac{\partial \rho }{\partial t}+ \sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$

How if $\rho$ is constant along trajectories, then LHS is 0 and the equation you have written follows directory.

So:

• a trajectory is any curve in 2N dimensional space described in $q_i$ and $p_i$ coordinates
• $\rho$ is a function of both time and $\vec q$ and $\vec p$
• whole concept is just an application of a chain rule.
• Thanks for answering; I was confused for the relation was proved taking a volume in the phase space and that the number of systems in the ensemble are constant. After the relation, the author without talking about the volume any more, just said $\rho$ is constant along trajectory. I mean he was talking about systems of ensemble in that volume and then he turned to trajectory. That's what I couldn't understood. Also, if the time-derivative of a function is zero, wouldn't that function be independent of time?
– user36790
Feb 18, 2016 at 2:32

Just throwing a little historical context in here, but Gibbs was the first to publish regarding this idea (1902). In my opinion there really isn't anything fundamental encoded in Liouville's theorem, other than rules of calculus and n-dimensional divergence theorem. When we go a little further with the Hamiltonian we can start to extract some interesting information from it.

I've been enjoying studying this concept for a number of years now and here's my explanation of it. First there's a probability distribution that is a function of all of the position and momenta in a system (and is an explicit function of time),

$$\rho \equiv \rho(p_1, q_1, ..., p_n, q_n, t)$$

It's the probability of "finding" the state with the above state variables. It is also a true probability in the sense that integrating it over the limit of all states gives 1, $$\int^{phase\ limits}... \int \rho\ dp_1 dq_1...dp_n dq_n dt = 1$$

Using the partial differential chain rule we can establish, $$\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t}\frac{dt}{dt}+\frac{\partial \rho}{\partial p_1}\frac{dp_1}{dt}+\frac{\partial \rho}{\partial q_1}\frac{dq_1}{dt}+ ...$$

If we assume the phase space is analogous to an incompressible fluid, we can use n-dimensional divergence theorem to arrive at the following (Gibbs showed this as well, but using older notation),

$$\frac{d\rho}{dt}=0$$

We get Liouville's Theorem,

$$\frac{\partial \rho}{\partial t}=-\sum^{n}_{i=1}\left( \frac{\partial \rho}{\partial p_i}\frac{dp_i}{dt}+\frac{\partial \rho}{\partial q_i}\frac{dq_i}{dt} \right)$$

Using Hamiltonian notation and rearranging slightly,

$$\frac{\partial \rho}{\partial t}=-\sum^{n}_{i=1}\left(\frac{\partial \rho}{\partial q_i}\frac{\partial H}{\partial p_i} -\frac{\partial \rho}{\partial p_i}\frac{\partial H}{\partial q_i} \right)$$

As per the interesting types of things you can do with Liouville's Theorem, we can actually derive the canonical ensemble using the condition of statistical equilibrium (the idea that the probability density function does not change with respect to time, considering all phase variables constant),

$$\frac{\partial \rho}{\partial t} = 0$$

And then we get, $$-\sum^{n}_{i=1}\left(\frac{\partial \rho}{\partial q_i}\frac{\partial H}{\partial p_i} -\frac{\partial \rho}{\partial p_i}\frac{\partial H}{\partial q_i} \right)=0$$

A solution of this partial differential equation, is the following (this is easy to verify with kinetic/potential energy functions in Newtonian mechanics),

$$\rho=Ae^{-H/kT}=e^{F/kt}e^{-H/kT}=e^{(F-H)/kT}$$

From which falls out the thermodynamic identity.

• The Liouville equation only suggests that $\rho$ may be a function of $H$, the exponential comes from other assumptions. Jun 10 at 17:29

Let's say, the phase-space distribution function at time $$t_0$$ is

$$\rho(t_{0},q(t_{0}),p(t_{0}))$$ Now, $$q$$ and $$p$$ will evolve according to Hamilton's equations and at time $$t_1$$, they will become $$q(t_{1})$$ and $$p(t_{1})$$.

Therefore, at time $$t_1$$ the phase-space distribution function will be

$$\rho(t_{1},q(t_{1}),p(t_{1}))$$ Notice that not only we are calculating $$\rho$$ at a different time but also at different $$q$$ and $$p$$, because q and p themselves are changing with time. In other words, we are calculating $$\rho$$ along the trajectory of a phase-point. According to Liouville's Theorem, $$\rho(t_{0},q(t_{0}),p(t_{0}))=\rho(t_{1},q(t_{1}),p(t_{1}))$$ That is what we mean when we say, "The phase-space distribution function is constant along the trajectories of the system"