3
$\begingroup$

I understand the proof of Liouville's theorem to the point where we conclude that Hamiltonian flow in phase-space is volume preserving as we flow in the phase space. Meaning the total derivative of any initial volume element is 0.

From here, how do we say that probability distribution function is constant as we flow in the phase-space?

What's the relation between phase space volume and the density function, which instantaneously tells us the probability of finding the system in a neighborhood in phase-space?

$\endgroup$
  • $\begingroup$ How is your volume defined? There are several different volumes, which one can transport along a trajectory in phase-space. $\endgroup$ – image Apr 2 '15 at 11:08
3
$\begingroup$

To obtain the result $\frac{\text d \rho }{\text d t}=0$ you need two facts: the first is that the hamiltonian flow preserves the volume of phase space. The second fact is the conservation of probability, that is, the probability that the system is found in a volume $U$ at time $t=0$ equals the probability of finding it within $\Phi _t U$ at time $t$, where $\Phi _t$ denotes the hamiltonian flow. This is a direct consequence of the deterministic nature of classical mechanics: the two propositions “$(p(0),q(0))\in U$” and “$(p(t),q(t))\in \Phi _t U$” are equivalent.

Using conservation of probability, for an arbitrary volume $U$ we can write an equation: $$\int _U \rho(p,q,0) \text d p \text d q=\int _{\Phi _t U} \rho(p,q,t)\text dp \text d q .$$ By Jacobi's theorem: $$\int _{\Phi_t U} \rho (p,q,t)\text d p \text d q=\int _U\rho (\Phi _t (p,q),t)\text J_{\Phi _t}d p \text d q.$$ The Jacobian $J_{\Phi _t}=1$, because the flow preserves volumes. It follows that: $$\int _U \rho (p,q,0)\text d p \text d q =\int _U \rho (\Phi _t (p,q),t)\text d p \text d q,$$ and, since the volume $U$ was arbitrary, $\rho (p,q,0)=\rho (\Phi _t (p,q),t)$, or $\text d\rho /\text d t=0$.

$\endgroup$
1
$\begingroup$

I do not think probability distributions are preserved by the Hamiltonian flow...consider a probability distribution that is a $\delta$-function on the phase space at initial time (you have just one point with probability one), so it is a particle with fixed coordinate and momentum. If you evolve in time by the Hamiltonian flow, you will find yourself at the phase space point corresponding to the evolved position and momentum of the particle. It corresponds again to a $\delta$-function probability distribution, but on a different point, so different from the starting one.

I think you should characterize the evolution of probability distributions with the aid of measures. A probability distribution has a mathematical meaning as a probability measure $\mu$ on the phase space $\mathscr{Z}$. You have that $\mu(\mathscr{Z})=1$ (total probability is one). Given that the initial distribution is the measure $\mu_0$, and calling $\Phi(t)$ the Hamiltonian flow, you should get the measure at time $t$ as the pushforward of the initial measure by the flow: $\mu_t=\Phi(t)_*\mu_0$.

However, I am not completely sure about that, I hope to get some feedback and eventual corrections from someone more expert on classical statistical mechanics ;-)

$\endgroup$
  • $\begingroup$ The Liouville theorem says that the phase-space density, $\rho(p,q)$, is preserved: $\partial\rho/\partial t=0$. The probability distribution is the phase-space integral of the density: $P=\int\rho\,d^np\,d^nq$. $\endgroup$ – Kyle Kanos Jul 22 '14 at 16:53
  • $\begingroup$ @KyleKanos $P$ is the probability, not the probability distribution...based on its definition it seems really a probability measure from the mathematical standpoint. But I am not sure about the time-evolved one being the push-forward of the initial. But the example is, in my opinion, convincing. $\endgroup$ – yuggib Jul 22 '14 at 17:15
  • $\begingroup$ @Kyle Kanos I think I'm a little clear now. I was thinking that phase-space density and probability distribution/density functions were different. Looks like they are one and the same. And I think you are wring in saying ∂ρ/∂t=0 it must be dρ/dt=0, right? $\endgroup$ – levitt Jul 22 '14 at 17:33
  • $\begingroup$ @yuggib: You are correct, distribution is misplaced in my comment. What I meant was that Liouville theorem says that the phase-space density is preserved, but this, I think, does not mean that the probability is preserved. $\endgroup$ – Kyle Kanos Jul 22 '14 at 17:33
1
$\begingroup$

The phase-space density $\varrho(p,q)$ tells us how many dynamical possible trajectories (DPT) pass through a given unit volume of phase space. Therefore it is a measure for the probability of finding a system in state $(p,q)$ since if there are more DPTs, it is more likely to find a system in this state (if we assume that every DPT is equally probable, that is, every initial condition is equally probable).

In this sense $\varrho(p,q)$ is the probability distribution

$\endgroup$
0
$\begingroup$

You can think about it as in the case of a simple fluid. You need to enforce conservation of probability (in the same way you could enforce mass conservation in fluid dynamics for instance or charge conservation in electrodynamics).

As far as I understand it, probability conservation is an additional premiss to the conservation of phase space volume under hamiltonian flow and this seems to go in the sense of yuggib's answer.

If you probability measure is characterized by a probability density on phase space $\rho(\mathbf{q},\mathbf{p})$ associated to the Lebesgue measure in phase space for instance, then you know that probability conservation leads to an equation like this:

\begin{equation} \frac{\partial \rho}{\partial t} + \textrm {div} (\rho \mathbf{v}) = 0 \end{equation}

Applying now the fact that $\textrm {div} (\rho \mathbf{v}) = \rho\, \textrm {div}(\mathbf{v}) + \mathbf{v} \cdot \textrm{grad} (\rho)$ and that $$\textrm {div}(\mathbf{v}) = \frac{\partial \dot{q}}{\partial q} + \frac{\partial \dot{p}}{\partial p} = \frac{\partial^2 H}{\partial q \partial p} - \frac{\partial^2 H}{\partial p \partial q} \equiv 0$$ (Liouville's theorem), you can use the other fact that $\mathbf{v} \cdot\textrm{grad}(\rho) = \{H, \rho \}$ which leads to the result you are looking for (I haven't checked the signs so there can be a minus wondering around somewhere):

\begin{equation} \frac{\partial \rho}{\partial t} = \{\rho, H \} \end{equation}

$\endgroup$
0
$\begingroup$

From here, how do we say that probability distribution function is constant as we flow in the phase-space?

More accurately, value of probability distribution function $f_t(p,q)$ at representative point $p^*(t),q^*(t)$ moving along any Hamiltonian trajectory in phase space is constant in time. The function itself generally changes in time. This value is

$$f_t(p^*(t),q^*(t)) =$$

$$ = \frac{\text{prob. per phase space element centered on the moving point}}{\text{volume of the element}}$$

This is constant in time because both numerator and denominator are constant in time.

The numerator is constant because the element evolves with the flow and although its boundary deforms, the element does not lose any probability.

The denominator is constant because of the Hamiltonian property of the flow (trajectories of representative pointw obey Hamilton's equations).

$\endgroup$

protected by Qmechanic Apr 4 '18 at 6:42

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.