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Suppose we have a classical statistical problem with canonical coordinates $\vec{q} = (q_1, q_2, \dots, q_n)$ and $\vec{p} = (p_1, p_2, \dots, p_n)$ such that they fulfill the usual Poisson brackets: \begin{align} \{ q_i, p_j \} & = \delta_{i,j} \\ \{ q_i, q_j \} & = 0 \\ \{ p_i, p_j \} & = 0 \\ \end{align} One can show that the time evolution of every dynamical quantity $f(\vec{q}, \vec{p}, t)$ is given by $$ \frac{d f}{d t} = \frac{\partial f}{\partial t} + \{ f, H \} = \frac{\partial f}{\partial t} + \sum_{i=1}^n \frac{\partial f}{\partial q_i} \dot{q_i} + \frac{\partial f}{\partial p_i} \dot{p_i} = \frac{\partial f}{\partial t} + (\nabla f) \cdot \vec{v} = \frac{\partial f}{\partial t} + \nabla (f \cdot \vec{v}) $$ with $\nabla = (\frac{\partial}{\partial q_1}, \dots, \frac{\partial}{\partial q_n}, \frac{\partial}{\partial p_1}, \dots, \frac{\partial}{\partial p_n})$, $\vec{v} = (\dot{q_1}, \dots, \dot{q_n}, \dot{p_1}, \dots, \dot{p_n})$ and $H$ the Hamiltonian of the system.

Liouville's theorem states that: $$ \int d^n p ~ ~ d^n q = \int d^n p' ~ ~ d^n q' $$ if $(\vec{q}~', \vec{p}~')$ and $(\vec{q}, \vec{p})$ are both canonical coordinates, e.g. related by a canonical transformation. So the phase space volume is a constant between systems which are described by canonical coordinates.

Now consider the phase space density $\varrho(\vec{q}, \vec{p}, t)$ which is the density of dynamically allowed trajectories at a given point $(\vec{q}, \vec{p})$ in phase space for a given instance of time $t$.

Liouville's equation reads: $$\frac{d \varrho}{d t} = 0$$ which (together with the equation for $f$) says that $\varrho$ is a (locally) conserved density in phase space. Because $\varrho \ge 0$ one can conclude that there are no sources of trajectories in phase space: trajectories do not start, end or cross.

Usually the Liouville equation is derived from Liouville's theorem. However I haven't seen such a derivation for which at some point it wasn't assumed that $\varrho$ is a (locally) conserved density. Hence, that reasoning is circular.

Do you know a non-circular derivation of Liouville's equation or is it indeed an axiom of classical statistical mechanics?

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    $\begingroup$ The Jacobian is the identity. $\endgroup$ – Jon May 22 '17 at 11:47
  • $\begingroup$ @WetSavannaAnimalakaRodVance $\omega^n$ is invariant under canonical transformations, e.g. phase space volume is conserved. Why is the number of states, enclosed in a finite volume transported along the trajectories constant? The enclosing volume is constant (not the shape) because dynamical transport can be expressed by a canonical transformation. However, why is the number of states in it constant? E.g. why are trajectories not allowed to flow out of it? If they aren't allowed (axiom?), surely a continuity equation holds (= Liouville equation). But why does it hold in the first place? $\endgroup$ – image May 22 '17 at 13:06
  • $\begingroup$ @image This happens due to Hamilton equations and the fact that you have an energy hypersurface that the system cannot leave, being energy a motion constant. $\endgroup$ – Jon May 22 '17 at 13:09
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    $\begingroup$ The uniqueness theorem for Hamilton's equations make sure that every point $(q,p)$ at instant $t$ comes from a dynamic evolution of a unique point $(q_0,p_0)$ at time $t_0$. In other words, during dynamic evolution, there is no representative points poping up or disappearing. The number of points in a given region is constant even though the region is deformed. Since Liouville theorem implies that the volume of a given region is constant during time evolution, so it is the number of points contained in it. $\endgroup$ – Diracology May 22 '17 at 13:45
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    $\begingroup$ Notwithstanding some comments, I think this is precisely the kind of question that makes Physics SE really worthwhile and helps spread knowledge to nonspecialists such as I am in a forum that lets users state and discuss their precise reasoning clearly. I learnt a great deal from thinking about your question carefully, so thanks very much for asking it. $\endgroup$ – WetSavannaAnimal May 25 '17 at 4:02
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Why There is a Need for a Further Axiom

To derive Liouville's equation, you indeed need another axiom further to your assumptions. Something like: "there is no nett creation or destruction of any particle of any species throughout the particle system state evolution". The easiest way to understand the need for this axiom is to cite a system wherefor Liouville's equation cannot hold, even though particles undergo dynamical evolution described by Hamilton's equations throughout their lifetimes: a system of particles undergoing a far-from-equilibrium chemical reaction. In such a system, reactant particle species are consumed by the reaction, and disappear from phase space. Reaction product particles appear in phase space in their place. Moreover, chemical energy is converted to kinetic energy (or contrariwise), so that a product species will "suddenly" appear at a different point in phase space from the one where the correspondingly consumed reactant species particles vanished. Liouville's equations would conceptually be replaced by a coupled system of equations, one for each species $j$, of the form:

$$\frac{\partial\,\rho_j(X,\,t)}{\partial\,t} = \{H,\,\rho_j\} + \sum\limits_k \int_\mathcal{P} M_{j\,k}(X,\,X^\prime)\,\rho_k(X^\prime,\,t)\,\mathrm{d}\Gamma^\prime$$

where the integral is over all phase space $\mathcal{P}$, $\Gamma$ is the measure defined by the volume form and the kernel $M_{j\,k}$ expresses detailed stochimetric balance between the chemically reacting species as well as other physical principles such as conservation of energy, momentum and strict increase with time of entropy. Note that I said "nett" creation or destruction: Liouville's equation would work if the reaction were at equilibrium.

Complete Axioms

The following axioms (1. and 2. are equivalent to yours) will get you Liouville's equation:

  1. Axiom 1: Phase space is a $2\,N$ dimensional $C^2$ manifold $\mathcal{P}$;
  2. Axiom 2: Points in phase space always and only evolve with a flow parameter $t$ through Hamilton's equations defined by a $C^2$ Hamiltonian $H:\mathcal{P}\times \mathbb{R}\to\mathbb{R}$, the latter possibly time varying (hence the domain $\mathcal{P}\times\mathbb{R}$);
  3. Axiom 3: The full states of particles are points in $\mathcal{P}$ evolving according to axiom 2 and there is no nett creation or destruction of any particle of any species throughout the particle system state evolution.

From Complete Axioms to Liouville's Equation

From these axioms, the chain of inference you need is as follows:

  1. Inference 1: From axioms 1. and 2., deduce that any $X\in T_p\,\mathcal{P};\;\forall p\in \mathcal{P}$ expressed in canonical co-ordinates (i.e. ones for which the Hamilton equations hold) that is Lie-dragged by the Hamiltonian flow evolves according to $\dot{X} = A(t)\,X$ where $A(t)\in\mathfrak{sp}(N,\,\mathbb{R})$, thus the symplectic 2-form $\omega(X,\,Y)\stackrel{def}{=} X^T\,\Omega\,Y$ where, for the special case of canonical co-ordinates, $\Omega =\left(\begin{array}{cc}0&-1_N\\1_N&0\end{array}\right)\;\forall p\in\mathcal{P}$ is conserved under the mapping $\mathcal{P}\mapsto \Phi(H,\,t)\,\mathcal{P},\,\forall t\in\mathbb{R}$ induced by the Hamiltonian flow. (Indeed, at any given point $p\in\mathcal{P}$ find $N$ different $C^2$ Hamiltonians such that the tangents to their flows span $T_p\,\mathcal{P}$ to deduce that the Lie derivative of $\omega$ in any direction is nought, thus $\mathrm{d}\omega=0$ from Cartan's formula relating Lie and Exterior derivatives, but this information is further to our immediate needs). Take heed that inference 1 holds whether or not the Hamiltonian be time-varying. In the latter case, the Hamiltonian is not constant along the flow, but the flow still conserves the symplectic form.
  2. Inference 2: From inference 1, we have immediately that the volume form $\Gamma = \omega^N$ ($N^{th}$ exterior power) is conserved under Hamiltonian flows. Thus deduce Liouville's theorem (as opposed to equation). Alternatively, the conservation of the symplectic form shown in Inference 1 implies that the Jacobi matrix of the transformation $\mathcal{P}\mapsto \Phi(H,\,t)\mathcal{P}$ is a symplectic matrix (member of $\mathrm{Sp}(N,\,\mathbb{R})$), which always has a unit determinant. Thus the volume form is conserved.
  3. Inference 3: But the volume form is also the Jacobian of the transformation $\mathcal{P}\mapsto \Phi(H,\,t)\,\mathcal{P}$ and $J(p,\,\Phi(H,\,0))=J(p,\,\mathrm{id})=1$. Since the volume form is conserved, the Jacobian $J(p,\,\Phi(H,\,t))=1,\forall p\in\mathcal{P},\forall t\in \mathbb{R}$. Thus $\Phi$ is everywhere a local bijection (inverse function theorem). Alternatively, we can make the same deduction this straight from Inference 1, which implies that the Jacobi matrix of the transformation $\mathcal{P}\mapsto \Phi(H,\,t)\mathcal{P}$ is a symplectic matrix (member of $SP(N,\,\mathbb{R})$), which is never singular and indeed always has a unit determinant.
  4. Inference 4: From Axioms 2 and 1, deduce that the distance function defined in canonical co-ordinates by $d(p_1,\,p_2) = (p_1-p_2)^T\,(p_1-p_2)$, zero iff $p_1=p_2$, between any pair of points $p_1,\,p_2\in\mathcal{P}$ must be a continuous function of the flow parameter $t$ (continuous with respect to the topology with basis of open balls defined by this distance function);
  5. Inference 5: From Inference 3, at any $p\in\mathcal{P}$ and $t\in \mathbb{R}$, there is an open set $\mathcal{U}_p$ small enough such that $\Phi(H,\,t):\mathcal{U}_p\to\Phi(H,\,t)(\mathcal{U}_p)$ is a bijection. The question now arises as to whether $\Phi(H,\,t)$ can map any point outside $\mathcal{U}_p$ into $\Phi(H,\,t)(\mathcal{U}_p)$ (which situation would make $\Phi(H,\,t)$ a local bijection but globally many to one for some $t\in\mathbb{R}$). However, if two or more points are mapped to one point $\tilde{p}\in\Phi(H,\,t)(\mathcal{U}_p)$, from inference 4. deduce that $\exists t$ small enough that the two chosen preimages of $\tilde{p}$ both lie in $\mathcal{U}_p$, thus contradicting local bijectivity. (Informally, from inference 4, multiple points of a function can only arise from mappings along connected "forked" flow lines, so zoom in near enough in on the fork point and thus contradict local bijectivity, showing that forks are impossible). Repeating the reasoning for $-t$ lets us deduce that multiple points are impossible and $\Phi(H,\,t):\mathcal{P}\to\mathcal{P}$ is a global bijection (indeed a symplectomophism in the light of inference 1, but, again, this information is further to our needs);
  6. Inference 6: From inference 5 and axiom 3, deduce that if there is some number $M$ of particles in any subset $\mathcal{V}\subseteq\mathcal{P}$, then there are precisely $M$ particles in $\Phi(H,\,t)(\mathcal{V})$. From inference 2. deduce that $\mathcal{V}$ and $\Phi(H,\,t)(\mathcal{V})$ have the same volumes. Therefore infer that the average particle density in any subset $\mathcal{V}\subseteq\mathcal{P}$ is constant if the particle states and subsets evolve by Hamiltonian flows;
  7. Inference 7: Apply inference 6 to a small open set that is shrunken according to an appropriate limiting process to deduce that the density function $\rho(p,\,t)$ at point $p$ and at time $t$ must be the same as the density at point $\Phi(H,\,-\mathrm{d}t)\,p$ at time $t-\mathrm{d}t$. Putting these words into symbols: $\mathcal{L}_{-X}\rho=\frac{\partial\,\rho}{\partial\,t}$, where $X$ is the vector field tangent to the Hamiltonian flow $\Phi$. This is, of course, $\{H,\,\rho\}=\frac{\partial\,\rho}{\partial\,t}$, or Liouville's equation.

Circularity of Other Proofs

Ultimately, I don't believe that proofs of Liouville's equation grounded on the divergence theorem are different from the above: I think that they are tacitly introducing Axiom 3 as "obvious" (even though I hope I have shown at the beginning of my answer that it doesn't always hold) and then the continuity equation and incompressible flows are simply an expression of this tacitly assumed axiom. So I don't think that these "proofs" are circular, just somewhat badly written in making use of tacit assumptions.

Summary

User Image sums all this up nicely (I was perhaps too brainfried to make the last step):

For Axiom 3 however, you showed that Axiom 3 $\Rightarrow \frac{d \varrho}{d t} = 0$. The other direction $\frac{d \varrho}{d t} = 0 \Rightarrow$ Axiom 3 is readily discussed in any textbook (trajectories do not start, end or cross etc.). So in fact we have Axiom 3 $\Leftrightarrow \frac{d \varrho}{d t} = 0$ when we are in the context of Axiom 1+2, e.g. classical mechanics. Hence, Liouville's equation is an axiom.

and indeed, in the presence of the other two, my axiom 3 is logically equivalent to Liouville's equation. My version is perhaps more physically transparent, but open to interpretation, and so the assertion of Liouville's equation as an axiom is perhaps more succinct and precise. So the answer to the title question is that Liouville's Equation must indeed be added as an axiom, and, in the presence of Axioms 1 and 2, it has the meaning that particle number of all species is conserved.

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    $\begingroup$ Thanks for this great answer! Maybe the term "circular" was a bit strong. What I was aiming at was: Axiom 1+2 can be taken as granted, since this is what defines classical mechanics. For Axiom 3 however, you showed that Axiom 3 $\Rightarrow \frac{d \varrho}{d t} = 0$. The other direction $\frac{d \varrho}{d t} = 0 \Rightarrow$ Axiom 3 is readily discussed in any textbook (trajectories do not start, end or cross etc.). So in fact we have Axiom 3 $\Leftrightarrow \frac{d \varrho}{d t} = 0$ when we are in the context of Axiom 1+2, e.g. classical mechanics. Hence, Liouville's equation is an axiom $\endgroup$ – image May 25 '17 at 9:59
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    $\begingroup$ @image Agreed - Liouville's equation can indeed be an axiom equivalent to Axiom 3, but I think it is more physically clear to begin with conservation of particles. I think we've both worked out a great deal together here. Thanks for the fun! $\endgroup$ – WetSavannaAnimal May 25 '17 at 22:21
  • $\begingroup$ @WetSavannaAnimalakaRodVance Correct me if I'm wrong, but net particle creation and destruction wouldn't imply that phase space changes dimension in time? In this case the manifold model wouldn't work. Moreover, as far as I am aware, this formulation is for isolated systems (microcanonical ensembles, aside from the fact that they might be far from equilibrium), and that implies that particle number is fixed. Unless I misunderstand your answer. $\endgroup$ – Bence Racskó May 26 '17 at 11:14
  • $\begingroup$ @Uldreth - Our discussion here applies to independent particles, whose states are distributed over a fixed dimension manifold. That's probably something else that should be made clear here (perhaps as axiom 0!). Alternatively, an equilibrium system with particle interactions would also work (because the distribution of states is then unchanging change and you can then calculate as though the particles are independent). You're probably right that these formulations usually are for microcanonical ensembles, but I think the point of the OP's question is to discover exactly what axioms are .... $\endgroup$ – WetSavannaAnimal May 26 '17 at 11:28
  • $\begingroup$ @Uldreth ... needed from these discussions to get to Liouville's equation. Many aspects of this discussion are broader than a microcanonical ensemble. BTW I warn that my knowledge of this stuff is mainly geometrical and, recently, has been applied to illumination optical systems, where you truly do have noninteracting particles (i.e. states of optical rays in optical phase space), so the application of this geometry to statistical mechanics is something I've only read about and not used. $\endgroup$ – WetSavannaAnimal May 26 '17 at 11:29
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The probability for the system to be in the phase cell $d\Gamma(t)$ at time $t$ is $$ P(t)=\rho(q,p,t)d\Gamma(t). $$ The time evolution of the trajectories and the possible explicit time dependence of $\rho$ is also considered into this. Now infinitesimal $dt$ time later $$ P(t+dt)=\rho(q+dq,p+dp,t+dt)d\Gamma(t+dt), $$ because probability is normed (eg. if we let the phase volume flow with the time evolution it should not change), these two must be equal, but by Liouville's theorem we have $d\Gamma(t)=d\Gamma(t+dt)$, the $\rho$ expressions must agree, so we have $$ 0=\rho(q+dq,p+dp,t+dt)-\rho(q,p,t)=\frac{\partial\rho}{\partial q}dq+\frac{\partial\rho}{\partial p}dp+\frac{\partial\rho}{\partial t}dt, $$ from which we have (by "dividing" by $dt$) $$ \frac{d\rho}{dt}=\frac{\partial\rho}{\partial q}\dot{q}+\frac{\partial\rho}{\partial p}\dot{p}+\frac{\partial\rho}{\partial t}=0,$$ which is equal to $$ \{H,\rho\}=\frac{\partial\rho}{\partial t}, $$ which is Liouville's equation.

Now, in equilibrium, we posulate $\rho$ to be explicitly time independent (as a definition of equilibrium), then we have $$ \{H,\rho\}=0, $$ so $\rho$ is a constant of motion.


Edit:

To justify the handwave-y step ($P(t)=P(t+dt)$), consider that for the entirety of phase space $\mathcal{P}$, we must have $$ 1=\int_{\mathcal{P}}\rho(q,p,t)d\Gamma(t). $$ By Liouville's theorem, the phase flow preserves $d\Gamma$, so it is time independent. Therefore, for the integral to be time-independent, the integrand must be so.

This is still a bit handwave-y, as the integral happens over phase space, whereas we take our time derive to be "along the time evolution", but I think, this can be further formalised by letting the phase flow act on the entire integral, eg. by first taking $$ 1=\int_{\mathcal{P}}\rho(q,p,t)d\Gamma, $$ and then taking $$ 1=\int_{\Phi_t(\mathcal{P})}\rho\ d\Gamma, $$ and comparing the two (where $\Phi_t$ is the phase flow).


Edit2:

Sorry for the mass edits, but I just checked, and my previous intuition seems correct. Some differential geometry shall be employed here.

The Liouville form is $d\Gamma$ (not that $d$ here is just a notation, not an exterior derivative), which in canonical (Darboux-) coordinates is given by $d\Gamma=dq^1\wedge...\wedge dq^n\wedge dp_1\wedge...\wedge dp_n$.

The phase density itself defines a (possibly) time-dependent $2n$-form as $\rho(q,p,t)d\Gamma$. Let $\Phi_t$ be the phase flow and consider $$ \frac{d}{dt}|_{t=0}\int_{\phi_t(\mathcal{P})}\rho d\Gamma=\frac{d}{dt}\int_{\mathcal{P}}(\Phi_t)^*(\rho d\Gamma)=\int_\mathcal{P}\frac{d}{dt}(\Phi^*_t\rho\Phi^*_td\Gamma)=\int_\mathcal{P}\left(\mathcal{L}_X\rho+\frac{\partial\rho}{\partial t}\right)d\Gamma+\rho\mathcal{L}_Xd\Gamma. $$ Here $X=d\Phi/dt|_{t=0}$ the Hamiltonian vector field, all time derivatives are taken at $t=0$, the partial time derivative appeared because $\rho$ has explicit time dependence too, and the last term is zero by Liouville's theorem.

Because integrals are diffeomorphism-invariant, this derivative must be zero, moreover, this must be true for any probability density $\rho$, hence the integrand itself must vanish, so we have $$ \mathcal{L}_X\rho+\frac{\partial\rho}{\partial t}=0, $$ and $X$ is the same as your $\vec{v}$, so this is actually Liouville's equation.

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  • $\begingroup$ Think I have to bother you again: The argument where you got rid of the integral in the end is a bit fishy. For instance what you have proven is essentially the following: "Every function $\varrho(q,p,t)$, whose phase space integral is a constant in time $d/dt \int \varrho ~ dq ~ dp = 0$ is already constant $d \varrho/dt = 0$". This is not correct though: An counterexample is $\varrho = \exp(-1/2 \cdot ((q-t)^2 + (p-t)^2))$ for which $\int \varrho ~ dq ~ dp = 2\pi$ but in general $d\varrho / dt \ne 0$. $\endgroup$ – image May 23 '17 at 13:41
  • $\begingroup$ The problem here is that you can only get rid of the integration if your function is arbitrary up to a Null set, which is not the case for the strong restriction $d/dt \int \varrho = 0$ $\endgroup$ – image May 23 '17 at 13:43
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind May 23 '17 at 14:00
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    $\begingroup$ Also, please do not let posts look like revision histories. $\endgroup$ – ACuriousMind May 23 '17 at 14:01
  • $\begingroup$ +1 ultimately I don't think your and my answer differ aside from the discussion of potentially disappearing and appearing particles (see my other comments to you under my answer): the differential geometry is the same. $\endgroup$ – WetSavannaAnimal May 27 '17 at 4:03

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