It can be proved that the size of an initial volume element in phase space remain constant in time even for time-dependent Hamiltonians. So I was wondering whether it is still true even when the Hamiltonian system is dissipative like a damped harmonic oscillator?


3 Answers 3


The interplay of Hamiltonian and Lagrangian theory is based on the following general identities, where $L$ is the Lagrangian function of the system, $$\dot{q}^k = \frac{\partial H}{\partial p_k}\:,\qquad(1)$$ $$\frac{\partial L}{\partial q^k} = -\frac{\partial H}{\partial q^k}\:.\qquad(2)$$ Above, the RH sides are functions of $t,q,p$ whereas the LH sides are functions of $t,q,\dot{q}$ and the two types of coordinates are related by means of the bijective smooth (with smooth inverse) relation, $$t=t\:,\quad q^k=q^k\:,\quad p_k = \frac{\partial L(t,q,\dot{q})}{\partial \dot{q}^k}\:.\qquad(3)$$ Finally, the Hamiltonian function is defined as follows $$H(t,q,p) = \sum_{k}\left.\frac{\partial L}{\partial \dot{q}^k}\right|_{(t,q,\dot{q}(t,q,p))}\dot{q}(t,q,p) - L(t,q,\dot{q}(t,q,p))\:.$$ Suppose that the following E-L hold, $$\frac{d}{dq}\left(\frac{\partial L}{\partial \dot{q}^k}\right) - \frac{\partial L}{\partial q^k} = Q_k(t,q,\dot{q})\:, \quad \frac{d q^k}{dt}= \dot{q}^k \quad (4)\:.$$ The functions $Q_k$ take the (e.g. dissipative) forces into account, which cannot be included in the Lagrangian. For a system of $N$ points of matter with positions $\vec{x}_i$, if the degrees of freedom of the system are described by coordinates $q^1,\ldots,q^n$ such that $\vec{x}_i= \vec{x}_i(t,q^1,\ldots,q^n)$, one has: $$Q_k = \sum_{i=1}^N \frac{\partial \vec{x}_i}{\partial q^k} \cdot \vec{f_i}$$ $\vec{f}_i$ being the total force, not described in the lagrangian, acting on the $i$th point.

One easily proves that, in view of the general identities (1) and (2), a curve $t \mapsto (t, q(t), \dot{q}(t))$ satisfies the EL equations (4), if and only if the corresponding curve $t \mapsto (t, q(t), p(t))$ (constructed out of the previous one via (3)), verifies the following equations: $$\frac{dq^k}{dt} = \frac{\partial H}{\partial p_k}\:, \quad \frac{dp_k}{dt} = -\frac{\partial H}{\partial q^k} + Q_k\:.\quad(5)$$ In the absence of the terms $Q_k$, these are the standard Hamilton equation. If $Q_k\equiv 0$, even if $H$ explicitely depend on time, the solutions of Hamilton equations preserve, in time, the canonical volume: $$dq^1 \wedge \cdots \wedge dq^n \wedge dp_1 \wedge \cdots \wedge d p_n\:.$$ In the presence of dissipative forces which cannot be included in the Lagrangian, the term $Q_k$ show up and the volume above generally fails to be preserved. However this is not the whole story. Let us consider the damped harmonic oscillator. In the absence of dissipative force, the Lagrangian reads $$L(x, \dot{x}) = \frac{m}{2} \dot{x}^2 -\frac{k}{2} x^2\:.$$ The dissipative force $-\gamma \dot{x}$ takes place in the EL equations due to the presence of the term: $$Q = - \gamma \dot{x}\:.$$ In this juncture, passing to the Hamiltonian formulation, the canonical volume is not preserved along the solutions of the equation of motion. However, sticking to the damped oscillator, there is a way to include the dissipative force in the Lagrangian function. As a matter of fact, this new Lagrangian produces the correct equation of motion of a damped oscillator $$L(t,q,\dot{q}) = e^{\gamma t/m}\left(\frac{m}{2} \dot{x}^2 -\frac{k}{2} x^2\right)\:.$$ In this case the Hamiltonian function turns out to be $$H(t,q,p) = e^{-\gamma t/m}\frac{p^2}{2m} + e^{\gamma t/m}\frac{k}{2}x^2\:.$$ As the general theory proves, the canonical volume is preserved by the solutions of Hamilton equations referred to that Hamiltonian function, regardless the fact that the system is dissipative.

It is important to notice that, with the second Lagrangian, $p$ ceases to be the standard momentum $mv$ differently from the first case.

  • $\begingroup$ @ V. Moretti- Thanks. But can you give an example of a system where this would not hold? In other words, I want to ask which kind of dissipative systems cannot be modeled (or difficult to model) as a Hamiltonian system? What would the shrinking of phase space volume imply physically? $\endgroup$
    – SRS
    Commented May 3, 2014 at 11:45
  • 1
    $\begingroup$ Do you mean as a pure Hamiltonian system, without dissipative terms in the rhs of H. equations? In general as far as I know the trick I illustrated does not work for more than one point with different masses. In general dissipative systems do not admit a pure Hamiltonian formulation. I do not know a precise meaning for the shrinking of canonical volume. Obviously statistical mechanics, for example, is difficult to formulate in that case. $\endgroup$ Commented May 3, 2014 at 14:30
  • $\begingroup$ @ValterMoretti Thanks for this summary! I only have one question: Your main conclusion was that when dissipative forces are present, the system is not closed anymore, energy is exchanged, thus the phase space distribution function cannot be a constant of motion anymore. So far I understand. But then you show a way of including the dissipative force in the Lagrangian and pass to Hamilton's equation of motion, then you conclude that the canonical volume is preserved. How can this be? Isn't this a contradiction? No matter how we write the Lagrangian, the considered system was dissipative! $\endgroup$
    – user929304
    Commented Nov 11, 2015 at 12:26
  • $\begingroup$ @user929304 The two Lagrangians, $m\dot{x}^2/2-kx^2/2$ and $e^{-\gamma t/m}(m\dot{x}^2/2-kx^2/2)$ respectively define two pair of canonical coordinates $(x,p)$ and $(X,P)$. These coordinate systems on the space of phases are related by $X=x$ and $P=e^{\gamma t/m}p$. This transformation is not canonical, so we are referring to two incompatible notions of phase space volume which explains the apparent contradiction. $\endgroup$ Commented Nov 11, 2015 at 12:52
  • $\begingroup$ @ValterMoretti Makes perfect sense, thanks for clarifying it. $\endgroup$
    – user929304
    Commented Nov 11, 2015 at 12:58

There is something you should be careful of regarding Liouville's theorem. If there are momentum-dependent forces, then Liouville's theorem changes because phase density is no longer incompressible.

Suppose we define $f_{s}$ = $f_{s}(\mathbf{x},\mathbf{p},t)$ $\equiv$ the particle distribution function of species $s$, which is non-negative, contains a finite amount of matter, and it exists in the space of positive times and $\mathbb{R}^{3}$ and $\mathbb{R}_{\mathbf{p}}^{3}$, where $\mathbb{R}_{\mathbf{p}}^{3}$ is the space of all momentum vectors. Another way to say this is that $f_{s}(\mathbf{x},\mathbf{p},t)$ $\equiv$ density of particles per unit $d^{3}x$ per unit $d^{3}p$ at fixed time $t$.

We can see that there are two ways to interpret $f$: (1) it can be an approximation of the true phase space density of a gas (large scale compared to inter-particle separations); or (2) it can reflect our ignorance of the true positions and velocities of the particles in the system. The first interpretation is deterministic while the second is probabilistic. The latter was used implicitly by Boltzmann [e.g., Villani, 2002, 2006].

The general equation of motion for the canonical phase space, $(\mathbf{q},\mathbf{p})$, is given by: $$ \frac{ \partial f }{ \partial t } = f \left[ \left( \frac{ \partial }{ \partial \textbf{q} } \frac{ d\textbf{q} }{ dt } \right) + \left( \frac{ \partial }{ \partial \textbf{p} } \frac{ d\textbf{p} }{ dt } \right) \right] + \left[ \frac{ d\textbf{q} }{ dt } \cdot \frac{ \partial f }{ \partial \textbf{q} } + \frac{ d\textbf{p} }{ dt } \cdot \frac{ \partial f }{ \partial \textbf{p} } \right] $$ If we simplify by letting terms dQ/dt $\rightarrow$ $\dot{Q}$ and let $\boldsymbol{\Gamma}$ $\equiv$ $(\mathbf{q},\mathbf{p})$, then we have: $$ \begin{align} \frac{ \partial f }{ \partial t } & = - f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} - \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \\ & = - \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \left( \dot{\boldsymbol{\Gamma}} f \right) \end{align} $$ and if we define the total time derivative as: $$ \frac{ d }{ dt } = \frac{ \partial }{ \partial t } + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial }{ \partial \boldsymbol{\Gamma} } $$ then we can show that the time rate of change of the distribution function is given by: $$ \begin{align} \frac{ d f }{ dt } & = \frac{ \partial f }{ \partial t } + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \\ & = - \left[ f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \right] + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \\ & = - f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} \\ & \equiv - f \Lambda\left( \boldsymbol{\Gamma} \right) \end{align} $$ where $\Lambda ( \boldsymbol{\Gamma} )$ is called the phase space compression factor [Evans and Morriss, 1990]. Note that these equations are different forms of Liouville's equation, which have been obtained without reference to the equations of motion and they do not require the existence of a Hamiltonian.

We can rewrite the last expression as: $$ \frac{ d }{ dt } \ln \lvert f \rvert = - \Lambda\left( \boldsymbol{\Gamma} \right) $$ which starts to look more familiar to the typical form of Liouville's theorem. If one can derive the equations of motion from a Hamiltonian, then $\Lambda ( \boldsymbol{\Gamma} )$ = 0, even in the presence of external fields that act to drive the system away from equilibrium. Note that the existence of a Hamiltonian is a sufficient, but not necessary condition for $\Lambda ( \boldsymbol{\Gamma} )$ = 0. For incompressible phase space, we recover the simple form of Liouville's equation: $$ \frac{ d f }{ dt } = 0 $$ However, Liouville's theorem can be violated by any of the following:

  • sources or sinks of particles;
  • existence of collisional, dissipative, or other forces causing $\nabla_{\mathbf{p}}$ $\cdot$ $\mathbf{F}$ $\neq$ 0;
  • boundaries which lead to particle trapping or exclusion, so that only parts of a distribution can be mapped from one point to another;
  • spatial inhomogeneities that lead to velocity filtering (e.g., particle drift velocities that prevent particles with smaller velocities from reaching the location they would have reached had they not drifted);
  • temporal variability at source or elsewhere which leads to non-simultaneous observation of oppositely-directed trajectories;
  • etc. [Paschmann and Daly 1998].

The existence of sources or sinks of particles, momentum-dependent forces, trapping or exclusion, velocity filtering, etc. can introduce compressibility to phase space and result in: $$ \frac{ d f }{ dt } \neq 0 $$


  1. Evans, D.J., and G. Morriss (1990), Statistical Mechanics of Nonequilibrium Liquids, 1st edition, Academic Press, London 1990.
  2. Paschmann, G., and P. W. Daly (1998), Analysis Methods for Multi-Spacecraft Data. ISSI Scientific Reports Series SR-001, ESA/ISSI, Vol. 1. ISBN 1608-280X, 1998, ISSI Sci. Rep. Ser., 1.
  3. Villani, C. (2002), Chapter 2 A review of mathematical topics in collisional kinetic theory, pp. 71-74, North-Holland, Washington, D.C., doi:10.1016/S1874-5792(02)80004-0.
  4. Villani, C. (2006), Entropy production and convergence to equilibrium for the Boltzmann equation, in XIVTH International Congress on Mathematical Physics, edited by J.-C. Zambrini, pp. 130-144, doi:10.1142/9789812704016_0011.
  • $\begingroup$ Note: I had previously posted this here but thought the answer fit better for this question. So I deleted the original post and moved it here. $\endgroup$ Commented Oct 31, 2014 at 19:04
  • $\begingroup$ I think so, yes. I think it's similar to the difference between Boltzmann's equation and the Vlasov equation, where the Vlasov version is supposed to be time-reversible (thus isentropic). However, after reading some of Villani's work, I have become somewhat confused as to the difference between irreversible and entropy-producing systems. See my question here. $\endgroup$ Commented Nov 11, 2015 at 18:10
  • $\begingroup$ Thanks for your reply. I was wondering, since I don't have a copy of Evans 1990 book, do you know of other introductory articles that I can read up and get acquainted with such ideas? (entropy production in irreversible systems, violation of Liouville's theorem, compressibility of phase space volume etc.) Any suggestions would be much appreciated. $\endgroup$
    – user929304
    Commented Nov 12, 2015 at 12:30
  • $\begingroup$ I think I found the Evans and Morriss book online (yep, I have an electronic version). I would recommend looking into the works by Villani, which are also online (good references within). You can also look up a paper by Oliver Penrose in 1979 in Rep. Prog. Phys. Vol. 42 called "Foundations of Statistical Mechanics," which I have found very useful. $\endgroup$ Commented Nov 12, 2015 at 16:27
  • $\begingroup$ The Penrose article can be found on SAO/NASA ADS at: http://adsabs.harvard.edu/abs/1979RPPh...42.1937P. $\endgroup$ Commented Nov 13, 2015 at 12:33

So I was wondering whether it is still true even when the system is dissipative like a damped harmonic oscillator?

It is true if the dissipative system is Hamiltonian, i.e. if the dissipative behaviour can be described by time-dependent Hamiltonian. For example, one oscillator connected to million oscillators can be described by the Hamiltonian $$ H = \frac{p^2}{2m} + \frac{1}{2}kx^2 - xF(t), $$ where $F(t)$ is the force acting on the oscilator due to other oscillators. Appropriate function $F(t)$ will make the system behave in a dissipative way, but since the system is described by Hamiltonian, the Liouville theorem is valid.


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