# Liouville's theorem for systems with dissipation described by a single hamiltonian

Following this link, one can treat dissipation by using a factor $$e^{\frac{t \beta}{ m}}$$ in addition to the Lagrangian $$L_0$$ of a system without disspation:

$$L[q, \dot{q}, t] = e^{\frac{t \beta}{ m}} L_0[q, \dot{q}] \, , \quad \text{where}\quad L_0[q, \dot{q}] = \frac{m}{2}\dot{q}^2 - V[x]\, .$$

The equation of motion associated to $$L[q, \dot{q}, t]$$ is:

$$e^{\frac{t \beta }{ m}} \frac{d}{d t} \frac{ \partial L_0}{\partial \dot{q}} - e^{\frac{t \beta}{ m}} \frac{ \partial L_0}{\partial q} = e^{\frac{t \beta}{ m}} \frac{\beta}{m} \frac{ \partial L_0}{\partial \dot{q}} \qquad \Rightarrow \qquad \frac{d}{d t} \frac{ \partial L_0}{\partial \dot{q}} - \frac{ \partial L_0}{\partial q} = \frac{\beta}{2} \dot{q}^2$$

This equation describes the motion of a particle in a potential under the effect of dissipation forces proportional to its velocity.

Question: If I use Hamiltonian mechanics, this system is described by a single Hamiltonian and the canonical equations of motion. That means that Liouville's theorem holds, and therefore the state density would be constant along the physical trajectories of the system in the phase-space. I can't imagine that this is possible for a system with dissipation, where I would expect the phase-space volume to shrink and the density to grow.

Consider for example a damped harmonic oscillator: The trajectories would be circles with exponentially decreasing radius in the phase space, so the phase-space-volume decreases.

Where is my mistake? I thought the only requirement for Liouville's theorem is that the system is described only by the Hamiltonian. So why does it hold where it clearly should not?

• There is a reason why the procedure is called "unconventional approach" in the original question. I do not believe that it is an actually correct model for physics. What it does is to exponentially re-scale the dynamics (including the volume of the phase space element, IMHO) so that things look like they are constant. In reality, however, a system that loses energy is subject to the dissipation-fluctuation theorem, i.e. the phase space element does shrink but only to an average size given by the thermodynamic noise that causes the dissipation, i.e. the above does not obey thermodynamics. Commented Dec 26, 2015 at 22:26
• I haven't done the calculation, but I would assume that with this approach you get different generalised coordinates (e.g. the momentum will be scaled by the exponential factor as well). Liouville's theorem applies to the "new" coordinates, not to the $p,q$ of the original Hamiltonian (those derived from $L_0$, which are the "familiar" momentum and position). Commented Dec 26, 2015 at 22:46
• You are right, this explains it. Indeed, the momenta will be rescaled with the exponential factor. This would also mean, that "rescaling" to the "old" momentum and location coordinates (p, q) is not a canonical transformation. Commented Dec 27, 2015 at 1:09
• The "exponential factor trick" to include dissipation: physics.stackexchange.com/a/483535/226902 and physics.stackexchange.com/a/89395/226902 Commented Oct 24, 2022 at 12:29
• Interesting question. Wikipedia says that "One of the assumptions of Liouville's Theorem is that the system obeys the conservation of energy, so that $\rho$ is constant on phase space surfaces of constant energy. If energy is not conserved (this is the case for time-dependent hamiltonians), we find that ρ also fails to be constant" en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian). However, see this answer on the time dependence of the Hamiltonian: physics.stackexchange.com/a/531112/226902 Commented Oct 24, 2022 at 13:57

This was bumped by Community; there is an accepted answer by Daniel in comments but since comments are fleeting, I will community wiki it:

I haven't done the calculation, but I would assume that with this approach you get different generalised coordinates (e.g. the momentum will be scaled by the exponential factor as well). Liouville's theorem applies to the "new" coordinates, not to the $$p,q$$ of the original Hamiltonian (those derived from $$L_0$$ , which are the "familiar" momentum and position).

You are right, this explains it. Indeed, the momenta will be rescaled with the exponential factor. This would also mean, that "rescaling" to the "old" momentum and location coordinates $$(p, q)$$ is not a canonical transformation.

In my opinion, we should define the Lagrangian only when the force is conservative. Eventhough the force is not conservative, we can write down the Lagrangian which is mathmatically correct. However, the Lagrangian and the Hamiltonian do not contain correct nature inside it. Only mathmatically, those work. So, the divergence is still zero, but this does not means that the density in the phase is constant.