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

$ L_0[q, \dot{q}] = \frac{m}{2}\dot{q}^2 - V[x] $

$ L[q, \dot{q}, t] = e^{\frac{t \beta}{ m}} L_0[q, \dot{q}] $

Then the equations of motion follow to be:

$ 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}} $

Und thus:

$ \frac{d}{d t} \frac{ \partial L_0}{\partial \dot{q}} - \frac{ \partial L_0}{\partial q} = \frac{\beta}{2} \dot{q}^2$ Which describes the motion of a particle in a potential under the effect of dissipation forces proportional to velocity.

So here is my 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 oszillator: 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 requirements for liouville's theorem is that the system is described only by the hamiltonian and that is äquivalent for description by one lagrangian. So why does it hold where it clearly should not?

  • $\begingroup$ 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. $\endgroup$ – CuriousOne Dec 26 '15 at 22:26
  • $\begingroup$ 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). $\endgroup$ – Daniel Dec 26 '15 at 22:46
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    $\begingroup$ 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. $\endgroup$ – Quantumwhisp Dec 27 '15 at 1:09

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