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?
[Physics] Liouville’s theorem and conservation of phase space volume
classical-mechanicsdissipationhamiltonian-formalismphase-spacevolume
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Actually, the Liouville theorem is more general - it is valid even if the distribution function depends on time, and even if the Hamiltonian depends on time.
http://en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian)
-> phase space volume preservation but no energy conservation: any Hamiltonian which depends on time, but you already know that. For example, system of free particles under action of prescribed time-dependent forces.
-> energy is conserved but Liouville's theorem doesn't hold : this is harder to find. Liouville theorem is valid for every normal Hamiltonian, so we have to look for non-Hamiltonian system which nevertheless has energy and this is conserved. The only thing that may have such behaviour that comes to my mind is non-holonomic system with some nasty moving constraints, like ball on a plane with no slipping. Based on what Goldstein says in the 2nd chapter of his book, I think that for such systems there may not be Hamiltonian, ergo no Liouville theorem. One has rather the basic Newtonian equations of motion and constraint equations and inequalities - energy then can be defined as sum of kinetic energies and may be conserved.
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 $\text d\rho /\text d t=0$.
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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.