classical mechanics - Liouville's theorem and conservation of phase space volume

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 system is dissipative like a damped harmonic oscillator?

Answer

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.