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I Liouville's theorem

  1. Nov 10, 2016 #1
    Liouville's theorem states that the total time-derivative of the distribution function is zero along a system trajectory in phase-space. Where the system follows a trajectory that satisfies the Hamilton's equations of motion.

    I have a hard time getting an inuitive understanding of this statement. For instance, what does this theorem tell me about a free-falling particle in a gravitational field?

    Edit: I assume that the distribution function for a free falling particle would be proportional to a product of delta functions.
     
    Last edited: Nov 10, 2016
  2. jcsd
  3. Nov 11, 2016 #2

    vanhees71

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    Liouville's theorem states that the volume of phase space is unchanged under Hamiltonian flow. That immediately follows from Hamilton's canonical equations. Suppose you have a system with initial conditions ##(q_0,p_0)## (where ##(q,p) \in \mathbb{R}^{2f}## are the phase-space variables) filling a certain infinitesimal volume elemen ##\mathrm{d} \Omega_0## in phase space. With time ##t## the trajectories in phase space are given by
    $$\dot{q}=\frac{\partial H}{\partial p}, \quad \dot{p}=-\frac{\partial H}{\partial q}.$$
    The volume element changes with the Jacobian of the transformation ##(q_0,p_0) \rightarrow (q,p)##. Now the time derivative of this determinant is
    $$\frac{\mathrm{d} \Omega}{\mathrm{d} t}=\frac{\partial (q,p)}{\partial(q_0,p_0)}=\mathrm{div}_q \dot{q}+\mathrm{div}_p \dot{p} = \mathrm{div}_q \frac{\partial H}{\partial p}-\mathrm{div}_p \frac{\partial H}{\partial q}=\sum_{j=1}^f \left (\frac{\partial}{\partial q^j} \frac{\partial H}{\partial p_j} - \frac{\partial}{\partial p_j} \frac{\partial H}{\partial q^j} \right)=0.$$
     
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