- #1

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in phase space, then this implies that \rho wil remain independent of time for all [itex]t>t_0[/itex] because by Liouville's theorem

[tex]\frac{\partial\rho}{\partial t}=-\sum(\frac{\partial\rho}{\partial q}\dot{q}+\frac{\partial\rho}{\partial p}\dot{p})=0[/tex]

I don't understand this because we take [itex]\frac{\partial\rho}{\partial q}=\frac{\partial\rho}{\partial p}=0[/itex] only at t=t_0 which means that [itex]\frac{\partial\rho}{\partial t}=0[/itex] only at t=t_0 it doesn't mean that [itex]\frac{\partial\rho}{\partial t}=0[/itex] for t>t_0 therefore we can not assert that [itex]\frac{\partial\rho}{\partial q}=0[/itex] and [itex]\frac{\partial\rho}{\partial q}=0[/itex] for [itex]t>t_0[/itex]