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How does one prove that for a system whose Hamiltonian is dependent explicitly on time ($H (q,p,t)$), the volume of an element in phase space is conserved i.e. $\frac{d V}{dt} = 0$ ?

In what follows I've dropped the subscripts of q and p to make things neater.

For autonomous case I used $\frac{d(\delta V)}{dt}=(\nabla \cdot \vec{f} )(\delta V)$ where $\vec{f}=(f_1, f_2)$ and $\dot{q}= f_1= \frac{\partial H}{\partial p} $and $\dot{p}= f_2= - \frac{\partial H}{\partial q} $ and the symmetry of the second derivatives.
I'm not sure whether this argument carries over to the non-autonomous case as f may be complicated in general. Does it??
This is how I progressed
$q(\delta t)=q(0)+\frac{\partial H}{\partial p}\delta t + O((\delta t)^2)$ and
$p(\delta t)=p(0)-\frac{\partial H}{\partial q}\delta t + O((\delta t)^2)$. Treating $q(t),p(t)$ as the new variables after transformation, we are done if we show the

$det(\frac{\partial (q(t),p(t))}{\partial (q(0),p(0))})=0$ but this requires

$\frac{\partial^2 H}{\partial p^2}\frac{\partial^2 H}{\partial q^2} = (\frac{\partial^2 H}{\partial q \partial p} )^2$ which I cnnot show to be true! I am stuck here.

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  • $\begingroup$ Are you sure that the statement you wish to prove is correct and doesn’t require further assumptions? $\endgroup$
    – Wrzlprmft
    Jul 23, 2017 at 8:20
  • $\begingroup$ I think it is sufficient. In autonomous case its fine: showing that the divergence of f =(f_1, f_2) = 0 suffices, where \frac{dq}{dt}=f_1 = \frac{\partial H}{\partial p} and \frac{dp}{dt}= f_2= - \frac{\partial H}{\partial p} respectively denotes the time. For the non autonomous case I proceeded by linearizing (dropping higher powers of t) but cannot proceed further. I'll just include in the original post how i progressed and what remains to be shown. $\endgroup$
    – Soumak B.
    Jul 23, 2017 at 13:35

1 Answer 1

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I rarely think about time-dependent Hamiltonians, so I may be wrong. But I think the standard proof of Liouville's theorem is valid for a time-dependent Hamiltonian. For a full derivation, See Sec. 3.2 of "Statistical physics of particles" by M. Kardar. There are PDFs available online. The gist is the following: You already got the total differential (to first order in $\delta t$) $$dq_\mu'=dq_\mu + \frac{\partial \dot{q}_\mu}{\partial q_\mu}dq_\mu\delta t,$$ and $$dp_\mu'=dp_\mu + \frac{\partial \dot{p}_\mu}{\partial p_\mu}dp_\mu\delta t,$$ so that the linearized phase space volume is now $$dq_\mu' dp_\mu'=dq_\mu dp_\mu\left[1+\left(\frac{\partial^2 H}{\partial q_\mu\partial p_\mu} - \frac{\partial^2 H}{\partial q_\mu\partial p_\mu} \right)\delta t + \mathcal{O}\{\delta t^2\} \right],$$ from Hamilton's equations. Since the phase-space volume differential remains unchanged, so does the full volume.

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  • $\begingroup$ omg, i had forgotten the \delta t term. thus the formidable stuff vanishes. $\endgroup$
    – Soumak B.
    Jul 23, 2017 at 16:00
  • $\begingroup$ Glad I could help! $\endgroup$ Jul 23, 2017 at 17:34

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