2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Are you sure that the statement you wish to prove is correct and doesn’t require further assumptions? $\endgroup$ – Wrzlprmft Jul 23 '17 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 '17 at 13:35
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ omg, i had forgotten the \delta t term. thus the formidable stuff vanishes. $\endgroup$ – Soumak B. Jul 23 '17 at 16:00
  • $\begingroup$ Glad I could help! $\endgroup$ – David Ruiz-Tijerina Jul 23 '17 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.