I have found a proof of Liouville's theorem on the internet, which fits me very well except one step I don't understand, the derivation is as follows:

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In the derivative, it must have used the relation $dq'_i=dq_i+\frac{\partial\dot{q}_i}{\partial q_i}dq_idt$ and $dp'_i=dp_i+\frac{\partial\dot{p}_i}{\partial p_i}dp_idt$ which I don't understand. Since in the Hamiltonian's formalism, the independent variables are $q$'s and $p$'s, why $dq_i'$ not equal to $dq_i+\sum_j\frac{\partial\dot{q}_i}{\partial q_j}dq_jdt + \sum_j\frac{\partial\dot{q}_i}{\partial p_j}dp_jdt$, etc?

  • 3
    $\begingroup$ I guess you do have the additional term which you listed at the end. However, as $d\Gamma$ is a volume form, such additional terms do not contribute. $\endgroup$
    – chichi
    Aug 14 '19 at 14:55
  • $\begingroup$ @chichi Please write your comment as an explicit answer if you have time. Three votes up to your answer, however I don't know what you are talking about. Those terms have the exactly same dimensionality $\endgroup$ Aug 14 '19 at 23:39
  • $\begingroup$ Please don't cut and paste on the internet without crediting the author. It's rude. $\endgroup$
    – user4552
    Jan 9 '20 at 21:12
  • $\begingroup$ We don't delete questions at the request of users, as that wastes the effort that other people have put into answering the question. In any case, if you have a pressing reason you think something should be deleted, you should flag it for moderator attention using the "flag" button under the post rather than vandalizing it. $\endgroup$
    – Chris
    Jan 10 '20 at 2:37
  • $\begingroup$ @Chris I know the rule, but I have the right to edit the question so no one knows what I am talking about. $\endgroup$ Jan 11 '20 at 1:23

Consider \begin{equation} d\Gamma' = \left( dq_1 + \frac{\partial\dot{q}_1}{\partial q_1}dq_1dt + \cdots + \frac{\partial\dot{q}_1}{\partial p_1}dp_1dt + \cdots \right) \left( dq_2 + \cdots \right) \cdots (dp_1+\cdots)(dp_{3N}+\cdots) \end{equation} where the omitted dots are the remaining terms in your correct expansion \begin{equation} dq'_i = dq_i + \sum_j \frac{\partial\dot{q}_i}{\partial q_j}dq_j dt + \sum_j \frac{\partial\dot{q}_i}{\partial p_j} dp_jdt. \end{equation} Now we want to expand the expression of $d\Gamma'$ up to the first order in $dt$. That means we can only choose one $dt$ term in each bracket. Consider we choose $dt$ factor from the first bracket. If we choose \begin{equation} \frac{\partial\dot{q}_1}{\partial q_1}dq_1dt \end{equation} from the first bracket, then all other brackets should contribute terms like $dq_2,\cdots$ and this leads to your (2.19).

Suppose we instead choose a term in linear $dt$ like \begin{equation} \frac{\partial\dot{q}_1}{\partial p_1}dp_1dt. \end{equation} Again, other brackets should contribute something without $dt$ and these can only be $dq_2,\cdots$ and also the term $dp_1$. However, the volume form is a top form, you can't have $dp_1$ show up twice, therefore, these terms do not contribute.

  • $\begingroup$ What if we instead choose a term in linear dt like $\frac{\partial\dot{q}_1}{\partial q_2}dq_2dt. $ $\endgroup$ Aug 15 '19 at 3:18
  • $\begingroup$ Then the $dq_2$ term from the second bracket will kill it. $\endgroup$
    – chichi
    Aug 15 '19 at 3:23
  • $\begingroup$ Why kill it .... $\endgroup$ Aug 15 '19 at 4:14
  • $\begingroup$ What prevent $(dq_2)^2$ from entering into the equation? $\endgroup$ Aug 15 '19 at 11:29
  • $\begingroup$ $(dq_2)^2 = 0$ because $dq_2 \wedge dq_2 = 0$. $\endgroup$
    – chichi
    Aug 15 '19 at 12:42

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