# A proof of Liouville’s theorem

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:

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?

• 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. – chichi Aug 14 at 14:55
• @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 – an offer can't refuse Aug 14 at 23:39

## 1 Answer

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

Suppose we instead choose a term in linear $$dt$$ like $$$$\frac{\partial\dot{q}_1}{\partial p_1}dp_1dt.$$$$ Again, other brackets should contribute something without $$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.

• What if we instead choose a term in linear dt like $\frac{\partial\dot{q}_1}{\partial q_2}dq_2dt.$ – an offer can't refuse Aug 15 at 3:18
• Then the $dq_2$ term from the second bracket will kill it. – chichi Aug 15 at 3:23
• Why kill it .... – an offer can't refuse Aug 15 at 4:14
• What prevent $(dq_2)^2$ from entering into the equation? – an offer can't refuse Aug 15 at 11:29
• $(dq_2)^2 = 0$ because $dq_2 \wedge dq_2 = 0$. – chichi Aug 15 at 12:42