In the book Equilibrium Statistical Physics: Phases of Matter and Phase Transitions by Marc Baus and Carlos F. Tejero at page 49 (English version) says

3.2 Liouville's Equation


and hence $$ a(\textbf{r},t)\equiv \int \text{d}q\int\text{d}p a(q,p;\textbf{r},t)\rho(q,p),$$ may be written as $$ a(\textbf{r},t) = \int \text{d}q \int \text{d}p \left[ \text{e}^{-L_N t} a(q,p; \textbf{r},0)\right] \rho (q,p) \overset{?}{=} \int \text{d}q\int\text{d}p a(q,p;\textbf{r},0) \left[ \text{e}^{L_N t} \rho (q,p)\right], $$ where to go from the first to the second expression an integration by parts has been performed (since $L_N$ is a linear differential operator), and it has been admitted that $\rho(q,p)$ and all of its derivatives vanish at the limits of integration.

How can I go from the first to the second expression? I don't understand how the book has made the integration by parts.

*In particular, if $$ \int x y' =xy-\int x' y.$$ Who is who?

Just like this: enter image description here from Statistical Mechanics of Nonequilibrium Liquids by Evans and Morriss. (Page 48)

  • 1
    $\begingroup$ Suggestion: if you expand in series $e^{L_N t}$ you will have terms that are even in $L_N$ and terms that are odd in $L_N$: if you transfer each term from $a$ to $\rho$ integrating by parts the odd ones will change sign, the even ones will not... $\endgroup$
    – DelCrosB
    Commented Nov 3, 2016 at 20:15

1 Answer 1


I think that the answer is: $$ \int \text{d}\Gamma \text{ }a(\Gamma)\text{ } L \text{ } \rho (\Gamma) = \int \text{ }\text{d}\Gamma \text{ } a(\Gamma) \text{ }\dot{\Gamma } \dfrac{\partial}{\partial \Gamma} \rho (\Gamma)\tag{1}$$ because of the Chain rule and the Hamilton's equations. The measure of the phase space $\Gamma$ is d$\Gamma =\text{d}q\text{ d}p.$ We know that $$ \int \text{d}\Gamma \text{ }a(\Gamma)\text{ } L \text{ } \rho (\Gamma) = \int \text{d}\Gamma \text{ }\rho(\Gamma)\text{ }\left[- L \right]\text{ } a(\Gamma) \tag{2}$$ The equation (2) is obtained by integrating by part once $$ \int \text{ }\text{d}\Gamma \text{ } a(\Gamma) \text{ }\dot{\Gamma } \dfrac{\partial}{\partial \Gamma} \rho (\Gamma) = \left. a(\Gamma) \text{ }\dot{\Gamma} \text{ } \rho(\Gamma) \right|_S-\int \text{d}\Gamma \rho(\Gamma) \dfrac{\partial}{\partial \Gamma}\left[ a(\Gamma)\dot{\Gamma}\right] = -\int \text{d}\Gamma \rho (\Gamma) \dot{\Gamma}\dfrac{\partial}{\partial \Gamma} a(\Gamma). \tag{3}$$ So, by induction, we have $$ \int \text{d}\Gamma \text{ }a(\Gamma)\text{ } L^n \text{ } \rho (\Gamma) = \int \text{ }\text{d}\Gamma \text{ } a(\Gamma) \text{ }(-1)^n L^n \rho (\Gamma),\quad \forall n \in \mathbb{N}\tag{4}.$$ Note that, in order to get the equation (4), we've been made $n$ integrations by parts: $$ \int \text{ }\text{d}\Gamma \text{ } a(\Gamma) \text{ }\dot{\Gamma }^n \dfrac{\partial^n}{\partial \Gamma^n} \rho (\Gamma) = \left. a(\Gamma) \text{ }\dot{\Gamma}^n \dfrac{\partial^{n-1}}{\partial \Gamma^{n-1}} \text{ } \rho(\Gamma) \right|_S-\int \text{d}\Gamma \dfrac{\partial^{n-1}}{\partial \Gamma^{n-1}}\rho(\Gamma) \dfrac{\partial}{\partial \Gamma}\left[ a(\Gamma)\dot{\Gamma}^n\right] $$ $$ =...= (-1)^n\int \text{d}\Gamma \rho (\Gamma) \dot{\Gamma}^n\dfrac{\partial^n}{\partial \Gamma^n} a(\Gamma). \tag{3}$$ Therefore, we can express the exponential function $\text{e}^{\pm Lt}$ as $$ e^{\pm Lt} = 1 \pm tL + \dfrac{t^2}{2}L^2 \pm \dfrac{t^3}{6} L^3 + O(L^4)$$ and apply the previous result.


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