A Liouville's theorem and time evolution of ensemble average

[Jeremy1986]

TL;DR Summary

Question about using Liouville's theorem to calculate time evolution of ensemble average

With the Liouville's theorem
$$\frac{{d\rho }}{{dt}} = \frac{{\partial \rho }}{{\partial t}} + \sum\limits_{a = 1}^{3N} {(\frac{{\partial \rho }}{{\partial {p_a}}}\frac{{d{p_a}}}{{dt}} + \frac{{\partial \rho }}{{\partial {q_a}}}\frac{{d{q_a}}}{{dt}})} = 0$$
when we calculate the time evolution of the ensemble average of a quantity $O(p,q)$ we have
$$\frac{{d\left\langle O \right\rangle }}{{dt}} = \int {d\Gamma \frac{{\partial \rho (p,q,t)}}{{\partial t}}O(p,q)} = \sum\limits_{a = 1}^{3N} {\int {d\Gamma } } O(p,q)(\frac{{\partial \rho }}{{\partial {p_a}}}\frac{{\partial H}}{{\partial {q_a}}} - \frac{{\partial \rho }}{{\partial {q_a}}}\frac{{\partial H}}{{\partial {p_a}}})$$
here $p,q$ represents a bunch of generalized coordinates and momentum ${p_a},{q_a},a = 1,...,3N$ .
Then by using the method of integration by parts, the above integration becomes
$$\frac{{d\left\langle O \right\rangle }}{{dt}} = - \sum\limits_{a = 1}^{3N} {\int {d\Gamma } } \rho [(\frac{{\partial O}}{{\partial {p_a}}}\frac{{\partial H}}{{\partial {q_a}}} - \frac{{\partial O}}{{\partial {q_a}}}\frac{{\partial H}}{{\partial {p_a}}}) + O(\frac{{{\partial ^2}H}}{{\partial {p_a}\partial {q_a}}} - \frac{{{\partial ^2}H}}{{\partial {q_a}\partial {p_a}}})$$Here comes my questions, I think the integration by parts uses $\int {d{p_a}} \frac{{\partial \rho }}{{\partial {p_a}}} = \int {d\rho }$ . However as
$\rho ~\rho (p,q,t)$ , should we have $\frac{{d\rho }}{{d{p_a}}} = \frac{{\partial \rho }}{{\partial {p_a}}} + \frac{{\partial \rho }}{{\partial t}}\frac{{dt}}{{d{p_a}}}$ .


I take that last relation for granted because when we calculate $\frac{{dy}}{{dx}}$ , if $y=y(x)$ determines implicitly by some relation $F(x,y)=0$ , we use
$$\frac{{\partial F(x,y)}}{{\partial x}} + \frac{{\partial F(x,y)}}{{\partial y}}\frac{{dy}}{{dx}} = 0$$ and get $$\frac{{dy}}{{dx}} = - \frac{{\frac{{\partial F(x,y)}}{{\partial x}}}}{{\frac{{\partial F(x,y)}}{{\partial y}}}}$$ If we differentiate $F(x,y)=0$ with y, in order to get the same value of $\frac{{dy}}{{dx}}$ , we need to have
$$\frac{{\partial F(x,y)}}{{\partial x}}\frac{{dx}}{{dy}} + \frac{{\partial F(x,y)}}{{\partial y}} = 0$$ where we consider $x$ ~$x(y)$ . So back to the $\rho$ case, in calculating $\frac{{d\rho }}{{d{p_a}}}$ , I think $\frac{{\partial \rho }}{{\partial t}}\frac{{dt}}{{d{p_a}}}$ term should be taken into account since $p_a$ ~$p_a(t)$ . Then the integration by parts seems to be wrong, so I think I have made a mistake.______________________________________________________________________

My second question concerns Liouville's theorem itself, if we have $\frac{{d\rho }}{{dt}}=0$ , then can we have $\frac{{d\rho }}{{d{q_a}}} = \frac{{d\rho }}{{dt}}\frac{{d{q_a}}}{{dt}} = 0$ ? It sounds ridiculous as it indicates that the probability density is everywhere the same in the phase space, regardless of what the system is. Then where did I make a mistake? Thanks for your patience reading my long questions!

 

[vanhees71]

Concerning the "integration by parts" question: In multidimensional integration "integration by parts" means to apply the generalized Gauss's theorem in arbitrary dimensions. E.g., for the first piece in your integral (Einstein summation convention applies!),
$$\int \mathrm{d} \Gamma O \frac{\partial \rho}{\partial p_a} \frac{\partial H}{\partial q_a} = \int \mathrm{d} \Gamma [\partial_{p_a} (O \rho \partial_{q_a} H)-\rho \partial_{p_a} (O \partial_{q_a} H)].$$
Now the first is a total divergence in $p$ space and the integral over $\mathrm{d}^{3N} p$ gives a hyper-surface integral due to Gauss's theorem. Since you assume that $\rho$ drops to 0 sufficiently fast at infinity of phase space, this contribution can be dropped, and you are left with the other phase-space integral. Combining everything leads to the desired result (just go on calculating; the result is very intuitive!).

The 2nd question doesn't make sense to me. What is a total derivative of $\rho$ wrt. $q_a$? Of course $\rho$ is not uniform across phase space. This wouldn't be a properly normalized distribution: Integrating over entire phase space should give the particle number, which is finite!

 

[Jeremy1986]

Concerning the "integration by parts" question: In multidimensional integration "integration by parts" means to apply the generalized Gauss's theorem in arbitrary dimensions. E.g., for the first piece in your integral (Einstein summation convention applies!),
$$\int \mathrm{d} \Gamma O \frac{\partial \rho}{\partial p_a} \frac{\partial H}{\partial q_a} = \int \mathrm{d} \Gamma [\partial_{p_a} (O \rho \partial_{q_a} H)-\rho \partial_{p_a} (O \partial_{q_a} H)].$$
Now the first is a total divergence in $p$ space and the integral over $\mathrm{d}^{3N} p$ gives a hyper-surface integral due to Gauss's theorem. Since you assume that $\rho$ drops to 0 sufficiently fast at infinity of phase space, this contribution can be dropped, and you are left with the other phase-space integral. Combining everything leads to the desired result (just go on calculating; the result is very intuitive!).

The 2nd question doesn't make sense to me. What is a total derivative of $\rho$ wrt. $q_a$? Of course $\rho$ is not uniform across phase space. This wouldn't be a properly normalized distribution: Integrating over entire phase space should give the particle number, which is finite!

Thank you very much for your nice reply! I think I start to understand my first question. What I was puzzled is actually the difference between $\frac{{d\rho }}{{dt}}$ and $\frac{{\partial \rho }}{{\partial t}}$, which I now realized defines change in $\rho$ along the path in phase space determined by Hamiltonian and defines the variation of a field "$\rho$" with time at a fixed point in phase space. For the case of ensemble average, the integration is just over phase space, it does not relates to any evolution of ${q,p}$ in the phase space, so the ${q,p}$ in the integration has nothing to do with time t. The only time dependence in $\left\langle O \right\rangle$ is the variation of $\rho$ at each point in phase space.