- #1

Jeremy1986

- 17

- 3

- 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!

$$\frac{{d\rho }}{{dt}} = \frac{{\partial \rho }}{{\partial t}} + \sum\limits_{a = 1}^{3N} {(\frac{{\partial \rho }}{{\partial {p_a}}}\frac{{d{p_a}}}{{dt}} + \frac{{\partial \rho }}{{\partial {q_a}}}\frac{{d{q_a}}}{{dt}})} = 0$$

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!