Time scales in the BBGKY Hierarchy

Question

In Kardar's or D. Tong's lectures on the BBGKY hierarchy, it is argued that the equation for the two-particle distribution function $f_2$,

$$ \left( \frac{\partial}{\partial t} + \frac{\vec{p}_1}{m} \cdot \frac{\partial}{\partial \vec{q}_1} + \frac{\vec{p}_2}{m} \cdot \frac{\partial}{\partial \vec{q}_2} - \frac{\partial U(|\vec{q}_1 - \vec{q}_2|)}{\partial \vec{q}_1} \cdot \left( \frac{\partial}{\partial \vec{p}_1} - \frac{\partial}{\partial \vec{p}_2} \right) \right) f_2 \simeq 0, $$

implies that $f_2$ evolves on a time scale of order $\tau_c$, the microscopic collision time.

I’m having trouble interpreting precisely what this statement means. Are we supposed to understand that

$$ \frac{\partial f_2}{\partial t} \sim \frac{f_2}{\tau_c}, $$

or more generally, that the full streaming operator acting on $f_2$ varies over a time scale $\tau_c$? But in that case, since the left-hand side is approximately zero (i.e., $\frac{d f_2}{dt} \approx 0$), what does it even mean to say that $f_2$ varies—let alone on a specific time scale?

By contrast, for the one-particle distribution function $f_1$, we have the equation

$$ \left( \frac{\partial}{\partial t} + \frac{\vec{p}_1}{m} \cdot \frac{\partial}{\partial \vec{q}_1} \right) f_1 = \int d^3 q_2 \, d^3 p_2 \, \frac{\partial U(|\vec{q}_1 - \vec{q}_2|)}{\partial \vec{q}_1} \cdot \frac{\partial f_2}{\partial \vec{p}_1}, $$

and here it is stated that $f_1$ varies on a time scale $\tau_U$, which makes sense, as the right-hand side gives a non-zero source term.

Definitions of time scales:

• $\tau_c \sim \frac{d}{v}$, where $d$ is the range of the interparticle interaction potential, and $v$ is a typical particle speed. This is the time over which two particles experience a significant collision.
• $\tau_U \sim \frac{L}{v}$, where $L$ is a macroscopic length scale, typically corresponding to the size of the system.

So my question is: what exactly does it mean to say that $f_2$ evolves on a time scale $\tau_c$, given that its total derivative is (approximately) zero? Are we talking about the scale over which its arguments vary? Or does it mean that if we had a non-zero collision term on the right-hand side (i.e., involving $f_3$), then $f_2$ would change on this time scale?

Any clarification would be appreciated.

Answer 1

So my question is: what exactly does it mean to say that $f_2$ evolves on a time scale $\tau_c$, given that its total derivative is (approximately) zero? Are we talking about the scale over which its arguments vary? Or does it mean that if we had a non-zero collision term on the right-hand side (i.e., involving $f_3$), then $f_2$ would change on this time scale?

One typical definition of a scale associated with varying quantity $x(t)$ is: $$ \tau \sim \left|\frac{x(t)}{\dot{x}(t)}\right|, $$ i.e., the ration of its magnitude to the speed with which this magnitude varies. Note that this is also applicable to spatial dependence or any other type of continuous variables. We also would probably want to average in respect to time, since $\left|\frac{x(t)}{\dot{x}(t)}\right|$ is of course time-dependent.

In case of $f_2$ another interpretation is possible - the correlation time, i.e., how soon $f_2$ for two particles becomes decomposed into a product of one-particle distribution functions: $$ f(p_1,q_1,p_2,q_2,t)\rightarrow f(p_1,q_1,t)f(p_2,q_2,t). $$ we could literally write a correlation of of any two quantities for two particles and plot it with respect to time: $$ \langle A(q_1,p_1,t)A(q_2,p_2,t) \rangle = \int dq_1dq_2dp_1dp_2 A(q_1,p_1,t)A(q_2,p_2,t) f(p_1,q_1,p_2,q_2,t)\propto e^{-\frac{t}{\tau_c}} $$ That is, after some time after the collision the two particles can be considered as moving independently of the velocities that had before and just after the collision. Indeed, after time $d/v$ they are likely to experience another collision, which a particle having unrelated velocity, after time $2d/v$ they will experience two collisions and so on. if the correlation between the particles decreases by factor $\alpha < 1$ with every collision, we get an exponential law: after $n$ collisions $\sim \alpha^n=e^{n\log\alpha}=e^{- t/\tau_c}$.

For one-particle distribution function the scale of change is determined by the potential in which it moves, as seen from the right-hand-side of the corresponding equation. $L$ is the scale on which this potential changes.

Same logic is applicable to higher_order distribution functions, which is why we can truncate the hierarchy at some order, assuming that the loss of correlation at higher orders occurs much faster.

However, this argument obviously disagree with the statement in the notes that the shortness of the correlation time follows from an already truncated equation for the two-particle correlation function. IMHO, if we solve this equation, the distribution function would evolve with the scale of the potential involved, i.e., $\tau_U$. But I am not an expert in Boltzmann equation, so I might be missing something.