Collision Term in the Classical Boltzmann Transport Equation

Question

I cannot get over the feeling that in the classical derivation of the collision term of Boltzmann's transport equation molecules that are already knocked out of a $(\textbf r, \textbf v)$ space volume are double-counted many times. In essence, my question really is: how can one multiply the number of "bullets" (incoming particles) by the number of "targets" (particles within an infinitesimal spatial-velocity element) to get the number of times a bullet hits a target? Isn't it that one "bullet" can only hit one "target"?

Please allow me to elaborate.

In the classical Boltzmann equation with the 2-body collision term for a gas of a single kind of molecule, the collision term is given as the sum of a gain through collision term and a loss through collision term.

Largely following Huang's Statistical Mechanics, the derivation of the loss term (the rate of decrease of particle density $f(\textbf r, \textbf v, t)$ owing to collisions) is reproduced as follows:

To a molecule in a given spatial volume $d^3 r$ about $\textbf r$, whose velocity lies in $d^3 v_1$ about $\textbf v_1$, other molecules of any given velocity $\textbf v_2$ in the same spatial volume pose as an incident beam. The flux of this incident beam is
$$I = |\textbf v_1 - \textbf v_2| f(\textbf r, \textbf v_2, t) d^3 v_2$$

So far so good. But the standard derivation goes on to say that the number of collisions happening in this spatial volume $d^3 r$ during $\delta t$ of the type $\{\textbf v_1, \textbf v_2\} \rightarrow \{\textbf v'_1, \textbf v'_2\}$ is given by
$$I \sigma(\textbf v_1, \textbf v_2 ; \textbf v'_1, \textbf v'_2) d^3 v'_1 d^3 v'_2 \delta t = |\textbf v_1 - \textbf v_2| f(\textbf r, \textbf v_2, t) \sigma(\textbf v_1, \textbf v_2 ; \textbf v'_1, \textbf v'_2) d^3 v_2 d^3 v'_1 d^3 v'_2 \delta t$$ A comment here from me before continuing on with the derviation: a single collision will knock the molecule with velocity $\textbf v_1$ out of the phase-space volume of $d^3 r d^3 v_1$ around $(\textbf r, \textbf v_1)$, how can it still be at the same location $\textbf r$ and velocity $\textbf v_1$ for the other collisions of the same type $\{\textbf v_1, \textbf v_2\} \rightarrow \{\textbf v'_1, \textbf v'_2\}$?

To help drive my point home, let's follow through with the standard derivation: The total rate of loss is obtained by integrating over all $\textbf v_2$, $\textbf v'_1$ and $\textbf v'_2$ and then multiplying the result by the number of molecules within the volume of $d^3 r d^3 v_1$ around $(\textbf r, \textbf v_1)$, i.e.,
$$df|_{\text{loss}} \equiv f(\textbf r, \textbf v_1, t+\delta t) d^3 r d^3 v_1 - f(\textbf r, \textbf v_1, t) d^3 r d^3 v_1 \\ = f(\textbf r, \textbf v_1, t) d^3 r d^3 v_1 \iiint |\textbf v_1 - \textbf v_2| f(\textbf r, \textbf v_2, t) \sigma(\textbf v_1, \textbf v_2 ; \textbf v'_1, \textbf v'_2) d^3 v_2 d^3 v'_1 d^3 v'_2 \delta t$$

To me, this further indicates that each molecule within $d^3 r d^3 v_1$ around $(\textbf r, \textbf v_1)$ undergoes $n$ number of collisions where $n = \iiint |\textbf v_1 - \textbf v_2| f(\textbf r, \textbf v_2, t) \sigma(\textbf v_1, \textbf v_2 ; \textbf v'_1, \textbf v'_2) d^3 v_2 d^3 v'_1 d^3 v'_2 \delta t$.

How are these multiple collisions possible? Isn't this double-counting?

The above is reproduced based on the derivation given in Huang's Statistical Mechanics, essentially same derivations can be found from multiple sources online, e.g., https://farside.ph.utexas.edu/teaching/plasma/Plasma/node33.html

Answer 1

The particles are not counted double. There are losses and gains to our phase space due to collisions which would force us to evaluate separate integrals (primed and non-primed velocity space). For elastic collisions the process is reversible and we can consider the reverse process.

Maybe it is easier to understand in the following way which can be found in the German-language book about molecular gas dynamics by Dieter Hänel. I will choose a nomenclature very similar to his as I try to keep my nomenclature consistent across my answers here on Stack Exchange.

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Stoßzahl

The Stoßzahl $Z_s$ is the number of collisions per time and space.

In order to determine it, we could integrate the flux of particles with velocity $\vec \xi_1$ that can be found in an area element perpendicular to the relative velocity

$$ \vec g = \vec \xi_1 - \vec \xi, $$

the so called differential cross-section $A_c$ around the particles with velocity $\vec \xi$. Putting this together we find

$$ d Z_s = | \vec g| f(\vec \xi_1) d \vec \xi_1 d A_c f(\vec \xi) d \vec \xi.$$

Now to determine the total number of collision we will have to integrate over all possible velocity combinations of $\vec \xi_1$ and $\vec \xi$ as well as the cross-section, which for central collisions of rigid spherical particles of equal size does not depends on $\vec \xi_1$ and $\vec \xi$ and is given by $ A_c = d^2 \pi $.

$$Z_s = \int\limits_{ \vec \xi } \int\limits_{ \vec \xi_1 } \int\limits_{ A_c } |\vec g| f( \vec \xi_1 ) f ( \vec \xi) d A_c d \vec \xi_1 d \vec \xi $$

The collision term in the Boltzmann equation

For determining the right-hand-side of the Boltzmann equation we look at losses and gains due to collisions in a similar way. Our phase-space volume is given by

$$\Delta V \Delta \vec \xi := \Delta x \Delta y \Delta z \Delta \xi_x \Delta \xi_y \Delta \xi_z$$

and correspondingly the number of molecules in this phase volume is

$$ \Delta N(\vec \xi, \vec r, t) = f(\vec \xi, \vec r, t) \Delta V \Delta \vec \xi$$

Obviously after collision the particles likely will leave the space volume of interest with velocities $\vec \xi'$ and $\vec \xi_1'$. We take a similar approach to determining the Stoßzahl but in this case we will not integrate over $\vec \xi$: $d \vec \xi$ will be replaced by $\Delta \vec \xi$. The corresponding specific loss flux is then given by

$$\frac{\Delta N_{\text{loss}}}{\Delta V \Delta t} = \int\limits_{ \vec \xi_1 } \int\limits_{ A_c } |\vec g| f( \vec \xi_1 ) f ( \vec \xi) d A_c d \vec \xi_1 \Delta \vec \xi. $$

Dividing by the velocity element we obtain the change of $f$ caused by collisions

$$df |_{\text{loss}} = \frac{\Delta N_{\text{loss}}}{\Delta V \Delta \vec \xi \Delta t} = \int\limits_{ \vec \xi_1 } \int\limits_{ A_c } |\vec g| f( \vec \xi_1 ) f ( \vec \xi) d A_c d \vec \xi_1. $$

Similarly our gains can be evaluated. As the elastic collisions are reversible this can be done by evaluating the inverse collision

$$\frac{\Delta N_{\text{gain}}}{\Delta V \Delta \vec \xi' \Delta t} = \int\limits_{ \vec \xi_1' } \int\limits_{ A_c } |\vec g'| f( \vec \xi_1' ) f ( \vec \xi') d A_c d \vec \xi_1' = \int\limits_{ \vec \xi_1' } \int\limits_{ A_c } |\vec g'| f_1' f' d A_c d \vec \xi_1'. $$

The two integrals can be combined with the laws for elastic collisions. As a consequence of the conservation of mass

$$ m_1 + m = m_1' + m', $$

momentum

$$ m_1 \vec \xi_1 + m \vec \xi = m_1' \vec \xi_1' + m' \vec \xi', $$

and energy

$$ \frac{m_1 \vec \xi_1^2}{2} + \frac{m \vec \xi^2}{2} = \frac{m_1' \vec \xi_1'^2}{2} + \frac{m' \vec \xi'^2}{2}, $$

before and after collisions we can determine the velocity of the common center of gravity as

$$\vec G = \frac{m_1 \vec \xi_1 + m \vec \xi}{m_1 + m} = \frac{m_1 \vec \xi_1' + m \vec \xi'}{m_1 + m} = \vec G'$$

and find that the relative velocity for elastic collisions behaves like

$$ \vec g = - \vec g' .$$

The transformation for the volume of absolute velocities can be transformed into the system of absolute and relative velocities by

$$d \vec \xi_1 d \vec \xi = \underbrace{det | J |}_1 d \vec g d \vec G$$

with the Jacobian matrix (you would have to rearrange the equation system given by the conservation equations to find the relations and that $det | J | = 1$, basically that the transformation conserves its volume during the transformation)

$$ J = \left( \begin{matrix} \frac{\partial \vec \xi}{\partial \vec g} & \frac{\partial \vec \xi}{\partial \vec G} \\ \frac{\partial \vec \xi_1}{\partial \vec g} & \frac{\partial \vec \xi_1}{\partial \vec G} \end{matrix} \right)$$

and therefore the transformation rule

$$ |\vec g'| d \vec \xi_1' \Delta \vec \xi' = |\vec g| d \vec \xi_1 \Delta \vec \xi $$

holds.

This leads to the gain term

$$df |_{\text{gain}} = \frac{\Delta N_{\text{gain}}}{\Delta V \Delta \vec \xi \Delta t} = \int\limits_{ \vec \xi_1 } \int\limits_{ A_c } |\vec g| f_1' f' d A_c d \vec \xi_1. $$

And as a consequence the entire collision term can be written as

$$df |_{\text{collision}} = \frac{\Delta N_{\text{gain}} - N_{\text{loss}}}{\Delta V \Delta \vec \xi \Delta t} = \int\limits_{ \vec \xi_1 } \int\limits_{ A_c } |\vec g| (f_1' f' - f f_1 ) d A_c d \vec \xi_1. $$