I am reading the textbook Transport Phenomena by Bird, Stewart & Lightfoot. In chapter 1 they derive expression for viscosity of gases at low density from the kinetic theory. They start by writing down four important expressions, one of which is mean free path $\lambda$ and the last being a "rough" estimate of the distance $a$ between a plane and the last molecular collision. This quantity is given by:

$a = \frac{2}{3} \lambda$

This derivation can be seen on Google Books, page 24.

I would like to understand what is the reasoning for taking such value for $a$? Why is it $\frac{2}{3}$ and why do they say that it is "roughly" given by that expression? How would one go about estimating $a$?

I understand the importance of $a$ and it seems to me that it is a very important quantity to estimate correctly in the derivation - later they use it to write momentum balance. It seems crucial to me that we take $a$ such that there are no more collisions between a plane at $y$ and at $y+a$, since every collision would be another opportunity to exchange momentum.


1 Answer 1


Distance until next collision is exponentially distributed, as is distance since last collision. The mean value of this distribution is $\lambda$. The direction from the last collision until the point at which the particle hits the plane is assumed to be distributed uniformly over the sphere. Together, this characterizes the full distribution of the location of the last collision, relative to the current location of the molecule.

Let the plane of interest be the $xy$ plane, and let the particle hit it at the origin. Then the location of the last collision is distributed with probability density proportional to $e^{-\frac{3\sqrt{x^2 + y^2 + z^2}}{\lambda}}$. The constant of proportionality is given by $$ \frac{1}{\int e^{-\frac{3\sqrt{x^2 + y^2 + z^2}}{\lambda}}} $$ since the distribution must be normalized. The distance from the plane is just $|z|$. The mean distance from the location of the last collision to the plane is then $$\frac{\int |z| e^{-\frac{3\sqrt{x^2 + y^2 + z^2}}{\lambda}}}{\int e^{-\frac{3\sqrt{x^2 + y^2 + z^2}}{\lambda}}} $$

If you want to try these integrals yourself, change to spherical coordinates and integrate the radial terms by parts. I find $\frac{1}{2}\lambda$, so it seems likely that something is wrong above.

Edit: The book you reference cites chapters II and III of this https://archive.org/details/in.ernet.dli.2015.1789/page/n151/mode/2up for a derivation. This book gives the $\frac{\lambda}{2}$ answer, and then proceeds to explain why it is wrong. It turns out that what we are interested is not the distance between a plane and the last collision for a molecule chosen uniformly at random. Instead, it refers to that distance for a random molecule chosen from all molecules that cross the plane over some short time. Some molecules are more likely to cross the plane than others, so this is not a (uniformly) random sample.

  • 1
    $\begingroup$ This ratio appear to give $3\lambda/2$ ... $\endgroup$
    – user26872
    Commented Jul 30, 2020 at 16:07
  • $\begingroup$ Good point, I also get that answer. I think this is because I've used $\lambda$ as the parameter of the exponential distribution, but that's only actually equal to the mean free path in 1D. In 3D, the mean free path is given by $$\lambda' = \int r p(r)r^2sin(\theta)dr d\theta d\phi = 3\lambda $$ However, changing out $\lambda$ for $\lambda/3$ in the above gives $a = \lambda/2$, so that still doesn't quite match. $\endgroup$
    – Daniel
    Commented Jul 30, 2020 at 20:09
  • $\begingroup$ I suspect the authors may have done something like approximating the exponential distribution as a dirac-delta with the same mean (i.e. just averaging over the surface of a sphere of radius $\lambda$). Does that approximation give $\frac{2}{3}$? $\endgroup$
    – Daniel
    Commented Jul 30, 2020 at 20:12
  • $\begingroup$ Why would we approximate an integral that can be done analytically with very little effort? Why would any reasonable approximation of $3\lambda/2$ result in $(4/9)3\lambda/2 = 2\lambda/3$? $\endgroup$
    – user26872
    Commented Jul 31, 2020 at 14:41
  • $\begingroup$ Physics textbooks do sometimes do that for pedagogical reasons, but these are reasonable questions. Do you think my answer has something else wrong with it? $\endgroup$
    – Daniel
    Commented Jul 31, 2020 at 18:55

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