Let us say we have a collection of $n$ stationary balls per square meter each of diameter $d$ and we roll an identical ball between them. Then what is the free mean path of this ball?

My calculations would indicate that it is: $$\lambda=\frac{1}{2dn}$$ However in the book I am working through (namely Thermodynamics, Kinetic theory and statistical thermodynamics, 3rd ed by Sears and Salinger problem 10-7*) they seem to have used from there numerical answers: $$\lambda =\frac{1}{dn}$$ Is my form correct, or the form I think they have used and why?

*Note the above question in block quotes aboveis not problem 10-7 but is a generalisation of it.


Here is how I get my result. Since all balls are radius $d/2$ there centres need to be within a distance of $d$ if they are going to collide. This means that if any stationary ball is within the area $2dL$ of the moving ball they will collide. Where $L$ is the distance travelled by and we have $2d$ since the centre of the stationary ball can be a distance $d$ either side of the moving for it to collide. When there is on average one ball in this area the area is given by: $$2dL=\frac{1}{n}$$ Meaning that our mean free path is: $$\lambda=\frac{1}{2dn}$$

  • 1
    $\begingroup$ can you give a more detailed analysis of your calculation? $\endgroup$
    – Bruce Lee
    Jan 13 '16 at 9:30
  • $\begingroup$ @BruceLee I have edited the question to include this detail $\endgroup$ Jan 13 '16 at 10:17
  • $\begingroup$ I agree with your result, and so does this answer. I'm now skeptical of your textbook's answer - they do sometimes get it wrong (rarely, but sometimes). $\endgroup$
    – Judge
    Jan 13 '16 at 12:40

The deviation of a factor 2 comes from the fact that you assumed other balls are stationary, which is not true. In fact other balls are also moving, and this increases the possibility of colliding with the moving ball you're considering. Similar correction can also be found in 3D case. You can refer to http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/menfre.html.


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