# Why does computing mean free path from an air molecule's reference frame seem to greatly overestimate path length?

I was interested in seeing if I could derive the mean free path of an "air molecule" by considering the reference frame of an individual molecule as other particles moved around it randomly. This is sort of the opposite of the simple model of a moving beam through a slab of stationary particles. My approach strikes me as kind of stupid but I had fun with it.

I considered a stationary molecule at the center of a sphere of radius $$R$$ ($$R$$ was chosen to match the typical number density of air). Random molecules would pass through the sphere. I calculated the probability $$P$$ that a molecule on a random path through the sphere would collide with the molecule at the center. Then to calculate the mean free path I take the expected number of random paths through the sphere to produce a collision $$(\frac{1}{P})$$, and multiply by mean random path length through the sphere (which if I am not mistaken is simply the mean chord length). Plugging in typical values I come up with a mean free path of about 1300 nm which is a good order of magnitude too large. $$mean\;free\;path: l = \frac{1}{P} \frac{4 R}{\pi}$$ Where here $$P$$, the probability of a random molecule through the sphere colliding with the molecule (both of diameter $$d$$) at center, is: $$P=\frac{1}{\pi}(ArcCos(1-2\frac{d^2}{R^2}) - 2\frac{d}{R}(1 - 2\frac{d^2}{R^2})\sqrt{1-\frac{d^2}{R^2}})$$ Which looks reasonable Using values of $$d=0.3 nm$$ and $$R=2.6 nm$$ produces $$l=1270 nm$$, much greater than the $$66 nm$$ quoted on wikipedia using the same number density of $$2.7 \times 10^{19} \frac{molecules}{cm^3}$$.

Derivation of $$P$$:

Two random points on the surface of the sphere of radius $$R$$ define a random path through the sphere. Without loss of generality we can assume the first point is at $$\mathbf p_i : (x=R,y=0,z=0)$$. The second point is then characterized by azimuth $$\phi$$ and elevation $$\theta$$: $$\mathbf p_f : (x=R Cos(\phi) Sin(\theta),\;y=R Sin(\phi) Sin(\theta),\;z=R Cos(\theta))$$. The path through the sphere can be parametrized as $$\mathbf r(t) = \mathbf p_i + \frac{(\mathbf p_f - \mathbf p_i) v t}{\Vert \mathbf p_f - \mathbf p_i \Vert}$$ By symmetry, the point of closest approach to the center of the sphere occurs at $$t=\frac{\Vert \mathbf p_f - \mathbf p_i \Vert}{2 v}$$ so that $$\mathbf r_{closest} = \frac{1}{2}(\mathbf p_i + \mathbf p_f)$$ After some basic trig identities and simplification we find $$r = \Vert \mathbf r_{closest} \Vert = \frac{1}{\sqrt{2}} R \sqrt{1 + Cos(\phi)Sin(\theta)}$$ To find $$P$$ we look at $$r < d$$ or $$2 \frac{d^2}{R} > R Cos(\phi) Sin(\theta) + R$$ $$2 \frac{d^2}{R} > X + R$$ This amounts to finding the area of the circle sector with $$h=2 \frac{d^2}{R}$$ divided by the area of the circle radius R. $$P=\frac{sector\;area}{cirlce\;area}$$ In doing so we recover the formula for $$P$$ that I quoted previously. As a final step I convert the number density $$n$$ into $$R$$. To do this we note that there should be be two molecules in the sphere so$$\frac{2}{n}=\frac{4}{3}\pi R^3$$ and $$R=(\frac{3}{2 \pi n})^{1/3}$$

I realized the issue. $$P=\frac{sector\;area}{circle\;area}$$ is incorrect. Since all point on the sphere map to the circle the proper equation is: $$P=\frac{arc}{circumference}$$ From there it's easy to see that $$h = R (1-Cos(\alpha))$$ $$\alpha=ArcCos(1-h/R), h=2\frac{d^2}{R}$$ $$L=2 R ArcCos(1-2\frac{d^2}{R^2})$$ $$P=\frac{1}{\pi}ArcCos(1-2\frac{d^2}{R^2})$$
This yields then for the mean free path: $$l=\frac{4R}{ArcCos(1-2\frac{d^2}{R^2})}$$
For the values $$d=0.3nm, R=2.6nm$$ this gives $$l=45nm$$ much closer to the actual value.