Mean Free Path — Question 9

Question 9

A nonuniform medium requires an integrated collision hazard

A probe moves along $0\le x\le L$ through stationary scatterers with density $\nu(x)=\nu_0(1+x/L)$. Cross section $\sigma$ is constant and $\sigma\nu_0L=2.00$. The survival obeys $dS/dx=-\sigma\nu(x)S$, with $S(0)=1$.

1. Derive $S(x)$ and find $S(L)$.

2. Find the position at which half the probes have collided. Compare final survival with estimates using only the inlet or outlet density.

Question 9 - Solution

A nonuniform medium requires an integrated collision hazard

(a) Method and calculation

$$\frac{dS}{S}=-\sigma\nu_0(1+x/L)dx\ \Rightarrow\ S(x)=\exp\left[-\sigma\nu_0\left(x+\frac{x^2}{2L}\right)\right].$$$$S(L)=e^{-3}=.0497871.$$

(b) Interpretation and further calculation

With $y=x/L$,$$2y+y^2=\ln2\ \Rightarrow\ y_{50}=-1+\sqrt{1+\ln2}\approx.301210.$$$$S_{\mathrm{inlet}}=e^{-2}=.135335,\quad S_{\mathrm{outlet}}=e^{-4}=.0183156.$$The inlet approximation overpredicts survival; the outlet approximation underpredicts it.

Physical check

Optical depth is an integral of local inverse free path, not an average of local free paths.