The random walk process

Suppose that a neutron is created at x,y,z,t=0. It suffers collisions with mean free path $\lambda$. The collisions are isotropic so that one can write, after n collisions:
$$\begin{align}x_{n} & =\sum_{i=0}^{n}\Delta_{xi}\nonumber\\ y_{n} & =\sum_{i=0}^{n}\Delta_{yi}\nonumber\\ z_{n} & =\sum_{i=0}^{n}\Delta_{zi}\nonumber \end{align}$$
Since the signs of the $\Delta$ are positive or negative with equal probability one has <x_n>=<y_n>=<z_n>=0. The average distance traveled by a neutron is $\lambda.$ The average of the square of the distance is $2\lambda^{2}$. Thus $<\Delta_{x}^{2}+\Delta_{y}^{2}+\Delta _{z}^{2}>=2\lambda^{2}$ so that $<x_{n}^{2}>={\frac{2n}{3}}\lambda^{2},$ with similar expressions for the other coordinates. It follows that, after n collisions, the probability distributions are given by a normal distribution(we only consider the x distribution):

$$P(n,x)={\frac{1}{\lambda\sqrt{{\frac{4\pi}{3}}n}}}e^{-{\frac{3x^{2}} {4n\lambda^{2}}}}$$ (4.41)

If one considers the distance r to the source one gets, for the density distribution:

$$P(n,r)={\frac{1}{\lambda^{3}\left( {\frac{4\pi}{3}}n\right) ^{3/2}} }e^{-{\frac{3r^{2}}{4n\lambda^{2}}}}$$ (4.42)

For very heavy scatterers it is a reasonable approximation to assume that the energy is decreased by a fixed relative amount, following each collision. This is given by :

$${\frac{\Delta E}{E}}\ =\xi\simeq\ \ \frac{1-\varpi}{2}\simeq\frac{2m}{M}$$ (4.43)

Since $\frac{\Delta E}{E}=\Delta\ln E$ the number of collisions required for the neutron to decrease its energy from E₀ to E is:

$$n\ =\ {\frac{1}{\xi}}\ln{\frac{E_{0}}{E}}$$ (4.44)

Thus the spatial distribution reads:

$$P(E,r)={\frac{1}{\lambda^{3}\left( {\frac{2\pi}{3}}\ {\frac{1... ...{2}}{4{\frac {\lambda^{2}}{\xi}}Log{\frac{E_{0}}{E}}}}}\right)$$ (4.45)

The root-mean square radius reads:

$$\sigma(r:E)=\lambda\sqrt{{\frac{2}{3\xi}}\ln{\frac{E_{0}}{E}}}$$ (4.46)

and the r probability distribution thus reads more simply:

$$P(E,r)={\frac{1}{(2\pi)^{3/2}\sigma^{3}(r:E)}}\left( e^{-{\frac{r^{2} }{2\sigma^{2}(r:E)}}}\right)$$ (4.47)

This result is, essentially, similar to that of the Fermi age theory, which reads[34]

$$q(r,\tau)=\frac{e^{-\frac{r^{2}}{4\tau}}}{\left( 4\pi\tau\right) ^{3/2}}$$ (4.48)

where $\tau$ is the Fermi age, $q(r,\tau)$ is the slowing down density, i.e. the number of neutrons at position r and age $\tau$ which cross the energy E. The Fermi age is defined:

$$\tau=\int_{E}^{E_{0}}\frac{D(E)}{\xi\Sigma_{s}(E)}\frac{dE}{E}$$

with $D(E)=\frac{1}{3\Sigma_{s}(E)(1-\overline{\mu})}$ and $\overline{\mu }=\frac{2}{3A}$. $\overline{\mu}$ will be neglected for large A, like in the case of lead. Thus, for D and $\Sigma_{s}$ independent of E

$$2\sigma^{2}(r:E)=\frac{4\lambda^{2}}{3\xi}\ln\frac{E_{0}}{E}=\frac{4} {3\xi\Sigma_{s}^{2}}\ln\frac{E_{0}}{E}=4\tau$$ (4.49)

which shows the identity of the age distribution with that obtained from the random walk calculation. Since the age equation accepts a dependence of D and $\Sigma_{s}$ on energy, it is more general than that we have just deduced, so that we shall keep it. The relation between $\tau$ and n is obtained for constant D and $\Sigma_{s}$:

$$\tau=\frac{\lambda^{2}}{3\xi}\ln\frac{E_{0}}{E}=\frac{n\lambda^{2}}{3}$$