The diffusion of neutrons; a uniform spherical source in a homogeneous medium
We take another example that gives the same kind of equation, this time having to do with diffusion. In Chapter 43 of Vol. I we considered the diffusion of ions in a single gas, and of one gas through another. This time, let’s take a different example—the diffusion of neutrons in a material like graphite. We choose to speak of graphite (a pure form of carbon) because carbon doesn’t absorb slow neutrons. In it the neutrons are free to wander around. They travel in a straight line for several centimeters, on the average, before being scattered by a nucleus and deflected into a new direction. So if we have a large block—many meters on a side—the neutrons initially at one place will diffuse to other places. We want to find a description of their average behavior—that is, their average flow. Let $N(x,y,z)\,\Delta V$ be the number of neutrons in the element of volume $\Delta V$ at the point $(x,y,z)$. Because of their motion, some neutrons will be leaving $\Delta V$, and others will be coming in. If there are more neutrons in one region than in a nearby region, more neutrons will go from the first region to the second than come back; there will be a net flow. Following the arguments of Chapter 43 in Vol. I, we describe the flow by a flow vector $\FLPJ$. Its $x$-component $J_x$ is the net number of neutrons that pass in unit time a unit area perpendicular to the $x$-direction. We found that \begin{equation} \label{Eq:II:12:19} J_x=-D\,\ddp{N}{x}, \end{equation}  where the diffusion constant $D$ is given in terms of the mean velocity $v$, and the mean-free-path $l$ between scatterings is given by \begin{equation} D=\frac{1}{3}\,lv.\notag \end{equation}  The vector equation for $\FLPJ$ is \begin{equation} \label{Eq:II:12:20} \FLPJ=-D\,\FLPgrad{N}. \end{equation}  The rate at which neutrons flow across any surface element $da$ is $\FLPJ\cdot\FLPn\,da$ (where, as usual, $\FLPn$ is the unit normal). The net flow out of a volume element is then (following the usual Gaussian argument) $\FLPdiv{\FLPJ}\,dV$. This flow would result in a decrease with time of the number in $\Delta V$ unless neutrons are being created in $\Delta V$ (by some nuclear process). If there are sources in the volume that generate $S$ neutrons per unit time in a unit volume, then the net flow out of $\Delta V$ will be equal to $(S-\ddpl{N}{t})\,\Delta V$. We have then that \begin{equation} \label{Eq:II:12:21} \FLPdiv{\FLPJ}=S-\ddp{N}{t}. \end{equation}  Combining (12.21) with (12.20), we get the neutron diffusion equation \begin{equation} \label{Eq:II:12:22} \FLPdiv{(-D\,\FLPgrad{N})}=S-\ddp{N}{t}. \end{equation}  In the static case—where $\ddpl{N}{t}=0$—we have Eq. (12.4) all over again!  We can use our knowledge of electrostatics to solve problems about the diffusion of neutrons. So let’s solve a problem. (You may wonder: Why do a problem if we have already done all the problems in electrostatics? We can do it faster this time because we have done the electrostatic problems!) Suppose we have a block of material in which neutrons are being generated—say by uranium fission—uniformly throughout a spherical region of radius $a$ (Fig. 12–7). We would like to know: What is the density of neutrons everywhere? How uniform is the density of neutrons in the region where they are being generated? What is the ratio of the neutron density at the center to the neutron density at the surface of the source region? Finding the answers is easy. The source density $S_0$ replaces the charge density $\rho$, so our problem is the same as the problem of a sphere of uniform charge density. Finding $N$ is just like finding the potential $\phi$. We have already worked out the fields inside and outside of a uniformly charged sphere; we can integrate them to get the potential. Outside, the potential is $Q/4\pi\epsO r$, with the total charge $Q$ given by $4\pi a^3\rho/3$. So \begin{equation} \label{Eq:II:12:23} \phi_{\text{outside}}=\frac{\rho a^3}{3\epsO r}. \end{equation}  For points inside, the field is due only to the charge $Q(r)$ inside the sphere of radius $r$, $Q(r)=4\pi r^3\rho/3$, so \begin{equation} \label{Eq:II:12:24} E=\frac{\rho r}{3\epsO}. \end{equation}  The field increases linearly with $r$. Integrating $E$ to get $\phi$, we have \begin{equation*} \phi_{\text{inside}}=-\frac{\rho r^2}{6\epsO}+\text{a constant}. \end{equation*}  At the radius $a$, $\phi_{\text{inside}}$ must be the same as $\phi_{\text{outside}}$, so the constant must be $\rho a^2/2\epsO$. (We are assuming that $\phi$ is zero at large distances from the source, which will correspond to $N$ being zero for the neutrons.) Therefore, \begin{equation} \label{Eq:II:12:25} \phi_{\text{inside}}=\frac{\rho}{3\epsO} \biggl(\frac{3a^2}{2}-\frac{r^2}{2}\biggr). \end{equation}  We know immediately the neutron density in our other problem. The answer is \begin{equation} \label{Eq:II:12:26} N_{\text{outside}}=\frac{Sa^3}{3Dr}, \end{equation}  and \begin{equation} \label{Eq:II:12:27} N_{\text{inside}}=\frac{S}{3D} \biggl(\frac{3a^2}{2}-\frac{r^2}{2}\biggr). \end{equation}  $N$ is shown as a function of $r$ in Fig. 12–7. Now what is the ratio of density at the center to that at the edge? At the center ($r=0$), it is proportional to $3a^2/2$. At the edge ($r=a$) it is proportional to $2a^2/2$, so the ratio of densities is $3/2$. A uniform source doesn’t produce a uniform density of neutrons. You see, our knowledge of electrostatics gives us a good start on the physics of nuclear reactors. There are many physical circumstances in which diffusion plays a big part. The motion of ions through a liquid, or of electrons through a semiconductor, obeys the same equation. We find again and again the same equations. 