Consider a plane of thickness $$\Delta z$$. The plane emits $$q$$ neutrons per second per unit area at a single energy and is comprised of a pure absorber. What is the angular flux at the right edge of the plane, $$\Phi(\Delta z, \mu)$$? Use your solution to give the angular flux emerging from an infinitely thin plane source. Hint: Use an integrating factor of $$\exp{(\sigma z / \mu)}$$.

The derivation of this solution is after the jump

The time- and energy-independent transport equation can be written as

\[\hat{\Omega } \cdot \nabla \Phi (\vec{r}, \hat{\Omega }) + \sigma \Phi (\vec{r}, \hat{\Omega }) = \int_{4\pi} d\Omega{}’ \: \sigma’f’\Phi(\vec{r}, \hat{\Omega }’\rightarrow \hat{\Omega }) + \frac{q}{4\pi \Delta z}\:.\]

Since the angular flux is being found for pure absorbing (no fissioning or scattering) slab, the scattering and azimuthal terms drop out.

\begin{equation}
\frac{\partial \Phi (\vec{r}, \hat{\Omega },E) }{\partial x} = \frac{\partial \Phi (\vec{r}, \hat{\Omega },E) }{\partial y}=\int_{4\pi} d\Omega{}’ \: \sigma’f’\Phi(\vec{r}, \hat{\Omega }’\rightarrow \hat{\Omega })= 0\:.
\end{equation} The streaming term is now 1-D,

\begin{equation}
\hat{\Omega} \cdot \frac{\partial \Phi (\vec{r}, \hat{\Omega },E) }{\partial z}\hat{z} = \mu \frac{\partial \Phi (\vec{r}, \hat{\Omega },E) }{\partial z}\:,
\end{equation}

If we think of $$\Phi(z,\mu)$$ as the angular flux per unit area, the factor of $$2\pi$$ can be removed from the source to get:
\[\mu \frac{\partial \Phi (z, \mu)}{\partial z} + \sigma \Phi (z, \mu) = \frac{q}{2 \Delta z}\]
with the following boundary conditions

\[\Phi (z=0, \mu) = 0 \quad \text{for} \: \mu > 0, \: \text{and} \quad \Phi (z=\Delta z, \mu) = 0 \quad \text{for} \: \mu < 0\:.\]
Next, the differential equation is solved.

\[\frac{e^{\frac{\sigma z}{\mu}}}{\mu}[\mu \frac{\partial \Phi (z, \mu)}{\partial z}+ \sigma \Phi (z, \mu)] = \frac{q}{2\Delta z}\frac{e^{\frac{\sigma z}{\mu}}}{\mu}\:.\]

\begin{equation}
\frac{\partial \Phi (z, \mu)e^{\frac{\sigma z}{\mu}}}{\partial z} = \frac{q}{2 \Delta z \mu}e^{\frac{\sigma z}{\mu}}\:.
\end{equation}

\begin{equation}
\Phi (z, \mu)e^{\frac{\sigma z}{\mu}} = \int \frac{q}{\mu}e^{\frac{\sigma z}{\mu}} \partial z = \frac{q}{2 \Delta z \sigma}e^{\frac{\sigma z}{\mu}} + C\:.
\end{equation}

\begin{equation}
\Phi (z, \mu) = \frac{q}{2 \Delta z \sigma} + C e^{-\frac{\sigma z}{\mu}}\:.
\end{equation}
Using the boundary condition at $$z = 0$$ gives that

\begin{equation}
C = -\frac{q}{2 \Delta z \sigma} \hspace{0.25in} \text{for} \hspace{0.05in} \mu > 0\:.
\end{equation}Thus, the equation for the angular flux at the right edge of the slab is

\begin{equation}
\boxed{\Phi (\Delta z, \mu) = \left\{\begin{matrix} \frac{q}{2 \Delta z \sigma}(1-e^{-\frac{\sigma \Delta z}{\mu}})
& \text{for} \hspace{0.05in} \mu > 0 \\
0 & \text{for} \hspace{0.05in} \mu < 0
\end{matrix} \right ] } \:.
\end{equation}Next, the limit is taken as $$\Delta z$$ approaches 0. A careful way of doing this is to expand the exponential

\begin{equation}
e^{-\frac{\sigma \Delta z}{\mu}} = 1+\frac{\sigma \Delta z}{\mu}+\frac{(\frac{\sigma \Delta z}{\mu})^2}{2!}+\frac{(\frac{\sigma \Delta z}{\mu})^3}{3!}+… +\frac{(\frac{\sigma \Delta z}{\mu})^N}{N!} ,
\end{equation} and then it can simply be shown that
\begin{equation}
\boxed{\Phi(0^+, \mu) = \lim_{\Delta z \rightarrow 0}\left [ \left\{\begin{matrix} \frac{q}{2 \Delta z \sigma}(1-e^{-\frac{\sigma \Delta z}{\mu}})
& \text{for} \hspace{0.05in} \mu > 0 \\
0 & \text{for} \hspace{0.05in} \mu < 0
\end{matrix}\right. \right ] = \left\{\begin{matrix} \frac{q}{2 \mu} & \text{for} \hspace{0.05in} \mu > 0 \\
0 & \text{for} \hspace{0.05in} \mu < 0
\end{matrix}\right. } \:,
\end{equation} which is the angular flux emerging from the right side of a plane source that is infinite in the x and y dimensions.