1. PUNEQE

Flux, Current, and Directional Current

This material is taken from (Lamarsh, 2002). Note that in neutronics, the integrals are often written backwards, so

\[ \begin{align}\begin{aligned}\int\overbrace{\mathscr{f}\left(x, y, \dots \right)}^{\text{integrand}} \underbrace{dx dy \dots}_{\text{differential}} \equiv \int \overbrace{dx dy \dots}^{\text{differential}} \underbrace{\mathscr{f}\left(x, y, \dots \right)}_{\text{integrand}}\\This can be confusing, but it's how it is.\end{aligned}\end{align} \]

Also, often the integration of the polar angle \(\theta\) is done by substitution, so

\[\int_{0}^{\pi}\mathscr{f}\left(\theta \right)\sin \theta d\theta \equiv \int_{-1}^{1} \mathscr{f} \left( \mu \right) d\mu\]

Given a differential flux of \(\psi = a + b \cos \theta\), find the flux, current, and each directions partial current.

To find the flux, we must integrate the differential flux over the solid angle \(\Omega\). In spherical coordinates, \(d\Omega = \sin \theta d\theta_{0}^{\pi} d\phi_{0}^{2\pi}\)

\[\phi\left(\vec{r}\right) = \int_{0}^{2\pi}\int_{0}^{\pi}\left(a + b \cos \theta \right) \sin \theta d\theta d\varphi\]
\[\phi\left(\vec{r}\right) = 2\pi \int_{-1}^{1}\left(a + b \mu \right) d \mu = 4\pi a\]

To find the current, we must integrate \(\phi \Omega\) over \(\Omega\), so we have

\[J=\int_{0}^{2\pi}\int_{0}^{\pi}\left( a + b \cos \theta\right) \sin \theta \Omega d\theta d\varphi\]

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