Lab Frame versus COM
An important distinction in neutronics is the lab frame versus the
center-of-mass frame, and how to convert between the two.
The central conversion is simple, defined as
\[\cos \theta_{s} = \frac{1 + A \cos \theta_{c}}{\sqrt{A^{2}+2A\cos \theta_{c}+1}}\]
However, applying this can prove more difficult.
For example, we can try to show that scattered neutrons are always
forward directional in the lab frame
(\(0\leq \theta_{s} \leq \nicefrac{\pi}{2}\)).
We start with the above correlation between lab and COM frame, and since
this is a neutron, \(A=1\), so
\[\cos \theta_{s} = \frac{1+\cos \theta_{c}}{\sqrt{2+2\cos \theta_{c}}} = \sqrt{\frac{1+\cos \theta_{c}}{2}}\]
And from there, we can square both sides and perform algebraic
simplification.
\[2\cos^{s} \theta_{s}=\cos \theta_{c} + 1 \therefore 2\cos^{2} \theta_{s} -1 = \cos \theta_{c}\]
and using identities,
\[\cos \left( 2\theta_{s}\right) = \cos \theta_{c} \therefore 2\theta_{s} = \theta_{c}\]
This leads to the conclusion that the lab frame angle is always half of
the center of mass angle. So, even if the the center of mass angle was
\(180^{\circ}\), the lab angle would only be \(90^{\circ}\),
which is to say that a neutron can never backscatter.
Another example is to find the scattering cross section as a function of
\(\mu_{c}=\cos \theta_{c}\). We start with the definition of the
differential cross section
\[\sigma_{s} \left( \theta_{c}, \phi_{c}\right) = \frac{\sigma_{s}}{4 \pi}\]
and then we integrate this over the whole range of \(\phi_{c}\)
\[ \begin{align}\begin{aligned}\sigma_{s} \left( \theta_{c}\right) = \int_{0}^{2\pi} \frac{\sigma_{s}}{4\pi}d\phi = 2\pi \frac{\sigma_{s}}{4\pi}\\And convert to :math:`\mu_{c}` (which requires no manipulation since\end{aligned}\end{align} \]
\(\theta_{c}\) does not appear in the expression). So, finally, we
have
\[\sigma_{s}\left( \mu_{c}\right)=\frac{\sigma_{s}}{2}\]
The last example is, using the above, to find the cross section in the
lab frame. We start with the conversion from
\(\mu_{c}\rightarrow \mu_{s}\).
\[\sigma_{s}\left( \mu_{s}\right) = \sigma_{s}\left(\mu_{c}\right) \left| \frac{d\mu_{c}}{d\mu_{s}}\right|\]
and, since
\[\mu_{s} = \sqrt{\frac{1+\mu_{c}}{2}}\]
\[ \begin{align}\begin{aligned}d\mu_{s}=\frac{1}{2\sqrt{2\left(1+\mu_{c}\right)}} d\mu_{c}\\we can now plug this in\end{aligned}\end{align} \]
\[\sigma_{s}\left( \mu_{s}\right)= \sigma_{s}\sqrt{2\left(1 + \mu_{c}\right)}=2\sigma_{s}\mu_{s}\]
