Question 3.2: Show that in Eq. (3.31) the normalization condition ∫0^Eo s(......
Show that in Eq. (3.31) the normalization condition ∫_0^{E_o} s(E)dE = 1 must be obeyed. Hint: Note that ∫_0^{E_o} p(E^′ → E)dE = 1 for E^′ ≤ E_0 .
Write \Sigma_{s}(E^{\prime}\to E)=\Sigma_{s}(E^{\prime})p(E^{\prime}\to E)
\Sigma_{t}(E)\varphi(E)=\int_0^{E_o}\Sigma_{s}(E^{\prime})p(E^{\prime}\to E)\varphi(E^{\prime})d E^{\prime}+s(E)q_{o}
Next integrate E between 0 and E_o:
\int_{0}^{E_{o}}\ \Sigma_{t}(E)\varphi(E)d E =\int_{0}^{E_{o}}\ \Sigma_{s}(E^{\prime})\biggl[\int_{0}^{E_{o}}\ p(E^{\prime}\to E)d E\biggr]\varphi(E^{\prime})d E^{\prime}+\int_{0}^{E_{o}}\ s(E)d E q_{o}Since the bracketed term is equal to one, (because there is no up-scatter past E_o )and \Sigma_{a}(E)=\Sigma_{t}(E)-\Sigma_{s}(E)\,, we have
\int_{0}^{E_{o}}\ \Sigma_{a}(E)\varphi(E)d E=\int_{0}^{E_{o}}\ s(E)d E q_{o} But all of the neutrons slowing down past E_o must be absorbed at lower energies. Thus \int_{0}^{E_{o}}\ \Sigma_{a}(E)\varphi(E)d E = q_o and hence we must have \int_{0}^{E_{o}}\,S(E)d E=1.