Question 3.7: Lethargy defined as u = ln(Eo / E) is often used in neutron ......
Lethargy defined as u = ln(E_o / E) is often used in neutron slowing down problems; lethargy increases as energy decreases. Noting the following transformations: φ(E)dE = – φ (u)du p(E → E^′) = -p(u → u^′)du^′ and Σ_x(E) = Σ_x(u) ,
a. Show that p(E → E^′) given by Eq. (2.47) becomes
p(u → u^′) = \Bigg \{ \begin {matrix}\frac{1}{1 – α} exp (u – u^′) ,& u ≤ u′ ≤ u + ln(1/α) \\ 0, & otherwise \end {matrix}
b. Express Eq. (3.22) in terms of u.
First note that E = E_o exp(-u) and thus d E/\,d u=-E_{o}\exp(-u)
a. p(u\to u^{\prime})=-p(E\to E^{\prime})d E^{\prime}/\,d u^{\prime}=p(E\to E^{\prime})E_{o}\exp(-u^{\prime}) Substituting from Eq. (2.47):
p(u\to u^{\prime})={\frac{1}{1-\alpha}}\cdot{\frac{1}{E}}E_{o}\exp(-u^{\prime})={\frac{1}{1-\alpha}}\exp(u-u^{\prime})
Finally when E^{\prime}=E~,~u^{\prime}=u~~\mathrm{and~when}~E^{\prime}=\alpha E\,.
u^{\prime}=\operatorname{ln}(E_{o}\ /\alpha E)=\operatorname{ln}(E_{o}\ /E)-\operatorname{ln}(\alpha)=u+\operatorname{ln}(1/\alpha)
b. First note that \varphi(E)=-\varphi(u)d u\;/\,d E=\varphi(u)\frac{1}{E_{o}}\mathrm{exp}(u)\;. Thus Eq. (3.22) may be written as
\Sigma_{s}(u)\varphi(u)\frac{1}{E_{o}}\mathrm{exp}(u) = \frac{1}{1-\alpha}\int_{E}^{E/\alpha}\;\frac{1}{E^{\prime}}\Sigma_{s}(E^{\prime})\varphi(E^{\prime})d E^{\prime}= -\frac{1}{1-\alpha}\int_{u}^{u-\mathrm{ln}(1 / \alpha)}\ \frac{\exp(u^{\prime})}{E_{o}}\Sigma_{s}(u^{\prime})\varphi(u^{\prime})d u^{\prime}
Thus \Sigma_{s}(u)\varphi(u)={\frac{1}{1-\alpha}}\int_{u-\operatorname{ln}(1 / \alpha)}^{u} {\exp}(u^{\prime}-u)\Sigma_{s}(u^{\prime})\varphi(u^{\prime})d u^{\prime}