Question 11.P.8: In an experiment, 650MeV π^0 pions are scattered from a heav...

(a) In the case of a totally absorbing nucleus, $\eta_{l}(k)=0$, the total elastic and inelastic cross sections, which are given by (11.113)

$\sigma_{e l}=4 \pi \sum_{l=0}^{\infty}(2 l+1)\left|f_{l}\right|^{2}=\frac{\pi}{k^{2}} \sum_{l}(2 l+1)\left(1+\eta_{l}^{2}-2 \eta_{l} \cos 2 \delta_{l}\right)$.           (11.113)

and (11.114),

$\sigma_{\text {inel }}=\frac{\pi}{k^{2}} \sum_{l=0}^{\infty}(2 l+1)\left(1-\eta_{l}^{2}(k)\right)$.           (11.114)

become equal:

$\sigma_{e l}=\frac{\pi}{k^{2}} \sum_{l=0}^{l_{\max }}(2 l+1)=\sigma_{\text {inel }}$.             (11.194)

This experiment can be viewed as a scattering of high-energy pions, E = 650MeV, from a black “disk” of radius a = 1.4 fm; thus, the number of partial waves involved in this scattering can be obtained from $l_{\max } \simeq k a$, ehere $k=\sqrt{2 m_{\pi^{0} E / \bar{h}^{2}}}$. Since the rest mass energy of a $\pi^{0}$ pion is $m_{\pi^{0} }c^{2} \simeq 135 MeV$ and since $\hbar c=197.33 MeV fm$, we have

$k \simeq \sqrt{\frac{2 m_{\pi^{0} E}}{\hbar^{2}}}=\sqrt{\frac{2(m_{\pi ^{0}}c^{2})}{(\bar{h}c )^{2}} } =\sqrt{\frac{2(135 MeV )(650 MeV )}{(197.33 MeV fm )^{2}}}=2.12 fm ^{-1}$;                (11.195)

hence $l_{\max }=k a \simeq\left(2.12 fm ^{-1}\right)(1.4 fm )=2.97 \simeq 3$. We can thus reduce (11.194) to

$\sigma_{e l}=\sigma_{\text {inel }}=\frac{\pi}{k^{2}} \sum_{l=0}^{3}(2 l+1)=\frac{16 \pi}{k^{2}} \simeq \frac{16 \pi}{\left(2.12 fm ^{-1}\right)^{2}}=40.1 fm ^{2}=0.40 \text { barn }$.               (11.196)

The total cross section

$\sigma_{\text {tot }}=\sigma_{e l}+\sigma_{\text {inel }}=\frac{32 \pi}{k^{2}}=0.80 \text { barn }$.                    (11.197)

(b) The scattering amplitude can be obtained from (11.112)

$f(\theta)=\frac{1}{2 k} \sum_{l=0}^{\infty}(2 l+1)\left[\eta_{l} \sin 2 \delta_{l}+i\left(1-\eta_{l} \cos 2 \delta_{l}\right)\right] P_{l}(\cos \theta)$.             (11.112)

with $\eta_{l}(k)=0$:

$=\frac{i}{2 k}\left[1+3 \cos \theta+\frac{5}{2}\left(3 \cos ^{2} \theta-1\right)+\frac{7}{2}\left(5 \cos ^{3} \theta-3 \cos \theta\right)\right]$,               (11.198)

where we have used the following Legendre polynomials: $P_{0}(u)=1, P_{1}(u)=u, P_{2}(u)=\frac{1}{2}\left(3 u^{2}-1\right), P_{3}(u)=\frac{1}{2}\left(5 u^{3}-3 u\right)$. The forward scattering amplitude (θ = 0) is

$f(0)=\frac{i}{2 k}\left[1+3+\frac{5}{2}(3-1)+\frac{7}{2}(5-3)\right]=\frac{8 i}{k}$.              (11.199)

Combining (11.197) and (11.199), we get the optical theorem: $\operatorname{Im} f(0)=(k / 4 \pi) \sigma_{\text {tot }}=8 / k$.

(c) From (11.198) the differential elastic cross section is

$\frac{d \sigma}{d \Omega}=|f(\theta)|^{2}=\frac{1}{4 k^{2}}\left[1+3 \cos \theta+\frac{5}{2}\left(3 \cos ^{2} \theta-1\right)+\frac{7}{2}\left(5 \cos ^{3} \theta-3 \cos \theta\right)\right]^{2}$.           (11.200)

As shown in Figure 11.9, the differential cross section displays an interference pattern due to the superposition of incoming and outgoing waves. The total elastic cross section is given by $\sigma_{e l}=\int_{0}^{\pi}|f(\theta)|^{2} \sin \theta d \theta \int_{0}^{2 \pi} d \varphi$ which, combined with (11.200), leads to

$\sigma_{e l}=\frac{2 \pi}{4 k^{2}} \int_{0}^{\pi}\left[1+3 \cos \theta+\frac{5}{2}\left(3 \cos ^{2} \theta-1\right)+\frac{7}{2}\left(5 \cos ^{3} \theta-3 \cos \theta\right)\right]^{2} \sin \theta d \theta=\frac{16 \pi}{k^{2}} .$(11.201)

This is the same expression we obtained in (11.196). Unlike the differential cross section, the total cross section displays no interference pattern because its final expression does not depend on any angle, since the angles were integrated over.